Mathematical Analysis
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MATHEMATICAL ANALYSIS
Elias Zakon University of Windsor
University of Windsor Book: Mathematical Analysis (Zakon)
n
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TABLE OF CONTENTS About the Author Licensing
1: Set Theory 1.1: Sets and Operations on Sets. Quantifiers 1.1.E: Problems in Set Theory (Exercises) 1.2: Relations. Mappings 1.2.E: Problems on Relations and Mappings (Exercises) 1.3: Sequences 1.4: Some Theorems on Countable Sets 1.4.E: Problems on Countable and Uncountable Sets (Exercises)
2: Real Numbers and Fields 2.1: Axioms and Basic Definitions 2.2: Natural Numbers. Induction 2.2.E: Problems on Natural Numbers and Induction (Exercises) 2.3: Integers and Rationals 2.4: Upper and Lower Bounds. Completeness 2.4.E: Problems on Upper and Lower Bounds (Exercises) 2.5: Some Consequences of the Completeness Axiom 2.6: Powers with Arbitrary Real Exponents. Irrationals 2.6.E: Problems on Roots, Powers, and Irrationals (Exercises) 2.7: The Infinities. Upper and Lower Limits of Sequences 2.7.E: Problems on Upper and Lower Limits of Sequences in \(E^{*}\) (Exercises)
3: Vector Spaces and Metric Spaces 3.1: The Euclidean n-Space, Eⁿ 3.1.E: Problems on Vectors in \(E^{n}\) (Exercises) 3.2: Lines and Planes in Eⁿ 3.2.E: Problems on Lines and Planes in \(E^{n}\) (Exercises) 3.3: Intervals in Eⁿ 3.3.E: Problems on Intervals in Eⁿ (Exercises) 3.4: Complex Numbers 3.4.E: Problems on Complex Numbers (Exercises) 3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces 3.5.E: Problems on Linear Spaces (Exercises) 3.6: Normed Linear Spaces 3.6.E: Problems on Normed Linear Spaces (Exercises) 3.7: Metric Spaces 3.7.E: Problems on Metric Spaces (Exercises) 3.8: Open and Closed Sets. Neighborhoods
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3.8.E: Problems on Neighborhoods, Open and Closed Sets (Exercises) 3.9: Bounded Sets. Diameters 3.9.E: Problems on Boundedness and Diameters (Exercises) 3.10: Cluster Points. Convergent Sequences 3.10.E: Problems on Cluster Points and Convergence (Exercises) 3.11: Operations on Convergent Sequences 3.11.E: Problems on Limits of Sequences (Exercises) 3.12: More on Cluster Points and Closed Sets. Density 3.12.E: Problems on Cluster Points, Closed Sets, and Density 3.13: Cauchy Sequences. Completeness 3.13.E: Problems on Cauchy Sequences
4: Function Limits and Continuity 4.1: Basic Definitions 4.1.E: Problems on Limits and Continuity 4.2: Some General Theorems on Limits and Continuity 4.2.E: More Problems on Limits and Continuity 4.3: Operations on Limits. Rational Functions 4.3.E: Problems on Continuity of Vector-Valued Functions 4.4: Infinite Limits. Operations in E* 4.4.E: Problems on Limits and Operations in \(E^{*}\) 4.5: Monotone Function 4.5.E: Problems on Monotone Functions 4.6: Compact Sets 4.6.E: Problems on Compact Sets 4.7: More on Compactness 4.8: Continuity on Compact Sets. Uniform Continuity 4.8.E: Problems on Uniform Continuity; Continuity on Compact Sets 4.9: The Intermediate Value Property 4.9.E: Problems on the Darboux Property and Related Topics 4.10: Arcs and Curves. Connected Sets 4.10.E: Problems on Arcs, Curves, and Connected Sets 4.11: Product Spaces. Double and Iterated Limits 4.11.E: Problems on Double Limits and Product Spaces 4.12: Sequences and Series of Functions 4.12.E: Problems on Sequences and Series of Functions 4.13: Absolutely Convergent Series. Power Series 4.13.E: More Problems on Series of Functions
5: Differentiation and Antidifferentiation 5.1: Derivatives of Functions of One Real Variable 5.1.E: Problems on Derived Functions in One Variable 5.2: Derivatives of Extended-Real Functions 5.2.E: Problems on Derivatives of Extended-Real Functions
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5.3: L'Hôpital's Rule 5.3.E: Problems on \(L^{\prime}\) Hôpital's Rule 5.4: Complex and Vector-Valued Functions on \(E^{1}\) 5.4.E: Problems on Complex and Vector-Valued Functions on \(E^{1}\) 5.5: Antiderivatives (Primitives, Integrals) 5.5.E: Problems on Antiderivatives 5.6: Differentials. Taylor’s Theorem and Taylor’s Series 5.6.E: Problems on Tayior's Theorem 5.7: The Total Variation (Length) of a Function f - E1 → E 5.7.E: Problems on Total Variation and Graph Length 5.8: Rectifiable Arcs. Absolute Continuity 5.8.E: Problems on Absolute Continuity and Rectifiable Arcs 5.9: Convergence Theorems in Differentiation and Integration 5.9.E: Problems on Convergence in Differentiation and Integration 5.10: Sufficient Condition of Integrability. Regulated Functions 5.10.E: Problems on Regulated Functions 5.11: Integral Definitions of Some Functions 5.11.E: Problems on Exponential and Trigonometric Functions
6: Differentiation on Eⁿ and Other Normed Linear Spaces 6.1: Directional and Partial Derivatives 6.1.E: Problems on Directional and Partial Derivatives 6.2: Linear Maps and Functionals. Matrices 6.2.E: Problems on Linear Maps and Matrices 6.3: Differentiable Functions 6.3.E: Problems on Differentiable Functions 6.4: The Chain Rule. The Cauchy Invariant Rule 6.4.E: Further Problems on Differentiable Functions 6.5: Repeated Differentiation. Taylor’s Theorem 6.5.E: Problems on Repeated Differentiation and Taylor Expansions 6.6: Determinants. Jacobians. Bijective Linear Operators 6.6.E: Problems on Bijective Linear Maps and Jacobians 6.7: Inverse and Implicit Functions. Open and Closed Maps 6.7.E: Problems on Inverse and Implicit Functions, Open and Closed Maps 6.8: Baire Categories. More on Linear Maps 6.8.E: Problems on Baire Categories and Linear Maps 6.9: Local Extrema. Maxima and Minima 6.9.E: Problems on Maxima and Minima 6.10: More on Implicit Differentiation. Conditional Extrema 6.10.E: Further Problems on Maxima and Minima
7: Volume and Measure 7.1: More on Intervals in Eⁿ. Semirings of Sets 7.1.E: Problems on Intervals and Semirings
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7.2: \(\mathcal{C}_{\sigma}\)-Sets. Countable Additivity. Permutable Series 7.2.E: Problems on \(\mathcal{C}_{\sigma}\) -Sets, \(\sigma\) -Additivity, and Permutable Series 7.3: More on Set Families 7.3.E: Problems on Set Families 7.4: Set Functions. Additivity. Continuity 7.4.E: Problems on Set Functions 7.5: Nonnegative Set Functions. Premeasures. Outer Measures 7.5.E: Problems on Premeasures and Related Topics 7.6: Measure Spaces. More on Outer Measures 7.6.E: Problems on Measures and Outer Measures 7.7: Topologies. Borel Sets. Borel Measures 7.7.E: Problems on Topologies, Borel Sets, and Regular Measures 7.8: Lebesgue Measure 7.8.E: Problems on Lebesgue Measure 7.9: Lebesgue–Stieltjes Measures 7.9.E: Problems on Lebesgue-Stieltjes Measures 7.10: Generalized Measures. Absolute Continuity 7.10.E: Problems on Generalized Measures 7.11: Differentiation of Set Functions 7.11.E: Problems on Vitali Coverings
8: Measurable Functions and Integration 8.1: Elementary and Measurable Functions 8.1.E: Problems on Measurable and Elementary Functions in \((S, \mathcal{M})\) 8.2: Measurability of Extended-Real Functions 8.2.E: Problems on Measurable Functions in \((S, \mathcal{M}, m)\) 8.3: Measurable Functions in \((S, \mathcal{M}, m)\) 8.3.E: Problems on Measurable Functions in \((S, \mathcal{M}, m)\) 8.4: Integration of Elementary Functions 8.4.E: Problems on Integration of Elementary Functions 8.5: Integration of Extended-Real Functions 8.5.E: Problems on Integration of Extended-Real Functions 8.6: Integrable Functions. Convergence Theorems 8.6.E: Problems on Integrability and Convergence Theorems 8.7: Integration of Complex and Vector-Valued Functions 8.7.E: Problems on Integration of Complex and Vector-Valued Functions 8.8: Product Measures. Iterated Integrals 8.8.E: Problems on Product Measures and Fubini Theorems 8.9: Riemann Integration. Stieltjes Integrals 8.9.E: Problems on Riemann and Stieltjes Integrals 8.10: Integration in Generalized Measure Spaces 8.10.E: Problems on Generalized Integration 8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition 8.11.E: Problems on Radon-Nikodym Derivatives and Lebesgue Decomposition
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8.12: Integration and Differentiation 8.12.E: Problems on Differentiation and Related Topics
9: Calculus Using Lebesgue Theory 9.1: L-Integrals and Antiderivatives 9.1.E: Problems on L-Integrals and Antiderivatives 9.2: More on L-Integrals and Absolute Continuity 9.2.E: Problems on L-Integrals and Absolute Continuity 9.3: Improper (Cauchy) Integrals 9.3.E: Problems on Cauchy Integrals 9.4: Convergence of Parametrized Integrals and Functions 9.4.E: Problems on Uniform Convergence of Functions and C-Integrals
Index Index Glossary Detailed Licensing
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About the Author Elias Zakon was born in Russia under the czar in 1908, and he was swept along in the turbulence of the great events of twentiethcentury Europe. Zakon studied mathematics and law in Germany and Poland, and later he joined his father’s law practice in Poland. Fleeing the approach of the German Army in 1941, he took his family to Barnaul, Siberia, where, with the rest of the populace, they endured five years of hardship. The Leningrad Institute of Technology was also evacuated to Barnaul upon the siege of Leningrad, and there he met the mathematician I. P. Natanson; with Natanson’s encouragement, Zakon again took up his studies and research in mathematics. Zakon and his family spent the years from 1946 to 1949 in a refugee camp in Salzburg, Austria, where he taught himself Hebrew, one of the six or seven languages in which he became fluent. In 1949, he took his family to the newly created state of Israel and he taught at the Technion in Haifa until 1956. In Israel he published his first research papers in logic and analysis. Throughout his life, Zakon maintained a love of music, art, politics, history, law, and especially chess; it was in Israel that he achieved the rank of chess master. In 1956, Zakon moved to Canada. As a research fellow at the University of Toronto, he worked with Abraham Robinson. In 1957, he joined the mathematics faculty at the University of Windsor, where the first degrees in the newly established Honors program in Mathematics were awarded in 1960. While at Windsor, he continued publishing his research results in logic and analysis. In this post-McCarthy era, he often had as his house-guest the prolific and eccentric mathematician Paul Erdös, who was then banned from the United States for his political views. Erdös would speak at the University of Windsor, where mathematicians from the University of Michigan and other American universities would gather to hear him and to discuss mathematics. While at Windsor, Zakon developed three volumes on mathematical analysis, which were bound and distributed to students. His goal was to introduce rigorous material as early as possible; later courses could then rely on this material. We are publishing here the latest complete version of the second of these volumes, which was used in a two-semester class required of all secondyear Honours Mathematics students at Windsor.
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Licensing A detailed breakdown of this resource's licensing can be found in Back Matter/Detailed Licensing.
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CHAPTER OVERVIEW 1: Set Theory 1.1: Sets and Operations on Sets. Quantifiers 1.1.E: Problems in Set Theory (Exercises) 1.2: Relations. Mappings 1.2.E: Problems on Relations and Mappings (Exercises) 1.3: Sequences 1.4: Some Theorems on Countable Sets 1.4.E: Problems on Countable and Uncountable Sets (Exercises)
This page titled 1: Set Theory is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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1.1: Sets and Operations on Sets. Quantifiers Sets and Operations on Sets A set is a collection of objects of any specified kind. Sets are usually denoted by capitals. The objects belonging to a set are called its elements or members. We write x ∈ A if x is a member of A , and x ∉ A if it is not. means that A consists of the elements a, b, c, . . .. In particular, A = {a, b} consists of consists of p alone. The empty or void set, ∅, has no elements. Equality (=) means logical identity. A = {a, b, c, . . . }
a
and b ;
A = {p}
If all members of A are also in B , we call A a subset of B (and B is a superset of A ), and write A ⊆ B or B ⊇ A . It is an axiom that the sets A and B are equal (A = B ) if they have the same members, i.e., A ⊆B
and
B ⊆ A.
(1.1.1)
If, however, A ⊆ B but B ⊈ A (i.e., B has some elements not in A), we call A a proper subset of B and write A ⊂ B or B ⊃ A . “ ⊆” is called the inclusion relation. Set equality is not affected by the order in which elements appear. Thus a, b = b, a . Not so for ordered pairs (a, b). For such pairs, (a, b) = (x, y)
iff
a =x
and
b = y,
(1.1.2)
but not if a = y and b = x . Similarly, for ordered n-tuples, (a1 , a2 , . . . , an ) = (x1 , x2 , . . . , xn )
iff
ak = xk , k = 1, 2, . . . , n.
(1.1.3)
We write x|P (x) for "the set of all x satisfying the condition P (x)." Similarly, (x, y)|P (x, y) is the set of all ordered pairs for which P (x, y) holds; x ∈ A|P (x) is the set of those x in A for which P (x) is true. For any sets A and B , we define their union product) A × B , as follows: A∪B
A∩B
A−B
A×B
A∪B
, intersection
A∩B
, difference
A =B
, and Cartesian product (or cross
is the set of all members of A and B taken together: {x|x ∈ A or x ∈ B}.
(1.1.4)
{x ∈ A|x ∈ B}.
(1.1.5)
{x ∈ A|x ∉ B}.
(1.1.6)
is the set of all common elements of A and B :
consists of those x ∈ A that are not in B :
is the set of all ordered pairs (x, y), with x ∈ A and y ∈ B : {(x, y) x ∈ A, y ∈ B}.
Similarly, A
1
× A2 ×. . . ×An
A × A×. . . ×A
is the set of all ordered n-tuples (x
1,
. . . , xn )
(1.1.7)
such that x
k
∈ Ak , k = 1, 2, . . . , n
. We write A for n
(n factors).
and B are said to be disjoint iff A ∩ B = ∅ (no common elements). Otherwise, we say that A meets B (A ∩ B ≠ ∅ ). Usually all sets involved are subsets of a "master set" S , called the space. Then we write −X for S − X , and call −X the complement of X ( in S ). Various other notations are likewise in use. A
Example 1.1.1 Let A = {1, 2, 3}, B = {2, 4}. Then A ∪ B = {1, 2, 3, 4},
A ∩ B − {2},
A − B = {1, 3},
A × B = {(1, 2), (1, 4), (2, 2), (2, 4), (3, 2), (3, 4)}.
(1.1.8) (1.1.9)
If N is the set of all naturals (positive integers), we could also write A = {x ∈ N |x < 4}.
1.1.1
(1.1.10)
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Theorem 1.1.1 a. A ∪ A = A; A ∩ A = A ; b. A ∪ B = B ∪ A, A ∩ B = B ∩ A ; c. (A ∪ B) ∪ C = A ∪ (B ∪ C ); (A ∩ B) ∩ C d. (A ∪ B) ∩ C = (A ∩ C ) ∪ (B ∩ C ) ; e. (A ∩ B) ∪ C = (A ∪ C ) ∩ (B ∪ C ) .
;
= A ∩ (B ∩ C )
Proof The proof of (d) is sketched out in Problem 1. The rest is left to the reader. Because of (c), we may omit brackets in A ∪ B ∪ C and A ∩ B ∩ C ; similarly for four or more sets. More generally, we may consider whole families of sets, i.e., collections of many (possibly infinitely many) sets. If M is such a family, we define its union, ⋃ M , to be the set of all elements x, each belonging to at least one set of the family. The intersection of M, denoted ⋂ M , consists of those x that belong to all sets of the family simultaneously. Instead, we also write ⋃{X|X ∈ M}
and
⋂{X|X ∈ M},
respectively.
(1.1.11)
Often we can number the sets of a given family: A1 , A2 , . . . , An , . . .
(1.1.12)
More generally, we may denote all sets of a family M by some letter (say, X) with indices i attached to it (the indices may, but need not, be numbers). The family M then is denoted by {X } or {X |i ∈ I, where i is a variable index ranging over a suitable set I of indices ("index notation"). In this case, the union and intersection of M are denoted by such symbols as i
i
⋃{ Xi |i ∈ I } = ⋃ Xi = ⋃ Xi = ⋃ Xi ; i
(1.1.13)
i∈I
⋂{ Xi |i ∈ I } = ⋂ Xi = ⋂ Xi = ⋂ Xi ; i
(1.1.14)
i∈I
If the indices are integers, we may write m
∞
m
⋃ Xn , ⋃ Xn , ⋃ Xn , etc. n=1
n=1
(1.1.15)
n=k
Theorem 1.1.1: De Morgan's Duality Laws For any sets S and A
i
(i ∈ I )
, the following are true: (i) S − ⋃ Ai = ⋂(S − A); i
(ii) S − ⋂ Ai = ∪i (S − Ai ).
i
(1.1.16)
i
(If S is the entire space, we may write −A for S − A , − ⋃ A for S − ⋃ A , etc. i
i
i
i
Before proving these laws, we introduce some useful notation.
Logical Quantifiers From logic we borrow the following abbreviations. "(∀x ∈ A) ..." means "For each member x of A , it is true that . . ." "(∃x ∈ A) ..." means "There is at least one x in A such that . . ." "(∃!x ∈ A) ..." means "There is a unique x in A such that . . ." The symbols "(∀x ∈ A) " and "(∃x ∈ A) " are called the universal and existential quantifiers, respectively. If confusion is ruled out, we simply write "(∀x)," "(∃x)," and "(∃!x)" instead. For example, if we agree that m, and n denote naturals, then " (∀n)(∃m) m > n "
1.1.2
(1.1.17)
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means "For each natural n , there is a natural m such that m > n ." We give some more examples. Let M = {A
i |i
∈ I}
be an indexed set family. By definition, x ∈ ⋃ A means that x is in at least one of the sets A ; in symbols, i
i
(∃i ∈ I ) x ∈ Ai .
(1.1.18)
Thus we note that x ∈ ⋃ Ai
iff [(∃i ∈ I )x ∈ Ai ].
(1.1.19)
iff [(∀i ∈ I )x ∈ Ai ].
(1.1.20)
i∈I
Similarly, x ∈ ⋂ Ai i
Also note that x ∉ ⋃ A iff x is in none of the A , i.e., i
i
(∀i) x ∉ Ai .
(1.1.21)
Similarly, x ∉ ⋂ A iff x fails to be in some A . i.e., i
i
(∃i) x ∉ Ai .
(W hy?)
(1.1.22)
We now use these remarks to prove Theorem 2(i). We have to show that S − ⋃ A has the same elements as ⋂(S − A ) , i.e., that x ∈ S − ⋃ A iff x ∈ ⋂(S − A ) . But, by our definitions, we have i
i
i
i
x ∈ S − ⋃ Ai
⟺
[x ∈ S, x ∉ ⋃ Ai ]
⟺
(∀i)[x ∈ S, x ∉ Ai ]
⟺
(∀i)x ∈ S − Ai
⟺
x ∈ ⋂(S − Ai ),
(1.1.23)
as required. One proves part (ii) of Theorem 2 quite similarly. (Exercise!) We shall now dwell on quantifiers more closely. Sometimes a formula additional property Q(x).
P (x)
holds not for all
x ∈ A
, but only for those with an
This will be written as (∀x ∈ A|Q(x)) P (x),
(1.1.24)
where the vertical stroke stands for "such that." For example, if N is again the naturals, then the formula (∀x ∈ N |x > 3) x ≥ 4
means "for each x ∈ N such that x ≥ 4 ." In other words, for naturals, (1) can also be written as (∀x ∈ N ) x > 3
x >3
⟹
(1.1.25) ⟹
x ≥4
(the arrow stands for "implies"). Thus
x ≥ 4.
(1.1.26)
In mathematics, we often have to form the negation of a formula that starts with one or several quantifiers. It is noteworthy, then, that each universal quantifier is replaced by an existential one (and vice versa), followed by the negation of the subsequent part of the formula. For example, in calculus, a real number p is called the limit of a sequence x , x , . . . , x , . . . iff the following is true: 1
2
n
For every real ϵ > 0 , there is a natural k (depending on ϵ) such that, for all natural n > k , we have |x
n
If we agree that lower case letters (possibly with subscripts) denote real numbers, and that n , sentence can be written as (∀ϵ > 0)(∃k)(∀n > k) | xn − p| < ϵ.
k
− p| < ϵ|
.
denote naturals
(n, k ∈ N)
, this
(1.1.27)
Here the expressions "(∀ϵ > 0) " and "(∀n > k) " stand for "(∀ϵ|ϵ > 0) " and "(∀n|n > k) ," respectively (such self-explanatory abbreviations will also be used in other similar cases).
1.1.3
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Now, since (2) states that "for all ϵ > 0 " something (i.e., the rest of (2)) is true, the negation of (2) starts with "there is an (for which the rest of the formula fails). Thus we start with "(∃ϵ > 0) ," and form the negation of what follows, i.e., of (∃k)(∀n > k) | xn − p| < ϵ.
ϵ>0
"
(1.1.28)
This negation, in turn, starts with "(∀k) ," etc. Step by step, we finally arrive at (∃ϵ > 0)(∀k)(∃n > k) | xn − p| ≥ ϵ.
(1.1.29)
Note that here the choice of n > k may depend on k. To stress it, we often write n for n . Thus the negation of (2) finally emerges as k
(∃ϵ > 0)(∀k)(∃nk > k) | xn
k
− p| ≥ ϵ.
(1.1.30)
The order in which the quantifiers follow each other is essential. For example, the formula (∀n ∈ N )(∃m ∈ N ) m > n
(1.1.31)
("each n ∈ N is exceeded by some m ∈ N ") is true, but (∃m ∈ N )(∀n ∈ N ) m > n
(1.1.32)
is false. However, two consecutive universal quantifiers (or two consecutive existential ones) may be interchanged. We briefly write " (∀x, y ∈ A) " for " (∀x ∈ A)(∀y ∈ A), "
(1.1.33)
" (∃x, y ∈ A) " for " (∃x ∈ A)(∃y ∈ A), " etc.
(1.1.34)
and
does not imply the existence of an x for which P (x) is true. It is only meant to imply that there is no x in A for which P (x) fails. The latter is true even if A = ∅ ; we then say that "(∀x ∈ A)
P (x)
" is vacuously true. For examplek the formula ∅ ⊆ B , i.e.,
(∀x ∈ ∅) x ∈ B
(1.1.35)
is always true (vacuously). This page titled 1.1: Sets and Operations on Sets. Quantifiers is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
1.1.4
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1.1.E: Problems in Set Theory (Exercises) Exercise 1.1.E. 1 Prove Theorem 1 (show that x is in the left-hand set iff it is in the right-hand set). For example, for (d), x ∈ (A ∪ B) ∩ C
⟺ [x ∈ (A ∪ B) and x ∈ C ] ⟺ [(x ∈ A or x ∈ B), and x ∈ C ] ⟺ [(x ∈ A, x ∈ C ) or (x ∈ B, x ∈ C )].
Exercise 1.1.E. 2 Prove that (i) −(−A) = A ; (ii) A ⊆ B iff −B ⊆ −A .
Exercise 1.1.E. 3 Prove that A − B = A ∩ (−B) = (−B) − (−A) = −[(−A) ∪ B].
(1.1.E.1)
Also, give three expressions for A ∩ B and A ∪ B, in terms of complements.
Exercise 1.1.E. 4 Prove the second duality law (Theorem 2(ii)).
Exercise 1.1.E. 5 Describe geometrically the following sets on the real line: (i) {x|x < 0};
(ii) {x||x| < 1};
(iii) {x||x − a| < ε};
(iv) {x|a < x ≤ b};
(1.1.E.2)
(v) {x||x| < 0}.
Exercise 1.1.E. 6 Let (a, b) denote the set {{a}, {a, b}}
(1.1.E.3)
(Kuratowski's definition of an ordered pair). (i) Which of the following statements are true? (a) a ∈ (a, b);
(b) {a} ∈ (a, b);
(c) (a, a) = {a};
(d) b ∈ (a, b);
(e) {b} ∈ (a, b);
(f) {a, b} ∈ (a, b).
(1.1.E.4)
(ii) Prove that (a, b) = (u, v) if a = u and b = v . [Hint: Consider separately the two cases a = b and a ≠ b, noting that {a, a} = {a}. Also note that {a} ≠ a. ]
1.1.E.1
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Exercise 1.1.E. 7 Describe geometrically the following sets in the xy-plane. (i) {(x, y)|x < y}; (ii) {(x, y)|x + y < 1} ; (iii) {(x, y)| max(|x|, |y|) < 1}; (iii) {(x, y)|y > x }; (iv) {(x, y)|y > x }; (vii) {(x, y)||x| + |y| < 4}; (vii) {(x, y)|(x − 2) + (y + 5) ≤ 9} ; (viii) {(x, y)|x − 2xy + y < 0} ; (ix) {(x, y)|x − 2xy + y = 0} . 2
2
2
2
2
2
2
2
2
2
Exercise 1.1.E. 8 Prove that (i) (A ∪ B) × C = (A × C ) ∪ (B × C ) ; (ii) (A ∩ B) × (C ∩ D) = (A × C ) ∩ (B × D) ; (iii) (X × Y ) − (X × Y ) = [(X ∩ X ) × (Y − Y )] ∪ [(X − X ) × Y ] ; [Hint: In each case, show that an ordered pair (x, y) is in the left-hand set iff it is in the right-hand set, treating element of the Cartesian product. ] ′
′
′
′
′
(x, y)
as one
Exercise 1.1.E. 9 Prove the distributive laws (i) A ∩ ∪X = ⋃ (A ∩ X ) ; (ii) A ∪ ∩X = ⋂ (A ∪ X ) ; (iii) (∩X ) − A = ∩ (X − A) ; (iv) (∪X ) − A = ∪ (X − A) ; (v) ∩X ∪ ∩Y = ∩ (X ∪ Y ) ; (vi) ∪X ∩ ∪Y = ∪ (X ∩ Y ) . i
i
i
i
i
i
i
i
i
j
i
i,j
j
i
i,j
j
i
j
Exercise 1.1.E. 10 Prove that (i) (∪A ) × B = ⋃ (A × B) ; (ii) (∩A ) × B = ∩ (A × B) ; (iii) (∩ A ) × (∩ B ) = ⋂ (A i
i
i
i
(iv) (∪
i
i
i Ai )
j
j
i,j
i
× Bi )
;
× (⋃j Bj ) = ⋃i,j (Ai × Bj )
.
1.1.E: Problems in Set Theory (Exercises) is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
1.1.E.2
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1.2: Relations. Mappings Relations In §1, we have already considered sets of ordered pairs, such as Cartesian products A × B or sets of the form {(x, y)|P (x, y)}(cf. §§1–3, Problem 7). If the pair (x, y) is an element of such a set R , we write (x, y) ∈ R
(1.2.1)
treating (x, y) as one thing. Note that this does not imply that x and y taken separately are members of R (in which case we would write x, y ∈ R). We call x, y the terms of (x, y). In mathematics, it is customary to call any set of ordered pairs a relation. For example, all sets listed in Problem 7 of §§1–3 are relations. Since relations are sets, equality R = S for relations means that they consist of the same elements (ordered pairs), i.e., that (x, y) ∈ R ⟺
(x, y) ∈ S
(1.2.2)
If (x, y) ∈ R, we call y an R -relative of x; we also say that y is R -related to x or that the relation R holds between x and y (in this order). Instead of (x, y) ∈ R, we also write xRy, and often replace “R ” by special symbols like between numbers are inverse to each other; so also are the relations (We may treat “⊆” as the name of the set of all pairs (X, Y ) such that X ⊆ Y in a given space.)
⊆
and
⊇
between sets.
If R contains the pairs (x, x'), (y, y'), (z, z'), . . ., we shall write x R =(
′
x
To obtain R
−1
y y
z
′
z
′
… );
e. g. ,
1
4
1
3
2
2
1
1
R =(
).
(1.2.5)
, we simply interchange the upper and lower rows in Equation 1.2.1.
Definition 1 The set of all left terms x of pairs (x, y) ∈ R is called the domain of R , denoted D . The set of all right terms of these pairs is called the range of R , denoted D . Clearly, x ∈ D iff xRy for some y . In symbols, R
′
R
R
′
x ∈ DR
⟺
(∃y) xRy;
similarly,
y ∈ D
R
⟺
(∃x) xRy.
(1.2.6)
In Equation 1.2.1, D is the upper row, and D is the lower row. Clearly, R
′
R
′
D
−1
R
=D
R
′
and D
−1
R
= DR .
(1.2.7)
For example, if R =(
1
4
1
2
2
1
),
(1.2.8)
then ′
DR = D
′ −1
R
= {1, 4}
and D
R
1.2.1
= DR−1 = {1, 2}.
(1.2.9)
https://math.libretexts.org/@go/page/19022
Definition 2 The image of a set A under a relation R (briefly, the R − image of A ) is the set of all R -relatives of elements of A , denoted R[A] . The inverse image of A under R is the image of A under the inverse relation, i.e., R [A] . If A consists of a single element, A = x , then R[A] and R [A] are also written R[x] and R [x], respectively, instead of R[x] and R [x]. −1
−1
−1
−1
Example 1.2.1 Let R =(
1
1
1
2
2
3
3
3
3
7
1
3
4
5
3
4
1
3
5
1
),
A = {1, 2},
B = {2, 4}.
(1.2.10)
Then R[1] = {1, 3, 4};
R[2] = {3, 5}; −1
R[5] = ∅; −1
R
−1
R
R
R[3] = {1, 3, 4, 5}; −1
[1] = {1, 3, 7};
R
[2] = ∅;
[4] = {1, 3};
R[A] = {1, 3, 4, 5};
(1.2.11)
−1
[3] = {1, 2, 3};
R
[A] = {1, 3, 7};
R[B] = {3, 5}.
By definition, R[x] is the set of all R -relatives of x. Thus y ∈ R[x]
iff (x, y) ∈ R;
i.e.,
xRy.
(1.2.12)
More generally, y ∈ R[A] means that (x, y) ∈ R for some x ∈ A . In symbols, y ∈ R[A]
⟺
(∃x ∈ A)(x, y) ∈ R.
(1.2.13)
Note that R[A] is always defined.
Mappings We shall now consider an especially important kind of relation.
Definition 3 A relation R is called a mapping (map), or a function, or a transformation, iff every element x ∈ D has a unique R -relative, so that R[x] consists of a single element. This unique element is denoted by R(x) and is called the function value at x (under R ). Thus R(x) is the only member of R[x]. R
If, in addition, different elements of D have different images, R is called a one-to-one (or one-one) map. In this case, R
x ≠ y (x, y ∈ DR ) implies R(x) ≠ R(y);
(1.2.14)
R(x) = R(y) implies x = y.
(1.2.15)
equivalently,
In other words, no two pairs belonging to R have the same left, or the same right, terms. This shows that R is one to one iff too, is a map. Mappings are often denoted by the letters f , g, h, F , ψ, etc. A mapping f is said to be “from A to B ” iff D
f
=A
and D
f : A → B
If, in particular, D
f
=A
and D
′ f
=B
′ f
⊆B
−1
R
,
; we then write
(" f maps A into B ")
(1.2.16)
, we call f a map of A onto B , and we write f : A ⟶ B
(" f maps A onto B ")
(1.2.17)
onto
If f is both onto and one to one, we write f : A ⟷ B
1.2.2
(1.2.18)
https://math.libretexts.org/@go/page/19022
(f
: A ⟷ B
means that f is one to one).
All pairs belonging to a mapping f have the form (x, f (x)) where f (x) is the function value at x, i.e., the unique f -relative of x, x ∈ D . Therefore, in order to define some function f , it suffices to specify its domain D and the function value f (x) for each x ∈ D . We shall often use such definitions. It is customary to say that f is defined on A (or “f is a function on A ”) iff A = D . f
f
f
f
Example 1.2.2 (a) The relation R = {(x, y)|x is the wife of y}
is a one-to-one map of the set of all wives onto the set of all husbands. R (= D ) onto the set of all wives (= D ) .
−1
(1.2.19)
is here a one-to-one map of the set of all husbands
′
R
R
(b) The relation f = {(x, y)|y is the father of x}
(1.2.20)
is a map of the set of all people onto the set of their fathers. It is not one to one since several persons may have the same father (f -relative), and so x ≠ x does not imply f (x) ≠ f (x ) . ′
′
(c) Let 1
2
3
4
2
2
3
8
g =(
Then g is a map of D
g
= {1, 2, 3, 4}
onto D
′ g
= {2, 3, 8},
)
(1.2.21)
with
g(1) = 2, g(2) = 2, g(3) = 3, g(4) = 8
(1.2.22)
(As noted above, these formulas may serve to define g. ) It is not one to one since g(1) = g(2), so g
−1
is not a map.
(d) Consider f : N → N , with f (x) = 2x for each x ∈ N
(1.2.23)
By what was said above, f is well defined. It is one to one since x ≠ y implies 2x ≠ 2y. Here D = N ( the naturals ), but D consists of even naturals only. Thus f is not onto N (it is onto a smaller set, the even naturals); f maps the even naturals onto all of N . f
−1
′
f
The domain and range of a relation may be quite arbitrary sets. In particular, we can consider functions f in which each element of the domain D is itself an ordered pair (x, y) or n -tuple (x , x , … , x ) . Such mappings are called functions of two (respectively, n) variables. To any n -tuple (x , … , x ) that belongs to D , the function f assigns a unique function value, denoted by f (x , … , x ) . It is convenient to regard x , x , … , x as certain variables; then the function value, too, becomes a variable depending on the x , … , x . Often D consists of all ordered n -tuples of elements taken from a set A , i.e., D = A (crossproduct of n sets, each equal to A). The range may be an arbitrary set B; so f : A → B . Similarly, f : A × B → C is a function of two variables, with D = A × B, D ⊆ C . f
1
1
1
n
n
2
n
f
1
2
n
n
1
n
f
f
n
f
′
f
Functions of two variables are also called (binary) operations. For example, addition of natural numbers may be treated as a map f : N × N → N , with f (x, y) = x + y.
Definition 4 A relation R is said to be (i) reflexive iff we have xRx for each x ∈ D ; (ii) symmetric iff xRy always implies yRx; (iii) transitive iff xRy combined with yRz always implies xRz. R
1.2.3
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is called an equivalence relation on a set A iff A = D and R has all the three properties (i), (ii), and (iii). For example, such is the equality relation on A (also called the identity map on A) denoted R
R
IA = {(x, y)|x ∈ A, x = y}
(1.2.24)
Equivalence relations are often denoted by special symbols resembling equality, such as ≡, ≈, ∼, etc. The formula xRy, where R is such a symbol, is read " x is equivalent (orR-equivalent) to y, "
(1.2.25)
and R[x] = {y|xRy} (i.e., the R -image of x) is called the R -equivalence class (briefly R -class) of x in A; it consists of all elements that are R -equivalent to x and hence to each other (for xRy and xRz imply first yRx, by symmetry, and hence yRz, by transitivity). Each such element is called a representative of the given R -class, or its generator. We often write [x] for R[x].
Example 1.2.3 (a') The inequality relation < between real numbers is transitive since x < y and y < z implies x < z
(1.2.26)
it is neither reflexive nor symmetric. (Why?) (b') The inclusion relation but it is not symmetric.
⊆
between sets is reflexive (for
A ⊆A
) and transitive
(
for
A ⊆B
and
B ⊆C
implies
(c') The membership relation ∈ between an element and a set is neither reflexive nor symmetric nor transitive A ∈ M does not imply x ∈ M) .
A ⊆ C ),
(x ∈ A
and
(d') Let R be the parallelism relation between lines in a plane, i.e., the set of all pairs (X, Y ), where X and Y are parallel lines. Writing ∥ for R, we have X∥X, X∥Y implies Y ∥X, and (X∥Y and Y \| Z) implies X∥Z, so R is an equivalence relation. An R -class here consists of all lines parallel to a given line in the plane. (e') Congruence of triangles is an equivalence relation. (Why?)
Theorem 1.2.1 If R (also written ≡) is an equivalence relation on A, then all R-classes are disjoint from each other, and A is their union. Proof Take two R-classes, [p] ≠ [q] . Seeking a contradiction, suppose they are not disjoint, so (∃x)
x ∈ [p] and x ∈ [q];
(1.2.27)
i.e., p ≡ x ≡ q and hence p ≡ q. But then, by symmetry and transitivity, y ∈ [p] ⇔ y ≡ p ⇔ y ≡ q ⇔ y ∈ [q];
(1.2.28)
i.e., [p] and [q] consist of the same elements y, contrary to assumption [q] ≠ [q] . Thus, indeed, any two (distinct) are disjoint.
R
-classes
Also, by reflexivity, (∀x ∈ A)
x ≡x
(1.2.29)
i.e., x ∈ [x]. Thus each x ∈ A is in some R-class (namely, in [x]); of A is in the union of such classes, A ⊆ ⋃ R[x]
(1.2.30)
x
Conversely, (∀x)
R[x] ⊆ A
(1.2.31)
since
1.2.4
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y ∈ R[x] ⇒ xRy ⇒ yRx ⇒ (y, x) ∈ R ⇒ y ∈ DR = A
(1.2.32)
by definition. Thus A contains all R[x], hence their union, and so A = ⋃ R[x]. □
(1.2.33)
x
This page titled 1.2: Relations. Mappings is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
1.2.5
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1.2.E: Problems on Relations and Mappings (Exercises) Exercise 1.2.E. 1 For the relations specified in Problem 7 of §§1-3, find D
R,
(a) A = {
1 2
′
D
R
};
,
and R
−1
. Also, find R[A] and R
−1
[A]
if
(b) A = {1}
(c) A = {0};
(d) A = ∅ ;
(e) A = {0, 3, −15};
(f) A = {3, 4, 7, 0, −1, 6}
(1.2.E.1)
(g) A = {x| − 20 < x < 5}
Exercise 1.2.E. 2 Prove that if A ⊆ B , then R[A] ⊆ R[B]. Disprove the converse by a counterexample.
Exercise 1.2.E. 3 Prove that (i) R[A ∪ B] = R[A] ∪ R[B] ; (ii) R[A ∩ B] ⊆ R[A] ∩ R[B] ; (iii) R[A − B] ⊇ R[A] − R[B] . Disprove reverse inclusions in (ii) and (iii) by examples. Do (i) and (ii) with { A |i ∈ I }.
A, B
replaced by an arbitrary set family
i
Exercise 1.2.E. 4 Under which conditions are the following statements true? (i) R[x] = ∅; (iii) R[A] = ∅;
−1
(ii) R
−1
(iv) R
[x] = ∅; (1.2.E.2) [A] = ∅;
Exercise 1.2.E. 5 Let f : N → N (N = { naturals }). For each of the following functions, specify f [N ], i.e., D one to one and onto N , given that for all x ∈ N ,
′ f
3
(i) f (x) = x ;
(ii) f (x) = 1;
,
and determine whether
f
is
(iii) f (x) = |x| + 3 (1.2.E.3)
2
(iv) f (x) = x ;
(v)f (x) = 4x + 5
Do all this also if N denotes (a) the set of all integers; (b) the set of all reals.
Exercise 1.2.E. 6 Prove that for any mapping f and any sets A, B, A (a) f [A ∪ B] = f [A] ∪ f [B] ; (b) f [A ∩ B] = f [A] ∩ f [B] ; (c) f [A − B] = f [A] − f [B] ; (d) f [⋃ A ] = ⋃ f [A ] ; (e) f [⋂ A ] = ⋂ f [A ] .
i (i
−1
−1
−1
−1
−1
−1
−1
−1
−1
∈ I)
,
−1
−1
i
i
i
i
i
i
−1
i
−1
i
1.2.E.1
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Compare with Problem 3. [Hint: First verify that x ∈ f
−1
[A]
iff x ∈ D and f (x) ∈ A. ] f
Exercise 1.2.E. 7 Let f be a map. Prove that (a) f [ f [A]] ⊆ A ; (b) f [ f [A]] = A if A ⊆ D ; (c) if A ⊆ D and f is one to one, A = f Is f [A] ∩ B ⊆ f [A ∩ f [B]]? −1
−1
′
f
f
−1
[f [A]]
/
−1
Exercise 1.2.E. 8 Is R an equivalence relation on the set J of all integers, and, if so, what are the R -classes, if (a) R = {(x, y)|x − y is divisible by a fixed n} ; (b) R = {(x, y)|x − y is odd } ; (c) R = {(x, y)|x − y is a prime } . (x, y, n denote integers.)
Exercise 1.2.E. 9 Is any relation in Problem 7 of §§1-3 reflexive? Symmetric? Transitive? 10. Show by examples that R may be (a) reflexive and symmetric, without being transitive; (b) reflexive and transitive without being symmetric. Does symmetry plus transitivity imply reflexivity? Give a proof or counterexample. 1.2.E: Problems on Relations and Mappings (Exercises) is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
1.2.E.2
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1.3: Sequences By an infinite sequence (briefly sequence) we mean a mapping (call it u) whose domain is N (all natural numbers 1, 2, 3, …); D may also contain 0.
u
A finite sequence is a map u in which D consists of all positive (or non-negative) integers less than a fixed integer p. The range D of any sequence u may be an arbitrary set B; we then call u a sequence of elements of B, or in B. For example, u
′ u
1
2
3
4
…
n
…
2
4
6
8
…
2n
…
u =(
)
(1.3.1)
is a sequence with Du = N = {1, 2, 3, …}
(1.3.2)
and with function values u(1) = 2, u(2) = 4, u(n) = 2n,
n = 1, 2, 3, …
(1.3.3)
Instead of u(n) we usually write u ("index notation"), and call u the n term of the sequence. If n is treated as a variable, u is called the general term of the sequence, and {u } is used to denote the entire (infinite) sequence, as well as its range D (whichever is meant, will be clear from the context). The formula {u } ⊆ B means that D ⊆ B, i.e., that u is a sequence in B . To determine a sequence, it suffices to define its general term u by some formula or rule. In (1) above, u = 2n . th
n
n
n
′ u
n
′ u
n
n
Often we omit the mention of D
u
=N
n
(since it is known) and give only the range
′ Du .
Thus instead of (1), we briefly write
2, 4, 6, … , 2n, …
(1.3.4)
u1 , u2 , … , un , …
(1.3.5)
or, more generally,
Yet it should be remembered that u is a set of pairs (a map). If all u are distinct (different from each other), u is a one-to-one map. However, this need not be the case. It may even occur that all u are equal (then u is said to be constant); e.g., u = 1 yields the sequence 1, 1, 1, … , 1, … ,i.e. n
n
n
u =(
1
2
3
…
n
…
1
1
1
…
1
…
)
(1.3.6)
Note that hereu is an infinite sequence (since D = N ), even though its range D has only one element, D = {1}. (In sets, repeated terms count as one element; but the sequence u consists of infinitely many distinct pairs (n, 1). ) If all u are real numbers, we call u a real sequence. For such sequences, we have the following definitions. ′ u
u
′ u
n
Definition 1 A real sequence {u
n}
is said to be monotone (or monotonic) iff it is either nondecreasing, i.e. (∀n)
un ≤ un+1
(1.3.7)
(∀n)
un ≥ un+1
(1.3.8)
or nonincreasing, i.e.,
Notation: {u } ↑ and {u } ↓, respectively. If instead we have the strict inequalities u call {u } strictly monotone (increasing or decreasing). n
n
n
< un+1
(respectively, u
n
> un+1 ),
we
n
A similar definition applies to sequences of sets.
1.3.1
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Definition 2 A sequence of sets A
1,
A2 , … , An , …
is said to be monotone iff it is either expanding, i.e., (∀n)
An ⊆ An+1
(1.3.9)
(∀n)
An ⊇ An+1
(1.3.10)
or contracting, i.e.,
Notation: {A } ↑ and {A } ↓, respectively. For example, any sequence of concentric solid spheres (treated as sets of points), with increasing radii, is expanding; if the radii decrease, we obtain a contracting sequence. n
n
Definition 3 Let {u
n}
be any sequence, and let n1 < n2 < ⋯ < nk < ⋯
be a strictly increasing sequence of natural numbers. Select from Then the sequence {u } so selected (with k th term equal to u subscripts n , k = 1, 2, 3, …. nk
(1.3.11)
those terms whose subscripts are n , n , … , n , … is called the subsequence of {u } , determined by the
{ un }
nk
),
1
2
k
n
k
Thus (roughly) a subsequence is any sequence obtained from {u } by dropping some terms, without changing the order of the remaining terms (this is ensured by the inequalities n < n < ⋯ < n < ⋯ where the n are the subscripts of the remaining terms). For example, let us select from (1) the subsequence of terms whose subscripts are primes (including 1). Then the subsequence is n
1
2
k
k
2, 4, 6, 10, 14, 22, …
(1.3.12)
u1 , u2 , u3 , u5 , u7 , u11 , …
(1.3.13)
i.e.,
All these definitions apply to finite sequences accordingly. Observe that every sequence arises by "numbering" the elements of its range (the terms): u is the first term, u is the second term, and so on. By so numbering, we put the terms in a certain order, determined by their subscripts 1, 2, 3, … (like the numbering of buildings in a street, of books in a library, etc. ). The question now arises: Given a set A, is it always possible to "number" its elements by integers? As we shall see in $4, this is not always the case. This leads us to the following definition. 1
2
Definition 4 A set A is said to be countable iff sequence).
A
is contained in the range of some sequence (briefly, the elements of
A
can be put in a
If, in particular, this sequence can be chosen finite, we call A a finite set. (The empty set is finite.) Sets that are not finite are said to be infinite. Sets that are not countable are said to be uncountable. Note that all finite sets are countable. The simplest example of an infinite countable set is N
.
= {1, 2, 3, …}
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1.3.2
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1.4: Some Theorems on Countable Sets We now derive some corollaries of Definition 4 in § 3.
COROLLARY 1.4.1 If a set A is countable or finite, so is any subset B ⊆ A . For if A ⊂ D for a sequence u, then certainly B ⊆ A ⊆ D ′ u
′ u
COROLLARY 1.4.2 If A is uncountable (or just infinite), so is any superset B ⊃ A . For, if B were countable or finite, so would be A ⊆ B, by Corollary 1
Theorem 1.4.1 If A and B are countable, so is their cross product A × B Proof If A or B is ∅, then A × B = ∅, and there is nothing to prove. Thus let A and B be nonvoid and countable. We may assume that they fill two infinite sequences, (repeat terms if necessary). Then, by definition, A × B is the set of all ordered pairs of the form (an , bm ) ,
Call n + m the rank of the pair (a
n,
bm ) .
n, m ∈ N
n,
bm )
(1.4.1)
For each r ∈ N , there are r − 1 pairs of rank r :
(a1 , br−1 ) , (a2 , br−2 ) , … , (ar−1 , b1 )
We now put all pairs (a
A = { an } , B = { bn }
(1.4.2)
in one sequence as follows. We start with (a1 , b1 )
(1.4.3)
(a1 , b2 ) , (a2 , b1 )
(1.4.4)
as the first term; then take the two pairs of rank three,
then the three pairs of rank four, and so on. At the (r − 1) st step, we take all pairs of rank r, in the order indicated in (1). Repeating this process for all ranks ad infinitum, we obtain the sequence of pairs (a1 , b1 ) , (a1 , b2 ) , (a2 , b1 ) , (a1 , b3 ) , (a2 , b2 ) , (a3 , b1 ) , …
in which u
1
= (a1 , b1 ) , u2 = (a1 , b2 ) ,
(1.4.5)
etc.
By construction, this sequence contains all pairs of all ranks r, hence all pairs that form the set A × B (for every such pair has some rank r and so it must eventually occur in the sequence). Thus A × B can be put in a sequence. □
COROLLARY 1.4.3 The set R of all rational numbers is countable. Proof Consider first the set Q of all positive rationals, i.e., n fractions
, with n, m ∈ N
(1.4.6)
m
1.4.1
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We may formally identify them with ordered pairs Theorem 1, we obtain the sequence 1
1 ,
1
2 ,
2
(n, m),
1 ,
1
2 ,
3
i.e., with 3
1
, 2
, 1
N ×N
2 ,
3 ,
4
n+m
the rank of
(n, m).
As in
4 ,
3
We call
2
,…
(1.4.7)
1
By dropping reducible fractions and inserting also 0 and the negative rationals, we put R into the sequence 1 0, 1, −1,
1 ,−
2
1 , 2, −2,
2
1 ,−
3
, 3, −3, … , as required. □
(1.4.8)
3
Theorem 1.4.2 The union of any sequence {A
n}
of countable sets is countable.
Proof As each A is countable, we may put n
An = { an1 , an2 , … , anm , …}
(1.4.9)
(The double subscripts are to distinguish the sequences representing different sets A . ) As before, we may assume that all sequences are infinite. Now, U A obviously consists of the elements of all A combined, i.e., all a (n, m ∈ N ). We call n + m the rank of a and proceed as in Theorem 1, thus obtaining n
n
n
n
nm
nm
⋃ An = { a11 , a12 , a21 , a13 , a22 , a31 , …}
(1.4.10)
n
Thus ∪
n An
can be put in a sequence. □
Note 1: Theorem 2 is briefly expressed as "Any countable union of countable sets is a countable set." (The term "countable union" means "union of a countable family of sets", i.e., a family of sets whose elements can be put in a sequence {A } . ) In particular, if A and B are countable, so are A ∪ B, A ∩ B, and A − B (by Corollary 1) . n
Note 2: From the proof it also follows that the range of any double sequence{a u whose domain D is N × N ; say, u : N × N → B. If n, m ∈ N , we write u
nm }
u
nm
is countable. (A double sequence is a function for u(n, m) here u = a . ) nm
nm
To prove the existence of uncountable sets, we shall now show that the interval [0, 1) = {x|0 ≤ x < 1}
(1.4.11)
of the real axis is uncountable. We assume as known the fact that each real number x ∈ [0, 1) has a unique infinite decimal expansion 0. x1 , x2 , … , xn , …
where the x are the decimal digits (possibly zeros), and the sequence {x
n}
n
(1.4.12)
does not terminate in nines (this ensures uniqueness)."
Theorem 1.4.3 The interval[0, 1) of the real axis is uncountable. Proof We must show that no sequence can comprise all of fraction; say,
[0, 1).
Indeed, given any
un = 0. an1 , an2 , … , anm , …
{ un } ,
write each term
un
as an infinite
(1.4.13)
Next, construct a new decimal fraction
1.4.2
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z = 0. x1 , x2 , … , xn , …
(1.4.14)
choosing its digits x as follows. n
If a (i.e., the n th digit of u ) is 0, put x = 1; if, however, a ≠ 0, put x = 0. Thus, in all cases, x ≠ a , i.e., z differs from each u in at least one decimal digit (namely, the n th digit). It follows that z is different from all u and hence is not in {u } , even though z ∈ [0, 1). nn
n
n
nn
n
n
n
nn
n
n
Thus, no matter what the choice of {u contains all of [0, 1). □
n}
was, we found some z ∈ [0, 1) not in the range of that sequence. Hence no {u
n}
Note 3: By Corollary 2, any superset of [0, 1), e.g., the entire real axis, is uncountable. Note 4: Observe that the numbers a used in the proof of Theorem 3 form the diagonal of the infinitely extending square composed of all a . Therefore, the method used above is called the diagonal process (due to G. Cantor). nn
nm
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1.4.3
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1.4.E: Problems on Countable and Uncountable Sets (Exercises) Exercise 1.4.E. 1 Prove that if A is countable but B is not, then B − A is uncountable. [Hint: If B − A were countable, so would be (B − A) ∪ A ⊇ B.
(Why?)
(1.4.E.1)
Use Corollary 1. ]
Exercise 1.4.E. 2 Let f be a mapping, and A ⊆ D . Prove that (i) if A is countable, so is f [A] ; (ii) if f is one to one and A is uncountable, so is f [A]. f
[ Hints: ( i) If A = { un } , then f [A] = {f (u1 ) , f (u2 ) , … , f (un ) , …}
(ii) If f [A] were countable, so would be f
−1
[f [A]],
(1.4.E.2)
by (i). Verify that f
−1
[f [A]] = A
(1.4.E.3)
here; cf. Problem 7 in §§4-7.]
Exercise 1.4.E. 3 Let a, b be real numbers (a < b). Define a map f on [0, 1) by f (x) = a + x(b − a).
Show that f is one to one and onto the interval [a, b) = {x|a ≤ x < b} . From Problem Hence, by Problem 1, so is(a, b) = {x|a < x < b} .
(1.4.E.4)
2,
deduce that
[a, b)
is uncountable.
Exercise 1.4.E. 4 Show that between any real numbers a, b(a < b) there are uncountably many irrationals, i.e., numbers that are not rational. [Hint: By Corollary 3 and Problems 1 and 3, the set (a, b) − R is uncountable. Explain in detail.
Exercise 1.4.E. 5 Show that every infinite set A contains a countably infinite set, i.e., an infinite sequence of distinct terms. [Hint: Fix any a ∈ A; A cannot consist of a alone, so there is another element 1
1
a2 ∈ A − { a1 } .
Again, A ≠ {a , a } , so there is an a { a } . Why are all a distinct? ] 1
n
2
3
∈ A − { a1 , a2 } .
(Why?)
(1.4.E.5)
(Why?) Continue thusly ad infinitum to obtain the required sequence
n
1.4.E.1
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Exercise 1.4.E. 6 From Problem 5, prove that if A is infinite, there is a map f : A → A that is one to one but not onto A. [Hint: With a as in Problem 5, define f (a ) = a . If, however, x is none of the a , put f (x) = x. Observe that f (x) = a is never true, so f is not onto A. Show, however, that f is one to one. n
n
n+1
n
1
Exercise 1.4.E. 7 Conversely (cf. Problem 6), prove that if there is a map f : A → A that is one to one but not onto infinite sequence {a } of distinct terms. [Hint: As f is not onto A, there is a ∈ A such that a ∉ f [A]. (Why?) Fix a and define
A,
then
A
contains an
n
1
1
1
a2 = f (a1 ) , a3 = f (a2 ) , … , an+1 = f (an ) , … ad infinitum.
To prove distinctness, show that each a is distinct from all a ∈ f [A](m > 1). Then proceed inductively.] n
am
with
m > n.
For
a1 ,
this is true since
(1.4.E.6)
a1 ∉ f [A],
whereas
m
1.4.E: Problems on Countable and Uncountable Sets (Exercises) is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
1.4.E.2
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CHAPTER OVERVIEW 2: Real Numbers and Fields 2.1: Axioms and Basic Definitions 2.2: Natural Numbers. Induction 2.2.E: Problems on Natural Numbers and Induction (Exercises) 2.3: Integers and Rationals 2.4: Upper and Lower Bounds. Completeness 2.4.E: Problems on Upper and Lower Bounds (Exercises) 2.5: Some Consequences of the Completeness Axiom 2.6: Powers with Arbitrary Real Exponents. Irrationals 2.6.E: Problems on Roots, Powers, and Irrationals (Exercises) 2.7: The Infinities. Upper and Lower Limits of Sequences 2.7.E: Problems on Upper and Lower Limits of Sequences in \(E^{*}\) (Exercises)
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1
2.1: Axioms and Basic Definitions Real numbers can be constructed step by step: first the integers, then the rationals, and finally the irrationals. Here, however, we shall assume the set of all real numbers, denoted E , as already given, without attempting to reduce this notion to simpler concepts. We shall also accept without definition (as primitive concepts) the notions of the sum (a + b) and the product, (a ⋅ b) or (ab), of two real numbers, as well as the inequality relation < (read "less than" ). Note that x ∈ E means "x is in E , i.e., "x is a real number." 1
1
1,
It is an important fact that all arithmetic properties of reals can be deduced from several simple axioms, listed (and named) below.
Axioms of Addition and Multiplication Definition 1. (closure laws) The sum x + y, and the product xy, any real numbers are real numbers themselves. In symbols, (∀x, y ∈ E
1
)
(x + y) ∈ E
1
and (xy) ∈ E
1
(2.1.1)
2. (commutative laws) (∀x, y ∈ E
1
)
x + y = y + x and xy = yx
(2.1.2)
3. (associative laws) (∀x, y, z ∈ E
1
)
(x + y) + z = x + (y + z) and (xy)z = x(yz)
(2.1.3)
4. (existence of neutral elements) (a) There is a (unique) real number, called zero (0), such that, for all real x, x + 0 = x . (b) There is a (unique) real number, called one (1), such that 1 1 ≠ 0 and, for all real x, x ⋅ 1 = x. In symbols, (a) (∃!0 ∈ E
1
) (∀x ∈ E
1
)
x + 0 = x;
(2.1.4)
x ⋅ 1 = x, 1 ≠ 0.
(2.1.5)
(b) (∃1 ∈ E
1
) (∀x ∈ E
1
)
(The real numbers 0 and 1 are called the neutral elements of addition and multiplication, respectively.) 5. (existence of inverse elements) (a) For every real x, there is a (unique) real, denoted −x, such that x + (−x) = 0 . (b) For every real x other than 0, there is a (unique) real, denoted x , such that x ⋅ x −1
−1
=1
.
In symbols, (a) (∀x ∈ E
1
) (∃! − x ∈ E
1
)
x + (−x) = 0;
(2.1.6)
(b) (∀x ∈ E
(The real numbers −x and x the reciprocal) of x. )
−1
1
−1
|x ≠ 0) (∃! x
∈ E
1
)
−1
xx
= 1.
(2.1.7)
are called, respectively, the additive inverse (or the symmetric) and the multiplicative inverse (or
6. (distributive law) (∀x, y, z ∈ E
1
)
(x + y)z = xz + yz
2.1.1
(2.1.8)
https://math.libretexts.org/@go/page/19028
Axioms of Order Definition 7. (trichotomy) For any real x and y, we have eitherx < y or y < x or x = y
(2.1.9)
but never two of these relations together. 8. (transitivity) (∀x, y, z ∈ E
1
)
x < y and y < z implies x < z
(2.1.10)
9. (monotonicity of addition and multiplication) For any x, y, z ∈ E , we have 1
(a) x < y implies x + z < y + z;
(2.1.11)
(b) \[x0 \text{ implies ] x z c} .
(3.3.E.2)
To fix ideas, let k = 1, i.e., cut the first edge. Then let ¯ ¯ ¯
p = (c, a2 , … , an ) and ¯ q¯ = (c, b2 , … , bn ) (see Figure 2),
and verify that P
¯ ¯ ¯¯ = (¯ a , q]
(3.3.E.3)
¯¯
and Q = (p, b]. Give a proof. ] ¯ ¯ ¯
Exercise 3.3.E. 6 In Problem 5, assume that A is closed, and make Q closed. (Prove it!)
Exercise 3.3.E. 7 In Problem 5 show that ( with k fixed ) the k th edge-lengths of P and Q equal i ≠ k the edge-length ℓ is the same in A, P , and Q, namely, ℓ = b − a . [Hint: If k = 1, define p and q as in Problem 5. ] i
¯ ¯ ¯
i
i
c − ak
and
bk − c,
respectively, while for
i
¯¯
3.3.E.1
https://math.libretexts.org/@go/page/22261
Exercise 3.3.E. 8 Prove that if an interval A is split into subintervals P and Q(P ∩ Q = ∅) , then vA = vP [Hint: Use Problem 7 to compute vA, vP , and vQ. Add up. ] Give an example. (Take A as in Problem 4 and split it by the plane x = 1. )
+ vQ.
4
Exercise 3.3.E. 9 *9. Prove the additivity of the volume of intervals, namely, if subintervals A , A , … , A inE , then
A
is subdivided, in any manner, into
m
mutually disjoint
n
1
2
m
m
vA = ∑ vAi .
(3.3.E.4)
i=1
(This is true also if some A contain common faces). [Proof outline: For m = 2, use Problem 8. Then by induction, suppose additivity holds for any number of intervals smaller than a certain m (m > 1). Now let i
m
A = ⋃ Ai
(Ai disjoint ) .
(3.3.E.5)
i=1
One of the A (say, A = [a, p]) must have some edge-length smaller than the corresponding edge-length of A (say, ℓ ) . Now cut all of A into P = [a, d] and Q = A − P ( Figure 4) by the plane x = c (c = p ) so that A ⊆ P while A ⊆ Q. For simplicity, assume that the plane cuts each A into two subintervals A and A . (One of them may be empty.) Then i
¯ ¯ ¯
1
¯ ¯ ¯
¯ ¯ ¯
1
¯ ¯ ¯
1
i
1
′
′′
i
i
m
1
2
m ′
′′
P = ⋃ A and Q = ⋃ A . i
Actually, however, P and our inductive assumption,
Q
are split into fewer than
m
(3.3.E.6)
i
i=1
i=1
(nonempty) intervals since
m
′′
A
1
′
=∅ =A
2
by construction. Thus, by
m ′
′′
vP = ∑ vA and vQ = ∑ vA , i
where vA
′′ 1
′
= 0 = vA , 2
and vA
i
′
′′
i
i
= vA + vA
(3.3.E.7)
i
i=1
i=1
by Problem 8. Complete the inductive proof by showing that m
vA = vP + vQ = ∑ vAi . ]
(3.3.E.8)
i=1
3.3.E: Problems on Intervals in Eⁿ (Exercises) is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
3.3.E.2
https://math.libretexts.org/@go/page/22261
3.4: Complex Numbers With all the operations defined in §§1-3, E (n > 1) is not yet a field because of the lack of a vector multiplication satisfying the field axioms. We shall now define such a multiplication, but only for E . Thus E will become a field, which we shall call the complex field, C . n
2
2
We make some changes in notation and terminology here. Points ofE , when regarded as elements of C , will be called complex numbers (each being an ordered pair of real numbers). We denote them by single letters (preferably z) without a bar or an arrow. For example, z = (x, y) . We preferably write (x, y) for (x , x ) . If z = (x, y), then x and y are called the real and imaginary parts of z, respectively, and z denotes the complex number (x, −y), called the conjugate of z (see Figure 5) . 2
1
1
2
¯ ¯ ¯
Complex numbers with vanishing imaginary part, (x, 0), are called real points of C . For brevity, we simply write x for (x, 0); for ¯ ¯ ¯
example, 2 = (2, 0). In particular, 1 = (1, 0) = θ is called the real unit in C . Points with vanishing real part, (0, y), are called (purely) imaginary numbers. In particular, θ = (0, 1) is such a number; we shall now denote it by i and call it the imaginary unit in C . Apart from these peculiarities, all our former definitions of §§1-3 remain valid in E = C . In particular, if z = (x, y) and z = (x , y ) , we have 1
¯ ¯ ¯
2
2
′
′
′
z±z
′
′
′
′
′
= (x, y) ± (x , y ) = (x ± x , y ± y ) ,
(3.4.1)
− −−−−−−−−−−−−−− − ′
2
′
ρ (z, z ) = √ (x − x )
′
+ (y − y )
2
, and
(3.4.2)
− −− −− − 2
|z| = √ x
+y
2
.
(3.4.3)
All theorems of §§1-3 are valid. We now define the new multiplication in C , which will make it a field.
Definition If z = (x, y) and z
′
′
′
= (x , y ) , then zz
′
′
′
′
′
= (x x − y y , x y + y x ) .
(3.4.4)
Theorem 3.4.1 E
2
=C
is a field, with zero element 0 = (0, 0) and unity 1 = (1, 0), under addition and multiplication as defined above.
Proof We only must show that multiplication obeys Axioms 1-6 of the field axioms. Note that for addition, all is proved in Theorem 1 of §§1-3. Axiom 1 (closure) is obvious from our definition, for if z and z are in C , so is zz . ′
′
To prove commutativity, take any complex numbers z = (x, y) and z
3.4.1
′
′
′
= (x , y )
(3.4.5)
https://math.libretexts.org/@go/page/19038
and verify that zz
′
′
= z z.
Indeed, by definition, zz
′
′
′
′
′
′
′
′
′
′
= (x x − y y , x y + y x ) and z z = (x x − y y, x y + y x) ;
(3.4.6)
but the two expressions coincide by the commutative laws for real numbers. Associativity and distributivity are proved in a similar manner. Next, we show that 1 = (1, 0) satisfies Axiom 4(b), i.e., that definition, and by axioms for E ,
for any complex number
1z = z
z = (x, y).
In fact, by
1
1z = (1, 0)(x, y) = (1x − 0y, 1y + 0x) = (x − 0, y + 0) = (x, y) = z.
It remains to verify Axiom 5(b), i.e., to show that each complex number z = (x, y) ≠ (0, 0) has an inverse z zz = 1. It turns out that the inverse is obtained by setting
(3.4.7) −1
such that
−1
z
−1
x
y
=(
,− 2
).
(3.4.8)
2
|z|
|z|
In fact, we then get 2
zz
−1
x =(
2
|z|
since x
2
+y
2
2
= |z| ,
y
2
xy
+
,− 2
|z|
2
yx +
2
x
2
|z|
+y
) =(
2
, 0) = (1, 0) = 1
(3.4.9)
2
|z|
|z|
by definition. This completes the proof. □
corollary 3.4.1 2
i
.
= −1; i. e. , (0, 1)(0, 1) = (−1, 0)
Proof By definition, (0, 1)(0, 1) = (0 ⋅ 0 − 1 ⋅ 1, 0 ⋅ 1 + 1 ⋅ 0) = (−1, 0) . Thus C has an element i whose square is −1, while E has no such element, by Corollary 2 in Chapter 2, 8{1 − 4. This is no contradiction since that corollary holds in ordered fields only. It only shows that C cannot be made an ordered field. 1
However, the "real points" in C form a subfield that can be ordered by setting ′
′
(x, 0) < (x , 0) iff x < x in E
1
.
(3.4.10)
Then this subfield behaves exactly like E . Therefore, it is customary not to distinguish between "real points in C and "real numbers," identifying (x, 0) with x. With this convention, E simply is a subset ( and a subfield ) of C . Henceforth, we shall simply say that "x is real" or "x ∈ E instead of "x = (x, 0) is a real point." We then obtain the following result. 1
′′
1
1′
Theorem 3.4.2 Every z ∈ C has a unique representation as z = x + yi,
(3.4.11)
z = x + yi iff z = (x, y).
(3.4.12)
where x and y are real and i = (0, 1). Specifically,
Proof By our conventions, x = (x, 0) and y = (y, 0), so x + yi = (x, 0) + (y, 0)(0, 1).
(3.4.13)
Computing the right-hand expression from definitions, we have for any x, y ∈ E that 1
3.4.2
https://math.libretexts.org/@go/page/19038
x + yi = (x, 0) + (y ⋅ 0 − 0 ⋅ 1, y ⋅ 1 + 0 ⋅ 1) = (x, 0) + (0, y) = (x, y).
Thus
(x, y) = x + yi
z = (x, y) = x + yi,
for any x, y ∈ E as required.
1
In particular, if
.
(x, y)
is the given number
z ∈ C
(3.4.14)
of the theorem, we obtain
To prove uniqueness, suppose that we also have ′
′
′
′
z = x + y i with x = (x , 0) and y
′
′
= (y , 0) .
Then, as shown above, z = (x , y ) . since also z = (x, y), we have (x, y) = (x , y and so x = x and y = y after all. □ ′
′
′
′
′
),
(3.4.15)
i.e., the two ordered pairs coincide,
′
Geometrically, instead of Cartesian coordinates (x, y), we may also use polar coordinates r, θ, where − −− −− − 2
r = √x
+y
2
= |z|
(3.4.16) →
and θ is the (counterclockwise) rotation angle from the x-axis to the directed line 0z; see Figure 6. Clearly, z is uniquely determined by r and θ but θ is not uniquely determined by z; indeed, the same point of E results if θ is replaced by θ + 2nπ(n = 1, 2, …) . (If z = 0, then θ is not defined at all.) The values r and θ are called, respectively, the modulus and argument of z = (x, y). By elementary trigonometry, x = r cos θ and y = r sin θ. Substituting in z = x + yi, we obtain the following corollary. 2
corollary 3.4.2 z = r(cos θ + i sin θ)(trigonometric or polar form of z).
(3.4.17)
This page titled 3.4: Complex Numbers is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
3.4.3
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3.4.E: Problems on Complex Numbers (Exercises) Exercise 3.4.E. 1 Complete the proof of Theorem 1 (associativity, distributivity, etc.).
Exercise 3.4.E. 1
′
Verify that the "real points" in C form an ordered field.
Exercise 3.4.E. 2 Prove that zz = |z|
2
¯ ¯ ¯
.
Deduce that z
−1
if z ≠ 0.
2
4
¯ ¯ =¯ z /|z|
Exercise 3.4.E. 3 Prove that ¯¯¯¯¯¯¯¯¯¯¯ ¯ ′
¯ ¯¯ ¯ ′
¯¯¯¯¯ ¯ ′
¯ ¯ =¯ z + z and zz
z+z
¯ ¯¯ ¯ ′
¯ ¯ =¯ z ⋅z
(3.4.E.1)
Hence show by induction that ¯ ¯¯ ¯ ¯ n
z
¯ ¯ ¯ n
= (z ) , n = 1, 2, … ,
and
¯¯¯¯¯¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ n k ∑ ak z k=1
n
k ¯ ¯ ¯ ¯ ¯ ¯ ak z
=∑
k=1
Exercise 3.4.E. 4 Define e
Describe e geometrically. Is ∣∣e θi
θi
θi
= cos θ + i sin θ.
(3.4.E.2)
∣ ∣ = 1?
Exercise 3.4.E. 5 Compute (a) ; (b) (1 + 2i)(3 − i); and (c) ,x ∈ E . Do it in two ways: (i) using definitions only and the notation (x, y) for x + yi; and ( ii) using all laws valid in a field. 1+2i 3−i
x+1+i
1
x+1−i
Exercise 3.4.E. 6 Solve the equation (2, −1)(x, y) = (3, 2) for x and y in E . 1
Exercise 3.4.E. 7 Let z = r(cos θ + i sin θ) z z
′
′′
′
′
′
= r (cos θ + i sin θ ) , and ′′
=r
′′
(cos θ
′′
+ i sin θ )
3.4.E.1
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as in Corollary 2. Prove that z = z
′
z
′′
if ′
′′
′
r = |z| = r r , i.e., | z z
′′
′
| = |z | |z
′′
′
′′
| , and θ = θ + θ .
(3.4.E.3)
− →
Discuss the following statement: To multiply z by z means to rotate 0z counterclockwise by the angle θ and to multiply it by the scalar r = |z | . Consider the cases z = i and z = −1 . [Hint: Remove brackets in ′
′′
′′
′′
′′
′
′′
′′
′
′
′
′′
r(cos θ + i sin θ) = r (cos θ + i sin θ ) ⋅ r
′′
(cos θ
′′
+ i sin θ )
(3.4.E.4)
and apply the laws of trigonometry.]
Exercise 3.4.E. 8 By induction, extend Problem 7 to products of z = r(cos θ + i sin θ), then z
n
complex numbers, and derive de Moivre's formula, namely, if
n
n
= r (cos(nθ) + i sin(nθ)).
(3.4.E.5)
Use it to find, for n = 1, 2, … n
1
n
(a)i ;
(b)(1 + i ) ;
(c)
n
.
(3.4.E.6)
(1 + i)
Exercise 3.4.E. 9 From Problem 8, prove that for every complex number z ≠ 0, there are exactly n complex numbers w such that w
n
= z;
(3.4.E.7)
they are called the n th roots of z [Hint: If ′
′
′
z = r(cos θ + i sin θ) and w = r (cos θ + i sin θ ) ,
the equation w
n
=z
(3.4.E.8)
yields, by Problem 8 ′
(r )
n
′
= r and nθ = θ,
(3.4.E.9)
and conversely. While this determines r uniquely, θ may be replaced by θ + 2kπ without affecting z. Thus ′
′
θ + 2kπ
θ =
,
k = 1, 2, …
(3.4.E.10)
n
Distinct points w result only from k = 0, 1, … , n − 1 (then they repeat cyclically). Thus n values of w are obtained.]
3.4.E.2
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Exercise 3.4.E. 10 Use Problem 9 to find in C (a) all cube roots of 1;
(b) all fourth roots of 1
(3.4.E.11)
Describe all n th roots of 1 geometrically. 3.4.E: Problems on Complex Numbers (Exercises) is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
3.4.E.3
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3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces I. We shall now follow the pattern of E to obtain the general notion of a vector space (just as we generalized E to define fields). n
1
Let V be a set of arbitrary elements (not necessarily n -tuples), called "vectors" or "points," with a certain operation (call it "addition," +) somehow defined in V . Let F be any field (e.g. , E or C ); its elements will be called scalars; its zero and unity will be denoted by 0 and 1, respectively. Suppose that yet another operation ("multiplication of scalars by vectors") has been defined that assigns to every scalar c ∈ F and every vector x ∈ V a certain vector, denoted cx or xc and called the c -multiple of x. Furthermore, suppose that this multiplication and addition in V satisfy the nine laws specified in Theorem 1 of §§1-3. That is, we have closure: 1
(∀x, y ∈ V )(∀c ∈ F )
x + y ∈ V and cx ∈ V
Vector addition is commutative and associative. There is a unique zero-vector,
→ 0 ,
(3.5.1)
such that
→ (∀x ∈ V )
x+ 0 =x
(3.5.2)
and each x ∈ V has a unique inverse, −x, such that → x + (−x) = 0 .
(3.5.3)
a(x + y) = ax + ay and (a + b)x = ax + bx.
(3.5.4)
1x = x
(3.5.5)
(ab)x = a(bx)
(3.5.6)
We have distributivity:
Finally, we have
and
(a, b ∈ F ; x, y ∈ V ).
In this case, V together with these two operations is called a vector space (or a linear space) over the field F ; F is called its scalar field, and elements of F are called the scalars of V .
Example 3.5.1 (a) E is a vector space over E (its scalar field). n
1
(a') R , the set of all rational points of E (i.e., points with rational coordinates is a vector space over R, the rationals in E . (Note that we could take R as a scalar field for all of E ; this would yield another vector space, E over R, not to be confused with E over E , i.e., the ordinary E . ) n
n
1
n
n
1
n
n
(b) Let F be any field, and let F be the set of all ordered n -tuples of elements of F , with sums and scalar multiples defined as in E (with F playing the role of E ). Then F is a vector space over F ( proof as in Theorem 1 of §§1-3). n
n
1
n
(c) Each field F is a vector space (over itself) under the addition and multiplication defined in F . Verify! (d) Let V be a vector space over a field F , and let W be the set of all possible mappings f : A → V
(3.5.7)
from some arbitrary set A ≠ ∅ into V . Define the sum f + g of two such maps by setting (f + g)(x) = f (x) + g(x) for all x ∈ A.
Similarly, given a ∈ F and f
∈ W,
(3.5.8)
define the map af by (af )(x) = af (x).
(3.5.9)
Vector spaces over E (respectively, C ) are called real (respectively, complex) linear spaces. Complex spaces can always be transformed into real ones by restricting their scalar field C to E (treated as a subfield of C ). 1
1
3.5.1
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II. An important example of a complex linear space is C
n
the set of all ordered n -tuples
,
x = (x1 , … , xn )
(3.5.10)
of complex numbers x (now treated as scalars), with sums and scalar multiples defined as in E . In order to avoid confusion with conjugates of complex numbers, we shall not use the bar notation x for a vector in this section, writing simply x for it. Dot products in C are defined by n
k
¯¯ ¯
n
n ¯ ¯ ¯
x ⋅ y = ∑ xk y k ,
(3.5.11)
k=1
where y is the conjugate of the complex number vectors with real components, ¯ ¯ ¯ k
yk
(see §8), and hence a scalar in
C.
Note that
¯ ¯ ¯ y k
= yk
if
yk ∈ E
1
. Thus, for
n
x ⋅ y = ∑ xk yk ,
(3.5.12)
k=1
as in E
n
.
The reader will easily verify (exactly as for E ) that, for x, y ∈ C and a, b ∈ C , we have the following properties: n
n
1. x ⋅ y ∈ C ; thus x ⋅ y is a scalar, not a vector. 2. x ⋅ x ∈ E
1
,
and x ⋅ x ≥ 0; moreover, x ⋅ x = 0 iff x =
→ 0 .
(Thus the dot product of a vector by itself is a real number ≥ 0. )
3. x ⋅ y = y ⋅ x (= conjugate of y ⋅ x). Commutativity fails in general. ¯ ¯¯¯¯¯¯¯ ¯
4. (ax) ⋅ (by) = (ab)(x ⋅ y). Hence (iv ) (ax) ⋅ y = a(x ⋅ y) = x ⋅ (ay) . ¯¯
′
5. (x + y) ⋅ z = x ⋅ z + y ⋅ z
¯ ¯ ¯
and (5 ) z ⋅ (x + y) = z ⋅ x + z ⋅ y . ′
Observe that (5') follows from (5) by (3). (Verify!) III. Sometimes (but not always) dot products can also be defined in real or complex linear spaces other than E or C , in such a manner as to satisfy the laws (1)-(5), hence also (5'), listed above, with C replaced by E if the space is real. If these laws hold, the space is called Euclidean. For example, E is a real Euclidean space and C is a complex one. n
n
1
n
n
In every such space, we define absolute values of vectors by − − − − |x| = √ x ⋅ x .
(This root exists in E by formula (ii).) In particular, this applies to obtain as before the following properties: 1
(a') |x| ≥ 0; and |x| = 0 iff x =
→ 0
E
n
(3.5.13)
and C
n
.
Then given any vectors
x, y
and a scalar
a,
we
.
(b') |ax| = |a||x|. (c') Triangle inequality: |x + y| ≤ |x| + |y| . (d') Cauchy-Schwarz inequality: |x ⋅ y| ≤ |x||y|, and |x ⋅ y| = |x||y| iff x∥y (i.e., x = ay or y = ax for some scalar a). We prove only (d') ;\) the rest is proved as in Theorem 4 of §§1-3. If
all is trivial, so let z = x ⋅ y = rc ≠ 0, where r = |x ⋅ y| and consider |tx + y | . By definition and (5),(3), and (4),
x ⋅ y = 0,
)t ∈ E
1
,
c
has modulus
1,
and let
y
′
= cy.
For any (variable
′
′
|tx + y |
2
′
′
= (tx + y ) ⋅ (tx + y ) ′
′
′
= tx ⋅ tx + y ⋅ tx + tx ⋅ y + y ⋅ y 2
′
′
′
′
′
= t (x ⋅ x) + t (y ⋅ x) + t (x ⋅ y ) + (y ⋅ y )
since t = t. Now, since c c = 1 , ¯
¯¯
x⋅y
′
¯ ¯ = x ⋅ (cy) = (¯ c x) ⋅ y = ¯ c rc = r = |x ⋅ y|.
(3.5.14)
Similarly,
3.5.2
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¯ ¯¯¯¯¯¯¯¯¯ ¯ ′
′
y ⋅x = x⋅y
2
′
¯ =¯ r = r = |x ⋅ y|, x ⋅ x = |x | , and y ⋅ y
′
2
= y ⋅ y = |y | .
(3.5.15)
Thus we obtain (∀t ∈ E
Here |x |
2
, 2|x ⋅ y|,
1
)
2
|tx + cy |
2
2
= t |x |
2
+ 2t|x ⋅ y| + |y | .
(3.5.16)
and |y| are fixed real numbers. We treat them as coefficients in t of the quadratic trinomial 2
2
2
f (t) = t |x |
2
+ 2t|x ⋅ y| + |y | .
(3.5.17)
Now if x and y are not parallel, then cy ≠ −tx, and so ′
|tx + cy| = |tx + y | ≠ 0
for any t ∈ E
1
.
(3.5.18)
Thus by (1), the quadratic trinomial has no real roots; hence its discriminant, 2
4|x ⋅ y |
2
− 4(|x||y| ) ,
(3.5.19)
is negative, so that |x ⋅ y| < |x||y|. If, however, x∥y, one easily obtains |x ⋅ y| = |x||y|, by (b ) . (Verify.) ′
Thus |x ⋅ y| = |x||y| or |x ⋅ y| < |x||y| according to whether x∥y or not. □ In any Euclidean space, we define distances by Thus ¯ ¯¯¯ ¯
ρ(x, y) = |x − y|.
line pq = {p + t(q − p)|t ∈ E
1
Planes, lines, and line segments are defined exactly as in
} (in real and complex spaces alike).
E
n
.
(3.5.20)
This page titled 3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
3.5.3
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3.5.E: Problems on Linear Spaces (Exercises) Exercise 3.5.E. 1 Prove that F in Example (b) is a vector space, i.e., that it satisfies all laws stated in Theorem 1 in §§1-3; similarly for W in Example (d). n
Exercise 3.5.E. 2 Verify that dot products in C obey the laws (i) − (v ) . Which of these laws would fail if these products were defined by n
′
n
n ¯ ¯ ¯
x ⋅ y = ∑ xk yk instead of x ⋅ y = ∑ xk y k ? k=1
(3.5.E.1)
k=1
How would this affect the properties of absolute values given in (a ) − (d )? ′
′
Exercise 3.5.E. 3 Complete the proof of formulas were restated as
′
′
(a ) − (d )
for Euclidean spaces. What change would result if property (ii) of dot products →
→
′′
" x ⋅ x ≥ 0 and 0 ⋅ 0 = 0 ?
(3.5.E.2)
Exercise 3.5.E. 4 Define orthogonality, parallelism and angles in a general Euclidean space following the pattern of §§1-3 (text and Problem 7 there). Show that u =
→ 0
iff u is orthogonal to all vectors of the space.
Exercise 3.5.E. 5 Define the basic unit vectors e in C exactly as in E , and prove Theorem 2 in §§1-3 for C ( replacing E by C ) . Also, do Problem 5(a) of §§1-3 for C . n
n
k
n
1
n
Exercise 3.5.E. 6 Define hyperplanes in C as in Definition 3 of §§4-6, and prove Theorem 1 stated there, for C . Do also Problems 4 − 6 there for C (replacing E by C ) and Problem 4 there for vector spaces in general (replacing E by the scalar field F ). n
n
n
1
1
Exercise 3.5.E. 7 Do Problem 3 of §§4-6 for general Euclidean spaces (real or complex). Note: Do not replace line and a line segment.
E
1
by C in the definition of a
Exercise 3.5.E. 8 A finite set of vectors B = {x
1,
… , xm }
in a linear space V over F is said to be independent iff m
(∀a1 , a2 , … , am ∈ F )
→
( ∑ ai xi = 0 ⟹ a1 = a2 = ⋯ = am = 0) .
(3.5.E.3)
i=1
Prove that if B is independent, then
3.5.E.1
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→
(i) 0 ∉ B ; (ii) each subset of B is independent (∅ counts as independent ); and (iii) if for some scalars a , b ∈ F , i
i
m
m
∑ ai xi = ∑ bi xi , i=1
then a
i
= bi , i = 1, 2, … , m
(3.5.E.4)
i=1
.
Exercise 3.5.E. 9 Let V be a vector space over F and let A ⊆ V . By the span of combinations" of vectors from A, i.e., all vectors of the form
A
in
V
, denoted
span(A),
is meant the set of all "linear
m
∑ ai xi ,
ai ∈ F , xi ∈ A, m ∈ N .
(3.5.E.5)
i=1
Show that span(A) is itself a vector space V ⊆ V (a subspace of V ) over the same field F , with the operations defined in V . (We say that A spans V . Show that in E and C , the basic unit vectors span the entire space. ′
′
n
n
3.5.E: Problems on Linear Spaces (Exercises) is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
3.5.E.2
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3.6: Normed Linear Spaces By a normed linear space (briefly normed space) is meant a real or complex vector space E in which every vector x is associated with a real number |x|, called its absolute value or norm, in such a manner that the properties (a ) − (c ) of §9 hold. That is, for any vectors x, y ∈ E and scalar a, we have ′
′
(i) |x| ≥ 0;
′
(i ) |x| = 0
iff x =
(ii) |ax| = |a||x|;
→ 0 ;
and
(iii) |x + y| ≤ |x| + |y| (triangle inequality).
Mathematically, the existence of absolute values in E amounts to that of a map (called a norm map) x → |x| on E, i.e., a map φ : E → E , with function values φ(x) written as |x|, satisfying the laws (i)-(iii) above. Often such a map can be chosen in many ways (not necessarily via dot products, which may not exist in E , thus giving rise to different norms on E. Sometimes we write ∥x∥ for |x| or use other similar symbols. 1
Note 1. From (iii), we also obtain |x − y| ≥ ||x| − |y|| exactly as in E
n
.
Example 3.6.1 (A) Each Euclidean space (§9) such as E or C n
n
,
is a normed space, with norm defined by − − − − |x| = √ x ⋅ x ,
as follows from formulas (a')-(c') in §9. In E and C n
n
,
(3.6.1)
one can also equivalently define − − − − − − − n
2
|x| = √∑ | xk | ,
(3.6.2)
k=1
where x = (x
1,
… , xn ) .
This is the so-called standard norm, usually presupposed in E
(B) One can also define other, "nonstandard," norms on E and C n
n
.
n
(C
n
).
For example, fix some real p ≥ 1 and put 1
n
|x |
p
p
p
= ( ∑ | xk | )
.
(3.6.3)
k=1
One can show that |x| so defined satisfies (i)−( iii) and thus is a norm (see Problems 5-7 below). p
(C) Let W be the set of all bounded maps f : A → E
(3.6.4)
from a set A ≠ ∅ into a normed space E, i.e., such that (∀t ∈ A)
|f (t)| ≤ c
(3.6.5)
for some real constant c > 0 (dependent on f but not on t). Define f + g and af as in Example (d) of §9 so that W becomes a vector space. Also, put ∥f ∥ = sup |f (t)|,
(3.6.6)
t∈A
i.e., the supremum of all |f (t)|, with t ∈ A. Due to boundedness, this supremum exists in E
1
,
so ∥f ∥ ∈ E
1
.
It is easy to show that ∥f ∥ is a norm on W . For example, we verify (iii) as follows. By definition, we have for f , g ∈ W and x ∈ A,
3.6.1
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|(f + g)(x)| = |f (x) + g(x)| ≤ |f (x)| + |g(x)| ≤ sup |f (t)| + sup |g(t)| t∈A
t∈A
= ∥f ∥ + ∥g∥.
(The first inequality is true because (iii) holds in the normed space E to which ∥f ∥ + ∥g∥ + ∥g∥ is an upper bound of all expressions |(f + g)(x)|, x ∈ A. Thus
f (x)
and
g(x)
∥f ∥ + ∥g∥ ≥ sup |(f + g)(x)| = ∥f + g∥.
belong.) By (1),
(3.6.7)
x∈A
Note 2. Formula (1) also shows that the map f + g is bounded and hence is a member of af ∈ W for any scalar a and f ∈ W . Thus we have the closure laws for W . The rest is easy.
W.
Quite similarly we see that
In every normed (in particular, in each Euclidean) space E, we define distances by ρ(x, y) = |x − y|
Such distances depend, of course, on the norm chosen for standard norm in E and C (Example (A)), we have n
E;
for all x, y ∈ E.
(3.6.8)
thus we call them norm-induced distances. In particular, using the
n
− − − − − − − − − − − n
ρ(x, y) = √∑ | xk − yk |
2
.
(3.6.9)
k=1
Using the norm of Example (B), we get 1
n
p
ρ(x, y) = ( ∑ | xk − yk |
p
)
(3.6.10)
k=1
instead. In the space W of Example (C) we have ρ(f , g) = ∥f − g∥ = sup |f (x) − g(x)|.
(3.6.11)
x∈A
Proceeding exactly as in the proof of Theorem 5 in §§1-3, we see that norm- induced distances obey the three laws stated there. (Verify!) Moreover, by definition, ρ(x + u, y + u) = |(x + u) − (y + u)| = |x − y| = ρ(x, y).
(3.6.12)
Thus we have ρ(x, y) = ρ(x + u, y + u) for norm-induced distances;
i.e., the distance ρ(x, y) does not change if both translation-invariant.
x
and
y
(3.6.13)
are "translated" by one and the same vector
u.
We call such distances
A more general theory of distances will be given in §§11ff. This page titled 3.6: Normed Linear Spaces is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
3.6.2
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3.6.E: Problems on Normed Linear Spaces (Exercises) Exercise 3.6.E. 1 Show that distances in normed spaces obey the laws stated in Theorem 5 of §§1-3.
Exercise 3.6.E. 2 Complete the proof of assertions made in Example (C) and Note 2.
Exercise 3.6.E. 3 Define |x| = x for about formula (2)? 1
x = (x1 , … , xn )
in
C
n
or
E
n
.
Is this a norm? Which (if any) of the laws (i) - (iii) does it obey? How
Exercise 3.6.E. 4 Do Problem 3 in §§4-6 for a general normed space E, with lines defined as in E (see also Problem 7 in §9). Also, show that contracting sequences of line segments in E are f -images of contracting sequences of intervals in E . Using this fact, deduce from Problem 11 in Chapter 2 §§8-9, an analogue for line segments in E , namely, if n
1
L [ an , bn ] ⊇ L [ an+1 , bn+1 ] ,
n = 1, 2, …
(3.6.E.1)
then ∞
⋂ L [ an , bn ] ≠ ∅.
(3.6.E.2)
n=1
Exercise 3.6.E. 5 Take for granted the lemma that 1/p
a
1/q
b
a ≤
b +
p
(3.6.E.3) q
if a, b, p, q ∈ E with a, b ≥ 0 and p, q > 0, and 1
1
1 +
p
(A proof will be suggested in Chapter 5, §6, Problem + = 1, then 1
1
p
q
= 1.
(3.6.E.4)
q
11. )
Use it to prove Hölder's inequality, namely, if
1
n
q
n
p
q
∑ | xk yk | ≤ ( ∑ | xk | ) k=1
and
1
p
n
p >1
( ∑ | yk | )
k=1
for any xk , yk ∈ C .
(3.6.E.5)
k=1
[Hint: Let 1
n
1
n
p
p
A = ( ∑ | xk | )
q
q
and B = ( ∑ | yk | )
k=1
.
(3.6.E.6)
k=1
3.6.E.1
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If A = 0 or B = 0, then all setting
xk
or all
yk
vanish, and the required inequality is trivial. Thus assume p
B ≠ 0.
Then,
| yk | and b =
p
and
q
| xk | a =
A ≠0
A
(3.6.E.7)
q
B
in the lemma, obtain p
| xk yk |
q
| xk | ≤
AB
| yk | +
p
pA
q
, k = 1, 2, … , n.
(3.6.E.8)
qB
Now add up these n inequalities, substitute the values of A and B, and simplify. ]
Exercise 3.6.E. 6 Prove the Minkowski inequality, 1
1
p
n
( ∑ | xk + yk |
p
p
)
≤ ( ∑ | xk | )
k=1
1
p
n
p
n p
+ ( ∑ | yk | )
k=1
(3.6.E.9)
k=1
for any real p ≥ 1 and x , y ∈ C . [Hint: If p = 1, this follows by the triangle inequality in C . If p > 1 , let k
k
n
A = ∑ | xk + yk |
p
≠ 0.
(3.6.E.10)
k=1
(If A = 0 , all is trivial.) Then verify (writing "∑" for "∑
n k=1
A = ∑ | xk + yk | | xk + yk |
p−1
" for simplicity)
≤ ∑ | xk | | xk + yk |
p−1
+ ∑ | yk | | xk + yk |
Now apply Hölder's inequality (Problem 5) to each of the last two sums, with 1/p = 1 − 1/q. Thus obtain 1
p
Then divide by A
1 q
= (∑ | xk + yk |
p
1
)
q
1
p
A ≤ (∑ | xk | )
(∑ | xk + yk |
p
1
q
)
q = p/(p − 1),
p
+ (∑ | yk | )
p−1
so that
(3.6.E.11)
(p − 1)q = p
and
1
p
(∑ | xk + yk |
p
q
)
.
(3.6.E.12)
and simplify. ]
Exercise 3.6.E. 7 Show that Example (B) indeed yields a norm for C and E . [Hint: For the triangle inequality, use Problem 6. The rest is easy. ] n
n
Exercise 3.6.E. 8 A sequence {x
m}
of vectors in a normed space E ( e.g. , in E (∃c ∈ E
i.e., iff sup
m
| xm |
1
n
or C
) (∀m)
n
)
is said to be bounded iff
| xm | < c,
(3.6.E.13)
is finite.
3.6.E.2
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Denote such sequences by single letters, x = {x
m}
, y = { ym } ,
etc. and define
x + y = { xm + ym } , and ax = {axm } for any scalar a.
(3.6.E.14)
|x| = sup | xm | .
(3.6.E.15)
Also let m
Show that, with these definitions, the set M of all bounded infinite sequences in E becomes a normed space (in which every such sequence is to be treated as a single vector, and the scalar field is the same as that of E ). 3.6.E: Problems on Normed Linear Spaces (Exercises) is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
3.6.E.3
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3.7: Metric Spaces I. In §§1-3, we defined distances ρ(x, y) for points x, y in E using the formula ¯¯ ¯
¯ ¯ ¯
¯¯ ¯
n
¯ ¯ ¯
− −−−−−−−−− − n
¯ ¯ ¯ ρ(¯¯ x ,¯ y ) = √∑ (xk − yk )
2
¯ ¯ ¯ = |¯¯ x −¯ y |.
(3.7.1)
k=1
This actually amounts to defining a certain function ρ of two variables x, y ∈ E Theorem 5 there. (We call them metric laws.) ¯¯ ¯
¯ ¯ ¯
n
.
We also showed that ρ obeys the three laws of
Now, as will be seen, such functions ρ can also be defined in other sets, using quite different defining formulas. In other words, given any set S ≠ ∅ of arbitrary elements, one can define in it, so to say, "fancy distances" ρ(x, y) satisfying the same three laws. It turns out that it is not the particular formula used to define ρ but rather the preservation of the three laws that is most important for general theoretical purposes. Thus we shall assume that a function ρ with the same three properties has been defined, in some way or other, for a set S ≠ ∅ , and propose to study the consequences of the three metric laws alone, without assuming anything else. (In particular, no operations other than ρ, or absolute values, or inequalities < need be defined in S. ) All results so obtained will, of course, apply to distances in E (since they obey the metric laws), but they will also apply to other cases where the metric laws hold. n
The elements of S (though arbitrary) will be called "points," usually denoted by p, q, x, y, z (sometimes with bars, etc. ); ρ is called a metric for S. We symbolize it by ρ : S ×S → E
since it is function defined on S × S (pairs of elements of S) into E
1
.
1
(3.7.2)
Thus we are led to the following definition.
Definition A metric space is a set S ≠ ∅ together with a function ρ : S ×S → E
1
(3.7.3)
(called a metric for S ) satisfying the metric laws (axioms): For any x, y, and z in S, we have i. ρ(x, y) ≥ 0, and (i ) ρ(x, y) = 0 iff x = y; ii. ρ(x, y) = ρ(y, x) (symmetry law); and iii. ρ(x, z) ≤ ρ(x, y) + ρ(y, z)( triangle law). ′
Thus a metric space is a pair (S, ρ), namely, a set S and a metric ρ for it. In general, one can define many different metrics ′
′′
ρ, ρ , ρ , …
(3.7.4)
for the same S. The resulting spaces ′
′′
(S, ρ), (S, ρ ) , (S, ρ ) , …
then are regarded as different. However, if confusion is unlikely, we simply write with metric ρ," and "A ⊆ (S, ρ) " for "A ⊆ S in (S, ρ)."
(3.7.5) S
for (S, ρ). We write "p ∈ (S, ρ)" for "p ∈ S
Example 3.7.1 (1) In E
n
,
we always assume ¯¯ ¯
¯ ¯ ¯
¯¯ ¯
¯ ¯ ¯
ρ(x, y ) = | x − y | (the "standard metric")
unless stated otherwise. By Theorem 5 in §§1-3, (E
n
, ρ)
(3.7.6)
is a metric space.
(2) However, one can define for E many other "nonstandard" metrics. For example, n
3.7.1
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1/p
n ′
¯¯ ¯
¯ ¯ ¯
ρ (x, y ) = ( ∑ | xk − yk |
p
)
for any real p ≥ 1
(3.7.7)
k=1
likewise satisfies the metric laws (a proof is suggested in §10, Problems 5-7; similarly for C
n
.
(3) Any set S ≠ ∅ can be "metrized" (i.e., endowed with a metric) by setting ρ(x, y) = 1 if x ≠ y, and ρ(x, x) = 0.
(3.7.8)
(Verify the metric laws!) This is the so-called discrete metric. The space (S, ρ) so defined is called a discrete space. (4) Distances ("mileages") on the surface of our planet are actually measured along circles fitting in the curvature of the globe (not straight lines). One can show that they obey the metric laws and thus define a (nonstandard) metric for S = (surface of the globe). (5) A mapping f
: A → E
1
is said to be bounded iff (∃K ∈ E
1
) (∀x ∈ A)
|f (x)| ≤ K.
(3.7.9)
For a fixed A ≠ ∅, let W be the set of all such maps (each being treated as a single "point" of W ). Metrize W by setting, for f, g ∈ W , ρ(f , g) = sup |f (x) − g(x)|.
(3.7.10)
x∈A
(Verify the metric laws; see a similar proof in §10.) II. We now define "balls" in any metric space (S, ρ).
Definition Given p ∈ (S, ρ) and a real denoted
ε > 0,
we define the open ball or globe with center
p
and radius
ε
(briefly "ε -globe about p",
Gp or Gp (ε) or G(p; ε),
(3.7.11)
ρ(x, p) < ε.
(3.7.12)
Gp = Gp (ε) = {x ∈ S|ρ(x, p) ≤ ε}.
(3.7.13)
Sp (ε) = {x ∈ S|ρ(x, p) = ε}.
(3.7.14)
to be the set of all x ∈ S such that
Similarly, the closed ε -globe about p is ¯ ¯¯ ¯
¯ ¯¯ ¯
The ε -sphere about p is defined by
Note. An open globe in E is an ordinary solid sphere (without its surface S 3
is a disc (the interior of a circle). In interval [p − ε, p + ε].
p (ε)),
E
1
, the globe
Gp (ε)
as known from geometry. In E
is simply the open interval
(p − ε, p + ε)
, while
¯ ¯¯ ¯
2
,
an open globe
Gp (ε)
is the closed
The shape of the globes and spheres depends on the metric ρ. It may become rather strange for various unusual metrics. For example, in the discrete space (Example (3)), any globe of radius < 1 consists of its center alone, while G (2) contains the entire space. (Why?) See also Problems 1 ,2, and 4. p
III. Now take any nonempty set A ⊆ (S, ρ). The distances ρ(x, y) in S are, of course, also defined for points of A (since A ⊆ S , and the metric laws remain valid in A. Thus A is likewise a (smaller) metric space under the metric ρ "inherited" from S; we only have to restrict the domain of ρ to A × A (pairs of points from A). The set A with this metric is called a subspace of S. We shall denote it by (A, ρ), using the same letter ρ or simply by A. Note that A with some other metric ρ is not called a subspace of (S, ρ). ′
3.7.2
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By definition, points in (A, ρ) have the same distances as in (S, ρ). However, globes and spheres in (A, ρ) must consist of points from A only, with centers in A. Denoting such a globe by ∗
Gp (ε) = {x ∈ A|ρ(x, p) < ε},
we see that it is obtained by restricting Thus
Gp (ε)
(the corresponding globe in
S)
(3.7.15)
to points of
A,
i.e., removing all points not in
∗
Gp (ε) = A ∩ Gp (ε);
similarly for closed globes and spheres. A ∩ G ρ(p, p) = 0 < ε, and p ∈ A.
p (ε)
A.
(3.7.16)
is often called the relativized (to A ) globe G
p (ε).
For example, let R be the subspace of E consisting of rationals only. Then the relativized globe G the interval 1
Note that
∗ p (ε)
∗
p ∈ Gp (ε)
since
consists of all rationals in
Gp (ε) = (p − ε, p + ε),
(3.7.17)
and it is assumed here that p is rational itself. IV. A few remarks are due on the extended real number system E (see Chapter 2, §§13) . As we know, E consists of all reals and two additional elements, ±∞, with the convention that −∞ < x < +∞ for all x ∈ E . The standard metric ρ does not apply to E . However, one can metrize E in various other ways. The most common metric ρ is suggested in Problems 5 and 6 below. Under that metric, globes turn out to be finite and infinite intervals in E . ∗
∗
1
∗
∗
′
∗
Instead of metrizing E
∗
,
we may simply adopt the convention that intervals of the form (a, +∞] and [−∞, a), a ∈ E
1
,
(3.7.18)
will be called "globes" about +∞ and −∞, respectively (without specifying any "radii"). Globes about finite points may remain as they are in E . This convention suffices for most purposes of limit theory. We shall use it often (as we did in Chapter 2, §13). 1
This page titled 3.7: Metric Spaces is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
3.7.3
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3.7.E: Problems on Metric Spaces (Exercises) The "arrowed" problems should be noted for later work.
Exercise 3.7.E. 1 Show that E becomes a metric space if distances ρ(x, y) are defined by (a) ρ(x, y) = |x − y | + |x − y | or (b) ρ(x, y) = max {|x − y | , |x − y |} , where x = (x , x ) and y = (y , y ) . In each case, describe G (1) and nonnegative coordinates. 2
¯¯ ¯
¯ ¯ ¯
¯¯ ¯
¯ ¯ ¯
¯¯ ¯
1
1
2
1
¯¯ ¯
1
2
2
1
¯ ¯ ¯
¯ ¯ ¯
2
1
2
¯ ¯ ¯
2
0
S¯¯¯ (1). 0
Do the same for the subspace of points with
Exercise 3.7.E. 2 Prove the assertions made in the text about globes in a discrete space. Find an empty sphere in such a space. Can a sphere contain the entire space?
Exercise 3.7.E. 3 Show that ρ in Examples (3) and (5) obeys the metric axioms.
Exercise 3.7.E. 4 Let M be the set of all positive integers together with the "point" ∞. Metrize M by setting 1 ∣ 1 ∣ 1 ρ(m, n) = ∣ − ∣ , with the convention that = 0. ∣m ∣ n ∞
Verify the metric axioms. Describe G
∞
(
1 2
) , S∞ (
1 2
),
and G
(3.7.E.1)
.
1 (1)
Exercise 3.7.E. 5 ⇒ 5.
Metrize the extended real number system E by ∗
′
ρ (x, y) = |f (x) − f (y)|,
(3.7.E.2)
where the function f : E
∗
⟶ [−1, 1]
(3.7.E.3)
onto
is defined by x f (x) =
if x is finite, f (−∞) = −1, and f (+∞) = 1.
(3.7.E.4)
1 + |x|
Compute ρ (0, +∞), ρ (0, −∞), ρ (−∞, +∞), ρ (0, 1), ρ (1, 2), and ρ (n, +∞). Describe G Verify the metric axioms (also when infinities are involved). ′
′
′
′
′
′
0 (1),
G+∞ (1),
and G
−∞
(
1 2
).
Exercise 3.7.E. 6 ⇒ 6.
In Problem 5, show that the function f is one to one, onto [−1, 1], and increasing; i.e. ′
′
′
x < x implies f (x) < f (x ) for x, x ∈ E
3.7.E.1
∗
.
(3.7.E.5)
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Also show that the f -image of an interval (a, b) ⊆ E is the interval (f (a), f (b)). Hence deduce that globes in E (with ρ as in Problem 5) are intervals in E (possibly infinite). [Hint: For a finite x, put ∗
∗
′
∗
x y = f (x) =
.
(3.7.E.6)
1 + |x|
Solving for x (separately in the cases x ≥ 0 and x < 0), show that (∀y ∈ (−1, 1))
x =f
−1
y (y) =
;
(3.7.E.7)
1 − |y|
thus x is uniquely determined by y, i.e., f is one to one and onto-each y ∈ (−1, 1) corresponds to some x ∈ E
1
.
(How about
±1?)
To show that f is increasing, consider separately the three cases x < 0 < x , x < x and x ). ] ′
′
0)(∃k)(∀m > k)
| xm ± ym − (q ± r)| < ε.
Thus we fix an arbitrary ε > 0 and look for a suitable k. since x
m
′
(∀m > k )
→ q
and y
m
→ r,
(3.11.1)
there are k and k such that ′
′′
ε | xm − q|
k )
(as ε is arbitrary, we may as well replace it by them, we obtain
1 2
ε).
(∀m > k)
ε | ym − r|
k, k = max (k , k ) .
| xm − q| + | ym − r| < ε.
Adding
(3.11.4)
Hence by the triangle law, | xm − q ± (ym − r)| < ε, i.e., | xm ± ym − (q ± r)| < ε for m > k,
(3.11.5)
as required. □ This proof of (i) applies to sequences of vectors as well, without any change. The proof of (ii) and (iii) is sketched in Problems 1-4 below. Note 1. By induction, parts (i) and (ii) hold for sums and products of any finite (but fixed) number of suitable convergent sequences. Note 2. The theorem does not apply to infinite limits q, r, a. Note 3. The assumption a ≠ 0 in Theorem 1( iii) is important. It ensures not only that q/a is defined but also that at most finitely many a can vanish (see Problem 3). Since we may safely drop a finite number of terms (see Note 2 in §14), we can achieve that noa is 0, so that x /a is defined. It is with this understanding that part (iii) of the theorem has been formulated. The next two theorems are actually special cases of more general propositions to be proved in Chapter 4, §§3 and 5. Therefore, we only state them here, leaving the proofs as exercises, with some hints provided. m
m
m
m
3.11.1
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Theorem 3.11.2 (componentwise convergence). We have x → p in E ( component of p, i.e., iff x → p , k = 1, 2, … , n, in E ¯¯ ¯ m
¯ ¯ ¯
mk
n
¯ ¯ ¯
∗
k
iff each of the n components of x (C ). (See Problem 8 for hints.)
C
1
n
¯¯ ¯ m
)
tends to the corresponding
Theorem 3.11.3 Every monotone sequence {x } ⊆ E has a finite or infinite limit, which equals sup_ x if {x { x } is monotone and bounded in E , its limit is finite (by Corollary 1 of Chapter 2, §13). ∗
n
n
n}
n
↑
and inf
if {x
n xn
n}
↓.
If
1
n
The proof was requested in Problem 9 of Chapter 2, §13. See also Chapter 4, §5, Theorem 1. An important application is the following.
Example 3.11.1 (the number e). Let x
= (1 +
n
1 n
n
)
in E
1
.
By the binomial theorem,
n(n − 1) xn
= 1 +1 +
n(n − 1)(n − 2) +
2
1 = 2 + (1 −
n
3!n
1
n!n
1
) n
n(n − 1) ⋯ (n − (n − 1)) +⋯ +
3
2!n
+ (1 −
2 ) (1 −
2!
n
1 )
1
2
+ ⋯ + (1 −
n
3!
) (1 − n
n−1 ) ⋯ (1 −
n
If n is replaced by n + 1, all terms in this expansion increase, as does their number. Thus x for n > 1 ,
n
1 2 < xn
1
1. Hence 2 < sup x ≤ 3; and by Theorem 3, sup x an important role in analysis. It can be shown that it is irrational, and (to within 10 any case, n
n}
n−1
1 1 −(
n−2
i.e., {x
2
1
2
< xn+1 ,
n!
1 +⋯ +
2
1
1 )
n
n
n
n
= lim xn .
−20
n
1 2 < e = lim (1 +
This limit, denoted by e, plays In
) e = 2.71828182845904523536 …
)
≤ 3.
(3.11.6)
n
n→∞
The following corollaries are left as exercises for the reader.
corollary 3.11.1 Suppose lim x
m
=p
(a) If p > q, then x
m
(b) If x
m
≤ ym
and lim y
m
> ym
=q
exist in E . ∗
for all but finitely many m.
for infinitely many m, then p ≤ q; i.e., lim x
m
≤ lim ym
.
This is known as passage to the limit in inequalities. Caution: The strict inequalities p ≤ q. For example, let
xm < ym
do not imply
p ym ;
(3.11.8)
= lim ym = 0.
corollary 3.11.2 Let x
→ p
m
in E
∗
,
and let c ∈ E (finite or not). Then the following are true: ∗
(a) If p > c (respectively,p < c) , we have x
m
(b) If x
m
≤c
(respectively, x
m
≥c
> c(xm < c)
for all but finitely many m.
) for infinitely many m, then p ≤ c
One can prove this from Corollary 1, with y
m
=c
(or x
m
=c
(p ≥ c)
.
) for all m.
corollary 3.11.3 (rule of intermediate sequence). If x z → p.
m
→ p
and y
m
→ p
in E and if ∗
xm ≤ zm ≤ ym
for all but finitely many
m,
then also
m
Theorem 3.11.4 (continuity of the distance function). If xm → p and ym → q in a metric space (S, ρ),
(3.11.9)
then ρ (xm , ym ) → ρ(p, q) in E
1
.
(3.11.10)
Proof Hint: Show that |ρ (xm , ym ) − ρ(p, q)| ≤ ρ (xm , p) + ρ (q, ym ) → 0
(3.11.11)
by Theorem 1.
This page titled 3.11: Operations on Convergent Sequences is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
3.11.3
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3.11.E: Problems on Limits of Sequences (Exercises) See also Chapter 2, §13.
Exercise 3.11.E. 1 Prove that if x
m
→ 0
and if {a
m}
is bounded in E or C , then 1
am xm → 0.
(3.11.E.1)
This is true also if the x are vectors and the a are scalars (or vice versa). [Hint: If {a } is bounded, there is a K ∈ E such that m
m
1
m
(∀m)
As x
m
→ 0
| am | < K.
(3.11.E.2)
, ε (∀ε > 0)(∃k)(∀m > k)
| xm |
0)(∃k)(∀m > k)
| am | ≥ ε.
(3.11.E.6)
(We briefly say that the a are bounded away from 0, for m > k. ) Hence prove the boundedness of { m
[Hint: For the first part, proceed as in the proof of Corollary 1 in §14, with x For the second part, the inequalities
m
(∀m > k)
1 ∣ 1 ∣ ∣ ∣ ≤ ∣ am ∣ ε
= am , p = a,
1 am
}
for m > k .
and q = 0.
(3.11.E.7)
lead to the desired result. ]
3.11.E.1
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Exercise 3.11.E. 4 Prove that if a
m
→ a ≠0
in E or C , then 1
1
1 →
am
.
(3.11.E.8)
a
Use this and Theorem 1( ii) to prove Theorem 1( iii), noting that xm am
1 = xm ⋅
.
(3.11.E.9)
am
[Hint: Use Note 3 and Problem 3 to find that (∀m > k)
where {
1 am
}
is bounded and
Hence, by Problem
1, ∣ ∣
1 am
−
1 |a| 1 a
| am − a| → 0.
∣ → 0. ∣
1∣ 1 1 ∣ 1 ∣ − ∣ = | am − a| , ∣ am a∣ |a| | am |
(3.11.E.10)
(Why?)
Proceed. ]
Exercise 3.11.E. 5 Prove Corollaries 1 and 2 in two ways: (i) Use Definition 2 of Chapter 2, §13 for Corollary 1(a), treating infinite limits separately; then prove (b) by assuming the opposite and exhibiting a contradiction to (a). (ii) Prove (b) first by using Corollary 2 and Theorem 3 of Chapter 2, §13; then deduce (a) by contradiction.
Exercise 3.11.E. 6 Prove Corollary 3 in two ways (cf. Problem 5).
Exercise 3.11.E. 7 Prove Theorem 4 as suggested, and also without using Theorem 1(i).
Exercise 3.11.E. 8 Prove Theorem 2. [Hint: If x → p, then ¯¯ ¯ m
¯ ¯ ¯
(∀ε > 0)(∃q)(∀m > q)
¯ ¯ ¯ ¯ ε > |¯¯ x m − p| ≥ | xmk − pk | .
(Why?)
(3.11.E.11)
Thus by definition x → p , k = 1, 2, … , n . Conversely, if so, use Theorem 1(i)( ii ) to obtain mk
k
n
n
⃗ → ∑ pk e k ⃗ , ∑ xmk e k k=1
(3.11.E.12)
k=1
with e ⃗ as in Theorem 2 of §§1-3]. k
3.11.E.2
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Exercise 3.11.E. 8
′
In Problem 8, prove the converse part from definitions. ( Fix ε > 0, etc. )
Exercise 3.11.E. 9 Find the following limits in E
1
,
in two ways: (i) using Theorem 1, justifying each step; (ii) using definitions only. m+1
(a) limm→∞
m 1
(c) limn→∞
2
1+n
;
3m+2
(b) limm→∞
;
(d) limn→∞
2m−1
(3.11.E.13)
n(n−1) 2
1−2n
[ Solution of (a) by the first method: Treat m +1
1 =1+
m
as the sum of x
m
=1
(3.11.E.14) m
(constant) and 1 ym =
§
→ 0 (proved in 14).
(3.11.E.15)
m
Thus by Theorem 1(i), m +1 = xm + ym → 1 + 0 = 1.
(3.11.E.16)
m
Second method: Fix ε > 0 and find k such that ∣ m +1 ∣ ∣ − 1 ∣ < ε. ∣ ∣ m
(∀m > k)
Solving for m, show that this holds if m >
1 ε
.
Thus take an integer k >
1 ε
,
(3.11.E.17)
so
∣ m +1 ∣ ∣ − 1 ∣ < ε. ∣ ∣ m
(∀m > k)
(3.11.E.18)
Caution: One cannot apply Theorem 1 (iii) directly, treating (m + 1)/m as the quotient of x = m + 1 and a = m, because x and a diverge in E . (Theorem 1 does not apply to infinite limits.) As a remedy, we first divide the numerator and denominator by a suitable power of m( or n). ] m
m
1
m
m
Exercise 3.11.E. 10 Prove that | xm | → +∞ in E
∗
1 iff
→ 0 xm
(xm ≠ 0) .
(3.11.E.19)
Exercise 3.11.E. 11 Prove that if xm → +∞ and ym → q ≠ −∞ in E
3.11.E.3
∗
,
(3.11.E.20)
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then xm + ym → +∞.
(3.11.E.21)
" +∞ + q = +∞ if q ≠ −∞. "
(3.11.E.22)
" −∞ + q = −∞ if q ≠ +∞. "
(3.11.E.23)
" (+∞) ⋅ q = +∞ if q > 0 "
(3.11.E.24)
" (+∞) ⋅ q = −∞ if q < 0. "
(3.11.E.25)
This is written symbolically as
Do also
Prove similarly that
and
[Hint: Treat the cases q ∈ E
1
, q = +∞,
and q = −∞ separately. Use definitions.]
Exercise 3.11.E. 12 ¯ ¯¯¯¯¯ ¯
Find the limit (or – lim and lim) of the following sequences in E –– (a) x = 2 ⋅ 4 ⋯ 2n = 2 n! ; (b) x = 5n − n ; (c) x = 2n − n − 3n − 1 ; (d) x = (−1) n! ;
∗
:
n
n
3
n
4
3
2
n
n
n
n
(−1)
(e) x = . [Hint for (b) : x n
n!
2
n
= n (5 − n ) ;
use Problem 11.]
Exercise 3.11.E. 13 Use Corollary 4 in §14, to find the following: n
(a) lim
(−1)
(b) lim
1−n+(−1)
n→∞
1+n2
; n
n→∞
.
2n+1
Exercise 3.11.E. 14 Find the following. (a) lim
1+2+⋯+n
n→∞
(b) lim
n→∞
(c) lim
n→∞
n2
n
∑
k=1
n
∑
k=1
;
2
k 3
n +1 3
k 4
; .
n −1 n m
[Hint: Compute ∑ k using Problem 10 of Chapter 2, §§5-6.] What is wrong with the following "solution" of (a) : → 0, → 0, etc.; hence the limit is 0? k=1
1
2
n
2
2
n
3.11.E.4
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Exercise 3.11.E. 15 For each integer m ≥ 0, let m
m
Smn = 1
+2
m
+⋯ +n
.
(3.11.E.26)
.
(3.11.E.27)
Prove by induction on m that Smn
lim n→∞
1 =
m+1
m +1
(n + 1)
[Hint: First prove that m−1 m+1
(m + 1)Smn = (n + 1 )
m +1
−1 − ∑ (
by adding up the binomial expansions of (k + 1)
m+1
) Smi
(3.11.E.28)
i
i=0
, k = 1, … , n. ]
Exercise 3.11.E. 16 Prove that lim q
n
= +∞ if q > 1;
n→∞
lim q
n
= 0 if |q| < 1;
n→∞
n
lim 1
= 1.
(3.11.E.29)
n→∞
[Hint: If q > 1, put q = 1 + d, d > 0. By the binomial expansion, q
If |q| < 1, then ∣∣
1 q
∣ > 1; ∣
so lim ∣∣
n
1 q
n
= (1 + d)
∣ ∣
n
= +∞;
= 1 + nd + ⋯ + d
n
> nd → +∞.
(Why?)
(3.11.E.30)
use Problem 10. ]
Exercise 3.11.E. 17 Prove that n lim n→∞
q
n
n
= 0 if |q| > 1, and lim n→∞
q
n
= +∞ if 0 < q < 1.
(3.11.E.31)
[Hint: If |q| > 1, use the binomial as in Problem 16 to obtain n
|q |
1 >
n
2
n(n − 1)d , n ≥ 2, so 2
n
2
1.
(3.11.E.34)
n→∞
[Hint: If r > 1 and a > 1, use Problem 17 with q = a
1/r
r
−n
0 1, put √c = 1 + dn , dn > 0. Expand c = (1 + dn )
n
to show that
c 0 < dn
1, so p = 1 . −− − In case 0 < c < 1, consider √1/c and use Theorem 1( iii) . n
Exercise 3.11.E. 24 Prove the existence of lim x and find it when x is defined inductively by – − − − (i) x = √2, x = √2x ; − − − − − − (ii) x = c > 0, x = √c + x ; (iii) x = c > 0, x = ; hence deduce that lim = 0. [Hint: Show that the sequences are monotone and bounded in E (Theorem 3). For example, in (ii) induction yields n
1
n+1
1
n
n
2
n+1
n
n
cxn
1
n+1
c
n→∞
n+1
n!
1
xn < xn+1 < c + 1.
Thus lim x
n
= lim xn+1 = p
( Verify! )
(3.11.E.41)
( given )
(3.11.E.42)
exists. To find p, square the equation − − − − − − 2
xn+1 = √ c
+ xn
and use Theorem 1 to get 2
p
2
=c
+ p.
(Why?)
(3.11.E.43)
Solving for p (noting that p > 0), obtain 1 p = lim xn =
− − − − − − 2 (1 + √ 4 c + 1 ) ;
(3.11.E.44)
2
similarly in cases (i) and (iii). ]
Exercise 3.11.E. 25 Find lim x in E or E (if any), given that (a) x = (n + 1) − n , 0 < q < 1 ; − − − − − − (b) x = √− n (√n + 1 − √n ) ; n
n
1
q
∗
q
n
3.11.E.7
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(c) x
n
(d) x
;
1
=
√n2 +k
with |c| < 1 ;
n
= n(n + 1)c ,
n
− − − − − − −
(e) x
= √∑ a k=1
(f) x
=
m
n
n
n k
3⋅5⋅7⋯(2n+1)
n
with a
,
2⋅5⋅8⋯(3n−1)
k
>0
;
.
[Hints: (a) 0 < x
q
n
=n
[(1 +
1 n
q
)
q
− 1] < n
(1 +
1 n
q−1
− 1) = n
− − − − −
(b) x
=
n
1 1+√1+1/n
,
where 1 < √1 +
1 n
0. Then
n ).
ε (∃k)(∀n > k)
ε
p−
< xn < p +
4
.
(3.11.E.47)
4
Adding n − k inequalities, get n
ε (n − k) (p −
) < 4
ε
∑ xi < (n − k) (p + i=k+1
).
(3.11.E.48)
4
With k so fixed, we thus have n−k (∀n > k)
ε
1
(p − n
) < 4
n
n−k (xk+1 + ⋯ + xn )
K )
ε ) K
′′
ε )
−
1
0) ,
then − − − − − − − − n − lim √ x1 x2 ⋯ xn = p.
(3.11.E.57)
n→∞
[Hint: Let first 0 < p < +∞. Given ε > 0, use density to fix δ > 1 so close to 1 that p p −ε
k)
4
4
< xn < p √δ.
(3.11.E.59)
√δ
Continue as in Problem 26, replacing ε by δ, and multiplication by addition (also subtraction by division, etc., as shown above). Find a similar solution for the case p = +∞. Note the result of Problem 20.]
3.11.E.9
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Exercise 3.11.E. 28 Disprove by counterexamples the converse implications in Problems 26 and 27. For example, consider the sequences 1, −1, 1, −1, …
(3.11.E.60)
and 1
1 , 2,
2
1 , 2,
2
, 2, …
(3.11.E.61)
2
Exercise 3.11.E. 29 Prove the following. (i) If {x } ⊂ E and lim (x − x ) = p in E , then → p. − − (ii) If {x } ⊂ E (x > 0) and if → p ∈ E , then √x → p . Disprove the converse statements by counterexamples. [Hint: For (i), let y = x and y = x − x , n = 2, 3, … Then y 1
n→∞
n+1
n
n
xn+1
1
n
xn
∗
n
n
∗
n
n
xn
1
1
n
n
n−1
n
n
1 n
∑ yi =
xn
→ p
and
,
(3.11.E.62)
n
i=1
so Problems 26 and 26 apply. For (ii), use Problem 27. See Problem 28 for examples. ] ′
Exercise 3.11.E. 30 From Problem 29 deduce that − − √n! = +∞ ; (a) lim (b) lim = 0; n
n→∞
n+1
n→∞
(c) lim
n!
n
n→∞
(d) lim (e) lim
n→∞
n→∞
−− n
√ 1 n
n
n!
=e
− − √n! = n
n − √n = 1
.
; 1 e
;
Exercise 3.11.E. 31 Prove that a + 2b lim xn =
n→∞
,
(3.11.E.63)
(xn + xn+1 ) .
(3.11.E.64)
3
given 1 x0 = a, x1 = b, and xn+2 =
[Hint: Show that the differences dn = x Then use the result of Problem 19. ]
n
− xn−1
2
form a geometric sequence, with ratio
3.11.E.10
q =−
1 2
,
and
n
xn = a + ∑
k=1
dk .
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Exercise 3.11.E. 32 ⇒ 32.
For any sequence {x
n}
⊆E
1
,
prove that n
n
i=1
i=1
1 ¯ ¯¯¯¯¯ ¯ 1 ¯ ¯¯¯¯¯ ¯ lim xn ≤ lim ∑ xi ≤ lim ∑ xi ≤ lim xn . ––– ––– n n
(3.11.E.65)
Hence find a new solution of Problems 26 and 26 . ′
¯ ¯¯¯¯¯ ¯
[Proof for lim: Fix any k ∈ N . Put k
c = ∑ xi and b = sup xi .
(3.11.E.66)
i≥k
i=1
Verify that (∀n > k)
xk+1 + xk+2 + ⋯ + xn ≤ (n − k)b.
(3.11.E.67)
Add c on both sides and divide by n to get n
1 (∀n > k) n
Now fix any ε > 0, and first let |b| < +∞. As
c n
→ 0
and
n−k n
b → b,
ε
nk )
c
∑ xi ≤
there is n
k
n−k and
2
(3.11.E.68)
n
>k
such that
ε b nk )
This clearly holds also if b = sup
i≥k
xi = +∞.
n
n
∑ xi ≤ ε + b.
(3.11.E.70)
i=1
Hence also 1 sup n≥nk
n
n
∑ xi ≤ ε + sup xi . i=1
(3.11.E.71)
i≥k
As k and ε were arbitrary, we may let first k → +∞, then ε → 0, to obtain n
1 ¯ ¯¯¯¯¯ ¯ lim ∑ xi ≤ lim sup xi = lim xn . ––– n k→∞ i≥k
( Explain! )]
(3.11.E.72)
i=1
Exercise 3.11.E. 33 ⇒ 33.
Given {x
n}
⊆E
1
, xn > 0,
prove that
¯ ¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯ ¯ − − − − − − − − − − − − − − − − n − n − lim xn ≤ lim √ x1 x2 ⋯ xn and lim √ x1 x2 ⋯ xn ≤ lim xn . ––– –––
(3.11.E.73)
Hence obtain a new solution for Problem 27. [Hint: Proceed as suggested in Problem 32, replacing addition by multiplication.]
3.11.E.11
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Exercise 3.11.E. 34 Given x
n,
1
yn ∈ E
(yn > 0) ,
with n
xn → p ∈ E
∗
and bn = ∑ yi → +∞,
(3.11.E.74)
i=1
prove that n
∑i=1 xi yi
lim n→∞
n
∑
i=1
= p.
(3.11.E.75)
yi
Note that Problem 26 is a special case of Problem 34 (take all y = 1) . [Hint for a finite However, before adding the n − k inequalities, multiply by y and obtain n
p :
Proceed as in Problem
26.
i
n
ε (p − 4
n
Put bn = ∑
i=1
n
) ∑ yi < i=k+1
n
bn
n
i=k+1
4
) ∑ yi .
(3.11.E.76)
i=k+1
and show that
yi
1
where b
n
ε
∑ xi yi < (p +
→ +∞( by assumption ),
k
1
∑ xi yi = 1 − i=k+1
∑ xi yi ,
bn
(3.11.E.77)
i=1
so k
1 bn
∑ xi yi → 0
(for a fixed k).
(3.11.E.78)
i=1
Proceed. Find a proof for p = ±∞. ]
Exercise 3.11.E. 35 ¯ ¯¯¯¯¯ ¯
Do Problem 34 by considering – lim and lim as in Problem 32. –– [ Hint: Replace
c n
by
n
c bn
, where bn = ∑
i=1
yi → +∞. ]
Exercise 3.11.E. 36 Prove that if u
n,
vn ∈ E
1
,
with {v
n}
↑
(strictly) and v
n
→ +∞,
un − un−1
lim n→∞
and if
=p
(p ∈ E
∗
),
(3.11.E.79)
vn − vn−1
then also lim n→∞
un
= p,
(3.11.E.80)
vn
[Hint: The result of Problem 34, with xn =
un − un−1 vn − vn−1
and yn = vn − vn−1 .
3.11.E.12
(3.11.E.81)
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leads to the final result. ]
Exercise 3.11.E. 37 From Problem 36 obtain a new solution for Problem 15. Also prove that lim ( n→∞
Smn
1
1
−
m+1
) =
.
m +1
n
(3.11.E.82)
2
[Hint: For the first part, put m+1
un = Smn and vn = n
.
(3.11.E.83)
For the second, put m+1
un = (m + 1)Smn − n
m
and vn = n
(m + 1). ]
(3.11.E.84)
Exercise 3.11.E. 38 − − = √ab
Let 0 < a < b < +∞. Define inductively: a
1
and b
1
=
1 2
(a + b)
;
1 − − − − an+1 = √an bn and bn+1 = (an + bn ) , n = 1, 2, … 2
Then a
n+1
< bn+1
(3.11.E.85)
for 1 bn+1 − an+1 =
2
1 − − − − − − − − 2 (an + bn ) − √an bn = (√bn − √an ) > 0. 2
(3.11.E.86)
Deduce that a < an < an+1 < bn+1 < bn < b,
so {a
n}
↑
and {b
n}
↓.
By Theorem 3, a
n
→ p
and b
n
→ q
for some p, q ∈ E
1
.
(3.11.E.87)
Prove that p = q, i.e.,
lim an = lim bn .
(This is Gauss's arithmetic-geometric mean of a and b. ) [Hint: Take limits of both sides in b = (a + b ) to get q = 1
n+1
2
n
n
1 2
(3.11.E.88)
(p + q). ]
Exercise 3.11.E. 39 Let 0 < a < b in E
1
.
Define inductively a
1
an+1 =
= a, b1 = b
2an bn an + bn
, 1
, and bn+1 =
2
(an + bn ) ,
n = 1, 2, …
(3.11.E.89)
Prove that − − √ab = lim an = lim bn . n→∞
(3.11.E.90)
n→∞
[Hint: Proceed as in Problem 38.]
3.11.E.13
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Exercise 3.11.E. 40 Prove the continuity of dot multiplication, namely, if ¯¯ ¯ xn
¯¯
¯ ¯ ¯
¯¯
→ q and y n → r in E
n
(3.11.E.91)
(*or in another Euclidean space; see §9), then ¯¯ ¯ xn
¯ ¯ ¯
¯¯
¯¯
⋅ y n → q ⋅ r.
(3.11.E.92)
3.11.E: Problems on Limits of Sequences (Exercises) is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
3.11.E.14
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3.12: More on Cluster Points and Closed Sets. Density This page is a draft and is under active development. I. The notions of cluster point and closed set (§§12, 14) can be characterized in terms of convergent sequences. We start with cluster points.
Theorem 3.12.1 (i) A sequence {x
m}
⊆ (S, ρ)
clusters at a point p ∈ S iff it has a subsequence {x
mn
(ii) A set A ⊆ (S, ρ) clusters at p ∈ S iff p is the limit of some sequence {x can be made distinct.
n}
}
converging to p.
of points of A other than p; if so, the terms x
n
Proof (i) If p = lim x , then by definition each globe about x . Thus p is a cluster point. n→∞
p
mn
contains all but finitely many
xm , n
hence infinitely many
m
Conversely, if so, consider in particular the globes 1 Gp (
By assumption, G
p (1)
contains some x
m.
),
n = 1, 2, …
(3.12.1)
n
Thus fix xm
1
∈ Gp (1).
(3.12.2)
) with m2 > m1 .
(3.12.3)
Next, choose a term 1 xm2 ∈ Gp (
(Such terms exist since G
p
(
1 2
)
2
contains infinitely many x
m.
)
Next, fix
1 xm3 ∈ Gp (
3
) , with m3 > m2 > m1 ,
(3.12.4)
and so on. Thus, step by step (inductively), select a sequence of subscripts m1 < m2 < ⋯ < mn < ⋯
(3.12.5)
that determines a subsequence (see Chapter 1, §8) such that 1 (∀n)
whence ρ (x
mn
, p) → 0,
or x
mn
→ p.
xm
n
∈ Gp (
n
1 ) , i.e., ρ (xm , p) < n
→ 0,
(3.12.6)
n
(Why?) Thus we have found a subsequence x
mn
→ p,
and assertion (i) is proved.
Assertion (ii) is proved quite similarly - proceed as in the proof of Corollary 6 in §§14; the inequalities m not needed here. □
1
< m2 < ⋯
are
Example 3.12.1 (a) Recall that the set R of all rationals clusters at each p ∈ E (§§14, Example (e)). Thus by Theorem 1(ii), each real p is the limit of a sequence of rationals. See also Problem 6 of §§12 for p in E . 1
¯ ¯ ¯
n
(b) The sequence
3.12.1
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0, 1, 0, 1, …
(3.12.7)
x2n = 1 → 1 and x2n−1 = 0 → 0.
(3.12.8)
has two convergent subsequences,
Thus by Theorem 1(i), it clusters at 0 and 1. Interpret Example (f) and Problem 10(a) in §14 similarly. As we know, even infinite sets may have no cluster points (take N in E ). However, a bounded infinite set or sequence in E (*or C ) must cluster. This important theorem (due to Bolzano and Weierstrass) is proved next. 1
n
n
Theorem 3.12.1 (Bolzano-Weierstrass). (i) Each bounded infinite set or sequence A in E (* or C ) has at least one cluster point p there (possibly outside A. n
n
¯ ¯ ¯
(ii) Thus each bounded sequence in E (* or C ) has a convergent subsequence. n
n
Proof Take first a bounded sequence {z
m}
⊆ [a, b]
in E
1
.
Let ¯ ¯¯¯¯¯ ¯
p = lim zm .
(3.12.9)
By Theorem 2(i) of Chapter 2, §13, {z } clusters at p. Moreover, as m
a ≤ zm ≤ b,
(3.12.10)
a ≤ inf zm ≤ p ≤ sup zm ≤ b
(3.12.11)
we have
by Corollary 1 of Chapter 2, §13. Thus p ∈ [a, b] ⊆ E
1
,
(3.12.12)
and so {z } clusters in E . 1
m
Assertion (ii) now follows - for E
1
by Theorem 1(i) above.
−
Next, take ¯ ¯ ¯
{ zm } ⊆ E
If {z
¯ ¯ ¯ m}
is bounded, all z
2
¯ ¯ ¯
, z m = (xm , ym ) ; xm , ym ∈ E
1
.
(3.12.13)
¯¯
are in some square [a, b]. (Why?) Let
¯ ¯ ¯ m
¯ ¯ ¯
¯¯
¯ ¯ ¯
a = (a1 , a2 ) and b = (b1 , b2 ) .
(3.12.14)
Then a1 ≤ xm ≤ b1 and a2 ≤ ym ≤ b2 in E
Thus by the first part of the proof, {x
m}
1
.
(3.12.15)
has a convergent subsequence xmk → p1 for some p1 ∈ [ a1 , b1 ] .
For simplicity, we henceforth write x for x with x → p , and a ≤ y ≤ b , as before. m
m
1
2
m
mk
, ym
m}
,
and z
¯ ¯ ¯ m
for z
¯ ¯ ¯ mk
. Thus z
¯ ¯ ¯ m
= (xm , ym )
is now a subsequence,
and obtain a subsubsequence ym
mi
mk
2
We now reapply this process to {y
The corresponding terms x
for y
(3.12.16)
i
→ p2 for some p2 ∈ [ a2 , b2 ] .
(3.12.17)
still tend to p by Corollary 3 of §14. Thus we have a subsequence 1
¯ ¯ ¯ z mi
= (xm , ym ) → (p1 , p2 ) i
i
3.12.2
in E
2
(3.12.18)
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by Theorem 2 in §15. Hence p = (p , p ) is a cluster point of theorem for sequences in E (hence in C ). ¯ ¯ ¯
1
¯ ¯ ¯
{ zm } .
2
Note that
¯ ¯ ¯
¯ ¯ ¯
¯¯
p ∈ [ a, b]
(see above). This proves the
2
The proof for E is similar; one only has to take subsequences n times. (*The same applies to replaced by complex ones.) n
C
n
with real components
Now take a bounded infinite set A ⊂ E ( C ) . Select from it an infinite sequence {z } of distinct points (see Chapter 1, §9, Problem 5). By what was shown above, {z } clusters at some point p, so each G contains infinitely many distinct points z ∈ A. Thus by definition, A clusters at p. □ n
∗
n
¯ ¯ ¯ m
¯ ¯ ¯ m
¯ ¯ ¯ m
¯ ¯ ¯
¯ ¯ ¯
p
¯ ¯ ¯
Note 1. We have also proved that if {z only.)
¯ ¯ ¯ m}
¯¯
¯ ¯ ⊆ [¯ a , b] ⊂ E
Note 2. The theorem may fail in spaces other than E cluster.
n
n
,
∗
( C
then {z
¯ ¯ ¯ m}
n
).
has a cluster point in
¯¯
¯ ¯ [¯ a , b].
(This applies to closed intervals
For example, in a discrete space, all sets are bounded, but no set can
II. Cluster points are closely related to the following notion.
Definition The closure of a set ¯ ¯¯ ¯
A ⊆ (S, ρ),
denoted
¯ ¯¯ ¯
A,
is the union of
and the set of all cluster points of
A
A
call it
′
A
. Thus
′
A = A∪A .
Theorem 3.12.1 ¯ ¯¯ ¯
We have p ∈ A in (S, ρ) iff each globe G
p (δ)
about p meets A , i. e., (∀δ > 0)
A ∩ Gp (δ) ≠ ∅.
(3.12.19)
¯ ¯¯ ¯
Equivalently, p ∈ A iff p = lim xn for some { xn } ⊆ A.
(3.12.20)
n→∞
Proof The proof is as in Corollary 6 of §14 and Theorem 1. (Here, however, the x need not be distinct or different from p. ) The details are left to the reader. n
This also yields the following new characterization of closed sets (cf. §12).
Theorem 3.12.1 A set A ⊆ (S, ρ) is closed iff one of the following conditions holds. (i) A contains all its cluster points (or has none); i.e., A ⊇ A . ′
¯ ¯¯ ¯
(ii) A = A . (iii) A contains the limit of each convergent sequence {x
n}
⊆A
(if any).
Proof Parts (i) and (ii) are equivalent since ′
′
¯ ¯¯ ¯
A ⊇ A ⟺ A = A ∪ A = A.
(Explain!)
(3.12.21)
Now let A be closed. If p ∉ A, then p ∈ −A; therefore, by Definition 3 in §12, some G fails to meet A (G ∩ A = ∅) . Hence no p ∈ −A is a cluster point, or the limit of a sequence {x } ⊆ A. (This would contradict Definitions 1 and 2 of §14.) Consequently, all such cluster points and limits must be in A, as claimed. p
p
n
3.12.3
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Conversely, suppose A is not closed, so −A is not open. Then entirely in −A. This means that each G meets A. Thus
−A
has a noninterior point
p;
i.e.,
p ∈ −A
but
noGp
is
p
¯ ¯¯ ¯
p ∈ A (by Theorem 3),
(3.12.22)
p = lim xn for some { xn } ⊆ A (by the same theorem),
(3.12.23)
and n→∞
even though p ∉ A( for p ∈ −A) . We see that (iii) and (ii), hence also (i), fail if A is not closed and hold if A is closed. (See the first part of the proof.) Thus the theorem is proved. □ ¯ ¯ ¯
Corollary 1. ∅ = ∅ . ¯ ¯¯ ¯
¯ ¯¯ ¯
Corollary 2. A ⊆ B ⟹ A ⊆ B . ¯ ¯¯ ¯
Corollary 3. A is always a closed set ⊇ A . ¯ ¯¯¯¯¯¯¯¯¯¯¯ ¯
Corollary 4. A ∪ B
¯ ¯¯ ¯
¯ ¯¯ ¯
= A∪B
¯ ¯¯ ¯
¯ ¯¯ ¯
(the closure of A ∪ B equals the union of A and B).
III. As we know, the rationals are dense in E (Theorem 3 of Chapter 2, §10). This means that every globe G (δ) = (p − δ, p + δ) in E contains rationals. Similarly (see Problem 6 in §12), the set R of all rational points is dense in E . We now generalize this idea for arbitrary sets in a metric space (S, ρ). 1
1
n
n
p
Definition Given A ⊆ B ⊆ (S, ρ), we say that A is dense in B iff each globe G p ∈ B is in A; i.e.,
p
p ∈ B,
meets A. By Theorem 3, this means that each
¯ ¯¯ ¯
p = lim xn n→∞
for some { xn } ⊆ A.
(3.12.24)
¯ ¯¯ ¯ 3
Equivalently, A ⊆ B ⊆ A. . Summing up, we have the following: 0
A is open iff A = A .
(3.12.25)
¯ ¯¯ ¯
′
A is closed iff A = A; equivalently, iff A ⊇ A . ¯ ¯¯ ¯
A is dense in B iff A ⊆ B ⊆ A. ′
A is perfect iff A = A .
(3.12.26)
(3.12.27) (3.12.28)
This page titled 3.12: More on Cluster Points and Closed Sets. Density is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
3.12.4
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3.12.E: Problems on Cluster Points, Closed Sets, and Density Exercise 3.12.E. 1 Complete the proof of Theorem 1( ii ) .
Exercise 3.12.E. 2 ¯ ¯¯ ¯
¯¯¯¯¯ ¯ n
Prove that R = E and R 1
=E
n
( Example (a))
.
Exercise 3.12.E. 3 Prove Theorem 2 for E
3
.
Prove it for E
n
∗
( and C
n
)
by induction on n.
Exercise 3.12.E. 4 Verify Note 2.
Exercise 3.12.E. 5 Prove Theorem 3.
Exercise 3.12.E. 6 Prove Corollaries 1 and 2.
Exercise 3.12.E. 7 Prove that (A ∪ B) = A ∪ B . [Hint: Show by contradiction that A ⊆ (A ∪ B) , etc. ] ′
′
′
′
′
′
p ∉ (A ∪ B )
excludes
′
p ∈ (A ∪ B) .
Hence
′
′
′
(A ∪ B) ⊆ A ∪ B .
Then show that
′
Exercise 3.12.E. 8 From Problem 7, deduce that A ∪ B is closed if A and B are. Then prove Corollary 4. By induction, extend both assertions to any finite number of sets.
Exercise 3.12.E. 9 From Theorem 4, prove that if the sets A
i (i
∈ I)
are closed, so is ⋂
i∈I
Ai
.
Exercise 3.12.E. 10 ¯ ¯¯ ¯ ¯ ¯¯ ¯
¯ ¯¯ ¯
Prove Corollary 3 from Theorem 3. Deduce that A = A and prove footnote 3. [Hint: Consider Figure 7 and Example (1) in §12 when using Theorem 3 (twice). ]
Exercise 3.12.E. 11 ¯ ¯¯ ¯
Prove that A is contained in any closed superset of A and is the intersection of all such supersets. [Hint: Use Corollaries 2 and 3. ]
3.12.E.1
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Exercise 3.12.E. 12 (i) Prove that a bounded sequence {x } ⊆ E ( C ) converges to p iff p is its only cluster point. (ii) Disprove it for (a) unbounded {x } and (b) other spaces. [Hint: For (i), if x → p fails, some G leaves out infinitely many x . These x form a bounded subsequence that, by Theorem 2, clusters at some q ≠ p. (Why? ) Thus q is another cluster point (contradiction!) For (ii), consider (a) Example (f) in §14 and (b) Problem 10 in §14, with (0,2] as a subspace of E . ] n
¯¯ ¯ m
∗
n
¯ ¯ ¯
¯ ¯ ¯
¯¯ ¯ m
¯¯ ¯ m
¯ ¯ ¯
¯¯ ¯ m
¯ ¯ ¯
p
¯¯
¯ ¯ ¯
¯¯ ¯ m
¯¯
1
Exercise 3.12.E. 13 ¯ ¯¯ ¯
In each case of Problem 10 in §14, find A . Is A closed? (Use Theorem 4.)
Exercise 3.12.E. 14 ¯ ¯¯ ¯
Prove that if {b } ⊆ B ⊆ A in (S, ρ), there is a sequence {a [Hint: Choose a ∈ G (1/n). ]
n}
n
n
⊆A
such that ρ (a
n,
bn ) → 0.
Hence a
n
→ p
iff b
n
→ p.
bn
Exercise 3.12.E. 15 We have, by definition, 0
p ∈ A iff (∃δ > 0)Gp (δ) ⊆ A;
(3.12.E.1)
p ∉ A iff (∀δ > 0)Gp (δ) ⊈ A, i.e., Gp (δ) − A ≠ ∅.
(3.12.E.2)
hence 0
¯ ¯¯ ¯
¯ ¯¯ ¯
(See Chapter 1, §§1 − 3. ) Find such quantifier formulas for p ∈ A, p ∉ A , p ∈ A , and p ∉ A . [Hint: Use Corollary 6 in §14, and Theorem 3 in §16. ] ′
′
Exercise 3.12.E. 16 Use Problem 15 to prove that (i) −(A) = (−A) and ¯ ¯¯ ¯
0
(ii) − (A
0
¯ ¯¯¯¯¯¯ ¯
) = −A
.
Exercise 3.12.E. 17 Show that ¯ ¯¯ ¯
¯ ¯¯¯¯¯¯ ¯
A ∩ (−A ) = bdA( boundary of A);
(3.12.E.3)
cf. §12, Problem 18. Hence prove again that A is closed iff A ⊇ bd A. [Hint: Use Theorem 4 and Problem 16 above. ]
Exercise 3.12.E. ∗18 ¯ ¯¯ ¯
A set A is said to be nowhere dense in (S, ρ) iff (A)
0
= ∅.
Show that Cantor's set P (§14, Problem 17) is nowhere dense.
¯ ¯¯ ¯
[ Hint: P is closed, so P = P . ]
3.12.E.2
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Exercise 3.12.E. ∗19 Give another proof of Theorem 2 for E . [Hint: Let A ⊆ [a, b]. Put 1
Q = {x ∈ [a, b]|x exceeds infinitely many points (or terms) of A}.
(3.12.E.4)
Show that Q is bounded and nonempty, so it has a glb, say, p = inf A. Show that A clusters at p. ]
Exercise 3.12.E. ∗20 For any set A ⊆ (S, ρ) define GA (ε) = ⋃ Gx (ε).
(3.12.E.5)
x∈A
Prove that ∞ ¯ ¯¯ ¯
A = ⋂ GA ( n=1
1 ).
(3.12.E.6)
n
Exercise 3.12.E. ∗21 Prove that ¯ ¯¯ ¯
A = {x ∈ S|ρ(x, A) = 0}; see $13, Note 3.
(3.12.E.7)
Hence deduce that a set A in (S, ρ) is closed iff (∀x ∈ S) ρ(x, A) = 0 ⟹ x ∈ A . 3.12.E: Problems on Cluster Points, Closed Sets, and Density is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
3.12.E.3
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3.13: Cauchy Sequences. Completeness This page is a draft and is under active development. A convergent sequence is characterized by the fact that its terms x become (and stay) arbitrarily close to its limit, as m → +∞. Due to this, however, they also get close to each other; in fact, ρ (x , x ) can be made arbitrarily small for sufficiently large m and n. It is natural to ask whether the latter property, in turn, implies the existence of a limit. This problem was first studied by Augustin-Louis Cauchy (1789 − 1857). Thus we shall call sequences Cauchy sequences. More precisely, we formulate the following. m
m
n
Definition A sequence {x } ⊆ (S, ρ) is called a Cauchy sequence (we briefly say that " {x } is Cauchy") iff, given any matter how small), we have ρ (x , x ) < ε for all but finitely many m and n. In symbols, m
m
m
ε >0
(no
n
(∀ε > 0)(∃k)(∀m, n > k)
ρ (xm , xn ) < ε.
(3.13.1)
Observe that here we only deal with terms x , x , not with any other point. The limit (if any) is not involved, and we do not have to know it in advance. We shall now study the relationship between property (1) and convergence. m
n
Theorem 3.13.1 Every convergent sequence {x
m}
⊆ (S, ρ)
is Cauchy.
Proof Let x
m
→ p.
Then given ε > 0, there is a k such that ε (∀m > k)
ρ (xm , p)
k, it also holds for any other term x with n > k . n
Thus ε (∀m, n > k)
ρ (xm , p)
k. Then m
m
(1),
there is
k
such that
n
(∀m > k)ρ (xm , xn ) < ε, i.e., xm ∈ Gxn (ε).
(3.13.5)
Thus the globe G (ε) contains all x except possibly the k terms x , … , x . To include them as well, we only have to take a larger radius r, greater than ρ (x , x ) , m = 1, … , k. Then all x are in the enlarged globe G (r). □ xn
m
m
1
n
k
m
3.13.1
xn
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Note 1. In E , under the standard metric, only sequences with finite limits are regarded as convergent. If x → ±∞, then {x } is not even a Cauchy sequence in E ( in view of Theorem 2); but in E , under a suitable metric (cf. Problem 5 in §11, it is convergent (hence also Cauchy and bounded). 1
n
1
n
∗
Theorem 3.13.3 If a Cauchy sequence {x
m}
clusters at a point p, then x
→ p
m
.
Proof We want to show that x
m
→ p,
i.e., that (∀ε > 0)(∃k)(∀m > k)
Thus we fix ε > 0 and look for a suitable k. Now as {x
m}
ρ (xm , p) < ε.
(3.13.6)
is Cauchy, there is a k such that ε
(∀m, n > k)
Also, as
is a cluster point, the globe
p
n > k (k as above). Then ρ (xn , p)
0)(∃δ > 0) (∀x ∈ A ∩ G¬p (δ))
f (x) ∈ Gq (ε).
(4.2.4)
Thus, given ε > 0, there is δ > 0 (henceforth fixed) such that f (x) ∈ Gq (ε) whenever x ∈ A, x ≠ p, and x ∈ Gp (δ).
(4.2.5)
We want to deduce (2'). Thus we fix any sequence { xm } ⊆ A − {p}, xm → p.
(4.2.6)
Then (∀m)
xm ∈ A and xm ≠ p,
and G (δ) contains all but finitely many x . Then these x all but finitely many m. As ε is arbitrary, this implies f (x (2'). Thus (2) ⟹ (2'). p
m
m
(4.2.7)
satisfy the conditions stated in (3). Hence f (x ) ∈ G (ε) for (by the definition of lim f (x )), as is required in m
m) → q
m→∞
q
m
Conversely, suppose (2) fails, i.e., its negation holds. (See the rules for forming negations of such formulas in Chapter 1, §§1-3.) Thus (∃ε > 0)(∀δ > 0) (∃x ∈ A ∩ G¬p (δ))
f (x) ∉ Gq (ε)
(4.2.8)
by the rules for quantifiers. We fix an ε satisfying (4), and let 1 δm =
,
m = 1, 2, …
(4.2.9)
m
By (4), for each δ there is x m
m
(depending on δm )
such that 1
xm ∈ A ∩ G¬p (
)
(4.2.10)
m = 1, 2, 3, …
(4.2.11)
m
and f (xm ) ∉ Gq (ε),
We fix these x
m.
As x
m
∈ A
and x
m
≠ p,
we obtain a sequence
4.2.1
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{ xm } ⊆ A − {p}.
(4.2.12)
Also, as x ∈ G ( ) , we have ρ (x , p) < 1/m → 0, and hence x → p . On the other hand, by (6), the image sequence {f (x )} canverge to q (why?), i.e., (2') fails. Thus we see that (2') fails or holds accordingly as (2) does. 1
m
p
m
m
m
m
This proves assertion (ii). Now, by setting continuity. □
q = f (p)
in (2) and (2'), we also obtain the first clause of the theorem, as to
Note 1. The theorem also applies to relative limits and continuity over a path B (just replace A by B in the proof), as well as to the cases p = ±∞ and q = ±∞ in E (for E can be treated as a metric space; see the end of Chapter 3, §11). ∗
∗
If the range space (T , ρ ) is complete (Chapter 3, §17), then the image sequences leads to the following corollary. ′
{f (xm )}
converge iff they are Cauchy. This
Corollary 1. Let (T , ρ ) be complete, such as E . Let a map f : A → T with A ⊆ (S, ρ) and a point p ∈ S be given. Then for f to have a limit at p, it suffices that {f (x )} be Cauchy in (T , ρ ) whenever {x } ⊆ A − {p} and x → p in (S, ρ). ′
n
′
m
m
m
Indeed, as noted above, all such {f (x )} converge. Thus it only remains to show that they tend to one and the same limit q, as is required in part (ii) of Theorem 1. We leave this as an exercise (Problem 1 below). m
Theorem 4.2.2 (Cauchy criterion for functions). With the assumptions of Corollary 1, the function f has a limit at p iff for each ε > 0, there is δ > 0 such that ′
′
′
ρ (f (x), f (x )) < ε for all x, x ∈ A ∩ G¬p (δ).
(4.2.13)
In symbols, ′
(∀ε > 0)(∃δ > 0) (∀x, x ∈ A ∩ G¬p (δ))
′
′
ρ (f (x), f (x )) < ε.
(4.2.14)
Proof Assume (7). To show that f has a limit at p, we use Corollary 1. Thus we take any sequence { xm } ⊆ A − {p} with xm → p
and show that {f (x
m )}
(4.2.15)
is Cauchy, i.e., ′
(∀ε > 0)(∃k)(∀m, n > k)
ρ (f (xm ) , f (xn )) < ε.
(4.2.16)
To do this, fix an arbitrary ε > 0. By (7), we have ′
′
(∀x, x ∈ A ∩ G¬p (δ))
for some δ > 0. Now as x
m
→ p,
m}
⊆ A − {p},
we even have x
m,
m )}
xm , xn ∈ Gp (δ).
xn ∈ A ∩ G¬p (δ). (∀m, n > k)
i.e., {f (x that limit.
(4.2.17)
there is k such that (∀m, n > k)
As {x
′
ρ (f (x), f (x )) < ε,
(4.2.18)
Hence by (7'),
′
ρ (f (xm ) , f (xn )) < ε;
(4.2.19)
is Cauchy, as required in Corollary 1, and so f has a limit at p . This shows that (7) implies the existence of
The easy converse proof is left to the reader. (See Problem 2.) □ II. Composite Functions. The composite of two functions f : S → T and g : T → U ,
(4.2.20)
denoted g∘ f
(in that order),
4.2.2
(4.2.21)
https://math.libretexts.org/@go/page/19043
is by definition a map of S into U given by (g ∘ f )(x) = g(f (x)),
x ∈ S.
(4.2.22)
Our next theorem states, roughly, that g ∘ f is continuous if g and f are. We shall use Theorem 1 to prove it.
Theorem 4.2.3 Let (S, ρ), (T , ρ ) , and (U , ρ ) be metric spaces. If a function f : S → T is continuous at a point p ∈ S, and if g : T continuous at the point q = f (p), then the composite function g ∘ f is continuous at p. ′
′′
→ U
is
Proof The domain of g ∘ f is S. So take any sequence { xm } ⊆ S with xm → p.
As f is continuous at f (p), we have
formula (1') yields
p,
where
f (xm ) → f (p),
f (xm )
(4.2.23)
is in
T = Dg .
Hence, as
g
is continuous at
g (f (xm )) → g(f (p)), i.e., (g ∘ f ) (xm ) → (g ∘ f )(p),
and this holds for any {x
m}
⊆S
with x
m
(4.2.24)
Thus g ∘ f satisfies condition (1') and is continuous at p. □
→ p.
Caution: The fact that lim f (x) = q and lim g(y) = r x→p
(4.2.25)
y→q
does not imply lim g(f (x)) = r
(4.2.26)
x→p
(see Problem 3 for counterexamples). Indeed, if {x } ⊆ S − {p} and x → p, we obtain, as before, f (x ) → q, but not f (x ) ≠ q. Thus we cannot re-apply formula (2') to obtain g (f (x )) → r since (2') requires that f (x ) ≠ q. The argument still works if g is continuous at q (then (1') applies) or if f (x) never equals q then f (x ) ≠ q . It even suffices that f (x) ≠ q for x in some deleted globe about p( (see §1, Note 4). Hence we obtain the following corollary. m
m
m
m
m
m
m
Corollary 2. With the notation of Theorem 3, suppose f (x) → q\)as\(x → p, and g(y) → r as y → q.
(4.2.27)
g(f (x)) → r as x → p,
(4.2.28)
Then
provided, however, that (i) g is continuous at q, or (ii) f (x) ≠ q for x in some deleted globe about p, or (iii) f is one to one, at least when restricted to some G
.
¬p (δ)
Indeed, (i) and (ii) suffice, as was explained above. Thus assume (iii). Then f can take the value q at most once, say, at some point
As x
0
≠ p,
0
(4.2.29)
= ρ (x0 , p) > 0.
(4.2.30)
let δ
Then x
x0 ∈ G¬p (δ).
′
∉ G¬p (δ ) ,
so f (x) ≠ q on G
¬p
′
(δ ) ,
′
and case (iii) reduces to (ii).
We now show how to apply Corollary 2.
4.2.3
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Note 2. Suppose we know that r = lim g(y) exists.
(4.2.31)
y→q
Using this fact, we often pass to another variable x, setting y = f (x) where f is such that q = lim f (x) for some p. We shall say that the substitution (or "change of variable") y = f (x) is admissible if one of the conditions (i), (ii), or (iii) of Corollary 2 holds. Then by Corollary 2, x→p
lim g(y) = r = lim g(f (x)) y→q
(4.2.32)
x→p
(yielding the second limit).
Example 4.2.1 (A) Let x
1 h(x) = (1 +
)
for |x| ≥ 1.
(4.2.33)
x
Then lim
h(x) = e.
(4.2.34)
x→+∞
For a proof, let n = f (x) = [x] be the integral part of x. Then for x > 1 , n
1 (1 +
)
n+1
1 ≤ h(x) ≤ (1 +
n+1
)
.
( Verify! )
(4.2.35)
n
As x → +∞, n tends to +∞ over integers, and by rules for sequences, n+1
1 lim (1 +
)
1 = lim (1 +
n
n→∞
n→∞
n
1 ) (1 +
n
)
n
1 = 1 ⋅ lim (1 +
n
)
= 1 ⋅ e = e,
(4.2.36)
n
n→∞
with e as in Chapter 3, §15. Similarly one shows that also n
1 lim (1 + n→∞
Thus (8) implies that also lim
x→+∞
h(x) = e
)
= e.
(4.2.37)
n+1
(see Problem 6 below).
Remark. Here we used Corollary 2(ii) with n
1 f (x) = [x], q = +∞, and g(n) = (1 +
)
.
(4.2.38)
n
The substitution n = f (x) is admissible since f (x) = n never equals +∞, its limit, thus satisfying Corollary 2(ii). (B) Quite similarly, one shows that also x
1 lim x→−∞
(1 +
)
= e.
(4.2.39)
x
See Problem 5. (C) In Examples (A) and (B), we now substitute x = 1/z. This is admissible by Corollary 2(ii) since the dependence between and z is one to one. Then 1 z =
+
→ 0
−
as x → +∞, and z → 0
as x → −∞.
x
(4.2.40)
x
Thus (A) and (B) yield
4.2.4
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1/z
lim (1 + z) +
1/z
= lim (1 + z)
= e.
(4.2.41)
−
z→0
z→0
Hence by Corollary 3 of §1, we obtain 1/z
lim(1 + z)
= e.
(4.2.42)
z→0
This page titled 4.2: Some General Theorems on Limits and Continuity is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
4.2.5
https://math.libretexts.org/@go/page/19043
4.2.E: More Problems on Limits and Continuity Exercise 4.2.E. 1 Complete the proof of Corollary 1. ′
[ Hint: Consider {f (xm )} and {f (xm )} , with ′
xm → p and xm → p.
(4.2.E.1)
By Chapter 3, §14, Corollary 4, p is also the limit of ′
′
1
2
x1 , x , x2 , x , … ,
(4.2.E.2)
so, by assumption, ′
f (x1 ) , f (x ) , … converges (to q, say ).
(4.2.E.3)
1
Hence {f (x
m )}
and {f (x
′ m )}
must have the same limit q. (Why?)]
Exercise 4.2.E. ∗2 Complete the converse proof of Theorem 2 (cf. proof of Theorem 1 in Chapter 3, §17).
Exercise 4.2.E. 3 Define f , g : E → E by setting (i) f (x) = 2; g(y) = 3 if y ≠ 2, and g(2) = 0; or (ii) f (x) = 2 if x is rational and f (x) = 2x otherwise; g as in (i). In both cases, show that 1
1
lim f (x) = 2 and lim g(y) = 3 but not lim g(f (x)) = 3. x→1
y→2
(4.2.E.4)
x→1
Exercise 4.2.E. 4 Prove Theorem 3 from "ε, δ " definitions. Also prove (both ways) that if f is relatively continuous on B, and g on f [B], then g ∘ f is relatively continuous on B .
Exercise 4.2.E. 5 Complete the missing details in Examples (A) and (B). [Hint for (B): Verify that −n−1
1 (1 −
)
n =(
n+1
−n−1
) n+1
n+1 =(
n+1
) n
1 = (1 + n
n
1 ) (1 +
)
→ e. ]
(4.2.E.5)
n
Exercise 4.2.E. 6 ⇒ 6.
Given f , g, h : A → E
∗
, A ⊆ (S, ρ),
with f (x) ≤ h(x) ≤ g(x)
4.2.E.1
(4.2.E.6)
https://math.libretexts.org/@go/page/22631
for x ∈ G
¬p (δ)
∩A
for some δ > 0. Prove that if lim f (x) = lim g(x) = q, x→p
(4.2.E.7)
x→p
also then lim h(x) = q.
(4.2.E.8)
x→p
Use Theorem 1. [Hint: Take any { xm } ⊆ A − {p} with xm → p.
Then f (x
m)
→ q, g (xm ) → q,
(4.2.E.9)
and (∀xm ∈ A ∩ G¬p (δ))
f (xm ) ≤ h (xm ) ≤ g (xm ) .
(4.2.E.10)
Now apply Corollary 3 of Chapter 3, §15. ]
Exercise 4.2.E. 7 Given f , g : A → E (i) If q > r, then ⇒ 7.
∗
, A ⊆ (S, ρ),
with f (x) → q and g(x) → r as x → p (p ∈ S), prove the following:
(∃δ > 0) (∀x ∈ A ∩ G¬p (δ))
f (x) > g(x).
(4.2.E.11)
f (x) ≤ g(x),
(4.2.E.12)
(ii) (Passage to the limit in inequalities.) If (∀δ > 0) (∃x ∈ A ∩ G¬p (δ))
then q ≤ r. (Observe that here A clusters at p necessarily, so the limits are unique.) [Hint: Proceed as in Problem 6; use Corollary 1 of Chapter 3, §15. ]
Exercise 4.2.E. 8 Do Problems 6 and 7 using only Definition 2 of §1. [ Hint: Here prove 7( ii ) first. ]
Exercise 4.2.E. 9 Do Examples (a) − (d) of §1 using Theorem 1. [Hint: For (c), use also Example (a) in Chapter 3, §16. ]
Exercise 4.2.E. 10 Addition and multiplication in E may be treated as functions 1
f, g : E
2
→ E
1
(4.2.E.13)
with
4.2.E.2
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f (x, y) = x + y and g(x, y) = xy.
(4.2.E.14)
Show that f and g are continuous on E (see footnote 2 in Chapter 3 §15). Similarly, show that the standard metric 2
ρ(x, y) = |x − y|
(4.2.E.15)
is a continuous mapping from E to E . 2
1
[ Hint: Use Theorems 1, 2, and, 4 of Chapter 3,
§15 and the sequential
criterion. ]
Exercise 4.2.E. 11 Using Corollary 2 and formula (9), find lim
for a fixed m ∈ N .
1/x
x→0
(1 ± mx )
Exercise 4.2.E. 12 ⇒ 12.
Let a > 0 in E
1
.
Prove that lim
x→0
x
a
=1
[ Hint: Let n = f (x) be the integral part of
1 x
.
(x ≠ 0). Verify that
−1/(n+1)
a
x
≤a
1/n
≤a
if a ≥ 1,
(4.2.E.16)
with inequalities reversed if 0 < a < 1. Then proceed as in Example (A), noting that 1/n
lim a
−1/(n+1)
= 1 = lim a
n→∞
(4.2.E.17)
n→∞
by Problem 20 of Chapter 3, §15. ( Explain! )]
Exercise 4.2.E. 13 ⇒ 13.
Given f , g : A → E
∗
, A ⊆ (S, ρ),
with f ≤g
for x in G¬p (δ) ∩ A.
(4.2.E.18)
Prove that (a) if lim f (x) = +∞, then also lim g(x) = +∞ ; (b) if lim g(x) = −∞, then also lim f (x) = −∞ . Do it it two ways: (i) Use definitions only, such as (2 ) in §1 . (ii) Use Problem 10 of Chapter 2, §13 and the sequential criterion. x→p
x→p
x→p
x→p
′
Exercise 4.2.E. 14 Prove that (i) if a > 1 in E , then ⇒ 14.
1
x
−x
a lim x→+∞
a = +∞ and
x
lim x→+∞
= 0;
(4.2.E.19)
= +∞;
(4.2.E.20)
x
(ii) if 0 < a < 1, then x
−x
a lim x→+∞
a = 0 and
x
lim x→+∞
4.2.E.3
x
https://math.libretexts.org/@go/page/22631
(iii) if a > 1 and 0 ≤ q ∈ E
1
,
then x
−x
a lim x→+∞
( iv )
if 0 < a < 1 and 0 ≤ q ∈ E
1
,
q
a = +∞ and
x
lim
= 0;
(4.2.E.21)
= +∞.
(4.2.E.22)
q
x
x→+∞
then x
−x
a lim x→+∞
q
a = 0 and
x
lim
q
x
x→+∞
[Hint: (i) From Problems 17 and 10 of Chapter 3, §15, obtain n
a lim
= +∞.
(4.2.E.23)
n
Then proceed as in Examples (A) − (C); (iii) reduces to (i) by the method used in Problem 18 of Chapter 3, §15. ]
Exercise 4.2.E. 15 For a map f : (S, ρ) → (T , ρ ) , show that the following statements are equivalent: is continuous on S . ′
⇒ ∗15.
(i) f
¯¯¯¯¯¯¯¯ ¯
¯ ¯¯ ¯
(ii) (∀A ⊆ S)f [A] ⊆ f [A] . ¯ ¯¯ ¯
¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ −1
(iii) (∀B ⊆ T )f [B] ⊇ f [B] . (iv) f [B] is closed in (S, ρ) whenever B is closed in (T , ρ ). (v) f [B] is open in (S, ρ) whenever B is open in (T , ρ ). [Hints: (i) ⟹ (ii) : Use Theorem 3 of Chapter } 3, §16 and the sequential criterion to show that −1
−1
′
−1
′
¯¯¯¯¯¯¯¯ ¯
¯ ¯¯ ¯
p ∈ A ⟹ f (p) ∈ f [A].
(ii) ⟹ ( iii ) : Let A = f
−1
[B].
(4.2.E.24)
Then f [A] ⊆ B, so by ( ii ) , ¯ ¯¯ ¯
¯¯¯¯¯¯¯¯ ¯
¯ ¯¯ ¯
f [ A] ⊆ f [A] ⊆ B.
(4.2.E.25)
Hence ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ −1
f
¯ ¯¯ ¯
[B] = A ⊆ f
−1
¯ ¯¯ ¯
[f [ A]] ⊆ f
−1
¯ ¯¯ ¯
[ B].
(Why?)
(4.2.E.26)
¯ ¯¯ ¯
(iii) ⟹ (iv) : If B is closed, B = B (Chapter 3, §16, Theorem 4(ii)), so by (iii), f
−1
[B] = f
−1
¯ ¯¯ ¯
¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯ −1
[ B] ⊇ f
[B]; deduce (iv) .
(4.2.E.27)
Pass to complements in (iv). Assume (v). Take any p ∈ S and use Definition 1 in §1. ]
(iv) ⟹ (v) : (v) ⟹ (i) :
Exercise 4.2.E. 16 Let f
: E
1
→ E
1
be continuous. Define g : E
1
→ E
2
by g(x) = (x, f (x)).
4.2.E.4
(4.2.E.28)
https://math.libretexts.org/@go/page/22631
Prove that (a) g and g are one to one and continuous; (b) the range of g, i.e., the set −1
′
Dg = {(x, f (x))|x ∈ E
1
},
(4.2.E.29)
is closed in E . [Hint: Use Theorem 2 of Chapter 3, §15, Theorem 4 of Chapter 3, §16, and the sequential criterion.] 2
4.2.E: More Problems on Limits and Continuity is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
4.2.E.5
https://math.libretexts.org/@go/page/22631
4.3: Operations on Limits. Rational Functions I. A function f : A → T is said to be real if its range D lies in E , complex if D ⊆ C , vector valued if D is a subset of E , and scalar valued if D lies in the scalar field of E . ( 'ln the latter two cases, we use the same terminology if E is replaced by some other (fixed) normed space under consideration.) The domain A may be arbitrary. ′
1
f
′
′
′
f
f
n
n
n
f
For such functions one can define various operations whenever they are defined for elements of their ranges, to which the function values f (x) belong. Thus as in Chapter 3, §9, we define the functions f ± g, f g, and f /g "pointwise," setting f (f ± g)(x) = f (x) ± g(x),
(f g)(x) = f (x)g(x), and (
f (x) ) (x) =
g
(4.3.1) g(x)
whenever the right side expressions are defined. We also define |f | : A → E by 1
(∀x ∈ A)
|f |(x) = |f (x)|.
(4.3.2)
In particular, f ± g is defined if f and g are both vector valued or both scalar valued, and f g is defined if f is vector valued while g is scalar valued; similarly for f /g. (However, the domain of f /g consists of those x ∈ A only for which g(x) ≠ 0. ) In the theorems below, all limits are at some (arbitrary, but fixed) point
p
of the domain space (S, ρ). For brevity, we often omit
" x → p. "
Theorem 4.3.1 For any functions f , g, h : A → E
1
(C ), A ⊆ (S, ρ),
we have the following:
i. (i) If f , g, h are continuous at p(p ∈ A), so are f ± g and fh. So also is f /h, provided h(p) ≠ 0; similarly for relative continuity over B ⊆ A . ii. (ii) If f (x) → q, g(x) → r, and h(x) → a( all, as x → p over B ⊆ A), then a. b. c.
f (x) ± g(x) → q ± r f (x)h(x) → qa; f (x) h(x)
→
q a
,
and
provided a ≠ 0
All this holds also if f and g are vector valued and h is scalar valued. For a simple proof, one can use Theorem 1 of Chapter 3, §15. (An independent proof is sketched in Problems 1-7 below.) We can also use the sequential criterion (Theorem 1 in §2). To prove (ii), take any sequence { xm } ⊆ B − {p}, xm → p
(4.3.3)
f (xm ) → q, g (xm ) → r, and h (xm ) → a
(4.3.4)
Then by the assumptions made,
Thus by Theorem 1 of Chapter 3, §15, f (xm ) ± g (xm ) → q ± r, f (xm ) g (xm ) → qa, and
f (xm )
q →
g (xm )
As this holds for any sequence {x (i).
m}
⊆ B − {p}
with x
m
→ p,
(4.3.5) a
our assertion (ii) follows by the sequential criterion; similarly for
Note 1. By induction, the theorem also holds for sums and products of any finite number of functions (whenever such products are defined). Note 2. Part (ii) does not apply to infinite limits q, r, a; but it does apply to limits at p = ±∞ (take E with a suitable metric for the space S) . ∗
Note 3. The assumption h(x) → a ≠ 0( as x → p over B) implies that h(x) ≠ 0 for Problem 5 below. Thus the quotient function f /h is defined on B ∩ G (δ) at least.
x
in
B ∩ G¬p (δ)
for some
δ > 0;
see
¬p
4.3.1
https://math.libretexts.org/@go/page/19044
II. If the range space of f is E complex) components, denoted
n
∗
( or C
n
),
then each function value
fk (x),
Here we may treat f as a mapping of A = D into E f (x). In this manner, each function k
1
f
f (x)
is a vector in that space; thus
n
real (* respectively,
k = 1, 2, … , n.
(∗ or C );
f : A → E
(4.3.6)
it carries each point n
∗
( C
n
x ∈ A
into f
k (x),
the
k
th component of
)
(4.3.7)
uniquely determines n scalar-valued maps fk : A → E
called the components of f . Notation: f
= (f1 , … , fn )
1
(C )
(4.3.8)
.
Conversely, given n arbitrary functions fk : A → E
one can define f
: A → E
n
∗
( C
n
)
1
(C ),
k = 1, 2, … , n,
(4.3.9)
by setting f (x) = (f1 (x), f2 (x), … , fn (x)) .
(4.3.10)
Then obviously f = (f , f , … , f ) . Thus the f in turn determine f uniquely. To define a function f give its n components f . Note that 1
2
n
k
: A → E
n
∗
( C
n
)
means to
k
n
n
¯ ¯ f (x) = (f1 (x), … , fn (x)) = ∑ ¯ e k fk (x),
¯ ¯ i.e., f = ∑ ¯ e k fk
k=1
(4.3.11)
k=1
where the e are the n basic unit vectors; see Chapter 3, §1-3, Theorem 2. Our next theorem shows that the limits and continuity of f reduce to those of the f . ¯ ¯ ¯ k
k
Theorem 4.3.2 (componentwise continuity and limits). For any function we have that
f : A → E
n
(∗ C
n
),
with
A ⊆ (S, ρ)
and with
f = (f1 , … , fn ) ,
(i) f is continuous at p(p ∈ A) iff all its components f are, and k
(ii) f (x) → q as x → p(p ∈ S) iff ¯¯
fk (x) → qk as x → p
(k = 1, 2, … , n),
(4.3.12)
i.e., iff each f has, as its limit at p, the corresponding component of q . ¯¯
k
Similar results hold for relative continuity and limits over a path B ⊆ A . We prove (ii). If f (x) → q as x → p then, by definition, ¯¯
− −−−−−−−−−−− − n
(∀ε > 0)(∃δ > 0) (∀x ∈ A ∩ G¬p (δ))
ε > |f (x) − ¯ q¯| = √∑ | fk (x) − qk |
2
;
(4.3.13)
k=1
in turn, the right-hand side of the inequality given above is no less than each | fk (x) − qk | ,
k = 1, 2, … , n.
(4.3.14)
Thus (∀ε > 0)(∃δ > 0) (∀x ∈ A ∩ G¬p (δ))
i.e., f
k (x)
| fk (x) − qk | < ε;
(4.3.15)
→ qk , k = 1, … , n.
Conversely, if each f
k (x)
→ qk ,
then Theorem 1 (ii) yields
4.3.2
https://math.libretexts.org/@go/page/19044
n
n ¯ ¯ ¯
¯ ¯ ¯
∑ ek fk (x) → ∑ ek qk . k=1
By formula continuity.
(1),
then,
¯¯
f (x) → q
(for
n
¯ ¯ ¯ ek qk
∑
k=1
(4.3.16)
k=1
Thus (ii) is proved; similarly for (i) and for relative limits and
¯¯
= q ).
Note 4. Again, Theorem 2 holds also for p = ±∞ (but not for infinite q). Note 5. A complex function f : A → C may be treated as f : A → E . Thus it has two real components: Traditionally, f and f are called the real and imaginary parts of f , also denoted by f and f , so 2
1
2
re
f = (f1 , f2 ) .
im
f = fre + i ⋅ fim .
By Theorem 2, f is continuous at p iff f
re
and f
im
(4.3.17)
are.
Example 4.3.1 The complex exponential is the function f
: E
1
defined by
→ C
f (x) = cos x + i ⋅ sin x, also written f (x) = e
xi
.
(4.3.18)
As we shall see later, the sine and the cosine functions are continuous. Hence so is f by Theorem 2. III. Next, consider functions whose domain is a set in E ( or C ) . We call them functions of (∗ or complex) variables, treating x = (x , … , x ) as a variable n -tuple. The range space may be arbitrary. n
¯¯ ¯
1
∗
n
n
real
n
In particular, a monomial in n variables is a map on E
n
∗
( or C
n
)
given by a formula of the form n
m1
¯ f (¯¯ x ) = ax
1
m2
x
2
mn
⋯ xn
mk
= a⋅∏x
k
,
(4.3.19)
k=1
where the m are fixed integers ≥ 0 and a ∈ E ( or a ∈ C ) . If a ≠ 0 , the sum m = ∑ m is called the degree of the monomial. Thus 1
∗
2
k
n
k=1
k
2
f (x, y, z) = 3 x y z
3
2
1
= 3x y z
3
(4.3.20)
defines a monomial of degree 6, in three real (or complex) variables x, y, z. (We often write x, y, z for x
1,
x2 , x3 . )
A polynomial is any sum of a finite number of monomials; its degree is, by definition, that of its leading term, i.e., the one of highest degree. (There may be several such terms, of equal degree.) For example, 2
f (x, y, z) = 3 x y z
3
− 2x y
7
(4.3.21)
defines a polynomial of degree 8 in x, y, z. Polynomials of degree 1 are sometimes called linear. A rational function is the quotient f /g of two polynomials f and g on E does not vanish. For example, 2
x
n
∗
( orC
n
. Its domain consists of those points at which g
)
− 3xy
h(x, y) =
(4.3.22) xy − 1
defines a rational function on points (x, y), with xy ≠ 1. Polynomials and monomials are rational functions with denominator 1.
Theorem 4.3.1 Any rational function (in particular, every polynomial) in one or several variables is continuous on all of its domain. Proof Consider first a monomial of the form ¯ f (¯¯ x ) = xk
(k fixed );
it is called the k th projection map because it "projects" each x ∈ E ¯¯ ¯
4.3.3
n
∗
( C
n
(4.3.23)
)
onto its k th component x . k
https://math.libretexts.org/@go/page/19044
Given any ε > 0 and p, choose δ = ε. Then ¯ ¯ ¯
− −−−−−−−− − n
¯ (∀¯¯ x ∈ G¯p ¯ ¯ (δ))
¯ ¯ ¯ |f (¯¯ x ) − f (¯ p )| = | xk − pk | ≤ √∑ | xi − pi |
2
¯ ¯ ¯ ¯ = ρ(¯¯ x ,p ) < ε.
(4.3.24)
i=1
Hence by definition, f is continuous at each p. Thus the theorem holds for projection maps. ¯ ¯ ¯
However, any other monomial, given by m1
¯¯ ¯
f (x) = ax
1
m2
x
2
mn
⋯ xn
,
(4.3.25)
is the product of finitely many (namely of m = m + m + … + m ) projection maps multiplied by a constant a . Thus by Theorem 1, it is continuous. So also is any finite sum of monomials (i.e., any polynomial), and hence so is the quotient f /g of two polynomials (i.e., any rational function) wherever it is defined, i.e., wherever the denominator does not vanish. 1
2
n
□
IV. For functions on E
n
∗
( or C
n
),
we often consider relative limits over a line of the form ¯¯ ¯
th
¯ ¯ ¯
⃗ (parallel to the k x = p + te k
¯ ¯ ¯
axis, through p);
(4.3.26)
see Chapter 3, §§4-6, Definition 1. If f is relatively continuous at p over that line, we say that f is continuous at p in the k th variable x (because the other components of x remain constant, namely, equal to those of p, as x runs over that line). As opposed to this, we say that f is continuous at p in all n variables jointly if it is continuous at p in the ordinary (not relative) sense. Similarly, we speak of limits in one variable, or in all of them jointly. ¯ ¯ ¯
¯¯ ¯
k
¯ ¯ ¯
¯ ¯ ¯
¯ ¯ ¯
¯¯ ¯
¯ ¯ ¯
since ordinary continuity implies relative continuity over any path, joint continuity in all n variables always implies that in each variable separately, but the converse fails (see Problems 9 and 10 below ); similarly for limits at p. ¯ ¯ ¯
This page titled 4.3: Operations on Limits. Rational Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
4.3.4
https://math.libretexts.org/@go/page/19044
4.3.E: Problems on Continuity of Vector-Valued Functions Exercise 4.3.E. 1 Give an "ε, δ " proof of Theorem 1 for f ± g . [Hint: Proveed as in Theorem 1 of Chapter 3, §15, replacing max (k , k f (x) → q and g(x) → r as x → p over B , then (∃δ , δ > 0) such that ′
′
′
,δ
′′
),
)
by δ = min (δ
ε
′
(∀x ∈ B ∩ G¬p (δ ))
Put δ = min (δ
′′
′
,δ
′′
)
. Thus fix ε > 0 and p ∈ S. If
′′
|f (x) − q| < 2
and (∀x ∈ B ∩ G¬p (δ
′′
ε ))
|g(x) − r|
0, f is bounded on B ∩ G
¬p (δ),
i.e.,
′
f [B ∩ G¬p (δ)] is a bounded set in (T , ρ ) .
Thus if T
= E,
(4.3.E.4)
there is K ∈ E such that 1
(∀x ∈ B ∩ G¬p (δ))
|f (x)| < K
(4.3.E.5)
(Chapter 3, §13, Theorem 2) .
Exercise 4.3.E. 4 Given f , h : A → E (C ) (or f : A → E, h : A → F ), prove that if one of other is bounded on B ∩ G (δ), then h(x)f (x) → 0(0). 1
f
and
h
has limit 0 (respectively,
¯ ¯ ¯
0),
while the
¯ ¯ ¯
¬p
Exercise 4.3.E. 5 Given h : A → E Prove that
1
(C ),
with h(x) → a as x → p over B, and a ≠ 0 .
(∃ε, δ > 0) (∀x ∈ B ∩ G¬p (δ))
i.e., h(x) is bounded away from 0 on B ∩ G
¬p (δ).
|h(x)| ≥ ε,
(4.3.E.6)
Hence show that 1/h is bounded on B ∩ G
¬p (δ).
§
[ Hint: Proceed as in the proof of Corollary 1 in 1, with q = a and r = 0. Then use
4.3.E.1
https://math.libretexts.org/@go/page/22633
∣ ∣
(∀x ∈ B ∩ G¬p (δ))
1
∣ 1 ∣ ≤ .]
∣ h(x) ∣
(4.3.E.7)
ε
Exercise 4.3.E. 6 Using Problems 1 to 5, give an independent proof of Theorem 1. [Hint: Proceed as in Problems 2 and 4 of Chapter 3, §15 to obtain Theorem 1(ii). Then use Corollary 2 of $1. ]
Exercise 4.3.E. 7 Deduce Theorems 1 and 2 of Chapter 3, §15 from those of the present section, setting A = B = N , S = E [Hint: See §1, Note 5.]
∗
,
and p = +∞ .
Exercise 4.3.E. 8 Redo Problem 8 of §1 in two ways: (i) Use Theorem 1 only. (ii) Use Theorem 3. . where h(x) = 1 (constant) and g(x) = x (identity map). As so is f by Theorem 1. Thus lim f (x) = f (1) = 1 + 1 = 2 . 2
[ Example for (i) : Find limx→1 (x
Here
2
f (x) = x
or
+ 1,
+ 1)
f = gg + h,
§
( 1, Examples ( a ) and (b)),
h
and
g
are continuous
2
x→1
Or, using Theorem 1( ii) , lim
2
x→1
(x
2
+ 1) = limx→1 x
+ limx→1 1, etc. ]
Exercise 4.3.E. 9 Define f
: E
2
→ E
by
1
2
x y f (x, y) =
(x4 + y 2 )
, with f (0, 0) = 0.
¯ ¯ ¯
(4.3.E.8)
Show that f (x, y) → 0 as (x, y) → (0, 0) along any straight line through 0, but not over the parabola y = x (then the limit is ). Deduce that f is continuous at 0 = (0, 0) in x and y separately, but not jointly. 2
¯ ¯ ¯
1 2
Exercise 4.3.E. 10 Do Problem 9, setting |y| f (x, y) = 0 if x = 0, and f (x, y) =
2
2
−|y|/x
⋅2
if x ≠ 0.
(4.3.E.9)
x
Exercise 4.3.E. 11 Discuss the continuity of f at 0, when
: E
2
→ E
1
in x and y jointly and separately,
¯ ¯ ¯
(a) f (x, y) =
2
x y 2
2
x +y
2
, f (0, 0) = 0
;
(b) f (x, y) = integral part of x + y ; (c) f (x, y) = x + if x ≠ 0, f (0, y) = 0 ; xy
|x|
(d) f (x, y) =
xy |x|
+ x sin
1 y
if xy ≠ 0, and f (x, y) = 0 otherwise;
(e) f (x, y) = sin(x + |xy|) if x ≠ 0, and f (0, y) = 0. [Hints: In (c) and (d), |f (x, y)| ≤ |x| + |y|; in (e), use | sin α| ≤ |α|⋅] . 1
2
x
4.3.E.2
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4.3.E.3
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4.4: Infinite Limits. Operations in E* As we have noted, Theorem 1 of §3 does not apply to infinite limits, even if the function values f (x), g(x), h(x) remain finite (i.e., inE ). Only in certain cases (stated below) can we prove some analogues. 1
There are quite a few such separate cases. Thus, for brevity, we shall adopt a kind of mathematical shorthand. The letter q will not necessarily denote a constant; it will stand for "a function f : A → E
1
, A ⊆ (S, ρ), such that f (x) → q ∈ E
1
as x → p. "
(4.4.1)
Similarly, "0" and "±∞ " will stand for analogous expressions, with q replaced by 0 and ±∞, respectively. For example, the "shorthand formula" (+∞) + (+∞) = +∞ means "The sum of two real functions, with limit + ∞ at p (p ∈ S), is itself a function with limit + ∞ at p. "
The point p is fixed, possibly ±∞ (if A ⊆ E
∗
).
(4.4.2)
With this notation, we have the following theorems.
Theorems 1. (±∞) + (±∞) = ±∞ . 2. (±∞) + q = q + (±∞) = ±∞ . 3. (±∞) ⋅ (±∞) = +∞ . 4. (±∞) ⋅ (∓∞) = −∞ . 5. | ± ∞| = +∞ . 6. (±∞) ⋅ q = q ⋅ (±∞) = ±∞ if q > 0 . 7. (±∞) ⋅ q = q ⋅ (±∞) = ∓∞ if q < 0 . 8. −(±∞) = ∓∞ . 9.
(±∞)
10.
= (±∞) ⋅
q q
(±∞)
=0
11. (+∞)
+∞
12. (+∞)
−∞
13. (+∞)
q
14. (+∞)
q
if q ≠ 0 .
1 q
. = +∞ =0
.
= +∞ =0
.
if q > 0 .
if q < 0 .
15. If q > 1, then q
+∞
= +∞
16. If 0 < q < 1, then q
+∞
and q
=0
−∞
and q
=0
−∞
.
= +∞
.
Proof We prove Theorems 1 and 2, leaving the rest as problems. (Theorems 11-16 are best postponed until the theory of logarithms is developed.) 1. Let f (x) and g(x) → +∞ as x → p. We have to show that f (x) + g(x) → +∞,
(4.4.3)
i.e., that (∀b ∈ E
1
) (∃δ > 0) (∀x ∈ A ∩ G¬p (δ))
f (x) + g(x) > b
(we may assume b > 0). Thus fix b > 0. As f (x) and g(x) → +∞, there are δ ′
′
,δ
(∀x ∈ A ∩ G¬p (δ )) f (x) > b and (∀x ∈ A ∩ G¬p (δ
4.4.1
′′
′′
>0
(4.4.4)
such that
)) g(x) > b.
(4.4.5)
https://math.libretexts.org/@go/page/19045
Let δ = min (δ
′
,δ
′′
).
Then (∀x ∈ A ∩ G¬p (δ))
f (x) + g(x) > b + b > b,
(4.4.6)
as required; similarly for the case of −∞ . 2. Let f (x) → +∞ and g(x) → q ∈ E g(x) > q − 1 . Also, given any b ∈ E
1
,
1
Then there is
.
δ
′
′
,δ
′′
).
such that for
))
f (x) > b − q + 1.
x
in
′
A ∩ G¬p (δ ) , |q − g(x)| < 1,
so that
there is δ such that ′′
(∀x ∈ A ∩ G−p (δ
Let δ = min (δ
>0
′′
(4.4.7)
Then (∀x ∈ A ∩ G¬p (δ))
f (x) + g(x) > (b − q + 1) + (q − 1) = b,
(4.4.8)
as required; similarly for the case of f (x) → −∞ . Caution: No theorems of this kind exist for the following cases (which therefore are called indeterminate expressions): ±∞ (+∞) + (−∞),
(±∞) ⋅ 0,
0 ,
,
±∞
0
0
(±∞) ,
0 ,
±∞
1
.
(4.4.9)
0
In these cases, it does not suffice to know only the limits of f and g. It is necessary to investigate the functions themselves to give a definite answer, since in each case the answer may be different, depending on the properties of f and g. The expressions (1*) remain indeterminate even if we consider the simplest kind of functions, namely sequences, as we show next.
Examples (a) Let um = 2m and vm = −m.
(4.4.10)
(This corresponds to f (x) = 2x and g(x) = −x. ) Then, as is readily seen, um → +∞, vm → −∞, and um + vm = 2m − m = m → +∞.
If, however, we take x
m
= 2m
and y
m
= −2m,
(4.4.11)
then xm + ym = 2m − 2m = 0;
thus x
m
+ ym
(4.4.12)
is constant, with limit 0 (for the limit of a constant function equals its value; see §1, Example (a)).
Next, let m
um = 2m and zm = −2m + (−1 )
.
(4.4.13)
Then again m
um → +∞ and zm → −∞, but um + zm = (−1 ) um + zm
;
(4.4.14)
"oscillates" from −1 to 1 as m → +∞, so it has no limit at all.
These examples show that (+∞) + (−∞) is indeed an indeterminate expression since the answer depends on the nature of the functions involved. No general answer is possible. (b) We now show that 1
+∞
Take first a constant {x
m}
is indeterminate. , xm = 1,
and let y
m
= m.
Then ym
xm → 1, ym → +∞, and xm
If, however, x = 1 + Chapter 3, §15), but m
1 m
and
ym = m,
then again
ym → +∞
4.4.2
m
=1
and
= 1 = xm → 1.
xm → 1
(4.4.15)
(by Theorem 10 above and Theorem 1 of
https://math.libretexts.org/@go/page/19045
ym
xm
m
1 = (1 +
)
(4.4.16)
m
does not tend to 1; it tends to e > 2, as shown in Chapter 3, §15. Thus again the result depends on {x
m}
and {y
m}
.
In a similar manner, one shows that the other cases (1*) are indeterminate. Note 1. It is often useful to introduce additional "shorthand" conventions. Thus the symbol functionf such that
∞
(unsigned infinity) might denote a
|f (x)| → +∞ as x → p;
we then also write f (x) → ∞. The symbol 0 (respectively, 0 +
−
)
(4.4.17)
denotes a function f such that
f (x) → 0 as x → p
(4.4.18)
f (x) > 0 (f (x) < 0, respectively) on some G¬p (δ).
(4.4.19)
and, moreover
We then have the following additional formulas: (i)
(±∞) +
0
(±∞)
= ±∞,
(ii) If q > 0, then (iii) (iv)
∞ 0 q ∞
=∞ =0
−
0 q
+
= ∓∞ = +∞
0
.
and
q −
0
= −∞
.
.
.
The proof is left to the reader. Note 2. All these formulas and theorems hold for relative limits, too. So far, we have defined no arithmetic operations in E . To fill this gap (at least partially), we shall henceforth treat Theorems 1-16 above not only as certain limit statements (in "shorthand") but also as definitions of certain operations in E . For example, the formula (+∞) + (+∞) = +∞ shall be treated as the definition of the actual sum of +∞ and +∞ in E , with +∞ regarded this time as an element of E (not as a function). This convention defines the arithmetic operations for certain cases only; the indeterminate expressions (1*) remain undefined, unless we decide to assign them some meaning. ∗
∗
∗
∗
In higher analysis, it indeed proves convenient to assign a meaning to at least some of them. We shall adopt these (admittedly arbitrary) conventions: 0
(±∞) + (∓∞) = (±∞) − (±∞) = +∞; 0
= 1;
{ 0 ⋅ (±∞) = (±∞) ⋅ 0 = 0 (even if 0 stands for the zero-vector ).
Caution: These formulas must not be treated as limit theorems (in "short-hand"). Sums and products of the form (2*) will be called "unorthodox." This page titled 4.4: Infinite Limits. Operations in E* is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
4.4.3
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4.4.E: Problems on Limits and Operations in E ∗ E∗ Exercise 4.4.E. 1 Show by examples that all expressions (1
∗
)
are indeterminate.
Exercise 4.4.E. 2 Give explicit definitions for the following "unsigned infinity" limit statements: (a) lim f (x) = ∞;
(b) lim f (x) = ∞;
x→p
x→p
(c) lim f (x) = ∞.
+
(4.4.E.1)
x→∞
Exercise 4.4.E. 3 Prove at least some of Theorems 1 − 10 and formulas (i) − (iv) in Note 1.
Exercise 4.4.E. 4 In the following cases, find accordingly.
lim f (x)
in two ways: (i) use definitions only; (ii) use suitable theorems and justify each step
(a) limx→∞
x(x−1)
1 x
(= 0).
(b) limx→∞
2
(c) limx→2+
x −2x+1 x2 −3x+2
2
1−3x
2
x −2x+1
(d) limx→2−
x2 −3x+2
(4.4.E.2)
2
(e) limx→2
x −2x+1 x2 −3x+2
(= ∞)
[Hint: Before using theorems, reduce by a suitable power of x.]
Exercise 4.4.E. 5 Let n
m k
k
f (x) = ∑ ak x and g(x) = ∑ bk x k=0
Find lim
x→∞
f (x) g(x)
(an ≠ 0, bm ≠ 0) .
(4.4.E.3)
k=0
if (i)n > m; ( ii )n < m; and (iii) n = m(n, m ∈ N ) .
Exercise 4.4.E. 6 Verify commutativity and associativity of addition and multiplication in E , treating Theorems 1 − 16 and formulas (2 ) as definitions. Show by examples that associativity and commutativity (for three terms or more) would fail if, instead of (2 ) , the formula (±∞) + (∓∞) = 0 were adopted. [Hint: For sums, first suppose that one of the terms in a sum is +∞; then the sum is + ∞. For products, single out the case where one of the factors is 0; then consider the infinite cases.] ∗
∗
∗
Exercise 4.4.E. 7 Continuing Problem 6, verify the distributive law (x + y)z = xz + yz in E , assuming that x and y have the same sign (if infinite), or that z ≥ 0 . Show by examples that it may fail in other cases; e.g., if x = −y = +∞, z = −1. ∗
4.4.E.1
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4.4.E: Problems on Limits and Operations in E is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts. ∗
4.4.E.2
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4.5: Monotone Function A function f
: A → E
∗
,
with A ⊆ E
∗
,
is said to be nondecreasing on a set B ⊆ A iff x ≤ y implies f (x) ≤ f (y) for x, y ∈ B.
(4.5.1)
x ≤ y implies f (x) ≥ f (y) for x, y ∈ B.
(4.5.2)
It is said to be nonincreasing on B iff
Notation: f
↑
and f
↓ ( on B),
respectively.
In both cases, f is said to be monotone or monotonic on B. If f is also one to one on B (i.e., when restricted to B ), we say that it is strictly monotone (increasing if f ↑ and decreasing if f ↓). Clearly, f is nondecreasing iff the function −f = (−1)f is nonincreasing. Thus in proofs, we need consider only the case f ↑. The case f ↓ reduces to it by applying the result to −f .
Theorem 4.5.1 If a function f
: A → E
In particular, if f
↑
∗
(A ⊆ E
∗
)
is monotone on A, it has a left and a right (possibly infinite) limit at each point p ∈ E . ∗
on an interval (a, b) ≠ ∅, then −
f (p
) =
sup
f (x) for p ∈ (a, b]
(4.5.3)
f (x) for p ∈ [a, b).
(4.5.4)
a 0 such that (∀k) (∃mk > k)
ρ (xmk , p) ≥ ε,
(4.8.9)
where we write “m ” for “m” to stress that the m may be different for different k . Thus by (1), we fix some m for each k so that (1) holds, choosing step by step, k
k
k
mk+1 > mk ,
k = 1, 2, …
(4.8.10)
Then the x form a subsequence of {x }, and the corresponding y = f (x ) form a subsequence of {y }. Henceforth, for brevity, let {x } and {y } themselves denote these two subsequences. Then as before, x ∈ B, y = f (x ) ∈ f [B] , and y → q, q = f (p) . Also,by(1), mk
m
m
m
m
m
mk
mk
m
m
m
4.8.2
https://math.libretexts.org/@go/page/20965
(∀m)
Now as {x
m}
⊆B
and B is compact, {x
m}
ρ (xm , p) ≥ ε (xm stands for xmk ) .
(4.8.11)
has a (sub)subsequence ′
′
xmi → p for some p ∈ B.
(4.8.12)
As f is relatively continuous on B, this implies ′
f (xmi ) = ymi → f (p )
However, the subsequence {y } must have the same limit as {y is one to one on B), so x → p = p .
(4.8.13)
, i.e., f (p). Thus f (p ) = f (p) whence p = p (for ′
m}
mi
′
f
′
mi
This contradicts (2), however, and thus the proof is complete. □
Example 4.8.2 (3) For a fixed n ∈ N , define f
: [0, +∞) → E
1
by n
f (x) = x .
(4.8.14)
Then f is one to one (strictly increasing) and continuous (being a monomial; see §3). Thus by Theorem 3, function) is relatively continuous on each interval n
f
n
f = [ a , b ].
−1
(the nth root
(4.8.15)
hence on [0, +∞). See also Example (a) in §6 and Problem 1 below. II. Uniform Continuity. If f is relatively continuous on B , then by definition, ′
(∀ε > 0)(∀p ∈ B)(∃δ > 0) (∀x ∈ B ∩ Gp (δ))
ρ (f (x), f (p)) < ε.
(4.8.16)
Here, in general, δ depends on both ϵ and p (see Problem 4 in §1); that is, given ϵ > 0 , some values of δ may fit a given p but fail (3) for other points. It may occur, however, that one and the same δ (depending on ϵ only) satisfies (3) for all p ∈ B simultaneously, so that we have the stronger formula (∀ε > 0)(∃δ > 0)(∀p, x ∈ B|ρ(x, p) < δ)
′
ρ (f (x), f (p)) < ε.
(4.8.17)
Definition If (4) is true, we say that f is uniformly continuous on B . Clearly, this implies (3), but the converse fails.
Theorem 4.8.4 If a function f : A → (T , ρ ) , A ⊆ (S, ρ) , is relatively continuous on a compact set continuous on B . ′
B ⊂A
, then
f
is also uniformly
Proof (by contradiction). Suppose f is relatively continuous on B, but (4) fails. Then there is an ϵ > 0 such that (∀δ > 0)(∃p, x ∈ B)
′
ρ(x, p) < δ, and yet ρ (f (x), f (p)) ≥ ε;
(4.8.18)
here p and x on δ . We fix such an ϵ and let 1 δ = 1,
1 ,…,
2
4.8.3
,…
(4.8.19)
m
https://math.libretexts.org/@go/page/20965
Then for each δ (i.e., each m), we get two points x
m,
pm ∈ B
with 1
ρ (xm , pm )
0 , we simply take δ = ε . Then ∀x, p ∈ A ′
ρ(x, p) < δ implies ρ (f (x), f (p)) ≤ ρ(x, p) < δ = ε,
(4.8.25)
as required in (3). (b) As a special case, consider the absolute value map (norm map) given by ¯ ¯ f (¯¯ x ) = |¯¯ x | on E
n
∗
( or another normed space ) .
(4.8.26)
It is uniformly continuous onE because n
′
¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ||¯¯ x | − |¯ p || ≤ |¯¯ x −¯ p |, i.e., ρ (f (¯¯ x ), f (¯ p )) ≤ ρ(¯¯ x ,p ),
(4.8.27)
which shows that f is a contraction map, so Example (a) applies. (c) Other examples of contraction maps are (1) constant maps (see §1, Example (a)) and (2) projection maps (see the proof of Theorem 3 in §3). Verify! (d) Define f
: E
1
→ E
1
by f (x) = sin x
By elementary trigonometry, | sin x| ≤ |x|. Thus (∀x, p ∈ E
1
(4.8.28)
)
|f (x) − f (p)| = | sin x − sin p| 1 1 ∣ ∣ = 2 ∣sin (x − p) ⋅ cos (x + p)∣ ∣ ∣ 2 2 ,
1 ∣ ∣ ≤ 2 ∣sin (x − p)∣ ∣ ∣ 2 1 ≤2⋅
|x − p| = |x − p| 2
and f is a contraction map again. Hence the sine function is uniformly continuous on E ; similarly for the cosine function. 1
(e) Given ∅ ≠ A ⊆ (S, ρ), define f
: S → E
1
by
4.8.4
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f (x) = ρ(x, A) where ρ(x, A) = inf ρ(x, y)
(4.8.29)
y∈A
It is easy to show that (∀x, p ∈ S)
ρ(x, A) ≤ ρ(x, p) + ρ(p, A)
(4.8.30)
i.e., f (x) ≤ ρ(p, x) + f (p), or f (x) − f (p) ≤ ρ(p, x)
(4.8.31)
Similarly, f (p) − f (x) ≤ ρ(p, x). Thus |f (x) − f (p)| ≤ ρ(p, x)
(4.8.32)
i.e., f is uniformly continuous (being a contraction map). (f) The identity map f
: (S, ρ) → (S, ρ),
given by f (x) = x
(4.8.33)
ρ(f (x), f (p)) = ρ(x, p) (a contraction map!)
(4.8.34)
is uniformly continuous on S since
However, even relative continuity could fail if the metric in the domain space the range space (e.g., make ρ discrete!)
S
were not the same as in
S
when regarded as
′
(g) Define f
: E
1
→ E
1
by f (x) = a + bx
(b ≠ 0).
(4.8.35)
|f (x) − f (p)| = |b||x − p|;
(4.8.36)
Then (∀x, p ∈ E
1
)
i.e., ρ(f (x), f (p)) = |b|ρ(x, p).
(4.8.37)
ρ(x, p) < δ ⟹ ρ(f (x), f (p)) = |b|ρ(x, p) < |b|δ = ε,
(4.8.38)
Thus, given ε > 0, take δ = ε/|b|. Then
proving uniform continuity. (h) Let 1 f (x) =
on B = (0, +∞).
(4.8.39)
x
Then f is continuous on B, but not uniformly so. Indeed, we can prove the negation of (4), i.e. (∃ε > 0)(∀δ > 0)(∃x, p ∈ B)
′
ρ(x, p) < δ and ρ (f (x), f (p)) ≥ ε.
(4.8.40)
Take ε = 1 and anyδ > 0. We look for x, p such that |x − p| < δ and |f (x) − f (p)| ≥ ε,
(4.8.41)
1∣ ∣ 1 ∣ − ∣ ≥ 1, ∣x p∣
(4.8.42)
i.e.,
This is achieved by taking 1 p = min (δ,
p ),x =
2
.
( Verify! )
(4.8.43)
2
4.8.5
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Thus (4) fails on B = (0, +∞), yet it holds on [a, +∞) for any a > 0 . (Verify!) This page titled 4.8: Continuity on Compact Sets. Uniform Continuity is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
4.8.6
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4.8.E: Problems on Uniform Continuity; Continuity on Compact Sets Exercise 4.8.E. 1 Prove that if f is relatively continuous on each compact subset of D, then it is relatively continuous on D. [Hint: Use Theorem 1 of §2 and Problem 7 in §6.]
Exercise 4.8.E. 2 Do Problem 4 in Chapter 3, §17, and thus complete the last details in the proof of Theorem 4.
Exercise 4.8.E. 3 Give an example of a continuous one-to-one map f such that f is not continuous. [Hint: Show that any map is continuous on a discrete space (S, ρ).] −1
Exercise 4.8.E. 4 Give an example of a continuous function f and a compact set D ⊆ [Hint: Let f be constant on E .]
′
(T , ρ )
such that f
−1
[D]
is not compact.
1
Exercise 4.8.E. 5 Complete the missing details in Examples (1) and (2) and (c) − (h) .
Exercise 4.8.E. 6 Show that every polynomial of degree one on E
n
(*or C
n
)
is uniformly continuous.
Exercise 4.8.E. 7 Show that the arcsine function is uniformly continuous on [−1, 1]. [Hint: Use Example (d) and Theorems 3 and 4. ]
Exercise 4.8.E. 8 Prove that if f is uniformly continuous on B, and if {x } ⊆ B is a Cauchy sequence, so is {f (x )} . (Briefly, preserves Cauchy sequences.) Show that this may fail if f is only continuous in the ordinary sense. (See Example (h).) ⇒ 8.
m
m
f
Exercise 4.8.E. 9 Prove that if f : S → T is uniformly continuous on B ⊆ S, and composite function g ∘ f is uniformly continuous on B .
g : T → U
is uniformly continuous on
f [B],
then the
Exercise 4.8.E. 10 Show that the functions f and f in Problem 5 of Chapter 3, §11 are contraction maps, 5 hence uniformly continuous. By Theorem 1, find again that (E , ρ ) is compact. −1
∗
′
4.8.E.1
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Exercise 4.8.E. 11 Let A be the set of all cluster points of A ⊆ (S, ρ). Let f : A → (T , ρ ) be uniformly continuous on complete. (i) Prove that lim f (x) exists at each p ∈ A . (ii) Thus define f (p) = lim f (x) for each p ∈ A − A, and show ′
′
A,
and let
′
(T , ρ )
be
′
x→p
′
x→p
¯ ¯¯ ¯
that f so extended is uniformly continuous on the set A = A ∪ A . (iii) Consider, in particular, the case A = (a, b) ⊆ E , so that ′
1
¯ ¯¯ ¯
′
A = A = [a, b].
[Hint: Take any sequence {x §2 to prove existence of lim
m}
x→p
(4.8.E.1)
As it is Cauchy (why?), so is {f (x )} by Problem 8. Use Corollary 1 in . For uniform continuity, use definitions; in case (iii), use Theorem 4 .] ′
⊆ A, xm → p ∈ A .
m
f (x)
Exercise 4.8.E. 12 Prove that if two functions f , g with values in a normed vector space are uniformly continuous on a set and af for a fixed scalar a. For real functions, prove this also for f ∨ g and f ∧ g defined by
B,
so also are
f ±g
(f ∨ g)(x) = max(f (x), g(x))
(4.8.E.2)
(f ∧ g)(x) = min(f (x), g(x)).
(4.8.E.3)
and
[Hint: After proving the first statements, verify that 1 max(a, b) =
1 (a + b + |b − a|) and min(a, b) =
2
(a + b − |b − a|)
(4.8.E.4)
2
and use Problem 9 and Example (b).]
Exercise 4.8.E. 13 Let f be vector valued and h scalar valued, with both uniformly continuous on B ⊆ (S, ρ). Prove that (i) if f and h are bounded on B , then hf is uniformly continuous on B ; (ii) the function f /h is uniformly continuous on B if f is bounded on B and h is "bounded away" from 0 on B , i.e., (∃δ > 0)(∀x ∈ B)
Give examples to show that without these additional conditions, 14 below).
|h(x)| ≥ δ.
hf
(4.8.E.5)
and f /h may not be uniformly continuous (see Problem
Exercise 4.8.E. 14 In the following cases, show that f is uniformly continuous on B ⊆ E , but only continuous (in the ordinary sense) on D, as indicated, with 0 < a < b < +∞ . (a) f (x) = ; B = [a, +∞); D = (0, 1) . (b) f (x) = x ; B = [a, b]; D = [a, +∞) . 1
1
x2
2
4.8.E.2
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(c) f (x) = sin ; B and D as in (a). (d) f (x) = x cos x; B and D as in (b). 1
x
Exercise 4.8.E. 15 Prove that if f is uniformly continuous on B, it is so on each subset A ⊆ B .
Exercise 4.8.E. 16 For nonvoid sets A, B ⊆ (S, ρ), define ρ(A, B) = inf{ρ(x, y)|x ∈ A, y ∈ B}.
(4.8.E.6)
Prove that if ρ(A, B) > 0 and if f is uniformly continuous on each of A and B, it is so on A ∪ B . Show by an example that this fails if ρ(A, B) = 0, even if A∩B = ∅ (e. g. , take A = [0, 1], B = (1, 2] in E , making f constant on each of A and B) . Note, however, that if A and B are compact, A ∩ B = ∅ implies ρ(A, B) > 0. (Prove it using Problem 13 in §6. ) Thus A ∩ B = ∅ suffices in this case. 1
Exercise 4.8.E. 17 Prove that if f is relatively continuous on each of the disjoint closed sets F1 , F2 , … , Fn ,
(4.8.E.7)
it is relatively continuous on their union n
F = ⋃ Fk ;
(4.8.E.8)
k=1
hence (see Problem 6 of §6) it is uniformly continuous on F if the F are compact. [Hint: Fix any p ∈ F . Then p is in some F , say, p ∈ F . As the F are disjoint, p ∉ F , … , F ; hence p also is no cluster point of any of F , … , F (for they are closed). Deduce that there is a globe G (δ) disjoint from each of F , … , F , so that F ∩ G (δ) = F ∩ G (δ). From this it is easy to show that relative continuity of f on F follows from relative continuity on F . ] k
k
2
1
k
2
p
n
p
2
n
p
1
p
1
Exercise 4.8.E. 18 \Rightarrow 18 .\) Let p Let
¯ ¯ ¯ ¯ ¯ ¯ , p1 , 0
¯ ¯ ¯
… , pm
be fixed points in E
n
∗
(
or in another normed space).
¯ ¯ ¯ ¯ ¯ ¯ f (t) = ¯ p + (t − k) (¯ p −¯ p ) k k+1 k
(4.8.E.9)
whenever k ≤ t ≤ k + 1, t ∈ E , k = 0, 1, … , m − 1 . Show that this defines a uniformly continuous mapping f of the interval [0, m] ⊆ E onto the "polygon" 1
1
m−1
⋃ L[ pk , pk+1 ].
(4.8.E.10)
k=0
In what case is f one to one? Is f uniformly continuous on each L [p , p [Hint: First prove ordinary continuity on [0, m] using −1
k
k+1 ]?
On the entire polygon? Theorem 1 of
§3.
(For
the
points 1, 2, … , m − 1, consider left and right limits.) Then use Theorems 1 − 4. ]
4.8.E.3
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Exercise 4.8.E. 19 Prove the sequential criterion for uniform continuity: A function f : A → T is uniformly continuous on a set B ⊆ A iff for any two (not necessarily convergent) sequences {x } and {y } in B, with ρ (x , y ) → 0, we have ρ (f (x ) , f (y )) → 0 (i.e., f preserves con-current pairs of sequences; see Problem 4 in Chapter 3, §17). m
′
m
m
m
m
m
4.8.E: Problems on Uniform Continuity; Continuity on Compact Sets is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
4.8.E.4
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4.9: The Intermediate Value Property This page is a draft and is under active development.
Definition: intermediate value property A function f : A → E is said to have the intermediate value property, or Darboux property, on a set B ⊆ A iff, together with any two function values f (p) and f (p ) (p, p ∈ B) , it also takes all intermediate values between f (p) and f (p ) at some points of B . ∗
1
1
1
1
In other words, the image set f [B] contains the entire interval between f (p) and f (p
1)
Note 1. It follows that f [B] itself is a finite or infinite interval in E
∗
,
in E
∗
.
with endpoints inf f [B] and sup f [B]. (Verify!)
Geometrically, if A ⊆ E , this means that the curve y = f (x) meets all horizontal lines y = q, for q between f (p) and f (p ) . For example, in Figure 13 in §1, we have a "smooth" curve that cuts each horizontal line y = q between f (0) and f (p ) ; so f has the Darboux property on [0, p ] . In Figures 14 and 15, there is a "gap" at p ; the property fails. In Example (f) of §1, the property holds on all of E despite a discontinuity at 0. Thus it does not imply continuity. 1
1
1
1
1
Intuitively, it seems plausible that a "continuous curve" must cut all intermediate horizontals. A precise proof for functions continuous on an interval, was given independently by Bolzano and Weierstrass (the same as in Theorem 2 of Chapter 3, §16). Below we give a more general version of Bolzano's proof based on the notion of a convex set and related concepts.
Definition: Convex ¯¯
¯¯
A set B in E (* or in another normed space) is said to be convex iff for each a, b ∈ B the line segment L[a, b] is a subset of B. n
¯ ¯ ¯
¯ ¯ ¯
¯¯
A polygon joining a and b is any finite union of line segments (a "broken line") of the form ¯ ¯ ¯
m−1 ¯ ¯ ¯
¯ ¯ ¯
¯ ¯ ¯
¯ ¯ ¯
¯ ¯ ¯
¯¯
⋃ L [ pi , pi+1 ] with p0 = a and pm = b.
(4.9.1)
i=0
The set B is said to be polygon connected (or piecewise convex) iff any two points a, b ∈ B can be joined by a polygon contained in B . ¯ ¯ ¯
4.9.1
¯¯
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Example Any globe in E (* or in another normed space) is convex, so also is any interval in E or in E . Figures 19 and 20 represent a convex set A and a polygon-connected set B in E is not convex; it has a "cavity"). n
n
∗
2
We shall need a simple lemma that is noteworthy in its own right as well.
Lemma 4.9.1 (principle of nested line segments) Every contracting sequence of closed line segments intersection; i.e., there is a point
n
(* or in any other normed space) has a nonvoid
¯ ¯ p ∈ ⋂ L[¯ p ,¯ q¯m ]. m
(4.9.2)
¯ ¯ ¯
¯¯
L [ pm , q m ]
in
E
∞ ¯ ¯ ¯
m=1
Proof Use Cantor's theorem (Theorem 5 of §6) and Example (1) in §8. □ We are now ready for Bolzano's theorem. The proof to be used is typical of so-called "bisection proofs." (See also §6, Problems 9 and 10 for such proofs.)
Theorem 4.9.1: Bolzano's theorem If f : B → E is relatively continuous on a polygon-connected set Darboux property on B. 1
in
B
E
n
(* or in another normed space), then
f
has the
In particular, if B is convex and if f (p) < c < f (q ) for some p, q ∈ B, then there is a point r ∈ L(p, q ) such that f (r) = c . ¯ ¯ ¯
¯¯
¯ ¯ ¯
¯¯
¯¯
¯ ¯ ¯
¯¯
¯¯
Proof First, let B be convex. Seeking a contradiction, suppose p, q ∈ B with ¯ ¯ ¯
¯¯
¯ ¯ f (¯ p ) < c < f (¯ q¯),
(4.9.3)
yet f (x) ≠ c for all x ∈ L(p, q ) . ¯¯ ¯
¯¯ ¯
¯ ¯ ¯
¯¯
Let P be the set of all those x ∈ L[p, q ] for which f (x) < c, i.e., ¯¯ ¯
¯ ¯ ¯
¯¯
¯¯ ¯
¯¯ ¯
¯ ¯ ¯
¯¯
¯¯ ¯
P = { x ∈ L[ p, q ]|f (x) < c},
(4.9.4)
¯ ¯ ¯ ¯¯ ¯ Q = {¯¯ x ∈ L[¯ p , q ]|f (¯¯ x ) > c}.
(4.9.5)
and let
Then p ∈ P , q ∈ Q, P ¯ ¯ ¯
¯¯
∩ Q = ∅,
and P
¯ ¯ ¯
¯¯
∪ Q = L[ p, q ] ⊆ B
. (Why?)
Now let ¯¯ r0
1 =
¯ ¯ ¯
¯¯
(p + q )
(4.9.6)
2
be the midpoint on L[p, q ]. Clearly, r is either in P or in Q. Thus it bisects must have its left endpoint in P and its right endpoint in Q. ¯ ¯ ¯
¯¯
¯¯ 0
We denote this particular closed segment by L [p
¯ ¯ ¯ ¯¯ , q1] 1
¯ ¯ ¯
¯¯
, p1 ∈ P , q 1 ∈ Q.
¯ ¯ ¯ ¯ ¯¯ L[¯ p ,¯ q¯1 ] ⊆ L[¯ p , q ] and | p1 − q1 | = 1
Now we bisect L[p
¯ ¯ ¯ ,¯ q¯1 ] 1
1
¯ ¯ ¯¯ L[¯ p , q]
into two subsegments, one of which
We then have
¯ ¯ |¯ p −¯ q¯|. (Verify!)
(4.9.7)
2
and repeat the process. Thus let ¯¯ r1
1 = 2
¯ ¯ (¯ p +¯ q¯1 ). 1
4.9.2
(4.9.8)
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By the same argument, we obtain a closed subsegment L [p
¯ ¯ ¯ ,¯ q¯2 ] 2
¯ ¯ ¯
1
¯¯
| p2 − q 2 | =
2
¯ ¯ ¯
¯ ¯ ⊆ L [¯ p ,¯ q¯1 ] 1
¯¯
| p1 − q 1 | =
1
¯ ¯ ¯
with p
¯ ¯ ¯ 2
∈ P,¯ q¯2 ∈ Q,
and
¯¯
| p − q |.
(4.9.9)
4
Next, we bisect L [p , q ] , and so on. Continuing this process indefinitely, we obtain an infinite contracting sequence of closed line segments L [p , q ] such that ¯ ¯ ¯ 2
¯¯ 2
¯ ¯ ¯ m
¯¯ m
¯ ¯ ¯ p m
(∀m)
∈ P,¯ q¯m ∈ Q,
(4.9.10)
and ¯ ¯ ¯
1
¯¯
| pm − q m | =
m
¯ ¯ ¯
¯¯
| p − q | → 0 as m → +∞.
(4.9.11)
2
By Lemma 1, there is a point ∞ ¯¯
¯ ¯ r ∈ ⋂ L [¯ p ,¯ q¯m ] . m
(4.9.12)
m=1
This implies that (∀m)
whence p
¯ ¯ ¯ m
¯¯
→ r.
Similarly, we obtain q
¯¯ m
Now since r ∈ L[p, q ] ⊆ B, the function criterion, then, ¯¯
¯ ¯ ¯
¯¯
¯¯
→ r f
¯ ¯ ¯ ¯ ¯ |¯ r −¯ p | ≤ |¯ p −¯ q¯m | → 0, m m
(4.9.13)
. is relatively continuous at
¯¯
r
over
B
(by assumption). By the sequential
¯ ¯ ¯ ¯ f (¯ p ) → f (¯ r ) and f (¯ q¯m ) → f (¯ r ). m
Moreover, f (p ) < c < f (q Corollary 1 ) and get ¯ ¯ ¯ m
¯¯ ) m
¯ ¯ ¯
¯¯
( for pm ∈ P and q m ∈ Q) .
Letting
(4.9.14)
m → +∞,
we pass to limits (Chapter 3, §15,
¯ ¯ f (¯ r ) ≤ c ≤ f (¯ r ),
(4.9.15)
so that r is neither in P nor in Q, which is a contradiction. This completes the proof for a convex B. ¯¯
The extension to polygon-connected sets is left as an exercise (see Problem 2 below). Thus all is proved. □ Note 2. In particular, the theorem applies if B is a globe or an interval. Thus continuity on an interval implies the Darboux property. The converse fails, as we have noted. However, for monotone functions, we obtain the following theorem.
Theorem 4.9.2 If a function f : A → E is monotone and has the Darboux property on a finite or infinite interval (a, b) ⊆ A ⊆ E continuous on (a, b). 1
1
,
then it is
Proof Seeking a contradiction, suppose f is discontinuous at some p ∈ (a, b) . For definiteness, let f ↑ on (a, b). Then by Theorems 2 and 3 in §5, we have either both, with no function values in between.
−
f (p
) < f (p)
or
+
f (p) < f (p
)
or
On the other hand, since f has the Darboux property, the function values f (x) for x in (a, b) fill an entire interval (see Note 1). Thus it is impossible for f (p) to be the only function value between f (p ) and f (p ) unless f is constant near p, but then it is also continuous at p, which we excluded. This contradiction completes the proof. □ −
+
Note 3. The theorem holds (with a similar proof) for nonopen intervals as well, but the continuity at the endpoints is relative (right at a, left at b).
4.9.3
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Theorem 4.9.3 If f f
is strictly monotone and continuous when restricted to a finite or infinite interval B ⊆ A ⊆ E has the same properties on the set f [B] (itself an interval, by Note 1 and Theorem 1).
: A → E
−1
1
1
,
then its inverse
Proof It is easy to see that f is increasing (decreasing) if f is; the proof is left as an exercise. Thus f is monotone on f [B] if f is so on B. To prove the relative continuity of f , we use Theorem 2, i.e., show that f has the Darboux property on −1
−1
−1
−1
f [B].
Thus let f (p) < c < f (q) for some p, q ∈ f [B]. We look for an r ∈ f [B] such that f (r) = c, i.e., r = f (c). Now since p, q ∈ f [B], the numbers f (p) and f (q) are in B, an interval. Hence also the intermediate value c is in B ; thus it belongs to the domain of f , and so the function value f (c) exists. It thus suffices to put r = f (c) to get the result. □ −1
−1
−1
−1
−1
ExampleS (a) Define f
: E
1
→ E
1
by n
f (x) = x for a fixed n ∈ N .
As
(4.9.16)
is continuous (being a monomial), it has the Darboux property on E . By Note 1, setting B = [0, +∞), we have . (Why?) Also, f is strictly increasing on B . Thus by Theorem 3, the inverse function f (i.e., the nth root function) exists and is continuous on f [B] = [0, +∞). 1
f
−1
f [B] = [0, +∞)
If n is odd, then f
−1
has these properties on all of E
1
by a similar proof; thus √− x exists for x ∈ E . 1
n
,
(b) Logarithmic functions. From the example in §5, we recall that the exponential function given by x
F (x) = a
is continuous and strictly monotone on E . Its inverse, F Theorem 3, it is continuous and strictly monotone on F [ E 1
To fix ideas, let a > 1, so F
↑
and (F
−1
−1
By Note 1, F
) ↑.
p = inf F [ E
1
1
,
(a > 0)
(4.9.17)
is called the logarithmic function to the base a, denoted log. By
].
[E
1
]
is an interval with endpoints p and r, where x
] = inf { a | − ∞ < x < +∞}
(4.9.18)
and r = sup F [ E
1
x
] = sup { a | − ∞ < x < +∞} .
(4.9.19)
Now by Problem 14 (iii) of §2 (with q = 0 ), lim
x
a
= +∞ and
x→+∞
As F
↑,
lim
x
a
(4.9.20)
we use Theorem 1 in §5 to obtain x
r = sup a
=
lim
x
a
= +∞ and p =
x→+∞
Thus
= 0.
x→−∞
] , i.e., the domain of log , is the interval it is called the logarithm of x to the base a .
F [E
(0, +∞);
1
a
lim
(p, r) = (0, +∞)
−1
1
lim x→+∞
= 0.
. It follows that
The range of log ( i.e. of F ) is the same as the domain of F , i.e., E . Thus if as x increases from 0 to +∞. Hence a
x
a
(4.9.21)
x→−∞
loga x
a > 1, log
loga x = +∞ and lim loga x = −∞,
a
is uniquely defined for
x
increases from
−∞
x
in
to +∞
(4.9.22)
x→0+
provided a > 1 . If 0 < a < 1, the values of these limits are interchanged (since F
4.9.4
↓
in this case), but otherwise the results are the same.
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If a = e, we write ln x or log x for log x, and we call ln x the natural logarithm of x. Its inverse is, of course, the exponential f (x) = e , also written exp(x). Thus by definition, ln e = x and a
x
x
x = exp(ln x) = e
(c) The power function g : (0, +∞) → E
1
ln x
(0 < x < +∞).
(4.9.23)
is defined by a
g(x) = x for a fixed real a.
(4.9.24)
If a > 0, we also define g(0) = 0. For x > 0, we have a
x
a
= exp(ln x ) = exp(a ⋅ ln x).
(4.9.25)
Thus by the rules for composite functions (Theorem 3 and Corollary 2 in §2), the continuity of g on (0, +∞) follows from that of exponential and log functions. If a > 0, g is also continuous at 0. (Exercise!) This page titled 4.9: The Intermediate Value Property is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
4.9.5
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4.9.E: Problems on the Darboux Property and Related Topics Exercise 4.9.E. 1 Prove Note 1.
Exercise 4.9.E. 1
′
Prove Note 3.
Exercise 4.9.E. 1
′′
Prove continuity at 0 in Example (c).
Exercise 4.9.E. 2 Prove Theorem 1 for polygon-connected sets. [Hint: If m−1 ¯ ¯ ¯
¯ ¯ ¯
B ⊇ ⋃ L [ pi , pi+1 ]
(4.9.E.1)
i=0
with ¯ ¯ ¯ ¯ f (¯ p ) < c < f (¯ p ), 0 m
show
that
for
at
least
one
i,
B in the theorem by the convex segment L
either ¯ ¯ ¯ ¯ ¯ ¯ [ pi , pi+1
¯ ¯ ¯
c = f (pi )
or
(4.9.E.2)
¯ ¯ ¯
¯ ¯ ¯
f (pi ) < c < f (pi+1 ) .
Then
replace
].]
Exercise 4.9.E. 3 Show that, if f is strictly increasing on B ⊆ E, then f decreasing functions.
−1
has the same property on f [B], and both are one to one; similarly for
Exercise 4.9.E. 4 For functions on B = [a, b] ⊂ E
1
,
Theorem 1 can be proved thusly: If f (a) < c < f (b),
(4.9.E.3)
P = {x ∈ B|f (x) < c}
(4.9.E.4)
let
and put r = sup P . Show that f (r) is neither greater nor less than c, and so necessarily f (r) = c. [Hint: If f (r) < c , continuity at r implies that f (x) < c on some G (δ) (§2, Problem 7), contrary to r = sup P . (Why?)] r
4.9.E.1
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Exercise 4.9.E. 5 Continuing Problem 4, prove Theorem 1 in all generality, as follows. Define ¯ ¯ ¯
¯¯
¯ ¯ ¯
g(t) = p + t(q − p),
0 ≤ t ≤ 1.
(4.9.E.5)
§ , and so is the composite function h = f ∘ g, on there is a t ∈ (0, 1) with h(t) = c. Put r = g(t), and show that f (r) = c .
Then g is continuous (by Theorem 3 in 3) B = [0, 1],
¯¯
[0, 1].
By Problem
4,
with
¯¯
Exercise 4.9.E. 6 Show that every equation of odd degree, of the form n k
f (x) = ∑ ak x
=0
(n = 2m − 1, an ≠ 0) ,
(4.9.E.6)
k=0
has at least one solution for x in E . [Hint: Show that f takes both negative and positive values as also take the intermediate value 0 for some x ∈ E . ] 1
x → −∞
or
x → +∞
; thus by the Darboux property,
f
must
1
Exercise 4.9.E. 7 Prove that if the functions f by
: A → (0, +∞)
and g : A → E
1
are both continuous, so also is the function h : A → E g(x)
h(x) = f (x )
[Hint: See Example (c)]
.
1
given
(4.9.E.7)
.
Exercise 4.9.E. 8 Using Corollary 2 in §2, and limit properties of the exponential and log functions, prove the "shorthand" Theorems 11 − 16 of §4.
Exercise 4.9.E. 8
′
Find lim
x→+∞
(1 +
1 x
)
√x
.
Exercise 4.9.E. 8
′′
Similarly, find a new solution of Problem 27 in Chapter 3, §15, reducing it to Problem 26.
Exercise 4.9.E. 9 Show that if f : E → E has the Darboux property on B(e. g. , if B is and if f is one to one on B , then f is necessarily strictly monotone on B . 1
∗
4.9.E.2
convex and f is relatively continuous on B)
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Exercise 4.9.E. 10 Prove that if two real functions f , g are relatively continuous on [a, b] (a < b) and f (x)g(x) > 0 for x ∈ [a, b],
(4.9.E.8)
(x − a)f (x) + (x − b)g(x) = 0
(4.9.E.9)
then the equation
has a solution between a and b; similarly for the equation f (x)
g(x) +
x −a
=0
(a, b ∈ E
1
).
(4.9.E.10)
x −b
Exercise 4.9.E. 10
′
Similarly, discuss the solutions of 2
9 +
x −4
1 +
x −1
= 0.
(4.9.E.11)
x −2
4.9.E: Problems on the Darboux Property and Related Topics is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
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4.10: Arcs and Curves. Connected Sets This page is a draft and is under active development. A deeper insight into continuity and the Darboux property can be gained by generalizing the notions of a convex set and polygonconnected set to obtain so-called connected sets. I. As a first step, we consider arcs and curves.
Definition A set A ⊆ (S, ρ) is called an arc iff mapping
A
is a continuous image of a compact interval
[a, b] ⊂ E
f : [a, b] ⟶ A.
1
,
i.e., iff there is a continuous
(4.10.1)
onto
If, in addition, f is one to one, A is called a simple arc with endpoints f (a) and f (b). If instead f (a) = f (b), we speak of a closed curve. A curve is a continuous image of any finite or infinite interval in E . 1
corollary 4.10.1 Each arc is a compact (hence closed and bounded) set (by Theorem 1 of §8).
Definition A set A ⊆ (S, ρ) is said to be arcwise connected iff every two points then also say the p and q can be joined by an arc in A. )
p, q ∈ A
are in some simple arc contained in
A.
(We
Example 4.10.1 (a) Every closed line segment Example (1) of §8). (b) Every polygon
¯ ¯ ¯
¯¯
L[ a, b]
in
E
n
∗
( or in any other normed space )
is a simple arc (consider the map
f
in
m−1 ¯ ¯ ¯ ¯ A = ⋃ L [¯ p ,¯ p i
i+1
]
(4.10.2)
i=0
is an arc (see Problem 18 in §8). It is a simple arc if the half-closed segments L [p , p ) do not intersect and the points p are distinct, for then the map f in Problem 18 of §8 is one to one. (c) It easily follows that every polygon-connected set is also arcwise connected; one only has to show that every polygon joining two points p , p can be reduced to a simple polygon (not a self-intersecting one). See Problem 2. However, the converse is false. For example, two discs in E connected by a parabolic arc form together an arcwise- (but not polygonwise-) connected set. (d) Let f , f , … , f be real continuous functions on an interval I ⊆ E . Treat them as components of a function f : I → E , ¯ ¯ ¯ i
¯ ¯ ¯ 0
¯ ¯ ¯ i+1
¯ ¯ ¯ i
¯ ¯ ¯ m
2
1
1
2
n
n
f = (f1 , … , fn ) .
(4.10.3)
Then f is continuous by Theorem 2 in §3. Thus the image set f [I ] is a
4.10.1
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curve in E ; it is an arc if I is a closed interval. Introducing a parameter t varying over I , we obtain the parametric equations of the curve, namely, n
xk = fk (t),
Then as t varies over I , the point x = (x , … , x f [I ]. This is the usual way of treating curves in E ¯¯ ¯
1
n) n
k = 1, 2, … , n.
(4.10.4)
describes the curve ). ∗
( and C
n
It is not hard to show that Theorem 1 in §9 holds also if B is only arcwise connected (see Problem 3 below). However, much more can be proved by introducing the general notion of a connected set. We do this next. II. For this topic, we shall need Theorems 2-4 of Chapter 3, §12, and Problem 15 of Chapter 4, §2. The reader is advised to review them. In particular, we have the following theorem.
Theorem 4.10.1 function f : (A, ρ) → (T , ρ ) is continuous on A iff f for open sets. ′
A
−1
is closed in
[B]
(A, ρ)
for each closed set B ⊆ (T , ρ ) ; similarly ′
Indeed, this is part of Problem 15 in §2 with (S, ρ) replaced by (A, ρ).
Definition A metric space (S, ρ) is said to be connected iff S is not the union P ∪ Q of any two nonvoid disjoint closed sets; it is disconnected otherwise. A set A ⊆ (S, ρ) is called connected iff (A, ρ) is connected as a subspace of (S, ρ); i.e., iff A is not a union of two disjoint sets P , Q ≠ ∅ that are closed (hence also open) in (A, ρ), as a subspace of (S, ρ). Note 1. By Theorem 4 of Chapter 3, §12, this means that P = A ∩ P1 and Q = A ∩ Q1
(4.10.5)
for some sets P , Q that are closed in (S, ρ). Observe that, unlike compact sets, a set that is closed or open in (A, ρ) need not be closed or open in (S, ρ). 1
1
Example 4.10.1 (a') ∅ is connected. (b') So is any one-point set {p}. (Why?) (c') Any finite set of two or more points is disconnected. (Why?) Other examples are provided by the theorems that follow.
Theorem 4.10.2 The only connected sets in E are exactly all convex sets, i.e., finite and infinite intervals, including E itself. 1
1
Proof The proof that such intervals are exactly all convex sets in E is left as an exercise. 1
Seeking a contradiction, suppose p ∉ A for some p ∈ (a, b), a, b ∈ A. Let P = A ∩ (−∞, p) and Q = A ∩ (p, +∞).
(4.10.6)
Then A = P ∪ Q, a ∈ P , b ∈ Q, and P ∩ Q = ∅. Moreover, (−∞, p) and (p, +∞) are open sets in E . (Why?) Hence P and Q are open in A, each being the intersection of A with a set open in E (see Note 1 above). As A = P ∪ Q, with P ∩ Q = ∅, it follows that A is disconnected. This shows that if A is connected in E , it must be convex. 1
1
1
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Conversely, let A be convex in E of §9, so we only briefly sketch it here.
1
.
The proof that A is connected is an almost exact copy of the proof given for Theorem 1
If A were disconnected, then A = P ∪ Q for some disjoint sets P , Q ≠ ∅ , both closed in A. Fix any p ∈ P and q ∈ Q. Exactly as in Theorem 1 of §9, select a contracting sequence of line segments (intervals) [p , q ] ⊆ A such that p ∈ P,q ∈ Q, and | p − q | → 0, and obtain a point m
m
m
m
m
m
∞
r ∈ ⋂ [ pm , qm ] ⊆ A
(4.10.7)
m=1
so that p → r, q → r, and r ∈ A. As the sets P and Q are closed in (A, ρ), Theorem 4 of Chapter 3, $16 shows that both P and Q must contain the common limit r of the sequences {p } ⊆ P and {q } ⊆ Q. This is impossible, however, since P ∩ Q = ∅, by assumption. This contradiction shows that A cannot be disconnected. Thus all is proved. □ m
m
m
m
Note 2. By the same proof, any convex set in a normed space is connected. In particular, connected themselves.
E
n
and all other normed spaces are
Theorem 4.10.3 If a function f in (T , ρ ).
′
: A → (T , ρ )
with A ⊆ (S, ρ) is relatively continuous on a connected set B ⊆ A, then f [B] is a connected set
′
Proof By definition (§1), relative continuity on B becomes ordinary continuity when f is restricted to B. Thus we may treat f as a mapping of B into f [B], replacing S and T by their subspaces B and f [B]. Seeking a contradiction, suppose f [B] is disconnected, i.e., f [B] = P ∪ Q
(4.10.8)
for some disjoint sets P , Q ≠ ∅ closed in (f [B], ρ ) . Then by Theorem 1, with T replaced by f [Q] are closed in (B, ρ). They also are nonvoid and disjoint (as are P and Q) and satisfy ′
f [B],
the sets
f
−1
[P ]
and
−1
B =f
−1
[P ∪ Q] = f
−1
[P ] ∪ f
−1
[Q]
(4.10.9)
\[ (see Chapter 1, §4-7, Problem 6). Thus B is disconnected, contrary to assumption. □
corollary 4.10.2 All arcs and curves are connected sets (by Definition 2 and Theorems 2 and 3).
lemma 4.10.1 set A ⊆ (S, ρ) is connected iff any two points p, q ∈ A are in some connected subset B ⊆ A. Hence any arcwise connected set is connected. A
Proof Seeking a contradiction, suppose the condition stated in Lemma 1 holds but disjoint sets P ≠ ∅, Q ≠ ∅ both closed in (A, ρ) . Pick any p ∈ P and (A, ρ), and let
q ∈ Q.
By assumption,
p
P
and ′
q
A
are in some connected ′
= B ∩ P and Q = B ∩ Q.
4.10.3
is disconnected, so
set B ⊆ A.
Treat
A =P ∪Q
(B, ρ)
for some
as a subspace of
(4.10.10)
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Then by Theorem 4 of Chapter 3, §12, P and Q are closed in B. Also, they are disjoint (for (for p ∈ P , q ∈ Q ), and ′
′
′
P
and Q are ) and nonvoid
′
B = B ∩ A = B ∩ (P ∪ Q) = (B ∩ P ) ∪ (B ∩ Q) = P
′
′
∪Q .
(4.10.11)
Thus B is disconnected, contrary to assumption. This contradiction proves the lemma (the converse proof is trivial). In particular, if A is arcwise connected, then any points p, q in A are in some arc B ⊆ A, a connected set by Corollary Thus all is proved. □
2.
corollary 4.10.3 Any convex or polygon-connected set connected.
(e. g. , a globe)
in
E
n
(or in any other normed space) is arcwise connected, hence
Proof Use Lemma 1 and Example (c) in part I of this section. □ Caution: The converse fails. A connected set need not be arcwise connected, let alone polygon connected (see Problem 17). However, we have the following theorem.
Theorem 4.10.4 Every open connected set A in E (* or in another normed space) is also arcwise connected and even polygon connected. n
Proof If A = ∅, this is "vacuously" true, so let A ≠ ∅ and fix a ∈ A . ¯ ¯ ¯
Let P be the set of all p ∈ A that can be joined with a by a polygon K ⊆ A Let Q = A − P . Clearly, a ∈ P , so P We shall show that P is open, i.e., that each p ∈ P is in a globe G ⊆ P . ¯ ¯ ¯
¯ ¯ ¯
¯ ¯ ¯
¯ ¯ ¯
≠∅
.
¯ ¯ ¯
p
Thus we fix any p ∈ P . As A is open and p ∈ A, there certainly is a globe G contained in A. Moreover, as G is convex, each point x ∈ G is joined with p by the line segment L[x, p] ⊆ G . Also, as p ∈ P , some polygon K ⊆ A joins p with a . Then ¯ ¯ ¯
¯¯ ¯
¯ ¯ ¯
¯ ¯ ¯
¯ ¯ ¯
p
¯ ¯ ¯
p
¯ ¯ ¯
¯¯ ¯
¯ ¯ ¯
p
¯ ¯ ¯
¯ ¯ ¯
p
¯ ¯ ¯
¯ ¯ ¯
¯¯ ¯
¯ ¯ ¯
K ∪ L[ x, p]
(4.10.12)
is a polygon joining x and a, and hence by definition x ∈ P . Thus each x ∈ G is in P , so that G P is open (also apen in A as a subspace). ¯¯ ¯
¯ ¯ ¯
¯¯ ¯
¯¯ ¯
¯ ¯ ¯
¯ ¯ ¯
p
p
⊆ P,
as required, and
Next, we show that the set Q = A − P is open as well. As before, if Q ≠ ∅ , fix any q ∈ Q and a globe G ⊆ A, and show that G ⊆ Q. Indeed, if some x ∈ G were not in Q, it would be in P , and thus it would be joined with a (fixed above) by a polygon K ⊆ A. Then, however, q itself could be so joined by the polygon ¯¯
¯¯ ¯
¯¯
q
¯¯
q
¯ ¯ ¯
¯¯
q
¯¯
¯ L[¯ q¯, ¯¯ x ] ∪ K,
implying that q ∈ P , not q ∈ Q. This shows that G ¯¯
Thus
with
¯¯
q
⊂Q
indeed, as claimed.
disjoint and open (hence clopen) in A. The connectedness of A then implies that noted.) Hence A = P . By the definition of P , then, each point b ∈ A can be by a polygon. As a ∈ A was arbitrary, A is polygon connected. □
A =P ∪Q
Q = ∅.
¯¯
P,Q
¯¯
(P is not empty, as has been
joined to a
¯ ¯ ¯
(4.10.13)
¯ ¯ ¯
Finally, we obtain a stronger version of the intermediate value theorem.
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Theorem 4.10.5 If a function f
: A → E
1
is relatively continuous on a connected set B ⊆ A ⊆ (S, ρ), then f has the Darboux property on B .
In fact, by Theorems 3 and 2, f [B] is a connected set in E
1
,
i.e., an interval. This, however, implies the Darboux property.
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4.10.E: Problems on Arcs, Curves, and Connected Sets Exercise 4.10.E. 1 ¯¯
Discuss Examples (a) and (b) in detail. In particular, verify that L[a, b] is a simple arc. (Show that the map f in Example of §8 is one to one.) ¯ ¯ ¯
(1)
Exercise 4.10.E. 2 Show that each polygon m−1 ¯ ¯ ¯ ¯ K = ⋃ L [¯ p ,¯ p ] i i+1
(4.10.E.1)
i=0
can be reduced to a simple polygon P (P ⊆ K) joining p and p . [Hint: First, show that if two line segments have two or more common points, they lie in one line. Then use induction on the number m of segments in K. Draw a diagram in E as a guide. 0
m
2
Exercise 4.10.E. 3 Prove Theorem 1 of §9 for an arcwise connected B ⊆ (S, ρ) . [Hint: Proceed as in Problems 4 and 5 in §9, replacing g by some continuous map f
: [a, b] ⟶ B. ]
Exercise 4.10.E. 4 Define f as in Example (f ) of §1. Let Gab = {(x, y) ∈ E
2
|a ≤ x ≤ b, y = f (x)} .
(4.10.E.2)
Prove the following: (i) If a > 0, then G is a simple arc in E . (ii) If a ≤ 0 ≤ b, G is not even arcwise connected. [Hints: (i) Prove that f is continuous on [a, b], a > 0, using the continuity of the sine function. Then use Problem 16 in §2, restricting f to [a, b]. (Gab is the graph of f over [a, b]. )
2
ab
ab
¯ ¯ ¯
¯ ¯ (ii) For a contradiction, assume 0 is joined by a simple arc to some ¯ p ∈ Gab . ]
Exercise 4.10.E. 5 Show that each arc is a continuous image of [0, 1]. [ Hint: First, show that any [a, b] ⊆ E is such an image. Then use a suitable composite mapping. 1
Exercise 4.10.E. ∗6 Prove that a function f
: B → E
1
on a compact set B ⊆ E must be continuous if its graph, 1
{(x, y) ∈ E
2
|x ∈ B, y = f (x)} ,
(4.10.E.3)
is a compact set (e.g., an arc) in E . [Hint: Proceed as in the proof of Theorem 3 of §8.] 2
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Exercise 4.10.E. ∗7 Prove that A is connected iff there is no continuous map f : A ⟶ {0, 1}.
(4.10.E.4)
onto
[Hint: If there is such a map, Theorem 1 shows that A is disconnected. (Why?) Conversely, if A = P ∪ Q(P , Q as in Definition 3), put f = 0 on P and f = 1 on Q. Use again Theorem 1 to show that so defined is continuous on A .]
f
Exercise 4.10.E. ∗8 Let B ⊆ A ⊆ (S, ρ). Prove that B is connected in S iff it is connected in (A, ρ).
Exercise 4.10.E. ∗9 Suppose that no two of the sets A (i ∈ I ) are disjoint. Prove that if all A are connected, so is A = ⋃ A . [Hint: If not, let A = P ∪ Q(P , Q as in Definition 3). Let P = A ∩ P and Q = A ∩ Q, so A = P ∪ Q At least one of the P , Q must be ∅ (why?); say, Q = ∅ for some j ∈ I . Then (∀i)Q = ∅, for Q ≠ ∅ implies P = ∅, whence i
i
i
i
i
i
i
i∈I
i
i
i
i
i
i,
i ∈ I.
j
i
i
Ai = Qi ⊆ Q ⟹ Ai ∩ Aj = ∅ ( since Aj ⊆ P ) ,
(4.10.E.5)
contrary to our assumption. Deduce that Q = ⋃ Qi = ∅. (Contradiction!) ] i
Exercise 4.10.E. ∗10 Prove that if {A
is a finite or infinite sequence of connected sets and if
n}
(∀n)
An ∩ An+1 ≠ ∅,
(4.10.E.6)
then A = ⋃ An
(4.10.E.7)
n
is connected. [Hint: Let B
A . Use Problem 9 and induction to show that the B are connected and no two are disjoint. Verify that and apply Problem 9 to the sets B . ]
n
A =⋃
n
Bn
n
=⋃
k
k=1
n
n
Exercise 4.10.E. ∗11 Given p ∈ A, A ⊆ (S, ρ), let A denote the union of all connected subsets of A that contain p (one of them is {p}); A called the p-component of A. Prove that (i) A is connected (use Problem 9) ; (ii) A is not contained in any other connected set B ⊆ A with p ∈ B ; (iii) (∀p, q ∈ A)A ∩ A = ∅ iff A ≠ A ; and (iv) A = ∪ {A |p ∈ A} . [Hint for (iii): If A ∩ A ≠ ∅ and A ≠ A , then B = A ∪ A is a connected set larger than A , contrary to (ii). ] p
p
is
p
p
p
q
p
q
p
p
q
p
q
p
q
4.10.E.2
p
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Exercise 4.10.E. ∗12 ¯ ¯¯ ¯
Prove that if A is connected, so is its closure (Chapter 3, §16 Definition 1 ), and so is any set D such that A ⊆ D ⊆ A . [Hints: First show that D is the "least" closed set in (D, ρ) that contains (Problem 11 in Chapter 3, §16 and Theorem 4 of Chapter 3, §12). Next, seeking a contradiction, D = P ∪ Q, P ∩ Q = ∅, P , Q ≠ ∅, clopen in D. Then A = (A ∩ P ) ∪ (A ∩ Q)
proves
A
disconnected,
for
if
A ∩ P = ∅,
say,
A
let
(4.10.E.8)
then
A ⊆Q ⊂D
(why?),
contrary
to
the minimality of D; similarly for A ∩ Q = ∅. ]
Exercise 4.10.E. ∗13 A set is said to be totally disconnected iff its only connected subsets are one-point sets and ∅. Show that R (the rationals) has this property in E . 1
Exercise 4.10.E. ∗14 Show that any discrete space is totally disconnected (see Problem 13).
Exercise 4.10.E. ∗15 From Problems 11 and 12 deduce that each component A is closed (A p
p
¯ ¯¯¯¯ ¯
= Ap ) .
Exercise 4.10.E. ∗16 Prove that a set A ⊆ (S, ρ) is disconnected iff A = P ∪ Q, with P , Q ≠ ∅, and each of P , Q disjoint from the closure of the other: P ∩ Q = ∅ = P ∩ Q. [Hint: By Problem 12, the closure of P in (A, ρ) (i.e., the least closed set in (A, ρ) that contains P ) is ¯ ¯¯ ¯
¯ ¯¯ ¯
¯ ¯¯ ¯
¯ ¯¯ ¯
¯ ¯¯ ¯
¯ ¯¯ ¯
A ∩ P = (P ∪ Q) ∩ P = (P ∩ P ) ∪ (Q ∩ P ) = P ∪ ∅ = P ,
(4.10.E.9)
so P is closed in A; similarly for Q. Prove the converse in the same manner.
Exercise 4.10.E. ∗17 Give an example of a connected set that is not arcwise connected. [Hint: The set G (a = 0) in Problem 4 is 0b
the
closure
of
¯ ¯ ¯
G0b − { 0}
(verify!),
and
the latter is connected (why?); hence so is G0b by Problem 12. ]
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4.11: Product Spaces. Double and Iterated Limits This page is a draft and is under active development. Given two metric spaces (X, ρ ) and (Y , ρ ) , we may consider the Cartesian product X × Y , suitably metrized. Two metrics for X × Y are suggested in Problem 10 in Chapter 3, §11. We shall adopt the first of them as follows. 1
2
Definition By the product of two metric spaces (X, ρ
1)
and (Y , ρ
2)
′
is meant the space (X × Y , ρ), where the metric ρ is defined by
′
′
′
ρ ((x, y), (x , y )) = max { ρ1 (x, x ) , ρ2 (y, y )}
for x, x
′
and y, y
∈ X
′
∈ Y
(4.11.1)
.
Thus the distance between (x, y) and (x , y ) is the larger of the two distances ′
′
′
′
ρ1 (x, x ) in X and ρ2 (y, y ) in Y .
(4.11.2)
The verification that ρ in (1) is, indeed, a metric is left to the reader. We now obtain the following theorem.
Theorem 4.11.1 (i) A globe G
(p,q) (ε)
in (X × Y , ρ) is the Cartesian product of the corresponding ε -globes in X and Y , G(p,q) (ε) = Gp (ε) × Gq (ε).
(ii) Convergence of sequences {(x
m,
ym )}
(4.11.3)
in X × Y is componentwise. That is, we have
(xm , ym ) → (p, q) in X × Y iff xm → p in X and ym → q in Y .
(4.11.4)
Proof Again, the easy proof is left as an exercise. In this connection, recall that by Theorem 2 of Chapter 3, §15, convergence in E is componentwise as well, even though the standard metric in E is not the product metric (1); it is rather the metric (ii) of Problem 10 in Chapter 3, §11. We might have adopted this second metric for X × Y as well. Then part (i) of Theorem 1 would fail, but part (ii) would still follow by making 2
2
ρ1 (xm , p)
0)(∃k)(∀m, n > k)
Note 4. Given any sequence {x sees that
ρ (umn , pn ) < ε.
we may consider the double limit lim
m→ ∞ n→ ∞
ρ (xm , xn )
in E
1
.
By using (8), one easily
lim ρ (xm , xn ) = 0
(4.11.37)
m→ ∞ n→ ∞
iff (∀ε > 0)(∃k)(∀m, n > k)
i.e., if f {x
m}
ρ (xm , xn ) < ε,
is a Cauchy sequence. Thus Cauchy sequences are those for which lim
n→ ∞ n→ ∞
(4.11.38)
ρ (xm , xn ) = 0
.
Theorem 4.11.3 In every metric space (S, ρ), the metric ρ : (S × S) → E
1
is a continuous function on the product space S × S .
Proof Fix any (p, q) ∈ S × S. By Theorem 1 of §2, \rho\) is continuous at (p, q) iff ρ (xm , ym ) → ρ(p, q) whenever (xm , ym ) → (p, q),
i.e., whenever x
m
→ p
and y
m
→ q.
(4.11.39)
However, this follows by Theorem 4 in Chapter 3, §15. Thus continuity is proved. □
This page titled 4.11: Product Spaces. Double and Iterated Limits is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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4.11.E: Problems on Double Limits and Product Spaces Exercise 4.11.E. 1 Prove Theorem 1(i). Prove Theorem 1( ii ) for both choices of ρ, as suggested.
Exercise 4.11.E. 2 Formulate Definitions 2 and 3 for the cases (i) p = q = s = +∞ ; (ii) p = +∞, q ∈ E , s = −∞ ; (iii) p ∈ E , q = s = −∞; and (iv) p = q = s = −∞ . 1
1
Exercise 4.11.E. 3 Prove Theorem 2 from Theorem 2 using Theorem 1 of §2. Give a direct proof as well. ′
Exercise 4.11.E. 4 Define f
: E
2
→ E
1
by xy f (x, y) =
x2 + y 2
if (x, y) ≠ (0, 0), and f (0, 0) = 0;
(4.11.E.1)
see §1, Example (g). Show that lim lim f (x, y) = 0 = lim lim f (x, y), y→0 x→0
(4.11.E.2)
x→0 y→0
but lim f (x, y) does not exist.
(4.11.E.3)
x→ 0 y→ 0
Explain the apparent failure of Theorem 2.
Exercise 4.11.E. 4
′
Define f
: E
2
→ E
1
by f (x, y) = 0 if xy = 0 and f (x, y) = 1 otherwise.
(4.11.E.4)
Show that f satisfies Theorem 2 at (p, q) = (0, 0), but lim
f (x, y)
(4.11.E.5)
(x,y)→(p,q)
does not exist.
4.11.E.1
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Exercise 4.11.E. 5 Do Problem 4, with f defined as in Problems 9 and 10 of §3.
Exercise 4.11.E. 6 Define f as in Problem 11 of §3. Show that for (c), we have lim
f (x, y) = lim f (x, y) = lim lim f (x, y) = 0, x→ 0
(x,y)→(0,0)
(4.11.E.6)
x→0 y→0
y→ 0
but lim
y→0
limx→0 f (x, y)
does not exist; for (d), lim lim f (x, y) = 0,
(4.11.E.7)
y→0 x→0
but the iterated limits do not exist; and for (e), lim
(x,y)→(0,0) f (x,
y)
fails to exist, but
lim f (x, y) = lim lim f (x, y) = lim lim f (x, y) = 0. y→0 x→0
x→ 0
(4.11.E.8)
x→0 y→0
y→ 0
Give your comments.
Exercise 4.11.E. 7 Find (if possible) the ordinary, the double, and the iterated limits of f at (0, 0) assuming that expressions below, and f is defined at those points of E where the expression has sense.
f (x, y)
is given by one of the
2
y sin xy
2
(i)
x 2
x +y
(iii)
2
x+2y x−y 2
2
2
2
x −y
(v)
;
x +y
(ii)
2
3
;
(iv)
x y 6
x +y
2
5
x +y
;
(vi)
2
( x +y −y
(vii)
2
x +y
y+x⋅2
2
4+x
(4.11.E.9)
4
2
2
)
2
;
(viii)
sin xy sin x⋅sin y
Exercise 4.11.E. 8 Solve Problem 7 with x and y tending to +∞ .
Exercise 4.11.E. 9 Consider the sequence u
mn
in E defined by 1
m + 2n umn =
.
(4.11.E.10)
lim umn = 1,
(4.11.E.11)
m +n
Show that lim
lim umn = 2 and lim
m→∞ n→∞
n→∞ m→∞
but the double limit fails to exist. What is wrong here? (See Theo rem 2 . ) ′
4.11.E.2
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Exercise 4.11.E. 10 Prove Theorem 2, with (i) replaced by the weaker assumption ("subuni-form limit") (∀ε > 0)(∃δ > 0) (∀x ∈ G¬p (δ)) (∀y ∈ G¬q (δ))
ρ(g(x), f (x, y)) < ε
(4.11.E.12)
and with iterated limits defined by s = lim lim f (x, y)
(4.11.E.13)
x→p y→q
iff (∀ε > 0) (∃δ
′
′
′′
′′
> 0) (∀x ∈ G¬p (δ )) (∃δx > 0) (∀y ∈ G¬q (δx ))
ρ(f (x, y), s) < ε.
(4.11.E.14)
Exercise 4.11.E. 11 Does the continuity of f on X × Y imply the existence of (i) iterated limits? (ii) the double limit? [Hint: See Problem 6.]
Exercise 4.11.E. 12 Show that the standard metric in E is equivalent to ρ of Problem 7 in Chapter 3, §11. 1
′
Exercise 4.11.E. 13 Define products of n spaces and prove Theorem 1 for such product spaces.
Exercise 4.11.E. 14 Show that the standard metric in similar problem for C . [Hint: Use Problem 13.]
E
n
is equivalent to the product metric for
E
n
treated as a product of
n
spaces
E
1
.
Solve a
n
Exercise 4.11.E. 15 Prove that {(x and Y are.
m,
ym )}
is a Cauchy sequence in X × Y iff {x
m}
and {y
m}
are Cauchy. Deduce that X × Y is complete iff X
Exercise 4.11.E. 16 Prove that X × Y is compact iff X and Y are. [Hint: See the proof of Theorem 2 in Chapter 3, §16, for E .] 2
Exercise 4.11.E. 17 (i) Prove the uniform continuity of projection maps P and P on X × Y , given by P (x, y) = x and P (x, y) = y. (ii) Show that for each open set G in X × Y , P [G] is open in X and P [G] is open in Y . [Hint: Use Corollary 1 of Chapter 3, {12. ] (iii) Disprove (ii) for closed sets by a counterexample. [Hint: Let X × Y = E . Let G be the hyperbola xy = 1. Use Theorem 4 of Chapter 3, §16 to prove that G is closed.] 1
2
1
1
2
2
2
4.11.E.3
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Exercise 4.11.E. 18 Prove that if X × Y is connected, so are X and Y . [Hint: Use Theorem 3 of §10 and the projection maps P and P of Problem 17.] 1
2
Exercise 4.11.E. 19 Prove that if X and Y are connected, so is X × Y under the product metric. [Hint: Using suitable continuous maps and Theorem 3 in §10, show that any two "lines" x = p and y = q are connected sets in X × Y . Then use Lemma 1 and Problem 10 in §10.]
Exercise 4.11.E. 20 Prove Theorem 2 under the weaker assumptions stated in footnote 1.
Exercise 4.11.E. 21 Prove the following: (i) If g(x) = lim f (x, y) and H = lim f (x, y)
(4.11.E.15)
x→ p
y→q
y→ q
exist for x ∈ G
¬p (r)
and y ∈ G
¬q
(r),
then lim lim f (x, y) = H .
(4.11.E.16)
x→p y→q
(ii) If the double limit and one iterated limit exist, they are necessarily equal.
Exercise 4.11.E. 22 In Theorem 2, add the assumptions h(y) = f (p, y)
for y ∈ Y − {q}
(4.11.E.17)
for x ∈ X − {p}.
(4.11.E.18)
and g(x) = f (x, q)
Then show that lim
f (x, y)
(4.11.E.19)
(x,y)→(p,q)
exists and equals the double limits. [Hint: Show that here (5) holds also for x = p and y ∈ G
¬q
(δ)
and for y = q and x ∈ G
¬p (δ).
]
Exercise 4.11.E. 23 From Problem 22 prove that a function f
: (X × Y ) → T
is continuous at (p, q) if
f (p, y) = lim f (x, y) and f (x, q) = lim f (x, y) x→p
(4.11.E.20)
y→q
4.11.E.4
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for (x, y) in some G
(p,q) (δ),
and at least one of these limits is uniform.
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4.12: Sequences and Series of Functions This page is a draft and is under active development. I. Let f1 , f2 , … , fm , …
(4.12.1)
be a sequence of mappings from a common domain A into a metric space (T , ρ ) . For each (fixed) x ∈ A, the function values ′
f1 (x), f2 (x), … , fm (x), …
form a sequence of points in the range space define a function f : B → T by setting
′
(T , ρ ) .
(4.12.2)
Suppose this sequence converges for each
f (x) =
x
in a set
B ⊆ A.
Then we can
lim fm (x) for all x ∈ B.
(4.12.3)
m→∞
This means that (∀ε > 0)(∀x ∈ B)(∃k)(∀m > k)
′
ρ (fm (x), f (x)) < ε.
(4.12.4)
Here k depends not only on ε but also on x, since each x yields a different sequence {f (x)} . However, in some cases (resembling uniform continuity), k depends on ε only; i.e., given ε > 0, one and the same k fits all x in B. In symbols, this is indicated by changing the order of quantifiers, namely, m
(∀ε > 0)(∃k)(∀x ∈ B)(∀m > k)
′
ρ (fm (x), f (x)) < ε.
(4.12.5)
Of course, (2) implies (1), but the converse fails (see examples below). This suggests the following definitions.
Definition 1 With the above notation, we call f the pointwise limit of a sequence of functions f on a set B(B ⊆ A) iff m
f (x) =
lim fm (x) for all x in B;
(4.12.6)
m→∞
i.e., formula (1) holds. We then write fm → f (pointwise) on B.
(4.12.7)
In case (2), we call the limit uniform (on B) and write fm → f (uniformly) on B.
II. If the f m , so
m
are real, complex, or vector valued (§3), we can also define s
(4.12.8)
m
m
=∑
k=1
fk
(= sum of the first m functions) for each
m
(∀x ∈ A)(∀m)
sm (x) = ∑ fk (x).
(4.12.9)
k=1
The s form a new sequence of functions on A. The pair of sequences m
({ fm } , { sm })
is called the (infinite) series with general term ∑ f , ∑ f (x), etc. m
fm ; sm
is called its
m
(4.12.10)
th partial sum. The series is often denoted by symbols like
m
Definition 2 The series ∑ f on A is said to converge (pointwise or uniformly) to a function f on a set B ⊆ A iff the sequence {s partial sums does as well.
m}
m
4.12.1
of its
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We then call f the sum of the series and write ∞
∞
f (x) = ∑ fk (x) or f = ∑ fm = lim sm k=1
(4.12.11)
m=1
(pointwise or uniformly) on B . Note that series of constants, ∑ c
m,
If the range space is E or E 1
∗
,
may be treated as series of constant functions f
m,
with f
m (x)
= cm
for x ∈ A.
we also consider infinite limits, lim fm (x) = ±∞.
(4.12.12)
m→∞
However, a series for which ∞
∑ fm = lim sm
(4.12.13)
m=1
is infinite for some x is regarded as divergent (i.e., not convergent) at that x. III. Since convergence of series reduces to that of sequences simple and useful test for uniform convergence of sequences f
{ sm } ,
we shall first of all consider sequences. The following is a ′
: A → (T , ρ ) .
m
Theorem 4.12.1 Given a sequence of functions f
′
: A → (T , ρ ) ,
m
let B ⊆ A and ′
Qm = sup ρ (fm (x), f (x)) .
(4.12.14)
x∈B
Then f
m
→ f (uniformly on B)
iff Q
m
→ 0
.
Proof If Q
→ 0,
m
then by definition (∀ε > 0)(∃k)(∀m > k)
However, Q is an upper bound of all distances ρ
′
m
Qm < ε.
(fm (x), f (x)) , x ∈ B.
(4.12.15)
Hence (2) follows.
Conversely, if (∀x ∈ B)
′
ρ (fm (x), f (x)) < ε,
(4.12.16)
then ′
ε ≥ sup ρ (fm (x), f (x)) ,
(4.12.17)
x∈B
i.e., Q
m
≤ ε.
Thus (2) implies (∀ε > 0)(∃k)(∀m > k)
and Q
m
Qm ≤ ε
(4.12.18)
→ 0. □
Examples (a) We have n
lim x
n
= 0 if |x| < 1 and lim x
n→∞
Thus, setting f
n (x)
n
=x ,
consider B = [0, 1] and C
(4.12.19)
.
= [0, 1)
We have f → 0 (pointwise) on C and f → f ( pointwise ) on limit is not uniform on C , let alone on B. Indeed, n
= 1 if x = 1.
n→∞
n
4.12.2
B,
with
f (x) = 0
for
x ∈ C
and
f (1) = 1.
However, the
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Qn = sup | fn (x) − f (x)| = 1 for each n.
(4.12.20)
x∈C
Thus Q does not tend to 0, and uniform convergence fails by Theorem 1. n
(b) In Example (a), let D = [0, a], 0 < a < 1. Then f
(uniformly) on D because, in this case,
→ f
n
n
Qn = sup | fn (x) − f (x)| = sup | x x∈D
n
− 0| = a
→ 0.
(4.12.21)
x∈D
(c) Let sin nx
2
fn (x) = x
+
,
x ∈ E
1
.
(4.12.22)
n
For a fixed x, 1 ∣ sin nx ∣ since ∣ ∣ ≤ → 0. ∣ ∣ n n
2
lim fn (x) = x
n→∞
Thus, setting f (x) = x
2
,
we have f
→ f
n
(pointwise) on E
1
.
(4.12.23)
Also,
1 ∣ sin nx ∣ | fn (x) − f (x)| = ∣ ∣ ≤ . ∣ ∣ n n
Thus (∀n)Q
n
≤
1
→ 0.
n
By Theorem 1, the limit is uniform on all of E
1
(4.12.24)
.
Theorem 4.12.2 Let
be
′
fm : A → (T , ρ )
a
sequence
of
functions
on A ⊆ (S, ρ). If continuous on B , then the limit
fm → f (uniformly on a set B ⊆ A, and if the fm are relatively (or uniformly)
function f has the same property. Proof Fix ε > 0. As f
m
→ f
(uniformly) on B, there is a k such that (∀x ∈ B)(∀m ≥ k)
Take any f
m
ε
′
ρ (fm (x), f (x))
k, and take any p ∈ B. By continuity, there is δ > 0, with (∀x ∈ B ∩ Gp (δ))
Also, setting x = p in (3) gives ρ
′
(fm (p), f (p))
0)(∃k)(∀x ∈ B)(∀m, n > k)
ρ (fm (x), fn (x)) < ε.
(4.12.28)
Proof If (5) holds then, for any (fixed) x ∈ B, {f (x)} is a Cauchy sequence of points in T , so by the assumed completeness of T , it has a limit f (x). Thus we can define a function f : B → T with m
f (x) =
lim fm (x) on B.
(4.12.29)
m→∞
To show that f → f (uniformly) on B, we use (5) again. Keeping ε, k, x, and m temporarily fixed, we let n → ∞ so that f (x) → f (x) . Then by Theorem 4 of Chapter 3, §15, ρ (f (x), f (x)) → p (f (x), f (x)) . Passing to the limit in (5), we thus obtain (2). m
′
n
m
′
n
m
The easy proof of the converse is left to the reader (cf. Chapter 3, §17, Theorem 1). □ IV. If the range space (T , ρ ) is E have ′
1
, C,
or E (*or another normed space), the standard metric applies. In particular, for series we n
′
ρ (sm (x), sn (x)) = | sn (x) − sm (x)| ∣
n
m
∣
= ∣∑ fk (x) − ∑ fk (x)∣ ∣ k=1
∣
k=1
n ∣ ∣ ∣ ∣ = ∑ fk (x) ∣ ∣ ∣k=m+1 ∣
Replacing here m by m − 1 and applying Theorem 3 to the sequence {s
m}
for m < n.
,
we obtain the following result.
Theorem 4.12.3
′
Let the range space of f , m = 1, 2, … , be converges uniformly on B iff m
E
1
, C,
or
E
n
(*or another complete normed space). Then the series
∣ (∀ε > 0)(∃q)(∀n > m > q)(∀x ∈ B)
n
∣
∣∑ fk (x)∣ < ε. ∣k=m
Similarly, via {s it!
m}
,
∑ fm
(4.12.30)
∣
Theorem 2 extends to series of functions. (Observe that the s
m
are continuous if the f
m
are.) Formulate
∞
V. If ∑ f exists on B, one may arbitrarily "group" the terms, i.e., replace every several consecutive terms by their sum. This property is stated more precisely in the following theorem. m=1
m
Theorem 4.12.4 Let ∞
f = ∑ fm (pointwise) on B.
(4.12.31)
m=1
Let m
< m2 < ⋯ < mn < ⋯
1
in N , and define g1 = sm , 1
(Thus g
n+1
= fmn +1 + ⋯ + fmn+1 . )
gn = sm
n
− sm
n−1
,
n > 1.
(4.12.32)
Then
4.12.4
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∞
f = ∑ gn (pointwise) on B as well;
(4.12.33)
n=1
similarly for uniform convergence. Proof Let n ′
sn = ∑ gk ,
n = 1, 2, …
(4.12.34)
k=1
Then s = s (verify!), so (pointwise); i.e., ′ n
mn
′
{ sn }
is a subsequence,
{ smn } ,
of
{ sm } .
Hence
sm → f ( pointwise )
implies
′
sn → f
∞
f = ∑ gn (pointwise).
(4.12.35)
n=1
For uniform convergence, see Problem 13 (cf. also Problem 19). □
This page titled 4.12: Sequences and Series of Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
4.12.5
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4.12.E: Problems on Sequences and Series of Functions Exercise 4.12.E. 1 Complete the proof of Theorems 2 and 3.
Exercise 4.12.E. 2 Complete the proof of Theorem 4.
Exercise 4.12.E. 2
′
In Example (a), show that f → +∞ (pointwise) on (1, +∞), but not uniformly so. Prove, however, that the limit is uniform on any interval [a, +∞), a > 1. (Define "lim f = +∞ (uniformly)" in a suitable manner.) n
n
Exercise 4.12.E. 3 Using Theorem 1, discuss lim f on B and C ( as in Example (a)) for each of the following. (i) f (x) = ; B = E ; C = [a, b] ⊂ E . (ii) f (x) = ;B =E . (iii) f (x) = ∑ x ; B = (−1, 1); C = [−a, a], |a| < 1 . (iv) f (x) = ; C = [0, +∞) . n→∞
x
n
n
1
1
n cos x+nx
n
n n
n
1
k
k=1
x
n
1+nx
[Hint: Prove that Qn = sup
(v) f
n (x)
n
= cos
(vi) f (x) = (vii) f (x) = n
n
2
sin
x; B = (0, nx
1+nx 1 1+xn
;B =E
1
π 2
1 n
(1 −
1 nx+1
),C = [
1 4
,
) = π 2
)
1 n
.]
.
.
; B = [0, 1); C = [0, a], 0 < a < 1
.
Exercise 4.12.E. 4 Using Theorems 1 and 2, discuss lim f on the sets given below, with f (x) as indicated and 0 < a < +∞. (Calculus rules for maxima and minima are assumed known in (v), (vi), and (vii).) (i) ; [a, +∞), (0, a). (ii) ; (a, +∞), (0, a) . n
n
nx
1+nx
nx
1+n3 x3
− − − (iii) √− cos x; (0, ) , [0, a], a < . (iv) ; (0, a), (0, +∞). (v) x e ; [0, +∞); E . (vi) nx e ; [a, +∞), (0, +∞). (vii) nx e ; [a, +∞), (0, +∞). [Hint: lim f cannot be uniform if the f are continuous on a set, but lim f is not. [For (v), f has a maximum at x = ; hence find Q .] n
π
π
2
2
x
n
−nx
1
−nx
2
−nx
n
n
n
1
n
n
n
Exercise 4.12.E. 5 Define f
n
: E
1
→ E
1
by ⎧ nx ⎪ ⎪
if 0 ≤ x ≤
fn (x) = ⎨ 2 − nx ⎪ ⎩ ⎪
Show
that
all
fn
and
lim fn
0
if
1 n
1 n
0)(∃δ > 0)(∀n) (∀x ∈ A ∩ Gp (δ))
Prove that if so, and if f
n
→ f
′
ρ (fn (x), fn (p)) < ε.
(4.12.E.17)
(pointwise) on A, then f is continuous at p.
4.12.E.4
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[Hint: "Imitate" the proof of Theorem 2 .] 4.12.E: Problems on Sequences and Series of Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
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4.13: Absolutely Convergent Series. Power Series This page is a draft and is under active development. I. A series ∑ f is said to be absolutely convergent on a set B iff the series ∑ |f f converges on B (pointwise or uniformly). Notation:
m (x)|
m
(briefly, ∑ |f
m |)
of the absolute values of
m
f = ∑ | fm | (pointwise or uniformly ) on B.
(4.13.1)
In general, ∑ f may converge while ∑ |f | does not (see Problem 12). In this case, the convergence of ∑ f is said to be conditional. (It may be absolute for some x and conditional for others.) As we shall see, absolute convergence ensures the commutative law for series, and it implies ordinary convergence (i.e., that of ∑ f ), if the range space of the f is complete. m
m
m
m
m
Note 1. Let m
σm = ∑ | fk | .
(4.13.2)
k=1
Then σm+1 = σm + | fm+1 | ≥ σm
i.e., the σ
m (x)
on B;
(4.13.3)
form a monotone sequence for each x ∈ B. Hence by Theorem 3 of Chapter 3, §15, ∞
lim σm = ∑ | fm | always exists in E
m→∞
converges iff ∑
∞
∑ | fm |
m=1
| fm | < +∞
∗
;
(4.13.4)
m=1
.
For the rest of this section we consider only complete range spaces.
Theorem 4.13.1 Let E
n
the
∗
(
range
defined on or another complete normed space). Then for B ⊆ A, we have the following:
(i) If ∑ |f
m|
space
of
the
functions
(all
fm
converges on B (pointwise or uniformly), so does ∑ f
m
∣
∞
∣
1
,
or
C,
itself. Moreover, on B.
(4.13.5)
m=1
(ii) (Commutative law for absolute convergence.) If ∑ |f | converges (pointwise or uniformly on ∑ | g | obtained by rearranging the f in any different order. Moreover, m
m
E
∞
∣
∣∑ fm ∣ ≤ ∑ | fm | ∣m=1
be
A)
B,
so does any series
m
∞
∞
∑ fm = ∑ gm m=1
Note 2. More precisely, a sequence {g
m}
( both exist on B)/
(4.13.6)
m=1
is called a rearrangement of {f
m}
(∀m ∈ N )
iff there is a map u : N
⟷onto N
gm = fu(m) .
such that (4.13.7)
Proof (i) If ∑ |f
m|
converges uniformly on B, then by Theorem 3 of §12, ′
(∀ε > 0)(∃k)(∀n > m > k)(∀x ∈ B) n
ε >∑
i=m
n
| fi (x)| ≥ | ∑
i=m
.
(4.13.8)
fi (x)| (triangle law)
However, this shows that ∑ f satisfies Cauchy's criterion (6) of §12, so it converges uniformly on B. n
Moreover, letting n → ∞ in the inequality
4.13.1
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∣
n
n
∣
∣∑ fm ∣ ≤ ∑ | fm | , ∣m=1
∣
(4.13.9)
m=1
we get ∣
∞
∞
∣
∣∑ fm ∣ ≤ ∑ | fm | < +∞ ∣m=1
∣
on B, as claimed.
(4.13.10)
m=1
By Note 1, this also proves the theorem for pointwise convergence. (ii) Again, if ∑ f | converges uniformly on B, the inequalities (1) hold for all f except (possibly) for f , f , … , f . Now when ∑ f is rearranged, these k functions will be renumbered as certain g . Let q be the largest of their new subscripts i. Then all of them (and possibly some more functions) are among g , g , … , g (so that q ≥ k). Hence if we exclude g , … , g , the inequalities (1) will certainly hold for the remaining g (i > q). Thus m
i
m
1
1
1
2
k
i
q
2
q
i
n
(∀ε > 0)(∃q)(∀n > m > q)(∀x ∈ B)
∣
n
∣
ε > ∑ | gi | ≥ ∣∑ gi ∣ . i=m
∣i=m
(4.13.11)
∣
By Cauchy's criterion, then, both ∑ g and ∑ |g | converge uniformly. i
i
Moreover, by construction, the two partial sums k
q ′
sk = ∑ fi and sq = ∑ gi i=1
(4.13.12)
i=1
can differ only in those terms whose original subscripts (before the rearrangement) were sum of such terms is less than ε in absolute value. Thus |s − s | < ε . ′ q
> k. By(1),
however, any finite
k
This argument holds also if k in (1) is replaced by a larger integer. (Then also q increases, since q ≥ k as noted above.) Thus we may let k → +∞( hence also q → +∞) in the inequality | s − s | < ε, with ε fixed. Then ′ q
k
∞
∞ ′
sk → ∑ fm and sq → ∑ gi , m=1
(4.13.13)
i=1
so ∣
∞
∞
∣
∣∑ gi − ∑ fm ∣ ≤ ε. ∣ i=1
m=1
(4.13.14)
∣
Now let ε → 0 to get ∞
∞
∑ gi = ∑ fm ; i=1
(4.13.15)
m=1
similarly for pointwise convergence. □ II. Next, we develop some simple tests for absolute convergence.
Theorem 4.13.2 (comparison test). Suppose (∀m)
| fm | ≤ | gm | on B.
(4.13.16)
Then (i) ∑
∞ m=1 ∞
(ii) ∑
m=1
∞
| fm | ≤ ∑
m=1
| fm | = +∞
| gm |
on B ; ∞
implies ∑
m=1
| gm | = +∞
on B; and
4.13.2
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(iii) If ∑ |g
m|
converges (pointwise or uniformly ) on B, so does ∑ |f |. m
Proof Conclusion (i) follows by letting n → ∞ in n
n
∑ | fm | ≤ ∑ | gm | . m=1
(4.13.17)
m=1
In turn, (ii) is a direct consequence of (i) . Also, by (i), ∞
∞
∑ | gm | < +∞ implies ∑ | fm | < +∞. m=1
(4.13.18)
m=1
This proves (iii) for the pointwise case (see Note 1). The uniform case follows exactly as in Theorem 1(i) on noting that n
n
∑ | fk | ≤ ∑ | gk | k=m
(4.13.19)
k=m
and that the functions |f | and |g | are real (so Theorem 3 in §12 does apply). □ ′
k
k
Theorem 4.13.3 (Weierstrass "M-test") If ∑ M is a convergent series of real constants M n
n
≥0
and if
(∀n)
on a set B, then ∑ |f
n|
| fn | ≤ Mn
(4.13.20)
converges uniformly on B. Moreover, ∞
∞
∑ | fn | ≤ ∑ Mn n=1
on B.
(4.13.21)
n=1
Proof Use Theorem 2 with |g
n|
= Mn ,
noting that ∑ |g
n|
converges uniformly since the |g
n|
are constant (§12, Problem 7). □
ExampleS (a) Let n
1 fn (x) = (
sin x) on E
1
.
(4.13.22)
2
Then (∀n) (∀x ∈ E
1
−n
)
| fn (x)| ≤ 2
,
and ∑ 2 converges (geometric series with ratio ; see §12, Problem 18). Thus, setting that the series ∑ ∣∣ sin x ∣∣ converges uniformly on E , as does ∑ ( sin x) ; moreover, −n
1
1
n
2
2
−n
Mn = 2
in Theorem 3, we infer
n
1
1
(4.13.23)
2
∞
∞ −n
∑ | fn | ≤ ∑ 2 n=1
= 1.
(4.13.24)
n=1
4.13.3
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Theorem 4.13.4 (necessary condition of convergence) If ∑ f
m
or ∑ |f
m|
converges on B (pointwise or uniformly), then |f
m|
→ 0
on B (in the same sense). ¯ ¯ ¯
Thus a series cannot converge unless its general term tends to 0 (respectively, 0). Proof If ∑ f
m
= f,
say, then s
and also s
→ f
m
m−1
Hence
→ f.
¯ ¯ ¯
sm − sm−1 → f − f = 0.
However, s
m
Thus f
− sm−1 = fm .
¯ ¯ ¯
m
→ 0,
and |f
m|
(4.13.25)
as claimed.
→ 0,
This holds for pointwise and uniform convergence alike (see Problem 14 in §12) . Caution: The condition |f as we show next.
m|
□
is necessary but not sufficient. Indeed, there are divergent series with general term tending to 0,
→ 0
Examples (Continued) (b) ∑
∞
1
n=1
n
= +∞
(the so-called harmonic series).
Indeed, by Note 1, ∞
1
∑
exists (in E
∗
),
(4.13.26)
n
n=1
so Theorem 4 of §12 applies. We group the series as follows: 1
1
∑
=1+ n
1 +(
2 1 ≥
3
1 +
2
1
Each bracketed expression now equals
) +( 4
1 +(
2
2
.
1 +
5
1 +
4 1
1
+
6
1 ) +(
4
1 + 7
1 +
8
1 + 8
1 +
8
1 ) +( 9
1 +
8
) +⋯ 16
1 ) +(
8
1 +⋯ +
16
) +⋯ . 16
Thus 1 ∑ n
As g does not tend to 0, ∑ g diverges, i.e., ∑
∞
m
1 +⋯ +
m
m=1
1 ≥ ∑ gm ,
gm
gm =
.
(4.13.27)
2
is infinite, by Theorem 4. A fortiori, so is ∑
∞
1
n=1
n
.
Theorem 4.13.5 (root and ratio tests) A series of constants ∑ a
n
(| an | ≠ 0)
converges absolutely if ¯ ¯¯¯¯¯ ¯
− − −
¯ ¯¯¯¯¯ ¯
n
lim √| an | < 1 or lim (
| an+1 | ) < 1.
(4.13.28)
) > 1.
(4.13.29)
| an |
It diverges if ¯ ¯¯¯¯¯ ¯
− − − n
lim √| an | > 1 or lim ( –––
| an+1 | | an |
It may converge or diverge if ¯ ¯¯¯¯¯ ¯
− − − n
lim √| an | = 1
(4.13.30)
or if
4.13.4
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| an+1 |
lim ( –––
¯ ¯¯¯¯¯ ¯
) ≤ 1 ≤ lim (
| an+1 |
| an |
).
(4.13.31)
| an |
(The a may be scalars or vectors.) n
Proof ¯ ¯¯¯¯¯ ¯
− − − n | < 1,
If lim √|a n
choose r > 0 such that ¯ ¯¯¯¯¯ ¯
− − − n
lim √| an | < r < 1.
− − − n √| an | < r
Then by Corollary 2 of Chapter 2, §13, (§12, Problem 17), we may assume that
(4.13.32)
for all but finitely many
Thus, dropping a finite number of terms
n.
n
| an | < r for all n.
(4.13.33)
As 0 < r < 1, the geometric series ∑ r converges. Hence so does ∑ |a n
by Theorem 2.
n|
In the case | an+1 |
¯ ¯¯¯¯¯ ¯
lim (
) < 1,
(4.13.34)
| an |
we similarly obtain (∃m)(∀n ≥ m) |a
n+1 |
< | an | r;
n−m
(∀n ≥ m)
Thus ∑ |a
n|
hence by induction,
| an | ≤ | am | r
.
(Verify!)
(4.13.35)
converges, as before.
− − − n lim √| an | > 1,
¯ ¯¯¯¯¯ ¯
If then by Corollary 2 of Chapter 2, §13,\left|a_{n}\right|>1\) for infinitely many tend to 0, and so ∑ a diverges by Theorem 4.
n.
Hence
| an |
cannot
n
Similarly, if | an+1 | lim ( –––
then |a
n+1 |
> | an |
) > 1,
(4.13.36)
| an |
for all but finitely many n, so |a | cannot tend to 0 again. n
□
Note 3. We have lim ( –––
| an+1 | | an |
− − −
¯ ¯¯¯¯¯ ¯
n
− − −
¯ ¯¯¯¯¯ ¯
n
) ≤ lim √| an | ≤ lim √| an | ≤ lim ( –––
| an+1 |
).
(4.13.37)
| an |
Thus ¯ ¯¯¯¯¯ ¯
lim (
| an+1 | | an |
lim ( –––
− − ¯ ¯¯¯¯¯ ¯ n − ) < 1 implies lim √| an | < 1; and
| an+1 | | an |
(4.13.38)
− − ¯ ¯¯¯¯¯ ¯ n − ) > 1 implies lim √| an | > 1.
Hence whenever the ratio test indicates convergence or divergence, so certainly does the root test. On the other hand, there are cases where the root test yields a result while the ratio test does not. Thus the root test is stronger (but the ratio test is often easier to apply).
Examples (continued) (c) Let a
n
−k
=2
if n = 2k − 1 (odd) and a
n
−k
=3
1 ∑ an =
1
2
if n = 2k (even). Thus
1 +
1
3
1 +
2
2
1 +
2
3
1 +
3
2
1 +
3
3
1 +
4
2
1 +
4
+⋯ .
(4.13.39)
3
Here
4.13.5
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−k
an+1 lim ( –––
3 ) = lim
an
k→∞
¯ ¯¯¯¯¯ ¯
−k
−k−1
an+1
= 0 and lim (
2 ) = lim
an
2
k→∞
= +∞,
−k
(4.13.40)
3
while − − − 2n−1 −n − n − lim √an = lim √2 =
1 – < 1. √2
¯ ¯¯¯¯¯ ¯
(Verify!)
(4.13.41)
Thus the ratio test fails, but the root test proves convergence. Note 4. The assumption |a
n|
≠0
is needed for the ratio test only.
III. Power Series. As an application, we now consider so-called power series, n
∑ an (x − p ) ,
where x, p, a
n
∈ E
1
(C );
(4.13.42)
the a may also be vectors. n
Theorem 4.13.6 For any power series ∑ a (x − p) , there is a unique r ∈ E (0 ≤ r ≤ +∞), called its convergence radius, such that the series converges absolutely for each x with |x − p| < r and does not converge (even conditionally) if |x − p| > r. n
∗
n
Specifically, 1 r = d
¯ ¯¯¯¯¯ ¯
− − − n
, where d = lim √| an |
(with r = +∞ if d = 0).
(4.13.43)
Proof Fix any x = x
0.
By Theorem 5, the series ∑ a
n (x0
− p)
n
¯ ¯¯¯¯¯ ¯
1 | x0 − p| < r
− − − n | | x0 − p| < 1,
converges absolutely if lim √|a
(r =
n
i.e., if
1
− − − lim √| an |
=
n
),
(4.13.44)
d
and diverges if |x − p| > r. (Here we assumed d ≠ 0; but if d = 0, the condition d |x − p| < 1 is trivial for anyx so r = +∞ in this case.) Thus r is the required radius, and clearly there can be only one such r. (Why?) □ 0
| an+1 |
Note 5. If lim
n→∞
− − − n √| an |,
exists, it equals lim
n→∞
| an |
0,
0
¯ ¯¯¯¯¯¯ ¯
by Note 3 (for lim and – lim coincide here). In this case, one can use the –––
ratio test to find | an+1 | d = lim n→∞
(4.13.45) | an |
and hence (if d ≠ 0) | an |
1 r =
= lim d
n→∞
.
(4.13.46)
| an+1 |
Theorem 4.13.7 If a power series ∑ a
n (x
n
− p)
¯ ¯¯ ¯
closed globe G
converges absolutely for some x = x
0
p (δ) δ = | x0 − p| .
So does ∑ a
n
n (x − p )
≠ p,
then ∑ |a
n (x
n
− p) |
converges uniformly on the
if the range space is complete (Theorem 1).
Proof Suppose ∑ |a
n (x0
− p)
n
|
converges. Let δ = | x0 − p| and Mn = | an | δ
n
;
(4.13.47)
thus ∑ M converges. n
4.13.6
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¯ ¯¯ ¯
Now if x ∈ G
p (δ),
then |x − p| ≤ δ, so n
| an (x − p ) | ≤ | an | δ
Hence by Theorem 3, ∑ |a
n (x
n
− p) |
¯ ¯¯ ¯
converges uniformly on G
n
= Mn .
p (δ).
(4.13.48)
□
Examples (Continued) (d) Consider ∑
n
x
n!
Here | an |
1 p = 0 and an =
, so n!
= n + 1 → +∞.
By Note 5, then, r = +∞; i.e., the series converges absolutely on all of E ¯ ¯¯ ¯
any G
0 (δ),
(4.13.49)
| an+1 | 1
.
Hence by Theorem 7, it converges uniformly on
hence on any finite interval in E . (The pointwise convergence is on all of E .) 1
1
This page titled 4.13: Absolutely Convergent Series. Power Series is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
4.13.7
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4.13.E: More Problems on Series of Functions Exercise 4.13.E. 1 Verify Note 3 and Example (c) in detail.
Exercise 4.13.E. 2 Show that the so-called hyperharmonic series of order p, 1 ∑
(p ∈ E
p
1
),
(4.13.E.1)
n
converges iff p > 1 . [Hint: If p ≤ 1 , ∞
∞
1
∑
p
n
n=1
1
≥∑ n=1
= +∞
( Example (b)).
(4.13.E.2)
n
If p > 1 , ∞
∑ n=1
1 p
1 = 1 +(
n
1 +
p
p
2
3
1 ≤ 1 +(
p
1 +⋯ +
p
p
1 ) +(
7
1 ) +(
2
4
p
1 +⋯ +
8
1 +⋯ +
p
1 ) +(
4
p
p
) +⋯
15 1 +⋯ +
8
p
) +⋯
8
1
=∑ p−1
n=0
p
4
1 +
p
2 ∞
1 ) +(
(2
.
n
)
a convergent geometric series. Explain each step.]
Exercise 4.13.E. 3 Prove the refined comparison test: (i) If two series of constants, ∑ |a | and ∑ |b ⇒ 3.
n|
n
,
are such that the sequence {|a
n|
∞
/ | bn |}
is bounded in E
1
,
then
∞
∑ | bn | < +∞ implies ∑ | an | < +∞. n=1
(4.13.E.3)
n=1
(ii) If 0 < lim n→∞
then ∑ |a What is
n|
converges if and only if ∑ |b
n|
| an |
< +∞,
(4.13.E.4)
| bn |
does.
lim n→∞
[Hint: If (\forall n)|a_{n}| / |b_{n}| \leq K\), then |a
n|
| an |
= +∞?
(4.13.E.5)
| bn |
≤ K| bn |.
]
4.13.E.1
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Exercise 4.13.E. 4 Test ∑ a for absolute convergence in each of the following. Use Problem 3 or Theorem 2 or the indicated references. (i) a = ( take b = ) ; n
n+1
n
1
n
√n4 +1
(ii) a
cos n
=
n
(−1)
(iii) a (iv) a (v) a
n
=
n
=n e
p
n 5
n
n
3 +1
1 √n3
;
; use Problem 2)
− − − − − − 1 (√n + 1 − √n ), p ∈ E
; (use Problem 18 of Chapter 3, §15);
−n
2 +n
=
n
( take bn =
√n3 −1 n
n
; n
(−1)
(vi) a
=
n
q
(log n)
;n ≥2
;
q
(log n)
(vii) a
=
n
2
n( n +1)
,q ∈ E
1
.
[Hint for (vi) and (vii): From Problem 14 in §2, show that y lim y→+∞
q
= +∞
(4.13.E.6)
(log y)
and hence q
(log n) lim n→∞
= 0.
(4.13.E.7)
n
Then select b
n.
Exercise 4.13.E. 5 Prove that ∑ [Hint: Show that n
n
∞
n
n=1
n! n
. does not tend to 0.]
= +∞ /n!
Exercise 4.13.E. 6 n
Prove that lim = 0. [Hint: Use Example (d) and Theorem 4.] x
n→∞
n!
Exercise 4.13.E. 7 Use Theorems 3, 5, 6, and 7 to show that ∑ |f | converges uniformly on with 0 < a < +∞ and b ∈ E . For parts ( ix ) − ( xii ), find M = max maxima are assumed known.) (i) ; [−a, b]. n
B,
1
n
x∈B
provided f (x) and B are as indicated below, | f (x)| and use Theorem 3. (Calculus rules for n
n
2n
x
(2n)!
(ii) (−1)
n+1
(iii) (iv) n
n
x
n
n
3
(v)
; [−a, a]
.
; [−a, b]
.
n
x ; [−a, a](a < 1)
sin nx 2
n
(vi) e (vii)
2n−1
x
(2n−1)!
−nx
;B =E
1
.
(use Problem 2)
.
.
sin nx; [a, +∞)
cos nx
√n3 +1
;B =E
1
.
(viii) a cos nx, with ∑ | a | < +∞; B = E . (ix) x e ; [0, +∞). (x) x e ; (−∞, ]. (xi) (x ⋅ log x ) , f (0) = 0; [− , ]. ∞
n
n
n
n=1
1
n
−nx
1
nx
2
n
n
3
3
2
2
4.13.E.2
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(xii) (
n
log x
.
) ; [1, +∞)
x
n
q(q−1)⋯(q−n+1)x
(xiii)
,q ∈ E
n!
1
; [−
1 2
,
1 2
]
.
Exercise 4.13.E. 8 (Summation by parts.) Let vector valued). Let f = h − h ⇒ 8.
n
fn , hn ,
n−1 (n
n
and
≥ 2).
g be real or complex functions (or let Verify that (∀m > n > 1) n
m
fn
and
hn
be scalar valued and
gn
be
m
∑
fk gk =
k=n+1
∑
(hk − hk−1 ) gk
k=n+1 m−1
= hm gm − hn gn+1 −
∑
hk (gk+1 − gk ) .
k=n+1
[Hint: Rearrange the sum.]
Exercise 4.13.E. 9 (Abel's test.) Let the f , g , and h be as in Problem 8, with h (i) the range space of the g is complete; (ii) |g | → 0 (uniformly) on a set B; and (iii) the partial sums h = ∑ f are uniformly bounded on B; i.e., ⇒ 9.
n
n
n
n
n
=∑
i=1
fi .
Suppose that
n
n
n
n
i=1
i
(∃K ∈ E
Then prove that ∑ f
k gk
1
) (∀n)
converges uniformly on B if ∑ |g
n+1
| hn | < K on B.
− gn |
(4.13.E.8)
does.
(This always holds if the gn are real and gn ≥ gn+1 on B. )
[Hint: Let ε > 0. Show that m
(∃k)(∀m > n > k)
∑ | gi+1 − gi | < ε and | gn | < ε on B.
(4.13.E.9)
i=n+1
Then use Problem 8 to show that ∣ m ∣ ∣ ∣ ∑ fi gi < 3Kε. ∣ ∣ ∣i=n+1 ∣
(4.13.E.10)
Apply Theorem 3 of §12.] ′
Exercise 4.13.E. 9
′
Prove that if ∑ a is a convergent series of constants then ∑ a b converges. [Hint: Let b → b . Write ′
⇒ 9 .
n
n
an ∈ E
1
and if
{ bn }
is a bounded monotone sequence in
E
1
,
n
n
an bn = an (bn − b) + an b
(4.13.E.11)
and use Problem 9 with fn = an and gn = bn − b. ]
4.13.E.3
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Exercise 4.13.E. 10 Prove the Leibniz test for alternating series: If (−1 ) b differs from s = ∑ (−1) b by b
⇒ 10. ∞
∑
n=1
n
n
n
n
k
k=1
k
and at most.
{ bn } ↓
n+1
bn → 0
in
E
1
,
then
n
∑(−1) bn
converges, and the sum
Exercise 4.13.E. 11 (Dirichlet test.) Let the and with ⇒ 11.
fn , gn ,
and h be as in Problem 8 with n
∞
∑
n=0
fn
uniformly convergent on
B
to a function
f,
∞
hn = − ∑ fi on B.
(4.13.E.12)
i=n+1
Suppose that (i) the range space of the g is complete; and (ii) there is K ∈ E such that n
1
∞
| g0 | + ∑ | gn+1 − gn | < K on B.
(4.13.E.13)
n=0
Show that ∑ f g converges uniformly on B . [Proof outline: We have n
n
n−1
∣
n−1
∣
| gn | = ∣g0 + ∑ (gi+1 − gi )∣ ≤ | g0 | + ∑ | gi+1 − gi | < K ∣
∣
i=0
by (ii).
(4.13.E.14)
i=0
Also, ∣
n
∣
| hn | = ∣∑ fi − f ∣ → 0 (uniformly) on B ∣ i=0
(4.13.E.15)
∣
by assumption. Hence (∀ε > 0)(∃k)(∀n > k)
| hn | < ε on B.
(4.13.E.16)
Using Problem 8, obtain (∀m > n > k)
∣ m ∣ ∣ ∣ ∑ fi gi < 2Kε. ] ∣ ∣ ∣i=n+i ∣
(4.13.E.17)
Exercise 4.13.E. 12 Prove that if 0 < p ≤ 1 , then ∑ [Hint: Use Problems 11 and 2 .]
n
(−1) p
n
converges conditionally.
Exercise 4.13.E. 13 ⇒ 13.
Continuing Problem 14 in §12, prove that if ∑ |f
n|
and ∑ |g
n|
converge on B (pointwise or uniformly), then so do the
series ∑ |afn + b gn | , ∑ | fn ± gn | , and ∑ |afn | .
4.13.E.4
(4.13.E.18)
https://math.libretexts.org/@go/page/23742
[ Hint : |afn + b gn | ≤ |a| | fn | + |b| | gn | . Use Theorem 2. ]
For the rest of the section, we define +
x
−
= max(x, 0) and x
= max(−x, 0).
(4.13.E.19)
Exercise 4.13.E. 14 show the following: (i) ∑ +∑ = ∑ |a | . (ii) If ∑ a < +∞ or ∑ a < +∞, then ∑ a (iii) If ∑ a converges conditionally, then ∑ a (iv) If ∑ |a | < +∞, then for any {b } ⊂ E , ⇒ 14.
Given {a
n}
+ an
⊂E
∗
− an
n
+ n
− n
+
n
.
−
= ∑ an − ∑ an
n
+ n
= +∞ = ∑
− an
.
1
n
n
+
∑ | an ± bn | < +∞ iff ∑ | bn | < ∞;
(4.13.E.20)
moreover, ∑ an ± ∑ bn = ∑ (an ± bn ) if ∑ bn exists.
(4.13.E.21)
−
+
§
−
[Hint: Verify that | an | = an + an and an = an − an . Use the rules of 4. ]
Exercise 4.13.E. 15 ⇒ 15.
(Abel's theorem.) Show that if a power series ∞ n
∑ an (x − p )
(an ∈ E, x, p ∈ E
1
)
(4.13.E.22)
n=0
converges for some x = x ≠ p, it converges uniformly on [p, x ] (or [x [Proof outline: First let p = 0 and x = 1. Use Problem 11 with 0
0,
0
p] if x0 < p)
.
0
n
fn = an and gn (x) = x
As f = a 1 = a (x − p) , the series constant. Verify that if x = 1 , then n
n
n
n
n
0
∑ fn
n
= (x − p ) .
(4.13.E.23)
converges by assumption. The convergence is uniform since the
fn
are
∞
∑ | gk+1 − gk | = 0,
(4.13.E.24)
k=1
and if 0 ≤ x < 1, then ∞
∞
∞ k
k
∑ | gk+1 − gk | = ∑ x |x − 1| = (1 − x) ∑ x k=0
k=0
=1
(a geometric series).
(4.13.E.25)
k=0
Also, |g (x)| = x = 1. Thus by Problem 11( with K = 2), ∑ f g converges uniformly on [0, 1], proving the theorem for p = 0 and x = 1. The general case reduces to this case by the substitution x − p = (x − p) y. Verify!] 0
0
n
n
0
0
Exercise 4.13.E. 16 Prove that if ¯ ¯¯¯¯¯ ¯
0 < lim an ≤ lim an < +∞, –––
4.13.E.5
(4.13.E.26)
https://math.libretexts.org/@go/page/23742
then the convergence radius of ∑ a
n (x
n
− p)
is 1.
Exercise 4.13.E. 17 Show that a conditionally convergent series ∑ a (a ∈ E ) can be rearranged so as to diverge, or to converge to any prescribed sum s . [Proof for s ∈ E : Using Problem 14(iii), take the first partial sum 1
n
n
1
+
a
1
+
+ ⋯ + am > s.
(4.13.E.27)
Then adjoin terms −
−a
1
−
, −a
2
−
, … , −an
(4.13.E.28)
until the partial sum becomes less than s. Then add terms a until it exceeds s . Then adjoin terms −a than s, and so on. As a → 0 and a → 0 (why?), the rearranged series tends to s. (Why?) Give a similar proof for s = ±∞ . Also, make the series oscillate, with no sum.] +
−
k
k
+
−
k
k
until it becomes less
Exercise 4.13.E. 18 Prove that if a power series δ ≤ | x − p| .
n
∑ an (x − p )
converges at some
x = x0 ≠ p
, it converges absolutely (pointwise) on
Gp (δ)
if
0
− →
[Hint: By Theorem 6, δ ≤ |x − p| ≤ r (r = convergence radius). Fix any x ∈ G (δ). Show that the line px, when extended, contains a point x such that |x − p| < |x − p| < δ ≤ r. By Theorem 6, the series converges absolutely at x , hence at x as well, by Theorem 7.] 0
1
p
1
1
4.13.E: More Problems on Series of Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
4.13.E.6
https://math.libretexts.org/@go/page/23742
CHAPTER OVERVIEW 5: Differentiation and Antidifferentiation 5.1: Derivatives of Functions of One Real Variable 5.1.E: Problems on Derived Functions in One Variable 5.2: Derivatives of Extended-Real Functions 5.2.E: Problems on Derivatives of Extended-Real Functions 5.3: L'Hôpital's Rule 5.3.E: Problems on \(L^{\prime}\) Hôpital's Rule 5.4: Complex and Vector-Valued Functions on \(E^{1}\) 5.4.E: Problems on Complex and Vector-Valued Functions on \(E^{1}\) 5.5: Antiderivatives (Primitives, Integrals) 5.5.E: Problems on Antiderivatives 5.6: Differentials. Taylor’s Theorem and Taylor’s Series 5.6.E: Problems on Tayior's Theorem 5.7: The Total Variation (Length) of a Function f - E1 → E 5.7.E: Problems on Total Variation and Graph Length 5.8: Rectifiable Arcs. Absolute Continuity 5.8.E: Problems on Absolute Continuity and Rectifiable Arcs 5.9: Convergence Theorems in Differentiation and Integration 5.9.E: Problems on Convergence in Differentiation and Integration 5.10: Sufficient Condition of Integrability. Regulated Functions 5.10.E: Problems on Regulated Functions 5.11: Integral Definitions of Some Functions 5.11.E: Problems on Exponential and Trigonometric Functions
This page titled 5: Differentiation and Antidifferentiation is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
1
5.1: Derivatives of Functions of One Real Variable In this chapter, " E " will always denote any one of E , E , C (the complex field), E , or another normed space. We shall consider functions f : E → E of one real variable with values in E . Functions f : E → E (admitting finite and infinite values) are said to be extended real. Thus f : E → E may be real, extended real, complex, or vector valued. 1
∗
n ∗
1
1
∗
1
Operations in E were defined in Chapter 4, §4. Recall, in particular, our conventions (2 ) there. Due to them, addition, subtraction, and multiplication are always defined in E (with sums and products possibly "unorthodox"). ∗
∗
∗
To simplify formulations, we shall also adopt the convention that f (x) = 0 unless defined otherwise. (" 0 " stands also for the zero-vector in E if E is a vector space.) Thus each
(5.1.1)
function
is defined on all of
f
E
1
. For
convenience, we call f (x) "finite" if f (x) ≠ ±∞( also if it is a vector ) .
Definition For each function f : E → E, we define its derived function f by setting, for every point p ∈ E , 1
′
: E
1
→ E
1
f (x)−f (p) ′
f (p) = {
limx→p
if this limit exists (finite or not);
x−p
(5.1.2)
0, otherwise.
Thus f
′
(p)
is always defined.
If the limit in (1) exists, we call it the derivative of f at p. If, in addition, this limit is finite, we say that f is differentiable at p. If this holds for each p in a set B ⊆ E function f thederivative of f on B .
1
,
we say that
f
has a derivative (respectively, is differentiable) on
B,
and we call the
′
If the limit in (1) is one sided (with x → p f .
−
or x → p
+
),
we call it a one-sided (left or right) derivative at p, denoted f or ′ −
′ +
Definition Given a function by induction:
f : E
1
→ E,
we define its n th derived function (or derived function of order
n),
denoted
f
(n)
: E
1
→ E,
′
f
Thus f : E
f 1
(0)
= f, f
(n+1)
= [f
(n)
] ,
is the derived function of f . By our conventions, f → E. We have f = f , and we write f for f , f for f (n+1)
(n)
(1)
′
n = 0, 1, 2, …
(n)
′′
(2)
(5.1.3)
is defined on all of E for each n and each function , etc. We say that f has n derivatives at a point p iff the 1
′′′
(3)
(x) − f
(k)
limits f
(k)
(q)
lim
(5.1.4) x −q
x→q
exist for all q in a neighborhood G of p and for k = 0, 1, … , n − 2, and also p
f
(n−1)
(x) − f
(n−1)
(p)
lim x→p
(5.1.5) x −p
exists. If all these limits are finite, we say that f is n times differentiable on I ; similarly for one-sided derivatives. It is an important fact that differentiability implies continuity.
5.1.1
https://math.libretexts.org/@go/page/19049
Theorem 5.1.1 If a function f
: E
1
→ E
is differentiable at a point p ∈ E
1
,
it is continuous at p, and f (p) is finite (even if E = E
∗
)
.
Proof Setting Δx = x − p and Δf
we have the identity
= f (x) − f (p),
∣ Δf ∣ |f (x) − f (p)| = ∣ ⋅ (x − p)∣ ∣ Δx ∣
for x ≠ p.
(5.1.6)
By assumption, Δf
′
f (p) = lim
(5.1.7)
x→p
Δx
exists and is finite. Thus as x → p, the right side of (2) (hence the left side as well) tends to 0, so lim |f (x) − f (p)| = 0, or lim f (x) = f (p) x→p
(5.1.8)
x→p
proving continuity at p . Also, f (p) ≠ ±∞, for otherwise |f (x) − f (p)| = +∞ for all x, and so |f (x) − f (p)| cannot tend to 0.
□
Note 1. Similarly, the existence of a finite left (right) derivative at p implies left (right) continuity at p. The proof is the same. Note 2. The existence of an infinite derivative does not imply continuity, nor does it exclude it. For example, consider the two cases (i) f (x) = (ii)
1 x
,
with f (0) = 0, and
3 − f (x) = √x
.
Give your comments for p = 0 . Caution: A function may be continuous on E without being differentiable anywhere (thus the converse to Theorem 1 fails). The first such function was indicated by Weierstrass. We give an example due to Olmsted (Advanced Calculus). 1
Example 5.1.1 (a) We first define a sequence of functions f
n
: E
1
→ E
−n
fn (x) = 0 if x = k ⋅ 4
Between k ⋅ 4 and (k ± uniformly on E . (Verify!) −n
1 2
−n
)⋅4
, fn
1
as follows. For each k = 0, ±1, ±2, … , let
(n = 1, 2, …) 1
, and fn (x) =
is linear (see Figure
−n
⋅4
1 if x = (k +
2
21),
−n
)⋅4
.
(5.1.9)
2
so it is continuous on
E
1
.
The series
∑ fn
converges
1
Let ∞
f = ∑ fn .
(5.1.10)
n=1
Then f is continuous on E
1
( why? yet it is nowhere differentiable.
To prove this fact, fix any p ∈ E
1
.
For each n, let −n−1
xn = p + dn , where dn = ±4
,
(5.1.11)
choosing the sign of d so that p and x are in the same half of a "sawtooth" in the graph of f (Figure 21) . Then n
n
n
fn (xn ) − fn (p) = ±dn = ± (xn − p) .
( Why ?)
(5.1.12)
Also, fm (xn ) − fm (p) = ±dn if m ≤ n
5.1.2
(5.1.13)
https://math.libretexts.org/@go/page/19049
but vanishes for m > n. (Why?) Thus, when computing f (x
n)
− f (p),
we may replace ∞
n
f = ∑ fm by f = ∑ fm . m=1
(5.1.14)
m=1
Since fm (xn ) − fm (p) = ±1 for m ≤ n.
(5.1.15)
xn − p
the fraction f (xn ) − f (p) (5.1.16) xn − p
is an integer, odd if n is odd and even if n is even. Thus this fraction cannot tend to a finite limit as d =4 → 0 and x = p + d → p. A fortiori, this applies to
n → ∞,
i.e., as
−n−1
n
n
n
f (x) − f (p) lim
.
(5.1.17)
x −p
x→p
Thus f is not differentiable at any p. The expressions f (x) − f (p) and x − p, briefly denoted Δf and Δx, and Δx, are called the increments of f and x (at p), respectively. 2 We now show that for differentiable functions, Δf and Δx are "nearly proportional' when x approaches p; that is, Δf = c + δ(x)
(5.1.18)
Δx
with c constant and lim
x→p
δ(x) = 0
.
Theorem 5.1.2 A function f limx→p
is differentiable at p, and f δ(x) = δ(p) = 0, and such that : E
1
→ E
′
(p) = c,
iff there is a finite c ∈ E and a function δ : E
Δf = [c + δ(x)]Δx
for all x ∈ E
1
.
1
→ E
such that
(5.1.19)
Proof If f is differentiable at p, put c = f
′
(p).
Define δ(p) = 0 and Δf δ(x) =
′
− f (p) for x ≠ p.
(5.1.20)
Δx
Then lim
x→p
′
′
δ(x) = f (p) − f (p) = 0 = δ(p).
Also, (3) follows.
Conversely, if (3) holds, then Δf = c + δ(x) → c as x → p( since δ(x) → 0).
(5.1.21)
Δx
Thus by definition, Δf c = lim x→p
′
′
= f (p) and f (p) = c is finite. □
(5.1.22)
Δx
5.1.3
https://math.libretexts.org/@go/page/19049
Theorem 5.1.3 (chain rule). Let the functions g : E → E ( real ) and f : E → E (real or not) be differentiable at where q = g(p). Then the composite function h = f ∘ g is differentiable at p, and 1
1
1
′
′
p
and
q,
respectively,
′
h (p) = f (q)g (p).
(5.1.23)
Proof Setting Δh = h(x) − h(p) = f (g(x)) − f (g(p)) = f (g(x)) − f (q).
(5.1.24)
we must show that Δh lim x→p
′
′
= f (q)g (p) ≠ ±∞.
Now as f is differentiable at q, Theorem 2 yields a function δ : E (∀y ∈ E
1
(5.1.25)
Δx 1
→ E
such that lim
x→q
δ(x) = δ(q) = 0
and such that
′
)
f (y) − f (q) = [ f (q) + δ(y)] Δy, Δy = y − q.
(5.1.26)
Taking y = g(x), we get (∀x ∈ E
1
)
′
f (g(x)) − f (q) = [ f (q) + δ(g(x))] [g(x) − g(p)],
(5.1.27)
where g(x) − g(p) = y − q = Δy and f (g(x)) − f (q) = Δh,
(5.1.28)
as noted above. Hence Δh
g(x) − g(p)
′
= [ f (q) + δ(g(x))] ⋅
Let x → p. Then we obtain h (p) = f ′
for all x ≠ p.
(5.1.29)
x −p
Δx ′
′
(q)g (p),
for, by the continuity of δ ∘ g at p (Chapter 4, §2, Theorem 3),
lim δ(g(x)) = δ(g(p)) = δ(q) = 0. □
(5.1.30)
x→p
The proofs of the next two theorems are left to the reader.
Theorem 5.1.4 If f , g, and h are real or complex and are differentiable at p, so are f f ± g, hf , and
(5.1.31) h
(the latter if h(p) ≠ 0), and at the point p we have (i) (f ± g)
′
(ii) (hf )
′
(iii) (
f h
=f
= hf ′
′
′
±g
′
+h f;
′
) =
;
′
′
hf −h f 2
h
and
.
All this holds also if f and g are vector valued and h is scalar valued. It also applies to infinite (even one-sided) derivatives, except when the limits involved become indeterminate (Chapter 4, §4). Note 3. By induction, if
f , g,
and
h
are
n
times differentiable at a point
p,
so are
f ±g
and
hf ,
and, denoting by
n (
) k
the
binomial coefficients, we have (i*) (f ± g)
(n)
=f
(n)
±g
(n)
;
and
5.1.4
https://math.libretexts.org/@go/page/19049
(ii*) (hf )
(n)
n
n
=∑
k=0
(k)
(
)h
f
(n−k)
.
k
Formula (ii ) is known as the Leibniz formula; its proof is analogous to that of the binomial theorem. It is symbolically written as (hf ) = (h + f ) , with the last term interpreted accordingly. (n)
n
Theorem 5.1.5 (componentwise differentiation). A function (f , … , f ) is, and then 1
f : E
1
→ E
n
∗
( C
n
is differentiable at
)
p
iff each of its
n
components
n
n ′
′
′
′
¯ ¯ f (p) = (f (p), … , fn (p)) = ∑ f (p)¯ e k, 1
(5.1.32)
k
k=1
with e as in Theorem 2 of Chapter 3, §§1-3. ¯ ¯ ¯ k
In particular, a complex function Chapter 4, §3, Note 5).
f : E
1
→ C
is differentiable iff its real and imaginary parts are, and
f
′
′
= fre + i ⋅ f
′
im
Example 5.1.2 (b) Consider the complex exponential f (x) = cos x + i ⋅ sin x = e
xi
( Chapter 4,
§3).
(5.1.33)
We assume the derivatives of cos x and sin x to be known (see Problem 8). By Theorem 5, we have 1
′
f (x) = − sin x + i ⋅ cos x = cos(x +
1 π) + i ⋅ sin(x +
2
(x+
π) = e
1 2
π)i
.
(5.1.34)
2
Hence by induction, f
(c) Define f
: E
1
→ E
3
(n)
(x+
(x) = e
1 2
nπ)i
, n = 1, 2, … . ( Verify! )
(5.1.35)
by f (x) = (1, cos x, sin x),
x ∈ E
1
.
(5.1.36)
Here Theorem 5 yields ′
f (p) = (0, − sin p, cos p),
For a fixed p = p
0,
p ∈ E
1
.
(5.1.37)
we may consider the line ¯¯ ¯
¯ ¯ ¯
x = a + tu⃗ ,
(5.1.38)
where ¯ ¯ ¯
a = f (p0 ) and u⃗ = f
′
(p0 ) = (0, − sin p0 , cos p0 ) .
This is, by definition, the tangent vector at p to the curve f [ E 0
More generally, if f : E f [ G ] to be the line
1
→ E
1
]
(5.1.39)
in E . 3
is differentiable at p and continuous on some globe about p, we define the tangent at p to the curve
p
′
¯¯ ¯
x = f (p) + t ⋅ f (p);
′
f (p) f [E
1
(5.1.40)
is its direction vector in E, while t is the variable real parameter. For real functions f ] but the curve y = f (x) in E , i.e., the set
: E
1
→ E
1
,
we usually consider not
2
{(x, y)|y = f (x), x ∈ E
The tangent to that curve at p is the line through (p, f (p)) with slope f
5.1.5
′
1
}.
(5.1.41)
.
(p)
https://math.libretexts.org/@go/page/19049
In conclusion, let us note that differentiation (i.e., taking derivatives) is a local limit process at some point p. Hence (cf. Chapter 4, §1, Note 4 ) the existence and the value of f (p) is not affected by restricting f to some globe G about p or by arbitrarily redefining f outside G . For one-sided derivatives, we may replace G by its corresponding "half." ′
p
p
p
This page titled 5.1: Derivatives of Functions of One Real Variable is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.1.6
https://math.libretexts.org/@go/page/19049
5.1.E: Problems on Derived Functions in One Variable Exercise 5.1.E. 1 Prove Theorems 4 and 5, including (i
∗
)
and (i i
∗
).
Do it for dot products as well.
Exercise 5.1.E. 2 Verify Note 2.
Exercise 5.1.E. 3 Verify Example (a).
Exercise 5.1.E. 3
′
Verify Example (b).
Exercise 5.1.E. 4 Prove that if f has finite one-sided derivatives at p, it is continuous at p.
Exercise 5.1.E. 5 Restate and prove Theorems 2 and 3 for one-sided derivatives.
Exercise 5.1.E. 6 Prove that if the functions f
i
: E
1
→ E
∗
(C )
are differentiable at p, so is their product, and m ′
′
(f1 f2 ⋯ fm ) = ∑ (f1 f2 ⋯ fi−1 f fi+1 ⋯ fm ) at p.
(5.1.E.1)
i
i=1
Exercise 5.1.E. 7 A function f
: E
1
→ E
is said to satisfy a Lipschitz condition (L) of order α(α > 0) at p iff (∃δ > 0) (∃K ∈ E
1
) (∀x ∈ G¬p (δ))
α
|f (x) − f (p)| ≤ K|x − p |
.
(5.1.E.2)
(i) This implies continuity at p but not conversely; take 1 f (x) =
,
f (0) = 0,
p = 0.
(5.1.E.3)
ln |x|
[Hint: For the converse, start with Problem 14 (iii) of Chapter 4,
§2. ]
(ii) L of order α > 1 implies differentiability at p, with f (p) = 0 . (iii) Differentiability implies L of order 1, but not conversely. (Take ′
1 f (x) = x sin
, f (0) = 0, p = 0;
(5.1.E.4)
x
then even one-sided derivatives fail to exist.)
5.1.E.1
https://math.libretexts.org/@go/page/23751
Exercise 5.1.E. 8 Let f (x) = sin x and g(x) = cos x.
Show that f and g are differentiable on E
1
,
(5.1.E.5)
with
′
′
f (p) = cos p and g (p) = − sin p for each p ∈ E
1
.
(5.1.E.6)
).
(5.1.E.7)
Hence prove for n = 0, 1, 2, … that f
(n)
nπ (p) = sin(p +
) and g
(n)
nπ (p) = cos(p +
2
2
[Hint: Evaluate Δf as in Example (d) of Chapter 4, §8. Then use the continuity of f and the formula sin z
z
lim
= lim z
z→0
z→0
= 1.
(5.1.E.8)
sin z
To prove the latter, note that | sin z| ≤ |z| ≤ | tan z|,
(5.1.E.9)
whence z
1
1 ≤
≤
→ 1;
sin z
(5.1.E.10)
| cos z|
similarly for g .]
Exercise 5.1.E. 9 Prove that if f is differentiable at p then f (x) − f (y) lim
′
exists, is finite, and equals f (p);
(5.1.E.11)
x −y
+ x→ p − y→ p
i.e., (∀ε > 0)(∃δ > 0)(∀x ∈ (p, p + δ))(∀y ∈ (p − δ, p)) ∣ f (x) − f (y) ∣
Show,
by
redefining
at
f
that
p,
∣
′
− f (p)∣ < ε. x −y
∣
(5.1.E.12)
∣
even
if
the
limit
exists,
may
f
not
be
differentiable (note that the above limit does not involve f (p)).
[Hint: If y < p < x then ∣ f (x) − f (y) ∣ ∣
x −y
′
∣
− f (p)∣ ∣
∣ f (x) − f (p) ≤∣ ∣
x −p −
x −y
′
∣
∣ f (p) − f (y)
f (p)∣ + ∣ x −y ∣ ∣
p −y −
x −y
′
∣
f (p)∣ x −y ∣
∣ f (x) − f (p) ∣ ∣ f (p) − f (y) ∣ ′ ′ ≤∣ − f (p)∣ + ∣ − f (p)∣ → 0. ] ∣
x −p
∣
5.1.E.2
∣
p −y
∣
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Exercise 5.1.E. 10 Prove that if f is twice differentiable at p, then f
′′
f (p + h) − 2f (p) + f (p − h) (x) = lim
≠ ±∞.
2
h→0
(5.1.E.13)
h
Does the converse hold (cf. Problem 9)?
Exercise 5.1.E. 11 In Example (c), find the three coordinate equations of the tangent line at p =
1 2
π.
Exercise 5.1.E. 12 Judging from Figure 22 in §2, discuss the existence, finiteness, and sign of the derivatives (or one-sided derivatives) of f at the points p indicated. i
Exercise 5.1.E. 13 Let f
: E
n
→ E
be linear, i.e., such that ¯ ¯ ¯ (∀¯¯ x ,¯ y ∈ E
n
) (∀a, b ∈ E
1
)
¯ ¯ ¯ ¯ ¯ ¯ f (a¯¯ x + b¯ y ) = af (¯¯ x ) + bf (¯ y ).
(5.1.E.14)
Prove that if g : E → E is differentiable at p, so is h = f ∘ g and h (p) = f (g (p)) . [Hint: f is continuous since f (x) = ∑ x f (e ) . See Problem 5 in Chapter 3, §§4-6.] 1
n
′
¯¯ ¯
n
k=1
k
′
¯ ¯ ¯ k
5.1.E: Problems on Derived Functions in One Variable is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.1.E.3
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5.2: Derivatives of Extended-Real Functions For a while (in §§2 and 3), we limit ourselves to extended-real functions. Below, f and g are real or extended real (f, g : E → E ). We assume, however, that they are not constantly infinite on any interval (a, b), a < b . 1
∗
Lemma 5.2.1 If f
′
(p) > 0
at some p ∈ E
1
,
then x
f (y) for x, y in some G
p (δ).
Proof If f
′
(p) > 0,
the "0" case in Definition 1 of §1, is excluded, so Δf
′
f (p) = lim x→p
Hence we must also have Δf /Δx > 0 for x in some G
p (δ)
It follows that Δf and Δx have the same sign in G
p (δ);
> 0.
(5.2.3)
Δx
.
i.e.,
f (x) − f (p) > 0 if x > p and f (x) − f (p) < 0 if x < p.
(5.2.4)
(This implies f (p) ≠ ±∞. Why? Hence x < p < y ⟹ f (x) < f (p) < f (y),
as claimed; similarly in case f
′
(p) < 0.
(5.2.5)
□
Corollary 5.2.1 If f (p) is the maximum or minimum value of f (x) for x in some G at all, at p.
p (δ),
For, by Lemma 1, f
′
(p) ≠ 0
then f
′
(p) = 0;
i.e., f has a zero derivative, or none
excludes a maximum or minimum at p. (Why?)
Note 1. Thus f (p) = 0 is a necessary condition for a local maximum or minimum at p. It is insufficient, however. For example, if f (x) = x , f has no maxima or minima at all, yet f (0) = 0. For sufficient conditions, see §6. ′
3
′
Figure 22 illustrates these facts at the points graph. Geometrically, f
′
(p) = 0
p2 , p3 , … , p11 .
Note that in Figure 22, the isolated points
P , Q, R
belong to the
means that the tangent at p is horizontal, or that a two-sided tangent does not exist at p.
Theorem 5.2.1 Let f : E → E be relatively continuous on an interval and f is signconstant there (possibly 0 at a and b), with f 1
′
∗
, with f ≠ 0 on (a, b). Then ≥ 0 if f ↑, and f ≤ 0 if f ↓.
[a, b] ′
′
f
is strictly monotone on
[a, b],
′
Proof By Theorem 2 of Chapter 4, §8, f attains a least value m, and a largest value M , at some points of [a, b]. However, neither can occur at an interior point p ∈ (a, b), for, by Corollary 1, this would imply f (p) = 0, contrary to our assumption. ′
5.2.1
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Thus
or M = f (b); for the moment we assume M = f (b) and m = f (a). We must have would make f constant on [a, b], implying f = 0. Thus m = f (a) < f (b) = M .
M = f (a)
m =M
m < M,
for
′
Now let a ≤ x < y ≤ b . Applying the previous argument to each of the intervals [a, x], [a, y], [x, y],and [x, b] (now using that m = f (a) < f (b) = M ) , we find that f (a) ≤ f (x) < f (y) ≤ f (b).
(Why?)
Thus a ≤ x < y ≤ b implies f (x) < f (y); i.e., f increases on [a, b]. Hence otherwise, by Lemma 1, f would decrease at p. Thus f ≥ 0 on [a, b].
′
f
(5.2.6)
cannot be negative at any
p ∈ [a, b],
for,
′
In the case M
= f (a) > f (b) = m,
we would obtain f
′
≤0
Caution: The function f may increase or decrease at p even if f See Note 1.
′
.
□
(p) = 0.
Corollary 5.2.2 (Rolle's theorem) If : E
1
→ E
∗
is relatively continuous on [a, b] and if f (a) = f (b), then f
′
(p) = 0
for at least one interior point p ∈ (a, b) .
For, if f ≠ 0 on all of (a, b), then by Theorem 1, f would be strictly monotone on [a, b], so the equality be impossible. ′
Figure 22 illustrates this on the intervals apparent failure on [0, p ].
[ p2 , p4 ]
and
[ p4 , p6 ] ,
with
f
′
(p3 ) = f
′
(p5 ) = 0.
f (a) = f (b)
would
A discontinuity at 0 causes an
2
Note 2. Theorem 1 and Corollary 2 hold even if f (a) and f (b) are infinite, if continuity is interpreted in the sense of the metric ρ of Problem 5 in Chapter 3, §11. (Weierstrass' Theorem 2 of Chapter 4, §8 applies to (E , ρ ) , with the same proof.)
′
∗
′
Theorem 5.2.2 (Cauchy's law of the mean) Let the functions f , g : E → E be relatively continuous and finite on never both infinite at the same point p ∈ (a, b). Then 1
∗
′
[a, b]
and have derivatives on
′
g (q)[f (b) − f (a)] = f (q)[g(b) − g(a)] for at least one q ∈ (a, b).
(a, b),
with
f
′
and
g
′
(5.2.7)
Proof Let A = f (b) − f (a) and B = g(b) − g(a). We must show that Ag (q) = Bf (q) for some q ∈ (a, b) . For this purpose, consider the function h = Ag − Bf . It is relatively continuous and finite on [a, b], as are g and f . Also, ′
h(a) = f (b)g(a) − g(b)f (a) = h(b).
5.2.2
′
(Verify!)
(5.2.8)
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Thus by Corollary 2, h (q) = 0 for some q ∈ (a, b). Here, by Theorem 4 of §1, h = (Ag − Bf ) = Ag − Bf . (This is legitimate, for, by assumption, f and g never both become infinite, so no indeterminate limits occur.) Thus h (q) = Ag (q) − Bf (q) = 0, and (1) follows. □ ′
′
′
′
′
′
′
′
′
′
Corollary 5.2.3 (Lagrange's law of the mean) If f
1
: E
→ E
1
is relatively continuous on [a, b] with a derivative on (a, b), then ′
f (b) − f (a) = f (q)(b − a) for at least one q ∈ (a, b).
(5.2.9)
Proof Take g(x) = x in Theorem 2, so g
′
=1
on E
1
.□
Note 3. Geometrically, f (b) − f (a) (5.2.10) b −a
is the slope of the secant through (a, f (a)) and (b, f (b)), and f (q) is the slope of the tangent line at q. Thus Corollary 3 states that the secant is parallel to the tangent at some intermediate point q; see Figure 23. Theorem 2 states the same for curves given parametrically: x = f (t), y = g(t) . ′
Corollary 5.2.4 Let f be as in Corollary 3. Then (i) f is constant on [a, b] iff f (ii) f
↑
(iii) f
on [a, b] iff f
↓
′
on [a, b] iff f
≥0 ′
≤0
′
=0
on (a, b);
on (a, b); and on (a, b).
Proof Let f
′
=0
on (a, b). If a ≤ x ≤ y ≤ b, apply Corollary 3 to the interval [x, y] to obtain ′
′
f (y) − f (x) = f (q)(y − x) for some q ∈ (a, b) and f (q) = 0.
(5.2.11)
Thus f (y) − f (x) = 0 for x, y ∈ [a, b], so f is constant. The rest is left to the reader.
□
5.2.3
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Theorem 5.2.3 (inverse functions) Let f
be relatively continuous and strictly monotone on an interval I ⊆ E . Let f (p) ≠ 0 at some interior point p ∈ I . Then the inverse function g = f (with f restricted to I ) has a derivative at q = f (p), and : E
1
→ E
1
1
′
−1
1
′
g (q) =
.
′
(5.2.12)
f (p)
(If f
′
(p) = ±∞,
then g
′
(q) = 0.)
Proof By Theorem 3 of Chapter 4, §9, g = f is strictly monotone and relatively continuous on interior to I , then q = f (p) is interior to f [I ]. (Why?) −1
f [I ],
itself an interval. If
p
is
Now if y ∈ f [I ], we set Δg = g(y) − g(q), Δy = y − q, x = f
−1
(y) = g(y), and f (x) = y
(5.2.13)
and obtain Δg
g(y) − g(q) =
x −p
Δx
=
Δy
y −q
=
for x ≠ p.
(5.2.14)
Δf
f (x) − f (p)
Now if y → q, the continuity of g at q yields g(y) → g(q); i.e., x → p. Also, x ≠ p iff y ≠ q , for f and g are one-to-one functions. Thus we may substitute y = f (x) or x = g(y) to get Δg
′
g (q) = lim
where we use the convention
1 ∞
=0
if f
′
Δx = lim
Δy
y→q
(p) = ∞.
1
1
=
=
Δf
x→p
limx→p (Δf /Δx)
,
′
(5.2.15)
f (p)
□
Examples (A) Let f (x) = loga |x| with f (0) = 0.
(5.2.16)
Let p > 0. Then (∀x > 0) Δf
= f (x) − f (p) = loga x − loga p = loga (x/p) p + (x − p) = loga
Δx = loga (1 +
p
). p
Thus Δf Δx
1/Δx
Δx = loga (1 +
)
.
(5.2.17)
p
Now let z = Δx/p. (Why is this substitution admissible?) Then using the formula 1/z
lim(1 + z)
=e
(see Chapter 4,
§2,
Example (C))
(5.2.18)
z→0
and the continuity of the log and power functions, we obtain
x→p
The same formula results also if f (0) = 0 by Definition 1 in §1.
1/p
Δf
′
f (p) = lim Δx
p < 0,
1/z
= lim loga [(1 + z)
i.e.,
]
z→0
|p| = −p.
At
p = 0, f
= loga e
1/p
1 = p
loga e.
has one-sided derivatives
(5.2.19)
(±∞)
only (verify!), so
′
(B) The inverse of the log function is the exponential g : E a
1
→ E
5.2.4
1
,
with
https://math.libretexts.org/@go/page/19050
y
g(y) = a
(a > 0, a ≠ 1).
(5.2.20)
By Theorem 3, we have (∀q ∈ E
1
1
′
)
g (q) =
q
, p = g(q) = a .
′
(5.2.21)
f (p)
Thus 1
′
g (q) =
1 p
q
p
a
=
= loga e
loga e
.
(5.2.22)
loga e
Symbolically, (log
a
In particular, if a = e, we have log
e
x
1
′
|x|) =
log
a
x
(a )
a
a
x
=
=a log
x = ln x;
(x ≠ 0)
ln a.
(5.2.23)
e
hence
1
′
(ln |x| ) =
′
a
and log
a =1
x
e(x ≠ 0);
x
and
(e )
′
=e
x
(x ∈ E
1
).
(5.2.24)
x
(C) The power function g : (0, +∞) → E
is given by
1
a
g(x) = x
= exp(a ⋅ ln x) for x > 0 and fixed a ∈ E
1
.
(5.2.25)
By the chain rule (§1, Theorem 3), we obtain a
′
g (x) = exp(a ⋅ ln x) ⋅
a
=x x
a ⋅
a−1
= a⋅x
.
(5.2.26)
x
Thus we have the symbolic formula a
(x )
′
a−1
= a⋅x
for x > 0 and fixed a ∈ E
1
.
(5.2.27)
Theorem 5.2.4 (Darboux) If f : E ) on I .
1
→ E
∗
is relatively continuous and has a derivative on an interval I, then f has the Darboux property (Chapter 4, §9 ′
Proof Let p, q ∈ I and f (p) < c < f (q). Put g(x) = f (x) − cx. Assume Theorem 1. Details are left to the reader. □ ′
′
g
′
≠0
on
(p, q)
and find a contradiction to
This page titled 5.2: Derivatives of Extended-Real Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.2.5
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5.2.E: Problems on Derivatives of Extended-Real Functions Exercise 5.2.E. 1 Complete the missing details in the proof of Theorems 1, 2, and 4, Corollary 4, and Lemma 1. [Hint for converse to Corollary 4(ii) : Use Lemma 1 for an indirect proof. ]
Exercise 5.2.E. 2 Do cases p ≤ 0 in Example (A).
Exercise 5.2.E. 3 Show that Theorems 1, 2, and 4 and Corollaries 2 to 4 hold also if f is discontinuous at a and b but f (a ) and f (b ) exist and are finite. (In Corollary 2, assume also f (a ) = f (b ) ; in Theorems 1 and 4 and Corollary 2, finiteness is unnecessary.) +
+
−
−
[Hint: Redefine f (a) and f (b). ]
Exercise 5.2.E. 4 Under the assumptions of Corollary 3, show that f cannot stay infinite on any interval (p, q), a ≤ p < q ≤ b. ′
[Hint: Apply Corollary 3 to the interval [p, q]. ]
Exercise 5.2.E. 5 Justify footnote 1. [Hint: Let 2
f (x) = x + 2 x
1 sin
2
with f (0) = 0.
(5.2.E.1)
x
At 0, find f from Definition 1 in §1. Use also Problem 8 of §1. Show that f is not monotone on any G ′
0 (δ).
]
Exercise 5.2.E. 6 Show that f need not be continuous or bounded on [a, b] (under the standard metric), even if f is differentiable there. ′
[Hint: Take f as in Problem 5. ]
Exercise 5.2.E. 7 With
as in Corollaries 3 and 4, prove that if (p, q) ≠ ∅, then f is strictly monotone on [a, b]. f
f
′
≥ 0 (f
′
≤ 0)
on
(a, b)
and if
f
′
is not constantly 0 on any subinterval
Exercise 5.2.E. 8 Let x = f (t), y = g(t), where t varies over an open interval I g have derivatives on I and f ≠ 0, then the function h = f curve at t equals g (t ) /f (t ) . ′
′
0
, define a curve in E parametrically. Prove that if f and has a derivative on f [I ], and the slope of the tangent to the
⊆E
−1
1
′
0
0
[Hint: The word "curve" implies that f and g are continuous on I (Chapter 4, h =f
−1
2
§10), so Theorems 1 and 3 apply, and
is a function. Also, y = g(h(x)). Use Theorem 3 of §1. ]
5.2.E.1
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Exercise 5.2.E. 9 Prove that if f is continuous and has a derivative on (a, b) and if f has a finite or infinite (even one-sided) limit at some p ∈ (a, b), then this limit equals f (p). Deduce that f is continuous at p if f (p ) and f (p ) exist. [Hint: By Corollary 3, for each x ∈ (a, b), there is some q between p and x such that ′
′
′
′
−
′
+
x
f
′
′
Δf (qx ) =
′
→ f (p) as x → p.
(5.2.E.2)
Δx
′
Set y = qx , so limy→p f (y) = f (p). ]
Exercise 5.2.E. 10 From Theorem 3 and Problem 8 in §1, deduce the differentiation formulas ′
(arcsin x ) =
1 −1 1 ′ ′ . − −−− − ; (arccos x ) = − −−− − ; (arctan x ) = 1 + x2 √ 1 − x2 √ 1 − x2
(5.2.E.3)
Exercise 5.2.E. 11 Prove that if (q, p), p ≠ q .
f
has a derivative at
p,
then
f (p)
[Hint: If f (p) = ±∞, each G has points at which p
is finite, provided Δf Δx
= +∞,
f
is not constantly infinite on any interval
as well as those x with
Δf Δx
(p, q)
or
= −∞. ]
5.2.E: Problems on Derivatives of Extended-Real Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.2.E.2
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5.3: L'Hôpital's Rule We shall now prove a useful rule for resolving indeterminate limits. Below, about ±∞ of the form (a, +∞) or (−∞, a). For one-sided limits, replace G
denotes a deleted globe by its appropriate "half."
G¬p
¬p
G¬p (δ)
in
E
1
,
or one
Theorem 5.3.1 (L'Hôpital's rule) Let
f, g : E
x → p
1
→ E
∗
be differentiable on
G¬p
, with
g
′
there. If
≠0
and
|f (x)|
|g(x)|
tend both to
1
+∞,
or both to
0,
as
and if ′
f (x) lim
= r exists in E
′
x→p
∗
,
(5.3.1)
g (x)
then also f (x) lim x→p
similarly for x → p
+
= r;
(5.3.2)
g(x)
or x → p . −
Proof It suffices to consider left and right limits. Both combined then yield the two-sided limit. First, let −∞ ≤ p < +∞ , ′
f (x) lim |f (x)| = lim |g(x)| = +∞ and lim x→p
+
x→p
Then given ε > 0, we can fix a > p (a ∈ G
¬p )
′
∣ f (x) ∣
+
x→p
= r (finite).
such that ∣
− r∣ < ε, for all x in the interval (p, a).
′
∣ g (x)
(5.3.4)
∣
Now apply Cauchy's law of the mean (§2, Theorem 2) to each interval some q ∈ (x, a) with ′
[x, a], p < x < a.
This yields, for each such
′
g (q)[f (x) − f (a)] = f (q)[g(x) − g(a)].
As g
′
≠0
(5.3.3)
g ′ (x)
+
x,
(5.3.5)
(by assumption), g(x) ≠ g(a) ≠ g(a) by Theorem 1, §2, so we may divide to obtain ′
f (x) − f (a)
f (q) =
g(x) − g(a)
,
′
where p < x < q < a.
(5.3.6)
g (q)
This combined with (1) yields ∣ f (x) − f (a) ∣
∣ − r∣ < ε,
∣ g(x) − g(a)
(5.3.7)
∣
or, setting 1 − f (a)/f (x) F (x) =
,
(5.3.8)
1 − g(a)/g(x)
we have ∣ f (x) ∣ ∣ g(x)
As
|f (x)|
b ∈ (p, a)
∣ ⋅ F (x) − r∣ < ε for all x inside (p, a).
(5.3.9)
∣
and |g(x)| → +∞ (by assumption), we have F (x) → 1 as x → p . Hence by rules for right limits, there is such that for all x ∈ (p, b) , both |F (x) − 1| < ε and F (x) > . (Why?) For such x , formula (2) holds as +
1 2
well. Also,
5.3.1
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1 < 2 and |r − rF (x)| = |r||1 − F (x)| < |r|ε.
(5.3.10)
|F (x)|
Combining this with (2), we have for x ∈ (p, b) ∣ f (x) ∣
∣
∣ f (x)
1
− r∣ =
∣ g(x)
∣
∣
∣ F (x) − rF (x)∣
|F (x)| ∣ g(x)
∣
∣ f (x) 0, we found b > p such that ∣ f (x) ∣
∣ − r∣ < 2ε(1 + |r|),
∣ g(x) f (x)
As ε is arbitrary, we have lim
x→p
The case lim
x→p
+
= r,
+
g(x)
x ∈ (p, b).
(5.3.11)
∣
as claimed.
f (x) = limx→p+ g(x) = 0
is simpler. As before, we obtain ∣ f (x) − f (a) ∣
∣ − r∣ < ε.
∣ g(x) − g(a)
(5.3.12)
∣
Here we may as well replace " a by any y ∈ (p, a). Keeping y fixed, let x → p . Then f (x) → 0 and g(x) → 0, so we get ′′
+
∣ f (y) ∣
∣ − r∣ ≤ ε for any y ∈ (p, a).
∣ g(y) f (y)
As ε is arbitrary, this implies lim
y→p
The cases r = ±∞ and x → p f (x) g(x)
g(y)
= r.
Thus the case x → p
+
is settled for a finite r.
are analogous, and we leave them to the reader.
−
Note 1. lim
+
′
f (x)
may exist even if lim
′
g (x)
(5.3.13)
∣
□
does not. For example, take f (x) = x + sin x and g(x) = x.
(5.3.14)
Then f (x) lim x→+∞
sin x =
lim
(1 +
) =1
( why? ),
(5.3.15)
x
x→+∞
g(x)
but ′
f (x) ′
= 1 + cos x
(5.3.16)
g (x)
does not tend to any limit as x → +∞ . Note 2. The rule fails if the required assumptions are not satisfied, e.g., if g has zero values in each G ′
¬p ;
see Problem 4 below.
Often it is useful to combine L'Hôpital's rule with some known limit formulas, such as 1/z
lim(1 + z)
x = e or lim
z→0
x→0
= 1 (see
§1,
Problem 8).
(5.3.17)
sin x
Examples ′
(a) lim
ln x
x→+∞
(b) lim
x→0
x
(ln x)
= limx→+∞
ln(1+x) x
1
= limx→+∞
1 x
= 0.
1/(1+x)
= limx→0
1
= 1.
5.3.2
https://math.libretexts.org/@go/page/19051
(c) lim
x−sin x
x→0
x3
= limx→0
1−cos x
sin x
= limx→0
3x2
6x
=
1 6
sin x
limx→0
x
=
1 6
.
(Here we had to apply L'Hôpital's rule repeatedly.) (d) Consider e
−1/x
lim
.
(5.3.18)
x
+
x→0
Taking derivatives (even n times), one gets e
−1/x
lim +
x→0
n+1
,
n = 1, 2, 3, … (always indeterminate!).
(5.3.19)
n!x
Thus the rule gives no results. In this case, however, a simple device helps (see Problem 5 below). (e) lim
n→∞
1/n
n
does not have the form
or
0 0
∞
,
∞
so the rule does not apply directly. Instead we compute
1/n
lim ln n n→∞
ln n = lim n→∞
= 0 (Example (a)).
(5.3.20)
n
Hence 1/n
n
1/n
= exp(ln n
) → exp(0) = e
0
=1
(5.3.21)
by the continuity of exponential functions. The answer is then 1. This page titled 5.3: L'Hôpital's Rule is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.3.3
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5.3.E: Problems on L ′ L′ Hôpital's Rule Elementary differentiation formulas are assumed known.
Exercise 5.3.E. 1 Complete the proof of L'Hôpital's rule. Verify that the differentiability assumption may be replaced by continuity plus existence of finite or infinite (but not both together infinite) derivatives f and g on G (same proof). ′
′
¬p
Exercise 5.3.E. 2 Show that the rule fails for complex functions. See, however, Problems 3, 7, and 8. [Hint: Take p = 0 with 2
f (x) = x and g(x) = x + x e
2
i/x
2
= x +x
1 (cos
1
2
+ i ⋅ sin
x
2
).
(5.3.E.1)
x
Then ′
f (x) lim x→0
Indeed, g
′
2
(x) − 1 = (2x − 2i/x)e
i/x
.
f (x) = 1, though lim
g(x)
x→0
1 = lim
′
g (x)
= 0.
′
(5.3.E.2)
g (x)
x→0
(Verify!) Hence ′
(for ∣ ∣e
| g (x)| + 1 ≥ |2x − 2i/x|
2
i/x
∣ = 1) , ∣
(5.3.E.3)
so 2
′
| g (x)| ≥ −1 +
.
( Why? )
(5.3.E.4)
x
Deduce that ∣
1
∣
x ∣ ∣ ∣ ≤∣ ∣ → 0. ] ∣ 2 −x ∣ ∣ g (x) ∣ ∣
(5.3.E.5)
′
Exercise 5.3.E. 3 Prove the "simplified rule of L Hôpital" for real or complex functions (also for vector-valued f and scalar-valued g) : If f and g are differentiable at p, with g (p) ≠ 0 and f (p) = g(p) = 0 , then ′
′
′
f (x) lim x→p
f (p) =
g(x)
.
′
(5.3.E.6)
g (p)
[Hint: f (x)
f (x) − f (p) =
g(x)
Δf =
g(x) − g(p)
Δx
5.3.E.1
′
Δg /
f (p) →
Δx
′
.
(5.3.E.7)
g (p)
https://math.libretexts.org/@go/page/23754
Exercise 5.3.E. 4 f (x)
Why does lim
x→+∞
g(x)
′
f (x)
not exist, though lim
x→+∞
f (x) = e
−2x
[Hint: g vanishes many times in each G ′
+∞ .
′
g (x)
does, in the following example? Verify and explain.
(cos x + 2 sin x),
g(x) = e
−x
(cos x + sin x).
(5.3.E.8)
Use the Darboux property for the proof. ]
Exercise 5.3.E. 5 Find lim
e
+
−1/x
x→0
.
x
1
[Hint: Substitute z =
x
→ +∞. Then use the rule. ]
Exercise 5.3.E. 6 Verify that the assumptions of L'Hôpital's rule hold, and find the following limits. x
e −e
(a) lim
x→0
x
−x
(b) lim
e −e
(c) lim (d) lim (e) lim (f) lim (g) lim
(1+x)
x→0
1/x
(x
+
x→0
q
(
(i) lim
x→+∞
(
−x
(x a 1
;
) , a > 1, q > 0 2
2
x
(
;
ln x) , q > 0
;
x
x→+∞
x→0
(x
x
+
x→0
x→0
;
ln x) , q > 0 −q
x→+∞
(j) lim
−e
x q
;
;
−2x
x−sin x
x→0
(h) lim
−x
ln(e−x)+x−1
− cotan
π 2
sin x x
x)
; 1/ ln x
− arctan x) 1/(1−cos x)
)
;
;
.
Exercise 5.3.E. 7 Prove L'Hôpital's rule for f
: E
1
→ E
n
(C )
and g : E
1
→ E
1
,
with
lim |f (x)| = 0 = lim |g(x)|, p ∈ E
∗
and r ∈ E
n
,
(5.3.E.9)
x→p
k→p
leaving the other assumptions unchanged. f
[Hint: Apply the rule to the components of
g
f
(respectively, to (
g
f
)
and (
re
g
)
).]
im
Exercise 5.3.E. 8 Let f and g be complex and differentiable on G
¬p ,
p ∈ E
1
Let
.
′
′
lim f (x) = lim g(x) = 0, lim f (x) = q, and lim g (x) = r ≠ 0. x→p
Prove that lim
x→p
f (x) g(x)
=
q r
x→p
x→p
(5.3.E.10)
x→p
.
[Hint: f (x)
f (x) =
g(x)
g(x) /
x −p
5.3.E.2
.
(5.3.E.11)
x −p
https://math.libretexts.org/@go/page/23754
Apply Problem 7 to find f (x) lim x→p
g(x) and lim
x −p
x→p
.]
(5.3.E.12)
x −p
Exercise 5.3.E. ∗9 Do Problem 8 for f
: E
1
→ C
n
and g : E
1
→ C
.
5.3.E: Problems on L Hôpital's Rule is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts. ′
5.3.E.3
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5.4: Complex and Vector-Valued Functions on E 1 E1 The theorems of §§2-3 fail for complex and vector-valued functions (see Problem 3 below and Problem 2 in §3). However, some analogues hold. In a sense, they even are stronger, for, unlike the previous theorems, they do not require the existence of a derivative on an entire interval I ⊆ E , but only on I − Q , where Q is a countable set, one contained in the range of a sequence, Q ⊆ { p } . (We henceforth presuppose §9 of Chapter 1.) 1
m
In the following theorem, due to N. Bourbaki, g : E → E is extended real while f may also be complex or vector valued. We call it the finite increments law since it deals with "finite increments" f (b) − f (a) and g(b) − g(a). Roughly, it states that | f | ≤ g implies a similar inequality for increments. 1
′
∗
′
Theorem 5.4.1 (finite increments law) Let f : E → E and g : E → E be relatively continuous and finite on a closed interval derivatives with |f | ≤ g , on I − Q where Q ⊆ {p , p , … , p , …} . Then 1
1
′
∗
′
1
2
I = [a, b] ⊆ E
1
,
and have
m
|f (b) − f (a)| ≤ g(b) − g(a).
(5.4.1)
The proof is somewhat laborious, but worthwhile. (At a first reading, one may omit it, however.) We outline some preliminary ideas. Given any x ∈ I , suppose first that x > p
m
for at least one p
m
In this case, we put
∈ Q.
−m
Q(x) = ∑ 2
;
(5.4.2)
pm 0)
1
,
we have
|f (b) − f (a)| ≤ g(b) − g(a) + Kε,
(5.4.4)
for then, letting ε → 0, we obtain (1). We choose K = b − a + Q(b), with Q(x) as above.
(5.4.5)
Temporarily fixing ε > 0, let us call a point r ∈ I "good" iff |f (r) − f (a)| ≤ g(r) − g(a) + [r − a + Q(r)]ε
(5.4.6)
and "bad" otherwise. We shall show that b is "good." First, we prove a lemma.
Lemma 5.4.1 Every "good" point r ∈ I (r < b) is followed by a whole interval (r, s), r < s ≤ b, consisting of "good" points only. Proof First let r ∉ Q, so by assumption, f and g have derivatives at r, with ′
′
| f (r)| ≤ g (r).
Suppose g (r, s), ,
′
(r) < +∞.
(5.4.7)
Then (treating g as a right derivative) we can find s > r ′
∣ g(x) − g(r) ∣ ∣
′
∣ ∣
5.4.1
such that, for all x in the interval
ε
− g (r)∣ < x −r
(s ≤ b)
(why?);
(5.4.8)
2
https://math.libretexts.org/@go/page/19052
similarly for f . Multiplying by x − r, we get ε
′
|f (x) − f (r) − f (r)(x − r)| < (x − r)
and 2 ε
′
|g(x) − g(r) − g (r)(x − r)| < (x − r)
, 2
and hence by the triangle inequality (explain!), ε
′
|f (x) − f (r)| ≤ | f (r)| (x − r) + (x − r)
(5.4.9) 2
and ε
′
g (r)(x − r) + (x − r)
< g(x) − g(r) + (x − r)ε.
(5.4.10)
2
Combining this with |f
′
′
(r)| ≤ g (r),
we obtain
|f (x) − f (r)| ≤ g(x) − g(r) + (x − r)ε whenever r < x < s.
(5.4.11)
Now as r is "good," it satisfies (2); hence, certainly, as Q(r) ≤ Q(x) , |f (r) − f (a)| ≤ g(r) − g(a) + (r − a)ε + Q(x)ε whenever r < x < s.
(5.4.12)
Adding this to (3) and using the triangle inequality again, we have |f (x) − f (a)| ≤ g(x) − g(a) + [x − a + Q(x)]ε for all x ∈ (r, s).
By definition, this shows that each with g (r) < +∞ .
x ∈ (r, s)
(5.4.13)
is "good," as claimed. Thus the lemma is proved for the case
r ∈ I − Q,
′
The cases g
′
(r) = +∞
and r ∈ Q are left as Problems 1 and 2.
□
We now return to Theorem 1. Proof of Theorem 1. Seeking a contradiction, suppose b is "bad," and let B ≠ ∅ be the set of all "bad" points in [a, b]. Let r = inf B,
r ∈ [a, b].
(5.4.14)
Then the interval [a, r) can contain only "good" points, i.e., points x such that |f (x) − f (a)| ≤ g(x) − g(a) + [x − a + Q(x)]ε.
(5.4.15)
As x < r implies Q(x) ≤ Q(r), we have |f (x) − f (a)| ≤ g(x) − g(a) + [x − a + Q(r)]ε for all x ∈ [a, r).
Note that [a, r) ≠ ∅, for by (2), contained in [a, r).
a
(5.4.16)
is certainly "good' (why?), and so Lemma 1 yields a whole interval
[a, s)
of "good" points
Letting x → r in (4) and using the continuity of f at r, we obtain (2). Thus r is "good" itself. Then, however, Lemma 1 yields a new interval (r, q) of "good" points. Hence [a, q) has no "bad" points, and so q is a lower bound of the set B of "bad" points in I , contrary to q > r = glb B . This contradiction shows that b must be "good," i.e., |f (b) − f (a)| ≤ g(b) − g(a) + [b − a + Q(b)]ε.
Now, letting ε → 0, we obtain formula (1), and all is proved.
(5.4.17)
□
Corollary 5.4.1 If f : E → E is relatively continuous and finite on such that 1
I = [a, b] ⊆ E
1
,
and has a derivative on
|f (b) − f (a)| ≤ M (b − a) and M ≤
sup
′
| f (t)| .
I − Q,
then there is a real
M
(5.4.18)
t∈I−Q
5.4.2
https://math.libretexts.org/@go/page/19052
Proof Let M0 =
sup
′
| f (t)| .
(5.4.19)
t∈I−Q
If M < +∞, put M = M formula (1) yields (5) since 0
0
′
≥ |f |
on
I − Q,
and take
g(x) = M x
in Theorem 1. Then
g
′
′
= M ≥ |f |
on
I − Q,
g(b) − g(a) = M b − M a = M (b − a).
If, however, M
0
= +∞,
so
(5.4.20)
let ∣ f (b) − f (a) ∣ M =∣
∣ < M0 .
b −a
∣
Then (5) clearly is true. Thus the required M exists always.
(5.4.21)
∣
□
Corollary 5.4.2 Let f be as in Corollary 1. Then f is constant on I iff f
′
=0
on I − Q.
Proof If
f
′
=0
on
I − Q,
then
[a, x](x ∈ I ), |f (x) − f (a)| ≤ 0;
Conversely, if so, then f
′
= 0,
M =0 in Corollary 1, so Corollary 1 yields, i.e., f (x) = f (a) for all x ∈ I . Thus f is constant on I .
even on all of I .
for
any
subinterval
□
Corollary 5.4.3 Let f , g : E → E be relatively continuous and finite on I iff f = g on I − Q. 1
′
= [a, b],
and differentiable on I − Q. Then f − g is constant on I
′
Proof Apply Corollary 2 to the function f − g.
□
We can now also strengthen parts (ii) and (iii) of Corollary 4 in §2.
Theorem 5.4.2 Let f be real and have the properties stated in Corollary 1. Then (i) f (ii) f
↑
on I
↓
= [a, b]
on I iff f
′
iff f
≤0
′
≥0
on I − Q; and
on I − Q .
Proof Let f ≥ 0 on I − Q. Fix any x, y ∈ I (x < y) and define g(t) = 0 on E . Then |g | = 0 ≤ f on I − Q. Thus g and satisfy Theorem 1 (with their roles reversed on I , and certainly on the subinterval [x, y]. Thus we have ′
1
′
′
f (y) − f (x) ≥ |g(y) − g(x)| = 0, i.e., f (y) ≥ f (x) whenever y > x in I ,
so f
↑
f
(5.4.22)
on I .
Conversely, if f ↑ on I , then for every p ∈ I , we must have f (p) ≥ 0, for otherwise, by Lemma 1 of §2, decrease at p. Thus f ≥ 0 , even on all of I , and (i) is proved. Assertion (ii) is proved similarly. □ ′
f
would
′
5.4.3
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5.4.4
https://math.libretexts.org/@go/page/19052
5.4.E: Problems on Complex and Vector-Valued Functions on E 1 E1 Exercise 5.4.E. 1 Do the case g (r) = +∞ in Lemma 1. [Hint: Show that there is s > r with ′
′
g(x) − g(r) ≥ (| f (r)| + 1) (x − r) ≥ |f (x) − f (r)| for x ∈ (r, s).
(5.4.E.1)
Such x are "good." ]
Exercise 5.4.E. 2 Do the case r = p ∈ Q in Lemma 1. [Hint: Show by continuity that there is s > r such that (∀x ∈ (r, s)) n
ε |f (x) − f (r)|
0
at some
p ∈ I,
then
f (a) < f (b).
Do it also with
f
′
https://math.libretexts.org/@go/page/23755
Exercise 5.4.E. 7 Let f , g : E → E be relatively continuous on I infinite) on I − Q . (i) Prove that if 1
1
′
for some fixed m, M
∈ E
1
,
and have right derivatives f and g (finite or infinite, but not both ′
= [a, b]
′
+
′
+
′
m g+ ≤ f+ ≤ M g+ on I − Q
(5.4.E.5)
m[g(b) − g(a)] ≤ f (b) − f (a) ≤ M [g(b) − g(a)].
(5.4.E.6)
then
[Hint: Apply Theorem 2 and Problem 4 to each of M g − f and f − mg. ]
(ii) Hence prove that m0 (b − a) ≤ f (b) − f (a) ≤ M0 (b − a),
(5.4.E.7)
where m0 = inf f
′
+
[Hint: Take g(x) = x if m0 ∈ E
1
or M0 ∈ E
′
[I − Q] and M0 = sup f
1
+
[I − Q] in E
∗
.
(5.4.E.8)
. The infinite case is simple. ]
Exercise 5.4.E. 8 (i) Let f
be finite, continuous, with a right derivative on (a, b). Prove that q = lim
: (a, b) → E
+
x→a
f
′
+
(x)
exists (finite) iff
f (x) − f (y) q =
lim
, +
x,y→a
(5.4.E.9)
x −y
i.e., iff (∀ε > 0)(∃c > a)(∀x, y ∈ (a, c)|x ≠ y)
∣ f (x) − f (y) ∣ ∣ − q ∣ < ε. ∣
[Hints: If so, let y → x
+
+
x→a
(5.4.E.10)
∣
(keeping x fixed) to obtain (∀x ∈ (a, c))
Conversely, if lim
x −y
′
f+ (x) = q,
|f
′
+
(x) − q| ≤ ε.
(Why?)
(5.4.E.11)
then (∀ε > 0)(∃c > a)(∀t ∈ (a, c))
|f
′
+
(t) − q| < ε.
(5.4.E.12)
Put ′
M = sup | f+ (t) − q| ≤ ε
( why ≤ ε?)
(5.4.E.13)
a 0 .
∫
b
a
f = 0, f ≥ 0
on
I,
yet
f (p) > 0
for some
p ∈ I −P
(P as above), so
′
Now if a ≤ p < b, Lemma 1 of §2 yields F (c) > F (p) for some c ∈ (p, b]. Then by (iii), b
∫
c
f ≥∫
a
contrary to ∫
b
a
f = 0;
similarly in case a < p ≤ b.
f = F (c) − F (p) > 0,
(5.5.39)
p
□
5.5.6
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Note 4. Hence b
∫
|f | = 0 implies f = 0 on [a, b] − P
(5.5.40)
a
(P countable), even for vector-valued functions (for |f | is always real, and so Theorem 3 applies). However, ∫
b
a
f =0
does not suffice, even for real functions (unless f is signconstant). For example, 2π
∫
sin xdx = 0, yet sin x ≢ 0 on any I − P .
(5.5.41)
0
See also Example (b).
Corollary 5.5.9 (first law of the mean) If f is real and ∫ f exists on [a, b], exact on (a, b), then b
∫
f = f (q)(b − a) for some q ∈ (a, b).
(5.5.42)
a
Proof Apply Corollary 3 in §2 to the function F
= ∫ f.
□
Caution: Corollary 9 may fail if ∫ f is inexact at some p ∈ (a, b). (Exactness on [a, b] − Q does not suffice, as it does not in Corollary 3 of §2, used here.) Thus in Example (b) above, ∫ f = 0. Yet for no q is f (q)(2 + 2) = 0, since f (q) = ±1. The reason is that ∫ f is inexact just at 0, an interior point of [−2, 2]. 2
−2
This page titled 5.5: Antiderivatives (Primitives, Integrals) is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.5.7
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5.5.E: Problems on Antiderivatives Exercise 5.5.E. 1 Prove in detail Corollaries 3, 4, 6, 7, 8, and 9 and Theorem 3(i ) and ( iv ). ′
Exercise 5.5.E. 2 In Examples (a) and (b) discuss continuity and differentiability of interval (−a, a). [Hint: Use Theorem 1.]
and
f
F
at
0.
In (a) show that
∫ f
does not exist on any
Exercise 5.5.E. 3 Show that Theorem 2 holds also if g is relatively continuous on I and differentiable on I − Q .
Exercise 5.5.E. 4 Under the assumptions of Theorem
2,
show that if
is one to one on
g
I,
then automatically
∫ f
is exact on
g[I − Q](Q countable).
[Hint: If F
=∫ f
on g[I ], then F
Let
Q =g
−1
[P ].
Use Problem 6 of Chapter
′
= f on g[I ] − P , P countable.
1,
(5.5.E.1)
§§4 − 7 and Problem 2 of Chapter 1 §9 to show that
Q
is countable and
g[I ] − P = g[I − Q].
Exercise 5.5.E. 5 Prove Corollary 5 for dot products f ⋅ g of vector-valued functions.
Exercise 5.5.E. 6 Prove that if ∫ f exists on [a, p] and [p, b], then it exists on [a, b]. By induction, extend this to unions of n adjacent intervals. [ Hint: Choose F = ∫ f on [a, p] and G = ∫ f on [p, b] such that F (p) = G(p) . (Why do such F , G exist?) Then construct a primitive H = ∫ f that is relatively continuous on all of [a, b]. ]
Exercise 5.5.E. 7 Prove the weighted law of the mean: If g is real and nonnegative on f : E → E, then there is a finite c ∈ E with
I = [a, b],
and if
∫ g
and
∫ gf
exist on
I
for some
1
b
∫
b
gf = c ∫
a
g.
(5.5.E.2)
a
(The value c is called a g -weighted mean of f .) [Hint: If ∫ g > 0, put b
a
b
c =∫
b
gf / ∫
a
5.5.E.1
g.
(5.5.E.3)
a
https://math.libretexts.org/@go/page/23756
If ∫
b
a
g = 0, use Theorem 3(i) and (iv) to show that also ∫
b
a
gf = 0, so any c will do. ]
Exercise 5.5.E. 8 In Problem 7, prove that if, in addition, f is real and has the Darboux property on I , then c = f (q) for some q ∈ I (the second law of the mean). [Hint: Choose c as in Problem 7. If ∫
b
a
g > 0,
put
m = inf f [I ] and M = sup f [I ], in E
so m ≤ f
≤M
∗
,
(5.5.E.4)
on I . Deduce that b
m∫
b
g ≤∫
a
b
gf ≤ M ∫
a
g,
(5.5.E.5)
a
whence m ≤ c ≤ M . If m < c < M , then f (x) < c < f (y) for some x, y ∈ I (why?), so the Darboux property applies. If c = m, then g ⋅ (f − c) ≥ 0 and Theorem 3( iv ) yields gf = gc on I − P . (Why?) Deduce that f (q) = c if g(q) ≠ 0 and q ∈ I − P . (Why does such a q exist?) What if c = M ?]
Exercise 5.5.E. 9 Taking g(x) ≡ 1 in Problem 8, obtain a new version of Corollary 9. State it precisely!
Exercise 5.5.E. 10 ⇒ 10.
Prove that if F
=∫ f
on I
= (a, b)
and f is right (left) continuous and finite at p ∈ I , then ′
′
f (p) = F+ (p) (respectively, F− (p)) .
(5.5.E.6)
Deduce that if f is continuous and finite on I , all its primitives on I are exact on I . [Hint: Fix ε > 0. If f is right continuous at p, there is c ∈ I (c > p), with |f (x) − f (p)| < ε for x ∈ [p, c).
(5.5.E.7)
Fix such an x. Let G(t) = F (t) − tf (p),
t ∈ E
1
.
(5.5.E.8)
Deduce that G (t) = f (t) − f (p) for t ∈ I − Q . By Corollary 1 of §4, ′
|G(x) − G(p)| = |F (x) − F (p) − (x − p)f (p)| ≤ M (x − p),
with M
≤ ε.
(5.5.E.9)
(Why?) Hence ∣ ΔF ∣ ∣ − f (p)∣ ≤ ε for x ∈ [p, c), ∣ Δx ∣
(5.5.E.10)
and so
5.5.E.2
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ΔF lim x→p
= f (p)
+
( why? );
(5.5.E.11)
Δx
similarly for a left-continuous f . ]
Exercise 5.5.E. 11 State and solve Problem 10 for the case I
= [a, b]
.
Exercise 5.5.E. 12 (i) Prove that if f is constant (f
= c ≠ ±∞)
on I − Q, then b
∫
f = (b − a)c
for a, b ∈ I .
(5.5.E.12)
a = a0 < a1 < ⋯ < an = b,
(5.5.E.13)
a
(ii) Hence prove that if f
= ck ≠ ±∞
on Ik = [ ak , ak+1 ) ,
then ∫ f exists on [a, b], and b
∫
n−1
f = ∑ (ak+1 − ak ) ck .
a
Show that this is true also if f
= ck ≠ ±∞
on I
k
(5.5.E.14)
k=0
− Qk
.
[Hint: Use Problem 6. ]
Exercise 5.5.E. 13 Prove that if ∫ f exists on each I
n
= [ an , bn ] ,
where
an+1 ≤ an ≤ bn ≤ bn+1 ,
n = 1, 2, … ,
(5.5.E.15)
then ∫ f exists on ∞
I = ⋃ [ an , bn ] ,
(5.5.E.16)
n=1
itself an interval with endpoints a = inf a and b = sup b [Hint: Fix some c ∈ I . Define
n,
n
a, b ∈ E
∗
.
1
t
Hn (t) = ∫
f on In , n = 1, 2, … .
(5.5.E.17)
c
Prove that (∀n ≤ m)
Thus H (t) is the same for all n such that t ∈ I all of I ; verify that I is, indeed, an interval.] n
n,
Hn = Hm on In ( since { In } ↑) .
so we may simply write H for H on I n
5.5.E.3
(5.5.E.18)
∞
= ⋃n=1 In .
Show that H = ∫ f on
https://math.libretexts.org/@go/page/23756
Exercise 5.5.E. 14 Continuing Problem 13, prove that ∫ f exists on an interval I iff it exists on each closed subinterval [a, b] ⊆ I . [Hint: Show that each I is the union of an expanding sequence I = [a , b ] . For example, if I = (a, b), a, b ∈ E n
1 an = a +
n
n
n
1
,
put
1 and bn = b −
for large n (how large?) ,
(5.5.E.19)
n
and show that I = ⋃ [ an , bn ] over such n. ]
(5.5.E.20)
n
5.5.E: Problems on Antiderivatives is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.5.E.4
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5.6: Differentials. Taylor’s Theorem and Taylor’s Series Recall (Theorem 2 of §1) that a function f is differentiable at p iff ′
Δf = f (p)Δx + δ(x)Δx,
with lim δ(x) = δ(p) = 0. It is customary to write (at p and x). Thus x→p
for f
df
′
and
(p)Δx
(5.6.1)
o(Δx)
for
δ(x)Δx; df
is called the differential of
Δf = df + o(Δx);
f
(5.6.2)
i.e., df approximates Δf to within o(Δx). More generally, given any function f
: E
d
n
1
→ E
f =d
n
and p, x ∈ E
f (p, x) = f
(n)
1
,
we define n
(p)(x − p ) ,
n = 0, 1, 2, … ,
(5.6.3)
where f is the n th derived function (Definition 2 in §1); d f is called the nth differential, or differential of order n, of f (at and x). In particular, d f = f (p)Δx = df . By our conventions, d f is always defined, as is f . (n)
n
1
′
n
As we shall see, good approximations of follows:
Δf
p
(n)
(suggested by Taylor) can often be obtained by using higher differentials (1), as 2
3
d f Δf = df +
d f +
d
n
f
+⋯ +
2!
3!
+ Rn ,
n!
n = 1, 2, 3, … ,
(5.6.4)
where n
Rn = Δf − ∑ k=1
k
d f (the "remainder term")
is the error of the approximation. Substituting the values of Δf and d ′
f (p) f (x) = f (p) +
f
′′
k
and transposing f (p), we have
f
(p)
(x − p) + 1!
2
(x − p )
f
(n)
(p)
+⋯ +
2!
n
(x − p )
+ Rn .
(5.6.6)
n!
Formula (3) is known as the nth Taylor expansion of f about p (with remainder term fixed and x as variable. Writing R (x) for R and setting n
(5.6.5)
k!
Rn
to be estimated). Usually we treat
p
as
n
n
f
(k)
(p)
Pn (x) = ∑ k=0
k
(x − p ) ,
(5.6.7)
k!
we have f (x) = Pn (x) + Rn (x).
(5.6.8)
The function P : E → E so defined is called the nth Taylor polynomial for f about p. Thus (3) yields approximations of f by polynomials P , n = 1, 2, 3, … . This is one way of interpreting it. The other (easy to remember) one is (2), which gives approximations of Δf by the d f . It remains, however, to find a good estimate for R . We do it next. 1
n
n
k
n
Theorem 5.6.1 (taylor) Let the function f : E → E and its first n derived functions be relatively continuous and finite on an interval I and differentiable on I − Q (Q countable). Let p, x ∈ I . Then formulas (2) and (3) hold, with 1
x
1 Rn =
∫ n!
f
(n+1)
n
(t) ⋅ (x − t) dt
("integral form of Rn ")
(5.6.9)
p
and n+1
|x − p| | Rn | ≤ Mn
for some real Mn ≤ (n + 1)!
sup
∣f (n+1) (t)∣ . ∣ ∣
(5.6.10)
t∈I−Q
5.6.1
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Proof By definition, R
n
or
= f − Pn ,
n
k
Rn = f (x) − f (p) − ∑ f
(k)
(x − p) (p)
.
We use the right side as a "pattern" to define a function replace p by a variable t. Thus we set
h : E
1
This time, we keep x fixed
→ E.
′
f (t) h(t) = f (a) − f (t) −
(5.6.11)
k!
k=1
f
(n)
(t)
(a − t) − ⋯ −
n
(a − t) for all t ∈ E
1!
1
( say , x = a ∈ I )
.
and
(5.6.12)
n!
Then h(p) = R and h(a) = 0. Our assumptions imply that h is relatively continuous and finite on on I − Q. Differentiating (4), we see that all cancels out except for one term n
I,
and differentiable
n
′
h (t) = −f
(n+1)
(a − t) (t)
,
t ∈ I − Q.
(Verify!)
(5.6.13)
n!
Hence by Definitions 1 and 2 of §5, a
1 −h(t) =
∫ n!
f
(n+1)
n
(s)(a − s) ds
on I
(5.6.14)
t
and a
1 ∫ n!
f
(n+1)
n
(t)(a − t) dt = −h(a) + h(p) = 0 + Rn = Rn
(for h(p) = Rn ) .
(5.6.15)
p
As x = a, (3') is proved. Next, let M =
sup
∣f (n+1) (t)∣ . ∣ ∣
(5.6.16)
t∈I−Q
If M
define
< +∞,
n+1
n+1
(t − a) g(t) = M
(a − t) for t ≥ a and g(t) = −M
for t ≤ a.
(n + 1)!
(5.6.17)
(n + 1)!
In both cases, n
′
|a − t|
g (t) = M
′
≥ | h (t)| on I − Q by (5).
(5.6.18)
n!
Hence, applying Theorem 1 in §4 to the functions h and g on the interval [a, p] (or [p, a]), we get |h(p) − h(a)| ≤ |g(p) − g(a)|,
(5.6.19)
or n+1
|a − p| | Rn − 0| ≤ M
.
(5.6.20)
(n + 1)!
Thus (3") follows, with M
n
Finally, if M
= +∞,
=M
.
we put (n + 1)! Mn = | Rn |
n+1
< M.
□
(5.6.21)
|a − p|
For real functions, we obtain some additional estimates of R . n
5.6.2
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Theorem 5.6.1
′
If f is real and n + 1 times differentiable on I , then for p ≠ x (x, p)) such that f Rn =
(n+1)
there are q
(p, x ∈ I ),
n,
(qn )
′
qn
in the interval
n+1
(x − p )
(p, x)
(respectively,
(5.6.22)
(n + 1)!
and f
(n+1)
Rn =
′
(qn )
′
(x − p) (x − qn )
n!
(Formulas (5') and (5") are known as the Lagrange and Cauchy forms of R
n
.
(5.6.23)
respectively.)
n,
Proof Exactly as in the proof of Theorem 1, we obtain the function h and formula (5). By our present assumptions, h is differentiable (hence continuous) on I , so we may apply to it Cauchy's law of the mean (Theorem 2 of §2) on the interval [a, p] (or [p, a] if p < a), where a = x ∈ I . For this purpose, we shall associate h with another suitable function there is a real q ∈ (a, p) (respectively, q ∈ (p, a)) such that ′
g
(to be specified later). Then by Theorem 2 of §2,
′
g (q)[h(a) − h(p)] = h (q)[g(a) − g(p)].
Here by the previous proof, h(a) = 0, h(p) = R
(5.6.24)
and
n,
f
′
(n+1) n
h (q) = −
(a − q ) .
(5.6.25)
n!
Thus f
′
(n+1)
(q)
g (q) ⋅ Rn =
n
(a − q ) [g(a) − g(p)].
(5.6.26)
n!
Now define g by g(t) = a − t,
t ∈ E
1
.
(5.6.27)
Then ′
g(a) − g(p) = −(a − p) and g (q) = −1,
so (6) yields (5") (with q
′ n
=q
(5.6.28)
and a = x) .
Similarly, setting g(t) = (a − t)
n+1
,
we obtain (5'). (Verify!) Thus all is proved.
□
Note 1. In (5') and (5"), the numbers q and q depend on n and are different in general (q of the function g . Since they are between p and x, they may be written as ′ n
n
n
′
′
≠ qn ) ,
′
qn = p + θn (x − p) and qn = p + θn (x − p),
where 0 < θ
n
0 if x > 0)
x
+ 1!
to within an error R
n
x
≈1+
+⋯ + 2!
(5.6.36) n!
with n+1
| Rn | < e
a
a
,
(5.6.37)
(n + 1)!
which tends to 0 as n → +∞. For a = 1 = x, we get 1 e =1+
1
1
+ 1!
e
+⋯ + 2!
1
+ Rn with 0 < Rn
p. The same argument then shows that f (x) < f (p) on one side of p and f (x) > f (p) on the other side; thus no local maximum or minimum can exist at p. This completes the proof. □ n
n
Examples (a') Let 2
1
f (x) = x on E
and p = 0.
(5.6.55)
Then ′
f (x) = 2x and f
′′
(x) = 2 > 0,
(5.6.56)
so ′
f (0) = 0 and f
By Theorem 2, f (p) = 0
2
′′
(0) = 2 > 0.
(5.6.57)
is a minimum value.
=0
It turns out to be a minimum on all of (−∞, 0) and increases on (0, +∞).
E
1
. Indeed,
′
f (x) > 0
for
x >0
, and
f
′
0
and f
for x in some Gp (ε), 0 < ε ≤ δ.
(5.6.61)
is the equation of the tangent at p, it follows that f (x) ≤ y; i.e., near p the curve lies below the 2
∈ CD
on G
p (δ)
implies that the curve near p lies above the tangent.
5.6.6
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This page titled 5.6: Differentials. Taylor’s Theorem and Taylor’s Series is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.6.7
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5.6.E: Problems on Tayior's Theorem Exercise 5.6.E. 1 Complete the proofs of Theorems 1, 1 , and 2. ′
Exercise 5.6.E. 2 Verify Note 1 and Examples (b) and (b ). ′′
Exercise 5.6.E. 3 Taking g(t) = (a − t)
s
in (6), find
, s > 0,
f Rn =
Obtain (5 ) and (5 ′
′′
)
(n+1)
(q)
s
n+1−s
(x − p ) (x − q )
( Schloemilch-Roche remainder ).
(5.6.E.1)
n!s
from it.
Exercise 5.6.E. 4 Prove that P (as defined) is the only polynomial of degree n such that n
f
(k)
(k)
(p) = Pn
(p),
k = 0, 1, … , n.
(5.6.E.2)
[Hint: Differentiate P n times to verify that it satisfies this property. For uniqueness, suppose this also holds for n
n k
P (x) = ∑ ak (x − p ) .
(5.6.E.3)
k=0
Differentiate P n times to show that P
(k)
(k)
(p) = f
(p) = ak k!,
(5.6.E.4)
so P = Pn . (Why?) ]
Exercise 5.6.E. 5 With P as defined, prove that if f is n times differentiable at p, then n
n
f (x) − Pn (x) = o ((x − p ) ) as x → p
(5.6.E.5)
(Taylor's theorem with Peano remainder term). [Hint: Let R(x) = f (x) − P (x) and n
R(x) δ(x) =
n
with δ(p) = 0.
(5.6.E.6)
(x − p)
Using the "simplified" L'Hôpital rule (Problem 3 in §3 ) repeatedly n times, prove that lim
x→p
5.6.E.1
δ(x) = 0. ]
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Exercise 5.6.E. 6 Use Theorem 1 with p = 0 to verify the following expansions, and prove that lim ′
n→∞
(a) sin x = x −
3
m
(−1)
5
x
+
3!
x
5!
−⋯ −
m
2m−1
x
(2m−1)!
(−1)
+
Rn = 0
.
2m+1
x
cos θm x
(2m+1)!
for all x ∈ E ; 1
2
(b) cos x = 1 −
x
2!
m
(−1)
4
+
x
−⋯ +
4!
m
2m
x
(−1)
−
(2m)!
2m+2
x
for
sin θm x
(2m+2)!
all x ∈ E . [Hints: Let f (x) = sin x and g(x) = cos x. Induction shows that 1
f
(n)
nπ (x) = sin(x +
) and g
(n)
nπ (x) = cos(x +
2
).
(5.6.E.7)
2
Using formula (5 ) , prove that ′
n+1
∣
∣
x
| Rn (x)| ≤ ∣
∣ → 0.
(5.6.E.8)
∣ (n + 1)! ∣
Indeed, x
n
/n!
is the general term of a convergent series n
x ∑
(see Chapter 4,
§13, Example (d)) .
(5.6.E.9)
n!
n
Thus x /n! → 0 by Theorem 4 of the same section. ]
Exercise 5.6.E. 7 For any s ∈ E and n ∈ N , define 1
s(s − 1) ⋯ (s − n + 1)
s (
) =
s with (
n!
n
) = 1.
(5.6.E.10)
0
Then prove the following. s
(i) lim
n→∞
n(
) =0
if s > 0 ,
n
(ii) lim
n→∞
s (
) =0
if s > −1 ,
n
(iii) For any fixed s ∈ E and x ∈ (−1, 1). 1
lim ( n→∞
s
n
) nx
= 0;
(5.6.E.11)
n
hence lim ( n→∞
s
n
)x
= 0.
(5.6.E.12)
n
∣ ∣ s [ Hints: (i) Let an = ∣n ( ) ∣ . Verify that ∣
n
∣ s ∣∣ s∣ s ∣ ∣ ∣ an = |s| 1 − 1− ⋯ ∣1 − ∣. ∣ ∣ 1∣∣ 2∣ n−1 ∣
5.6.E.2
(5.6.E.13)
https://math.libretexts.org/@go/page/24085
If s > 0, {a Prove that
n}
↓
for n > s + 1, so we may put L = lim a
n
a2n
= lim a2n ≥ 0.
s ∣n ∣ − 1− → e ∣ 2n ∣
0) ,
−
1 2
s
−
1 2
s
+ ε) an .
Then with ε → 0, obtain Le
+ ε) L.
1 2
s
(5.6.E.15)
≤L
.
this implies L = 0, as claimed.
(ii) For s > −1, s + 1 > 0, so by (i), s+1
s
(n + 1) (
) → 0; i.e., (s + 1) ( n+1
) → 0.
( Why? )
(5.6.E.16)
n
(iii) Use the ratio test to show that the series ∑ (
s
n
) nx
converges when |x| < 1.
n Then apply Theorem 4 of Chapter 4,
§13. ]
Exercise 5.6.E. 8 Continuing Problems 6 and 7, prove that n s
(1 + x )
= ∑( k=0
s
k
)x
+ Rn (x),
(5.6.E.17)
k
where R (x) → 0 if either |x| < 1, or x = 1 and s > −1, or x = −1 and s > 0. [Hints: (a) If 0 ≤ x ≤ 1, use (5 ) for n
′
s Rn−1 (x) = (
n
)x
(1 + θn x)
s−n
,
0 < θn < 1. (Verify!)
(5.6.E.18)
n
Deduce that |R
n−1 (x)|
∣ ∣ s n ≤ ∣( ) x ∣ → 0. ∣ ∣ n
(b) If −1 ≤ x < 0, write (5
′′
)
Use Problem 7( iii) if |x| < 1 or Problem 7( ii ) if x = 1 .
as n−1
′
s Rn−1 (x) = (
n
) nx
′
−1
(1 + θn x) s
(
n
1 − θn 1+
)
′ θn x
. (Check!)
(5.6.E.19)
As −1 ≤ x < 0, the last fraction is ≤ 1. (Why?) Also, ′
(1 + θn x)
Thus,
with
x
fixed,
these
s−1
s−1
≤ 1 if s > 1, and ≤ (1 + x )
expressions
are
bounded,
while
if s ≤ 1.
s (
(5.6.E.20)
n
) nx
→ 0
by
Problem
7(i)
n or ( iii ). Deduce that Rn−1 → 0, hence Rn → 0. ]
5.6.E.3
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Exercise 5.6.E. 9 Prove that n
k
k+1
x
ln(1 + x) = ∑(−1 )
+ Rn (x),
k
k=1
(5.6.E.21)
where lim R (x) = 0 if −1 < x ≤ 1 . [Hints: If 0 ≤ x ≤ 1, use formula (5 ). If −1 < x < 0, use formula (6) with g(t) = ln(1 + t) to obtain n→∞
n
′
ln(1 + x) Rn (x) =
(−1)n
n
1 − θn
(
⋅ x)
.
(5.6.E.22)
1 + θn x
Proceed as in Problem 8. ]
Exercise 5.6.E. 10 Prove that if f
: E
1
→ E
∗
is of class CD on [a, b] and if −∞ < f 1
′′
(x0 − a) + f (a);
b −a
(5.6.E.23)
i.e., the curve y = f (x) lies above the secant through (a, f (a)) and (b, f (b)). [Hint: This formula is equivalent to f (x0 ) − f (a)
f (b) − f (a) >
,
x0 − a
i.e.,
the
average
of
f
′
on
[a, x0 ]
is
(5.6.E.24)
b −a
strictly
greater
than
the
average
of
f
′
on
[a, b],
′
which follows because f decreases on (a, b). ( Explain! )]
Exercise 5.6.E. 11 Prove that if a, b, r, and s are positive reals and r + s = 1, then r
s
a b
≤ ra + sb.
(5.6.E.25)
(This inequality is important for the theory of so-called L -spaces.) [Hints: If a = b , all is trivial. Therefore, assume a < b . Then p
a = (r + s)a < ra + sb < b.
Use Problem 10 with x
0
= ra + sb ∈ (a, b)
(5.6.E.26)
and f (x) = ln x, f
′′
1 (x) = −
x2
< 0.
(5.6.E.27)
Verify that x0 − a = x0 − (r + s)a = s(b − a)
5.6.E.4
(5.6.E.28)
https://math.libretexts.org/@go/page/24085
and r ⋅ ln a = (1 − s) ln a; hence deduce that r ⋅ ln a + s ⋅ ln b − ln a = s(ln b − ln a) = s(f (b) − f (a)).
(5.6.E.29)
After substitutions, obtain r
s
f (x0 ) = ln(ra + sb) > r ⋅ ln a + s ⋅ ln b = ln(a b ). ]
(5.6.E.30)
Exercise 5.6.E. 12 Use Taylor's theorem (Theorem 1') to prove the following inequalities: − − − − − (a) √1 + x < 1 + if x > −1, x ≠ 0 . (b) cos x > 1 − x if x ≠ 0 . (c) < arctan x < x if x > 0 . x
3
3
1
2
2
x
2
1+x
(d) x > sin x > x −
1 6
3
x
if x > 0 .
5.6.E: Problems on Tayior's Theorem is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.6.E.5
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5.7: The Total Variation (Length) of a Function f - E1 → E This page is a draft and is under active development. The question that we shall consider now is how to define reasonably (and precisely) the notion of the length of a curve (Chapter 4, §10) described by a function f : E → E over an interval I = [a, b], i.e., f [I ]. 1
We proceed as follows (see Figure 24).
Subdivide [a, b] by a finite set of points P
= { t0 , t1 , … , tm } ,
with
a = t0 ≤ t1 ≤ ⋯ ≤ tm = b; P
(5.7.1)
is called a partition of [a, b]. Let qi = f (ti ) ,
i = 1, 2, … , m,
(5.7.2)
and, for i = 1, 2, … , m, Δi f = qi − qi−1 = f (ti ) − f (ti−1 ) .
We also define m
m
S(f , P ) = ∑ | Δi f | = ∑ | qi − qi−1 | . i=1
Geometrically, |Δ f | = |q the length of the polygon i
i
− qi−1 |
(5.7.3)
i=1
is the length of the line segment L [q
i−1
, qi ]
in E, and S(f , P ) is the sum of such lengths, i.e.,
m
W = ⋃ L [ qi−1 , qi ]
(5.7.4)
i=1
inscribed into f [I ]; we denote it by ℓW = S(f , P ).
(5.7.5)
ti−1 ≤ c ≤ ti .
(5.7.6)
Now suppose we add a new partition point c, with
Then we obtain a new partition
5.7.1
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Pc = { t0 , … , ti−1 , c, ti , … , tm } ,
(5.7.7)
called a refinement of P , and a new inscribed polygon W in which L [q , q ] is replaced by two segments, L [q, q ] , where q = f (c); see Figure 24. Accordingly, the term | Δ f | = | q − q | in S(f , P ) is replaced by c
i−1
i
i
i
| qi − q| + |q − qi−1 | ≥ | qi − qi−1 |
i
L [ qi−1 , q]
and
i−1
(triangle law).
(5.7.8)
It follows that S(f , P ) ≤ S (f , Pc ) ; i.e., ℓW ≤ ℓ Wc .
(5.7.9)
Hence we obtain the following result.
Corollary 5.7.1 The sum S(f , P ) = ℓW cannot decrease when P is refined. Thus when new partition points are added, S(f , P ) grows in general; i.e., it approaches some supremum value (finite or not). Roughly speaking, the inscribed polygon W gets "closer" to the curve. It is natural to define the desired length of the curve to be the lub of all lengths ℓW , i.e., of all sums S(f , P ) resulting from the various partitions P . This supremum is also called the total variation of f over [a, b], denoted V [a, b]. f
Definition 1 Given any function f
: E
1
→ E,
and I
= [a, b] ⊂ E
1
,
we set m
Vf [I ] = Vf [a, b] = sup S(f , P ) = sup ∑ |f (ti ) − f (ti−1 )| ≥ 0 in E P
where the supremum is over all partitions Briefly, we denote it by V .
P
P = { t0 , … , tm }
of
∗
,
(5.7.10)
i=1
I.
We call
Vf [I ]
the total variation, or length, of
f
on
I.
f
Note 1. If f is continuous on [a, b], the image set A = f [I ] is an arc (Chapter 4, §10). It is customary to call V [I ] the length of that arc, denoted ℓ A or briefly ℓA. Note, however, that there may well be another function g, continuous on an interval J, such that g[J] = A but V [I ] ≠ V [J], and so ℓ A ≠ ℓ A. Thus it is safer to say "the length of A as described by f on I." Only for simple arcs (where f is one to one), is "ℓA" unambiguous. (See Problems 6-8.) f
f
f
g
f
g
In the propositions below, f is an arbitrary function, f
Theorem 5.7.1 (additivity of V
f
: E
1
→ E
.
)
If a ≤ c ≤ b, then Vf [a, b] = Vf [a, c] + Vf [c, b];
(5.7.11)
i.e., the length of the whole equals the sum of the lengths of the parts. Proof Take any partition P
= { t0 , … , tm }
of [a, b]. If c ∉ P , refine P to Pc = { t0 , … , ti , c, ti , … , tm } .
(5.7.12)
Then by Corollary 1, S(f , P ) ≤ S (f , P ) . c
Now P splits into partitions of [a, c] and [c, b], namely, c
P
′
= { t0 , … , ti−1 , c} and P
′′
= {c, ti , … , tm } ,
(5.7.13)
so that ′
S (f , P ) + S (f , P
′′
) = S (f , Pc ) . (Verify!)
5.7.2
(5.7.14)
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Hence (as V
f
is the lub of the corresponding sums)
,
Vf [a, c] + Vf [c, d] ≥ S (f , Pc ) ≥ S(f , P ).
(5.7.15)
As P is an arbitrary partition of [a, b], we also have Vf [a, c] + Vf [c, b] ≥ sup S(f , P ) = Vf [a, b].
(5.7.16)
Thus it remains to show that, conversely, Vf [a, b] ≥ Vf [a, c] + Vf [c, b].
The latter is trivial if V [a, b] = +∞. Thus assume V [a, b] = K < +∞. Let [c, b], respectively. Then P =P ∪P is a partition of [a, b], and f
f
∗
′
(5.7.17)
P
′
and
P
′′
be any partitions of
[a, c]
and
′′
′
S (f , P ) + S (f , P
′′
) = S (f , P
∗
) ≤ Vf [a, b] = K,
(5.7.18)
whence ′
S (f , P ) ≤ K − S (f , P
Keeping P fixed and varying P ′′
′
,
we see that K − S (f , P
′′
)
′′
).
(5.7.19)
is an upper bound of all S (f , P
Vf [a, c] ≤ K − S (f , P
′′
′
)
over [a, c]. Hence
)
(5.7.20)
) ≤ K − Vf [a, c].
(5.7.21)
or S (f , P
Similarly, varying P
′′
,
′′
we now obtain Vf [c, b] ≤ K − Vf [a, c]
(5.7.22)
Vf [a, c] + Vf [c, b] ≤ K = Vf [a, b],
(5.7.23)
or
as required. Thus all is proved.
□
Corollary 5.7.2 (monotonicity of V
f
)
If a ≤ c ≤ d ≤ b, then Vf [c, d] ≤ Vf [a, b].
(5.7.24)
Proof By Theorem 1, Vf [a, b] = Vf [a, c] + Vf [c, d] + Vf [d, b] ≥ Vf [c, d].
□
(5.7.25)
Definition 2 If V
f
[a, b] < +∞,
we say that f is of bounded variation on I
= [a, b],
and that the set f [I ] is rectifiable (by f on I ).
Corollary 5.7.3 For each t ∈ [a, b] , |f (t) − f (a)| ≤ Vf [a, b].
(5.7.26)
Hence if f is of bounded variation on [a, b], it is bounded on [a, b]. Proof
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If t ∈ [a, b], let P
= {a, t, b},
so
|f (t) − f (a)| ≤ |f (t) − f (a)| + |f (b) − f (t)| = S(f , P ) ≤ Vf [a, b],
(5.7.27)
proving our first assertion. Hence (∀t ∈ [a, b])
This proves the second assertion.
|f (t)| ≤ |f (t) − f (a)| + |f (a)| ≤ Vf [a, b] + |f (a)|.
(5.7.28)
□
Note 2. Neither boundedness, nor continuity, nor differentiability of does. (See Problems 1 and 3.)
on
f
[a, b]
implies
Vf [a, b] < +∞,
but boundedness of
f
′
Corollary 5.7.4 A function f is finite and constant on [a, b] iff V
f
[a, b] = 0
.
The proof is left to the reader. (Use Corollary 3 and the definitions.
Theorem 5.7.2 Let f , g, h be real or complex (or let f and g be vector valued and h scalar valued). Then on any interval I (i) V
|f |
≤ Vf
(ii) V
f ±g
(iii) V
hf
= [a, b],
we have
;
≤ Vf + Vg ;
and
≤ sVf + rVh ,
with r = sup
t∈I
|f (t)|
and s = sup
t∈I
|h(t)|
.
Hence if f , g, and h are of bounded variation on I , so are f ± g, hf , and |f |. Proof We first prove (iii). Take any partition P
= { t0 , … , tm }
of I . Then
| Δi hf | = |h (ti ) f (ti ) − h (ti−1 ) f (ti−1 )| ≤ |h (ti ) f (ti ) − h (ti−1 ) f (ti )| + |h (ti−1 ) f (ti ) − h (ti−1 ) f (ti−1 )| = |f (ti ) ∥ Δi h| + |h (ti−1 )∥ Δi f | ≤ r | Δi h| + s | Δi f | .
Adding these inequalities, we obtain S(hf , P ) ≤ r ⋅ S(h, P ) + s ⋅ S(f , P ) ≤ rVh + sVf .
(5.7.29)
As this holds for all sums S(hf , P ), it holds for their supremum, so Vhf = sup S(hf , P ) ≤ rVh + sVf ,
(5.7.30)
||f (ti ) | − |f (ti−1 ) || ≤ |f (ti ) − f (ti−1 )| .
(5.7.31)
as claimed. Similarly, (i) follows from
The analogous proof of (ii) is left to the reader. Finally, (i)-(iii) imply that V
f
, Vf ±g
, and V
hf
are finite if V
f
, Vg ,
and V are. This proves our last assertion. h
□
Note 3. Also f /h is of bounded variation on I if f and h are, provided h is bounded away from 0 on I ; i.e., (∃ε > 0)
|h| ≥ ε on I .
(5.7.32)
(See Problem 5.)
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Special theorems apply in case the range space E is E or E 1
n
∗
( or C
n
)
.
Theorem 5.7.3 (i) A real function f is of bounded variation on I
= [a, b]
iff f
= g−h
for some nondecreasing real functions g and h on I .
(ii) If f is real and monotone on I , it is of bounded variation there. Proof We prove (ii) first. Let f
↑
on I . If P
= { t0 , … , tm } ,
then ti ≥ ti−1 implies f (ti ) ≥ f (ti−1 ) .
Hence Δ
if
≥ 0;
i.e., |Δ
if|
= Δi f .
(5.7.33)
Thus m
m
m
S(f , P ) = ∑ | Δi f | = ∑ Δi f = ∑ [f (ti ) − f (ti−1 )] i=1
i=1
i=1
= f (tm ) − f (t0 ) = f (b) − f (a)
for any P . (Verify!) This implies that also Vf [I ] = sup S(f , P ) = f (b) − f (a) < +∞.
(5.7.34)
Thus (ii) is proved. Now for (i), let f = g − h with g ↑ and h ↑ on I . By (ii), by Theorem 2 (last clause). Conversely, suppose V
f
[I ] < +∞.
g
and h are of bounded variation on
I.
Hence so is
Then define g(x) = Vf [a, x], x ∈ I , and h = g − f on I ,
so f
= g − h,
f = g−h
(5.7.35)
and it only remains to show that g ↑ and h ↑.
To prove it, let a ≤ x ≤ y ≤ b. Then Theorem 1 yields Vf [a, y] − Vf [a, x] = Vf [x, y];
(5.7.36)
i.e., g(y) − g(x) = Vf [x, y] ≥ |f (y) − f (x)| ≥ 0
(by Corollary 3).
(5.7.37)
Hence g(y) ≥ g(x). Also, as h = g − f , we have h(y) − h(x)
= g(y) − f (y) − [g(x) − f (x)] = g(y) − g(x) − [f (y) − f (x)] ≥0
by (2).
Thus h(y) ≥ h(x). We see that a ≤ x ≤ y ≤ b implies g(x) ≤ g(y) and h(x) ≤ h(y), so h ↑ and g ↑, indeed.
□
Theorem 5.7.4 (i) A function f
: E
1
→ E
n
∗
( C
n
)
is of bounded variation on I
(ii) If this is the case, then finite limits f (p
+
)
and f (q
−
)
= [a, b]
iff all of its components (f
1,
f2 , … , fn )
are.
exist for every p ∈ [a, b) and q ∈ (a, b].
Proof (i) Take any partition P
= { t0 , … , tm }
of I . Then n
| fk (ti ) − fk (ti−1 )|
2
≤ ∑ | fj (ti ) − fj (ti−1 )|
2
= |f (ti ) − f (ti−1 )|
2
;
(5.7.38)
j=1
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i.e., |Δ
i fk |
≤ | Δi f | , i = 1, 2, … , m.
Thus (∀P )
and V
fk
≤ Vf
S (fk , P ) ≤ S(f , P ) ≤ Vf ,
(5.7.39)
follows. Thus Vf < +∞ implies Vf
k
The converse follows by Theorem 2 since f
n
=∑
k=1
(ii) For real monotone functions, f (p ) and f (q of bounded variation, for by Theorem 3, +
−
< +∞,
⃗ . fk e k
k = 1, 2, … , n.
(5.7.40)
(Explain!)
exist by Theorem 1 of Chapter 4, §5. This also applies if f is real and
)
f = g − h with g ↑ and h ↑ on I ,
(5.7.41)
and so +
f (p
+
) = g (p
+
) − h (p
) and f (q
−
) = g (q
−
) − h (q
−
) exist.
(5.7.42)
The limits are finite since f is bounded on I by Corollary 3. Via components (Theorem 2 of Chapter 4, §3), this also applies to functions f applies to complex functions (treat C as E (*and so it extends to functions f : E 2
1
: E
1
→ C
. (Why?) In particular, (ii) as well). □
→ E n
.
n
We also have proved the following corollary.
Corollary 5.7.5 A complex function f 5).
: E
1
→ C
is of bounded variation on [a, b] iff its real and imaginary parts are. (See Chapter 4, §3, Note
This page titled 5.7: The Total Variation (Length) of a Function f - E1 → E is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.7.6
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5.7.E: Problems on Total Variation and Graph Length Exercise 5.7.E. 1 In
the
following
cases
show
that
though
Vf [I ] = +∞,
I . ( In case (iii), f is continuous, and in case (iv), it is even differentiable
(i) For I
1
if x ∈ R( rational ), and
0
if x ∈ E
= [a, b](a < b), f (x) = {
1
is
f
bounded
on
on I . )
− R.
(ii) f (x) = sin ; f (0) = 0; I = [a, b], a ≤ 0 ≤ b, a < b . (iii) f (x) = x ⋅ sin ; f (0) = 0; I = [0, 1] . (iv) f (x) = x sin ; f (0) = 0; I = [0, 1] . [Hints: (i) For any m there is P , with 1
x
π
2x 1
2
x2
| Δi f | = 1,
i = 1, 2, … , m,
(5.7.E.1)
so S(f , P ) = m → +∞ . (iii) Let 1 Pm = {0,
m
Prove that S (f , Pm ) ≥ ∑k=1
1 k
1 ,
m
1 ,…,
, 1} .
(5.7.E.2)
("piecewise monotone") .
(5.7.E.3)
m −1
2
→ +∞. ]
Exercise 5.7.E. 2 Let f
: E
1
→ E
1
be monotone on each of the intervals [ ak−1 , ak ] ,
k = 1, … , n
Prove that n
Vf [ a0 , an ] = ∑ |f (ak ) − f (ak−1 )| .
(5.7.E.4)
k=1
In particular, show that this applies if f (x) = ∑ c x (polynomial), with c ∈ E . [Hint: It is known that a polynomial of degree n has at most n real roots. Thus it is piecewise monotone, for its derivative vanishes at finitely many points (being of degree n − 1). Use Theorem 1 in §2. ] n
i=1
i
i
i
1
Exercise 5.7.E. 3 Prove that if then ⇒
f
is finite and relatively continuous on
I = [a, b],
with a bounded derivative,
Vf [a, b] ≤ M (b − a).
However, we may have V
f
[I ] < +∞,
and yet |f
′
| = +∞
′
|f | ≤ M ,
on
§
I − Q( see 4),
(5.7.E.5)
at some p ∈ I .
− [Hint: Take f (x) = √x on [−1, 1]. ] 3
5.7.E.1
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Exercise 5.7.E. 4 Complete the proofs of Corollary 4 and Theorems 2 and 4.
Exercise 5.7.E. 5 Prove Note 3. [Hint: If |h| ≥ ε on I , show that ∣
∣ | Δi h| 1 1 ∣ − ∣ ≤ 2 ∣ h (ti ) h (ti−1 ) ∣ ε
(5.7.E.6)
and hence 1 S(
S(h, P ) ,P) ≤
h
Deduce that
1 h
ε
≤
2
Vh ε
2
.
(5.7.E.7)
is of bounded variation on I if h is. Then apply Theorem 2( iii) to
1 h
⋅ f. ]
Exercise 5.7.E. 6 Let g : E → E (real) and and b = g(d). Let 1
1
f : E
1
→ E
be relatively continuous on
J = [c, d]
and
I = [a, b],
respectively, with
h = f ∘ g.
a = g(c)
(5.7.E.8)
Prove that if g is one to one on J, then (i) g[J] = I , so f and h describe one and the same arc A = f [I ] = h[J] ; (ii) V [I ] = V [J]; i.e., ℓ A = ℓ A . [Hint for (ii): Given P = {a = t , … , t = b} , show that the points s = g (t ) form a partition P of J = [c, d], with S (h, P ) = S(f , P ). Hence deduce V [I ] ≤ V [J]. Then prove that V [J] ≤ V [I ], taking an arbitrary P = {c = s , … , s = d} , f
h
f
h
0
m
′
f
h
−1
i
′
i
h
′
f
0
m
and defining P = { t0 , … , tm } , with ti = g (si ) . What if g(c) = b, g(d) = a?]
Exercise 5.7.E. 7 Prove that if f , h : E
1
→ E
are relatively continuous and one to one on I
= [a, b]
and J = [c, d], respectively, and if
f [I ] = h[J] = A
(i.e., f and h describe the same simple arc A),
(5.7.E.9)
then ℓf A = ℓh A.
Thus for simple arcs, ℓ A is independent of f . [Hint: Define g : J → E by g = f ∘ h. Use Problem 6 and Chapter also if g(c) = b and g(d) = a, i.e., g ↓ on J. ]
(5.7.E.10)
f
1
−1
4,
§9, Theorem
3.
First check that Problem 6 works
Exercise 5.7.E. 8 Let I
= [0, 2π]
and define f , g, h : E
1
→ E
2
(C )
by
5.7.E.2
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f (x) = (sin x, cos x), g(x) = (sin 3x, cos 3x), h(x) = (sin
1 x
, cos
Show that f [I ] = g[I ] = h[I ] (the unit circle; call it A),
1 x
(5.7.E.11)
) with h(0) = (0, 1).
yet
ℓf A = 2π
,
ℓg A = 6π,
while
Vh [I ] = +∞.
(Thus the result
of Problem 7 fails for closed curves and nonsimple arcs. )
Exercise 5.7.E. 9 In Theorem 3, define two functions G, H : E
1
→ E
1
by 1
G(x) = 2
[ Vf [a, x] + f (x) − f (a)]
(5.7.E.12)
and H (x) = G(x) − f (x) + f (a).
(G and H are called, respectively, the positive and negative variation functions for f . )
(5.7.E.13)
Prove that
(i) G ↑ and H ↑ on [a, b]; (ii) f (x) = G(x) − [H (x) − f (a)] (thus the functions f and g of Theorem 3 are not unique); (iii) V [a, x] = G(x) + H (x) ; (iv) if f = g − h, with g ↑ and h ↑ on [a, b], then f
VG [a, b] ≤ Vg [a, b], and VH [a, b] ≤ Vh [a, b] ;
(5.7.E.14)
(v) G(a) = H (a) = 0 .
Exercise 5.7.E. 10∗ Prove that if f
: E
1
→ E
n
∗
( C
n
)
is of bounded variation on I
= [a, b],
then f has at most countably many discontinuities in
I.
[Hint: Apply Problem 5 of Chapter 4, §5. Proceed as in the proof of Theorem 4 in §
§
7. Finally, use Theorem 2 of Chapter 1, 9. ]
5.7.E: Problems on Total Variation and Graph Length is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.7.E.3
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5.8: Rectifiable Arcs. Absolute Continuity This page is a draft and is under active development. If a function f
: E
1
→ E
is of bounded variation (§7) on an interval I
= [a, b],
we can define a real function v on I by f
vf (x) = Vf [a, x](= total variation of f on [a, x]) for x ∈ I ;
(5.8.1)
is called the total variation function, or length function, generated by f on I . Note that v ↑ on I . (Why?) We now consider the case where f is also relatively continuous on I , so that the set A = f [I ] is a rectifiable arc (see §7, Note 1 and Definition 2). vf
f
Definition 1 A function f
: E
1
→ E
is (weakly) absolutely continuous on I
= [a, b]
iff V
f
[I ] < +∞
and f is relatively continuous on I .
Theorem 5.8.1 The following are equivalent: (i) f is (weakly) absolutely continuous on I
= [a, b]
;
(ii) v is finite and relatively continuous on I ; and f
(iii) (∀ε > 0) (∃δ > 0) (∀x, y ∈ I |0 ≤ y − x < δ) V
f
[x, y] < ε
.
Proof We shall show that (ii) ⇒ (iii) ⇒ (i) ⇒ (ii). (ii) ⇒ (iii). As I
= [a, b]
is compact, (ii) implies that v is uniformly continuous on I (Theorem 4 of Chapter 4, §8). Thus f
(∀ε > 0) (∃δ > 0) (∀x, y ∈ I |0 ≤ y − x < δ)
vf (y) − vf (x) < ε.
(5.8.2)
However, vf (y) − vf (x) = Vf [a, y] − Vf [a, x] = Vf [x, y]
(5.8.3)
by additivity (Theorem 1 in §7). Thus (iii) follows. (iii) ⇒ (i). By Corollary 3 of §7, |f (x) − f (y)| ≤ V
f
[x, y].
Therefore, (iii) implies that
(∀ε > 0) (∃δ > 0) (∀x, y ∈ I ||x − y| < δ)
|f (x) − f (y)| < ε,
(5.8.4)
and so f is relatively (even uniformly) continuous on I . Now with ε and δ as in (iii), take a partition P
= { t0 , … , tm }
ti − ti−1 < δ,
Then (∀i)V
f
[ ti−1 , ti ] < ε.
of I so fine that
i = 1, 2, … , m.
(5.8.5)
Adding up these m inequalities and using the additivity of V
f
,
we obtain
m
Vf [I ] = ∑ Vf [ ti−1 , ti ] < mε < +∞.
(5.8.6)
i=1
Thus (i) follows, by definition. That (i) ⇒ (ii) is given as the next theorem.
□
5.8.1
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Theorem 5.8.2 If V v .
f
[I ] < +∞
and if f is relatively continuous at some p ∈ I (over I
= [a, b]),
then the same applies to the length function
f
Proof We consider left continuity first, with a < p ≤ b . Let ε > 0. By assumption, there is δ > 0 such that ε |f (x) − f (p)|
0)|h| ≥ ε on I .
Exercise 5.8.E. 5 Prove that (i) If f is finite and ≠ 0 on I ′
= [a, b],
so also is v
′ f
( with one-sided
derivatives at the endpoints of the interval); moreover,
∣ f′ ∣ ∣ ∣ = 1 on I . ′ ∣ vf ∣
′
§
′
Thus show that f / v is the tangent unit vector (see 1) f
(5.8.E.1)
.
(ii) Under the same assumptions, F = f ∘ v is differentiable on J = [0, v (b)] (with one-sided derivatives at the endpoints of the interval) and F [J] = f [I ]; i.e., F and f describe the same simple arc, with V [I ] = V [I ]. Note that this is tantamount to a change of parameter. Setting s = v (t), i.e., t = v (s), we have −1
f
f
F
f
−1
f
−1
f (t) = f (v
f
(s)) = F (s),
f
with the arclength s as parameter.
5.8.E: Problems on Absolute Continuity and Rectifiable Arcs is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.8.E.1
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5.9: Convergence Theorems in Differentiation and Integration This page is a draft and is under active development. Given ′
Fn = ∫
fn or Fn = fn ,
n = 1, 2, … ,
(5.9.1)
′
what can one say about ∫ lim f or (lim F ) if the limits exist? Below we give some answers, for complete range spaces E (such as E ). Roughly, we have lim F = (lim F ) on I − Q if n
n
n
′
′ n
(a) {F
n (p)}
n
converges for at least one p ∈ I and
(b) {F
converges uniformly.
Here
is a finite or infinite interval in
′ n}
I
I −Q ⊆ I
0
E
1
and
Q
is countable. We include in
Q
the endpoints of
I
(if any), so
(= interior of I ).
Theorem 5.9.1 Let F
: E
n
1
→ E (n = 1, 2, …)
(a) lim
Fn (p)
n→∞
(b) F
′ n
be finite and relatively continuous on I and differentiable on I − Q. Suppose that
exists for some p ∈ I ;
→ f ≠ ±∞(uniformly)
on J − Q for each finite subinterval J ⊆ I ;
(c) E is complete. Then (i) lim
n→∞
(ii) F
Fn = F
=∫ f
(iii) (lim F
exists uniformly on each finite subinterval J ⊆ I ;
on I ; and
′ n)
=F
′
′
= f = limn→∞ Fn
on I − Q .
Proof Fix ε > 0 and any subinterval J ⊆ I of length δ < ∞, with p ∈ J (p as in (a)). By (b), F there is a k such that for m, n > k ,
′ n
→ f
(uniformly) on J − Q, so
ε
′
| Fn (t) − f (t)|
k, x ∈ J and | Fm (x) − Fn (x) − Fm (p) + Fn (p)| ≤ ε|x − p| ≤ εδ.
(5.9.5)
As ε is arbitrary, this shows that the sequence { Fn − Fn (p)}
(5.9.6)
satisfies the uniform Cauchy criterion (Chapter 4, §12, Theorem 3). Thus as E is complete, uniformly on J. So does {F } , for {F (p)} converges, by (a). Thus we may set n
{ Fn − Fn (p)}
converges
n
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F = lim Fn (uniformly) on J,
(5.9.7)
proving assertion (i). Here by Theorem 2 of Chapter 4, §12, F is relatively continuous on each such J ⊆ I , hence on all of I . Also, letting m → +∞ (with n fixed), we have F → F in (3), and it follows that for n > k and x ∈ G (δ) ∩ I . m
p
|F (x) − Fn (x) − F (p) + Fn (p)| ≤ ε|x − p| ≤ εδ.
(5.9.8)
Having proved (i), we may now treat p as just any point in I . Thus formula (4) holds for any globe G show that
p (δ),
F
′
= f on I − Q; i.e., F = ∫
Indeed, if p ∈ I − Q, each F is differentiable at Thus for each n, there is δ > 0 such that n
p
p ∈ I.
f on I .
(by assumption), and
p ∈ I
0
We now
(5.9.9)
(since I − Q ⊆ I
0
by our convention).
n
∣ Fn (x) − Fn (p) ∣ ∣ ΔFn ∣ ε ′ ′ ∣ − Fn (p)∣ = ∣ − Fn (p)∣ < ∣ Δx ∣ x −p 2 ∣ ∣
for all x ∈ G
¬p
(5.9.10)
.
(δn ) ; Gp (δn ) ⊆ I
By assumption (b) and by (4), we can fix n so large that ε
′
| Fn (p) − f (p)|
0, there is δ > 0 such that ∣ ΔF ∣ ∣ ΔF ΔFn ∣ ∣ ΔFn ∣ ′ ′ ∣ − f (p)∣ ≤ ∣ − ∣+∣ − Fn (p)∣ + | Fn (p) − f (p)| < 2ε, x ∈ Gp (δ). ∣ Δx ∣ ∣ Δx ∣ Δx ∣ Δx ∣
(5.9.13)
This shows that ΔF lim x→p
= f (p) for p ∈ I − Q,
(5.9.14)
Δx
i.e., F = f on I − Q, with f finite by assumption, and F finite by (4). As F is also relatively continuous on I , assertion (ii) is proved, and (iii) follows since F = lim F and f = lim F . □ ′
′ n
n
Note 1. The same proof also shows that F → F (uniformly) on each closed subinterval J ⊆ I if F → f (uniformly) on for all such J (with the other assumptions unchanged). In any case, we then have F → F (pointwise) on all of I . ′ n
n
J −Q
n
We now apply Theorem 1 to antiderivatives.
Theorem 5.9.2 Let the functions f : E → E, n = 1, 2, … , have antiderivatives, F = ∫ f , on I . Suppose (uniformly) on each finite subinterval J ⊆ I , with f finite there. Then ∫ f exists on I , and 1
n
n
x
∫ p
x
f =∫ p
n
E
is complete and
fn → f
x
lim fn = lim ∫
n→∞
n→∞
fn for any p, x ∈ I .
(5.9.15)
p
Proof Fix any p ∈ I . By Note 2 in §5, we may choose
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x
Fn (x) = ∫
fn for x ∈ I .
(5.9.16)
p
Then F
n (p)
=∫
p
p
and so lim
fn = 0,
exists, as required in Theorem 1(a).
Fn (p) = 0
n→∞
Also, by definition, each F is relatively continuous and finite on countable sets Q need not be the same, so we replace them by n
and differentiable, with
I
′
Fn = fn ,
on
I − Qn .
The
n
∞
Q = ⋃ Qn
(5.9.17)
n=1
(including in
Q
I − Q ⊆ I − Qn ,
also the endpoints of I , if any. Then Q is countable (see Chapter 1, §9, Theorem 2), and so all F are differentiable on I − Q, with F = f there. ′ n
n
n
Additionally, by assumption, fn → f (uniformly)
(5.9.18)
Fn → f (uniformly) on J − Q
(5.9.19)
on finite subintervals J ⊆ I . Hence ′
for all such J, and so the conditions of Theorem 1 are satisfied. By that theorem, then, F =∫
f = lim Fn exists on I
(5.9.20)
and, recalling that x
Fn (x) = ∫
fn ,
(5.9.21)
p
we obtain for x ∈ I x
∫
x
f = F (x) − F (p) = lim Fn (x) − lim Fn (p) = lim Fn (x) − 0 = lim ∫
p
fn .
(5.9.22)
p
As p ∈ I was arbitrary, and f
= lim fn
(by assumption), all is proved.
□
Note 2. By Theorem 1, the convergence x
∫
x
fn → ∫
p
f
( i.e. , Fn → F )
(5.9.23)
p
is uniform in x (with p fixed), on each finite subinterval J ⊆ I . We now "translate" Theorems 1 and 2 into the language of series.
Corollary 5.9.1 Let E and the functions F
n
: E
1
→ E
be as in Theorem 1. Suppose the series ∑ Fn (p)
(5.9.24)
converges for some p ∈ I , and ′
∑ Fn
(5.9.25)
converges uniformly on J − Q, for each finite subinterval J ⊆ I . Then ∑ F converges uniformly on each such J, and n
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∞
F = ∑ Fn
(5.9.26)
n=1
is differentiable on I − Q, with ′
∞ ′
F
∞ ′
(5.9.27)
n = 1, 2, … ,
(5.9.28)
= ( ∑ Fn ) = ∑ Fn there. n=1
n−1
In other words, the series can be differentiated termwise. Proof Let n
sn = ∑ Fk , k=1
be the partial sums of ∑ F . From our assumptions, it then follows that the (Verify!) Thus the conclusions of Theorem 1 hold, with F replaced by s . n
n
We have F
= limsn
and F
′
′
= (lim sn ) = lim
′ sn ,
sn
satisfy all conditions of Theorem 1.
n
whence (7) follows.
□
Corollary 5.9.2 If E and the f are as in Theorem 2 and if ∑ f converges uniformly to f on each finite interval J ⊆ I , then ∫ f exists on and n
n
x
∞
x
∫
f =∫
p
∞
x
∑ fn = ∑ ∫
p
n=1
I,
fn for any p, x ∈ I .
(5.9.29)
p
n=1
Briefly, a uniformly convergent series can be integrated termwise.
Theorem 5.9.3 (Power Series) Let r be the convergence radius of n
∑ an (x − p ) ,
an ∈ E, p ∈ E
1
.
(5.9.30)
Suppose E is complete. Set ∞ n
f (x) = ∑ an (x − p )
on I = (p − r, p + r).
(5.9.31)
n=0
Then the following are true: (i) f is differentiable and has an exact antiderivative on I . (ii) f
′
∞
(x) = ∑
n=1
n−1
nan (x − p )
and ∫
x
p
an
∞
f =∑
n=0
n+1
n+1
(x − p )
,x ∈ I
.
(iii) r is also the convergence radius of the two series in (ii). (iv) ∑
∞ n=0
n
an (x − p )
is exactly the Taylor series for f (x) on I about p.
Proof We prove (iii) first. By Theorem 6 of Chapter 4, §13, r = 1/d, where ¯ ¯¯¯¯¯ ¯ − − n d = lim √ an .
Let r be the convergence radius of ∑ na ′
n (x
n−1
− p)
,
(5.9.32)
so
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1
′
r = d
′
with d
¯ ¯¯¯¯¯ ¯ − − n − = lim √nan .
′
(5.9.33)
However, lim √− n = 1 (see §3, Example (e)). It easily follows that n
d
Hence r
′
= 1/ d
′
= 1/d = r
¯ ¯¯¯¯¯ ¯ ¯ ¯¯¯¯¯ ¯ 2 − − − n − n − = lim √ nan = 1 ⋅ lim √ an = d .
′
(5.9.34)
.
The convergence radius of ∑
an n+1
is determined similarly. Thus (iii) is proved.
n+1
(x − p )
Next, let an
n
fn (x) = an (x − p ) and Fn (x) =
Then the F are differentiable on I , with F
′ n
n
n+1
(x − p )
, n = 0, 1, 2, … .
(5.9.35)
n+1
there. Also, by Theorems 6 and 7 of Chapter 4, §13, the series
= fn
n
′
∑ Fn = ∑ an (x − p )
(5.9.36)
converges uniformly on each closed subinterval J ⊆ I = (p − r, p + r). Thus the functions Corollary 1, with Q = ∅, and the f satisfy Corollary 2. By Corollary 1, then,
Fn
satisfy all conditions of
n
∞
F = ∑ Fn
(5.9.37)
n=1
is differentiable on I , with ∞
∞
′
n
′
F (x) = ∑ Fn (x) = ∑ an (x − p ) n=1
= f (x)
(5.9.38)
n=1
for all x ∈ I . Hence F is an exact antiderivative of f on I , and (8) yields the second formula in (ii). Quite similarly, replacing F and follows. This proves (i) as well. n
F
by
fn
and
one shows that
f,
is differentiable on
f
I,
and the first formula in (ii)
Finally, to prove (iv), we apply (i)-(iii) to the consecutive derivatives of f and obtain ∞
f
(k)
n−k
(x) = ∑ n(n − 1) ⋯ (n − k + 1)an (x − p )
(5.9.39)
n=k
for each x ∈ I and k ∈ N . If x = p, all terms vanish except the first term (n = k), i.e., k!a
k.
f
(n)
Thus f
(k)
(p) = k! ak , k ∈ N .
We may rewrite it as
(p)
an =
,
n = 0, 1, 2, … ,
(5.9.40)
n!
since f
(0)
(p) = f (p) = a0 .
Assertion (iv) now follows since ∞
∞ n
f (x) = ∑ an (x − p ) n=0
Note 3. If ∑ a
n (x
n
− p)
f
(n)
(p)
=∑ n=0
n
(x − p ) ,
x ∈ I , as required.
□
(5.9.41)
n!
converges also for x = p − r or x = p + r, so does the integrated series n+1
(x − p) ∑ an
(5.9.42) n+1
since we may include such x in I . However, the derived series example (see §6, Problem 9), the expansion
n−1
∑ nan (x − p )
2
5.9.5
(Why?) For
x +
2
x.
3
x ln(1 + x) = x −
need not converge at such
−⋯
(5.9.43)
3
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converges for x = 1 but the derived series 2
1 −x +x
−⋯
(5.9.44)
does not. In this respect, there is the following useful rule for functions f
: E
1
→ E
m
(*C ). m
Corollary 5.9.3 Let a function f
: E
1
→ E
m
∗
( C
m
)
be relatively continuous on [p, x ] (or [x
0,
0
,
p]) x0 ≠ p.
If
∞ n
f (x) = ∑ an (x − p ) for p ≤ x < x0 (respectively, x0 < x ≤ p),
(5.9.45)
n=0
and if ∑ a
n (x0
− p)
n
converges, then necessarily ∞
f (x0 ) = ∑ an (x0 − p)
n
.
(5.9.46)
n=0
Proof The proof is sketched in Problems 4 and 5. Thus in the above example, f (x) = ln(1 + x) defines a continuous function on [0, 1], with ∞
n
n−1
x
f (x) = ∑(−1 )
on [0, 1].
We gave a direct proof in §6, Problem 9. However, by Corollary 3, it suffices to prove this for the convergence of ∞
[0, 1),
which is much easier. Then
n−1
(−1)
∑ n=1
(5.9.47)
n
n=1
(5.9.48) n
yields the result for x = 1 as well. This page titled 5.9: Convergence Theorems in Differentiation and Integration is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.9.6
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5.9.E: Problems on Convergence in Differentiation and Integration Exercise 5.9.E. 1 Complete all proof details in Theorems 1 and 3, Corollaries 1 and 2, and Note 3.
Exercise 5.9.E. 2 Show that assumptions (a) and (c) in Theorem 1 can be replaced by F applies to incomplete spaces E as well.) [Hint: F → F ( pointwise e), together with formula ( 3) , implies F
n
n
→ F
n
→ F
(pointwise) on I . (In this form, the theorem (uniformly) on I . ]
Exercise 5.9.E. 3 Show that Theorem 1 fails without assumption (b), even if F → F (uniformly) and if F is differentiable on I . [Hint: For a counterexample, try F (x) = sin nx, on any nondegenerate I . Verify that F → 0 (uniformly), yet (b) and assertion (iii) fail. n
1
n
n
n
Exercise 5.9.E. 4 Prove Abel's theorem (Chapter 4, §13, Problem 15) for series n
∑ an (x − p ) ,
(5.9.E.1)
with all a in E ( or in C ) but with x, p ∈ E . [Hint: Split a (x − p) into components.] m
∗
m
1
n
n
n
Exercise 5.9.E. 5 Prove Corollary 3. [Hint: By Abel's theorem (see Problem 4),
we may put ∞ n
∑ an (x − p )
= F (x)
(5.9.E.2)
n=0
This implies that Show that
uniformly on [p, x0 ] (respectively, [ x0 , p]) .
assumption. Also f
=F
on [p, x
0)
((x0 , p]) .
F
is relatively continuous at
f (x0 ) = lim f (x) = lim F (x) = F (x0 )
x0 .
(Why?) So is
f,
by
(5.9.E.3)
as x → x from the left (right).] 0
Exercise 5.9.E. 6 In the following cases, find the Taylor series of F about 0 by integrating the series of F . Use Theorem 3 and Corollary 3 to find the convergence radius r and to investigate convergence at −r and r. Use (b) to find a formula for π. (a) F (x) = ln(1 + x) ; (b) F (x) = arctan x; (c) F (x) = arcsin x. ′
5.9.E.1
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Exercise 5.9.E. 7 Prove that x
ln(1 − t)
∫ 0
∞
dt = ∑ t
n=1
n
x
2
for x ∈ [−1, 1].
(5.9.E.4)
n
[Hint: Use Theorem 3 and Corollary 3. Take derivatives of both sides.] 5.9.E: Problems on Convergence in Differentiation and Integration is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.9.E.2
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5.10: Sufficient Condition of Integrability. Regulated Functions This page is a draft and is under active development. In this section, we shall determine a large family of functions that do have antiderivatives. First, we give a general definition, valid for any range space (T , p) (not necessarily E). The domain space remains E . 1
Definition 1 A function f : E → (T , p) is said to be regulated on an interval I ⊆ E , with endpoints a < b, iff the limits f (p ) and f (p ) , other than ±∞, exist at each p ∈ I . However, at the endpoints a, b, if in I , we only require f (a ) and f (b ) to exist. 1
1
−
+
+
−
Examples (a) If f is relatively continuous and finite on I , it is regulated. (b) Every real monotone function is regulated (see Chapter 4, §5, Theorem 1). (c) If f
: E
1
→ E
n
∗
( C
n
)
has bounded variation on I , it is regulated (§7, Theorem 4).
(d) The characteristic function of a set B, denoted C
B,
is defined by
CB (x) = 1 if x ∈ B and CB = 0 on − B.
For any interval J ⊆ E
1
, CJ
(5.10.1)
is regulated on E . 1
(e) A function f is called a step function on I iff I can be represented as the union, I = ⋃ I , of a sequence of disjoint intervals I such that f is constant and ≠ ±∞ on each I . Note that some I may be singletons, {p}. k
k
k
k
k
If the number of the I is finite, we call f a simple step function. k
When the range space T is E, we can give the following convenient alternative definition. If, say, f f = ∑ ak CI
k
on I ,
= ak ≠ ±∞
on I , then k
(5.10.2)
k
where C is as in (d). Note that ∑ Ik
k
ak CIk (x)
always exists for disjoint I . (Why?) k
Each simple step function is regulated. (Why?)
Theorem 5.10.1 Let the functions f , g, h be real or complex (or let f , g be vector valued and h scalar valued). If they are regulated on I , so are f ± g, f h, and |f |; so also is f /h if h is bounded away from 0 on I , i.e., (∃ε > 0)|h| ≥ ε on I.
Proof The proof, based on the usual limit properties, is left to the reader. We shall need two lemmas. One is the famous Heine-Borel lemma.
lemma 5.10.1 (Heine-Borel) If a closed interval A = [a, b] in E (or E 1
n
)
is covered by open sets G
i (i
A ⊆ ⋃ Gi ,
∈ I ),
i.e., (5.10.3)
i∈I
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then A can be covered by a finite number of these G . i
Proof The proof was sketched in Problem 10 of Chapter 4, §6. Note 1. This fails for nonclosed intervals A. For example, let A = (0, 1) ⊆ E
1
1 and Gn = (
, 1) .
(5.10.4)
n
Then ∞
m
A = ⋃ Gn ( verify! ), but not A ⊆ ⋃ Gn n=1
(5.10.5)
n=1
for any finite m. (Why?) The lemma also fails for nonopen sets G . For example, cover A by singletons {x}, x ∈ A. Then none of the {x} can be dropped! i
lemma 5.10.2 If a function f
: E
1
is regulated on I
→ T
= [a, b],
then f can be uniformly approximated by simple step functions on I .
That is, for any ε > 0, there is a simple step function g, with ρ(f , g) ≤ ε on I ; hence sup ρ(f (x), g(x)) ≤ ε.
(5.10.6)
x∈I
Proof By assumption, f (p
−
)
exists for each p ∈ (a, b], and f (p
+
exists for p ∈ [a, b), all finite.
)
Thus, given ε > 0 and any p ∈ I , there is G (δ ) (δ depending on p ) such that ρ(f (x), r) < ε whenever r = f (p x ∈ (p − δ, p), and ρ(f (x), s) < ε whenever s = f (p ) and x ∈ (p, p + δ); x ∈ I
−
p
)
and
+
We choose such a G (δ) for every p ∈ I . Then the open globes Lemma 1, I is covered by a finite number of such globes, say, p
Gp = Gp (δ)
cover the closed interval
I = [a, b],
so by
n
I ⊆ ⋃ Gp
k
(δk ) ,
a ∈ Gp , a ≤ p1 < p2 < ⋯ < pn ≤ b. 1
(5.10.7)
k=1
We now define the step function g on I as follows. If x = p
k,
we put g(x) = f (pk ) ,
If x ∈ [a, p
1)
,
k = 1, 2, … , n.
then −
g(x) = f (p
1
If x ∈ (p
1,
(5.10.8)
p1 + δ1 ) ,
).
(5.10.9)
).
(5.10.10)
then +
g(x) = f (p
1
More generally, if x is in G
¬pk
(δk )
but in none of the G
pi
(δi ) , i < k, −
g(x) = f (p
k
we put
)
if x < pk
(5.10.11)
)
if x > pk .
(5.10.12)
and +
g(x) = f (p
k
Then by construction, ρ(f , g) < ε on each G
pk
,
hence on I .
□
5.10.2
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*Note 2. If T is complete, we can say more: f is regulated on I on I . (See Problem 2.)
= [a, b]
iff f is uniformly approximated by simple step functions
Theorem 5.10.2 If f : E f in I .
1
→ E
is regulated on an interval I
⊆E
1
and if E is complete, then ∫ f exists on I , exact at every continuity point of
0
In particular, all continuous maps f
: E
1
→ E
n
∗
( C
n
)
have exact primitives.
Proof In view of Problem 14 of §5, it suffices to consider closed intervals. Thus let I = [a, b], a < b, in E . Suppose first that f is the characteristic function C of a subinterval J ⊆ I with endpoints c and d (a ≤ c ≤ d ≤ b), so f = 1 on J and f = 0 on I − J. We then define F (x) = x on J, F = c on [a, c], and F = d on [d, b] (see Figure 25). Thus F is continuous (why?), and F = f on I − {a, b, c, d} (why?). Hence F = ∫ f on I ; i.e., characteristic functions are integrable. 1
J
′
Then, however, so is any simple step function m
f = ∑ ak CI ,
(5.10.13)
k
k=1
by repeated use of Corollary 1 in §5. Finally, let f be any regulated function on I . Then by Lemma 2, for any that
εn =
1 n
,
there is a simple step function
gn
such
1 sup | gn (x) − f (x)| ≤ x∈I
,
n = 1, 2, … .
(5.10.14)
n
As → 0, this implies that g → f (uniformly) on I (see Chapter 4, §12, Theorem 1). Also, by what was proved above, the step functions g have antiderivatives, hence so has f (Theorem 2 in §9); ; i.e., F = ∫ f exists on I , as claimed. Moreover, ∫ f is exact at continuity points of f in I (Problem 10 in §5). □ 1
n
n
n
0
In view of the sufficient condition expressed in Theorem 2, we can now replace the assumption "∫ f exists" in our previous theorems by "f is regulated" (provided E is complete). For example, let us now review Problems 7 and 8 in §5.
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Theorem 5.10.3 (weighted law of the mean) Let f
: E
1
(E complete) and g : E
→ E
1
→ E
be regulated on I
1
= [a, b],
with g ≥ 0 on I . Then the following are true: b
(i) There is a finite c ∈ E (called the "g-weighted mean of f on I ") such that ∫
a
gf = c ∫
b
a
g
.
(ii) If f , too, is real and has the Darboux property on I , then c = f (q) for some q ∈ I . Proof Indeed, as f and g are regulated, so is gf by Theorem 1. Hence by Theorem 2, ∫ f and ∫ gf exist on I . The rest follows as in Problems 7 and 8 of §5. □
Theorem 5.10.4 (second law of the mean) Suppose f and g are real, f is monotone with f
=∫ f
on I , and g is regulated on I ; I
′
b
q
∫
f g = f (a) ∫
a
= [a, b].
Then
b
g + f (b) ∫
a
g for some q ∈ I .
(5.10.15)
q
Proof To fix ideas, let f
↑;
i.e., f
′
≥0
on I .
The formula f = ∫ f means that countable). As g is regulated, ′
is relatively continuous (hence regulated) on
f
and differentiable on
I
I −Q
(Q
x
∫
g = G(x)
(5.10.16)
a a
does exist on I , so G has similar properties, with G(a) = ∫
a
By Theorems 1 and 2, ∫ f G
′
= ∫ fg
exists on I . (Why?) Hence by Corollary 5 in §5, so does ∫ Gf
b
∫
.
g =0
b
a
b b
′
fg = ∫
f G = f (x)G(x)|a − ∫
a
′
= f (b)G(b) − ∫
and we have
′
Gf .
(5.10.17)
a
Now G has the Darboux property on I (being relatively continuous), and f Problems 7 and 8 in §5, b
,
b
Gf
a
∫
′
′
≥ 0.
Also, ∫ G and ∫ Gf exist on I . Thus by ′
b
Gf
′
= G(q) ∫
a
f
′
b
= G(q)f (x)| , a
q ∈ I.
(5.10.18)
a
Combining all, we obtain the required result (1) since b
∫
f g = f (b)G(b) − ∫
′
Gf
a
= f (b)G(b) − f (b)G(q) + f (a)G(q) b
= f (b) ∫
q
g + f (a) ∫
q
We conclude with an application to infinite series. Given f ∞
∫ a
: E
1
→ E,
x
f =
lim x→+∞
∫
g.
□
a
we define
a
f and ∫
a
−∞
a
f =
lim x→−∞
∫
f
(5.10.19)
x
if these integrals and limits exist. We say that ∫
∞
a
f
and ∫
a
−∞
f
converge iff they exist and are finite.
5.10.4
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Theorem 5.10.5 (integral test of convergence) If f
: E
1
→ E
1
is nonnegative and nonincreasing on I
then
= [a, +∞),
∞
∞
∫
f converges iff ∑ f (n) does.
a
(5.10.20)
n=1
Proof As f
↓, f
is regulated, so ∫ f exists on I
= [a, +∞).
We fix some natural k ≥ a and define x
F (x) = ∫
f for x ∈ I .
(5.10.21)
k
By Theorem 3(iii) in §5, F
↑
on I . Thus by monotonicity, x
lim
F (x) =
lim
x→+∞
exists in E
∗
;
so does ∫
k
a
f.
x→+∞
x
k
f
k
f
(5.10.22)
k
k
f =∫
a
a
f =∫
Since ∫
where ∫
∞
∫
x
f +∫
a
f,
(5.10.23)
k
is finite by definition, we have ∞
∫
∞
f < +∞ iff ∫
a
f < +∞.
(5.10.24)
k
Similarly, ∞
∞
∑ f (n) < +∞
iff ∑ f (n) < +∞.
n=1
(5.10.25)
n=k
Thus we may replace "a " by "k ." Let In = [n, n + 1),
and define two step functions, g and h, constant on each I
n,
n = k, k + 1, … ,
(5.10.26)
by
h = f (n) and g = f (n + 1) on In , n ≥ k.
Since f
↓,
we have g ≤ f
≤h
on all I
n,
(5.10.27)
hence on J = [k, +∞). Therefore, x
∫
x
g ≤∫
k
x
f ≤∫
k
h for x ∈ J.
(5.10.28)
k
Also, m
∫
m−1
n+1
h = ∑∫
k
n=k
m−1
h = ∑ f (n),
n
(5.10.29)
n=k
since h = f (n) (constant) on [n, n + 1), and so n+1
∫
n+1 n+1
h(x)dx = f (n) ∫
n
1dx = f (n) ⋅ x|n
= f (n)(n + 1 − n) = f (n).
(5.10.30)
n
Similarly, m
∫ k
m−1
m
g = ∑ f (n + 1) = n=k
∑
f (n).
(5.10.31)
n=k+1
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Thus we obtain m
m
∑
f (n) = ∫
m
g ≤∫
k
n=k+1
m−1
m
f ≤∫
k
h = ∑ f (n),
k
(5.10.32)
n=k
or, letting m → ∞ , ∞
f (n) ≤ ∫
∞
k
f
is finite iff ∑
∞ n=1
f (n)
f ≤ ∑ f (n).
k
n=k+1
Hence ∫
∞
∞
∑
is, and all is proved.
(5.10.33)
n=k
□
Examples (continued) (f) Consider the hyperharmonic series 1 ∑
p
(Problem 2 of Chapter 4,
§13).
(5.10.34)
n
Let 1 f (x) =
p
,
x ≥ 1.
(5.10.35)
x
If p = 1, then f (x) = 1/x, so ∫
x
1
f = ln x → +∞
as x → +∞. Hence ∑ 1/n diverges.
If p ≠ 1, then ∞
∫ 1
so ∫
∞
1
f
x
f =
lim x→+∞
∫
1−p
x f =
1
lim x→+∞
∣
x
∣ ,
(5.10.36)
1 −p ∣ 1
converges or diverges according as p > 1 or p < 1, and the same applies to the series ∑ 1/n
p
.
(g) Even nonregulated functions may be integrable. Such is Dirichlet's function (Example (c) in Chapter 4, §1). Explain, using the countability of the rationals. This page titled 5.10: Sufficient Condition of Integrability. Regulated Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.10.6
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5.10.E: Problems on Regulated Functions Exercise 5.10.E. 1 Complete all details in the proof of Theorems 1 − 3 .
Exercise 5.10.E. 1
′
Explain Examples (a) − (g) .
Exercise 5.10.E. 2∗ Prove Note 2. More generally, assuming T to be complete, prove that if gn → f ( uniformly ) on I = [a, b]
(5.10.E.1)
and if the g are regulated on I , so is f . [Hint: Fix p ∈ (a, b]. Use Theorem 2 of Chapter 4, §11 with n
X = [a, p], Y = N ∪ {+∞}, q = +∞, and F (x, n) = gn (x).
(5.10.E.2)
Then show that −
f (p
) = lim x→p
+
similarly for f (p
−
lim gn (x) exists;
(5.10.E.3)
n→∞
).]
Exercise 5.10.E. 3 Given f , g : E → E , define f ∨ g and f ∧ g as in Problem 12 of Chapter 4, §8. Using the hint given there, show that f ∨ g and f ∧ g are regulated if f and g are. 1
1
Exercise 5.10.E. 4 Show that the function g ∘ f need not be regulated even if g and f are. [Hint: Let 1 f (x) = x ⋅ sin
x , g(x) =
, and f (0) = g(0) = 0 with I = [0, 1].
x
(5.10.E.4)
|x|
Proceed.]
Exercise 5.10.E. 5 ⇒
Given f
: E
1
→ (T , ρ),
regulated on I , put −
j(p) = max {ρ (f (p), f (p
+
)) , ρ (f (p), f (p
−
)) , ρ (f (p
+
) , f (p
))} ;
(5.10.E.5)
call it the jump at p. (i) Prove that f is discontinuous at p ∈ I iff j(p) > 0, i.e., iff 0
5.10.E.1
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1 (∃n ∈ N )
j(p) >
.
(5.10.E.6)
n
(ii) For a fixed n ∈ N , prove that a closed subinterval J ⊆ I contains at most finitely many x with j(x) > 1/n . [Hint: Otherwise, there is a sequence of distinct points x ∈ J, j (x ) > , hence a subsequence x → p ∈ J. (Why?) Use Theorem 1 of Chapter 4, §2, to show that f (p ) or f (p ) fails to exist. ] 1
m
−
m
mk
n
+
Exercise 5.10.E. 6 Show that if f : E → (T , ρ) is regulated on I , then it has at most countably many discontinuities in I ; all are of the "jump" type (Problem 5). [Hint: By Problem 5, any closed subinterval J ⊆ I contains, for each n, at most finitely many discontinuities x with j(x) > 1/n. Thus for n = 1, 2, … , obtain countably many such x. ] 1
⇒
Exercise 5.10.E. 7 Prove that if E is complete, all maps f : E → E, with V [I ] < +∞ on I = [a, b], are regulated on I . [Hint: Use Corollary 1, Chapter 4, §2, to show that f (p ) and f (p ) exist. Say, 1
f
−
+
xn → p with xn < p
but {f (x
n )}
is not Cauchy. Then find a subsequence, {x
nk
} ↑,
(xn , p ∈ I ) ,
and ε > 0 such that
|f (xnk+1 ) − f (xnk )| ≥ ε,
Deduce a contradiction to V
f
(5.10.E.7)
k = 1, 3, 5, …
(5.10.E.8)
[I ] < +∞.
Provide a similar argument for the case xn > p. ]
Exercise 5.10.E. 8 Prove that if f : E → (T , ρ) is regulated on compact subset of I . 1
I,
then
¯¯¯¯¯¯¯¯ ¯
f [B]
(the closure
¯¯¯¯¯¯¯¯ ¯
of f [B])
[Hint: Given {z } in f [B], find {y } ⊆ f [B] such that ρ (z , y ) → 0 (use "imitate" the proof of Theorem 1 in Chap ter 4, §8 suitably. Distinguish the cases: (i) all but finitely many x are < p ; (ii) infinitely many x exceed p; or (iii) infinitely many x equal p.] m
m
m
m
is compact in
(T , ρ)
whenever
Theorem 3 of Chapter 3,
B
is a
§16). Then
m
m
m
5.10.E: Problems on Regulated Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.10.E.2
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5.11: Integral Definitions of Some Functions This page is a draft and is under active development. By Theorem 2 in §10, ∫ f exists on I whenever the function f such an f is given, we can define a new function F by setting
: E
1
→ E
is regulated on
I,
and E is complete. Hence whenever
x
F =∫
f
(5.11.1)
a
on I for some a ∈ I . This is a convenient method of obtaining new continuous functions, differentiable on I − Q (Q countable). We shall now apply it to obtain new definitions of some functions previously defined in a rather strenuous step-by-step manner. I. Logarithmic and Exponential Functions. From our former definitions, we proved that x
1
ln x = ∫
dt,
x > 0.
(5.11.2)
t
1
Now we want to treat this as a definition of logarithms. We start by setting 1 f (t) =
,
t ∈ E
1
, t ≠ 0,
(5.11.3)
t
and f (0) = 0 . Then f is continuous on I = (0, +∞) and J = (−∞, 0), so it has an exact primitive on I and J separately (not on E ). Thus we can now define the log function on I by 1
x
1
∫ 1
dt = log x (also written ln x) for x > 0.
(5.11.4)
t
By the very definition of an exact primitive, the log function is continuous and differentiable on I f . Thus we again have the symbolic formula
= (0, +∞)
; its derivative on I is
1
′
(log x ) =
,
x > 0.
(5.11.5)
x
If x < 0, we can consider log(−x). Then the chain rule (Theorem 3 of §1) yields 1
′
(log(−x)) =
.
(Verify!)
(5.11.6)
x
Hence ′
1
(log |x| ) =
for x ≠ 0.
(5.11.7)
x
Other properties of logarithms easily follow from (1). We summarize them now.
Theorem 5.11.1 (i) log 1 = ∫
1
1
1 t
dt = 0
.
(ii) log x < log y whenever 0 < x < y . (iii) lim
x→+∞
log x = +∞
and lim
+
x→0
log x = −∞
.
(iv) The range of log is all of E . 1
(v) For any positive x, y ∈ E , 1
x log(xy) = log x + log y and log(
) = log x − log y.
(5.11.8)
y
5.11.1
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(vi) log a
r
= r ⋅ log a, a > 0, r ∈ N
(vii) log e = 1, where e = lim
n→∞
.
(1 +
n
1
)
n
.
Proof (ii) By (2), (log x )
′
>0
on I
= (0, +∞),
so log x is increasing on I .
(iii) By Theorem 5 in §10, ∞
lim
1
log x = ∫
x→+∞
dt = +∞
(5.11.9)
t
1
since ∞
1
∑
= +∞
§13,
(Chapter 4,
Example (b)).
(5.11.10)
n
n=1
Hence, substituting y = 1/x, we obtain 1 lim log y = +
lim
log
.
(5.11.11)
x
x→+∞
y→0
However, by Theorem 2 in §5 (substituting s = 1/t), 1/x
1 log x
x
1
=∫
1
dt = − ∫ t
1
ds = − log x.
(5.11.12)
s
1
Thus 1 lim log y = +
1 x
log
=− x
x→+∞
y→0
as claimed. (We also proved that log
lim
lim
log x = −∞
(5.11.13)
x→+∞
)
= − log x.
(iv) Assertion (iv) now follows by the Darboux property (as in Chapter 4, §9, Example (b)). (v) With x, y fixed, we substitute t = xs in xy
1
∫
dt = log xy
(5.11.14)
t
1
and obtain xy
y
1
log xy = ∫
1
dt = ∫ t
1 1
y
1
=∫ 1/x
ds
1/x
s
1
ds + ∫ s
ds s
1
1 = − log
+ log y x
= log x + log y.
Replacing y by 1/y here, we have x log
1 = log x + log
y
= log x − log y.
(5.11.15)
y
Thus (v) is proved, and (vi) follows by induction over r . (vii) By continuity, n
1 log e = lim log x = lim log (1 + x→e
n→∞
) n
log(1 + 1/n) = lim n→∞
,
(5.11.16)
1/n
where the last equality follows by (vi). Now, L'Hôpital's rule yields
5.11.2
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log(1 + x)
1
lim
= lim x
x→0
Letting x run over
1 n
→ 0,
we get (vii).
= 1.
(5.11.17)
1 +x
x→0
□
Note 1. Actually, (vi) holds for any r ∈ E , with a as in Chapter 2, §§11-12. One uses the techniques from that section to prove it first for rational r, and then it follows for all real r by continuity. However, we prefer not to use this now. 1
r
Next, we define the exponential function ("exp") to be the inverse of the log function. This inverse function exists; it is continuous (even differentiable) and strictly increasing on its domain (by Theorem 3 of Chapter 4, §9 and Theorem 3 of Chapter 5, §2) since the log function has these properties. From (log x ) = 1/x we get, as in 2, ′
′
(exp x ) = exp x
§2,
(cf.
Example (B)).
(5.11.18)
The domain of the exponential is the range of its inverse, i.e., E (cf. Theorem 1(iv)). Thus exp x is defined for all range of exp is the domain of log, i.e., (0, +∞). Hence exp x > 0 for all x ∈ E . Also, by definition, 1
x ∈ E
1
.
The
1
exp(log x) = x for x > 0 exp 0 = 1 (cf. Theorem 1(i)), and r
exp r = e for r ∈ N .
Indeed, by Theorem 1(vi) and (vii), log e = r ⋅ log e = r. Hence (6) follows. If the definitions and rules of Chapter 2, §§11-12 are used, this proof even works for any r by Note 1. Thus our new definition of exp agrees with the old one. r
Our next step is to give a new definition of a , for any a, r ∈ E r
r
a
r
log a
r
r
r
=a b ;
rs
a
(a > 0).
We set
= exp(r ⋅ log a) or = r ⋅ log a
In case r ∈ N , (8) becomes Theorem 1(vi). Thus for natural also obtain, for a, b > 0 , (ab )
1
r
r,
(r ∈ E
1
).
our new definition of
s
r+s
= (a ) ;
r
a
s
=a a ;
r
a
is consistent with the previous one. We
(r, s ∈ E
1
).
(5.11.19)
The proof is by taking logarithms. For example, r
log(ab)
= r log ab = r(log a + log b) = r ⋅ log a + r ⋅ log b r
= log a
Thus (ab)
r
r
r
=a b .
r
+ log b
r
r
= log(a b ).
Similar arguments can be given for the rest of (9) and other laws stated in Chapter 2, §§11-12.
We can now define the exponential to the base a(a > 0) and its inverse, log , as before (see the example in Chapter 4, §5 and Example (b) in Chapter 4, §9). The differentiability of the former is now immediate from (7), and the rest follows as before. a
II. Trigonometric Functions. These shall now be defined in a precise analytic manner (not based on geometry). We start with an integral definition of what is usually called the principal value of the arcsine function, x
arcsin x = ∫ 0
1 − − − − − √ 1 − t2
dt.
(5.11.20)
We shall denote it by F (x) and set 1 f (x) =
(F = f = 0
on E
1
−I
.) Thus by definition, F
=∫ f
− −−− − √ 1 − x2
on I = (−1, 1).
(5.11.21)
on I .
Note that ∫ f exists and is exact on I since f is continuous on I . Thus ′
F (x) = f (x) =
and so F is strictly increasing on I . Also, F (0) = ∫
0
0
f =0
1 − −−− − >0 √ 1 − x2
for x ∈ I ,
(5.11.22)
.
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We also define the number π by setting − − 1
π
= 2 arcsin √
2
− − 1 . 2
c
= 2F (c) = 2 ∫ 2
c =√
f,
0
(5.11.23)
Then we obtain the following theorem.
Theorem 5.11.2 F
has the limits −
F (1
π ) =
π
+
and F (−1
) =−
.
2
Thus F becomes relatively continuous on I
¯ ¯ ¯
= [−1, 1]
(5.11.24)
2
if one sets π
F (1) =
π and F (−1) = −
2
,
(5.11.25)
2
i.e., π
π
arcsin 1 =
and arcsin(−1) = − 2
.
(5.11.26)
2
Proof We have x
c
F (x) = ∫
f =∫
0
− − − − −
By substituting s = √1 − t
2
Now as x → 1
−
x
c
1 − − − − − √ 1 − t2
c
− −−− −
,
c
− −−− − 2 ,
f =∫
c
0
we have y = √1 − x
2
→ 0,
− − − − − √ 1 − s2
y
−
Similarly, one gets F
+
(−1
y→0
) = −π/2.
we obtain
ds = F (c) − F (y).
(Verify!)
(5.11.28)
and hence F (y) → F (0) = 0 (for F is continuous at 0). Thus
) = lim F (x) = lim ( ∫ x→1
(5.11.27)
1
dt = ∫
c −
F (1
c =√
f,
in the last integral and setting, for brevity, y = √1 − x
x
∫
− − 1 . 2
x
f +∫
c
f +∫
0
c
f) = ∫
y
c
f + F (c) = 2 ∫
0
π f =
.
(5.11.29)
2
0
□
The function F as redefined in Theorem 2 will be denoted by F F (x) = ∫ f , −1 ≤ x ≤ 1, and we may now write
0.
¯ ¯ ¯
It is a primitive of f on the closed interval I (exact on I ). Thus
x
0
0
1
π =∫ 2
0
f and π = ∫
0
1
f +∫
−1
1
f =∫
0
f.
(5.11.30)
−1
Note 2. In classical analysis, the last integrals are regarded as so-called improper integrals, i.e., limits of integrals rather than integrals proper. In our theory, this is unnecessary since F is a genuine primitive of f on I . ¯ ¯ ¯
0
For each integer n (negatives included), we now define F
n
: E
1
→ E
1
by ¯ ¯ ¯
n
Fn (x) = nπ + (−1 ) F0 (x) for x ∈ I = [−1, 1] ¯ ¯ ¯
Fn = 0
on − I .
is called the n th branch of the arcsine. Figure 26 shows the graphs of following theorem. Fn
5.11.4
F0
and
F1
( that of
F1
is dotted). We now obtain the
https://math.libretexts.org/@go/page/21247
Theorem 5.11.3 (i) Each F is differentiable on I n
= (−1, 1)
¯ ¯ ¯
and relatively continuous on I
= [−1, 1]
.
(ii) F is increasing on I if n is even, and decreasing if n is odd. ¯ ¯ ¯
n
n
(iii) F
′ n (x)
(−1)
=
(iv) F
n (−1)
√1−x2
on I . n π
= Fn−1 (−1) = nπ − (−1 )
2
n π
; Fn (1) = Fn−1 (1) = nπ + (−1 )
2
.
Proof
The proof is obvious from (12) and the properties of F . Assertion (iv) ensures that the graphs of the F add up to one curve. By (ii), each F is one to one (strictly monotone) on I . Thus it has a strictly monotone inverse on the interval J =F [[−1, 1]], i.e., on the F -image of I . For simplicity, we consider only 0
n
¯ ¯ ¯
n
¯¯¯¯ ¯ n
¯ ¯ ¯
n
n
¯ ¯¯¯ ¯
π
J0 = [−
π ,
2
2
π ] and J1 = [
3π ,
2
],
(5.11.31)
2
¯ ¯¯¯ ¯
as shown on the Y -axis in Figure 26. On these, we define for x ∈ J
0
sin x = F
−1
0
(x)
(5.11.32)
and − − − − − − − − 2 cos x = √ 1 − sin x ,
5.11.5
(5.11.33)
https://math.libretexts.org/@go/page/21247
and for x ∈ J
¯ ¯¯¯ ¯ 1
sin x = F
−1
1
(x)
(5.11.34)
and − − − − − − − − 2 cos x = −√ 1 − sin x .
On the rest of E
1
,
(5.11.35)
we define sin x and cos x periodically by setting sin(x + 2nπ) = sin x and cos(x + 2nπ) = cos x,
n = 0, ±1, ±2, … .
(5.11.36)
Note that by Theorem 3(iv), F
π
−1
(
0
) =F
−1
1
2
π (
) = 1.
(5.11.37)
2
Thus (13) and (14) both yield sin π/2 = 1 for the common endpoint π/2 of J and J We also have ¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
0
π sin(−
1,
so the two formulas are consistent.
3π ) = sin(
2
) = −1,
(5.11.38)
2
in agreement with (15). Thus the sine and cosine functions (briefly, s and c) are well defined on E . 1
Theorem 5.11.4 The sine and cosine functions (s and c ) are differentiable, hence continuous, on all of c = −s; that is,
E
1
,
with derivatives
′
s =c
and
′
′
′
(sin x ) = cos x and (cos x ) = − sin x.
(5.11.39)
Proof It suffices to consider the intervals the rest of E .
¯ ¯¯¯ ¯
J0
¯ ¯¯¯ ¯
and J
1,
for by (15), all properties of
s
and c repeat themselves, with period
2π,
on
1
By (13), s =F
−1
0
where F is differentiable on I that 0
= (−1, 1).
π
¯ ¯¯¯ ¯
on J0 = [−
π ,
2
],
(5.11.40)
2
Thus Theorem 3 of §2 shows that s is differentiable on J
0
= (−π/2, π/2)
and
1
′
s (q) =
′
whenever p ∈ I and q = F0 (p);
(5.11.41)
F (p) 0
i.e., q ∈ J and p = s(q). However, by Theorem 3(iii), 1
′
F (p) = 0
− − − − − 2 √1 − p
.
(5.11.42)
Hence, − − − − − − − − ′
2
s (q) = √ 1 − sin
q = cos q = c(q),
q ∈ J.
(5.11.43)
This proves the theorem for interior points of J as far as s is concerned. ¯ ¯¯¯ ¯ 0
As − − − − − 2 2 c = √ 1 − s = (1 − s )
1 2
on J0 (by (13)),
(5.11.44)
we can use the chain rule (Theorem 3 in §1) to obtain
5.11.6
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1
′
c =
2
−
(1 − s )
1 2
′
(−2s)s = −s
(5.11.45)
2
on noting that s
′
1
2
= c = (1 − s )
2
on J
0.
Similarly, using (14), one proves that s
′
=c
and c
′
= −s
¯ ¯¯¯ ¯
on J (interior of J ) . 1
1
Next, let q be an endpoint, say, q = π/2. We take the left derivative s(x) − s(q)
′
s− (q) = lim x→q
,
−
x ∈ J0 .
x −q
(5.11.46)
By L'Hôpital's rule, we get ′
s (x)
′
s− (q) = lim x→q
since s
′
=c
on J
0.
However, s = F
−1
0
−
= lim c(x) 1
x→q
− − − − − 2 .
is left continuous at q (why?); hence so is c = √1 − s ′
s− (q) = lim c(x) = c(q), x→q
Similarly, one shows that s
′ +
(q) = c(q).
(5.11.47)
−
as required.
(Why?) Therefore, (5.11.48)
−
Hence s (q) = c(q) and c (q) = −s(q) as before. ′
′
□
The other trigonometric functions reduce to s and c by their defining formulas sin x tan x =
cos x , cot x =
cos x
1 , sec x =
sin x
1 , and csc x =
cos x
,
(5.11.49)
sin x
so we shall not dwell on them in detail. The various trigonometric laws easily follow from our present definitions; for hints, see the problems below. This page titled 5.11: Integral Definitions of Some Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
5.11.7
https://math.libretexts.org/@go/page/21247
5.11.E: Problems on Exponential and Trigonometric Functions Exercise 5.11.E. 1 Verify formula (2).
Exercise 5.11.E. 2 Prove Note 1, as suggested (using Chapter 2,
§§11 − 12) .
Exercise 5.11.E. 3 Prove formulas (1) of Chapter 2, §§11 − 12 from our new definitions.
Exercise 5.11.E. 4 Complete the missing details in the proofs of Theorems 2 − 4 .
Exercise 5.11.E. 5 Prove that (i) sin 0 = sin(nπ) = 0 ; (ii) cos 0 = cos(2nπ) = 1 ; (iii) sin = 1 ; (iv) sin(− ) = −1 ; (v) cos(± ) = 0 ; (vi) | sin x| ≤ 1 and | cos x| ≤ 1 for x ∈ E . π 2
π
2 π 2
1
Exercise 5.11.E. 6 Prove that (i) sin(−x) = − sin x and (ii) cos(−x) = cos x for x ∈ E . [Hint: For (i), let h(x) = sin x + sin(−x). 1
h = q on E
1
Show
that
′
h = 0;
hence
h
is
constant,
say,
. Substitute x = 0 to find q. For (ii), use (13) − (15). ]
Exercise 5.11.E. 7 Prove the following for x, y ∈ E : (i) sin(x + y) = sin x cos y + cos x sin y; hence sin(x + ) = cos x . (ii) cos(x + y) = cos x cos y − sin x sin y; hence cos(x + ) = − sin x . [Hint for (i) : Fix x, y and let p = x + y. Define h : E → E by 1
π 2
π 2
1
1
h(t) = sin t cos(p − t) + cos t sin(p − t),
t ∈ E
1
.
(5.11.E.1)
n
is odd. How about the cosine? Find
Proceed as in Problem 6. Then let t = x. ]
Exercise 5.11.E. 8 ¯¯¯¯ ¯
¯¯¯¯ ¯
With J as in the text, show that the sine increases on J if n is even and decreases if the endpoints of J . n
n
¯¯¯¯ ¯ n
5.11.E.1
https://math.libretexts.org/@go/page/24107
5.11.E: Problems on Exponential and Trigonometric Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
5.11.E.2
https://math.libretexts.org/@go/page/24107
CHAPTER OVERVIEW 6: Differentiation on Eⁿ and Other Normed Linear Spaces 6.1: Directional and Partial Derivatives 6.1.E: Problems on Directional and Partial Derivatives 6.2: Linear Maps and Functionals. Matrices 6.2.E: Problems on Linear Maps and Matrices 6.3: Differentiable Functions 6.3.E: Problems on Differentiable Functions 6.4: The Chain Rule. The Cauchy Invariant Rule 6.4.E: Further Problems on Differentiable Functions 6.5: Repeated Differentiation. Taylor’s Theorem 6.5.E: Problems on Repeated Differentiation and Taylor Expansions 6.6: Determinants. Jacobians. Bijective Linear Operators 6.6.E: Problems on Bijective Linear Maps and Jacobians 6.7: Inverse and Implicit Functions. Open and Closed Maps 6.7.E: Problems on Inverse and Implicit Functions, Open and Closed Maps 6.8: Baire Categories. More on Linear Maps 6.8.E: Problems on Baire Categories and Linear Maps 6.9: Local Extrema. Maxima and Minima 6.9.E: Problems on Maxima and Minima 6.10: More on Implicit Differentiation. Conditional Extrema 6.10.E: Further Problems on Maxima and Minima
This page titled 6: Differentiation on Eⁿ and Other Normed Linear Spaces is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
1
6.1: Directional and Partial Derivatives In Chapter 5 we considered functions f
: E
Now we take up functions f
where both E and E are normed spaces.
: E
′
→ E
1
of one real variable.
→ E
′
The scalar field of both is always assumed the same: E or C (the complex field). The case assumed finite. 1
E =E
∗
is excluded here; thus all is
We mostly use arrowed letters p ,⃗ q ,⃗ … , x⃗ , y ,⃗ z ⃗ for vectors in the domain space E , and nonarrowed letters for those in E and for scalars. ′
As before, we adopt the convention that f is defined on all of E , with f (x⃗ ) = 0 if not defined otherwise. ′
Note that, if p ⃗ ∈ E
′
,
one can express any point x⃗ ∈ E as ′
x⃗ = p ⃗ + tu⃗ ,
(6.1.1)
with t ∈ E and u⃗ a unit vector. For if x⃗ ≠ p ,⃗ set 1
⃗ t = | x⃗ − p | and u⃗ =
1
⃗ (x⃗ − p );
(6.1.2)
t
and if x⃗ = p ,⃗ set t = 0, and anyu⃗ will do. We often use the notation t ⃗ = Δx⃗ = x⃗ − p ⃗ = tu⃗
(t ∈ E
1
′
, t ,⃗ u⃗ ∈ E ) .
(6.1.3)
First of all, we generalize Definition 1 in Chapter 5, §1.
Definition 1 Given f at p ⃗ by
: E
′
→ E
and p ,⃗ u⃗ ∈ E
′
→ (u⃗ ≠ 0 ),
we define the directional derivative of
⃗ = lim Du⃗ f (p ) t→0
1
f
along u⃗ (or u⃗ -directed derivative of
⃗ [f (p ⃗ + tu⃗ ) − f (p )],
f)
(6.1.4)
t
if this limit exists in E (finite). We also define the u⃗ -directed derived function, Du⃗ f : E
′
→ E,
(6.1.5)
as follows. For any p ⃗ ∈ E , ′
Duf ⃗ ( p ) ⃗ = {
1
limt→0
t
⃗ [f (p ⃗ + tu⃗ ) − f (p )]
(6.1.6)
0
Thus D
otherwise.
is always defined, but the name derivative is used only if the limit (1) exists (finite). If it exists for each p ⃗ in a set we call D f (in classical notation ∂f /∂ u⃗ ) the u⃗ -directed derivative of f on B .
f
u⃗
′
B ⊆E ,
if this limit exists,
u⃗
Note that, as t → 0, x⃗ tends to p ⃗ over the line x⃗ = p ⃗ + tu⃗ . Thus D f (p )⃗ can be treated as a relative limit over that line. Observe that it depends on both the direction and the length of u⃗ . Indeed, we have the following result. u⃗
Corollary 6.1.1 Given f
: E
→
′
→ E, u⃗ ≠ 0 ,
and a scalar s ≠ 0, we have Dsu⃗ f = sDu⃗ f .
Moreover, D
f (p ) ⃗
su⃗
is a genuine derivative iff D
f (p ) ⃗
u⃗
(6.1.7)
is.
Proof
6.1.1
https://math.libretexts.org/@go/page/19197
Set t = sθ in (1) to get ⃗ = lim sDu⃗ f (p ) θ→0
1
⃗ = D ⃗ [f (p ⃗ + θsu⃗ ) − f (p )] f (p ). su⃗
θ
□
(6.1.8)
In particular, taking s = 1/|u⃗ |, we have | u⃗ |
1
|su⃗ | =
= 1 and Du⃗ f =
| u ⃗|
Thus all reduces to the case does not simplify matters.
Dv ⃗ f ,
where
′
n
n
scalar variables
n
(6.1.9)
is a unit vector. This device, called normalization, is often used, but actually it
v ⃗ = su⃗
If E = E (C ) , then f is a function of example leads us to the following definition.
Dsu⃗ f .
s
xk (k = 1, … , n)
and
E
has the
′
n
basic unit vectors
⃗ . ek
This
Definition 2 If in formula (1), E to x , denoted
′
=E
n
(C
n
)
and u⃗ = e ⃗ for a fixed k ≤ n, we call D k
f
u⃗
the partially derived function for f , with respect
k
∂f Dk f or
,
(6.1.10)
∂xk
and the limit (1) is called the partial derivative of f at p ,⃗ with respect to x
k,
⃗ or Dk f (p ),
If it exists for all p ⃗ ∈ B, we call D
kf
=E
3
(C
3
),
we often write x, y, z for x
1,
x2 , x3 , ∂f
,
=E
1
,
(6.1.11)
∂y
scalars are also "vectors," and
n
(C
n
).
and
∂f ,
∂x
′
are always defined on all of E
= 1, … , n)
∂f
Note 1. If E Explain!
∂f ∣ ∣ . ∂xk ∣x= ⃗ p ⃗
k
k f (k
′
⃗ or f (p ),
the partial derivative (briefly, partial) of f on B, with respect to x .
In any case, the derived functions D If E
∂ ∂xk
denoted
∂z
D1 f
for Dk f
(k = 1, 2, 3).
coincides with
f
′
(6.1.12)
as defined in Chapter 5, §1 (except where
f
′
= ±∞
).
Note 2. As we have observed, the u⃗ -directed derivative (1) is obtained by keeping x⃗ on the line x⃗ = p ⃗ + tu⃗ . If u⃗ = e ⃗ , the line is parallel to the k th axis; so all coordinates of x⃗ , except x , remain fixed (x = p , i ≠ k) , and f behaves like a function of one variable, x . Thus we can compute D f by the usual rules of differentiation, treating all x (i ≠ k) as constants and x as the only variable. k
k
k
i
k
i
i
k
For example, let f (x, y) = x
2
y.
Then ∂f ∂x
Note 3. More generally, given p ⃗ and u⃗ ≠
∂f = D1 f (x, y) = 2xy and
→ 0 ,
∂y
2
= D2 f (x, y) = x .
(6.1.13)
set h(t) = f (p ⃗ + tu⃗ ),
t ∈ E
1
.
(6.1.14)
⃗ so Then h(0) = f (p );
6.1.2
https://math.libretexts.org/@go/page/19197
⃗ = lim Du⃗ f (p ) t→0
1
⃗ [f (p ⃗ + tu⃗ ) − f (p )]
t h(t) − h(0)
= lim t −0
t→0 ′
= h (0)
if the limit exists. Thus all reduces to a function h of one real variable. For functions
f : E
1
the existence of a finite derivative ("differentiability") at
→ E,
Chapter 5, §1). But in the general case, f
: E
′
→ E,
this may fail even if D
p
implies continuity at
exists for all u⃗ ≠
⃗ f (p )
u⃗
→ 0
p
(Theorem 1 of
.
Examples (a) Define f
: E
2
→ E
1
by 2
x y f (x, y) =
4
x
Fix a unit vector u⃗ = (u
1,
u2 )
in E
2
Let p ⃗ = (0, 0). To find D
.
,
2
+y
f (0, 0) = 0.
f (p),
u⃗
(6.1.15)
use the h of Note 3 : 2
tu u2 1
h(t) = f (p ⃗ + tu⃗ ) = f (tu⃗ ) = f (tu1 , tu2 ) =
2
4
t u
1
and h = 0 if u
2
= 0.
2
if u2 ≠ 0,
(6.1.16)
+u
2
Hence 2
u
′
⃗ = h (0) = Du⃗ f (p )
and h (0) = 0 if u ′
2
= 0.
Thus D
→ ( 0 )
u⃗
1
u2
if u2 ≠ 0,
exists for all u⃗ . Yet f is discontinuous at
(6.1.17)
→ 0
(see Problem 9 in Chapter 4, §3).
(b) Let x +y
if xy = 0,
1
otherwise.
f (x, y) = {
Then f (x, y) = x on the x-axis; so D
1 f (0,
0) = 1
(6.1.18)
.
→
Yet f is discontinuous at 0 (even relatively so) over any line y = ax so f (x, y) → 1 but f (0, 0) = 0 + 0 = 0 . Thus continuity at
→ 0
(a ≠ 0).
For on that line, f (x, y) = 1 if (x, y) ≠ (0, 0);
fails. (But see Theorem 1 below!)
Hence, if differentiability is to imply continuity, it must be defined in a stronger manner. We do it in §3. For now, we prove only some theorems on partial and directional derivatives, based on those of Chapter 5.
Theorem 6.1.1 If f
: E
′
→ E
has a u⃗ -directed derivative at p ⃗ ∈ E
′
,
then f is relatively continuous at p ⃗ over the line →
x⃗ = p ⃗ + tu⃗
′
( 0 ≠ u⃗ ∈ E ) .
(6.1.19)
Proof Set h(t) = f (p ⃗ + tu⃗ ), t ∈ E . 1
By Note 3, our assumption implies that h (a function on E ) is differentiable at 0. 1
By Theorem 1 in Chapter 5, §1, then, h is continuous at 0; so ⃗ lim h(t) = h(0) = f (p ),
(6.1.20)
t→0
6.1.3
https://math.libretexts.org/@go/page/19197
i.e., ⃗ lim f (p ⃗ + tu⃗ ) = f (p ).
(6.1.21)
t→0
But this means that f (x⃗ ) → f (p )⃗ as x⃗ → p ⃗ over the line x⃗ = p ⃗ + tu⃗ , for, on that line, x⃗ = p ⃗ + tu⃗ . Thus, indeed, f is relatively continuous at p ,⃗ as stated. Note that we actually used the substitution (Corollary 2(iii) of Chapter 4, §2). Why?
x⃗ = p ⃗ + tu⃗ .
□
This is admissible since the dependence between
x
and
t
is one-to-tone
Theorem 6.1.2 Let E
′
→ ∋ u⃗ = q ⃗ − p ,⃗ u⃗ ≠ 0
If f : E then
′
→ E
.
is relatively continuous on the segment
I = L[ p ,⃗ q ]⃗
and has a u⃗ -directed derivative on
⃗ − f (p )| ⃗ ≤ sup | D ⃗ f (x⃗ )| , |f (q ) u
I −Q
(Q countable),
x⃗ ∈ I − Q.
(6.1.22)
Proof Set again h(t) = f (p ⃗ + tu⃗ ) and g(t) = p ⃗ + tu⃗ . Then h = f ∘ g, and g is continuous on E As f is relatively continuous on I (1)). Now fix t
0
∈ J.
If x⃗
0
1
.
(Why?)
⃗ = L[ p ,⃗ q ],
= p ⃗ + t0 u⃗ ∈ I − Q,
so is h = f ∘ g on the interval J = [0, 1] ⊂ E (cf. Chapter 4, §8, Example 1
our assumptions imply the existence of
D ⃗ f (x⃗ 0 ) = lim u
1
[f (x⃗ 0 + tu⃗ ) − f (x⃗ 0 )]
t
t→0
1 = lim
[f (p ⃗ + t0 u⃗ + tu⃗ ) − f (p ⃗ + t0 u⃗ )]
t
t→0
1 = lim t→0
t
[h (t0 + t) − h (t0 )]
′
= h (t0 ) .
This can fail for at most a countable set Q of points t ′
0
(Explain!)
∈ J
(those for which x⃗
0
∈ Q).
Thus h is differentiable on J − Q ; and so, by Corollary 1 in Chapter 5, §4, ′
|h(1) − h(0)| ≤
′
sup
| h (t)| = ′
t∈J−Q
sup
| D ⃗ f (x⃗ )| .
(6.1.23)
u
⃗ x∈I−Q
→
⃗ Now as h(1) = f (p ⃗ + u⃗ ) = f (q )⃗ and h( 0 ) = f (p ), formula (2) follows.
□
Theorem 6.1.3 ⃗ then If in Theorem 2, E = E and if f has a u⃗ -directed derivative at least on the open line segment L(p ,⃗ q ), 1
⃗ − f (p ) ⃗ = D ⃗ f (x⃗ 0 ) f (q ) u
for some x⃗
0
⃗ ∈ L(p ,⃗ q )
(6.1.24)
.
Proof The proof is as in Theorem 2, based on Corollary 3 in Chapter 5, §2 (instead of Corollary 1 in Chapter 5, §4). Theorems 2 and 3 are often used in "normalized" form, as follows.
6.1.4
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Corollary 6.1.2 If in Theorems 2 and 3, we set r = | u⃗ | = | q ⃗ − p |⃗ ≠ 0 and v ⃗ =
1
u⃗ ,
(6.1.25)
x⃗ ∈ I − Q,
(6.1.26)
r
then formulas (2) and (3) can be written as ⃗ − f (p )| ⃗ ≤ | q ⃗ − p |⃗ sup | D ⃗ f (x⃗ )| , |f (q ) v
and
for some x⃗
0
⃗ ∈ L(p ,⃗ q )
⃗ − f (p ) ⃗ = | q ⃗ − p |⃗ D ⃗ f (x⃗ 0 ) f (q ) v
(6.1.27)
Du⃗ f = rDv ⃗ f = | q ⃗ − p |⃗ Dv ⃗ f ;
(6.1.28)
.
For by Corollary 1,
so (2') and (3') follow. This page titled 6.1: Directional and Partial Derivatives is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
6.1.5
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6.1.E: Problems on Directional and Partial Derivatives Exercise 6.1.E. 1 Complete all missing details in the proof of Theorems 1 to 3 and Corollaries 1 and 2.
Exercise 6.1.E. 2 Complete all details in Examples (a) and (b). Find Note 3; (ii) use Definition 2 only.
D1 f (p ) ⃗
and D
⃗ 2 f (p )
also for
p ⃗ ≠ 0.
Do Example (b) in two ways: (i) use
Exercise 6.1.E. 3 In Examples (a) and (b) describe D f : E → E . Compute it for \(\vec{u =(1,1)=\vec{p}.\) In (b), show that f has no directional derivatives D f (p )⃗ except if u⃗ ∥e ⃗ or u⃗ ∥e ⃗ . Give two proofs: (i) use Theorem 1; (ii) use definitions only. 2
1
u⃗
u⃗
1
2
Exercise 6.1.E. 4 Prove that if f : E (C ) → E has a zero partial derivative, D for x⃗ ∈ A. (Use Theorems 1 and 2.) n
n
kf
= 0,
on a convex set A, then f (x⃗ ) does not depend on x
k,
Exercise 6.1.E. 5 Describe D f and D f (x, y) equals: 1
2f
on the various parts of E
(i)
xy x2 +y 2
2
,
;
xy
(iii)
|x|
and discuss the relative continuity of f over lines through
+ x sin
=0
0 ,
given that
(ii) the integral part of x + y; 1 y
(v) sin(y cos x);
(Set f
→
2
2
2
2
x −y
;
(iv) xy
x +y
(6.1.E.1)
;
y
(vi) x .
wherever the formula makes no sense.)
Exercise 6.1.E. 6 Prove that if f : E → E has a local maximum or minimum at p ⃗ ∈ E [Hint: Use Note 3, then Corollary 1 in Chapter 5, §2. ⇒
′
1
′
,
then D
⃗ = 0 f (p )
u⃗
for every vector u⃗ ≠
→ 0
in E
′
.
Exercise 6.1.E. 7 State and prove the Finite Increments Law (Theorem 1 of Chapter 5, §4) for directional derivatives. [Hint: Imitate Theorem 2 using two auxiliary functions, h and k .]
Exercise 6.1.E. 8 State and prove Theorems 4 and 5 of Chapter 5, §1, for directional derivatives. 6.1.E: Problems on Directional and Partial Derivatives is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.1.E.1
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6.2: Linear Maps and Functionals. Matrices For an adequate definition of differentiability, we need the notion of a linear map. Below, over the same scalar field, E or C .
′
E ,E
′′
,
and
E
denote normed spaces
1
Definition 1 A function f
: E
′
→ E
is a linear map if and only if for all x⃗ , y ⃗ ∈ E and scalars a, b ′
⃗ = af (x⃗ ) + bf (y ); ⃗ f (ax⃗ + b y )
(6.2.1)
⃗ = f (x) + f (y) and f (ax⃗ ) = af (x⃗ ). (Verify!) f (x⃗ + y )
(6.2.2)
equivalently, iff for all such x⃗ , y ,⃗ and a
If E = E
′
,
such a map is also called a linear operator.
If the range space E is the scalar field of E , (i.e., E or C , ) the linear f is also called a (real or complex) linear functional on ′
1
′
E .
Note 1. Induction extends formula (1) to any "linear combinations": m
m
f ( ∑ ai x⃗ i ) = ∑ ai f (x⃗ i ) i=1
for all x⃗
i
∈ E
′
(6.2.3)
i=1
and scalars a . i
Briefly: A linear map f preserves linear combinations. →
Note 2. Taking a = b = 0 in (1), we obtain f ( 0 ) = 0 if f is linear.
Examples (a) Let E
′
=E
n
(C
n
).
Fix a vector v ⃗ = (v
1,
… , vn )
in E and set ′
′
(∀x⃗ ∈ E )
f (x⃗ ) = x⃗ ⋅ v ⃗
(6.2.4)
(inner product; see Chapter 3, §§1-3 and §9). Then ⃗ f (ax⃗ + b y )
⃗ ⋅ v ⃗ = (ax⃗ ) ⋅ v ⃗ + (b y ) ⃗ + b(y ⃗ ⋅ v) ⃗ = a(x⃗ ⋅ v) ⃗ = af (x⃗ ) + bf (y );
so f is linear. Note that if E
′
=E
n
,
then by definition, n
n
f (x⃗ ) = x⃗ ⋅ v ⃗ = ∑ xk vk = ∑ vk xk .
If, however, E
′
=C
n
,
k=1
k=1
n
n
(6.2.5)
then ¯ ¯ ¯ ¯ ¯ f (x⃗ ) = x⃗ ⋅ v ⃗ = ∑ xk ¯ v k = ∑ vk xk , k=1
(6.2.6)
k=1
where v is the conjugate of the complex number v . ¯ ¯ ¯ k
k
By Theorem 3 in Chapter 4, §3, f is continuous (a polynomial!). Moreover, f (x⃗ ) = x⃗ ⋅ v⃗ is a scalar (in E or C ). Thus the range of f lies in the scalar field of E ; so f is a linear functional on 1
′
′
E .
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(b) Let I = [0, 1]. Let E be the set of all functions there (Theorem 2 of Chapter 4, §8). ′
u : I → E
that are of class
∞
CD
(Chapter 5, §6) on I , hence bounded
As in Example (C) in Chapter 3, §10, E is a normed linear space, with norm ′
∥u∥ = sup |u(x)|.
(6.2.7)
x∈I
Here each function u ∈ E is treated as a single "point" in E . The distance between two such points, u and v, equals ∥u − v∥, by definition. ′
′
Now define a map D on E by setting D(u) = u (derivative of u on I ). As every u ∈ E is of class C D ′
Thus D(u) = u
′
′
∈ E ,
′
and so D : E
′
→ E
′
′
∞
,
so is u . ′
is a linear operator. (Its linearity follows from Theorem 4 in Chapter 5, §1.)
(c) Let again I = [0, 1]. Let E be the set of all functions u : I → E that are bounded and have antiderivatives (Chapter 5, §5) on I . With norm ∥u∥ as in Example (b), E is a normed linear space. ′
′
Now define ϕ : E
′
by
→ E
1
ϕ(u) = ∫
u,
(6.2.8)
0
with ∫ u as in Chapter 5, §5. (Recall that ∫ map of E into E . (Why?)
1
0
u
is an element of E if u : I
→ E.
) By Corollary 1 in Chapter 5, §5, ϕ is a linear
′
(d) The zero map f
=0
on E is always linear. (Why?) ′
Theorem 6.2.1 A linear map f : E real c > 0 such that
′
→ E
is continuous (even uniformly so) on all of ′
(∀x⃗ ∈ E )
E
′
iff it is continuous at
|f (x⃗ )| ≤ c| x⃗ |.
→ 0 ;
equivalently, iff there is a
(6.2.9)
(We call this property linear boundedness.) Proof Assume that f is continuous at
→ 0 .
Then, given ε > 0, there is δ > 0 such that →
whenever |x⃗ −
→ 0 | = | x⃗ | < δ
Now, for any x⃗ ≠
→ 0 ,
|f (x⃗ ) − f ( 0 )| = |f (x⃗ )| ≤ ε
(6.2.10)
∣ δx⃗ ∣ δ ∣ ∣ = < δ. 2 ∣ | x⃗ | ∣
(6.2.11)
.
we surely have
Hence →
∣
(∀x⃗ ≠ 0 )
δx⃗
∣f ( ∣
∣ ) ∣ ≤ ε,
2| x⃗ |
(6.2.12)
∣
or, by linearity, δ
|f (x⃗ )| ≤ ε,
(6.2.13)
2| x⃗ |
i.e.,
6.2.2
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|f (x⃗ )| ≤
2ε
| x⃗ |.
(6.2.14)
δ
By Note 2, this also holds if x⃗ =
→ 0
.
Thus, taking c = 2ε/δ, we obtain ′
(∀x⃗ ∈ E )
f (x⃗ ) ≤ c| x⃗ |
(linear boundedness).
(6.2.15)
Now assume (3). Then ′
(∀x⃗ , y ⃗ ∈ E )
⃗ ≤ c| x⃗ − y |; ⃗ |f (x⃗ − y )|
(6.2.16)
⃗ ≤ c| x⃗ − y |. ⃗ |f (x⃗ ) − f (y )|
(6.2.17)
or, by linearity, ′
(∀x⃗ , y ⃗ ∈ E )
Hence f is uniformly continuous (given ε > 0, take δ = ε/c). This, in turn, implies continuity at equivalent, as claimed. □ A linear map need not be continuous. But, for E and C n
n
,
→ 0 ;
so all conditions are
we have the following result.
Theorem 6.2.2 (i) Any linear map on E or C is uniformly continuous. n
n
(ii) Every linear functional on E
n
(C
n
)
has the form f (x⃗ ) = x⃗ ⋅ v ⃗
for some unique vector v ⃗ ∈ E
n
(C
n
),
(dot product)
(6.2.18)
dependent on f only.
Proof Suppose f
: E
n
→ E
is linear; so f preserves linear combinations.
But every x⃗ ∈ E is such a combination, n
n
⃗ x⃗ = ∑ xk e k
(Theorem 2 in Chapter 3,
§§1-3).
(6.2.19)
k=1
Thus, by Note 1, n
n
⃗ ) = ∑ xk f (e k ⃗ ) . f (x⃗ ) = f ( ∑ xk e k k=1
(6.2.20)
k=1
Here the function values f (e ⃗ ) are fixed vectors in the range space E, say, k
⃗ ) = vk ∈ E, f (e k
(6.2.21)
so that n
n
⃗ ) = ∑ xk vk , f (x⃗ ) = ∑ xk f (e k k=1
vk ∈ E.
(6.2.22)
k=1
Thus f is a polynomial in n real variables x , hence continuous (even uniformly so, by Theorem 1). k
In particular, if E = E (i.e., f is a linear functional) then all v in (5) are real numbers; so they form a vector 1
k
v ⃗ = (v1 , … , vk ) in E
n
,
(6.2.23)
and (5) can be written as f (x⃗ ) = x⃗ ⋅ v.⃗
6.2.3
(6.2.24)
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The vector v ⃗ is unique. For suppose there are two vectors, u⃗ and v,⃗ such that (∀x⃗ ∈ E
n
)
f (x⃗ ) = x⃗ ⋅ v ⃗ = x⃗ ⋅ u⃗ .
(6.2.25)
x⃗ ⋅ (v ⃗ − u⃗ ) = 0.
(6.2.26)
Then (∀x⃗ ∈ E
By Problem 10 of Chapter 3, §§1-3, this yields v ⃗ − u⃗ =
n
)
→
or v ⃗ = u⃗ . This completes the proof for E = E
0 ,
It is analogous for C ; only in (ii) the v are complex and one has to replace them by their conjugates vector v ⃗ to obtain f (x⃗ ) = x⃗ ⋅ v ⃗ . Thus all is proved. □ n
k
Note 3. Formula (5) shows that a linear map f If further E = E
m
(C
m
: E
n
(C
n
¯ ¯ ¯ v k
n
when forming the
is uniquely determined by the n function values v
) → E
.
k
⃗ ) = f (e k
.
the vectors v are m -tuples of scalars,
),
k
vk = (v1k , … , vmk ) .
(6.2.27)
We often write such vectors vertically, as the n "columns" in an array of m "rows" and n "columns": v11
v12
…
v1n
⎜ v21 ⎜ ⎜ ⎜ ⎜ ⋮
v22
…
⋮
⋱
v2n ⎟ ⎟ ⎟. ⎟ ⎟ ⋮
vm2
…
vmn
⎛
⎝
vm1
⎞
(6.2.28)
⎠
Formally, (6) is a double sequence of mn terms, called an m × n matrix. We denote it by [f ] = (v
ik )
⃗ ) = vk = (v1k , … , vmk ) . f (e k
Thus linear maps f
: E
n
→ E
m
(or f
: C
n
→ C
m
)
,
where for k = 1, 2, … , n, (6.2.29)
correspond one-to-one to their matrices [f ].
The easy proof of Corollaries 1 to 3 below is left to the reader.
Corollary 6.2.1 If f , g : E
′
are linear, so is
→ E
h = af + bg
(6.2.30)
for any scalars a, b. If further E
′
=E
n
(C
n
)
and E = E
m
(C
m
),
with [f ] = (v
ik )
and [g] = (w
[h] = (avik + b wik ) .
ik )
, then (6.2.31)
Corollary 6.2.2 A map f
: E
n
(C
n
) → E
is linear iff n
f (x⃗ ) = ∑ vk xk ,
(6.2.32)
k=1
where v
k
⃗ ) = f (e k
.
Hint: For the "if," use Corollary 1. For the "only if," use formula (5) above.
Corollary 6.2.3 If f
: E
′
→ E
′′
and g : E
′′
→ E
are linear, so is the composite h = g ∘ f .
Our next theorem deals with the matrix of the composite linear map g ∘ f
6.2.4
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Theorem 6.2.3 Let f
: E
′
→ E
′′
and g : E
′′
→ E
be linear, with E
If [f ] = (v
ik )
and [g] = (w
ji )
,
′
=E
n
(C
n
),E
′′
=E
m
(C
m
) , and E = E
r
(C
r
).
(6.2.33)
then [h] = [g ∘ f ] = (zjk ) ,
(6.2.34)
where m
zjk = ∑ wji vik ,
j = 1, 2, … , r, k = 1, 2, … , n.
(6.2.35)
i=1
Proof Denote the basic unit vectors in E by ′
′
′
(6.2.36)
e , … , em ,
′′
(6.2.37)
e1 , … , er .
(6.2.38)
e , … , en , 1
those in E by ′′
′′ 1
and those in E by
Then for k = 1, 2, … , n, m
r
′
′′
′
i
k
f (e ) = vk = ∑ vik e and h (e ) = ∑ zjk ej , k
i=1
(6.2.39)
j=1
and for i = 1, … m , r ′′
g (e ) = ∑ wji ej .
(6.2.40)
i
j=1
Also, m ′
′
k
k
m
m
′′
r
′′
h (e ) = g (f (e )) = g ( ∑ vik e ) = ∑ vik g (e ) = ∑ vik ( ∑ wji ej ) . i
i
i=1
i=1
i=1
(6.2.41)
j=1
Thus r
r
m
′
h (e ) = ∑ zjk ej = ∑ ( ∑ wji vik ) ej . k
j=1
j=1
(6.2.42)
i=1
But the representation in terms of the e is unique (Theorem 2 in Chapter 3, §§1-3), so, equating coefficients, we get (7). j
□
Note 4. Observe that z (an m × n matrix).
jk
is obtained, so to say, by "dot-multiplying" the j th row of [g] (an r × m matrix) by the k th column of
[f ]
It is natural to set [g][f ] = [g ∘ f ],
(6.2.43)
(wji ) (vik ) = (zjk ) ,
(6.2.44)
or
6.2.5
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with z
jk
as in (7).
Caution. Matrix multiplication, so defined, is not commutative.
Definition 2 The set of all continuous linear maps f If E = E
′
,
: E
′
→ E
(for fixed E and E ) is denoted L(E ′
′
, E).
we write L(E) instead.
For each f in L (E
′
, E) ,
we define its norm by ∥f ∥ = sup |f (x⃗ )|.
(6.2.45)
⃗ | x|≤1
Note that ∥f ∥ < +∞, by Theorem 1.
Theorem 6.2.4 ′
L(E , E)
is a normed linear space under the norm defined above and under the usual operations on functions, as in Corollary
1. Proof Corollary 1 easily implies that L(E
′
, E)
is a vector space. We now show that ∥ ⋅ ∥ is a genuine norm.
The triangle law, ∥f + g∥ ≤ ∥f ∥ + ∥g∥,
(6.2.46)
follows exactly as in Example (C) of Chapter 3, §10. (Verify!) Also, by Problem 5 in Chapter 2, §§8-9, sup |af (x⃗ )| = |a| sup |f (x⃗ )|. Hence ∥af ∥ = |a|∥f ∥ for any scalar a. As noted above, 0 ≤ ∥f ∥ < +∞ . It remains to show that ∥f ∥ = 0 iff f is the zero map. If ∥f ∥ = sup |f (x⃗ )| = 0,
(6.2.47)
⃗ | x|≤1
then |f (x⃗ )| = 0 when |x⃗ | ≤ 1. Hence, if x⃗ ≠
→ 0
, x⃗ f(
1 ) =
| x⃗ |
f (x⃗ ) = 0.
(6.2.48)
| x⃗ |
→
As f ( 0 ) = 0, we have f (x⃗ ) = 0 for all x⃗ ∈ E . ′
Thus ∥f ∥ = 0 implies f
= 0,
Note 5. A similar proof, via f (
x⃗ ⃗ | x|
and the converse is clear. Thus all is proved.
)
□
and properties of lub, shows that ∣ f (x⃗ ) ∣ ∣
∥f ∥ = sup ∣ ⃗ x≠0
∣
| x⃗ |
(6.2.49)
∣
and ′
(∀x⃗ ∈ E )
|f (x⃗ )| ≤ ∥f ∥| x⃗ |.
(6.2.50)
|f (x⃗ )| ≤ c| x⃗ |.
(6.2.51)
It also follows that ∥f ∥ is the least real c such that ′
(∀x⃗ ∈ E )
Verify. (See Problem 3'.)
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As in any normed space, we define distances in L(E
′
, E)
by
ρ(f , g) = ∥f − g∥,
(6.2.52)
making it a metric space; so we may speak of convergence, limits, etc. in it.
Corollary 6.2.4 If f
′
∈ L(E , E
′′
)
and g ∈ L(E
′′
, E),
then ∥g ∘ f ∥ ≤ ∥g∥∥f ∥.
(6.2.53)
Proof By Note 5, ′
(∀x⃗ ∈ E )
|g(f (x⃗ ))| ≤ ∥g∥|f (x⃗ )| ≤ ∥g∥∥f ∥| x⃗ |.
(6.2.54)
Hence →
∣ (g ∘ f )(x⃗ ) ∣
(∀x⃗ ≠ 0 )
∣ ∣
∣ ≤ ∥g∥∥f ∥, | x⃗ |
(6.2.55)
∣
and so |(g ∘ f )(x⃗ )| ∥g∥∥f ∥ ≥ sup ¯ ¯ ¯
⃗ 0 x≠
= ∥g ∘ f ∥.
□
(6.2.56)
| x⃗ |
This page titled 6.2: Linear Maps and Functionals. Matrices is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
6.2.7
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6.2.E: Problems on Linear Maps and Matrices Exercise 6.2.E. 1 Verify Note 1 and the equivalence of the two statements in Definition 1.
Exercise 6.2.E. 2 In Examples (b) and (c) show that fn → f (uniformly) on I iff ∥ fn − f ∥ → 0,
(6.2.E.1)
i.e., f → f in E . [Hint: Use Theorem 1 in Chapter 4, §2.] Hence deduce the following. (i) If E is complete, then the map ϕ in Example (c) is continuous. [Hint: Use Theorem 2 of Chapter 5, §9, and Theorem 1 in Chapter 4, §12.] (ii) The map D of Example (b}) is not continuous. [Hint: Use Problem 3 in Chapter 5, §9.] ′
n
Exercise 6.2.E. 3 Prove Corollaries 1 to 3.
Exercise 6.2.E. 3
′
Show that |f (x⃗ )| ∥f ∥ = sup |f (x⃗ )| = sup |f (x⃗ )| = sup ⃗ | x|≤1
[Hint: From linearity of f deduce that computing sup |f (x⃗ )|. Why?]
→
⃗ | x|=1
|f (x⃗ )| ≥ |f (cx)|
if
⃗ 0 x≠
|c| < 1.
.
(6.2.E.2)
| x⃗ |
Hence one may disregard vectors of length
0, there is δ > 0 such that, setting Δf 1
we have
⃗ ⃗ = f (p ⃗ + t ) − f (p )
⃗ ⃗ |Δf − ϕ(t )| < ε whenever 0 < | t | < δ;
(6.3.17)
⃗ |t |
or, by the triangle law, ⃗ ⃗ ⃗ ⃗ |Δf | ≤ |Δf − ϕ(t )| + |ϕ(t )| ≤ ε| t | + |ϕ(t )|,
⃗ 0 < | t | < δ.
(6.3.18)
Now, by Definition 1, ϕ is linear and continuous; so → ⃗ lim |ϕ(t )| = |ϕ( 0 )| = 0.
(6.3.19)
→ ⃗ t→ 0
Thus, making t ⃗ →
→
in (5), with ε fixed, we get
0
lim |Δf | = 0.
(6.3.20)
→ ⃗ t→ 0
As t ⃗ is just another notation for Δx⃗ = x⃗ − p ,⃗ this proves assertion (i). Next, fix any u⃗ ≠
→
in E , and substitute tu⃗ for t ⃗ in (4). ′
0
In other words, t is a real variable, 0 < t < δ/|u⃗ |, so that t ⃗ = tu⃗ satisfies 0 < |t |⃗ < δ . Multiplying by |u⃗ |, we use the linearity of ϕ to get ⃗ ∣ Δf ϕ(tu⃗ ) ∣ ∣ f (p ⃗ + tu⃗ ) − f (p ) ∣ ∣ Δf ∣ ε| u⃗ | > ∣ − ∣ =∣ − ϕ(u⃗ )∣ = ∣ − ϕ(u⃗ )∣ . t t ∣ t ∣ t ∣ ∣ ∣ ∣
(6.3.21)
As ε is arbitrary, we have ϕ(u⃗ ) = lim t→0
But this is simply D
⃗ f (p ),
u⃗
Thus D ′
⃗ = ϕ(u⃗ ) = df (p ;⃗ u⃗ ), f (p )
=E
n
(C
n
⃗ [f (p ⃗ + tu⃗ ) − f (p )].
(6.3.22)
t
by Definition 1 in §1.
u⃗
Note 2. If E
1
proving (ii). □
Theorem 2(iii) shows that if f is differentiable at p ,⃗ it has the n partials
),
Dk f (p ) ⃗ = df (p ;⃗ e k ⃗ ) ,
But the converse fails: the existence of the we have the following result.
⃗ Dk f (p )
k = 1, … , n.
(6.3.23)
does not even imply continuity, let alone differentiability (see §1). Moreover,
Corollary 6.3.1 If E
′
=E
n
(C
n
)
and if f
: E
′
→ E
is differentiable at p ,⃗ then n
n
⃗ = ∑ t D f (p ) ⃗ = ∑ tk df (p ;⃗ t ) k k k=1
where t ⃗ = (t
1,
… , tn )
k=1
∂
⃗ f (p ),
(6.3.24)
∂xk
.
Proof By definition, ϕ = df (p ;⃗ ⋅) is a linear map for a fixed p .⃗ If E
′
=E
n
or C
n
,
we may use formula (3) of §2, replacing f and x⃗ by ϕ and t ,⃗ and get n
n
⃗ ⃗ ⃗ ) = ∑ tk Dk f (p ) ⃗ ϕ(t ) = df (p ;⃗ t ) = ∑ tk df (p ;⃗ e k k=1
6.3.3
(6.3.25)
k=1
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by Note 2.
□
Note 3. In classical notation, one writes Δx or dx for t in (6). Thus, omitting p ⃗ and t ,⃗ formula (6) is often written as k
k
k
∂f
∂f
df = ∂x1
In particular, if n = 3, we write x, y, z for x
1,
x2 , x3 .
dx1 +
∂f dx2 + ⋯ +
∂x2
∂xn
dxn .
(6.3.26)
This yields ∂f
df =
∂f dx +
∂x
∂f dy +
∂y
dz
(6.3.27)
∂z
(a familiar calculus formula). Note 4. If the range space E in Corollary 1 is E In case f
: E
n
→ E
1
,
1
(C ),
then the D
⃗
k f (p )
form an n -tuple of scalars, i.e., a vector in E
n
(C
n
).
we denote it by n
⃗ = (D1 f (p ), ⃗ … , Dn f (p )) ⃗ = ∑ e k ⃗ Dk f (p ). ⃗ ∇f (p )
(6.3.28)
k=1
In case f
: C
n
we replace the D
→ C,
⃗
¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
k f (p )
by their conjugates D
⃗
k f (p )
and set
n ¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
⃗ = ∑ e k ⃗ Dk f (p ) ⃗ . ∇f (p )
(6.3.29)
k=1
The vector ∇f (p )⃗ is called the gradient of f ("grad f ") at p .⃗ From (6) we obtain n
⃗ = ∑ t D f (p ) ⃗ = t ⃗ ⋅ ∇f (p ) ⃗ df (p ;⃗ t ) k k
(6.3.30)
k=1
⃗ (dot product of t ⃗ by ∇f (p )), provided f
: E
n
→ E
1
(or f
: C
n
→ C)
is differentiable at p .⃗
This leads us to the following result.
Corollary 6.3.2 A function f
: E
n
→ E
1
(or f
: C
n
→ C
) is differentiable at p ⃗ iff 1 lim ¯ ¯ ¯
⃗ 0 t→
for some v ⃗ ∈ E
n
(C
n
)
⃗ ⃗ − t ⃗ ⋅ v|⃗ = 0 |f (p ⃗ + t ) − f (p )
(6.3.31)
| t |⃗
.
⃗ t ⃗ ∈ E In this case, necessarily v ⃗ = ∇f (p )⃗ and t ⃗ ⋅ v ⃗ = df (p ;⃗ t ),
n
(C
n
)
.
Proof If f is differentiable at p ,⃗ we may set ϕ = df (p ;⃗ ⋅) and v ⃗ = ∇f (p )⃗ Then by (7), ⃗ = df (p ;⃗ t ) ⃗ = t ⃗ ⋅ v;⃗ ϕ(t )
(6.3.32)
so by Definition 1, (8) results. Conversely, if some v ⃗ satisfies (8), set ϕ(t )⃗ = t ⃗ ⋅ v.⃗ Then (8) implies (2), and ϕ is linear and continuous. Thus by definition, f is differentiable at p ;⃗ so (7) holds. Also, ϕ is a linear functional on E
n
(C
Thus by (7), v ⃗ = ∇f (p )⃗ necessarily.
n
).
By Theorem 2(ii) in §2, the v ⃗ in ϕ(t )⃗ = t ⃗ ⋅ v ⃗ is unique, as is ϕ.
□
6.3.4
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Corollary 6.3.3 (law of the mean) If f
: E
n
→ E
⃗ p ⃗ ≠ q ,⃗ and differentiable on L(p ,⃗ q ), ⃗ then (real) is relatively continuous on a closed segment L[p ,⃗ q ],
1
⃗ − f (p ) ⃗ = (q ⃗ − p ) ⃗ ⋅ ∇f (x⃗ 0 ) f (q )
for some x⃗
0
⃗ ∈ L(p ,⃗ q )
(6.3.33)
.
Proof Let ⃗ r = | q ⃗ − p |,
v ⃗ =
1
⃗ and rv ⃗ = (q ⃗ − p ). ⃗ (q ⃗ − p ),
(6.3.34)
r
By (7) and Theorem 2(ii), ⃗ = v ⃗ ⋅ ∇f (x⃗ ) Dv ⃗ f (x⃗ ) = df (x⃗ ; v)
(6.3.35)
⃗ Thus by formula (3') of Corollary 2 in §1, for x⃗ ∈ L(p ,⃗ q ). ⃗ − f (p ) ⃗ = rD f (x⃗ ) = rv ⃗ ⋅ ∇f (x⃗ ) = (q ⃗ − p ) ⃗ ⋅ ∇f (x⃗ ) f (q ) 0 0 0 v ⃗
for some x⃗
0
⃗ ∈ L(p ,⃗ q ).
(6.3.36)
□
As we know, the mere existence of partials does not imply differentiability. But the existence of continuous partials does. Indeed, we have the following theorem.
Theorem 6.3.3 Let E
′
=E
n
(C
n
)
.
If f : E → E has the partial derivatives D some p ⃗ ∈ A, then f is differentiable at p .⃗ ′
k f (k
= 1, … , n)
on all of an open set A ⊆ E
′
,
and if the D
kf
are continuous at
Proof With p ⃗ as above, let n
n
′ ⃗ ⃗ ⃗ ⃗ ∈ E . ϕ(t ) = ∑ tk Dk f (p ) with t = ∑ tk e k k=1
(6.3.37)
k=1
Then ϕ is continuous (a polynomial!) and linear (Corollary 2 in §2). Thus by Definition 1, it remains to show that lim
⃗ = 0; |Δf − ϕ(t )|
(6.3.38)
→ ⃗ ⃗ t → 0 |t |
that is; 1 lim → ⃗ t∈ 0
n ∣ ⃗ ⃗ − ∑ tk Dk f (p ) ⃗ ∣ = 0. ∣f (p ⃗ + t ) − f (p )
∣
⃗ |t | ∣
To do this, fix ε > 0. As A is open and the simultaneously (explain this!) (∀x⃗ ∈ G ⃗ (δ)) p
Hence for any set I
Dk f
are continuous at
(6.3.39)
∣
k=1
p ⃗ ∈ A
⃗ < | Dk f (x⃗ ) − Dk f (p )|
there is a
δ >0
such that
G ⃗ (δ) ⊆ A p
and
ε , k = 1, … , n.
(6.3.40)
n
⊆ G ⃗ (δ) p
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⃗ ≤ sup | Dk f (x⃗ ) − Dk f (p )| ⃗ x∈I
Now fix any t ⃗ ∈ E
′
⃗ , 0 < | t | < δ,
and let p ⃗
0
ε .
(W hy?)
(6.3.41)
n
,
= p ⃗
k
⃗ = p ⃗ + ∑ t e , pk i i
k = 1, … , n.
(6.3.42)
i=1
Then n
⃗ = p ⃗ + ∑ ti e i⃗ = p ⃗ + t ,⃗ pn
(6.3.43)
i=1
⃗ − p ⃗ |p k | = | tk | , k−1
and all p ⃗ lie in G
(δ),
p ⃗
k
for
− −−−− − − − −−− − k n ∣ k ∣ 2 2 ⃗ − p |⃗ = ∣∑ ti ei ∣ = ∑ | ti | ≤ √∑ | ti | = | t |⃗ < δ, |p k ∣ ∣ ⎷ ∣ i=1 ∣ i=1 i=1
(6.3.44)
as required. As G (δ) is convex (Chapter 4, §9), the segments partials there. p
all lie in
⃗ ⃗ ] Ik = L [ p k−1 , pk
and by assumption,
G ⃗ (δ) ⊆ A; p
f
has all
Hence by Theorem 1 in §1, f is relatively continuous on all I . k
All this also applies to the functions g
k,
defined by ′
(∀x⃗ ∈ E )
⃗ gk (x⃗ ) = f (x⃗ ) − xk Dk f (p ),
k = 1, … , n.
(6.3.45)
(Why?) Here ⃗ Dk gk (x⃗ ) = Dk f (x⃗ ) − Dk f (p ).
(6.3.46)
(Why?) Thus by Corollary 2 in §1, and (11) above, ⃗ ) − gk (p ⃗ | gk (p k )| k−1
⃗ − p ⃗ ⃗ ≤ |p k | sup | Dk f (x⃗ ) − Dk f (p )| k−1 x∈Ik
ε ≤
ε | tk | ≤
n
⃗ | t |,
n
since ⃗ ⃗ − p ⃗ ⃗ | ≤ | t |, |p k | = | tk e k k−1
(6.3.47)
by construction. Combine with (12), recalling that the k th coordinates x , for p ⃗ and p ⃗ k
| gk (p ⃗ ) − gk (p ⃗ k
k−1
)|
k
k−1
= |f (p ⃗ ) − f (p ⃗ k
ε ≤
differ by t
k−1
k;
so we obtain
⃗ ) − tk Dk f (p )|
⃗ | t |.
n
Also, n
⃗ ) − f (p ⃗ ⃗ ) − f (p ⃗ ) ∑ [f (p k )] = f (p n k−1 0 k=1
⃗ − f (p ) ⃗ = Δf (see above). = f (p ⃗ + t )
Thus,
6.3.6
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n
∣
∣
⃗ ∣ ∣Δf − ∑ tk Dk f (p ) ∣
∣
∣
k=1
n
∣
⃗ ) − f (p ⃗ ⃗ ∣ = ∣∑ [f (p k ) − tk Dk f (p )] k−1 ∣ k=1
∣
ε
⃗ ⃗ | t | = ε| t |.
≤ n⋅ n
As ε is arbitrary, (10) follows, and all is proved.
□
Theorem 6.3.4 If f
: E
n
→ E
m
(or f
: C
n
→ C
m
is differentiable at p ,⃗ with f
)
′
⃗ = [ Dk fi (p )] ⃗ , [ f (p )]
= (f1 , … , fm ) ,
then [f
′
⃗ (p )]
is an m × n matrix,
i = 1, … , m, k = 1, … , n.
(6.3.48)
Proof By definition, [f
′
is the matrix of the linear map ϕ = df (p ;⃗ ⋅),
⃗ (p )]
ϕ = (ϕ1 , … , ϕm ) .
Here
n
⃗ ⃗ ϕ(t ) = ∑ tk Dk f (p )
(6.3.49)
k=1
by Corollary 1. As f
we can compute D
= (f1 , … , fm ) ,
⃗
k f (p )
componentwise by Theorem 5 of Chapter 5, §1, and Note 2 in §1 to get
⃗ = (Dk f1 (p ), ⃗ … , Dk fm (p )) ⃗ Dk f (p ) m ′
⃗ = ∑ e Dk fi (p ),
k = 1, 2, … , n,
i
i=1
where the e are the basic vectors in E ′
m
i
(C
m
).
(Recall that the e ⃗ are the basic vectors in E k
n
(C
n
).)
Thus m
⃗ ′ ⃗ ϕ(t ) = ∑ e ϕi (t ).
(6.3.50)
i
i=1
Also, n
m
m
n
⃗ ′ ⃗ = ∑ e′ ∑ tk Dk fi (p ). ⃗ ϕ(t ) = ∑ tk ∑ e Dk fi (p ) i
k=1
(6.3.51)
i
i=1
i=1
k=1
The uniqueness of the decomposition (Theorem 2 in Chapter 3, §§1-3) now yields n
⃗ ⃗ ϕi (t ) = ∑ tk Dk fi (p ),
i = 1, … , m,
n n ⃗ t ∈ E (C ) .
(6.3.52)
k=1
If here t ⃗ = e ⃗ , then t k
k
= 1,
while t
j
=0
for j ≠ k. Thus we obtain
⃗ ) = Dk fi (p ), ⃗ ϕi (e k
i = 1, … , m, k = 1, … , n.
(6.3.53)
Hence, ⃗ ) = (v1k , v2k , … , vmk ) , ϕ (e k
(6.3.54)
where ⃗ ) = D f (p ). ⃗ vik = ϕi (e k k i
But by Note 3 of §2, v , … , v (14) results indeed. □ 1k
mk
(6.3.55)
(written vertically) is the k th column of the m × n matrix [ϕ] = [f
In conclusion, let us stress again that while D
⃗ f (p )
u⃗
′
⃗ . (p )]
Thus formula
is a constant, for a fixed p ,⃗ df (p ;⃗ ⋅) is a mapping
6.3.7
https://math.libretexts.org/@go/page/19199
′
ϕ ∈ L (E , E) ,
(6.3.56)
especially "tailored" for p .⃗ The reader should carefully study at least the "arrowed" problems below. This page titled 6.3: Differentiable Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
6.3.8
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6.3.E: Problems on Differentiable Functions Exercise 6.3.E. 1 Complete the missing details in the proofs of this section.
Exercise 6.3.E. 2 Verify Note 1. Describe [f
′
(p )] ⃗
for f
: E
1
→ E
m
,
too. Give examples.
Exercise 6.3.E. 3 ⇒ A map f : E
′
→ E
is said to satisfy a Lipschitz condition (L) of order α > 0 at p ⃗ iff (∃δ > 0) (∃K ∈ E
1
α
) (∀x⃗ ∈ G¬p ⃗ (δ))
⃗ ≤ K| x⃗ − p |⃗ |f (x⃗ ) − f (p )|
.
(6.3.E.1)
Prove the following. (i) This implies continuity at p ⃗ (but not conversely; see Problem 7 in Chapter 5, §1). (ii) L of order > 1 implies differentiability at p ,⃗ with df (p ;⃗ ⋅) = 0 on E . (iii) Differentiability at p ⃗ implies L of order 1 (apply Theorem 1 in §2 to ϕ = df ). (iv) If f and g are differentiable at p ,⃗ then ′
1 lim
|Δf ||Δg| = 0.
⃗ x→ p ⃗
(6.3.E.2)
|Δx⃗ |
Exercise 6.3.E. 4 For the functions of Problem 5 in §1, find those p ⃗ at which f is differentiable. Find ′
⃗ df (p ;⃗ ⋅), and [ f (p )] ⃗ . ∇f (p ),
(6.3.E.3)
[Hint: Use Theorem 3 and Corollary 1.]
Exercise 6.3.E. 5 Prove the following statements. (i) If f : E → E is constant on an open globe G ⊂ E ⇒
′
′
,
it is differentiable at each p ⃗ ∈ G, and df (p ,⃗ ⋅) = 0 on E
(ii) If the latter holds for each p ⃗ ∈ G − Q (Q countable), then continuous there. [Hint: Given p ,⃗ q ⃗ ∈ G, use Theorem 2 in §1 to get f (p )⃗ = f (q )⃗ .]
f
is constant on
G
(even on
¯ ¯¯ ¯
G
) provided
f
′
.
is relatively
Exercise 6.3.E. 6 Do Problem 5 in case G is any open polygon-connected set in E . (See Chapter 4, §9.) ′
Exercise 6.3.E. 7 Prove the following. (i) If f , g : E → E are differentiable at p ,⃗ so is ⇒
′
h = af + bg,
6.3.E.1
(6.3.E.4)
https://math.libretexts.org/@go/page/24088
for any scalars a, b (if f and g are scalar valued, a and b may be vectors; moreover, d(af + bg) = adf + bdg,
(6.3.E.5)
i.e., ′ t ⃗ ∈ E .
⃗ = adf (p ;⃗ t ) ⃗ + bdg(p ;⃗ t ), ⃗ dh(p ;⃗ t )
(ii) In case f , g : E
m
→ E
1
or C
m
→ C,
(6.3.E.6)
deduce also that ⃗ = a∇f (p ) ⃗ + b∇g(p ). ⃗ ∇h(p )
(6.3.E.7)
Exercise 6.3.E. 8 ⇒
Prove that if f , g : E
′
→ E
1
are differentiable at p ,⃗ then so are
(C )
g h = gf and k =
.
(6.3.E.8)
f
⃗ (the latter, if f (p )⃗ ≠ 0). Moreover, with a = f (p )⃗ and b = g(p ), show that (i) dh = adg + bdf and (ii) dk = (adg − bdf )/a . If further E = E (C ) , verify that (iii) ∇h(p )⃗ = a∇g(p )⃗ + b∇f (p )⃗ and ⃗ (iv) ∇k(p )⃗ = (a∇g(p )⃗ − b∇f (p ))/ a . Prove (i) and (ii) for vector-valued g, too. [Hints: (i) Set ϕ = adg + bdf , with a and b as above. Verify that 2
′
n
n
2
⃗ = g(p )(Δf ⃗ + f (p )(Δg ⃗ + (Δf )(Δg). Δh − ϕ(t ) ⃗ − df (t )) ⃗ − dg(t ))
Use Problem 3(iv) and Definition 1. ⃗ Show that dF (ii) Let F (t )⃗ = 1/f (t ).
2
= −df / a .
(6.3.E.9)
Then apply (i) to gF .]
Exercise 6.3.E. 9 Let f : E → E (C ) , f = (f , … , f ) . Prove that (i) f is linear iff all its m components f are; (ii) f is differentiable at p ⃗ iff all f are, and then df = (df , … , df ⇒
′
m
m
1
m
k
k
1
m)
. Hence if f is complex, df
= dfre + i ⋅ dfim .
Exercise 6.3.E. 10 Prove the following statements. (i) If f ∈ L (E , E) then f is differentiable on E , and df (p ;⃗ ⋅) = f , p ⃗ ∈ E (ii) Such is any first-degree monomial, hence any sum of such monomials. ′
′
′
.
Exercise 6.3.E. 11 Any rational function is differentiable in its domain. [Hint: Use Problems 10(i), 7, and 8. Proceed as in Theorem 3 in Chapter 4, §3.]
6.3.E.2
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Exercise 6.3.E. 12 Do Problem 8(i) in case g is only continuous at p ,⃗ and f (p )⃗ = 0. Find dh.
Exercise 6.3.E. 13 Do Problem 8(i) for dot products h = f ⋅ g of functions f , g : E
′
→ E
m
(C
m
).
Exercise 6.3.E. 14 Prove the following. ⃗ with v ⃗ as in §2, Theorem 2(ii). (i) If ϕ ∈ L (E , E ) or ϕ ∈ L (C , C ) , then ∥ϕ∥ = |v|, (ii) If f : E → E (f : C → C ) is differentiable at p ,⃗ then n
1
n
n
1
n
1
⃗ ∥df (p ;⃗ ⋅)∥ = |∇f (p )|.
Moreover, in case f
: E
n
→ E
1
(6.3.E.10)
, ⃗ ≥ D ⃗ f (p ) ⃗ |∇f (p )| u
if | u⃗ | = 1
(6.3.E.11)
and ⃗ = D ⃗ f (p ) ⃗ |∇f (p )|
⃗ ∇f (p )
when u⃗ =
u
(6.3.E.12) ⃗ |∇f (p )|;
thus ⃗ = max D ⃗ f (p ). ⃗ |∇f (p )| u
(6.3.E.13)
⃗ | u|=1
[Hints: Use the equality case in Theorem 4(c') of Chapter 3, §§1-3. Use formula (7), Corollary 2, and Theorem 2(ii).]
Exercise 6.3.E. 15 Show that Theorem 3 holds even if (i) D f is discontinuous at p ,⃗ and (ii) f has partials on A − Q only (Q countable, p ⃗ ∉ Q), provided f is continuous on A in each of the last n − 1 variables. [Hint: For k = 1, formula (13) still results by definition of D f , if a suitable δ has been chosen.] 1
1
Exercise 6.3.E. 16∗ Show that Theorem 3 and Problem 15 apply also to any f : E → E where E is n -dimensional with basis {u⃗ , … , u⃗ } (see Problem 12 in §2) if we write D f for D f . [Hints: Assume |u⃗ | = 1, 1 ≤ k ≤ n (if not, replace u⃗ by u⃗ / |u⃗ | ; show that this yields another basis). Modify the proof so that the p ⃗ are still in G (δ). Caution: The standard norm of E does not apply here.] ′
′
1
k
n
⃗ uk
k
k
k
k
n
k
p ⃗
Exercise 6.3.E. 17 Let f
k
: E
1
→ E
1
be differentiable at p
k (k
= 1, … , n).
For x⃗ = (x
1,
n
… , xn ) ∈ E
n
,
set
n
F (x⃗ ) = ∑ fk (xk ) and G(x⃗ ) = ∏ fk (xk ) . k=1
(6.3.E.14)
k=1
6.3.E.3
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Show that F and G are differentiable at p ⃗ = (p , … , p ) . Express ∇F (p )⃗ and ∇G(p )⃗ in terms of the f (p ). ⃗ "imitate" Problem 6 [Hint: In order to use Problems 7 and 8, replace the f by suitable functions defined on E . For ∇G(p ), in Chapter 5, §1.] 1
′
n
k
k
n
k
6.3.E: Problems on Differentiable Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.3.E.4
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6.4: The Chain Rule. The Cauchy Invariant Rule To generalize the chain rule (Chapter 5, §1), we consider the composite g : E → E, with E , E , and E as before. ′′
′
h = g∘ f
of two functions,
f : E
′
→ E
′′
and
′′
Theorem 6.4.1 (chain rule) If f : E
′
→ E
′′
and g : E
′′
→ E
(6.4.1)
⃗ respectively, then are differentiable at p ⃗ and q ⃗ = f (p ), h = g∘ f
(6.4.2)
dh(p ;⃗ ⋅) = dg(q ;⃗ ⋅) ∘ df (p ;⃗ ⋅).
(6.4.3)
is differentiable at p ,⃗ and
Briefly: "The differential of the composite is the composite of differentials." Proof Let U
= df (p ;⃗ ⋅), V = dg(q ;⃗ ⋅),
and ϕ = V
∘U
.
As U and V are linear continuous maps, so is ϕ. We must show that ϕ = dh(p ;⃗ ⋅). ⃗ Here it is more convenient to write Δx⃗ or x⃗ − p ⃗ for the "t ⃗ " of Definition 1 in §3. For brevity, we set (with q ⃗ = f (p )) w(x⃗ )
⃗ − ϕ(x⃗ − p ), ⃗ = Δh − ϕ(Δx⃗ ) = h(x⃗ ) − h(p )
u(x⃗ )
⃗ − U (x⃗ − p ), ⃗ = Δf − U (Δx⃗ ) = f (x⃗ ) − f (p )
⃗ v(y )
⃗ = g(y ) ⃗ − g(q ) ⃗ − V (y ⃗ − q ), ⃗ = Δg − V (Δy )
′
x⃗ ∈ E , ′
x⃗ ∈ E , y ⃗ ∈ E
′′
.
Then what we have to prove (see Definition 1 in §3) reduces to w(x⃗ ) lim
= 0,
⃗ x→ p ⃗
(6.4.4)
| x⃗ − p |⃗
while the assumed existence of df (p ;⃗ ⋅) = U and dg(q ;⃗ ⋅) = V can be expressed as u(x⃗ ) lim
= 0,
⃗ p ⃗ x→
(6.4.5)
| x⃗ − p |⃗
and ⃗ v(y ) lim y →q ⃗
¯ ¯ ¯
= 0,
⃗ q ⃗ = f (p ).
(6.4.6)
| y ⃗ − q |⃗
From (2) and (3), recalling that h = g ∘ f and ϕ = V
∘ U,
we obtain
⃗ − V (U (x⃗ − p )) ⃗ w(x⃗ ) = g(f (x⃗ )) − g(q ) ⃗ − V (f (x⃗ ) − f (p ) ⃗ − u(x⃗ )). = g(f (x⃗ )) − g(q )
Using (4), with y ⃗ = f (x⃗ ), and the linearity of V , we rewrite (6) as w(x⃗ )
⃗ − V (f (x⃗ ) − f (p )) ⃗ − V (u(x⃗ )) = g(f (x⃗ )) − g(q ) = v(f (x⃗ )) + V (u(x⃗ )).
(Verify!) Thus the desired formula (5) will be proved if we show that V (u(x⃗ )) lim ⃗ p ⃗ x→
=0
(6.4.7)
| x⃗ − p |⃗
and
6.4.1
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v(f (x⃗ )) lim
= 0.
(6.4.8)
| x⃗ − p |⃗
⃗ p ⃗ x→
Now, as V is linear and continuous, formula (5') yields (6'). Indeed, V (u(x⃗ ))
u(x⃗ )
lim ⃗ p ⃗ x→
= lim V ( | x⃗ − p |⃗
) = V (0) = 0
(6.4.9)
| x⃗ − p |⃗
⃗ p ⃗ x→
by Corollary 2 in Chapter 4, §2. (Why?) Similarly, (5") implies (6") by substituting y ⃗ = f (x⃗ ), since ⃗ ≤ K| x⃗ − p |⃗ |f (x⃗ ) − f (p )|
by Problem 3(iii) in §3. (Explain!) Thus all is proved.
(6.4.10)
□
Note 1 (Cauchy invariant rule). Under the same assumptions, we also have ⃗ ⃗ dh(p ;⃗ t ) = dg(q ;⃗ s )
(6.4.11)
dh(p ;⃗ ⋅) = ϕ = V ∘ U .
(6.4.12)
⃗ = U (t ), ⃗ s ⃗ = df (p ;⃗ t )
(6.4.13)
⃗ = ϕ(t ) ⃗ = V (U (t )) ⃗ = V (s ) ⃗ = dg(q ;⃗ s ), ⃗ dh(p ;⃗ t )
(6.4.14)
⃗ ⃗ if s ⃗ = df (p ;⃗ t ), t ∈ E . ′
For with U and V as above,
Thus if
we have
proving (7). Note 2. If E
′
=E
n
(C
n
),E
′′
=E
m
(C
m
) , and E = E
r
(C
r
)
(6.4.15)
then by Theorem 3 of §2 and Definition 2 in §3, we can write (1) in matrix form, ′
′
′
[ h (p )] ⃗ = [ g (q )] ⃗ [ f (p )] ⃗ ,
(6.4.16)
resembling Theorem 3 in Chapter 5, §1 (with f and g interchanged). Moreover, we have the following theorem.
Theorem 6.4.2 With all as in Theorem 1, let E
′
=E
n
(C
n
),E
′′
=E
m
(C
m
),
(6.4.17)
and f = (f1 , … , fm ) .
(6.4.18)
Then m
⃗ = ∑ Di g(q ) ⃗ Dk fi (p ); ⃗ Dk h(p )
(6.4.19)
i=1
or, in classical notation, ∂
m
∂
⃗ = ∑ h(p ) ∂xk
i=1
∂ ⃗ ⋅ g(q )
∂yi
∂xk
6.4.2
⃗ fi (p ),
k = 1, 2, … , n.
(6.4.20)
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Proof Fix any basic vector e ⃗ in E and set k
′
⃗ ) , s ⃗ = df (p ;⃗ e k
s ⃗ = (s1 , … , sm ) ∈ E
′′
.
(6.4.21)
As f is differentiable at p ,⃗ so are its components f (Problem 9 in §3), and i
⃗ ) = Dk fi (p ) ⃗ si = dfi (p ;⃗ e k
(6.4.22)
by Theorem 2(ii) in §3. Using also Corollary 1 in §3, we get m
m
⃗ = ∑ si Di g(q ) ⃗ = ∑ Dk fi (p ) ⃗ Di g(q ). ⃗ dg(q ;⃗ s ) i=1
(6.4.23)
i=1
But as s ⃗ = df (p ;⃗ e ⃗ ) , formula (7) yields k
⃗ = dh (p ;⃗ e ⃗ ) = D h(p ) ⃗ dg(q ;⃗ s ) k k
by Theorem 2(ii) in §3. Thus the result follows.
(6.4.24)
□
Note 3. Theorem 2 is often called the chain rule for functions of several variables. It yields Theorem 3 in Chapter 5, §1, if m =n =1 . In classical calculus one often speaks of derivatives and differentials of variables mappings. Thus Theorem 2 is stated as follows. Let u = g (y
1,
… , ym )
y = f (x1 , … , xn )
rather than those of
be differentiable. If, in turn, each yi = fi (x1 , … , xn )
is differentiable for i = 1, … , m, then formula (8)) we have
u
(6.4.25)
is also differentiable as a composite function of the m
∂u
∂u
=∑ ∂xk
i=1
∂yi
,
k = 1, 2, … , n.
n
variables
xk ,
and ("simplifying"
(6.4.26)
∂yi ∂xk
It is understood that the partials ∂u and ∂xk ⃗ while the ∂u/∂y are at q ⃗ = f (p ), where cause ambiguities (see the next example). i
∂yi
′
are taken at some p ⃗ ∈ E ,
(6.4.27)
∂xk
f = (f1 , … , fm ) .
This "variable" notation is convenient in computations, but may
Example Let u = g(x, y, z), where z depends on x and y : z = f3 (x, y).
Set f
1 (x,
y) = x, f2 (x, y) = y, f = (f1 , f2 , f3 ) ,
(6.4.28)
and h = g ∘ f ; so h(x, y) = g(x, y, z).
(6.4.29)
By (8'), ∂u
∂u ∂x =
∂x
∂u ∂y +
∂x ∂x
∂u ∂z +
∂y ∂x
.
(6.4.30)
∂z ∂x
Here ∂x = ∂x
∂f1
∂y = 1 and
∂x
= 0,
(6.4.31)
∂x
6.4.3
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for f does not depend on x. Thus we obtain 2
∂u
∂u =
∂x
∂u ∂z +
∂x
.
(6.4.32)
∂z ∂x
(Question: Is (∂u/∂z)(∂z/∂x) = 0? ) The trouble with (9) is that the variable u "poses" as both g and h. On the left, it is h; on the right, it is g. To avoid this, our method is to differentiate well-defined mappings, not "variables." Thus in (9), we have the maps g : E
with f
1,
f2 , f3
3
→ E and f : E
2
→ E
3
,
(6.4.33)
as indicated. Then if h = g ∘ f , Theorem 2 states (9) unambiguously as ⃗ = D1 g(q ) ⃗ + D3 g(q ) ⃗ ⋅ D1 f (p ), ⃗ D1 h(p )
(6.4.34)
⃗ = (p1 , p2 , f3 (p )) ⃗ . q ⃗ = f (p )
(6.4.35)
where p ⃗ ∈ E and 2
(Why?) In classical notation, ∂h
∂g =
∂x
+ ∂x
∂g ∂f3 ∂z
(6.4.36)
∂x
(avoiding the "paradox" of (9)). Nonetheless, with due caution, one may use the "variable" notation where convenient. The reader should practice both (see the Problems). Note 4. The Cauchy rule (7), in "variable" notation, turns into m
i=1
where dx
k
= tk
and dy
i
⃗ = dfi (p ;⃗ t )
n
∂u
du = ∑ ∂yi
dyi = ∑ k=1
∂u ∂xk
dxk ,
(6.4.37)
.
Indeed, by Corollary 1 in §3, n
m
⃗ ⃗ ⋅ tk and dg(q ;⃗ s ) ⃗ = ∑ Di g(q ) ⃗ ⋅ si . dh(p ;⃗ t ) = ∑ Dk h(p ) k=1
(6.4.38)
i=1
Now, in (7), ⃗ s ⃗ = (s1 , … , sm ) = df (p ;⃗ t );
(6.4.39)
so by Problem 9 in §3, ⃗ = s , dfi (p ;⃗ t ) i
i = 1, … , m.
(6.4.40)
Rewriting all in the "variable" notation, we obtain (10). The "advantage" of (10) is that du has the same form, independently of whether u is treated as a function of the x or of the (hence the name "invariant" rule). However, one must remember the meaning of dx and dy , which are quite different. k
k
yi
i
The "invariance" also fails completely for differentials of higher order (§5). The advantages of the "variable" notation vanish unless one is able to "translate" it into precise formulas. This page titled 6.4: The Chain Rule. The Cauchy Invariant Rule is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
6.4.4
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6.4.E: Further Problems on Differentiable Functions Exercise 6.4.E. 1 For E = E [Hint: Find
r
(C
r
)
prove Theorem 2 directly. ⃗ Dk hj (p ),
j = 1, … , r,
(6.4.E.1)
from Theorem 4 of §3, and Theorem 3 of §2. Verify that r
r
⃗ = ∑ ej Dk hj (p ) and ⃗ ⃗ = ∑ ej Di gj (q ), ⃗ Dk h(p ) Di g(q ) j=1
where the e are the basic unit vectors in E j
r
.
(6.4.E.2)
j=1
Proceed.]
Exercise 6.4.E. 2 Let g(x, y, z) = u, x = f
1 (r,
and
θ), y = f2 (r, θ), z = f3 (r, θ),
f = (f1 , f2 , f3 ) : E
2
→ E
3
.
(6.4.E.3)
Assuming differentiability, verify (using "variables") that ∂u du =
∂u dx +
∂x
∂u dy +
∂y
∂u dz =
∂z
∂u dr +
∂r
dθ
(6.4.E.4)
∂θ
⃗ by computing derivatives from (8'). Then do all in the mapping notation for H = g ∘ f , dH (p ;⃗ t ).
Exercise 6.4.E. 3 For the specific functions f , g, h, and k of Problems 4 and 5 of §2, set up and solve problems analogous to Problem 2, using (a) k ∘ f ;
(b) g ∘ k;
(c) f ∘ h;
(d) h ∘ g.
(6.4.E.5)
Exercise 6.4.E. 4 For the functions of Problem 5 in §1, find the formulas for Describe it for a chosen p .⃗
⃗ df (p ;⃗ t ).
At which
p ⃗
does
df (p ;⃗ ⋅)
exist in each given case?
Exercise 6.4.E. 5 From Theorem 2, with E = E
1
(C ),
find n
∇h(p ) ⃗ = ∑ Dk g(q )∇ ⃗ fk (p ). ⃗
(6.4.E.6)
k=1
Exercise 6.4.E. 6 Use Theorem 1 for a new solution of Problem 7 in §3 with E = E [Hint: Define F on E and G on E (C ) by ′
2
1
(C ).
2
⃗ = ay1 + b y2 . F (x⃗ ) = (f (x⃗ ), g(x⃗ )) and G(y )
6.4.E.1
(6.4.E.7)
https://math.libretexts.org/@go/page/24089
Then h = af + bg = G ∘ F . (Why?) Use Problems 9 and 10(ii) of §3. Do all in "variable" notation, too.]
Exercise 6.4.E. 7 Use Theorem 1 for a new proof of the "only if " in Problem 9 in §3. [Hint: Set f = g ∘ f , where g(x⃗ ) = x (the ith "projection map") is a monomial. Verify!] i
i
Exercise 6.4.E. 8 Do Problem 8(I) in §3 for the case E
′
=E
2
(C
2
),
with
f (x⃗ ) = x1 and g(x⃗ ) = x2 .
(6.4.E.8)
(Simplify!) Then do the general case as in Problem 6 above, with ⃗ = y1 y2 . G(y )
(6.4.E.9)
Exercise 6.4.E. 9 Use Theorem 2 for a new proof of Theorem 4 in Chapter 5, §1. (Proceed as in Problems 6 and 8, with D h = h ). Do it in the "variable" notation, too.
E
′
=E
1
,
so that
′
1
Exercise 6.4.E. 10 Under proper differentiability assumptions, use formula (8') to express the partials of u if (i) u = g(x, y), x = f (r)h(θ), y = r + h(θ) + θf (r) ; (ii) u = g(r, θ), r = f (x + f (y)), θ = f (xf (y)) ; (iii) u = g (x , y , z ) . Then redo all in the "mapping" terminology, too. y
z
x+y
Exercise 6.4.E. 11 Let the map g : E → E be differentiable on E h = g ∘ f and (i) f (x⃗ ) = ∑ x , x⃗ ∈ E ; (ii) f (x⃗ ) = |x⃗ | , x⃗ ∈ E . 1
1
n
k=1
1
.
⃗ if Find |∇h(p )|
n
k
2
n
Exercise 6.4.E. 12 (Euler's theorem.) A map f
: E
n
→ E
1
(or C
n
→ C
) is called homogeneous of degree m on G iff
(∀t ∈ E
1
m
f (tx⃗ ) = t
(C ))
f (x⃗ )
(6.4.E.10)
when x⃗ , tx⃗ ∈ G. Prove the following statements. (i) If so, and f is differentiable at p ⃗ ∈ G (an open globe), then ⃗ = mf (p ). ⃗ p ⃗ ⋅ ∇f (p )
*(ii) Conversely, if the latter holds for all p ⃗ ∈ G and if
→ 0 ∉ G,
(6.4.E.11)
then f is homogeneous of degree m on G.
→
(iii) What if 0 ∈ G? [Hints: (i) Let g(t) = f (tp ). ⃗ Find g
′
(1).
(iii) Take f (x, y) = x
2
y
2
6.4.E.2
if x ≤ 0, f
=0
if x > 0, G = G
0 (1).
]
https://math.libretexts.org/@go/page/24089
Exercise 6.4.E. 13 Try Problem 12 for f
: E
′
→ E,
replacing p ⃗ ⋅ ∇f (p )⃗ by df (p ;⃗ p )⃗ .
Exercise 6.4.E. 14 With all as in Theorem 1, prove the following. (i) If E
′
=E
1
and s ⃗ = f
′
→ (p) ≠ 0 ,
then h (p) = D ′
⃗ g(q )
s ⃗
(ii) If u⃗ and v ⃗ are nonzero in E and au⃗ + bv ⃗ ≠ ′
→ 0
.
for some scalars a, b, then
⃗ = aD ⃗ f (p ) ⃗ + b D ⃗ f (p ). ⃗ Dau+b f (p ) ⃗ v ⃗ u v
(iii) If f is differentiable on a globe G
,
p ⃗
and u⃗ ≠
→ 0
(6.4.E.12)
in E , then ′
⃗ = lim D ⃗ (p ). ⃗ D ⃗ f (p ) u
⃗ u⃗ x→
x
(6.4.E.13)
[Hints: Use Theorem 2(ii) from §3 and Note 1.]
Exercise 6.4.E. 15 Use Theorem 2 to find the partially derived functions of f , if (i) f (x, y, z) = (sin(xy/z)) ; (ii) f (x, y) = log | tan(y/x)|. (Set f = 0 wherever undefined.) x
x
6.4.E: Further Problems on Differentiable Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.4.E.3
https://math.libretexts.org/@go/page/24089
6.5: Repeated Differentiation. Taylor’s Theorem In §1 we defined u⃗ -directed derived functions, D
f
u⃗
for any f
: E
′
→ E
and any u⃗ ≠
→
in E
0
′
.
→
Thus given a sequence {u⃗ } ⊆ E − { 0 }, we can first form D f , then D (D f ) (the u⃗ -directed derived function of D f ), then the u⃗ -directed derived function of D (D f ) , and so on. We call all functions so formed the higher-order directional derived functions of f . ′
i
⃗ u1
3
⃗ u2
⃗ u2
⃗ u1
2
⃗ u1
If at each step the limit postulated in Definition 1 of §1 exists for all p ⃗ in a set B ⊆ E derivatives of f (on B ). If all u⃗ are basic unit vectors in E i
We also define D
1
n
(C
n
k
k−1
⃗ u2 ⃗ …uk ⃗ u1
⃗ u2 ⃗ …uk ⃗ u1
f
f = Du⃗ (D
¯¯ ¯
⃗ u2 …uk−1 ⃗ u1
k
f) ,
k = 2, 3, … ,
⃗ u⃗ ⃗ uk …u1 k−1
k
i
f
u⃗
1
∂
2
D12 f = D
⃗ e 2 ⃗ e1
f =f
u⃗
f. )
instead.
For partially derived functions, we simplify this notation, writing 12 … for e ⃗ e ⃗ notation):
0
(6.5.1)
a directional derived function of order k. (Some authors denote it by D
If all u⃗ equal u⃗ , we write D
We also set D
we call them the higher-order directional
,
we say "partial' instead of "directional."
),
D
k
′
and
f = Du⃗ f
u⃗
and call D
⃗ u1
2
f
f =
, ∂ x1 ∂ x2
…
2
and omitting the "k " in D (except in classical k
∂
2
D11 f = D
⃗ e 1 ⃗ e1
2
f =
f 2
, etc.
(6.5.2)
∂x
1
for any vector u⃗ .
Example (A) Define f
: E
2
→ E
1
by 2
xy (x f (0, 0) = 0,
f (x, y) =
2
−y )
2
+y
2
2
x
.
2
(6.5.3)
Then 4
y (x
∂f ∂x
whence D
1 f (0,
y) = −y
+ 4x y
= D1 f (x, y) = (x2 + y 2 )
4
−y ) ,
(6.5.4)
(Verify!)
(6.5.5)
2
if y ≠ 0; and also f (x, 0) − f (0, 0) D1 f (0, 0) = lim
= 0. x
x→0
Thus D
1 f (0,
y) = −y
always, and so D
12 f (0,
y) = −1; D12 f (0, 0) = −1 4
x (x
2
− 4x y
D2 f (x, y) =
2
(x
if x ≠ 0 and D
2 f (0,
0) = 0.
Thus (∀x)D
2 f (x,
0) = x
2
2
+y )
Similarly, 4
−y ) (6.5.6)
2
and so
D21 f (x, 0) = 1 and D21 f (0, 0) = 1 ≠ D12 f (0, 0) = −1.
The previous example shows that we may well have the following theorem.
D12 f ≠ D21 f ,
6.5.1
or more generally,
2
D
(6.5.7) 2
f ≠D
⃗ ⃗ uv
f.
⃗ ⃗ vu
However, we obtain
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Theorem 6.5.1 Given nonzero vectors u⃗ and v⃗ in E , suppose f ′
: E
′
has the derivatives
→ E
2
Du⃗ f , Dv ⃗ f , and D
f
(6.5.8)
⃗ ⃗ uv
on an open set A ⊆ E . ′
If D
2
f
⃗ ⃗ uv
is continuous at some p ⃗ ∈ A, then the derivative D
2
⃗ f (p )
⃗ ⃗ vu
also exists and equals D
2
⃗ f (p )
⃗ ⃗ uv
.
Proof ⃗ (Why?) By Corollary 1 in §1, all reduces to the case |u⃗ | = 1 = |v|.
Given ε > 0, fix δ > 0 so small that G = G
(δ) ⊆ A
p ⃗
and simultaneously
2 2 ⃗ ∣ ≤ ε sup ∣ f (x⃗ ) − D f (p ) ∣Duv ∣ ⃗ ⃗ ⃗ ⃗ uv
(6.5.9)
⃗ x∈G
(by the continuity of D
2
f
⃗ ⃗ uv
Now (∀s, t ∈ E
1
)
at p )⃗ .
define H
t
: E
1
→ E
by ⃗ Ht (s) = Du⃗ f (p ⃗ + tu⃗ + sv).
(6.5.10)
Let δ I = (−
If s, t ∈ I , the point x⃗ = p ⃗ + tu⃗ + sv ⃗ is in G
δ ,
2
).
(6.5.11)
2
since
(δ) ⊆ A,
p ⃗
| x⃗ − p |⃗ = |tu⃗ + sv|⃗
2. Thus the derivative D f is independent of the order in which the u⃗ follow each other if it exists and is continuous on an open set A ⊆ E , along with appropriate derivatives of order < k. ⃗ u2 ⃗ …uk ⃗ u1
′
i
If E
′
=E
n
(C
n
),
this applies to partials as a special case.
For E and C only, we also formulate the following definition. n
n
Definition 1 Let E = E (C ) . We say that differentiable at p .⃗ ′
n
n
f : E
If this holds for all p ⃗ in a set B ⊆ E
′
,
′
→ E
is
m
times differentiable at
p ⃗ ∈ E
′
iff
f
and all its partials of order
0)(∃δ > 0) (∀x⃗ ∈ Gp ⃗ (δ))
(6.5.E.8)
with norm ∥ as in Definition 2 in §2. (Does it apply?) [Hint: If f ∈ C D , use Theorem 2 in §3. For the converse, verify that 1
∣
n
∣
1 1 ⃗ ⃗ ⃗ ⃗ ∣ ⃗ ⃗ ε ≥∣ ∣d f (p ; t ) − d f (x; t )∣ = ∣∑ [ Dk f (p ) − Dk f (x)] tk ∣
∣ k=1
if x⃗ ∈ G
(δ)
p ⃗
and |t |⃗ ≤ 1. Take t ⃗ = e ⃗ , to prove continuity of D k
kf
6.5.E.2
(6.5.E.9)
∣
⃗ at p .]
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Exercise 6.5.E. 11 Prove the following. (i) If ϕ : E → E is linear and [ϕ] = (v n
m
ik )
,
then 2
2
∥ϕ∥
≤ ∑ | vik | .
(6.5.E.10)
i,k
(ii) If f
: E
n
→ E
m
is differentiable at p ,⃗ then 2
∥df (p ;⃗ ⋅)∥
2
⃗ ≤ ∑ | Dk fi (p )| .
(6.5.E.11)
i,k
(iii) Hence find a new converse proof in Problem 10 for f : E → E . Consider f : C → C , too. [Hints: (i) By the Cauchy-Schwarz inequality, |ϕ(x⃗ )| ≤ |x⃗ | ∑ |v | n
n
m
m
2
2
2
ik
i,k
.
(Why?) (ii) Use part (i) and Theorem 4 in §3.]
Exercise 6.5.E. 12 (i) Find d u for the functions of Problem 10 in §4, in the "variable" and "mapping" notations. (ii) Do it also for 2
2
u = f (x, y, z) = (x
+y
2
2
+z )
−
1 2
(6.5.E.12)
and show that D f + D f + D f = 0 . (iii) Does the latter hold for u = arctan ? 11
22
33
y
x
Exercise 6.5.E. 13 Let u = g(x, y), x = r cos θ, y = r sin θ (passage to polars). Using "variables" and then the "mappings" notation, prove that if g is differentiable, then (i) = cos θ + sin θ and ∂u
∂u
∂r
∂x
(ii) |∇g(x, y)|
2
∂u ∂y
=(
∂u ∂r
2
)
(iii) Assuming g ∈ C D
2
,
+(
1
∂u
r
∂θ
express
2
) ∂
2
u
∂r∂θ
. ,
∂
2
u 2
∂r
,
and
∂
2
u 2
∂θ
as in (i).
Exercise 6.5.E. 14 Let f , g : E → E be of class C D on E (i) D h = a D h if a ∈ E (fixed) and 1
1
2
11
2
1
.
Verify (in "variable" notation, too) the following statements.
1
22
h(x, y) = f (ax + y) + g(y − ax).
(ii) x
2
2
D11 h(x, y) + 2xy D12 h(x, y) + y D22 h(x, y) = 0
if y
h(x, y) = xf (
y )+g(
x
(iii) D
1h
⋅ D21 h = D2 h ⋅ D11 h
(6.5.E.13)
).
(6.5.E.14)
x
if h(x, y) = g(f (x) + y)
6.5.E.3
(6.5.E.15)
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Find D
12 h,
too.
Exercise 6.5.E. 15 Assume
and E = E (C ) . Let ⃗ ∈ E q ⃗ = f (p ) , respectively, and set h = g ∘ f . Show that h is twice differentiable at p ,⃗ and E
′
=E
n
(C
n
′′
)
m
m
f : E
′
→ E
′′
and
g : E
′′
→ E
be twice differentiable at
p ⃗ ∈ E
′
and
′′
2 ⃗ = d 2 g(q ;⃗ s ) ⃗ + dg(q ;⃗ v), ⃗ d h(p ;⃗ t )
where t ⃗ ∈ E
′
⃗ , s ⃗ = df (p ;⃗ t ),
and v ⃗ = (v
1,
… , vm ) ∈ E
′′
(6.5.E.16)
satisfies
2 ⃗ vi = d fi (p ;⃗ t ),
i = 1, … , m.
(6.5.E.17)
Thus the second differential is not invariant in the sense of Note 4 in §4. [Hint: Show that m
m
m
⃗ = ∑ ∑ Dij g(q ) ⃗ Dk fi (p ) ⃗ Dl fj (p ) ⃗ + ∑ Di g(q ) ⃗ Dkl fi (p ). ⃗ Dkl h(p ) j=1
i=1
(6.5.E.18)
i=1
Proceed.]
Exercise 6.5.E. 16 Continuing Problem 15, prove the invariant rule: r
r
⃗ = d g(q ;⃗ s ), ⃗ d h(p ;⃗ t )
(6.5.E.19)
if f is a first-degree polynomial and g is r times differentiable at q .⃗ [Hint: Here all higher-order partials of f vanish. Use induction.] 6.5.E: Problems on Repeated Differentiation and Taylor Expansions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.5.E.4
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6.6: Determinants. Jacobians. Bijective Linear Operators We assume the reader to be familiar with elements of linear algebra. Thus we only briefly recall some definitions and well-known rules.
Definition Given a linear operator ϕ : E
n
→ E
n
( or ϕ : C
n
→ C
n
with matrix
),
[ϕ] = (vik ) ,
i, k = 1, … , n,
(6.6.1)
we define the determinant of [ϕ] by ∣ v11
v12
…
v1n ∣
v22
…
∣ v2n ∣
⋮
⋱
vn2
…
∣ ∣ v21 det[ϕ] = det(vik ) = ∣
∣
∣
⋮
∣
∣ vn1
⋮
∣ ∣
vnn ∣
λ
= ∑(−1 ) v1k1 v2k2 … vnkn
where the sum is over all ordered n -tuples (k
1,
… , kn )
of distinct integers k
j
(1 ≤ kj ≤ n) ,
0
if ∏
1
if ∏j 0 (for ϕ
−1
6.6.5
is not the zero map, being one-to-one).
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Thus we may set 1 ε =
Clearly x⃗ = ϕ
−1
⃗ (y )
−1
∥ ∥ϕ
,
−1
∥
−1
⃗ ∣ (y ) ∣ ≤
∥ϕ
∥ ∥ =
1 .
(6.6.40)
ε
if y ⃗ = ϕ(x⃗ ). Also, ∣ ∣ϕ
1
| y |⃗
(6.6.41)
⃗ ∣ (y ) ∣,
(6.6.42)
ε
by Note 5 in §2, Hence −1
| y |⃗ ≥ ε ∣ ∣ϕ
i.e., |ϕ(x⃗ )| ≥ ε| x⃗ |
(6.6.43)
|(ϕ − θ)(x⃗ )| ≤ ∥ϕ − θ∥| x⃗ | = σ| x⃗ |.
(6.6.44)
for all x⃗ ∈ E and y ⃗ ∈ E . ′
Now suppose ϕ ∈ L (E
′
, E)
and ∥θ − ϕ∥ = σ < ε .
Obviously, θ = ϕ − (ϕ − θ), and by Note 5 in §2,
Thus for every x⃗ ∈ E , ′
|θ(x⃗ )| ≥ |ϕ(x⃗ )| − |(ϕ − θ)(x⃗ )| ≥ |ϕ(x⃗ )| − σ| x⃗ | ≥ (ε − σ)| x⃗ |
by (2). Therefore, given p ⃗ ≠ r ⃗ in E and setting x⃗ = p ⃗ − r ⃗ ≠ ′
→
we obtain
0 ,
⃗ − θ(r )| ⃗ = |θ(p ⃗ − r )| ⃗ = |θ(x⃗ )| ≥ (ε − σ)| x⃗ | > 0 |θ(p )
(6.6.45)
(since σ < ε) . ⃗ We see that p ⃗ ≠ r ⃗ implies θ(p )⃗ ≠ θ(r ); so θ is one-to-one, indeed.
Also, setting θ(x⃗ ) = z ⃗ and x⃗ = θ
−1
⃗ (z )
in (3), we get −1
| z |⃗ ≥ (ε − σ) ∣ ∣θ
⃗ ∣ (z ) ∣;
(6.6.46)
that is, −1
∣ ∣θ
for all z ⃗ in the range of θ (domain of θ
−1
Thus θ
−1
)
−1
⃗ ∣ (z ) ∣ ≤ (ε − σ )
| z |⃗
(6.6.47)
.
is linearly bounded (by Theorem 1 in §2), hence uniformly continuous, as claimed.
□
Corollary 6.6.3 If E
′
=E =E
n
(C
n
)
in Theorem 2 above, then for given ϕ and δ > 0, there always is δ ′
−1
∥θ − ϕ∥ < δ implies ∥ ∥θ
In other words, the transformation ϕ → ϕ
−1
−1
−ϕ
is continuous on L(E), E =
E
n
∥ ∥ < δ. (C
n
′
>0
such that (6.6.48)
).
Proof First, since E
′
=E =E
n
(C
n
),θ
is bijective by Theorem 1(iii), so θ
−1
∈ L(E)
.
As before, set ∥θ − ϕ∥ = σ < ε .
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By Note 5 in §2, formula (5) above implies that −1
∥ ∥θ
∥ ∥ ≤
1 .
(6.6.49)
ε−σ
Also, −1
ϕ
−1
∘ (θ − ϕ) ∘ θ
−1
=ϕ
−1
−θ
(6.6.50)
(see Problem 11). Hence by Corollary 4 in §2, recalling that ∥ ∥ϕ
−1
∥ ∥ = 1/ε,
we get
∥θ−1 − ϕ−1 ∥ ≤ ∥ϕ−1 ∥ ⋅ ∥θ − ϕ∥ ⋅ ∥θ−1 ∥ ≤ ∥ ∥ ∥ ∥ ∥ ∥
σ → 0 as σ → 0.
□
(6.6.51)
ε(ε − σ)
This page titled 6.6: Determinants. Jacobians. Bijective Linear Operators is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
6.6.7
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6.6.E: Problems on Bijective Linear Maps and Jacobians Exercise 6.6.E. 1 (i) Can a functional determinant f = det(v ) (see Note 1) be continuous or differentiable even if the functions v are not? (ii) Must a Jacobian map J be continuous or differentiable if f is? Give proofs or counterexamples. ik
ik
f
Exercise 6.6.E. 2 ⇒
Prove rule (b) on determinants. More generally, show that if f (x⃗ ) = x⃗ on an open set A ⊆ E
n
(C
n
),
then J
f
=1
on A .
Exercise 6.6.E. 3 Let f : E → E (or C → C ), f = (f Suppose each f depends on x only, i.e., n
n
n
k
n
1,
… , fn )
.
k
fk (x⃗ ) = fk (y ) if ⃗ xk = yk ,
regardless of the other coordinates x , y . Prove that J [Hint: Show that D f = 0 if i ≠ k .] i
k
i
f
n
=∏
k=1
Dk fk
(6.6.E.1)
.
i
Exercise 6.6.E. 4 In Corollary 1, show that n
⃗ = ∏ Dk fk (p ) ⃗ ⋅ Jg (q ) ⃗ Jh (p )
(6.6.E.2)
k=1
if f also has the property specified in Problem 3. Then do all in "variables," with y
k
= yk (xk )
instead of f . k
Exercise 6.6.E. 5 Let E = E in Note 1. Prove that if all the v are differentiable at p, then f from det (v ) , by replacing the terms of one column by their derivatives. [Hint: Use Problem 6 in Chapter 5, §1.] ′
1
ik
′
(p)
is the sum of
n
determinants, each arising
ik
Exercise 6.6.E. 6 Do Problem 5 for partials of f , with E = E (C ) , and for directionals D f , in any normed space E formulas analogous to Problem 6 in Chapter 5, §1; use Note 3 in §1.) Finally, do it for the differential, df (p ;⃗ ⋅). ′
n
n
u⃗
′
.
(First, prove
Exercise 6.6.E. 7 In Note 1 of §4, express the matrices in terms of partials (see Theorem 4 in §3). Invent a "variable" notation for such matrices, imitating Jacobians (Corollary 3).
Exercise 6.6.E. 8 (i) Show that
6.6.E.1
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∂(x, y, z)
2
= −r
sin α
(6.6.E.3)
y = r sin θ sin α, and
(6.6.E.4)
∂(r, θ, α)
if x = r cos θ,
z = r cos α
(This transformation is passage to polars in E ; see Figure 27, where r = OP , ∠XOA = θ, and ∠AOP (ii) What if x = r cos θ, y = r sin θ, and z = z remains unchanged (passage to cylindric coordinates)? (iii) Same for x = e cos θ, y = e sin θ, and z = z . 3
r
= α. )
r
Exercise 6.6.E. 9 Is f = (f , f ) : E → E one-to-one or bijective, and is J (i) f (x, y) = e cos y and f (x, y) = e sin y ; (ii) f (x, y) = x − y and f (x, y) = 2xy? 2
1
2
2
f
x
if
≠ 0,
x
1
2
2
2
1
2
Exercise 6.6.E. 10 Define f
: E
3
→ E
3
(or C
3
→ C
3
) x⃗
f (x⃗ ) =
3
1 +∑
k=1
(6.6.E.5) xk
on 3
A = { x⃗ | ∑ xk ≠ −1}
(6.6.E.6)
k=1
→
and f = 0 on −A. Prove the following. (i) f is one-to-one on A (find f !). (ii) J (x⃗ ) = . −1
1
f
(1+∑
3 k=1
4
xk )
(iii) Describe −A geometrically.
6.6.E.2
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Exercise 6.6.E. 11 Given any sets A, B and maps f , g : A → E , h : E (i) (f ± g) ∘ k = f ∘ k ± g ∘ k, and (ii) h ∘ (f ± g) = h ∘ f ± h ∘ g if h is linear. Use these distributive laws to verify that ′
−1
ϕ
′
and k : B → A, prove that
→ E,
−1
∘ (θ − ϕ) ∘ θ
−1
=ϕ
−1
−θ
(6.6.E.7)
In Corollary 3. [Hint: First verify the associativity of mapping composition.]
Exercise 6.6.E. 12 Prove that if ϕ : E
′
→ E
is linear and one-to-one, so is ϕ
−1
: E
′′
′
→ E ,
where E
′′
′
= ϕ [E ] .
Exercise 6.6.E. 13 Let v ⃗ , … , v ⃗ be the column vectors in det[ϕ]. Prove that det[ϕ] turns into (i) c ⋅ det[ϕ] if one of the v ⃗ is multiplied by a scalar c ; (ii) − det[ϕ], if any two of the v ⃗ are interchanged (consider λ in formula (1)). Furthermore, show that (iii) det[ϕ] does not change if some v ⃗ is replaced by v ⃗ + cv ⃗ (i ≠ k) ; 1
n
k
k
k
(iv) det[ϕ] = 0 if some v ⃗ is k
→ 0 ,
k
i
or if two of the v ⃗ are the same. k
6.6.E: Problems on Bijective Linear Maps and Jacobians is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.6.E.3
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6.7: Inverse and Implicit Functions. Open and Closed Maps I. "If f ∈ C D at p ,⃗ then f resembles a linear map (namely df ) at p .⃗ " Pursuing this basic idea, we first make precise our notion ⃗ of "f ∈ C D at p ." 1
1
Definition 1 A map f
: E
′
→ E
is continuously differentiable, or of class C D (written f 1
1
∈ C D ),
at p ⃗ iff the following statement is true:
Given any ε > 0, there is δ > 0 such that f is differentiable on the ¯ ¯¯ ¯
¯¯¯¯¯¯¯¯¯¯¯ ¯
globe G = Gp ⃗ (δ), with
(6.7.1) ¯ ¯¯ ¯
∥df (x⃗ ; ⋅) − df (p ;⃗ ⋅)∥ < ε for all x⃗ ∈ G.
By Problem 10 in §5, this definition agrees with Definition 1 §5, but is no longer limited to the case Problems 1 and 2 below.
E
′
=E
n
(C
n
).
See also
We now obtain the following result.
Theorem 6.7.1 Let E and E be complete. If f globe G = G (δ). ′
¯ ¯¯ ¯
: E
′
→ E
is of class C D at p ⃗ and if df (p ;⃗ ⋅) is bijective (§6), then f is one-to-one on some 1
¯¯¯¯¯ ¯
p ⃗
Thus f "locally" resembles df (p ;⃗ ⋅) in this respect. Proof Set ϕ = df (p ;⃗ ⋅) and −1
∥ ∥ϕ
∥ ∥ =
1 (6.7.2) ε
(cf. Theorem 2 of §6). ¯ ¯¯ ¯
¯¯¯¯¯¯¯¯¯¯¯ ¯
By Definition 1, fix δ > 0 so that for x⃗ ∈ G = G
(δ)
p ⃗
.
∥df (x⃗ ; ⋅) − ϕ∥
0) (∃δ
′′
> 0)
∥U − W ∥ < δ
′′
⇒ ∥ ∥U
⃗ = dg(y )
and W
n
the bijectivity of ϕ = df (p ;⃗ ⋅) is equivalent to
(C
n
),
−1
⃗ = dg(y ⃗ + Δy ),
−1
−W
−1
′
∥ ∥ 0} ;
(6.7.E.2)
x
use Theorem 4(iii) in Chapter 3, §16 and continuity to show that D is closed in E However, f is open on all of E by Problem 8. (Verify!)]
2
,
but f [D] = (0, +∞) is not closed in E
1
.
2
Exercise 6.7.E. 10 Continuing Problem 9(ii), define f : E → E (or Show that f is open, but not closed, on E (C ). n
1
n
C
n
→ C)
by
f (x⃗ ) = xk
for a fixed
k ≤n
(the "k th projection map").
n
Exercise 6.7.E. 11 (i) In Example (a), take (p, q) = (5, 0) or (−5, 0). Are the conditions of Theorem 4 satisfied? Do the conclusions hold? (ii) Verify Example (b).
Exercise 6.7.E. 12 (i) Treating z as a function of x and y, given implicitly by f (x, y, z) = z
3
+ xz
2
discuss the choices of P and Q that satisfy Theorem 4. Find (ii) Do the same for f (x, y, z) = e
xyz
−1 = 0
− yz = 0,
∂z ∂x
and
∂z ∂y
f : E
3
→ E
1
,
(6.7.E.3)
.
.
Exercise 6.7.E. 13 Given f : E (C ) → E (C ) , n > m, prove that if f ∈ C D on a globe G, f cannot be one-to-one. [Hint for f : E → E : If, say, D f ≠ 0 on G, set F (x, y) = (f (x, y), y).] n
n
2
m
m
1
1
1
Exercise 6.7.E. 14 Suppose that f satisfies Theorem 1 for every p ⃗ in an open set A ⊆ E , and is one-to-one on A (cf. Problem 4). Let (restrict f to A and take its inverse). Show that f and g are open and of class C D on A and f [A], respectively. ′
g =f
−1
A
1
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Exercise 6.7.E. 15 Given v⃗ ∈ E and a scalar c ≠ 0, define T
v ⃗
: E → E
("translation by v⃗ ") and M
c
("dilation by c "), by setting
: E → E
⃗ Tv ⃗ (x⃗ ) = x⃗ + v and Mc (x⃗ ) = c x⃗ .
(6.7.E.4)
Prove the following. (i) T and T (= T ) are bijective, continuous, and "clopen" on E; so also are M and M (= M ) . (ii) Similarly for the Lipschitz property on E . (iii) If G = G (δ) ⊂ E, then T [G] = G (δ), and M [G] = G (|cδ|). (iv) If f : E → E is linear, and v ⃗ = f (p )⃗ for some p ⃗ ∈ E , then T ∘ f = f ∘ T and M ∘ f = f ∘ M , where T and M −1
v ⃗
−1
−v ⃗
v ⃗
c
q ⃗
v ⃗
⃗ v ⃗ q+
c
′
c
1/c
cq ⃗
′
′
v ⃗
′ c
c
p ⃗
′
′ c
p ⃗
are the corresponding maps on E . If, further, f is continuous at p ,⃗ it is continuous on all of E . ⃗ g = T ∘ f ∘ T [Hint for (iv): Fix any x⃗ ∈ E . Set v ⃗ = f (x⃗ − p ), . Verify that g = f , T (x⃗ ) = p ,⃗ and g is continuous at ′
′
′
′
v ⃗
′
⃗ x⃗ p−
⃗ x⃗ p−
.]
x⃗
Exercise 6.7.E. 16 Show that if f : E → E is linear and if f [G ] is open in E for some G = G (δ) ⊆ E , then (i) f is open on all of E ; (ii) f is onto E . [Hints: (i) By Problem 8, it suffices to show that the set f [G] is open, for any globe G (why?). First take ′
∗
∗
′
p ⃗
′
G = G→ (δ).
use Problems 7 and 15(i)-(iv), with suitable v ⃗ and c. (ii) To prove E = f [E ] , fix any y ⃗ ∈ E. As \(f=G_{\overrightarrow{0} (\delta\) is open, it contains a globe ′
For small c, cy ⃗ ∈ G
′
′
G = G→ (r). 0
Hence y ⃗ ∈ f [E ] (Problem 10 in §2).]
′
Then
0
′
⊆ f [E ] .
Exercise 6.7.E. 17 Continuing Problem 16, show that if f is also one-to-one on G , then ∗
f : E
′
f ∈ L (E , E) , f
−1
⟷ onto E,
(6.7.E.5)
is clopen on E , and f is so on E . is one-to-one on E , let f (x⃗ ) = y ⃗ for some x⃗ , x⃗ ∈ E . Show that ′
′
∈ L (E, E ) , f
[Hints: To prove that f
′
−1
′
′
(∃c, ε > 0)
′
′
∗
⃗ = f (c x⃗ + p ) ⃗ ∈ f [ G ⃗ (δ)] = f [ G ] . c y ⃗ ∈ G→ (ε) ⊆ f [G→ (δ)] and f (c x⃗ + p ) p 0
′
(6.7.E.6)
0
′
Deduce that cx⃗ + p ⃗ = cx⃗ + p ⃗ and x⃗ = x⃗ . Then use Problem 15(v) in Chapter 4, §2, and Note 1.]
Exercise 6.7.E. 18 A map ⟷
′
f : (S, ρ) onto (T , ρ )
is said to be bicontinuous, or a homeomorphism, (from S onto T ) iff both f and f following. (i) x → p in S iff f (x ) → f (p) in T ; (ii) A is closed (open, compact, perfect) in S iff f [A] is so in T ; n
(6.7.E.7)
−1
are continuous. Assuming this, prove the
n
¯ ¯¯ ¯
¯¯¯¯¯¯¯¯ ¯
(iii) B = A in S iff f [B] = f [A] in T ; (iv) B = A in S iff f [B] = (f [A]) in T ; 0
0
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¯ ¯¯ ¯
(v) A is dense in B (i.e., A ⊆ B ⊆ A ⊆ S ) in (S, ρ) iff f [A] is dense in f [B] ⊆ (T , ρ ) . [Hint: Use Theorem 1 of Chapter 4, §2, and Theorem 4 in Chapter 3, §16, for closed sets; see also Note 1.] ′
Exercise 6.7.E. 19 Given A, B ⊆ E, v⃗ ∈ E and a scalar c, set A + v ⃗ = { x⃗ + v|⃗ x⃗ ∈ A} and cA = {c x⃗ | x⃗ ∈ A}.
(6.7.E.8)
Assuming c ≠ 0, prove that (i) A is closed (open, compact, perfect) in E iff cA + v⃗ is; ¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
¯ ¯¯ ¯
(ii) B = A iff cB + v ⃗ = cA + v ⃗ ; (iii) B = A iff cB + v ⃗ = (cA + v)⃗ ; (iv) A is dense in B iff cA + v ⃗ is dense in cB + v ⃗ . [Hint: Apply Problem 18 to the maps T and M of Problem 15, noting that A + v ⃗ = T 0
0
v ⃗
c
[A]
v ⃗
and cA = M
c [A].
]
Exercise 6.7.E. 20 Prove Theorem 2, for a reduced δ, assuming that only one of E and E is E (C ) , and the other is just complete. [Hint: If, say, E = E (C ) , then f [G] is compact (being closed and bounded), and so is G = f [f [G]]. (Why?) Thus the Lemma works out as before, i.e., f [G] ⊇ G (α) . Now use the continuity of f to obtain a globe G = G (δ ) ⊆ G such that f [G ] ⊆ G (α). Let g = f , further restricted to G (α). Apply Problem 15(v) in Chapter 4, §2, to g, with S = G (α), T = E .] ′
n
n
n
n
¯ ¯¯ ¯
¯ ¯¯ ¯
¯ ¯¯ ¯
−1
¯ ¯ ¯ q
′
′
−1
′
p ⃗
q ⃗
G
′
q ⃗
q ⃗
6.7.E: Problems on Inverse and Implicit Functions, Open and Closed Maps is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.7.E.4
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6.8: Baire Categories. More on Linear Maps We pause to outline the theory of so-called sets of Category I or Category II, as introduced by Baire. It is one of the most powerful tools in higher analysis. Below, (S, ρ) is a metric space.
Definition 1 ¯ ¯¯ ¯
¯ ¯¯ ¯ 0
A set A ⊆ (S, ρ) is said to be nowhere dense (in S) iff its closure A has no interior points (i.e., contains no globes): (A) Equivalently, the set A is nowhere dense iff every open set G
∗
≠∅
=∅
.
¯ ¯¯ ¯
in S contains a globe G disjoint from A. (Why?)
Definition 2 A set A ⊆ (S, ρ) is meagre, or of Category I (in S), iff ∞
A = ⋃ An ,
(6.8.1)
n=1
for some sequence of nowhere dense sets A . n
Otherwise, A is said to be nonmeagre or of Category II. A
is residual iff −A is meagre, but A is not.
Examples (a) ∅ is nowhere dense. (b) Any finite set in a normed space E is nowhere dense. (c) The set N of all naturals in E is nowhere dense. 1
(d) So also is Cantor's set P (Problem 17 in Chapter 3, §14); indeed, P is closed (P ¯ ¯¯ ¯ 0
so (P )
=P
0
=∅
¯ ¯¯ ¯
=P)
and has no interior points (verify!),
.
(e) The set R of all rationals in E is meagre; for it is countable (see Chapter 1, §9), hence a countable union of nowhere dense singletons {r }, r ∈ R. But R is not nowhere dense; it is even dense in E , since R = E (see Definition 2, in Chapter 3, §14). Thus a meagre set need not be nowhere dense. (But all nowhere dense sets are meagre why?) 1
1
n
¯ ¯¯ ¯
1
n
Examples (c) and (d) show that a nowhere dense set may be infinite (even uncountable). Yet, sometimes nowhere dense sets are treated as "small" or "negligible," in comparison with other sets. Most important is the following theorem.
Theorem 6.8.1 (Baire) In a complete metric space (S, ρ), every open set G
∗
≠∅
is nonmeagre. Hence the entire space S is residual.
Proof Seeking a contradiction, suppose G is meagre, i.e., ∗
∞ ∗
G
= ⋃ An
(6.8.2)
n=1
for some nowhere dense sets A
n.
Now, as A is nowhere dense, G contains a closed globe ∗
1
¯ ¯¯ ¯
¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
G1 = Gx1 (δ1 ) ⊆ −A1 .
(6.8.3)
Again, as A is nowhere dense, G contains a globe 2
1
¯ ¯¯ ¯
1
¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
G2 = Gx
2
(δ2 ) ⊆ −A2 ,
6.8.1
with 0 < δ2 ≤
2
δ1 .
(6.8.4)
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By induction, we obtain a contracting sequence of closed globes 1
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
¯ ¯¯ ¯
Gn = Gxn (δn ) ,
with 0 < δn ≤
δ1 → 0.
n
2
(6.8.5)
¯ ¯¯ ¯
As S is complete, so are the G (Theorem 5 in Chapter 3, §17). Thus, by Cantor's theorem (Theorem 5 of Chapter 4, §6), there is n
∞ ¯ ¯¯ ¯
p ∈ ⋂ Gn .
(6.8.6)
n=1
¯ ¯¯ ¯
As G
∗
we have p ∈ G
∗
⊇ Gn ,
¯ ¯¯ ¯
But, as G
.
n
we also have (∀n)p ∉ A ; hence
⊆ −An ,
n
∞ ∗
p ∉ ⋃ An = G
(6.8.7)
n=1
(the desired contradiction!).
□
We shall need a lemma based on Problems 15 and 19 in §7. (Review them!)
lemma Let f
′
complete. Let then G ⊆ f [G]. ′
∈ L (E , E) , E
G0 = G0 (r) ⊂ E,
be the unit globe in
G = G0 (1)
′
E .
If
¯¯¯¯¯¯¯¯ ¯
f [G]
(closure of
f [G]
in
E
) contains a globe
0
Note. Recall that we "arrow" only vectors from E (e.g., ′
→
but not those from E (e.g., 0).
0 ),
Proof ¯ ¯¯ ¯
Let A = f [G] ∩ G ⊆ G . We claim that A is dense in G ; i.e., G ⊆ A. Indeed, by assumption, any q ∈ G is in f [G]. Thus by Theorem 3 in Chapter 3, §16, any G meets f [G] ∩ G = A if q ∈ G . Hence 0
0
0
0
q
(∀q ∈ G0 ) ¯ ¯¯ ¯
i.e., G
0
0
0
0
¯ ¯¯ ¯
q ∈ A,
(6.8.8)
as claimed.
⊆ A,
Now fix any q
0
∈ G0 = G0 (r)
and a real c(0 < c < 1). As A is dense in G , 0
A ∩ Gq (cr) ≠ ∅;
(6.8.9)
0
so let q
1
∈ A ∩ Gq (cr) ⊆ f [G]. 0
Then | q1 − q0 | < cr,
As q ∈ f [G], we can fix some p ⃗ ∈ G = G is dense in cG + q = G (cr) . But q ∈ G 1
0 (1),
1
0
1
q1
0
q1
∈ Gq
0
2
(c r) ∩ (cA + q1 ) ,
so q
∈ Gq
0
2
Inductively, we fix for each n > 1 some q
n
with f (p ⃗ ) = q . Thus
1.
(c r) ∩ (cA + q1 ) ≠ ∅;
2
(c r) ,
(6.8.11)
with
(c r) ,
0
cA + q1
etc.
n
∈ Gq
Also, by Problems 19(iv) and 15(iii) in §7,
2
0
2
(6.8.10)
1
1
(cr)
Gq
so let q
q0 ∈ Gq (cr).
n−1
qn ∈ c
A + qn−1 ,
(6.8.12)
i.e., n−1
qn − qn−1 ∈ c
As A ⊆ f [G
0 (1)]
,
A.
(6.8.13)
linearity yields n−1
qn − qn−1 ∈ f [ c
n−1
G0 (1)] = f [ G0 (c
6.8.2
)] ,
n > 1.
(6.8.14)
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Thus for each n > 1, there is p ⃗ 0 1.
(6.8.39)
By the continuity of the norm ||, we can choose r so small that 1
¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
(∀x⃗ ∈ Gx⃗ (r1 ) ) 1
Again by (4), we fix f
2
∈ N
and x⃗
2
∈ Gx⃗ (r1 ) 1
such that |f
2|
|f (x⃗ )| > 1.
>2
(6.8.40)
on some globe
¯ ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
Gx⃗ (r2 ) ⊆ Gx⃗ (r1 ) , 2
with 0 < r
2
< 1/2.
Inductively, we thus form a contracting sequence of closed globes ¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
G
⃗ xn
and a sequence {f
(6.8.41)
1
n}
⊆N,
(rn ) ,
1 0 < rn
n on G
⃗ xn
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ ¯
′
Gx⃗ (rn ) ⊆ E . n
′
(rn ) ⊆ E .
Also,
0 < rn
}.
(6.8.E.4)
k
m=n
6.8.E.2
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By the continuity of k = 1, 2, …
Also, as f
n
′
ρ , fn
and
fm , Akm
is open in
S.
(Why?) So by Problem 12,
∞
⋃
m=1
¯ ¯¯¯¯¯¯¯ ¯
(Akm − Akm )
is meagre for
.
→ f
on S, ⋂
∞
Akm = ∅.
m=1
(Verify!) Thus ∞
(∀k)
m=1
(Why?) Hence the set Q = ⋃ Moreover, S − Q = ⋂ ∪
∞
∞
k=1
∞
∞
k=1
m=1
⋂
m=1
∞ ¯ ¯¯¯¯¯¯¯ ¯
¯ ¯¯¯¯¯¯¯ ¯
⋂ Akm ⊆ ⋃ (Akm − Akm ) .
(6.8.E.5)
m=1
¯ ¯¯¯¯¯¯¯ ¯
is meagre in S. by Problem 16 in Chapter 3, §16. Deduce that if p ∈ S − Q, then
Akm
(−Akm )
0
(∀ε > 0) (∃m0 ) (∃Gp ) (∀n, m ≥ m0 ) (∀x ∈ G0 )
′
ρ (fm (x), fn (x)) < ε.
(6.8.E.6)
Keeping m fixed, let n → ∞ to get (∀ε > 0) (∃m0 ) (∃Gp ) (∀m ≥ m0 ) (∀x ∈ Gp )
′
ρ (fm (x), f (x)) ≤ ε.
(6.8.E.7)
Now modify the proof of Theorem 2 of Chapter 4, §12, to show that this implies the continuity of f at each p ∈ S − Q .] 6.8.E: Problems on Baire Categories and Linear Maps is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.8.E.3
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6.9: Local Extrema. Maxima and Minima We say that f : E → E has a local maximum (minimum) at about p ;⃗ more precisely, iff ′
1
(∀x⃗ ∈ G)
p ⃗ ∈ E
′
iff
is the largest (least) value of
⃗ f (p )
f
on some globe
⃗ < 0(> 0). Δf = f (x⃗ ) − f (p )
We speak of an improper extremum if we only have Δf
G
(6.9.1)
on G. In any case, all depends on the sign of Δf .
≤ 0(≥ 0)
From Problem 6 in §1, recall the following necessary condition.
Theorem 6.9.1 If f
: E
′
→ E
In the case E
1
=E
1
n
(C
n
),
this means that d
⃗ = ∑n f (p ;⃗ t )
k=1
⃗ tk . Dk f (p )
1
f (p ;⃗ ⋅) = 0
→
for all u⃗ ≠
⃗ = 0 f (p )
u⃗
′
(Recall that d
has a local extremum at p ⃗ then D
in E
0
′
.
on E . ′
It vanishes if the D
⃗
k f (p )
do.
Note 1. This condition is only necessary, not sufficient. For example, if f (x, y) = xy, then d at
→ 0 .
1
→ f ( 0 ; ⋅) = 0;
yet f has no extremum
(Verify!)
Sufficient conditions were given in Theorem 2 of §5, for E
′
=E
1
We now take up E
.
′
=E
2
.
Theorem 6.9.2 Let f : E → E be of class C D on a globe G = G and C = D f (p )⃗ . 2
1
2
Suppose d
(δ).
p ⃗
1
f (p ;⃗ ⋅) = 0
on E
2
.
Set A = D
⃗
11 f (p ),
⃗ B = D12 f (p ),
22
Then the following statements are true. (i) If AC A B ,
has a maximum or minimum at p ,⃗ according to whether
or A > 0.
(ii) If AC
2
B ,
2
2
2
+ (AC − B ) u , 2
2
2
2
+ (AC − B ) u . 2
the right-side expression in (3) is > 0; so AD > 0 , i.e., D has the same sign as A.
Hence if A < 0, we also have Δf p .⃗
0, then Δf
Now let AC < B . We claim that no matter how small extremum at p .⃗ 2
Indeed, we have x⃗ = p ⃗ + u⃗ , u⃗ = (u
1,
→
If u
u2 ) ≠ 0 .
2
G = G ⃗ (δ), Δf p
changes sign as
and f has a minimum at
> 0,
x⃗
varies in
G,
and so
f
has no
(3) shows that D and Δf have the same sign as A(A ≠ 0).
= 0,
But if u ≠ 0 and u = −Bu /A (assuming A ≠ 0), then D and Δf have the sign opposite to that of A; and x⃗ is still in G if u is small enough (how small?). 2
1
2
2
One proceeds similarly if C
≠0
(interchange A and C , and use (3').
Finally, if A = C = 0, then by (2), D = 2Bu u and B ≠ 0 (since does; so f has no extremum at p .⃗ Thus all is proved. □ 1
2
2
AC < B )
. Again
D
and
Briefly, the proof utilizes the fact that the trinomial (2) is sign-changing iff its discriminant ∣A ∣ ∣B
B∣ ∣ 0
on all of E
n
→ −{ 0 }
iff the following n determinants A are positive: k
∣ a11
a12
…
a1k ∣
a22
…
∣ a2k ∣
………
…
a2k
ak2
…
akk ∣
∣ Ak =
∣ a21 ∣ ∣
…
∣ ak1
(ii) We have P
0 f : E
n
∣
,
k = 1, 2, … , n.
(6.9.7)
∣
for k = 1, 2, … , n.
→ E
1
. (This will also cover
6.9.2
f : C
n
→ E
1
,
treated as
f : E
2n
→ E
1
.)
The
https://math.libretexts.org/@go/page/21630
Theorem 6.9.3 Let
f : E
aij = Dij
be of class C D on some G = G (δ). Suppose ⃗ i, j, k ≤ n Then the following statements hold. f (p ), n
→ E
1
2
df (p ;⃗ ⋅) = 0
p ⃗
(i) f has a local minimum at p ⃗ if A
k
>0
(ii) f has a local maximum at p ⃗ if (−1)
k
on
E
n
.
Define the
as in (4), with
Ak
for k = 1, 2, … , n. Ak > 0
for k = 1, … , n .
(iii) f has no extremum at p ⃗ if the expression n
n
P (u⃗ ) = ∑ ∑ aij ui uj j=1
(6.9.8)
i=1
is > 0 for some u⃗ ∈ E and < 0 for others (i.e., P changes sign on E n
n
)
.
Proof Let again x⃗ ∈ G, u⃗ = x⃗ − p ⃗ ≠
→
and use Taylor's theorem to obtain
0 ,
1
⃗ = R1 = Δf = f (x⃗ ) − f (p )
2
n
n
2
⃗ ui uj , d f (s ;⃗ u⃗ ) = ∑ ∑ Dij f (s ) j=1
(6.9.9)
i=1
with s ⃗ ∈ L(x⃗ , p )⃗ . As f ∈ C D , the partials D f are continuous on G. Thus we can make G so small that the sign of the last double sum does not change if s ⃗ is replaced by p .⃗ Hence the sign of Δf on G is the same as that of P (u⃗ ) = ∑ ∑ a u u , with the a as stated in the theorem. 2
ij
n
n
j=1
i=1
ij
i
j
ij
The quadratic form P is symmetric since a = a by Theorem 1 in §5. Thus by Sylvester's theorem stated above, one easily obtains our assertions (i) and (ii). Indeed, they are immediate from clauses (i) and (ii) of that theorem. ij
ji
⃗ Now, for (iii), suppose P (u⃗ ) > 0 > P (v), i.e., n
n
n
n
for some u⃗ , v ⃗ ∈ E
∑ ∑ aij ui uj > 0 > ∑ ∑ aij vi vj j=1
i=1
j=1
n
→ − { 0 }.
(6.9.10)
i=1
⃗ ≠ 0), then u If here u⃗ and v ⃗ are replaced by tu⃗ and tv(t
i uj
and v
i vj
2
turn into t
2
ui uj
and t
2
vi vj ,
respectively. Hence
2
⃗ = P (tv). ⃗ P (tu⃗ ) = t P (u⃗ ) > 0 > t P (v)
Now, for any
t ∈ (0, δ/| u⃗ |),
the point
x⃗ = p ⃗ + tu⃗
lies on the
-directed line through
u⃗
(6.9.11) p ,⃗
inside
G = G ⃗ (δ). p
(Why?)
′
Similarly for the point x⃗ = p ⃗ + tv.⃗ Hence for such ′
′
s ⃗ ∈ L (p ,⃗ x⃗ )
x⃗
and
′
Taylor's theorem again yields formulas analogous to (5) for some
x⃗ ,
s ⃗ ∈ L(p ,⃗ x⃗ )
and
lying on the same two lines. It again follows that for small δ , ′
⃗ > 0 > f (x⃗ ) − f (p ), ⃗ f (x⃗ ) − f (p )
(6.9.12)
just as P (u⃗ ) > 0 > P (v)⃗ . Thus Δf changes sign on G
(δ),
p ⃗
and (iii) is proved.
□
→
Note 3. Still unresolved are cases in which P (u⃗ ) vanishes for some u⃗ ≠ 0 , without changing its sign; e.g., P (u⃗ ) = (u + u + u ) = 0 for u⃗ = (1, 1, −2) . Then the answer depends on higher-order terms of the Taylor formula. In ⃗ particular, if d f (p ;⃗ ⋅) = d f (p ;⃗ ⋅) = 0 on E , then Δf = R = d f (p ;⃗ s ), etc. 2
1
2
1
3
2
1
n
2
3
6
Note 4. The largest or least value of f on a set A (sometimes called the absolute maximum or minimum) may occur at sominterior (e.g., boundary) point p ⃗ ∈ A, and then fails to be among the local extrema (where, by definition, a globe G ⊆ A is presupposed). Thus to find absolute extrema, one must also explore the behaviour of f at noninterior points of A. p ⃗
6.9.3
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By Theorem 1, local extrema can occur only at so-called critical points p ,⃗ i.e., those at which all directional derivatives vanish (or fail to exist, in which case D f (p )⃗ = 0 by convention). u⃗
In practice, to find such points in E (C ) , one equates the partials D considerations to determine whether an extremum really exists. n
n
kf
to 0. Then one uses Theorems 2 and 3 or other
(k ≤ n)
Examples (A) Find the largest value of f (x, y) = sin x + sin y − sin(x + y)
(6.9.13)
on the set A ⊆ E bounded by the lines x = 0, y = 0 and x + y = 2π . 2
We have D1 f (x, y) = cos x − cos(x + y) and D2 f (x, y) = cos y − cos(x + y).
(6.9.14)
– √3.
Inside the triangle A, both partials vanish only at the point ( , ) at which f = On the boundary of A (i.e., on the lines x = 0, y = 0 and x + y = 2π), f = 0. Thus even without using Theorem 2, it is evident that f attains its largest value, 2π
2π
2π
3
3
3
2
2π
f (
, 3
3 ) =
3
– √3,
(6.9.15)
2
at this unique critical point. (B) Find the largest and the least value of 2
2
2
f (x, y, z) = a x
on the condition that x
2
As z
2
2
= 1 −x
+y
2
+z
2
+b y
2
2
− (ax
+ by
−y ,
2
2
2
+ cz )
,
(6.9.16)
and a > b > c > 0 .
=1
2
2
2
2
−c )x
+ (b
2
−c )y
¯ ¯¯ ¯
¯ ¯¯ ¯
(Explain!) For F , we seek the extrema on the disc G = G x + y + z = 1) . 2
2
we can eliminate z from f (x, y, z) and replace f by F
2
F (x, y) = (a
2
2
+c z
2
0 (1)
2
+c
⊂E
: E
2
2
− [(a − c)x
2
,
where x
2
→ E
1
:
+ (b − c)y
+y
2
≤1
2
2
+ c]
.
(6.9.17)
(so as not to violate the condition
2
Equating to 0 the two partials 2
D1 F (x, y) = 2x(a − c) {(a + c) − 2 [(a − c)x
+ (b − c)y
2
2
+ c]
} = 0, (6.9.18)
2
D2 F (x, y) = 2y(b − c) {(b + c) − 2 [(a − c)x
+ (b − c)y
2
+ c]
2
} =0
and solving this system of equations, we find these critical points inside G : (1) x = y = 0 (F
= 0)
(2) x = 0, y = ±2
−
(3) x = ±2
−
;
1 2
(F =
1 2
, y = 0 (F =
1 4 1 4
and
2
(b − c ) ) ; 2
(a − c ) )
.
(Verify!) ¯ ¯¯ ¯
Now, for the boundary of G, i.e., the circle F (x, y), thus reducing it to 2
h(x) = (a
on the interval [−1, 1] ⊂ E
1
.
2
x
+y
2
2
−b )x
2
= 1,
2
+b
repeat this process: substitute
2
+ [(a − b)x
2
+ b]
,
h : E
1
y
→ E
2
1
,
2
= 1 −x
in the formula for
(6.9.19)
In (−1, 1) the derivative ′
2
h (x) = 2(a − b)x (1 − 2 x )
(6.9.20)
vanishes only when
6.9.4
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(4) x = 0 (h = 0), and (5) x = ±2
−
1 2
(h =
1 4
2
(a − b ) )
.
Finally, at the endpoints of [−1, 1], we have (6) x = ±1 (h = 0) . Comparing the resulting function values in all six cases, we conclude that the least of them is 0, while the largest is (a − c) . These are the desired least and largest values of f , subject to the conditions stated. They are attained, respectively, at the points 1
2
4
(0, 0, ±1), (0, ±1, 0), (±1, 0, 0), and (±2
−
1 2
, 0, ±2
−
1 2
).
(6.9.21)
Again, the use of Theorems 2 and 3 was redundant. However, we suggest as an exercise that the reader test the critical points of F by using Theorem 2. Caution. Theorems 1 to 3 apply to functions of independent variables only. In Example (B), the imposed equation 2
x
(which geometrically limits all to the surface of Theorems 1 to 3 be used.
G→ (1)
+y
in
2
E
+z 3
),
2
=1
so that one of them,
x, y, z
were made interdependent by
(6.9.22)
z,
could be eliminated. Only then can
0
This page titled 6.9: Local Extrema. Maxima and Minima is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
6.9.5
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6.9.E: Problems on Maxima and Minima Exercise 6.9.E. 1 Verify Note 1.
Exercise 6.9.E. 1
′
Complete the missing details in the proof of Theorems 2 and 3.
Exercise 6.9.E. 2 Verify Examples (A) and (B). Supplement Example (A) by applying Theorem 2.
Exercise 6.9.E. 3 Test f for extrema in E if f (x, y) is 2
(i)
y
2
x
+
2p x
−
2p 2 2
(p > 0, q > 0)
2q y
2
(ii) (iii) y (iv) y
2
2
2q
(p > 0, q > 0)
4
+x
3
+x
; ;
; .
Exercise 6.9.E. 4 (i) Find the maximum volume of an interval A ⊂ E (see Chapter 3, §7) whose edge lengths x +y +z = a . (ii) Do the same in E and in E ; show that A is a cube. (iii) Hence deduce that 3
4
x, y, z
have a prescribed sum:
n
− − − − − − − − n − √ x1 x2 ⋯ xn ≤
1
n
∑ xk n
(xk ≥ 0) ,
(6.9.E.1)
1
i.e., the geometric mean of n nonnegative numbers is ≤ their arithmetic mean.
Exercise 6.9.E. 5 Find the minimum value for the sum f (x, y, z, t) = x + y + z + t of four positive numbers on the condition that (constant). [Answer: x = y = z = t = c; f = 4c .]
4
xyzt = c
max
Exercise 6.9.E. 6 Among all triangles inscribed in a circle of radius R, find the one of maximum area. [Hint: Connect the vertices with the center. Let x, y, z be the angles at the center. Show that the area of the triangle = R (sin x + sin y + sin z), with z = 2π − (x + y) .] 1
2
2
Exercise 6.9.E. 7 Among all intervals A ⊂ E inscribed in the ellipsoid 3
6.9.E.1
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2
x
2
y +
2a
2b
,
√3
,
√3
2c √3
2
z +
2
a
find the one of largest volume. [Answer: the edge lengths are
2
b
=1
2
(6.9.E.2)
c
.]
Exercise 6.9.E. 8 Let P = (a ⋅ b ) , i = 1, 2, 3, be 3 points in E forming a triangle in which one angle (say, Find a point P = (x, y) for which the sum of the distances, 2
i
i
i
3
easureangleP1 )
is ≥ 2π/3.
− −−−−−−−−−−−−−− −
P P1 + P P2 + P P3 = ∑ √ (x − ai )
2
+ (y − bi )
2
,
(6.9.E.3)
i=1
is the least possible.
− −−−−−−−−−−−−−− − 3
[Outline: Let f (x, y) = ∑
√(x − a ) i
i=1
2
+ (y − bi )
2
.
Show that f has no partial derivatives at P , P , or P (and so P , P , and P are critical points at which an extremum may occur), while at other points P , partials do exist but never vanish simultaneously, so that there are no other critical points. Indeed, prove that D f (P ) = 0 = D f (P ) would imply that 1
1
2
3
1
2
3
2
3
3
∑ cos θi = 0 = ∑ sin θI , i=1
(6.9.E.4)
1
¯ ¯¯¯¯¯¯¯ ¯
where θ is the angle between P P and the x-axis; hence i
i
sin(θ1 − θ2 ) = sin(θ2 − θ3 ) = sin(θ3 − θ1 )
(why?),
(6.9.E.5)
and so θ − θ = θ − θ = θ − θ = 2π/3, contrary to ∠ P ≥ 2π/3. (Why?) From geometric considerations, conclude that f has an absolute minimum at P . (This shows that one cannot disregard points at which f has no partials.)] 1
2
2
3
3
1
1
1
Exercise 6.9.E. 9 Continuing Problem 8, show that if none of ∠ P , ∠ P , and ∠P is ≥ 2π/3, then f attains its least value at some P (inside the triangle) such that ∠ P P P = ∠ P P P = ∠ P P P = 2π/3 . [Hint: Verify that D f = 0 = D f at P . Use the law of cosines to show that P P > P P + P P and P P > P P + P P . Adding, obtain P P + P P > P P + P P + P P , i.e., f (P ) > f (P ). Similarly, f (P ) > f (P ) and f (P ) > f (P ). Combining with Problem 8, obtain the result.] 1
1
2
2
1
3
2
3
3
1
2
1
1
1
3
1
2
1
2
2
2
1
1
2
1
3
3
1
3
2
1
2
3
Exercise 6.9.E. 10 In a circle of radius R inscribe a polygon with n + 1 sides of maximum area. [Outline: Let x , x , … , x be the central angles subtended by the sides of the polygon. Then its area A is 1
2
n+1
1
n+1 2
R 2
with x
n+1
n
= 2π − ∑
k=1
xk .
∑ sin xk ,
(6.9.E.6)
k=1
(Why?) Thus all reduces to maximizing
6.9.E.2
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rk
n
f (x1 , … , xn ) = ∑ sin xk + sin(2π − ∑ xk ), k=1
(6.9.E.7)
k=1
on the condition that 0 ≤ x and ∑ x ≤ 2π. (Why?) These inequalities define a bounded set D ⊂ E (called a simplex). Equating all partials of f to 0, show that the only critical point interior to D is x⃗ = (x , … , x ) , with x = , k ≤ n (implying that x = , too). For that x⃗ , we get n
k
k=1
k
n
2π
1
n
k
2π
n+1
n+1
n+1
f (x⃗ ) = (n + 1) sin[2π/(n + 1)].
(6.9.E.8)
This value must be compared with the "boundary" values of f , on the "faces" of the simplex D (see Note 4). Do this by induction. For n = 2, Problem 6 shows that f (x⃗ ) is indeed the largest when all x equal . Now let D be the "face" of D, where x = 0. On that face, treat f as a function of only n − 1 variables, x , … , x . By the inductive hypothesis, the largest value of f on D is n sin(2π/n). Similarly for the other "faces." As n sin(2π/n) < (n + 1) sin 2π/(n + 1), the induction is complete. Thus, the area A is the largest when the polygon is regular, for which 2π
k
n
1
n+1
n
n−1
n
1 A =
2π
2
R (n + 1) sin 2
.]
(6.9.E.9)
n+1
Exercise 6.9.E. 11 Among all triangles of a prescribed perimeter 2p, find the one of maximum area. [Hint: Maximize p(p − x)(p − y)(p − z) on the condition that x + y + z = 2p .]
Exercise 6.9.E. 12 Among all triangles of area A, find the one of smallest perimeter.
Exercise 6.9.E. 13 Find the shortest distance from a given point p ⃗ ∈ E
n
to a given plane u⃗ ⋅ x⃗ = c (Chapter 3, §§4-6). Answer: u⃗ ⋅ p ⃗ − c ±
.
(6.9.E.10)
| u⃗ |
[Hint: First do it in E
3
,
writing (x, y, z) for x⃗ .]
6.9.E: Problems on Maxima and Minima is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.9.E.3
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6.10: More on Implicit Differentiation. Conditional Extrema I. Implicit differentiation was sketched in §7. Under suitable assumptions (Theorem 4 in §7), one can differentiate a given system of equations, gk (x1 , … , xn , y1 , … , ym ) = 0,
k = 1, 2, … , n,
(6.10.1)
treating the x as implicit functions of the y without seeking an explicit solution of the form j
i
xj = Hj (y1 , … , ym ) .
This yields a new system of equations from which the partials D
i Hj
=
∂xj ∂yi
(6.10.2)
can be found directly.
We now supplement Theorem 4 in §7 (review it!) by showing that this new system is linear in the partials involved and that its determinant is ≠ 0. Thus in general, it is simpler to solve than (1). As in Part IV of §7, we set ⃗ = (x1 , … , xn , y1 , … , ym ) and g = (g1 , … , gn ) , (x⃗ , y )
(6.10.3)
replacing the f of §7 by g. Then equations (1) simplify to → ⃗ = 0 , g(x⃗ , y )
where g : E
n+m
→ E
n
(or g : C
n+m
→ C
n
)
(6.10.4)
.
Theorem 6.10.1 (implicit differentiation) Adopt all assumptions of Theorem 4 in §7, replacing f by g and setting H = (H
1,
⃗ = ajk , Dj gk (p ,⃗ q )
j ≤ n + m,
… , Hn )
,
k ≤ n.
(6.10.5)
Then for each i = 1, … , m, we have n linear equations, n
⃗ = −an+i,k , ∑ ajk Di Hj (q )
k ≤ n,
(6.10.6)
j=1
with det(ajk ) ≠ 0,
that uniquely determine the partials D
i Hj (q
) ⃗
(j, k ≤ n),
(6.10.7)
for j = 1, 2, … , n .
Proof As usual, extend the map H : Q → P of Theorem 4 in §7 to H : E Also, define σ : E
m
→ E
n+m
(C
m
→ C
n+m
)
m
→ E
n
(or C
m
→ C
n
)
by setting H =
→ 0
on −Q .
by
⃗ = (H (y ), ⃗ y ) ⃗ = (H1 (y ), ⃗ … , Hn (y ), ⃗ y1 , … , ym ) , σ(y )
y ⃗ ∈ E
m
(C
m
).
(6.10.8)
Then σ is differentiable at q ⃗ ∈ Q, as are its n + m components. (Why?) since x⃗ = H (y )⃗ is a solution of (2), equations (1) ⃗ Also, σ(q ) ⃗ = (H (q ), ⃗ q ) ⃗ = (p ,⃗ q ) ⃗ and (2) become identities when x⃗ is replaced by H (y ). since H (q )⃗ = p ⃗ . Moreover, → ⃗ = g(H (y ), ⃗ y ) ⃗ = 0 for y ⃗ ∈ Q; g(σ(y ))
i.e., g ∘ σ =
→ 0
(6.10.9)
on Q .
⃗ so the chain rule (Theorem 2 in §4) applies, with f , p ,⃗ q ,⃗ n, and m replaced by Now, by assumption, g ∈ C D at (p ,⃗ q ); ⃗ m, and n + m, respectively. σ, q ,⃗ (p ,⃗ q ), 1
→
As h = g ∘ σ = 0 component of σ,
on
Q,
an open set, the partials of
h
vanish on
6.10.1
Q.
So by Theorem 2 of §4, writing
σj
for the j th
https://math.libretexts.org/@go/page/21631
n+m
→
⃗ ⋅ Di σj (q ), ⃗ 0 = ∑ Dj g(p ,⃗ q )
i ≤ m.
(6.10.10)
j=1
By (4), σ
j
j = n + i,
if j ≤ n, and σ (y )⃗ = y if j = n + i. Thus D and D σ = 0 otherwise. Hence by (5),
= Hj
j
i
i σj
i
= Di Hj
j ≤ n;
but for j > n, we have
Di σj = 1
if
j
n
→
⃗ ⋅ D H (q ) ⃗ + D ⃗ ⃗ 0 = ∑ Dj g(p ,⃗ q ) i j n+i g(p , q ),
i = 1, 2, … , m.
(6.10.11)
j=1
As g = (g
1,
… , gn ) ,
each of these vector equations splits into n scalar ones: n
⃗ ⋅ Di Hj (q ) ⃗ + Dn+i gk (p ,⃗ q ), ⃗ 0 = ∑ Dj gk (p ,⃗ q )
i ≤ m, k ≤ n.
(6.10.12)
j=1
With D
this yields (3), where det(a
⃗ q ) ⃗ = ajk ,
j gk (p ,
Thus all is proved.
jk )
⃗ ≠ 0 = det(Dj gk (p ,⃗ q ))
by hypothesis (see Theorem 4 in §7).
□
⃗ ≠ 0 for all (x⃗ , y ) ⃗ in a sufficiently small neighborhood of (p ,⃗ q ). ⃗ Note 1. By continuity (Note 1 in §6), we have det D g (x⃗ , y )) Thus Theorem 1 holds also with (p ,⃗ q )⃗ replaced by such (x⃗ , y )⃗ . In practice, one does not have to memorize (3), but one obtains it by implicitly differentiating equations (1). j
k
II. We shall now apply Theorem 1 to the theory of conditional extrema.
Definition 1 We say that f
: E
n+m
→ E
1
has a local conditional maximum (minimum) at p ⃗ ∈ E
n+m
,
with constraints
→ g = (g1 , … , gn ) = 0 (g : E
n+m
→ E
n
(6.10.13)
iff in some neighborhood G of p ⃗ we have
)
⃗ ≤ 0 Δf = f (x⃗ ) − f (p )
for all x⃗ ∈ G for which g(x⃗ ) =
→ 0
(≥ 0, respectively)
(6.10.14)
. →
In §9 (Example (B) and Problems), we found such conditional extrema by using the constraint equations g = 0 to eliminate some variables and thus reduce all to finding the unconditional extrema of a function of fewer (independent) variables. Often, however, such elimination is cumbersome since it involves solving a system (1) of possibly nonlinear equations. It is here that implicit differentiation (based on Theorem 1) is useful. Lagrange invented a method (known as that of multipliers) for finding the critical points at which such extrema may exist; to wit, we have the following: Given f
: E
n+m
→ E
1
,
set n
F = f + ∑ ck gk ,
(6.10.15)
k=1
where the constants c are to be determined and g are as above. k
k
Then find the partials D
jF
(j ≤ n + m)
and solve the system of 2n + m equations
Dj F (x⃗ ) = 0,
for the 2n + m "unknowns" x
j (j
≤ n + m)
j ≤ n + m,
and c
k (k
≤ n),
and
gk (x⃗ ) = 0,
k ≤ n,
(6.10.16)
the c originating from (7). k
Any x⃗ satisfying (8), with the c so determined is a critical point (still to be tested). The method is based on Theorem 2 below, where we again write (p ,⃗ q )⃗ for p ⃗ and (x⃗ , y )⃗ for x⃗ (we call it "double notation"). k
6.10.2
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Theorem 6.10.2 (Lagrange multipliers) Suppose f
: E
n+m
→ E
1
is differentiable at ⃗ = (p1 , … , pn , q1 , … , qm ) (p ,⃗ q )
(6.10.17)
and has a local extremum at (p ,⃗ q )⃗ subject to the constraints → g = (g1 , … , gn ) = 0 ,
with g as in Theorem 1, g : E
n+m
→ E
n
(6.10.18)
Then
.
n
⃗ = −Dj f (p ,⃗ q ), ⃗ ∑ ck Dj gk (p ,⃗ q )
j = 1, 2, … , n + m,
(6.10.19)
k=1
for certain multipliers c (determined by the first n equations in (9)). k
Proof These n equations admit a unique solution for the c
k,
as they are linear, and
⃗ ≠ 0 det(Dj gk (p ,⃗ q ))
(j, k ≤ n)
(6.10.20)
by hypothesis. With the c so determined, (9) holds for j ≤ n. It remains to prove (9) for n < j ≤ n + m. k
Now, since f has a conditional extremum at (p ,⃗ q )⃗ as stated, we have ⃗ − f (p ,⃗ q ) ⃗ ≤ 0 f (x⃗ , y )
for all (x⃗ , y )⃗ ∈ P
×Q
with g(x⃗ , y )⃗ =
→ 0 ,
( or
≥ 0)
(6.10.21)
provided we make the neighborhood P
×Q
small enough.
Define H and σ as in the previous proof (see (4)); so x⃗ = H (y )⃗ is equivalent to g(x⃗ , y )⃗ = ⃗ with x⃗ = H (y ), ⃗ Then, for all such (x⃗ , y ), we surely have g(x⃗ , y )⃗ =
→ 0
→ 0
for (x⃗ , y )⃗ ∈ P
×Q
and also
⃗ = f (H (y ), ⃗ y ) ⃗ = f (σ(y )). ⃗ f (x⃗ , y )
Set h = f ∘ σ, h : E
m
→ E
1
.
.
(6.10.22)
Then (10) reduces to ⃗ − h(q ) ⃗ ≤ 0( or ≥ 0) h(y )
for all y ⃗ ∈ Q.
(6.10.23)
This means that h has an unconditional extremum at q ,⃗ an interior point of Q. Thus, by Theorem 1 in §9, ⃗ = 0, Di h(q )
i = 1, … , m.
(6.10.24)
Hence, applying the chain rule (Theorem 2 of §4) to h = f ∘ σ, we get, much as in the previous proof, n+m
⃗ Di σj (q ) ⃗ 0 = ∑ Dj f (p ,⃗ q ) j=1 n
⃗ Di Hj (q ) ⃗ + Dn+i f (p ,⃗ q ), ⃗ = ∑ Dj f (p ,⃗ q )
i ≤ m.
j=1
(Verify!) Next, as g by hypothesis satisfies Theorem 1, we get equations (3) or equivalently (6). Multiplying (6) by combining with (11), we obtain n
∑
n
j=1
⃗ + ∑ [ Dj f (p ,⃗ q )
k=1
ck ,
adding and
⃗ Di Hj (q ) ⃗ ck Dj gk (p ,⃗ q )] (6.10.25)
n
⃗ + ∑ ⃗ = 0, +Dn+i f (p ,⃗ q ) ck Dn+i gk (p ,⃗ q ) k=1
i ≤ m.
(Verify!) But the square-bracketed expression is 0; for we chose the c so as to satisfy (9) for j ≤ n. Thus all simplifies to k
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n
⃗ = −D ⃗ ⃗ ∑ ck Dn+i gk (p ,⃗ q ) n+i f (p , q ),
i = 1, 2, … , m.
(6.10.26)
k=1
Hence (9) holds for n < j ≤ n + m, too, and all is proved.
□
Remarks. Lagrange's method has the advantage that all variables (the x and dependent ones. Thus in applications, one uses only F , i.e., f and g (not H ). k
One can also write only).
for
x⃗ = (x1 , … , xn+m )
⃗ = (x1 , … , xn , y1 , … , ym ) (x⃗ , y )
yi
) are treated equally, without singling out the
(the "double" notation was good for the proof
On the other hand, one still must solve equations (8). Theorem 2 yields only a necessary condition (9) for extrema with constraints. There also are various sufficient conditions, but mostly one uses geometric and other considerations instead (as we did in §9). Therefore, we limit ourselves to one proposition (using "single" notation this time).
Theorem 6.10.3 (sufficient conditions) Let n
F = f + ∑ ck gk ,
(6.10.27)
k=1
with f
: E
n+m
→ E
1
,g : E
n+m
→ E
n
,
and c as in Theorem 2. k
Then f has a maximum (minimum) at p ⃗ = (p , … , p ) (with constraints ⃗ fortiori, this is the case if F has an unconditional extremum at p .) 1
n+m
→ g = (g1 , … , gn ) = 0
whenever
F
does. (A
Proof Suppose F has a maximum at p ,⃗ with constraints g =
→ 0 .
Then n
⃗ = f (x⃗ ) − f (p ) ⃗ + ∑ ck [ gk (x⃗ ) − gk (p )] ⃗ 0 ≥ F (x⃗ ) − F (p )
(6.10.28)
k=1
for those x⃗ near p ⃗ (including x⃗ = p )⃗ for which g(x⃗ ) = But for such x⃗ , g
→ 0
.
⃗ = g (p ) ⃗ = 0, c [ g (x⃗ ) − g (p )] ⃗ = 0, k k k k
k (x)
and so
⃗ = f (x⃗ ) − f (p ). ⃗ 0 ≥ F (x⃗ ) − F (p )
(6.10.29)
Hence f has a maximum at p ,⃗ with constraints as stated. Similarly, ΔF
= Δf
in case F has a conditional minimum at p .⃗
□
Example 1 Find the local extrema of f (x, y, z, t) = x + y + z + t
(6.10.30)
on the condition that 4
g(x, y, z, t) = xyzt − a
= 0,
(6.10.31)
with a > 0 and x, y, z, t > 0. (Note that inequalities do not count as "constraints" in the sense of Theorems 2 and 3.) Here one can simply eliminate t = a /(xyz), but it is still easier to use Lagrange's method. 4
Set F (x, y, z, t) = x + y + z + t + cxyzt. (We drop a since it will anyway disappear in differentiation.) Equations (8) then read 4
6.10.4
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4
0 = 1 + cyzt = 1 + cxzt = 1 + cxyt = 1 + cxyz,
Solving for x, z, t and c, we get c = −a
−3
3
,
= 0.
(6.10.32)
.
,x =y =z =t =a
Thus F (x, y, z, t) = x + y + z + t − xyzt/a
xyzt − a
and the only critical point is p ⃗ = (a, a, a, a). (Verify!)
⃗ By Theorem 3, one can now explore the sign of F (x⃗ ) − F (p ), where x⃗ = (x, y, z, t). For x⃗ near p ,⃗ it agrees with the sign of d F (p ;⃗ ⋅). (See proof of Theorem 2 in §9.) We shall do it below, using yet another device, to be explained now. 2
Elimination of dependent differentials. If all partials of (briefly dF ≡ 0 ). Conversely, if d
1
f (p ;⃗ ⋅) = 0
on a globe G
,
p ⃗
vanish at
F
p ⃗
(e.g., if
p ⃗
satisfies (9), then
1
d F (p ;⃗ ⋅) = 0
on
E
n+m
for some function f on n independent variables, then ⃗ = 0, Dk f (p )
k = 1, 2, … , n,
since d f (p ;⃗ ⋅) (a polynomial!) vanishes at infinitely many points if its coefficients the variables are interdependent.) 1
(6.10.33) Dk f (p ) ⃗
vanish. (The latter fails, however, if
Thus, instead of working with the partials, one can equate to 0 the differential dF or df . Using the "variable" notation and the invariance of df (Note 4 in §4), one then writes dx, dy, … for the "differentials" of dependent and independent variables alike, and tries to eliminate the differentials of the dependent variables. We now redo Example 1 using this method.
Example 2 With f and g as in Example 1, we treat t as the dependent variable, i.e., an implicit function of x, y, z, 4
t = a /(xyz) = H (x, y, z),
and differentiate the identity xyzt − a
4
(6.10.34)
to obtain
=0
0 = yztdx + xztdy + xytdz + xyzdt;
(6.10.35)
so dx dt = −t (
dy +
Substituting this value of dt in df
t
).
y
t
t
) dx + (1 − x
(6.10.36)
z
(the equation for critical points), we eliminate dt and find:
= dx + dy + dz + dt = 0
(1 −
dz +
x
) dy + (1 − y
) dz ≡ 0.
As x, y, z are independent variables, this identity implies that the coefficients of above. Thus t
t
1−
=1− x
(6.10.37)
z dx, dy,
and
dz
must vanish, as pointed out
t =1−
y
= 0.
(6.10.38)
z
Hence x = y = z = t = a . (Why?) Thus again, the only critical point is p ⃗ = (a, a, a, a). Now, returning to Lagrange's method, we use formula (5) in §5 to compute 2
2
d F =−
(dxdy + dxdz + dzdt + dxdt + dydz + dydt).
(6.10.39)
a
(Verify!) We shall show that this expression is sign-constant (if xyzt = a ) , near the critical point x = y = z = t = a in (12), we get dt = −(dx + dy + dz), and (13) turns into 4
2 −
p .⃗
Indeed, setting
2
[dxdy + dxdz+ dydz − (dx + dy + dz) ] a 1 =
2
[dx
+ dy
2
+ dz
2
2
2
+ (dx + dy + dz) ] = d F .
a
6.10.5
https://math.libretexts.org/@go/page/21631
This expression is > 0 (for dx, dy, and dz are not all 0). Thus f has a local conditional minimum at p ⃗ = (a, a, a, a). Caution; here we cannot infer that f (p )⃗ is the least value of f under the imposed conditions: x, y, z > 0 and xyzt = a
4
.
The simplification due to the Cauchy invariant rule (Note 4 in §4) makes the use of the "variable" notation attractive, though caution ismandatory. Note 2. When using Theorem 2, it suffices to ascertain that some n equations from (9) admit a solution for the c ; for then, renumbering the equations, one can achieve that these become the first n equations, as was assumed. This means that the n × (n + m) matrix (D g (p ,⃗ q )) ⃗ must be of rank n, i.e., contains an n × n -submatrix (obtained by deleting some columns), with a nonzero determinant. k
j
k
In the Problems we often use r, s, t, … for Lagrange multipliers. This page titled 6.10: More on Implicit Differentiation. Conditional Extrema is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
6.10.6
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6.10.E: Further Problems on Maxima and Minima Exercise 6.10.E. 1 Fill in all details in Examples 1 and 2 and the proofs of all theorems in this section.
Exercise 6.10.E. 2 Redo Example (B) in §9 by Lagrange's method. [Hint: Set F (x, y, z) = f (x, y, z) − r (x + y + z points.] 2
2
2
2
) , g(x, y, z) = x
+y
2
2
+z
. Compare the values of f at all critical
−1
Exercise 6.10.E. 3 An ellipsoid 2
x
y +
2
2
2
a
2
z +
b
=1
2
(6.10.E.1)
c
is cut by a plane ux + vy + wz = 0. Find the semiaxes of the section-ellipse, i.e., the extrema of 2
ρ
under the constraints g = (g
→
1,
g2 ) = 0 ,
2
2
= [f (x, y, z)]
=x
+y
2
+z
2
(6.10.E.2)
where 2
y
x g1 (x, y, z) = ux + vy + wz and g2 (x, y, z) =
+
a2
2
b2
z +
2
c2
− 1.
(6.10.E.3)
Assume that a > b > c > 0 and that not all u, v, w = 0 . [Outline: By Note 2, explore the rank of the matrix 2
2
x/a
2
y/b
z/c
(
). u
v
(6.10.E.4)
z
(Why this particular matrix?) Seeking a contradiction, suppose all its 2 × 2 determinants vanish at all points of the section-ellipse. Then the upper and lower entries in (14) are proportional (why?); so x /a + y /b + z /c = 0 (a contradiction!). Next, set 2
2
2
2
2
2
2
2
F (x, y, z) = x
+y
2
+z
2
x +r(
y +
2
a
2
2
z +
2
b
2
) + 2s(ux + vy + wz).
(6.10.E.5)
c
Equate dF to 0 : rx x+
2
ry + su = 0,
y+
a
2
rz + sv = 0,
2
2
a
2
−ρ
→
obtain r = −ρ
,
z =
2
y =
(6.10.E.6)
2
b
2
−ρ
6.10.E.1
;
so, by (15), for a, b, c ≠ ρ ,
2
−svb ,
+ sw = 0.
0 ,
2
−sua
2
c
Multiplying by x, y, z, respectively, adding, and combining with g = x =
z+
b
−swc 2
c
2
.
(6.10.E.7)
−ρ
https://math.libretexts.org/@go/page/24102
Find s, x, y, z, then compare the ρ-values at critical points.]
Exercise 6.10.E. 4 Find the least and the largest values of the quadratic form n
f (x⃗ ) = ∑ aik xi xk
(aik = aki )
(6.10.E.8)
i,k=1
on the condition that g(x⃗ ) = |x⃗ | − 1 = 0 (f , g : E [Outline: Let F (x⃗ ) = f (x⃗ ) − t (x + x + … + x 2
2
2
1
2
n
2 n)
) . Equating dF to 0, obtain
→ E .
1
(a11 − t) x1 + a12 x2 + … + a1n xn = 0, a21 x1 + (a22 − t) x2 + … + a2n xn = 0,
(6.10.E.9)
⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯ an1 x1 + an2 x2 + … + (ann − t) xn = 0.
Using Theorem 1(iv) in §6, derive the so-called characteristic equation of f , ∣ a11 − t ∣ ∣ ∣ ∣ ∣
of degree
…
a1n
a21
a22 − t
…
a2n
…
…
…
…
an1
a2n
…
∣ ∣ ∣ ∣
= 0,
(6.10.E.10)
∣
ann − t ∣
If t is one of its n roots (known to be real), then equations (16) admit a nonzero solution for x⃗ = (x , … , x ) ; by replacing x⃗ by x⃗ /| x⃗ | if necessary, x⃗ satisfies also the constraint equation g(x⃗ ) = | x⃗ | − 1 = 0. (Explain!) Thus each root t of (17) yields a critical point x⃗ = (x , … , x ) . Now, to find f (x⃗ ) , multiply the k th equation in (16) by x , k = 1, … , n, and add to get n
in
a12
t.
2
1
n
t
t
1
n
k
n
n 2
0 = ∑ aik xi xk − t ∑ x
k
i,k=1
= f (x⃗ t ) − t.
(6.10.E.11)
k=1
Hence f (x⃗ ) = t . Thus the values of f at the critical points x⃗ are simply the roots of (17). The largest (smallest) root is also the largest (least) value of f on S = {x⃗ ∈ E ||x⃗ | = 1} (Explain!)] t
t
n
Exercise 6.10.E. 5 Use the method of Problem 4 to find the semiaxes of (i) the quadric curve in E
2
,
centered at
→ 0 ,
3
given by ∑
2 i,k=1
aik xi xk = 1;
and
(ii) the quadric surface ∑ a x x = 1 in E , centered at (\overrightarrow{0}\). Assume a = a . [Hint: Explore the extrema of f (x⃗ ) = |x⃗ | on the condition that i,k=1
ik
3
ik
i
k
kI
2
g(x⃗ ) = ∑ aik xi xk − 1 = 0.]
(6.10.E.12)
i,k
6.10.E.2
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Exercise 6.10.E. 6 Using Lagrange's method, redo Problems 4, 5, 6, 7, 11, 12, and 13 of §9.
Exercise 6.10.E. 7 In E
2
,
find the shortest distance from
→ 0
to the parabola y
2
= 2(x + a)
.
Exercise 6.10.E. 8 In E , find the shortest distance from 3
with u⃗ and v ⃗ different from
→ 0 .
→ 0
to the intersection line of two planes given by the formulas
u⃗ ⋅ x⃗ = a
and
v ⃗ ⋅ x⃗ = b
(Rewrite all in coordinate form!)
Exercise 6.10.E. 9 In E
n
,
find the largest value of |a⃗ ⋅ x⃗ | if |x⃗ | = 1. Use Lagrange's method.
Exercise 6.10.E. 10∗ (Hadamard's theorem.) If A = det(x
ik )(i,
then
k ≤ n),
n
|A| ≤ ∏ | x⃗ i | ,
(6.10.E.13)
i=1
where x⃗ = (x , x , … , x ) . [Hints: Set a = |x⃗ | . Treat A as a function of n variables. Using Lagrange's method, prove that, under the n constraints | x⃗ | − a = 0, A cannot have an extremum unless A = det(y ), with y = 0 (if i ≠ k ) and \(y_{ii}=a_{i}^{2}.] i
i1
i2
in
2
i
2
i
2 i
i
2
ik
Ik
6.10.E: Further Problems on Maxima and Minima is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
6.10.E.3
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CHAPTER OVERVIEW 7: Volume and Measure I. Our theory of set families leads quite naturally to a generalization of metric spaces. As we know, in any such space (S, ρ), there is a family G of open sets, and a family F of all closed sets. In Chapter 3, §12, we derived the following two properties. (i) G is closed under any (even uncountable) unions and under finite intersections (Chapter 3, §12, Theorem 2). Moreover, ∅ ∈ G and S ∈ G.
(7.1)
(ii) F has these properties, with "unions" and "intersections" interchanged (Chapter 3, §12, Theorem 3). Moreover, by definition, A ∈ F iff − A ∈ G.
(7.2)
Now, quite often, it is not so important to have distances (i.e., a metric) defined in S, but rather to single out two set families, G and F , with properties (i) and (ii), in a suitable manner. For examples, see Problems 1 to 4 below. Once G and F are given, one does not need a metric to define such notions as continuity, limits, etc. (See Problems 2 and 3.) This leads us to the following definition. 7.1: More on Intervals in Eⁿ. Semirings of Sets 7.1.E: Problems on Intervals and Semirings 7.2: \(\mathcal{C}_{\sigma}\)-Sets. Countable Additivity. Permutable Series 7.2.E: Problems on \(\mathcal{C}_{\sigma}\) -Sets, \(\sigma\) -Additivity, and Permutable Series 7.3: More on Set Families 7.3.E: Problems on Set Families 7.4: Set Functions. Additivity. Continuity 7.4.E: Problems on Set Functions 7.5: Nonnegative Set Functions. Premeasures. Outer Measures 7.5.E: Problems on Premeasures and Related Topics 7.6: Measure Spaces. More on Outer Measures 7.6.E: Problems on Measures and Outer Measures 7.7: Topologies. Borel Sets. Borel Measures 7.7.E: Problems on Topologies, Borel Sets, and Regular Measures 7.8: Lebesgue Measure 7.8.E: Problems on Lebesgue Measure 7.9: Lebesgue–Stieltjes Measures 7.9.E: Problems on Lebesgue-Stieltjes Measures 7.10: Generalized Measures. Absolute Continuity 7.10.E: Problems on Generalized Measures 7.11: Differentiation of Set Functions 7.11.E: Problems on Vitali Coverings
This page titled 7: Volume and Measure is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
1
7.1: More on Intervals in Eⁿ. Semirings of Sets I. As a prologue, we turn to intervals in E (Chapter 3, §7). n
Theorem 7.1.1 If A and B are intervals in E
n
,
then
(i) A ∩ B is an interval (∅ counts as an interval); (ii) A − B is the union of finitely many disjoint intervals (but need not be an interval itself). Proof The easy proof for E is left to the reader. 1
An interval in E is the cross-product of two line intervals. 2
Let A = X × Y and B = X
where X, Y , X , and Y are intervals in E ′
′
1
.
′
×Y
′
,
(7.1.1)
Then (see Figure 29)
and ′
′
A − B = [(X − X ) × Y ] ∪ [(X ∩ X ) × (Y − Y
′
)] ;
(7.1.2)
see Problem 8 in Chapter 1, §§1-3. As the theorem holds in E , 1
′
X ∩ X and Y ∩ Y
are intervals in E
1
,
′
(7.1.3)
while X −Y
′
and Y − Y
′
(7.1.4)
are finite unions of disjoint line intervals. (In Figure 29 they are just intervals, but in general they are not.) It easily follows that A ∩ B is an interval in theorem holds in E .
E
2
,
while A − B splits into finitely many such intervals. (Verify!) Thus the
2
Finally, for E , use induction. An interval in E is the cross-product of an interval in E theorem holds in E , the same argument shows that it holds in E , too. (Verify!) n
n
n−1
This completes the inductive proof.
n−1
by a line interval. Thus if the
n
□
7.1.1
https://math.libretexts.org/@go/page/19204
Actually, Theorem 1 applies to many other families of sets (not necessarily intervals or sets in E name.
n
).
We now give such families a
Definition 1 A family C of arbitrary sets is called a semiring iff (i) ∅ ∈ C (∅ is a member), and (ii) for any sets A and B from C, we have A ∩ B ∈ C, while A − B is the union of finitely many disjoint sets from C. Briefly: C is a semiring iff it satisfies Theorem 1. Note that here C is not just a set, but a whole family of sets. Recall (Chapter §§1-3) that a set family (family of sets) is a set whose members are other sets. If A is a member of M, we call A an M-set and write A ∈ M (not A ⊆ M) .
M
Sometimes we use index notation: M = { Xi |i ∈ I } ,
(7.1.5)
M = { Xi } ,
(7.1.6)
briefly
where the X are M-sets distinguished from each other by the subscripts i varying over some index set I . i
A set family M = {X
i}
and its union ⋃ Xi
(7.1.7)
i
are said to be disjoint iff Xi ∩ Xj = ∅ whenever i ≠ j.
(7.1.8)
⋃ Xi (disjoint).
(7.1.9)
Notation:
In our case, A ∈ C means that A is a C -set (a member of the semiring C). The formula (∀A, B ∈ C)
A∩B ∈ C
(7.1.10)
means that the intersection of two C -sets in a C -set itself. Henceforth, we will often speak of semirings this case in mind!
C
in general. In particular, this will apply to the case
C =
{intervals}. Always keep
Note 1. By Theorem 1, the intervals in E form a semiring. So also do the half-open and the half-closed intervals separately (same proof!), but not the open (or closed) ones. (Why?) n
Caution. The union and difference of two C -sets need not be a C -set. To remedy this, we now enlarge C.
Definition 2 We say that a set A (from C or not) is C -simple and write ′
A ∈ Cs
(7.1.11)
iff A is a finite union of disjoint C -sets (such as A − B in Theorem 1). Thus C is the family of all C -simple sets. ′
s
Every C -set is also a C -set, i.e., a C -simple one. (Why?) Briefly: ′
s
7.1.2
https://math.libretexts.org/@go/page/19204
′
C ⊆ Cs .
(7.1.12)
If C is the set of all intervals, a C -simple set may look as in Figure 30.
Theorem 7.1.2 If C is a semiring, and if A and B are C -simple, so also are A ∩ B, A − B, and A ∪ B.
(7.1.13)
In symbols, ′
(∀A, B ∈ C ) s
′
′
′
s
s
s
A ∩ B ∈ C , A − B ∈ C , and A ∪ B ∈ C .
(7.1.14)
Proof We give a proof outline and suggest the proof as an exercise. Before attempting it, the reader should thoroughly review the laws and problems of Chapter 1, §§1-3. (1) To prove A ∩ B ∈ C
′
s
,
let m
n
A = ⋃ Ai (disjoint) and B = ⋃ Bk (disjoint), i=1
with A
i,
Bk ∈ C.
(7.1.15)
k=1
Verify that n
m
A ∩ B = ⋃ ⋃ (Ai ∩ Bk ) (disjoint),
(7.1.16)
k=1 i=1
and so A ∩ B ∈ C . ′
s
(2) Next prove that A − B ∈ C if A ∈ C and B ∈ C . ′
′
s
s
Indeed, if m
A = ⋃ Ai (disjoint),
(7.1.17)
i=1
then m
m
A − B = ⋃ Ai − B = ⋃ (Ai − B) (disjoint). i=1
(7.1.18)
i=1
Verify and use Definition 2. (3) Prove that ′
(∀A, B ∈ Cs )
′
A − B ∈ Cs ;
(7.1.19)
we suggest the following argument.
7.1.3
https://math.libretexts.org/@go/page/19204
Let n
B = ⋃ Bk ,
Bk ∈ C.
(7.1.20)
k=1
Then n
n
A − B = A − ⋃ Bk = ⋂ (A − Bk ) k=1
(7.1.21)
k=1
by duality laws. But A − B is C -simple by step (2). Hence so is k
m
A − B = ⋂ (A − Bk )
(7.1.22)
k=1
by step (1) plus induction. (4) To prove A ∪ B ∈ C
′
s
,
verify that A ∪ B = A ∪ (B − A),
where B − A ∈ C
′
s
(7.1.23)
by (3).
,
Note 2. By induction, Theorem 2 extends to any finite number of C -sets. It is a kind of "closure law." ′
s
We thus briefly say that C is closed under finite unions, intersections, and set differences. Any (nonempty) set family with these properties is called a set ring (see also §3). ′
s
Thus Theorem 2 states that if C is a semiring, then C is a ring. ′
s
Caution. An infinite union of C -simple sets need not be C -simple. Yet we may consider such unions, as we do next. In Corollary 1 below, C may be replaced by any set ring M. ′
s
Corollary 7.1.1 If {A } is a finite or infinite sequence of sets from a semiring sequence of C -simple sets (or M-sets) B ⊆ A such that
C
n
n
(or from a ring
M
such as
′
Cs
), then there is a disjoint
n
⋃ An = ⋃ Bn . n
(7.1.24)
n
Proof Let B
1
= A1
and for n = 1, 2, …, n
Bn+1 = An+1 − ⋃ Ak ,
Ak ∈ C.
(7.1.25)
k=1
By Theorem 2, the B are C -simple (as are A a contradiction) and verify that ⋃ A = ⋃ B n
and ⋃ A ). Show that they are disjoint (assume the opposite and find If x ∈ ⋃ A , take the least n for which x ∈ A . Then n > 1 and n
n+1
n
n
:
k
k=1
n
n
n−1
x ∈ An − ⋃ Ak = Bn ,
(7.1.26)
k=1
or n = 1 and x ∈ A
1
Note 3. In Corollary 1, B
n
= B1 .
′
∈ Cs ,
□
i.e., B
n
mn
=⋃
i=1
Cni
for some disjoint sets C
ni
∈ C.
Thus
mn
⋃ An = ⋃ ⋃ Cni n
n
7.1.4
(7.1.27)
i=1
https://math.libretexts.org/@go/page/19204
is also a countable disjoint union of C -sets. II. Recall that the volume of intervals is additive (Problem 9 in Chapter 3, §7). That is, if A ∈ C is split into finitely many disjoint subintervals, then vA (the volume of A) equals the sum of the volumes of the parts. We shall need the following lemma.
lemma 1 Let X
1,
X2 , … , Xm ∈ C
m
(i) ⋃
i=1
(ii) ⋃
m i=1
m
implies ∑
Xi ⊆ Y ∈ C p
Xi ⊆ ⋃
k=1
(intervals in E i=1
n
).
If the X are mutually disjoint, then i
vXi ≤ vY ;
Yk (with Yk ∈ C)
and
implies ∑
m
p
i=1
vXi ≤ ∑
vYk
k=1
.
Proof (i) By Theorem 2, the set m
Y − ⋃ Xi
(7.1.28)
i=1
is C -simple; so m
q
Y − ⋃ Xi = ⋃ Cj i=1
(7.1.29)
j=1
for some disjoint intervals C . Hence j
Y = ⋃ Xi ∪ ⋃ Cj (all disjoint).
(7.1.30)
Thus by additivity, q
m
m
vY = ∑ vXi + ∑ vCj ≥ ∑ vXi , i=1
j=1
(7.1.31)
i=1
as claimed. (ii) By set theory (Problem 9 in Chapter 1, §§1-3), p
Xi ⊆ ⋃ Yk
(7.1.32)
k=1
implies p
p
Xi = Xi ∩ ⋃ Yk = ⋃ (Xi ∩ Yk ) . k=1
(7.1.33)
k=1
If it happens that the Y are mutually disjoint also, so certainly are the smaller intervals X k
i
∩ Yk ;
so by additivity,
p
vXi = ∑ v (Xi ∩ Yk ) .
(7.1.34)
k=1
Hence m
p
m
p
m
∑ vXi = ∑ ∑ v (Xi ∩ Yk ) = ∑ [ ∑ v (Xi ∩ Yk )] . i=1
i=1
k=1
k=1
(7.1.35)
i=1
But by (i), m
∑ v (Xi ∩ Yk ) ≤ vYk (why?);
(7.1.36)
i=1
7.1.5
https://math.libretexts.org/@go/page/19204
so p
m
∑ vXi ≤ ∑ vYk , i=1
(7.1.37)
k=1
as required. If, however, the Y are not disjoint, Corollary 1 yields k
⋃ Yk = ⋃ Bk (disjoint),
(7.1.38)
with mk
Yk ⊇ Bk = ⋃ Ckj (disjoint),
Ckj ∈ C.
(7.1.39)
j=1
By (i), mk
∑ vCkj ≤ vYk .
(7.1.40)
j=1
As p
m
p
p
mk
⋃ Xi ⊆ ⋃ Yk = ⋃ Bk = ⋃ ⋃ Ckj (disjoint), i=1
k=1
all reduces to the previous disjoint case.
k=1
(7.1.41)
k=1 j=1
□
Corollary 7.1.2 Let A ∈ C (C = intervals in E ). If ′
n
s
p
m
A = ⋃ Xi (disjoint) = ⋃ Yk (disjoint) i=1
with X
i,
Yk ∈ C,
(7.1.42)
k=1
then p
m
∑ vXi = ∑ vYk . i=1
(7.1.43)
k=1
(Use part (ii) of the lemma twice.) Thus we can (and do) unambiguously define vA to be either of these sums. This page titled 7.1: More on Intervals in Eⁿ. Semirings of Sets is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
7.1.6
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7.1.E: Problems on Intervals and Semirings Exercise 7.1.E. 1 Complete the proof of Theorem 1 and Note 1.
Exercise 7.1.E. 1
′
Prove Theorem 2 in detail.
Exercise 7.1.E. 2 Fill in the details in the proof of Corollary 1.
Exercise 7.1.E. 2
′
Prove Corollary 2.
Exercise 7.1.E. 3 Show that, in the definition of a semiring, the condition ∅ ∈ C is equivalent to C ≠ ∅ . m
[Hint: Consider ∅ = A − A = ∪
i=1
Ai (A, Ai ∈ C) to get ∅ = Ai ∈ C. ]
Exercise 7.1.E. 4 Given a set S, show that the following are semirings or rings. (a) C = { all subsets of S} ; (b) C = { all finite subsets of S} ; (c) C = {∅} ; (d) C = {∅ and all singletons in S} . Disprove it for C = {∅ and all two − point sets in S}, S = {1, 2, 3, …} . In (a) − (c), show that C = C. Disprove it for (d). ′
s
Exercise 7.1.E. 5 Show that the cubes in E
n
(n > 1)
do not form a semiring.
Exercise 7.1.E. 6 Using Corollary 2 and the definition thereafter, show that volume is additive for C -simple sets. That is, m
m
if A = ⋃ Ai (disjoint) then vA = ∑ vAi i=1
′
(A, Ai ∈ Cs ) .
(7.1.E.1)
i=1
Exercise 7.1.E. 7 Prove the lemma for C -simple sets. [Hint: Use Problem 6 and argue as before. ]
7.1.E.1
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Exercise 7.1.E. 8 Prove that if C is a semiring, then C
′
s
(C -simple sets ) = Cs ,
the family of all finite unions of C -sets (disjoint or not).
[Hint: Use Theorem 2. ]
7.1.E: Problems on Intervals and Semirings is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
7.1.E.2
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7.2: C σ Cσ -Sets. Countable Additivity. Permutable Series We now want to further extend the definition of volume by considering countable unions of intervals, called semiring of all intervals in E ).
Cσ
-sets (C being the
n
We also ask, if A is split into countably many such sets, does additivity still hold? This is called countable additivity or σ-additivity (the σ is used whenever countable unions are involved). We need two lemmas in addition to that of §1.
lemma 1 If B is a nonempty interval in E
n
,
then given ε > 0, there is an open interval C and a closed one A such that A ⊆B ⊆C
(7.2.1)
vC − ε < vB < vA + ε.
(7.2.2)
and
Proof Let the endpoints of B be ¯¯
¯ ¯ ¯
a = (a1 , … , an ) and b = (b1 , … , bn ) .
(7.2.3)
For each natural number i, consider the open interval C , with endpoints i
1 ( a1 −
i
1 , a2 −
i
1 , … , an −
i
1 ) and ( b1 +
i
1 , b2 +
i
1 , … , bn +
).
(7.2.4)
i
Then B ⊆ C and i
n
1
vCi = ∏ [bk + k=1
i
n
1 − ( ak −
i
)] = ∏ ( bk − ak + k=1
2 ).
(7.2.5)
i
Making i → ∞, we get n
lim vCi = ∏ (bk − ak ) = vB. i→∞
(7.2.6)
k=1
(Why?) Hence by the sequential limit definition, given ε > 0, there is a natural i such that vCi − vB < ε,
(7.2.7)
vCi − ε < vB.
(7.2.8)
or
As C is open and ⊇ B, it is the desired interval C . i
Similarly, one finds the closed interval A ⊆ B. (Verify!)
□
lemma 2 Any open set G ⊆ E
n
is a countable union of open cubes A and also a disjoint countable union of half-open intervals. k
(See also Problem 2 below.) Proof If G = ∅, take all A
k
=∅
.
If G ≠ ∅, every point p ∈ G has a cubic neighborhood
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Cp ⊆ G,
(7.2.9)
centered at p (Problem 3 in Chapter 3, §12). By slightly shrinking this C , one can make its endpoints rational, with p still in it (but not necessarily its center), and make C open, half-open, or closed, as desired. (Explain!) p
p
Choose such a cube C for every p ∈ G; so p
G ⊆ ⋃ Cp .
(7.2.10)
p∈G
But by construction, G contains all C , so that p
G = ⋃ Cp .
(7.2.11)
p∈G
Moreover, because the coordinates of the endpoints of all C are rational, the set of ordered pairs of endpoints of the C is countable, and thus, while the set of all p ∈ G is uncountable, the set of distinct C is countable. Thus one can put the family of all C in a sequence and rename it {A }: p
p
p
p
k
∞
G = ⋃ Ak .
(7.2.12)
k=1
If, further, the A are half-open, we can use Corollary 1 and Note 3, both from §1, to make the union disjoint (half-open intervals form a semiring!). □ k
Now let C be the family of all possible countable unions of intervals in unions). Thus A ∈ C means that A is a C -set, i.e., σ
σ
E
n
,
such as
G
in Lemma 2 (we use
Cs
for all finite
σ
∞
A = ⋃ Ai
(7.2.13)
i=1
for some sequence of intervals {A
i}
.
Such are all open sets in E
We can always make the sequence {A
i}
n
,
but there also are many other C -sets. σ
infinite (add null sets or repeat a term!).
By Corollary 1 and Note 3 of §1, we can decompose any C -set A into countably many disjoint intervals. This can be done in many ways. However, we have the following result. σ
Theorem 7.2.1 If ∞
∞
A = ⋃ Ai (disjoint) = ⋃ Bk (disjoint) i=1
for some intervals A
i,
Bk
in E
n
,
(7.2.14)
k=1
then ∞
∞
∑ vAi = ∑ vBk . i=1
(7.2.15)
k=1
Thus we can (and do) unambiguously define either of these sums to be the volume vA of the C -set A. σ
Proof We shall use the Heine-Borel theorem (Problem 10 in Chapter 4, §6; review it!). Seeking a contradiction, let (say) ∞
∞
∑ vAi > ∑ vBk , i=1
(7.2.16)
k=1
so, in particular,
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∞
∑ vBk < +∞.
(7.2.17)
k=1
As ∞
m
∑ vAi = i=1
lim ∑ vAi ,
m→∞
(7.2.18)
i=1
there is an integer m for which m
∞
∑ vAi > ∑ vBk . i=1
(7.2.19)
k=1
We fix that m and set m
∞
2ε = ∑ vAi − ∑ vBk > 0. i=1
Dropping "empties" (if any), we assume A
i
Then Lemma 1 yields open intervals Y
k
≠∅
and B
k
(7.2.20)
k=1
.
≠∅
with
⊇ Bk ,
ε vBk > vYk −
,
k = 1, 2, … ,
(7.2.21)
k
2
and closed ones X
i
⊆ Ai ,
with ε vXi +
> vAi ;
m
(7.2.22)
so m
∞
m
2ε = ∑ vAi − ∑ vBk i=1
k=1
∞
ε
< ∑ (vXi +
m
i=1 m
ε
) − ∑ (vYk −
k
)
2
k=1
∞
= ∑ vXi − ∑ vYk + 2ε. i=1
k=1
Thus m
∞
∑ vXi > ∑ vYk . i=1
(7.2.23)
k=1
(Explain in detail!) Now, as ∞
∞
Xi ⊆ Ai ⊆ A = ⋃ Bk ⊆ ⋃ Yk , k=1
(7.2.24)
k=1
each of the closed intervals X is covered by the open sets Y . i
k
By the Heine-Borel theorem, ⋃
m i=1
Xi
is already covered by a finite number of the Y , say, k
p
m
⋃ Xi ⊆ ⋃ Yk . i=1
(7.2.25)
k=1
The X are disjoint, for even the larger sets A are. Thus by Lemma 1(ii) in §1, i
i
m
p
∞
∑ vXi ≤ ∑ vYk ≤ ∑ vYk , i=1
contrary to (1). This contradiction completes the proof.
k=1
(7.2.26)
k=1
□
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Corollary 7.2.1 If ∞
A = ⋃ Bk (disjoint)
(7.2.27)
k=1
for some intervals B , then k
∞
vA = ∑ vBk .
(7.2.28)
k=1
Indeed, this is simply the definition of vA contained in Theorem 1. Note 1. In particular, Corollary 1 holds if A is an interval itself. We express this by saying that the volume of intervals is σ-additive or countably additive. This also shows that our previous definition of volume (for intervals) agrees with the definition contained in Theorem 1 (for C -sets). σ
Note 2. As all open sets are C -sets (Lemma 2), volume is now defined for any open set A ⊆ E σ
n
(in particular, for A = E ). n
Corollary 7.2.2 If A
i,
Bk
are intervals in E
n
,
with ∞
∞
⋃ Ai ⊆ ⋃ Bk , i=1
(7.2.29)
k=1
then provided the A are mutually disjoint, i
∞
∞
∑ vAi ≤ ∑ vBk . i=1
(7.2.30)
k=1
Proof The proof is as in Theorem 1 (but the B need not be disjoint here). k
Corollary 7.2.3 ("σ-subadditivity" of the volume) If ∞
A ⊆ ⋃ Bk ,
(7.2.31)
k=1
where A ∈ C and the B are intervals in E σ
k
n
,
then ∞
vA ≤ ∑ vBk .
(7.2.32)
k=1
Proof Set ∞
A = ⋃ Ai ( disjoint ), Ai ∈ C,
(7.2.33)
i=1
and use Corollary 2.
□
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Corollary 7.2.4 ("monotonicity") If A, B ∈ C
σ,
with A ⊆ B,
(7.2.34)
vA ≤ vB.
(7.2.35)
then
("Larger sets have larger volumes.") This is simply Corollary 3, with ⋃
k
Bk = B
.
Corollary 7.2.5 The volume of all of E is ∞ (we write ∞ for +∞ ). n
Proof We have A ⊆ E for any interval A. n
Thus, by Corollary 4, vA ≤ vE . n
As vA can be chosen arbitrarily large, vE must be infinite. n
□
Corollary 7.2.6 For any countable set A ⊂ E
n
, vA = 0.
In particular, v∅ = 0 .
Proof First let A = {a} be a singleton. Then we may treat A as a degenerate interval [a, a]. As all its edge lengths are 0, we have vA = 0 . ¯ ¯ ¯
Next, if A = {a
¯ ¯ ¯ ¯ ¯ ¯ 1 , a2 ,
¯ ¯ ¯
…}
¯ ¯ ¯
is a countable set, then ¯ ¯ A = ⋃ {¯ a k} ;
(7.2.36)
k
so ¯ ¯ vA = ∑ v {¯ a k} = 0
(7.2.37)
k
by Corollary 1. Finally, ∅ is the degenerate open interval (a, a); so v∅ = 0. ¯ ¯ ¯
¯ ¯ ¯
□
Note 3. Actually, all these propositions hold also if all sets involved are intervals!).
Cσ
-sets, not just intervals (split each
Cσ
-set into disjoint
Permutable Series. Since σ-additivity involves countable sums, it appears useful to generalize the notion of a series. We say that a series of constants, ∑ an ,
(7.2.38)
is permutable iff it has a definite (possibly infinite) sum obeying the general commutative law: Given any one-one map onto
u : N ⟷ N
7.2.5
(7.2.39)
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(N
the naturals), we have
=
∑ an = ∑ au ,
(7.2.40)
n
n
where u
n
= u(n)
n
.
(Such are all positive and all absolutely convergent series in a complete space the sum does not depend on the choice of the map u. Thus, given any u : N
onto
⟷ J
E;
see Chapter 4, §13.) If the series is permutable,
(where J is a countable index set) and a set { ai |i ∈ J} ⊆ E
(7.2.41)
(where E is E or a normed space), we can define ∗
∞
∑ ai = ∑ au
(7.2.42)
n
i∈J
if ∑
n
aun
n=1
is permutable.
In particular, if J = N ×N
(7.2.43)
∑ ai
(7.2.44)
(a countable set, by Theorem 1 in Chapter 1, §9), we call
i∈J
a double series, denoted by symbols like ∑ akn
(k, n ∈ N ).
(7.2.45)
n,k
Note that ∑ | ai |
(7.2.46)
i∈J
is always defined (being a positive series). If ∑ | ai | < ∞,
(7.2.47)
i∈J
we say that ∑
i∈J
ai
converges absolutely.
For a positive series, we obtain the following result.
Theorem 7.2.2 (i) All positive series in E are permutable. ∗
(ii) For positive double series in E
∗
,
we have ∞
∞
∞
∞
∞
∑ ank = ∑ ( ∑ ank ) = ∑ ( ∑ ank ) . n,k=1
n=1
k=1
k=1
(7.2.48)
n=1
Proof (i) Let ∞
m
s = ∑ an and sm = ∑ an n=1
(an ≥ 0) .
(7.2.49)
n=1
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Then clearly sm+1 = sm + am+1 ≥ sm ;
i.e., {s
m}
(7.2.50)
, and so
↑
s =
lim sm = sup sm
m→∞
(7.2.51)
m
by Theorem 3 in Chapter 3, §15. Hence s certainly does not exceed the lub of all possible sums of the form ∑ ai ,
(7.2.52)
i∈I
where I is a finite subset of N (the partial sums s are among them). Thus m
s ≤ sup ∑ ai ,
(7.2.53)
i∈I
over all finite sets I
⊂N
.
On the other hand, every such ∑ and so
i∈I
ai
is exceeded by, or equals, some s . Hence in (4), the reverse inequality holds, too, m
s = sup ∑ ai .
(7.2.54)
i∈I
But sup ∑ a clearly does not depend on any arrangement of the a_{i}. Therefore, the series assertion (i) is proved. i
i∈I
∑ an
is permutable, and
Assertion (ii) follows similarly by considering sums of the form ∑ a where I is a finite subset of N × N , and showing that the lub of such sums equals each of the three expressions in (3). We leave it to the reader. □ i∈I
i
A similar formula holds for absolutely convergent series (see Problems). This page titled 7.2: C -Sets. Countable Additivity. Permutable Series is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. σ
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7.2.E: Problems on C σ Cσ -Sets, σ σ -Additivity, and Permutable Series Exercise 7.2.E. 1 Fill in the missing details in the proofs of this section.
Exercise 7.2.E. 1
′
Prove Note 3.
Exercise 7.2.E. 2 Show that every open set A ≠ ∅ in E is a countable union of disjoint half-open cubes. [Outline: For each natural m, show that E is split into such cubes of edge length 2 by the hyperplanes n
n
−m
i xk =
m
i = 0, ±1, ±2, … ; k = 1, 2, … , n,
(7.2.E.1)
2
and that the family C of such cubes is countable. For m > 1, let C , C , … be the sequence of those cubes from m
m1
m2
Cm
(if any) that lie in
A
but not in any cube
Csj
with
s < m. As A is open, x ∈ A iff x ∈ some Cmj . ]
Exercise 7.2.E. 3 Prove that any open set A ⊆ E is a countable union of disjoint (possibly infinite) open intervals. [Hint: By Lemma 2, A = ⋃ (a , b ) . If, say, (a , b ) overlaps with some (a , b ), replace both by their union. Continue inductively. 1
n
n
n
1
1
m
m
Exercise 7.2.E. 4 Prove that C is closed under finite intersections and countable unions. σ
Exercise 7.2.E. 5 (i) Find A, B ∈ C such that A − B ∉ C (ii) Show that C is not a semiring. [Hint: Try A = E , B = R (the rationals).] Note. In the following problems, J is countably infinite, a σ
σ
σ
1
i
∈ E(E complete).
Exercise 7.2.E. 6 Prove that ∑ | ai | < ∞
(7.2.E.2)
i∈J
iff for every ε > 0, there is a finite set F ⊂J
(F ≠ ∅)
(7.2.E.3)
such that
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∑ | ai | < ε
(7.2.E.4)
i∈I
for every finite I ⊂ J − F . [Outline: By Theorem 2, fix u : N onto
⟺
J
with ∞
∑ | ai | = ∑ | au | .
(7.2.E.5)
n
i∈J
n=1
By Cauchy's criterion, ∞
∑ | au | < ∞
(7.2.E.6)
n
n=1
iff n
(∀ε > 0)(∃q)(∀n > m > q)
∑ | au | < ε. k
(7.2.E.7)
k=m
Let F
= { u1 , … , uq } .
If I is as above, (∃n > m > q)
{ um , … , un } ⊇ I ;
(7.2.E.8)
so n
∑ | ai | ≤ ∑ | auk | < ε. ] i∈I
(7.2.E.9)
k=m
Exercise 7.2.E. 7 Prove that if ∑ | ai | < ∞,
(7.2.E.10)
i∈J
then for every ε > 0, there is a finite F
⊂ J(F ≠ ∅)
such that ∣ ∣ ∣ ∣ ∑ ai − ∑ ai < ε ∣ ∣ ∣ i∈J ∣ i∈K
for each finite K ⊃ F (K ⊂ J) . [Hint: Proceed as in Problem 6, with I
= K −F
(7.2.E.11)
and q so large that
∣ ∣ 1 ∣ ∣ ∑ ai − ∑ ai < ε ∣ ∣ 2 ∣ i∈J ∣ i∈F
and
∣ ∣ 1 ∣ ∣ ∑ ai < ε. ] ∣ ∣ 2 ∣ i∈F ∣
(7.2.E.12)
Exercise 7.2.E. 8 Show that if
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∞
J = ⋃ In (disjoint),
(7.2.E.13)
n=1
then ∞
∑ | ai | = ∑ bn , where bn = ∑ | ai | . i∈J
n=1
(7.2.E.14)
i∈In
(Use Problem 8' below.)
Exercise 7.2.E. 8
′
Show that ∑ | ai | = sup ∑ | ai | F
i∈J
over all finite sets F
⊂ J(F ≠ ∅)
(7.2.E.15)
i∈F
.
[Hint: Argue as in Theorem 2. ]
Exercise 7.2.E. 9 Show that if ∅ ≠ I
⊆ J,
then ∑ | ai | ≤ ∑ | ai | . i∈I
′
(7.2.E.16)
i∈J
[Hint: Use Problem 8 and Corollary 2 of Chapter 2,
§§8 − 9. ]
Exercise 7.2.E. 10 Continuing Problem 8, prove that if ∞
∑ | ai | = ∑ bn < ∞, i∈J
(7.2.E.17)
n=1
then ∞
∑ ai = ∑ cn with cn = ∑ ai . i∈J
n=1
(7.2.E.18)
i∈In
[Outline: By Problem 9, (∀n) ∑ | ai | < ∞;
(7.2.E.19)
i∈In
so cn = ∑ ai
(7.2.E.20)
i∈In
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and ∞
∑ cn
(7.2.E.21)
n=1
converge absolutely. Fix ε and F as in Problem 7. Choose the largest q ∈ N with F ∩ Iq ≠ ∅
(7.2.E.22)
(why does it exist?), and fix any n > q. By Problem 7, (∀k ≤ n) (∀k ≤ n) (∃ finite Fk |J ⊇ Fk ⊇ F ∩ Iq )
(∀ finite Hk | Ik ⊇ Hk ⊇ Fk )
∣ ∣
n
∑ ai − ∑ ck
∣ ∣i∈Hk
k=1
∣ ∣ ∣ ∣
1
0 (take rationals r ↘ r ). By translation, m = cm on half-open intervals. Proceed as in Problem 13 of §8. (iii) See Problems 4 to 6 in §8. Note that, by Theorem 2, one may assume m = m (a translation-invariant LS measure). As m = cm on half-open intervals, Lemma 2 in §2 yields m = cm on G (open sets). Use G -regularity to prove m = cm and M = M .] 1
n
k
k
i
′
′
α
α
∗ α
α
∗
∗
∗
α
Exercise 7.9.E. 9∗ (LS measures in E .) Let n
C
For any map G : E
n
→ E
1
and any (a, b] ∈ C ¯¯
¯ ¯ ¯
¯ ¯ ¯
∗
∗
= {alf-open intervals in E
,
n
}.
(7.9.E.5)
set
¯¯
Δk G(a, b] = G (x1 , … , xk−1 , bk , xk+1 , … , xn ) − G (x1 , … , xk−1 , ak , xk+1 , … , xn ) ,
Given α : E
n
→ E
1
,
1 ≤ k ≤ n.
set ¯ ¯ ¯
¯¯
¯ ¯ ¯
¯¯
sα (a, b] = Δ1 (Δ2 (⋯ (Δn α(a, b]) ⋯)) .
(7.9.E.6)
sα (a, b] = α (b1 , b2 ) − α (b1 , a2 ) − [α (a1 , b2 ) − α (a1 , a2 )] .
(7.9.E.7)
For example, in E , 2
Show that s is additive on C . Check that the order in which the Δ are applied is immaterial. Set up a formula for s in E . [Hint: First take two disjoint intervals ∗
3
α
k
¯ ¯ ¯
¯¯
¯ ¯ ¯
¯¯
α
¯ ¯ ¯
¯¯
(a, q ] ∪ (p, b] = (a, b],
(7.9.E.8)
as in Figure 2 in Chapter 3, §7. Then use induction, as in Problem 9 of Chapter 3, §7.]
Exercise 7.9.E. 10∗ If s in Problem 9 is nonnegative, and α is right continuous in each variable x separately, we call and s is called the α -induced LS premeasure in E ; the LS outer measure m and measure α
k
n
α
a distribution function,
∗ α
α
∗
mα : Mα → E
∗
(7.9.E.9)
in E (obtained from s as shown in } §§5 and 6) are said to be induced by α. For s , m , and m so defined, redo Problems 1-3 above. n
α
α
∗ α
α
7.9.E: Problems on Lebesgue-Stieltjes Measures is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
7.9.E.2
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7.10: Generalized Measures. Absolute Continuity 7.10: Generalized Measures. Absolute Continuity is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
7.10.1
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7.10.E: Problems on Generalized Measures Exercise 7.10.E. 1 Complete the proofs of Theorems 1,4, and 5.
Exercise 7.10.E. 1
′
Do it also for the lemmas and Corollary 3.
Exercise 7.10.E. 2 Verify the following. (i) In Definition 2, one can equivalently replace "countable {X }" by "finite {X }." (ii) If M is a ring, Note 1 holds for finite sequences {X }. (iii) If s : M → E is additive on M, a semiring, so is v . [Hint: Use Theorem 1 from §4.] i
i
i
s
Exercise 7.10.E. 3 For any set functions s, t on M, prove that (i) v = v , and (ii) v ≤ av , provided st is defined and |s|
s
st
t
a = sup{|sX||X ∈ M}.
(7.10.E.1)
Exercise 7.10.E. 4 Given s, t : M → E, show that (i) v ≤ v + v ; (ii) v = |k|v ((\k\) as in Corollary 2); and (iii) if E = E (C ) and s+t
s
ks
t
s
n
n
n ¯ ¯ s = ∑ sk ¯ e k,
(7.10.E.2)
k=1
then n
vs
k
≤ vs ≤ ∑ vsk .
(7.10.E.3)
k=1
[Hints: (i) If A ⊇ ⋃ Xi (disjoint),
with A
i,
Xi ∈ M,
(7.10.E.4)
verify that |(s + t)Xi | ≤ |sXi | + |tXi | ,
(7.10.E.5)
∑ |(s + t)Xi | ≤ vs A + vt A, etc.;
(ii) is analogous.
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(iii) Use (ii) and (i), with |e
¯ ¯ ¯ k|
=1
.]
Exercise 7.10.E. 5 If
, can one define the signed LS measure s by simply setting s = m )? [Hint: the domains of m and m may be different. Give an example. How about taking their intersection?] g ↑, h ↑,
and
α = g−h
on
E
1
α
α
g
(assuming
− mh
mh < ∞
g
h
Exercise 7.10.E. 6 Find an LS measure m such that α is continuous and one-to-one, but m is not m-finite (m = Lebesgue measure). [Hint: Take α
α
3
x
α(x) = {
|x|
,
x ≠ 0, (7.10.E.6)
0,
x = 0,
and ∞
1
A = ⋃ (n, n +
2
].]
(7.10.E.7)
n
n=1
Exercise 7.10.E. 7 Construct complex and vector-valued LS measures s
∗
α
: Mα → E
n
(C
n
)
in E
1
.
Exercise 7.10.E. 8 Show that if s : M → E (C ) is additive and bounded on M, a ring, so is v . [Hint: By Problem 4(iii), reduce all to the real case. Use Problem 2. Given a finite disjoint sequence {X } ⊆ M, let U (U sX ≥ 0(sX < 0, respectively). Show that n
n
s
+
−
i
i
)
be the union of those
Xi
for which
i
∑ sXi = sU
+
− sU
−
≤ 2 sup |s| < ∞. ]
(7.10.E.8)
Exercise 7.10.E. 9 For any s : M → E
∗
and A ∈ M, set +
s
A = sup{sX|A ⊇ X ∈ M}
(7.10.E.9)
and −
s
A = sup{−sX|A ⊇ X ∈ M}.
Prove that if s is additive and bounded on M, a ring, so are s and s +
+
s
−
;
furthermore,
1 = 2
s
−
(7.10.E.10)
(vs + s) ≥ 0,
1 = 2
(vs − s) ≥ 0,
+
s =s
+
vs = s
−
−s
−
+s
, and .
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[Hints: Use Problem 8. Set 1
′
s = 2
(vs + s) .
(7.10.E.11)
Then (∀X ∈ M|X ⊆ A) 2sX = sA + sX − s(A − X)
≤ sA + (|sX| + |s(A − X)|) ′
≤ sA + vs A = 2 s A.
Deduce that s A ≤ s A . To prove also that s A ≤ s +
′
′
+
A,
let ε > 0. By Problems 2 and 8, fix {X
i}
⊆M
, with
n
A = ⋃ Xi (disjoint)
(7.10.E.12)
i=1
and n
vs A − ε < ∑ |sXi | .
(7.10.E.13)
i=1
Show that n ′
2 s A − ε = vs A + sA − ε ≤ sU
+
− sU
−
+ s ⋃ Xi = 2sU
+
(7.10.E.14)
i=1
and +
2s
A ≥ 2sU
+
′
≥ 2 s A − ε. ]
(7.10.E.15)
Exercise 7.10.E. 10 Let K = {compact sets in a topological space (S, G)}
(7.10.E.16)
(adopt Theorem 2 in Chapter 4, §7, as a definition). Given s : M → E,
S
M ⊆2
,
(7.10.E.17)
we call s compact regular (CR) iff (∀ε > 0) (∀A ∈ M)(∃F ∈ K)(∃G ∈ G) F,G
∈ M, F ⊆ A ⊆ G, and vs G − ε ≤ vs A ≤ vs F + ε.
Prove the following. (i) If s, t : M → E are CR, so are s ± t and ks (k as in Corollary 2). (ii) If s is additive and CR on M, a semiring, so is its extension to the ring M (Theorem 1 in §4 and Theorem 4 of §3). (iii) If E = E (C ) and v < ∞ on M, a ring, then s is CR iff its components s are, or in the case E = E , iff s and s are (see Problem 9). [Hint for (iii): Use (i) and Problem 4(iii). Consider v (G − F ) .] s
n
n
1
s
+
−
k
s
7.10.E.3
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Exercise 7.10.E. 11 (Aleksandrov.) Show that if s : M → E is CR (see Problem 10) and additive on M, a ring in a topological space S, and if v < ∞ on M, then v and s are σ-additive, and v has a unique σ-additive extension v to the σ-ring N generated by M. The latter holds for s, too, if S ∈ M and E = E (C ) . [Proof outline: The σ-additivity of v results as in Theorem 1 of §2 (first check Lemma 1 in §1 for v ). For the σ-additivity of s, let s
s
¯ ¯ ¯ s
s
n
n
s
s
∞
A = ⋃ Ai (disjoint),
A, Ai ∈ M;
(7.10.E.18)
i=1
then r−1
∣
∣
∞
∣sA − ∑ sAi ∣ ≤ ∑ vs Ai → 0 ∣
∣
i=1
(7.10.E.19)
i=r
as r → ∞, for ∞
∞
∑ vs Ai = vs ⋃ Ai < ∞. i=1
(7.10.E.20)
i=1
(Explain!) Now, Theorem 2 of §6 extends v to a measure on a σ-field s
∗
M
⊇N ⊇M
(7.10.E.21)
(use the minimality of N ). Its restriction to N is the desired v (unique by Problem 15 in §6). A similar proof holds for s, too, if s : M → [0, ∞). The case s : M → E (C ) results via Theorem 5 and Problem 10(iii) provided S ∈ M; for then by Corollary 1, v S < ∞ ensures the finiteness of v , s , and s even on N .] ¯ ¯ ¯ s
n
n
+
s
−
s
Exercise 7.10.E. 12 Do Problem 11 for semirings M. [Hint: Use Problem 10(ii).] 7.10.E: Problems on Generalized Measures is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
7.10.E.4
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7.11: Differentiation of Set Functions 7.11: Differentiation of Set Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
7.11.1
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7.11.E: Problems on Vitali Coverings Exercise 7.11.E. 1 Prove Theorem 1 for globes, filling in all details. [Hint: Use Problem 16 in §8.]
Exercise 7.11.E. 2 Show that any (even uncountable) union of globes or nondegenerate cubes J ⊂ E is L-measurable. [Hint: Include in K each globe (cube) that lies in some J . Then Theorem 1 represents ∪J as a countable union plus a null set.] ⇒
i
i
n
I
Exercise 7.11.E. 3 Supplement Theorem 1 by proving that m
∗
(A − ⋃ I
o
k
) =0
(7.11.E.1)
and ∗
m A =m
here I
o
=
∗
(A ∩ ⋃ I
o
k
);
(7.11.E.2)
interior of I .
Exercise 7.11.E. 4 ¯ ¯¯ ¯
Fill in all proof details in Lemmas 1 and 2. Do it also for K = {globes}.
Exercise 7.11.E. 5 Given mZ = 0 and ε > 0, prove that there are open globes ∗
G
k
⊆E
n
,
(7.11.E.3)
∗
(7.11.E.4)
with ∞
Z ⊂ ⋃ G
k
k=1
and ∞ ∗
∑ mG
k
< ε.
(7.11.E.5)
k=1
[Hint: Use Problem 3(f) in §5 and Problem 16(iii) from §8.]
7.11.E.1
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Exercise 7.11.E. 6 Do Problem 3 in §5 for (i) C = {open globes}, and (ii) C = {all globes in E } . [Hints for (i): Let m = outer measure induced by v ′
′
n
′
′
: C
′
→ E
(∀A ⊆ E
To prove m A ≤ m ′
∗
n
1
.
From Problem 3(e) in §5, show that ′
)
∗
m A ≥ m A.
(7.11.E.6)
also, fix ε > 0 and an open set G ⊇ A with
A
∗
m A + ε ≥ mG (Theorem 3 of
§8).
(7.11.E.7)
Globes inside G cover A in the V -sense (why?); so A ⊆ Z ∪ ⋃ Gk (disjoint)
(7.11.E.8)
for some globes G and null set Z. With G as in Problem 5, ∗
k
k
′
∗
∗
m A ≤ ∑ (m Gk + m G ) ≤ mG + ε ≤ m A + 2ε. ]
(7.11.E.9)
k
Exercise 7.11.E. 7 Suppose f
: E
n
onto
⟷ E
n
is an isometry, i.e., satisfies ¯ ¯ ¯ ¯ ¯ ¯ |f (¯¯ x ) − f (¯ y )| = |¯¯ x −¯ y |
¯ ¯ ¯ for ¯¯ x ,¯ y ∈ E
n
.
(7.11.E.10)
Prove that (i) (∀A ⊆ E ) m A = m f [A], and (ii) A ∈ M iff f [A] ∈ M . [Hints: If A is a globe of radius r, so is f [A] (verify!); thus Problems 14 and 16 in §8 apply. In the general case, argue as in Theorem 4 of §8 , replacing intervals by globes (see Problem 6). Note that f is an isometry, too.] n
∗
∗
∗
∗
−1
Exercise 7.11.E. 7
′
From Problem 7 infer that Lebesgue measure in f (p) = p .) ¯ ¯ ¯
E
n
is rotation invariant. (A rotation about
¯ ¯ ¯
p
is an isometry
f
such that
¯ ¯ ¯
Exercise 7.11.E. 8 A V -covering K of A ⊆ E
n
¯ ¯ ¯
is called normal iff
(i) (∀I ∈ K)0 < mI = mI , and (ii) for every p ∈ A, there is some c ∈ (0, ∞) and a sequence o
¯ ¯ ¯
¯ ¯ ¯
Ik → p
({ Ik } ⊆ K)
(7.11.E.11)
such that (∀k) (∃ cube Jk ⊇ Ik )
∗
c ⋅ m Ik ≥ m Jk .
(7.11.E.12)
(We then say that p and {I } are normal; specifically, c -normal.) Prove Theorems 1 and 2 for any normal K. ¯ ¯ ¯
k
7.11.E.2
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[Hints: By Problem 21 of Chapter 3, §16, dI = dI . First, suppose K is uniformly normal, i.e., all p ∈ A are c -normal for the same c. In the general case, let ¯ ¯ ¯
¯ ¯ ¯
¯¯ ¯
¯¯ ¯
Ai = { x ∈ A| x is i-normal},
so K is uniform for A . Verify that A ↗ A . Then select, step by step, as in Theorem 1, a disjoint sequence {I i
i = 1, 2, … ;
(7.11.E.13)
i
k}
⊆K
and naturals n
1
ni
(∀i)
m
∗
< n2 < ⋯ < ni < ⋯
such that
1
( Ai − ⋃ Ik ) < k=1
.
(7.11.E.14)
i
Let ∞
U = ⋃ Ik .
(7.11.E.15)
k=1
Then (∀i)
m
∗
1 (Ai − U )
Ds) –– ––
(7.11.E.21)
and ∗ ¯ ¯¯ ¯
′
sI
′
K = {I ∈ K |I ⊆ G ,
> v}
(7.11.E.22)
mI
to prove a.e. that – D –
∗
s ≤ Ds; ––
similarly for – Ds ≤ D –
∗
Throughout assume that s : M
′
→ E
′
∗
¯ ¯¯ ¯
s
.
∗ ¯ ¯¯ ¯
(M ⊇ K ∪ K )
is a measure in E
n
,
¯ ¯¯ ¯
∗ ¯ ¯¯ ¯
finite on K ∪ K .]
Exercise 7.11.E. 10 Continuing Problems 8 and 9, verify that (a) \(\overline{\mathcal{K}}=\{\text {nondegenerate cubes \}\}\) is a normal and universal V -covering of E ; n
o ¯ ¯¯ ¯
(b) so also is K
= {all globes inE
n
;
}
¯ ¯ ¯
(c) C = {nondegenerate intervals} is normal. Note that C is not universal. ¯ ¯ ¯
Exercise 7.11.E. 11 Continuing Definition 3, we call q a derivate of s, and write q ∼ Ds(p), iff ¯ ¯ ¯
q = lim k→∞
for some sequence I Set
k
with I
¯ ¯ ¯
→ p,
k
¯ ¯¯ ¯
∈ K
sIk
(7.11.E.23)
mIk
.
D¯p ¯ ¯ = {q ∈ E
∗
¯ ¯ ¯
|q ∼ Ds(p)}
(7.11.E.24)
and prove that ¯¯¯ ¯
¯ ¯ ¯
¯ ¯ ¯
Ds(p) = min D¯p ¯ ¯ and Ds(p) = max D¯ ¯ ¯. p ––
(7.11.E.25)
Exercise 7.11.E. 12 Let K be a normal V -covering of E (see Problem 8). Given a measure s in E ∗
n
∗
¯ ¯ ¯
q ∼ D s(p)
n
,
finite on K
∗
¯ ¯¯ ¯
∪ K,
write (7.11.E.26)
iff q = lim k→∞
for some normal sequence I Set
k
¯ ¯ →¯ p ,
with I
k
∗
∈ K
sIk
(7.11.E.27)
mIk
. ∗
D¯¯¯ = {q ∈ E p
∗
∗
¯ ¯ |q ∼ D s(¯ p )} ,
(7.11.E.28)
and then
7.11.E.4
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∗
∗ ¯¯¯ ¯
∗
∗
¯ ¯ ¯ ¯ D s(¯ p ) = inf D¯¯¯ and D s(¯ p ) = sup D¯¯¯ . p p ––
(7.11.E.29)
Prove that ∗ ¯¯¯ ¯
∗
¯¯¯ ¯
Ds = D s = D s = Ds a.e. on E –– ––
[Hint: E
n
∞
=⋃
i=1
Ei ,
i,
∗
K
.
(7.11.E.30)
where ¯ Ei = {¯¯ x ∈ E
On each E
n
is uniformly normal. To prove – Ds = D – ––
∗
s
n ¯¯ ¯
| x is i-normal} .
(7.11.E.31)
a.e. on E , "imitate" Problem 9(c). Proceed.] i
7.11.E: Problems on Vitali Coverings is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
7.11.E.5
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CHAPTER OVERVIEW 8: Measurable Functions and Integration 8.1: Elementary and Measurable Functions 8.1.E: Problems on Measurable and Elementary Functions in \((S, \mathcal{M})\) 8.2: Measurability of Extended-Real Functions 8.2.E: Problems on Measurable Functions in \((S, \mathcal{M}, m)\) 8.3: Measurable Functions in \((S, \mathcal{M}, m)\) 8.3.E: Problems on Measurable Functions in \((S, \mathcal{M}, m)\) 8.4: Integration of Elementary Functions 8.4.E: Problems on Integration of Elementary Functions 8.5: Integration of Extended-Real Functions 8.5.E: Problems on Integration of Extended-Real Functions 8.6: Integrable Functions. Convergence Theorems 8.6.E: Problems on Integrability and Convergence Theorems 8.7: Integration of Complex and Vector-Valued Functions 8.7.E: Problems on Integration of Complex and Vector-Valued Functions 8.8: Product Measures. Iterated Integrals 8.8.E: Problems on Product Measures and Fubini Theorems 8.9: Riemann Integration. Stieltjes Integrals 8.9.E: Problems on Riemann and Stieltjes Integrals 8.10: Integration in Generalized Measure Spaces 8.10.E: Problems on Generalized Integration 8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition 8.11.E: Problems on Radon-Nikodym Derivatives and Lebesgue Decomposition 8.12: Integration and Differentiation 8.12.E: Problems on Differentiation and Related Topics
This page titled 8: Measurable Functions and Integration is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
1
8.1: Elementary and Measurable Functions From set functions, we now return to point functions ′
f : S → (T , ρ )
(8.1.1)
whose domain D consists of points of a set S. The range space T will mostly be E, i.e., E , E , C , E , or another normed space. We assume f (x) = 0 unless defined otherwise. (In a general metric space T , we may take some fixed element q for 0. ) Thus D is all of S , always. 1
∗
n
f
f
We also adopt a convenient notation for sets: " A(P ) " for " {x ∈ A|P (x)}. "
(8.1.2)
Thus A(f ≠ a) = {x ∈ A|f (x) ≠ a}, A(f = g) = {x ∈ A|f (x) = g(x)}, A(f > g) = {x ∈ A|f (x) > g(x)}, etc.
Definition A measurable space is a set S ≠ ∅ together with a set ring M of subsets of S, denoted (S, M) . Henceforth, (S, M) is fixed.
Definition An M-partition of a set A is a countable set family P = {A
i}
such that
A = ⋃ Ai (disjoint),
(8.1.3)
i
with A, A
i
∈ M
.
We briefly say "the partition A = ⋃ A . " i
An M-partition P
′
= { Bik }
is a refinement of P = {A
i}
′
( or P refines
(∀i)
P,
or P is finer than P) iff ′
Ai = ⋃ Bik
(8.1.4)
k
i.e., each B is contained in some A . ik
The intersection P
i
′
∩P
′′
of P
′
= { Ai }
and P
′′
= { Bk }
is understood to be the family of all sets of the form
Ai ∩ Bk ,
i, k = 1, 2, …
(8.1.5)
It is an M -partition that refines both P and P . ′
′′
Definition A map (function) f : S → T is elementary, or M-elementary, on a set A ∈ M iff there is an M-partition P = {A that f is constant (f = a ) on each A .
i}
i
If P = {A
1,
… , Aq }
is finite, we say that f is simple, or M-simple, on A.
If the A are intervals in E i
of A such
i
n
,
we call f a step function; it is a simple step function if P is finite.
The function values a are elements of T (possibly vectors). They may be infinite if T of course. i
8.1.1
=E
∗
.
Any simple map is also elementary,
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Definition A map f
′
: S → (T , ρ )
is said to be measurable (or M -measurable ) on a set A in (S, M) iff f =
lim fm
( pointwise ) on A
(8.1.6)
m→∞
for some sequence of functions f
m
all elementary on A. (See Chapter 4, §12 for "pointwise.")
: S → T,
Note 1. This implies A ∈ M, as follows from Definitions 2 and 3. (Why ? )
Corollary 8.1.1 If f
′
: S → (T , ρ )
is elementary on A, it is measurable on A.
Proof Set f
m
= f , m = 1, 2, … ,
in Definition 4. Then clearly f
on A. square
→ f
m
Corollary 8.1.2 If f is simple, elementary, or measurable on A in (S, M), it has the same property on any subset B ⊆ A with B ∈ M . Proof Let f be simple on A; so f
= ai
on A
i,
i = 1, 2, … , n,
for some finite M -partition, A = ⋃
n i=1
Ai
.
If A ⊇ B ∈ M, then {B ∩ Ai } ,
is a finite M -partition of B( why? ), and f
= ai
i = 1, 2, … , n,
on B ∩ A
(8.1.7)
so f is simple on B.
i;
For elementary maps, use countable partitions. Now let f be measurable on A, i.e., f =
lim fm
(8.1.8)
m→∞
for some elementary maps measurable on B. □
fm
on
A.
As shown above, the
fm
are elementary on
B,
too, and
fm → f
on
B;
so
f
is
Corollary 8.1.3 If f is elementary or measurable on each of the (countably many ) sets A in (S, M), it has the same property on their union A =⋃ A . n
n
n
Proof Let f be elementary on each A (so A n
n
∈ M
by Note 1) .
By Corollary 1 of Chapter 7, §1, A = ⋃ An = ⋃ Bn
for some disjoint sets B
n
⊆ An (Bn ∈ M)
By Corollary 2, f is elementary on each B
(8.1.9)
.
n;
i.e., constant on sets of some M -partition {B
ni }
of B . i
All B combined (for all n and all i) form an M-partition of A, ni
A = ⋃ Bn = ⋃ Bni . n
8.1.2
(8.1.10)
n,i
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As f is constant on each B
ni ,
it is elementary on A.
For measurable functions f , slightly modify the method used in Corollary 2. □
Corollary 8.1.4 If f : S → (T , ρ ) is measurable on continuous on f [A]. ′
A
in
(S, M),
so is the composite map
g ∘ f,
provided
′′
g : T → (U , ρ )
is relatively
Proof By assumption, f =
lim fm (pointwise)
(8.1.11)
m→∞
for some elementary maps f
m
on A.
Hence by the continuity of g , g (fm (x)) → g(f (x)),
i.e., g ∘ f
→ g∘ f
m
Moreover, all g ∘ f
m
(8.1.12)
(pointwise) on A. are elementary on A (for g ∘ f
m
is constant on any partition set, if f
m
is).
Thus g ∘ f is measurable on A, as claimed. □
Theorem 8.1.1 If the maps f , g, h : S → E (C ) are simple, elementary, or measurable on a ≠ 0) and f /h (if h ≠ 0 on A). 1
A
in
(S, M),
so are
a
f ± g, f h, |f |
(for real
Similarly for vector-valued f and g and scalar-valued h . Proof First, let f and g be elementary on A. Then there are two M-partitions, A = ⋃ Ai = ⋃ Bk ,
such that f
= ai
The sets A
∩ Bk
on A and g = b on B , say. i
k
k
(for all i and k) then form a new M -partition of A( why? ) , such that both f and g are constant on each ( why?); hence so is f ± g . i
Ai ∩ Bk
(8.1.13)
Thus f ± g is elementary on A. Similarly for simple functions. Next, let f and g be measurable on A; so f = lim fm and g = lim gm (pointwise) on A
for some elementary maps f
m,
By what was shown above, f
m
gm
(8.1.14)
.
± gm
is elementary for each m. Also, fm ± gm → f ± g( pointwise ) on A,
(8.1.15)
Thus f ± g is measurable on A. The rest of the theorem follows quite similarly. □ If the range space is E following theorem.
n
( or C
n
),
then f has n real (complex) components f
1,
8.1.3
… , fn ,
as in Chapter 4,§3 (Part II). This yields the
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Theorem 8.1.2 A function
f : S → E
f1 , f2 , … , fn
n
(C
n
is simple, elementary, or measurable on a set
)
A
in
(S, M)
iff all its
n
component functions
are.
Proof For simplicity, consider f
: S → E
2
, f = (f1 , f2 )
.
If f and f are simple or elementary on A then (exactly as in Theorem 1) , one can achieve that both are constant on sets A ∩B of one and the same M-partition of A. Hence f = (f , f ) , too, is constant on each A ∩ B , as required. 1
i
2
k
1
2
i
k
Conversely, let ¯¯
f = ci = (ai , bi ) on Ci
(8.1.16)
A = ⋃ Ci .
(8.1.17)
for some M-partition
Then by definition, f
1
In the general case (E
= ai n
and f
2
or C
n
),
= bi
on C ; so both are elementary (or simple) on A. i
the proof is analogous.
For measurable functions, the proof reduces to limits of elementary maps (using Theorem 2 of Chapter 3, §15). The details are left to the reader. □ Note 2. As C
=E
2
a complex function f
,
is simple, elementary, or measurable on A iff its real and imaginary parts are.
: S → C
By Definition 4, a measurable function is a pointwise limit of elementary maps. However, if M is a σ-ring, one can make the limit uniform. Indeed, we have the following theorem.
Theorem 8.1.3 If M is a σ-ring, and f
′
: S → (T , ρ )
is M-measurable on A, then f =
lim gm (uniformly) on A
(8.1.18)
m→∞
for some finite elementary maps g . m
Proof Thus given ε > 0, there is a finite elementary map g such that ρ (f , g) < ε on A. ′
Theorem 8.1.4 If M is a σ-ring in S, if fm → f (pointwise) on A ′
(fm : S → (T , ρ )) ,
(8.1.19)
and if all f are M -measurable on A, so also is f . m
Briefly: A pointwise limit of measurable maps is measurable (unlike continuous maps; cf. Chapter 4, §12). Proof By the second clause of Theorem 3, each f ε = 1/m, m = 1, 2, … ,
is uniformly approximated by some elementary map g
m
m
1
′
ρ (fm (x), gm (x))
0 , ′
(∃k)(∀m > k)
ρ (f (x), fm (x)) < δ.
(8.1.21)
Take k so large that, in addition, 1 (∀m > k)
< δ.
(8.1.22)
m
Then by the triangle law and by (1), we obtain for m > k that ′
′
′
ρ (f (x), gm (x)) ≤ ρ (f (x), fm (x)) + ρ (fm (x), gm (x)) .
1 n) ⋅]
Exercise 8.1.E. 4 . Let f : S → T be M-elementary on A, with M a σ -ring in S. Show the following. (i) A(f = a) ∈ M, A(f ≠ a) ∈ M . (ii) If T = E , then A(f < a), A(f ≥ a), A(f > a), and A(f ≥ a) are in M, too. (iii) (∀B ⊆ T )A ∩ f [B] ∈ M . [Hint: If ⇒ 4
∗
−1
∞
A = ⋃ Ai
(8.1.E.4)
i−1
and f
= ai on Ai , then A(f = a) is the countable union of those Ai for which ai = a. ]
8.1.E.1
https://math.libretexts.org/@go/page/24952
Exercise 8.1.E. 5 Do Problem 4(i) for measurable f . [Hint: If f = lim f for elementary maps f
m,
m
then H = A(f = a) = A (fm → a) .
(8.1.E.5)
Express H as in Problem 3, with 1 Amn = A ( hm
n) ∈ M
(8.2.3)
n=1
by (i
∗
).
And if a ∈ E , 1
∞
1
A(f ≥ a) = ⋂ A (f > a −
).
(8.2.4)
n
n=1
(Verify!) By (i ), ∗
1 A (f > a −
) ∈ M;
(8.2.5)
n
so A(f ∗
≥ a) ∈ M( a σ-ring! )
For (i
∗
∗
(ii ) ⇒ (iii ) .
.
and A ∈ M imply
)
A(f < a) = A − A(f ≥ a) ∈ M. ∗
(8.2.6)
If a ∈ E ,
∗
1
(iii ) ⇒ (iv ) .
∞
1
A(f ≤ a) = ⋂ A (f < a +
) ∈ M.
(8.2.7)
n
n=1
What if a = ±∞? ∗
∗
(i v ) ⇒ (i ) .
Indeed, (iv
∗
)
and A ∈ M imply A(f > a) = A − A(f ≤ a) ∈ M.
Thus, indeed, each of (i
∗
)
to (iv
∗
)
(8.2.8)
implies the others. To finish, we need two lemmas that are of interest in their own right.
Lemma 8.2.1 If the maps f
m
: S → E
∗
(m = 1, 2, …)
satisfy conditions (i
∗
∗
) − (iv ) ,
so also do the functions
¯ ¯¯¯¯¯ ¯
sup fm , inf fm , lim fm , and lim fm , –––
(8.2.9)
defined pointwise, i.e., (sup fm ) (x) = sup fm (x),
8.2.1
(8.2.10)
https://math.libretexts.org/@go/page/19212
and similarly for the others. Proof Let f
= sup fm .
Then ∞
A(f ≤ a) = ⋂ A (fm ≤ a)
(Why?)
(8.2.11)
m=1
But by assumption, A (fm ≤ a) ∈ M ∗
(fm satisfies (i v )) .
Thus sup f
Hence A(f
satisfies (i
∗
m
So does inf f
m;
≤ a) ∈ M( for M is a σ-ring )
(8.2.12)
.
∗
) − (i v ) .
for ∞
A (inf fm ≥ a) = ⋂ A (fm ≥ a) ∈ M.
(8.2.13)
m=1
(Explain!) ¯ ¯¯¯¯¯ ¯
So also do – lim f and lim f ––
m;
m
for by definition, lim tfm = sup gk , –––
(8.2.14)
gk = inf fm
(8.2.15)
k
where m≥k
satisfies (i
∗
as was shown above; hence so does sup g
∗
) − (i v ) ,
Similarly for lim f ¯ ¯¯¯¯¯ ¯
m.
k
.
= lim fm –––
□
Lemma 8.2.2 If f satisfies (i
∗
∗
) − (iv ) ,
then f =
lim fm (uniformly) on A
(8.2.16)
m→∞
for some sequence of finite functions f
m,
Moreover, if f
≥0
all M -elementary on A .
on A, the f can be made nonnegative, with {f
m}
m
↑
on A .
Proof Let H = A(f
= +∞), K = A(f = −∞),
and k−1 Amk = A (
m
2
k ≤f
0)
m
on A, and all is proved. □
Theorem 8.2.1 (Restated) function f them): A
: S → E
∗
is measurable on a set A ∈ M iff it satisfies one of the following equivalent conditions (hence all of ∗
(i ) (∀a ∈ E ∗
∗
∗
) A(f > a) ∈ M;
(iii ) (∀a ∈ E
∗
(ii ) (∀a ∈ E
) A(f < a) ∈ M;
∗
∗
(iv ) (∀a ∈ E
) A(f ≥ a) ∈ M;
∗
(8.2.29) ) A(f ≤ a) ∈ M.
Proof If f is measurable on A, then by definition, f
= lim fm
(pointwise) for some elementary maps f
m
8.2.3
on A.
https://math.libretexts.org/@go/page/19212
By Problem 4 (ii) in §1, all f
m
satisfy (i*)-(iv*). Thus so does f by Lemma 1, for here f
¯ ¯¯¯¯¯ ¯
= lim fm = lim fm
.
The converse follows by Lemma 2. This completes the proof. □ Note 1. Lemmas 1 and 2 prove Theorems 3 and 4 of $1, for f f : S → E (C ) . Verify! n
: S → E
∗
. By using also Theorem 2 in §1, one easily extends this to
n
Corollary 8.2.1 If f
: S → E
∗
is measurable on A, then (∀a ∈ E
∗
)
A(f = a) ∈ M and A(f ≠ a) ∈ M.
(8.2.30)
Indeed, A(f = a) = A(f ≥ a) ∩ A(f ≤ a) ∈ M
(8.2.31)
A(f ≠ a) = A − A(f = a) ∈ M.
(8.2.32)
and
Corollary 8.2.2 If f
′
: S → (T , ρ )
is measurable on A in (S, M), then A∩f
for every globe G = G
q (δ)
−1
[G] ∈ M
(8.2.33)
in (T , ρ ). ′
Proof Define h : S → E
1
by ′
h(x) = ρ (f (x), q).
(8.2.34)
Then h is measurable on A by Problem 6 in §1. Thus by Theorem 1, A(h < δ) ∈ M.
(8.2.35)
But as is easily seen, ′
A(h < δ) = {x ∈ A| ρ (f (x), q) < δ} = A ∩ f
−1
[ Gq (δ)] .
(8.2.36)
Hence the result. □
Definition Given f , g : S → E
∗
,
we define the maps f ∨ g and f ∧ g on S by (f ∨ g)(x) = max{f (x), g(x)}
(8.2.37)
(f ∧ g)(x) = min{f (x), g(x)};
(8.2.38)
and
similarly for f ∨ g ∨ h, f ∧ g ∧ h, etc. We also set f
Clearly, f
+
≥0
and f
−
≥0
on S. Also, f
=f
+
+
−f
= f ∨ 0 and f
−
−
and |f | = f
8.2.4
+
= −f ∨ 0.
+f
−
(8.2.39)
.
https://math.libretexts.org/@go/page/19212
(Why?) We now obtain the following theorem.
Theorem 8.2.2 If the functions |f | (a ≠ 0) .
f, g : S → E
∗
are simple, elementary, or measurable on
A,
so also are
f ± g, f g, f ∨ g, f ∧ g, f
+
,f
−
,
and
a
Proof If f and g are finite, this follows by Theorem 1 of §1 on verifying that 1 f ∨g =
(f + g + |f − g|)
(8.2.40)
(f + g − |f − g|)
(8.2.41)
2
and 1 f ∧g = 2
on S. (Check it!) Otherwise, consider A(f = +∞), A(f = −∞), A(g = +∞), and A(g = −∞).
(8.2.42)
By Theorem 1, these are M-sets; hence so is their union U . On each of them f ∨ g and f ∧ g equal f or g; so by Corollary 3 in §1, f ∨ g and f ∧ g have the desired properties on So also have f = f ∨ 0 and f = −f ∨ 0( take g = 0) . +
We claim that the maps above, hence on U . For example, on A(f
f ±g
= +∞)
and
fg
are simple (hence elementary and measurable) on each of the four sets mentioned
, f ± g = +∞( constant )
by our conventions (2 ) in Chapter 4, §4. For f g, split A , g < 0 on A , and g = 0 on A ; so f g = +∞ on A A(f = +∞) . ∗
1
For
2
a
|f | ,
use
A − U ∈ M.
U.
−
into three sets A , A , A ∈ M, with g > 0 on on A , and f g = 0 on A . Hence f g is simple on
A(f = +∞)
1,
3
(8.2.43)
f g = −∞
1
2
Again, the theorem holds on U , and also on Thus it holds on A = (A − U ) ∪ U by Corollary 3 in §1. □
U = A(|f | = ∞).
2
3
3
A − U,
since
f
and
g
are finite on
Note 2. Induction extends Theorem 2 to any finite number of functions. Note 3. Combining Theorem 2 with f We also obtain the following result.
=f
+
−f
−
,
we see that f
: S → E
∗
is simple (elementary, measurable) iff f
+
and f
−
are.
Theorem 8.2.3 If the functions A(f ≠ g) ∈ M
f , g : S → E∗
are measurable on
A ∈ M,
then
A(f ≥ g) ∈ M, A(f < g) ∈ M, A(f = g) ∈ M,
and
.
This page titled 8.2: Measurability of Extended-Real Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
8.2.5
https://math.libretexts.org/@go/page/19212
8.2.E: Problems on Measurable Functions in (S,M,m) (S,M,m) Exercise 8.2.E. 1 In Theorem 1, give the details in proving the equivalence of (i
∗
∗
) − (i v )
.
Exercise 8.2.E. 2 Prove Note 1.
Exercise 8.2.E. 2
′
Prove that f
=f
+
−f
−
and |f | = f
+
+f
−
.
Exercise 8.2.E. 3 Complete the proof of Theorem 2, in detail.
Exercise 8.2.E. 4 . Prove Theorem 3. [Hint: By our conventions, A(f ≥ g) = A(f − g ≥ 0) even if g or f is ±∞ for some x ∈ A. (Verify all cases!) By Theorems 1 and 2, A(f − g ≥ 0) ∈ M; so A(f ≥ g) ∈ M, and A(f < g) = A − A(f ≥ g) ∈ M. Proceed. ] ⇒ 4
Exercise 8.2.E. 5 Show that the measurability of |f | does not imply that of f . [Hint: Let f = 1 on Q and f = −1 on A − Q for some Q ∉ M(Q ⊂ A); e.g., use Q of Problem 6 in Chapter 7, §8. ]
Exercise 8.2.E. 6 . Show that a function f ≥ 0 is measurable on A iff f ↗ f (pointwise) on A for some finite simple maps . [Hint: Modify the proof of Lemma 2, setting H = A(f ≥ m) and f = m on H , and defining the A for 1 ≤ k ≤ m2 only.] ⇒ 6
m
fm ≥ 0, { fm } ↑
m
m
m
m
mk
Exercise 8.2.E. 7 . Prove Theorem 3 in §1. [Outline: By Problems 7 and 8 in ξ1, there are q ⇒ 7
i
∈ T
such that ∞
(∀n)
1
f [A] ⊆ ⋃ Gqi ( i=1
).
(8.2.E.1)
n
Set Ani = A ∩ f
by Corollary 2; so ρ (f (x), q ) < By Corollary 1 in Chapter 7, §1 ′
i
1 n
−1
1 [Gq ( i
)] ∈ M
(8.2.E.2)
n
on A . ni
8.2.E.1
https://math.libretexts.org/@go/page/25010
∞
∞
A = ⋃ Ani = ⋃ Bni (disjoint) i=1
for
some
sets
Now
Bni ∈ M, Bni ⊆ Ani .
Bni , hence on A. By Theorem 1 in Chapter 4,
(8.2.E.3)
i=1
define
§12, gn
on
gn = qi
so
Bni ;
′
ρ (f , gn )
a) ∩ A(f < b).
(8.2.E.5)
Then by Lemma 2 of Chapter 7, 2, R ⊇ G, hence R ⊇ B. (Why?) Conversely, if so, (a, ∞) ∈ R ⇒ A ∩ f
−1
[(a, ∞)] ∈ M ⇒ A(f > a) ∈ M. ]
(8.2.E.6)
Exercise 8.2.E. 9 ⇒ 9
. Do Problem 8 for f
[Hint: If f
: S → E
n
. ¯¯
and B = (a, b) ⊂ E ¯ ¯ ¯
= (f1 , … , fn )
n
¯ = (a with a
,
1,
… , an )
and ¯b = (b
1,
… , bn ) ,
show that
n
f
−1
[B] = ⋂ f
−1
k
[(ak , bk )] .
(8.2.E.7)
k=1
Apply Problem 8 to each f
k
: S → E
1
and use Theorem 2 in §1. Proceed as in Problem 8. ]
Exercise 8.2.E. 10 Do Problem 8 for f
: S → C
n
,
treating C as E n
2n
.
Exercise 8.2.E. 11 Prove that f : S → (T , ρ ) is measurable on A in (S, M) iff (i) A ∩ f [G] ∈ M for every open globe G ⊆ T , and (ii) f [A] is separable in T ( Problem 7 in §1). [Hint: If so, proceed as in Problem 7 (without assuming measurability of f ) to show that maps g on A. For the converse, use Problem 7 in §1 and Corollary 2 in §2 .] ′
−1
f = lim gn
for some elementary
n
Exercise 8.2.E. 12 (i) Show that if all of T is separable (Problem 7 in §1), there is a sequence of globes G set B ⊆ T is the union of some of these G .
k
⊆T
such that each nonempty open
k
8.2.E.2
https://math.libretexts.org/@go/page/25010
(ii) Show that E and C are separable. [Hints: (i) Use the G ( ) of Problem 8 in §1, putting them in one sequence. (ii) Take D = R ⊂ E in Problem 7 of §1. ] n
n
1
q
n
i
n
n
Exercise 8.2.E. 13 Do Problem 11 with "globe G ⊆ T replaced by "Borel set B ⊆ T ." [Hints: Treat f as f : A → T , T = f [A], noting that ′′
′
′
A∩f
−1
[B] = A ∩ f
−1
′
[B ∩ T ] .
By Problem 12, if B ≠ ∅ is open in T , then B ∩ T is a countable union of "globes" Chapter 3, §12. Proceed as in Problem 8, replacing E by T .] ′
(8.2.E.8)
Gq ∩ T
′
in (T
′
′
,ρ );
see Theorem 4 in
1
Exercise 8.2.E. 14 A map g : (T , ρ ) → (U , ρ ) is said to be of Baire class 0 (g ∈ B ) on D ⊆ T iff g is relatively continuous on D. Inductively, g is of Baire class n (g ∈ B , n ≥ 1) iff g = lim g (pointwise) on D for some maps g ∈ B . Show by induction that Corollary 4 in §1 holds also if g ∈ B on f [A] for some n. ′
′′
0
n
m
m
n−1
n
8.2.E: Problems on Measurable Functions in LibreTexts.
(S, M, m)
is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by
8.2.E.3
https://math.libretexts.org/@go/page/25010
8.3: Measurable Functions in (S,M,m) (S,M,m) I. Henceforth we shall presuppose not just a measurable space (§1) but a measure space measure on a σ -ring M ⊆ 2 .
(S, M, m),
where
m : M → E
∗
is a
S
We saw in Chapter 7 that one could often neglect sets of Lebesgue measure zero on E − if a property held everywhere except on a set of Lebesgue measure zero, we said it held "almost everywhere." The following definition generalizes this usage. n
Definition We say that a property P (x) holds for almost all x ∈ A (with respect to the measure m) or almost everywhere (a.e. (m)) on A iff it holds on A − Q for some Q ∈ M with mQ = 0 . Thus we write fn → f (a. e. ) or f = lim fn (a. e. (m)) on A
iff f
n
→ f ( pointwise )
(8.3.1)
on A − Q, mQ = 0. Of course, "pointwise" implies " a. e. " ( take Q = ∅), but the converse fails.
Definition We say that f
′
: S → (T , ρ )
is almost measurable on A iff A ∈ M and f is M -measurable on A − Q, mQ = 0 .
We then also say that f is m -measurable (m being the measure involved ) as opposed to M-measurable. Observe that we may assume Q ⊆ A here (replace Q by A ∩ Q) . *Note 1. If m is a generalized measure (Chapter 7, §11), replace Definitions 1 and 2 and in the following proofs.
mQ = 0
by
vm Q = 0 (vm = total variation of m)
in
Corollary 8.3.1 If the functions ′
fn : S → (T , ρ ) ,
n = 1, 2, …
(8.3.2)
are m-measurable on A, and if fn → f (a. e. (m))
(8.3.3)
on A, then f is m-measurable on A . Proof By assumption, f
n
→ f ( pointwise )
on A − Q
0,
m Q0 = 0.
Also, f is M-measurable on
A − Qn , m Qn = 0,
n
n = 1, 2, …
(8.3.4)
(The Q need not be the same.) n
Let ∞
Q = ⋃ Qn ;
(8.3.5)
n=0
so ∞
mQ ≤ ∑ m Qn = 0.
(8.3.6)
n=0
By Corollary 2 in §1, all f are M-measurable on A − Q (why?), and f (pointwise) on A − Q, as A − Q ⊆ A − Q . n
n
→ f
0
8.3.1
https://math.libretexts.org/@go/page/19213
Thus by Theorem 4 in §1, f is M -measurable on A − Q. As mQ = 0 , this is the desired result. □
Corollary 8.3.2 If f
on A and f is m-measurable on A, so is g .
= g(a. e. (m))
Proof By assumption, f Let Q = Q
1
=g
∪ Q2 .
on A − Q and f is M-measurable on A − Q , with mQ 1
2
1
= m Q2 = 0
.
Then mQ = 0 and g = f on A − Q. (Why? )
By Corollary 2 of §1, f is M-measurable on A − Q . Hence so is g , as claimed. □
Corollary 8.3.3 If f
is m -measurable on A, then
′
: S → (T , ρ )
f = lim fn (uniformly) on A − Q(mQ = 0),
(8.3.7)
n→∞
for some maps f
n,
all elementary on A − Q .
Proof Add proof here and it will automatically be hidden (Compare Corollary 3 with Theorem 3 in §1). Quite similarly all other propositions of §1 carry over to almost measurable (i.e., m -measurable) functions. Note, however, that the term "measurable" in §§1 and 2 always meant " M -measurable." This implies m-measurability (take Q = ∅), but the converse fails. (See Note 2, however.) We still obtain the following result.
Corollary 8.3.4 If the functions fn : S → E
∗
(n = 1, 2, …)
(8.3.8)
are m-measurable on a set A, so also are ¯ ¯¯¯¯¯ ¯
sup fn , inf fn , lim fn , and lim fn –––
(8.3.9)
(Use Lemma 1 of §2). Similarly, Theorem 2 in §2 carries over to m-measurable functions. Note 2. If m is complete (such as Lebesgue measure and LS measures) then, for measurability coincide (see Problem 3 below).
f : S → E
∗
(E
n
,C
n
) , m−
and
M
-
II. Measurability and Continuity. To study the connection between these notions, we first state two lemmas, often treated as definitions.
Lemma 8.3.1 A map f : S → E
n
(C
n
)
is M-measurable on A iff A∩f
−1
[B] ∈ M
8.3.2
(8.3.10)
https://math.libretexts.org/@go/page/19213
for each Borel set (equivalently, open set) B in E
n
(C
n
)
.
Proof See Problems 8 − 10 in §2 for a sketch of the proof.
Lemma 8.3.2 is relatively continuous on A ⊆ S iff for any open set B ⊆ (T , ρ ) , the set A ∩ f (A, ρ) as a subspace of (S, ρ). (This holds also with "open" replaced by "closed.") ′
′
A map f : (S, ρ) → (T , ρ )
−1
[B]
is open in
Proof By Chapter 4, §1, footnote 4, f is relatively continuous on A iff its restriction to A (call it g : A → T ) is continuous in the ordinary sense. Now, by Problem 15(iv)(v) in Chapter 4, §2, with S replaced by A, this means that g [B] is open (closed) in(A, ρ) when B is so in (T , ρ ) . But −1
′
g
−1
[B] = {x ∈ A|f (x) ∈ B} = A ∩ f
−1
[B].
(8.3.11)
(Why?) Hence the result follows. □
Theorem 8.3.1 Let m : M → E be a topological measure in (S, ρ). If f measurable on A . ∗
: S → E
n
(C
n
)
is relatively continuous on a set A ∈ M, it is M -
Proof Let B be open in E
n
(C
n
).
By Lemma 2, A∩f
−1
[B]
(8.3.12)
[B] = A ∩ U
(8.3.13)
is open in(A, ρ). Hence by Theorem 4 of Chapter 3, §12, A∩f
−1
for some open set U in (S, ρ) . Now, by assumption, A is in M. So is U , as M is topological (M ⊇ G) . Hence A∩f
for any open B ⊆ E
n
(C
n
).
−1
[B] = A ∩ U ∈ M
(8.3.14)
The result follows by Lemma 1. □
Note 3. The converse fails. For example, the Dirichlet function (Example (c) in Chapter 4, §1) is L-measurable (even simple) but discontinuous everywhere. Note 4. Lemma 1 and Theorem 1 hold for a map f : S → (T , ρ ) , too, provided f [A] is separable, i.e., ′
¯¯¯ ¯
f [A] ⊆ D
(8.3.15)
for a countable set D ⊆ T (cf. Problem 11 in §2).
8.3.3
https://math.libretexts.org/@go/page/19213
*III. For strongly regular measures (Definition 5 in Chapter 7, §7), we obtain the following theorem.
*Theorem 8.3.2 (Luzin). Let m : M → E be a strongly regular measure in (S, ρ). Let f Then given ε > 0, there is a closed set F ⊆ A(F ∈ M) such that ∗
′
: S → (T , ρ )
be m-measurable on A .
m(A − F ) < ε
(8.3.16)
and f is relatively continuous on F . (Note that if T = E , ρ is as in Problem 5 of Chapter 3, §11.) ∗
′
Proof By assumption, f is M-measurable on a set H = A − Q, mQ = 0;
(8.3.17)
so by Problem 7 in §1, f [H ] is separable in T . We may safely assume that f is M-measurable on S and that all of T is separable. (If not, replace S and T by H and f [H ], restricting f to H , and m to M-sets inside H ; see also Problems 7 and 8 below.) Then by Problem 12 of §2, we can fix globes G , G , … in T such that 1
2
each open set B ≠ ∅ in T is the union of a subsequence of { Gk } .
(8.3.18)
Now let ε > 0, and set Sk = S ∩ f
By Corollary 2 in §2, S
k
∈ M.
−1
[ Gk ] = f
−1
[ Gk ] ,
k = 1, 2, …
As m is strongly regular, we find for each S an open set k
Uk ⊇ Sk ,
with U
k
∈ M
(8.3.19)
(8.3.20)
and ε m (Uk − Sk )
a).
in
§2,
and
Q(f > a) ∈ M
(8.3.E.1)
if
m
is
complete.
For
§
) , use Theorem 2 of 1. ]
Exercise 8.3.E. 4 *4. Show that if m is complete and measurable on A. [Hint: Use Problem 13 in §2. ]
′
f : S → (T , ρ )
is
m
-measurable on
A
with
f [A]
separable in
T,
then
f
is
M
-
Exercise 8.3.E. 5 *5. Prove Theorem 1 for f
′
: S → (T , ρ ) ,
assuming that f [A] is separable in T .
Exercise 8.3.E. 6 Given f
n
→ f ( a.e. )
on A, prove that f
n
→ g( a.e. )
on A iff f
= g( a.e. )
on A.
Exercise 8.3.E. 7 Given A ∈ M in (S, M, m), let m be the restriction of m to A
MA = {X ∈ M|X ⊆ A}.
(8.3.E.2)
Prove that (i) (A, M , m ) is a measure space (called a subspace of (S, M, m)) ; (ii) if m is complete, topological, σ-finite or (strongly) regular, so is m . A
A
A
8.3.E.1
https://math.libretexts.org/@go/page/25013
Exercise 8.3.E. 8 ¯ ¯ (i) Show that if D ⊆ K ⊆ (T , ρ ) , then the closure of D in the subspace (K, ρ ) is K ∩ D , where D is the closure of ′
′
D
in
′
(T , ρ ) .
[Hint: Use Problem 11 in Chapter 3, §16. ] (ii) Prove that if B ⊆ K and if B is separable in (T , ρ ) , it is so in (K, ρ ) . [Hint: Use Problem 7 from ξ1.] ′
′
Exercise 8.3.E. 9 *9. Fill in all proof details in Lemma 4.
Exercise 8.3.E. 10 Simplify the proof of Theorem 2 for the case mA < ∞. [Outline: (i) First, let f be elementary, with f = a on A Given ε > 0 i
i
n
(∃n)
Ai
has a closed subset
n
F =⋃
i=1
Fi ∈ M
with
i
= mA < ∞
.
1
mA − ∑ m Ai < i=1
Each
(disjoint), ∑ mA
∈ M, A = ∪i Ai
m (Ai − Fi ) < ε/2n.
ε.
(8.3.E.3)
2
(Why?) Now use Problem 17 in Chapter 4, §8, and set
Fi .
(ii) If f is M -measurable on H = A − Q, mQ = 0, then by Theorem 3 in ξ1, f → f (uniformly) on H for some elementary maps f . By (i), each f is relatively continuous on a closed M-set with mH − mF < ε/2 ; so all f are relatively continuous on F = ⋂ F . Show that F is the required set. n
n
Fn ⊆ H ,
n
∞
n
n
n
n=1
n
Exercise 8.3.E. 11 Given f : S → (T , ρ ) , n = 1, 2, … , we say that (i) f → f almost uniformly on A ⊆ S iff ′
n
n
(∀δ > 0)(∃D ∈ M|mD < δ)
(ii) f
n
→ f
fn → f (uniformly) on A − D;
(8.3.E.4)
in measure on A iff (∀δ, σ > 0)(∃k)(∀n > k) (∃Dn ∈ M|m Dn < δ) ′
ρ (f , fn ) < σ on A − Dn .
Prove the following. (a) f → f (uniformly) n
implies
fn → f
(almost
uniformly),
and
the
latter
implies
both
fn → f ( in measure) and fn → f (a. e. ).
(b) Given f → f (almost uniformly), we have f → g (almost uniformly) iff measure. (c) If f and f are M -measurable on A, then f → f in measure on A iff n
n
n
f = g( a.e. );
similarly for convergence in
n
(∀σ > 0)
′
lim mA (ρ (f , fn ) ≥ σ) = 0.
(8.3.E.5)
n→∞
8.3.E.2
https://math.libretexts.org/@go/page/25013
Exercise 8.3.E. 12 Assuming that f : S → (T , ρ ) is m -measurable on A for n = 1, 2, … , that mA < ∞, and that f → f (a. e. ) on A, prove the following. (i) Lebesgue's theorem: f → f (in measure) on A (see Problem 11 ). (ii) Egorov's theorem: f → f (almost uniformly) on A . [Outline: (i) f and f are M -measurable on H = A − Q, mQ = 0 (Corollary 1) , with f → f (pointwise) on H . For all i, k, set ′
n
n
n
n
n
n
∞
1
′
Hi (k) = ⋂ H ( ρ (fn , f ) < n=i
by Problem 6 in §1. Show that ( ∀k)H
i (k)
) ∈ M
(8.3.E.6)
k
; hence
↗ H
lim m Hi (k) = mH = mA < ∞;
(8.3.E.7)
i→∞
so δ (∀δ > 0)(∀k) (∃ik )
m (A − Hi (k)) < k
,
(8.3.E.8)
k
2
proving (i), since (∀n > ik )
(ii) Continuing, set (∀k)D
k
= Hik (k)
1
′
ρ (fn , f )
∫
+G
and G ≥ g such that
¯ ¯¯¯ ¯
F and v > ∫ A
As f + g ≤ F
≥f
G
(8.5.E.11)
A
on A, Theorem 1(c) of §5 and Problem 6 of §4 show that ¯ ¯¯¯ ¯
∫
(f + g) ≤ ∫ A
(F + G) = ∫
A
F +∫
A
G < u + v,
(8.5.E.12)
A
contrary to ¯ ¯¯¯ ¯
u +v ≤ ∫
(f + g).
(8.5.E.13)
A
Similarly prove clause (ii).]
8.5.E.2
https://math.libretexts.org/@go/page/25014
Exercise 8.5.E. 6 Continuing Problem 5, prove that ¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
∫
(f + g) ≥ ∫ A
provided ∣∣∫
– –A
∣ g −∞.
A
(8.5.E.15)
A
(Why?) Apply Problems 5 and 4(a) to ¯ ¯¯¯ ¯
∫
((f + g) + (−g)).
(8.5.E.16)
A
Use Theorem 1 (e ) . ] ′
Exercise 8.5.E. 7 Prove the following. (i) ¯ ¯¯¯ ¯
∫
¯ ¯¯¯ ¯
|f | < ∞ iff − ∞ < ∫ A
¯ ¯ ¯
(ii) If f
A
¯¯ ¯
|f | < ∞
and ∫
A
|g| < ∞,
––
f ≤∫
f < ∞.
(8.5.E.17)
A
A
then ∣
¯ ¯¯¯ ¯
∣∫ ∣
¯ ¯¯¯ ¯
A
¯ ¯¯¯ ¯
∣
f −∫
g∣ ≤ ∫ ∣
A
|f − g|
(8.5.E.18)
|f − g|.
(8.5.E.19)
A
and ∣
∣
∣∫ ∣ ––
f −∫ A
––
¯ ¯¯¯ ¯
g∣ ≤ ∫ A
∣
A
[Hint: Use Problems 5 and 6. ]
Exercise 8.5.E. 8 Show that any signed measure s¯
f
(Note 4)
is the difference of two measures: s¯
f
= s ¯f + − s ¯f −
.
8.5.E: Problems on Integration of Extended-Real Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
8.5.E.3
https://math.libretexts.org/@go/page/25014
8.6: Integrable Functions. Convergence Theorems I. Some important theorems apply to integrable functions.
Theorem 8.6.1 (linearity of the integral) If f , g : S → E are integrable on a set A ∈ M in (S, M, m), so is ∗
pf + qg
for any p, q ∈ E
1
,
(8.6.1)
and ∫
(pf + qg) = p ∫
A
f +q ∫
A
g;
(8.6.2)
A
in particular, ∫
(f ± g) = ∫
A
f ±∫
A
g.
(8.6.3)
A
Proof By Problem 5 in §5, ¯ ¯¯¯ ¯
∫
¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
f +∫
g ≥∫
A
(f + g) ≥ ∫
A
A
––
(f + g) ≥ ∫ A
––
f +∫ A
––
g.
(8.6.4)
A
(Here ¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
∫
f, ∫ A
––
f, ∫
g, and ∫ A
A
g
(8.6.5)
A
––
are finite by integrability; so all is orthodox.) As ¯ ¯¯¯ ¯
∫
¯ ¯¯¯ ¯
f = ∫ A
––
f and ∫
g = ∫ A
A
g,
(8.6.6)
(f + g).
(8.6.7)
––
A
the inequalities turn into equalities, so that ¯ ¯¯¯ ¯
∫ A
f +∫
g =∫
A
(f + g) = ∫ A
––
A
Using also Theorem 1(e)(e') from §5, we obtain the desired result for any p, q ∈ E
1
.
□
Theorem 8.6.2 A function f
: S → E
∗
is integrable on A in (S, M, m) iff
(i) it is m-measurable on A, and ¯¯ ¯
(ii) ∫
A
¯¯ ¯
f
(equivalently ∫
A
) is finite.
|f |
Proof If these conditions hold, f is integrable on A by Theorem 3 of §5. Conversely, let ¯ ¯¯¯ ¯
∫
f = ∫ A
––
f ≠ ±∞.
(8.6.8)
A
8.6.1
https://math.libretexts.org/@go/page/19216
Using Lemma 2 in §5, fix measurable maps g and h (g ≤ f ∫
g =∫
A
) on A, with
≤h
f =∫
A
h ≠ ±∞.
(8.6.9)
A
By Theorem 3 in §5, g and h are integrable on A; so by Theorem 1, ∫
(h − g) = ∫
A
h −∫
A
g = 0.
(8.6.10)
A
As h − g ≥ h − f ≥ 0,
(8.6.11)
we get ∫
(h − f ) = 0,
(8.6.12)
A
and so by Theorem 1(h) of §5, h − f
=0
a.e. on A.
Hence f is almost measurable on A, and ∫
f ≠ ±∞
(8.6.13)
A
by assumption. From formula (1), we then get ∫
f
+
and ∫
A
−
f
< ∞,
(8.6.14)
A
and hence ∫
|f | = ∫
A
(f
+
+f
−
) =∫
A
f
A
by Theorem 1 and by Theorem 2 of §2. Thus all is proved.
+
+∫
f
−
p ∈ E
∗
,
(8.6.30)
A
there is an elementary and nonnegative map g on A such that ∫
g > p,
(8.6.31)
A
and g < f on A except only at those x ∈ A (if any) at which f (x) = g(x) = 0.
(8.6.32)
(We then briefly write g ⊂ f on A.) Proof By Lemma 1 in §5, there is an elementary and nonnegative map G ≤ f on A, with ∫
f ≥∫
G > p.
(8.6.33)
A
A
––
For the rest, proceed as in Problem 7 of §4, replacing f by G there.
□
Theorem 8.6.4 (monotone convergence) If 0 ≤ f
n
↗ f
(a.e.) on A ∈ M, i.e., 0 ≤ fn ≤ fn+1
and f
n
→ f
(∀n),
(8.6.34)
(a.e.) on A, then ¯ ¯¯¯ ¯
∫
¯ ¯¯¯ ¯
fn ↗ ∫ A
f.
(8.6.35)
A
Proof for M -measurable f and f on A. n
By Corollary 2 in §5, we may assume that f
n
By Theorem 1(c) of §5, 0 ≤ f
n
↗ f
↗ f
(pointwise) on A (otherwise, drop a null set).
implies 0 ≤∫
fn ≤ ∫
A
8.6.4
f,
(8.6.36)
A
https://math.libretexts.org/@go/page/19216
and so lim ∫ n→∞
The limit, call it p, exists in E
∗
,
as {∫
A
fn ≤ ∫
A
f.
(8.6.37)
A
It remains to show that
fn } ↑.
¯ ¯¯¯ ¯
p ≥∫
f = ∫ A
––
f.
(8.6.38)
A
(We know that ¯ ¯¯¯ ¯
∫
f = ∫ A
––
f,
(8.6.39)
A
by the assumed measurability of f ; see Theorem 3 in §5.) Suppose ∫ ––
f > p.
(8.6.40)
A
Then Lemma 1 yields an elementary and nonnegative map g ⊂ f on A, with p 0, then f (x) > g(x), so that f
n (x)
> g(x)
for large n; hence x ∈ A
n.
By Note 4 in §5, the set function s = ∫ g is a measure, hence continuous by Theorem 2 in Chapter 7, §4. Thus ∫
g = sA = lim sAn = lim ∫ n→∞
A
But as g ≤ f on A
n,
n
n→∞
g.
(8.6.44)
An
we have ∫
g ≤∫
An
fn ≤ ∫
An
fn .
(8.6.45)
A
Hence ∫
g = lim ∫
A
contrary to p < ∫
A
g.
g ≤ lim ∫
An
This contradiction completes the proof.
fn = p,
(8.6.46)
A
□
Lemma 8.6.2 (Fatou) If f
n
≥0
on A ∈ M(n = 1, 2, …), then ¯ ¯¯¯ ¯
∫ A
¯ ¯¯¯ ¯
lim fn ≤ lim ∫ ––– –––
fn .
(8.6.47)
A
Proof
8.6.5
https://math.libretexts.org/@go/page/19216
Let gn = inf fk ,
n = 1, 2, … ;
(8.6.48)
k≥n
so f
n
≥ gn ≥ 0
and {g
n}
↑
on A. Thus by Theorem 4, ¯ ¯¯¯ ¯
∫
¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
lim gn = lim ∫ A
A
gn = lim ∫ –––
A
¯ ¯¯¯ ¯
gn ≤ lim ∫ –––
fn .
(8.6.49)
A
But lim gn = sup gn = sup inf fk = lim fn .
n→∞
n
n
(8.6.50)
n
k≥n
Hence ¯ ¯¯¯ ¯
∫ A
as claimed.
¯ ¯¯¯ ¯
lim fn = ∫ –––
A
¯ ¯¯¯ ¯
lim gn ≤ lim ∫ –––
fn ,
(8.6.51)
A
□
Theorem 8.6.5 (dominated convergence) Let f
n
: S → E
be m-measurable on A ∈ M(n = 1, 2, …). Let fn → f (a.e.) on A.
(8.6.52)
lim ∫
(8.6.53)
Then
n→∞
provided that there is a map g : S → E
1
| fn − f | = 0,
A
such that ∫
g 0 such that ¯ ¯¯¯ ¯
∫ X
whenever mX < δ
(A ⊇ X, X ∈ M).
(8.6.77)
Proof By Lemma 2 in §5, fix h ≥ |f |, measurable on A, with ¯ ¯¯¯ ¯
∫
h =∫
A
|f | < ∞.
(8.6.78)
A
Neglecting a null set, we assume that |h| < ∞ on A (Corollary 1 of §5). Now, (∀n) set h(x),
x ∈ An = A(h < n),
0,
x ∈ −An .
gn (x) = {
Then g
n
Also, g
n
≤n
and g is measurable on A. (Why?)
≥0
and g
(8.6.79)
n
n
→ h
(pointwise) on A.
For let ε > 0, fix x ∈ A, and find k > h(x). Then (∀n ≥ k)
h(x) ≤ n and gn (x) = h(x).
8.6.8
(8.6.80)
https://math.libretexts.org/@go/page/19216
So (∀n ≥ k)
Clearly, g
n
≤ h.
| gn (x) − h(x)| = 0 < ε.
(8.6.81)
Hence by Theorem 5 lim ∫ n→∞
|h − gn | = 0.
(8.6.82)
A
Thus we can fix n so large that 1 ∫
(h − gn )
−∞.
(8.6.E.12)
A
(ii) If f
n
↘ f ( a.e. )
on A, then ∫ ––
fn ↘ ∫ A
––
f,
(8.6.E.13)
A
provided (∃n)
∫ ––
[Hint: Replace f , f by F , F as in Problem 1 (e ) in §5. (All is orthodox; why?)] n
n
4.
fn < ∞.
(8.6.E.14)
A
Then apply Problem 3 to
Fn ;
thus obtain (i). For (ii), use (i) and Theorem
′
Exercise 8.6.E. 6 Show by examples that (i) the conditions ¯ ¯¯¯ ¯
∫
fn > −∞ and ∫ A
––
8.6.E.2
fn < ∞
(8.6.E.15)
A
https://math.libretexts.org/@go/page/25028
in Problem 5 are essential; and (ii) Problem 5(i) fails for lower integrals. What about 5(ii)? [Hints: (i) Let A = (0, 1) ⊂ E , m = Lebesgue measure, f = −∞ on (0, ) , f = 1 elsewhere. (ii) Let M = {E , ∅} , mE = 1, m∅ = 0, f = 1 on (−n, n), f = 0 elsewhere. If f = 1 on A = E not 1
1
1
n
n
n
1
n
n
∫
fn → ∫ A
––
––
f.
1
,
then
fn → f ,
but
(8.6.E.16)
A
Explain!]
Exercise 8.6.E. 7 Given f
n
: S → E
and A ∈ M, let
∗
gn = inf fk and hn = sup fk k≥n
(n = 1, 2, …).
(8.6.E.17)
k≥n
Prove that ¯¯ ¯
(i) ∫
¯¯ ¯
A
(ii) ∫
lim fn ≤ lim ∫ fn ––– ––– – –A ¯ ¯¯¯¯¯ ¯
– –A
¯ ¯¯¯¯¯ ¯
lim fn ≤ lim ∫
– –A
provided (∃n) ∫
A
fn provided(∃n) ∫
– –A
[Hint: Apply Problem 5 to g and h (iii) Give examples for which
n.
n
and
gn > −∞;
.
hn < ∞
]
¯ ¯¯¯ ¯
¯ ¯¯¯ ¯ ¯ ¯¯¯¯¯ ¯
∫ A
lim fn ≠ limA ∫ –––
¯ ¯¯¯¯¯ ¯
fn and ∫ A
––
lim fn ≠ lim ∫ –––
A
fn .
––
(8.6.E.18)
A
(See Note 2).
Exercise 8.6.E. 8 Let f ≥ 0 on A ∈ M and f Prove the following. (i) If n
n
→ f ( a.e. )
on A. Let A ⊇ X, X ∈ M.
¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
∫
fn → ∫ A
f < ∞,
(8.6.E.19)
A
then ¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
∫
fn → ∫ X
f.
(8.6.E.20)
X
(ii) This fails for sign-changing f . [Hints: If (i) fails, then n
¯ ¯¯¯ ¯
lim ∫ –––X
¯ ¯¯¯ ¯
fn < ∫ X
X
¯ ¯¯¯ ¯
f or lim ∫ –––X
¯ ¯¯¯ ¯
fn > ∫ X
f.
(8.6.E.21)
X
Find a subsequence of
8.6.E.3
https://math.libretexts.org/@go/page/25028
¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
{∫
fn } or { ∫ X
contradicting Lemma 2. (ii) Let m = Lebesgue measure; A = (0, 1), X = (0,
fn }
(8.6.E.22)
A−X
1
,
)
2
1
n
on (0,
−n
on (1 −
fn = {
],
2n
(8.6.E.23)
1 2n
, 1[ .
Exercise 8.6.E. 9 ⇒ 9
. (i) Show that if f and g are m-measurable and nonnegative on A, then (∀a, b ≥ 0)
∫
(af + bg) = a ∫
A
(ii) If, in addition, ∫ f < ∞ or ∫ [Hint: Proceed as in Theorem 1.] A
A
g < ∞,
f +b ∫
A
this formula holds for any a, b ∈ E
g.
(8.6.E.24)
A
1
.
Exercise 8.6.E. 10 ⇒ 10
. If ∞
f = ∑ fn ,
(8.6.E.25)
n=1
with all f measurable and nonnegative on A, then n
∞
∫
f = ∑∫
A
n=1
fn .
(8.6.E.26)
A
[Hint: Apply Theorem 4 to the maps n
gn = ∑ fk ↗ f .
(8.6.E.27)
k=1
Use Problem 9.]
Exercise 8.6.E. 11 If ∞
q = ∑∫ n=1
| fn | < ∞
(8.6.E.28)
A
and the f are m-measurable on A, then n
∞
∑ | fn | < ∞(a. e. ) on A
(8.6.E.29)
n=1
8.6.E.4
https://math.libretexts.org/@go/page/25028
and f
∞
=∑
n=1
fn
is m-integrable on A, with ∞
∫
f = ∑∫
A
[Hint: Let g = ∑
∞ n=1
| fn | .
fn .
(8.6.E.30)
| fn | = q < ∞;
(8.6.E.31)
A
n=1
By Problem 10, ∞
∫
g = ∑∫
A
n=1
A
so g < ∞(a. e. ) on A. (Why?) Apply Theorem 5 and Note 1 to the maps n
gn = ∑ fk ;
(8.6.E.32)
k=1
note that |g
n|
≤ g. ]
Exercise 8.6.E. 12 (Convergence in measure; see Problem 11(ii) of §3). (i) Prove Riesz' theorem: If f → f in measure on uniformly), hence (a.e.), on A . [Outline: Taking n
A ⊆S
, there is a subsequence
−k
σk = δk = 2
,
{ fnk }
such that
fnk → f
(almost
(8.6.E.33)
pick, step by step, naturals n1 < n2 < ⋯ < nk < ⋯
and sets D
k
∈ M
(8.6.E.34)
such that (∀k) −k
m Dk < 2
(8.6.E.35)
and ′
−k
ρ (fnk , f ) < 2
on A − D
k.
(8.6.E.36)
(Explain!) Let ∞
En = ⋃ Dk ,
(8.6.E.37)
k=n
1−n
m En < 2
.(
Why?) Show that (∀n)(∀k > n)
on A − E . Use Problem 11 in §3. ] (ii) For maps f : S → E and g : S → E
′
1−n
ρ (fnk , f ) < 2
(8.6.E.38)
n
n
1
deduce that if fn → f
8.6.E.5
(8.6.E.39)
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in measure on A and (∀n)
| fn | ≤ g( a.e. ) on A,
(8.6.E.40)
then |f | ≤ g( a.e. ) on A.
(8.6.E.41)
fn → f
(8.6.E.42)
[ Hint: fnk → f (a. e. ) on A. ]
Exercise 8.6.E. 13 Continuing Problem 12( ii ), let
in measure on A ∈ M (f
n
: S → E)
and (∀n)
| fn | ≤ g(a. e.) on A,
(8.6.E.43)
with ¯ ¯¯¯¯ ¯
∫
g < ∞.
(8.6.E.44)
A
Prove that ¯ ¯¯¯ ¯
lim ∫
| fn − f | = 0.
n→∞
(8.6.E.45)
A
Does ¯ ¯¯¯ ¯
¯ ¯¯¯ ¯
∫
fn → ∫
f?
A
(8.6.E.46)
A
[Outline: From Corollary 1 of §5, infer that g = 0 on A − C , where ∞
C = ⋃ Ck (disjoint),
(8.6.E.47)
k=1
m Ck < ∞.
(We may assume gM -measurable on A. Why?) Also, ∞
∞ >∫
g =∫
A
g+∫
A−C
g = 0 +∑∫
C
k=1
g;
(8.6.E.48)
g,
(8.6.E.49)
Ck
so the series converges. Hence p
(∀ε > 0)(∃p)
∫
g−ε < ∑∫
A
k=1
8.6.E.6
Ck
g =∫ H
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where p
H = ⋃ Ck ∈ M
(8.6.E.50)
k=1
and mH < ∞. As |f
n
− f | ≤ 2g( a.e. ),
we get ¯ ¯¯¯ ¯
(1) ∫ ––
¯ ¯¯¯ ¯
| fn − f | ≤ ∫
| fn − f | + ∫
A
A
¯¯¯¯¯ ¯
| fn − f | ≤ ∫ H
2g < ∫
A−H
| fn − f | + 2ε.
(8.6.E.51)
H
(Explain!) As mH < ∞, we can fix σ > 0 with σ ⋅ mH < ε.
(8.6.E.52)
Also, by Theorem 6, fix δ such that 2∫
g n0 )
m Dn < δ
(8.6.E.54)
and | fn − f | < σ on An = H − Dn .
(We may use the standard metric, as |f | and |f
n|
¯¯¯¯¯ ¯
| fn − f | A
a.e. Why?) Thus from (1), we get
0) (∃k) (∀n > k)
| fn − f | < ε on A;
(8.7.10)
so (∀n > k)
∫
| fn − f | ≤ ∫
A
As ε is arbitrary, the result follows.
(ε) = ε ⋅ mA < ∞.
(8.7.11)
A
□
Lemma 8.7.2 If ∫
|f | < ∞
(mA < ∞)
(8.7.12)
A
and f = lim fn (uniformly) on A − Q (mQ = 0)
(8.7.13)
n→∞
for some elementary maps f on A, then all but finitely many f are elementary and integrable on A, and n
n
lim ∫ n→∞
fn
(8.7.14)
A
exists in E; further, the latter limit does not depend on the sequence {f
.
n}
Proof By Lemma 1, (∀ε > 0) (∃q) (∀n, k > q)
∫
| fn − f | < ε and ∫
A
| fn − fk | < ε.
(8.7.15)
A
(The latter can be achieved since
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lim ∫ k→∞
| fn − fk | = ∫
A
| fn − f | < ε. )
(8.7.16)
A
Now, as | fn | ≤ | fn − f | + |f |,
(8.7.17)
Problem 7 in §5 yields (∀n > k)
∫
| fn | ≤ ∫
A
| fn − f | + ∫
A
|f | < ε + ∫
A
|f | < ∞.
(8.7.18)
A
Thus f is elementary and integrable for n > k, as claimed. Also, by Theorem 2 and Corollary 1(ii), both in §4, n
∣ ∣ ∣ ∣ ∣∫ fn − ∫ fk ∣ = ∣∫ (fn − fk )∣ ≤ ∫ | fn − fk | < ε. ∣ A ∣ ∣ A ∣ A A
(∀n, k > q)
Thus {∫
A
fn }
(8.7.19)
is a Cauchy sequence. As E is complete, lim ∫
fn ≠ ±∞
(8.7.20)
A
exists in E, as asserted. Finally, suppose g → f (uniformly) on above, lim ∫ g exists, and
A−Q
n
A
for some other elementary and integrable maps
gn .
By what was shown
n
∣ ∣ ∣ ∣ ∣lim ∫ gn − lim ∫ fn ∣ = ∣lim ∫ (gn − fn )∣ ≤ lim ∫ | gn − fn − 0| = 0 ∣ ∣ ∣ ∣ A A A A
by Lemma 1, as g
n
− fn → 0
(uniformly) on A. Thus lim ∫
gn = lim ∫
A
and all is proved.
(8.7.21)
fn ,
(8.7.22)
A
□
This leads us to the following definition.
Definition If f
: S → E
is integrable on A ∈ M
(mA < ∞),
∫
we set
f =∫
A
for any elementary and integrable maps f such that f n
f dm = lim ∫ n→∞
A
n
→ f
fn
(8.7.23)
A
(uniformly) on A − Q, mQ = 0 .
Indeed, such maps exist by Theorem 3 of §1, and Lemma 2 excludes ambiguity. *Note 1. If f itself is elementary and integrable, Definition 2 agrees with that of §4. For, choosing f
n
∫
f =∫
A
= f (n = 1, 2, …),
fn
we get (8.7.24)
A
(the latter as in §4). *Note 2. We may neglect sets on which f = 0 on A − B in Definition 2. Then
f = 0,
along with null sets. For if
f =0
on A − B
(A ⊇ B, B ∈ M),
we may choose
n
∫ A
f = lim ∫
fn = lim ∫
A
B
fn = ∫
f.
(8.7.25)
B
Thus we now define
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∫
f =∫
A
even if mA = ∞, provided f
=0
f,
(8.7.26)
B
on A − B, i.e., f = f CB on A
(CB =
(8.7.27)
characteristic function of B), with A ⊇ B, B ∈ M, and mB < ∞ .
If such a B exists, we say that f has m-finite support in A . *Note 3. By Corollary 1 in §5, ∫
|f | < ∞
(8.7.28)
A
implies that A(f
≠ 0)
is σ-finite. Neglecting A(f
= 0),
we may assume that
A = ⋃ Bn , m Bn < ∞, and { Bn } ↑
(if not, replace B by ∪
n
n
Bk
k=1
); so B
n
↗ A
(8.7.29)
.
Lemma 8.7.3 Let ϕ : S → E be integrable on A . Let B
n
↗ A, m Bn < ∞
and set
fn = ϕCBn ,
Then f
→ ϕ
n
n = 1, 2, … .
(8.7.30)
(pointwise) on A, all f are integrable on A, and n
lim ∫ n→∞
fn
(8.7.31)
A
exists in E. Furthermore, this limit does not depend on the choice of {B
.
n}
Proof Fix any x ∈ A. As B
n
↗ A = ∪Bn
, (∃n0 ) (∀n > n0 )
By assumption, f
n
=ϕ
on B
n.
n
→ ϕ
(8.7.32)
fn (x) = ϕ(x);
(8.7.33)
Thus (∀n > n0 )
so f
x ∈ Bn .
(pointwise) on A.
Moreover, f
n
= ϕCBn
is m-measurable on A (as ϕ and C
Bn
are); and
| fn | = |ϕ| CB
(8.7.34)
n
implies ∫
| fn | ≤ ∫
A
|ϕ| < ∞.
(8.7.35)
A
Thus all f are integrable on A. n
As f
n
=0
on A − B
n (mB
< ∞)
, ∫
fn
(8.7.36)
A
is defined. Since f
n
→ ϕ
(pointwise) and |f
n|
≤ |ϕ|
on A, Theorem 5 in §6, with g = |ϕ|, yields ∫
| fn − ϕ| → 0.
(8.7.37)
A
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The rest is as in Lemma 2, with our present Theorem 2 below (assuming m-finite support of f and g), replacing Theorem 2 of §4. Thus all is proved. □
Definition If ϕ : S → E is integrable on A ∈ M, we set ∫
ϕ =∫
A
ϕdm = lim ∫ n→∞
A
fn ,
(8.7.38)
A
with the f as in Lemma 3 (even if ϕ has no m-finite support). n
Theorem 8.7.2 (linearity) If f , g : S → E are integrable on A ∈ M, so is pf + qg
(8.7.39)
for any scalars p, q. Moreover, ∫
(pf + qg) = p ∫
A
f +q ∫
A
g.
(8.7.40)
A
Furthermore if f and g are scalar valued, p and q may be vectors in E . Proof For the moment, f , g denotes mappings with m-finite support in A. Integrability is clear since pf + qg is measurable on A (as f and g are), and |pf + qg| ≤ |p||f | + |q||g|
(8.7.41)
yields ∫
|pf + qg| ≤ |g| ∫
A
|f | + |q| ∫
A
|g| < ∞.
(8.7.42)
A
Now, as noted above, assume that f = f CB and g = gCB 1
for some B
1,
B2 ⊆ A(m B1 + m B2 < ∞).
Let B = B
1
(8.7.43)
2
so
∪ B2 ;
f = g = pf + qg = 0 on A − B;
(8.7.44)
additionally, ∫
f =∫
A
f, ∫
B
A
g =∫
g, and ∫
(pf + qg) = ∫
B
A
f = lim ∫
fn and ∫
(pf + qg).
(8.7.45)
B
Also, mB < ∞; so by Definition 2, ∫ B
B
B
g = lim ∫
gn
(8.7.46)
B
for some elementary and integrable maps fn → f (uniformly) and gn → g (uniformly) on B − Q, mQ = 0.
(8.7.47)
p fn + q gn → pf + qg (uniformly) on B − Q.
(8.7.48)
Thus
But by Theorem 2 and Corollary 1(vii), both of §4 (for elementary and integrable maps),
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∫
(p fn + q gn ) = p ∫
B
fn + q ∫
B
gn .
(8.7.49)
B
Hence ∫
(pf + qg) = ∫
A
(pf + qg) = lim ∫
B
(p fn + q gn )
B
= lim (p ∫
fn + q ∫
B
gn ) = p ∫
B
f +q ∫
B
g =p∫
B
A
f +q ∫
g.
A
This proves the statement of the theorem, provided f and g have m-finite support in A. For the general case, we now resume the notation f , g, … for any functions, and extend the result to any integrable functions. Using Definition 3, we set ∞
A = ⋃ Bn , { Bn } ↑, m Bn < ∞,
(8.7.50)
n=1
and fn = f CB , gn = gCB , n
n = 1, 2, … .
n
(8.7.51)
Then by definition, ∫
f = lim ∫ n→∞
A
fn and ∫
A
g = lim ∫ n→∞
A
gn ,
(8.7.52)
gn ) .
(8.7.53)
A
and so p∫
f +q ∫
A
As f
n,
gn
g = lim (p ∫ n→∞
A
fn + q ∫
A
A
have m-finite supports, the first part of the proof yields p∫
fn + q ∫
A
gn = ∫
A
(p fn + q gn ) .
(8.7.54)
A
Thus as claimed, p∫
f +q ∫
A
g = lim ∫
A
(p fn + q gn ) = ∫
A
(pf + qg).
□
(8.7.55)
A
Similarly, one extends Corollary 1(ii)(iii)(v) of §4 first to maps with parts of that corollary need no new proof. (Why?)
m
-finite support, and then to all integrable maps. The other
Theorem 8.7.3 (additivity) (i) If f
: S → E
is integrable on each of n disjoint M-sets A
k,
it is so on their union n
A = ⋃ Ak ,
(8.7.56)
k=1
and n
∫ A
f = ∑∫ k=1
f.
(8.7.57)
Ak
(ii) This holds for countable unions, too, if f is integrable on all of A. Proof Let f have m-finite support: f
= f CB
on A, mB < ∞. Then
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∫
f =∫
A
f and ∫
f,
(8.7.58)
k = 1, 2, … , n.
(8.7.59)
B
f =∫
Ak
Bk
where Bk = Ak ∩ B,
By Definition 2, fix elementary and integrable maps B − Q (hence also on B − Q ), with
fi
(on
A
) and a set
such that
Q (mQ = 0)
fi → f
(uniformly) on
k
∫
f =∫
A
f = lim ∫ i→∞
B
fi
and ∫
B
f = lim ∫ i→∞
Ak
fi ,
k = 1, 2, … , n.
(8.7.60)
Bk
As the f are elementary and integrable, Theorem 1 in §4 yields i
n
∫
fi = ∫
A
n
fi = ∑ ∫
B
k=1
fi = ∑ ∫
Bk
k=1
fi .
(8.7.61)
Ak
Hence n
∫
f = lim ∫ i→∞
A
n
fi = lim ∑ ∫ i→∞
B
Bk
k=1
n
fi = ∑ ( lim ∫ i→∞
k=1
fi ) = ∑ ∫
Ak
k=1
f.
(8.7.62)
Ak
Thus clause (i) holds for maps with m-finite support. For other functions, (i) now follows quite similarly, from Definition 3. (Verify!) As for (ii), let f be integrable on ∞
A = ⋃ Ak (disjoint),
Ak ∈ M.
(8.7.63)
k=1
In this case, set g
n
= f CBn ,
where B
n
n
=⋃
k=1
Ak , n = 1, 2, …
. By clause (i), we have
n
∫
gn = ∫
A
since g
n
=f
on each A
⊆ Bn
k
Also, as is easily seen, |g
n|
n
gn = ∑ ∫
Bn
k=1
gn = ∑ ∫
Ak
k=1
f,
(8.7.64)
Ak
.
≤ |f |
on A and g
n
→ f
(pointwise) on A (proof as in Lemma 3). Thus by Theorem 5 in §6, ∫
| gn − f | → 0.
(8.7.65)
A
As ∣ ∣ ∣ ∣ ∣∫ gn − ∫ f ∣ = ∣∫ (gn − f )∣ ≤ ∫ | gn − f | , ∣ A ∣ ∣ A ∣ A A
(8.7.66)
we obtain ∫ A
and the result follows by (3).
f = lim ∫ n→∞
gn ,
(8.7.67)
A
□
8.7: Integration of Complex and Vector-Valued Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
8.7.7
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8.7.E: Problems on Integration of Complex and Vector-Valued Functions Exercise 8.7.E. 1 Prove Corollary 1( iii )− (vii) in §4 componentwise for integrable maps f
: S → E
n
(C
n
).
Exercise 8.7.E. 2 Prove Theorems 2 and 3 componentwise for E = E
n
(C
n
)
.
Exercise 8.7.E. 2
′
Do it for Corollary 3 in §6.
Exercise 8.7.E. 3 Prove Theorem 1 with ∫
|f | < ∞
(8.7.E.1)
A
replaced by ∫
| fk | < ∞,
k = 1, … , n.
(8.7.E.2)
A
Exercise 8.7.E. 4 Prove that if f
: S → E
n
(C
n
)
is integrable on A, so is |f |. Disprove the converse.
Exercise 8.7.E. 5 Disprove Lemma 1 for mA = ∞ .
Exercise 8.7.E. ∗6 Complete the proof of Lemma 3.
Exercise 8.7.E. ∗7 Complete the proof of Theorem 3.
Exercise 8.7.E. ∗8 Do Problem 1 and 2 for f ′
: S → E
.
Exercise 8.7.E. ∗9 Prove formula (1) from definitions of Part II of this section.
8.7.E.1
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Exercise 8.7.E. 10 ⇒ 10
. Show that ∣ ∣∫ ∣
∣ f∣ ≤ ∫ ∣
A
|f |
(8.7.E.3)
A
for integrable maps f : S → E. See also Problem 14. [Hint: If mA < ∞, use Corollary 1( ii ) of §4 and Lemma 1. If mA = ∞, us imitate" the proof of Lemma 3. ]
Exercise 8.7.E. 11 Do Problem 11 in §6 for f
n
: S → E.
Do it componentwise for E =
E
n
(C
n
).
Exercise 8.7.E. 12 Show that if f , g : S → E
1
(C )
are integrable on A, then ∣ ∣∫ ∣
A
∣ f g∣ ∣
2 2
≤∫
|f |
A
⋅∫
2
|g| .
(8.7.E.4)
A
In what case does equality hold? Deduce Theorem 4 (c ) in Chapter 3, §§1-3, from this result. [Hint: Argue as in that theorem. Consider the case ′
(∃t ∈ E
1
)
∫
|f − tg| = 0. ]
(8.7.E.5)
A
Exercise 8.7.E. 13 Show that if f
: S → E
1
(C )
is integrable on A and ∣ ∣ ∣∫ f∣ = ∫ |f |, ∣ A ∣ A
(8.7.E.6)
then (∃c ∈ C )
[Hint: Let a = ∫
A
f.
cf = |f |
a.e. on A.
(8.7.E.7)
The case a = 0 is trivial. If a ≠ 0, let |a| c =
; |c| = 1; ca = |a|.
(8.7.E.8)
a
Let r = (cf )
re .
Show that r ≤ |cf | = |f | , ∣ ∣ ∣ ∣ ∣∫ f∣ = ∫ cf = ∫ r ≤∫ |f | = ∣∫ f∣ , ∣ A ∣ ∣ A ∣ A A A ∫
|f | = ∫
A
r =∫
A
(cf )re,
A
(cf )re = |cf |(a. e.), and cf = |cf | = |f | a.e. on A. ]
8.7.E.2
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Exercise 8.7.E. 14 Do Problem 10 for E = C using the method of Problem 13.
Exercise 8.7.E. 15 Show that if f
: S → E
is integrable on A, it is integrable on each M-set B ⊆ A. If, in addition, ∫
f =0
(8.7.E.9)
B
for all such B, show that f = 0 a.e. on A. Prove it for E = E first. [Hint for E = E : A = A(f > 0) ∪ A(f ≤ 0). Use Theorems 1(h) and 2 from §5. ] n
∗
Exercise 8.7.E. 16 In Problem 15, show that s =∫
f
(8.7.E.10)
is a σ-additive set function on MA = {X ∈ M|X ⊆ A}.
(8.7.E.11)
(Note 4 in §5); s is called the indefinite integral of f in A. 8.7.E: Problems on Integration of Complex and Vector-Valued Functions is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
8.7.E.3
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8.8: Product Measures. Iterated Integrals Let (X, M, m) and (Y , N , n) be measure spaces, with X ∈ M and Y
∈ N.
Let C be the family of all "rectangles," i.e., sets
A × B,
(8.8.1)
with A ∈ M, B ∈ N , mA < ∞, and nB < ∞ . Define a premeasure s : C → E
1
by s(A × B) = mA ⋅ nB,
A × B ∈ C.
(8.8.2)
Let p be the s -induced outer measure in X × Y and ∗
∗
p : P
→ E
∗
(8.8.3)
the p -induced measure ("product measure," p = m × n ) on the σ-field P of all p -measurable sets in X × Y (Chapter 7, §§5-6). ∗
∗
We consider functions f
: X ×Y → E
∗
(extended-real).
∗
I. We begin with some definitions.
Definitions (1) Given a function f
: X → Y → E
∗
(of two variables x, y), let f or f (x, ⋅) denote the function on Y given by x
fx (y) = f (x, y);
(8.8.4)
it arises from f by fixing x. Similarly, f or f (⋅, y) is given by f y
(2) Define g : X → E
∗
y
.
(x) = f (x, y)
by g(x) = ∫
f (x, ⋅)dn,
(8.8.5)
Y
and set ∫
∫
X
f dndm = ∫
Y
gdm,
(8.8.6)
X
also written ∫
dm(x) ∫
X
f (x, y)dn(y).
(8.8.7)
Y
This is called the iterated integral of f on Y and X, in this order. Similarly, h(y) = ∫
f
y
dm
(8.8.8)
X
and ∫
∫
Y
f dmdn = ∫
X
hdn.
(8.8.9)
Y
Note that by the rules of §5, these integrals are always defined. (3) With f , g, h as above, we say that f is a Fubini map or has the Fubini properties (after the mathematician Fubini) iff (a) g is m-measurable on X and h is n -measurable on Y ; (b)
-measurable on − Q , nQ = 0; and
fx
y ∈ Y
is
n
′
Y
for almost all
x
(i.e., for
x ∈ X −Q
,
mQ = 0); f
y
is
m
-measurable on
X
for
′
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(c) the iterated integrals above satisfy ∫
∫
X
f dndm = ∫
Y
∫
Y
f dmdn = ∫
X
f dp
(8.8.10)
X×Y
(the main point). For monotone sequences fk : X × Y → E
∗
(k = 1, 2, …),
(8.8.11)
we now obtain the following lemma.
Lemma 8.8.1 If 0 ≤ f
↗ f
k
(pointwise) on X × Y and if each f has Fubini property (a), (b), or (c), then f has the same property. k
Proof For k = 1, 2, … , set gk (x) = ∫
fk (x, ⋅)dn
(8.8.12)
fk (⋅, y)dm.
(8.8.13)
Y
and hk (y) = ∫ X
By assumpsion, 0 ≤ fk (x, ⋅) ↗ f (x, ⋅)
(8.8.14)
pointwise on Y . Thus by Theorem 4 in §6, ∫
fk (x, ⋅) ↗ ∫
Y
i.e., g
k
↗ g
f (x, ⋅)dn,
(8.8.15)
Y
(pointwise) on X, with g as in Definition 2.
Again, by Theorem 4 of §6, ∫
gk dm ↗ ∫
X
gdm;
(8.8.16)
X
or by Definition 2, ∫
∫
X
f dndm = lim ∫ k→∞
Y
∫
X
fk dndm.
(8.8.17)
Y
Similarly for ∫
∫
Y
f dmdn
(8.8.18)
f dp.
(8.8.19)
X
and ∫ X×Y
Hence f satisfies (c) if all f do. k
Next, let f have property (b); so (∀k)f k
k (x,
⋅)
is n -measurable on Y if x ∈ X − Q (mQ k
k
=0
). Let
∞
Q = ⋃ Qk ;
(8.8.20)
k=1
8.8.2
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so mQ = 0, and all f
k (x,
⋅)
are n -measurable on Y , for x ∈ X − Q. Hence so is f (x, ⋅) = lim fk (x, ⋅).
(8.8.21)
k→∞
Similarly for f (⋅, y). Thus f satisfies (b). Property (a) follows from g
k
→ g
and h
→ h.
k
□
Using Problems 9 and 10 from §6, the reader will also easily verify the following lemma.
Lemma 8.8.2 (i) If f and f are nonnegative, p-measurable Fubini maps, so is af 1
2
1
+ b f2
for a, b ≥ 0 .
(ii) If, in addition, ∫
f1 dp < ∞ or ∫
X×Y
then f
1
f2 dp < ∞,
(8.8.22)
X×Y
is a Fubini map, too
− f2
Lemma 8.8.3 Let f
∞
=∑
i=1
fi
(pointwise), with f
i
≥0
on X × Y .
(i) If all f are p-measurable Fubini maps, so is f . i
(ii) If the f have Fubini properties (a) and (b), then i
∞
∫
∫
X
Y
f dndm = ∑ ∫
∫
X
i=1
fi dndm
(8.8.23)
fi dmdn.
(8.8.24)
Y
and ∞
∫
∫
Y
X
f dmdn = ∑ ∫ i=1
∫
Y
X
II. By Theorem 4 of Chapter 7, §3, the family C (see above) is a semiring, being the product of two rings, {A ∈ M|mA < ∞} and {B ∈ N |nB < ∞}.
(Verify!) Thus using Theorem 2 in Chapter 7, §6, we now show that p is an extension of s : C → E
(8.8.25) 1
.
Theorem 8.8.1 The product premeasure s is σ-additive on the semiring C. Hence (i) C ⊆ P
∗
and p = s < ∞ on C, and
(ii) the characteristic function C of any set D ∈ C is a Fubini map. D
Proof Let D = A × B ∈ C; so CD (x, y) = CA (x) ⋅ CB (y).
(Why?) Thus for a fixed x, C
D (x,
⋅)
is just a multiple of the N -simple map C
B,
g(x) = ∫
CD (x, ⋅)dn = CA (x) ⋅ ∫
Y
so g = C
A
⋅ nB
(8.8.26)
hence n -measurable on Y . Also,
CB dn = CA (x) ⋅ nB;
(8.8.27)
Y
is M-simple on X, with
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∫
∫
X
Similarly for C
D (⋅,
y),
CD dndm = ∫
Y
gdm = nB ∫
X
CA dm = nB ⋅ mA = sD.
(8.8.28)
X
and h(y) = ∫
CD (⋅, y)dm.
(8.8.29)
X
Thus C has Fubini properties (a) and (b), and for every D ∈ C D
∫
∫
X
CD dndm = ∫
Y
∫
Y
CD dmdn = sD.
(8.8.30)
X
To prove σ -additivity, let ∞
D = ⋃ Di (disjoint), Di ∈ C;
(8.8.31)
i=1
so ∞
CD = ∑ CDi .
(8.8.32)
i=1
(Why?) As shown above, each C
Di
has Fubini properties (a) and (b); so by (1) and Lemma 3, ∞
sD = ∫ X
∫
∞
CD dndm = ∑ ∫
Y
∫
X
i=1
CDi dndm = ∑ sDi ,
Y
(8.8.33)
i=1
as required. Assertion (i) now follows by Theorem 2 in Chapter 7, §6. Hence sD = pD = ∫
CD dp;
(8.8.34)
X×Y
so by formula (1), C also has Fubini property (c), and all is proved. D
□
Next, let P be the σ-ring generated by the semiring C (so C ⊆ P ⊆ P ). ∗
Lemma 8.8.4 P
is the least set family R such that
(i) R ⊇ C ; (ii) R is closed under countable disjoint unions; and (iii) H − D ∈ R if D ∈ R and D ⊆ H , H ∈ C . This is simply Theorem 3 in Chapter 7, §3, with changed notation.
Lemma 8.8.5 If D ∈ P (σ-generated by C), then C is a Fubini map. D
Proof Let R be the family of all D ∈ P such that C is a Fubini map. We shall show that R satisfies (i)-(iii) of Lemma 4, and so D
P ⊆ R.
(ii) Let
8.8.4
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∞
D = ⋃ Di (disjoint),
Di ∈ R.
(8.8.35)
i=1
Then ∞
CD = ∑ CD ,
(8.8.36)
i
i=1
and each C
Di
is a Fubini map. Hence so is C by Lemma 3. Thus D ∈ R , proving (ii). D
(iii) We must show that C
H −D
is a Fubini map if C is and if D ⊆ H , H ∈ C. Now, D ⊆ H implies D
CH −D = CH − CD .
(8.8.37)
CH dp = pH = sH < ∞,
(8.8.38)
(Why?) Also, by Theorem 1, H ∈ C implies ∫ X×Y
and C is a Fubini map. So is C by assumption. So also is H
D
CH −D = CH − CD
(8.8.39)
by Lemma 2(ii). Thus H − D ∈ R, proving (iii). By Lemma 4, then, P ⊆ R. Hence (∀D ∈ P)C
is a Fubini map.
D
□
We can now establish one of the main theorems, due to Fubini.
Theorem 8.8.2 (Fubini I) Suppose f (i) f
≥0
: X ×Y → E
∗
is P -measurable on X × Y (P as above) rom. Then f is a Fubini map if either
on X × Y , or
(ii) one of ∫
|f |dp, ∫
X×Y
∫
X
|f |dndm, or ∫
Y
∫
Y
|f |dmdn
(8.8.40)
X
is finite. In both cases, ∫ X
∫
f dndm = ∫
Y
∫
Y
f dmdn = ∫
X
f dp.
(8.8.41)
X×Y
Proof First, let ∞
f = ∑ ai CD
i
(ai ≥ 0, Di ∈ P) ,
(8.8.42)
i=1
i.e., f is P -elementary, hence certainly p -measurable. (Why?) By Lemmas 5 and 2, each a is f (Lemma 3). Formula (2) is simply Fubini property (c).
i CDi
Now take any P -measurable f
≥ 0.
is a Fubini map. Hence so
By Lemma 2 in §2, f = lim fk on X × Y
(8.8.43)
k→∞
for some sequence {f } ↑ of P -elementary maps, Lemma 1. This settles case (i). k
fk ≥ 0.
As shown above, each
8.8.5
fk
is a Fubini map. Hence so is
f
by
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Next, assume (ii). As f is Fubini maps by case (i). So is f
=f
+
−f
−
P
-measurable, so are
by Lemma 2(ii), since f
+
f
+
≤ |f |
, f− ,
and
|f |
(Theorem 2 in §2). As they are nonnegative, they are
implies
∫
f
+
dp < ∞
(8.8.44)
X×Y
by our assumption (ii). (The three integrals are equal, as |f | is a Fubini map.) Thus all is proved.
□
III. We now want to replace P by P in Lemma 5 and Theorem 2. This works only under certain σ-finiteness conditions, as shown below. ∗
Lemma 8.8.6 Let D ∈ P be σ-finite, i.e., ∗
∞
D = ⋃ Di (disjoint)
(8.8.45)
i=1
for some D
i
∈ P
∗
,
with pD
i
< ∞ (i = 1, 2, …).
Then there is aK ∈ P such that p(K − D) = 0 and D ⊆ K . Proof As P is a σ -ring containing C, it also contains C
σ.
Thus by Theorem 3 of Chapter 7, §5, p is P -regular. ∗
For the rest, proceed as in Theorems 1 and 2 in Chapter 7, §7.
□
Lemma 8.8.7 If D ∈ P is σ-finite (Lemma 6), then C is a Fubini map. ∗
D
Proof By Lemma 6, (∃K ∈ P)
Let Q = K − D, so pQ = 0, and C
Q
= CK − CD ;
p(K − D) = 0, D ⊆ K.
that is, C
∫
D
= CK − CQ
(8.8.46)
and
CQ dp = pQ = 0.
(8.8.47)
X×Y
As K ∈ P, C
K
is a Fubini map. Thus by Lemma 2(ii), all reduces to proving the same for C
Q.
Now, as pQ = 0, Q is certainly σ -finite; so by Lemma 6, (∃Z ∈ P)
Q ⊆ Z, pZ = pQ = 0.
(8.8.48)
Again C is a Fubini map; so Z
∫ X
As Q ⊆ Z, we have C
Q
≤ CZ ,
∫ Y
CZ dndm = ∫
CZ dp = pZ = 0.
(8.8.49)
X×Y
and so
8.8.6
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∫ X
∫
CQ dndm = ∫
Y
[∫
X
≤∫
CQ (x, ⋅)dn] dm
Y
[∫
X
CZ (x, ⋅)dn] dm = ∫
Y
CZ dp = 0.
X×Y
Similarly, ∫
∫
Y
CQ dmdn = ∫
X
[∫
Y
CQ (⋅, y)dm] dn = 0.
(8.8.50)
X
Thus setting g(x) = ∫
CQ (x, ⋅)dn and h(y) = ∫
Y
CQ (⋅, y)dm,
(8.8.51)
X
we have ∫
gdm = 0 = ∫
X
hdn.
(8.8.52)
Y
Hence by Theorem 1(h) in §5, g = 0 a.e. on X, and h = 0 a.e. on Y . So g and h are "almost" measurable (Definition 2 of §3); i.e., C has Fubini property (a). Q
Similarly, one establishes (b), and (3) yields Fubini property (c), since ∫ X
as required.
∫
CQ dndm = ∫
Y
∫
Y
CQ dmdn = ∫
X
CQ dp = 0,
(8.8.53)
X×Y
□
Theorem 8.8.3 (Fubini II) Suppose f
: X ×Y → E
∗
is P -measurable on X × Y and satisfies condition (i) or (ii) of Theorem 2. ∗
Then f is a Fubini map, provided f has σ-finite support, i.e., f vanishes outside some σ-finite set H ⊆ X × Y . Proof First, let ∞
f = ∑ ai CDi
(ai > 0, Di ∈ P
∗
),
(8.8.54)
i=1
with f
on −H (as above).
=0
As f = a ≠ 0 on A , we must have D and so is f . (Why?) i
i
i
⊆ H;
so all D are σ -finite. (Why?) Thus by Lemma 7, each C i
Di
is a Fubini map,
If f is P -measurable and nonnegative, and f = 0 on −H , we can proceed as in Theorem 2, making all f vanish on −H . Then the f and f are Fubini maps by what was shown above. ∗
k
k
Finally, in case (ii), f
=0
on −H implies f
Thus f
+
,f
−
,
+
=f
−
= |f | = 0 on − H .
and f are Fubini maps by part (i) and the argument of Theorem 2.
(8.8.55) □
Note 1. The σ-finite support is automatic if f is p-integrable (Corollary 1 in §5), or if p or both m and n are σ-finite (see Problem 3). The condition is also redundant if f is P -measurable (Theorem 2; see also Problem 4). Note 2. By induction, our definitions and Theorems 2 and 3 extend to any finite number q of measure spaces (Xi , Mi , mi ) ,
8.8.7
i = 1, … , q.
(8.8.56)
https://math.libretexts.org/@go/page/22219
One writes p = m1 × m2
(8.8.57)
m1 × m2 × ⋯ × mq+1 = (m1 × ⋯ × mq ) × mq+1 .
(8.8.58)
if q = 2 and sets
Theorems 2 and 3 with similar assumptions then state that the order of integrations is immaterial. Note 3. Lebesgue measure in E can be treated as the product of q one- dimensional measures. Similarly for LS product measures (but this method is less general than that described in Problems 9 and 10 of Chapter 7, §9). q
IV. Theorems 2(ii) and 3(ii) hold also for functions f : X ×Y → E
n
(C
n
)
(8.8.59)
if Definitions 2 and 3 are modified as follows (so that they make sense for such maps): In Definition 2, set g(x) = ∫
fx dn
(8.8.60)
Y
if f is n -integrable on Y , and g(x) = 0 otherwise. Similarly for h(y). In Definition 3, replace "measurable" by "integrable." x
For the proof of the theorems, apply Theorems 2(i) and 3(i) to |f |. This yields ∫ Y
∫
|f |dmdn = ∫
X
∫
X
|f |dndm = ∫
Y
|f |dp.
(8.8.61)
X×Y
Hence if one of these integrals is finite, f is p-integrable on X × Y , and so are its q components. The result then follows on noting that f is a Fubini map (in the modified sense) iff its components are. (Verify!) See also Problem 12 below. V. In conclusion, note that integrals of the form ∫
f dp
(D ∈ P
∗
)
(8.8.62)
D
reduce to ∫
f ⋅ CD dp.
(8.8.63)
X×Y
Thus it suffices to consider integrals over X × Y . 8.8: Product Measures. Iterated Integrals is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
8.8.8
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8.8.E: Problems on Product Measures and Fubini Theorems Exercise 8.8.E. 1 Prove Lemmas 2 and 3.
Exercise 8.8.E. 1
′
Show that {A ∈ M|mA < ∞} is a set ring.
Exercise 8.8.E. 2 Fill in all proof details in Theorems 1 to 3.
Exercise 8.8.E. 2
′
Do the same for Lemmas 5 to 7.
Exercise 8.8.E. 3 Prove that if m and n are σ-finite, so is p = m × n. Disprove the converse by an example. [Hint: (∪ A ) × (U B ) = U (A × B ) . Verify!] i
i
j
j
i,j
i
j
Exercise 8.8.E. 4 Prove the following. (i) Each D ∈ P (as in the text) is (p) σ-finite. (ii) All P -measurable maps f : X × Y → E have σ-finite support. [Hints: (i) Use Problem 14(b) from Chapter 7, §3. (ii) Use (i) for P -elementary and nonnegative maps first. ] ∗
Exercise 8.8.E. 5 (i) Find D ∈ P and x ∈ X such that C (x, ⋅) is not n -measurable on Y . Does this contradict Lemma 7? [Hint: Let m = n = Lebesgue measure in E ; D = {x} × Q, with Q non-measurable. ] (ii) Which C -sets have nonzero measure if X = Y = E , m is as in Problem 2(b) of Chapter 7, §5( with S = X), and n is Lebesgue measure? ∗
D
1
1
∗
Exercise 8.8.E. 5
′
Let m = n = Lebesgue measure in [0, 1] = X = Y . Let 1
on (
0
elsewhere.
k+1
,
1
k(k + 1) fk = {
k
] and (8.8.E.1)
Let ∞
f (x, y) = ∑ [ fk (x) − fk+1 (x)] fk (y);
(8.8.E.2)
k=1
the series converges. (Why?) Show that (i) (∀k) ∫ f = 1 ; X
k
8.8.E.1
https://math.libretexts.org/@go/page/25046
(ii) ∫ ∫ f dndm = 1 ≠ 0 = ∫ ∫ What is wrong? Is f P -measurable? [Hint: Explore X
Y
Y
X
.
f dmdn
∫ X
∫
|f |dndm. ]
(8.8.E.3)
Y
Exercise 8.8.E. 6 Let X = Y = [0, 1], m as in Example (c) of Chapter 7, §6, (S = X) and n = Lebesgue measure in Y . (i) Show that p = m × n is a topological measure under the standard metric in E . (ii) Prove that D = {(x, y) ∈ X × Y |x = y} ∈ P . (iii) Describe C . [Hints: (i) Any subinterval of X × Y is in P ; (ii) D is closed. Verify!] 2
∗
∗
Exercise 8.8.E. 7 Continuing Problem 6, let f (i) Show that
= CD
.
∫ Y
∫
f dndm = 0 ≠ 1 = ∫
X
Y
∫
f dmdn.
(8.8.E.4)
X
What is wrong? [Hint: D is not σ-finite; for if ∞
D = ⋃ Di ,
(8.8.E.5)
i=1
at least one D is uncountable and has no finite basic covering values (why?), so p (ii) Compute p {(x, 0)|x ∈ X} and p {(0, y)|y ∈ Y }.
∗
i
∗
Di = ∞.
]
∗
Exercise 8.8.E. 8 Show that D ∈ P is σ-finite iff ∗
∞
D ⊆ ⋃ Di (disjoint)
(8.8.E.6)
i=1
for some sets D [Hint: First let p
i ∗
∈ C
.
D < ∞. Use Corollary 1 from Chapter 7,
§1. ]
Exercise 8.8.E. 9 Given D ∈ P, a ∈ X, and b ∈ Y , let Da = {y ∈ Y |(a, y) ∈ D}
(8.8.E.7)
and b
D
= {x ∈ X|(x, b) ∈ D}.
8.8.E.2
(8.8.E.8)
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(See Figure 34 for X = Y = E Prove that (i) D ∈ N , D ∈ M ; (ii) C (a, ⋅) = C , nD = ∫ [Hint: Let
1
.)
b
a
D
Da
a
Y
b
CD (a, ⋅)dn, m D
=∫
X
CD (⋅, b)dm
H = {(x, y) ∈ E
Show that R is a σ-ring ⊇ C . Hence R ⊇ P; D ∈ R; D
2
|0 ≤ y < f (x)}
(8.8.E.9)
Similariy for D .] b
∈ N.
a
.
Exercise 8.8.E. 10 ⇒ 10
. Let m = n = Lebesgue measure in E
1
Let f
=X =Y .
H = {(x, y) ∈ E
: E
2
1
→ [0, ∞)
be m-mensurable on X. Let
|0 ≤ y < f (x)}
(8.8.E.10)
and G = {(x, y) ∈ E
2
|y = f (x, y)}
(8.8.E.11)
f dm
(8.8.E.12)
(the "graph" of f ). Prove that (i) H ∈ P and ∗
pH = ∫ X
(="the area under f") (ii) G ∈ P and pG = 0 . [Hints: (i) First take f = C , and elementary and nonnegative maps. Then use Lemma 2 in §2 (last clause). Fix elementary and nonnegative maps f ↗ f , assuming f < f (if not, replace f by (1 − ) f ) . Let ∗
D
1
k
Show that H (ii) Set
k
↗ H ∈ P
∗
k
k
k
k
Hk = {(x, y)|0 ≤ y < fk (x)} .
(8.8.E.13)
ϕ(x, y) = y − f (x).
(8.8.E.14)
.
Using Corollary 4 of §1, show that assume G ∈ P. By Problem 9 (ii),
ϕ
is p-measurable on
(∀x ∈ E
1
)
∫
E
2
;
so
G=E
2
(ϕ = 0) ∈ P
CG (x, ⋅)dn = nGx = 0,
∗
. Dropping a null set (Lemma
6),
(8.8.E.15)
Y
as G
x
= {f (x)}, a singleton. ]
Exercise 8.8.E. 11 Let f (x, y) = ϕ1 (x)ϕ2 (y).
8.8.E.3
(8.8.E.16)
https://math.libretexts.org/@go/page/25046
Prove that if ϕ is m-integrable on X and ϕ is n -integrable on Y , then f is p-integrable on X × Y and 1
2
∫
f dp = ∫
X×Y
ϕ1 ⋅ ∫
X
ϕ2 .
(8.8.E.17)
Y
Exercise 8.8.E. ∗12 Prove Theorem 3(ii) for f : X × Y → E(E complete) . [Outline: If f is P -simple, use Lemma 7 above and Theorem 2 in §7. If ∗
∞
f = ∑ ak CDk ,
Dk ∈ P
∗
,
(8.8.E.18)
k=1
let k
Hk = ⋃ Di
(8.8.E.19)
i=1
and f
k
= f CH , k
so the f are P -simple (hence Fubini maps), and f ∗
k
k
∫
→ f
(point-wise) on X × Y , with |f
k|
|f |dp < ∞
≤ |f |
and
(8.8.E.20)
X×Y
(by assumption). Now use Theorem 5 from §6. Let now f be P -measurable; so ∗
f = lim fk (uniformly)
(8.8.E.21)
k→∞
for some
P
gk = gk CH .
∗
§ By assumption, f = f C (H σ-finite); so we may assume are Fubini maps. So is f by Lemma 1 in §7 (verify!), provided H ⊆ D for some
-elementary maps gk (Theorem 3 in 1) .
Then as shown above, all
gk
H
D ∈ C.
In the general case, by Problem 8 , H ⊆ ⋃ Di (disjoint), Di ∈ C.
(8.8.E.22)
i
Let H
i
= H ∩ Di .
By the previous step, each f C
Hi
is a Fubini map; so is k
fk = ∑ f CHi
(8.8.E.23)
i=1
(why?), hence so is f
= limk→∞ fk ,
by Theorem 5 of §6. (Verify!)]
Exercise 8.8.E. 13 Let m = Lebesgue measure in E
1
,p =
Lebesgue measure in E
s
, X = (0, ∞),
s
¯ ∈ E || y ¯| = 1} . Y = {y
8.8.E.4
and (8.8.E.24)
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Given x ¯ ∈ E
s
¯ ¯ ¯
− { 0},
let x ¯ r = |x ¯| and u ¯ =
∈ Y .
(8.8.E.25)
r
¯ the polar coordinates of x ¯ ≠ 0. Call r and u If D ⊆ Y , set ¯ ¯ ¯
∗
∗
n D = s ⋅ p {ru ¯| u ¯ ∈ D, 0 < r ≤ 1}.
(8.8.E.26)
Show that n is an outer measure in Y ; so it induces a measure n in Y . Then prove that ∗
∫
f dp = ∫
E
if f is p-measurable and nonnegative on E [Hint: Start with f = C ,
s
s
s−1
r
dm(r) ∫
X
¯)dn(u ¯) f (ru
(8.8.E.27)
Y
.
A
A = {ru ¯| u ¯ ∈ H , a < r < b},
for A ⊆P
some ∗
open
set
s
H ⊆ Y (subspace of E ) .
Next,
(8.8.E.28)
let
A ∈ B( Borel set in Y );
then
. Then let f be p-elementary, and so on. ]
8.8.E: Problems on Product Measures and Fubini Theorems is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
8.8.E.5
https://math.libretexts.org/@go/page/25046
8.9: Riemann Integration. Stieltjes Integrals I. In this section, C is the family of all intervals in function v (Chapter 7, §§1-2).
E
n
and
,
is an additive finite premeasure on
m
C
(or
Cs
), such as the volume
By a C -partition of A ∈ C (or A ∈ C ), we mean a finite family s
P = { Ai } ⊂ C
(8.9.1)
A = ⋃ Ai (disjoint).
(8.9.2)
such that
i
As we noted in §5, the Riemann integral, R∫
f = R∫
A
of f
: E
n
→ E
1
f dm,
(8.9.3)
A
can be defined as its Lebesgue counterpart, ∫
f,
(8.9.4)
A
with elementary maps replaced by simple step functions ("C -simple" maps.) Equivalently, one can use the following construction, due to J. G. Darboux.
Definitions (a) Given f
: E
n
→ E
∗
and a C -partition P = { A1 , … , Aq }
(8.9.5)
¯¯ ¯
of A, we define the lower and upper Darboux sums, S and S , of f over P (with respect to m) by – –
q
q ¯¯ ¯
S (f , P) = ∑ m Ai ⋅ inf f [ Ai ] and S (f , P) = ∑ m Ai ⋅ sup f [ Ai ] . – – i=1
(8.9.6)
i=1
(b) The lower and upper Riemann integrals ("R-integrals") of f on A (with respect to m) are R∫
f =R∫
– –A
– –A
¯¯ ¯
¯¯ ¯
R∫
A
f = R∫
f dm = sup P S (f , P) and – –
}
(8.9.7)
¯¯ ¯
A
f dm = infP S (f , P),
where the "inf" and "sup" are taken over all C -partitions P of A . (c) We say that f is Riemann-integrable ("R-integrable") with respect to m on A iff f is bounded on A and ¯ ¯¯¯ ¯
R∫ ––
f = R∫
f.
(8.9.8)
A
A
We then set ¯ ¯¯¯ ¯
R∫ A
f =R∫ ––
f = R∫
f dm = R ∫ A
A
f dm
(8.9.9)
A
and call it the Riemann integral ("R-integral") of f on A. "Classical" notation: R∫
¯ ¯ f (¯¯ x )dm(¯¯ x ).
(8.9.10)
A
If A = [a, b] ⊂ E
1
,
we also write
8.9.1
https://math.libretexts.org/@go/page/22250
b
R∫
b
f = R∫
a
f (x)dm(x)
(8.9.11)
a
instead. If m is Lebesgue measure (or premeasure) in E
1
,
we write "dx" for "dm(x)."
For Lebesgue integrals, we replace "R " by "L," or we simply omit "R." If f is R-integrable on A, we also say that R∫
f
(8.9.12)
A
exists (note that this implies the boundedness of f ); note that ¯ ¯¯¯ ¯
R∫ ––
f and R ∫
f
(8.9.13)
A
A
are always defined in E . ∗
Below, we always restrict f to a fixed A ∈ C (or A ∈ C ); P, P s
′
,P
′′
,P
∗
and P denote C -partitions of A. k
We now obtain the following result for any additive m : C → [0, ∞) .
Corollary 8.9.1 If P refines P (§1), then ′
¯¯ ¯
′
¯¯ ¯
′
S (f , P ) ≤ S (f , P) ≤ S (f , P) ≤ S (f , P ) . – – – –
(8.9.14)
Proof Let P
′
= { Ai } , P = { Bik } ,
and (∀i)
Ai = ⋃ Bik .
(8.9.15)
k
By additivity, m Ai = ∑ m Bik .
(8.9.16)
k
Also, B
ik
⊆ Ai
implies f [ Bik ] ⊆ f [ Ai ] ; sup f [ Bik ] ≤ sup f [ Ai ] ; and inf f [ Bik ] ≥ inf f [ Ai ] .
So setting ai = inf f [ Ai ] and bik = inf f [ Bik ] ,
(8.9.17)
we get ′
S (f , P ) = ∑ ai m Ai = ∑ ∑ ai m Bik – – i
i
k
≤ ∑ bik m Bik = S (f , P). – – i,k
Similarly, ¯¯ ¯
′
¯¯ ¯
S (f , P ) ≤ S (f , P),
(8.9.18)
and
8.9.2
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¯¯ ¯
S (f , P) ≤ S (f , P) – –
is obvious from (1).
(8.9.19)
□
Corollary 8.9.2 For any P and P , ′
′′
¯¯ ¯
′
S (f , P ) ≤ S (f , P – –
′′
).
(8.9.20)
Hence ¯ ¯¯¯ ¯
R∫ ––
f ≤ R∫
f.
(8.9.21)
A
A
Proof Let P = P
′
∩P
′′
(see §1). As P refines both P and P , Corollary 1 yields ′
′′
¯¯ ¯
′
¯¯ ¯
S (f , P ) ≤ S (f , P) ≤ S (f , P) ≤ S (f , P – – – –
Thus, indeed, no lower sum –S (f , P –
′
)
¯¯ ¯
exceeds any upper sum S (f , P
′′
)
′′
).
(8.9.22)
.
Hence also, ¯¯ ¯
′
sup S (f , P ) ≤ inf S (f , P – – ′′ ′
′′
),
(8.9.23)
P
P
i.e., ¯ ¯¯¯ ¯
R∫ ––
as claimed.
f ≤ R∫
f,
(8.9.24)
A
A
□
Lemma 8.9.1 A map f
: A → E
1
is R -integrable iff f is bounded and, moreover, ¯¯ ¯
(∀ε > 0) (∃P)
S (f , P) − S (f , P) < ε. – –
(8.9.25)
Proof By formulas (1) and (2), ¯ ¯¯¯ ¯
S (f , P) ≤ R ∫ – – ––
¯¯ ¯
f ≤ R∫
f ≤ S (f , P).
(8.9.26)
A
A
Hence (3) implies ∣
¯ ¯¯¯ ¯
∣R ∫ ∣
∣ f −R∫
A
f ∣ < ε. A
––
(8.9.27)
∣
As ε is arbitrary, we get ¯ ¯¯¯ ¯
R∫
f =R∫ A
––
f;
(8.9.28)
A ––
so f is R-integrable. Conversely, if so, definitions (b) and (c) imply the existence of P and P such that ′
8.9.3
′′
https://math.libretexts.org/@go/page/22250
1
′
S (f , P ) > R ∫ – –
f −
ε
(8.9.29)
ε.
(8.9.30)
2
A
and ¯¯ ¯
S (f , P
′′
1 ) < R∫
f + 2
A
Let P refine both P and P ′
′′
.
Then by Corollary 1,
¯¯ ¯
¯¯ ¯
S (f , P) − S (f , P) ≤ S (f , P – –
′′
′
) − S (f , P ) – – 1
< (R ∫
f +
A
as required.
1 ε) − (R ∫
2
f −
ε) = ε, 2
A
□
Lemma 8.9.2 Let f be C -simple; say, f
= ai
on A for some C -partition P
∗
i
= { Ai }
of A (we then write
f = ∑ ai CA
(8.9.31)
i
i
on A; see Note 4 of §4). Then ¯ ¯¯¯ ¯
R∫ ––
f = R∫ A
A
f = S (f , P – –
∗
¯¯ ¯
) = S (f , P
∗
) = ∑ ai m Ai .
(8.9.32)
i
Hence any finite C -simple function is R-integrable, with R ∫
A
f
as in (4).
Proof Given any C -partition P = {B
k}
of A, consider P
As f
= ai
on A
i
∩ Bk
∗
\acdotP = { Ai ∩ Bk } .
(8.9.33)
(even on all of A ), i
ai = inf f [ Ai ∩ Bk ] = sup f [ Ai ∩ Bk ] .
(8.9.34)
A = ⋃ (Ai ∩ Bk ) (disjoint)
(8.9.35)
Also,
i,k
and (∀i)
Ai = ⋃ (Ai ∩ Bk ) ;
(8.9.36)
k
so m Ai = ∑ m (Ai ∩ Bk )
(8.9.37)
k
and S (f , P) = ∑ ∑ ai m (Ai ∩ Bk ) = ∑ ai m Ai = S (f , P – – – – i
k
∗
)
(8.9.38)
i
for any such P . Hence also
8.9.4
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∑ ai m Ai = sup S (f , P) = R ∫ – – P
i
¯¯ ¯
Similarly for R ∫
A
f.
––
f.
(8.9.39)
A
This proves (4).
If, further, f is finite, it is bounded (by max |a |) since there are only finitely many a ; so f is R-integrable on A, and all is proved. □ i
i
¯¯ ¯
Note 1. Thus S and S are integrals of C -simple maps, and definition (b) can be restated: – –
¯ ¯¯¯ ¯
R∫ ––
f = sup R ∫ g
A
g and R ∫
f = inf R ∫
A
h
A
h,
(8.9.40)
A
taking the sup and inf over all C -simple maps g, h with g ≤ f ≤ h on A.
(8.9.41)
(Verify by properties of glb and lub!) Therefore, we can now develop R-integration as in §§4-5, replacing elementary maps by particular, Problem 5 in §5 works out as before.
C
-simple maps, with
S =E
n
.
In
Hence linearity (Theorem 1 of §6) follows, with the same proof. One also obtains additivity (limited to C -partitions). Moreover, the R-integrability of f and g implies that of f g, f ∨ g, f ∧ g, and |f |. (See the Problems.)
Theorem 8.9.1 If f
i
→ f
(uniformly) on A and if the f are R-integrable on A , so also is f . Moreover, i
lim R ∫ i→∞
|f − fi | = 0 and lim R ∫ i→∞
A
fi = R ∫
A
f.
(8.9.42)
A
Proof As all f are bounded (definition (c)), so is f , by Problem 10 of Chapter 4, §12. i
Now, given ε > 0, fix k such that ε (∀i ≥ k)
|f − fi |
0. If g(p) = h(p), then k→∞
so (∃k)
As G
p
(δk ) ⊆ Akp ,
| hk (p) − gk (p)| = sup f [ Akp ] − inf f [ Akp ] < ε.
(8.9.82)
we get (∀x ∈ Gp (δk ))
|f (x) − f (p)| ≤ sup f [ Akp ] − inf f [ Akp ] < ε,
(8.9.83)
proving continuity (clause (ii)). For (i), given ε > 0, choose δ > 0 so that (∀x, y ∈ A ∩ Gp (δ))
|f (x) − f (y)| < ε.
(8.9.84)
Because (∀δ > 0) (∃k0 ) (∀k > k0 )
for k > k
0,
Akp ⊆ Gp (δ).
| Pk | < δ
(8.9.85)
Deduce that (∀k > k0 )
| hk (p) − gk (p)| ≤ ε.
8.9.9
□
(8.9.86)
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Note 5. The Lebesgue measure of B in Lemma 4 is zero; for B consists of countably many "faces" (degenerate intervals), each of measure zero.
Theorem 8.9.3 A map f : A → E is R-integrable on A (with m = Lebesgue measure) iff f is bounded on A and continuous on A − Q for some Q with mQ = 0 . 1
Note that relative continuity on A − Q is not enough-take f
of Note 2.
= CR
Proof If these conditions hold, choose {P
k}
as in Lemma 4.
Then by the assumed continuity, g = h on A − Q, mQ = 0 . Thus ∫
g =∫
A
h
(8.9.87)
A
(Corollary 2 in §5). Hence by formula (6), f is R-integrable on A. Conversely, if so, use Lemma 1 with 1 ε = 1,
1 ,…,
2
,…
(8.9.88)
k
to get for each k some P such that k
1 ¯¯ ¯ S (f , Pk ) − S (f , Pk ) < → 0. – – k
By Corollary 1, this will hold if we refine each Then Lemmas 3 and 4 apply.
Pk ,
(8.9.89)
step by step, so as to achieve properties (i) and (ii) of Note 4 as well.
As ¯¯ ¯
S (f , Pk ) − S (f , Pk ) → 0, – –
(8.9.90)
formula (6) show that ∫ A
¯¯ ¯
g = lim S (f , Pk ) = lim S (f , Pk ) = ∫ – k→∞ – k→∞
h.
(8.9.91)
A
As h and g are integrable on A, ∫
(h − g) = ∫
A
h −∫
A
Also h − g ≥ 0; so by Theorem 1(h) in §5, h = g on A − Q , mQ is continuous on ′
g = 0.
(8.9.92)
A
′
=0
′
A − Q − B,
(under Lebesgue measure). Hence by Lemma 4, f
(8.9.93)
with mB = 0 (Note 5). Let Q = Q
′
∪ B.
Then mQ = 0 and ′
A − Q = A − Q − B;
so f is continuous on A − Q. This completes the proof.
(8.9.94)
□
8.9.10
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Note 6. The first part of the proof does not involve requires that mB = 0 .
and thus works even if
B
m
is not the Lebesgue measure. The second part
Theorem 3 shows that R-integrals are limited to a.e. continuous functions and hence are less flexible than L-integrals: Fewer functions are R-integrable, and convergence theorems (§6, Theorems 4 and 5) fail unless R ∫ f exists. A
III. Functions f : E → E is R-integrable on A iff all f n
s
k
(C
s
).
(k ≤ s)
For such functions, R-integrals are defined componentwise (see §7). Thus f are, and then
= (f1 , … , fs )
s ¯ ¯ ¯
R∫
f = ∑ ek R ∫
A
A complex function f is R-integrable iff f and f re
fk .
(8.9.95)
A
k=1
are, and then
im
R∫
f = R∫
A
fre + iR ∫
A
fim .
(8.9.96)
A
Via components, Theorems 1 to 3, Corollaries 3 and 4, additivity, linearity, etc., apply. IV. Stieltjes Integrals. Riemann used Lebesgue premeasure v only. But as we saw, his method admits other premeasures, too. Thus in E 7, §9).
1
,
we may let m be the LS premeasure s or the LS measure m where α ↑ (Chapter 7, §5, Example (b), and Chapter α
α
Then R∫
f dm
(8.9.97)
A
is called the Riemann-Stieltjes (RS) integral of f with respect to α, also written b
R∫
f dα
or
R∫
A
f (x)dα(x)
(8.9.98)
a
(the latter if A = [a, b] ); f and α are called the integrand and integrator, respectively. If α(x) = x, m becomes the Lebesgue measure, and α
R∫
f (x)dα(x)
(8.9.99)
turns into R∫
f (x)dx.
(8.9.100)
Our theory still remains valid; only Theorem 3 now reads as follows.
Corollary 8.9.4 If f is bounded and a.e. continuous on A = [a, b] (under an LS measure m ) then α
b
R∫
f dα
(8.9.101)
a
exists. The converse holds if α is continuous on A . For by Notes 5 and 6, the "only if" in Theorem 3 holds if m B = 0. Here consists of countably many endpoints of partition subintervals. But (see Chapter §9) m {p} = 0 if α is continuous at p. Thus the later implies m B = 0 . α
α
α
RS-integration has been used in many fields (e.g., probability theory, physics, etc.), but it is superseded by LS-integration, i.e., Lebesgue integration with respect to m , which is fully covered by the general theory of §§1-8. α
Actually, Stieltjes himself used somewhat different definitions (see Problems 10-13), which amount to applying the set function σ of Problem 9 in Chapter 7, §4, instead of s or m . We reserve the name "Stieltjes integrals," denoted
α
α
α
8.9.11
https://math.libretexts.org/@go/page/22250
b
S∫
f dα,
(8.9.102)
a
for such integrals, and "RS-integrals" for those based on m or s (this terminology is not standard). α
α
Observe that σ need not be ≥ 0. Thus for the first time, we encounter integration with respect to sign-changing set functions. A much more general theory is presented in §10 (see Problem 10 there). α
8.9: Riemann Integration. Stieltjes Integrals is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
8.9.12
https://math.libretexts.org/@go/page/22250
8.9.E: Problems on Riemann and Stieltjes Integrals Exercise 8.9.E. 1 Replacing "M" by "C, " and "elementary and integrable" or "elementary and nonnegative" by "C -simple," price Corollary 1(ii) (iv)(vii) and Theorems 1(i) and 2(ii), all in §4, and do Problems 5-7 in §4, for R-integrals.
Exercise 8.9.E. 2 Verify Note 1.
Exercise 8.9.E. 2
′
Do Problems 5 − 7 in §5 for R-integrals.
Exercise 8.9.E. 3 Do the following for R-integrals. (i) Prove Theorems 1(a) − (g) and 2, both in §5(C-partitions only ) . (ii) Prove Theorem 1 and Corollaries 1 and 2, all in §6. (iii) Show that definition (b) can be replaced by formulas analogous to formulas (1 ) , (1 [Hint: Use Problems 1 and 2 . ] ′
′′
),
and ( 1) of Definition 1 in §5.
′
Exercise 8.9.E. 4 Fill in all details in the proof of Theorem 1, Lemmas 3 and 4, and Corollary 4.
Exercise 8.9.E. 5 For f , g : E → E (C ) , via components, prove the following. (i) Theorems 1 − 3 and (ii) additivity and linearity of R-integrals. Do also Problem 13 in §7 for R-integrals. n
s
s
Exercise 8.9.E. 6 Prove that if f
: A → E
s
(C
s
)
is bounded and a.e. continuous on A, then R∫ A
∣ ∣ |f | ≥ ∣R ∫ f∣ . ∣ ∣ A
(8.9.E.1)
For m = Lebesgue measure, do it assuming R-integrability only.
Exercise 8.9.E. 7 Prove that if f , g : A → E are R-integrable, then (i) so is f , and (ii) so is f g. [Hints: (i) Use Lemma 1. Let h = |f | ≤ K < ∞ on A. Verify that 1
2
2
(inf h [ Ai ])
= inf f
2
[ Ai ] and (sup h [ Ai ])
8.9.E.1
2
= sup f
2
[ Ai ] ;
(8.9.E.2)
https://math.libretexts.org/@go/page/25047
so sup f
2
[ Ai ] − inf f
2
[ Ai ]
= (sup h [ Ai ] + inf h [ Ai ]) (sup h [ Ai ] − inf h [ Ai ]) ≤ (sup h [ Ai ] − inf h [ Ai ]) 2K.
(ii) Use 1 fg =
2
[(f + g)
2
− (f − g) ] .
(8.9.E.3)
4
(iii) For m = Lebesgue measure, do it using Theorem 3.]
Exercise 8.9.E. 8 Prove that if
m =
the volume function
v
(or LS function
sα
for a continuous
α
), then in formulas ( 1) and
(2),
one may
¯ ¯¯ ¯
replace A by A (closure of A ) . [Hint: Show that here mA = mA , i
i
i
¯ ¯¯ ¯
R∫
f = R∫
f,
(8.9.E.4)
¯ ¯¯ ¯
A
A
and additivity works even if the A have some common "faces" (only their interiors being disjoint).] i
Exercise 8.9.E. 9 (Riemann sums.) Instead of –S and S¯ , Riemann used sums – S(f , P) = ∑ f (xi ) dm Ai ,
(8.9.E.5)
i
¯¯¯¯ ¯
where m = v (see Problem 8) and x is arbitrarily chosen from A . For a bounded f , prove that i
i
r = R∫
f dm
(8.9.E.6)
A
exists on A = [a, b] iff for every ε > 0, there is P such that ε
|S(f , P) − r| < ε
(8.9.E.7)
P = { Ai }
(8.9.E.8)
for every refinement
¯¯¯¯ ¯
of P and any choice of x ∈ A . [Hint: Show that by Problem 8, this is equivalent to formula ( 3 ).] ε
i
i
8.9.E.2
https://math.libretexts.org/@go/page/25047
Exercise 8.9.E. 10 Replacing m by the σ of Problem 9 of Chapter 7, §4, write definition of the Stieltjes integral, α
S(f , P, α)
b
S∫
in Problem
9,
treating Problem 9 as a
( or S ∫
f dσα ) .
(8.9.E.9)
a
Here f , α : E → E (monotone or not; even f , α : E Prove that if α : E → E is continuous and α ↑, then 1
1
S(f , P)
b
f dα
a
1
for
1
→ C
will do).
1
b
S∫
b
f dα = R ∫
a
f dα,
(8.9.E.10)
a
the RS -integral.
Exercise 8.9.E. 11 (Integration by parts.) Continuing Problem 10, prove that b
S∫
f dα
(8.9.E.11)
αdf
(8.9.E.12)
a
exists iff b
S∫ a
does, and then b
S∫
b
f dα + S ∫
a
αdf = K,
(8.9.E.13)
a
where K = f (b)α(b) − f (a)α(a).
[Hints: Take any C -partition P = {A
i}
(8.9.E.14)
of [a, b], with ¯¯¯¯ ¯
say. For any x
¯ ¯¯ ¯
i
Ai = [ yi−1 , yi ] ,
(8.9.E.15)
S(f , P, α) = ∑ f (xi ) [α (yi ) − α (yi−1 )] = ∑ f (xi ) α (yi ) − ∑ f (xi ) α (yi−1 )
(8.9.E.16)
K = ∑ f (xi ) α (yi ) − ∑ f (xi−1 ) α (yi−1 ) .
(8.9.E.17)
∈ Ai ,
verify that
and
Deduce that ′
K − S(f , P, α) = S (α, P , f ) = ∑ α (xi ) [f (xi ) − f (yi )] − ∑ α (xi−1 ) [f (yi ) − f (xi−1 )] ;
8.9.E.3
(8.9.E.18)
https://math.libretexts.org/@go/page/25047
here P results by combining the partition points x and y , so it refines P . Now, if S ∫ αdf exists, fix P as in Problem 9 and show that ′
i
i
b
ε
a
b
∣ ∣K − S(f , P, α) − S ∫ ∣
whenever P refines P
ε.
∣ αdf ∣ < ε
(8.9.E.19)
∣
a
]
Exercise 8.9.E. 12 If α : E
1
→ E
1
is of class C D on [a, b] and if 1
b
S∫
f dα
(8.9.E.20)
a
exists (see Problem 10), it equals b
R∫
′
f (x)α (x)dx.
(8.9.E.21)
a
[Hints: Set ϕ = f α , P = {A ′
i}
¯¯¯¯ ¯
, Ai = [ ai−1 , ai ]
. Then ¯¯¯¯ ¯
′
S(ϕ, P) = ∑ f (xi ) α (xi ) (ai − ai−1 ) ,
xi ∈ Ai
(8.9.E.22)
and (Corollary 3 in Chapter 5, §2) ′
S(f , P, α) = ∑ f (xi ) [α (ai ) − α (ai−1 )] = ∑ f (xi ) α (qi ) ,
qi ∈ Ai .
(8.9.E.23)
As f is bounded and α is uniformly continuous on [a, b] (why?), deduce that ′
(∀ε > 0) (∃Pε ) (∀Pε ) (∀P refining Pε ) 1 |S(ϕ, P) − S(f , P, α)|
0)
as in Problem 17(iii) of Chapter 7, §8.
F (x) = (x − an )
2
(x − bn )
2
1 sin
if x ∉ P .
(9.1.E.14)
(bn − an ) (x − an ) (x − bn )
Prove that F has a bounded derivative f , yet f is not R-integrable on A; so Theorem 2 applies, but Corollary 1 does not. [Hints: If p ∉ P , compute F (p) as in calculus. If p ∈ P and x → p+ over A − P , then x is always in some (a , b ) , p ≤ a < x. (Why?) Deduce that Δx = x − p > x − a and ′
n
n
n
n
∣ ΔF ∣ 2 2 ∣ ∣ ≤ (x − an ) (b − a) ≤ |Δx|(b − a) ; ∣ Δx ∣
so F (p) = 0. (What if x → p+ over P ?) Similarly, show that F = 0 on P . Prove however that F (x) oscillates from 1 to −1 as x → a + or x → b − , hence also as discontinuous on all of P , with mP > 0. Now use Theorem 3 in Chapter 8, §9.] ′
(9.1.E.15)
′
+
−
′
n
n
x → p ∈ P
(why?); so
F
′
is
Exercise 9.1.E. 8 ⇒ 8
. If Q ⊆ A = [a, b]
and mQ = 0, find a continuous map g : A → E
1
, g ≥ 0, g ↑,
g
′
with
= +∞
[Hints: By Theorem 2 of Chapter 7, §8, fix ( ∀n ) an open G
n
(9.1.E.16)
on Q.
⊇ Q,
(9.1.E.17)
with
−n
m Gn < 2
.
(9.1.E.18)
Set gn (x) = m (Gn ∩ [a, x])
(9.1.E.19)
and ∞
g = ∑ gn
(9.1.E.20)
n=1
9.1.E.3
https://math.libretexts.org/@go/page/25152
on A; ∑ g converges uniformly on A. (Why?) By Problem 4 in Chapter 7, §9, and Theorem 2 of Chapter 7, §4, each g (hence g) is continuous. (Why?) If [p, x] ⊆ G that n
n
gn (x) = gn (p) + (x − p),
n,
show
(9.1.E.21)
so Δgn
=1
(9.1.E.22)
Δx
and Δg
∞
=∑ Δx
Δgn
→ ∞. ]
(9.1.E.23)
Δx
n=1
Exercise 9.1.E. 9 (i) Prove Corollary 4. (ii) State and prove earlier analogues for Corollary 5 of Chapter 5, §5, and Theorems 3 and 4 from Chapter 5, §10. [Hint for (i): For primitives, this is Problem 3 in Chapter 5, §5. As g[Q] is countable (Problem 2 in Chapter 1, §9) and bounded on g[A] − g[Q] ⊆ g[A − Q],
f
is
(9.1.E.24)
′
f satisfies Theorem 2 on g[A], with P = g[Q], while (f ∘ g)g satisfies it on A. ]
Exercise 9.1.E. 10 ⇒ 10
. Show that if h : E
1
→ E
∗
is L-integrable on A = [a, b], and x
(∀x ∈ A)
L∫
h = 0,
(9.1.E.25)
a
then h = 0 a.e. on A . [Hints: Let K = A(h > 0) and H = A − K, with, say, mK = ε > 0. Then by Corollary 1 in Chapter 7, §1 and Definition 2 of Chapter 7, §5, H ⊆ ⋃ Bn (disjoint)
(9.1.E.26)
n
for some intervals B
n
⊆ A,
with ∑ m Bn < mH + ε = mH + mK = mA.
(9.1.E.27)
n
(Why?) Set B = ∪
n Bn ;
so ∫ B
h = ∑∫ n
h =0
(9.1.E.28)
Bn
(for L ∫ h = 0 on intervals B ). Thus n
9.1.E.4
https://math.libretexts.org/@go/page/25152
∫
h =∫
A−B
h −∫
A
h = 0.
(9.1.E.29)
B
But B ⊇ H ; so A − B ⊆ A − H = K,
(9.1.E.30)
where h > 0, even though m(A − B) > 0. (Why?) Hence find a contradiction to Theorem 1(h) of Chapter 8, §5. Similarly, disprove that mA(h < 0) = ε > 0. ]
Exercise 9.1.E. 11 . Let F prove that ⇒ 11
on A = [a, b], |F | < ∞, with derived function
↑
F
′
= f.
Taking Theorem 3 from Chapter 7, §10, for granted,
x
L∫
f ≤ F (x) − F (a),
x ∈ A,
(9.1.E.31)
a
[Hints: With f as in (3), F and f are bounded on A and measurable by Theorem 1 of Chapter 8, §2. (Why?) Deduce that f → f (a.e.) on A. Argue as in Lemma 1 using Fatou's lemma (Chapter 8, §6, Lemma 2).] n
n
n
Exercise 9.1.E. 12 ("Truncation.") Prove that if g : S → E is m-integrable on A ∈ M in a measure space (S, M, m), then for any ε > 0, there is a bounded, M -measurable and integrable on A map g : S → E such that 0
∫
|g − g0 | dm < ε.
(9.1.E.32)
A
[Outline: Redefine g = 0 on a null set, to make gM -measurable on A. Then for n = 1, 2, … set g
on A(|g| < n), and
0
elsewhere.
gn = {
(9.1.E.33)
(The function g is called the n th truncate of g .) Each g is bounded and M -measurable on A (why?), and n
n
∫
|g|dm < ∞
(9.1.E.34)
A
by integrability. Also, |g | ≤ |g| and g → g(pointwise) on A. (Why?) Now use Theorem 5 from Chapter 8, §6, to show that one of the g may serve as the desired g n
n
o.
n
]
Exercise 9.1.E. 13 Fill in all proof details in Lemma 2. Prove it for unbounded g . [Hints: By Problem 12, fix a bounded g (|g | ≤ B) , with o
o
1 L∫ A
|g − go |
n);
gn = {
so g
n
→ f
(pointwise), g
n
This makes each g
n
L
≤ f , gn ≤ n,
and |g
n|
(9.2.32)
.
≤ |f |
-integrable on A. Thus as before, by Theorem 5 of Chapter 8, §6, x
x
lim L ∫ n→∞
gn = L ∫
a
f,
x ∈ A.
(9.2.33)
a
Now, set x
Fn (x) = F (x) − L ∫
gn .
(9.2.34)
a
Then by Theorem 3 of §1 (already proved), ′
(since g
n
≤f
x
d
′
Fn (x) = F (x) −
L∫ dx
gn = f (x) − gn (x) ≥ 0
a.e. on A
(9.2.35)
a
).
Thus F has solely nonnegative derivates on A − Q(mQ = 0). Also, as g n
≤ n,
n
we get
x
1 L∫ x −p
gn ≤ n,
(9.2.36)
a
even if x < p. (Why?) Hence ΔFn
ΔF ≥
− n,
Δx
(9.2.37)
Δx
as x
Fn (x) = F (x) − L ∫
gn .
(9.2.38)
a
Thus none of the F -derivates on A can be −∞ . n
By Lemma 1, then, F is monotone (↑) on A; so F
n (x)
n
≥ Fn (a),
i.e.,
x
F (x) − L ∫
a
gn ≥ F (a) − L ∫
a
gn = F (a),
(9.2.39)
a
or x
F (x) − F (a) ≥ L ∫
gn ,
x ∈ A, n = 1, 2, … .
(9.2.40)
a
Hence by (1), x
F (x) − F (a) ≥ L ∫
f,
x ∈ A.
(9.2.41)
a
For the reverse inequality, apply the same formula to −f . Thus we obtain the desired result: x
F (x) = F (a) + L ∫
f
for x ∈ A.
□
(9.2.42)
a
Note 1. Formula (2) is equivalent to F
= L∫ f
on A (see the last part of §1). For if (2) holds, then
9.2.4
https://math.libretexts.org/@go/page/19219
x
F (x) = c + L ∫
f,
(9.2.43)
f,
(9.2.44)
a
with c = F (a); so F
by definition.
= L∫ f
Conversely, if x
F (x) = c + L ∫ a
set x = a to find c = F (a) . II. A bsolute continuity redefined.
Definition A map f
: E
1
is absolutely continuous on an interval I
→ E
⊆E
r
1
iff for every ε > 0, there is δ > 0 such that
r
∑ (bi − ai ) < δ implies ∑ |f (bi ) − f (ai )| < ε i=1
for any disjoint intervals (a
i,
bi ) ,
with a
i,
(9.2.45)
i=1
bi ∈ I
.
From now on, this replaces the "weaker" definition given in Chapter 5, §8. The reader will easily verify the next three "routine" propositions.
Theorem 9.2.1 If f , g, h : E
1
→ E
∗
(C )
are absolutely continuous on A = [a, b] so are f ± g, hf , and |f |.
(9.2.46)
So also is f /h if (∃ε > 0)
All this also holds if f , g : E
1
→ E
|h| ≥ ε on A.
(9.2.47)
are vector valued and h is scalar valued. Finally, if E ⊆ E f ∨ g, f ∧ g, f
+
, and f
∗
,
then
−
(9.2.48)
are absolutely continuous along with f and g.
Corollary 9.2.1 A function F
: E
1
→ E
n
(C
Hence a complex function F
n
)
is absolutely continuous on A =
: E
1
→ C
[a, b]
iff all its components F
1,
… , Fn
is absolutely continuous iff its real and imaginary parts, F
re
are.
and F
im ,
are.
Corollary 9.2.2 If Vf
is absolutely continuous on [a, b] < ∞ all on A.
f : E
1
→ E
A = [a, b],
it is bounded, is uniformly continuous, and has bounded variation,
Lemma 9.2.2 If F
: E
1
→ E
n
(C
n
)
is of bounded variation on A = [a, b], then
(i) F is a.e. differentiable on A, and (ii) F is L-integrable on A . ′
Proof
9.2.5
https://math.libretexts.org/@go/page/19219
Via components (Theorem 4 of Chapter 5, §7), all reduces to the real case, F Then since V
F
[A] < ∞,
: E
1
→ E
1
.
we have F = g−h
(9.2.49)
for some nondecreasing g and h (Theorem 3 in Chapter 5, §7). Now, by Theorem 3 from Chapter 7, §10, g and h are a.e. differentiable on A. Hence so is g−h = F.
Moreover, g
′
and h
′
≥0
≥0
(9.2.50)
since g ↑ and h ↑.
Thus for the L-integrability of F , proceed as in Problem 11 in §1, i.e., show that F is measurable on A and that ′
′
b
L∫
b ′
F
a
is finite. This yields the result.
b ′
= L∫
g −L∫
a
′
h
(9.2.51)
a
□
Theorem 9.2.2 (Lebesgue) If F
: E
1
→ E
n
(C
n
)
is absolutely continuous on A = [a, b], then the following are true:
(i*) F is a.e. differentiable, and F is L-integrable, on A . ′
(ii*) If, in addition, F
′
=0
a.e. on A, then F is constant on A .
Proof Assertion (i*) is immediate from Lemma 2, since any absolutely continuous function is of bounded variation by Corollary 2. (ii*) Now let F
′
=0
a.e. on A. Fix any B = [a, c] ⊆ A
and let Z consist of all p ∈ B at which the derivative F
′
=0
(9.2.52)
.
Given ε > 0, let K be the set of all closed intervals [p, x], p < x, such that ∣ F (x) − F (p) ∣ ∣ ΔF ∣ ∣ ∣ =∣ ∣ < ε. ∣ Δx ∣ x −p ∣ ∣
(9.2.53)
By assumption, ΔF lim x→p
and m(B − Z) = 0; B = [a, c] ∈ M
∗
.
=0
(p ∈ Z),
(9.2.54)
Δx
If p ∈ Z, and x − p is small enough, then ∣ ΔF ∣ ∣ ∣ < ε, ∣ Δx ∣
(9.2.55)
i.e., [p, x] ∈ K . It easily follows that K covers Z in the Vitali sense (verify!); so for any δ > 0, Theorem 2 of Chapter 7, §10 yields disjoint intervals Ik = [ pk , xk ] ∈ K, Ik ⊆ B,
(9.2.56)
with q
m
∗
(Z − ⋃ Ik ) < δ,
(9.2.57)
k=1
9.2.6
https://math.libretexts.org/@go/page/19219
hence also q
m (B − ⋃ Ik ) < δ
(9.2.58)
k=1
(for m(B − Z) = 0 ). But q
q−1
B − ⋃ Ik
= [a, c] − ⋃ [ pk , xk ]
k=1
k=1 q−1
= [a, p1 ) ∪ ⋃ [ xk , pk+1 ) ∪ [ xq , c]
( if xk < pk < xk+1 ) ;
k=1
so q
q−1
m (B − ⋃ Ik ) = (p1 − a) + ∑ (pk+1 − xk ) + (c − xq ) < δ. k=1
(9.2.59)
k=1
Now, as F is absolutely continuous, we can choose δ > 0 so that (3) implies q−1
|F (p1 ) − F (a)| + ∑ |F (pk+1 ) − F (xk )| + |F (c) − F (xq )| < ε.
(9.2.60)
k=1
But I
k
∈ K
also implies |F (xk ) − F (pk )| < ε (xk − pk ) = ε ⋅ m Ik .
(9.2.61)
Hence ∣
q
q
∣
∣∑ [F (xk ) − F (pk )] ∣ < ε ∑ m Ik ≤ ε ⋅ mB = ε(c − p). ∣ k=1
∣
(9.2.62)
k=1
Combining with (4), we get |F (c) − F (a)| ≤ ε(1 + c − a) → 0 as ε → 0;
so F (c) = F (a). As c ∈ A was arbitrary, F is constant on A, as claimed.
(9.2.63)
□
Note 2. This shows that Cantor's function (Problem 6 of Chapter 4, §5) is not absolutely continuous, even though it is continuous and monotone, hence of bounded variation on [0, 1]. Indeed (see Problem 2 in §1), it has a zero derivative a.e. on [0, 1] but is not constant there. Thus absolute continuity, as now defined, differs from its "weak" counterpart (Chapter 5, §8).
Theorem 9.2.3 A map F
: E
1
→ E
1
(C
n
)
is absolutely continuous on A =
[a, b]
F = L∫
f
iff on A
(9.2.64)
for some function f ; and then x
F (x) = F (a) + L ∫
f,
x ∈ A.
(9.2.65)
a
Briefly: Absolutely continuous maps are exactly all L-primitives. Proof If F
= L ∫ f,
then by Theorem 1 of §1, F is absolutely continuous on A, and by Note 1,
9.2.7
https://math.libretexts.org/@go/page/19219
x
F (x) = F (a) + L ∫
f,
x ∈ A.
(9.2.66)
a
Conversely, if F is absolutely continuous, then by Theorem 2, it is a.e. differentiable and A ). Let
F
′
=f
is L-integrable (all on
x
H (x) = L ∫
f,
x ∈ A.
(9.2.67)
a
Then H , too, is absolutely continuous and so is F
− H.
Also, by Theorem 3 of §1,
H
′
′
=f =F ,
(9.2.68)
and so ′
(F − H ) = 0
By Theorem 2, F
a.e. on A.
(9.2.69)
i.e.,
− H = c;
x
F (x) = c + H (x) = c + L ∫
f,
(9.2.70)
a
and so F
= L∫ f
on A, as claimed.
□
Corollary 9.2.3 If f , F
: E
1
→ E
∗
(E
n
,C
n
we have
),
F = L∫
on an interval I
⊆E
1
iff F is absolutely continuous on I and F
′
f
=f
(9.2.71)
a.e. on I .
(Use Problem 3 in §1 and Theorem 3.) Note 3. This (or Theorem 3) could serve as a definition. Comparing ordinary primitives F =∫
f
(9.2.72)
with L-primitives F = L∫
f,
(9.2.73)
we see that the former require F to be just relatively continuous but allow only a countable "exceptional" set require absolute continuity but allow Q to even be uncountable, provided mQ = 0 .
Q,
while the latter
The simplest and "strongest" kind of absolutely continuous functions are so-called Lipschitz maps (see Problem 6). See also Problems 7 and 10. III. We conclude with another important idea, due to Lebesgue.
Definition We call p ∈ E a Lebesgue point ("L-point") of f 1
(i) f is L-integrable on some G
: E
1
→ E
iff
;
p (δ)
(ii) q = f (p) is finite; and (iii) lim
x→p
1 x−p
L∫
x
p
|f − q| = 0
.
The Lebesgue set of f consists of all such p.
9.2.8
https://math.libretexts.org/@go/page/19219
Corollary 9.2.4 Let F = L∫
f
on A = [a, b].
(9.2.74)
If p ∈ A is an L-point of f , then f (p) is the derivative of F at p (but the converse fails). Proof By assumption, x
F (x) = c + L ∫
f,
x ∈ Gp (δ),
(9.2.75)
p
and 1
x
∣ ∣L ∫
|Δx| ∣
∣
x
1
(f − q)∣ ≤ ∣
p
L∫ |Δx|
|f − q| → 0
(9.2.76)
p
as x → p. (Here q = f (p) and Δx = x − p .) Thus with x → p, we get ∣ F (x) − F (p) ∣ ∣
∣ − q∣
x −p
∣
∣L ∫ |x − p| ∣
∣ x
∣L ∫ |x − p| ∣
∣ f − (x − p)q ∣
p
∣
1 =
as required.
x
∣
1 =
x
f −L∫
p
p
∣ (q)∣ → 0, ∣
□
Corollary 9.2.5 Let f
: E
1
→ E
n
(C
n
).
Then p is an L-point of f iff it is an L-point for each of the n components, f
1,
… , fn ,
of f .
Proof (Exercise!)
Theorem 9.2.4 If f
: E
1
→ E
∗
(E
n
,C
n
)
is L-integrable on A = [a, b], then almost all p ∈ A are Lebesgue points of f .
Note that this strengthens Theorem 3 of §1. Proof By Corollary 5, we need only consider the case f For any r ∈ E
1
, |f − r|
: E
1
→ E
∗
.
is L-integrable on A; so by Theorem 3 of §1, setting x
Fr (x) = L ∫
|f − r|,
(9.2.77)
|f − r| = |f (p) − r|
(9.2.78)
a
we get x
1
′
Fr (p) = lim x→p
L∫ |x − p|
p
for almost all p ∈ A .
9.2.9
https://math.libretexts.org/@go/page/19219
Now, for each r, let A be the set of those p ∈ A for which (5) fails; so mA in E . Let r
r
Let {r
= 0.
k}
be the sequence of all rationals
1
∞
Q = ⋃ Ark ∪ {a, b} ∪ A∞ ,
(9.2.79)
k=1
where A∞ = A(|f | = ∞);
(9.2.80)
so mQ = 0. (Why?) To finish, we show that all p ∈ A − Q are L-points of f . Indeed, fix any p ∈ A − Q and any ε > 0. Let q = f (p). Fix a rational r such that ε |q − r|
0 such that |x − p| < δ implies ∣
x
1
∣( ∣
L∫ |x − p|
∣
ε
|f − r|) −∣ f (p) − r|| < ∣
p
.
(9.2.85)
3
As ε |f (p) − r| = |q − r|
0, there is some b ∈ [a, ∞) = A
(9.3.41)
such that ∣ ∣∫ ∣ a
x
∣ 1 f dm − r∣ < ε ∣ 2
for x ≥ b.
(9.3.42)
(Why may we use the standard metric here?) Taking x = b, we get (2'). Also, if a ≤ b ≤ v ≤ x, we have x
∣ ∣∫ ∣ a
∣ 1 f dm − r∣ < ε ∣ 2
(9.3.43)
and ∣ ∣r − ∫ ∣ a
ν
∣ 1 f dm ∣ < ε. ∣ 2
(9.3.44)
Hence by the triangle law, (2) follows also. Thus this b satisfies (2). Conversely, suppose such a define F : A → E by
b
exists for every given
ε > 0.
Fixing
b,
we thus have (2) and (2'). Now, with
A = [a, ∞),
x
F (x) = ∫
f dm,
(9.3.45)
a
so ∞
C ∫
f = lim F (x)
(9.3.46)
x→∞
a
if this limit exists. By (2), ∣ |F (x)| = ∣∫ ∣ a
x
∣ ∣ f dm ∣ ≤ ∣∫ ∣ ∣ a
b
∣ ∣ f dm ∣ + ∣∫ ∣ b ∣
x
∣ ∣ f dm ∣ < ∣∫ ∣ ∣ a
b
∣ f dm ∣ + ε
(9.3.47)
∣
if x ≥ b. Thus F is finite on [b, ∞), and so we may again use the standard metric ∣ ρ(F (x), F (v)) = |F (x) − F (v)| = ∣∫ ∣ a
x
9.3.5
v
f dm − ∫ a
∣ ∣ f dm ∣ ≤ ∣∫ ∣ ∣ v
x
∣ f dm ∣ < ε ∣
(9.3.48)
https://math.libretexts.org/@go/page/19220
if x, v ≥ b. The existence of ∞
C ∫
f dm = lim F (x) ≠ ±∞
(9.3.49)
x→∞
a
now follows by Theorem 2 of Chapter 4, §2. (We shall henceforth presuppose this "starred" theorem.) Thus all is proved.
□
Corollary 9.3.1 Under the same assumptions as in Theorem 2, the convergence of ∞
C ∫
|f |dm
(9.3.50)
f dm.
(9.3.51)
a
implies that of ∞
C ∫ a
Indeed, ∣ ∣∫ ∣
x
v
x
∣ f∣ ≤ ∫ ∣
|f |
(9.3.52)
v
(Theorem 1(g) of Chapter 8, §5, and Problem 10 in Chapter 8, §7). Note 4. We say that C ∫ f converges absolutely iff C ∫ |f | converges.
Corollary 9.3.2 (comparison test) If |f | ≤ |g| a.e. on A = [a, ∞) for some f , g : E
1
→ E,
then
∞
C ∫
∞
|f | ≤ C ∫
a
|g|;
(9.3.53)
a
so the convergence of ∞
C ∫
|g|
(9.3.54)
|f |.
(9.3.55)
a
implies that of ∞
C ∫ a
For as |f |, |g| ≥ 0, Theorem 1 reduces all to Theorem 1(c) of Chapter 8, §5. Note 5. As we see, absolutely convergent C-integrals coincide with proper (finite) Lebesgue integrals of nonnegative or measurable maps. For conditional (i.e., nonabsolute) convergence, see Problems 6-9, 13, and 14.
m
-
Iterated C-Integrals. Let the product space X × Y of Chapter 8, §8 be E
1
×E
1
=E
2
,
(9.3.56)
and let p = m × n, where m and n are Lebesgue measure or LS measures in E . Let 1
A = [a, b], B = [c, d], and D = A × B.
(9.3.57)
Then the integral ∫ B
∫ A
f dmdn = ∫ Y
9.3.6
∫
f CD dmdn
(9.3.58)
X
https://math.libretexts.org/@go/page/19220
is also written d
∫
b
∫
f dmdn
(9.3.59)
f (x, y)dm(x)dn(y).
(9.3.60)
c
a
or d
∫
b
∫
c
a
As usual, we write "dx" for "dm(x)" if m is Lebesgue measure in E
1
similarly for n.
;
We now define ∞
C ∫ a
∞
∫
b
f dndm = lim ∫ b→∞
c
d
( lim ∫ d→∞
a ∞
=C ∫
f (x, y)dn(y)) dm(x)
c
∞
∫
a
f (x, y)dn(y)dm(x),
c
provided the limits and integrals involved exist. If the integral (3) is finite, we say that it converges. Again, convergence is absolute if it holds also with conditional otherwise. Similar definitions apply to ∞
C ∫
∞
∫
c
b
f dmdn, C ∫
a
f
replaced by
|f |,
and
∞
∫
−∞
f dndm, etc.
(9.3.61)
c
Theorem 9.3.3 Let f
: E
2
→ E
be p-measurable on E (p, m, n as above). Then we have the following.
∗
2
(i*) The Cauchy integrals ∞
C ∫
∞
∫
−∞
∞
|f |dndm and C ∫
−∞
∞
∫
−∞
|f |dmdn
(9.3.62)
−∞
exist (≤ ∞), and both equal ∫
|f |dp.
E
(9.3.63)
2
(ii*) If one of these three integrals is finite, then ∞
C ∫ −∞
∞
∫
∞
f dndm and C ∫
−∞
−∞
∞
∫
f dmdn
(9.3.64)
−∞
converge, and both equal ∫
f dp.
E
(Similarly for C ∫
∞
a
∫
b
−∞
f dndm,
(9.3.65)
2
etc.)
Proof As m and n are σ -finite (finite on intervals!), f surely has σ -finite support. As |f | ≥ 0, clause (i*) easily follows from our present Theorem 1(i) and Theorem 3(i) of Chapter 8, §8. Similarly, clause (ii*) follows from Theorem 3(ii) of the same section.
9.3.7
□
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Theorem 9.3.4 (passage to polars) Let p = Lebesgue measure in E
2
.
Suppose f
: E
2
→ E
∗
is p-measurable on E
F (r, θ) = f (r cos θ, r sin θ),
2
.
Set
r > 0.
(9.3.66)
Then ∞
(a) C ∫
−∞
(b) C ∫
∞
0
∫
∞
−∞
∫
∞
f dxdy = C ∫
f dxdy = C ∫
0
∞
0
∞
0
rdr ∫
2π
0
π/2
rdr ∫
0
F dθ,
and
F dθ,
provided f is nonnegative or p-integrable on E (for (a)) or on (0, ∞) × (0, ∞) (for (b)). 2
Proof Outline First let f
= CD ,
with D a "curved rectangle" {(r, θ)| r1 < r ≤ r2 , θ1 < θ ≤ θ2 }
for some r
1
< r2
in X = (0, ∞) and θ
1
in Y
< θ2
By elementary geometry (or calculus), the area
= [0, 2π). 1
2
pD =
(r
2
2
(9.3.67)
2
− r ) (θ2 − θ1 )
(9.3.68)
1
(the difference between two circular sectors). For f
= CD ,
formulas (a) and (b) easily follow from pD = L ∫
CD dp.
E
(9.3.69)
2
(Verify!) Now, curved rectangles behave like half-open intervals (r1 , r2 ] × (θ1 , θ2 ]
(9.3.70)
in E , since Theorem 1 in Chapter 7, §1, and Lemma 2 of Chapter 7, §2, apply with the same proof. Thus they form a semiring generating the Borel field in E . 2
2
Hence show (as in Chapter 8, §8 that Theorem 4 holds for f = C (D ∈ B ). Then take D ∈ M . Next let f be elementary and nonnegative, and so on, as in Theorems 2 and 3 in Chapter 8, §8. □ ∗
D
Examples (continued) (D) Let ∞ 2
J = L∫
e
−x
dx;
(9.3.71)
0
so ∞
J
2
∞ 2
= (C ∫
e
−x
dx) (C ∫
0
2
dy)
∞ 2
0
−y
0
∞
=C ∫
e
∫
e
−(x +y
2
)
dxdy.
(Why?)
0
Set 2
f (x, y) = e
in Theorem 4(b). Then F (r, θ) = e
2
−r
;
−(x +y
2
)
(9.3.72)
hence
9.3.8
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π
∞
J
2
=C ∫
2
2
rdr ( ∫
0
e
dθ)
0 ∞
π
2
=C ∫
−r
re
−r
=− 2
0
∞
1
dr ⋅
πe
−t
4
∣ ∣ ∣
0
1 =
π. 4
(Here we computed 2
∫
by substituting r
2
=t
re
−r
dr
(9.3.73)
.) Thus ∞
∞ 2
C ∫ 0
e
−x
2
dx = L ∫
e
−x
0
−− − 1 1 − π = √π . 4 2
dx = √
(9.3.74)
This page titled 9.3: Improper (Cauchy) Integrals is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
9.3.9
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9.3.E: Problems on Cauchy Integrals Exercise 9.3.E. 1 Fill in all proof details in Theorems 1 − 3. Verify also at least some of the cases other than ∫ integrals (footnote 6).
∞
f.
a
Check the validity for
LS
-
Exercise 9.3.E. 2 Prove Theorem 4 in detail.
Exercise 9.3.E. 2
′
Verify Notes 2 and 3 and examples (A)-(D).
Exercise 9.3.E. 3 Assuming a > 0, verify the following: (i) ∫ e dt ≤ ∫ e dt = . [Hint: Use Corollary 2.] (ii) ∫ e dt = . (iii) ∫ e dt = . (iv) ∫ e sin btdt = . ∞
1
1
∞
−t
∞
1
−t
1
t
e
−at
1
∞
−a
a 1
−at
0 ∞
e
a
−at
b
2
0
a2 +b
Exercise 9.3.E. 4 Verify the following: (i) ∫ ∫ e dydx = ∫ e dx ≤ ( converges, by 3(i)) . (ii) ∫ ∫ e dydx ≥ ∫ ∫ e dydx = ∫ (1 − e ) dx ≥ ∫ Does this contradict formula (4) in the text, or Problem 5, which follows? ∞
1
∞
0
∞
1
∞
∞
−xy
1
−xy
0
∞
1
1
x
−x
∞
1 e
−xy
0
∞
1
1
∞
−x
x
1
1
(
x
−e
−x
) dx = ∞
.
Exercise 9.3.E. 5 Let f (x, y) = e
−xy
and 1
g(x) = L ∫
e
−xy
dy;
(9.3.E.1)
0
so g(0) = 1. (Why?) (i) Is g R-integrable on A = [0, 1]? Is f so on A × A? (ii) Find g(x) using Corollary 1 in §1. (iii) Find the value of 1
R∫ 0
1
∫
1
e
−xy
dydx = R ∫
0
g
(9.3.E.2)
0
to within 1/10. [Hint: Reduce it to Problem 6(b) in §1.]
9.3.E.1
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Exercise 9.3.E. 6 . Let f , g : E (i) If ⇒ 6
1
∗
→ E
be m-measurable on A = [a, b), b ≤ ∞. Prove the following: b−
C ∫
b−
f
+
< ∞ or C ∫
a
then C ∫
b−
a
f
b−
b−
f
< ∞,
(9.3.E.3)
f dm( proper ).
(9.3.E.4)
b−
f
+
−C ∫
a
a
−
exists and equals C ∫
(ii) If ∫
f
a
f
−
=∫
a
A
converges conditionally only, then b−
∫
b−
f
+
=∫
a
(iii) In case C ∫
b−
a
|f | < ∞,
f
−
= +∞.
(9.3.E.5)
a
we have b−
C ∫
|f ± g| = ∞
(9.3.E.6)
a
iff C ∫
b−
a
|g| = ∞;
also, b−
C ∫
b−
(f ± g) = C ∫
a
if C ∫
b−
a
g
b−
f ±C ∫
a
g
(9.3.E.7)
a
exists (finite or not).
Exercise 9.3.E. 7 ⇒ 7
. Suppose f
: E
1
→ E
∗
is m-integrable and sign-constant on each An = [ an , an+1 ) ,
but changes sign from A to A n
n+1 ,
n = 1, 2, …
(9.3.E.8)
with ∞
⋃ An = [a, ∞)
(9.3.E.9)
n=1
and {a } ↑ fixed. Prove that if n
∣
∣
∣∫ ∣
f dm ∣ ↘ 0
(9.3.E.10)
∣
An
as n → ∞, then ∞
c∫
f
(9.3.E.11)
a
9.3.E.2
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converges. [Hint: Use Problem 10 in Chapter 4, §13.]
Exercise 9.3.E. 8 ⇒ 8
. Let sin x f (x) =
,
f (0) = 1.
(9.3.E.12)
x
Prove that ∞
C ∫
f (x)dx
(9.3.E.13)
0
converges conditionally only. [Hints: Use Problem 7. Show that ∞
C ∫
∞
|f | = L ∫
0
|f | = L ∫
(0,∞)
∞
f
+
= L∫
0
f
−
= ∞. ]
(9.3.E.14)
0
Exercise 9.3.E. 9 ⇒ 9
. (Additivity.) Given f
: E
1
, suppose that
→ E(E complete) and a < b < c ≤ ∞ x
∫
f dm ≠ ±∞
(9.3.E.15)
a
(proper) exists for each x ∈ [a, c). Prove the following: (a) C ∫ f and C ∫ f converge. (b) If b−
a
b
a+
c−
C ∫
f
(9.3.E.16)
b
converges, so does c−
C ∫ a
b−
f =C ∫
c−
f +C ∫
a
f.
(9.3.E.17)
b
(c) Countable additivity does not necessarily hold for C-integrals. [Hint: Use Problem 8 suitably splitting [0, ∞).]
Exercise 9.3.E. 10 (Refined comparison test.) Given f , g : E (i) If for some a < b and k ∈ E ,
1
→ E(E complete) and b ≤ ∞,
prove the following:
1
|f | ≤ |kg|
on [a, b)
(9.3.E.18)
then
9.3.E.3
https://math.libretexts.org/@go/page/25155
b−
∫
b−
|g| < ∞ implies ∫
a
|f | < ∞.
(9.3.E.19)
a
(ii) Such a, k ∈ E do exist if 1
|f (t)| lim t→b−
−1 ; (iii) ∫ t dt = ∞ . ∞
p
1
1
p
0+ ∞
p
0+
Exercise 9.3.E. 12 Use Problems 10 and 11 to test for convergence of the following: (a) ∫ ; 3/2
∞
t
0
2
∞
(b) ∫
dt
1
2 t√1+t
;
P (t)
∞
(c) ∫
dt
1+t
a
dt
Q(t)
(Q, P polynomials of degree s and r, s > r; Q ≠ 0 for t ≥ a)
(d) ∫
1−
(e) ∫
1
dt
0
√1−t4 p
t
0+ 1−
(f) ∫
0 π
(g) ∫
2
ln tdt
dt ln t
;
;
−
0+
;
;
p
tan
tdt
.
Exercise 9.3.E. 13 . (The Abel-Dirichlet test.) Given f , g : E (a) f ↓, with lim f (t) = 0 ; (b) g is L-measurable on A = [a, ∞); and; (c) (∃K ∈ E ) (∀x ∈ A) ∣∣L ∫ g∣∣ < K . Then C ∫ f (x)g(x)dx converges. [Outline: Set ⇒ 13
1
→ E
1
,
suppose that
t→∞
1
∞
x
a
a
x
G(x) = ∫
g;
(9.3.E.22)
a
so |G| < K on A. By Lemma 2 of §1, f g is L-integrable on each [u, v] ⊂ A, and (∃c ∈ [u, v]) such that
9.3.E.4
https://math.libretexts.org/@go/page/25155
∣ ∣L ∫ ∣ u
v
∣ ∣ f g∣ = ∣f (u) ∫ ∣ ∣ u
c
∣ g∣ = |f (u)[G(c) − G(u)]| < 2Kf (u). ∣
(9.3.E.23)
Now, by (a), ε (∀ε > 0)(∃k ∈ A)(∀u ≥ k)
|f (u)|
0)(∃b > a)(∀x ≥ b)
x
a
∣ f (t, u)dm(t) − F (u)∣ < ε, ∣
(9.4.6)
so |F | < ∞ on B . Here b depends on both ε and u (convergence is "pointwise"). However, it may occur that one and the same b fits all u ∈ B, so that b depends on ε alone. We then say that ∞
C ∫
f (t, u)dm(t)
(9.4.7)
a
converges uniformly on B (i.e., for u ∈ B ), and write ∞
F (u) = C ∫
f (t, u)dm(t) (uniformly) on B.
(9.4.8)
a
Explicitly, this means that ∣ ∣∫ ∣ a
(∀ε > 0)(∃b > a)(∀u ∈ B)(∀x ≥ b)
x
∣ f (t, u)dm(t) − F (u)∣ < ε. ∣
(9.4.9)
Clearly, this implies (1), but not conversely. We now obtain the following.
Theorem 9.4.1 (Cauchy criterion) Suppose x
∫
f (t, u)dm(t)
(9.4.10)
a
exists for x ≥ a and u ∈ B ⊆ E
1
.
(This is automatic if E ⊆ E
∗
;
9.4.1
see Chapter 8, §5.)
https://math.libretexts.org/@go/page/19221
Then ∞
C ∫
f (t, u)dm(t)
(9.4.11)
a
converges uniformly on B iff for every ε > 0, there is b > a such that ∣ ∣∫ ∣ v
(∀v, x ∈ [b, ∞))(∀u ∈ B)
x
∣ f (t, u)dm(t)∣ < ε, ∣
(9.4.12)
and b
∣ ∣∫ ∣
∣ f (t, u)dm(t)∣ < ∞.
(9.4.13)
∣
a
Proof The necessity of (3) follows as in Theorem 2 of §3. (Verify!) To prove sufficiency, suppose the desired b exists for every ε > 0. Then for each (fixed) u ∈ B , ∞
C ∫
f (t, u)dm(t)
(9.4.14)
a
satisfies Theorem 2 of §3. Hence x
F (u) = lim ∫ x→∞
f (t, u)dm(t) ≠ ±∞
(9.4.15)
a
exists for every u ∈ B (pointwise). Now, from (3), writing briefly ∫ f for ∫ f (t, u)dm(t), we obtain ∣ ∣∫ ∣
x
v
x
∣ ∣ f ∣ = ∣∫ ∣ ∣
v
f −∫
a
a
∣ f∣ < ε ∣
(9.4.16)
for all u ∈ B and all x > v ≥ b . Making x → ∞ (with u and v temporarily fixed), we have by (4) that ∣ ∣F (u) − ∫ ∣ a
v
∣ f∣ ≤ ε ∣
(9.4.17)
whenever v ≥ b . But by our assumption, convergence of
b
depends on
ε
alone (not on
u
). Thus unfixing
u
, we see that (5) establishes the uniform
∞
∫
f,
(9.4.18)
f (t, u)dm(t)
(9.4.19)
|f (t, u)|dm(t)
(9.4.20)
a
as required.
□
Corollary 9.4.1 Under the assumptions of Theorem 1, ∞
C ∫ a
converges uniformly on B if ∞
C ∫ a
does.
9.4.2
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Indeed, ∣ ∣∫ ∣ v
x
∣ f∣ ≤ ∫ ∣ v
x
|f | < ε.
(9.4.21)
Corollary 9.4.2 (comparison test) Let f
: E
2
→ E
and M
: E
2
→ E
∗
satisfy |f (t, u)| ≤ M (t, u)
(9.4.22)
for u ∈ B ⊆ E and t ≥ a . 1
Then ∞
C ∫
|f (t, u)|dm(t)
(9.4.23)
M (t, u)dm(t)
(9.4.24)
a
converges uniformly on B if ∞
C ∫ a
does. Indeed, Theorem 1 applies, with ∣ ∣∫ ∣ v
x
∣ f∣ ≤ ∫ ∣ v
x
M < ε.
(9.4.25)
Hence we have the following corollary.
Corollary 9.4.3 ("M -test") Let f
: E
2
→ E
and M
: E
1
→ E
∗
satisfy |f (t, u)| ≤ M (t)
(9.4.26)
for u ∈ B ⊆ E and t ≥ a. Suppose 1
∞
C ∫
M (t)dm(t)
(9.4.27)
|f (t, u)|dm(t)
(9.4.28)
a
converges. Then ∞
C ∫ a
converges (uniformly) on B. So does ∞
C ∫
f (t, u)dm(t)
(9.4.29)
a
by Corollary 1. Proof Set h(t, u) = M (t) ≥ |f (t, u)|.
(9.4.30)
Then Corollary 2 applies (with M replaced by h there). Indeed, the convergence of C ∫
h =C ∫
9.4.3
M
(9.4.31)
https://math.libretexts.org/@go/page/19221
is trivially "uniform" for u ∈ B, since M does not depend on u at all. Note 1. Observe also that, if equivalent.
h(t, u)
does not depend on
u,
□
then the (pointwise) and (uniform) convergence of
C ∫ h
are trivially
We also have the following result.
Corollary 9.4.4 Suppose ∞
C ∫
f (t, u)dm(t)
(9.4.32)
a
converges (pointwise) on B ⊆ E
1
.
Then this convergence is uniform iff ∞
lim C ∫ ν →∞
f (t, u)dm(t) = 0 (uniformly) on B,
(9.4.33)
v
i.e., iff (∀ε > 0)(∃b > a)(∀u ∈ B)(∀v ≥ b)
∣ ∣C ∫ ∣ v
∞
∣ f (t, u)dm(t)∣ < ε. ∣
(9.4.34)
Proof The proof (based on Theorem 1) is left to the reader, along with that of the following corollary.
Corollary 9.4.5 Suppose b
∫
f (t, u)dm(t) ≠ ±∞
(9.4.35)
a
exists for each u ∈ B ⊆ E . 1
Then ∞
C ∫
f (t, u)dm(t)
(9.4.36)
f (t, u)dm(t)
(9.4.37)
a
converges (uniformly) on B iff ∞
C ∫ b
does. II. The Abel-Dirichlet tests for uniform convergence of series (Problems 9 and 11 in Chapter 4, §13) have various analogues for Cintegrals. We give two of them, using the second law of the mean (Corollary 5 in §1). First, however, we generalize our definitions, "unstarring" some ideas of Chapter 4, §11. Specifically, given H : E
2
→ E (E complete),
we say that H (x, y) converges to F (y), uniformly on B, as x → q (q ∈ E
∗
),
(9.4.38)
and write
lim H (x, y) = F (y) (uniformly) on B
(9.4.39)
x→q
iff we have (∀ε > 0) (∃G¬q ) (∀y ∈ B) (∀x ∈ G¬q )
9.4.4
|H (x, y) − F (y)| < ε;
(9.4.40)
https://math.libretexts.org/@go/page/19221
hence |F | < ∞ on B . If here q = ∞, the deleted globe G
¬q
has the form (b, ∞). Thus if x
H (x, u) = ∫
f (t, u)dt,
(9.4.41)
a
(6) turns into (2) as a special case. If (6) holds with " (∃G only.
¬q
As in Chapter 8, §8, we denote by f (⋅, y), or f
y
,
and " (∀y ∈ B) " interchanged, as in (1), convergence is pointwise
) "
the function of x alone (on E ) given by 1
f
y
(x) = f (x, y).
(9.4.42)
fx (y) = f (x, y).
(9.4.43)
Similarly,
Of course, we may replace f (x, y) by f (t, u) or H (t, u), etc. We use Lebesgue measure in Theorems 2 and 3 below.
Theorem 9.4.2 Assume f , g : E (i) C ∫
∞
a
2
→ E
g(t, u)dt
(ii) each g
u
(iii) each f
satisfy
converges (uniformly) on B ;
(u ∈ B)
u
1
is L-measurable on A = [a, ∞) ;
(u ∈ B)
is monotone (↓ or ↑) on A; and
(iv) |f | < K ∈ E (bounded) on A × B . 1
Then ∞
C ∫
f (t, u)g(t, u)dt
(9.4.44)
a
converges uniformly on B . Proof Given ε > 0, use assumption (i) and Theorem 1 to choose b > a so that x
∣ ∣L ∫ ∣ v
∣ ε g(t, u)dt∣ < , ∣ 2K
(9.4.45)
written briefly as ∣ ∣L ∫ ∣ v
x
g
∣ ε ∣ < , ∣ 2K
u
(9.4.46)
for all u ∈ B and x > v ≥ b, with K as in (iv). Hence by (ii), each g (u ∈ B) is L-integrable on any interval [v, x] ⊂ A , with x > v ≥ b. Thus given such we can use (iii) and Corollary 5 from §1 to find that u
x
L∫
c
f
u
g
u
=f
u
v
(v)L ∫
u
and [v, x],
x
g
u
+f
u
(x)L ∫
v
g
u
(9.4.47)
c
for some c ∈ [v, x]. Combining with (7) and using (iv), we easily obtain ∣ ∣L ∫ ∣
v
x
∣ f (t, u)g(t, u)dt∣ < ε ∣
9.4.5
(9.4.48)
https://math.libretexts.org/@go/page/19221
whenever u ∈ B and x > v ≥ b. (Verify!) Our assertion now follows by Theorem 1.
□
Theorem 9.4.3 (Abel-Dirichlet test) Let f , g : E
2
(a) lim
f (t, u) = 0
t→∞
(b) each f (c) each g
∗
satisfy (uniformly) for u ∈ B ;
(u ∈ B)
is nonincreasing (↓) on A = [0, ∞) ;
(u ∈ B)
is L-measurable on A; and
u
u
→ E
(d) (∃K ∈ E
1
) (∀x ∈ A)(∀u ∈ B) ∣ ∣L ∫
x
a
.
g(t, u)dt∣ ∣ v ≥ a . Then use assumption (a) to fix k so that ε |f (t, u)|
k and u ∈ B.
□
Note 2. Via components, Theorems 2 and 3 extend to the case g : E
2
→ E
n
(C
n
).
Note 3. While Corollaries 2 and 3 apply to absolute convergence only, Theorems 2 and 3 cover conditional convergence, too (a great advantage!). The theorems also apply if f or g is independent of u (see Note 1). This supersedes Problems 13 and 14 in §3.
Examples (A) The integral ∞
sin tu
∫
dt
converges uniformly on B
δ
= [δ, ∞)
(9.4.52)
t
0
if δ > 0, and pointwise on B = [0, ∞) .
Indeed, we can use Theorem 3, with g(t, u) = sin tu
(9.4.53)
and 1 f (t, u) =
, f (0, u) = 1,
(9.4.54)
t
say. Then the limit
9.4.6
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1 lim
=0
t→∞
(9.4.55)
t
is trivially uniform for u ∈ B , as f is independent of u. Thus assumption (a) is satisfied. So is (d) because δ
∣ ∣∫ ∣ 0
x
xu
∣ ∣ 1 sin tudt∣ = ∣ ∫ ∣ ∣u 0
∣ 1 sin θdθ∣ ≤ ⋅ 2. ∣ δ
(9.4.56)
(Explain!) The rest is easy. Note that Theorem 2 fails here since assumption (i) is not satisfied. (B) The integral ∞
1
∫
e
−tu
sin atdt
(9.4.57)
t
0
converges uniformly on B = [0, ∞). It does so absolutely on B
= [δ, ∞),
δ
if δ > 0.
Here we shall use Theorem 2 (though Theorem 3 works, too). Set f (t, u) = e
−tu
(9.4.58)
and sin at g(t, u) =
, g(0, u) = a.
(9.4.59)
t
Then ∞
∫
g(t, u)dt
(9.4.60)
0
converges (substitute x = at in Problem 8 or 15 in §3). Convergence is trivially uniform, by Note 1. Thus assumption (i) holds, and so do the other assumptions. Hence the result. For absolute convergence on B , use Corollary 3 with δ
M (t) = e
so M
≥ |f g|
−δt
,
(9.4.61)
.
Note that, quite similarly, one treats C-integrals of the form ∞
∫
∞
e
−tu
2
g(t)dt, ∫
a
e
−t u
g(t)dt, etc.,
(9.4.62)
a
provided ∞
∫
g(t)dt
(9.4.63)
a
converges (a ≥ 0) . In fact, Theorem 2 states (roughly) that the uniform convergence of C ∫ g implies that of C ∫ f g, provided f is monotone (in t ) and bounded. III. We conclude with some theorems on uniform convergence of functions LS (or Lebesgue) measure in E ; the deleted globe G is fixed. 1
H : E
2
→ E
(see (6)). In Theorem 4,
m
is again an
∗ ¬q
Theorem 9.4.4 Suppose lim H (x, y) = F (y) (uniformly)
(9.4.64)
x→q
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for y ∈ B ⊆ E (i) If all H
x
1
Then we have the following:
.
are continuous or m-measurable on B, so also is F .
∗
(x ∈ G¬q )
(ii) The same applies to m-integrability on B, provided mB < ∞; and then lim ∫ x→q
| Hx − F | = 0;
(9.4.65)
B
hence lim ∫ x→q
Hx = ∫
B
F =∫
B
( lim Hx ) .
B
(9.4.66)
x→q
Formula (8') is known as the rule of passage to the limit under the integral sign. Proof (i) Fix a sequence x
k
→ q (xk
in the deleted globe G
∗ ¬q
),
and set
Hk = Hx
(k = 1, 2, …).
k
(9.4.67)
The uniform convergence H (x, y) → F (y)
(9.4.68)
is preserved as x runs over that sequence (see Problem 4). Hence if all H are continuous or measurable, so is F (Theorem 2 in Chapter 4, §12 and Theorem 4 in Chapter 8, §1. Thus clause (i) is proved. k
(ii) Now let all H be m-integrable on B; let x
mB < ∞.
(9.4.69)
Then the H are m-measurable on B, and so is F , by (i). Also, by (6), k
(∀ε > 0) (∃G¬q ) (∀x ∈ G¬q )
∫
| Hx − F | ≤ ∫
B
(ε) = εmB < ∞,
(9.4.70)
B
proving (8). Moreover, as ∫
| Hx − F | < ∞,
(9.4.71)
B
Hx − F
is m-integrable on B, and so is F = Hx − (Hx − F ) .
(9.4.72)
∣ ∣ ∣ ∣ ∣∫ Hx − ∫ F ∣ = ∣∫ (Hx − F )∣ ≤ ∫ | Hx − F | → 0, ∣ B ∣ ∣ B ∣ B B
(9.4.73)
Hence
as x → q, by (8). Thus (8') is proved, too.
□
Quite similarly (keeping E complete and using sequences), we obtain the following result.
Theorem 9.4.5 Suppose that (i) all H set Q;
x
(ii) lim
x→q
(iii) lim
∗
(x ∈ G
x→q
−q
)
are continuous and finite on a finite interval B ⊂ E , and differentiable on B − Q, for a fixed countable
H (x, y0 ) ≠ ±∞
1
exists for some y
D2 H (x, y) = f (y)
0
∈ B;
and
(uniformly) exists on B − Q .
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Then f , so defined, has a primitive F on B, exact on B − Q (so F
′
=f
on B − Q); moreover,
F (y) = lim H (x, y) (uniformly) for y ∈ B.
(9.4.74)
x→y
Outline of Proof Note that d D2 H (x, y) =
Hx (y).
(9.4.75)
dy
Use Theorem 1 of Chapter 5, §9, with F
n
= Hxn , xn → q.
□
Note 4. If x → q over a path P (clustering at q), one must replace G 4 and 5.
¬q
and G
∗ ¬q
by P
∩ G¬q
and P
∗
∩ G¬q
in (6) and in Theorems
This page titled 9.4: Convergence of Parametrized Integrals and Functions is shared under a CC BY 3.0 license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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9.4.E: Problems on Uniform Convergence of Functions and C-Integrals Exercise 9.4.E. 1 Fill in all proof details in Theorems 1 − 5, Corollaries 4 and 5, and examples (A) and (B).
Exercise 9.4.E. 1
′
Using (6), prove that lim H (x, y) (uniformly)
(9.4.E.1)
x→q
exists on B ⊆ E iff 1
′
′
(∀ε > 0) (∃G¬q ) (∀y ∈ B) (∀x, x ∈ G¬q )
|H (x, y) − H (x , y)| < ε.
(9.4.E.2)
Assume E complete and |H | < ∞ on G × B. [Hint: "Imitate" the proof of Theorem 1, using Theorem 2 of Chapter 4, §2.] ¬q
Exercise 9.4.E. 2 State formulas analogous to ( 1) and ( 2) for ∫
a
−∞
,∫
b−
a
,
and ∫
b
a+
.
Exercise 9.4.E. 3 State and prove Theorems 1 to 3 and Corollaries 1 to 3 for a
b−
∫
,∫
−∞
b
, and ∫
a
.
(9.4.E.3)
a+
In Theorems 2 and 3 explore absolute convergence for b−
∫
b
and ∫
a
.
(9.4.E.4)
a+
Do at least some of the cases involved. [Hint: Use Theorem 1 of §3 and Problem 1', if already solved.]
Exercise 9.4.E. 4 Prove that lim H (x, y) = F (y) (uniformly)
(9.4.E.5)
x→q
on B iff lim H (xn , ⋅) = F ( uniformly )
(9.4.E.6)
n→∞
on B for all sequences x → q (x ≠ q) . [Hint: "Imitate" Theorem 1 in Chapter 4, §2. Use Definition 1 of Chapter 4, §12.] n
n
9.4.E.1
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Exercise 9.4.E. 5 Prove that if lim H (x, y) = F (y) (uniformly)
(9.4.E.7)
x→q
on A and on B, then this convergence holds on A ∪ B. Hence deduce similar propositions on C -integrals.
Exercise 9.4.E. 6 Show that the integrals listed below violate Corollary 4 and hence do not converge uniformly on P = (0, δ) though proper Lintegrals exist for each u ∈ P . Thus show that Theorem 1 (ii) does not apply to uniform convergence. (a) ∫ ; 1
0+
(b) ∫
1
0+
2
2
1
0+
(c) ∫
udt
2
t −u
2
u −t 2
2
2
dt
;
(t +u ) 2
2
tu(t −u ) 2
2
(t +u )
2
dt
.
[ Hint for (b) : To disprove uniform convergence, fix any ε, v > 0. v
2
u
2
−t
∫ 2
(t
0
as u → 0. Thus if v
> ε. ]
(9.4.E.9)
2v
Exercise 9.4.E. 7 Using Corollaries 3 to 5, show that the following integrals converge (uniformly) on U (as listed) but only pointwise on P (for the latter, proceed as in Problem 6 ). Specify P and M (t) in each case where they are not given. (a) ∫ e dt; U = [δ, ∞); P = (0, δ) . ∞
2
−ut
0
[ Hint: Set M (t) = e
(b) ∫
∞
(c) ∫
1
0
0+ 1
e
−ut
u−1
t
a
t
−δt
for t ≥ 1 (Corollaries 3 and 5) . ]
cos tdt(a ≥ 0); U = [δ, ∞)
dt; U = [δ, ∞)
.
.
(d) ∫ t sin tdt; U = [0, δ], 0 < δ < 2; P [Hint: Fix v so small that −u
0+
.
1−δ
= [δ, 2); M (t) = t
sin t
1
(∀t ∈ (0, v))
> t
.
(9.4.E.10)
2
Then, if u → 2 , v
∫
−u
t
v
1 sin tdt ≥
0
∫ 2
0
dt u−1
→ ∞. ]
(9.4.E.11)
t
Exercise 9.4.E. 8 In example (A), disprove uniform convergence on P [Hint: Proceed as in Problem 6. ]
= (0, ∞)
.
9.4.E.2
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Exercise 9.4.E. 9 Do example (B) using Theorem 3 and Corollary 5. Disprove uniform convergence on B.
Exercise 9.4.E. 10 Show that ∞
sin tu
∫
cos tdt
(9.4.E.12)
t
0+
converges uniformly on any closed interval U , with ±1 ∉ U . [Hint: Transform into ∞
1
1
∫ 2
{sin[(u + 1)t] + sin[(u − 1)t]}dt. ]
(9.4.E.13)
t
0+
Exercise 9.4.E. 11 Show that ∞ 3
∫
t sin t
sin tudt
(9.4.E.14)
0
converges (uniformly) on any finite interval U . [Hint: Integrate y 3
∫
t sin t
sin tudt
(9.4.E.15)
x
by parts twice. Then let y → ∞ and x → 0. ]
Exercise 9.4.E. 12 Show that ∞
∫
e
−tu
cos t ta
0+
converges (uniformly) for u ≥ 0. [Hints: For t → 0+, use M (t) = t
−a
dt
(0 < a < 1)
(9.4.E.16)
. For t → ∞, use example (B) and Theorem 2. ]
Exercise 9.4.E. 13 Prove that ∞
cos tu
∫
a
dt
(0 < a < 1)
(9.4.E.17)
t
0+
converges (uniformly) for u ≥ δ > 0, but (pointwise) for u > 0. [Hint: Use Theorem 3 with g(t, u) = cos tu and ∣ ∣∫ ∣ 0
x
∣ 1 ∣ sin xu ∣ g∣ = ∣ ∣ ≤ . ∣ ∣ ∣ u δ
9.4.E.3
(9.4.E.18)
https://math.libretexts.org/@go/page/25154
For u > 0 , ∞
∞
cos tu
∫
ta
v
a−1
dt = u
cos z
∫
dz → ∞
(9.4.E.19)
z
vu
if v = 1/u and u → 0. Use Corollary 4. ]
Exercise 9.4.E. 14 Given A, B ⊆ E (mA < ∞) and f : E → E, suppose that (i) each f (x, ⋅) = f (x ∈ A) is relatively (or uniformly) continuous on B; and (ii) each f (⋅, y) = f (y ∈ B) is m-integrable on A . Set 1
2
x
y
F (y) = ∫
f (x, y)dm(x),
y ∈ B.
(9.4.E.20)
A
Then show that F is relatively (or uniformly) continuous on B. [Hint: We have (∀x ∈ A)(∀ε > 0) (∀y0 ∈ B)( ∃δ > 0) (∀y ∈ B ∩ Gy (δ)) 0
ε |F (y) − F (y0 )| ≤ ∫
|f (x, y) − f (x, y0 )| dm(x) ≤ ∫
A
A
(
) dm = ε. mA
Similarly for uniform continuity.]
Exercise 9.4.E. 15 Suppose that (a) C ∫ f (t, y)dm(t) = F (y)( uniformly ) on B = [b, d] ⊆ E ; (b) each f (x, ⋅) = f (x ≥ a) is relatively continuous on B; and (c) each f (⋅, y) = f (y ∈ B) is m -integrable on every [a, x] ⊂ E , x ≥ a. Then show that F is relatively continuous, hence integrable, on B and that ∞
1
a
x y
1
∫
F = lim ∫ x→∞
B
Hx ,
(9.4.E.21)
B
where x
H (x, y) = ∫
f (t, y)dm(t).
(9.4.E.22)
a
(Passage to the limit under the ∫ -sign.) [Hint: Use Problem 14 and Theorem 4; note that ∞
C ∫ 0
f (t, y)dm(t) = lim H (x, y)(uniformly). ]
(9.4.E.23)
x→∞
9.4.E: Problems on Uniform Convergence of Functions and C-Integrals is shared under a CC BY 1.0 license and was authored, remixed, and/or curated by LibreTexts.
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Index A
D
arc
Lebesgue Theory 9: Calculus Using Lebesgue Theory
Darboux property
4.10: Arcs and Curves. Connected Sets
4.9: The Intermediate Value Property
de Morgan's Duality Law
B
1.1: Sets and Operations on Sets. Quantifiers
Bolzano's theorem
Differentiable function 6.3: Differentiable Functions
4.9: The Intermediate Value Property
Borel Measures 7.7: Topologies. Borel Sets. Borel Measures
Borel Sets
M Metric Spaces 3.7: Metric Spaces
Monotone Function 4.5: Monotone Function
E
P
Euclidean Spaces
7.7: Topologies. Borel Sets. Borel Measures
3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces
parametric equation of the line 3.2: Lines and Planes in Eⁿ
C
F
Cauchy criterion
finite sequence
9.3: Improper (Cauchy) Integrals
Cauchy integrals 9.3: Improper (Cauchy) Integrals
Cauchy Invariant Rule
3.10: Cluster Points. Convergent Sequences
Compact Sets 4.6: Compact Sets
Compactness 4.7: More on Compactness
Completeness 3.13: Cauchy Sequences. Completeness
Complex Numbers 3.4: Complex Numbers
convergent sequences 3.10: Cluster Points. Convergent Sequences
Convex polygons 4.9: The Intermediate Value Property
8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition
I improper integrals 9.3: Improper (Cauchy) Integrals
Intermediate Value Property
3.13: Cauchy Sequences. Completeness
Cluster Point
Radon–Nikodym theorem
1.3: Sequences
6.4: The Chain Rule. The Cauchy Invariant Rule
Cauchy sequences
R
4.9: The Intermediate Value Property
Irrational Numbers 2.6: Powers with Arbitrary Real Exponents. Irrationals
J
S saddle point 1.3: Sequences
singularity 9.3: Improper (Cauchy) Integrals
U Uniform Continuity
Jacobian matrix 6.3: Differentiable Functions
4.8: Continuity on Compact Sets. Uniform Continuity
V
L Lebesgue decomposition 8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition
vector spaces 3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces
Lebesgue Measure 7.8: Lebesgue Measure
1
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Index A
D
arc
Darboux property
4.10: Arcs and Curves. Connected Sets
Lebesgue Theory 4.9: The Intermediate Value Property
de Morgan's Duality Law
B
1.1: Sets and Operations on Sets. Quantifiers
Bolzano's theorem
Differentiable function 6.3: Differentiable Functions
4.9: The Intermediate Value Property
Borel Measures 7.7: Topologies. Borel Sets. Borel Measures
Borel Sets
9: Calculus Using Lebesgue Theory
M Metric Spaces 3.7: Metric Spaces
Monotone Function 4.5: Monotone Function
E Euclidean Spaces
7.7: Topologies. Borel Sets. Borel Measures
3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces
P parametric equation of the line 3.2: Lines and Planes in Eⁿ
C
F
Cauchy criterion
finite sequence
9.3: Improper (Cauchy) Integrals
Cauchy integrals 9.3: Improper (Cauchy) Integrals
Cauchy Invariant Rule
1.3: Sequences
Cauchy sequences
improper integrals 9.3: Improper (Cauchy) Integrals
Intermediate Value Property
3.13: Cauchy Sequences. Completeness
Cluster Point 3.10: Cluster Points. Convergent Sequences
Compact Sets 4.6: Compact Sets
Compactness 4.7: More on Compactness
Completeness 3.13: Cauchy Sequences. Completeness
Complex Numbers 3.4: Complex Numbers
convergent sequences 3.10: Cluster Points. Convergent Sequences
Convex polygons 4.9: The Intermediate Value Property
Radon–Nikodym theorem 8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition
I
6.4: The Chain Rule. The Cauchy Invariant Rule
R
4.9: The Intermediate Value Property
Irrational Numbers 2.6: Powers with Arbitrary Real Exponents. Irrationals
J
S saddle point 1.3: Sequences
singularity 9.3: Improper (Cauchy) Integrals
U Uniform Continuity
Jacobian matrix 6.3: Differentiable Functions
4.8: Continuity on Compact Sets. Uniform Continuity
V
L Lebesgue decomposition 8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition
Lebesgue Measure 7.8: Lebesgue Measure
vector spaces 3.5: Vector Spaces. The Space Cⁿ. Euclidean Spaces
Glossary Sample Word 1 | Sample Definition 1
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5.8: Rectifiable Arcs. Absolute Continuity - CC BY 3.0
4.8.E: Problems on Uniform Continuity; Continuity on Compact Sets - CC BY 1.0
5.8.E: Problems on Absolute Continuity and Rectifiable Arcs - CC BY 1.0
4.9: The Intermediate Value Property - CC BY 3.0 4.9.E: Problems on the Darboux Property and Related Topics - CC BY 1.0
5.9: Convergence Theorems in Differentiation and Integration - CC BY 3.0
4.10: Arcs and Curves. Connected Sets - CC BY 3.0
5.9.E: Problems on Convergence in Differentiation and Integration - CC BY 1.0
4.10.E: Problems on Arcs, Curves, and Connected Sets - CC BY 1.0
5.10: Sufficient Condition of Integrability. Regulated Functions - CC BY 3.0
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5.10.E: Problems on Regulated Functions - CC BY 1.0
7.1: More on Intervals in Eⁿ. Semirings of Sets - CC BY 3.0
5.11: Integral Definitions of Some Functions - CC BY 3.0
7.1.E: Problems on Intervals and Semirings - CC BY 1.0
5.11.E: Problems on Exponential and Trigonometric Functions - CC BY 1.0
7.2: \(\mathcal{C}_{\sigma}\)-Sets. Countable Additivity. Permutable Series - CC BY 3.0
6: Differentiation on Eⁿ and Other Normed Linear Spaces - CC BY 3.0
7.2.E: Problems on \(\mathcal{C}_{\sigma}\) Sets, \(\sigma\) -Additivity, and Permutable Series - CC BY 1.0
6.1: Directional and Partial Derivatives - CC BY 3.0
7.3: More on Set Families - CC BY 3.0
6.1.E: Problems on Directional and Partial Derivatives - CC BY 1.0
7.3.E: Problems on Set Families - CC BY 1.0
6.2: Linear Maps and Functionals. Matrices - CC BY 3.0
7.4: Set Functions. Additivity. Continuity - CC BY 3.0 7.4.E: Problems on Set Functions - CC BY 1.0
6.2.E: Problems on Linear Maps and Matrices CC BY 1.0
7.5: Nonnegative Set Functions. Premeasures. Outer Measures - CC BY 3.0
6.3: Differentiable Functions - CC BY 3.0
7.5.E: Problems on Premeasures and Related Topics - CC BY 1.0
6.3.E: Problems on Differentiable Functions - CC BY 1.0
7.6: Measure Spaces. More on Outer Measures - CC BY 3.0
6.4: The Chain Rule. The Cauchy Invariant Rule - CC BY 3.0
7.6.E: Problems on Measures and Outer Measures - CC BY 1.0
6.4.E: Further Problems on Differentiable Functions - CC BY 1.0
7.7: Topologies. Borel Sets. Borel Measures - CC BY 3.0
6.5: Repeated Differentiation. Taylor’s Theorem - CC BY 3.0
7.7.E: Problems on Topologies, Borel Sets, and Regular Measures - CC BY 1.0
6.5.E: Problems on Repeated Differentiation and Taylor Expansions - CC BY 1.0
7.8: Lebesgue Measure - CC BY 3.0
6.6: Determinants. Jacobians. Bijective Linear Operators - CC BY 3.0
7.8.E: Problems on Lebesgue Measure - CC BY 1.0
6.6.E: Problems on Bijective Linear Maps and Jacobians - CC BY 1.0
7.9: Lebesgue–Stieltjes Measures - CC BY 1.0 7.9.E: Problems on Lebesgue-Stieltjes Measures CC BY 1.0
6.7: Inverse and Implicit Functions. Open and Closed Maps - CC BY 3.0
7.10: Generalized Measures. Absolute Continuity CC BY 1.0
6.7.E: Problems on Inverse and Implicit Functions, Open and Closed Maps - CC BY 1.0
7.10.E: Problems on Generalized Measures - CC BY 1.0
6.8: Baire Categories. More on Linear Maps - CC BY 3.0
7.11: Differentiation of Set Functions - CC BY 1.0
6.8.E: Problems on Baire Categories and Linear Maps - CC BY 1.0
7.11.E: Problems on Vitali Coverings - CC BY 1.0 8: Measurable Functions and Integration - CC BY 3.0
6.9: Local Extrema. Maxima and Minima - CC BY 3.0
8.1: Elementary and Measurable Functions - CC BY 3.0
6.9.E: Problems on Maxima and Minima - CC BY 1.0
8.1.E: Problems on Measurable and Elementary Functions in \((S, \mathcal{M})\) - CC BY 1.0
6.10: More on Implicit Differentiation. Conditional Extrema - CC BY 3.0
8.2: Measurability of Extended-Real Functions - CC BY 3.0
6.10.E: Further Problems on Maxima and Minima - CC BY 1.0
8.2.E: Problems on Measurable Functions in \((S, \mathcal{M}, m)\) - CC BY 1.0
7: Volume and Measure - CC BY 3.0
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8.3: Measurable Functions in \((S, \mathcal{M}, m)\) - CC BY 3.0
8.10.E: Problems on Generalized Integration - CC BY 1.0
8.3.E: Problems on Measurable Functions in \((S, \mathcal{M}, m)\) - CC BY 1.0
8.11: The Radon–Nikodym Theorem. Lebesgue Decomposition - CC BY 1.0
8.4: Integration of Elementary Functions - CC BY 3.0
8.11.E: Problems on Radon-Nikodym Derivatives and Lebesgue Decomposition - CC BY 1.0
8.4.E: Problems on Integration of Elementary Functions - CC BY 1.0
8.12: Integration and Differentiation - CC BY 1.0
8.5: Integration of Extended-Real Functions - CC BY 3.0
8.12.E: Problems on Differentiation and Related Topics - CC BY 1.0
8.5.E: Problems on Integration of Extended-Real Functions - CC BY 1.0
9: Calculus Using Lebesgue Theory - CC BY 3.0 9.1: L-Integrals and Antiderivatives - CC BY 3.0
8.6: Integrable Functions. Convergence Theorems CC BY 3.0
9.1.E: Problems on L-Integrals and Antiderivatives - CC BY 1.0
8.6.E: Problems on Integrability and Convergence Theorems - CC BY 1.0
9.2: More on L-Integrals and Absolute Continuity CC BY 3.0
8.7: Integration of Complex and Vector-Valued Functions - CC BY 1.0
9.2.E: Problems on L-Integrals and Absolute Continuity - CC BY 1.0
8.7.E: Problems on Integration of Complex and Vector-Valued Functions - CC BY 1.0
9.3: Improper (Cauchy) Integrals - CC BY 3.0 9.3.E: Problems on Cauchy Integrals - CC BY 1.0
8.8: Product Measures. Iterated Integrals - CC BY 1.0
9.4: Convergence of Parametrized Integrals and Functions - CC BY 3.0
8.8.E: Problems on Product Measures and Fubini Theorems - CC BY 1.0
9.4.E: Problems on Uniform Convergence of Functions and C-Integrals - CC BY 1.0
8.9: Riemann Integration. Stieltjes Integrals - CC BY 1.0
Back Matter - CC BY 1.0
8.9.E: Problems on Riemann and Stieltjes Integrals - CC BY 1.0
Index - CC BY 1.0 Index - Undeclared Glossary - CC BY 1.0 Detailed Licensing - Undeclared
8.10: Integration in Generalized Measure Spaces - CC BY 1.0
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