Distribution Theory Applied to Differential Equations [1 ed.] 9783030671587, 9783030671594

Table of contents :
Contents
1 Introduction
1.1 Initial Remarks
References
2 Preliminaries
2.1 Introduction
2.2 Test Functions and Regularization
2.3 Seminorms and Locally Convex Spaces
2.3.1 Locally Convex Spaces
2.3.2 Convex and Balanced Sets
2.3.3 Absorbing Sets
2.4 Duals
2.4.1 Reflexive Spaces
2.5 The Inductive Limit Topology
References
3 Convex and Lower-Semicontinuous Functions
3.1 Introduction
3.2 Convex Functions
3.3 Lower Semicontinuous Functions
3.4 Convexity and Lower Semicontinuity
References
4 The Subdifferential of a Convex Function
4.1 Introduction
4.2 The Conjugate Function
4.3 The Additivity of the Subdifferential
References
5 Evolution Equations
5.1 Introduction
5.2 The Resolvent and the Yosida Approximation
References
6 Distributions
6.1 Fundamental Spaces in the Theory of Distributions
6.1.1 On Some Properties of the Spaces Cm(Ω) and Cinfty(Ω)
6.2 The Space of Distributions
6.3 The Dual of Cinfty
6.4 The Derivative of a Distribution
6.5 Distributions as Generalized Functions
6.6 On Some Spaces of Distributions
6.7 The Primitive of a Distribution
6.7.1 Structure Theorems
6.8 Extras
6.8.1 Higher-Order Primitives
6.8.2 The Local Structure of Distributions
6.9 Convolutions
6.9.1 The Direct Product of Distributions
6.9.2 Convolution of Distributions
6.9.3 Convolution of Functions and Distributions: Regularization
6.9.4 Convolution Maps
References
7 Tempered Distributions
7.1 The Schwartz Space of Infinitely Differentiable Functions …
7.2 Tempered Distributions
7.3 The Fourier Transform in mathcalS(mathbbRn)
7.3.1 The Inverse Fourier Transform
7.3.2 Properties of the Fourier Transform
7.4 Fourier Transform of Tempered Distributions
7.5 The Fourier Transform of a Distribution with Compact Support
7.6 The Product of a Distribution by a Cinfty Function
7.7 The Space of Multipliers of mathcalS'(mathbbRn)
7.8 Some Results on Convolutions with Tempered Distributions
7.9 The Paley-Wiener-Schwartz Theorem
7.10 A Result on the Fourier Transform of a Convolution of Two Distributions
References
8 Differential Equations in Distributions
8.1 Ordinary Differential Equations
8.1.1 Linear Differential Equations with Constant Coefficients
8.1.2 An Application
8.2 Partial Differential Equations
8.2.1 The Direct Product
8.2.2 Hyperbolic Partial Differential Equations
8.2.3 Parabolic Partial Differential Equations
8.2.4 Elliptic Partial Differential Equations
8.2.5 The Cauchy Problem
8.2.6 An Application
References
9 Sobolev Spaces
9.1 The Sobolev Space H1(Ω)
9.2 The Sobolev Space Hm(Ω)
9.3 The Sobolev Space Wk,p(Ω)
9.4 The Sobolev Spaces Hs(mathbbRn)
9.5 Besov Spaces
9.5.1 The Nonhomogeneous Littlewood-Paley Decomposition
9.5.2 Definition and Properties
9.5.3 The Homogeneous Littlewood-Paley Decomposition and the Homogeneous Besov Spaces
References
10 Variational Problems
10.1 Introduction
10.1.1 The Stokes System
10.1.2 The Elasticity System
10.1.3 The Plate Equation
10.2 The Approximation of Variational Problems
10.2.1 The Galerkin Method
10.2.2 The Finite Element Method
References
11 On Some Spaces of Distributions
11.1 The Spaces mathcalDLp
11.2 The Space mathcalO'C
References
12 On Some Differential Operators
12.1 Local and Pseudolocal Operators
12.2 Hypoelliptic Partial Differential Operators
12.3 Existence of Fundamental Solutions
References
Index
Index

Citation preview

Adina Chirilă Marin Marin Andreas Öchsner

Distribution Theory Applied to Differential Equations

Distribution Theory Applied to Differential Equations

Adina Chirilă Marin Marin Andreas Öchsner •

•

Distribution Theory Applied to Differential Equations

123

Adina Chirilă Department of Mathematics and Computer Sciences Transilvania University of Braşov Braşov, Romania

Marin Marin Department of Mathematics and Computer Sciences Transilvania University of Braşov Braşov, Romania

Andreas Öchsner Faculty of Mechanical Engineering Esslingen University of Applied Sciences Esslingen am Neckar, Baden-Württemberg Germany

ISBN 978-3-030-67158-7 ISBN 978-3-030-67159-4 https://doi.org/10.1007/978-3-030-67159-4

(eBook)

© The Editor(s) (if applicable) and The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 This work is subject to copyright. All rights are solely and exclusively licensed by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

Contents

1 1 2

1

Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Initial Remarks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2

Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Introduction . . . . . . . . . . . . . . . . . . . . . 2.2 Test Functions and Regularization . . . . . 2.3 Seminorms and Locally Convex Spaces . 2.3.1 Locally Convex Spaces . . . . . . . 2.3.2 Convex and Balanced Sets . . . . 2.3.3 Absorbing Sets . . . . . . . . . . . . . 2.4 Duals . . . . . . . . . . . . . . . . . . . . . . . . . . 2.4.1 Reflexive Spaces . . . . . . . . . . . 2.5 The Inductive Limit Topology . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Convex and Lower-Semicontinuous Functions 3.1 Introduction . . . . . . . . . . . . . . . . . . . . . . 3.2 Convex Functions . . . . . . . . . . . . . . . . . . 3.3 Lower Semicontinuous Functions . . . . . . . 3.4 Convexity and Lower Semicontinuity . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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The Subdifferential of a Convex Function . 4.1 Introduction . . . . . . . . . . . . . . . . . . . 4.2 The Conjugate Function . . . . . . . . . . 4.3 The Additivity of the Subdifferential . References . . . . . . . . . . . . . . . . . . . . . . . . .

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Evolution Equations . . . . . . . . . . . . . . . . . . . . . . . 5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 The Resolvent and the Yosida Approximation References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

6

Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.1 Fundamental Spaces in the Theory of Distributions . . 6.1.1 On Some Properties of the Spaces C m ðXÞ and C1 ðXÞ . . . . . . . . . . . . . . . . . . . . . . . . 6.2 The Space of Distributions . . . . . . . . . . . . . . . . . . . 6.3 The Dual of C 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 The Derivative of a Distribution . . . . . . . . . . . . . . . 6.5 Distributions as Generalized Functions . . . . . . . . . . . 6.6 On Some Spaces of Distributions . . . . . . . . . . . . . . . 6.7 The Primitive of a Distribution . . . . . . . . . . . . . . . . 6.7.1 Structure Theorems . . . . . . . . . . . . . . . . . . 6.8 Extras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8.1 Higher-Order Primitives . . . . . . . . . . . . . . . 6.8.2 The Local Structure of Distributions . . . . . . 6.9 Convolutions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.9.1 The Direct Product of Distributions . . . . . . . 6.9.2 Convolution of Distributions . . . . . . . . . . . . 6.9.3 Convolution of Functions and Distributions: Regularization . . . . . . . . . . . . . . . . . . . . . . 6.9.4 Convolution Maps . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Tempered Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.1 The Schwartz Space of Infinitely Differentiable Functions Rapidly Decreasing at Infinity . . . . . . . . . . . . . . . . . . . . . . . 7.2 Tempered Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.3 The Fourier Transform in SðRn Þ . . . . . . . . . . . . . . . . . . . . . 7.3.1 The Inverse Fourier Transform . . . . . . . . . . . . . . . . 7.3.2 Properties of the Fourier Transform . . . . . . . . . . . . . 7.4 Fourier Transform of Tempered Distributions . . . . . . . . . . . . 7.5 The Fourier Transform of a Distribution with Compact Support . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 The Product of a Distribution by a C 1 Function . . . . . . . . . 7.7 The Space of Multipliers of S 0 ðRn Þ . . . . . . . . . . . . . . . . . . . 7.8 Some Results on Convolutions with Tempered Distributions . 7.9 The Paley-Wiener-Schwartz Theorem . . . . . . . . . . . . . . . . . . 7.10 A Result on the Fourier Transform of a Convolution of Two Distributions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Differential Equations in Distributions . . . . . . . . . . . . . . 8.1 Ordinary Differential Equations . . . . . . . . . . . . . . . . 8.1.1 Linear Differential Equations with Constant Coefficients . . . . . . . . . . . . . . . . . . . . . . . . 8.1.2 An Application . . . . . . . . . . . . . . . . . . . . . . 8.2 Partial Differential Equations . . . . . . . . . . . . . . . . . . 8.2.1 The Direct Product . . . . . . . . . . . . . . . . . . . 8.2.2 Hyperbolic Partial Differential Equations . . . 8.2.3 Parabolic Partial Differential Equations . . . . 8.2.4 Elliptic Partial Differential Equations . . . . . . 8.2.5 The Cauchy Problem . . . . . . . . . . . . . . . . . 8.2.6 An Application . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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Sobolev Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.1 The Sobolev Space H 1 ðXÞ . . . . . . . . . . . . . . . . . . . . . . . . . 9.2 The Sobolev Space H m ðXÞ . . . . . . . . . . . . . . . . . . . . . . . . . 9.3 The Sobolev Space W k;p ðXÞ . . . . . . . . . . . . . . . . . . . . . . . . 9.4 The Sobolev Spaces H s ðRn Þ . . . . . . . . . . . . . . . . . . . . . . . . 9.5 Besov Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.1 The Nonhomogeneous Littlewood-Paley Decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9.5.2 Definition and Properties . . . . . . . . . . . . . . . . . . . . 9.5.3 The Homogeneous Littlewood-Paley Decomposition and the Homogeneous Besov Spaces . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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10 Variational Problems . . . . . . . . . . . . . . . . . . . . . 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . 10.1.1 The Stokes System . . . . . . . . . . . . . 10.1.2 The Elasticity System . . . . . . . . . . . 10.1.3 The Plate Equation . . . . . . . . . . . . . 10.2 The Approximation of Variational Problems . 10.2.1 The Galerkin Method . . . . . . . . . . . 10.2.2 The Finite Element Method . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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11 On Some Spaces of Distributions 11.1 The Spaces DLp . . . . . . . . . 11.2 The Space O0C . . . . . . . . . . References . . . . . . . . . . . . . . . . . .

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12 On Some Differential Operators . . . . . . . . . . . 12.1 Local and Pseudolocal Operators . . . . . . . 12.2 Hypoelliptic Partial Differential Operators 12.3 Existence of Fundamental Solutions . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . .

Contents

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Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 275

Chapter 1

Introduction

Abstract We give a brief motivation for the concept of distributions. We explain why this concept is useful in applied mathematics. We present some landmarks in the history of this subject. Then we present the main topics and results in the theory of distributions.

1.1 Initial Remarks The concept of distributions is a relatively recent one since it was used for the first time in 1951 in a study of the French mathematician L. Schwarz [43]. Distributions are a generalization of the concept of functions, hence some mathematicians call them today “generalized functions”. This generalization was motivated by practical situations. For instance, a boundary value problem from the theory of differential equations (or even a Cauchy problem) has a solution if the right-hand side of the problem is at least of class C 1 . But in the phenomena modelled by these problems these conditions are not satisfied. In most of these phenomena, the functions are not even continuous, very often having discontinuities of the second kind. We will see that the same problem, formulated in the context of the theory of distributions will not impose restrictions on the right-hand side, these being automatically satisfied. We anticipate that every distribution is differentiable (in a sense that we will discuss and which does not differ too much from the differentiation of a function) of every order. In the context of the theory of distributions the regularity conditions are weaker. For example, the functions do not need to be continuous [1, 2, 4]. The theory of distributions, as it is structured today, has the disadvantage that it employs very sophisticated mathematical concepts, see for example [13, 18, 20, 21, 31, 35, 42]. Therefore, this theory is less accessible to the researchers in mathematics. Most of the concepts used in the theory of distributions are at the intersection of other subjects of mathematics. The fundamentals of the theory of distributions are based on advanced topics in functional analysis [3, 5, 6, 9, 24, 30, 36, 37, 47, 49], topology, mathematical physics [7, 10, 11, 14, 16, 17, 19, 22, 23, 25, 32, 34, 38–40, 45, 46, 48] differential geometry © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_1

1

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1 Introduction

and so on [8, 15, 26–29, 33, 44]. We may consider that the origins of this theory coincide with Sobolev’s trial to define generalized solutions for hyperbolic partial differential equations in the 1940s. Sobolev’s idea was employed by L. Schwarz in the 1960s, when the concept of distributions was used for the first time. In 1971 Schwarz [43] published a book in which he laid the foundations of the unified theory of distributions by using the method of linear functionals. Afterwards, Sikorski [41] introduced a theory of distributions by using fundamental sequences of continuous functions. This method is similar to that used by Dedekind [12] to introduce the real numbers as cuts in the set of natural numbers. The way in which Schwarz introduced the concept of distributions seems more natural and for this reason we use it in the sequel. Since the contributions of Schwarz intersect with the ones of Sobolev, many concepts and results from the theory of distributions use the names of these two mathematicians, even though the theory as it is today used, is almost completely different from that initiated by these two scientists. The main qualitative improvement expected from this theory is the simplification of the mathematical theory in order to make it more accesible. Starting with the initiators of this theory, the concepts have become more complex.

References 1. R.A. Adams, J.J.F. Fournier, Sobolev Spaces (Elsevier, 2003) 2. J.J. Alibert, G. Bouchitte, Non-uniform integrability and generalized Young measures. J. Convex Anal. 4(1), 129–147 (1997) 3. H.W. Alt, Lineare Funktionalanalysis (Springer, Berlin Heidelberg, 2006) 4. L. Ambrosio, N. Gigli, G. Savare, Gradient Flows in Metric Spaces and in the Space of Probability Measures (Birkhäuser, Basel, 2005) 5. V. Barbu, T. Precupanu, Convexity and Optimization in Banach Spaces, 4th edn. (Springer, 2012) 6. J. Borwein, Convex Functions: Constructions, Characterizations and Counterexamples (Cambridge University Press, New York, 2010) 7. R.I. Bot, S.M. Grad, G. Wanka, Fenchel’s duality theorem. J. Optim. Theory Appl. 132, 509– 515 (2007) 8. D. Braess, Finite Elemente: Theorie, schnelle Löser und Anwendungen in der Elastizitätstheorie (Springer, 2007) 9. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, 2011) 10. M.G. Crandall, Semigroups of nonlinear contractions and dissipative sets. J. Funct. Anal. 3, 376–418 (1969) 11. C. De Lellis, L. Szekelyhidi, Jr., The Euler equations as a differential inclusion. Ann. Math. (2) 170(3), 1417–1436 (2009) 12. R. Dedekind, Stetigkeit und irrationale Zahlen (Braunschweig University, 1872) 13. J.J. Duistermaat, J.A.C. Kolk, Distributions: Theory and Applications (Birkhäuser, 2010) 14. D.G. Ebin, J. Marsden, Groups of diffeomorphisms and the motion of an incompressible fluid. Ann. Math. 2(92), 102–163 (1970) 15. H. Elman, D. Silvester, A. Wathen, Finite Elements and Fast Iterative Solvers: With Applications in Incompressible Fluid Dynamics, 2nd edn. (Oxford University Press, 2014)

References

3

16. L.C. Evans, Partial Differential Equations, 4th edn. (American Mathematical Society, 2008) 17. I. Fonseca, G. Leoni, Modern Methods in the Calculus of Variations: L p Spaces (Springer, 2007) 18. F.G. Friedlander, M. Joshi, Introduction to the Theory of Distributions, 2nd edn. (Cambridge University Press, 1999) 19. D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order, 2nd edn. (Springer, 2001) 20. D.D. Haroske, H. Triebel, Distributions, Sobolev Spaces, Elliptic Equations (European Mathematical Society, 2008) 21. L. Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, 2nd edn. (Springer, 1990) 22. T. Kato, Nonstationary flows of viscous and ideal fluids in R3 . J. Funct. Anal. 9, 296–305 (1972) 23. T. Kato, On the eigenfunction of many-particle systems in quantum mechanics. Comm. Pure Appl. Math. 10, 151–177 (1957) 24. Y. Komura, On linear topological spaces. Kumamoto J. Sci. A 5, 148–157 (1962) 25. L. Lichtenstein, Grundlagen der Hydromechanik (Springer, 1929) 26. M. Marin, H. Arabnia, Equations of Evolution (Elliot & Fitzpatrick, Athens, USA, 2010) 27. M. Marin, A. Öchsner, Complements of Higher Mathematics (Springer, 2018) 28. M. Marin, A. Öchsner, Essentials of Partial Differential Equations (Springer, 2019) 29. S.D. Micu, Introduction to Finite Element Method (Romanian), Publications of the Centre for Nonlinear Analysis and its Applications 30. G. Minty, Monotone operators in Hilbert space. Duke Math. J. 29, 341–346 (1962) 31. D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis (Springer, 2013) 32. L. Onsager, Statistical hydrodynamics. Nuovo Cimento (9), 6(Supplemento, 2(Convegno Internazionale di Meccanica Statistica)), 279–287 (1949) 33. A. Öchsner, Computational Statics and Dynamics: An Introduction Based on the Finite Element Method (Springer, Singapore, 2020) 34. N.S. Papageorgiou, Weak solutions of differential equations in Banach spaces. Bull. Austral. Math. Soc. 33, 407–418 (1986) 35. I. Richards, H. Youn, Theory of Distributions: A Non-technical Introduction (Cambridge University Press, 1995) 36. R.T. Rockafellar, Convex Analysis, No. 28 (Princeton University Press, Princeton, N.J., 1970) 37. W. Rudin, Functional analysis (Tata-McGraw Hill Publishing, 1974) 38. V. Scheffer, An inviscid flow with compact support in space-time. J. Geom. Anal. 3(4), 343–401 (1993) 39. A.I. Shnirelman, On the nonuniqueness of weak solution of the Euler equation. Comm. Pure Appl. Math. 50(12), 1261–1286 (1997) 40. R.E. Showalter, Hilbert space methods for partial differential equations. Electron. J. Diff. Eqns., Monograph 01 (1994) 41. R. Sikorski, The elementary theory of distributions. Bull. Acad. Pol., Warszawa (1957) 42. R. Strichartz, A Guide to Distribution Theory and Fourier Transforms (CRC Press, 1994) 43. L. Schwartz, Theorie des Distributions (Herman & Cie, Paris, 1971) 44. E. Stein, Singular Integral and Differentiability Properties of Functions (Princeton University Press, 1970) 45. L. Szekelyhidi, Weak solutions of the Euler equations: non-uniqueness and dissipation. Journees Equations aux derivees partielles, Roscoff, 1-5 juin 2015 46. L. Szekelyhidi Jr., E. Wiedemann, Young measures generated by ideal incompressible fluid flows. Arch. Rational Mech. Anal. (2012) 47. H. Triebel, Theory of Function Spaces (Birkhäuser, 1983) 48. E. Wiedemann, Existence of weak solutions for the incompressible Euler equations. Ann. Inst. H. Poincare Anal. Non Lineaire 28(5), 727–730 (2011) 49. L.C. Young, Lecture on the Calculus of Variations and Optimal Control Theory (American Mathematical Society, 1980)

Chapter 2

Preliminaries

Abstract This chapter presents some concepts that are used in the next chapters. One section is dedicated to seminorms and locally convex spaces, as well as to convex, balanced and absorbing sets. Another section presents duals and reflexive spaces. Finally, the inductive limit topology is presented.

2.1 Introduction Let X be a set. We denote by 2 X the set of all subsets of X . This includes the empty set. Definition 2.1.1 Let X be a set and T be a subset of 2 X . Then the pair (X, T ) is a topological space if it satisfies the following properties: (i) ∅ ∈ T , X ∈ T ; (ii) if U ∈ T , V ∈ T , then U ∩ V ∈ T ; (iii) if  is an arbitrary set and Uλ ∈ T for all λ ∈ , then ∪λ∈ Uλ ∈ T . Then we call T a topology on X [6]. Definition 2.1.2 A topological space (X, T ) is called Hausdorff if for different points there exist disjoint open neighbourhoods, that is, x = y ⇒ ∃Ux , U y ∈ T such that x ∈ Ux , y ∈ U y and Ux ∩ U y = ∅.

(2.1.1)

Proposition 2.1.1 Let (X, T ) be a Hausdorff space. Then every sequence has at most one limit point. Definition 2.1.3 We call a pair (X,d) a metric space if d : X × X → [0, ∞) has the three properties below: (i) d(x, y) = 0 ⇔ x = y; (ii) d(y, x) = d(x, y), ∀x, y ∈ X ; (iii) d(x, y) ≤ d(x, z) + d(z, y), ∀x, y, z ∈ X .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_2

5

6

2 Preliminaries

Then a map d : X × X → [0, ∞) satisfying these three properties (of definiteness, symmetry and the triangle inequality) is called a metric on X [2, 7]. Definition 2.1.4 Let (X, d) be a metric space. Then a sequence x : N → X is called a Cauchy sequence if ∀ε > 0, ∃k0 ∈ N such that ∀k, j ≥ k0 , d(x j , xk ) < ε.

(2.1.2)

Definition 2.1.5 Let (X, d) be a metric space. Then it is called complete if every Cauchy sequence converges. Definition 2.1.6 We call a pair (X,  · ) a normed space if X is a K vector space and  ·  : X → [0, ∞) is a map satisfying the three properties below: (i) x = 0 ⇒ x = 0; (ii) αx = |α|x, ∀α ∈ K, x ∈ X . (iii) x + y ≤ x + y. Then a map  ·  : X → [0, ∞) satisfying these three properties (of definiteness, homogeneity and the triangle inequality) is called a norm on X [2, 7]. Definition 2.1.7 Let (X,  · ) be a normed space. Then it is called a Banach space if it is complete under the metric d : X × X → [0, ∞) defined by d(x, y) = x − y, ∀x, y ∈ X.

(2.1.3)

Definition 2.1.8 Let X be a K vector space. We call a map (x, y) → (x, y) X from X × X to K a sesquilinear form if for all x, y, x1 , y1 , x2 , y2 ∈ X and all α ∈ K, the following holds true: ¯ y) X ; (i) (αx, y) X = α(x, y) X , (x, αy) X = α(x, (ii) (x1 + x2 , y) X = (x1 , y) X + (x2 , y) X , (x, y1 + y2 ) X = (x, y1 ) X + (x, y2 ) X . Definition 2.1.9 Let X be a K vector space. Then a sesquilinear form which is symmetric, i.e., (i) ∀x, y ∈ X, (y, x) X = (x, y) X and positive definite, i.e., (ii) ∀x ∈ X, (x, x) X ≥ 0 and (x, x) = 0 ⇔ x = 0 is called a scalar product. Definition 2.1.10 Let X be a K vector space. The pair (X, (·, ·) X ), where (·, ·) X is a scalar product, is called a pre-Hilbert space. Definition 2.1.11 A pre-Hilbert space (X, (·, ·) X ) which is complete under the norm induced by the scalar product is called a Hilbert space. Below we present the theorem of Hahn-Banach.

2.1 Introduction

7

Theorem 2.1.1 Let X be an R vector space. We assume that (i) p : X → R is sublinear, that is, ∀x, y ∈ X p(x + y) ≤ p(x) + p(y), p(αx) = α p(x), ∀α ≥ 0;

(2.1.4)

(ii) f : Y → R is linear, where Y is a subspace of X ; (iii) f (x) ≤ p(x), ∀x ∈ Y . Then there is F : X → R linear with F(x) = f (x), ∀x ∈ Y, F(x) ≤ p(x), ∀x ∈ X.

(2.1.5)

Below we present another version of the Hahn-Banach theorem Theorem 2.1.2 Let X be a normed K vector space. Let Y be a subspace of X with the norm induced by X . Then for every y  ∈ Y  , there exists x  ∈ X  such that x  = y  on Y, x   X  = y  Y  .

(2.1.6)

2.2 Test Functions and Regularization Definition 2.2.1 Let  ⊂ Rn be open. Let f be a complex-valued function defined on . We call support of f , and denote it by supp f , the closure in  of the set {x ∈  : f (x) = 0}.

(2.2.1)

Observation 2.2.1 The support of f is the smallest relative closed set outside of which f is identically zero. By definition, a test function on  is a C ∞ function on  which has compact support in . By Cc∞ () we denote the vector space of all test functions on . Note that this vector space endowed with an appropriate topology is important in the definition of distributions on . We introduce the function  1 e |x|2 −1 if |x| < 1 β(x) = . (2.2.2) 0 if |x| ≥ 1 This is a C ∞ function for which the support is the closed unit ball. We can divide by a constant, namely Rn β(x)d x. Then we obtain another C ∞ function, which we denote by α. Its support is the closed unit ball and we have

8

2 Preliminaries

 Rn

α(x)d x = 1.

Then we introduce ∀ε > 0 the function αε (x) = ε−n α

x  ε

(2.2.3)

.

(2.2.4)

Note that αε (x) ∈ Cc∞ (Rn ). Moreover, its support is Bε (0), the closed ball with center at the origin and radius ε. We have  Rn

αε (x)d x = 1.

(2.2.5)

Note that the family of test functions (αε )ε>0 can be used to regularize discontinuous functions, like integrable ones or L p functions [1, 4, 5]. In other words, we can show that such functions can be approximated by test functions. Definition 2.2.2 A function f defined in  is locally integrable if f is integrable (in the sense of Lebesgue [3]) on every compact subset K ⊂ . We denote by χ K the characteristic function of K , which is equal to 1 on K and 0 outside of K . Hence, f is locally integrable in  if, for every compact set K ⊂ , the product f · χ K is integrable on . Definition 2.2.3 We consider that u is a locally integrable function on Rn . Then the convolution of u and αε is defined by  u ε (x) =

 Rn

u(x − y)αε (y)dy =

Rn

u(y)αε (x − y)dy.

(2.2.6)

Note that other possible notations are (u ∗ αε )(x) or (αε ∗ u)(x). Theorem 2.2.1 Let u be a locally integrable function in Rn . We have 1. The convolution u ε is a C ∞ function in Rn . 2. If u has compact support K (and in the case u is integrable in Rn ), the support of u ε is contained in the ε-neighbourhood of K . 3. If u is a continuous function, then u ε converges uniformly to u on compact subsets of Rn . 4. If u ∈ L p (Rn ), 1 ≤ p < +∞, then u ε → u in L p (Rn ). Proof Step 1 We want to show that u ε ∈ C ∞ (Rn ). Note that both integrals in (2.2.6) are being evaluated over compact subsets of Rn . Hence, it suffices to apply the classical theorem about differentiation inside the integral sign. Step 2 By definition, the ε-neighbourhood of K is the set Kε =

 x∈K

Bε (x)

(2.2.7)

2.2 Test Functions and Regularization

9

of all closed balls with center x ∈ K and radius ε. If x ∈ Rn and d(x, K ) > ε, then x∈ / K ε so that αε (x − y) = 0 for all y ∈ K . Hence, the second integral in (2.2.6) is equal to zero. We deduce that the support of u ε is contained in K ε . Step 3 Let u be a continuous function. We assume that L is an arbitrary fixed compact subset of Rn . We evaluate  u ε (x) − u(x) =

Rn

[u(x − y) − u(x)] αε (y)dy.

(2.2.8)

Note that u is uniformly continuous on L. Therefore, given σ > 0, there is δ > 0 such that |u(x − y) − u(x)| < σ (2.2.9) for all x ∈ L and for all |y| < δ. By considering that ε ≤ δ, we obtain  |u ε (x) − u(x)| ≤

|u(x − y) − u(x)| αε (y)dy < σ,

Rn

(2.2.10)

for all x ∈ L. This implies that u ε → u, uniformly on L, as ε → 0. Step 4 Let u ∈ L p (Rn ), 1 ≤ p < ∞. We recall that u can be approximated in L p by continuous functions with compact support. Moreover, if u ∈ L p , then u ε ∈ L p . This can be derived from Minkowski’s inequality [3] in its integral form  ||u ε || p = ≤

Rn

 =

Rn





Rn

|u ε (x)| d x p

Rn

1p

 =

|u(x − y)αε (y)| p d x

1p



αε (y)

|u(x − y)| d x p

Rn

R

1p





n

R

p 1p

u(x − y)αε (y)dy

d x ≤ n

dy =

(2.2.11)

dy = ||u|| p .

Given σ > 0, we consider a continuous function v with compact support, such that ||u − v|| p
0 is the set ¯ 0 , r ) = {x ∈ E| p(x − x0 ) ≤ r }. B(x

(2.3.6)

2.3 Seminorms and Locally Convex Spaces

13

A neighbourhood of x0 is a set V containing a ball with center x0 . Let x0 ∈ E be arbitrary. Note that the set of all open (closed) balls with center x0 defines a fundamental system of neighbourhoods of x0 of a topology on E which is compatible with the vector space structure of E. Let a ∈ E be fixed and λ ∈ K , λ = 0 be fixed. The following maps x → a + x, x → λx

(2.3.7)

are called translation and homothety and are homeomorphisms on E. Hence, it suffices to have a fundamental system of neighbourhoods of the origin in order to have a fundamental system of neighbourhoods of a point a ∈ E. Let ( pi )i∈I be a family of seminorms defined on E. We define for every x0 ∈ E, ε ∈ R, ε > 0 and F a finite part of I V (x0 , ε, F) = {x ∈ E| pi (x − x0 ) < ε, i ∈ F}.

(2.3.8)

The set V (x0 , ε, F) is the intersection of the balls with center x0 and radius ε, corresponding to the seminorms pi with i ∈ F. The family of all sets V (x0 , ε, F) when ε runs through the set of all positive real numbers and F runs through all finite subsets of I defines a fundamental system of neighbourhoods of x0 of a topology on E compatible with the vector space structure of E. The space E endowed with this topology is called a locally convex topological vector space. If E is a locally convex topological vector space and ( pi )i∈I is a family of seminorms defining its topology, then E is a Hausdorff space [3] if and only if to any pair x, y ∈ E with x = y we associate a seminorm pk such that pk (x) = pk (y).

2.3.2 Convex and Balanced Sets Definition 2.3.2 Let E be a vector space over K . A subset A of E is convex if, given two points x, y ∈ A, the segment αx + β y with α, β ∈ R+ and α + β = 1 is contained in A. Example 1 The whole space E is convex. The empty set is convex. Example 2 Balls are convex sets. Example 3 Segments are convex sets. Example 4 If (Ai )i∈I is a family of convex sets of E, then the intersection A = ∩i∈I Ai is a convex set. Definition 2.3.3 Let A be a subset of E. The smallest convex subset of E containing A is called the convex hull of A.

14

2 Preliminaries

Note that, given a set A, its convex hull is always well defined. Clearly, it suffices to consider the intersection of the family of all convex sets which contain A. Note that this family is nonempty because it contains the whole space E. The convex hull of A will be denoted by (A). This can be characterized as the set of all x ∈ E which can be represented as a finite sum x = i∈F αi xi with xi ∈ A, αi ∈ R+ , i∈F αi = 1, where F is a finite set of indices which depend on x. Definition 2.3.4 Let A be a subset of E. Then it is called balanced if λA ⊂ A for all λ ∈ K such that |λ| ≤ 1. Example 1 Balls with center at the origin are balanced sets. Example 2 Let E = R. The set [0, 1] is convex, but it is not balanced. Let A be a subset of E. We assume that it is convex and balanced. Then for every pair of points x, y ∈ A we have αx + β y ∈ A for all α, β ∈ K with |α| + |β| ≤ 1. Note that the intersection of a family of convex and balanced subsets of E is a convex and balanced subset of E. Definition 2.3.5 Let A be a subset of E. The smallest balanced convex subset b (A) of E containing A is called the balanced convex hull of A. Note that the balanced convex hull of A is the intersection of all balanced convex sets containing A. Moreover, we have the following characterization x ∈ b (A) ⇔ x =



αi xi ,

(2.3.9)

i∈F

where xi ∈ A,

i∈F

|αi | ≤ 1 and F is a finite set of indices.

2.3.3 Absorbing Sets Definition 2.3.6 Let V be a subset of a vector space E. V is called absorbing if given x ∈ E, there is λ ∈ R, λ > 0 such that λx ∈ V . Example 1 The set {−1, 0, 1} is an absorbing set on the real line R. Example 2 Balls with center at the origin are absorbing sets. More generally, every neighbourhood of the origin in a topological vector space is an absorbing set. Let E be a vector space endowed with a seminorm p. The unit ball U = {x ∈ E| p(x) < 1}

(2.3.10)

is a balanced convex set. Moreover, it is absorbing. We can show that if V is a balanced convex and absorbing subset of E, then q(x) = inf{λ ≥ 0|x ∈ λV }

(2.3.11)

2.3 Seminorms and Locally Convex Spaces

15

is a seminorm on E. Clearly, q is well defined on E and q : E → R+ because V is absorbing. If x, y ∈ E, let λ > 0 and μ > 0 be such that x ∈ λV and y ∈ μV . Since V is convex, we deduce that   μ λ V+ V ⊂ (λ + μ)V. (2.3.12) x + y ∈ λV + μV = (λ + μ) λ+μ λ+μ Hence, we obtain q(x + y) ≤ q(x) + q(y).

(2.3.13)

Since V is balanced, we deduce that q(λx) = |λ|q(x).

(2.3.14)

Therefore, we conclude that q is a seminorm on E. Moreover, we have {x ∈ E|q(x) < 1} ⊂ V ⊂ {x ∈ E|q(x) ≤ 1}.

(2.3.15)

Theorem 2.3.1 Let E be a topological vector space. The following conditions are equivalent: (i) E is locally convex; (ii) there is a fundamental system of convex neighbourhoods of the origin; (iii) there is a fundamental system of absorbing balanced convex neighbourhoods of the origin. Proof Step 1. We show that (i) implies (ii). If the topology of E is defined by a family of seminorms ( pi )i∈I , then the sets V (F, ε) = {x ∈ E| pi (x) ≤ ε, i ∈ F}

(2.3.16)

with F a finite part of I and 0 < ε < 1, form a fundamental system of convex neighbourhoods of the origin. Step 2. We show that (iii) implies (i). Let V be an absorbing balanced convex subset of E which belongs to a fundamental system of neighbourhoods of the origin. For every such V we have a corresponding seminorm pV . Hence, the family of seminorms obtained in this way defines the topology of E. Step 3. We show that (ii) implies (iii). Let V be a convex neighbourhood of the origin. It suffices to prove that the set U = ∩|λ|=1 λV

(2.3.17)

is an absorbing balanced convex neighbourhood of the origin. Note that the map (μ, x) → μx is continuous at the point (0, 0). Hence, there are ε > 0 and V  a neighbourhood of zero in E such that

16

2 Preliminaries

μx ∈ V, ∀ |μ| ≤ ε, ∀x ∈ V  .

(2.3.18)

Or there is a neighbourhood of zero W such that μW ⊂ V, ∀ |μ| ≤ 1.

(2.3.19)

μW ⊂ V, ∀ |μ| = 1.

(2.3.20)

Hence, we have This implies that W ⊂ λV , ∀ |λ| = 1. Therefore, U is a neighbourhood of zero in E. It is clear that U is a convex set since it is the intersection of convex sets. Moreover, it is absorbing. We will show that U is balanced. If x ∈ U , then the segment [0, x] is contained in U , that is λx ∈ U, ∀ λ such that 0 ≤ λ ≤ 1.

(2.3.21)

By the definition of U , if x ∈ U , then λx ∈ U , ∀ |λ| = 1. Hence, if μ = 0 and |μ| ≤ 1, then we have μ (2.3.22) μx = |μ| x ∈ U. |μ| This shows that U is balanced.



In the sequel, we give examples of locally convex spaces. Example 1 A seminormed space is a locally convex space. In particular, a normed space is a locally convex space. Example 2 Let X be a locally compact topological space. We consider that C(X ) is the space of all complex-valued continuous functions on X . Moreover, we consider that K is the collection of all compact subsets of X . Let us consider the family of seminorms ( pk )k∈K with pk ( f ) = supt∈K | f (t)|. This family of seminorms defines a Hausdorff locally convex topology on C(X ), see [3]. Note that a sequence ( f j ) j of functions of C(X ) converges to zero in this topology if and only if f j (x) converges to zero uniformly on each compact subset K of X . Therefore, this locally convex topology is called the topology of uniform convergence on compact subsets of X . Let X =  ⊂ Rn be open. Then such a topology can be defined by a sequence of seminorms. Clearly, it suffices to consider an increasing sequence (K j ) j∈N of compact subsets contained in  whose union is  and the corresponding sequence of seminorms ( p K j ) j∈N . Therefore, the topology of C() is defined by a countable fundamental system of neighbourhoods of the origin. Then the Hausdorff locally convex space C() is a metrizable space. Furthermore, C() is a complete space because the uniform limit on compact subsets of  of continuous functions is a continuous function. Therefore, the topology just defined on C() is the natural topology of C().

2.3 Seminorms and Locally Convex Spaces

17

Definition 2.3.7 A Frechet space is a Hausdorff locally convex, metrizable and complete space [3]. p

Example 3 Let  ⊂ Rn be open. Let 1 ≤ p < ∞. We consider that L loc () is the space of classes of measurable functions [1, 4, 5] such that on every compact subset K of  we have 1  | f (x)| p d x

p

< ∞.

(2.3.23)

K p

The elements of L loc () are called pth power locally integrable functions. We consider the seminorm  1p | f (x)| p d x . (2.3.24) pk ( f ) = K

The family of seminorms ( pk ), where k runs over all compact subsets of , defines a p locally convex topology on L loc (). Such a topology can be defined by a countable p family of seminorms. Then we can prove that L loc is a Frechet space.

2.4 Duals Let E be a vector space over a field K . Its algebraic dual E ∗ is the vector space of all linear maps x ∗ : E ∗ → K . We denote by x ∗ (x) or (x, x ∗ ) the value of x ∗ on x ∈ E. Let E be a topological vector space over a field K . Then the dual E  of E is the subspace of E ∗ consisting of all continuous linear maps (or functionals) on E. For every x ∗ ∈ E ∗ we set

 

px ∗ (x) = x, x ∗ .

(2.4.1)

Then px ∗ defines a seminorm on E and the family ( px ∗ )x ∗ ∈E ∗ defines a locally convex topology on E, which we denote by σ(E, E ∗ ). When E is a topological vector space, the topology σ(E, E  ) defined on E by the family of seminorms ( px  )x  ∈E  is called the weak topology of E. Clearly, the weak topology is coarser than the given topology of E and also coarser than σ(E, E ∗ ). Similarly, we can consider the weak topology σ(E  , E) on the dual E  of E. Then a sequence (x j ) converges weakly to zero in E  if and only if, for every x ∈ E, the sequence (x j (x)) converges to zero in x. Hence, the weak topology on E  coincides with the pointwise convergence topology. Moreover, on the dual E  of a topological vector space E we can consider the strong topology of E  , which is also a locally convex topology. Definition 2.4.1 Let E be a topological vector space. Let V be a neighbourhood of zero in E. Then A ⊂ E is bounded if there is λ > 0 such that λA ⊂ V .

18

2 Preliminaries

Let E be a locally convex topological vector space. Then every neighbourhood of zero contains a balanced neighbourhood of zero. Therefore, A is bounded in E if to every neighbourhood of zero V there is ε > 0 such that λA ⊂ V for all |λ| ≤ ε. Note that every topological vector space has a fundamental system of balanced neighbourhoods of zero. Example 1 Finite subsets of E are bounded sets. Example 2 In a seminormed space, balls are bounded sets. Example 3 Every relatively compact subset A of a locally convex space E is bounded. Let V be a neighbourhood of zero in E. Then there is an open neighbourhood of zero W in E such that W + W ⊂ V and μW ⊂ W for all |μ| ≤ 1. Note that A is relatively compact. Then we can find a finite subset (x j )1≤ j≤ p of elements of A such that the open sets (x j + W )1≤ j≤ p cover A. Note that the set (x j )1≤ j≤ p is bounded in E. Then we can find λ ∈ R, 0 < λ < 1 such that λx j ⊂ W , 1 ≤ j ≤ p. Then, we have p (2.4.2) λA ⊂ ∪ j=1 λ(x j + W ) ⊂ W + W ⊂ V. Therefore, A is bounded. Definition 2.4.2 Let B be a subset of a topological vector space E. The polar set B ◦ of B is the subset of E  defined as

 

B ◦ = {x  ∈ E  | x, x  ≤ 1, ∀x ∈ B}.

(2.4.3)

Let A ⊂ E be bounded. We will prove that its polar set A◦ is an absorbing, balanced and convex subset of E  . Let x  , y  ∈ A◦ and α, β ∈ R+ such that α + β = 1. Then we obtain





 





x, αx  + β y  ≤ α x, x  + β x, y  ≤ 1.

(2.4.4)

Therefore, A◦ is convex. Let x  ∈ A◦ and λ ∈ K , |λ| ≤ 1. Then



 



x, λx  = |λ| x, x  ≤ 1.

(2.4.5)

Therefore, we have λx  ∈ A◦ . Hence, A◦ is balanced. We consider z  ∈ E  and let V be a neighbourhood of zero in E defined as

 

V = {x ∈ E| x, z  ≤ 1}.

(2.4.6)

Note that A ⊂ E is bounded. Then there is λ > 0 such that λA ⊂ V . Therefore, we have

   

x, λz  = λx, z  ≤ 1, ∀x ∈ A. (2.4.7) This implies that A◦ is an absorbing set in E  .

2.4 Duals

E

19

Hence, to every bounded set A of E we can associate the following seminorm on p A◦ (x  ) = inf{λ ≥ 0|x  ∈ λA◦ }.

(2.4.8)

Let B denote the family of all bounded subsets of E. Then the family of seminorms ( p A◦ ) A∈B defines a Hausdorff locally convex topology on E  , called the strong topology of E  . Note that a sequence (x j ) converges strongly to zero in E  if and   only if the sequence x j (x) converges uniformly to zero on every bounded set of E. Hence, the strong topology on E  is also called the topology of uniform convergence on bounded sets of E. We denote by E b the dual E  endowed with the strong topology. Example Let E be a normed space. Let E  be its dual endowed with the norm

 

x   E  = sup x, x  .

(2.4.9)

x E ≤1

Then E  is a Banach space [3]. Therefore, the strong topology on E  coincides with the one defined by the above norm.

2.4.1 Reflexive Spaces Let E  be the dual of the Banach space E  . We consider the following norm

 

x   E  = sup x  , x 

x   E  ≤1

(2.4.10)

on E  . Therefore, E  becomes a Banach space. Every element x ∈ E defines a continuous linear functional x˜ on E    x(x ˜  ) = x, x  , ∀x  ∈ E  .

(2.4.11)

Note that the map x → x˜ is an isometry from E into E  . Moreover, E is called a reflexive space if this isometry is onto. We consider that E is a Hausdorff topological vector space and that E b is its strong dual. The dual of E b is the bidual E  of E. Then we consider the map x ∈ E → x˜ ∈ E  . If this map represents an isomorphism from E onto E b , then the topological vector space E is reflexive. Example 1 The n-dimensional Euclidean space Rn is reflexive. Moreover, finitedimensional Hausdorff topological vector spaces are reflexive. Example 2 Hilbert spaces are reflexive spaces. Example 3 The Banach spaces L p (), 1 < p < ∞ are reflexive. The spaces 1 L () and L ∞ () are not reflexive.

20

2 Preliminaries

2.5 The Inductive Limit Topology In the sequel, we consider the inductive limit topology. We consider that (E i )i∈N is an increasing sequence of locally convex spaces such that the identity map E i → E i+1 is continuous for every i. Moreover, we define ∞ Ei . E = ∪i=1

(2.5.1)

Then we consider on E the finest locally convex topology for which the identity map E i → E is continuous for every i = 1, 2, . . .. This topology represents the inductive limit topology of E defined by the subspaces E i . Moreover, the space E endowed with this topology is called the inductive limit of the spaces (E i )i∈N . Let V be a convex set. We want V to be a neighbourhood of zero in the inductive limit topology. Then it is necessary and sufficient to consider that every intersection V ∩ E i is a neighbourhood of zero in E i for all i = 1, 2, . . .. Moreover, by considering all convex hulls  ∞  (2.5.2) Vi V =  ∪i=1 we obtain a fundamental system of neighbourhoods of the origin in E. Note that every Vi runs over a fundamental system of convex neighbourhoods of zero in E i , i = 1, 2, . . .. Proposition 2.5.1 We consider that E is the inductive limit of (E i )i∈N and that F is any locally convex space. Then a linear map u : E → F is continuous if and only if the restriction u i of u is continuous from E i into F for all i. Proof Step 1. Let u be continuous. Since the identity map E i → E is continuous by definition, every restriction u i is continuous. Step 2. We assume that every u i is a continuous map from E i into F. Let U be a convex neighbourhood of zero in F. Then there is a convex neighbourhood of zero Vi in E i such that u i (Vi ) ⊂ U . Hence,  ∞  Vi V =  ∪i=1

(2.5.3)

is a neighbourhood of zero in E. In fact, u(V ) ⊂ U . This implies that u is continuous from E into F.  Lemma 2.5.1 We assume that F is a locally convex space and G is a closed subspace of F. Let V be a balanced convex open neighbourhood of zero in G. We consider x ∈ F such that x ∈ / G. Therefore, there is a balanced convex open neighbourhood of zero W in F such that x ∈ / W and W ∩ G = V . Proof Note that G is closed in F. Then there is a balanced convex open neighbourhood of zero V0 in F such that (x + V0 ) ∩ G = ∅ and V0 ∩ G ⊂ V.

(2.5.4)

2.5 The Inductive Limit Topology

21

We consider that W is the balanced convex hull of V ∪ V0 . Note that W is open. We will show that W ∩ G = V . In fact, we have W ∩ G ⊃ V . Let w ∈ W ∩ G. Therefore, we have (2.5.5) w = αv + βv0 , where v ∈ V , v0 ∈ V0 and |α| + |β| ≤ 1. If β = 0, there is nothing to show. Hence, we consider that β = 0. Then we have v0 ∈ V0 ∩ G ⊂ V and w ∈ V . We assume by contradiction that x ∈ W . Therefore, we have x = y + z, where y ∈ G and z ∈ V0 . So, we obtain (2.5.6) y = x − z ∈ (x + V0 ) ∩ G, 

which is impossible.

Theorem 2.5.1 Let E be a vector space. We assume that E is the union of an increasing sequence (E i )i∈N of locally convex spaces with the following properties: (i) the identity map E i → E i+1 is continuous for every i; (ii) the topology induced by E i+1 on E i coincides with the topology of E i for every i; (iii) E i is a closed subspace of E i+1 for every i. Therefore, the following hold true: (iv) the inductive limit topology of E induces on every E i its original topology; (v) a subset A is bounded in the inductive limit E if and only if there is an index j such that A is contained and it is bounded in E j . Proof Step 1. We will prove that the topology induced by E on every E i coincides with the given topology of E i . Let Vi be a balanced convex neighbourhood of zero in E i . It suffices to prove that there is a neighbourhood of zero V in E such that Vi = V ∩ E i , ∀i.

(2.5.7)

Note that there is a sequence (Vi+k ), k = 0, 1, 2, . . . of balanced convex neighbourhoods of zero in E i+k such that

We consider

Vi+k−1 = Vi+k ∩ E i+k−1 , k = 1, 2, . . . .

(2.5.8)

V = ∪∞ k=0 Vi+k .

(2.5.9)

Note that V is a neighbourhood of zero in E and Vi = V ∩ E i . Step 2. We consider A ⊂ E to be bounded. We assume by contradiction that there is no index i such that A is contained in E i . Hence, there is an increasing sequence of indices (i n ) and a sequence (xn ) of elements of E such that / E in−1 . xn ∈ A ∩ E in and xn ∈

(2.5.10)

22

2 Preliminaries

Note that there is a sequence (Vn ) of balanced convex open neighbourhoods of zero in E in such that / nVn and Vn ∩ E in−1 = Vn−1 . (2.5.11) xn ∈ We consider V = ∪∞ n=1 Vn . Hence, V is a neighbourhood of zero in E such that / nV. V ∩ E in = Vn and xn ∈

(2.5.12)

Finally, we have a contradiction with the assumption that A is bounded in E.



Example 1 Let  ⊂ Rn be open. We consider (K i ) an increasing sequence of compact subsets of  such that  = ∪i K i . Moreover, we consider E = Cc () to be the space of all complex-valued continuous functions defined in  and with compact support. Then we consider E i = Cc (; K i ) to be the subspace of Cc () which consists of all functions having support in K i . We have Cc () = ∪i Cc (; K i ).

(2.5.13)

We consider on Cc (; K i ) the topology of uniform convergence on K i . This topology is locally convex. Moreover, it is defined by the seminorm p K i ( f ) = sup | f (x)|.

(2.5.14)

x∈K i

Note that p K i is a norm on Cc (; K i ). Furthermore, the space Cc (; K i ) endowed with this norm is a Banach space. Note that the embedding Cc (; K i ) → Cc (; K i+1 )

(2.5.15)

is continuous. Moreover, Cc (; K i ) is a closed subspace of Cc (; K i+1 ). Then we consider on Cc () the inductive limit topology of the spaces Cc (; K i ). This topology is the natural topology of Cc (). Note that a sequence (ϕ j ) converges to zero in Cc () if and only if the following conditions hold true: (i) There is K ⊂  compact such that supp ϕ j ⊂ K , ∀ j; (ii) the sequence (ϕ j ) converges to zero uniformly on K . Definition 2.5.1 A continuous linear functional on Cc () is a Radon measure on , see [3]. We will denote by M() the space of all Radon measures on . This is the topological dual of Cc (). Note that a linear map μ : Cc () → C is a Radon measure if and only if

(2.5.16)

2.5 The Inductive Limit Topology

23

(i) for every K ⊂  compact, the linear map μ restricted to Cc (; K ) is continuous; (ii) or for every K ⊂  compact there is a constant Mk such that |μ(ϕ)| ≤ Mk sup |ϕ(x)| , ∀ϕ ∈ Cc (; K ).

(2.5.17)

x∈K

Note that every locally integrable function f in  (in particular, every continuous function) defines a Radon measure on  by considering  μ f (ϕ) =



f ϕ, ∀ϕ ∈ Cc ().

(2.5.18)

1 Let L loc () denote the vector space of all classes of locally integrable functions on 1 () into M(). Clearly, if . Then the map f → μ f gives an embedding of L loc μ f ≡ 0 on Cc (), then f must be zero almost everywhere in . Therefore, it defines 1 (). the zero element of L loc p Example 2 In the sequel, we discuss the spaces L c (), with 1 ≤ p ≤ ∞. We n consider that  ⊂ R is open and that K ⊂  is compact and arbitrary. We denote by L p (K ) with 1 ≤ p < ∞ the space of pth power integrable functions with compact support contained in K endowed with its natural norm. If p = ∞, then L ∞ (K ) is the space of essentially bounded functions with compact support in K , endowed with its natural norm. The spaces L p (K ) with 1 ≤ p ≤ ∞ are Banach spaces. If K 1 ⊂ K 2 , then the embedding L p (K 1 ) → L p (K 2 ) is continuous and the topology p induced by L p (K 2 ) on L p (K 1 ) coincides with the topology of L p (K 1 ). Let L c () be p the union of L (K ) for all compact subsets K of . If (K i ) is an increasing sequence p of compact subsets of  whose union is , then L c () and L p (K i ), i = 1, 2, . . ., p satisfy all the assumptions of the previous theorem. We can consider on L c () the p inductive limit topology relative to its subspaces L (K i ), i = 1, 2, . . ..

References 1. R.A. Adams, Sobolev Spaces (Academic Press, 1975) 2. H.W. Alt, Lineare Funktionalanalysis (Springer, 2006) 3. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, 2011) 4. L.C. Evans, Partial Differential Equations (American Mathematical Society, 1998) 5. D. Gilbarg, N.S. Trudinger, Elliptic Partial Differential Equations of Second Order (Springer, 2001) 6. Y. Komura, On linear topological spaces. Kumamoto J. Sci. A 5, 148–157 (1962) 7. W. Rudin, Functional Analysis (Tata-McGraw Hill Publishing, 1974)

Chapter 3

Convex and Lower-Semicontinuous Functions

Abstract In this chapter, one section is about convex functions and their properties. Another section is about lower-semicontinuous functions and their properties in compact topological spaces and in Banach spaces. The final section presents some properties of functions that are both convex and lower-semicontinuous. More precisely, some conditions are discussed that are needed for such a function to take a minimum value on a Banach space.

3.1 Introduction Let X be a Banach space. We denote by X ∗ the dual space of the space X , that is, the set of all linear and continuous functionals defined on X . By means of the dual space X ∗ we can define on the space X a family of seminorms { px ∗ }x ∗ ∈X ∗ by (3.1.1) px ∗ : X → R, px ∗ (x) = (x, x ∗ ), ∀x ∈ X. Formula (3.1.1) is defined for every x ∗ ∈ X ∗ . This family induces on the space X a topology of a locally convex space. The Banach space X already has a topology (which induces the structure of a Banach space). In order to distinguish between the two topologies, the topology induced by the seminorms is called the weak topology on X . In this topology, we can define the convergence by weakly

strongly

xn  x ⇔ (xn , x ∗ ) −→ (x, x ∗ ), ∀x ∗ ∈ X ∗ ⇔ def

strongly

⇔ px ∗ (xn ) −→ px ∗ (x), ∀x ∗ ∈ X ∗ .

(3.1.2)

Similarly, by means of the space X we can define on the dual space X ∗ a family of seminorms { px }x∈X by px : X ∗ → R, px (x ∗ ) = (x, x ∗ ), ∀x ∗ ∈ X ∗ .

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_3

(3.1.3)

25

26

3 Convex and Lower-Semicontinuous Functions

Formula (3.1.3) is defined for every x ∈ X . This family induces on the space X ∗ a topology of a locally convex space, which is called the weak topology in order to distinguish it from the initial topology on the space X ∗ . Hence, we can talk about the weak convergence of a sequence of functionals weakly

strongly

xn∗  x ∗ ⇔ (xn∗ , x) −→ (x ∗ , x), ∀x ∈ X ⇔ def

strongly

⇔ px (xn∗ ) −→ px (x ∗ ), ∀x ∈ X.

(3.1.4)

Let us denote by X ∗∗ the bidual of the Banach space X . We have X ⊂ X ∗∗ . If the other inclusion X ∗∗ ⊂ X holds true, then we have an equality X = X ∗∗ and the space X is called a reflexive space. Every Hilbert space is reflexive. The L p spaces are reflexive for p < 1. In fact, if X = L p (), then X ∗ = L q (), where q is chosen such that 1 1 + = 1. p q

(3.1.5)

Then X ∗∗ = L s (), where s is chosen such that 1 1 + = 1 ⇒ s = p ⇒ X ∗∗ = X. q s

(3.1.6)

In the L p () spaces we have the norm  ||y|| p =

|y(x)| d x p



 1p (3.1.7)

and the weak convergence in these spaces is  yn  y ⇔



 yn (x)z(x)d x →



y(x)z(x)d x, ∀z ∈ L q ().

(3.1.8)

Theorem 3.1.1 (Compactness) A Banach space X is reflexive if and only if any bounded set in X is weakly compact.

3.2 Convex Functions Definition 3.2.1 A set C ⊂ X is convex if it satisfies the condition λx + (1 − λ)y ∈ C, ∀x, y ∈ C, ∀λ ∈ [0, 1].

(3.2.1)

3.2 Convex Functions

27

Theorem 3.2.1 (Mazur) For a convex set, the strong closure coincides with the weak closure. ¯ = (−∞, ∞] is called convex [2, 4] if it Definition 3.2.2 The function ϕ : X → R satisfies the condition ϕ (λx + (1 − λ)y) ≤ λϕ(x) + (1 − λ)ϕ(y), ∀x, y ∈ X, ∀λ ∈ [0, 1].

(3.2.2)

Note that the function ϕ is called strictly convex if the inequality is strict for every λ ∈ (0, 1) and for all x, y ∈ X, x = y with ϕ(x) < ∞ and ϕ(y) < ∞. Note that if ϕ is convex, then the inequality  ϕ

n 

 λi xi

≤

i=1

n 

λi ϕ(xi ), λi ≥ 0,

i=1

n 

λi = 1

(3.2.3)

i=1

holds for all x1 , . . . , xn ∈ X , for which the right-hand side makes sense. ¯ is convex, then the level sets Observation 3.2.1 If ϕ : X → R {x ∈ X |ϕ(x) ≤ λ}

(3.2.4)

¯ On the other hand, the functions which have this property are convex, where λ ∈ R. are not necessarily convex and they are called quasi-convex. Let C be a nonempty subset of X . Then the function IC on X , defined by  IC (x) =

0, if x ∈ C, +∞, if x ∈ /C

(3.2.5)

is called the indicator function of C. Proposition 3.2.1 The subset C of X is convex if and only if its indicator function IC is convex. ¯ be any extended real-valued function on X . The set Let ϕ : X → R epi ϕ = {(x, α)|x ∈ X, α ∈ R, ϕ(x) ≤ α}

(3.2.6)

is called the epigraph of ϕ. The set hypo ϕ = {(x, α)|x ∈ X, α ∈ R, ϕ(x) ≥ α}

(3.2.7)

is called the hypograph of ϕ. ¯ is convex if and only if its epigraph is a Proposition 3.2.2 A function ϕ : X → R convex subset of X × R [1].

28

3 Convex and Lower-Semicontinuous Functions

Proof Sufficiency. We assume that ϕ is convex and (x, α), (y, β) ∈ epi ϕ and λ ∈ [0, 1]. Let us consider w = (1 − λ)x + λy and t = (1 − λ)α + λβ. We have ϕ(w) ≤ (1 − λ)ϕ(x) + λϕ(y) ≤ t,

(3.2.8)

so we obtain (w, t) ∈ epi ϕ. Hence, epi ϕ is a convex subset of X × R. Necessity. Let us assume that epi ϕ is convex. The proof is by contradiction. Let x, y ∈ X and λ ∈ [0, 1] be such that ϕ(w) = ϕ ((1 − λ)x + λy) > (1 − λ)ϕ(x) + λϕ(y)

(3.2.9)

holds. This inequality shows that 0 < λ < 1 and that neither ϕ(x) nor ϕ(y) can be +∞. Hence, there exist α, β ∈ R such that (x, α), (y, β) ∈ epi ϕ. Therefore, for every x, y and λ, we have inf {(1 − λ)α + λβ|(x, α), (y, β) ∈ epi ϕ} = (1 − λ)ϕ(x) + λϕ(y).

(3.2.10)

Since the epigraph of ϕ is convex, we obtain ϕ(w) = inf {t|(w, t) ∈ epi ϕ} ≤ (1 − λ)ϕ(x) + λϕ(y) < ϕ(w).

(3.2.11)

Hence, we arrive at a contradiction.



In the sequel, we present a continuity result for convex functions. ¯ be a proper Theorem 3.2.2 Let X be a topological linear space and let ϕ : X → R convex function on X . Then, the function ϕ is continuous on Int D(ϕ) if and only if ϕ is bounded from above on a neighbourhood of an interior point of D(ϕ). In the sequel, we consider the case when X is a finite-dimensional space. Proposition 3.2.3 Let X be a finite-dimensional separated topological linear space. Then every proper convex function ϕ on X is continuous on the interior of its effective domain.

3.3 Lower Semicontinuous Functions ¯ is called lower semicontinuous on X if Definition 3.3.1 The function ϕ : X → R it satisfies the condition lim inf ϕ(y) ≥ ϕ(x), ∀x ∈ X. y→x

Observation 3.3.1 Clearly, we have

(3.3.1)

3.3 Lower Semicontinuous Functions

29

lim inf ϕ(y) ≤ ϕ(x), ∀x ∈ X. y→x

(3.3.2)

Therefore, we may say that the function ϕ is called lower semicontinuous on X if it satisfies the condition (3.3.3) lim inf ϕ(y) = ϕ(x), ∀x ∈ X. y→x

We recall that lim inf ϕ(y) = sup inf ϕ(y).

(3.3.4)

lim sup ϕ(y) = inf sup ϕ(y).

(3.3.5)

y→x

V ∈V(x) y∈V

Similarly, we have V ∈V(x) y∈V

y→x

We may give an equivalent formulation of the definition of a lower semicontinuous function. We say that the function ϕ is lower semicontinuous on X if the set {x ∈ X |ϕ(x) ≤ λ} is closed for any fixed λ, where λ > 0. Proposition 3.3.1 The upper-envelope of a family {ϕi }i∈I of lower-semicontinuous functions is also a lower-semicontinuous function. Proof We note that

 {x ∈ X |ϕi (x) ≤ λ} , x ∈ X | sup ϕi (x) ≤ λ = i∈I

(3.3.6)

i∈I

and we consider the closeness of the level sets.



Example. We will give an example of a lower semicontinuous function. Let C be ¯ given by a closed subset of the Banach space X . Then the function ϕ : X → R  ϕ(x) =

0, x ∈C +∞, x ∈ /C

(3.3.7)

is lower semicontinuous since we clearly have {x ∈ X |ϕ(x) ≤ λ} = C, ∀λ > 0

(3.3.8)

and the set C is assumed to be closed. Below we give an important property of lower-semicontinuous functions which can be proven by the characterization by closed level sets. Theorem 3.3.1 (Weierstrass) Let X be a compact topological space. Then a lowersemicontinuous function ϕ on X takes a minimum value on X . Moreover, if it takes only finite values, then it is bounded from below.

30

3 Convex and Lower-Semicontinuous Functions

Definition 3.3.2 (i) If X is a Banach space, then the function ϕ : X → R is called a proper function if there exists at least one point x ∈ X such that ϕ(x) < ∞. (ii) The effective domain of the function ϕ is denoted by D(ϕ) and is the set D(ϕ) = {x ∈ X |ϕ(x) < ∞}.

(3.3.9)

An important property of a lower semicontinuous function on a Banach space is given in the following theorem, which we present without a proof. ¯ be a function. If ϕ is Theorem 3.3.2 Let X be a Banach space and let ϕ : X → R lower semicontinuous on X , then it is bounded from below by an affine function, i.e. ∃α ∈ R and x0∗ ∈ X ∗ such that ϕ(x) ≥ (x, x0∗ ) + α, ∀x ∈ X.

(3.3.10)

3.4 Convexity and Lower Semicontinuity Proposition 3.4.1 Let X be a locally convex space. A proper convex function ϕ : ¯ is lower-semicontinuous on X if and only if it is lower-semicontinous with X →R respect to the weak topology on X . ¯ be a function Theorem 3.4.1 Let X be a reflexive Banach space and let ϕ : X → R which is assumed to be proper, convex and lower-semicontinuous on X . Then ϕ takes a minimum value on every bounded, convex and closed subset M of X . So, there exists x0 ∈ M such that (3.4.1) ϕ(x0 ) = inf{ϕ(x)|x ∈ M}. Proof The proof follows by applying the theorem of Weierstrass to the space X endowed with the weak topology. Note that every closed and bounded subset of a reflexive Banach space is weakly compact.  Observation 3.4.1 If we assume that ϕ is strictly convex in the previous theorem, then the minimum point x0 is unique. Observation 3.4.2 In the previous theorem, the condition that M is bounded may be replaced by the coercivity condition lim

||x||→+∞,x∈M

ϕ(x) = +∞.

(3.4.2)

In the following theorem we will see that a lower semicontinuous function has a behaviour which is similar to that of a continuous function. ¯ be a function which Theorem 3.4.2 Let X be a Banach space and let ϕ : X → R is assumed to be convex and lower semicontinuous on X . Moreover, if ϕ has the property that

3.4 Convexity and Lower Semicontinuity

31

lim ϕ(x) = +∞,

(3.4.3)

||x||→∞

then ϕ attains its infimum on the space X . Proof We will show that ∃x0 ∈ X such that ϕ(x0 ) = inf ϕ(x).

(3.4.4)

x∈X

We denote this infimum with d. Let us show that d is finite. The proof is by contradiction and we assume that d is infinite. Clearly, d can not be ∞. If the infimum is ∞, then the other values of the function ϕ are infinite, that is, the function ϕ is improper. Hence, we can only assume that d = −∞. According to the definition of the infimum, we can deduce that there exists a sequence {xn } ⊂ X such that lim ϕ(xn ) = −∞.

(3.4.5)

n→∞

The sequence {xn } is bounded. If it was not bounded, then we would have ||xn || → ∞ and ϕ(xn ) → −∞, which is in contradiction with the assumption (3.4.3). If the sequence {xn } is bounded, then ϕ(xn ) is bounded, which contradicts the assumption that d = −∞. In conclusion, d is a finite number and by the definition of the infimum we can write 1 (3.4.6) d ≤ ϕ(xn ) ≤ d + . n Since the sequence {xn } is bounded, we deduce that we can consider a subsequence which is convergent xn k  x0 and d ≤ ϕ(xn k ) ≤ d +

1 . nk

(3.4.7)

We consider the set A = {x ∈ X |ϕ(x) ≤ d + ε}.

(3.4.8)

Clearly, A is a closed set since the inequality is not strict. Let us prove that A is convex, that is, let us show that ∀x, y ∈ A and λ ∈ [0, 1] ⇒ λx + (1 − λ)y ∈ A,

(3.4.9)

which is equivalent to ϕ(x) ≤ d + ε, ϕ(y) ≤ d + ε ⇒ ϕ (λx + (1 − λ)y) ≤ d + ε. Indeed, since ϕ is a convex function, we have

(3.4.10)

32

3 Convex and Lower-Semicontinuous Functions

ϕ (λx + (1 − λ)y) ≤ λϕ(x) + (1 − λ)ϕ(y) ≤ ≤ λ(d + ε) + (1 − λ)(d + ε) = d + ε.

(3.4.11)

Hence, A is a convex and closed set. Since X is a reflexive Banach space, we deduce that A is weakly closed and then along with a sequence of elements from the set it also contains the limit of the sequence. Clearly, xn k ∈ A and then x0 ∈ A, so ϕ(x0 ) ≤ d. Hence, we have ϕ(x0 ) = d since if ϕ(x0 ) < d, then we have a contradiction with the fact that d is the infimum.  In the following theorem we recall a result that is used very often in functional analysis. Theorem 3.4.3 (Beer) Every closed, convex, absorbing and equilibrated set from a Banach space contains a ball with the center in the origin. The following theorem provides yet another proof that lower semicontinuous functions are not too far from continuous functions [3]. ¯ is convex and lower semicontinuous on Theorem 3.4.4 If the function ϕ : X → R the Banach space X , then ϕ is continuous on D(ϕ), which is its effective domain. Proof Let ∀x0 ∈ D(ϕ) and let us show that the function ϕ is continuous in x0 , that is, it can be interchanged with the limit sign lim ϕ(x) = ϕ(x0 ).

x→x0

(3.4.12)

In the first step, let us show that we can restrict our considerations to x0 = 0 and ϕ(x0 ) = 0. In fact, if we consider instead of the function ϕ the function ψ given by ψ(x) = ϕ(x + x0 ) − ϕ(x0 ),

(3.4.13)

then we obtain ψ(0) = ϕ(x0 ) − ϕ(x0 ). Let us show that the function ψ is convex and lower semicontinuous. For the convexity, let ∀x, y ∈ X and λ ∈ [0, 1] and let us prove that ψ (λx + (1 − λ)y) ≤ λψ(x) + (1 − λ)ψ(y).

(3.4.14)

In fact ψ (λx + (1 − λ)y) = ϕ (λx + (1 − λ)y + x0 ) − ϕ(x0 ) = = ϕ (λ(x + x0 ) + (1 − λ)(y + x0 ) + x0 − λx0 − (1 − λ)x0 ) − − ϕ(x0 ) = ϕ (λ(x + x0 ) + (1 − λ)(y + x0 )) − ϕ(x0 ) ≤ ≤ λϕ(x + x0 ) + (1 − λ)ϕ(y + x0 ) − ϕ(x0 ) = = λ (ϕ(x + x0 ) − ϕ(x0 )) + λϕ(x0 ) + (1 − λ)· · (ϕ(y + x0 ) − ϕ(x0 )) + (1 − λ)ϕ(x0 ) − ϕ(x0 ) = = λψ(x) + (1 − λ)ψ(y).

(3.4.15)

3.4 Convexity and Lower Semicontinuity

33

In the sequel we prove that ψ is a lower semicontinuous function, that is lim inf ψ(y) ≥ ψ(x), ∀x ∈ X. y→x

(3.4.16)

In fact, we have lim inf ψ(y) = lim inf ϕ(y + x0 ) − lim inf ϕ(x0 ) ≥ y→x

y→x

y→x

≥ ϕ(x + x0 ) − ϕ(x0 ) = ψ(x).

(3.4.17)

We define the set K by K = {x ∈ X |ϕ(x) ≤ M} ∩ {x ∈ X |ϕ(−x) ≤ M}, M > 0.

(3.4.18)

We choose M sufficiently large to have M = ∅. We will show that the set K is closed, convex, equilibrated and absorbing. In order to be closed, a set needs to satisfy the following condition: along with a sequence it contains its limit. Let {xn }n∈N ⊂ K be a sequence such that xn → x. Let us prove that x ∈ K : xn ∈ K ⇒ ϕ(xn ) ≤ M, ϕ(−xn ) ≤ M ⇒ ⇒ lim inf ϕ(xn ) ≤ lim inf M, xn →x

xn →x

lim inf ϕ(−xn ) ≤ lim inf M ⇒ xn →x

(3.4.19)

xn →x

⇒ ϕ(x) ≤ M, ϕ(−x) ≤ M ⇒ x ∈ K . Let us show that the set K is convex, that is ∀x, y ∈ K and ∀λ ∈ [0, 1] ⇒ λx + (1 − λ)y ∈ K .

(3.4.20)

So if ϕ(x) ≤ M, ϕ(−x) ≤ M and ϕ(y) ≤ M, ϕ(−y) ≤ M ⇒ ϕ(λx + (1 − λ)y) ≤ M and ϕ (−λx − (1 − λ)y) ≤ M. In fact, we have ϕ (λx + (1 − λ)y) ≤ λϕ(x) + (1 − λ)ϕ(y) ≤ ≤ λM + (1 − λ)M = M

(3.4.21)

ϕ (−λx − (1 − λ)y) = ϕ (λ(−x) − (1 − λ)(−y)) ≤ ≤ λϕ(−x) + (1 − λ)ϕ(−y) ≤ λM + (1 − λ)M = M.

(3.4.22)

Note that the set K is equilibrated since if x ∈ K then −x ∈ K . We only have to prove that K is absorbing. To this end, we have to show that for all y ∈ X , ∃t > 0 such that t y ∈ K . Let us consider the function g(t)ϕ(t y). Since 0 ∈ D(ϕ), we deduce that ϕ(0) < ∞ and then g(0) < ∞. This means that g(t) is finite for t sufficiently small

34

3 Convex and Lower-Semicontinuous Functions

|g(t)| ≤ M ⇔ g(t) ≤ M and g(−t) ≤ M ⇒ t y ∈ K .

(3.4.23)

From the relations above we deduce that the set K has all the properties that are required by Beer’s theorem. Therefore, we have ∃S(0, ρ) ⊂ K |ϕ(x) ≤ M and ϕ(−x) ≤ M, ||x|| ≤ ρ.

(3.4.24)

Let us show that ϕ is bounded on this sphere, that is, |ϕ(x)| ≤ M. Clearly, we have ϕ(x) ≤ M. Then

(3.4.25)

   1 1 x−x =ϕ x + (−x) ⇒ 0 = ϕ(0) = ϕ 2 2 2 1 1 ⇒ 0 ≤ ϕ(x) + ϕ(−x) ⇒ −ϕ(−x) ≤ ϕ(x) ⇒ 2 2 ⇒ −ϕ(−x) ≤ M ⇒ ϕ(−x) ≥ −M. 

(3.4.26)

In the last inequality we replace x by −x and we obtain ϕ(x) ≥ −M.

(3.4.27)

By (3.4.25) and (3.4.27), we obtain |ϕ(x)| ≤ M, ∀x ∈ X such that ||x|| ≤ ρ.

(3.4.28)

We write x in the form x=

  ||x|| x ||x|| ρ+ 1− ·0 ||x|| ρ ρ

and we use the convexity of the function ϕ by taking λ to be equal to

(3.4.29) ||x|| ρ

    x ||x|| ||x|| ϕ ρ + 1− ϕ(0) ⇒ ρ ||x|| ρ   x ||x|| ||x|| ϕ ρ ≤ M, ⇒ ϕ(x) ≤ ρ ||x|| ρ

(3.4.30)

x ρ ∈ S(0, ρ). ||x||

(3.4.31)

M ||x||. ρ

(3.4.32)

ϕ(x) ≤

where we used

Finally, we have ϕ(x) ≤

3.4 Convexity and Lower Semicontinuity

35

Then, we have  1 1 x + (−x) ≤ 2 2 1 M 1 ||x|| ⇒ ≤ ϕ(x) + ϕ(−x) ⇒ −ϕ(−x) ≤ ϕ(x) ≤ 2 2 ρ M M ⇒ ϕ(−x) ≥ −ϕ(x) ≥ − ||x|| ⇒ ϕ(−x) ≥ − || − x||. ρ ρ 

0 = ϕ(0) = ϕ

x−x 2





=ϕ

Hence, we obtain

(3.4.33)

M ||x|| ρ

(3.4.34)

M ||x||. ρ

(3.4.35)

ϕ(x) ≥ − since || − x|| = ||x||. By (3.4.32) and (3.4.34), we obtain |ϕ(x)| ≤ Therefore, we obtain

lim ϕ(x) = 0 = ϕ(0),

x→0

(3.4.36)

so the function ϕ can be interchanged with the limit sign. Hence, it is continuous in  the point x0 ∈ D(ϕ).

References 1. V. Barbu, T. Precupanu, Convexity and Optimization in Banach Spaces, 4th edn. (Springer, 2012) 2. J. Borwein, Convex Functions: Constructions, Characterizations and Counterexamples (Cambridge University Press, New York, 2010) 3. R.I. Bot, S.M. Grad, G. Wanka, Fenchel’s duality theorem. J. Optim. Theory Appl. 132, 509–515 (2007) 4. R.T. Rockafellar, Convex Analysis, No. 28 (Princeton University Press, Princeton, N.J., 1970)

Chapter 4

The Subdifferential of a Convex Function

Abstract In this chapter, the first section presents the definitions of Gateaux differentiable functions and of Frechet differentiable functions and the concept of subdifferentiability. Monotone and maximal monotone operators are defined. Minty’s theorem is proved. The subdifferential is shown to be a maximal monotone operator. The conjugate function is used to transform a minimization problem into a maximization problem and conversely. Finally, the additivity of the subdifferential is studied.

4.1 Introduction First we recall the definition of a Gateaux differentiable function and of a Frechet differentiable function, see [1, 7, 10]. ¯ is Gateaux differentiable in the point Definition 4.1.1 The function ϕ : X → R x0 ∈ X if ∃grad ϕ in x0 , which is a linear and continuous functional on X , hence an element from the dual space X , which is usually denoted by X ∗ and lim

λ→0

ϕ(x0 + λx) − ϕ(x0 ) = (grad ϕ(x0 ), x) , ∀x ∈ X, λ

(4.1.1)

where by (·, ·) we denoted the scalar product. ¯ is Frechet differentiable in Definition 4.1.2 We say that the function ϕ : X → R the point x0 ∈ X if ∃grad ϕ in x0 and we have ϕ(x0 + x) − ϕ(x0 ) = (grad ϕ(x0 ), x) + ω(x), ∀x ∈ X,

(4.1.2)

where the function ω(x) has the property lim ω(x) = 0.

||x||→0

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_4

(4.1.3)

37

38

4 The Subdifferential of a Convex Function

¯ be a convex and lower semi-continuous function Definition 4.1.3 (i) Let ϕ : X → R on X . The subgradient of the function ϕ in the point x0 ∈ X is denoted by ∂ϕ(x0 ) and is the set ∂ϕ(x0 ) = {x ∗ ∈ X ∗ |ϕ(x0 ) − ϕ(x) ≤ (x ∗ , x0 − x), ∀x ∈ X }.

(4.1.4)

(ii) We say that the function ϕ is subdifferentiable in the point x0 if ∂ϕ(x0 ) = ∅. Note that ∂ϕ(x0 ) is a closed convex subset of X ∗ . Note that if ϕ is a proper convex function on X , then the (global) minimum of ϕ over X is attained at the point x ∈ X if and only if 0 ∈ ∂ϕ(x). In the following theorem, we give the relationship between Gateaux differentiability and subdifferentiability. Theorem 4.1.1 If the convex function ϕ is Gateaux differentiable in the point x0 , then ϕ is subdifferentiable in the point x0 and we have the following equality of sets ∂ϕ(x0 ) = grad ϕ(x0 ).

(4.1.5)

Proof Since ϕ is Gateaux differentiable in x0 , we deduce that ∃ grad ϕ in x0 and lim

λ→0

ϕ(x0 + λx) − ϕ(x0 ) = (grad ϕ(x0 ), x), ∀x ∈ X. λ

(4.1.6)

But the function ϕ is convex, so we have ϕ (x0 + λ(x − x0 )) − ϕ(x0 ) ≤ λϕ(x) + (1 − λ)ϕ(x0 ) ⇒ ⇒ ϕ (x0 + λ(x − x0 )) − ϕ(x0 ) ≤ λ (ϕ(x) − ϕ(x0 )) ⇒ ϕ (x0 + λ(x − x0 )) − ϕ(x0 ) ≤ ϕ(x) − ϕ(x0 ) ⇒ ⇒ λ ⇒ (grad ϕ(x0 ), x − x0 ) ≤ ϕ(x) − ϕ(x0 ) ⇒

(4.1.7)

⇒ ϕ(x0 ) − ϕ(x) ≤ (grad ϕ(x0 ), x − x0 ), ∀x ∈ X. Hence, we deduce that x ∗ = grad ϕ(x0 ) is an element of ∂ϕ(x0 ), so ∂ϕ(x0 ) = ∅, that is, the function ϕ is subdifferentiable in the point x0 . We can show that grad ϕ(x0 ) is the only element of ∂ϕ(x0 ) and then we can state  that grad ϕ(x0 ) = ∂ϕ(x0 ). Observation 4.1.1 Conversely, if ϕ is continuous at x0 and if ∂ϕ(x0 ) contains a single element, then ϕ is Gateaux differentiable at x0 and grad ϕ(x0 ) = ∂ϕ(x0 ). ¯ which is convex and lower semi-continuous We recall that a function ϕ : X → R on X is subdifferentiable in the point x0 if the subgradient of the function ϕ in the point x0 ∈ X is a non-empty set [2, 3]. The subgradient of ϕ in the point x0 ∈ X is the set ∂ϕ(x0 ) = {x ∗ ∈ X ∗ |ϕ(x0 ) − ϕ(x) ≤ (x ∗ , x0 − x), ∀x ∈ X }.

(4.1.8)

4.1 Introduction

39

Example 1. We consider the function ϕ : R → R, ϕ(x) = |x|. Clearly, this function is not differentiable in x0 = 0. We will prove that it is subdifferentiable and we will compute the subdifferential of the function. We have ∂ϕ(0) = {x ∗ ∈ X ∗ |ϕ(0) − ϕ(x) ≤ (x ∗ , 0 − x), ∀x ∈ R}.

(4.1.9)

In this case, X ∗ = R and the scalar product on R is the usual product (x ∗ , x) = x ∗ x. We replace this in the previous relation ∂ϕ(0) = {x ∗ ∈ R| − |x| ≤ −x ∗ x, ∀x ∈ R}.

(4.1.10)

If x > 0, then |x| = x and the previous inequality becomes x ≥ x ∗ x ⇒ x ∗ ≤ 1. If x < 0, then |x| = −x and the previous inequality becomes −x ≥ x ∗ x ⇒ x ∗ ≥ −1. By considering these two cases, we obtain ∂ϕ(0) = [−1, 1]. Finally, we have ⎧ ⎪ x 0.

(4.1.11)

Example 2. Let K be a closed convex subset of X . The normal cone N K (x) to K at a point x ∈ K consists, by definition, of all the normal vectors to half-spaces that support K at x, that is, N K (x) = {x ∗ ∈ X ∗ |(x ∗ , x − u) ≥ 0 for all u ∈ K }.

(4.1.12)

This is a closed convex cone containing the origin and, in terms of the indicator function I K of K , we can write it as N K (x) = ∂ I K (x), x ∈ K .

(4.1.13)

Example 3. If X is a Banach space, then we recall the duality mapping of X [2, 3], which we denote by F(x) and it is the set F(x) = {x ∗ ∈ X ∗ |(x ∗ , x) = ||x||2X = ||x ∗ ||2X , ∀x ∈ X }.

(4.1.14)

Note that F(x) is a convex set of X ∗ for every x ∈ X . Clearly, x ∗ ∈ F(x), with ¯ ||x||), x ∗ = 0, if and only if the element x maximizes x ∗ on the closed ball S(0, or equivalently if and only if x ∗ (u) = ||x||2 , u ∈ X , is the equation of a closed ¯ ||x||). supporting hyperplane to S(0, We define the function ϕ : X → R by ϕ(x) =

1 ||x||2 . 2

(4.1.15)

40

4 The Subdifferential of a Convex Function

Proposition 4.1.1 The subdifferential of the function ϕ is the duality mapping of the Banach space X , that is  ∂ϕ(x) = ∂

1 ||x||2 2

 = F(x).

(4.1.16)

Proof We fix an arbitrary point x0 ∈ X and we prove the equality of the sets ∂ϕ(x0 ) = F(x0 ).

(4.1.17)

Step 1. We prove that ∂ϕ(x0 ) ⊂ F(x0 ). We consider an arbitrary x ∗ ∈ ∂ϕ(x0 ) and we show that x ∗ ∈ F(x0 )  x ∗ ∈ ∂ϕ(x0 ) ⇒ ϕ(x0 ) − ϕ(u) ≤ x ∗ , x0 − u , ∀u ∈ X ⇒  1 1 ⇒ ||x0 ||2 − ||u||2 ≤ x ∗ , x0 − u , ∀u ∈ X. 2 2

(4.1.18)

We consider u = x0 + λv and the previous relation becomes  1 1 ||x0 ||2 − ||x0 + λv||2 ≤ x ∗ , −λv ⇒ 2 2 1 ∗ − λ(v, x ) ≥ − ||x0 ||2 − ||x0 + λv||2 ⇒ 2  ∗ 1 − λ x , v ≥ − 2λ(x0 , v) + λ2 ||v||2 , ∀λ ∈ R, ∀v ∈ X ⇒ 2 λ ∗ (x , v) ≤ (x0 , v) + ||v||2 , ∀λ ∈ R, ∀v ∈ X ⇒ 2 λ (x ∗ − x0 , v) ≤ ||v||2 , ∀λ ∈ R, ∀v ∈ X. 2

(4.1.19)

We pass to the limit with λ → 0 and we obtain (x ∗ − x0 , v) ≤ 0, ∀v ∈ X ⇒ x ∗ = x0 .

(4.1.20)

Then, we have F(x0 ) = (x ∗ , x0 ) = ||x0 ||2 = ||x ∗ ||2 ⇒ x ∗ ∈ F(x0 ).

(4.1.21)

Step 2. We prove that F(x0 ) ⊂ ∂ϕ(x0 ). We consider an arbitrary point x ∗ ∈ F(x0 ) and we show that x ∗ ∈ ∂ϕ(x0 ). Since x ∗ ∈ F(x0 ), we deduce that (x ∗ , x) = ||x||2X = ||x ∗ ||2X ∗ . We have

(4.1.22)

4.1 Introduction

41

(x0 − u, x ∗ ) = (x0 , x ∗ ) − (u, x ∗ ) ≥ ||x0 ||2 − ||u|| · ||x ∗ ||2 .

(4.1.23)

Based on the following inequality x2 − xy ≥

1 2 (x − y 2 ), 2

(4.1.24)

we deduce that ||x0 ||2 − ||u|| · ||x ∗ || ≥ Finally, we obtain

1 ||x0 ||2 − ||u||2 = ϕ(x0 ) − ϕ(u). 2

(x0 − u, x ∗ ) ≥ ϕ(x0 ) − ϕ(u)

(4.1.25)

(4.1.26)

and based on the definition of the subdifferential we deduce that x ∗ ∈ ∂ϕ(x0 ). ¯ be a convex and lower Theorem 4.1.2 Let X be a Banach space and let ϕ : X → R semi-continuous function. Then, we have D(ϕ) ⊂ D(∂ϕ),

(4.1.27)

where D(ϕ) is the effective domain of the function ϕ. Proof First we recall the Hahn-Banach separation theorem, see [5]. If a set H is convex and closed and has a non-empty interior, then every point on the boundary of H can be separated from the rest of the set H by a separation hyperplane. We define the set H by H = {(x, λ) ∈ X × R, ϕ ≤ λ},

(4.1.28)

which is called the epigraph of the function ϕ. Since the inequality from the definition of the set H is not strict, it follows that H is a closed set. Let us prove that H is a convex set by using the convexity of the function ϕ. We have to show that ∀(x1 , λ1 ), (x2 , λ2 ) ∈ H, ∀t ∈ [0, 1] ⇒ ⇒ t (x1 , λ1 ) + (1 − t)(x2 , λ2 ) ∈ H ⇒ ⇒ (t x1 + (1 − t)x2 , tλ1 + (1 − t)λ2 ) ∈ H ⇔ ⇔ ϕ (t x1 + (1 − t)x2 ) ≤ tλ1 + (1 − t)λ2 .

(4.1.29)

In fact, we have (x1 , λ1 ), (x2 , λ2 ) ∈ H ⇒ ϕ(x1 ) ≤ λ1 and ϕ(x2 ) ≤ λ2 .

(4.1.30)

42

4 The Subdifferential of a Convex Function

Since the function ϕ is convex, we have ϕ (t x1 + (1 − t)x2 ) ≤ tϕ(x1 ) + (1 − t)ϕ(x2 ) ≤ tλ1 + (1 − t)λ2 .

(4.1.31)

Let (x0 , ϕ(x0 )) ∈ Fr H , which is the boundary of the set H . Then (x0 , ϕ(x0 ) + ε) ∈ Int H , where ε is sufficiently small. Since the function ϕ is lower semi-continuous, we deduce that ϕ is continuous in x0 ∈ Int D(ϕ). Clearly, we have ϕ(x0 ) < ϕ(x0 ) + ε. We consider y ∈ Vx0 and λ ∈ Vϕ(x0 )+ε and by the continuity of the function ϕ in the point x0 ∈ Int D(ϕ) we deduce that ϕ(y) ≤ λ, which proves that Int H = ∅. In conclusion, the epigraph H satisfies the three conditions from the Hahn-Banach theorem and then the point (x0 , ϕ(x0 )) can be separated from the rest of the set H by a separation hyperplane. We consider the hyperplane of equation (x, y0∗ ) + λ = ρ.

(4.1.32)

Let us assume, without loss of generality, that H is on the side of the hyperplane that satisfies (4.1.33) (x, y0∗ ) + λ ≥ ρ. Then, since the point (x0 , ϕ(x0 )) is on the other side of the hyperplane, we have (x0 , y0∗ ) + ϕ(x0 ) ≤ ρ.

(4.1.34)

By the equations (4.1.33) and (4.1.34) we have ϕ(x0 ) + (x0 , y0∗ ) ≤ (x, y0∗ ) + λ, ∀(x, λ) ∈ H.

(4.1.35)

On the other hand, since (x, λ) ∈ H , we have ϕ(x) ≤ λ and then the last inequality leads to  (4.1.36) ϕ(x0 ) − ϕ(x) ≤ x0 − x, y0∗ , ∀x ∈ X. This proves that

− y0∗ ∈ ∂ϕ(x0 ),

(4.1.37)

that is, the function ϕ is subdifferentiable in x0 and then x0 ∈ Int (D(∂ϕ)), which finishes the proof.  Definition 4.1.4 If X and Y are Banach spaces, then we say that the operator A : X → Y is monotone if (y1 − y2 , x1 − x2 ) ≥ 0, ∀x1 , x2 ∈ D(A), y1 ∈ A(x1 ), y2 ∈ A(x2 ), if the operator A is multi-valued. If A is a single-valued operator, then A is called monotone if

(4.1.38)

4.1 Introduction

43

(Ax1 − Ax2 , x1 − x2 ) ≥ 0, ∀x1 , x2 ∈ D(A).

(4.1.39)

Definition 4.1.5 The operator A : X → Y is called maximal monotone if it is monotone in the sense of the previous definition and if it does not admit a proper extension that is monotone. So ˜ ∀x ∈ D(A) ⊂ D( A). ˜  A˜ = A such that Ax ∈ Ax,

(4.1.40)

Note that if A : X → X ∗ is maximal monotone, then, for each x ∈ X , Ax is a closed convex subset of X ∗ since Ax = {x ∗ ∈ X ∗ |(x ∗ − v, x − u) ≥ 0, ∀u ∈ D(A), v ∈ A(u)}.

(4.1.41)

In order to characterize the maximal monotony of an operator defined on a Hilbert space, we present the following theorem, which was introduced by G. Minty [8]. Theorem 4.1.3 Let H be a Hilbert space and let A : D(A) ⊂ H → H be an operator. The necessary and sufficient condition for the operator A to be maximal monotone is ∀λ > 0 ⇒ R(I + λ A) = H, (4.1.42) where I is the identity operator. Proof Necessity. The proof is very technical, so we do not present it [8]. Sufficiency. We will prove that if R(I + λA) = H , for λ > 0 arbitrarily fixed, then the operator A is maximal monotone. The proof is by contradiction and we assume that A is not a maximal monotone ˜ which is a monotone operator, operator. Therefore, it admits a proper extension A, ˜ = Ax, ∀x = x0 and Ax ˜ 0 = y0 , i.e. there exists at least one x0 ∈ A˜ such that Ax / Ax0 . So, we have y0 ∈ ˜ y0 ∈ Ax. ˜ (y − y0 , x − x0 ) ≥ 0, ∀x ∈ D(A), y ∈ Ax, x0 ∈ D( A),

(4.1.43)

Since x0 + λy0 ∈ and R(I + λA) = H , we deduce that there exists the pair (x1 , y1 ) with x1 ∈ D(A) and y1 ∈ Ax1 such that x1 + λy1 = x0 + λy0 .

(4.1.44)

In (4.1.43) we consider x = x1 and y = y1 and then (y1 − y0 , x1 − x0 ) ≥ 0.

(4.1.45)

By taking (4.1.44) into consideration, we obtain (y1 − y0 , λ(x1 − x0 )) ≥ 0 ⇒ −λ (y1 − y0 , y1 − y0 ) ≥ 0 ⇒ ⇒ ||y1 − y0 ||2 ≤ 0 ⇒ y1 = y0 ⇒ x1 = x0 ,

(4.1.46)

44

4 The Subdifferential of a Convex Function

where we used again (4.1.44) in the last equality. But (x1 , y1 ) ∈ A, so y1 ∈ Ax1 and then y0 ∈ Ax0 , which contradicts the assump/ Ax0 . This finishes the proof.  tion y0 ∈ In the following theorem we give an example of a maximal monotone operator, i.e. the subdifferential. Theorem 4.1.4 The subdifferential of a convex and lower semi-continuous function is a (multi-valued) maximal monotone operator on its effective domain. ¯ be a convex and lower semiProof Let X be a Banach space and let ϕ : X → R continuous function. We prove that the operator A = ∂ϕ, A : X → X ∗ is maximal monotone. Step 1. First, we prove that ∂ϕ is a monotone operator, that is (y1 − y2 , x1 − x2 ) ≥ 0, ∀x1 , x2 ∈ D(∂ϕ), y1 ∈ ∂ϕ(x1 ), y2 ∈ ∂ϕ(x2 ).

(4.1.47)

By the definition of the subdifferential, we deduce that y1 ∈ ∂ϕ(x1 ) ⇒ ϕ(x1 ) − ϕ(u) ≤ (y1 , x1 − u), ∀u ∈ H y2 ∈ ∂ϕ(x2 ) ⇒ ϕ(x2 ) − ϕ(v) ≤ (y2 , x2 − v), ∀v ∈ H.

(4.1.48)

If in these two relations we consider u = x2 and v = x1 , respectively, then we obtain ϕ(x1 ) − ϕ(x2 ) ≤ (y1 , x1 − x2 ) ϕ(x2 ) − ϕ(x1 ) ≤ (y2 , x2 − x1 ).

(4.1.49)

We add side by side these two relations and we obtain 0 ≤ (y1 , x1 − x2 ) + (y2 , x2 − x1 ) ⇒ (y1 − y2 , x1 − x2 ) ≥ 0,

(4.1.50)

which proves that ∂ϕ is a monotone operator. Step 2. In order to prove that ∂ϕ is a maximal monotone operator, we will use the sufficiency from Minty’s theorem [8] in the particular case λ = 1. So, let us show that R(I + A) = H , that is, ∀y ∈ H the equation x + ∂ϕ(x) = y

(4.1.51)

has at least one solution in the unknown x. To this end, let us recall another theorem concerning lower-semicontinuous functions, which was introduced by Minty [8]. If ϕ is a convex and lower semi-continuous function, then ϕ is bounded from below by an affine function, i.e. ∃y0∗ ∈ H and λ0 ∈ R such that (4.1.52) ϕ(x) ≥ (y0∗ , x) + λ0 , ∀x ∈ H. We define the function ψ by

4.1 Introduction

45

1 ψ(x) = ϕ(x) + ||x||2 − (y, x), x ∈ H. 2

(4.1.53)

The function ψ is lower semi-continuous since the norm and the scalar product are continuous, while the function ϕ is assumed to be lower semi-continuous. Let us show that ψ is a convex function ψ (λx1 + (1 − λ)x2 ) ≤ λψ(x1 ) + (1 − λ)ψ(x2 ),

(4.1.54)

∀λ ∈ [0, 1], x1 , x2 ∈ H. By the definition (4.1.53), we obtain ψ (λx1 + (1 − λ)x2 ) = ϕ (λx1 + (1 − λ)x2 ) + 1 + ||λx1 + (1 − λ)x2 ||2 − (y, λx1 + (1 − λ)x2 ) ≤ 2 ≤ λϕ(x1 ) + (1 − λ)ϕ(x2 )+ 1 + (λx1 + (1 − λ)x2 , λx1 + (1 − λ)x2 ) − 2 − λ(y, x1 ) − (1 − λ)(y, x2 ) = 

1 = λ ϕ(x1 ) + ||x1 ||2 − (y, x1 ) + 2 

1 + (1 − λ) ϕ(x2 ) + ||x2 ||2 − (y, x2 ) = 2 = λψ(x1 ) + (1 − λ)ψ(x2 ).

(4.1.55)

Since the function ψ is lower semi-continuous and convex, it satisfies the assumptions of Minty’s theorem, hence it is bounded from below by an affine function, i.e. ∃y0∗ ∈ H and λ0 ∈ R such that ψ(x) ≥ (y0∗ , x) + λ0 , ∀x ∈ H.

(4.1.56)

By this inequality, we deduce that lim ψ(x) = +∞.

||x||→∞

(4.1.57)

With this condition, the function ψ is, as we have already proven, convex and lower semi-continuous and it attains its minimum. Therefore, ∃x0 ∈ H such that ψ(x0 ) ≤ ψ(x), ∀x ∈ H.

(4.1.58)

We will prove that this x0 is exactly the solution of the Eq. (4.1.51). By (4.1.53), the inequality ψ(x0 ) ≤ ψ(x) becomes

46

4 The Subdifferential of a Convex Function

1 1 ϕ(x0 ) + ||x0 ||2 − (y, x0 ) ≤ ϕ(x) + ||x||2 − (y, x) ⇔ 2 2 1 1 ⇔ ϕ(x0 ) − ϕ(x) ≤ (y, x0 − x) + ||x||2 − ||x0 ||2 . 2 2

(4.1.59)

On the other hand, by the inequality of Schwarz for the scalar product [10], we obtain 1 1 (y, x0 − x) + ||x||2 − ||x0 ||2 ≤ (y − x, x0 − x). 2 2 In fact, we have

1 1 (y, x0 ) − (y, x) + ||x||2 − ||x0 ||2 ≤ 2 2 ≤ (y, x0 ) − (y, x) − (x, x0 ) + ||x||2 ⇔ 1 1 ⇔ ||x||2 + ||x0 ||2 ≥ (x, x0 ). 2 2

(4.1.60)

(4.1.61)

This last inequality holds true since it corresponds to the elementary inequality 1 2 a + b2 ≥ ab. 2

(4.1.62)

By the inequalities (4.1.59) and (4.1.60), we obtain ϕ(x0 ) − ϕ(x) ≤ (y − x, x0 − x), ∀x ∈ H.

(4.1.63)

Formula (4.1.63) is very close to the desired relation. The proof of the theorem is finished by (4.1.64) ϕ(x0 ) − ϕ(x) ≤ (y − x0 , x0 − x) , ∀x ∈ H, since this relation, according to the definition of the subdifferential, implies that y − x0 ∈ ∂ϕ(x0 ),

(4.1.65)

that is, x0 is the solution of the equation (4.1.51). In the sequel, we write the relation (4.1.63) for x = t x0 + (1 − t)v, where t is arbitrary, t ∈ [0, 1] and v is arbitrary, v ∈ H and we obtain (y − x, x0 − x) ≤ (y − x0 , x0 − v) + (1 − t)||x0 − v||2 , ∀x ∈ H. In fact, we have

(4.1.66)

4.1 Introduction

47

(y − t x0 − (1 − t)x0 , x0 − t x0 − (1 − t)v) ≤ ≤ (y − x0 , x0 − v) + (1 − t)||x0 − v||2 ⇔ ⇔ (y − x0 + x0 − t x0 − (1 − t)x0 , (1 − t)(x0 − v)) ≤ ≤ (y − x0 , x0 − v) + (1 − t)||x0 − v||2 ⇔ ⇔ (y − x0 , (1 − t)(x0 − v)) + (1 − t)2 ||x0 − v||2 ≤

(4.1.67)

≤ (y − x0 , x0 − v) + (1 − t)||x0 − v||2 ⇔

⇔ (1 − t) (y − x0 , x0 − v) + (1 − t)||x0 − v||2 ≤ ≤ (y − x0 , x0 − v) + (1 − t)||x0 − v||2 ⇔ 1 − t ≤ 1, where the last inequality is obvious. We pass to the limit in (4.1.66) with t → 0 and we obtain (4.1.68) (y − v, x0 − v) ≤ (y − x0 , x0 − v), ∀v ∈ H, because if t → 0, then x → v. By the last inequality, (4.1.63) becomes ϕ(x0 ) − ϕ(v) ≤ (y − x0 , x0 − v), ∀v ∈ H. Based on these relations, the proof of the theorem is finished.

(4.1.69) 

Definition 4.1.6 Let X be a Banach space. The operator A : X → X ∗ is called coercive if it satisfies the condition lim

||x||→∞

(Ax, x) = +∞. ||x||

(4.1.70)

The properties of maximal monotone operators prove useful in showing existence results. The following theorem, which was introduced by Crandall [6], is such an example. Theorem 4.1.5 If the operator A : X → X ∗ is maximal monotone and coercive and, moreover, if 0 ∈ D(A), then A is a surjective operator, so R(A) = X ∗ . In the case of a single-valued operator, this means that ∀y ∈ X ∗ ∃x ∈ X such that Ax = y. For multi-valued operators, surjectivity means that ∀y ∈ X ∗ ∃x ∈ X such that Ax  y. We do not present the proof since it is very technical [6]. ¯ is lower semi-continuous and ϕ satisfies the If the convex function ϕ : X → R coercivity assumptions (ϕ(x), x) = +∞, (4.1.71) lim ||x||→+∞ ||x|| then ∂ϕ is a surjective operator since it is a maximal operator and we can use the previous theorem: ∀y ∈ X ∗ , ∃x ∈ X such that ∂ϕ(x) = y.

48

4 The Subdifferential of a Convex Function

4.2 The Conjugate Function The goal of this function is to transform a minimization problem into a maximization problem and conversely. ¯ be a convex and lower Definition 4.2.1 Let X be a Banach space and let ϕ : X → R semi-continuous function. Then its conjugate function, which is denoted by ϕ∗ , is defined by ¯ ϕ∗ (x ∗ ) = sup{(x ∗ , x) − ϕ(x), x ∈ X }. (4.2.1) ϕ∗ : X ∗ → R, ¯ be a convex and lower semi-continuous function. Theorem 4.2.1 Let ϕ : X → R ∗ Then its conjugate function ϕ is a proper function, that is, ϕ∗ = ∅. Proof Since ϕ is a convex and lower semi-continuous function, we can apply Minty’s theorem [8] and we deduce that ϕ is bounded from below by an affine function, that is, ∃x ∗ ∈ X ∗ and α ∈ R such that ϕ(x) ≥ (x0∗ , x) + α, ∀x ∈ X.

(4.2.2)

We can write this relation in the equivalent form − ϕ(x) ≤ −(x0∗ , x) − α, ∀x ∈ X ⇔ ⇔ (x0∗ , x) − ϕ(x) ≤ −α, ∀x ∈ X.

(4.2.3)

In this last relation, we pass to the supremum   sup (x0∗ , x) − ϕ(x) ≤ −α ⇒ x∈X

⇒ ϕ∗ (x0∗ ) ≤ −α ⇒ ϕ∗ (x0∗ ) < +∞. So, ϕ∗ = ∅, hence the function ϕ∗ is proper, which finishes the proof.

(4.2.4) 

In the following theorem we prove other properties of the conjugate function, which are useful for passing from a minimum problem to a maximum problem and conversely. ¯ has the properties Theorem 4.2.2 The conjugate function ϕ∗ : X ∗ → R (i) ϕ∗ (x ∗ ) + ϕ(x) ≥ (x ∗ , x), ∀x ∈ X, ∀x ∗ ∈ X ∗ . The equality holds true if and only if x ∗ ∈ ∂ϕ(x). (ii) ϕ∗∗ (x) ≤ ϕ(x), ∀x ∈ X . The equality holds true if and only if x ∗ ∈ ∂ϕ(x). Proof (i) We use the definition of the conjugate function. For x ∗ arbitrarily fixed in X ∗ , we have ϕ∗ (x ∗ ) = sup{(x ∗ , x) − ϕ(x), x ∈ X } ⇒ ⇒ ϕ∗ (x ∗ ) ≥ (x ∗ , x) − ϕ(x), ∀x ∈ X ⇒ (4.2.5) ⇒ ϕ∗ (x ∗ ) + ϕ(x) ≥ (x ∗ , x), ∀x ∈ X.

4.2 The Conjugate Function

49

For the second assertion from (i) we consider an arbitrary x ∗ ∈ ∂ϕ(x). According to the definition of the subdifferential, we obtain ϕ(x) − ϕ(u) ≤ (x ∗ , x − u), ∀u ∈ X ⇔ ⇔ ϕ(x) − ϕ(u) ≤ (x ∗ , x) − (x ∗ , u), ∀u ∈ X ⇔ ⇔ (x ∗ , u) − ϕ(u) ≤ (x ∗ , x) − ϕ(x), ∀u ∈ X.

(4.2.6)

We pass to the supremum in the last inequality sup{(x ∗ , u) − ϕ(u), u ∈ X } ≤ (x ∗ , x) − ϕ(x) ⇒ ⇒ ϕ∗ (x ∗ ) ≥ (x ∗ , x) − ϕ(x), ∀x ∈ X ⇒

(4.2.7)

⇒ ϕ∗ (x ∗ ) + ϕ(x) ≤ (x ∗ , x), ∀x ∈ X. Since the converse inequality holds true without the assumption that x ∗ ∈ ∂ϕ(x), we deduce that (4.2.8) ϕ∗ (x ∗ ) + ϕ(x) = (x ∗ , x), ∀x ∗ ∈ X ∗ , ∀x ∈ X. (ii) For the conjugate of the conjugate function, we adapt the definition of the conjugate (4.2.9) ϕ∗∗ (x) = sup{(x ∗ , x) − ϕ∗ (x ∗ ), x ∗ ∈ X ∗ }. On the other hand, by (i) we have ϕ∗ (x ∗ ) + ϕ(x) ≥ (x ∗ , x), ∀x ∗ ∈ X ∗ ⇔ ⇔ ϕ(x) ≥ (x ∗ , x) − ϕ∗ (x ∗ ), ∀x ∗ ∈ X ∗ .

(4.2.10)

Then ϕ(x) is greater than the largest value from the right-hand side of the last inequality, that is, ϕ(x) ≥ sup{(x ∗ , x) − ϕ∗ (x ∗ ), x ∗ ∈ X ∗ } ⇒ ϕ(x) ≥ ϕ∗∗ (x).

(4.2.11)

For the second assertion from (ii) we consider an arbitrary point x ∗ ∈ ∂ϕ(x). According to (i), we have ϕ∗ (x ∗ ) + ϕ(x) = (x ∗ , x), ∀x ∈ X ⇒ ⇒ ϕ(x) = (x ∗ , x) − ϕ∗ (x ∗ ), ∀x ∈ X ⇒ ⇒ ϕ(x) = sup{(x ∗ , x) − ϕ∗ (x ∗ ), x ∈ X } = ϕ∗∗ (x), which finishes the proof.

(4.2.12)



In the following theorem, we show the relation between the subdifferential of a convex function and its conjugate function.

50

4 The Subdifferential of a Convex Function

Theorem 4.2.3 The subdifferential of the conjugate function is equal to the inverse of the subdifferential. Proof We will work on a Hilbert space H since every Hilbert space is reflexive and then we have H ∗ = H . So we have the function ϕ : H → R and then ∂ϕ : H → R. We will only prove the inclusion (∂ϕ)−1 ⊂ ∂ϕ∗

(4.2.13)

since the operator (∂ϕ)−1 is maximal monotone and it does not admit proper extensions which are monotone (we proved that the subdifferential ∂ϕ∗ is a maximal monotone operator). Then we deduce that (∂ϕ)−1 = ∂ϕ∗ .

(4.2.14)

Let ∀y ∈ (∂ϕ)−1 (x) ⇒ x ∈ ∂ϕ(y). By the definition of the subdifferential, we have ϕ(y) − ϕ(v) ≤ (x, y − v), ∀v ∈ H ⇔ ⇔ (x, v) − ϕ(v) ≤ (x, y) − ϕ(y), ∀v ∈ H.

(4.2.15)

We pass to the supremum in the last inequality sup{(x, v) − ϕ(v), v ∈ H } ≤ (x, y) − ϕ(y) ⇔ ⇔ ϕ∗ (x) ≤ (x, y) − ϕ(y).

(4.2.16)

According to Theorem 4.2.2, since x ∈ ∂ϕ(y), we deduce that ϕ∗ (x) = (x, y) − ϕ(y).

(4.2.17)

ϕ∗ (z) = sup{(z, u) − ϕ(u), u ∈ H } ⇒ ⇒ ϕ∗ (z) ≥ (z, u) − ϕ(u), ∀u ∈ H ⇒ ⇒ −ϕ∗ (z) ≤ −(z, u) + ϕ(u), ∀u ∈ H.

(4.2.18)

For z arbitrary in H , we have

If in the last inequality we take u = y, then we obtain − ϕ∗ (z) ≤ −(z, y) + ϕ(y).

(4.2.19)

We add up the relations (4.2.17) and (4.2.19) and we obtain ϕ∗ (x) − ϕ∗ (z) ≤ (y, x − z), ∀z ∈ H ⇒ y ∈ ∂ϕ∗ (x). This finishes the proof of the inclusion (∂ϕ)−1 ⊂ ∂ϕ∗ .

(4.2.20) 

4.2 The Conjugate Function

51

Example [Conjugate function] Let K be a closed and convex set. We denote by I K the indicator function of the set K , that is  0, x∈K (4.2.21) I K (x) = +∞, x ∈ / K. Then the conjugate of this function, I K∗ , which we denote by HK and is called the support function of the set K , is defined by I K∗ (x ∗ ) = HK (x ∗ ) = sup{(x ∗ , x) − I K (x), x ∈ K } = sup{(x ∗ , x), x ∈ K }. (4.2.22) In the following theorem, we pass from a minimization problem to a maximization problem by using the conjugate function. The result is due to Fenchel and we present it without a proof, see [4, 9]. ¯ be convex and lower Theorem 4.2.4 Let X be a Banach space and let ϕ, ψ : X → R semi-continuous functions. If Int D(ϕ) ∩ D(ψ) = ∅ or D(ϕ) ∩ Int D(ψ) = ∅,

(4.2.23)

then a minimum problem is transformed into a maximum problem by inf {ϕ(x) + ψ(x)} = sup {−ϕ∗ (x ∗ ) − ψ ∗ (−x ∗ )}.

x∈X

x ∗ ∈X ∗

(4.2.24)

The indicator function of the set K is used for the restrictions. So, if the set K contains the restrictions of a problem, then we can write a minimum problem with restrictions of the form inf {ϕ(x), x ∈ K } = inf {ϕ(x) + I K (x)}.

x∈X

x∈X

(4.2.25)

If Int D(ϕ) ∩ K = ∅ or D(ϕ) ∩ Int K = ∅, then we can use Fenchel’s theorem [4] inf {ϕ(x) + I K (x)} = sup {−ϕ∗ (x ∗ ) − HK (−x ∗ )},

x∈X

x ∗ ∈X ∗

(4.2.26)

where HK is the conjugate function of the indicator function. Example [on the real line] Let us consider the function ϕ(x) = (x − 1)2 and the set of restrictions K = {x, x ≤ 0}. Clearly, the function ϕ is convex and continuous, so it is lower semi-continuous and K is a convex and closed set. The conjugate function of the function ϕ is ϕ∗ (x ∗ ) = sup{(x ∗ , x) − ϕ(x), x ∈ K } = sup{(x ∗ , x) − (x − 1)2 , x ∈ K }. (4.2.27) On the real axis, the scalar product (x ∗ , x) becomes x ∗ x, hence

52

4 The Subdifferential of a Convex Function

ϕ∗ (x ∗ ) = x ∗ x − (x − 1)2 .

(4.2.28)

The extremal points are among the roots of the derivative, so we differentiate with respect to x x∗ + 2 ⇒ 2  ∗ 2 x +2 x∗ + 2 (x ∗ )2 + 4x ∗ − −1 = . ⇒ ϕ∗ (x ∗ ) = x ∗ 2 2 4

x ∗ − 2(x − 1) = 0 ⇒ x =

(4.2.29)

The support function becomes HK (x ∗ ) = sup{(x ∗ , x) − I K (x), x ∈ K } =  0, x∗ ≥ 0 = sup x ∗ x = (x ∗ )2 , x ∗ < 0.

(4.2.30)

Then the maximum problem becomes   (x ∗ )2 + 4x ∗ − (x ∗ )2 = sup{−ϕ∗ (x ∗ ) − HK (−x ∗ )} = sup − 4 x∗ x∗   5 ∗ 2  1 = . = sup − (x ) − x ∗ = − 4 4a 5 x∗

(4.2.31)

4.3 The Additivity of the Subdifferential We ask if the subdifferential is additive, like the differential and if it is true that ∂(ϕ + ψ) = ∂ϕ + ∂ψ.

(4.3.1)

The answer is that this is generally not true. But, under specific conditions (which we will emphasize), the subdifferential can become additive. Theorem 4.3.1 If the functions ϕ and ψ are convex and lower semi-continuous, then we have ∂ϕ + ∂ψ ⊂ ∂(ϕ + ψ). (4.3.2) Proof We prove the set inclusion because the subdifferential is a multi-valued operator. We consider an arbitrary point y ∈ ∂ϕ(x) + ∂ψ(x) and we have to prove that y ∈ ∂(ϕ + ψ)(x). We will write y = y1 + y2 , where y1 ∈ ∂ϕ(x) and y2 ∈ ∂ψ(x). According to the definition of the subdifferential, we have

4.3 The Additivity of the Subdifferential

(y1 , x − u) ≥ ϕ(x) − ϕ(u), ∀u ∈ H (y2 , x − u) ≥ ψ(x) − ψ(u), ∀u ∈ H.

53

(4.3.3)

We add these two relations side by side and we obtain (y1 + y2 , x − u) ≥ (ϕ + ψ)(x) − (ϕ + ψ)(u), ∀u ∈ H ⇒ ⇒ (y, x − u) ≥ (ϕ + ψ)(x) − (ϕ + ψ)(u), ∀u ∈ H ⇒

(4.3.4)

⇒ y ∈ ∂(ϕ + ψ)(x), 

which ends the proof.

In the following theorem, which was introduced by Rockafellar [9], we impose sufficient conditions for the subdifferential to be additive. ¯ = Theorem 4.3.2 Let X be a Banach space and let the functions ϕ, ψ : X → R (−∞, ∞] be convex and lower semi-continuous. If the functions ϕ and ψ satisfy one of the following two conditions Int D(ϕ) ∩ D(ψ) = ∅ or D(ϕ) ∩ Int D(ψ) = ∅,

(4.3.5)

then the subdifferential is additive, that is, the following equality holds true ∂(ϕ + ψ) = ∂ϕ + ∂ψ.

(4.3.6)

Proof We only have to prove the inclusion ∂(ϕ + ψ) ⊂ ∂ϕ + ∂ψ

(4.3.7)

since by Theorem 4.3.1 the inclusion ∂ϕ + ∂ψ ⊂ ∂(ϕ + ψ)

(4.3.8)

holds true without the extra conditions from the theorem. Let x0 ∈ D (∂(ϕ + ψ)) and y0 ∈ ∂(ϕ + ψ)(x0 ) and let us prove that y0 ∈ ∂ϕ(x0 ) + ∂ψ(x0 ). We can restrict our considerations to the case x0 = 0 and y0 = 0 since instead of the functions ϕ and ψ we can consider the functions ϕ(x) ˜ = ϕ(x + x0 ) − ϕ(x0 ), ˜ ψ(x) = ψ(x + x0 ) − ψ(x0 ).

(4.3.9)

In a previous proof, we showed that the functions built in this way are convex and lower semi-continuous, so they can replace the functions ϕ and ψ. Moreover, we have ϕ(0) ˜ = ϕ(x0 ) − ϕ(x0 ) = 0, (4.3.10) ˜ ψ(0) = ψ(x0 ) − ψ(x0 ) = 0.

54

4 The Subdifferential of a Convex Function

So, based on the assumption 0 ∈ ∂(ϕ + ψ)(0), let us prove that 0 ∈ ∂ϕ(0) + ∂ψ(0). But 0 ∈ ∂(ϕ + ψ)(0) based on the definition of the subdifferential, so (ϕ + ψ)(0) − (ϕ + ψ)(x) ≤ (0, 0 − x) ⇔ ⇔ ϕ(0) + ψ(0) − ϕ(x) + ψ(x) ≤ (0, 0 − x) = 0 ⇔

(4.3.11)

⇔ ϕ(x) + ψ(x) ≥ 0, since ϕ(0) = 0 and ψ(0) = 0 based on the translations mentioned above. We assume that the first hypothesis from the statement of the theorem is satisfied (the proof is analogous if the second hypothesis from the theorem holds true), that is, Int D(ϕ) ∩ D(ψ) = ∅. This means that ∃x 0 ∈ Int D(ϕ), that is, Int D(ϕ) is nonempty. We proved that if ϕ is a convex and lower semi-continuous function, then ϕ is continuous on the interior of the effective domain. So ϕ is continuous in the point x 0 . By the translations mentioned above, we can take x 0 = 0, so ϕ is continuous in 0. We consider the epigraphs of the functions ϕ and ψ K 1 = {(x, λ) ∈ X × R|ϕ(x) ≤ λ − ε}, K 2 = {(x, λ) ∈ X × R|ψ(x) ≤ −λ}.

(4.3.12)

Since ϕ is continuous in the point 0 ∈ Int D(ϕ), we deduce that Int K 1 is non-empty. Moreover, we have K 1 ∩ K 2 = ∅. Conversely, if we had K 1 ∩ K 2 = ∅, then ∃x ∈ K 1 and ∃x ∈ K 2 and we would have ϕ(x) ≤ λ − ε and ψ(x) ≤ −λ,

(4.3.13)

which we can add up side by side in order to obtain ϕ(x) + ψ(x) ≤ −ε.

(4.3.14)

This inequality contradicts (4.3.11), which proves that K 1 ∩ K 2 = ∅. Moreover, the sets K 1 and K 2 are convex and closed, hence all three conditions of the Hahn-Banach separation theorem [5] are satisfied. According to this, there exists a hyperplane, called a separation hyperplane, which we denote by H and which separates the two sets. We assume that the equation of the hyperplane H is (x ∗ , x) + λ = ρ, ∀(x, λ) ∈ H.

(4.3.15)

Then for the two sets we have (x ∗ , x) + λ ≥ ρ, ∀(x, λ) ∈ K 1 , (x ∗ , x) + λ ≤ ρ, ∀(x, λ) ∈ K 2 .

(4.3.16)

Since ψ(0) = 0, we deduce that (0, 0) ∈ K 2 so in the relations above we can take ρ = 0. Hence, we have

4.3 The Additivity of the Subdifferential

(x ∗ , x) + λ ≥ 0, ∀(x, λ) ∈ K 1 , (x ∗ , x) + λ ≤ 0, ∀(x, λ) ∈ K 2 .

55

(4.3.17)

Now we can prove that 0 ∈ ∂ϕ(0) + ∂ψ(0), so we show that there exists x ∗ such that (4.3.18) x ∗ ∈ ∂ϕ(0) and − x ∗ ∈ ∂ψ(0) and by adding these two relations we obtain x ∗ − x ∗ ∈ ∂ϕ(0) + ∂ψ(0).

(4.3.19)

If in (4.3.17)1 we consider λ = ϕ(x) and in (4.3.17)2 we consider λ = −ψ(x), then we obtain (x ∗ , x) ≥ −ϕ(x), (4.3.20) (x ∗ , x) + λ ≤ ψ(x). We can write these relations in the form ϕ(0) − ϕ(x) ≤ (−x ∗ , −x) ψ(0) − ψ(x) ≤ (x ∗ , −x).

(4.3.21)

− (ϕ(x) + ψ(x)) ≤ (0, 0 − x),

(4.3.22)

We add up these relations

which proves that 0 ∈ ∂ϕ(0) + ∂ψ(0). This finishes the proof.



References 1. H.W. Alt, Lineare Funktionalanalysis (Springer, Berlin, Heidelberg, 2006) 2. V. Barbu, T. Precupanu, Convexity and Optimization in Banach Spaces, 4th edn. (Springer, 2012) 3. J. Borwein, Convex Functions: Constructions, Characterizations and Counterexamples (Cambridge University Press, New York, 2010) 4. R.I. Bot, S.M. Grad, G. Wanka, Fenchel’s duality theorem. J. Optim. Theory Appl. 132, 509–515 (2007) 5. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, 2011) 6. M.G. Crandall, Semigroups of nonlinear contractions and dissipative sets. J. Funct. Anal. 3, 376–418 (1969) 7. V. Kadets, Course in Functional Analysis and Measure Theory (Springer, Cham, 2018) 8. G. Minty, Monotone operators in Hilbert space. Duke Math. J. 29, 341–346 (1962) 9. R.T. Rockafellar, Convex Analysis, No. 28 (Princeton University Press, Princeton, N.J., 1970) 10. W. Rudin, Real and Complex Analysis (McGraw-Hill, New York, 1974)

Chapter 5

Evolution Equations

Abstract This chapter presents some results about evolution equations. Moreover, it presents the definition and the main properties of the resolvent, the Yosida approximation and the principal section of a maximal monotone operator. A stability result is shown for the solution of the Cauchy problem associated to an evolution equation based on the inequality of Gronwall.

5.1 Introduction Definition 5.1.1 Let X be a Banach space, I a real interval, f a function, f : I → X , A an operator defined on X , A : X → X ∗ and u a function, u : I → X . An equation of the form du(t) + Au(t) = f (t) (5.1.1) dt is called an evolution equation. Their name suggests that such equations model phenomena which evolve in time. In fact, evolution equations are ordinary differential equations or partial differential equations on infinite dimensional spaces [2]. Definition 5.1.2 (i) A classical solution of the Eq. (5.1.1) is a function ϕ : I → X , ϕ ∈ C 1 (I ), which when replaced in (5.1.1) transforms it into an identity. (ii) A Caratheodory solution of the Eq. (5.1.1) is a function ϕ : I → X , which is absolutely continuous, so differentiable for almost every t ∈ I and which verifies the Eq. (5.1.1) for almost every t ∈ I . We can introduce other types of generalized solutions for the Eq. (5.1.1) under the condition that every generalized solution can be approximated by classical solutions. For the most concrete problems which are modelled by equations of the form (5.1.1) we impose an initial condition of the form u(0) = u 0 ,

(5.1.2)

where u 0 is a given number and the interval I is necessarily of the form I = [0, T ]. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_5

57

58

5 Evolution Equations

The problem formed with the evolution equation (5.1.1) and the initial condition (5.1.2) is the Cauchy problem [3]. Another condition that can be attached to the Eq. (5.1.1) is the periodicity condition, which has the form u 0 = u(0) = u(T ) = u T .

(5.1.3)

In the sequel, we only discuss the Cauchy problem (5.1.1)+ (5.1.2). We anticipate that if the operator A is maximal monotone and X is a Hilbert space, then the Cauchy problem (5.1.1)+(5.1.2) has a solution in the Caratheodory sense and this is unique [6]. First we present the proof in the case when X is a Hilbert space, which we denote by H . We will assume that the operator A is Lipschitz on H and we will write A ∈ Li p(H ), that is, ∃L > 0 such that |Au − Av| ≤ L|u − v|, ∀u, v ∈ H.

(5.1.4)

We denote by L 1 (0, T ; H ) the set of functions defined on the interval [0, T ] with values in the Hilbert space H and which have an integrable norm (computed in the sense of the space H ). In the sequel, we denote by (·, ·) and | · | the scalar product and the norm, respectively, on the Hilbert space H . Theorem 5.1.1 Let A : H → H be an operator. If A ∈ Li p(H ), then the Cauchy problem (5.1.1)+(5.1.2) has the solution u ∈ C 1 (0, T ; H ), ∀u 0 ∈ H, ∀ f ∈ L 1 (0, T ; H ), where C 1 (0, T ; H ) represents the set of differentiable functions with continuous derivative defined on the interval [0, T ] with values in the Hilbert space H . Proof Let us prove first that a function u is a solution for the Cauchy problem (5.1.1)+(5.1.2) if and only if u is a solution for the integral equation 



t

u(t) = u 0 −

t

Au(s)ds +

0

f (s)ds.

(5.1.5)

0

In fact, if u is a solution of the Cauchy problem (5.1.1)+(5.1.2), we have u(0) = u 0 , du(t) + Au(t) = f (t). dt

(5.1.6)

We integrate the Eq. (5.1.6)2 and we obtain 

t

u(t) − u(0) + 0



t

Au(s)ds =

f (s)ds.

(5.1.7)

0

If in this relation we consider the initial condition (5.1.6)1 , then we obtain the Eq. (5.1.5). Now we prove the converse relation, that is, we assume that the function u satisfies the Eq. (5.1.5) and we prove that u satisfies the Cauchy problem

5.1 Introduction

59

(5.1.1)+(5.1.2). If we differentiate in (5.1.5) by the rule of differentiation for parameter integrals, then we obtain du(t) = −Au(t) + f (t), dt

(5.1.8)

which is the Eq. (5.1.1). Now we replace t by 0 and we obtain  0  0 u(0) = u 0 − ··· + ⇒ u(0) = u 0 , 0

(5.1.9)

0

meaning that the initial condition (5.1.2) is satisfied. Owing to the equivalence proven above, instead of the Cauchy problem (5.1.1)+(5.1.2) we will consider the integral equation (5.1.5). We consider the space C(0, T ; H ), which has a clear meaning by the considerations above. On this space we define the norm ||u|| = sup eαt |u(t)|, α < 0,

(5.1.10)

0≤t≤T

where | · | is the usual norm on the Hilbert space H . We define the operator  by  : C(0, T ; H ) → C(0, T ; H ),  t  Au(s)ds + (u)(t) = u 0 − 0

t

f (s)ds.

(5.1.11)

0

Let us prove that  is a contractive operator (a contraction) on the space C(0, T ; H )  t  t u(t) − v(t) = − Au(s)ds + Av(s)ds = 0 0  t =− [Au(s) − Av(s)] ds ⇒ 0  t |Au(s) − Av(s)|ds ≤ ⇒ |u(t) − v(t)| ≤ 0  t  t L|u(s) − v(s)|ds = L e−αs eαs |u(s) − v(s)|ds ≤ ≤ 0 0  −αs   t e |t = e−αs ||u − v||ds = L||u − v|| ≤L −α 0 0     1 e−αt L = L||u − v|| − = − ||u − v|| e−αt − 1 ≤ α α α L ≤ − ||u − v||e−αt , α

(5.1.12)

60

so

5 Evolution Equations

L |u(t) − v(t)| ≤ − ||u − v||e−αt . α

(5.1.13)

In this inequality we take α = −2L < 0 and we obtain 1 ||u − v||e2Lt ⇔ 2 1 ⇔ e−2Lt |u(t) − v(t)| ≤ ||u − v||. 2

|u(t) − v(t)| ≤

(5.1.14)

In the last inequality we pass to the supremum for t ∈ [0, T ] and by considering the norm in (5.1.10), we obtain ||u − v|| ≤

1 ||u − v||, 2

(5.1.15)

which proves that the operator  is a contraction. So we can apply the Banach fixed point theorem [8], according to which the equation u(t) = u(t),

(5.1.16)

which is exactly the Eq. (5.1.5), has a solution and this is unique. Based on the equivalence from the beginning of the proof, we deduce that the Cauchy problem (5.1.1)+(5.1.2) has a unique solution and the proof of the theorem is finished.  The main result is the following theorem, which has three authors: Komura, Crandall, Kato, see [4, 5]. Theorem 5.1.2 Let A : D(A) ⊂ H → H be a maximal monotone operator. If u 0 ∈ D(A) and ∈ L 1 (0, T ; H ), then the Cauchy problem has a unique solution u ∈ C(0, T ; H ), which is absolutely continuous and differentiable almost everywhere on [0, T ], u(t) ∈ D(A), ∀t ∈ [0, T ] and the right derivative of the function u satisfies the equation d+ u(t) + A0 u(t) = f (t), ∀t ∈ [0, T ], dt

(5.1.17)

where A0 is called the principal section of the multi-valued operator A and is defined by (5.1.18) A0 u = inf{|y|, y ∈ Au}, that is, the element of minimal norm from the set Au. For the proof of the theorem we need some preparations. The goal of the principal section is to pass from multi-valued operators to singlevalued operators. In the case of a single-valued operator, the principal section of the operator coincides with the operator, that is A0 ≡ A.

5.2 The Resolvent and the Yosida Approximation

61

5.2 The Resolvent and the Yosida Approximation Let H be a Hilbert space and A : D(A) ⊂ H → H be a maximal monotone operator [6]. Besides the principal section of the operator A, denoted by A0 and defined by A0 u = inf{|y|, y ∈ Au},

(5.2.1)

we introduce two new operators defined by means of A. Definition 5.2.1 The resolvent of the operator A is the operator Jλ defined by Jλ = (I + λA)−1 , λ > 0. def

(5.2.2)

Definition 5.2.2 The Yosida approximation [1] of the operator A is the operator Aλ defined by I − Jλ def , λ > 0. (5.2.3) Aλ = (I − Jλ )λ−1 = λ In the following theorem, we formulate and prove the main properties of the resolvent, the Yosida approximation and the principal section of a maximal monotone operator. Theorem 5.2.1 Let H be a Hilbert space and A : D(A) ⊂ H → H be a maximal monotone operator. The resolvent operator, the Yosida approximation and the principal section associated to the operator A have the properties (i) |Jλ x − Jλ y| ≤ |x − y|, ∀x, y ∈ H , ∀λ > 0 (ii) Aλ (x) ∈ A Jλ (x), ∀x ∈ H (iii) Aλ is a Lipschitz operator on H (iv) |Aλ (x)| ≤ A0 x, ∀x ∈ D(A) (v) limλ→0 Jλ (x) = x, ∀x ∈ D(A) (vi) limλ→0 Aλ (x) = A0 (x). Proof (i) We use the notations xλ = Jλ (x), yλ = Jλ (y).

(5.2.4)

According to the definition of the resolvent Jλ , we have xλ + Axλ (x) = x, yλ + Ayλ (y) = y.

(5.2.5)

We subtract these relations side by side and we obtain xλ − yλ + λ A(xλ − yλ ) = x − y. We multiply this relation by xλ − yλ

(5.2.6)

62

5 Evolution Equations

|xλ − yλ |2 + λ (Axλ − Ayλ , xλ − yλ ) − (x − y, xλ − yλ ).

(5.2.7)

Now we consider that A is a monotone operator (even if A is maximal monotone) and λ > 0 such that (5.2.8) |xλ − yλ |2 ≤ (x − y, xλ − yλ ). By the inequality of Schwarz we obtain |xλ − yλ |2 ≤ |x − y||xλ − yλ | ⇒ |xλ − yλ | ≤ |x − y| ⇒ ⇒ |Jλ (x) − Jλ (y)| ≤ |x − y|, ∀x, y ∈ H.

(5.2.9)

(ii) We use the definition of the Yosida approximation Aλ x =

 1 x − (I + λA)−1 x ∈ λ

 1 (I + λA)(I + λA)−1 x − (I + λA)−1 x = λ 1 = (I + λ A)−1 [(I + λA)x − x] = A(I + λA)−1 x = A Jλ x. λ ∈

(5.2.10)

(iii) Based on (i), we deduce that the resolvent Jλ is a Lipschitz operator with the constant L = 1. Clearly, the identity operator is a Lipschitz operator with the constant L = 1. This helps us find the Lipschitz constant for the Yosida approximation 1 [(x − y) + Jλ y − Jλ x] ⇒ λ 1 2 ⇒ |Aλ x − Aλ y| ≤ (|x − y| + |Jλ x − Jλ y|) ≤ |x − y|, λ λ

Aλ x − Aλ y =

(5.2.11)

so Aλ is a Lipschitz operator with the constant L = λ2 . (iv) By using (ii) we have Aλ (x) ∈ A Jλ (x) ⇒ 1 ⇒ Aλ (x) ∈ [Jλ (I + λ A)x − Jλ x] ⇒ λ 1 ⇒ |Aλ (x)| ≤ |(I + λA)x − x| ⇒ λ ⇒ |Aλ (x)| ≤ |y|, ∀y ∈ Ax ⇒

(5.2.12)

⇒ |Aλ (x)| ≤ A0 x, ∀x ∈ D(A). (v) By the definition of Jλ and by (iv) we obtain I − Jλ = λAλ ⇒ |x − Jλ x| = λ|Aλ x| ≤ λA0 x, ∀x ∈ D(A).

(5.2.13)

5.2 The Resolvent and the Yosida Approximation

63

We pass to the limit with λ → 0 and we use the boundedness of A0 in order to obtain (5.2.14) lim Jλ (x) = x, ∀x ∈ D(A). λ→0

Let us now prove the result for x ∈ D(A). We consider x ∈ D(A) \ D(A) and by the definition of the closure we deduce that ∃xn ∈ D(A) such that lim xn = x. n→∞

(5.2.15)

By the definition of Jλ , we have |x − Jλ x| = |x − xn + xn − Jλ xn + Jλ xn − Jλ x| ⇒ ⇒ |x − Jλ x| ≤ |x − xn | + |Jλ xn − Jλ x| + |xn − Jλ xn | ⇒ ⇒ |x − Jλ x| ≤ 2|x − xn | + |Jλ xn − xn |.

(5.2.16)

But x ∈ D(A) and then by the first part of the proof we have lim Jλ (xn ) = xn ⇔ |Jλ xn − xn | ≤

λ→0

Then lim xn = x ⇔ |xn − x| ≤

n→∞

ε , ∀ε > 0. 2

ε , ∀ε > 0. 4

(5.2.17)

(5.2.18)

The last two inequalities are introduced in (5.2.16) and we obtain ε ε |Jλ x − x| ≤ 2 + = ε ⇔ lim Jλ (x) = x, ∀x ∈ D(A) \ D(A). λ→0 4 2

(5.2.19)

(v) By (iv) we have |Aλ x| ≤ A0 x

(5.2.20)

and this inequality shows that the sequence Aλ x is bounded with respect to λ. Hence, there exists at least one subsequence which is weakly convergent {Aλn } ⊂ {Aλ } such that Aλn x  y if λn → 0.

(5.2.21)

Aλn x ∈ A Jλn x and Jλn x → x

(5.2.22)

On the other hand,

and every maximal monotone operator is semi-closed, that is, closed in the weak topology, so y ∈ Ax ⇒ y = A0 x since |y| ≤ A0 x ⇔ Aλ x → A0 x.

(5.2.23)

64

5 Evolution Equations



The proof of the theorem is finished.

Theorem 5.2.2 (Gronwall) Let ψ be an integrable function on [a, b] and nonnegative on [a, b], that is, ψ(t) ≥ 0, ∀t ∈ [a, b]. We consider the continuous function h ∈ C[a, b] which satisfies the inequality 1 2 1 h (t) ≤ C 2 + 2 2



b

ψ(s)|h(s)|ds, ∀t ∈ [a, b],

(5.2.24)

a

where C is a real constant. Then h satisfies the inequality  |h(t)| ≤ |C| +

t

ψ(s)ds, ∀t ∈ [a, b].

(5.2.25)

a

Proof Step 1. In the first part we will present the proof by using the extra assumption C = 0. We introduce the notation  y(t) = C 2 + 2

t

ψ(s)|h(s)|ds

(5.2.26)

a

and then (5.2.24) becomes h 2 (t) ≤ y(t), ∀t ∈ [a, b].

(5.2.27)

We differentiate in (5.2.26) by using the rule of differentiation of the parameter integral (5.2.28) y  (t) = 2ψ(t)|h(t)| ⇒ y  (t) ≥ 0, from where we deduce that y is an increasing function. Then y(a) = C 2 > 0, so we deduce that y(t) is positive ∀t ∈ [a, b] and then we can extract the square root in (5.2.27) (5.2.29) |h(t)| ≤ y(t) ⇒ 2ψ(t)|h(t)| ≤ 2 y(t)ψ(t), ∀t ∈ [a, b]. By (5.2.28) and (5.2.29) we have y  (t) ≤ 2 y(t)ψ(t), ∀t ∈ [a, b] ⇒ y  (t) ⇒ √ ≤ ψ(t), ∀t ∈ [a, b] ⇒ 2 y(t) 

y(t) ≤ ψ(t), ∀t ∈ [a, b]. ⇒ We integrate this last inequality on the interval [a, t]

(5.2.30)

5.2 The Resolvent and the Yosida Approximation



y(t) −

65

 t √ C2 ≤ ψ(s)ds.

(5.2.31)

a

By the last inequality and by (5.2.27) we deduce that  |h(t)| ≤ |C| +

t

ψ(s)ds, ∀t ∈ [a, b],

(5.2.32)

a

which is the conclusion of the theorem. Step 2. If C = 0, then we rewrite the proof from Step 1 with ε instead of C and finally, we pass to the limit with ε → 0 and the proof of the theorem is finished.  Now we can prove the main result, that is, the existence and the uniqueness of the solution of the Cauchy problem dx + Ax  f dt x(0) = x0 ,

(5.2.33)

where the operator A is maximal monotone and it is defined on the Hilbert space H . Theorem 5.2.3 If the operator A : D(A) ⊂ H → H is maximal monotone, the functions f and f  ∈ L 2 (0, T ; H ) and x0 ∈ D(A), then the Cauchy problem (5.2.33) has a solution x which is (i) unique; (ii) continuous, that is, x ∈ C ((0, T ); H ); (iii) differentiable almost everywhere on the interval [0, T ]; (iv) x  ∈ L ∞ ((0, T ); H ); (v) x verifies the Eq. (5.2.33)1 for almost every t ∈ [0, T ]; (vi) x satisfies the equation d + x(t) + A0 x(t) = f (t), ∀t ∈ [0, T ]. dt

(5.2.34)

Proof We will present the proof in three steps. In the first step we prove the existence of the solution for an approximative problem, that is, for a problem which approximates the Cauchy problem (5.2.33). In the second step we will make some qualitative evaluations of the solution of the approximative problem found in the first step. Finally, in the third step, based on the estimates from the second step, we will be able to pass to the limit in the approximative solution and we will obtain the solution of the exact problem (5.2.33). Step 1. We consider the approximative problem d xλ + Aλ x λ = f dt xλ (0) = x0 ,

(5.2.35)

66

5 Evolution Equations

where Aλ is the Yosida approximation [1] of the operator A. The problem (5.2.35) approximates the Cauchy problem (5.2.33) since we proved that for λ → 0 we have Aλ → A. In Theorem 5.2.2 we proved that Aλ is a Lipschitz operator and then we can apply Caratheodory’s theorem to deduce that the approximative problem has a solution, which is unique and continuous, xλ ∈ C 1 (0, T ; H ). We will prove that for λ → 0, the solution xλ of the problem (5.2.35) tends to the solution x of the problem (5.2.33). Step 2. Since Aλ ∈ Li p(H ), according to a theorem of Lipschitz [7], we deduce that Aλ xλ is absolutely continuous, hence, differentiable almost everywhere. By assumption, the function f is differentiable almost everywhere and by Caratheodory’s theorem, xλ ∈ C 1 (0, T ; H ), so we can differentiate in the Eq. (5.2.35)1 side by side d d 2 xλ + (Aλ xλ ) = f  almost everywhere ⇔ dt 2 dt ⇔ xλ + (Aλ xλ ) = f  almost everywhere.

(5.2.36)

We scalarly multiply by xλ   (xλ , xλ ) + (Aλ xλ ) , xλ = ( f  , xλ ).

(5.2.37)

We have the following equality  1 |y(t)|2 = |y  (t)||y(t)|. 2

(5.2.38)

In fact, by the definition of the derivative, we have  1 |y(t + h)|2 − |y(t)|2 1 |y(t)|2 = lim = h→0 2 2 h (5.2.39) |y(t + h)| − |y(t)| |y(t + h)| + |y(t)| · lim = |y  (t)| · |y(t)|. = lim h→0 h→0 h 2 On the other hand, by the monotonicity of the operator Aλ , we have  lim

h→0

Aλ xλ (t + h) − Aλ xλ (t) xλ (t + h) − xλ (t) , h h

 ≥ 0.

(5.2.40)

We use (5.2.40) and (5.2.38) in (5.2.37) and we obtain  1  |xλ (t)|2 ≤ | f  (t)| · |xλ (t)| ⇒ 2 ⇒ |xλ (t)| ≤ | f  (t)| almost everywhere. We integrate on the interval [0, t]

(5.2.41)

5.2 The Resolvent and the Yosida Approximation

|xλ (t)| ≤ |xλ (0)| +

67



t

| f  (s)|ds =

0



t

= | f (0) − Aλ x0 | +

| f  (s)|ds ≤

(5.2.42)

0

 ≤ | f (0)| + |A0 x0 | +

t

| f  (s)|ds.

0

In order to obtain the last inequality from (5.2.42), we used |Aλ x0 | ≤ |A0 x0 |. By considering the assumption that f and f  ∈ L 2 (0, T ; H ) and x0 ∈ D(A) ⇒ |A0 x0 | < ∞,

(5.2.43)

we deduce that all terms on the right-hand side of the inequality (5.2.42) are bounded and then we have (5.2.44) |xλ (t)| ≤ M, ∀t ∈ [0, T ]. Clearly, we can write 

t

f  (s)ds, t ∈ [0, T ] ⇒  t | f  (s)|ds, t ∈ [0, T ]. ⇒ | f (t)| ≤ | f (0)| +

f (t) = f (0) +

0

(5.2.45)

0

By (5.2.44) and (5.2.45), we deduce that |Aλ xλ (t)| ≤ C, ∀t ∈ [0, T ].

(5.2.46)

Then we use the definition of the Yosida approximation Aλ x λ =

xλ − Jλ xλ λ

(5.2.47)

and the inequality (5.2.46) can be rewritten as |xλ (t) − Jλ xλ (t)| ≤ Cλ, ∀t ∈ [0, T ].

(5.2.48)

Step 3. We will use the evaluations derived in the second step in order to show that xλ tends to a solution of the Cauchy problem (5.2.33). Firstly, we show that xλ is a fundamental sequence. We have xλ + Aλ xλ  f, xμ + Aμ xμ  f. We substract side by side these two relations and we obtain

(5.2.49)

68

5 Evolution Equations

(xλ − xμ ) + Aλ xλ − Aμ xμ = 0.

(5.2.50)

We scalarly multiply this relation by xλ − xμ and we deduce    1 (xλ − xμ )2 + Aλ xλ − Aμ xμ , xλ − xμ = 0. 2 We can write     Aλ xλ − Aμ xμ , xλ − xμ = Aλ xλ − Aμ xμ , Jλ xλ − Jμ xμ +   + Aλ xλ − Aμ xμ , xλ − Jλ xλ − (xμ − Jμ xμ ) .

(5.2.51)

(5.2.52)

Based on Aλ x ∈ A Jλ x, we deduce that 

 Aλ xλ − Aμ xμ , Jλ xλ − Jμ xμ ≥ 0.

(5.2.53)

Moreover, owing to the linearity of the scalar product, we can write 

 Aλ xλ − Aμ xμ , xλ − Jλ xλ − (xμ − Jμ xμ ) =     = Aλ xλ − Aμ xμ , xλ − Jλ xλ − Aλ xλ − Aμ xμ , xμ − Jμ xμ .

(5.2.54)

If we consider (5.2.46) and (5.2.48), then the relation (5.2.51) becomes  1 |xλ − xμ |2 ≥ 2C 2 (λ + μ), 2

(5.2.55)

and by integration on the interval [0, t] we obtain

so

|xλ − xμ |2 ≤ 4C 2 (λ + μ)t ≤ 4C 2 (λ + μ)T,

(5.2.56)

√ |xλ − xμ | ≤ 2C T λ + μ.

(5.2.57)

The relation (5.2.57) proves that the sequence {xλ }λ satisfies the Cauchy condition of uniform convergence. So there exists limλ→0 xλ and this limit is uniform. We denote this limit by x(t) and this is defined ∀t ∈ [0, T ]. Since it is the uniform limit of a sequence of continuous functions, we deduce that the function x(t) is continuous. We will prove now that there exists x  (t) and x  ∈ L ∞ (0, T ; H ). Clearly, we can write |x(t) − x(s)| ≤ |xλ (t) − xλ (s)| + |xλ (t) − x(t)| + |xλ (s) − x(s)|. If we consider (5.2.44), then we obtain the inequality

(5.2.58)

5.2 The Resolvent and the Yosida Approximation



t

|xλ (t) − xλ (s)| ≤ s

|xλ (τ )|dτ ≤ M(t − s).

69

(5.2.59)

Then (5.2.58) leads to |x(t) − x(s)| ≤ M|t − s| + |xλ (t) − x(t)| + |xλ (s) − x(s)|.

(5.2.60)

Then we consider that limλ→0 xλ = x and the previous inequality becomes |x(t) − x(s)| ≤ M|t − s|.

(5.2.61)

Hence, we deduce that the function x(t) is a Lipschitz function and every Lipschitz function is absolutely continuous. But every absolutely continuous function is differentiable almost everywhere. So there exists x  (t) almost everywhere. By passing to the limit in |xλ (t)| ≤ M, we obtain |x  (t)| ≤ M, so x  ∈ L ∞ (0, T ; H ), that is, the function x(t) is essentially bounded. We will prove now that the function x(t) is a solution for the equation from the Cauchy problem, which is (5.2.33)1 . Let x 0 be arbitrarily fixed, x 0 ∈ D(A) and y 0 ∈ Ax 0 . We multiply the equation from the Cauchy problem by (xλ (t) − x 0 ) and we obtain 2      1 d xλ (t) − x 0 + Aλ xλ (t), xλ (t) − x 0 = f (t), xλ (t) − x 0 . 2 dt

(5.2.62)

We integrate this equality on the interval [s, t] ⊂ [0, T ] and we deduce 2 2  1 xλ (t) − x 0 − xλ (s) − x 0 + 2  t   + Aλ xλ (τ ), xλ (τ ) − x 0 dτ = s  t   = f (τ ), xλ (τ ) − x 0 dτ .

(5.2.63)

s

But we have 2 2    1 xλ (t) − x 0 − xλ (s) − x 0 ≥ xλ (t) − xλ (s), xλ (t) − x 0 . 2

(5.2.64)

By this inequality, the relation (5.2.63) becomes 

 xλ (t) − xλ (s), xλ (t) − x 0 ≤



t



 f (τ ) − Aλ xλ (τ ), xλ (τ ) − x 0 dτ .

s

We pass to the limit with λ → 0 and we obtain

(5.2.65)

70

5 Evolution Equations



 x(t) − x(s), x(t) − x 0 ≤



t



 f (τ ) − Ax(τ ), x(τ ) − x 0 dτ .

(5.2.66)

s

We assume that s is a point in which the function x is differentiable. We divide (5.2.66) by t − s and then we pass to the limit with t → s   x(t) − x(s) 0 lim , x(t) − x ≤ t→s t −s  t   1 ≤ lim f (τ ) − Ax(τ ), x(τ ) − x 0 dτ . t→s t − s s

(5.2.67)

On the right-hand side we use the mean-value theorem [9] and we deduce 

   x  (t), x(t) − x 0 ≤ f (t) − Ax, x(t) − x 0 ,

(5.2.68)

which can be rewritten as 

 f (t) − Ax − x  (t), x(t) − x 0 ≤ 0.

(5.2.69)

In this inequality, the second term of the scalar product has changing sign, so we deduce (5.2.70) x  (t) + Ax = f (t). Hence, the function x is the solution of the Eq. (5.2.33)1 . In all the computations above we assumed that the operator A is single-valued. In the case that A is a multi-valued operator, the computations are similar, but more laborious. We only have to prove that the Cauchy problem (5.2.33) has a unique solution. The proof is by contradiction and we assume that the problem has two solutions, which correspond to the same initial condition x0 ∈ D(A) and the same free term, that is x  + Ax(t)  f (5.2.71) x(0) = x0 and

y  + Ay(t)  f y(0) = x0 ,

(5.2.72)

respectively. We substract side by side these equations, then we scalarly multiply by x(t) − y(t) 

 x  − y  , x(t) − y(t) + (Ax(t) − Ay(t), x(t) − y(t)) = 0.

(5.2.73)

If we consider that the operator A is maximal monotone (in fact, it suffices that is monotone), then the previous inequality leads to

5.2 The Resolvent and the Yosida Approximation

71

 1 d  |x(t) − y(t)|2 ≤ 0. 2 dt

(5.2.74)

Then we integrate on the interval [0, t] and we obtain |x(t) − y(t)|2 − |x(0) − y(0)|2 ≤ 0, ∀t ∈ [0, T ] ⇒ ⇒ |x(t) − y(t)|2 ≤ 0, ∀t ∈ [0, T ] ⇒ x(t) = y(t), ∀t ∈ [0, T ],

(5.2.75)

which proves the uniqueness of the solution and the proof of the theorem is finished.  Now we will prove a stability result for the solution of the Cauchy problem (5.2.33), based on the inequality of Gronwall from Theorem 5.2.2. Theorem 5.2.4 If x(t) and y(t) are two solutions of the Cauchy problem (5.2.33) which correspond to (x0 , f ) ∈ D(A) × L 1 (0, T ; H ) and (y0 , g) ∈ D(A) × L 1 (0, T ; H ), respectively, then these functions satisfy the inequality  |x(t) − y(t)| ≤ |x0 − y0 | +

t

| f (τ ) − g(τ )|dτ ,

(5.2.76)

0

for almost every t ∈ [0, T ]. Proof The functions x(t) and y(t) verify the Eq. (5.2.33)1 , so we write x  + Ax(t)  f (t) for almost every t ∈ [0, T ], y  + Ay(t)  g(t) for almost every t ∈ [0, T ].

(5.2.77)

We substract these relations side by side and then we scalarly multiply by x(t) − y(t) 

 x  − y  , x(t) − y(t) + (Ax(t) − Ay(t), x(t) − y(t)) =

= ( f (t) − g(t), x(t) − y(t)) for almost every t ∈ [0, T ].

(5.2.78)

But A is a monotone operator and then we have (Ax(t) − Ay(t), x(t) − y(t)) ≥ 0,

(5.2.79)

so we can deduce from (5.2.78)  1 d  |x(t) − y(t)|2 ≤ ( f (t) − g(t), x(t) − y(t)) 2 dt

(5.2.80)

for almost every t ∈ [0, T ]. We integrate this inequality on the interval [0, t] and on the right-hand side we use the inequality of Schwarz, so we obtain the inequality

72

5 Evolution Equations

1 1 |x(t) − y(t)|2 ≤ |x(t) − y(t)|2 + 2 2



t

| f (τ ) − g(τ )||x(τ ) − y(τ )|dτ .

0

(5.2.81) We have the conditions of the inequality of Gronwall (Theorem 5.2.2) and then we obtain the conclusion of the theorem  t | f (τ ) − g(τ )|dτ (5.2.82) |x(t) − y(t)| ≤ |x0 − y0 | + 0

for almost every t ∈ [0, T ], which finishes the proof.



Observation 5.2.1 In fact, the result of the Theorem 5.2.4 is one of continuous dependence of the solution on the initial data and the right-hand side of the equation. Indeed, if the initial data are sufficiently close |x0 − y0 | < 2ε with ε > 0 arbitrarily small and if the terms on the right-hand side are sufficiently close | f (t) − g(t)| ≤ 2Tε , then the two solutions of the Cauchy problem are sufficiently close  t | f (τ ) − g(τ )|dτ ≤ |x(t) − y(t)| ≤ |x0 − y0 | + 0  t ε ε ε ε T ⇒ dτ ≤ + ≤ + 2 2T 0 2 2T ⇒ |x(t) − y(t)| ≤ ε.

(5.2.83)

The stability result is similar when T → ∞.

References 1. L. Ambrosio, N. Gigli, G. Savare, Gradient Flows in Metric Spaces and in the Space of Probability Measures (Birkhäuser, Basel, 2005) 2. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, Berlin, 2011) 3. M.G. Crandall, Semigroups of nonlinear contractions and dissipative sets. J. Funct. Anal. 3, 376–418 (1969) 4. T. Kato, On the eigenfunction of many-particle systems in quantum mechanics. Commun. Pure Appl. Math. 10, 151–177 (1957) 5. Y. Komura, On linear topological spaces. Kumamoto J. Sci. A 5, 148–157 (1962) 6. G. Minty, Monotone operators in Hilbert space. Duke Math. J. 29, 341–346 (1962) 7. T. Precupanu, Spa¸tii liniare topologice s¸i elemente de analiz˘a convex˘a, Ed. Acad. Române, Bucure¸sti (1992) 8. W. Rudin, Functional Analysis (Tata-McGraw Hill Publication, New York, 1974) 9. W. Rudin, Real and Complex Analysis (McGraw-Hill, New York, 1974)

Chapter 6

Distributions

Abstract First of all, we introduce some fundamental spaces that are needed in the theory of distributions. Then we introduce the space of distributions. Then we discuss about the derivative of a distribution, the primitive of a distribution and higher-order primitives. Finally, we present the direct product of distributions, the convolution of distributions and their properties.

6.1 Fundamental Spaces in the Theory of Distributions Let  be an open set on Rn , which is not necessarily bounded. Then C ∞ () = {ϕ = ϕ(x)|D k ϕ(x) ∈ C 0 (), ∀k, k = (k1 , k2 , . . . , kn )}, where D k ϕ(x) =

∂ |k| ϕ(x)

, |k| = kn

∂x1k1 ∂x2k2 · · · ∂xn

n 

ki .

(6.1.1)

(6.1.2)

i=1

The support of the function ϕ is the set supp ϕ(·) = {x|x ∈ , ϕ(x) = 0}.

(6.1.3)

A subset of the set C ∞ () is the set Cc∞ () defined by Cc∞ () = {ϕ(·) ∈ C ∞ ()|supp ϕ(·) = compact set in }.

(6.1.4)

If  is a bounded set, then supp ϕ is a bounded set since supp ϕ is a subset of . But supp ϕ is a closed set, hence supp ϕ is a compact set (a bounded and closed set is a compact set). In this case the sets C ∞ () and Cc∞ () coincide. Similarly, we introduce the following function spaces   C m () = ϕ(x)|D k ϕ(x) ∈ C 0 (), |k| ≤ m , © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_6

(6.1.5) 73

74

6 Distributions

  Ccm () = ϕ(x)|ϕ ∈ C m (), supp ϕ = compact set in  .

(6.1.6)

In particular, m can be zero and then we obtain the sets C 0 () and Cc0 (). These notations are usual in classical analysis or in functional analysis. In the theory of distributions we use the notations D() = D∞ () = Cc∞ (), Dm () = Ccm (), D0 () = Cc0 ().

(6.1.7)

Clearly, these spaces are vector spaces with respect to the usual function addition and the scalar multiplication of functions. We also use the linearity of differentiation. We want to show that these vector spaces are topological spaces. In order to introduce the topology on these spaces we use an algorithm which is similar to that for locally convex spaces. We define a sequence of compact sets from , which is ordered and increasing, such that  Q 1 ⊆ Q 2 ⊆ · · · ⊆ Q n ⊆ · · · ⊆ ; (6.1.8) n Qi .  = ∪i=1 On every compact set Q i we define the seminorm |ϕ| Q i, j =

sup

x∈Q i ,|k|≤ j

  k  D ϕ(x) ,

(6.1.9)

where i and j are fixed, i, j = 1, 2, . . . , ∀ϕ ∈ Dm (). We can easily verify that the seminorm from (6.1.9) is in fact a norm, but in the sequel it is enough to assume that it is a seminorm (as in the case of locally convex spaces). With the help of the seminorm we can introduce the following metric for i fixed m  1 |ϕ − ψ| Q i, j . di (ϕ, ψ) = j 1 + |ϕ − ψ| 2 Q i, j j=1

(6.1.10)

Clearly, j is the order of differentiation. For a fixed compact set, by means of (6.1.10), we define the topology of the metric space. The neighbourhoods are defined by means of open balls and the balls are defined by means of the metric (6.1.10). We can compare the seminorms defined on different compact sets by |ϕ| Q i+1, j ≥ |ϕ| Q i, j ,

Q i+1 ⊇ Q i .

(6.1.11)

Clearly, for any two compact sets we can establish an inclusion relation. Therefore, we can consider that the restriction of the topology defined on Q i+1 to Q i coincides with the topology defined on Q i . On the other hand, Dm () can be represented as a union of functions of class C m on the compact sets (6.1.8). Then we obtain the inductive limit topology defined with the seminorms (6.1.9) on .

6.1 Fundamental Spaces in the Theory of Distributions

75

By definition, a neighbourhood in Dm () is any subset of Dm () with the property that its intersection with each locally convex subspace corresponding to Q i , with fixed i, is a neighbourhood of that subspace. Observation 6.1.1 The topology defined above is the inductive limit topology because (i) Dm () can be represented as the union of the subspaces corresponding to each Q i , with fixed i. (ii) The restriction of the topology defined by the seminorms corresponding to Q i+1 to the space corresponding to Q i , coincides with the topology of Q i . In the particular case when m = 0, we have D0 () and the seminorms correspond only to the compact sets Q i , hence there is no ordering by the order of differentiation. Another particular case is m = ∞. In this case, the family of seminorms is similar to that from (6.1.9) and the metric is similar as well (the sum becomes a series) di (ϕ, ψ) =

∞  1 |ϕ − ψ| Q i, j . 2 j 1 + |ϕ − ψ| Q i, j j=1

(6.1.12)

Hence, the topology on D() is also the inductive limit topology. Let us define the convergence in such spaces. Definition 6.1.1 The sequence (ϕν (x))ν∈N ∈ D() converges to ϕ(x) ∈ D() if and only if by definition (i) there is a compact set for all indices ν such that supp ϕν (x) ⊂ Q ⊂ . uniformly (ii) D k ϕν (x) −→ D k ϕ(x), ∀k = (k1 , k2 , . . . , kn ). Similarly, we can define the convergence on the space Dm (), where m is a fixed natural number. The space D() with the topology above is called the fundamental space of distributions. The space Dm () with the topology above is called the space of distributions of order m. The space D0 () with the topology above is called the fundamental space of distributions of order 0 or measures. These spaces are called spaces of test functions of order ∞, of order m and of order 0, respectively.

6.1.1 On Some Properties of the Spaces C m () and C ∞ () Theorem 6.1.1 The locally convex space C m () is complete.

76

6 Distributions

Proof Let (ϕν ) be a Cauchy sequence in C m (). Note that, in particular, (ϕν ) is a Cauchy sequence in C(). We know that C() is a complete space. It follows that ϕν → φ, where φ ∈ C(). Moreover, ∀1 ≤ k ≤ n, the sequence (∂k ϕν ) is also a Cauchy sequence in C(). It follows that ∂k ϕν → φk in C(), 1 ≤ k ≤ n. We can conclude that φk = ∂k φ, 1 ≤ k ≤ n. Using mathematical induction, we deduce that ∀|α| ≤ m, we have D α ϕν → φα ,  where φα ∈ C() and φα = D α φ. Proposition 6.1.1 The space C m () is a Frechet space [2]. Below we give a characterization of bounded sets in C ∞ (). Theorem 6.1.2 Let A be a subset of C ∞ (). Then A is bounded if and only if, given m ∈ N and Q ⊂  compact, there is C > 0 a constant such that   k  D ϕ(x) ≤ C,

(6.1.13)

∀|k| ≤ m, ∀x ∈ Q, ∀ϕ ∈ A. Proof We assume that A satisfies the condition from the statement of the theorem. We want to prove that A is bounded. To this end, we have to show that, given V a neighbourhood of zero in C ∞ (), we can find λ > 0 such that λA ⊂ V . Clearly, we can suppose that the neighbourhood V is given by V = {ϕ ∈ C ∞ ()||ϕ| Q i, j ≤ ε}.

(6.1.14)

From the fact that A satisfies the condition from the statement of the theorem, we deduce that there is a constant C > 0 such that   k  D ϕ(x) ≤ C,

(6.1.15)

∀|k| ≤ m, ∀x ∈ Q i , ∀ϕ ∈ A. Let λ ∈ R, λ > 0, with λC ≤ ε. It follows from the last inequality that   λ  D k ϕ(x) ≤ ε, ∀|k| ≤ m, ∀x ∈ Q i , ∀ϕ ∈ A. Hence, λA ⊂ V , so A is bounded in C ∞ .

(6.1.16) 

Note that a subset of a topological space is relatively compact if its closure is compact. In the sequel, we will present a charaterization of relatively compact subsets of C m () based on Ascoli’s theorem. Then it will follow that in C ∞ () relatively compact and bounded subsets do coincide. Definition 6.1.2 A subset A of C() is called equicontinuous at a point x0 ∈  if, given ε > 0, there is a neighbourhood V of x0 such that

6.1 Fundamental Spaces in the Theory of Distributions

|ϕ(x) − ϕ(x0 )| ≤ ε,

77

(6.1.17)

∀x ∈ V, ∀ϕ ∈ A. Then A is equicontinuous in  if it is equicontinuous at every point of . Below we present the theorem of Ascoli [1]. Theorem 6.1.3 Let A be a subset of C(). Then A is relatively compact if and only if (i) A is equicontinuous; (ii) ∀x ∈ , the subset {ϕ(x)|ϕ ∈ A} is relatively compact in C. Lemma 6.1.1 Let B ⊂ C 1 () be bounded. Then B is an equicontinuous subset of C(). Proof By applying the mean value theorem, it follows that |ϕ(x + h) − ϕ(x)| ≤ sup ϕ (x + θh) · |h|,

(6.1.18)

0≤θ≤1

where h = (h 1 , . . . , h n ). Note that ϕ is the differential of ϕ and ϕ denotes its norm as a linear map from Rn into C. We assume that B is a bounded subset in C 1 (). Then there is M > 0 a constant such that (6.1.19) sup ϕ (x + θh) ≤ M, 0≤θ≤1

∀ϕ ∈ B. Therefore, we obtain |ϕ(x + h) − ϕ(x)| ≤ M · |h|, ∀ϕ ∈ B. It follows that B is an equicontinuous set.

(6.1.20) 

Theorem 6.1.4 Let A be a subset of C m (). Then A is relatively compact if and only if (i) A is bounded in C m (); (ii) ∀ p = ( p1 , . . . , pn ) with | p| = m, the image of A by the differential operator D p , i.e. D p A, is equicontinuous. Proof Step 1. We assume that A is a relatively compact subset of C m (). It follows that A is a bounded subset of C m (). By using Ascoli’s theorem [1], we deduce that D p A is equicontinuous ∀| p| = m. Step 2. We assume that conditions (i) and (ii) hold true. We consider p = ( p1 , . . . , pn ) with | p| = m and the set D p A. By assumption, D p A is bounded and equicontinuous. By using Ascoli’s theorem, we deduce that D p A is a relatively compact set.

78

6 Distributions

Then we consider q = (q1 , . . . , qn ) with |q| = m − 1 and B = D q A. Clearly, B is a bounded subset of C 1 (). It follows that B is an equicontinuous set. By using Ascoli’s theorem, we deduce that B = D q A is a relatively compact set. By using mathematical induction, we deduce that D p A is a relatively compact  subset of C(), ∀| p| ≤ m. Then A is a relatively compact subset of C m (). Corollary 6.1.1 Let A be a bounded subset of C m+1 (). Then A is relatively compact in C m (). Theorem 6.1.5 Let A be a subset of C ∞ (). Then A is relatively compact if and only if A is bounded in C ∞ (). Definition 6.1.3 A Hausdorff locally convex space E is called a Montel space if every bounded subset of E is relatively compact. [2] The space C ∞ () endowed with its natural topology is a Montel space. Every Montel space is a reflexive space. It follows that C ∞ () is a reflexive space. Proposition 6.1.2 The inductive limit of Montel spaces is a Montel space. Proof We consider a sequence E i of Montel spaces. Let E be its inductive limit. We consider that A is a bounded set in E. Then there is an index i 0 such that A ⊂ E i0 and A is bounded in E i0 . We know that E i0 is a Montel space. Hence, A is relatively compact in E i0 . Since the identity map from E i0 into E is continuous, A is relatively compact in E. We can conclude that E is a Montel space.  Let Q ⊂  be compact. Let Cc∞ (; Q) be the subspace of Cc∞ () which includes all functions with support contained in Q. On this space we consider the topology induced by C ∞ () that coincides with the locally convex topology defined by the sequence of norms |ϕ| j =

sup

|D k ϕ(x)|, j ∈ N.

(6.1.21)

x∈Q,|k|≤ j

Theorem 6.1.6 The space Cc∞ (; Q) is a Frechet space. Proof Note that Cc∞ (; Q) is a metrizable space since the topology of Cc∞ (; Q) is defined by a sequence of norms. Clearly, Cc∞ (; Q) is a closed subspace of C ∞ ().  Therefore, Cc∞ (; Q) is complete. Theorem 6.1.7 Let Q and L be two compact subsets of  such that Q ⊂ L ⊂ . Then (i) the identity map Cc∞ (; Q) → Cc∞ (; L) is a continuous one; (ii) Cc∞ (; Q) is a closed subspace of Cc∞ (; L). Note that

Cc∞ () = ∪i Cc∞ (; Q i ),

(6.1.22)

where (Q i ) is an increasing sequence of compact sets such that Q i ⊂  and the union of Q i is .

6.1 Fundamental Spaces in the Theory of Distributions

79

Theorem 6.1.8 The identity map Cc∞ () → C ∞ () is continuous. It follows that every bounded subset of Cc∞ () is a bounded subset of C ∞ (). Theorem 6.1.9 The space Cc∞ () is a Montel space. Corollary 6.1.2 The space Cc∞ () is a reflexive space. Note also that Cc∞ () is a complete space. Proposition 6.1.3 The space Cc∞ () is sequentially complete. Proof Let (ϕν ) be a Cauchy sequence in Cc∞ (). Then it is bounded. Moreover, it is a Cauchy sequence in some space Cc∞ (; Q). Since Cc∞ (; Q) is complete, it  follows that (ϕν ) converges in this space. Hence, (ϕν ) converges in Cc∞ ().

6.2 The Space of Distributions Definition 6.2.1 Let  be an open subset of Rn . A distribution on  is a continuous linear functional on Cc∞ (). The vector space of all distributions on  will be denoted by D () [4–7, 11–13]. It is then the topological dual of Cc∞ (). D () = { f : D() → R, f = linear and continuous functional}

(6.2.1)

By the continuity of the functional we mean the continuity of a function over a space that is equipped with the topology of a linear topological space. We use the convention that a distribution is a distribution of order ∞. A distribution of order m is a linear and continuous functional defined on Dm (). The set of all distributions of order m is the dual of Dm (), hence    D m () = f : Dm () → R, f = linear and continuous functional .

(6.2.2)

The distributions of order 0 or measures are linear and continuous functionals defined on D0 ():    D 0 () = f : D0 () → R, f = linear and continuous functional .

(6.2.3)

We use the notation ( f, ϕ) to denote the value of f in ϕ, where ϕ is arbitrarily fixed in D(). We use the notation f (ϕ) = ( f, ϕ) as well. D

We have the equality of two distributions from D (), i.e. f = g if ( f, ϕ) = (g, ϕ), ∀ϕ ∈ D(). 



Dm

For f, g ∈ D m (), we have f = g if ( f, ϕ) = (g, ϕ), ∀ϕ ∈ Dm ().

80

6 Distributions 

D0



For f, g ∈ D 0 (), we have f = g if ( f, ϕ) = (g, ϕ), ∀ϕ ∈ D0 (). An example of a test function is  ϕ(x, ε) =

where r = |O x| =



n i=1

2

− ε2ε−r 2

e 0

if r < ε if r ≥ ε,

(6.2.4)

xi2 . Note that Dxk ϕ(x, ε) ∈ C 0 (Rn ), ∀k = (k1 , k2 , . . ., kn ),

hence ϕ(x, ε)∈C ∞ (Rn). Moreover, supp ϕ(x, ε) = B(0, ε), that is supp ϕ(x, ε) is a compact set and then ϕ(x, ε) ∈ D(Rn ) = D∞ (Rn ). We consider the sequence ϕν (x) = ν1 ϕ(x, ε), ν = 1, 2, . . .. Note that ϕν (x) → 0 in the sense of the topology of D(Rn ). The compact set that is suitable for all ν exists since ν does not appear in the expression of ϕ(x, ε). Moreover, the derivatives with respect to x do not depend on ν.

If we choose the sequence ψν (x) = ν1 ϕ νx , ε , then the derivatives converge uniformly, but on different compact sets, so there does not exist a compact set that is suitable for all ν. Hence, the sequence ψν (x) is not convergent in the sense of the topology of D(). Proposition 6.2.1 Let μ be a Radon measure on  [2]. Then it defines a distribution on . Proof It suffices to show that if μ ∈ M() is such that μ(ϕ) = 0, ∀ϕ ∈ Cc∞ (),

(6.2.5)

then μ is the Radon measure identically zero, that is, μ(ψ) = 0, ∀ψ ∈ Cc (). Let ψ be an element of Cc (). It follows that ψε = αε ∗ ψ belongs to Cc∞ (). Moreover, we have ψε → ψ in Cc () as ε → 0. It follows by continuity that μ(ψ) = 0.  The space of distributions D () is a linear space with respect to the following usual operations of addition of distributions and scalar multiplication of a distribution: def ( f + g, ϕ) = ( f, ϕ) + (g, ϕ), ∀ f, g ∈ D () and ∀ϕ ∈ D(), def (λ f, ϕ) = ( f, λϕ) = λ( f, ϕ), ∀ f ∈ D (), ∀λ ∈ R and ∀ϕ ∈ D().

(6.2.6)

Similarly, we can equip with a structure of linear space, the space of distributions  of order m D m () and the space of distributions of order 0 (the space of measures)  D 0 (). We want to define an algebra structure on the space of distributions. In a classical algebra we define the usual product of the elements of the linear space. In the case of the linear space of distributions of order ∞ we should define the product of two distributions.

6.2 The Space of Distributions

81

Schwartz showed that the product of two distributions is not necessarily a distribution [14]. Hence, the product of two distributions is not an internal operation, so it can not induce an algebra structure. Instead of the usual product of distributions we introduce another product, namely the product of a distribution with a function of class C ∞ (). Definition 6.2.2 If f is a distribution, f ∈ D () and a(x) is a function of class C ∞ (), then we define the product of f with a as the new distribution a f given by def (a f, ϕ) = ( f, aϕ), ∀ϕ ∈ D(), ∀ f ∈ D () and ∀a ∈ C ∞ ().

(6.2.7)

Proposition 6.2.2 The product a f defined by the relation (6.2.7) is a distribution. Proof Step 1. First we show that the product a f is well defined. Since a ∈ C ∞ (), we deduce that aϕ ∈ C ∞ () for a test function ϕ ∈ C ∞ (). Moreover, we have supp(aϕ) ⊂ suppϕ. Hence, supp(aϕ) is a compact set because ϕ has a compact support. Therefore, aϕ ∈ Cc∞ () = D(), that is, the function aϕ is a test function and f is a distribution, so it is defined for any test function. Step 2. Now we prove the linearity of the functional from (6.2.7). Let ∀λ1 , λ2 ∈ R and ϕ1 , ϕ2 ∈ D(). By the definition from (6.2.7), we have (a f, λ1 ϕ1 + λ2 ϕ2 ) = ( f, a(λ1 ϕ1 + λ2 ϕ2 )) = = ( f, aλ1 ϕ1 + aλ2 ϕ2 ) = ( f, aλ1 ϕ1 ) + ( f, aλ2 ϕ2 ) =

(6.2.8)

= λ1 ( f, aϕ1 ) + λ2 ( f, aϕ2 ) = λ1 (a f, ϕ1 ) + λ2 (a f, ϕ2 ). We used the linearity of the functional f and the definition from (6.2.7). Step 3. Now we prove the continuity of the functional (6.2.7). Let {ϕν }ν∈N be an arbitrary sequence of test functions, ϕν ∈ D(), ∀ν = 1, 2, . . . such that ϕν → ϕ, ϕ ∈ D().

(6.2.9)

We will prove that the functional a f can be interchanged with the limit sign, so we deduce it is continuous. We have  lim (a f, ϕν ) = lim ( f, aϕν ) = f, lim (aϕν ) = ν→∞ ν→∞ ν→∞  (6.2.10) = ( f, aϕ) = (a f, ϕ) = a f, lim ϕν . ν→∞

We used the fact that f is a distribution, so it is a continuous functional and it can be interchanged with the limit sign.  In the sequel, we define the convergence in D (). So we consider a sequence of distributions ( f ν )ν∈N , f ν ∈ D ().

82

6 Distributions

Definition 6.2.3 We say that the sequence f ν converges, in the sense of distributions, to the distribution f ∈ D () if for any test function ϕ ∈ D() we have R

( f ν , ϕ) → ( f, ϕ), ∀ϕ ∈ D(). So, we have

D  ()

R

f ν −→ f ⇔ ( f ν , ϕ) −→ ( f, ϕ), ∀ϕ ∈ D().

(6.2.11)

(6.2.12)

Moreover, for a sequence of distributions ( f ν )ν∈N we define the series of distributions and use the classical notation ∞ 

f ν , f ν ∈ D (), ∀ν ∈ N.

(6.2.13)

ν=1

We say that the series (6.2.13) is convergent if the sequence of partial sums is convergent in the sense of distributions, i.e. ∞  ν=1

f ν is convergent ⇔

∞ 

 R

f ν , ϕ −→ ( f, ϕ), ∀ϕ ∈ D().

(6.2.14)

ν=1

We make the convention that for the convergent series (6.2.13), the distribution f is the sum of the series, i.e. ∞  fν . (6.2.15) f = ν=1

Let us define the translation of a distribution. If f is an arbitrary distribution, then its translation is defined by ( f (x + x0 ), ϕ(x)) = ( f (x), ϕ(x − x0 )) , ∀ϕ ∈ D().

(6.2.16)

In the one-dimensional case, for  ⊂ R, we define D() and D () and the translation ( f (t + t0 ), ϕ(t)) = ( f (t), ϕ(t − t0 )) , ∀ϕ ∈ D(),  ⊂ R.

(6.2.17)

Proposition 6.2.3 The translation of a distribution is a distribution. Proof We will prove that the translation of a distribution defined as in relation (6.2.17) is well defined, linear and continuous. Step 1. Let ϕ be a test function. Then ϕ(t − t0 ) is also a test function since it is the composition of the function ϕ with the function ψ(t) = t − t0 . Clearly, the function ψ is linear and ψ ∈ Cc∞ (). Since f is a distribution, it makes sense for any test function, including the test function ϕ(t − t0 ), so the right-hand side from (6.2.17) makes sense.

6.2 The Space of Distributions

83

Step 2. We prove the linearity of the functional from (6.2.17). To this end, we consider two arbitrary test functions ϕ1 and ϕ2 and two arbitrary scalars λ1 and λ2 . We have to show that ( f (t + t0 ), λ1 ϕ1 (t) + λ2 ϕ2 (t)) = λ1 ( f (t + t0 ), ϕ1 ) + λ2 ( f (t + t0 ), ϕ2 ) . (6.2.18) Clearly, by the definition from (6.2.17) and the linearity of the functional f (which is a distribution), we have ( f (t + t0 ), λ1 ϕ1 (t) + λ2 ϕ2 (t)) = ( f (t), (λ1 ϕ1 + λ2 ϕ2 )(t − t0 )) = = ( f (t), λ1 ϕ1 (t − t0 ) + λ2 ϕ2 (t − t0 )) = = ( f (t), λ1 ϕ1 (t − t0 )) + ( f (t), λ2 ϕ2 (t − t0 )) =

(6.2.19)

= λ1 ( f (t), ϕ1 (t − t0 )) + λ2 ( f (t), ϕ2 (t − t0 )) = = λ1 ( f (t + t0 ), ϕ1 ) + λ2 ( f (t + t0 ), ϕ2 ) . Step 3. We will prove the continuity of the functional defined in (6.2.17). Let {ϕν }ν∈N be an arbitrary sequence of test functions, ϕν ∈ D(), ∀ν = 1, 2, . . . such that (6.2.20) ϕν → ϕ, ϕ ∈ D(). We will show that the translation can be interchanged with the limit sign and so we deduce that it is continuous. We have lim ( f (t + t0 ), ϕν (t)) = lim ( f (t), ϕν (t − t0 )) = ν→∞  = f (t), lim ϕν (t − t0 ) = ( f (t), ϕ(t − t0 )) = ν→∞  = ( f (t + t0 ), ϕ(t)) = f (t + t0 ), lim ϕν (t) .

ν→∞

(6.2.21)

ν→∞

We used the fact that f is a distribution, so it is a continuous functional and it can be interchanged with the limit sign.  Theorem 6.2.1 A linear functional f is a distribution on  if and only if for every compact set Q ⊂  there is C > 0 a constant and m ∈ N such that |( f, ϕ)| ≤ C

sup x∈,|k|≤m

  k  D ϕ(x) , ∀ϕ ∈ C ∞ (; Q). c

(6.2.22)

Proof Step 1. Let f ∈ D (). Note that for every compact subset Q ⊂ , f is a continuous linear functional on Cc∞ (; Q). Then there is a neighbourhood of zero V = V (Q, m, ε) = {ϕ ∈ Cc∞ (; Q)||ϕ| Q,m ≤ ε}, with

(6.2.23)

84

6 Distributions

|ϕ| Q,m =

|D k ϕ(x)|,

sup

(6.2.24)

x∈Q,|k|≤m

such that |( f, ϕ)| ≤ 1, ∀ϕ ∈ V.

(6.2.25)

Moreover, if ϕ ∈ Cc∞ (; Q) is such that ϕ = 0, we deduce that εϕ ∈ V. |ϕ| Q,m Hence, we have | ( f, ϕ) |
0 a constant, m ∈ N and a compact subset Q ⊂  such that |( f, ϕ)| ≤ C

sup

  k  D ϕ(x) , ∀ϕ ∈ C ∞ ().

(6.3.1)

x∈Q,|k|≤m

Proof Let f ∈ E  (). Then there is a neighbourhood of zero in C ∞ () V = {ϕ ∈ C ∞ ()||ϕ| Q,m ≤ ε}

(6.3.2)

|( f, ϕ)| ≤ 1, ∀ϕ ∈ V.

(6.3.3)

such that We consider ϕ ∈ C ∞ () such that |ϕ| Q,m = 0. Then we have εϕ ∈V |ϕ| Q,m

(6.3.4)

|( f, ϕ)| ≤ C|ϕ| Q,m ,

(6.3.5)

and it follows that for C = 1ε . On the other hand, if ϕ ∈ C ∞ () is such that |ϕ| Q,m = 0, then we have ( f, ϕ) = 0. Clearly, such a ϕ belongs to V . The same holds true for any constant multiple λϕ of ϕ. If ( f, ϕ) were different from zero, then |( f, λϕ)| would become as large as we please. But this would contradict (6.3.3). Therefore, we can conclude that (6.3.1)  holds true for all ϕ ∈ C ∞ (). Theorem 6.3.2 A sequence ( f ν ) converges strongly to zero in E  () if and only if the numerical sequence (( f ν , ϕ)) converges to zero uniformly on bounded sets of C ∞ (). Definition 6.3.1 Let U be an open subset of . A distribution f ∈ D () is zero on U if (6.3.6) ( f, ϕ) = 0, ∀ϕ ∈ Cc∞ (U ). Definition 6.3.2 Two distributions f, g ∈ D () coincide on an open subset U of  if f − g is equal to zero on U . We will write f = g on U .

86

6 Distributions

Lemma 6.3.1 Let (Vi )i∈I be a family of open subsets of . We assume that f ∈ D () is zero on every subset Vi . Then f is zero on the union V = ∪i∈I Vi . Proof Let ϕ ∈ Cc∞ (V ). We denote by Q the support of ϕ. Since (Vi )i∈I is an open covering of the compact subset Q, we can select a finite subcovering Vi1 , . . . , Vir of Q. We consider that (ψ j )1≤ j≤r is a C ∞ partition of unity subordinated to this covering. Then we have ψ j ∈ Cc∞ (Vi j ), 0 ≤ ψ j ≤ 1 and

r 

ψ j = 1.

(6.3.7)

j=1

This implies that ϕ=

r 

ϕ · ψj.

(6.3.8)

j=1

Note that ϕ · ψ j ∈ Cc∞ (Vi j ) and f is zero on Vi j . Then f (ϕ) =

r 

f (ϕ · ψ j ) = 0.

(6.3.9)

j=1

Since ϕ is an arbitrary element of Cc∞ (V ), it follows that f is zero on V .



Definition 6.3.3 Let f ∈ D (). The support of f is the complement in  of the largest open subset of  where f is zero. We denote it by supp f . Note that a point belongs to the support of f if and only if there is no open neighbourhood of it on which f is zero. Moreover, the support of f is the smallest closed subset of  outside of which the distribution f is zero. Proposition 6.3.1 If ϕ ∈ Cc∞ () and f ∈ D () are such that supp ϕ ∩ supp f = ∅, then ( f, ϕ) = 0. Proposition 6.3.2 Let f ∈ D (). The support of f is the smallest closed subset F of U such that if ϕ, ψ ∈ Cc∞ () and ϕ = ψ on a neighbourhood of F, then ( f, ϕ) = ( f, ψ). Below we present the relation between the two spaces of distributions E  () and D (). Theorem 6.3.3 Let  ⊂ Rn be open. We have (i) E  () ⊂ D () and the identity map is continuous with respect to the strong topologies; (ii) the elements of E  () are distributions with compact support in .

6.3 The Dual of C ∞

87

Proof Step 1. (i) We will show that Cc∞ () is a dense subspace of C ∞ (). We consider an increasing sequence (Q i ) of compact subsets contained in  and such that their union is . Moreover, we consider a sequence of functions βi ∈ Cc∞ () such that βi = 1 on a neighbourhood of Q i . For ϕ ∈ C ∞ (), we consider ϕi = βi ϕ ∈ Cc∞ (). Clearly, we have ϕi → ϕ in Cc∞ (). Step 2. We consider f ∈ E  (). Note that the identity map Cc∞ () → C ∞ () is continuous. It follows that f defines a distribution on . We want to show that E  () can be identified with a subspace of D (). It suffices to prove that if f ∈ E  () is such that ( f, ϕ) = 0 for all ϕ ∈ Cc∞ (), then f is the distribution identically zero. Note that this follows from the density of Cc∞ () in C ∞ (). We want to show that the identity map from E  () endowed with the strong topology into D () endowed with its strong topology is continuous. To this end, it suffices to note that the embedding Cc∞ () → C ∞ () is continuous. It follows that every bounded set of Cc∞ () is a bounded set of C ∞ (). Step 3. (ii) We will prove that every f ∈ E  () has a compact support contained in . Note that if f ∈ E  (), then there is ε > 0, m ∈ N and a compact subset Q of  such that ∀ϕ ∈ C ∞ () with the property that |ϕ| Q,m ≤ ε, we have |( f, ϕ)| ≤ 1. Note also that ( f, ϕ) = 0, ∀ϕ ∈ C ∞ () such that |ϕ| Q,m = 0. Since every ψ ∈ Cc∞ ( − Q) satisfies this condition, we deduce that f must be zero on  \ Q. Hence, the support of f must be contained in Q.  Example 1. The Dirac measure has compact support equal to {0}. Example 2. The elements of C  () are Radon measures (hence distributions) with compact support in .

6.4 The Derivative of a Distribution Let  be an open set from Rn , which is not necessarily bounded. On this set we define the fundamental spaces D() and D (). For any distribution f ∈ D () we define the correspondence ∂f (6.4.1) f −→ ∂x j by



   ∂ϕ ∂f , ∀ϕ ∈ D(). , ϕ = f, − ∂x j ∂x j

(6.4.2)

Hence, we have ∂f ∂x j

D()  ϕ −→



∂ϕ f, − ∂x j

 , ∀ϕ ∈ D(),  ⊂ Rn .

(6.4.3)

88

6 Distributions

Note that in the one-dimensional case the derivative of a distribution is defined by 

   df dϕ def , ϕ = f, − , ∀ϕ ∈ D(),  ⊂ R. dt dt

(6.4.4)

Proposition 6.4.1 If f is a distribution, f ∈ D (), then the correspondence from (6.4.2) is a distribution as well, i.e. f ∈ D () ⇒

∂f ∈ D (). ∂x j

(6.4.5)

Proof We have to prove that the correspondence from (6.4.2) is well defined, linear and continuous. ∂ϕ Step 1. Let ϕ be a test function. Then it is infinitely differentiable and then ∂x j is infinitely differentiable, so this derivative is also a test function. Since f is a ∂ϕ . distribution, it makes sense for any test function, so also for ∂x j Step 2. For the linearity we have to show that ∀λ1 , λ2 ∈ R, ∀ϕ1 , ϕ2 ∈ D(),  ⊂ Rn ⇒       ∂f ∂f ∂f , λ 1 ϕ1 + λ 2 ϕ2 = λ 1 , ϕ1 + λ 2 , ϕ2 . ⇒ ∂x j ∂x j ∂x j

(6.4.6)

Clearly, we have ∀λ1 , λ2 ∈ R, ∀ϕ1 , ϕ2 ∈ D(),  ⊂ Rn ⇒     ∂ ∂f , λ1 ϕ1 + λ2 ϕ2 = f, − (λ1 ϕ1 + λ2 ϕ2 ) = ⇒ ∂x j ∂x j       ∂ϕ1 ∂ϕ2 ∂ϕ1 ∂ϕ2 = f, −λ1 + f, −λ2 = − λ2 = f, −λ1 ∂x j ∂x j ∂x j ∂x j         ∂ϕ1 ∂ϕ2 ∂f ∂f + λ2 f, − = λ1 = λ1 f, − , ϕ1 + λ 2 , ϕ2 . ∂x j ∂x j ∂x j ∂x j (6.4.7) Step 3. We prove the continuity of the functional defined in (6.4.2). To this end, we show that this correspondence can be interchanged with the limit sign. Let {ϕν }ν∈N be an arbitrary sequence of test functions, ϕν ∈ D(), ∀ν = 1, 2, . . . such that ϕν → ϕ, ϕ ∈ D(). 

Let us show that lim

ν→∞

Clearly, we have

∂f , ϕν ∂x j



 =

 ∂f , lim ϕν . ∂x j ν→∞

(6.4.8)

(6.4.9)

6.4 The Derivative of a Distribution

89

      ∂f ∂ϕν ∂ϕν = f, − lim = lim , ϕν = lim f, − ν→∞ ∂x j ν→∞ ν→∞ ∂x j ∂x j       ∂f ∂ϕ ∂f = = f, − ,ϕ = , lim ϕν . ∂x j ∂x j ∂x j ν→∞

(6.4.10)



This finishes the proof.

Observation 6.4.1 We deduce that any distribution is differentiable. Moreover, the derivative of a distribution is again a distribution, which is also differentiable. Hence, we deduce that any distribution is infinitely differentiable. Clearly, we have a result of Schwarz type for mixed derivatives. This result is based on the fact that the test functions are infinitely differentiable and satisfy the result of Schwarz. We can prove a more general result ∂ k1 +k2 f ∂x1k1 ∂x2k2

=

∂ k2 +k1 f ∂x2k2 ∂x1k1

.

(6.4.11)

This result is based on a similar property of the test functions. We have the following result for higher-order derivatives of distributions. Proposition 6.4.2 Given an arbitrary multi-index k = (k1 , k2 , . . . , kn ) and an arbitrary distribution f ∈ D (), the derivative of order k has the expression



D k f, ϕ = f, (−1)k1 +k2 +···+kn D k ϕ , ∀ϕ ∈ D(),  ⊂ Rn ,

(6.4.12)

where the multi-index derivative is defined by Dk f =

∂ |k| f ∂xk11 ∂xk22

· · · ∂xknn

, |k| = k1 + k2 + · · · + kn .

(6.4.13)

Proof We prove the result by mathematical induction



D f, ϕ =

∂ k1 +k2 +···+kn f



,ϕ = ∂x1k1 ∂x2k2 · · · ∂xnkn

 k1 k

∂ k2 +k3 +···+kn f ∂ ϕ = D f, ϕ = , (−1)k1 k1 = ∂x1 ∂x2k2 ∂x3k3 · · · ∂xnkn

 k3 +k4 +···+kn k1 +k2 ∂ f ∂ ϕ , (−1)k1 +k2 k1 k2 = · · · = = kn k3 k4 ∂x1 ∂x2 ∂x3 ∂x4 · · · ∂xn

k1 +k2 +···+kn k1 +k2 +···+kn D ϕ . = f, (−1) k

(6.4.14)

90

6 Distributions

6.5 Distributions as Generalized Functions The first distributions were the ones generated by functions and they are called generalized functions. 1 (), i.e. f (x) is an integrable function Let f (x) be a function from the space L loc in the sense of Lebesgue on any compact set in . With the help of f we build a functional that we denote again by f through the relation  ( f, ϕ) =



f (x)ϕ(x)d x, ∀ϕ ∈ D().

(6.5.1)

Proposition 6.5.1 The correspondence from (6.5.1) defines a distribution. Proof According to the definition of a distribution, we have to prove that f is well defined, linear and continuous. Step 1. The functional f is well defined if its value for any test function is a finite 1 (), we deduce that f is integrable on any compact set, so its number. Since f ∈ L loc integral on any compact set from  is a finite number. Since ϕ ∈ D(), we deduce that ϕ is a test function, so it is infinitely differentiable and hence, it is integrable. Then suppϕ = K = compact set ⊂ , so the values of ϕ outside the compact set K are null, hence   f (x)ϕ(x)d x = f (x)ϕ(x)d x. (6.5.2) 

K

Clearly, the right-hand side from (6.5.2) is a finite number by the assumptions on the functions f and ϕ. So the functional f is well defined. Step 2. We show that the functional f is linear, that is, ∀λ1 , λ2 ∈ R and ∀ϕ1 , ϕ2 ∈ D() ⇒ ⇒ ( f, λ1 ϕ1 + λ2 ϕ2 ) = λ1 ( f, ϕ1 ) + λ2 ( f, ϕ2 ).

(6.5.3)

Clearly, by the linearity of the integral and the formula (6.5.1) we deduce that  f (x) (λ1 ϕ1 (x) + λ2 ϕ2 (x)) d x = ( f, λ1 ϕ1 + λ2 ϕ2 ) =    = λ1 f (x)ϕ1 (x)d x + λ2 f (x)ϕ2 (x)d x = 

(6.5.4)



= λ1 ( f, ϕ1 ) + λ2 ( f, ϕ2 ). Step 3. We will prove the continuity of the functional f . We will use the property that a functional can be interchanged with the limit sign. Let us consider a sequence of test functions {ϕν }ν∈N , ϕν ∈ D(), ν = 1, 2, . . . such that (6.5.5) ϕν → ϕ, ϕ ∈ D().

6.5 Distributions as Generalized Functions

91

We have to prove that lim ( f, ϕν ) =

ν→∞



f, lim ϕν . ν→∞

(6.5.6)

Clearly, since the integral can be interchanged with the limit sign and by the definition (6.5.1), we have   lim ( f, ϕν ) = lim f (x)ϕν (x)d x = f (x) lim ϕν (x)d x = ν→∞ ν→∞  ν→∞   (6.5.7)  = f (x)ϕ(x)d x = ( f, ϕ) = f, lim ϕν . 

ν→∞

We can conclude that f defined by the relation (6.5.1) is a distribution.



Hence f ∈ D () and it is called a distribution generated by a locally integrable function or distribution of function type. Then we ask if every distribution is generated by a function. The answer is negative and we will present a counterexample. So there exist distributions which are not generated by functions. A simple example is the Dirac distribution [9], denoted by δ and which is defined by (δ, ϕ) = ϕ(0), ∀ϕ ∈ D(). (6.5.8) Proposition 6.5.2 The functional δ defined by the correspondence (6.5.8) is a distribution. Proof We have to show that δ is well defined, linear and continuous, that is, it can be interchanged with the limit sign. Step 1. Note that the right-hand side from the relation (6.5.8) is a finite number since it is the value of a test function in a given point, so δ : D() → R.

(6.5.9)

Step 2. For the linearity, let us consider ∀λ1 , λ2 ∈ R and ∀ϕ1 , ϕ2 ∈ D(). We have to prove that (δ, λ1 ϕ1 + λ2 ϕ2 ) = λ1 (δ, ϕ1 ) + λ2 (δ, ϕ2 ) .

(6.5.10)

(δ, λ1 ϕ1 + λ2 ϕ2 ) = (λ1 ϕ1 + λ2 ϕ2 ) (0) = = λ1 ϕ1 (0) + λ2 ϕ2 (0) = λ1 (δ, ϕ1 ) + λ2 (δ, ϕ2 ).

(6.5.11)

Clearly, we have

Step 3. For the continuity, we consider a sequence of test functions {ϕν }ν∈N , ϕν ∈ D(), ν = 1, 2, . . . such that ϕν → ϕ, ϕ ∈ D().

(6.5.12)

92

6 Distributions

We have to show that δ can be interchanged with the limit sign, that is  lim (δ, ϕν ) = δ, lim ϕν .

ν→∞

(6.5.13)

ν→∞

We have  lim (δ, ϕν ) = lim ϕν (0) = ϕ(0) = (δ, ϕ) = δ, lim ϕν .

ν→∞

ν→∞

ν→∞

(6.5.14)

Hence, δ is a distribution.



Now we can state that not every distribution is a generalized function, that is, we prove that the δ distribution is not of function type. Proposition 6.5.3 There is no locally integrable function that generates the Dirac distribution, i.e. δ is not a distribution of function type. Proof The proof is by contradiction. We assume that there exists a locally integrable function that generates the Dirac distribution δ, i.e.  1 ∃ f (x) ∈ L loc () such that (δ, ϕ) =



f (x)ϕ(x)d x, ∀ϕ ∈ D().

(6.5.15)

Since this equality holds true for any test function, it holds true for the following particular test function  ϕ(x, ε) =

e−ε 0,

2

/(ε2 −r 2 )

, r 0.

(6.7.7)

98

6 Distributions

1 Clearly, the function θ is locally integrable, θ ∈ L loc (R), hence it generates a distribution of function type that we denote again by θ

 (θ(t), ϕ(t)) =

∞

−∞

 θ(t)ϕ(t)dt =

∞

ϕ(t)dt.

(6.7.8)

0

Proposition 6.7.1 The functional θ defined by the correspondence (6.7.8) is a distribution. Proof (i) The functional θ is well defined since supp ϕ =compact set in R and ∞ ϕ ∈ Cc∞ (R), so the number 0 ϕ(t)dt is a finite number. (ii) The linearity of the functional θ follows from the linearity of the integral. For ∀λ1 , λ2 ∈ R and ∀ϕ1 , ϕ2 ∈ D(R) we should have (θ, λ1 ϕ1 + λ2 ϕ2 ) = λ1 (θ, ϕ1 ) + λ2 (θ, ϕ2 ).

(6.7.9)

Indeed, we have  ∞ (λ1 ϕ1 + λ2 ϕ2 ) (t)dt = (θ, λ1 ϕ1 + λ2 ϕ2 ) = 0  ∞  ∞ = λ1 ϕ1 (t)dt + λ2 ϕ2 (t) = λ1 (θ, ϕ1 ) + λ2 (θ, ϕ2 ). 0

(6.7.10)

0

(iii) The continuity of the functional θ follows from the fact that it can be interchanged with the limit sign. We consider an arbitrary sequence of test functions (ϕν )ν∈N , ϕν ∈ D(R), ν = 1, 2, . . . such that

We will prove that

ϕν → ϕ, ϕ ∈ D(R).

(6.7.11)

 lim (θ, ϕν ) = θ, lim ϕν .

(6.7.12)

ν→∞

ν→∞

Indeed, by the definition (6.7.8), we have 



∞

∞

ϕν (t)dt = lim ϕν (t)dt = lim (θ, ϕν ) = lim ν→∞ ν→∞ 0 0 ν→∞  ∞  = ϕ(t)dt = (θ, ϕ) = θ, lim ϕν . 0

(6.7.13)

ν→∞

By the properties (i), (ii) and (iii) we deduce that θ is a distribution.



Clearly, in the classical sense of functions we have dθ(t) = 0, ∀t = 0. dt

(6.7.14)

6.7 The Primitive of a Distribution

99

We will see that in the sense of distributions we have a different result from the case of functions. Proposition 6.7.2 The derivative in the sense of distributions of the Heaviside distribution is equal to the Dirac distribution dθ D (R) = δ⇔ dt



 dθ , ϕ = (δ, ϕ), ∀ϕ ∈ D(R). dt

(6.7.15)

Proof First we recall the definition of the Dirac distribution [9] (δ, ϕ) = ϕ(0), ∀ϕ ∈ D().

(6.7.16)

In this case we have  ⊂ R. By using the definition of the distribution of function type, the integration by parts formula and the definition of the Heaviside distribution, we obtain   ∞  dθ dθ ,ϕ = ϕ(t)dt = θ(t)ϕ(t)|∞ −∞ − dt −∞ dt  ∞  ∞  ∞ (6.7.17)   θ(t)ϕ (t)dt = − θ(t)ϕ (t)dt = − ϕ (t)dt = − −∞

−∞

0

= −ϕ(t)|∞ 0 = ϕ(0) − ϕ(∞) = ϕ(0) = (δ, ϕ) . In the relations above, we used twice the fact that supp ϕ = compact set, so it is a closed interval and then we have ϕ(∞) = 0.  Observation 6.7.2 The example given in the Proposition 6.7.2 can be used conversely. If the derivative of the Heaviside distribution is the Dirac distribution, then, conversely, the primitive of the Dirac distribution is the Heaviside distribution. Moreover, the example from the Proposition 6.7.2 shows that in the case of primitives, the set of distributions is larger than the set of classical functions.

6.7.1 Structure Theorems The goal of these theorems is to prove that it is not essential to impose the restriction that primitives of distributions are defined for those test functions that are derivatives of other test functions. We introduce the notation   d ϕ(t) ˜ . (6.7.18) 0 = ϕ0 ∈ D(R) | ∃ϕ˜ ∈ D(R) such that ϕ0 (t) = dt We have the following theorem for the characterization of the set 0 .

100

6 Distributions

Theorem 6.7.1 The necessary and sufficient condition for a test function ψ0 ∈ D(R) to be in the set 0 is that  ∞

−∞

ψ0 (τ )dτ = 0.

(6.7.19)

Proof Necessity. We assume that ψ0 ∈ 0 and we prove that formula (6.7.19) holds true. Since ψ0 ∈ 0 , we have d ϕ˜ ∃ ϕ˜ ∈ D(R) such that ψ0 (t) = ⇒ dt  ∞  ∞ d ϕ˜ dτ = ϕ(τ ˜ )|∞ ψ0 (τ )dτ = ⇒ −∞ = 0. dτ −∞ −∞

(6.7.20)

The last equality holds true since supp ϕ˜ = compact set ⊂ R. Sufficiency. We assume that ψ0 verifies relation (6.7.19) and we prove that ψ0 ∈ 0 . We have to find a test function ϕ˜ ∈ D(R) such that d ϕ˜ = ψ0 . dt

(6.7.21)

The proof is constructive. We will build the function ϕ˜ such that  ϕ(t) ˜ =

t −∞

ψ0 (τ )dτ .

(6.7.22)

By (6.7.22) we have ϕ(−∞) ˜ = 0. By (6.7.19) we have ϕ(∞) ˜ = 0. Hence, from the last two statements we obtain supp ϕ˜ = compact set ⊂ R. On the other hand, based on the properties of integrals with parameter we know that ϕ˜ is infinitely differentiable, hence ϕ˜ is a test function, i.e. ϕ˜ ∈ D(R). Moreover, based on the properties of integrals with parameter we obtain d ϕ˜ = ψ0 (t), dt

(6.7.23)

which leads to ψ0 ∈ 0 .



In the following theorem we establish the relationship between the sets D(R) and 0 . Theorem 6.7.2 Let ϕ1 be a test function, ϕ1 ∈ D(R), which satisfies the condition 

∞ −∞

ϕ1 (τ )dτ = 1.

(6.7.24)

Then for any test function ϕ ∈ D(R), there exists a function ϕ0 ∈ 0 such that

6.7 The Primitive of a Distribution

101

 ϕ(t) = ϕ1 (t)

∞

−∞

ϕ(τ )dτ + ϕ0 (t).

(6.7.25)

Proof The function ϕ0 is unique and it is called the projection of the test function ϕ on the set 0 . We prove that the set of test functions that satisfies condition (6.7.24) is non-empty by showing that at least one test function satisfies this condition. We consider a test function ϕ2 which does not belong to 0 , so it satisfies the condition 

∞ −∞

ϕ2 (τ )dτ = 0.

(6.7.26)

Then we can build the function ϕ1 as ϕ2 (t) . −∞ ϕ2 (τ )dτ

de f

ϕ1 (t) =  ∞

(6.7.27)

function ϕ1 is a test function since ϕ2 is a test function and the integral The ∞ ϕ −∞ 2 (τ )dτ is a scalar and the set of test functions is a linear space. The function ϕ1 built in (6.7.27) satisfies (6.7.24) since 

∞ −∞

 ϕ1 (t)dt =

1 = ∞ −∞ ϕ2 (τ )dτ

∞

ϕ2 (t) dt = −∞ ϕ2 (τ )dτ

∞

−∞  ∞

−∞

(6.7.28)

ϕ2 (t)dt = 1.

With ϕ1 built as in (6.7.27) we try to verify relation (6.7.25). To this end, we define the function ϕ0 by  ϕ0 (t) = −ϕ1 (t)

∞

−∞

ϕ(τ )dτ + ϕ(t).

(6.7.29)

Since the functions ϕ1 and ϕ are test functions, we can prove that the function ϕ0 from (6.7.29) is a test function. We want to prove that ϕ0 ∈ 0 . Based on Theorem 6.7.1, we have to prove that  ∞

−∞

ϕ0 (t)dt = 0.

(6.7.30)

By using (6.7.29), we have 

∞ −∞



=

 ∞   ∞ ϕ(t)dt − ϕ1 (t) ϕ(τ )dτ dt = −∞ −∞ −∞    ∞ ϕ(τ )dτ 1 − ϕ1 (t)dt = 0, 

ϕ0 (t)dt = ∞

−∞

∞

−∞

(6.7.31)

102

6 Distributions

where we used the hypothesis (6.7.24). We can rewrite relation (6.7.29) in order to obtain relation (6.7.25).



Theorem 6.7.3 The primitives of 0 (the constant zero) are constants in the sense of distributions, i.e. any primitive of the null distribution is a distribution generated by a constant function. Proof Let 0 ∈ D (R) and y ∈ D (R) such that dy =0 dt in the sense of distributions, that is   dy , ϕ = (0, ϕ), ∀ϕ ∈ D (R). dt

(6.7.32)

(6.7.33)

By considering the definition of the derivative of a distribution, we have (y, −ϕ ) = 0.

(6.7.34)

The relation (6.7.34) shows that the distribution y is defined for those test functions that are derivatives of other test functions, i.e. y is defined for test functions from the set 0 from Theorem 6.7.1. Moreover, for these test functions the distribution y is null (6.7.35) (y, ψ0 ) = 0, ∀ψ0 ∈ 0 . We consider an arbitrary test function ϕ ∈ D(R) and we apply the decomposition from Theorem 6.7.2 in order to obtain    ∞ ϕ(τ )dτ = (y, ϕ(t)) = y, ϕ0 (t) + ϕ1 (t) −∞    ∞ ϕ(τ )dτ = = (y, ϕ0 (t)) + y, ϕ1 (t) −∞   ∞  ∞    ∞ (6.7.36) ϕ(τ )dτ = y(t)ϕ1 (t) ϕ(τ )dτ dt = = y, ϕ1 (t) −∞ −∞ −∞  ∞  ∞  ∞ ϕ(τ )dτ y(t)ϕ1 (t)dt = C1 ϕ(τ )dτ = = −∞

−∞

−∞

= (C1 , ϕ), ∀ϕ ∈ D(R). Hence, we obtain

D  (R)

y = C1 ,

(6.7.37)

6.7 The Primitive of a Distribution

103

where C1 = (y, ϕ1 ) is a constant since it is the value of the distribution y for the fixed test function ϕ1 . By choosing another fixed test function ϕ1 , we obtain another constant.  Theorem 6.7.4 Let us consider the equation dy = f, f ∈ D (R). dt

(6.7.38)

Then the solution of this equation is of the form y = y0 + C,

(6.7.39)

where the distribution y0 is a particular solution of the equation, i.e. a particular primitive of the distribution f , and C is the distribution generated by a constant function. Proof The statement of the theorem can be reformulated in the following way: For an arbitrarily fixed distribution, any two primitives (which are distributions) differ by a constant (i.e. the distribution of function type generated by a constant function). In the context of distributions, the equation from the statement of the theorem becomes   dy (6.7.40) , ϕ = ( f, ϕ) ⇔ (y, −ϕ ) = ( f, ϕ), ∀ϕ ∈ D(R). dt Therefore, the distribution y, which is the primitive of the distribution f , is defined only for test functions from the set 0 . We define the distribution y0 by (y0 , ϕ) = (y, ϕ0 ),

(6.7.41)

where the test function ϕ0 is the projection of the arbitrary test function ϕ on the set 0 . We use the decomposition of the test function ϕ from Theorem 6.7.2 and we obtain  ∞ ϕ(τ )dτ ⇒ ϕ(t) = ϕ0 (t) + ϕ1 (t) −∞    ∞ ϕ(τ )dτ = ⇒ (y, ϕ) = (y, ϕ0 ) + y, ϕ1 (t) −∞ (6.7.42)  ∞ ϕ(τ )(y, ϕ1 )dτ = = (y0 , ϕ) + −∞  ∞ C1 ϕ(τ )dτ = (y0 , ϕ) + (C1 , ϕ) = (y0 + C1 , ϕ). = (y0 , ϕ) + −∞

104

6 Distributions

Therefore, we obtain y = y0 + C1 since the test function ϕ is arbitrary. The proof of the theorem is finished if we prove that the functional y0 defined in (6.7.41) is a distribution. (i) The distribution is well defined because y is a distribution and ϕ0 is a test function (even if it is a particular one), hence (y, ϕ0 ) is a finite number. (ii) Linearity follows from (y0 , λ1 ϕ1 + λ2 ϕ2 ) = (y, λ1 ϕ01 + λ2 ϕ02 ),

(6.7.43)

where ϕ01 and ϕ02 are the projections of the test functions ϕ1 and ϕ2 , respectively, on the set 0 . Since y is a distribution, it is linear, so y0 is linear. (iii) In order to prove continuity, let us consider an arbitrary sequence of test functions {ϕν }ν∈N , ϕν ∈ D(R) such that ϕν → ϕ, ϕ ∈ D(R).

(6.7.44)

We will prove that limν→∞ (y0 , ϕν ) = (y0 , limν→∞ ϕν ). Since y is a distribution, it is continuous and it can be interchanged with the limit sign. By definition (6.7.41), we have

 lim (y0 , ϕν ) = lim y, ϕ0ν = y, lim ϕ0ν = ν→∞ ν→∞ ν→∞  (6.7.45)

0 = y, ϕ = (y0 , ϕ) = y0 , lim ϕν . ν→∞

The proof of the theorem is finished.



6.8 Extras We will discuss some results (some of them without proof) about the distributions of function type and about the distributions of order zero (or measures). Theorem 6.8.1 Let I = (α, β) be an interval on the real line and let g be a distribution, g ∈ D (I ). The necessary and sufficient condition for the derivative in the sense of distributions of the distribution g to be a measure (or a distribution of order zero) is that g is generated by a function of bounded variation (i.e. a BV function) on every compact set of the interval I . Theorem 6.8.2 Let I = (α, β) be an interval on the real line and let g be a distribution, g ∈ D (I ). The necessary and sufficient condition for the derivative in the 1 is that g is generated by an sense of distributions of the distribution g to be in L loc absolutely continuous function on every compact subinterval of the interval I . Moreover, if g is generated by an absolutely continuous function on every compact subinterval of the interval I , then the classical derivative of g, g  (computed in the sense of functions), generates the distribution g  .

6.8 Extras

105

We do not show the proofs of these two theorems because they are technical and based on elements of BV functions and absolutely continous functions, respectively. For details see [10]. Definition 6.8.1 Let us consider the function ϕ : [a, b] → R and the sum vτ =

n 

|ϕ(tk ) − ϕ(tk+1 )|,

(6.8.1)

k=1

where τ is a division of the interval [a, b]. If the set of the sums vτ corresponding to all the divisions τ of the interval [a, b] is bounded, then ϕ is called a function of bounded variation. For a function of bounded variation, the quantity Vab ϕ = sup vτ τ

(6.8.2)

is called the total variation of the function ϕ. Definition 6.8.2 If A is a set of an algebra, then the function ϕ is absolutely continuous on A if μ(A) = 0 ⇒ ϕ(A) = 0, (6.8.3) where μ is a measure, i.e. a set function that is positive and additively countable. Definition 6.8.3 The Stieltjes measure is the measure generated by a function of bounded variation [16]. In the sequel, we present a theorem that is due to Vulih [15]. Theorem 6.8.3 The general form of linear and continuous functionals on the space C[a, b] is given by the Stieltjes integral  f (x) =

b

x(t)dg(t),

(6.8.4)

a

where g(t) is a function of bounded variation. Moreover, || f || = Vab g.

(6.8.5)

We can prove that the function g defines in a unique way the functional f . A distribution that is very often encountered in quantum mechanics [8] is the distribution generated by the Cauchy principal value of the function x1 , denoted by 1 Vp . x

(6.8.6)

106

6 Distributions

It is clear that the function f (x) = x1 , x = 0 is not locally integrable since it is not integrable on compact sets including the origin. Therefore, in mathematical analysis we use the Cauchy principal value of this function, which is by definition  Vp

b −a

1 d x = lim ε0 x



−ε

−a

1 dx + x



b ε

 1 d x , a, b > 0. x

(6.8.7)

We obtain directly 

  b

1 1 b lim dx + d x = lim ln |x||−ε −a + ln |x||ε = ε0 ε0 −a x ε x   ε b b b = lim ln = ln . = lim ln + ln ε0 ε0 a ε a a −ε

(6.8.8)

We can easily obtain the following two particular cases of the Cauchy principal value  Vp

a

−a

1 d x = 0, V p x



∞ −∞

1 d x = 0. x

(6.8.9)

According to the definition of the Cauchy principal value, it is locally integrable, so it can generate a distribution of function type that we denote by V p x1 and that we define by    1 ϕ(x) d x, ∀ϕ ∈ D(R). (6.8.10) Vp , ϕ = Vp x x R Proposition 6.8.1 The functional V p x1 defined in (6.8.10) is a distribution. Proof First of all, we have 

 −ε    ∞  ∞ 1 ϕ(x) ϕ(x) ϕ(x) Vp , ϕ = Vp d x = lim dx + d x . (6.8.11) ε0 x x x x −∞ −∞ ε

(i) The distribution is well defined because the test function ϕ is of class C0∞ , hence it is integrable and it has compact support and x does not reach the origin. (ii) The linearity follows from the linearity of the integral. (iii) Now we prove continuity. We consider an arbitrary sequence of test functions {ϕν }ν∈N , ϕν ∈ D(R) such that ϕν → ϕ, ϕ ∈ D(R).

(6.8.12)

Let us prove that the Cauchy principal value can be interchanged with the limit sign, that is     1 1 (6.8.13) lim V p , ϕν = V p , lim ϕν . ν→∞ x x ν→∞

6.8 Extras

107



  ∞ 1 ϕν (x) V p , ϕν = lim V p dx = ν→∞ ν→∞ x x −∞  −ε   ∞ ϕν (x) ϕν (x) dx + dx = = lim lim ν→∞ ε0 x x −∞ ε  −ε   ∞ ϕν (x) ϕν (x) dx + dx = = lim lim ε0 ν→∞ x x −∞ ε  −ε   ∞ ϕ(x) ϕ(x) dx + dx = = lim ε0 x x −∞ ε     1 1 = V p , ϕ = V p , lim ϕν , x x ν→∞

We have

lim

(6.8.14)



which finishes the proof.

As we have already proven, the product of a distribution with a function of class C ∞ is a distribution. The function g(x) = x is of class C ∞ and by multiplying it with the Cauchy principal value we obtain a distribution. Hence, x Vp

1 x

(6.8.15)

is a distribution. In the following proposition we want to prove that the Cauchy principal value V p x1 is close to the function that generates it, that is x1 . Proposition 6.8.2 In the set of distributions we have the following equality x Vp

1 D (R) = 1. x

(6.8.16)

Proof By using the definition of the Cauchy principal value and the definition of the product of a distribution with a function of class C ∞ , we obtain directly 

    ∞  ∞ 1 1 1 x V p , ϕ = V p , xϕ = V p xϕ(x)d x = V p ϕ(x)d x = x x −∞ x −∞  ∞  ∞ ϕ(x)d x = 1 · ϕ(x)d x = (1, ϕ), ∀ϕ ∈ D(R). = −∞

−∞

(6.8.17) Since the test function ϕ is arbitrary, we obtain the equality from the statement of the proposition.  Observation 6.8.1 We can state that the distribution x V p x1 is of function type and it is generated by the constant function 1.

108

6 Distributions

In the sequel we want to establish a relationship between the derivative in the sense of distributions and the derivative in the classical sense of functions. We study this problem for a function that is neither differentiable, nor continuous. We will restrict our considerations to a function that is continuous except for a point x0 in which the function f has a discontinuity of the first kind. We denote by s0 the jump of the function f in the point x0 , that is s0 = ld (s0 ) − ls (s0 ) = f (x0 + 0) − f (x0 − 0).

(6.8.18)

Theorem 6.8.4 The relationship between the derivative in the sense of distributions and the derivative in the classical sense of functions is given by f  (x) = f˜ (x) + s0 δ(x + x0 ),

(6.8.19)

where by f˜ we denote the derivative in the classical sense of functions and δ(x + x0 ) is the translation of the Dirac distribution. Proof We consider an arbitrary test function and we compute the derivative in the sense of distributions: ( f  (x), ϕ(x)) = ( f (x), −ϕ (x)) = −( f (x), ϕ (x)) =  x0 −ε   ∞   = − lim f (x)ϕ (x)d x + f (x)ϕ (x)d x = ε0 −∞ x0 +ε   x0 −ε x0 −ε = − lim f (x)ϕ(x)|−∞ − f˜ (x)ϕ(x)d x + f (x)ϕ(x)|∞ x0 +ε − ε0 −∞   ∞ − f˜ (x)ϕ(x)d x = − lim [ f (x0 − ε)ϕ(x0 − ε)−

(6.8.20)

ε0

x0 +ε



 ∞ f˜ (x)ϕ(x)d x + f˜ (x)ϕ(x)d x = x0 −∞  x0 = [ f (x0 + 0) − f (x0 − 0)] ϕ(x0 ) + f˜ (x)ϕ(x)d x+ −∞  ∞  ∞ f˜ (x)ϕ(x)d x = s0 ϕ(x0 ) + f˜ (x)ϕ(x)d x = + x0 −∞  = f˜ (x), ϕ(x) + s0 (δ(x), ϕ(x − x0 )) =  = f˜ (x), ϕ(x) + s0 (δ(x + x0 ), ϕ(x)) =  = f˜ (x) + s0 δ(x + x0 ), ϕ(x) . − f (x0 + ε)ϕ(x0 + ε)] +

x0

(6.8.21)

In conclusion, we have



 f  (x), ϕ(x) = f˜ (x) + s0 δ(x + x0 ), ϕ(x) .

(6.8.22)

6.8 Extras

109

Since the test function ϕ is arbitrary, we derive the equality in the statement of the theorem, which finishes the proof.  We can conclude that the derivative in the sense of distributions is the sum of the derivative in the sense of functions and the product of the jump of the function f in the discontinuity point x0 with the Dirac distribution that is concentrated at x0 . In particular, in the case of differentiable functions, the derivative in the sense of distributions coincides with the derivative in the sense of functions.

6.8.1 Higher-Order Primitives Definition 6.8.4 We say that the distribution F is a primitive of order m, m ∈ N of the arbitrarily fixed distribution f if D  (R)

F (m) =



f ⇔ F (m) , ϕ = ( f, ϕ), ∀ϕ ∈ D(R).

(6.8.23)

We have the following result for higher-order primitives. Theorem 6.8.5 (i) Every distribution has a primitive of any order, where the order is a natural number. (ii) Two primitives of the same order m corresponding to the same distribution differ by a polynomial (in the sense of distributions) of degree m − 1. Proof (i) The existence of the primitive F which verifies the definition above can be shown by induction. For m = 1 we have already proven that every distribution has a primitive. Since the primitive F is a distribution, it has also a primitive, which is a second-order primitive for the initial distribution f . This second-order primitive is a distribution, so it has a primitive, and so on. So we can assume that the distribution f has a primitive of order m − 1, which we denote by G D  (R)

G (m−1) =



f ⇔ G (m−1) , ϕ = ( f, ϕ) ∀ϕ ∈ D(R).

(6.8.24)

Since G is a distribution, it has a primitive that we denote by F, hence D  (R)

D  (R)

D  (R)

F  = G ⇒ F (m) = G (m−1) ⇒ F (m) =

f.

(6.8.25)

(ii) For the second part of the proof, we have already shown that two primitives of the same distribution differ by a constant, which we consider to be a polynomial of degree zero. We assume that there are two primitives of order m − 1 of the distribution f , which differ by a polynomial of degree m − 2. We have to prove the statement for m, i.e. two primitives of order m of the distribution f differ by a polynomial of order m − 1. Let G 1 and G 2 be the primitives of order m − 1 that differ by a polynomial of order m − 2, which we denote by Pm−2 .

110

6 Distributions



D  (R)

G 1 − G 2 = Pm−2 ⇔ (G 1 − G 2 , ϕ) =

∞

−∞

Pm−2 (x)ϕ(x)d x, ∀ϕ ∈ D(R).

(6.8.26) We denote by F1 and F2 the primitives of order 1 of the distributions G 1 and G 2 , that is D  (R) D  (R) (6.8.27) F1 = G 1 , F2 = G 2 . Then we can obtain directly  ∞ [G 1 (x) − G 2 (x)] ϕ(x)d x = (G 1 − G 2 , ϕ) = −∞  ∞  ∞    F1 (x) − F2 (x) ϕ(x)d x = = [F1 (x) − F2 (x)] ϕ(x)d x = −∞ −∞  ∞ − = [F1 (x) − F2 (x)] ϕ(x)|∞ [F1 (x) − F2 (x)] ϕ (x)d x = −∞ −∞  ∞ =− [F1 (x) − F2 (x)] ϕ (x)d x.

(6.8.28)

−∞

In conclusion, we have  (G 1 − G 2 , ϕ) = −

∞ −∞

[F1 (x) − F2 (x)] ϕ (x)d x.

(6.8.29)

On the other hand, we have 

∞

Pm−2 (x)ϕ(x)d x = (G 1 − G 2 , ϕ) = (Pm−2 , ϕ) = −∞  ∞  Pm−1 (x)ϕ(x)d x = Pm−1 (x)ϕ(x)|∞ = −∞ − −∞  ∞  ∞ Pm−1 (x)ϕ (x)d x = − Pm−1 (x)ϕ (x)d x. − −∞

(6.8.30)

−∞

Hence, we have

 (G 1 − G 2 , ϕ) = −

∞ −∞

Pm−1 (x)ϕ (x)d x.

(6.8.31)

If we compare formulae (6.8.29) and (6.8.31), then the expressions on the right-hand side have to be equal 

∞

−∞

[F1 (x) − F2 (x)] ϕ (x)d x =



∞

−∞

Pm−1 (x)ϕ (x)d x ⇒

⇒ (F1 − F2 , ϕ) = (Pm−1 , ϕ) , ∀ϕ ∈ D(R).

(6.8.32)

6.8 Extras

111

The last equality is valid only for test functions from the set 0 . By applying the second structure Theorem 6.7.2, the result is valid for any test function.  As an application of higher-order primitives, we will discuss an important result regarding the solution of a differential equation in the set of distributions. Proposition 6.8.3 The following differential equation, considered in the set of distributions d n y D (R) = δ (6.8.33) dxn has the solution yn =

1 x n−1 x+n−1 = θ(x) , (n − 1)! (n − 1)!

(6.8.34)

where δ is the Dirac distribution and θ is the Heaviside distribution. Proof We will use mathematical induction. For n = 1 the equation becomes dy dx

D  (R)

= δ

(6.8.35)

and the solution is y = θ(x). But we have already proven that the derivative of the Heaviside distribution is the Dirac distribution: dθ D (R) = δ. dx

(6.8.36)

We assume that the result is true for n and we prove it for n + 1. So we assume that the Eq. (6.8.33) has the solution (6.8.34). We will show that the equation d n+1 y D (R) = δ d x n+1 has the solution yn+1 = θ(x) We will show that

xn . n!

D  (R)

(n+1) yn+1 = δ

by direct computations and using the induction hypothesis:

(6.8.37)

(6.8.38)

(6.8.39)

112

6 Distributions

   x n (n) θ(x) = = = n!

 (n)    x n−1 x x (n)  θ(x) = = yn = (n − 1)! n n   x d x dn  x  + · · · + Cnn yn n = = Cn0 yn(n) + Cn1 yn(n−1) n dx n dx n  

 x = δ(x) + yn(n−1) = yn(n−1) . n

(n+1) yn+1



(n) yn+1



(6.8.40)

We used the Leibniz formula for higher-order derivatives and the fact that the derivatives of order greater than two of nx are zero. Finally, by considering the definition of the product of a distribution with a function of class C ∞ , we have δ(x) since

x D (R) = δ=0 n

x

 x x δ, ϕ = δ, ϕ = ϕ (0) = 0 · ϕ(0) = 0. n n n

(6.8.41)

(6.8.42)

Based on the induction hypothesis, we have (n−1) x (n−1) = θ(x) = (n − 1)!

 n−1 (n−1)  x , x >0 1, x > 0 (n−1)! = = θ(x). = 0, x ≤ 0 0, x ≤0

yn(n−1)

Finally, we obtain





yn(n−1)



D  (R)

D  (R)

= θ (x) = δ,

which ends the proof of the proposition.

(6.8.43)

(6.8.44) 

6.8.2 The Local Structure of Distributions We know that every distribution has derivatives of all orders. We will show that locally every distribution is the derivative (in the sense of distributions) of a bounded function. Theorem 6.8.6 Let f ∈ D () and let U be a relatively compact open set such that U¯ ⊂ . Then we can find a function g ∈ L ∞ (U ) and an integer m ≥ 0 such that

6.8 Extras

113

f =

∂ mn g on U. · · · ∂xnm

(6.8.45)

∂x1m

Proof Step 1. Let Q = U¯ . Note that f is a continuous linear functional on Cc∞ (; Q) since f is a distribution. It follows that, given ε > 0, we can find a neighbourhood     V = V (Q, j, η) = ϕ ∈ Cc∞ (; Q)|  D k ϕ(x) ≤ η, ∀|k| ≤ j, ∀x ∈ Q (6.8.46) such that | f (ϕ)| ≤ ε, ∀ϕ ∈ V.

(6.8.47)

∂ ∂ Step 2. Let m ∈ N. We denote by ∂x m the partial derivative ∂x m ···∂x m . n 1 We consider that E is the subspace of all functions of the form m

ψ=

mn

∂ j+1 ϕ , ∀ϕ ∈ Cc∞ (; Q). ∂x j+1

(6.8.48)

Note that the correspondence between ϕ and ψ from (6.8.48) is one-to-one because the functions have compact support. We consider that L 1 (Q) is the Banach space of integrable functions on Q. We will consider on E the topology induced by L 1 (Q). j+1 ϕν converges to zero in We will show that is a sequence of functions ψν = ∂∂x j+1 1 L (Q), then the sequence f (ϕν ) converges to zero. We consider that L is a hypercube with length side l ≥ 1, which contains Q. For any η > 0, there is an index ν0 such that 

  j+1   ∂ ϕν  η   ···  ∂x j+1  d x1 . . . d xn ≤ l ( j+1)η Q

(6.8.49)

for all ν ≥ ν0 . If we write ∂ j ϕν (x) = ∂x j then it follows that



x1

−∞

 ···

xn −∞

∂ j+1 ϕν dt1 . . . dtn , ∂t j+1

 j   ∂ ϕν  η    ∂x j (x) ≤ l jη

(6.8.50)

(6.8.51)

because l ≥ 1. This inequality leads to   k  D ϕν (x) ≤ η, ∀x ∈ Q, ∀|k| ≤ j, ∀ν ≥ ν0 .

(6.8.52)

This implies that ϕν ∈ V , ∀ν ≥ ν0 . It follows that | f (ϕν )| ≤ ε, ∀ν ≥ ν0 . Step 3. On E we consider the linear functional P(ψ) = f (ϕ). Note that P is a continuous linear functional on E endowed with the topology induced by L 1 (Q).

114

6 Distributions

Clearly, E is a proper linear subspace of L 1 (Q). Then by the Hahn-Banach theorem, the linear functional P can be extended to a continuous linear functional P˜ on L 1 (Q). It follows that there is φ ∈ L ∞ (Q) such that   ˜ φ · ψ, ∀ψ ∈ E. (6.8.53) P(ψ) = ··· Q

Hence, for every ϕ ∈ Cc∞ (; Q) and for every ϕ ∈ Cc∞ (U ), we obtain  f (ϕ) = P(ψ) =

  j+1 φ ∂ j+1 ϕ ( j+1)n ∂ ··· φ· = (−1) ,ϕ . ∂x j+1 ∂x j+1 Q 

By setting m = j + 1 and g = (−1)mn φ, it follows that f =

∂m g ∂x m

on U .

(6.8.54) 

Theorem 6.8.7 Let f ∈ D (). We consider that U is a relatively compact open set with U¯ ⊂ . Then the distribution f coincides on U with the derivative of a continuous function which has its support in an arbitrary neighbourhood of U¯ . Theorem 6.8.8 Let f ∈ E  () be a distribution. Then it can be written as f =



D k gk

(6.8.55)

|k|≤ j

on , where gk are continuous functions with compact support contained in an arbitrary neighbourhood ω of Q, which is the support of f . Proof We consider that U is a relatively compact open set with the property that Q ⊂ U ⊂ U¯ ⊂ ω ⊂ . Then there is a continuous function with compact support in ω such that f = D j g on U , that is ( f, ψ) = (−1)| j|



 ···



g · D j ψ, ∀ψ ∈ Cc∞ (U ).

(6.8.56)

Then we consider β ∈ Cc∞ (U ), which is equal to one on a neighbourhood of Q. It follows that for every ϕ ∈ C ∞ () we obtain ( f, ϕ) = ( f, βϕ) = (−1)

| j|



 ···



g · D j (βϕ).

(6.8.57)

If we apply the Leibniz formula, then we obtain D j (β · ϕ) =

 k≤ j

Therefore, we obtain

j! D j−k β · D k ϕ. k!( j − k)!

(6.8.58)

6.8 Extras

115

( f, ϕ) = (−1)| j|

 k≤ j

j! k!( j − k)!

We consider that gk = (−1)| j| + |k|



 ···



g · D j−k β · D k ϕ.

j! g · D j−k β. k!( j − k)!

(6.8.59)

(6.8.60)

Then we integrate by parts and we deduce that  ( f, ϕ) =

 ···



⎛ ⎞  ⎝ D k gk ⎠ ϕ

(6.8.61)

k≤ j

for all ϕ ∈ C ∞ ().



Corollary 6.8.1 Let f be a distribution with compact support. Then f is of finite order. Proof Let g be a locally integrable function in Rn . If f = D k g, then f defines a |k|  continuous linear functional on Cc (Rn ). It follows that f is of finite order. Theorem 6.8.9 Let f be a distribution with compact support Q ⊂ . We assume that f has an order less than or equal to m. Therefore, we have ( f, ϕ) = 0 for all ϕ ∈ C ∞ () such that D k ϕ = 0 on Q, for all |k| ≤ m. Proof Step 1. We consider that Q ε is the ε-neighbourhood of Q. Note that all derivatives D k ϕ with |k| = m become null on Q. Therefore, given δ > 0, we can find ε > 0 sufficiently small so that Q ε ⊂  and   k  D ϕ(x) ≤ δ, ∀x ∈ Q ε .

(6.8.62)

For every x ∈ Q, we consider that x0 ∈ Q is such that the distance between x0 and x is at most ε. Note that D k ϕ(x0 ) = 0. Hence, if |k| = m − 1, then we deduce that 

1

D k ϕ(x) = 0

It follows that

n 

∂ j D k ϕ(x0 + t (x − x0 )) · (x j − x0, j )dt.

(6.8.63)

j=1

 k  √  D ϕ(x) ≤ ε nδ.

(6.8.64)

By using mathematical induction, we have   k √  D ϕ(x) ≤ (ε n)m−|k| · δ, ∀|k| ≤ m.

(6.8.65)

Step 2. There is a function βε such that βε ∈ Cc∞ (), supp βε ⊂ Q ε , βε = 1 on Q ε/4 and

116

6 Distributions

  k  D βε (x) ≤ C(k, n)ε−|k| , ∀x.

(6.8.66)

If we choose C > 0 sufficiently large, then we obtain   k  D βε (x) ≤ C · ε−|k| , ∀x, ∀|k| ≤ m.

(6.8.67)

Step 3. We consider that φε = βε ϕ. It follows from the formula of Leibniz that D k φε =

 j≤k

k! D k− j βε D j ϕ. j!(k − j)!

(6.8.68)

By using the inequalities (6.8.65) and (6.8.67), we derive the following upper bound  k  √  D φε (x) ≤ Cε−|k|+| j| · (ε n)m−| j| · δ ≤ C · δ

(6.8.69)

for all x and for all |k| ≤ m, where C = C(m, n) is a suitable constant which does not depend on k. Step 4. Note that φε = ϕ on a neighbourhood of Q, the support of f . Hence, we have (6.8.70) ( f, φε ) = ( f, ϕ) . Since f ∈ Dm , it follows that there is C1 > 0 a constant such that |( f, ψ)| ≤ C1

sup x∈,|k|≤m

|D k ψ|, ∀ψ ∈ Cc∞ (, Q ε ).

(6.8.71)

If we replace ψ by φε and we consider the relations (6.8.69) and (6.8.70), then we obtain |( f, ϕ)| ≤ C1 Cδ. (6.8.72) Since δ is an arbitrary positive number, it follows that ( f, ϕ) = 0



In the sequel, we present a characterization of distributions whose support is the origin. Theorem 6.8.10 Let f be a distribution whose support is the origin. Then it can be represented in a unique way as a finite linear combination of derivatives of the Dirac measure. Proof Note that f has compact support. This implies that f is a distribution of finite order, that is, f ∈ Dm for some m ∈ N. Note that every ϕ ∈ C ∞ can be represented as ϕ(x) =

 D k ϕ(0) x k + Rm (x), k! |k|≤m

(6.8.73)

6.8 Extras

117

where Rm is a C ∞ function with the property that D k Rm (0) = 0, ∀|k| ≤ m. We deduce that  D k ϕ(0)

f, x k . (6.8.74) ( f, ϕ) = k! |k|≤m

(−1)|k| f, x k ck = k!

If we set

then we obtain f =



ck D k δ.

(6.8.75)

(6.8.76)

|k|≤m



Note that if f = 0, then f, x k = 0, which implies that ck = 0, ∀k. It follows that the decomposition is unique. 

6.9 Convolutions 6.9.1 The Direct Product of Distributions Since the product of two distributions is not necessarily a distribution, we introduced the product of a distribution with a function of class C ∞ , which is a distribution. Hence, we have an algebraic structure on the space of distributions. If f ∈ D () and a ∈ C ∞ , then (a f, ϕ) = ( f, aϕ), ∀ϕ ∈ D(). (6.9.1) We will introduce two more products of distributions. First, let us introduce the direct product of two functions. If f : Rn → R, g : Rm → R, then we define their direct product, denoted by f ⊗g (6.9.2) f ⊗ g : Rn+m → R, ( f ⊗ g)(x, y) = f (x)g(y). The direct product is commutative, associative and distributive over addition, which are the properties of the usual product of functions: (i) f ⊗ g = g ⊗ f ; (ii)( f ⊗ g) ⊗ h = f ⊗ (g ⊗ h);

(6.9.3)

(iii) f ⊗ (g + h) = ( f ⊗ g) + ( f ⊗ h). In particular, if f is a locally integrable function on Rn and g is a locally integrable function on Rm , then f ⊗ g is a locally integrable function on Rn+m .

118

6 Distributions

Observation 6.9.1 Let  (respectively  ) be an open subset of Rn (respectively Rm ). The algebraic tensor product Cc∞ () ⊗ Cc∞ ( ) is the vector space of all functions u(x, y) that can be represented as finite sums  u(x, y) = φ j (x) · ψ j (y), (6.9.4) ∞  Cc∞ () ⊗ Cc∞ ( ) is a vector subwhere φ j ∈ C c∞ () and

ψ j ∈ Cc ( ). Clearly, ∞  ∞ space of Cc  ×  , the space of all C functions in the variables (x, y) = (x1 , · · · , xn , y1 , · · · , ym ) with compact support in  ×  . It can be proven that restrictions to  ×  of polynomials in (x, y) are dense in C ∞ ( ×  ) equipped with its natural topology. It then follows that the tensor product Cc∞ () ⊗ Cc∞ ( ) is dense in Cc∞ ( ×  ). Indeed, if for every u ∈ Cc∞ ( ×  ) there is a sequence of polynomials (Pk (x, y)) converging to u in C ∞ ( ×  ) then, taking α ∈ Cc∞ () and β ∈ Cc∞ ( ) such that α(x)β(y) = 1 on a neighbourhood of the support of u, the sequence (α(x)β(y) Pk (x, y)) belongs to Cc∞ () ⊗ Cc∞ ( ) and converges to u in Cc∞ ( ×  ). As a consequence, it follows that every distribution T ∈ D ( ×  ) is well determined by its values on functions φ ⊗ ψ, where φ ∈ Cc∞ () and ψ ∈ Cc∞ ( ).

We will define the direct product of two distributions as the distribution generated by the direct product of two functions. We can consider that the direct product of two functions coincides with the usual product of functions, but this is not true for the direct product of two distributions since the usual product of two distributions is not necessarily a distribution. 1 1 Let f ∈ L loc (Rn ) and g ∈ L loc (Rm ) be two functions that generate the distri n  butions F ∈ D (R ) and G ∈ D (Rm ), respectively. We consider an arbitrary test function ϕ ∈ D(Rn+m ) and we define the direct product of these two distributions as  ( f ⊗ g)(x, y)ϕ(x, y)d yd x. (6.9.5) ((F ⊗ G) (x, y), ϕ(x, y)) = Rn+m

Proposition 6.9.1 The direct product of two distributions can be written as ((F ⊗ G) (x, y), ϕ(x, y)) = (F(x), (G(y), ϕ(x, y))) .

(6.9.6)

Proof We employ the definition of the direct product of distributions  ( f ⊗ g)(x, y)ϕ(x, y)d yd x = ((F ⊗ G)(x, y), ϕ(x, y)) = Rn+m     f (x)g(y)ϕ(x, y)d yd x = f (x)g(y)ϕ(x, y)dy d x = = n m (6.9.7) Rn+m   R R  f (x) g(y)ϕ(x, y)dy d x = f (x) (G(y), ϕ(x, y)) d x = = Rn

Rm

= (F(x), (G(y), ϕ(x, y))) .

Rn



6.9 Convolutions

119

Proposition 6.9.2 The direct product of two distributions satisfies the following commutative property ((F ⊗ G) (x, y), ϕ(x, y)) = (G(y), (F(x), ϕ(x, y))) .

(6.9.8)

Proof By direct computations, we obtain  ( f ⊗ g)(x, y)ϕ(x, y)d yd x = ((F ⊗ G)(x, y), ϕ(x, y)) = Rn+m   = f (x)g(y)ϕ(x, y)d yd x = f (y)g(x)ϕ(y, x)d xd y = n+m n+m (6.9.9)   R  R g(x) f (y)ϕ(y, x)d x dy = g(x) (F(y), ϕ(y, x)) d x = = Rm

Rn

Rm

= (G(x), (F(y), ϕ(y, x))) = (G(y), (F(x), ϕ(x, y))) . We could change the order of integration based on the theorem of Fubini [3].



Based on the results from Proposition 6.9.1 and Proposition 6.9.2, we have the following definition of the direct product of distributions: (i) ((F ⊗ G) (x, y), ϕ(x, y)) = (F(x), (G(y), ϕ(x, y))) = (G(y), (F(x), ϕ(x, y))) ;     (6.9.10) (ii) supp F(·) ⊗ G(·) = supp F(·) × supp G(·) .

Observation 6.9.2 If f is a distribution, then its support supp f is defined (as for functions) as the complement of the reunion of open sets on which the distribution f is null. The direct product of distributions has the usual commutative, associative and distributie properties (over the usual addition of distributions). In the sequel, we will discuss other properties of the direct product of distributions which are useful for applications. Proposition 6.9.3 The Dirac distribution satisfies the following property D

δ(x) ⊗ δ(y) = δ(x, y).

(6.9.11)

Proof By the definition of the direct product of distributions and by the definition of the Dirac distribution, we have (δ(x) ⊗ δ(y), ϕ(x, y)) = (δ(x), (δ(y), ϕ(x, y))) = = (δ(x), ϕ(x, 0)) = ϕ(0, 0). On the other hand, we have

(6.9.12)

120

6 Distributions

(δ(x, y), ϕ(x, y)) = ϕ(0, 0).

(6.9.13)

Note that for the last two equalities the expressions on the right-hand side are equal. Hence, it follows that the corresponding expressions on the left-hand side are equal, which finishes the proof.  Observation 6.9.3 We can generalise the result from Proposition 6.9.3 by mathematical induction D

δ(x1 ) ⊗ δ(x2 ) ⊗ · · · ⊗ δ(xn ) = δ(x1 , x2 , . . . , xn ).

(6.9.14)

In the following proposition we present a property of the derivative of the direct product of distributions, which differs from the corresponding property for the derivative of functions. Proposition 6.9.4 The mixed derivative of the direct product of distributions is equal to the direct product of derivatives, i.e. dG(y) D  ∂2 D  d F(x) ⊗ = F (x) ⊗ G  (y). (F(x) ⊗ G(y)) = ∂x∂ y dx dy

(6.9.15)

More generally, we have p ∂ p+q d q G(y) D  d F(x) = ⊗ , ∀ p, q ∈ N. ⊗ G(y)) (F(x) ∂x p ∂ y q dx p dy q

(6.9.16)

Proof In order to prove the first statement we use the definition of the direct product of distributions and the rule of differentiation for the distributions. Therefore, we obtain  2    ∂ ∂ 2 ϕ(x, y) = (F(x) ⊗ G(y)) , ϕ(x, y) = F(x) ⊗ G(y), ∂x∂ y ∂x∂ y       ∂ϕ(x, y) ∂ 2 ϕ(x, y) = F(x), − G  (y), = = F(x), G(y), ∂x∂ y ∂x       ∂ϕ(x, y) d F(x) = − G  (y), − , ϕ(x, y) = = − G  (y), − F(x), ∂x dx      d F(x) dG(y) dG(y) d F(x) , , ϕ(x, y) = ⊗ , ϕ(x, y) . = dy dx dx dy (6.9.17) The second statement of the proposition follows immediately by mathematical induction.  In the sequel, we present an application with practical importance of Proposition 6.9.3 and Proposition 6.9.4. More precisely, the direct product of distributions for the Heaviside distribution satisfies a property which is similar to the one satisfied by the Dirac distribution.

6.9 Convolutions

121

Proposition 6.9.5 The Heaviside distribution satisfies the following relation D

θ(x) ⊗ θ(y) = θ(x, y).

(6.9.18)

Proof We use the following relation satisfied by the Dirac distribution D

δ(x) ⊗ δ(y) = δ(x, y).

(6.9.19)

By using Proposition 6.9.4 and the fact that the derivative of the Heaviside distribution is the Dirac distribution, we obtain D

δ(x) ⊗ δ(y) =

dθ(x) dθ(y) D ∂ 2 = ⊗ (θ(x) ⊗ θ(y)) . dx dy ∂x∂ y

(6.9.20)

On the other hand, since the derivative of the Heaviside distribution is equal to the Dirac distribution, we have D

δ(x, y) =

∂2 θ(x, y). ∂x∂ y

(6.9.21)

Based on the relation (6.9.19), we deduce that the expressions on the left-hand side of the relations (6.9.20) and (6.9.21) are equal. Hence, it follows that the expressions on the right-hand side have to be equal, which proves the equality from the statement of the proposition.  Observation 6.9.4 We can generalise the result from Proposition 6.9.5 by mathematical induction D

θ(x1 ) ⊗ θ(x2 ) ⊗ · · · ⊗ θ(xn ) = θ(x1 , x2 , · · · , xn ).

(6.9.22)

6.9.2 Convolution of Distributions We recall the definition of the convolution of two functions. If f and g are integrable functions on R, then their convolution is defined as  f (y)g(x − y)dy = f (x) ∗ g(x). (6.9.23) ( f ∗ g)(x) = R

The convolution of two functions has the commutative, associative and distributive property (over the usual addition of functions). Moreover, if the convolution of two functions is null, then at least one of the functions is identically null. Moreover, if the functions f and g are continuous, then their convolution is a continuous function.

122

6 Distributions

In the sequel, we present conditions under which the convolution of two functions is well defined. Theorem 6.9.1 The convolution of two functions is well defined if one of the following conditions is satisfied: (i) the function f is integrable on R and g is bounded on R; (ii) supp f ⊂ [0, ∞), supp g ⊂ [0, ∞); (iii) supp f and supp g are compact sets. The criterion that is used the most for the existence of the convolution of two functions is item (ii). Therefore, we have the following result. Proposition 6.9.6 If supp f ⊂ [0, ∞) and supp g ⊂ [0, ∞), then their convolution has the form 

x

( f ∗ g)(x) = f (x) ∗ g(x) =



x

f (y)g(x − y)dy =

0

g(y) f (x − y)dy.

0

(6.9.24)

Proof The result follows from the additivity of the Riemann integral over the domain of integration. We have  ( f ∗ g)(x) =  =

0

−∞

∞ −∞

f (y)g(x − y)dy = 

x

f (y)g(x − y)dy +



∞

f (y)g(x − y)dy +

0

f (y)g(x − y)dy.

x

(6.9.25) The first integral on the right-hand side is null since y < 0. Then f (y) = 0 because supp f ⊂ [0, ∞). Similarly, the last integral on the right-hand side is null since y > x. This implies that x − y < 0 ⇒ g(x − y) = 0 since supp g ⊂ [0, ∞). Hence, we obtain the desired result.  We shall mention for future reference the following important results on the convolution of functions: (i) Let p, q and r be real numbers such that 1 ≤ p, q, r ≤ +∞ and r1 = If f ∈ L p (Rn ) and g ∈ L q (Rn ), then f ∗ g ∈ L r (Rn ) and || f ∗ g||r ≤ || f || p · ||g||q .

1 p

+

1 q

− 1.

(6.9.26)

As a consequence, we have the following. (ii) Let p = 1. For every f ∈ L 1 , the map g ∈ Lq → f ∗ g ∈ L p is a continuous linear one whose norm is ≤ || f ||.

(6.9.27)

6.9 Convolutions

123

(iii) The bilinear map L 1 × L 1  ( f, g) → f ∗ g ∈ L 1

(6.9.28)

is continuous and we have || f ∗ g||1 ≤ || f ||1 · ||g||1 . In the sequel, we define the convolution of distributions. Definition 6.9.1 If f and g are distributions on Rn , f, g ∈ D (),  ⊂ Rn , then their convolution is defined by 

 ( f (x) ∗ g(x), ϕ(x)) =

Rn

Rn

 f (y)g(x − y)dy ϕ(x)d x.

(6.9.29)

In the sequel, we present another form of the convolution of distributions. Proposition 6.9.7 The convolution of two distributions has the equivalent form  ( f (x) ∗ g(x), ϕ(x)) =

R2n

f (y)g(z)ϕ(y + z)dydz.

(6.9.30)

Proof In formula (6.9.29), we make the change of variables x − y = z with respect to the variable x and we consider that y is a parameter. We obtain 

 ( f ∗ g, ϕ) =

Rn

Rn

  f (y)g(z)dy ϕ(y + z)dz =

R2n

f (y)g(z)ϕ(y + z)dydz,

which ends the proof.

(6.9.31) 

We can easily prove some properties of the convolution of distributions by means of formulae (6.9.29) or (6.9.30). Proposition 6.9.8 The convolution of distributions has the following properties: D

(i) f ∗ g = g ∗ f ; D

(ii) ( f ∗ g) ∗ h = f ∗ (g ∗ h); D

(iii) f ∗ (g + h) = ( f ∗ g) + ( f ∗ h); D

D

D

(iv) f ∗ g = 0 ⇔ f ≡ 0 or g ≡ 0. In the following theorem we establish the relationship between the direct product of distributions and their convolution. Theorem 6.9.2 We have the following relation between the direct product of two distributions and their convolution ( f (x) ∗ g(x), ϕ(x)) = ( f (x) ⊗ g(y), ϕ(x + y)) .

(6.9.32)

124

6 Distributions

Proof For the left-hand side we use Proposition 6.9.7 and we obtain  ( f (x) ∗ g(x), ϕ(x)) =

f (y)g(z)ϕ(y + z)dydz.

R2n

(6.9.33)

For the right-hand side we have ( f (x) ⊗ g(y), ϕ(x + y)) = ( f (x), (g(y), ϕ(x + y))) =     f (x) (g(y), ϕ(x + y)) d x = f (x) g(y)ϕ(x + y)dy d x = = n Rn Rn R = f (x)g(y)ϕ(x + y)d xd y. R2n

(6.9.34)

Clearly, we can write this equality in the form  ( f (x) ⊗ g(y), ϕ(x + y)) =

R2n

f (y)g(z)ϕ(y + z)dydz.

(6.9.35)

By (6.9.33) and (6.9.35) we obtain the equality from the statement of the theorem.  By using the relationship established in Theorem 6.9.2 between the convolution of two distributions and their direct product, we can prove easily other properties of the convolution. Theorem 6.9.3 The convolution of distributions has the following properties: D

(i) δ ∗ f = f ; D

(ii) δ  ∗ f = f  ;

D

D

D

(iii) D p ( f ∗ g) = (D p f ) ∗ g = f ∗ (D p g) = D k1 f ∗ D k2 g, where k1 + k2 = p. Proof (i) This relation shows that the Dirac distribution is the unit element of the convolution product. By the result from Theorem 6.9.2 we have (δ(x) ∗ f (x), ϕ(x)) = (δ(x) ⊗ f (y), ϕ(x + y)) = = (δ(x), ( f (y), ϕ(x + y))) = ( f (y), (δ(x), ϕ(x + y))) =

(6.9.36)

D

= ( f (y), ϕ(y)) , ∀ϕ ⇒ δ ∗ f = f. (ii) We use the relation that gives us the expression of the convolution in terms of the direct product 



δ (x) ∗ f (x), ϕ(x) = δ  (x), ( f (y), ϕ(x + y)) =



= f (y), δ  (x), ϕ(x + y) = f (y), δ(x), −ϕ (x + y) =



= δ(y), f (x), −ϕ (y + x) = δ(y), f  (x), ϕ(y + x) =



= f  (x), (δ(y), ϕ(y + x)) = f  (x), ϕ(x) .

(6.9.37)

6.9 Convolutions

125

(iii) We use the definition of the derivative of a distribution and the relation that gives us the expression of the convolution in terms of the direct product in order to obtain p





D ( f ∗ g), ϕ = f ∗ g, (−1) p D p ϕ = (−1) p f ∗ g, D p ϕ =



= (−1) p f (x), g(y), D p ϕ(x + y) = f (x), D p g(y), ϕ(x + y) =

D = f (x) ∗ D p g(x), ϕ(x) , ∀ϕ ⇒ D P ( f ∗ g) = (D p f ) ∗ g. (6.9.38) We can obtain the other equalities from (iii) by the commutative property of the convolution.  Proposition 6.9.9 Let f, g ∈ D (Rn ) and suppose that at least one has compact support. We have supp ( f ∗ g) ⊂ supp f + supp g. (6.9.39) Proof Let A = supp f and B = supp g. Since A and B are closed and at least one is compact, the set A + B = {x + y|x ∈ A, y ∈ B} (6.9.40) is closed. Let us show that f ∗ g is equal to zero on the open set  = (A + B)c . Indeed, if ϕ ∈ Cc∞ (), the support of ϕ(x + y) is contained in the open set {(x, y) ∈ Rn × Rn |x + y ∈ }.

(6.9.41)

On the other hand, the support of f ⊗ g is, as we have seen, A × B. Since (x, y) ∈ A × B implies x + y ∈ A + B, the support of f ⊗ g does not intersect the support of ϕ(x + y). Consequently, we have ( f (x) ∗ g(x), ϕ(x)) = ( f (x) ⊗ g(y), ϕ(x + y)) , for all ϕ ∈ Cc∞ ().

(6.9.42) 

6.9.3 Convolution of Functions and Distributions: Regularization Let ϕ ∈ Cc∞ (Rn ) and let f ∈ D (Rn ). Then the convolution of f and ϕ is given by the formula ( f ∗ ϕ)(x) = ( f (y), ϕ(x − y)) . (6.9.43) In fact, for every ψ ∈ Cc∞ (Rn ) we have

126

6 Distributions

(( f ∗ ϕ)(x), ψ(x)) = ( f (x) ⊗ φ(y), ψ(x + y)) = = ( f (y), (φ(y), ψ(x + y))) .  φ(y)ψ(x + y)dy = (φ(y), ψ(x + y)) = Rn  = φ(ξ − x)ψ(ξ)dξ = (φ(ξ − x), ψ(ξ)) .

But

(6.9.44)

(6.9.45)

Rn

Making the replacement, we get (( f ∗ φ)(x), ψ(x)) = ( f (y), (φ(x − y), ψ(x))) = = (( f (y), φ(x − y)) , ψ(x)) .

(6.9.46)

˜ φ(x) = φ(−x).

(6.9.47)

Let us set

Using this notation, (6.9.43) can be written as ( f ∗ φ) (x) = We also have ( f, φ) =





f, τx φ˜ .

f ∗ φ˜ (0).

(6.9.48)

(6.9.49)

Theorem 6.9.4 Let φ ∈ Cc∞ (Rn ) and let f ∈ D (Rn ). Then (i) ( f ∗ φ)(x) ∈ C ∞ (Rn ). (ii) The bilinear map (φ, f ) → f ∗ φ

(6.9.50)

from Cc∞ (Rn ) × D (Rn ) into C ∞ (Rn ) is separately continuous. Proof (i) Since ( f ∗ φ) (x) = ( f (y), φ(x − y))

(6.9.51)

then it follows that ( f ∗ φ)(x) is a continuous function of x. Taking derivatives, we get that, for all p = ( p1 , . . . , pn ), ∂ p ( f ∗ φ) = f ∗ ∂ p φ

(6.9.52)

is a continuous function. Therefore, f ∗ φ ∈ C ∞ . (ii) To show that the map φ ∈ Cc∞ → f ∗ φ ∈ C ∞

(6.9.53)

6.9 Convolutions

127

is a continuous one, it suffices to show that for every compact subset K ⊂ Rn , the map (6.9.54) φ ∈ Cc∞ (Rn ; K ) → f ∗ φ ∈ C ∞ (Rn ) is a continuous one. In other words, for all p ∈ Nn , for all L being a compact subset of Rn , there is a constant C > 0 and an integer m > 0 such that sup |∂ p ( f ∗ φ) (x)| ≤ C x∈L

sup

|q|≤m,z∈K

|∂ q φ(z)|.

(6.9.55)



But when x ∈ L and y ∈ K , it is clear that τ y φ (x) = φ(x − y) belongs to Cc∞ (Rn ; L − K ). Since f ∈ D , there is a constant C > 0 and an integer l ≥ 0 such that |( f, ψ)| ≤ C

sup

|r |≤l,y∈Rn



|∂ r ψ(y)|, ∀ψ ∈ Cc∞ Rn ; L − K .

(6.9.56)

Since ∂ p ( f ∗ φ)(x) = ( f ∗ ∂ p φ) (x) = ( f (y), ∂ p φ(x − y)), we get sup |∂ p ( f ∗ φ)(x)| = sup | f (y), ∂ p φ(x − y)| ≤ x∈L

x∈L

≤ C sup x∈L

sup

|r |≤l,y∈Rn

|∂ ry ∂xp φ(x − y)| ≤ C

sup

z∈Rn ,|q|≤l+| p|

|∂ q φ(z)|,

(6.9.57)

as we wanted to prove. (iii) Let φ be a fixed element of Cc∞ (Rn ). It is easy to see that if x belongs to a ˜ ∈ K } is bounded in C ∞ (Rn ). compact subset K of Rn the set of functions {τx φ|x c Hence, if the distributions f j converge strongly to zero, then



f j ∗ φ (x) = f j , τx φ˜

(6.9.58)

converge to zero uniformly on K . By taking derivatives, we also get that for all p ∈ Nn the functions



(6.9.59) ∂ p f j ∗ φ (x) = f j ∗ ∂ p φ (x) converge to zero uniformly on K an arbitrary compact subset of Rn . Therefore  f j ∗ φ → 0 in C ∞ (Rn ). Regularization of distribution Let 1  x , αε (x) = n α − ε ε

(6.9.60)

where α is the test function defined in Sect. 1. We have the following lemma.

128

6 Distributions

Lemma 6.9.1 If ψ ∈ Cc∞ (Rn ), the functions αε ∗ ψ of Cc∞ (Rn ) converge to ψ in Cc∞ (Rn ) when ε → 0. Moreover, the convergence is uniform on bounded sets of Cc∞ (Rn ). Proof Let B be a bounded set in Cc∞ (Rn ). There is a compact subset L of Rn such that supp ψ ⊂ L, ∀ψ ∈ B and such that B is a bounded subset of Cc∞ (Rn ; L). On the other hand, there is a compact subset K of Rn such that L ⊂ K and such that supp (αε ∗ ψ) ⊂ K

(6.9.61)

for all ψ ∈ B and 0 < ε < 1. Since, by assumption, B is bounded in Cc∞ (Rn ), we can prove that for every p = ( p1 , . . . , pn ) the set (6.9.62) ∂ p B = {∂ p ψ|ψ ∈ B} is uniformly continuous on K . Hence, given m a nonnegative integer and σ > 0 a real number, there is δ > 0 such that |∂ p ψ(x − y) − ∂ p ψ(x)| ≤ 0

(6.9.63)

∀|y| ≤ δ, ∀ψ ∈ B, ∀x ∈ K and ∀| p| ≤ m. Consider now the neighbourhood of zero W (m, σ) = {φ ∈ Cc∞ (Rn ; K )||∂ p φ(x)| ≤ σ, ∀x ∈ K , ∀| p| ≤ m}

(6.9.64)

in Cc∞ (Rn ; K ). We can write

αε ∗ ∂ p ψ − ∂ p ψ (x) =

 Rn



αε (y) ∂ p ψ(x − y) − ∂ p ψ(x) dy.

(6.9.65)

Let ε0 be small enough so that for all ε < ε0 the function αε has its support contained in the ball of center at the origin and of radius δ. It follows that

| αε ∗ ∂ p ψ − ∂ p ψ (x)| ≤

 Rn



αε (y)| ∂ p ψ(x − y) − ∂ p ψ(x) |dy ≤ σ (6.9.66)

for all ψ ∈ B, | p| ≤ m and ε ≤ ε0 . In other words, αε ∗ ψ − ψ ∈ W (m, σ), ∀ε ≤ ε0 , ∀ψ ∈ B,

(6.9.67)

which proves that αε ∗ ψ → ψ when ε → 0, uniformly on bounded sets of Cc∞ (Rn ).  Observation 6.9.5 Lemma 6.9.1 holds true if we replace Cc∞ (Rn ) by C ∞ (Rn ). Theorem 6.9.5 If f ∈ D (Rn ), the functions f ∗ αε of C ∞ (Rn ) converge strongly to f in D (Rn ) when ε → 0.

6.9 Convolutions

129

Proof Let f be a fixed element of D (Rn ). We have   ( f ∗ αε − f, ψ) = ( f ∗ αε ) ∗ ψ˜ − f ∗ ψ˜ (0) =     ˜ − f ∗ ψ˜ (0) = f ∗ (αε ∗ ψ˜ − ψ) ˜ (0) = = f ∗ (αε ∗ ψ)   = f, α˜ ε ∗ ψ −ψ .

(6.9.68)

By Lemma 6.9.1, α˜ ∗ ψ − ψ → 0 uniformly when ψ belongs to a bounded subset of Cc∞ (Rn ), as ε → 0. This implies that 

 f, α˜ ε ∗ ψ −ψ →0

and therefore that f ∗ αε converges strongly to f in D (Rn ) when ε → 0.

(6.9.69) 

Corollary 6.9.1 The family of functions αε converges to the Dirac measure δ in the strong topology of D (Rn ). Proof It suffices to apply Theorem 6.9.5 to f = δ, recalling that δ is the unit element with respect to the convolution product.  Observation 6.9.6 We called (αε ) a regularizing family of functions and α j (x) = j n α( j x), j = 1, 2, . . .

(6.9.70)

a regularizing sequence of functions. We proved that integrable functions can be approximated in suitable topologies by smooth functions. Theorem 6.9.5 extends this result to distributions. It allows us to replace a distribution (which in general has singularities) by a converging family (or sequence) of functions.

6.9.4 Convolution Maps Let f ∈ D (Rn ). The continuous linear map L f : φ ∈ Cc∞ (Rn ) → f ∗ φ ∈ C ∞ (Rn )

(6.9.71)

is said to be a convolution map. To every f ∈ D (Rn ) we can then associate a convolution map. The following theorem characterizes such maps. Theorem 6.9.6 Let f ∈ D (Rn ). The convolution map L f is a continuous linear map from Cc∞ (Rn ) into C ∞ (Rn ) which commutes with translations. Conversely, if L : Cc∞ (Rn ) → C ∞ (Rn ) is a continuous linear map such that L ◦ τh = τh ◦ L ,

(6.9.72)

130

6 Distributions

for every h ∈ Rn , there is a unique f ∈ D (Rn ) such that L(φ) = f ∗ φ, ∀φ ∈ Cc∞ (Rn ).

(6.9.73)

Proof Let f ∈ D (Rn ). We have [ f ∗ (τh φ)] (x) = ( f (y), (τh φ)(x − y)) = ( f (y), φ(x − y − h)) = = ( f ∗ φ)(x − h) = [τh ( f ∗ φ)] (x),

(6.9.74)

which proves that convolution maps commute with translations. Conversely, suppose that L : Cc∞ (Rn ) → C ∞ (Rn ) is a continuous linear map commuting with translations. It is easy to see that the map φ ∈ Cc∞ (Rn ) → (Lφ)(0) ∈ C

(6.9.75)

defines a continous linear functional on Cc∞ (Rn ). Hence, there is a unique f ∈ D (Rn ) such that  (Lφ)(0) = f, φ˜ . (6.9.76) We get (Lφ)(0) = ( f ∗ φ)(0).

(6.9.77)

Since L commutes with translations, we have for every x ∈ Rn       (Lφ)(x) = τ−x (Lφ) (0) = L(τ−x φ) (0) = f ∗ (τ−x φ) (0) = = ( f (y), (τ−x φ) (−y)) = ( f (y), φ(x − y)) = ( f ∗ φ)(x),

(6.9.78)

which shows that L is a convolution map.

References 1. H.W. Alt, Lineare Funktionalanalysis (Springer, Berlin, Heidelberg, 2006) 2. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, 2011) 3. E. DiBenedetto, Real Analysis (Birkhäuser, Boston, 2002) 4. J.J. Duistermaat, J.A.C. Kolk, Distributions: Theory and Applications (Birkhäuser, 2010) 5. F.G. Friedlander, M. Joshi, Introduction to the Theory of Distributions, 2nd edn. (Cambridge University Press, 1999) 6. D.D. Haroske, H. Triebel, Distributions, Sobolev Spaces, Elliptic Equations (European Mathematical Society, 2008) 7. L. Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, 2nd edn. (Springer, 1990) 8. F.W. King, Hilbert Transforms (Cambridge University Press, Cambridge, 2009)

References

131

9. D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis (Springer, 2013) 10. T. Precupanu, Spa¸tii liniare topologice s¸i elemente de analiz˘a convex˘a, Ed. Acad. Române, Bucure¸sti (1992) 11. I. Richards, H. Youn, Theory of Distributions: A Non-technical Introduction (Cambridge University Press, 1995) 12. R. Sikorski, The elementary theory of distributions. Bull. Acad. Pol., Warszawa (1957) 13. R. Strichartz, A Guide to Distribution Theory and Fourier Transforms (CRC Press, 1994) 14. L. Schwartz, Theorie des Distributions (Herman & Cie, Paris, 1971) 15. Z. Vulih, Introduction to the Theory of Partially Ordered Spaces (Woltors–Noordhoff, Groningen, 1967) 16. J. Yeh, Real Analysis. Theory of Measure and Integration (World Scientific, New Jersey, London, Singapore, 2006)

Chapter 7

Tempered Distributions

Abstract First of all, we introduce the Schwartz space of infinitely differentiable functions that are rapidly decreasing at infinity. Then we define tempered distributions. We consider the Fourier transform and its properties. Finally, the Paley-WienerSchwartz theorem is discussed.

7.1 The Schwartz Space of Infinitely Differentiable Functions Rapidly Decreasing at Infinity Definition 7.1.1 A function φ ∈ C ∞ (Rn ) is rapidly decreasing at infinity if   lim x α ∂ p φ(x) = 0

|x|→+∞

(7.1.1)

for all multi-indices α ∈ Nn and all p ∈ Nn . The set of all smooth functions rapidly decreasing at infinity is a vector space over C denoted by S(Rn ) and it is called the Schwartz space of functions [8]. Observation 7.1.1 The condition (7.1.1) is equivalent to the following conditions (i) for all α, p ∈ Nn , x α ∂ p φ(x) is uniformly bounded on Rn ; (ii) for all integer k ≥ 0 and for all p ∈ Nn , (1 + r 2 )k/2 ∂ p φ(x) is uniformly bounded on Rn , where r = |x|. Example 1 The space Cc∞ (Rn ) is a vector subspace of S(Rn ). |x|2

Example 2 The function e− 2 belongs to S(Rn ). Example 3 Let α ∈ Cc∞ (Rn ) be such that 0 ≤ α ≤ 1, supp α ⊂ B1 and α(0) = 1. Let (x j ) be a sequence of elements of Rn such that |x j | + 2 ≤ |x j+1 |. We introduce the function ∞  α(x − x j ) γ (x) = . (7.1.2) (1 + |x j |2 ) j j=1

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_7

133

134

7 Tempered Distributions

Note that the supports of the functions α(x − x j ) are disjoint. Therefore, the sum is well defined. Let k ∈ N and p ∈ Nn . Then we obtain (1 + |x|2 )k ∂ p γ (x) =

(1 + |x|2 )k ∂ p α(x − x j ) , (1 + |x j |2 )k (1 + |x j |2 ) j−k

(7.1.3)

2

1+|x| for |x j | − 1 ≤ |x| ≤ |x j | + 1. Moreover, we have 1+|x 2 ≤ C, with a suitable conj| stant and     (7.1.4) sup ∂ p α(x − x j ) = sup ∂ p α(x) . x∈Rn

x∈Rn

Hence, we deduce that γ ∈ S(Rn ). Observation 7.1.2 If a ∈ C ∞ and φ ∈ S, then the function a(x)φ(x) does not necessarily belong to S. Example If a(x) = e|x| and φ(x) = e−|x| , then their product 2

2

a(x)φ(x) = e|x| e−|x| = 1 ∈ / S. 2

2

(7.1.5)

But if the function a ∈ C(Rn ) and all its derivatives of any order do not grow too much as x → ∞ (less than a polynomial), then a(x)φ(x) ∈ S. Proposition 7.1.1 If P(x) is a polynomial with constant coefficients and Q(∂) is a partial differential operator with constant coefficients, then the following conditions are equivalent: (1) φ ∈ S; (2) ∀P(x) and ∀Q(∂), P(x)Q(∂)φ ∈ S; (3) ∀P(x) and Q(∂), ∀Q(∂) (P(x)φ(x)) ∈ S. Proof Note that P(x)Q(∂)φ can be written as a linear combination of terms of the form described in the Observation 7.1.1. Hence, Condition (1) clearly implies Condition (2). It is obvious that Condition (2) implies Condition (1). By the Leibniz formula, we deduce that Q(∂) (P(x)φ(x)) is a linear combination of terms of the form described in the Observation 7.1.1. Therefore, Condition (1) implies Condition (3).  We introduce on S(Rn ) the following seminorms   γα, p (φ) = sup x α ∂ p φ(x) , α, p ∈ Nn .

(7.1.6)

x∈Rn

  The family of seminorms γα, p is countable and defines a Hausdorff locally convex topology on S(Rn ) [8]. We can show that this is a metrizable and complete topology. Therefore, S(Rn ) is a Frechet space. Note that we can also use the sequence of seminorms

7.1 The Schwartz Space of Infinitely Differentiable Functions …

  γk,m (φ) = sup sup (1 + r 2 )k/2 ∂ p φ(x) , k, m ∈ N | p|≤m x∈Rn

135

(7.1.7)

in order to define the topology of S(Rn ). Proposition 7.1.2 S is a Montel space. Proof Let B be a bounded set of S. Since the embedding S → C ∞ is continuous, it follows that B is a bounded set in C ∞ . Recall that C ∞ is a Montel space. Therefore, B is a relatively compact set in C ∞ . We want to show that B is relatively compact in S. It suffices to consider a sequence of elements of B, (φ j ) and show that if φ j → φ in C ∞ , then φ ∈ S and φ j → φ in S. Step 1. We show that φ ∈ S. Note that B is bounded in S, which implies that for all k ∈ N and for all p ∈ Nn , there is a constant Ck, p such that   sup (1 + r 2 )k/2 ∂ p f (x) ≤ Ck, p , f ∈ B.

(7.1.8)

x∈Rn

From this inequality it follows that, given ε > 0, there is M > 0 constant such that   (1 + r 2 )k/2 ∂ p f (x) ≤ ε, r = |x| > M, f ∈ B.

(7.1.9)

Since φ j → φ in C ∞ , we deduce from the last inequality that   (1 + r 2 )k/2 ∂ p φ(x) ≤ ε, r > M.

(7.1.10)

Therefore, we have φ ∈ S. Step 2. We show that φ j → φ in S.   We have φ j → φ in C ∞ , which implies that ∂ p φ j converges uniformly to ∂ p φ on the compact set {x ∈ Rn ||x| ≤ M}. Then it follows that, given ε > 0, we can find an integer j0 such that   (1 + r 2 )k/2 ∂ p φ j (x) − ∂ p φ(x) ≤ ε,

(7.1.11)

for all |x| ≤ M and for all j ≥ j0 . From the last three inequalities it follows that   sup (1 + r 2 )k/2 ∂ p φ j (x) − ∂ p φ(x) ≤ ε,

(7.1.12)

x∈Rn

for all j ≥ j0 . Hence, we have φ j → φ in S.



Proposition 7.1.3 S is a reflexive space. This is a direct consequence of the Proposition 7.1.2. Theorem 7.1.1 Let (φ j ) be a sequence in S. Then φ j → 0 in S if and only if one of the following equivalent conditions holds true:

136

7 Tempered Distributions

1. For all α, p ∈ Nn

x α ∂ p φ j (x) → 0

(7.1.13)

uniformly on Rn as j → ∞. 2. For all polynomials P(x) with constant coefficients and for all partial differential operators with constant coefficients Q(∂), P(x)Q(∂)φ j → 0

(7.1.14)

uniformly on Rn as j → ∞. 3. For all P(x) and all Q(∂) as before,   Q(∂) P(x)φ j (x) → 0

(7.1.15)

uniformly on Rn as j → ∞. The proof follows easily from the definition of the seminorms γα, p and Proposition 7.1.1. We can define a metric on S as follows d(φ, ψ) =

 p,α∈Nn

2−| p|−|α|

γα, p (φ − ψ) . 1 + γα, p (φ − ψ)

(7.1.16)

Note that the sum converges since 

2−|α| =

···

α1 =0

α∈Nn



∞ 

2

−| p|−|α|

∞  αn =0

2−α1 · · · 2−αn =

n  ∞ 

2−α j = 2n ,

j=1 α j =0

(7.1.17)

=4 . n

p,α∈Nn

Theorem 7.1.2 The space S is complete in the metric d(φ, ψ). Theorem 7.1.3 The following inclusions hold true Cc (Rn ) ⊂ S(Rn ) ⊂ C ∞ (Rn )

(7.1.18)

and the embeddings are continuous. Furthermore, Cc (Rn ) is a dense subspace of S(Rn ) and S(Rn ) is a dense subspace of C ∞ (Rn ). Proof Note that Cc∞ (Rn ) is dense in C ∞ (Rn ), which implies that S(Rn ) is dense in C ∞ (Rn ). Let β j ∈ Cc∞ (Rn ) be such that β j = 1 on the closed  ball with center at the origin and radius j. Let φ ∈ S(Rn ). Therefore, the sequence β j φ of functions of Cc∞ (Rn ) converges to φ in S(Rn ).

7.1 The Schwartz Space of Infinitely Differentiable Functions …

137

Step 1. We want to show that the identity map from Cc∞ (Rn ) into S(Rn ) is continuous. It suffices to prove that for every compact subset K of Rn the identity map from Cc∞ (Rn ; K ) into S(Rn ) is a continuous one. Note that they are both metrizable spaces. Hence, it suffices to prove that every sequence (φ j ) converging to zero in Cc∞ (Rn ; K ) converges to zero in S(Rn ). If φ j → 0 in Cc∞ (Rn ; K ) as j → ∞, then for all p ∈ Nn ∂ p φ j (x) → 0 uniformly on K as j → ∞.

(7.1.19)

Since supp φ j ⊂ K , ∀ j, this implies that for all α, p ∈ Nn x α ∂ p φ j → 0 uniformly on Rn as j → ∞.

(7.1.20)

Step 2. Assume that (φ j ) is a sequence such that φ j → 0 in S(Rn ) as j → ∞. Then for all p ∈ Nn ∂ p φ j → 0 uniformly on Rn as j → ∞.

(7.1.21)

In particular, the sequence converges to zero uniformly on every compact subset of  Rn . Note that the space S is larger than the space D and it extends D by characterizing the behaviour at infinity of the functions φ. 2 2 2 Example 1 The function φ(x) = e−(x1 +x2 +···+xn ) , x ∈ Rn belongs to S, but it does not belong to D. Example 2 The function φ(x) = e x , x ∈ R does not belong to S since φ(x)  0 as x → ∞. D S The convergence in D implies the convergence in S, i.e. φk → φ ⇒ φk → φ. Note that S(Rn ) ⊂ D (Rn ) and the embedding is continuous. Then S(Rn ) is a normal space of distributions. Therefore, its dual S (Rn ) is a subspace of D (Rn ).

7.2 Tempered Distributions Definition 7.2.1 The elements of S (Rn ) are called tempered distributions [6]. Hence a tempered distribution is a linear and continuous functional defined on S [2–4, 7, 9, 10] f : S → R, f (φ) = ( f, φ) . (7.2.1) Therefore, if f is a tempered distribution, then 1. ∀φ ∈ S, f (φ) ∈ R; 2. ∀α, β ∈ R and ∀φ, ψ ∈ S ⇒ ( f, αφ + βψ) = α ( f, φ) + β ( f, ψ);

138

7 Tempered Distributions S

3. f is a continuous functional on S, i.e. if φk → φ as k → ∞, then ( f, φk ) → ( f, φ) as k → ∞, ∀φ ∈ S. S

The convergence in S implies the convergence in D . Indeed, if f k → f as k → D

∞, then ( f k , φ) → ( f, φ) as k → ∞, ∀φ ∈ D ⊂ S ⇒ f k → f as k → ∞. Example 1 If f ∈ S , then D p f ∈ S , ∀ p ∈ N since (D p f, φ) = (−1)| p| ( f, D p φ), ∀φ ∈ S. Example 2 If f ∈ S and det A = 0, then f (Ax + b) ∈ S since  ( f (Ax + b), φ) =

 φ(A−1 (x − b)) . f, det |A|

(7.2.2)

Example 3 If f ∈ S (Rn ) and g ∈ S (Rm ), then f (x) ⊗ g(x) ∈ S (Rn+m ), where f (x) ⊗ g(y), φ(x, y)) = ( f (x), (g(y), φ(x, y))). ( Example 4 If f ∈ S and it has bounded support, then f ∗ g ∈ S and ( f ∗ g, φ) = ( f (x) ⊗ g(y), η(y)φ(x + y)), where η is a function from D which is equal to 1 in a neighbourhood V (supp g). Example 5 By Theorem 7.1.3, we infer that every distribution with compact support is a tempered one. Moreover, the space of all distributions with compact support is a subspace of S (Rn ). To be more precise, we have the continuous embeddings E (Rn ) → S (Rn ) → D (Rn ) in the sense of the strong topologies. Example 6 Let us consider a function f ∈ L p (Rn ), 1 ≤ p ≤ ∞. Then we can introduce a tempered distribution by means of ( f, φ) =

f · φd x, ∀φ ∈ S(Rn ).

Rn

(7.2.3)

Clearly, it follows from Hölder’s inequality that |( f, φ)| ≤ || f || p · ||φ||q ,

(7.2.4)

where 1p + q1 = 1. Note that if a sequence φ j → 0 in S as j → ∞, then φ j → 0 in L q as j → ∞. Hence every f ∈ L p defines a continuous linear functional on S. Furthermore, note that L p can be identified with a vector subspace of S . Example 7 Let P(x) be a polynomial with constant coefficients. Then it defines a tempered distribution. Clearly, we can set α

(x , φ) =

Rn

x α φ(x)d x, ∀φ ∈ S

(7.2.5)

and then prove that every monomial x α defines a tempered distribution. Definition 7.2.2 Let f be a continuous function. Then f (x) is slowly increasing at infinity if ∃k ∈ N such that (1 + r 2 )−k/2 f (x) is bounded in Rn or if ∃P(x) a polynomial such that | f (x)| ≤ |P(x)|, x ∈ Rn .

7.2 Tempered Distributions

139

Example 8 Let f be a continuous function which is slowly increasing at infinity. Then it defines a tempered distribution. Clearly, we can set ( f, φ) =

Rn

f (x)φ(x)d x, ∀φ ∈ S.

(7.2.6)

It follows that |( f, φ)| ≤ | f (x)φ(x)| d x = Rn   (1 + r 2 )−k/2 f (x) · (1 + r 2 )k/2 φ(x) d x ≤ = n R   (1 + r 2 )k/2 φ(x) d x. ≤C

(7.2.7)

Rn

Note that if a sequence φ j → 0 in S as j → ∞ then, for every k ≥ 0, (1 + r 2 )k/2 φ j → 0 in L 1 as j → ∞. Hence, f defines an element of S . Example 9 Let f be a continuous function which is slowly increasing at infinity. Then any of its derivatives (in the sense of distributions) defines a tempered distribution. We consider T = ∂ α f and introduce (T, φ) = (−1)

|α|

Rn

f (x) · ∂ α φ(x)d x, ∀φ ∈ S.

(7.2.8)

It follows that |(T, φ)| ≤ | f (x)||∂ α φ(x)|d x = n R    (1 + r 2 )−k/2 f (x)(1 + r 2 )k/2 ∂ α φ(x) d x ≤ = Rn   (1 + r 2 )k/2 ∂ α φ(x) d x. ≤C

(7.2.9)

Rn

Note that if a sequence φ j → 0 in S as j → ∞, then for every k ≥ 0 and every α ∈ Nn , the sequence (1 + r 2 )k/2 ∂ α φ j → 0 in L 1 as j → ∞. Therefore, T = ∂ α f defines a tempered distribution. On the other hand, we can show that every tempered distribution is the derivative of a continuous function slowly increasing at infinity. In fact, the name tempered given to the distributions belonging to S is motivated by precisely this property. Theorem 7.2.1 The necessary and sufficient condition for the linear functional f : S → R to be continuous ( f ∈ S ) is that there exist C > 0 and α ≥ 0, α ∈ Z+ such that |( f, φ)| ≤ C||φ||α , (7.2.10) where

140

7 Tempered Distributions

||φ||α =

  sup (1 + |x|)α  D p φ(x) .

(7.2.11)

p≤α,x∈Rn

Proof Necessity. We will show that ∃C > 0, α ≥ 0 such that ∀φ ∈ S the condition (7.2.10) holds true. The proof is by contradiction. We assume that there are no functions C and α which satisfy this condition and that there is a sequence (φk ) ∈ S such that (7.2.12) ( f, φk ) ≥ k||φk ||k . The sequence

φk (x) , k = 1, 2, . . . ψk = √ k||φk ||k

(7.2.13)

converges to zero in S because for |k| > | p| and k ≥ β we have    x β D p φk (x)  β p 1 x D ψk (x) = √ ≤√ . k||φk ||k k

(7.2.14)

This result and the continuity of f on S imply that ( f, ψk ) → 0 as k → ∞. On the other hand, the inequality (7.2.12) leads us to √ 1 |( f, φk )| ≥ k, |( f, ψk )| = √ k||φk ||k

(7.2.15)

which is a contradiction. Sufficiency. Let f : S → R be a linear functional that satisfies the condition S

(7.2.10) for C > 0, α ≥ 0. We will show that f ∈ S . Let φk → 0 as k → ∞. Then  ||φk ||α → 0 as k → ∞ and |( f, φk )| ≤ C||φk ||α .

7.3 The Fourier Transform in S(Rn ) Definition 7.3.1 Let f ∈ S(Rn ) be a function. Then the Fourier transform of f is defined by the integral e−ix,ξ  f (x)d x, (7.3.1) fˆ(ξ ) = Rn

where x, ξ  = x1 ξ1 + · · · + xn ξn . Note that the integral (7.3.1) is absolutely convergent since f ∈ S. We will denote the Fourier transform of f by F f . Example The Fourier transform of the function e− is equal to the function (2π )n/2 e−

|ξ |2 2

.

|x|2 2

(7.3.2)

7.3 The Fourier Transform in S (Rn )

141

Proposition 7.3.1 We consider that f ∈ S(Rn ). Then the Fourier transform of the ∂ fˆ . product i x j f is equal to the partial derivative − ∂ξ j Proof We differentiate both sides of (7.3.1) with respect to ξ j . Note that we can move the derivative inside the integral sign. Hence, we obtain  ∂ fˆ(ξ ) ∂  −ix,ξ  e = f (x) d x = ∂ξ j Rn ∂ξ j   =− e−ix,ξ  i x j f (x) d x = −i xj f .

(7.3.3)

Rn

 Proposition 7.3.2 The Fourier transform of the partial derivative product iξ j fˆ.

∂f ∂x j

is equal to the

Proof We integrate by parts. Hence, we obtain

∂f ∂f (ξ ) = e−ix,ξ  (x)d x = n ∂x j ∂ xj R = iξ j e−ix,ξ  f (x)d x = iξ j fˆ.

(7.3.4)

Rn

 We introduce the notation Dj =

1 ∂ j , 1 ≤ j ≤ n. i

(7.3.5)

Then by Propositions 7.3.1 and 7.3.2 we obtain  x j f = −D j fˆ and  D j f = ξ j fˆ.

(7.3.6)

More generally, for α = (α1 , . . . , αn ), we obtain  x α f = (−D)α fˆ and  D α f = ξ α fˆ.

(7.3.7)

Moreover, for the partial differential operator with constant coefficients P(D) = it follows that



ap D p,

f = P(ξ ) · fˆ(ξ ). P(D)

(7.3.8)

(7.3.9)

142

7 Tempered Distributions

Theorem 7.3.1 The Fourier transform defines a continuous linear map from S(Rn ) into S(Rn ). Proof Step 1. We show that fˆ ∈ S(Rn ). The results in Propositions 7.3.1 and 7.3.2 imply that   α p ˆ e−ix,ξ  D α (−x) p f (x) d x. (7.3.10) ξ D f (ξ ) = Rn

By Proposition 7.1.1, D α ((−x) p f (x)) belongs to S(Rn ). Therefore,    (1 + r 2 )k  D α (−x) p f (x) 

(7.3.11)

is uniformly bounded in Rn , ∀k ∈ N. By choosing k such that Rn

1 d x = C < ∞, (1 + r 2 )k

(7.3.12)

we are led to the following inequality       α p ˆ   D α x p f (x)  d x = ξ D f (ξ ) ≤ Rn    1 = (1 + r 2 )k  D α x p f (x)  d x ≤ 2 k Rn (1 + r )    ≤ C sup (1 + r 2 )k  D α x p f (x)  d x.

(7.3.13)

x∈Rn

This shows that ∀α and ∀ p, ξ α D p fˆ(ξ ) is uniformly bounded in Rn . Therefore, we have fˆ ∈ S(Rn ). Step 2. We prove continuity of the linear map. Taking into account Theorem 7.1.1, then from the same inequality it follows that if the sequence f j → 0 in S(Rn ), then the sequence of Fourier transforms fˆj → 0 in S(Rn ). Hence, F : S(Rn ) → S(Rn )

(7.3.14) 

is a continuous linear map.

7.3.1 The Inverse Fourier Transform Let g ∈ S(Rn ). Then the integral (2π )

−n



eix,ξ  g(x)d x Rn

(7.3.15)

7.3 The Fourier Transform in S (Rn )

143

is absolutely convergent and defines a function of the variables (ξ1 , . . . , ξn ) which belongs to S(Rn ). If we denote by 

 F −1 g (ξ )

(7.3.16)

the integral (7.3.15), then we can show, as in Theorem 7.3.1, that F −1 defines a continuous linear map from S(Rn ) into S(Rn ). We can easily check the following relations Fg = (2π )n F −1 g¯ and F g¯ = (2π )n F −1 g.

(7.3.17)

Theorem 7.3.2 (Fourier’s inversion formula) We have f = F −1 (F f ), ∀ f ∈ S.

(7.3.18)

Proof The following relation follows from Fubini’s theorem [1]  g(ξ )eix,ξ  e−iy,ξ  f (y)dy dξ = fˆ(ξ )g(ξ )eix,ξ  dξ = Rn Rn Rn = eix−y,ξ  g(ξ ) f (y)dξ dy = g(y ˆ − x) f (y)dy,



Rn

Rn

(7.3.19)

Rn

∀ f, g ∈ S(Rn ). The relation (7.3.19) can be rewritten by a change of variables as follows Rn

fˆ(ξ )g(ξ )eix,ξ  dξ =

Rn

g(y) ˆ f (x + y)dy. gˆ

y . Then we do ε

g(y) ˆ f (x + εy)dy.

(7.3.21)

We replace g(ξ ) by g(εξ ), whose Fourier transform is equal to a change of variables and we obtain Rn

fˆ(ξ )g(εξ )eix,ξ  dξ =

(7.3.20)

1 εn

Rn

In the last relation, we let ε → 0 and we get g(0) Rn

Let g(x) = e−

|x|2 2

fˆ(ξ )eix,ξ  dξ = f (x)

Rn

g(y)dy. ˆ

(7.3.22)

. Note that g(y) ˆ = (2π ) 2 e− n

and that (Gauss integral)

|y|2 2

(7.3.23)

144

7 Tempered Distributions



e−

|x|2 2

Rn

n

d x = (2π ) 2 .

Hence, it follows that (2π )n f (x) =

Rn

(7.3.24)

eix,ξ  fˆ(ξ )dξ.

(7.3.25) 

The property below follows from Theorems 7.3.1 and 7.3.2. Corollary 7.3.1 The Fourier transform F defines a topological isomorphism from S(Rn ) onto S(Rn ).

7.3.2 Properties of the Fourier Transform Property 1 Let f, g ∈ S(Rn ). Then we have Rn

fˆ · g =

Rn

f · g. ˆ

(7.3.26)

Clearly, it suffices to set x = 0 in (7.3.20).



Property 2 (Parseval’s formula) [10] Let f, g ∈ S(Rn ). Then we have Rn

f · g¯ = (2π )−n

Rn

¯ˆ fˆg.

(7.3.27)

Clearly, we have to employ the Property 1, the relations (7.3.17) and Fourier’s inversion formula. Then it follows that F f · Fg = f · F(Fg) = (2π )n f · F −1 (Fg) = (2π )n f · g. ¯ Rn

Rn

Rn

Rn

(7.3.28) 

Property 3 (The Fourier transform of a convolution) Let f and g belong to S(Rn ). Then we have  f ∗ g = fˆ · g. ˆ (7.3.29) Proof We use Fubini’s theorem in order to obtain the following

7.3 The Fourier Transform in S (Rn )

 e−ix,ξ  ( f ∗ g)(x)d x = f ∗ g(ξ ) = Rn  −ix,ξ  e f (y)g(x − y)dy d x = = n Rn R = e−ix−y,ξ  e−iy,ξ  f (y)g(x − y)d xd y = Rn Rn  −iy,ξ  −ix−y,ξ  e e g(x − y)d x f (y)dy = = Rn Rn = g(ξ ˆ ) e−iy,ξ  f (y)dy = fˆ(ξ ) · g(ξ ˆ ).

145

(7.3.30)

Rn

 Property 4 (The Fourier transform of a product) Let f, g ∈ S(Rn ). Then we have  f · g = (2π )−n fˆ ∗ g. ˆ

(7.3.31)

Proof From Fourier’s inversion formula and Fubini’s theorem, it follows that  f · g(ξ ) =



e−ix,ξ  f (x) · g(x)d x =  −n ix,ξ  ix,η ˆ e g(x) e f (η)dη d x = = (2π ) n Rn R = (2π )−n e−ix,ξ −η g(x) fˆ(η)d xdη = Rn Rn  −n −ix,ξ −η e g(x)d x · fˆ(η)dη = = (2π ) Rn Rn

 = (2π )−n g(ξ ˆ − η) fˆ(η)dη = (2π )−n fˆ ∗ gˆ (ξ ). Rn

Rn

(7.3.32)

Rn



7.4 Fourier Transform of Tempered Distributions We recall that the Fourier transform F is an isomorphism of S(Rn ) onto S(Rn ). It follows that we can define the Fourier transform of tempered distributions by duality. Definition 7.4.1 We consider that T ∈ S (Rn ). Then its Fourier transform is the tempered distribution F T defined by (F T, φ) = (T, Fφ) , ∀φ ∈ S(Rn ).

(7.4.1)

146

7 Tempered Distributions

Clearly, the linear map from S (Rn ) into S (Rn ) defined by (7.4.1) is the transpose of the isomorphism F : S(Rn ) → S(Rn ). Note that we will employ the same F to characterize either the Fourier transform of elements of S or the Fourier transform of tempered distributions. The reason is that when (7.4.2) T = ψ ∈ S(Rn ) then, by Property 1, we observe that the definition in (7.4.1) coincides with that in (7.3.1). Note that F : S(Rn ) → S(Rn ) is continuous. Hence, F : S (Rn ) → S (Rn ) is also a continuous map in the sense of the strong topology of S (Rn ). As in the case of the Fourier transforms of functions of S(Rn ), the Fourier transform F T of T ∈ S (Rn ) shall be denoted by Tˆ . Similarly, we introduce the inverse Fourier transform of tempered distributions by duality     −1 (7.4.3) F T, φ = T, F −1 φ , ∀φ ∈ S(Rn ). Clearly, F −1 is a continuous linear map from S (Rn ) into S (Rn ) equipped with the strong topologies. Moreover, the Fourier inversion formula T = F −1 (F T ) , ∀T ∈ S (Rn )

(7.4.4)

holds true. Hence, F is an isomorphism from S (Rn ) onto S (Rn ). Let f ∈ L 2 (Rn ). From the classical theory of Fourier transforms we know that fˆ(ξ ) =

Rn

e−ix,ξ  f (x)d x

(7.4.5)

exists, belongs to L 2 (Rn ) and the map f ∈ L 2 (Rn ) → fˆ ∈ L 2 (Rn )

(7.4.6)

is an isomorphism. Since L 2 (Rn ) is a subspace of S (Rn ), it follows that the Fourier transform of square integrable functions also exists in the sense of distributions. By Property 1, which holds true in the classical case, we deduce that these two notions are actually the same. The results below are shown within the framework of tempered distributions. Theorem 7.4.1 We consider that f ∈ L 2 (Rn ) and that fˆ is its Fourier transform in the sense of distributions. Then the following hold true 1. fˆ ∈ L 2 (Rn ); n 2. || fˆ|| L 2 (Rn ) = (2π ) 2 || f || L 2 (Rn ) (Parseval’s formula). Proof Let φ ∈ S. Then the Property 2 implies that

7.4 Fourier Transform of Tempered Distributions

147

           ˆ  f , φ  =  f, φˆ  =  f · φˆ  ≤

(7.4.7)

ˆ L 2 (Rn ) ≤ (2π ) || f || L 2 (Rn ) · ||φ|| L 2 (Rn ) . ≤ || f || L 2 (Rn ) · ||φ|| n 2

From this inequality it follows that fˆ ∈ L 2 (Rn ) and that n || fˆ|| L 2 (Rn ) ≤ (2π ) 2 || f || L 2 (Rn ) , ∀ f ∈ L 2 (Rn ).

(7.4.8)

Similarly, we can prove that ||F −1 f || L 2 (Rn ) ≤ (2π )− 2 || f || L 2 (Rn ) , ∀ f ∈ L 2 (Rn ). n

(7.4.9)

From inequalities (7.4.8) and (7.4.9), it follows that || f || L 2 (Rn ) = ||F −1 (F f )|| L 2 (Rn ) ≤ (2π )− 2 ||F f || L 2 (Rn ) ≤ || f || L 2 (Rn ) . n

Therefore, we obtain

|| fˆ|| L 2 (Rn ) = (2π ) 2 || f || L 2 (Rn ) . n

(7.4.10)

(7.4.11) 

Below we recall the following classical result about Fourier transforms of integrable functions [10]. Proposition 7.4.1 Let f ∈ L 1 (Rn ). Then fˆ(ξ ) =

Rn

e−ix,ξ  f (x)d x

(7.4.12)

is a well defined continuous function of ξ ∈ Rn . Proof Note that the integral is absolutely convergent. Hence, fˆ(ξ ) is well defined. We estimate   e−ix,ξ0  e−ix,ξ −ξ0  − 1 f (x)d x, (7.4.13) fˆ(ξ ) − fˆ(ξ0 ) = Rn

so we obtain the inequality      ˆ e−ix,ξ −ξ0  − 1 | f (x)|d x ≤  f (ξ ) − fˆ(ξ0 ) ≤ Rn    x, ξ − ξ0    | f (x)|d x ≤ sin ≤2   2 Rn    x, ξ − ξ0    | f (x)|d x + 2  sin ≤2 | f (x)|d x.   2 |ξ |≤A |ξ |>A

(7.4.14)

148

7 Tempered Distributions

    It follows that  fˆ(ξ ) − fˆ(ξ0 ) can be made arbitrarily small provided that we choose |ξ − ξ0 | sufficiently small. 

7.5 The Fourier Transform of a Distribution with Compact Support The goal of this section is to prove that the Fourier transform of every distribution with compact support is a C ∞ function in Rn that can be extended to the complex space Cn as an entire analytic function. a variable element of Cn , where ζ j = We consider that ζ = (ζ1 , . . . , ζn ) denotes √ ξ j + iη j (1 ≤ j ≤ n), ξ j , η j ∈ Rn , i = −1. We also consider     ∂ ∂ ∂ ∂ 1 ∂ 1 ∂ , , = −i = +i ∂ζ j 2 ∂ξ j ∂η j 2 ∂ξ j ∂η j ∂ ζ¯ j     ∂ ∂ ∂ ∂ ∂ ∂ , . = = ,··· , ,··· , ∂ζ ∂ζ1 ∂ζn ∂ ζ¯ ∂ ζ¯1 ∂ ζ¯n For p ∈ Nn , we set

∂ ∂ζ

p

=

∂ ∂ζ1

 p1

···

∂ ∂ζn

 pn

(7.5.1)

.

Definition 7.5.1 Let U ⊂ Cn be open. Let f : U → C be a function. Then f is holomorphic at a point ζ0 ∈ U if it can be represented as a power series in ζ − ζ0 f (ζ ) =



a p (ζ − ζ0 ) p ,

(7.5.2)

p

the series being convergent in some neighbourhood of ζ0 . The function f is holomorphic in U if it is holomorphic at every point of U . An entire function is a holomorphic function in Cn . In an equivalent way, a function f of class C 1 on U is holomorphic on U if it satisfies the Cauchy-Riemann equations ∂f = 0, 1 ≤ j ≤ n. ∂ ζ¯ j

(7.5.3)

The coefficient a p in the series expansion above can be represented as ap =

1 p!



∂ ∂ζ

p f (ζ0 ).

(7.5.4)

We can further employ the Cauchy’s integral formulas. Let ζ0 ∈ U and consider the polydisk

7.5 The Fourier Transform of a Distribution with Compact Support

149

D(r1 , . . . , rn ) = {ζ ∈ Cn ||ζ j − ζ j0 | ≤ r j , 1 ≤ j ≤ n},

(7.5.5)

which we assume to be contained in U . Then, ∀ p ∈ Nn , we obtain  ∂ p f (ζ0 ) = ∂ζ 1 f (ζ )dζ1 . . . dζn = · · · . 0 p1 +1 0 (2πi)n |ζ1 −ζ10 |=r1 (ζ − ζ · · · (ζn − ζn0 ) pn +1 1 |ζn −ζn |=rn 1)

1 p!



(7.5.6)

Definition 7.5.2 Let U ⊂ Cn be open and let E be a topological vector space. A function f : U → E is holomorphic in U if, for every ζ0 ∈ U , f can be represented as a power series in ζ − ζ0 with coefficients in E, the series being convergent in some neighbourhood of ζ0 . A holomorphic function with values in a topological vector space E is often called a vector-valued holomorphic function [5]. When U = Cn , f is called a vector-valued entire function. Let f : U → E be a vector-valued holomorphic function. Then for every element e belonging to the dual E of E, the complex valued function f e defined on U by f e (ζ ) =  f (ζ ), e  is holomorphic on U . Clearly, we can apply the operator relation. Moreover, note that ∂ ∂f  f (ζ ), e  =  (ζ ), e . ¯ ∂ζ j ∂ ζ¯ j

(7.5.7) ∂ ∂ ζ¯ j

to both sides of the last

(7.5.8)

Since by assumption, f is holomorphic, we obtain ∂∂ fζ¯e = 0, 1 ≤ j ≤ n, ∀e ∈ E . j Let f : U → E be a vector-valued function. If f is such that, for every e ∈ E , the complex-valued function f e is holomorphic in U , then it is called a scalarly holomorphic function in U . We already showed that every vector-valued holomorphic function is scalarly holomorphic. On the other hand, if E is a complete topological vector space every scalarly holomorphic function in U with values in E is a vector-valued holomorphic function. Example 1 We consider x ∈ R, ζ = ξ + iη ∈ C. The function ζ ∈ C → e−i xζ ∈ C ∞ (R)

(7.5.9)

is an entire function of ζ with values in the locally convex space C ∞ (R). Clearly, ∀x ∈ R, (−i xζ )n −i xζ + ··· + + ··· (7.5.10) e−i xζ = 1 + 1! n!

150

7 Tempered Distributions

is an entire function of ζ . Furthermore, the series converges uniformly on every compact subset of R. The same holds true for all its derivatives with respect to the variable x. Example 2 More generally, if x = (x1 , . . . , xn ) ∈ Rn , ζ = (ζ1 , . . . , ζn ) ∈ Cn and x, ζ  = x1 ζ1 + · · · + xn ζn , the function ζ ∈ Cn → e−ix,ζ  ∈ C ∞ (Rn )

(7.5.11)

is a C ∞ (Rn )-valued entire function. Theorem 7.5.1 Let T ∈ E (Rn ). Then its Fourier transform is represented as   Tˆ (ξ ) = T (x), e−ix,ξ 

(7.5.12)

and is a C ∞ function in Rn . Furthermore, Tˆ (ξ ) can be extended to the complex space Cn as an entire analytic function given by   Tˆ (ζ ) = T (x), e−ix,ζ  .

(7.5.13)

Proof Step 1. Note that T is a distribution with compact support and e−ix,ξ  is a C ∞ function of x. Therefore, the right-hand side of (7.5.12) is well defined. Moreover, the right-hand side of (7.5.12) is a C ∞ function of ξ ∈ Rn . Step 2. We show that relation (7.5.12) holds. Let (α j ) be a regularizing sequence. Then α j ∗ T → T as j → ∞ in E . Therefore, we have α j ∗ T → T in S . Let α˜ j (x) = α j (−x). Then   α˜ j ∗ e−i·,ξ  (x) → e−ix,ξ  in C ∞ (Rn ).

(7.5.14)

Since T ∈ E , we have 

     T (x), α˜ j ∗ e−i·,ξ  (x) → T (x), e−ix,ξ  .

(7.5.15)

If we can show that       T ∗ α j (ξ ) = (T ∗ α j )(x), e−ix,ξ  = T (x), α˜ j ∗ e−i·,ξ  (x)

(7.5.16)

then (7.5.12) will follow by letting j → ∞. Hence, it suffices to show (7.5.16). We recall that T ∗ α j is ∀ j, a C ∞ function with compact support. Therefore, its Fourier transform is given by T ∗ α j (ξ ) =

Rn

  e−ix,ξ  (T ∗ α j )(x)d x = (T ∗ α j )(x), e−ix,ξ  .

  But (T ∗ α j )(x) = T (y), α j (x − y) , so

(7.5.17)

7.5 The Fourier Transform of a Distribution with Compact Support

    (T ∗ α j )(x), e−ix,ξ  = T (y), α j (x − y) , e−ix,ξ  =        = T (y), α j (x − y), e−ix,ξ  = T (y), α˜ j ∗ e−i·,ξ  (y) ,

151



(7.5.18)

which implies (7.5.16). Step 3. Note that the function ζ ∈ Cn → e−ix,ζ  ∈ C ∞ (Rn )

(7.5.19)

is a vector-valued entire analytic function. Moreover, T is an element of the dual of C ∞ (Rn ). Therefore, the complex-valued function   ζ ∈ Cn → Tˆ (ζ ) = T (x), e−ix,ζ  ∈ C

(7.5.20) 

is an entire analytic function.

Definition 7.5.3 Let T ∈ E (Rn ) be a distribution. Then the Fourier-Laplace transform of T is the entire analytic function   Tˆ (ζ ) = T (x), e−ix,ζ  .

(7.5.21)

7.6 The Product of a Distribution by a C ∞ Function Definition 7.6.1 We consider T ∈ D (Rn ) and α ∈ C ∞ (Rn ). We introduce the product of α by T as the distribution αT with (αT, φ) = (T, αφ) , ∀φ ∈ Cc∞ (Rn ).

(7.6.1)

Note that if T is defined by a locally integrable function, then the product αT coincides with the usual product of functions. Note that it is not possible, in general, to define the product of two distributions S and T . Clearly, if S = f and T = g are locally integrable functions, then f · g is not necessarily locally integrable. Therefore, it does not define a distribution. Below we present some consequences of Definition 7.6.1. Consequence 1 The support of αT is contained in the intersection of the support of α and the support of T . Consequence 2 We have ∀1 ≤ j ≤ n ∂ j (αT ) = ∂ j α · T + α · ∂ j T. Clearly, we have

(7.6.2)

152

7 Tempered Distributions

      ∂ j (αT ), φ = − αT, ∂ j φ = − T, α∂ j φ =     = − T, ∂ j (αφ) + T, ∂ j αφ =     = ∂ j T, αφ + T, ∂ j αφ =   = α∂ j T + ∂ j αT, φ , ∀φ ∈ Cc∞ (Rn ).

(7.6.3)

 Consequently, we derive the Leibniz formula ∂ p (αT ) =



p! r s ∂ α∂ T. r !s! r +s= p

(7.6.4)

Consequence 3 The bilinear map C ∞ (Rn ) × D (Rn )  (α, T ) → αT ∈ D (Rn )

(7.6.5)

is separately continuous. We assume that (T j ) converges strongly to zero in D . Let B be a bounded set in ∞ C (Rn ). Hence, the set α B = {αφ|φ ∈ B} is also bounded in C ∞ (Rn ). Therefore, we have     (7.6.6) αT j , φ = T j , αφ → 0 uniformly when φ ∈ B. Hence, αT j → 0 strongly in D (Rn ). Note that αT is continuous with respect to the first variable. Consequence 4 Let α ∈ Cc∞ and T ∈ D . Therefore, αT ∈ E and the map (α, T ) ∈ Cc∞ × D → αT ∈ E

(7.6.7)

is separately continuous.

7.7 The Space of Multipliers of S  (Rn ) We consider T ∈ S . Assume that we ask for conditions on the C ∞ function α such 2 that αT ∈ S . Indeed, this holds true for α ∈ Cc∞ (Rn ). But for α(x) = e|x| , it is

not true that αT ∈ S . The reason is that an exponential function grows very fast at infinity. Definition 7.7.1 Let O M (Rn ) be the space of all φ ∈ C ∞ (Rn ) such that for every p ∈ Nn there is a polynomial Pp (x) for which   p ∂ φ(x) ≤ |Pp (x)|, ∀x ∈ Rn . O M is called the space of C ∞ functions slowly increasing at infinity.

(7.7.1)

7.7 The Space of Multipliers of S (Rn )

153

Proposition 7.7.1 We consider φ ∈ C ∞ (Rn ). We have the equivalent conditions: 1. For every p = ( p1 , . . . , pn ) ∈ Nn there is a polynomial Pp (x) such that   p ∂ φ(x) ≤ |Pp (x)|, ∀x ∈ Rn .

(7.7.2)

2. For all f ∈ S the product φ f ∈ S. 3. For every n-tuple p = ( p1 , . . . , pn ) and for every f ∈ S the function ∂ p φ · f is bounded in Rn . Proof (1)⇒(2) We consider p ∈ Nn . By the Leibniz formula, it follows that ∂ p (φ · f ) =

 q≤ p

p! ∂ p φ · ∂ p−q f. q!( p − q)!

(7.7.3)

Therefore, we obtain (1 + r 2 )k ∂ p (φ · f ) =

 q≤ p

p! ∂ q φ · (1 + r 2 )k ∂ p−q f, q!( p − q)!

(7.7.4)

with k being a positive real number. Since φ satisfies condition 1, ∃N ∈ N∗ such that   q ∂ φ(x) ≤ C(1 + r 2 ) N , ∀x ∈ Rn , ∀q ≤ p.

(7.7.5)

By an appropriate replacement, we obtain     sup sup (1 + r 2 )k ∂ p (φ · f )(x) ≤ C sup sup (1 + r 2 )k+N ∂ s f (x) ,

x∈Rn | p|≤m

x∈Rn |s|≤m

(7.7.6)

which implies, since f ∈ S, that f φ ∈ S. (2)⇒(3) We have ∂ j φ · f = ∂ j (φ · f ) − φ · ∂ j f, 1 ≤ j ≤ n.

(7.7.7)

Since f ∈ S then, by condition 2, φ · ∂ j f ∈ S. Therefore, we have ∂ j φ · f ∈ S. Using the Leibniz formula and an induction argument we are led to ∂ p φ · f ∈ S, ∀ p ∈ Nn . Therefore, condition 3 holds. (3)⇒(1) Assume, by contradiction, that condition 1 does not hold. Hence, for some p ∈ Nn , |∂ p φ| is not bounded by any polynomial. Therefore, by induction, we can find a sequence (x j ) of elements of Rn with |x j+1 | ≥ |x j | + 2 and such that |∂ p φ(x j )| > (1 + |x j |2 ) j . We consider γ ∈ S to be the function defined by γ (x) =

∞  α(x − x j )   , 2 j j=1 1 + |x j |

(7.7.8)

154

7 Tempered Distributions

where α ∈ Cc∞ (Rn ) is chosen such that 0 ≤ α ≤ 1, supp α ⊂ B1 and α(0) = 1. Clearly, we have   γ (x j )∂ p φ(x j ) > α(0) = 1, ∀ j = 1, 2, . . . ,

(7.7.9) 

which contradicts condition 3.

The Topology of O M This is the locally convex topology defined by the family of seminorms   (7.7.10) γ f, p (φ) = sup  f (x) · ∂ p φ(x) , x∈Rn

where f ∈ S(Rn ) and p ∈ Nn . Clearly, such topology does not have a countable basis. Moreover, O M is a complete space. A sequence (or filter) (φ j ) converges to zero in O M if and only if ∀ f ∈ S and  ∀ p ∈ Nn , f (x)∂ p φ j (x) converges to zero uniformly on Rn . In an equivalent way, ∀ f ∈ S, ( f φ j ) converges to zero in S. From Definition 7.7.1 and Proposition 7.7.1 we deduce that a set B is bounded in O M if and only if ∀ p ∈ Nn , ∃Pp (x) a polynomial such that   p ∂ φ(x) ≤ Pp (x), ∀x ∈ Rn , ∀φ ∈ B.

(7.7.11)

Proposition 7.7.2 The bilinear map O M × S  (φ, f ) → φ f ∈ S

(7.7.12)

is separately continuous. Proof Step 1. Fix φ ∈ O M . From the inequality (7.7.6) it follows that the linear map S  f → φf ∈ S

(7.7.13)

is continuous. Step 2. Fix f ∈ S and let (φ j ) be functions of O M which converge to zero in O M . Therefore, ∀g ∈ S and ∀ p ∈ Nn , we have   γg, p (φ) = sup g(x)∂ p φ j (x) → 0.

(7.7.14)

x∈Rn

Let k be an integer and q a n-tuple of nonnegative integers. Using the Leibniz formula, we obtain  q! q



∂q ( f · φ j ) = ∂ f · ∂q φ j . (7.7.15)



q !q ! q +q

=q Therefore, we have

7.7 The Space of Multipliers of S (Rn )

sup

x∈Rn ,|q|≤m

≤



155

  (1 + r 2 )k ∂ q ( f · φ j ) ≤

Cq q

sup

x∈Rn ,|q |≤m,|q

|≤m

 



  (1 + r 2 )k ∂ q f · ∂ q φ j  .

(7.7.16)



Since (1 + r 2 )k ∂ q f ∈ S, every term in the right-hand side converges to zero as  φ j → 0 in O M . Hence, we have f · φ j → 0 in S(Rn ). Note that if φ ∈ C ∞ is such that φ f ∈ S, ∀ f ∈ S, then, by Proposition 7.7.1, φ ∈ OM . Theorem 7.7.1 We have the following inclusions S(Rn ) → O M (Rn ) → S (Rn )

(7.7.17)

and the embeddings are continuous. Furthermore, S is dense in O M . Proof Let φ ∈ O M . Then we can introduce a tempered distribution by (φ, f ) =

Rn

φ f d x, ∀ f ∈ S.

(7.7.18)

Clearly, we have S ⊂ O M . Let (β j ) be a sequence of elements of Cc∞ (Rn ) such that β j = 1 on K j , where (K j ) j=1,2,... is an increasing sequence of compact subsets of Rn whose union is Rn . Hence, we have β j φ → φ in O M for all φ ∈ O M . Therefore,  S is dense in O M . Moreover, the embeddings are continuous. We deduce that the space O M is a normal space of distributions. Definition 7.7.2 Let φ ∈ O M and T ∈ S . Then we define the product φT by (φT, f ) = (T, φ f ) , ∀ f ∈ S.

(7.7.19)

As a consequence of Proposition 7.7.2, the product φT belongs to S . Theorem 7.7.2 The bilinear map O M × S  (φ, T ) → φT ∈ S

(7.7.20)

is separately continuous. From the reflexivity of the space S we deduce that O M is the space of all multipliers of S . Proposition 7.7.3 Let φ ∈ C ∞ be such that φT ∈ S for all T ∈ S . Then φ ∈ O M .

156

7 Tempered Distributions

Proof By assumption, we deduce that ∀ f ∈ S the map T → (φT, f )

(7.7.21)

is a continuous linear functional on S . Note that S is a reflexive space. Therefore, ∃g ∈ S such that (7.7.22) (φT, f ) = (T, g) , ∀T ∈ S . In particular,

which implies that

(φα, f ) = (α, g) , ∀α ∈ Cc∞ ,

(7.7.23)

(α, φ f ) = (α, g) , ∀α ∈ Cc∞ .

(7.7.24)

Therefore, we have φ f = g ∈ S, ∀ f ∈ S. By the remark following the proof of  Proposition 7.7.2, we deduce that φ ∈ O M .

7.8 Some Results on Convolutions with Tempered Distributions We already defined the convolution of a C ∞ function (respectively a C ∞ function with compact support) and a distribution with compact support (respectively a distribution). The same definition extends to the case of a C ∞ function rapidly decreasing at infinity and a tempered distribution. Let φ ∈ S(Rn ) and T ∈ S (Rn ). Then we define (T ∗ φ) (x) = (T (y), φ(x − y)) .

(7.8.1)

We can prove that T ∗ φ ∈ C ∞ (Rn ) and that the bilinear map S × S  (φ, T ) → T ∗ φ ∈ C ∞

(7.8.2)

is separately continuous. In general, it is not true that T ∗ φ ∈ S. Indeed, consider T to be the function identically equal to one on Rn . Then we have (T ∗ φ)(x) = (1(y), φ(x − y)) =

Rn

φ(y)dy = C,

(7.8.3)

with C being a constant. Therefore, we have T ∗ φ ∈ / S. But we can derive the result below by using the structure of tempered distributions. Theorem 7.8.1 Let φ ∈ S and T ∈ S . Then T ∗ φ ∈ O M .

7.8 Some Results on Convolutions with Tempered Distributions

Proof Let T ∈ S . Then

 k T = ∂ p (1 + r 2 ) 2 f ,

157

(7.8.4)

where f is a bounded continuous function on Rn . By definition, we have (T ∗ φ)(x) = (T (y), φ(x − y)) =   p  ∂ 2 k2 (1 + |y| ) f (y) , φ(x − y) = = ∂yp ∂ pφ k = (−1)| p| (1 + |y|2 ) 2 f (y) p (x − y)dy. ∂y Rn

(7.8.5)

Using the inequality k k k  (1 + |y|2 ) 2 ≤ C(1 + |x|2 ) 2 1 + |x − y|2 2

(7.8.6)

it follows that k

|T ∗ φ(x)| ≤ C(1 + |x|2 ) 2

Rn

 k 1 + |x − y|2 2

 p  ∂ φ    dy ≤ (x − y)  ∂x p 

(7.8.7)

k 2

≤ C (1 + |x| ) . 2

In a similar way, it follows that   q  k ∂ (T ∗ φ) ≤ Cq 1 + |x|2 2 , ∀q ∈ Nn . Therefore, by Proposition 7.7.1, we deduce that T ∗ φ ∈ O M .

(7.8.8) 

Theorem 7.8.2 Let S ∈ S and T ∈ E . Therefore, we have S ∗ T ∈ S . Furthermore, the bilinear map S × E  (S, T ) → S ∗ T ∈ S

(7.8.9)

is separately continuous. The proof is based on the following lemma. Lemma 7.8.1 Let T ∈ E and φ ∈ S. Then the function ψ(ξ ) = (T (η), φ(ξ + η))

(7.8.10)

belongs to S. Moreover, if φ → 0 in S, then ψ → 0 in S. Proof Let T be a distribution with compact support. Then it can be written as a finite sum of derivatives of continuous functions with compact support contained in

158

7 Tempered Distributions

an arbitrary neighbourhood of the support of T . Hence, it suffices to show the lemma for (7.8.11) T = ∂ q G, where G is a continuous function in Rn such that its support is contained in an arbitrary neighbourhood of the support of T . Note that ψ(ξ ) is a C ∞ function in Rn . Therefore, it suffices to prove that ψ(ξ ) is rapidly decreasing at infinity. We have ψ(ξ ) = (T (η), φ(ξ + η)) = (−1)|q|

G(η) Rn

∂q φ (ξ + η)dη. ∂ηq

(7.8.12)

Therefore, we obtain   k ∂ pψ 1 + |ξ |2 2 (ξ ) = (−1)|q| ∂ξ p

Rn

 k ∂ p+q φ  G(η) 1 + |ξ |2 2 (ξ + η)dη. (7.8.13) ∂ξ p ∂ηq

If we employ the inequality    1 + |ξ |2 ≤ 2 · 1 + |η|2 1 + |ξ + η|2 ,

(7.8.14)

then we can estimate the last expression in the following way     k p  1 + |ξ |2 2 ∂ ψ (ξ ) ≤   p ∂ξ  k k  ≤2· 1 + |η|2 2 |G(η)| 1 + |ξ + η|2 2 Rn

 p+q  ∂  φ   dη. (ξ + η)  ∂ξ p ∂ηq 

(7.8.15)

Note that G is a continuous function with compact support. Therefore, we deduce the inequality     k p 2 2 ∂ ψ  sup sup  1 + |ξ | (ξ ) ≤ p ∂ξ | p|≤m ξ ∈Rn   (7.8.16)   k ∂sφ  , ≤ C sup sup  1 + |ζ |2 2 (ζ )  ∂ζ s |s|≤l ζ ∈Rn which proves at the same time that ψ ∈ S and that it depends continuously on φ in S.  Observation 7.8.1 Note that we have (T ∗ φ) (x) = (T (y), φ(x − y)) .

(7.8.17)

Then we can prove, similarly as in the previous lemma, that if T ∈ E and φ ∈ S, then T ∗ φ ∈ S.

7.8 Some Results on Convolutions with Tempered Distributions

159

Proof of Theorem 7.8.2. Step 1. We consider S ∈ S , T ∈ E and φ ∈ S. It can be proven that (S(ξ ), φ(ξ + η))

(7.8.18)

is an infinitely differentiable function of η depending continuously on φ ∈ S. Hence, (T (η), (S(ξ ), φ(ξ + η)))

(7.8.19)

is well defined and depends continuously on φ ∈ S. By Lemma 7.8.1 the function ψ(ξ ) = (T (η), φ(ξ + η))

(7.8.20)

belongs to S and depends continuously on φ ∈ S. Since S ∈ S , we deduce that (S(ξ ), (T (η), φ(ξ + η)))

(7.8.21)

is also well defined and it depends continuously on φ ∈ S. Moreover, since (7.8.19) and (7.8.21) do coincide when φ ∈ Cc∞ (see the definition of the convolution product) and Cc∞ is dense in S (Theorem 7.1.3), they coincide everywhere in S. Then it follows that (S ∗ T, φ) = (S(ξ ) ⊗ T (η), φ(ξ + η)) = (7.8.22) = (S(ξ ), (T (η), φ(ξ + η))) = (T (η), (S(ξ ), φ(ξ + η))) for all φ ∈ S. Hence, we have S ∗ T ∈ S . Step 2. We fix T ∈ E (Rn ). Moreover, we assume that the distributions S j converge to zero strongly in S (Rn ). It follows that we have ∀φ ∈ S 



    S j ∗ T, φ = S j ∗ T ∗ φ˜ (0) = T ∗ S j ∗ φ˜ (0).

(7.8.23)

We recall that the convolution product of a function on S and a distribution of S is separately continuous. Therefore, we have S j ∗ φ˜ → 0 in C ∞ . Moreover, we can prove that S j ∗ φ˜ converges uniformly to zero whenever φ belongs to a bounded set in S. Therefore, we deduce that T ∗ S j ∗ φ˜ → 0 in C ∞ uniformly whenever φ belongs to a bounded set of S. If we take into account (7.8.23), then it follows that S j ∗ T → 0 strongly in S as S j → 0 strongly in S . Similarly, we can show the continuity of S ∗ T with respect to T .  The result below follows by means of Fourier transforms. Theorem 7.8.3 Let φ ∈ S and T ∈ S . Then φ ∗ T = φˆ · Tˆ .

(7.8.24)

160

7 Tempered Distributions

Proof We have φ ∗ T ∈ O M and by Theorem 7.7.1 we have φ ∗ T ∈ S . It follows that the Fourier transform φ ∗ T is well defined and it belongs to S . Moreover, φˆ ∈ S ⊂ O M and Tˆ ∈ S . Therefore, we have φˆ Tˆ ∈ S . We only have to prove that both sides of (7.8.24) are equal. For all ψ ∈ S we have

   ˆ + η) = φ ∗ T , ψ = φ ∗ T, ψˆ = φ(ξ ) ⊗ T (η), ψ(ξ



 ˆ + η) = T (η), φ(ξ − η), ψ(ξ ˆ ) , = T (η), φ(ξ ), ψ(ξ

(7.8.25)

 ˆ + η) . Moreover, we have after a change of variable inside the integral φ(ξ ), ψ(ξ

 ˆ ) = φ(ξ − η), ψ(ξ

Rn

It follows easily that

 ˆ )dξ = φ˜ ∗ ψˆ (η). φ(ξ − η)ψ(ξ



ˆˆ φ˜ = F F −1 φ˜ = (2π )−n φ,

(7.8.26)

(7.8.27)

where φˆˆ = F(Fφ). By an appropriate replacement, we obtain

 

   −n ˆˆ ˆ ˆ ˆ φ(ξ − η), ψ(ξ ) = (2π ) φ ∗ ψ = φ · ψ (η).

Hence, we deduce that

 

  φ ∗ T , ψ = φ ∗ T, ψˆ = T (η), φ˜ ∗ ψˆ (η) =        ˆ = T (η), φ · ψ (η) = Tˆ , φˆ · ψ = φˆ Tˆ , ψ .

(7.8.28)

(7.8.29)



7.9 The Paley-Wiener-Schwartz Theorem Theorem 7.9.1 1. Let T be a distribution with compact support in Rn . Then the Fourier-Laplace transform of T is an entire function F(ζ ) in Cn satisfying the following property (P1 ) ∃C, A constants and N ∈ N such that |F(ζ )| ≤ C (1 + |ζ |) N e A|I mζ | , ∀ζ ∈ Cn .

(7.9.1)

7.9 The Paley-Wiener-Schwartz Theorem

161

On the other hand, every entire function in Cn satisfying Property (P1 ) is the FourierLaplace transform of a distribution belonging to E (Rn ). 2. Let f be an infinitely differentiable function with compact support in Rn . Then the Fourier-Laplace transform of f is an entire function F(ζ ) in Cn satisfying the following property: (P2 ) ∃A > 0 constant such that ∀N ∈ N we can find a constant C such that |F(ζ )| ≤ C (1 + |ζ |)−N e A|I mζ | , ∀ζ ∈ Cn .

(7.9.2)

On the other hand, every entire function in Cn satisfying Property (P2 ) is the FourierLaplace transform of a C ∞ function with compact support in Rn . Proof (i) Let T ∈ E (Rn ) be a distribution. We assume that   F(ζ ) = T (x), e−ix,ζ 

(7.9.3)

is the Fourier-Laplace transform of T . Then ∃C > 0 constant, ∃N ∈ N and ∃K ⊂ Rn compact such that |(T, φ)| ≤ C ·

sup

|α|≤N ,x∈K

|D α φ|, ∀φ ∈ C ∞ (Rn ).

(7.9.4)

Let A be a positive real number such that the compact subset K is contained in the closed ball (7.9.5) {x ∈ Rn ||x| ≤ A} and let

We have

φζ (x) = e−ix,ζ  .

(7.9.6)

  α  D φζ (x) ≤ |ζ ||α| ex,η .

(7.9.7)

It follows that sup

|α|≤N ,|x|≤A

  α  D φζ (x) ≤

sup

|α|≤N ,|x|≤A

|ζ ||α| ex,η ≤ (1 + |ζ |) N e A|η| .

(7.9.8)

The inequalities (7.9.4) and (7.9.8), together with the definition of F(ζ ), imply the inequality (7.9.1). (ii) Let φ ∈ Cc∞ (Rn ) be a function. We assume that F(ζ ) is the Fourier-Laplace transform of φ. Let A > 0 be such that the support of φ is contained in the ball with center at the origin and radius A. For every n-tuple α = (α1 , . . . , αn ) of nonnegative integers we have α ˆ e−ix,ζ  D α φ(x)d x. (7.9.9) ζ φ(ζ ) = Rn

162

7 Tempered Distributions

From this relation we deduce the inequality    αˆ  ζ φ(ζ ) ≤ c · e A|η| . It follows that

  ˆ  φ(ζ ) ≤ C · e A|η| ,

 N 1 + |ζ |2 2

(7.9.10)

(7.9.11)

for every integer N ≥ 0. Finally, by means of the inequality 0 < c0 ≤

N  1 + |ζ |2 2 (1 + |ζ |) N

≤ C0

(7.9.12)

it follows that (7.9.11) is equivalent to (7.9.2). (iii) Conversely, we assume that Property (P2 ) holds. Then we consider f (x) = (2π )−n



F(ξ )eix,ξ  dξ,

Rn

(7.9.13)

i.e., f (x) is the inverse Fourier transform of F(ξ ), ξ ∈ Rn . Because of (7.9.2) the last integral is absolutely convergent. Moreover, from the same inequality (7.9.2) it follows that the integral α

D f (x) = (2π )

−n

Rn

F(ξ )ξ α eix,ξ  dξ

(7.9.14)

is absolutely convergent. Hence, we deduce that f (x) is an infinitely differentiable function in Rn . We will prove that f has a compact support. Let η = (η1 , . . . , ηn ) be an arbitrarily fixed point in Rn . Note that, by assumption, F(ζ ) is an entire function satisfying the inequality (7.9.2). Therefore, it follows that f (x) = (2π )−n

Rn

F(ξ + iη)eix,ξ +iη dξ,

(7.9.15)

where the integral is an absolutely convergent one. Take N = n + 1. By using (7.9.2) we deduce that −n −x,η+A|η| | f (x)| ≤ (2π ) C · e (7.9.16) (1 + |ξ |)−n−1 dξ. Rn

It follows that

| f (x)| ≤ C e A|η|−x,η .

We set η = t x in the last inequality. Hence, we deduce that

(7.9.17)

7.9 The Paley-Wiener-Schwartz Theorem

| f (x)| ≤ C e−t (|x|

163 2

−A|x|)

.

(7.9.18)

Letting t → ∞, it follows that f (x) must be zero for all x ∈ Rn with norm greater than A. As a consequence, the support of f (x) is contained in the closed ball with center at the origin and radius A. (iv) We assume that F(ζ ) is an entire function in Cn satisfying Property (P1 ). Then, from (7.9.1) we deduce that F(ξ ), ξ ∈ Rn is a function slowly increasing at infinity. Therefore, it defines a tempered distribution in Rn whose inverse Fourier transform T is an element of S (Rn ). We will show that T has a compact support in Rn . To this end, we consider that (α j ), j = 1, 2, . . . is a regularizing sequence in Cc∞ (Rn ) such that the support of α j is contained in the ball with center at the origin and radius 1j . We have, by Theorem 7.8.3, (7.9.19) T ∗ α j = Tˆ · αˆ j . By using (7.9.1), which, by assumption, holds for Tˆ = F(ζ ) and (7.9.2), which holds for every αˆ j , it follows that every Tˆ · αˆ j is an entire function in Cn satisfying an inequality of the type (7.9.2), where the constant A is replaced by the constant A + 1j . Then we deduce that T ∗ α j is an infinitely differentiable function with compact support contained in the ball of center at the origin of radius A + 1j . Hence, the distribution T which is the limit of the sequence of functions (T ∗ α j ) must have its support contained in the ball of center at the origin and radius A.  Corollary 7.9.1 Let T ∈ E (Rn ). Then its Fourier transform is an infinitely differentiable function slowly increasing at infinity. Proof The assertion is a trivial consequence of (7.9.1), letting |Imζ | = 0. Corollary 7.9.2 Let T ∈ S (Rn ). The following are equivalent conditions: (i) T has support contained in the closed set 

 x ∈ Rn ||x j | ≤ A j , 1 ≤ j ≤ n .

(7.9.20)

(ii) The Fourier-Laplace transform F(ζ ) of T is an entire function in Cn such that for any ε > 0 there are a constant Cε and an integer N ≥ 0 such that |F(ζ )| ≤ Cε (1 + |ξ |)n e(A1 +ε)|η1 |+·+(An +ε)|ηn | .

(7.9.21)

Definition 7.9.1 An entire function f (ξ ) in Cn is said to be of an exponential type (A1 , . . . , An ) if for every ε > 0 there is a constant Cε > 0 such that | f (ζ )| ≤ Cε e(A1 +ε)|ζ1 |+···+(An +ε)|ζn |

(7.9.22)

for ζ ∈ Cn . The Fourier-Laplace transform of a distribution with compact support is then an entire function of exponential type.

164

7 Tempered Distributions

7.10 A Result on the Fourier Transform of a Convolution of Two Distributions The Fourier transform F transforms the convolution of two functions of S into the product of their Fourier transforms. Moreover, Theorem 7.8.3 shows that the Fourier transform of φ ∗ T , with φ ∈ S and T ∈ S equals the product φˆ · Tˆ . We are going to extend this result to the convolution product of a tempered distribution and a distribution with compact support with the following theorem. Theorem 7.10.1 Let S ∈ S and T ∈ E . Then ˆ T ∗ S = Tˆ · S.

(7.10.1)

Proof By Theorem 7.8.2, we have T ∗ S ∈ S . Since T ∈ E , by the Paley-WienerSchwartz theorem [6], we deduce that Tˆ ∈ O M . Therefore, Tˆ · Sˆ is well defined and belongs to S . In order to show that both sides of (7.10.1) are equal, we shall use a regularizing sequence. Let (α j ) be a sequence in Cc∞ (Rn ) converging to δ in E (Rn ) as j → ∞. Then the sequence of functions (7.10.2) φ j = α j ∗ T ∈ Cc∞ converges to T in E . By Theorem 7.8.2, we deduce that φ j ∗ S → T ∗ S in S . Therefore, by taking Fourier transforms, we obtain ˆ ˆ T ∗ S = lim φ j ∗ S = lim φ j · S,

(7.10.3)

the last relation being a consequence of Theorem 7.8.3. Moreover, since φ j = α j ∗ T → T in E , it also converges in S , because the embedding of E into S is a continuous one (Theorem 7.1.3). Therefore, by Fourier transform, we obtain φˆ j → Tˆ in S . But the functions φˆ j and Tˆ do belong to O M and φˆ j → Tˆ in O M . Clearly, it suffices to prove that for all fˆ ∈ T , φˆ j · fˆ → Tˆ · fˆ in S or, equivalently, by taking Fourier transforms, that φ j ∗ f → T ∗ f in S, which is a trivial consequence of the remark following Lemma 7.8.1. Hence, we have ˆ Tˆ · Sˆ = lim φˆ j · S,

(7.10.4)

the limit being taken in S . The relations (7.10.3) and (7.10.4) imply (7.10.1).



Observation 7.10.1 Note that the result of Theorem 7.10.1 is not the best one in that direction. In fact, it is possible to show, without using the Paley-Wiener-Schwartz theorem, a more general theorem. We shall define the space Oc which plays with respect to the convolution with tempered distributions the same role that the space O M plays with respect to the product with tempered distributions. We shall see that F is a one-to-one map from O M onto Oc and vice versa and that F transforms the

7.10 A Result on the Fourier Transform of a Convolution of Two Distributions

165

convolution of a distribution in Oc with a tempered distribution into a product of an element of O M by a tempered distribution.

References 1. E. DiBenedetto, Real Analysis (Birkhuser, Boston, 2002) 2. F.G. Friedlander, M. Joshi, Introduction to the Theory of Distributions, 2nd edn. (Cambridge University Press, Cambridge, 1999) 3. D.D. Haroske, H. Triebel, Distributions, Sobolev Spaces, Elliptic Equations (European Mathematical Society, 2008) 4. L. Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, 2nd edn. (Springer, Berlin, 1990) 5. A.I. Markushevich, Theory of Functions of a Complex Variable, 2nd edn. (American Mathematical Society, New York, 2005) 6. D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis (Springer, 2013) 7. I. Richards, H. Youn, Theory of Distributions: a Non-technical Introduction (Cambridge University Press, Cambridge, 1995) 8. L. Schwartz, Theorie des Distributions (Herman & Cie, Paris, 1971) 9. R. Sikorski, The Elementary Theory of Distributions (Bull. Acad. Pol, Warszawa, 1957) 10. E. Stein, R. Shakarchi, Fourier Analysis: An Introduction (Princeton University Press, Princeton, 2003) 11. R. Strichartz, A Guide to Distribution Theory and Fourier Transforms (CRC Press, 1994)

Chapter 8

Differential Equations in Distributions

Abstract This chapter presents both ordinary and partial differential equations in distributions. Linear differential equations with constant coefficients are discussed in the framework of the theory of distributions. Hyperbolic, parabolic and elliptic partial differential equations are solved by means of the Fourier transform. The Cauchy problem is discussed.

8.1 Ordinary Differential Equations In this chapter we want to extend the concept of solution for ordinary differential equations and partial differential equations. This extension is possible if we consider differential equations in the framework of the theory of distributions.

8.1.1 Linear Differential Equations with Constant Coefficients For simplicity, we first consider a linear differential equation with constant coefficients of the form y (n) + a1 y (n−1) + · · · + an−1 y  + an y = f,

(8.1.1)

where y = y(x), f = f (x), x ∈ I = (a, b) ⊂ R and f is a distribution, f ∈ D (I ). Definition 8.1.1 A solution of the Eq. (8.1.1) is a distribution y ∈ D (I ) for which (8.1.1) becomes an equality in the sense of distributions.   We denote by P ddx y the polynomial from the left-hand side of the Eq. (8.1.1)  P

d dx



y = y (n) + a1 y (n−1) + · · · + an−1 y  + an y.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_8

(8.1.2)

167

168

8 Differential Equations in Distributions

Hence, we can rewrite the Eq. (8.1.1) as  P

d dx

 y = f.

(8.1.3)

We emphasize that the distribution y is a solution for the Eq. (8.1.1) or for its equivalent form (8.1.3) in the following sense     d P y, ϕ = ( f, ϕ), ∀ϕ ∈ D(I ). dx

(8.1.4)

If we consider the definition of the derivative of a distribution, then we can write (8.1.4) in the form 

 y, (−1)n ϕ(n) + a1 (−1)n−1 ϕ(n−1) + · · · + an−1 ϕ + an ϕ = ( f, ϕ),

(8.1.5)

where ϕ ∈ D(I ). The considerations above can be easily reformulated for the case of linear systems of first-order differential equations dyi (x) = ai j (x)y j (x) + f i (x), i = 1, n, dx

(8.1.6)

where we used the Einstein summation convention (for repeating indices, we consider the sum over all possible values of that index). In the case of the system (8.1.6), the index j takes the values 1, 2, . . . , n. In order to write the system (8.1.6) in matrix form, we use the notations   A(x) = ai j (x) i, j=1,n , f (x) = ( f 1 (x), f 2 (x), . . . , f n (x)) , y(x) = (y1 (x), y2 (x), y3 (x), . . . , yn (x))T .

(8.1.7)

Hence, we can rewrite the system (8.1.6) as dy(x) = A(x)y(x) + f (x), f ∈ D (I ). dx

(8.1.8)

We say that the distribution y = (y1 , y2 , . . . , yn ) ∈ D (I ) is a solution of the system (8.1.8) if   dy , ϕ = (Ay + f, ϕ) , ∀ϕ ∈ D(I ). (8.1.9) dx If we use the definition of the derivative of a distribution, then we obtain − (y, ϕ) = (Ay, ϕ) + ( f, ϕ) , ∀ϕ ∈ D(I ).

(8.1.10)

8.1 Ordinary Differential Equations

169

If we use the matrix notations (8.1.7), then we can rewrite (8.1.8) as − (yi , ϕ) =

n  

 y j , ai j ϕ + ( f i , ϕ) , ∀ϕ ∈ D(I ).

(8.1.11)

j=1

By the usual notations y n−i = yi (x), i = 1, n

(8.1.12)

(with the convention that y (0) = y), we can write the Eq. (8.1.1) in the form of a system of first-order linear differential equations, which is similar to (8.1.6) ⎧ y1 = y ⎪ ⎪ ⎪ ⎨y = y 2 ⎪· · · ⎪ ⎪ ⎩ yn = y (n−1) and from which we have ⎧  ⎪ ⎪ ⎪ y1 = y2 ⎪  ⎪ ⎪ ⎨ y2 = y3 ··· ⎪ ⎪  ⎪ = yn yn−1 ⎪ ⎪ ⎪ ⎩ y  = −a y − a y 1 n 2 n−1 − . . . − an y1 . n

(8.1.13)

(8.1.14)

Clearly, the distribution y ∈ D (I ) is a solution of the Eq. (8.1.1) if and only if   the distribution y = y, y , y  , . . . , y (n−1) is a solution in the sense of distributions for the system (8.1.14). We make the following convention: we call classical solutions the solutions that are functions of class C n . If f is considered to be a function and it is continuous on the interval I , then we know that the set of classical solutions of the Eq. (8.1.1) is given by C1 y1 + C2 y2 + · · · + Cn yn + y˜ ,

(8.1.15)

where (y1 , y2 , . . . , yn ) is a fundamental system of solutions for the homogeneous equation associated to the Eq. (8.1.1) and y˜ is a particular solution of the inhomogeneous equation. We will show that the only solutions in the sense of distributions of the Eq. (8.1.1) are the ones from (8.1.15) Theorem 8.1.1 The only solutions of the Eq. (8.1.1) and of the system (8.1.6), respectively, are the classical solutions if f is a continuous function on the interval I = (a, b).

170

8 Differential Equations in Distributions

Proof First we consider the equation y = 0

(8.1.16)

and we show that the only solutions are constants, in the sense of distributions generated by constant functions. Hence, the equation has only classical solutions. Let y ∈ D (I ) be a distribution that verifies the Eq. (8.1.16) (y  , ϕ) = 0, ∀ϕ ∈ D(I ), that is



(8.1.17)

 y, ϕ = 0.

(8.1.18)

If we consider the decomposition theorem, then we have (y, ψ) = 0, ∀ψ ∈ 0 . Since ψ ∈ 0 , we have

(8.1.19)

 R

ψ(x)d x = 0.

(8.1.20)

We fix an arbitrary function ϕ1 ∈ D(R) such that  R

ϕ1 (x)d x = 1.

(8.1.21)

Then ∀ϕ ∈ D(I ) we have   ϕ(x) = ϕ1 (x) ϕ(x)d x + ϕ(x) − ϕ1 (x) ϕ(x)d x R

(8.1.22)

R

and we replace in formula (8.1.19)      y, ϕ1 ϕ(x)d x + y, ϕ − ϕ1 ϕ(x)d x = R   R = (y, ϕ1 ) ϕ(x)d x + (y, 0) = (y, ϕ1 ) ϕ(x)d x. 

(y, ϕ) =

R

(8.1.23)

R



Since ϕ − ϕ1

R

ϕ(x)d x ∈ 0 ,

(8.1.24)

based on (8.1.19) we obtain 



 y, ϕ − ϕ1

R

ϕ(x)d x

= 0.

(8.1.25)

8.1 Ordinary Differential Equations

171

Hence, we obtain 

 (y, ϕ) = C

R

ϕ(x)d x =

Cϕ(x)d x = (C, ϕ) , ∀ϕ ∈ D(R).

R

(8.1.26)

Therefore, we have y = C,

(8.1.27)

where C = (y, −ϕ1 ). This is indeed constant since the test function ϕ1 was fixed. The first step is proven. In the sequel, let us consider the inhomogeneous form of the Eq. (8.1.16), that is (8.1.28) y  = f, where the distribution f is generated by a continuous function on the interval I . We define G(x) by  x

G(x) =

f (τ )dτ , x0 ∈ I

(8.1.29)

x0

and then we can write the Eq. (8.1.28) in the equivalent form (y − G) = 0.

(8.1.30)

According to the proof from the first step, we infer that the solution is y = G + C,

(8.1.31) 

where C is an arbitrary constant.

The results that we obtain for the differential equation y  = f can be transferred to systems of differential equations by considering the equivalence between the general Eq. (8.1.1) and a system of differential equations. Hence, if f is a vectorial function with the components from C(I ) and the vectorial distribution y with the components from D (I ) is a solution of the system y  = f, 

then we obtain y=

x

f (τ )dτ + C,

(8.1.32)

(8.1.33)

x0

where C is a constant vector. Definition 8.1.2 We say that the inhomogeneous linear differential equation with constant coefficients   d P y = f, f ∈ D (R) (8.1.34) dx

172

8 Differential Equations in Distributions

has as elementary solution or as fundamental solution the distribution E(x) ∈ D (R) if this verifies the equation  P

d dx

 E(x) = δ(x).

(8.1.35)

Theorem 8.1.2 The differential equation with constant coefficients  P

d dx



y = f, f ∈ D (R)

(8.1.36)

has a solution of the form y(x) = E(x) ∗ f (x),

(8.1.37)

where E(x) is a fundamental solution. Proof The convolution product from (8.1.37) exists if the distribution f has bounded support in (a, b) or if the fundamental solution E has bounded support. In order to prove the theorem, we will show that the distribution y from (8.1.8) transforms the equation into an identity    d d y(x) = P [E(x) ∗ f (x)] = dx dx

  d E(x) ∗ f = δ(x) ∗ f (x) = f (x). = P dx 

P

(8.1.38)

We used two properties of the convolution product:     (i) ddx ( f ∗ g) = ddx f ∗ g = f ∗ ddx g ; (ii) δ(x) ∗ f (x) = f (x). Finally, we emphasize that any differential equation with constant coefficients of the form   d y = f, f ∈ D (R) (8.1.39) P dx has infinitely many solutions which differ by a solution of the homogeneous equation based on the proof of Theorem 8.1.1. Definition 8.1.3 We call fundamental particular solution and denote by E + (x) any particular solution of the inhomogeneous equation. Observation 8.1.1 If we denote by E 0 (x) the general solution of the homogeneous equation, then the general solution of the inhomogeneous equation is E(x) = E 0 (x) + E + (x).

(8.1.40)

8.1 Ordinary Differential Equations

173

Applications 1. We showed that the inhomogeneous differential equation with constant coefficients dn y =δ (8.1.41) dxn has the fundamental particular solution E +n (x) =

1 x n−1 x+n−1 = θ(x) . (n − 1)! (n − 1)!

(8.1.42)

2. We consider the differential equation that models the elastic motion with friction of a material point of mass m to which we apply a momentum of conventional measure 1 at the initial moment my  + by  + ω 2 y = δ, (8.1.43) where b and ω are material constants with precise physical significance. For instance, such an equation can model the pinching of an elastic chord which vibrates freely. Owing to the initial momentum, represented in (8.1.43) by the Dirac distribution δ, we have a wave that is superimposed on the initial unperturbed motion of the chord. Let y be a distribution, y ∈ D (R), which is a solution of the Eq. (8.1.43), i.e. it verifies it in the sense of distributions    my + by  + ω 2 y, ϕ = (δ, ϕ) , ∀ϕ ∈ D(R),

(8.1.44)

m(y, ϕ ) − b(y, ϕ ) + ω 2 (y, ϕ) = ϕ(0), ∀ϕ ∈ D(R).

(8.1.45)

or, equivalently

This relation can be written in the form    m y(x)ϕ (x)d x − b y(x)ϕ (x)d x+ R R  + ω2 y(x)ϕ(x)d x = ϕ(0), ∀ϕ ∈ D(R).

(8.1.46)

R

If in formula (8.1.46) we consider the test function ϕ ∈ D(R+ ), then ϕ ∈ D(R− ), we can conclude that the distribution y verifies the equation my  − by  + ω 2 y = 0 on R − {0}.

(8.1.47)

According   to Theorem 8.1.1, y is a classical solution, i.e. it is of class C 2 R+ ∪ R− . We integrate by parts the first integral in (8.1.46)

174

8 Differential Equations in Distributions







0



∞

y(x)ϕ (x)d x = y(x)ϕ (x)d x + y(x)ϕ (x)d x = −∞ 0 R     = y(0− ) − y(0+ ) ϕ (0) + y  (0+ ) − y  (0− ) ϕ(0)+  ∞  0  y (x)ϕ(x)d x + y  (x)ϕ(x)d x. + −∞



(8.1.48)

0

Then we integrate by parts the second integral in (8.1.46) 

  y(x)ϕ (x)d x = y(0− ) − y(0+ ) ϕ(0)− R  ∞  0 y  (x)ϕ(x)d x + y  (x)ϕ(x)d x. − −∞

(8.1.49)

0

If we consider (8.1.46) and (8.1.47), then we obtain     m y(0− ) − y(0+ ) ϕ (0) + m y  (0+ ) − y  (0− ) ϕ(0)+   + b y(0− ) − y(0+ ) ϕ(0) = ϕ(0), ∀ϕ ∈ D(R).

(8.1.50)

We identify the coefficients and obtain y(0− ) = y(0+ ) and y  (0+ ) = y  (0− ) +

1 , m

(8.1.51)

i.e. y is a continuous function and its derivative y  is discontinuous at the origin, where it has a jump equal to m1 .

8.1.2 An Application In the sequel, we follow [8] in order to give an application for ordinary differential equations in distributions. We consider that X is a Banach space. We denote by X ∗ it topological dual. We consider that A is a nonempty, bounded subset of X . We introduce Pw (X ) = {B ⊆ X |B = ∅ and B is weakly compact}.

(8.1.52)

Let B1 be the closed unit ball in X . As in [8] and following [3], we define the following measure of noncompactness α(A) = inf{t > 0|∃K ∈ Pw (X ) such that A ⊆ K + t B1 }. We denote by A˜ the weak closure of A.

(8.1.53)

8.1 Ordinary Differential Equations

175

Lemma 8.1.1 Let A and B be bounded subsets of X . Then (a) A ⊆ B implies α(A) ≤ α(B); ˜ (b) α(A) = α( A); (c) α(A) = 0 if and only if A˜ is weakly compact; (d) α(A ∪ B) = max[α(A), α(B)]; (e) α(A) = α(conv A); (f) α(A + B) ≤ α(A) + α(B); (g) α(x + A) = α(A) for all x ∈ X ; (h) α(λA) = |λ|α(A) for all λ ∈ R;   (i) α ∪0≤t≤t0 t A = t0 α(A). Lemma 8.1.2 Let E ⊆ C X ([0, b]) be bounded and equicontinuous for the strong topology. Then (8.1.54) α(E) = sup α(E(t)) = α(E([0, b])), t∈[0,b]

where E(t) = {h(t)|h(·) ∈ E, t ∈ [0, b]}.

(8.1.55)

Proof We will prove that α (E([0, b])) ≤ supt∈[0,b] α (E([0, b])) = λ.

(8.1.56)

Note that by assumption, E is an equicontinuous family. Then given any ε > 0, we can find a η > 0 such that if |t − t  | < η, then h(t) − h(t  ) < ε, for all h(·) ∈ E. We consider that {tk }nk=0 is a partition of [0, b] such that for all k ∈ {0, 1, . . . , n − 1} we have |tk+1 − tk | < η. Moreover, we consider Q k ∈ Pw (X ) such that E(tk ) ⊆ Q k + (λ + ε)B1 .

(8.1.57)

E ([0, b]) ⊆ ∪nk=0 Q k + (λ + 2ε)B1 .

(8.1.58)

We will prove that To this end, let x ∈ E([0, b]). We suppose that x = h(t), t ∈ / {tk }nk=0 . Note that otherwise there is nothing to show. It follows that t ∈ (tk , tk+1 ) for some k ∈ {0, 1, . . . , n − 1}. Then we have x ∈ h(tk ) + (h(t) − h(tk )) ∈ Q k + (λ + ε)B1 + εB1 = Q k + (λ + 2ε)B1 . (8.1.59) This proves the claim. Now we consider ε ↓ 0. Then α (E([0, b])) ≤ λ = sup α (E([0, b])) .

(8.1.60)

t∈[0,b]

Moreover, we have α(E(t)) ≤ α (E([0, b])) for all t ∈ [0, b]. It follows that

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8 Differential Equations in Distributions

sup α (E(t)) ≤ α (E([0, b])) .

(8.1.61)

t∈[0,b]



This concludes the lemma.

We will consider that X w is the Banach space X with the weak topology. We will also consider subintervals [t, t + r ] of [0, b]. Definition 8.1.4 A Kamke function [8] is a function f : [0, b] × R+ → R+ such that (i) for all y ∈ R+ , t → f (t, y) is measurable and f (t, y) ≤ ϕ(t) with ϕ(·) ∈ L 1+ ; (ii) for all t ∈ [0, b], y → f (t, y) is continuous; t (iii) y(t) ≡ 0 is the only solution of y(t) ≤ 0 f (s, y(s)) ds, y(0) = 0. Let g : [0, b] × X → X be a weakly continuous vector field. We will consider the following Cauchy problem 

y˙ (x) = g (x, y(x)) y(0) = y0

.

(8.1.62)

A solution of the problem (8.1.62) is a strongly continuous, once weakly differentiable function y : [0, b] → X which satisfies (8.1.62), where y˙ (·) denotes the weak derivative. In this case, y(·) is almost everywhere strongly differentiable and satisfies (8.1.62) with y˙ (·) being the strong derivative. In the sequel, we present an existence result. Theorem 8.1.3 Let g : [0, b] × X → X be a vector field such that (i) g(·, ·) is continuous from [0, b] × X w into X w (i.e., g(·, ·) is weakly continuous); (ii) for all (x, y), we have g(x, y) ≤ N ; (iii) for all A ⊆ X nonempty and bounded we have lim α (g([t, t + r ] × A)) ≤ f (t, α(A)) , r ↓0

(8.1.63)

where f (·, ·) is a Kamke function. Then (8.1.62) admits a solution. Proof Let Lip N ([0, b]) = {y(·) ∈ C X ([0, b])|y(·) is N-Lipschitz}. We introduce the nonlinear integral operator ϕ : Lip N ([0, b]) → Lip N ([0, b]) by  (ϕy)(x) = y0 +

x

g (z, y(z)) dz.

(8.1.64)

0

We will show that it is weakly-weakly sequentially continuous. w−C X

We assume that yn (·) → y(·). We have [C X ([0, b])]∗ = M X ∗ ([0, b]), which includes bounded, regular, vector measures from [0, b] into X ∗ , which are of bounded variation. Hence, for all m(·) ∈ M X ∗ ([0, b]), we obtain

8.1 Ordinary Differential Equations

177

(m, yn (·) − y(·)) → 0, as n → ∞.

(8.1.65)

We consider that m = x ∗ δx , where x ∗ ∈ X ∗ , x ∈ [0, b] and δx is the Dirac measure concentrated on x. It follows that 

 x ∗ , yn (x) − y(x) → 0, as n → ∞.

(8.1.66)

Therefore, we have w

yn (x) → y(x) as n → ∞, ∀x ∈ [0, b].

(8.1.67)

By the weak continuity of the vector field g(·, ·) and by the Lebesgue dominated convergence theorem, it follows that 

x

w



x

g (z, yn (z)) dz →

0

g (z, y(z)) dz, ∀x ∈ [0, b].

(8.1.68)

0

Then for every m(·) ∈ M X ∗ ([0, b]), we obtain 

 (m, ϕyn − ϕy) =

[0,b]

x

(g(z, yn (z)) − g(z, y(z))) dz dm(x).

(8.1.69)

0

x w x Note that for all x ∈ [0, b], we have 0 g(z, yn (z))dz → 0 g(z, y(z))dz as n → ∞. Moreover, if we approximate m(·), uniformly on Lip N ([0, b]), by linear combinations of Dirac measures, then it follows that (m, ϕyn − ϕy) → 0, as n → ∞.

(8.1.70)

Therefore, ϕ(·) is weakly-weakly sequentially continuous. Then we consider the classical Caratheodory approximations [8]  yn (x) =

y0 y0 +



for 0 ≤ x ≤ x− n1 0

g (z, yn (z)) dz for

1 n

1 n

≤x ≤b

(8.1.71)

Clearly, for all n ≥ 1, we have yn (·) ∈ C X ([0, b]). Moreover, we have  x yn (x) − ϕyn (x) = y0 − y0 − g(z, yn (z))dz ≤ 0  x 1 ≤ g(z, yn (z))dz ≤ N x for 0 ≤ x ≤ n 0 and

(8.1.72)

178

8 Differential Equations in Distributions

 x− 1  x n yn (x) − ϕyn (x) = y0 + g(z, yn (z))dz − y0 − g(z, yn (z))dz ≤ 0 0  x 1 1 ≤ g(z, yn (z))dz ≤ N , for ≤ x ≤ b. 1 n n x− n (8.1.73) Hence, we have yn − ϕyn  → 0 as n → ∞. We consider that Q = {yn (·)}n≥1 and L = ϕ(Q) = {ϕyn (·)}n≥1 . Note that α(Q − L) = α((I − ϕ)(Q)) = 0.

(8.1.74)

Moreover, for all x ∈ [0, b], we have Q(x) ⊆ Q(x) − L(x) + L(x) = (I − ϕ)(Q(x)) + L(x).

(8.1.75)

Hence, we obtain α (Q(x)) ≤ α ((I − ϕ)(Q(x))) + α(L(x)).

(8.1.76)

Since α ((I − ϕ)(Q)) = 0, we obtain α ((I − ϕ)(Q(x))) = 0. It follows that α(Q(x)) ≤ α(L(x)).

(8.1.77)

For all x ∈ [0, b], we also have L(x) ⊆ L(x) − Q(x) + Q(x) = Q(x) − (I − ϕ)(Q(x)).

(8.1.78)

Hence, we obtain α(L(x)) ≤ α(Q(x)) + α ((I − ϕ)(Q(x))) .

(8.1.79)

Since α ((I − ϕ)(Q(x))) = 0, we obtain α(Q(x)) ≤ α(Q(x)).

(8.1.80)

By (8.1.77) and (8.1.80), we obtain α(Q(x)) = α(L(x)) = p(x).

(8.1.81)

We will show that L = ϕ(Q) is an equicontinuous set. We consider x, x  ∈ [0, b]. Then we have

8.1 Ordinary Differential Equations

179

   x  x     ϕy(x ) − ϕy(x) =  y0 + g(z, y(z))dz − y0 − g(z, y(z))dz  ≤   0 0     x    ≤ g(z, y(z))dz  ≤ N |x  − x|.  x  (8.1.82) This implies that L is equicontinuous. We also have (8.1.83) | p(x  ) − p(x)| ≤ α(B1 )m L (x  − x), 

where m L (·) is the modulus of equicontinuity of the family L, i.e, m L (r ) = sup{h(x  ) − h(x)|x, x  ∈ [0, b], |x  − x| ≤ r, h(·) ∈ L}.

(8.1.84)

We also have α(B1 ) ≤ 1. This implies that   | p(x  ) − p(x)| ≤ m L |x  − x| ≤ N |x  − x|.

(8.1.85)

It follows that p(·) is absolutely continuous, hence differentiable at all x ∈ [0, b] \ N , λ(N ) = 0. We fix x ∈ [0, b] \ N . We consider that ε > 0 is given. Then we can find η > 0 such that |s − p(x)| < η implies | f (x, s) − f (x, p(x))| < ε.

(8.1.86)

This is possible since f is a Kamke function, so f (x, ·) is continuous. Let r > 0 be such that Nr < η and t + r ≤ b. Let Q r = {y(z)|y(·) ∈ Q, x ≤ z ≤ x + r }. Then we can find q with 0 < q < r such that α (g([x, x + q] × Q r )) ≤ h (x, α(Q r )) + ε.

(8.1.87)

We have α(Q r ) =

sup

z∈[x,x+r ]

α(Q(z)) = p(x) ˜

(8.1.88)

for some x˜ ∈ [x, x + r ]. It follows that 0 ≤ α(Q r ) − p(x) ≤ m L (|x˜ − x|) ≤ N |x˜ − x| ≤ Nr < η.

(8.1.89)

By the choice of η > 0, we obtain | f (x, α(Q r )) − f (x, p(x))| < ε.

(8.1.90)

Note that g(·, ·) is weakly continuous and Pettis integrable. It follows that for all y(·) ∈ Q and for v < q, we obtain

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8 Differential Equations in Distributions



x+v

ϕy(x + v) = ϕy(x) +

g(z, y(z))dz ⊆

x

(8.1.91)

⊆ ϕy(x) + vconv ∪z∈[x,x+v] g(z, y(z)). Hence, we obtain L(x + v) ⊆ Q(x) + v · convg([x, x + v] × Q r ).

(8.1.92)

It follows that α(L(t + v)) ≤ α(L(x)) + α (v · convg([x, x + v] × Q r )) .

(8.1.93)

By the properties of α(·), we obtain α (v · convg([x, x + v] × Q r )) = v · α (convg([x, x + v] × Q r ) = = v · α (g([x, x + v] × Q r )) ≤ v · f (x, α(Q r )) + v · ε ≤

(8.1.94)

≤ v · f (x, p(x)) + 2v · ε. It follows that p(x + v) ≤ p(x) + v · f (x, p(x)) + 2v · ε.

(8.1.95)

p(x + v) − p(x) ≤ f (x, p(x)) + 2ε. v

(8.1.96)

Hence, we have

We pass to the limit as v → 0+ . Since x ∈ [0, b] \ N , we obtain

Let ε ↓ 0. Then we have

p(x) ˙ ≤ f (x, p(x)) + 2ε.

(8.1.97)

p(x) ˙ ≤ f (x, p(x)).

(8.1.98)

This implies that



x

p(x) ≤

f (z, p(z))dz,

(8.1.99)

0

while p(0) = α(Q(0)) = α(L(0)) = α({y0 }) = 0. Since f (·, ·) is a Kamke function, we have for all x ∈ [0, b] p(x) = 0. (8.1.100) Therefore, we obtain α(Q(x)) = 0.

(8.1.101)

We deduce that α(Q) = 0. which implies that K˜ is w-compact in C X ([0, b]).

8.1 Ordinary Differential Equations

181

By the Eberlein-Smulian theorem [8] and by passing to a subsequence if necessary, we may assume that w−C X

yn (·) → y(·) ∈ C X ([0, b]).

(8.1.102) w−C X

Note that ϕ(·) is w-sequentially continuous. Therefore, we have ϕyn → ϕy. Hence, we obtain w−C X yn − ϕyn → y − ϕy as n → ∞. (8.1.103) Since the norm is w-lower semicontinuous, we obtain lim yn − ϕyn ∞ ≥ y − ϕy∞ .

n→∞

(8.1.104)

Since we showed that yn − ϕyn ∞ → 0 as n → ∞, it follows that y − ϕy∞ = 0. Hence, we obtain



(8.1.105)

x

y(x) = y0 +

g(z, y(z))dz.

(8.1.106)

0

Since g(·, ·) is weakly continuous, it follows that y(·) is weakly differentiable and the weak derivative y˙ (·) satisifies the Cauchy problem. Then we conclude that y(·) is the desired weak solution. 

8.2 Partial Differential Equations 8.2.1 The Direct Product Let us consider the differential operator P(x, D) =

n 

aα (x)D α , aα ∈ C ∞ ,

(8.2.1)

α=0

where we used the notation  D≡

∂ ∂ ∂ , ,··· , ∂x1 ∂x2 ∂xn

 .

(8.2.2)

The notation P(x, D) shows that the operator is polynomial and has variable coefficients. For constant coefficients we use the notation P(D). We consider the

182

8 Differential Equations in Distributions

equation

P(x, D)u(x) = f (x), f ∈ D (Rn ), x ∈ R.

(8.2.3)

Definition 8.2.1 A generalized solution of the Eq. (8.2.3), in the domain  ⊂ Rn is any distribution u(x) ∈ D(Rn ) which verifies the equation in the domain under consideration in the sense of distributions, that is (P(x, D)u(x), ϕ(x)) = ( f (x), ϕ(x)) , ∀ϕ(x) ∈ D(Rn ).

(8.2.4)

Every classical solution of the Eq. (8.2.3) is a generalized solution. The converse statement is not always true. If f is a continuous function on the domain  and the function u(x) ∈ C m (), then u(x) verifies the equation both in the classical sense and in the sense of distributions. In this case, the derivatives of order ≤ m in the classical sense and in the sense of distributions coincide. The function [P(x, D)u(x) − f (x)]

(8.2.5)

is null for ∀x ∈  ⊂ Rn and we have P(x, D)u(x) − f (x) = 0

(8.2.6)

in the sense of distributions as well since (P(x, D)u(x) − f (x), ϕ(x)) = 0, ∀ϕ ∈ D ().

(8.2.7)

Definition 8.2.2 A fundamental solution of the differential operator with constant coefficients P(D) is a distribution E(x1 , x2 , . . . , xn ) ∈ D (),  ⊂ Rn which verifies the equation P(D)E(x1 , x2 , . . . , xn ) = δ(x1 , x2 , . . . , xn ),

(8.2.8)

where δ(x1 , x2 , . . . , xn ) is the n-dimensional Dirac distribution. Observation 8.2.1 The Eq. (8.2.8) has infinitely many fundamental solutions in the sense that if E 0 (x1 , x2 , . . . , xn ) is a solution of the homogeneous equation and E + (x1 , x2 , . . . , xn ) is a particular solution of the Eq. (8.2.8), then E(x1 , x2 , . . . , xn ) = E 0 (x1 , x2 , . . . , xn ) + E + (x1 , x2 , . . . , xn )

(8.2.9)

is a solution of the Eq. (8.2.8). In order to determine the particular fundamental solution we apply an integral transform and we obtain P (−is1 , −is2 , . . . , −isn ) F [E(x)] = 1.

(8.2.10)

8.2 Partial Differential Equations

183

Then we can derive the transform of the particular fundamental solution   F E + (x) =

1 . P (−is1 , −is2 , . . . , −isn )

(8.2.11)

Since P is a polynomial, the fraction on the right-hand side in (8.2.11) can be decomposed in simple fractions and we can derive the preimages, i.e. we can determine the particular fundamental solution E + (x). If E(x) verifies the Eq. (8.2.8), then E(x) ∗ f (x) is a solution of the Eq. (8.2.1) written for constant coefficients P(D) [E(x) ∗ f (x)] = P(D)E(x) ∗ f (x) = δ(x) ∗ f (x) = f (x),

(8.2.12)

owing to the properties of the convolution product. Therefore, we obtain the following theorem. Theorem 8.2.1 If E(x) ∈ D (Rn ) is a fundamental solution of the operator P(D) and the distribution f ∈ D (Rn ) has the property that there exists the convolution product E(x) ∗ f (x), then E(x) ∗ f (x) represents a solution of the equation P(D)u(x) = f (x), f ∈ D (Rn ), x ∈ Rn .

(8.2.13)

Note that the notion of fundamental solution makes sense for particular classes of equations with variable coefficients and we can state a theorem that is analogous to the preceding one.

8.2.2 Hyperbolic Partial Differential Equations Hyperbolic partial differential equations have the general form a u(x1 , x2 , . . . , xn , t) = δ(x1 , x2 , . . . , xn , t),

(8.2.14)

where we used the notation a =

∂2 ∂2 ∂2 1 ∂2 + + · · · + − . ∂xn2 a 2 ∂t 2 ∂x12 ∂x22

(8.2.15)

The Eq. (8.2.14) is a generalization of the wave equation. For n = 1 we have ∂ 2 E(x, t) 1 ∂ 2 E(x, t) − = δ(x, t), ∂x 2 a2 ∂t 2 where the Dirac distribution δ(x, t) is the direct product

(8.2.16)

184

8 Differential Equations in Distributions

δ(x, t) = δ(x) ⊗ δ(t).

(8.2.17)

We apply the Fourier transform with respect to x and we obtain − α2 Fx [E(x, t)] −

1 d2 Fx [E(x, t)] = 1[α] ⊗ δ(t), a 2 dt 2

(8.2.18)

where α is a complex variable. The operator on the left-hand side of the relation (8.2.18) leads us to a fundamental solution of the form Fx [E(x, t)] = −

θ(t)a sin (aαt) , a > 0. α

(8.2.19)

We apply the inverse transform in (8.2.19) and we obtain a E(x, t) = − θ(t)θ (at − |x|) , a > 0, 2

(8.2.20)

which can be written in the form  −a/2, |x| < at, t > 0, a > 0 E(x, t) = 0, otherwise.

(8.2.21)

In the sequel, we apply the Laplace transform with respect to the variable x and with respect to the variable t. If we denote by p1 the variable corresponding to x and with p2 the variable corresponding to t, then by the Laplace transform we obtain p12 L t [L x [E(x, t)]] −

a 2 p L t [L x [E(x, t)]] = 1. 2 2

Hence, we obtain L t [L x [E(x, t)]] =

a2 . − p22

(8.2.22)

(8.2.23)

a 2 p12

But the right-hand side of this relation can be decomposed in simple fractions a2 a = 2 2 p1 a 2 p1 − p22



1 1 + ap1 + p2 ap1 − p2

 (8.2.24)

and then we obtain a L t [L x [E(x, t)]] = 2 p1



1 1 + ap1 + p2 ap1 − p2

 .

(8.2.25)

In this form we can apply the inverse transform with respect to t and we obtain

8.2 Partial Differential Equations

185

L x [E(x, t)] = −

 a  ap1 t e − e−ap1 t . 2 p1

(8.2.26)

We apply the inverse transform with respect to x and we obtain a E(x, t) = − θ(t)θ(at − |x|), a > 0. 2

(8.2.27)

8.2.3 Parabolic Partial Differential Equations Parabolic partial differential equations have the following general form a u(x1 , x2 , . . . , xn , t) = δ(x1 , x2 , . . . , xn , t),

(8.2.28)

where the operator  has the expression a =

∂2 ∂2 ∂2 1 ∂ . + 2 + ··· + 2 − 2 ∂xn a ∂t ∂x1 ∂x2

(8.2.29)

The Eq. (8.2.28) is a generalization of the heat equation. For n = 1 we have ∂ 2 u(x, t) 1 ∂u(x, t) − = δ(x, t), ∂x 2 a ∂t

(8.2.30)

where the Dirac distribution δ(x, t) is the direct product δ(x, t) = δ(x) ⊗ δ(t).

(8.2.31)

We apply the Laplace transform with respect to the variable t and then the Fourier transform with respect to the variable x and we obtain − α2 L t [Fx [u]] −

p L t [Fx [u]] = 1(α) ⊗ 1( p) = 1, α

(8.2.32)

where α and p are complex variables that correspond by the two transforms to the variables x and t, respectively. Clearly, the last relation can be written as L t [Fx [u]] = −

a . p + aα2

(8.2.33)

We apply the inverse Laplace transform and we obtain Fx [u] = −aθ(t)e−aα t , a > 0. 2

(8.2.34)

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8 Differential Equations in Distributions

In this relation we apply the inverse Fourier transform and we obtain θ(t) √ − x 2 ae 4at , a > 0. E(x, t) = u(x, t) = − √ 2 πt

(8.2.35)

Observation 8.2.2 If in the Eq. (8.2.30) the constant a is negative, then we obtain the inverse equation of heat conduction. If a is a purely imaginary number, then we have Schrödinger’s equation from quantum mechanics for a material point situated in a field of conservative forces.

8.2.4 Elliptic Partial Differential Equations Elliptic partial differential equations have the general form ∂2u ∂2u ∂2u + 2 + · · · + 2 = f (x1 , x2 , . . . , xn ) , 2 ∂xn ∂x1 ∂x2

(8.2.36)

where the unknown is the distribution u u = u (x1 , x2 , . . . , xn ) , u ∈ D (Rn )

(8.2.37)

and f is the free term and it is a distribution f = f (x1 , x2 , . . . , xn ) , f ∈ D (Rn ).

(8.2.38)

In the particular case n = 3, we have the generalized Poisson equation ∂ 2 E(x, y, z) ∂ 2 E(x, y, z) ∂ 2 E(x, y, z) + + − k 2 E(x, y, z) = δ(x, y, z), ∂x 2 ∂ y2 ∂z 2 (8.2.39) where δ(x, y, z) is the 3-dimensional Dirac distribution and can be expressed by the direct product δ(x, y, z) = δ(x) ⊗ δ(y) ⊗ δ(z). (8.2.40) In (8.2.39) we apply the Fourier transform with respect to the three variables x, y and z and we obtain 1 , (8.2.41) F [E(x, y, z)] = − 2 α + β2 + γ2 where the variables α, β, γ correspond to the variables x, y and z by the Fourier transform. In (8.2.41) we apply the inverse Fourier transform and we obtain

8.2 Partial Differential Equations

187

E(x, y, z) =

1 −k R e , 4π R

(8.2.42)

where we used the notation R=



x 2 + y2 + z2.

(8.2.43)

1 . 4π R

(8.2.44)

In particular, for k = 0 we obtain E(x, y, z) = − If we replace k by −k, then we obtain 1 kR ¯ e , E(x, y, z) = − 4π R

(8.2.45)

where E¯ is the conjugate of E. We combine the last two expressions and we consider that ch(k R) =

 1  kR e + e−k R , 2

(8.2.46)

1 ch(k R). 4π R

(8.2.47)

 1  kR e − e−k R , 2

(8.2.48)

so we obtain E(x, y, z) = − If we consider that sh(k R) =

then we conclude that an integral of the homogeneous equation u − k 2 u = 0 can be written in the form u(x, y, z) =

1 sh(k R). R

(8.2.49)

(8.2.50)

If instead of the real constant k we consider the complex constant ik, then for the inhomogeneous equation u + k 2 u = δ(x, y, z) (8.2.51) we obtain solutions of the form E(x, y, z) = −

1 −ik R e 4π R

(8.2.52)

188

8 Differential Equations in Distributions

and

1 ik R ¯ E(x, y, z) = − e . 4π R

(8.2.53)

For the homogeneous equation u + k 2 u = 0

(8.2.54)

we obtain a solution of the form u(x, y, z) =

1 sin(k R). R

(8.2.55)

8.2.5 The Cauchy Problem Let us consider a linear partial differential equation with constant coefficients  P

∂ ∂ , ∂x ∂t

 u(x, t) = f (x, t), x ∈ Rn ,

(8.2.56)

where f (x, t) is a distribution, f (x, t) ∈ D (Rn ). For the Eq. (8.2.56) we search for a solution in the set of distributions, u(x, t) ∈ D (Rn ). The fundamental solution E(x, t) of the Eq. (8.2.56) verifies the equation  P

∂ ∂ , ∂x ∂t

 E(x, t) = δ(x, t), x ∈ Rn .

(8.2.57)

If in (8.2.56) we set f (x, t) ≡ 0, then we obtain the homogeneous equation  P

∂ ∂ , ∂x ∂t

 u(x, t) = 0, x ∈ Rn .

(8.2.58)

Definition 8.2.3 The Cauchy problem associated to the homogeneous equation (8.2.58) is the problem of determining the distribution u(x, t) ∈ D (Rn ) which verifies the Eq. (8.2.58) in the sense of distributions and which along with its derivatives, up to the order m − 1 verifies the conditions ∂ ∂ m−1 u(x, 0+ ) = u 1 (x), . . . , m−1 u(x, 0+ ) = u m−1 (x), ∂t ∂t (8.2.59) x ∈ Rn , where u 0 (x), u 1 (x), . . . , u m−1 (x) are given distributions. A fundamental solution of the Cauchy problem (8.2.58)–(8.2.59) is a distribution E(x, t) which satisfies the homogeneous equation (8.2.58) and the initial conditions u(x, 0+ ) = u 0 (x),

8.2 Partial Differential Equations

E(x, 0) = 0,

189

∂ m−2 ∂ ∂ m−1 E(x, 0) = 0, . . . , m−2 E(x, 0) = 0, m−1 E(x, 0) = δ(x), ∂t ∂t ∂t (8.2.60)

x ∈ Rn . Theorem 8.2.2 If the distribution E(x, t) is such that the convolution product E(x, t) ∗ u m−1 (x) makes sense, then the solution of the Cauchy problem for the homogeneous equation (8.2.58) and the initial conditions ∂ m−2 ∂ ∂ m−1 u(x, 0) = 0, . . . , m−2 u(x, 0) = 0, m−1 u(x, 0) = u m−1 (x), ∂t ∂t ∂t (8.2.61) x ∈ Rn , can be written in the form u(x, 0) = 0,

u(x, t) = E(x, t) ∗ u m−1 (x).

(8.2.62)

Proof We have  P

∂ ∂ , ∂x ∂t



 

∂ ∂ u(x, t) = P , E(x, t) ∗ u m−1 = 0 ∂x ∂t

(8.2.63)

since E(x, t) verifies the homogeneous equation (8.2.58). Moreover, we have ∂k ∂ k E(x, t) ∂k u(x, t) = ∗ u(x, t), t) ∗ u(x, t)] = [E(x, ∂t k ∂t k ∂t k

(8.2.64)

for k = 0, m − 1. If we consider (8.2.60) then we note that u(x, t) from (8.2.62) verifies the conditions (8.2.61), which concludes the proof of the theorem.  Observation 8.2.3 (i) If we apply several times the method from the Theorem 8.2.2, then we can determine the solution of the Cauchy problem for the homogeneous equation (8.2.58) with the initial conditions (8.2.59). (ii) In the considerations above, we can choose as the initial moment any t0 = 0 without changing the methods of finding the solution of the problem. (iii) In order to solve the Cauchy problem we can consider the function u(x, ¯ t) = θ(t)u(x, t).

(8.2.65)

We use the differentiation rule and the fact that for t > 0 ⇒ θ(t) = 1 ⇒ u(x, ¯ t) = u(x, t) in order to obtain from the Eq. (8.2.58)

(8.2.66)

190

8 Differential Equations in Distributions

 P

∂ ∂ , ∂x ∂t

 u(x, ¯ t) = g(x, t),

(8.2.67)

where the function g(x, t) can be expressed by means of the initial conditions. The fundamental solution of the Eq. (8.2.67) is known, so we can express the solution of the Cauchy problem as u(x, ¯ t) = E(x, t) ∗ g(x, t), (8.2.68) where the convolution product refers to all n + 1 variables.

8.2.6 An Application In the sequel, we follow [13] in order to present weak solutions of the incompressible Euler equations, which describe the motion of a perfect incompressible fluid. In Eulerian variables, we can write the continuum equations associated with the conservation of momentum and mass of arbitrary fluid regions as ∂t v + (v · ∇) v + ∇ p = 0, div v = 0.

(8.2.69)

Note that v = v(x, t) is the velocity and p = p(x, t) is the pressure. In the sequel, we will consider the three-dimensional case with periodic boundary conditions, as discussed in [13]. More precisely, the spatial domain will be the flat three-dimensional torus T3 = R3 \ Z3 . A on a given time interval [0, T ] is by definition a pair (v, p) ∈  classical solution  C 1 T3 × [0, T ] . As far as smooth solutions are concerned, existence and uniqueness for short time (or local well-posedness) were proven in Hölder spaces C 1,α , α > 0 [6] or Sobolev spaces H s , s > 5/2, see [4, 5]. Taking into consideration the physical applications, several weaker notions of solution have been proposed in the literature. First, a notion of solution was introduced that allows discontinuities in the vorticity and in the velocity. More precisely, weak solutions of (8.2.69) are defined as a   pair (v, p) ∈ C T3 × [0, T ] such that, for any connected region  ⊂ T3 with C 1 boundary and any t ∈ (0, T ) 





v(x, t)d x −

∂



v(x, 0)d x +

 t 0

∂

v · nd A(x) = 0.

Note that n is the unit outward normal to .

v(v · n) + pd Ads = 0, (8.2.70)

8.2 Partial Differential Equations

191

Clearly, if (v, p) ∈ C 1 is a solution of (8.2.70), then it is a classical solution of (8.2.69). In fact, we obtain (8.2.70) from the principles of continuum mechanics and the conservation laws of momentum and mass applied to arbitrary fluid regions . If in addition (v, p) ∈ C 1 , then the divergence theorem and a standard localization argument leads to (8.2.69). Note that this definition includes the pressure. Moreover, the pressure can be recovered uniquely, up to an additive constant from (8.2.69) via the equation − p = div div (v ⊗ v).

(8.2.71)

Hence, we can remove the pressure from the equation by projecting the first equation of (8.2.69) onto divergence-free fields. In the sequel, we will define distributional solutions. Using div v = 0, we employ the following identity (v · ∇v)k =

  i

∂ vi ∂xi

 vk =

 ∂ (vi vk ) = [div (v ⊗ v)]k ∂xi i

(8.2.72)

for any k = 1, 2, 3. Then it follows from (8.2.69) that  0

T



 T3

∂t ϕ · v + ∇ϕ : v ⊗ vd xdt +

T3

ϕ(x, 0) · v0 (x)d x = 0

(8.2.73)

  for all ϕ ∈ C ∞ T3 × [0, T ); R3 with div ϕ = 0. Accordingly, the  weakest possible  notion of solution of (8.2.69) is given by a vectorfield v ∈ L 2 T3 × (0, T ) with div v = 0 in the sense of distributions such that (8.2.73) holds. Other types of weak solutions are based on the concept of Young measure. We consider a sequence of velocity fields vk (x, t). Then from classical Young measure theory [16], it follows that there exists a subsequence (not relabeled) and a parametrized probability measure νx,t on R3 such that for all bounded continuous functions f, we have  ∗  f (vk (x, t))  νx,t , f (8.2.74)   weakly ∗ in L ∞ T3 × (0, T ) , where (·, ·) denotes the duality bracket for C0∗ (R3 ) = M(R3 ). The measure νx,t can be interpreted as the probability distribution of the velocity field at the point x at time t when the sequence (vk ) exhibits faster and faster oscillations as k → ∞. We can also employ the generalized Young measure (ν, λ, ν ∞ ), see [1]. In this triple, ν = νx,t is a parametrized probability measure on R3 as before (the oscillation measure), λ is a Radon measure on T3 × (0, T ) (the concentration mea∞ sure) and ν ∞ = νx,t is a parametrized probability measure on S 2 defined λ-a.e. (the concentration-angle measure).

192

8 Differential Equations in Distributions

Hence, (8.2.74) can be replaced by   ∗ f (vk )d xdt  (ν, f ) d xdt + ν ∞ , f ∞ λ

(8.2.75)

in the sense of measures for every continuous f : R3 → R that possesses an L 2 recession function f ∞ . This means that f ∞ (θ) = lims→∞ s −2 f (sθ) exists and is continuous. Clearly, for bounded f , the relation (8.2.75) becomes (8.2.74) since f ∞ = 0 in this case. Note that (ν, λ, ν ∞ ) is able to record oscillations and concentrations in the quadratic term v ⊗ v of the Euler equations (8.2.73). We consider that id is the identity map ξ → ξ and that σ(ξ) = ξ ⊗ ξ, ξ ∈ R3 . Since σ ∞ = σ, a measure-valued solution of the Euler equations is defined to be a generalized Young measure (ν, λ, ν ∞ ) such that div (ν, id) = 0 in the sense of distributions and   T   ∂t ϕ · (ν, id) + ∇ϕ : (ν, σ) d xdt + ∇ϕ : ν ∞ , σ λ(d xdt) = T3 T3 ×(0,T ) 0  =− ϕ(x, 0)v0 (x)d x T3

(8.2.76)   for all ϕ ∈ Cc∞ T3 × [0, T ); R3 with div ϕ = 0. Note that (8.2.76) is a constraint on the first and second moments of the generalized Young measure, that is, on     ∞   , σ λ(d xdt). v¯ = νx,t , id , v ⊗ v = νx,t , σ + νx,t

(8.2.77)

We present a result that follows from [2, 9, 11]. Theorem exist infinitely many non-trivial weak   8.2.3 There v ∈ L ∞ T3 × R of (8.2.73) which have compact support in time.

solutions

The next theorem follows from [15]. Theorem 8.2.4 For any solenoidal v0 ∈ L 2 (T3 ) there exist infinitely many global weak solutions v ∈ L ∞ 0, ∞; L 2 (T3 ) of (8.2.73). Below we present a result that follows from [14]. Theorem 8.2.5 For any measure-valued   solution of (8.2.76) there exists a sequence of weak solution vk ∈ L 2 T3 × (0, T ) of (8.2.73) generating this measure-valued solution, in the sense that (8.2.75) holds true. Note that C 1 solutions of (8.2.69) satisfy the conservation of the kinetic energy in a local form  2   |v| |v|2 + div + p v = 0. (8.2.78) ∂t 2 2

8.2 Partial Differential Equations

193

Clearly, this is a consequence of v · (v · ∇)v =

 k,i

   ∂ v2 ∂ |v|2 k . vk vi vk = vi = div v ∂xi ∂xi 2 2 k,i

(8.2.79)

If we integrate (8.2.78) in space, then we obtain the conservation of the total kinetic energy  d |v(x, t)|2 d x = 0. (8.2.80) dt T3 We denote the energy by 1 E(t) := 2

 T3

|v(x, t)|2 d x.

(8.2.81)

Below we present a result of weak-strong uniqueness.   Theorem 8.2.6 We consider that v ∈ L ∞ [0, T ), L 2 (T3 ) is a solution of (8.2.73) such that ∇v + ∇v T ∈ L ∞ . We suppose that (ν, λ, ν ∞ ) is a solution of (8.2.76) which satisfies the energy inequality 1 2

  T3

R3

|ξ|2 dνx,t (ξ)d x +

1 2



 T3

dλt (x) ≤

T3

|v0 |2 (x)d x for a.e. t.

(8.2.82)

It follows that (ν, λ, ν ∞ ) coincides with v as long as the latter exists, that is νx,t = δv(x,t) for a.a (x, t) ∈ T3 × (0, T ) and λ ≡ 0 on T3 × (0, T ).

(8.2.83)

In 1949, L. Onsager [7] stated the following conjecture. Conjecture Let (v, p) be solutions of (8.2.70) which satisfy the Hölder condition |v(x, t) − v(x  , t)| ≤ C|x − x  |θ ,

(8.2.84)

where C is a constant that does not depend on x, x  ∈ T3 and t. Then (i) if θ > 13 , then any solution (v, p) of (8.2.70) which satisfies (8.2.84) conserves the energy; (ii) for any θ < 13 there exist solutions (v, p) of (8.2.70) which satisfy (8.2.84) that does not conserve the energy. Below we present some estimates on convolutions. We consider that φ ∈ Cc∞ (R3 )  is a symmetric, non-negative mollifying kernel such that φ = 1. We set φk (x) =   k −3 φ xk . Let v ∈ C(T3 ). Then we introduce  vk (x) := v ∗ φk (x) =

R3

v(x − y)φk (y)dy.

(8.2.85)

194

8 Differential Equations in Distributions

Lemma 8.2.1 Let v ∈ C θ (T3 ). We have v − vk 0 ≤ Ck θ [v]θ ,

(8.2.86)

∇vk 0 ≤ Ck θ−1 [v]θ ,

(8.2.87)

(v ⊗ v)k − vk ⊗ vk 0 ≤ Ck 2θ [v]2θ .

(8.2.88)

Proof Step 1. We have     |vk (x) − v(x)| =  φk (x − y) (v(y) − v(x)) dy  ≤     ≤  φk (x − y)|y − x|θ dy  [v]θ .

(8.2.89)

Step 2. We have  ∇vk (x) =

 ∇φk (x − y)v(y)dy =

∇φk (x − y) (v(y) − v(x)) dy.

(8.2.90)

Step 3. We have  (v ⊗ v)k (x) = φk (x − y)v(y) ⊗ v(y)dy =  = φk (x − y) (v(y) − v(x)) ⊗ (v(y) − v(x)) dy+

(8.2.91)

+ v(x) ⊗ vk (x) + vk (x) ⊗ v(x) − v(x) ⊗ v(x). It follows that  (v ⊗ v)k (x) − vk (x) ⊗ vk (x) =

φk (x − y) (v(y) − v(x)) ⊗ (v(y) − v(x)) dy−

− (v(x) − vk (x)) ⊗ (v(x) − vk (x)) . (8.2.92)  We consider that (v, p) is a Hölder-continuous solution of (8.2.70). Then divvk = 0 and (8.2.93) ∂t vk + vk · ∇vk + ∇ pk = −divRk , where Rk = (v ⊗ v)k − vk ⊗ vk . We have the energy balance

(8.2.94)

8.2 Partial Differential Equations

195

d Ek = dt where E k =

1 2



 ∇vk : Rk d x,

(8.2.95)

|vk (x)|2 d x. Then for any T > 0, we have  |E k (T ) − E k (0)| ≤ 0

T

 T3

k 3θ−1 [v(t)]3θ d xdt.

(8.2.96)

As a consequence, if we pass to the limit as k → 0, then E(T ) = E(0), under the condition that  T [v(t)]3θ d x < ∞ for some θ > 1/3. (8.2.97) 0

This proves the part (i) of the conjecture.

References 1. J.J. Alibert, G. Bouchitte, Non-uniform integrability and generalized Young measures. J. Convex Anal. 4(1), 129–147 (1997) 2. C. De Lellis, L. Szekelyhidi Jr., The Euler equations as a differential inclusion. Ann. Math. (2) 170(3), 1417–1436 (2009) 3. F. DeBlasi, On a property of the unit sphere in a Banach space. Bull. Math. Soc. Sci. Math. R. S. Roumanie 21, 259–62 (1977) 4. D.G. Ebin, J. Marsden, Groups of diffeomorphisms and the motion of an incompressible fluid. Ann. Math. 2(92), 102–163 (1970) 5. T. Kato, Nonstationary flows of viscous and ideal fluids in R3 . J. Funct. Anal. 9, 296–305 (1972) 6. L. Lichtenstein, Grundlagen der Hydromechanik (Springer, Berlin, 1929) 7. L. Onsager, Statistical hydrodynamics. Nuovo Cimento (9), 6(Supplemento, 2(Convegno Internazionale di Meccanica Statistica)), 279–287 (1949) 8. N.S. Papageorgiou, Weak solutions of differential equations in Banach spaces. Bull. Austral. Math. Soc. 33, 407–418 (1986) 9. V. Scheffer, An inviscid flow with compact support in space-time. J. Geom. Anal. 3(4), 343–401 (1993) 10. L. Schwartz, Theorie des Distributions (Herman & Cie, Paris, 1971) 11. A.I. Shnirelman, On the nonuniqueness of weak solution of the Euler equation. Comm. Pure Appl. Math. 50(12), 1261–1286 (1997) 12. R. Sikorski, The Elementary Theory of Distributions (Bull. Acad. Pol, Warszawa, 1957) 13. L. Szekelyhidi, Weak solutions of the Euler equations: non-uniqueness and dissipation. Journees Equations aux derivees partielles, Roscoff, 1-5 juin (2015) 14. L. Szekelyhidi Jr., E. Wiedemann, Young measures generated by ideal incompressible fluid flows. Arch. Rational Mech. Anal. (2012) 15. E. Wiedemann, Existence of weak solutions for the incompressible Euler equations. Ann. Inst. H. Poincare Anal. Non Lineaire 28(5), 727–730 (2011) 16. L.C. Young, Lecture on the Calculus of Variations and Optimal Control Theory (American Mathematical Society, 1980)

Chapter 9

Sobolev Spaces

Abstract We define some Sobolev spaces and study some of their properties, such as the inequality of Poincare, trace theorems or embedding results. Then we present the Littlewood-Paley decomposition and Besov spaces.

9.1 The Sobolev Space H 1 () An important notion in the theory of partial differential equations is that of the Sobolev space [2, 4]. Definition 9.1.1 We define the following Sobolev space  H () = 1

 ∂f 2 f ∈ L ()| ∈ L (), i = 1, 2, . . . , N , ∂xi 2

(9.1.1)

where the partial derivatives are in the sense of distributions. Note that

∂f ∂xi

∈ L 2 () if there exists gi ∈ L 2 () such that

 

∂ϕ f (x) (x)d x = − ∂xi

 

gi (x)ϕ(x)d x, ∀ϕ ∈ D().

(9.1.2)

We introduce the following scalar product on H 1 ()  ( f, g) H 1 () =



f (x)g(x)d x +

N   ∂f ∂g (x) (x)d x, ∀ f, g ∈ H 1 () ∂x ∂x i i i=1  (9.1.3)

and the norm || f || H 1 () =



( f, f ) H 1 () .

(9.1.4)

Theorem 9.1.1 The Sobolev space H 1 () is a Hilbert space with respect to the scalar product (9.1.3). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_9

197

198

9 Sobolev Spaces

Proof Clearly, (9.1.3) defines a scalar product on H 1 (). We only have to show that H 1 () is complete with respect to this scalar product. Let (h n )n≥1 be a sequence in H 1 () which   is a Cauchy sequence with respect to n , 1 ≤ i ≤ N are Cauchy sequences the norm (9.1.4). Therefore, (h n )n≥1 and ∂h ∂xi n≥1

in L 2 (). Since L 2 () is complete, there exist h, h i ∈ L 2 (), 1 ≤ i ≤ N such that lim h n = h in L 2 (),

n→∞

(9.1.5)

∂h n = h i in L 2 () for 1 ≤ i ≤ N . n→∞ ∂x i lim

∂h If we show that ∂x = h i in the sense of distributions for every 1 ≤ i ≤ N , then i h ∈ H 1 () and the proof is finished. Let 1 ≤ i ≤ N and ϕ ∈ D(). We have



  ∂ϕ ∂h ∂ϕ ,ϕ = − h d x = − lim hn dx = n→∞ ∂xi ∂xi  ∂x i    ∂h n ϕd x = h i ϕd x = (h i , ϕ) . = lim n→∞  ∂x i 

∂h ∂xi

Therefore,

(9.1.6)

= h i in the sense of distributions and the proof is finished.



Theorem 9.1.2 The space H 1 (a, b) is a subset of C[a, b]. Proof In the sequel, we will denote by dtd the classical derivative and by  the derivative in the sense of distributions. Let f ∈ H 1 (a, b). Then f  ∈ L 2 (a, b). Moreover, the function g defined by  x g(x) = f  (s)ds (9.1.7) a

belongs to C[a, b], is differentiable almost everywhere in [a, b] and almost everywhere in [a, b]. Then for every ϕ ∈ D(a, b) 

 g  , ϕ = − g, ϕ = − 

=− a



b

g(x)ϕ (x)d x = −

a x

f  (s)dsϕ(x)|ab +

 a

 a

b

b



x

dg (x) dx

= f  (x)

f  (s)dsϕ (x)d x =

a

 f  (x)ϕ(x)d x = f  , ϕ .

(9.1.8) Therefore, we obtain g  = f  , so f = g + C, where C is a real constant. Since g is a continuous function, then f is also continuous and the proof is finished. 

9.1 The Sobolev Space H 1 ()

199

Definition 9.1.2 We introduce the Sobolev space H01 () = D()

||·|| H 1 ()

.

(9.1.9)

Theorem 9.1.3 We have H01 (R N ) = H 1 (R N ).

(9.1.10)

In the sequel, we present the inequality of Poincaré, see [2]. Theorem 9.1.4 If  is an open and bounded set, then there exists a constant C = C() > 0 such that  || f || L 2 () ≤ C



||∇

f ||2L 2 ()

21

, ∀ f ∈ H01 ().

(9.1.11)

Observation 9.1.1 By Theorem 9.1.4, we obtain H01 ()  H 1 () if  is bounded. Clearly, every constant function which is different from zero belongs to H 1 (), but it does not verify the inequality (9.1.11), so it cannot belong to H01 (). Let us consider R+N = {x = (x  , x N ) ∈ R N |x  ∈ R N −1 , x N > 0},

(9.1.12)

R−N = {x = (x  , x N ) ∈ R N |x  ∈ R N −1 , x N < 0}

(9.1.13)

and  their boundary  N = {x = (x  , x N ) ∈ R N |x  ∈ R N −1 , x N = 0}.

(9.1.14)

Lemma 9.1.1 The space D(R+N ) is dense in H 1 (R+N ). Proof Let f ∈ H 1 (R+N ). We introduce the function g defined by reflection

g(x) =

f (x  , x N ), f (x  , −x N ),

and we show that g ∈ H 1 (R N ). Since   2 |g(x)| d x = 2 RN

we obtain g ∈ L 2 (R N ).

N R+

if x N > 0 if x N < 0

(9.1.15)

|g(x)|2 d x

(9.1.16)

200

9 Sobolev Spaces

On the other hand, if we define

∂g ∂xi ∂g ∂xi

gi =

(x  , x N ), (x  , −x N ),

if x N > 0 if x N < 0

(9.1.17)

for 1 ≤ i ≤ N − 1 and

gN =

∂g (x  , x N ), ∂x N ∂g − ∂x (x  , −x N ), N

if x N > 0 if x N < 0

(9.1.18)

we obtain gi ∈ L 2 (R N ) for 1 ≤ i ≤ N . Moreover, for every ϕ ∈ D(R N ), we obtain

 ∂ϕ ∂ϕ g, = g(x) (x)d x = ∂xi ∂xi RN   ∂ϕ  ∂ϕ    g(x , x N ) (x , x N )d x d x N + g(x  , −x N ) (x , x N )d x  d x N = = N N ∂xi ∂xi R+ R−   ∂g  ∂g  =− (x , x N )ϕ(x  , x N )d x  d x N − (x , −x N )ϕ(x  , x N )d x  d x N = N ∂x i N ∂x i R+ R−  gi (x)ϕ(x)d x = − (gi , ϕ) . =− RN

(9.1.19) ∂g 1 N Therefore, ∂x = g and g ∈ H (R ). Then there exists a sequence (ϕ i k )k≥1 ⊂ i N D(R ) such that (9.1.20) ϕk → g in H 1 (R N ). If ψk is the restriction of ϕk to R+N , then we obtain (ψk )k≥1 ⊂ D(R+N ) and ψk → f in H 1 (R+N )

(9.1.21)

and the proof is finished. Lemma 9.1.2 For every f ∈ D(R+N ), we have || f (·, 0)|| L 2 (R N −1 ) ≤ || f || H 1 (R+N ) .

(9.1.22)

Therefore, the restriction to  N of f ∈ D(R+N ) → f (·, 0) ∈ D(R N −1 )

(9.1.23)

can be extended uniquely by density and by continuity to a linear and continuous function (9.1.24) γ : H 1 (R+N ) → L 2 ( N ).

9.1 The Sobolev Space H 1 ()

201

Definition 9.1.3 For every function f ∈ H 1 (R+N ), γ( f ) is called the trace of f on N . Note that if f is regular (for instance, f ∈ C(R+N )), then we can consider the restriction of f to  N and γ( f ) = f | N . Definition 9.1.4 A set  ⊂ R N is m-regular if it is bounded and there exists a finite number of open sets (Oi )0≤i≤M and for each i = 1, 2, . . . , M there is an application ϕi : Oi → BR N (0, 1) such that M ¯ ⊂ ∪i=0 1. O0 ⊂ ,  Oi ; 2. ϕi is bijective, ϕi and ϕi−1 are m times continuously differentiable; 3. ϕi (Oi ∩ ) = BR N (0, 1) ∩ R+N = {y = (y  , y N ) ∈ R N ||y| < 1, y N > 0}; ϕi (Oi ∩ ∂) = BR N (0, 1) ∩ R N −1 = {y = (y  , y N ) ∈ R N ||y| < 1, y N = 0}.

Lemma 9.1.3 If  is 1 − r egular , then there exists a linear and continuous operator P : H 1 () → H 1 (R N ) such that, for every f ∈ H 1 (), we have P f = f almost everywhere in .

(9.1.25)

¯ is dense in H 1 (). Lemma 9.1.4 If  is 1-regular, then D() ¯ we denote by γ( f ) the restriction of f to the boundary  For every f ∈ D(), of  γ( f ) = f | . (9.1.26)

Lemma 9.1.5 If  is 1-regular, then there exists a constant C > 0 such that ¯ ||γ( f )|| L 2 () ≤ C|| f || H 1 () , ∀ f ∈ D().

(9.1.27)

In the sequel, we present a trace theorem [4]. Theorem 9.1.5 Let  be a 1-regular open set. There exists a unique application γ with the following properties: 1. γ : H 1 () → L 2 (); 2. γ is linear and continuous ¯ 3. γ( f ) = f | , ∀ f ∈ D(). Definition 9.1.5 For every function f ∈ H 1 (), γ( f ) will be called the trace of f on . If  is a 1-regular set, then we denote by ν the exterior unit normal vector in every point of the boundary  of . By νi we denote the component i of ν.

202

9 Sobolev Spaces

Theorem 9.1.6 Let  be a 1-regular set. If f and g are two functions from H 1 (), then    ∂f ∂g gd x = − f dx + f gνi dσ. (9.1.28)  ∂x i  ∂x i  In the sequel, we will present a useful characterization of the functions from H01 (). Theorem 9.1.7 Let  be an open and 1-regular set. For every f ∈ H 1 () we define the extension by zero of f to R N

fˆ(x) =

f (x) if x ∈  0 if x ∈ R N \ .

(9.1.29)

The following properties are equivalent: 1. f ∈ H01 (); 2. fˆ ∈ H 1 (R N ). In the sequel, we will characterize the elements of H01 () as those functions in H () which become null on the boundary of . 1

Theorem 9.1.8 If  is a 1-regular set, then H01 () = { f ∈ H 1 ()|γ( f ) = 0}.

(9.1.30)

Proof Step 1. We prove the inclusion H01 () ⊂ { f ∈ H 1 ()|γ( f ) = 0}.

(9.1.31)

Let f ∈ H01 (). There exists a sequence ( f k )k≥1 ⊂ D() which converges to f in H 1 (). By the continuity of the trace application we deduce that γ( f ) is the limit of the sequence (γ( f k ))k≥1 when k tends to infinity. Since γ( f k ) = 0 for every k, we deduce that γ( f ) = 0. Step 2. We prove the inclusion { f ∈ H 1 ()|γ( f ) = 0} ⊂ H01 ().

(9.1.32)

Let f ∈ H 1 () be such that f | = 0 and let fˆ be the extension by zero to R N . If we prove that fˆ ∈ H 1 (R N ), then we obtain f ∈ H01 (). ∂f ∈ Clearly, fˆ ∈ L 2 (R N ). Let fˆi be the extension by zero to R N \  of f i = ∂x i L 2 (). Then we have fˆi ∈ L 2 (R N ). Moreover, for every ϕ ∈ D(R N ), by applying Green’s formula, we obtain

9.1 The Sobolev Space H 1 ()

203

   ∂f fˆi , ϕ = (x)ϕ(x)d x = fˆi (x)ϕ(x)d x = RN  ∂x i    ∂ϕ ∂ϕ f (x) (x)d x + f (x)ϕ(x)νi (x)d x = − f (x) (x)d x, =− ∂xi ∂xi    (9.1.33) where the last relation holds true since γ( f ) = 0. Therefore, we obtain

    ∂ϕ ∂ϕ ∂ϕ . (x) f (x)d x = − (x) fˆ(x)d x = − fˆ, fˆi , ϕ = − ∂xi  ∂x i R N ∂x i (9.1.34) ∂ fˆ 2 1 N ˆ ˆ We obtain ∂xi = f i ∈ L (), so f ∈ H (R ). 

9.2 The Sobolev Space H m () Definition 9.2.1 For every integer m ≥ 1, we define the Sobolev space of exponent m in  (9.2.1) H m () = { f ∈ L 2 ()|D α f ∈ L 2 (), |α| ≤ m}. In H m () we define the scalar product  ( f, g) H m () =

⎛ ⎝





⎞ Dα f Dα g⎠ d x

(9.2.2)

|α|≤m

and we denote by || f || H m () =



( f, f ) H m () .

(9.2.3)

Theorem 9.2.1 The Sobolev space H m () is a Hilbert space with respect to the scalar product from (9.2.2). Theorem 9.2.2 Let  be a 1-regular set. If f ∈ H 2 () and g ∈ H 1 (), then  −

 

 f gd x =

 

∇ f ∇gd x −



∂f gdσ. ∂ν

(9.2.4)

9.3 The Sobolev Space W k, p () In the sequel, we consider that  ⊂ Rn is open, that k ∈ N and that 1 ≤ p ≤ ∞.

204

9 Sobolev Spaces

Definition 9.3.1 W k, p () is the space of all distributions f ∈ D () such that D α f ∈ L p (), ∀|α| ≤ k.

(9.3.1)

This space is endowed with the norm ⎛

f W k, p () = ⎝

 |α|≤k



⎞ 1p |D α f (x)| p d x ⎠ .

(9.3.2)

Observation 9.3.1 1. For k = 0 we have W 0, p () = L p (). 2. If k ≥ l, then W k, p () → W l, p () and the embedding is continuous. Theorem 9.3.1 W k, p () is a Banach space, see [2]. Proof We will prove that W k, p () is complete. We consider a Cauchy sequence ( f j ) j∈N in W k, p (). For every α = (α1 , . . . , αn ) with |α| ≤ k, we note that (D α f j ) j∈N is a Cauchy sequence in L p (). This space is complete, so D α f j → gα in L p (), ∀|α| ≤ k.

(9.3.3)

In particular, we have f j → g0 in L p (). Therefore, we obtain f j → g0 in D (). On the other hand, D α is, for every α, a continuous linear operator from D () into D (). Therefore, we obtain D α f j → D α g0 , ∀|α| ≤ k.

(9.3.4)

Since the limit is unique, we obtain gα = D α g0 . Hence, we obtain g0 ∈ W k, p ()  and g j → g0 in W k, p (). Definition 9.3.2 We denote by W0 () the closure of Cc∞ () in W k, p (). k, p

k, p

Note that W0 () is a Banach space since it is a closed subspace of a Banach space. Note that, by definition, we have Cc∞ () ⊂ W0 () ⊂ D (), k, p

(9.3.5)

where the embedding is continuous and Cc∞ () is dense in W0 (). Hence,   k, p k, p W0 () is a normal space of distributions. Its dual is denoted by W0 () and it is a subspace of D ().   k, p Theorem 9.3.2 The dual W0 () coincides with the space of all distributions T ∈ D () such that k, p

9.3 The Sobolev Space W k, p ()

205

T =



∂ α fα ,

(9.3.6)

|α|≤k

where f α ∈ L q () and

1 p

+

1 q

= 1.

Proof Step 1. We consider T to be of the form (9.3.6). Then we define for all ϕ ∈ Cc∞ () the following linear functional (T, ϕ) =



(−1)|α| ( f α , ∂ α ϕ) =

|α|≤k



(−1)|α|



|α|≤k



fα

∂αϕ d x. ∂x α

(9.3.7)

By Hölder’s inequality [2], we obtain |(T, ϕ)| ≤



f α L q ∂ α ϕ L p ≤ C ϕ W k, p () .

(9.3.8)

|α|≤k

Since Cc∞ () is dense in W0 (), this proves that T defines a continuous linear k, p functional on W0 ().   k, p Step 2. Now we prove the converse statement. We assume that T ∈ W0 () . We consider that N is the number of n-tuples α = (α1 , . . . , αn ) such that |α| ≤ k. Let us denote by (L p ()) N = L p () × · · · × L p () (N times) the product of N copies of L p () endowed with the product topology. We define the map k, p

 N J : W k, p ()  ϕ → (∂ α ϕ)|α|≤k ∈ L p () .

(9.3.9)

Clearly, this is a canonical isometry from W k, p () into (L p ()) N . In particular, W k, p () can be identified, by J , with a subspace E of (L p ()) N . We consider on k, p E = J (W0 ()) the following linear functional  k, p F (∂ α ϕ)|α|≤k = (T, ϕ) , ∀ϕ ∈ W0 ().

(9.3.10)

k, p

F is continuous on E since T is continuous on W0 (). By the Hahn-Banach theorem [2], we can extend F to a continuous linear functional on (L p ()) N . But the dual of (L p ()) N can be identified with the product (L q ()) N . Therefore, there are functions gα ∈ L q (), |α| ≤ k, such that    α ∂αϕ k, p gα α d x, ∀ϕ ∈ W0 (), (T, ϕ) = F (∂ ϕ)|α|≤k = ∂x |α|≤k 

(9.3.11)

in particular, for all ϕ ∈ Cc∞ (). By the definition of the derivative in the sense of distributions, the last relation leads to

206

9 Sobolev Spaces

(T, ϕ) =



∂ α f α ϕ, ∀ϕ ∈ Cc∞ (),

(9.3.12)

|α|≤k

where f α = (−1)|α| gα .



Theorem 9.3.3 Let k ≥ 0 and 1 ≤ p < ∞. Then Cc∞ (Rn ) is a dense subspace of W k, p (Rn ). Proof Step 1. We will show that every element of W k, p (Rn ) can be approximated, in the sense of the norm · W k, p (Rn ) , by a sequence of elements of W k, p (Rn ) with compact support. We consider f ∈ W k, p (Rn ). Let (βk )k be a sequence of functions which belong to Cc∞ and such that βk = 1 on the ball B(0, k) and βk = 0 outside the ball B(0, k + 1). We consider that f k = βk f , k = 1, 2, . . .. Clearly, f k ∈ L p and f k → f in L p . For all 1 ≤ j ≤ n, we have D j f k = βk · D j f + D j βk · f.

(9.3.13)

We have D j f ∈ L p , so βk D j f → D j f in L p . By the definition of βk and since f ∈ L p , we have D j β j · f ∈ L p and D j βk · f → 0 in L p as k → ∞. By induction, it follows that D α f k → D α f in L p as k → ∞. Step 2. Let f be an element of W k, p (Rn ) with compact support. We consider (α j ) j to be a sequence of functions of Cc∞ (Rn ) converging to δ in D (Rn ). The function ϕ j = α j ∗ f belongs to Cc∞ (Rn ) for all j = 1, 2, . . .. On the other hand, if |α| ≤ k, we have (9.3.14) D α ϕ j = α j ∗ D α f. Since D α f ∈ L p , ∀|α| ≤ k, we have D α ϕ j → D α f in L p as j → ∞, ∀|α| ≤ k.

(9.3.15)

Hence, ϕ j → f in W k, p (Rn ).

 k, p

As a consequence, if  = Rn and 1 ≤ p < ∞, then W0 (Rn ) = W k, p (Rn ).

9.4 The Sobolev Spaces H s (Rn ) In the sequel, we will describe the structure of Sobolev spaces by using the Fourier transform [5, 6]. Recall that f ∈ H m (Rn ) if and only if D α f ∈ L 2 (Rn ), ∀|α| ≤ m. By the properties of the Fourier transform, this condition is equivalent to (i)

ξ α fˆ(ξ) ∈ L 2 (Rn ), ∀|α| ≤ m;

(9.4.1)

9.4 The Sobolev Spaces H s (Rn )

207

(ii) for every polynomial P(ξ) with constant coefficients and degree less or equal to m, we have P(ξ) fˆ ∈ L 2 (Rn ); (iii) m (9.4.2) (1 + |ξ|2 ) 2 fˆ(ξ) ∈ L 2 (Rn ). These results motivate the following definition [5]. Definition 9.4.1 Let s ∈ R. We denote by H s (Rn ) the space of tempered distributions f ∈ S  (Rn ) such that s (1 + |ξ|2 ) 2 fˆ(ξ) ∈ L 2 (Rn ).

(9.4.3)

This space is endowed with the scalar product  ( f, g)s =

Rn

s (1 + |ξ|2 ) 2 fˆ(ξ)g(ξ)dξ, ˆ

(9.4.4)

which induces the s-norm 

f s =

Rn

 s 1 + |ξ|2 | fˆ(ξ)|2 dξ

21

.

(9.4.5)

Theorem 9.4.1 H s (Rn ) is a Hilbert space ∀s ∈ R. Proof Let ( f j ) be a Cauchy sequence in H s (Rn ). By definition, we deduce that   s (1 + |ξ|2 ) 2 fˆj (ξ)

(9.4.6)

is a Cauchy sequence in L 2 (Rn ). Since this space is complete, the sequence   2 2s ˆ (1 + |ξ| ) f j (ξ) converges to gˆ in L 2 (Rn ), hence in S  (Rn ). We consider that − s  ˆ fˆ(ξ) = 1 + |ξ 2 | 2 g(ξ).

(9.4.7)

Clearly, we have fˆ ∈ S  (Rn ) and f j → f in H s (Rn ).



Theorem 9.4.2 Let s, t ∈ R such that s ≥ t. The following inclusions hold true S(Rn ) ⊂ H s (Rn ) ⊂ H t (Rn ) ⊂ S  (Rn )

(9.4.8)

and the embeddings are continuous. Moreover, S is dense in H s for all s. Proof By the definition of the space H s , we have the following embedding H s → H t (s ≥ t).

(9.4.9)

208

9 Sobolev Spaces

Let f ∈ H s (Rn ). Recall that S is dense in L 2 . Let ε > 0. Then we can find ϕ ∈ S such that s (9.4.10) ˆ

(1 + |ξ|2 ) 2 fˆ(ξ) − ϕ(ξ) L 2 < ε. On the other hand,

s ˆ ˆ ψ(ξ) = (1 + |ξ|2 )− 2 ϕ(ξ)

(9.4.11)

belongs to S. Hence, we obtain ˆ

(1 + |ξ|2 ) 2 ( fˆ(ξ) − ψ(ξ)) L2 < ε

(9.4.12)

f − ψ s < ε.

(9.4.13)

S → H s and H s → S 

(9.4.14)

s

and Therefore, S is dense in H s . Note also that the embeddings



are continuous.

Theorem 9.4.3 The strong dual of H s (Rn ) can be identified, in the algebraic and topological senses, with H −s (Rn ). Proof Step 1. Note that S ⊂ H s ⊂ S  with S dense in H s for all s. Hence the dual (H s ) of H s is a subspace of S  . Let f ∈ (H s ) . Then there is a constant C > 0 such that |( f, ϕ)| ≤ C

(9.4.15)

for all ϕ ∈ S such that ϕ s ≤ 1. By using the Fourier transform, we obtain

We consider

    ˆ  f , ϕˆ  ≤ C1 , ∀ ϕ s ≤ 1.

(9.4.16)

s  ˆ ˆ ψ(ξ) = 1 + |ξ|2 2 ϕ(ξ).

(9.4.17)

Hence, ϕ ∈ S with ϕ s ≤ 1 implies that ψ ∈ S with ψ L 2 ≤ 1 and conversely. From the last inequality we obtain     ψˆ   ˆ  < C1 , ∀ψ ∈ S with ψ L 2 ≤ 1 ⇔  f, s 2  (1 + |ξ| ) 2    s   ⇔  (1 + |ξ|2 )− 2 fˆ, ψˆ  ≤ C1 , ∀ψ ∈ S with ψ L 2 ≤ 1.

(9.4.18)

(9.4.19)

9.4 The Sobolev Spaces H s (Rn )

209

s Therefore, we obtain (1 + |ξ|2 )− 2 fˆ ∈ L 2 , which proves that f ∈ H −s . Step 2. We prove the converse statement. Let f ∈ H −s . Let us consider

 ( f, ϕ) =

Rn

 − s s  1 + |ξ|2 2 fˆ(ξ) 1 + |ξ|2 2 ϕ(ξ)dξ ˆ

(9.4.20)

for all ϕ ∈ S. By applying Hölder’s inequality, we obtain |( f, ϕ)| ≤ f −s ϕ s .

(9.4.21)

Therefore, f ∈ (H s ) . Finally, the (−s)-norm is equivalent to the dual norm

f = sup |( f, ϕ)| .

ϕ s ≤1

(9.4.22)

This implies that on H −s the strong topology of (H s ) coincides with the topology defined by the s-norm.  Theorem 9.4.4 Let m ∈ N. Then every f ∈ H −m can be represented as a finite sum of derivatives of order less or equal to m of square integrable functions. Proof If f ∈ H −m (Rn ), then, by definition, we obtain  − m 1 + |ξ|2 2 fˆ(ξ) ∈ L 2 (Rn ),

(9.4.23)

which is equivalent to g(ξ) ˆ =

1 + |ξ1

|m

fˆ(ξ) ∈ L 2 (Rn ). + · · · + |ξn |m

(9.4.24)

Therefore, we obtain fˆ(ξ) = g(ξ) ˆ +

n  j=1

= g(ξ) ˆ +

n 



ξ mj

j=1

= g(ξ) ˆ +

n 

|ξ j |m g(ξ) ˆ =  |ξ j |m g(ξ) ˆ = ξ mj

(9.4.25)

ξ mj gˆ j (ξ),

j=1



with gˆ j (ξ) =

|ξ j |m ξ mj

 g(ξ) ˆ ∈ L 2 (Rn ).

(9.4.26)

210

9 Sobolev Spaces

Hence, we obtain f =g+

n 

D mj g j .

(9.4.27)

j=1

 In the sequel, we will present a Sobolev embedding theorem. Theorem 9.4.5 If s > n2 , then H s (Rn ) ⊂ C 0 (Rn ) and the embedding is continuous. Proof Let s > n2 . Therefore, we have − s  1 + |ξ|2 2 ∈ L 2 (Rn ).

(9.4.28)

− s  s  fˆ(ξ) = 1 + |ξ|2 2 1 + |ξ|2 2 fˆ(ξ) ∈ L 1 (Rn ).

(9.4.29)

Let f ∈ H s . Then

Hence, we obtain f ∈ C 0 (Rn ). We assume that f j → 0 in H s . Then, by definition,  s 1 + |ξ|2 2 fˆj (ξ) → 0 in L 2 and by (9.4.29) we have fˆj → 0 in L 1 (Rn ). Therefore,  we obtain f j → 0 in C 0 (Rn ). Corollary 9.4.1 Let s > embedding is continuous.

n 2

+ k, where k ∈ N. Then H s (Rn ) ⊂ C k (Rn ) and the

Proof Let |α| ≤ k and let f ∈ H s . Hence, we obtain D α f ∈ H s−|α| . We have s −  |α| ≥ s − k ≥ n2 . Therefore, we have D α f ∈ C 0 , ∀|α| ≤ k. In the sequel, we define the vector spaces H ∞ (Rn ) = ∩s H s (Rn ),

(9.4.30)

H −∞ (Rn ) = ∪s H −s (Rn ).

(9.4.31)

As a consequence of the Corollary 9.4.1, H ∞ (Rn ) is a subspace of C ∞ (Rn ). Clearly, H ∞ coincides with the space D L 2 of all functions ϕ ∈ C ∞ (Rn ) such that D α ϕ ∈ L 2 (Rn ) for all α. On H ∞ we define the coarsest locally convex topology for which the identity maps ∞ H → H s , ∀s, are continuous. On H −∞ we define the inductive limit topology of H s , ∀s. Then, we have H ∞ ⊂ · · · ⊂ H s ⊂ H t ⊂ · · · ⊂ H −∞ for s ≥ t and the embeddings are continuous. In the sequel, we present a result about the structure of H −∞ .

(9.4.32)

9.4 The Sobolev Spaces H s (Rn )

211

Theorem 9.4.6 A distribution f belongs to H −∞ (Rn ) if and only if it is a finite sum of derivatives of functions from L 2 (Rn ). Example 1. The Dirac measure δ belongs to H s for all s < − n2 . In fact, we have s  δˆ = 1 and 1 + |ξ|2 2 ∈ L 2 (Rn ) if s < − n2 . Example 2. Let m ∈ N∗ such that −m < − n2 . Then the Dirac measure δ can be represented as a finite sum of derivatives of order less or equal to m of functions belonging to L 2 (Rn ). Example 3. Let x  (9.4.33) αε (x) = ε−n α ε be test functions. We know that αε → δ in E  (Rn ) as ε → 0. Hence, we have αˆ ε → 1 in S  as ε → 0. Clearly, we have αˆ ε (ξ) = ε

−n

 e

−i(x,ξ)

Rn

α

x  ε

 dx =

Letting ε → 0, we obtain

Rn

e−i(y,εξ) α(y)dy = α(εξ). ˆ

αˆ ε (ξ) → 1

(9.4.34)

(9.4.35)

pointwise in Rn . Let s < − n2 . Then we can write 

αε − δ 2s =

Rn

2  s  ˆ − 1 dξ. 1 + |ξ|2 α(εξ)

(9.4.36)

Note that the functions appearing inside the integral converge pointwise to zero as ε → 0 and they are all bounded by an integrable function. Hence, by the Lebesgue dominated convergence theorem [3], we have

αε − δ 2s → 0 as ε → 0.

(9.4.37)

Finally, for all s < − n2 , αε → δ in H s . Example 4. We present a fundamental solution of 1 − . Let  be the Laplace  ∂2 operator nj=1 ∂x 2 . Our goal is to find a tempered distribution E such that (1 − j

)E = δ. By using the Fourier transform we obtain (1 + |ξ|2 ) · E = 1. Hence Eˆ = is such that (1 + |ξ|2 )

s+2 2

1 1 + |ξ|2

n · Eˆ ∈ L 2 (Rn ), ∀s < − . 2

(9.4.38)

(9.4.39)

Therefore, the operator 1 −  possesses a fundamental solution E ∈ H s+2 (Rn ) with s < − n2 . Similarly, there is a fundamental solution E of the operator (1 − )k belong-

212

9 Sobolev Spaces

ing to H s+2k . We can choose k to be an integer sufficiently large that (1 − )k possesses a fundamental solution belonging to C 0 (Rn ). In the sequel, we present multiplication and convolution operations in H s (Rn ). Lemma 9.4.1 Let K (x, y) be a continuous function on Rn × Rn . Assume that there is a constant C > 0 such that  |K (x, y)|d x ≤ C uniformly on y (9.4.40) Rn



and

Rn

|K (x, y)|dy ≤ C uniformly on x.

(9.4.41)



Then A f (x) =

Rn

K (x, y) f (y)dy

(9.4.42)

defines a continuous linear operator from L 2 (Rn ) into L 2 (Rn ). Proof For all f, g ∈ L 2 (Rn ) we have          |(A f, g)| =  A f (x)g(x)d x  =  K (x, y) f (y)g(x)d xd y  ≤ Rn Rn ×Rn  1 1 ≤ |K (x, y)| 2 | f (y)||K (x, y)| 2 |g(x)|d xd y ≤ Rn ×Rn

  ≤

21   |K (x, y)|| f (y)| d xd y · 2

Rn ×Rn

|K (x, y)||g(x)| d xd y 2

Rn ×Rn

21

≤

≤ C · f L 2 · g L 2 . This proves that A : L 2 → L 2 is a continuous linear operator.

(9.4.43) 

Below we prove Peetre’s inequality. Lemma 9.4.2 We have ∀t ∈ R

1 + |ξ|2 1 + |η|2

t

 |t| ≤ 2|t| 1 + |ξ − η|2

(9.4.44)

with ξ, η ∈ Rn . Proof For all ξ, η ∈ Rn , we have 1 + |ξ − ζ|2 = 1 + |ξ|2 − 2ξζ + |ζ|2 ≤   ≤ 1 + 2|ξ|2 + 2|ζ|2 ≤ 2 1 + |ξ|2 1 + |ζ|2 By setting η = ξ − ζ, we obtain

(9.4.45)

9.4 The Sobolev Spaces H s (Rn )

213

  1 + |η|2 ≤ 2 1 + |ξ|2 1 + |ξ − η|2 .

(9.4.46)

We discuss two cases. If t < 0, it suffices to raise both sides of the last inequality to the power −t to get Peetre’s inequality. If t > 0, by interchanging ξ and η and raising the inequality to the power t, we again obtain Peetre’s inequality.  Theorem 9.4.7 If ϕ ∈ S and f ∈ H s , the product ϕ f belongs to H s . Moreover, the bilinear map (9.4.47) S × H s  (ϕ, f ) → ϕ f ∈ H s is separately continuous. Proof Step 1. Let ϕ ∈ S and f ∈ H s . We have    −n −n ˆ  ϕ f (ξ) = (2π) ϕˆ ∗ f (ξ) = (2π)

Rn

ϕ(ξ ˆ − η) f (η)dη.

(9.4.48)

We will show that ϕ f ∈ H s . It suffices to show that s  f (ξ) ∈ L 2 (Rn ). 1 + |ξ|2 2 ϕ

(9.4.49)

Let us write  s f (ξ) = (2π)−n 1 + |ξ|2 2 ϕ −n

= (2π)

 Rn



2 2

1 + |ξ| 1 + |η|2

s

 Rn

 s 1 + |ξ|2 2 ϕ(ξ − η) f (η)dη =

s  ϕ(ξ ˆ − η) 1 + |η|2 2 fˆ(η)dη

and set

K (ξ, η) =

1 + |ξ|2 1 + |η|2

2s

ϕ(ξ ˆ − η).

(9.4.50)

(9.4.51)

By Peetre’s inequality, we obtain |s| |s|  ˆ − η)| = |K (ξ, η)| ≤ 2 2 1 + |ξ − η|2 2 |ϕ(ξ |s| −n  |s|  +n = 2 2 1 + |ξ − η|2 2 |ϕ(ξ ˆ − η)| 1 + |ξ − η|2 ≤  2 −n ≤ C 1 + |ξ − η|

(9.4.52)

since ϕ ∈ S. Therefore, K (ξ, η) satisfies the assumptions of Lemma 9.4.1 and cons f (ξ) ∈ L 2 (Rn ). sequently (1 + |ξ|2 ) 2 ϕ Step 2. The following inequality holds true

ϕ f s ≤ C f s .

(9.4.53)

214

9 Sobolev Spaces

This implies the continuity of the map (ϕ, f ) → ϕ f with respect to f ∈ H s . We assume that ϕ j → 0 in S and let   |s| +n  C j = sup 1 + |ζ|2 2 ϕˆ j (ζ) ζ∈Rn

(9.4.54)

be the corresponding constant appearing in (9.4.52). We obtain s 

ϕ j f s = 1 + |ξ|2 2  ϕ j f (ξ) L 2 ≤ C j C  f s .

(9.4.55)

Since ϕ j → 0, hence C j → 0, the last inequality implies the continuity of the product ϕ f with respect to ϕ ∈ S.   Corollary 9.4.2 Let P = P(x, D) = |α|≤m aα (x)D α be a partial differential operator of order less or equal to m and coefficients belonging to S(Rn ). Then P defines a continuous linear map from H s (Rn ) into H s−m (Rn ), ∀s ∈ R. Proof It suffices to note that D α maps H s continuously into H s−|α| and to apply Theorem 9.4.7. Theorem 9.4.8 Let ϕ ∈ S and f ∈ H s . The convolution ϕ ∗ f ∈ H s and the bilinear map (9.4.56) (S × H s )  (ϕ, f ) → ϕ ∗ f ∈ H s is separately continuous. Moreover, ϕ ∗ f ∈ H ∞ . Proof Let ϕ ∈ S and f ∈ H s . Then ϕ ∗ f is a C ∞ function. By assumption, we s s have (1 + |ξ|2 ) 2 fˆ(ξ) ∈ L 2 . Since ϕˆ ∈ S, it follows that (1 + |ξ|2 ) 2 fˆ(ξ)ϕ(ξ) ˆ ∈ L 2. s Therefore, we obtain ϕ ∗ f ∈ H . The inequality 

ϕ ∗ f s =

2 (1 + |ξ| ) | fˆ(ξ)|2 |ϕ(ξ)| ˆ dξ 2 s

Rn

21

≤ sup |ϕ(ξ)| ˆ · f s ξ∈Rn

(9.4.57)

implies the separate continuity of the convolution product ϕ ∗ f . k Finally, we know that if ϕ ∈ S, then (1 + |ξ|2 ) 2 ϕ ∈ S. Therefore, we obtain (1 + |ξ|2 )

s+k 2

ϕ(ξ) ˆ fˆ(ξ) ∈ L 2

for every k, which implies that ϕ ∗ f ∈ H ∞ .

(9.4.58) 

Definition 9.4.2 A continuous linear operator L : Cc∞ (Rn ) → C ∞ (Rn ) is of order less or equal to m if it can be extended to a continuous linear operator from H s (Rn ) into H s−m (Rn ). Example 1. Let ϕ ∈ S and define Mϕ : Cc∞ → C ∞ by

9.4 The Sobolev Spaces H s (Rn )

215

Mϕ ( f ) = ϕ · f.

(9.4.59)

The multiplication operator Mϕ has order less or equal to 0. Example 2. Let ϕ ∈ S and define L ϕ : Cc∞ → C ∞ by L ϕ ( f ) = ϕ ∗ f.

(9.4.60)

The convolution operator L ϕ has order −∞.

9.5 Besov Spaces We denote by C(0, R1 , R2 ) the set {ξ ∈ Rn |R1 ≤ |ξ| ≤ R2 }. Below we state the Bernstein lemma [1]. Lemma 9.5.1 Let k ∈ N and 0 < R1 < R2 . There exists a constant C = C(R1 , R2 , n) such that for all 1 ≤ a ≤ b ≤ ∞ and u ∈ L a , we have supp uˆ ⊂ B(0, R1 λ) ⇒ sup ∂ α u L b ≤ C k+1 λk+n ( a − b ) u L a , 1

1

|α|=k

(9.5.1)

supp uˆ ⊂ C(0, R1 λ, R2 λ) ⇒ C −k−1 λk u L a ≤ sup ∂ α u L a ≤ C k+1 λk u L a . |α|=k

(9.5.2) Proof Step 1. By using rescaling, we may assume without loss of generality that λ = 1. We fix a smooth function φ with compact support with the property that φ ≡ 1 ˆ Then we denote by in a neighbourhood of the ball B(0, R1 ). Note that uˆ = φu. g := F −1 φ. Hence, it follows that for all multiindices α we have α



∂ u(x) =

∂ α g(y)u(x − y)dy.

(9.5.3)

By applying Young’s inequality, it follows that

∂ α u L b ≤ ∂ α g L c u L a , where

1 c

=1+

1 b

(9.5.4)

− a1 . The first inequality is easily proven since

∂ α g L c ≤ ∂ α g L ∞ + ∂ α g L 1 ≤ C k+1 .

(9.5.5)

Step 2. We want to prove the second inequality. We consider a smooth function ϕ˜ n property that ϕ˜ ≡ 1 in a neighbourhood with compact support in R \ {0},αwith the of C(0, R1 , R2 ). Since |α|=k (iξ) (−iξ)α = |ξ|2k , it follows that

216

9 Sobolev Spaces

u=



gα ∗ ∂ α u,

(9.5.6)

|α|=k

where gˆ α (ξ) := (iξ)α |ξ|−2k ϕ(ξ). ˜ The result follows by applying Young’s inequality. 

9.5.1 The Nonhomogeneous Littlewood-Paley Decomposition Let α > 1. We consider that (ϕ, χ) is a pair of smooth functions with values in [0, 1] such that (i) ϕ has the support in {ξ ∈ Rn |α−1 ≤ |ξ| ≤ 2α}; (ii) χ has the support in the ball {ξ ∈ Rn ||ξ| ≤ α}; (iii) ∀ξ ∈ Rn , χ(ξ) + q∈N ϕ(2−q ξ) = 1. Let u ∈ S  . If A is a smooth function with polynomial growth at infinity, then we set A(D)u := F −1 (AFu), where Fu denotes the Fourier transform of u. Then we define nonhomogeneous dyadic blocks in the following way: q u := 0, if q ≤ −2, −1 u := χ(D)u = h˜ ∗ u, with h˜ := F −1 χ,  q u := ϕ(2−q D)u = 2qn h(2q y)u(x − y)dy with h = F −1 ϕ, if q ≥ 0. We have

u=



(9.5.7) 

 j u in S (R ) n

(9.5.8)

j∈Z

for all tempered distributions u. We say that the right-hand side is the nonhomogeneous Littlewood-Paley decomposition of u [1]. We define the following low frequency cut-off   p u. (9.5.9) Sq u := p≤q−1

Note that S0 u = −1 u. Since ϕ(ξ) = χ(ξ/2) − χ(ξ), ∀ξ ∈ Rn , it follows that 

Sq u = χ 2

−q



D u=



˜ q y)u(x − y)dy, ∀q ∈ N. h(2

(9.5.10)

Proposition 9.5.1 Let α = 4/3. Let u ∈ S  (Rn ), v ∈ S  (Rn ). Then  p q u ≡ 0 if | p − q| ≥ 2,  q S p−1 u p v ≡ 0 if | p − q| ≥ 5.

(9.5.11)

9.5 Besov Spaces

217

9.5.2 Definition and Properties Let (aq )q∈Z be a non-negative sequence. Then we make the convention that the  1 r r notation a means supq aq in the case r = ∞. q q Definition 9.5.1 Let s ∈ R and 1 ≤ p, r ≤ ∞. For u ∈ S  (Rn ), we set

u B sp,r

 1  r r qs := . 2 q u L p

(9.5.12)

q

The Besov space B sp,r is the set of tempered distributions u such that u B sp,r < ∞. We introduce the following notation

u α,S =

sup

x∈Rn ,| p|≤α

(1 + |x|)α |∂ p u(x)|, ∀u ∈ S, α ∈ N.

(9.5.13)

Proposition 9.5.2 Let 1 ≤ p, r ≤ ∞ and s ∈ R. The space B sp,r is a Banach space which is continuously embedded in S  . Proof Step 1. We show that B sp,r is continuously embedded in S  . Note that B sp,r is a subspace of S  by definition. Therefore, it suffices to show that there exist a constant C and an integer α such that for any φ ∈ S we have | (u, φ) | ≤ C u B sp,r φ α,S .

(9.5.14)

With the specific notations from the second step of the proof of the lemma of Bernstein, it follows that we obtain after rescaling u q = 2−qk



2qn gα (2q ·) ∗ ∂ α u q .

(9.5.15)

|α|=k

Hence, for any test function φ ∈ S, we deduce that 

 u q , φ = (−1)k 2−qk u q , 2qn gˇ α (2q ·) ∗ ∂ α φ ,

(9.5.16)

|α|=k

where g(z) ˇ = g(−z). In this relation, we replace u q by q u. Hence, it follows that ∀k ∈ N, there exist a constant Ck and an integer αk such that  |(q u, φ)| ≤ Ck 2−q 2q(1−k) q u L ∞ φ αk ,S . By the lemma of Bernstein, we have

(9.5.17)

218

9 Sobolev Spaces n

q u L ∞ ≤ C2q p q u L p .

(9.5.18)

If we choose k so large as to satisfy s − n/ p ≥ 1 − k, then it follows that  | q u, φ | ≤ Ck 2−q u B sp,r φ αk ,S .

(9.5.19)

After summation  on q, we find the desired inequality. Step 2. Let u (n) n∈N be a Cauchy sequence in B sp,r . From the inequality (9.5.14)  it follows that for any test function φ ∈ S, the sequence u (n) , φ n∈N is a Cauchy sequence in R. Hence, the following  (u, φ) := lim u (n) , φ n→∞

(9.5.20)

defines a tempered distribution.  From the definition of the norm of B sp,r , it follows that the sequence q u (n) n∈N is  a Cauchy sequence in L p for any q. Hence, there exists u q ∈ L p such that q u (n) n∈N converges to u q in L p . Since the sequence (u (n) )n∈N converges to u in S  , it follows that q u = u q . We fix Q ∈ N and ε > 0. Since ∀q ≥ −1, q u (n) tends to q u in L p , it follows that ∀n ∈ N, we have ⎛

⎞ r1 ⎛ ⎞ r1    r r ⎝ 2qs q (u (n) − u) L p ⎠ = lim ⎝ 2qs q (u (n) − u (m) ) L p ⎠ . m→∞

q≤Q

q≤Q

(9.5.21) Note that the argument of the limit on the right-hand side is bounded by

u (n) −  (n) (m) s u B sp,r . Moreover, u n∈N is a Cauchy sequence in B p,r . Hence, it follows that there exists n 0 (independent of Q) such that for all n ≥ n 0 , we have ⎛

⎞ r1   r ⎝ 2qs q u (n) − u L p ⎠ ≤ ε.

(9.5.22)

q≤Q

 By letting Q → ∞, we deduce that u (n) n∈N converges to u in B sp,r .



Proposition 9.5.3 Let 1 ≤ p, r ≤ ∞ and s ∈ R. The space Cc∞ of smooth functions with compact support is dense in B sp,r if and only if p and r are finite. Proof Step 1. Let r be finite. We consider u ∈ B sp,r and ε > 0. Since r is finite, there exists an integer q such that

u − Sq u B sp,r
Q

2(Q+1)(1−θ)(˜s −s) 2−Qθ(˜s −s) ˜ + u B sp,∞ . ≤ u B sp,∞ (1−θ)(˜s −s) 2 −1 1 − 2−θ(˜s −s)

2−qθ(˜s −s) ≤ (9.5.40)

To finish the proof, it suffices to choose Q such that ˜

u B sp,∞

u B sp,∞

≤ 2 Q(˜s −s) < 2

˜

u B sp,∞

u B sp,∞

.

(9.5.41) 

Observation 9.5.1 We obtained the Besov spaces by considering first the L p norm on each dyadic block and then taking a weighted l r norm. If we consider first the

222

9 Sobolev Spaces

weighted l r norm and then the L p norm over Rn , then we obtain new Banach spaces s and called Triebel-Lizorkin spaces [1, 7]. denoted by F p,r

9.5.3 The Homogeneous Littlewood-Paley Decomposition and the Homogeneous Besov Spaces Definition 9.5.2 Let s ∈ R and 1 ≤ p, r ≤ ∞. Let u be a tempered distribution. We consider ⎛ ⎞1

u B˙ sp,r

:= ⎝



r

rqs

2

˙ q u rL p ⎠ .



(9.5.42)

q∈Z

Theorem 9.5.1 There exists C a constant such that ∀s ∈ R we have r1 ≤ r2 ⇒ u B˙ sp,r ≤ u B˙ sp,r , 2

p1 ≤ p2 ⇒ u

1

( B˙ p2 ,r s−n

1 1 p1 − p2

) ≤ C u B˙ sp1 ,r .

(9.5.43)

For any θ ∈ (0, 1) and s < s˜ , we have the following interpolation inequalities 1−θ θ s ≤ u ˙ s u s˜ ,

u B˙ θs+(1−θ)˜ B B˙ p,r p,r

u B˙ θs+(1−θ)˜s p,1

p,r

C

u θB˙ s u 1−θ ≤ . ˜ B˙ sp,∞ p,∞ θ(1 − θ)(˜s − s)

(9.5.44)

In the sequel, we consider the invariance by dilation of homogeneous Besov spaces. Let λ > 0. For any tempered distribution u we consider the tempered distribution u λ defined for all x ∈ Rn by u λ (x) := u(λx). Proposition 9.5.8 If u B˙ sp,r is finite, then u λ B˙ sp,r is finite. Moreover, we have n

u λ B˙ sp,r ≈ λs− p u B˙ sp,r .

(9.5.45)

Equality holds true if λ = 2m for some m ∈ Z. Proof We consider β ∈ R, β > 0. We have  ϕ β −1 D u λ (x) = β n

 h (β(x − y)) u(λy)dy.

We make the change of variables z = λy. This implies that

(9.5.46)

9.5 Besov Spaces

223

 ϕ β −1 D u λ (x) = β n λ−n



   h βx − βλ−1 z u(z)dz = ϕ(λβ −1 D)u (βx).

 Let q ∈ Z. Then we consider vq := ϕ 2[log2 λ]−lo f2 λ−q D u λ . In the equality above, we consider β = 2q−[log2 λ] λ. This implies that n

2qs vq L p ≈ λs− p 2(q−[log2 λ])s · q−[log2 λ] u L p . The equality follows if log2 λ is an integer. It follows that  r1   n ≈ λs− p u ·B sp,r . 2qs vq L p

(9.5.47)

(9.5.48)

(9.5.49)

q

 Note that supp vˆq ⊂ C 0, 43 2q , 16 2q . Therefore, we obtain 3 ˙ q uλ = 



· p u q .

(9.5.50)

| p−q|≤2

The result follows from straightforward computations.







Proposition 9.5.9 Let s ∈ R and 1 ≤ p, r ≤ ∞. Then the space B˙ sp,r , · B sp,r is a normed space. If r is finite, then Cc∞ ∩ B˙ sp,r is densely embedded in B˙ sp,r . Proof Note that · B˙ sp,r is a semi-norm. We will suppose that u B˙ sp,r = 0 for some u ∈ Sh . It follows that supp uˆ ⊂ {0}. Hence, for any j ∈ Z, we obtain S˙ j u = u. Since u ∈ Sh , we have lim j→−∞ S˙ j u = 0 and we deduce that u = 0.  Let r be finite and u ∈ B˙ sp,r . Clearly, the sequence of general term |q|≤n q u belongs to C ∞ ∩ B˙ sp,r and converges to u in B˙ sp,r . Then we can find a sequence of  functions of Cc∞ ∩ B˙ sp,r that converges to u in B˙ sp,r .   Theorem 9.5.2 If s < np or s = np and r = 1, then B˙ sp,r , · B˙ sp,r is a Banach space that is continuously embedded in S  . Proof Step 1. We will show that B˙ sp,r is continuously embedded in S  .  ˙ j u is convergent in L ∞ . Therefore, the case r = 1 and Note that the series   n ˙ j u, we obtain u ∈ L ∞ . Moreover, the s = p follows immediately. Since u = j  following embeddings hold true n

p 0 → B˙ ∞,1 → L ∞ → S  . B˙ p,1

(9.5.51)

In the sequel, we will suppose that s < np . Since we have the embedding B˙ sp,r →

s− n

p B˙ ∞,∞ , it follows that we can find a large integer α such that for all nonnegative j we obtain

224

9 Sobolev Spaces

    ˙ j u, φ  ≤ 2− j u

s− np

B∞,∞

φ α,S .

(9.5.52)

For negative j, we deduce that for large enough α, we have   n     ˙ j u, φ   2 j p −s u 

2

j

n p −s



s− n

p B˙ ∞,∞

φ L 1  (9.5.53)

u B˙ sp,r φ α,S . 

References 1. H. Bahouri, J.Y. Chemin, R. Danchin, Littlewood-Paley Theory, in: Fourier Analysis and Nonlinear Partial Differential Equations, vol. 343, Grundlehren der mathematischen Wissenschaften (Springer, Berlin, Heidelberg, 2011) 2. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, Berlin, 2011) 3. V. Kadets, Course in Functional Analysis and Measure Theory (Springer, Cham, 2018) 4. S.D. Micu, Introduction to Finite Element Method (Romanian). Publications of the Centre for Nonlinear Analysis and its Applications 5. T. Runst, W. Sickel, Sobolev Spaces of Fractional Order, Nemytskij Operators, and Nonlinear Partial Differential Equations (Walter de Gruyter, Berlin, New York, 1996) 6. E. Stein, R. Shakarchi, Fourier Analysis: An Introduction (Princeton University Press, Princeton, 2003) 7. H. Triebel, Theory of Function Spaces II, vol. 84, Monographs in Mathematics (Birkhäuser, Basel, 1992)

Chapter 10

Variational Problems

Abstract We present the existence and uniqueness of the solutions of a large class of variational problems. We apply the Lax-Milgram theorem for the variational formulations of the Stokes system, the elasticity system and the plate equation. Then we discuss the approximation of variational problems by means of the Galerkin method and by means of the finite element method.

10.1 Introduction We will study the existence and uniqueness of the solutions of a large class of variational problems. The variational methods that we introduce in this section are important for studying the approximation by the finite element method [1, 4]. Let V be a Hilbert space with the scalar product (·, ·) and the norm || · ||. On V , we consider a bilinear, continuous and coercive form a. To be more precise, a : V × V → R, is an application with the following properties 1. bilinearity: for any α, β ∈ R and u, v, w ∈ V , we have  a(αu + βv, w) = αa(u, w) + βa(v, w) . (10.1.1) a(w, αu + βv) = αa(w, u) + βa(w, v) 2. continuity: there exists a positive constant M > 0 such that, for any u, v ∈ V , we have (10.1.2) |a(u, v)| ≤ M||u|| · ||v||. 3. coercivity: there exists a positive constant m > 0 such that, for any u ∈ V m||u||2 ≤ a(u, u).

(10.1.3)

On V we also consider a linear and continuous form L, that is, an application L : V → R with the following properties 1. linearity: for any α, β ∈ R and u, v ∈ V , we have L(αu + βv) = αL(u) + β L(v). © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_10

(10.1.4) 225

226

10 Variational Problems

2. continuity: there exists a positive constant C > 0 such that, for any u ∈ V , we have |L(u)| ≤ C||u||. (10.1.5) We study the following variational problem: find u ∈ V such that a(u, v) = L(v), ∀v ∈ V.

(10.1.6)

In the following, we present the Lax-Milgram theorem [3]. Theorem 10.1.1 Let V be a Hilbert space. If a is a bilinear, continuous and coercive form and L is a linear and continuous form on V , then the problem (10.1.6) has a unique solution u ∈ V . Proof Let L be linear and continuous on V . Then by the Riesz representation theorem [3] it follows that there exists f ∈ V such that L(v) = ( f, v), ∀v ∈ V.

(10.1.7)

Let a be bilinear and continuous on V . Then for every w ∈ V , the map a(w, ·) : V → R

(10.1.8)

is linear and continuous on V . We apply again the Riesz representation theorem. Then for every w ∈ V there exists a unique element A(w) ∈ V such that a(w, v) = (A(w), v) , ∀v ∈ V.

(10.1.9)

Therefore, we defined the operator A : V → V , which is linear. Moreover, we have A(w)2 = (A(w), A(w)) = a(w, A(w)) ≤ MwA(w).

(10.1.10)

Hence, A is also continuous in V and A ≤ M. Therefore, we only have to show that for every f ∈ V there exists a unique element u ∈ V such that A(u) = f. (10.1.11) To this end, we define the map T : V → V T (w) = w − μ (A(w) − f ) ,

(10.1.12)

where μ ∈ R, μ > 0 will be chosen later. Note that A(u) = f if and only if T (u) = u. It suffices to prove that T is a contraction on V for a suitable μ. We have

10.1 Introduction

227

T (w1 ) − T (w2 )2 = w1 − w2 − μA(w1 − w2 )2 = = w1 − w2 2 + μ2 A(w1 − w2 )2 − 2μ (A(w1 − w2 ), w1 − w2 ) = = w1 − w2 2 + μ2 A(w1 − w2 )2 − 2μa (w1 − w2 , w1 − w2 ) ≤   ≤ 1 + μ2 M 2 − 2mμ w1 − w2 2 . If we choose μ =

m , M2

(10.1.13)

then

  m2 T (w1 ) − T (w2 ) ≤ 1 − 2 w1 − w2 2 M 2

(10.1.14)

and T is a contraction on V . Then by the theorem of Banach, T has a unique fixed point.  In the sequel, we assume that the bilinear, continuous and coercive form a is also symmetric a(u, v) = a(v, u), ∀u, v ∈ V. (10.1.15) We introduce the functional J : V → R J (v) =

1 a(v, v) − L(v). 2

(10.1.16)

Theorem 10.1.2 Let V be a Hilbert space. If a is a bilinear, continuous, coercive and symmetric form and L is a linear and continuous form on V , then the problem (10.1.6) has a unique solution u ∈ V which verifies J (u) = min J (v). v∈V

(10.1.17)

Proof From the properties of a it follows that v, w → a(v, w)

(10.1.18)

defines a scalar product on V . Let L be linear and continuous with respect to the norm induced by this scalar product. By the Riesz representation theorem, it follows that there exists a unique u ∈ V such that the problem (10.1.6) is verified. We will show that u minimizes the function J . Note that     1 1 a(v, v) − L(v) − a(u, u) − L(u) = J (v) − J (u) = 2 2   (10.1.19) 1 1 1 a(v − u, v) − L(v) + L(v) + a(u, u) = = 2 2 2

228

10 Variational Problems

 = =

   1 1 1 1 a(v − u, v) − L(v) + a(u − v, u) + L(v) = 2 2 2 2

1 a(v − u, v − u) ≥ mv − u2 . 2

Therefore, we obtain J (u) = minv∈V J (v).



Observation 10.1.1 The relation (10.1.17) has a mechanical interpretation. It shows that the energy associated to the equilibrium position is minimal. Theorem 10.1.3 Let X be a reflexive space and J : X → R a convex and lower semicontinuous functional which satisfies the coercivity property lim J (x) = ∞.

||x||→∞

(10.1.20)

Then there exists u ∈ X with the property that J (u) = min J (v). v∈X

(10.1.21)

Let  be a 1-regular set from R N . The exterior unit normal vector to the boundary ∂ of  will be denoted by ν. We will assume that ∂ = 0 ∪ 1 , where 0 and 1 are two disjoint sets of ∂ and 0 is non-empty. We introduce the elliptic operator of order two Au = −

  N  ∂ ∂ ai j u + a0 u, ∂x j ∂xi i, j=1

(10.1.22)

where ai j , a0 ∈ L ∞ () are functions defined on  with the following properties: 1. ai j = a ji , 1 ≤ i, j ≤ N ,  2. i,N j=1 ai j (x)ξ j ξi ≥ α|ξ|2 , ∀ξ ∈ R N almost everywhere in , 3. a0 (x) ≥ α0 almost everywhere in , and α and α0 are two positive constants. We also consider the following operator N  ∂u ∂u = ai j νi . ∂ν A ∂x j i, j=1

(10.1.23)

In the sequel, we consider the following second-order elliptic equation ⎧ ⎪ ⎨ Au = f u=0 ⎪ ⎩ ∂u =0 ∂ν A

in  on 0 on ∂ \ 0 ,

(10.1.24)

10.1 Introduction

229

with Dirichlet boundary conditions on 0 and Neumann boundary conditions on 1 and with the non-homogeneous term f ∈ L 2 (). The Eq. (10.1.24) may represent the equilibrium condition of an elastic membrane on which acts the force f and for which the boundary on 0 is fixed and the boundary on 1 is free. In the sequel, we introduce the variational formulation of (10.1.24) and the concept of weak solution, which will prove useful for the numerical treatment of the equation. Let us assume that (10.1.24) has a solution u. If we multiply the equation by a regular function ϕ and we integrate on , then we obtain ⎞

∂u ∂ϕ ⎝ ai j + a0 uϕ⎠ d x = f ϕd x. ∂x j ∂xi  i, j=1 

⎛

N 

(10.1.25)

We assume that u, ϕ ∈ H 1 () and f ∈ L 2 (). Therefore, we may study the existence of a function u ∈ H 1 () which verifies (10.1.25) for any ϕ ∈ H 1 (). Since we consider Dirichlet boundary conditions, we will assume that u belongs to the space (10.1.26) V = H0 () = {u ∈ H 1 ()|u|0 = 0}. On V we introduce the bilinear form a and the linear form L ⎞ ⎛

N  ∂u ∂ϕ ⎝ ai j + a0 uϕ⎠ d x a(u, ϕ) = ∂x j ∂x i  i, j=1

f ϕd x. L(ϕ) =

(10.1.27)



Then we can write (10.1.25) in an equivalent form a(u, ϕ) = L(ϕ), ∀ϕ ∈ V.

(10.1.28)

Definition 10.1.1 A function u ∈ H0 () is a weak solution of (10.1.24) if the relation (10.1.28) holds true for any ϕ ∈ H0 (). The existence and uniqueness of the solution of the problem (10.1.28) can be shown by the theorem of Lax-Milgram [1]. Theorem 10.1.4 There exists a unique weak solution of the problem (10.1.24), that is, u ∈ H0 (). Observation 10.1.2 The variational formulation (10.1.28) has a mechanical interpretation. If we consider that (10.1.24) represents the equation of the equilibrium position of an elastic membrane, then (10.1.28) shows that the virtual work is zero for any possible virtual displacement from the equilibrium position.

230

10 Variational Problems

By the rules of distributional differentiation, we have

Au, ϕ =



f ϕd x, ∀ϕ ∈ D().

(10.1.29)

Therefore, u verifies the Eq. (10.1.24) in the sense of distributions. Since f ∈ L 2 (), we have Au ∈ L 2 () and the Eq. (10.1.24) holds true in L 2 (), that is Au(x) = f (x) almost everywhere in .

(10.1.30)

The Dirichlet boundary condition is verified in the sense of trace, that is, u ∈ H0 () implies u|0 = 0. ¯ For the Neumann boundary conditions, we assume that the functions ai j ∈ C 1 (). N ∂u 1 Then j=1 ai j ∂x j belongs to H (). By Green’s formula, we have



f ϕd x = a(u, ϕ) =



Auϕd x +

1

∂u ϕdσ. ∂ν A

(10.1.31)

By (10.1.30), we have

1

∂u ϕ = 0, ∀ϕ ∈ H0 (). ∂ν A

(10.1.32)

If the space {ϕ|1 |ϕ ∈ H0 ()} is dense in L 2 (1 ), then by (10.1.32) we obtain ∂u =0 ∂ν A

(10.1.33)

and the Neumann boundary condition is verified almost everywhere on 1 . Note that the Dirichlet boundary condition is included in the definition of the solution space, while the Neumann condition is included in the variational formulation of the problem.

10.1.1 The Stokes System Let  ⊂ R2 be open and bounded. Let us consider the system ⎧ ⎪ ⎨−μu i + div u = 0 ⎪ ⎩ u = 0

∂p ∂xi

= f i in , i = 1, 2 , in  on ∂

(10.1.34)

10.1 Introduction

231

where u = (u 1 , u 2 ) and f = ( f 1 , f 2 ) are two vectorial functions and p is a real function defined on . This system describes the stationary flow of a viscous and incompressible fluid. We assume that the viscosity is sufficiently large to neglect the nonlinear transport terms in Euler’s equation. Note that this is the stationary version of the linearized Navier-Stokes equations [3]. From a physical point of view, u is the fluid velocity, p represents the pressure and μ ∈ R, μ > 0 represents the viscosity coefficient. We define the space 2  V = { v ∈ H01 () |div v = 0}.

(10.1.35)

2  Note that V is a closed subspace of H01 () , so it is a Hilbert space with respect to the induced scalar product. In the sequel, we will consider the variational formulation of the Stokes system. Find u ∈ V such that

 2  i, j=1

 ∂u i ∂vi dx = ∂x j ∂x j i=1 2

μ

f i vi d x, ∀ v ∈ V.



(10.1.36)

Definition 10.1.2 A function u ∈ V is a weak solution of (10.1.34) if it verifies (10.1.36) for all v ∈ V . The existence and uniqueness of the system (10.1.34) follows from the theorem of Lax-Milgram [1]. Theorem 10.1.5 The system (10.1.34) has a unique weak solution u ∈ V . Proof We define on V the bilinear form a( u , v ) =

 2

and the linear form L( v) =

 i, j=1

μ

2  i=1



∂u i ∂vi dx ∂x j ∂x j

(10.1.37)

f i vi d x.

(10.1.38)

Note that a is bilinear, continuous, symmetric and coercive (by the inequality of Poincaré), while L is linear and continuous. Hence, the conditions in the theorem of Lax-Milgram are satisfied and (10.1.36) has a unique solution u ∈ V .  Proposition 10.1.1 Let  ⊂ R2 be an open, connected and 1-regular set. Let L be  2 a linear and continuous form on H01 () . Then L = 0 in V if and only if there exists ϕ ∈ L 2 () such that

232

10 Variational Problems

L( v) =



 2 ϕdiv v d x, ∀v ∈ H01 () .

(10.1.39)

Theorem 10.1.6 Let  ⊂ R2 be an open, connected and 1-regular set. Then there exists a unique function u ∈ V and there exists p ∈ L 2 () unique up to a constant such that

 2 a( u , v ) − pdiv v d x = L( v ), ∀ v ∈ H01 () (10.1.40) 

and u is the solution of (10.1.36). Proof Let u ∈ V be the solution of (10.1.36). Then we define the map v → L( v ) = a( u , v ) − L( v ).

(10.1.41)

 2 Note that L is a linear and continuous map on H01 () with the property that L = 0 in V . There exists p ∈ L 2 () unique up to a constant such that

L( v) =



Finally, ( u , p) is the solution of (10.1.40).

pdiv v d x.

(10.1.42) 

Therefore, the solution ( u , p) of (10.1.40) verifies (10.1.34) in the sense of dis 2 tributions on . Moreover, if u ∈ H 2 () , then p ∈ H 1 () and the Eq. (10.1.34) is verified in L 2 ().

10.1.2 The Elasticity System Let  ⊂ R3 be an open and 1-regular set. Let 0 be a subset of the boundary ∂ of strictly positive measure and 1 = ∂ \ 0 . We consider v = (v1 , v2 , v3 ) to be a vectorial function defined on  and with values in R3 . Then we define   ∂v j 1 ∂vi v) = + εi j ( , 1 ≤ i, j ≤ 3, 2 ∂x j ∂xi  3  (10.1.43)  v) = λ εkk ( v ) δi j + 2μεi j ( v ), 1 ≤ i, j ≤ 3, σi j ( k=1

where λ ≥ 0 and μ > 0 are two real constants. In the sequel, we consider a problem of the three-dimensional elasticity. If f =  3 ( f 1 , f 2 , f 3 ) ∈ L 2 () , find u = (u 1 , u 2 , u 3 ) a solution of the system

10.1 Introduction

233

⎧ 3 ∂ u ) + f i = 0 in , 1 ≤ i ≤ 3 ⎪ ⎨ j=1 ∂x j σi j ( on 0 ui = 0 ⎪ ⎩3 u )ν j = 0 on 1 . j=1 σi j (

(10.1.44)

These equations describe the small deformations u of an elastic solid  which is homogeneous and isotropic and is under the action of an exterior force f . This solid is fixed on 0 and free on 1 . The tensor σi j represents the constraints, εi j represents the tensor of linear deformations and the constants μ and λ are the Lamé coefficients [3]. In the following, we consider the variational formulation of the problem (10.1.44). Therefore, we define the space 3  v = 0 on 0 } V = { v ∈ H 1 () | and the norms  v  H 1 () =

 3 

 21 vi 2H 1 ()

i=1

 v  L 2 () =

 3 

(10.1.45)

, (10.1.46)

 21 vi 2L 2 ()

.

i=1

The following result is based on the inequality of Korn [3]. Theorem 10.1.7 Let  be an open and 1-regular set in R N . Then there exists a positive constant C such that 3 

εi j ( v )2L 2 () ≥ C v 2H 1 () , ∀ v ∈ V.

(10.1.47)

i, j=1

We define the bilinear form a : V × V → R a( u , v ) =

3  i, j=1 

σi j ( u )εi j ( v )d x

(10.1.48)

and the linear form L : V → R L( v) =

3  i=1



vi f i d x.

(10.1.49)

Theorem 10.1.8 There exists a unique u ∈ V which verifies the variational formulation a( u , v ) = L( v ), ∀ v ∈ V. (10.1.50)

234

10 Variational Problems

Proof Note that a is bilinear, symmetric, continuous and coercive (by Theorem 10.1.7), while L is linear and continuous. The result follows by the theorem of Lax-Milgram [1].  Note that a is symmetric. Then we can find the solution u of (10.1.50) by minimizing the functional 1 J ( v ) = a( v , v ) − L( v ). (10.1.51) 2 Note that v ). J ( u ) = min J ( v ∈V

(10.1.52)

Definition 10.1.3 The solution u of the problem (10.1.50) is a weak solution of the system (10.1.44). In the sequel, we will analyse how the weak solution u verifies the initial system (10.1.44). By the symmetry of the tensor σi j , we have a( u , v ) =

3  i, j=1 

σi j ( u)

∂vi d x. ∂x j

(10.1.53)

We set v ∈ (D())3 . From (10.1.50) it follows that −

3  ∂ σi j ( u ) = f i in , 1 ≤ i ≤ 3 ∂x j j=1

(10.1.54)

3  in the sense of distributions. If u ∈ H 2 () , then (10.1.54) holds true in L 2 (). In the following, we study the boundary conditions. Note that u ∈ V , so these are satisfied in the sense of trace on 0 . On the other hand, the conditions on 1 are incorporated in the variational formulation (10.1.50). To this end, we consider 3  u ∈ H 2 () and apply Green’s formula in (10.1.50). Hence, we have 3  i=1

This leads to

1

3 

⎛ ⎞ 3  ⎝ σi j ( u )ν j ⎠ vi dσ = 0, ∀ v ∈ V.

(10.1.55)

j=1

σi j ( u )ν j = 0 on 1 , 1 ≤ i ≤ 3.

(10.1.56)

j=1

The variational formulation (10.1.50) has a physical interpretation. Note that V is the space of admissible deformations v . Then (10.1.50) describes the principle of virtual work. More precisely, if v ∈ V is an admissible deformation, then the work

10.1 Introduction

235

of deformation of the elastic solid a( u , v ) =

3  i, j=1 

σi j ( u )εi j ( v )d x

(10.1.57)

is equal to the work of the external forces L( v) =

3  i=1



vi f i d x.

(10.1.58)

The real deformation u is the one for which the two expressions of the work are equal. Hence, (10.1.50) holds true. The formulation (10.1.52) shows that the real deformation is the admissible deformation which minimizes the potential energy J . Note that J is composed of a part associated to the deformation energy 3 1  σi j ( v )εi j ( v )d x 2 i, j=1 

(10.1.59)

and a part associated to the energy of the external forces 3  i=1



vi f i d x.

(10.1.60)

10.1.3 The Plate Equation Let  ⊂ R N be an open and bounded set with 2-regular boundary. Let f ∈ L 2 () be a function. Then we consider  −2 u = f in  (10.1.61) ∂u = 0 on ∂. u = ∂ν This equation describes the equilibrium position u of a thin plate (or of a bar if N = 1) which is under the action of a distributed force f . We have N  ∂4u . (10.1.62) 2 u = (u) = ∂ 2 xi ∂ 2 x j i, j=1 We define the space

236

10 Variational Problems

  ∂v = 0 on ∂ . V = v ∈ H 2 ()|v = 0, ∂ν

(10.1.63)

We introduce the bilinear form a and the linear form L by

a(u, v) = uvd x,

 L(v) = f vd x.

(10.1.64)



Note that a is bilinear, symmetric and continuous on V , while L is linear and continuous on V . Theorem 10.1.9 Let  ⊂ R N be an open and 1-regular set. Then there exists a positive constant C such that

ϕ2H 2 ()

≤C



|ϕ|2 d x, ∀ϕ ∈ H02 ().

(10.1.65)

Proof Since D() is dense in H02 (), it suffices to prove the inequality for functions ϕ ∈ D(). By Green’s formula, we have for all ϕ ∈ D()

 

∂2ϕ ∂xi ∂x j

2

dx = −



∂ 3 ϕ ∂ϕ dx = ∂xi2 ∂x j ∂x j



∂2ϕ ∂2ϕ d x. ∂xi2 ∂x 2j

(10.1.66)

Hence, we have |ϕ|2H 2 () =



|ϕ|2 d x.

(10.1.67) 

In the sequel, we consider the variational formulation a(u, v) = L(v), ∀v ∈ V.

(10.1.68)

Theorem 10.1.10 There exists a unique solution u ∈ V of the problem (10.1.68). Proof The result follows by the theorem of Lax-Milgram.



Definition 10.1.4 The solution u ∈ V of the problem (10.1.68) is called a weak solution of the Eq. (10.1.61).

10.2 The Approximation of Variational Problems

237

10.2 The Approximation of Variational Problems Let V be a Hilbert space. Let us consider a sequence of vector spaces (Vh )h>0 with the following properties: 1. for any h, Vh is a vector subspace of V Vh ⊂ V.

(10.2.1)

2. for any h > 0, Vh has a finite dimension dim(Vh ) = Ih < ∞.

(10.2.2)

We will assume that the dimension of the space V is infinite dim(V ) = ∞.

(10.2.3)

The spaces Vh are approximations of finite dimension of the infinite dimensional space V . We will assume that the dimensions of the spaces Vh tend to the dimension of the space V as h tends to zero. Therefore, we will suppose that lim Ih = ∞.

h→0

(10.2.4)

In the space Vh , we consider the following problem. Find u h ∈ Vh such that a(u h , ϕ) = L(ϕ), ∀ϕ ∈ Vh .

(10.2.5)

Since Vh is a finite dimensional subspace of V , it is a Hilbert space with respect to the scalar product in V . Therefore, the existence and uniqueness of the solution u h ∈ Vh of the problem (10.2.5) can be proven by the theorem of Lax-Milgram. In the sequel, we present a different argument which takes into consideration that the problem is formulated in a finite dimensional space. Theorem 10.2.1 If Vh ⊂ V is a finite dimensional subspace of V with dimension Ih , then the problem (10.2.5) has a unique solution u h ∈ Vh . Proof Let {ϕ1 , ϕ2 , . . . , ϕ Ih } be a basis of the space Vh . We search for a solution of the problem (10.2.5) of the form uh =

Ih 

j

uh ϕ j ,

(10.2.6)

j=1 j

where (u h )1≤ j≤Ih are coefficients of u h in the basis (ϕ j )1≤ j≤Ih . By (10.2.6) and the linearity of the bilinear form a, the problem (10.2.5) can be rewritten as

238

10 Variational Problems Ih 

j

u h a(ϕ j , ϕ) = L(ϕ), ∀ϕ ∈ Vh .

(10.2.7)

j=1

If we consider the test function of the form ϕ=

Ih 

αi ϕi ,

(10.2.8)

i=1

then we deduce that the existence and uniqueness of u h is equivalent to the existence j and uniqueness of the scalars (u h )1≤ j≤Ih which verifies Ih 

⎛ αi ⎝

i=1

Ih 

⎞ u h a(ϕ j , ϕi ) − L(ϕi )⎠ = 0, j

(10.2.9)

j=1

for any choice of (αi )1≤i≤Ih . Therefore, u h of the form (10.2.6) is a solution of the problem (10.2.5) if and only if Ih 

j

u h a(ϕ j , ϕi ) = L(ϕi ), 1 ≤ i ≤ Ih .

(10.2.10)

j=1

Let R be a square matrix of dimension Ih with the elements Ri j = a(ϕ j , ϕi ), F the vector of dimension Ih with the elements Fi = L(ϕi ) and U the vector of j dimension Ih with the elements U j = u h . The relation (10.2.10) can be written in matrix form RU = F. (10.2.11) In order to find the solution u h , we have to solve the algebraic linear system of Eq. (10.2.11). Clearly, the matrix R is positive definite. If V = (v j )1≤ j≤Ih ∈ R Ih is an arbitrary vector, then by the coercivity property of a, we obtain V t RV =

Ih Ih   i=1 j=1

⎞ ⎛ Ih Ih   a(ϕ j , ϕi )v i v j = a ⎝ v jϕj, v i ϕi ⎠ ≥

     Ih  Ih    j     i · ≥ μ v ϕ v ϕ . j i      j=1  i=1

j=1

i=1

(10.2.12)

 Ih i Therefore, we obtain V t RV ≥ 0 and V t RV = 0 if and only if i=1 v ϕi = 0, which is equivalent to V = 0. Hence, we proved that R is positive definite. By the positive definiteness of R, the system (10.2.11) has a unique solution U ∈ R Ih . Therefore, (10.2.5) has a unique solution u h given by (10.2.6).

10.2 The Approximation of Variational Problems

239

Note that if the bilinear form a is symmetric, then the matrix R is also symmetric.  In the sequel, we will prove a convergence criterion. To this end, let u ∈ V be a solution of the problem: find u ∈ V such that a(u, ϕ) = L(ϕ), ∀ϕ ∈ V.

(10.2.13)

Let u h ∈ Vh be a solution of the problem: find u h ∈ Vh such that a(u h , ϕ) = L(ϕ), ∀ϕ ∈ Vh .

(10.2.14)

Our goal is to approximate u by u h . Then we define the error eh = u − u h ∈ V.

(10.2.15)

In the sequel, we will give estimates for the norm of eh in V . Theorem 10.2.2 If u ∈ V is a solution of the problem (10.2.13) and u h ∈ Vh is a solution of the problem (10.2.14), then (i) a(eh , ϕ) = 0, ∀ϕ ∈ Vh ; (ii) a(eh , eh ) ≤ a(u − ϕ, u − ϕ), ∀ϕ ∈ Vh ; (iii) a(u − u h , u − u h ) = minϕ∈Vh a(u − ϕ, u − ϕ). Proof (i) Since u ∈ V is a solution of the problem (10.2.13) and u h ∈ Vh is a solution of the problem (10.2.14), we have a(u, ϕ) = L(ϕ), a(u h , ϕ) = L(ϕ), ∀ϕ ∈ Vh .

(10.2.16)

The result is obtained by subtracting these two relations. (ii) Let ϕ ∈ Vh . We have a(u − ϕ, u − ϕ) = a(eh + u h − ϕ, eh + u h − ϕ) = = a(eh , eh ) + a(u h − ϕ, u h − ϕ) + 2a(eh , u h − ϕ).

(10.2.17)

Since u h − ϕ ∈ Vh , by the previous result, we have a(u − ϕ, u − ϕ) = a(eh , eh ) + a(u h − ϕ, u h − ϕ) ≥ ≥ a(eh , eh ) + μ||u h − ϕ||2 ≥ a(eh , eh ). (iii) This follows from the previous result by considering that u h ∈ Vh .

(10.2.18) 

In the sequel, we present Cea’s lemma [3]. Lemma 10.2.1 If eh = u − u h is the error obtained by approximating u by u h , then there exists a constant C > 0 which is independent of h and such that

240

10 Variational Problems

eh  ≤ C inf u − ϕ. ϕ∈Vh

(10.2.19)

Proof Let us consider ϕ ∈ Vh and wh = ϕ − u h ∈ Vh . Then we have a(u − u h , wh ) = 0

(10.2.20)

a(u − u h , u − u h ) = a(u − u h , u − ϕ).

(10.2.21)

so

We deduce that μu − u h 2 ≤ a(u − u h , u − u h ) = a(u − u h , u − ϕ) ≤ Mu − u h  · u − ϕ (10.2.22)  and the conclusion follows with C = Mμ . Definition 10.2.1 The sequence (Vh )h>0 is an internal approximation of V if (i) for any h > 0, Vh ⊂ V is a subspace of V of finite dimension; (ii) for any ϕ ∈ V and h > 0, there exists ϕh ∈ Vh such that lim ϕ − ϕh  = 0.

h→0

(10.2.23)

Theorem 10.2.3 If (Vh )h>0 is an internal approximation of the space V , u is the solution of the problem (10.2.13) in V and u h is the solution of the problem (10.2.14) in Vh , then (10.2.24) lim u − u h  = 0. h→0

Proof Let ε > 0 be arbitrary. By the approximation property of the spaces (Vh )h>0 we deduce that for any h > 0 there exists vh ∈ Vh such that lim u − vh  = 0.

h→0

(10.2.25)

Therefore, there exists h ε > 0 with the property that, for any h < h ε , we have u − vh 
0 converges to u. We assume that V is a separable Hilbert space. Then there exists an orthonormal basis (vi )i≥1 of V which is at most countable. Definition 10.2.2 (vi )i≥1 ⊂ V is an orthonormal basis of V if (i) span(vi )i≥1 = V ; (ii) (vi , v j ) = δi j , ∀i, j ≥ 1.   For every h > 0 we choose N = h1 . Let us consider Vh = span{v1 , . . . , v N }.

(10.2.28)

Then the sequence (Vh )h>0 is an internal approximation of V . Clearly, Vh ⊂ V has a finite dimension. In the following, we will verify the second condition in the definition of the internal approximation. Let us consider ϕ ∈ V and h > 0. We define ϕh =

N  (ϕ, vi )vi ∈ Vh .

(10.2.29)

i=1

Since (vi )i≥1 is an orthonormal basis of V , we have ϕ=

∞ 

(ϕ, vi )vi , ϕ2 =

i=1

∞ 

|(ϕ, vi )|2 < ∞.

(10.2.30)

i=1

Then we have ϕ − ϕh 2 =

∞ 

|(ϕ, vi )|2 .

(10.2.31)

i=N +1

Since the last term tends to zero as N tends to infinity, we conclude that the second condition from the definition of the internal approximation is verified. Let u be the solution of the problem (10.2.13) in V and let u h be the solution of the problem (10.2.14) in Vh given by (10.2.28). Since lim u − u h  = 0,

h→0

the method converges.

(10.2.32)

242

10 Variational Problems

10.2.2 The Finite Element Method In the sequel, we will discuss an example of discretization by the finite element method, see [1, 2, 4]. Let us consider V = H01 (0, 1),

1

1   p(x)u (x)ϕ (x)d x + q(x)u(x)ϕ(x)d x, a(u, ϕ) = 0 0

1 L(ϕ) = f (x)ϕ(x)d x, ∀ϕ ∈ V,

(10.2.33)

0

where f ∈ L 2 (0, 1), p, q ∈ L ∞ (0, 1) with p(x) ≥ p0 > 0 and q(x) ≥ 0 almost everywhere in (0, 1). The corresponding variational problem a(u, ϕ) = L(ϕ), ∀ϕ ∈ V

(10.2.34)

has a unique solution u ∈ V . We will determine the approximation u h of u in the case of the problem (10.2.34) by the finite element method. Let N ∈ N∗ and h = N 1+1 . We split the interval [0, 1] in N + 1 equal parts K i = [xi , xi+1 ] by the points xi = i h, 0 ≤ i ≤ N + 1. Hence, we obtain a partition of the interval [0, 1]. We define the space Vh = {v ∈ C[0, 1]|v(0) = v(1) = 0, v| Ki = αx + β, 1 ≤ i ≤ N }.

(10.2.35)

Cleary, Vh is a linear space included in V . Moreover, it has the finite dimension N . For the proof, let us define for any i ∈ {1, 2, . . . , N } the function ϕi by

ϕi (x) =

⎧ x−x i−1 ⎪ ⎨ h xi+1 −x ⎪ h

⎩

0

if x ∈ [xi−1 , xi ) if x ∈ [xi , xi+1 ] . otherwise

(10.2.36)

Note that ϕi (x j ) = δi j for any 1 ≤ i, j ≤ N . We obtain that ϕi ∈ Vh and (ϕi )1≤i≤N forms a basis in Vh . Clearly, (ϕi )1≤i≤N is linear independent since  1≤i≤N

 αi ϕi = 0 ⇒



 αi ϕi (x j ) = 0, ∀1 ≤ j ≤ N ⇒

1≤i≤N

⇒ α j = 0, ∀1 ≤ j ≤ N . This is also a system of generators since for any v ∈ Vh we have

(10.2.37)

10.2 The Approximation of Variational Problems



v=

243

v(xi )ϕi .

(10.2.38)

1≤i≤N

Therefore, we have a finite dimensional subspace Vh of V . Moreover, the basis (ϕi )1≤i≤N does not depend on the bilinear form from the problem (10.2.34). On Vh we consider the problem

1 p(x)u h (x)ϕ (x)d x + q(x)u h (x)ϕ(x)d x = 0 0

1 = f (x)ϕ(x)d x, ∀ϕ ∈ Vh ,

1

(10.2.39)

0

which has a unique solution u h ∈ Vh . We have uh =

N 

u ih ϕi ,

(10.2.40)

i=1

where U = (u ih )1≤i≤N is the solution of the system of linear equations RU = F.

(10.2.41)

The stiffness matrix R has the elements Ri j = a(ϕ j , ϕi ) and the vector F has the components F j = L(ϕ j ). Since the supports of the functions ϕi and ϕ j intersect in at most one point when |i − j| > 1, the matrix R is tridiagonal and it has many null elements. While its dimension is N (which can be very big), the number of elements different from zero on each line is at most equal to three. Therefore, we obtain a system of linear equations which is easily solvable from a numerical point of view. In the sequel, we will analyze the error obtained by approximating u by u h and will discuss the convergence of the method. In order to estimate the error eh = u − u h , it is sufficient to evaluate in the norm V the difference between the exact solution u and an element from the subspace Vh . We will search for an element Uh ∈ Vh such that the difference u − Uh is small. Let Uh be the interpolant of u in Vh , which is the unique function belonging to Vh which coincides with u in all points xi . Uh is given by Uh =

N 

u(i h)ϕi .

i=1

Clearly, Uh ∈ Vh . Moreover, we have Uh (i h) = u(i h), 1 ≤ i ≤ N .

(10.2.42)

244

10 Variational Problems

Theorem 10.2.4 Let u ∈ H 2 (0, 1) ∩ H01 (0, 1) be a function and Uh be given by (10.2.42). Then there exists a positive constant c which is independent of h and such that u − Uh  L 2 (0,1) ≤ ch 2 u   L 2 (0,1) , (10.2.43) u − Uh  H01 (0,1) ≤ chu   L 2 (0,1) . Proof Step 1. Let us consider an interval [xi , xi+1 ] and a function (x) = u(x) − Uh (x) which becomes null for xi and xi+1 . By using Fourier series, we obtain (x) =

∞ 

 an sin

n=1

 nπ(x − xi ) . h

(10.2.44)

  i) in L 2 (xi , xi+1 ) we deduce By the orthogonality of the functions sin nπ(x−x h that

xi+1 ∞ h  2 n2 π2 ( )2 d x = a , 2 n=1 n h 2 xi (10.2.45)

xi+1 ∞ h  2 n4 π4  2 ( ) d x = a . 2 n=1 n h 4 xi By the two equalities above and since (Uh ) (x) = 0 on each interval [xi , xi+1 ], we deduce that

xi+1 h 2 xi+1  2 h 2 xi+1  2  2 ( ) d x ≤ 2 ( ) d x = 2 (u ) d x. (10.2.46) π xi π xi xi In conclusion, we have

1

( )2 d x =

0

N  i=0

2

h ≤ 2 π

N xi+1  xi

i=0

xi+1

( )2 d x ≤

xi

h2 (u ) d x = 2 π  2

1

(10.2.47)  2

(u ) d x

0

and the second inequality from (10.2.43) is shown. Step 2. We will prove the first inequality. Note that

xi+1 xi

Similarly, we obtain

()2 d x =

∞ h 2 a . 2 n=1 n

(10.2.48)

10.2 The Approximation of Variational Problems

1

()2 d x =

0

h4 ≤ 4 π

N  i=0

xi+1 xi

xi+1

245

()2 d x ≤

xi

N h 4  xi+1  2 h 4 1  2 ( ) d x = 4 (u ) d x = 4 (u ) d x. π i=0 xi π 0

(10.2.49)

 2



This concludes the proof. Theorem 10.2.5 If u ∈ C 2 [0, 1] and Uh is given by (10.2.42), then h2 max |u  (x)|, x∈[0,1] 8 x∈[0,1] max |u  (x) − Uh (x)| ≤ h max |u  (x)|. max |u(x) − Uh (x)| ≤

x∈[0,1]

(10.2.50)

x∈[0,1]

Proof Step 1. Let us consider x ∈ [0, 1] and i ∈ {0, 1, . . . , n} such that x ∈[xi , xi+1]. Let us define the function (x) = u(x) − Uh (x) on the interval [xi , xi+1 ]. Note that it becomes null for xi and xi+1 . Then there exists z ∈ [xi , xi+1 ] with the property that  (z) = 0 and we have

x

 (x) =

 (t)dt.

(10.2.51)

z

Since  = u  on the interval [xi , xi+1 ], we obtain   | (x)| = 

x z

     (t)dt  = 

x z

  u  (t)dt  ≤ h max |u  (x)| x∈[0,1]

(10.2.52)

and the second inequality is shown. Step 2. For the first inequality, let us consider xm ∈ [xi , xi+1 ] with the property that (10.2.53) max |(x)| = |(xm )|. x∈[xi ,xi+1 ]

If (xm ) = 0, then the inequality is true. If (xm ) = 0, then xm ∈ (xi , xi+1 ) and  (xm ) = 0. We will use the Taylor series (x) = (xm ) + (xm − x) (xm ) + (xm − x)2   (ξ) = (xm ) + 2

(xm − x)2   (ξ) = 2

(10.2.54)

for any x ∈ [xi , xi+1 ] and where ξ lies between x and xm . We consider that x is the endpoint of the interval [xi , xi+1 ] which is the closest to the point xm . Hence, we obtain

246

10 Variational Problems

|(xm )| ≤

h 2  | (ξ)|, 8

(10.2.55) 

which proves the first inequality.

In the sequel, we present a convergence result. To be more precise, we estimate the error from the finite element method in the norm from H 1 . Theorem 10.2.6 If the solution of the problem (10.2.34) is u ∈ H 2 (0, 1) ∩ H01 (0, 1) and the solution of the problem (10.2.39) is u h , then u − u h  H01 ≤ chu   L 2 ,

(10.2.56)

where c > 0 is a constant which does not depend on h. Proof We have u − u h  H01 ≤ C inf u − ϕ H01 .

(10.2.57)

ϕ∈Vh

Since Uh ∈ Vh , we obtain inf u − ϕ H01 ≤ u − Uh  H01 ≤ chu   L 2

(10.2.58)

ϕ∈Vh



and the proof is finished.

Theorem 10.2.7 If the solution of the problem (10.2.34) is u ∈ H 2 (0, 1) ∩ H01 (0, 1) and the solution of the problem (10.2.39) is u h , then u − u h  L 2 ≤ ch 2 u   L 2 ,

(10.2.59)

where c > 0 is a constant which does not depend on h. Proof Let z be the solution of the problem (10.2.34) with f = eh = u − u h . Taking eh as a test function, we have a(z, eh ) = eh 2L 2

(10.2.60)

where a is the bilinear form from (10.2.34). On the other hand, since a(eh , ϕ) = 0 for any ϕ ∈ Vh , we have a(z − ϕ, eh ) = a(z, eh ) = eh 2L 2 , ∀ϕ ∈ Vh .

(10.2.61)

By Hölder’s inequality, we obtain for any ϕ ∈ Vh 1

1

eh 2L 2 = a(z − ϕ, eh ) ≤ [a(z − ϕ, z − ϕ)] 2 [a(eh , eh )] 2 . We also have

(10.2.62)

10.2 The Approximation of Variational Problems

247

[a(eh , eh )] 2 ≤ Ceh  H01 ≤ Chu   L 2 . 1

(10.2.63)

Moreover, if z h is the approximation of z given by (10.2.39), then we obtain [a(z − z h , z − z h )] 2 ≤ Chz   L 2 1

(10.2.64)

By taking ϕ = z h in (10.2.62), we obtain eh 2L 2 ≤ Ch 2 u   L 2 z   L 2 ≤ Ch 2 u   L 2 z H 2 .

(10.2.65)

Since the application that associates to each term the corresponding solution u of the Eq. (10.2.34) is continuous from L 2 into H 2 , we have z   L 2 ≤ z H 2 ≤ Ceh  L 2

(10.2.66) 

and the proof is finished.

In the sequel, we will discuss the pointwise convergence of the finite element method by considering only the vertices of the triangulation xi , 1 ≤ i ≤ N and the model equation

1

u  (x)ϕ (x)d x =

0

0

1

f (x)ϕ(x)d x, ∀ϕ ∈ H01 (0, 1),

(10.2.67)

which has the unique solution u ∈ H01 (0, 1). After discretization by the finite element method as before, we obtain an approximation of the solution u of the problem (10.2.67) by u h , which is the solution of the equation

1

1 u h (x)ϕ (x)d x = f (x)ϕ(x)d x, ∀ϕ ∈ Vh . (10.2.68) 0

0

After subtracting the relations (10.2.67) and (10.2.68), we obtain

1

(u − u h ) (x)ϕ (x)d x = 0, ∀ϕ ∈ Vh .

(10.2.69)

0

Let ϕi be a basis in Vh as before and u h (x) =

N 

j

u h ϕ j (x).

(10.2.70)

j=1 j

j

Note that u h is the value of u h in x j and u(x j ) ≈ u h (x j ) = u h , 1 ≤ j ≤ N . In the sequel, we will show that the exact solution coincides with the approximative solution in the nodes x j . This is the phenomenon of superconvergence since in these

248

10 Variational Problems

points the error is much smaller (in fact, equal to zero) than the error in the general case. Proposition 10.2.1 If u is the solution of (10.2.67) and u h is the solution of (10.2.68), then for any N ≥ 1, we have j

u(x j ) = u h (x j ) = u h , 1 ≤ j ≤ N .

(10.2.71)

Proof Let y ∈ (0, 1) and let us consider the equation 

g  + δ y = 0 g(0) = g(1) = 0

,

(10.2.72)

where δ y is the Dirac delta. The variational formulation of (10.2.72) is

1 0

g  (x)ϕ (x)d x = ϕ(y), ∀ϕ ∈ H01 (0, 1),

(10.2.73)

which has the unique solution  (1 − y)x x < y g(x) = (1 − x)y x ≥ y.

(10.2.74)

Note that if y = xi , then g ∈ Vh . Therefore, g can be used as a test function in (10.2.69). By (10.2.73) and by (10.2.69), we obtain

u(xi ) − u h (xi ) =

1

g  (x)(u − u h ) (x)d x = 0

(10.2.75)

0

and the proof is finished.



References 1. D. Braess, Finite Elemente: Theorie, schnelle Löser und Anwendungen in der Elastizitätstheorie (Springer, 2007) 2. H. Elman, D. Silvester, A. Wathen, Finite Elements and Fast Iterative Solvers: With Applications in Incompressible Fluid Dynamics, 2nd edn. (Oxford University Press, Oxford, 2014) 3. S.D. Micu, Introduction to Finite Element Method (Romanian), Publications of the Centre for Nonlinear Analysis and its Applications 4. A. Öchsner, Computational Statics and Dynamics: An Introduction Based on the Finite Element Method (Springer, Singapore, 2020)

Chapter 11

On Some Spaces of Distributions

Abstract We present the spaces D L p and their properties. We also present their duals and their properties. Then we ask which is the most general space of distributions operating continuously on S  by convolution and such that the Fourier transform maps convolution products into the product of the corresponding Fourier transforms. So we present the space OC and its properties.

11.1 The Spaces D L p In the sequel, we present the spaces D L p and their duals [2–10]. Definition 11.1.1 Let p ∈ R such that 1 ≤ p ≤ ∞. We denote by D L p (Rn ) the space of all functions ϕ ∈ C ∞ (Rn ) such that ∂ α ϕ ∈ L p (Rn ) for all α ∈ Nn endowed with the coarsest locally convex topology for which the maps ∂ α : D L p (Rn ) → L p (Rn ) are continuous for all α ∈ Nn . The topology of D L p coincides with the one defined by the countable family of norms ⎛ ⎞ 1p  p ϕm, p  = ⎝ ∂ α ϕ L p ⎠ , m ∈ N. (11.1.1) |α|≤m

Clearly, D L p is a Frechet space [1]. The following embeddings hold true Cc∞ (Rn ) ⊂ D L p (Rn ) ⊂ D (Rn ), 1 ≤ p < ∞

(11.1.2)

and the injections are continuous. Moreover, if 1 ≤ p < ∞, then Cc∞ is a dense subspace of D L p . Therefore, D L p is a normal space of distributions in Rn . But Cc∞ is not dense in D L ∞ . We denote by D˙ L ∞ the subspace of all functions in D L ∞ which converge to zero at infinity. Clearly, D˙ L ∞ is a closed subspace of D L ∞ . Therefore, it is a Frechet space. © The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_11

249

250

11 On Some Spaces of Distributions

Moreover, we have Cc∞ ⊂ D L ∞ ⊂ D and the injections are continuous. Furthermore, Cc∞ is dense in D˙ L ∞ . Hence, D˙ L ∞ is a normal space of distributions. In the sequel, we present some properties on the structure of the spaces D L p . Recall that if α ∈ Nn , then rankα = max1≤ j≤n α j . Theorem 11.1.1 Let A, a ∈ R such that 0 < a < A. We define B A and B A−a to be the concentric balls of radius A and A − a, respectively. If a distribution T ∈ D (Rn ) is such that all its derivatives of rank≤ 1 are functions such that their L p norms on B A are bounded by a constant M, then on B A−a , T is a function bounded by C(A, n)a −n M, where C(A, n) is a constant independent of T and of a. Proof Note that there is a function ψ ∈ Cc∞ (Rn ) such that suppψ ⊂ B A , ψ(x) = 1 on B A−a and for all α ∈ Nn we have |∂ α ψ(x)| ≤ C(α, n)a −|α| . Let us consider the Heaviside function in Rn  1 if x1 ≥ 0, . . . , xn ≥ 0 Y (x) = 0 otherwise. This is a fundamental solution of ψT = δ ∗ (ψT ) =

∂n . ∂x1 ···∂xn

(11.1.3)

(11.1.4)

We can write

∂n Y ∂ n (ψT ) ∗ (ψT ) = Y ∗ . ∂x1 · · · ∂xn ∂x1 · · · ∂xn

(11.1.5)

Therefore, T is a function in B A−a . For all x = (x1 , . . . , xn ) ∈ B A−a , we have  T (x) = (ψT )(x) =

x1

−∞

 ···

xn −∞

∂ n (ψT ) dy. ∂ y1 · · · ∂ yn

(11.1.6)

By the formula of Leibniz, we obtain ∂ n (ψT ) = ∂ y1 · · · ∂ yn

 |α|+|β|=n

Cα,β

∂αψ ∂β T · , ∂ yα ∂ yβ

(11.1.7)

where rankα ≤ 1 and rankβ ≤ 1. By the inequality of Hölder, we have α β   α β ∂ ψ ∂ T ∂ ψ ∂ T ··· ∂ y α · ∂ y β dy ≤ · · · α · ∂ y β dy ≤ −∞ −∞ BA ∂ y

  α q  q1   β p 1p ∂ ψ ∂ T ≤ ··· · ··· ≤ ∂ y α dy β dy BA BA ∂ y



x1



xn

≤ C(α, n)a −|α| M,

(11.1.8)

11.1 The Spaces D L p

251



which implies the desired inequality.

Corollary 11.1.1 If T ∈ D (Rn ) is such that all its derivatives of rank ≤ 1 are functions belonging to L P (Rn ), then we have the following estimate 

sup |T (x)| ≤ C ·

x∈Rn

∂ α T  L p .

(11.1.9)

rankα≤1

Proof Let x be a variable element in Rn . It suffices to apply the theorem to balls with center at x and radius 2 and 1.  Corollary 11.1.2 If ϕ ∈ D L p (1 ≤ p ≤ ∞), then ϕ is bounded in Rn . Moreover, if 1 ≤ p < ∞, then ϕ converges to zero at infinity. Proof ϕ ∈ D L p is bounded by Corollary 11.1.1. If ϕ ∈ D L p , 1 ≤ p < ∞, then it is possible to find a number R > 0 so large that 

α ∂ ϕ ··· α dx < ε |x|>R ∂x 

for all α with rank ≤ 1. Then |ϕ(x)| is arbitrarily small for large |x|.

(11.1.10) 

Corollary 11.1.3 If 1 ≤ p ≤ q ≤ ∞, then D L p ⊂ D L q and the embedding is continuous. Proof If q = ∞, then the result follows from Corollary 11.1.2. We assume that q < ∞. Let ϕ(β) = ∂ β ϕ. By Corollary 11.1.1, we obtain  Rn

|ϕ(β) (x)|q d x ≤ sup |ϕ(β) (x)|q− p

≤C



x∈Rn

 Rn

|ϕ(β) (x)| p d x ≤

∂ α ϕ(β)  L p · ϕβ  L p . q− p

p

(11.1.11)

rankα≤1

 Observation 11.1.1 The spaces D L p , 1 ≤ p ≤ ∞ are related to the Sobolev spaces H m, p (Rn ). Clearly, by the Sobolev embedding theorem H m, p (Rn ) is a space of continuous functions provided that m is sufficiently large. Therefore, we have m, p (Rn ) D L p (Rn ) = ∩∞ m=0 H

(11.1.12)

and the topology of D L p coincides with the coarsest one for which the embeddings D L p → H m, p are continuous. Since the spaces D L p , 1 ≤ p < ∞ and D L ∞ are normal spaces of distributions, their duals are subspaces of D (Rn ).

252

11 On Some Spaces of Distributions

Definition 11.1.2 We denote by DL p , 1 < p ≤ ∞, the dual of D L q , with We denote by DL 1 the dual of D L ∞ .

1 p

+

1 q

= 1.

These duals are subspaces of D (Rn ). We can define the strong topology of DL p and prove that the embedding DL p → D is strongly continuous. Since the spaces L p are reflexive, 1 < p < ∞, we deduce that the spaces D L p are reflexive. Since (11.1.13) D L q  ⊂ D L p , ∀q  ≤ p  , Cc∞ (Rn ) is dense in D L q  , 1 ≤ q  < ∞ and in D L ∞ , it follows that DL p ⊂ DL q , ∀ p ≤ q.

(11.1.14)

In the sequel, we present a result about the structure of DL p . Theorem 11.1.2 A distribution T ∈ DL p if and only if there is an integer m = m(T ) > 0 such that  T = ∂ α fα , (11.1.15) |α|≤m

where f α ∈ L p . Proof We only consider the case 1 < p ≤ ∞. If T ∈ D (Rn ) is of the form (11.1.15), then T defines a continuous linear functional on D L q . Therefore, we have T ∈ DL p . Conversely, we assume that T ∈ DL p . Then there is a neighbourhood of zero W (m, δ) in D L q such that |(T, ϕ)| ≤ 1, ∀ϕ ∈ W (m, δ).

(11.1.16)

Hence, there is C > 0 a constant such that |(T, ϕ)| ≤ C sup ∂ α ϕ L q , ∀ϕ ∈ D L q . |α|≤m

(11.1.17)

Let us denote by N the number of n-tuples α such that |α| ≤ m. We set (L q ) N = L q × · · · × L q (N times) and define the map J : D L p ϕ → (∂ α ϕ)|α|≤m ∈ (L q ) N .

(11.1.18)

Clearly, J is a one-to-one map. Then we can identify J D L p with a proper subspace of (L q ) N . On J D L q , we define the linear functional F : J D L q → C as follows   F (∂ α ϕ)|α|≤m = (T, ϕ) , ∀ϕ ∈ D L q .

(11.1.19)

By (11.1.17), F is a continuous linear functional on J D L q endowed with the topology induced by (L q ) N . By the Hahn-Banach theorem, F extends to a continuous linear

11.1 The Spaces D L p

253

functional on (L q ) N . Since the dual of the product (L q ) N can be identified with the product (L p ) N , there are functions gα ∈ L p , |α| ≤ m such that     (T, ϕ) = F (∂ α ϕ)|α|≤m = |α|≤m

Rn

gα

∂αϕ dx ∂x α

(11.1.20)

for all ϕ ∈ D L q . Therefore, we obtain T =



∂ α fα ,

(11.1.21)

|α|≤m

where f α = (−1)|α| gα .



Note that the space D L 2 coincides, in the algebraic and topological senses, with H ∞ . Then its dual D (L 2 ) will coincide with H −∞ . Corollary 11.1.4 For every distribution T ∈ DL p , there is an integer m = m(T ) > 0 such that  ∂ α fα (11.1.22) T = |α|≤m

with f α being bounded continuous functions belonging to L p . Furthermore, if p < ∞, then each f α converges to zero at infinity.  Proof We have T = |β|≤l ∂ β gβ , where gβ ∈ L p . Let E be a fundamental solution of (1 − )k and choose k so large that E is a continuous function. Let γ ∈ Cc∞ (Rn ) be such that γ = 1 on a neighbourhood of the origin in Rn and set F = γ E. We have (1 − )k F = δ − ϕ,

(11.1.23)

with ϕ ∈ Cc∞ (Rn ). The distribution F is called a parametrix of the partial differential operator (1 − )k . From (11.1.23) it follows that every distribution g ∈ D (Rn ) can be written as follows (11.1.24) g = (1 − )k (γ E ∗ g) + ϕ ∗ g. Moreover, if g ∈ L p , since γ E and ϕ are continuous functions with compact support, it follows that γ E ∗ g and ϕ ∗ g are bounded continuous functions belonging to L p and converging to zero at infinity if p < ∞. Therefore, every g ∈ L p can be written as a finite sum of derivatives (in the sense of distributions) of bounded continuous functions belonging to L p and converging to zero at infinity if p < ∞. We apply these facts to the functions gβ appearing in the representation of T and we obtain the desired result.  In the sequel, we will present some results about the Fourier transform of elements of DL p in the cases p = 1 and p = 2.

254

11 On Some Spaces of Distributions

Theorem 11.1.3 Let ϕ ∈ D L 1 . Then its Fourier transform is a continuous function rapidly decreasing at infinity. Let T ∈ DL 1 . Then its Fourier transform is a continuous function slowly increasing at infinity. Proof Step 1. We assume that ϕ ∈ D L 1 . Then for all α ∈ Nn , we have D α ϕ ∈ L 1 . ˆ is a continuous bounded function on Rn . Hence, for every polyTherefore, ξ α ϕ(ξ) nomial P(ξ), the product P(ξ)ϕ(ξ) ˆ is a bounded function on Rn . This implies that ϕˆ is a continuous function rapidly decreasing at infinity.  Step 2. Let T ∈ DL 1 . Then T = |α|≤m D α f α , with f α being bounded continuous  functions belonging to L 1 (Rn ). This implies that Tˆ (ξ) = |α|≤m ξ α fˆα , with fˆα being bounded continuous functions. Therefore, Tˆ (ξ) can be written as a product of a polynomial and a bounded continuous function.  Theorem 11.1.4 A distribution T belongs to DL 2 if and only if its Fourier transform is the product of a polynomial and an L 2 function.  Proof Let T ∈ DL 2 . Then T = |α|≤m D α f α , where fˆα ∈ L 2 (Rn ). Therefore, we have  ξ α fˆα , (11.1.25) T (ξ) = |α|≤m

where fˆα ∈ L 2 . Note that the function  ˆ h(ξ) =

|α|≤m

ξ α fˆα

(11.1.26)

m

(1 + |ξ|2 ) 2

ˆ belongs to L 2 . Therefore, we obtain Tˆ (ξ) = (1 + |ξ|2 ) 2 h(ξ). m



Recall that a derivative of a continuous function slowly increasing at infinity defines a tempered distribution. We will prove that, conversely, every tempered distribution is the derivative of a continuous function slowly increasing at infinity. Lemma 11.1.1 Let T be a tempered distribution. Then there is k > 0 such that k (1 + r 2 )− 2 T ∈ DL ∞ . Proof Let T ∈ S  . Then we can find a neighbourhood of zero   k V (k, m, ε) = ϕ ∈ S| (1 + r 2 ) 2 ∂ p ϕ(x) ≤ ε, ∀| p| ≤ m, ∀x ∈ Rn

(11.1.27)

in S such that |(T, ϕ)| ≤ 1, ∀ϕ ∈ V . We claim that there is a neighbourhood of zero W in D L 1 such that for all ψ ∈ m S ∩ W , ϕ = (1 + r 2 )− 2 ψ ∈ V . Clearly, let   α ∂ ϕ W (m + n, η) = ψ ∈ D L 1 | sup ∂x α (x) d x ≤ η , |α|≤m+n 

(11.1.28)

11.1 The Spaces D L p

255

where η will be determined later. Note that for every n-tuple s ∈ Nn , we have k k s ∂ (1 + r 2 )− 2 ≤ Cs,k (1 + r 2 )− 2 .

(11.1.29)

Let p ∈ Nn be such that | p| ≤ m. By the formula of Leibniz, we have 

∂ pϕ =

s+q= p

 p! s  k ∂ (1 + r 2 )− 2 ∂ q ψ. s!q!

(11.1.30)

Therefore, we obtain |∂ p ϕ| ≤ C p,k (1 + r 2 )− 2 k



|∂ p ψ|.

(11.1.31)

 ∂q ψ (t) dt1 · · · dtn , ∂t q

(11.1.32)

q≤ p

We also have  ∂ q ψ(x) = which leads to

x1 −∞

 ···

xn

−∞

∂n ∂t1 · · · ∂tn

  sup ∂ q ψ(x) ≤  

x∈Rn



∂n ∂t1 · · · ∂tn



 ∂q ψ   ≤η ∂t q  L 1

(11.1.33)

since |q| + n ≤ m + n. Then we obtain k (1 + r 2 ) 2 ∂ p ϕ ≤ C p,k · η.

(11.1.34)

Let us choose η such that C p,k · η ≤ ε. Hence, we have ϕ ∈ V whenever ψ ∈ S ∩ W . Then for all ψ ∈ T ∩ W we obtain     k k (11.1.35) (1 + r 2 )− 2 T, ψ = T, (1 + r 2 )− 2 ψ ≤ 1. Note that S is dense in D L 1 . Therefore, (1 + r 2 )− 2 T is a continuous linear functional  on D L 1 . k

Theorem 11.1.5 A distribution T ∈ S  if and only if it can be represented as a finite sum  (11.1.36) T = ∂ α fα with f α being continuous functions slowly increasing at infinity. Proof Note that by Lemma 11.1.1 there is a number k > 0 such that (1 + r 2 )− 2 T ∈ DL ∞ . We have  k ∂γ hγ , hγ ∈ L ∞. (11.1.37) (1 + r 2 )− 2 T = k

256

11 On Some Spaces of Distributions



We set



x1

gγ (x) =

···

0

xn

h γ (t)dt1 · · · dtn .

(11.1.38)

0

We have |gγ (x)| ≤ |x1 | · · · |xn | sup |h γ (x)|.

(11.1.39)

x∈Rn

Therefore, gγ is a continuous function slowly increasing at infinity. We obtain T =



(1 + r 2 ) 2 ∂ β gβ . k

(11.1.40)

Note that every term (1 + r 2 ) 2 ∂ β gβ can be written as a finite sum k



∂ α (1 + r 2 ) 2 gα , k

(11.1.41)

with gα being a bounded continuous function.



Note that the function f α appearing in (11.1.36) is a continuous function slowly increasing at infinity. Therefore, it can be written as lα

f α = (1 + r 2 ) 2 gα

(11.1.42)

with gα being a bounded continuous function in Rn . We can also reduce the sum (11.1.36) to a single derivative. We assume that   k l T = (1 + r 2 ) 2 f + ∂ j (1 + r 2 ) 2 f j

(11.1.43)

with f and f j being bounded continuous functions. We set 

xj

g=

k

(1 + r 2 ) 2 f dt j ,

(11.1.44)

0

h=

g (1 + r 2 ) 2 +1 k

.

(11.1.45)

k

Clearly, we have ∂ j g = (1 + r 2 ) 2 f and h is a bounded continuous function. Therefore, we obtain   m T = ∂ j (1 + r 2 ) 2 F (11.1.46) with a suitable m, where F is a bounded continuous function. The following result can be obtained by induction. Corollary 11.1.5 A distribution T belongs to S  if and only if it can be represented as

11.1 The Spaces D L p

257

  k T = ∂ α (1 + r 2 ) 2 f (x) ,

(11.1.47)

where f (x) is a bounded continuous function on Rn .

11.2 The Space OC We recall that the elements of S operate on S  by convolution and that E  operates on S  by convolution. In both cases, the operation is separately continuous and the Fourier transform operator maps the convolution product into the product of the corresponding Fourier transforms. It is then natural to ask which is the most general space of distributions operating continuously on S  by convolution and such that the Fourier transform maps convolution products into the product of the corresponding Fourier transforms. Definition 11.2.1 Let OC (Rn ) be the space of all distributions T ∈ D (Rn ) such that for all k ∈ R we have k (11.2.1) (1 + r 2 ) 2 T ∈ DL ∞ . In other words, T ∈ OC if and only if for all k, (1 + r 2 ) 2 T is a finite sum of derivatives of functions belonging to L ∞ . Equivalently, T ∈ OC if and only if for all polynomials P(x) we have P(x)T ∈ DL ∞ . Examples. 1. Every distribution with compact support defines an element of OC . 2. Every element of OC defines a tempered distribution. More precisely, the space OC can be defined as a dual of a normal space of distributions, namely the space OC of all C ∞ functions ϕ on Rn such that there is an k integer k such that (1 + r 2 ) 2 ∂ α ϕ vanishes at infinity for all α ∈ N. The space OC can be endowed with a Hausdorff locally convex topology such that it becomes a normal space of distributions whose dual consists of all distributions on Rn satisfying the condition of Definition 11.2.1. On OC we can define its strong topology and the embedding OC → D is strongly continuous. We also have E  ⊂ OC and OC ⊂ S  and the embeddings are continuous. From Definition 11.2.1 we can deduce that distributions T j ∈ OC converge to k zero if and only if for all k the distributions (1 + r 2 ) 2 T j converge to zero in DL ∞ . Note that this corresponds to strong convergence in OC . k

Theorem 11.2.1 A distribution T ∈ OC if and only if for every k > 0 there is an integer m = m(k) such that  ∂ α fα , (11.2.2) T = |α|≤m

where f α are continuous functions such that (1 + r 2 ) 2 f α ∈ L ∞ . k

Proof Let T ∈ OC . Then by definition

258

11 On Some Spaces of Distributions k

(1 + r 2 ) 2 T =



∂ β gβ ,

(11.2.3)

β

where gβ is a bounded function. Therefore, we can write T =



(1 + r 2 )− 2 ∂ β gβ . k

(11.2.4)

β

By the formula of Leibniz, we obtain   k k (1 + r 2 )− 2 ∂ β gβ = ∂ β (1 + r 2 )− 2 gβ −    k − cs,t ∂ s (1 + r 2 )− 2 ∂ t gβ .

(11.2.5)

0≤|t|≤|β|

We will show by induction on |t| that   k ∂ s (1 + r 2 )− 2 ∂ t gβ

(11.2.6) k

is a finite sum of derivatives of continuous functions whose product by (1 + r 2 ) 2 are bounded in Rn . Clearly, if t = 0, we set

We have

  k h = ∂ s (1 + r 2 )− 2 gβ .

(11.2.7)

|h| ≤ C(1 + r 2 )− 2 |gβ |.

(11.2.8)

k

Therefore, we obtain (1 + r 2 ) 2 h ∈ L ∞ . In the induction step, we assume that the result is true for |t| = l. Let |t| = l + 1. We obtain k

       k k ∂ s (1 + r 2 )− 2 ∂ t gβ = ∂ j ∂ s (1 + r 2 )− 2 ∂ t gβ −    k − ∂ j ∂ s (1 + r 2 )− 2 ∂ t gβ .

(11.2.9)

Therefore, the left-hand side is a sum of derivatives of continuous functions whose k product with (1 + r 2 ) 2 belongs to L ∞ . Conversely, we assume that a distribution T can be represented by (11.2.2). For every ϕ ∈ D L 1 and every |α| ≤ m we set    2 2k α |α| (1 + r ) ∂ f α , ϕ = (−1)

Rn

  k f α ∂ α (1 + r 2 ) 2 ϕ d x.

(11.2.10)

11.2 The Space OC

259

  k k  Note that ∂ α (1 + r 2 ) 2 ϕ is bounded by a sum (1 + r 2 ) 2 |l|≤|α| ∂ l ϕ with ∂ l ϕ ∈ L 1 and (1 + r 2 ) 2 f α ∈ L ∞ . Then, we obtain k

    k k (1 + r 2 ) 2 ∂ α f α , ϕ ≤ C (1 + r 2 ) 2 f α ∂ l ϕ d x ≤ |l|≤|α|

k 2

≤ C(1 + r ) f α  L ∞ · 2



(11.2.11)

∂ l ϕ L 1 .

|l|≤|α|

Therefore, we obtain (1 + r 2 ) 2 T ∈ DL ∞ . k



Corollary 11.2.1 A distribution T belongs to OC if and only if for every k ≥ 0 there is an integer m = m(k) such that T =



∂ α fα ,

(11.2.12)

|α|≤m

where f α is a continuous function on Rn such that k

lim (1 + r 2 ) 2 | f α (x)| = 0.

(11.2.13)

r →∞

Corollary 11.2.2 If T ∈ OC , then T ∈ DL 1 . Proof In Theorem 11.2.1 we take k such that (1 + r 2 )− 2 is integrable in Rn . Then  the functions f α appearing in (11.2.2) belong to L 1 (Rn ). k

Note that DL p ⊂ DL q , ∀ p ≤ q. Therefore, it follows from Corollary 11.2.2 that if T ∈ OC , then T ∈ DL p , ∀ p. In the sequel, we will show that the space OC operates on S  by convolution.   k Lemma 11.2.1 If S = ∂ α (1 + r 2 ) 2 f (x) ∈ S  with f (x) a bounded continuous function on Rn and if ϕ ∈ S, then   F(x) = S y , ϕ(x + y)

(11.2.14)

is a C ∞ function on Rn such that (1 + |x|2 )− 2 F(x) ∈ D L 1 for all l > k + n. Moreover, if ϕ j → 0 in S, then F j → 0 in D L 1 . l

Proof We have   F(x) = S y , ϕ(x + y) = (−1)|α| Therefore, we obtain



 k ∂αϕ 1 + |y|2 2 f (y) α (x + y)dy. ∂y Rn (11.2.15)

260

11 On Some Spaces of Distributions

∂β F = (−1)|α| ∂x β



k  ∂ α+β 1 + |y|2 2 f (y) β α ϕ(x + y)dy. ∂x ∂ y

(11.2.16)

Note that f is bounded on Rn . Hence, we have β  ∂ F   k ∂ α+β ≤ C(1 + |x|2 ) 2k 1 + |x + y|2 2 (x) ϕ(x + y)dy = ∂x β ∂x β ∂ y α Rn  k k = C(1 + |x|2 ) 2 (1 + |u|2 ) 2 ∂ α+β ϕ(u)du ≤ Rn

k 2

≤ C(1 + |x| ) γm  ,k  (ϕ), ∀β ∈ Nn , 2

(11.2.17) where m  and k  are suitable integers and γm  ,k  is one of the seminorms defining the topology of S. If l > k + n, the last inequality implies β l ∂ F k−l (1 + |x|2 )− 2 β (x) ≤ Cγm  ,k  (1 + |x|2 ) 2 . ∂x

(11.2.18)

Therefore, the left-hand side of (11.2.18) is integrable, ∀β ∈ Nn . Hence, we obtain l (1 + |x|2 )− 2 F(x) ∈ D L 1 . The last part of the theorem follows from (11.2.18).    k Let T ∈ OC and S ∈ S  . Then we have S = ∂ α (1 + r 2 ) 2 f (x) , where f (x) is a bounded continuous function on Rn . We also have by definition (1 + |x|2 ) 2 T ∈ DL ∞ , with l > k + n. We consider      l l (11.2.19) Tξ , Sη , ϕ(ξ + η) = (1 + |ξ|2 ) 2 Tξ , (1 + |ξ|2 )− 2 F(ξ) . l

Note that if we take, in particular, T ∈ E  ⊂ OC and ϕ ∈ Cc ⊂ S, then the left-hand side of (11.2.19) coincides with (S ∗ T, ϕ). By the Lemma 11.2.1, the right-hand side of (11.2.19) is well defined for all T ∈ OC , S ∈ S  and ϕ ∈ S. It can be shown that it depends continuously on each variable when the other two remain bounded. Clearly, if ϕ j → 0 in S then, by Lemma 11.2.1, F j → 0 in D L 1 . Therefore, the right-hand side of (11.2.19) converges to zero. If T j → 0 in OC then, l by the definition of convergence in this space, we have (1 + |ξ|2 ) 2 T j → 0 in DL ∞ and by Lemma 11.2.1, the right-hand side of (11.2.19) converges to zero. Finally, if S j → 0 in S  then (11.2.19) converges to zero. Therefore, we have sketched the proof of the following theorem. Theorem 11.2.2 Let T ∈ OC and S ∈ S  . Then S ∗ T is well defined by (11.2.19) and belongs to S  . Furthermore, the map S  × OC (S, T ) → S ∗ T ∈ S  is separately continuous.

(11.2.20)

11.2 The Space OC

261

Note that O M and OC are subspaces of S  . Therefore, the Fourier transform F is well defined. Theorem 11.2.3 The Fourier transform F is a one-to-one map from O M onto OC . Moreover, F is a one-to-one map from OC onto O M . Proof Since the inverse Fourier transform F −1 has the same properties as F, it suffices to show that F maps O M into OC and OC into O M . Let ϕ ∈ O M and let m be an integer such that m > n2 . Then for every k ∈ N there is an integer l sufficiently large that (1 − )k ϕ(x) ≤ C(1 + r 2 )l−m , ∀x ∈ Rn . We set h(x) =

(1 − )k ϕ(x) . (1 + r 2 )l

(11.2.21)

(11.2.22)

Clearly, by the choice of the integer m, it follows that h is an integrable function. By taking Fourier transforms, we obtain ˆ ˆ = (1 − ξ )l h(ξ), (1 + |ξ|2 )k ϕ(ξ)

(11.2.23)

where the right-hand side is a finite sum of derivatives of bounded functions. Hence, ˆ ∈ DL ∞ , which shows that ϕˆ ∈ OC . for every k ∈ N, we have (1 + |ξ|2 )k ϕ(ξ)  n α If T ∈ OC , then for all α ∈ N , x T ∈ DL . Then x α T = Dξα Tˆ (ξ)

(11.2.24)

is a continuous function slowly increasing at infinity, ∀α ∈ Nn . Therefore, we have Tˆ ∈ O M .  Theorem 11.2.4 If T ∈ OC and α ∈ S, then T ∗ α ∈ S. Proof It suffices to show that the Fourier transform T ∗ α ∈ S. We have OC ⊂ S  . Therefore, we obtain T ∗ α ∈ O M ⊂ S  . Hence, T ∗ α is well defined and T ∗ α = Tˆ · α. ˆ Then Tˆ ∈ O M and Tˆ · αˆ ∈ S.  Theorem 11.2.5 If T ∈ OC and S ∈ S  , then

Proof We have

S ∗ T = Sˆ · Tˆ .

(11.2.25)

S = D α ϕ,

(11.2.26)

where ϕ is a continuous function slowly increasing at infinity. By considering (11.2.19), we obtain for all ϕ ∈ S

262

11 On Some Spaces of Distributions

     ˆ + η) . (S ∗ T, ϕ) = S ∗ T, ϕˆ = Tξ , Sη , ϕ(ξ

(11.2.27)

We also have 



|α|



Sη , ϕ(ξ ˆ + η) = (−1) ϕ(η)Dηα ϕ(ξ ˆ + η)dη = Rn  = (−1)|α| ϕ(η − ξ)Dηα ϕ(η)dη ˆ = n R    = ϕ(η − ξ) y α ϕ(η)dη = ϕˇ ∗  y α ϕ (ξ).

(11.2.28)

Rn

We can easily verify that

ˆˆ ϕˇ = (2π)−n ϕ,

(11.2.29)

ϕ ∗ y α ϕ = (2π)−n ϕˆˆ ∗  ∗ y α ϕ. y α ϕ = ϕˆ

(11.2.30)

Hence, we have

     ˆ + η) = Tξ , ϕ · y α ϕ(ξ) = (S ∗ T, ϕ) = Tξ , Sη , ϕ(ξ         = Tˆ , ϕˆ · y α ϕ = Tˆ ,  Dηα ϕ · ϕ = Tˆ , Sˆ · ϕ = Sˆ · Tˆ , ϕ

since Tˆ ∈ O M and Sˆ ∈ S  .

(11.2.31)



Theorem 11.2.6 If T ∈ O M and S ∈ S  , then S · T = (2π)−n Sˆ ∗ Tˆ .

(11.2.32)

References 1. H. Brezis, Functional Analysis, Sobolev Spaces and Partial Differential Equations (Springer, 2011) 2. J.J. Duistermaat, J.A.C. Kolk, Distributions: Theory and Applications (Birkhäuser, 2010) 3. F.G. Friedlander, M. Joshi, Introduction to the Theory of Distributions, 2nd edn. (Cambridge University Press, Cambridge, 1999) 4. D.D. Haroske, H. Triebel, Distributions, Sobolev Spaces, Elliptic Equations (European Mathematical Society, 2008) 5. L. Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, 2nd edn. (Springer, Berlin, 1990) 6. D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis (Springer, 2013) 7. I. Richards, H. Youn, Theory of Distributions: A Non-Technical Introduction (Cambridge University Press, Cambridge, 1995) 8. L. Schwartz, Theorie des Distributions (Herman & Cie, Paris, 1971) 9. R. Sikorski, The Elementary Theory of Distributions (Bull. Acad. Pol, Warszawa, 1957) 10. R. Strichartz, A Guide to Distribution Theory and Fourier Transforms (CRC Press, 1994)

Chapter 12

On Some Differential Operators

Abstract We present local and pseudolocal operators, give examples and study their properties. Then we discuss the hypoelliptic partial differential operators and their properties. Finally, we study the existence of fundamental solutions.

12.1 Local and Pseudolocal Operators  Let P(x, D) = | p|≤m a p (x)D p be a partial differential operator with C ∞ coefficients. Clearly, we have supp P T ⊂ supp T, ∀T ∈ D .

(12.1.1)

Therefore, a partial differential operator shrinks the support of a distribution. Definition 12.1.1 Let L be a continuous linear operator from Cc∞ () into C ∞ (). L is called a local operator if it can be extended to a continuous linear operator from E  () into D () and it is such that supp L T ⊂ supp T, ∀T ∈ E  ().

(12.1.2)

Hence, every partial differential operator with C ∞ coefficients is a local operator. In the sequel, we will discuss about pseudolocal operators [5]. To this end, we will need the notion of the singular support of a distribution. Definition 12.1.2 The singular support of a distribution T is the complement of the largest open set where T is a C ∞ function. Clearly, we have sing supp T ⊂ supp T.

(12.1.3)

Note that in the complement of its support, T coincides with the function identically zero.

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4_12

263

264

12 On Some Differential Operators

Examples 1. The singular support of any function in C ∞ (Rn ) is the empty set. 2. The singular support of the Dirac measure δ is the origin. In this case, the singular support coincides with the support. 3. We consider the characteristic function of an open interval (a, b). Its support is the closed interval [a, b]. Its singular support is the set {a} ∪ {b}. Definition 12.1.3 Let L be a continuous linear operator from Cc∞ () into C ∞ (). Then L is called a pseudolocal operator if the following conditions hold true: (i) L can be extended to a continuous linear operator from E  () into D (); (ii) for every T ∈ E  () we have sing supp L T ⊂ sing supp T.

(12.1.4)

Examples 1. Let ϕ ∈ C ∞ (Rn ). We introduce the operator Mϕ (ψ) = ϕ · ψ, ∀ψ ∈ Cc∞ (Rn ).

(12.1.5)

This operator is pseudolocal. Clearly, it is also local. 2. Let ϕ ∈ C ∞ (Rn ). The operator L ϕ (ψ) = ϕ ∗ ψ is pseudolocal. Clearly, we have (12.1.6) sing supp (ϕ ∗ T ) = ∅, ∀T ∈ E  (Rn ). Let  ⊂ Rn be open. For y ∈ Rn we define the translation of  by y  − y = {x − y|x ∈ }. Note that the set  − y is open. Let A ⊂ Rn . The set  − A = ∪ y∈A  − y

(12.1.7)

(12.1.8)

is open. If A ⊂ B, then  − A ⊂  − B. Let ε > 0 be a real number and let c be the complement of  in Rn . The set ε = {x ∈ |d(x, c ) > ε}

(12.1.9)

is open. Let Bε denote the ball with center at the origin and radius ε. Then  = ε + Bε . Lemma 12.1.1 Let T ∈ E  (Rn ). We consider that K is the support of T . Let  ⊂ Rn be open. Let E ∈ D (Rn ) be a distribution. If E is zero on  \ K , then the convolution product E ∗ T is zero on . Proof We have supp T = K and supp E ⊂ ( − K )c . Moreover, we have

(12.1.10)

12.1 Local and Pseudolocal Operators

supp (E ∗ T ) ⊂ K + ( − K )c .

265

(12.1.11)

Note that ∀x ∈ K and ∀y ∈ ( − K )c we have x + y ∈ c . Hence, we have K + ( − K )c ⊂ c . Therefore, E ∗ T is zero on .

(12.1.12) 

Corollary 12.1.1 We consider the same assumptions as in the Lemma 12.1.1. Moreover, we assume that the distribution E is zero on ( − K )ε . Then the convolution product E ∗ T is zero on ε . Proof We have K + ( − K )c ⊂ c and (ε )c = c + Bε . Therefore, we obtain K + ( − K )c + Bε ⊂ c + Bε = (ε )c .

(12.1.13)

K + ( − K )c ⊂ (ε )c .

(12.1.14)

Hence, we have

This implies that E ∗ T is zero on ε .



Lemma 12.1.2 We consider the same assumptions as in the Lemma 12.1.1. Moreover, we assume that the distribution E is a C ∞ function on  − K . Then E ∗ T is a C ∞ function on . Proof Note that differentiability is a local property. Then we can assume that the set  is bounded. We consider that β is a C ∞ function equal to one on ( − K )ε , which ε has its support contained in ( − K ) 2 . We set E 1 = β E. Therefore, E 1 is a C ∞ function with compact support in Rn . This implies that the convolution product E 1 ∗ T is a C ∞ function in Rn . Moreover, E 1 coincides with E on ( − K )ε . Therefore, E 1 ∗ T coincides with E ∗ T on ε , so E ∗ T is a C ∞ function on ε . We let ε → 0 and we deduce that E ∗ T is C ∞ on .  Theorem 12.1.1 Let E ∈ D (Rn ). We assume that E is a C ∞ function in Rn \ {0}, which is the complement of the origin in Rn . We introduce the operator L E (ϕ) = E ∗ ϕ, ∀ϕ ∈ Cc∞ (Rn ).

(12.1.15)

Then L E is pseudolocal. Proof Let  be an open subset of Rn . Let T ∈ E  (Rn ). It suffices to prove that if T is a C ∞ function on , then the convolution E ∗ T is a C ∞ function on . We consider that ω is a relatively compact open set such that ω¯ ⊂ . Moreover, ¯ let α ∈ Cc∞ () be such that α = 1 on a neighbourhood of ω. We have E ∗ T = E ∗ αT + E ∗ (1 − α)T. (12.1.16)

266

12 On Some Differential Operators

Note that αT ∈ Cc∞ (Rn ). Then E ∗ αT ∈ C ∞ (Rn ). Let K denote the support of (1 − α)T . Therefore, we obtain ω¯ ∩ K = ∅ and ω − K ⊂ Rn − {0}. Note that E is a C ∞ function on Rn − {0}, so on ω − K . This implies that E ∗ (1 − α)T is a C ∞ function on ω. Hence, E ∗ T is a C ∞ function on ω. Note that ω is an arbitrary relatively compact open subset of . Therefore, E ∗ T  is a C ∞ function on .

12.2 Hypoelliptic Partial Differential Operators Definition 12.2.1 We consider the partial differential operator 

P = P(x, D) =

a p (x)D p

(12.2.1)

| p|≤m

with C ∞ coefficients on . Then P is hypoelliptic in  if it satisfies the following condition: (H) for every U ⊂  open and every distribution T in , we deduce from P T ∈ C ∞ (U ) that T ∈ C ∞ (U ). By the definition above, all solutions u of the equation Pu = f belong to C ∞ (U ) whenever f ∈ C ∞ (U ). In the sequel, we will consider only partial differential operators with constant coefficients [1–4, 6–9]. Therefore, if the operator P is hypoelliptic and has a fundamental solution, then E is a C ∞ function off the origin. Theorem 12.2.1 We consider a partial differential operator P with constant coefficients. We assume that P has a fundamental solution E which is a C ∞ function on Rn − {0}. Then P is hypoelliptic in Rn . Proof We will prove that Condition (H) is satisfied. We consider T ∈ D (Rn ). We assume that the distribution S = P T is a C ∞ function on some open set . Let Bε be a ball with center at the origin and radius ε. We consider α ∈ Cc∞ (Bε ) such that α is equal to one on a neighbourhood of the origin. We set F = αE. By the rule of Leibniz, we deduce that PF =

 | p|≤m

a p D p (αE) = αP E +



a p cr s Dr α · D s E = δ + β,

(12.2.2)

|r |>0

where β ∈ Cc∞ (Bε ). The distribution F is called a parametrix of P [5]. We have T = δ ∗ T = P F ∗ T − β ∗ T = F ∗ P T − β ∗ T = F ∗ S − β ∗ T. Note that β ∈ Cc∞ (B). Then the convolution β ∗ T belongs to C ∞ (Rn ).

(12.2.3)

12.2 Hypoelliptic Partial Differential Operators

267

Let K = supp F ⊂ B. We have ε − K ⊂ ε − Bε ⊂ . Note that S is a C ∞ function on , so on ε − K . Therefore, F ∗ S is a C ∞ function on ε . We let  ε → 0. Then T is a C ∞ function on . In the sequel, we will show that every partial differential operator with constant coefficients has a fundamental solution [1–4]. Then we can deduce that a partial differential operator with constant coefficients is hypoelliptic if and only if it has a fundamental solution which is a C ∞ function off the origin.

12.3 Existence of Fundamental Solutions Let E be a fundamental solution of P = P(D). Let f be a distribution. Then a solution of the inhomogeneous equation Pu = f

(12.3.1)

is given by u = E ∗ f whenever the convolution product is defined [6–9]. We assume that P = Pm is a homogeneous elliptic partial differential operator of degree m with constant coefficients. We will construct its fundamental solution. To this end, we will show that the homogeneous function of degree −m U (ξ) =

1 Pm (ξ)

(12.3.2)

defines a tempered distribution whose inverse Fourier transform is the fundamental solution of the operator P. Definition 12.3.1 Let λ ∈ C. Then a function f (x) defined on Rn is called homogeneous of degree λ if (12.3.3) f (t x) = t λ f (x), ∀t > 0. Let  be the unit sphere in Rn . We consider r = |x| and ω = (ω1 , . . . , ωn ) ∈ . Then f (x) = r λ g(ω). We will only assume that λ is an integer. Moreover, we will consider that f is a C ∞ function off the origin. Let (12.3.4) f (x) = r −m g(ω), where m is an integer and g is a C ∞ function on . We claim that the function f defines a tempered distribution on Rn . We will discuss the cases m < n and m ≥ n. We assume that m < n. Then we define for every ϕ ∈ S(Rn ) the function  u ϕ (r ) =



g(ω)ϕ(r ω)dω.

(12.3.5)

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12 On Some Differential Operators

Clearly, u ϕ (r ) ∈ S(R) for all ϕ ∈ S(Rn ). Then we introduce 

∞

( f, ϕ) =

r −m+n−1 u ϕ (r )dr, ∀ϕ ∈ S(Rn ).

(12.3.6)

0

Note that ϕ ∈ S and m < n. Therefore, the integral (12.3.6) is absolutely convergent. Moreover, (12.3.6) depends continuously on ϕ ∈ S. Hence, f defines a tempered distribution. We assume that m ≥ n. In this case, the integral in formula (12.3.6) is not well defined. Therefore, we shall employ the notion of finite part of a divergent integral in order to define f as a tempered distribution. First, we assume that m = n and we consider the following integral 

∞

ε

r −1 u ϕ (r )dr =



1

ε

r −1 u ϕ (r )dr +



∞

r −1 u ϕ (r )dr.

(12.3.7)

1

Note that the second integral on the right-hand side of (12.3.7) is convergent. Moreover, we will rewrite the first integral on the right-hand side of (12.3.7) in the following way:  1  1   r −1 u ϕ (r )dr = r −1 u ϕ (r ) − u ϕ (0) dr + u ϕ (0)r −1 dr = ε ε ε  1   r −1 u ϕ (r ) − u ϕ (0) dr − u ϕ (0)logε. =



1

(12.3.8)

ε

By replacing these expression in (12.3.7), we obtain  ε

∞

r −1 u ϕ (r )dr =

 ε

1

  r −1 u ϕ (r ) − u ϕ (0) dr +



∞

r −1 u ϕ (r )dr − u ϕ (0)logε.

1

(12.3.9)

We introduce  F(ε) =

ε

1

  r −1 u ϕ (r ) − u ϕ (0) dr +



∞

r −1 u ϕ (r )dr

(12.3.10)

1

and I (ε) = −u ϕ (0)logε.

(12.3.11)

Note that the integral (12.3.9) decomposes into two parts: F(ε), which has a finite limit as ε → 0 and I (ε), which tends to infinity as ε → 0. By definition, the limit of F(ε) as ε → 0 is the finite part of the integral (12.3.6) with m = n. We have 

∞

FP 0

r −1 u ϕ (r )dr =



1 0

  r −1 u ϕ (r ) − u ϕ (0) dr +

 1

∞

r −1 u ϕ (r )dr. (12.3.12)

12.3 Existence of Fundamental Solutions

269

Similarly, we can prove that when m > n, the finite part of the integral (12.3.6) is equal to 

∞

FP

r 0

 +

∞

−m+n−1

 u ϕ (r )dr =

 r −m+n−1 u ϕ (r ) −

0



1

r

−m+n−1

0 m−n−1  s=0

u ϕ (r ) − 

m−n  s=0

 1 ∂ s u ϕ (0) s r dr + s! ∂r s

1 ∂ s u ϕ (0) s r dr. s! ∂r s

(12.3.13) Clearly, when m < n, the finite part of (12.3.6) coincides with the value of the integral. Note that if ϕ j → 0 in S(Rn ), then u ϕ j → 0 in S(Rn ). As a consequence, the right-hand sides of (12.3.12) and (12.3.13) converge to 0. In summary, we can say that if f (x) is a homogeneous function of degree −m, which is C ∞ outside the origin, it defines a tempered distribution by 

∞

( f, ϕ) = F P

r −m+n−1 u ϕ (r )dr, ϕ ∈ S.

(12.3.14)

0

The symbol F P is not necessary for m < n. Note that f ∈ S  . Then its Fourier transform fˆ is well defined, belongs to S  and has the following homogeneity properties: (i) if m < n, then fˆ is a C ∞ function outside the origin and homogeneous of degree m − n; (ii) if m ≥ n, then fˆ decomposes into the sum fˆ = H + P ln |x|,

(12.3.15)

where H is a homogeneous function of degree m − n which is C ∞ outside the origin and P is a homogeneous polynomial of degree m − n. Observation 12.3.1 Let P(D) =



ap D p

(12.3.16)

| p|=m

be a homogeneous elliptic operator of order m which has constant coefficients. Then P(D) has a tempered fundamental solution E such that (i) if m < n, then E is a homogeneous function of degree m − n, which is C ∞ outside the origin; (ii) if m ≥ n, then E = H (x) + P(x) ln |x|, where H is a homogeneous function of degree m − n, which is C ∞ outside the origin and P is a homogeneous polynomial of degree m − n. In the sequel, we will present two lemmas that will be useful in proving Malgrange’s theorem on the existence of fundamental solutions [5].

270

12 On Some Differential Operators

Lemma 12.3.1 We consider that f (z) is an analytic function of a complex variable on the unit disk {z ∈ C||z| ≤ 1}. Let p(z) be a polynomial with leading coefficient a. Then the following inequality holds true 1 |a f (0)| ≤ 2π



2π

f (eiθ ) p(eiθ ) dθ.

(12.3.17)

0

Proof We consider that m is the degree of p(z). Then we can write p(z) = az m + bz m−1 + · · · .

(12.3.18)

If we conjugate the coefficients of p, then we obtain the polynomial ¯ m−1 + · · · . p(z) ¯ = az ¯ m + bz

(12.3.19)

 1 q(z) = z p¯ . z

(12.3.20)

iθ ¯ ) . q(0) ¯ = a¯ and p(eiθ ) = q(e

(12.3.21)

We set

m

Clearly, we have

By applying Cauchy’s formula, it follows that f (0)q(0) ¯ =

1 2πi

 |z|=1

f (z)q(z) ¯ dz. z

(12.3.22) 

This yields (12.3.17).

Lemma 12.3.2 We consider that f (z) is an entire function in C. Let p(z) be a polynomial with leading coefficient a. Then for all z 0 ∈ C, the following inequality holds true |a f (z 0 )| ≤ sup | f (z) p(z)| . (12.3.23) |z−z 0 |≤1

Proof We apply the Lemma 12.3.1 to f (z 0 + z) and p(z + z 0 ). Hence, we obtain |a f (z 0 )| ≤ This yields (12.3.23).

1 2π



2π

f (z 0 + eiθ ) p(z 0 + eiθ ) dθ.

(12.3.24)

0



Theorem 12.3.1 Let D(n+1) (Rn ) be the space of distributions of order n + 1. Then every partial differential operator with constant coefficients in Rn has a fundamental solution E belonging to D(n+1) (Rn ).

12.3 Existence of Fundamental Solutions

271

Proof Step 1. We consider that P = P(D) = operator with constant coefficients. Let t

P = t P(D) =



 | p|≤m

a p D p is a partial differential

(−1)| p| a p D p

(12.3.25)

| p|≤m

be the transpose of P. This is defined by   (P T, ϕ) = T, t Pϕ , T ∈ D , ∀ϕ ∈ Cc∞ .

(12.3.26)

We assume that E is a fundamental solution of P. If we replace T by E in (12.3.26) and if we consider that P E = δ, then it follows that 

 E, t Pϕ = ϕ(0), ∀ϕ ∈ Cc∞ .

(12.3.27)

We consider that t PCc∞ is the image of Cc∞ by t P. Clearly, we have t PCc∞ ⊂ Cc∞ . Note that if ψ ∈ t PCc∞ , then there is a unique ϕ ∈ Cc∞ such that ψ = t Pϕ. Clearly, if t Pϕ = 0 for some ϕ ∈ Cc∞ , then by taking the Fourier transform it follows that t

P(ξ) · ϕ(ξ) ˆ = 0, ∀ξ ∈ Rn .

(12.3.28)

Since t P(ξ) is a polynomial, it follows that ϕˆ is identically zero. Hence, if E is a fundamental solution, it follows that the linear functional E : ψ ∈ t PC ∞ → (E, ψ) = ϕ(0),

(12.3.29)

where t Pϕ = ψ, is well defined on t PC ∞ . Moreover, it is continuous with respect to the topology induced on t PCc∞ by the natural topology of Cc∞ . Conversely, if we can show that the above linear functional is continuous on t PCc∞ endowed with the topology induced by Cck for some k, then by the Hahn-Banach theorem we can extend it to a continuous linear functional E on the whole space Cck . Therefore, we have E ∈ Dk and P E = δ, which proves that E is a fundamental solution of P. Step 2. Note that to prove the theorem it suffices to show that there is an integer k, sufficiently large, such that the linear functional ψ ∈ t PCc∞ → ϕ(0) ∈ C,

(12.3.30)

where t Pϕ = ψ, is continuous with respect to the topology of Cck . Let Ph be a differential operator on D1 , . . . , Dn−1 . By changing variables, if necessary, we can assume that the operator t P(D) can be written in the following way m  t P(D) = Dnm + Ph (D1 , . . . , Dn−1 )Dnm−h . (12.3.31) h=1

272

12 On Some Differential Operators

We will denote the elements of the space Rn by x = (x  , t) with x  = (x1 , . . . , xn−1 ) and t = xn . Moreover, we set D = (Dx  , Dt ) with Dx  = (D1 , . . . , Dn−1 ) and Dt = Dxn . We will also denote the elements of the space Cn by ζ = (ζ  , τ ), with ζ  = (ζ1 , . . . , ζn−1 ), ζ j = ξ j + iη j , 1 ≤ j ≤ n − 1 and τ = μ + iσ. Hence, by using these notations, we can write t

P(D) = Dtm +

m 

Ph (Dx  )Dtm−h .

(12.3.32)

h=1

ˆ denote its Fourier-Laplace transform. By the Paley-Wiener For ϕ ∈ Cc∞ , let ϕ(ξ) ˆ ∈ S(Rn ). From Fourier’s theorem, ϕ(ξ) ˆ is an entire function on Cn and such that ϕ(ξ) inversion formula, it follows that the following inequality holds true |ϕ(0)| ≤ (2π)−n

 Rn

ϕ(ξ ˆ  , μ) dξ  dμ.

(12.3.33)

The last integral can be estimated in the following way  Rn

ϕ(ξ ˆ  , μ) dξ  dμ ≤ A ·

 Rn

dξ  dμ , 1 + |ξ1 |n+1 + · · · + |ξn−1 |n+1 + |μ|n+1 (12.3.34)

with   ˆ . A = sup 1 + |ξ1 |n+1 + · · · + |ξn−1 |n+1 + |μ|n+1 ϕ(ξ) ξ∈Rn

(12.3.35)

We set −n

M = (2π)

 Rn

dξ  dμ . 1 + |ξ1 |n+1 + · · · + |ξn−1 |n+1 + |μ|n+1

(12.3.36)

Hence, it follows that |ϕ(0)| ≤ A · M.

(12.3.37)

Step 3. We will estimate A. We consider the following entire functions of a complex variable τ = μ + iσ: ϕ(ξ ˆ  , τ ), ξ n+1 ϕ(ξ ˆ  , τ ), 1 ≤ j ≤ n − 1 and τ n+1 ϕ(ξ ˆ , τ ) j t   and the polynomial P(ξ , τ ) at the point μ. We assume that ξ is a fixed element in Rn−1 . Hence, we obtain ϕ(ξ ˆ  , μ) ≤ sup t P(ξ  , τ )ϕ(ξ ˆ  , τ ) , |τ −μ|≤1

n+1 t ˆ  , μ) ≤ sup ξ n+1 P(ξ  , τ )ϕ(ξ ˆ  , τ ) , 1 ≤ j ≤ n − 1, ξ j ϕ(ξ j |τ −μ|≤1

(12.3.38)

(12.3.39)

12.3 Existence of Fundamental Solutions

273

n+1 μ ϕ(ξ ˆ  , μ) ≤ sup τ n+1t P(ξ  , τ )ϕ(ξ ˆ  , τ ) . |τ −μ|≤1

(12.3.40)

Moreover, we have t







P(ξ , τ )ϕ(ξ ˆ , τ) =

Rn

    ex p − i (x  , ξ  ) + tτ t P(D)ϕ (x  , t)d x  dt. (12.3.41)

Therefore, it follows that the following inequality holds true ˆ  , τ ) ≤ sup sup t P(ξ  , τ )ϕ(ξ

|τ −μ|≤1



|σ|≤1 Rn

e|tσ| (t P(D)ϕ)(x  , t) d x  dt.

(12.3.42)

In a similar way, we obtain t P(ξ  , τ )ϕ(ξ ˆ , τ ) = ξ n+1 j

 Rn

    n+1 t ex p − i (x  , ξ  ) + tτ D j P(D)ϕ (x  , t)d x  dt, (12.3.43)

for 1 ≤ j ≤ n − 1 and ˆ , τ ) = τ n+1t P(ξ  , τ )ϕ(ξ



Rn

   n+1 t  ex p − i (x  , ξ  ) + tτ Dt P(D)ϕ (x  , t)d x  dt. (12.3.44)

Then we derive the inequalities  t   ≤ sup sup ξ n+1 P(ξ , τ ) ϕ(ξ ˆ , τ ) j

|τ −μ|≤1

|σ|≤1 Rn

  t e|tσ| D n+1 P(D)ϕ (x  , t)d x  dt, j (12.3.45)

for 1 ≤ j ≤ n − 1 and ˆ  , τ ) ≤ sup sup τ n+1t P(ξ  , τ )ϕ(ξ

|τ −μ|≤1



|σ|≤1 Rn

e|tσ| (Dtn+1 t P(D)ϕ)(x  , t) d x  dt.

(12.3.46) By using all these inequalities, it follows that we have the following estimate  A ≤ sup

|σ|≤1 Rn

⎤ n−1  t e|tσ| ⎣|t Pϕ| + Pϕ + Dtn+1 t Pϕ ⎦ d x  dt. D n+1 j ⎡

(12.3.47)

j=1

Therefore, it follows that  |ϕ(0)| < M sup

|σ|≤1 Rn

⎤ n−1  n+1 t e|tσ| ⎣|t Pϕ| + D j Pϕ + Dtn+1 t Pϕ ⎦ d x  dt. ⎡

j=1

(12.3.48)

274

12 On Some Differential Operators

Step 4. We consider on t PCc∞ the topology induced by Ccn+1 . If t Pϕ j → 0 in then all t Pϕ j have support contained in a fixed compact set of Rn and all the derivatives of order ≤ n + 1 of t Pϕ j converge uniformly to zero. Therefore, the right-hand side of (12.3.48) must converge to zero. This yields ϕ j (0) → 0. This proves that the linear functional

Ccn+1 ,

ψ ∈ t PCc∞ → ϕ(0) ∈ C

(12.3.49)

where t Pϕ = ψ, is continuous on t PCc∞ endowed with the topology induced by  Ccn+1 .

References 1. J.J. Duistermaat, J.A.C. Kolk, Distributions: Theory and Applications (Birkhäuser, 2010) 2. F.G. Friedlander, M. Joshi, Introduction to the Theory of Distributions, 2nd edn. (Cambridge University Press, Cambridge, 1999) 3. D.D. Haroske, H. Triebel, Distributions, Sobolev Spaces, Elliptic Equations (European Mathematical Society, 2008) 4. L. Hörmander, The Analysis of Linear Partial Differential Operators I: Distribution Theory and Fourier Analysis, 2nd edn. (Springer, Berlin, 1990) 5. D. Mitrea, Distributions, Partial Differential Equations, and Harmonic Analysis (Springer, 2013) 6. I. Richards, H. Youn, Theory of Distributions: A Non-Technical Introduction (Cambridge University Press, 1995) 7. L. Schwartz, Theorie des Distributions (Herman & Cie, Paris, 1971) 8. R. Sikorski, The Elementary Theory of Distributions (Bull. Acad. Pol, Warszawa, 1957) 9. R. Strichartz, A Guide to Distribution Theory and Fourier Transforms (CRC Press, 1994)

Index

A Absorbing set, 14

Gateaux differentiable function, 37 Gronwall’s inequality, 64

B Balanced set, 14 Beer’s theorem, 32 Besov space, 215

H Hahn-Banach’s theorem, 6 Hausdorff space, 5 Hyperbolic partial differential equations, 183 Hypoelliptic operator, 266 Hypograph, 27

C Caratheodory solution, 57 Cauchy sequence, 6 Conjugate function, 48 Convex function, 27 D Derivative of a distribution, 87 Distribution, 79 E Elasticity system, The, 232 Elliptic partial differential equations, 186 Epigraph, 27 Evolution equation, 57

I Indicator function, 27 Inductive limit topology, 20 K Kamke function, 176 L Littlewood-Paley decomposition, 216 Local operator, 263 Lower semicontinuous function, 28

F Finite element method, The, 242 Fourier transform, 140 Frechet differentiable function, 37 Frechet space, 17

M Maximal monotone operator, 43 Mazur, 27 Metric space, 5 Monotone operator, 42 Montel space, 78

G Galerkin method, The, 241

N Normed space, 6

© The Author(s), under exclusive license to Springer Nature Switzerland AG 2021 A. Chiril˘a et al., Distribution Theory Applied to Differential Equations, https://doi.org/10.1007/978-3-030-67159-4

275

276 P Paley-Wiener-Schwartz theorem, The, 160 Parabolic partial differential equations, 185 Plate equation, The, 235 Primitive of a distribution, 96 Pseudolocal operator, 264

R Resolvent of an operator, 61

S Seminorm, 11 Sobolev space, 197

Index Stokes system, The, 230 Subdifferentiable function, 38

T Tempered distributions, 137 Topology, 5

W Weierstrass, 29

Y Yosida approximation of an operator, 61