Lectures on quasiconformal and quasisymmetric mappings

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Le tures on quasi onformal and quasisymmetri mappings Pekka Koskela De ember 4, 2009

Contents 1

The metri denition

3

2

From lo al to global

6

2.1

Spe ial ase

6

2.2

Relaxing the regularity assumption

. . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.3

Covering theorems

2.4

The maximal fun tion

. . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

2.5

Upper gradients and Poin aré inequalities

2.6

Proof of Theorem 2.1.

8 10 11

. . . . . . . . . . .

14

. . . . . . . . . . . . . . . . . . . . . .

19

3

Quasisymmetri mappings

4

Gehring's lemma and regularity of quasi onformal mappings 27 4.1

The volume derivative

23

. . . . . . . . . . . . . . . . . . . . . .

27

4.2

The maximal stre hing . . . . . . . . . . . . . . . . . . . . . .

30

4.3

Gehring's lemma

. . . . . . . . . . . . . . . . . . . . . . . . .

34

4.4

Ap -weights

. . . . . . . . . . . . . . . . . . . . . . . . . . . . .

43

4.5

Dierentiability almost everywhere

. . . . . . . . . . . . . . .

45

5

The analyti denition

51

6

K -quasi onformal

60

7

Sobolev spa es and onvergen e of quasi onformal mappings 68

8

On 1-quasi onformal mappings

mappings

1

74

9

Mapping theorems

78

10 Examples of quasi onformal mappings

84

11 Appendix

93 . . . . . . .

93

11.2 Some linear algebra . . . . . . . . . . . . . . . . . . . . . . . . p 11.3 L -spa es . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

11.1 Conformal mappings of a square onto a re tangle

94

11.4 Regularity of

p-harmoni

fun tions

. . . . . . . . . . . . . . .

97 99

11.5 Fixed point theorem and related results . . . . . . . . . . . . . 100 Quasi onformal mappings appeared perhaps for the rst time in 1928 in a work by Grötzs h under the name most nearly onformal mappings. He essentially onsidered the problem of mapping a planar square to a planar (non-square) re tangle by a dieomorphism that sends the verti es of the square to the orner points of the re tangle.

Even though these two do-

mains are onformally equivalent, the given boundary ondition annot be realized by any onformal mapping. For a onformal mapping f , the ra|f ′ (z)|2 /Jf (z), is identi ally one by the Cau hy-Riemann equations, and

tio

one was then lead to try to minimize the maximum of this quantity under dieomorphisms with the given boundary ondition. Similar questions were

subsequently onsidered by Tei hmüller in the 1930's. The term quasifonformal mapping was oined by Ahlfors in 1935. He relaxed the regularity 1,2 assumption and onsidered homeomorphisms in the lo al Sobolev lass W for whi h

|Df (z)|2 ≤ KJf (z)

K ≥ 1. The restri tion K ≥ 1 n × n-matrix A,

almost everywhere for some onstant from simple linear algebra: for ea h

omes

det A ≤ |A|n , where

det A

refers to the determinant of

A

and

|A| = sup |Ah| = sup |Ah| |h|≤1

|h|=1

is the operator norm of the linear transformation asso iated with

A.

In 1938, Morrey proved a powerful existen e theorem, alled the measurable Riemann mapping theorem. This essentially states that, in the plane, 2 ∞ quasi onformal mappings with any pres ribed ratio |Df (z)| /Jf (z) ∈ L and any given dire tion for the the maximal dire tional derivative an be found. Other important developers of the theory in lude Lavrantiev and Bojarski. Planar quasi onformal mappings have sin e then been applied to many entirely dierent problems.

Let us simply here list the following: 2

Kleinian

groups, Nevanlinna theory, surfa e topology, omplex dynami s, partial differential equations, inverse problems and ondu tivity. Higher dimensional quasi onformal mappings were already introdu ed by Lavrantiev in 1938. The theory began to ourish around 1960 when important works by Loewner, Gehring, and Väisälä appeared.

Other signi ant

ontributors in lude Callender, Shabat, and Reshetnyak. Subsequently, these mappings were introdu ed also in non-Eu lidean settings by Mostow, who proved his elebrated rigidity theorem in 1968. Another elebrated result is the reverse Hölder inequality of Gehring's from 1972. In higher dimensions, the theory of and te hniques introdu ed to study quasi onformal mappings have been su

essfully applied in dierential geometry, topology, harmoni analysis, partial dierential equations, and non-linear elasti ity, among other elds. The purpose of these notes is to give an introdu tion to the theory. The sele ted approa h has been inuen ed by re ent advan es in the metri setting, but the framework is mostly that of a Eu lidean spa e. The on ept of quasisymmetry will be ru ial in our approa h.

We have tried to make

the notes as self- ontained as possible. The reader is nevertheless assumed to Lp -spa es. The topi s

know the basi s of the Lebesgue integration theory and

overed ree t the personal taste of the author. Naturally many important aspe ts must have been left untou hed. For further reading, we re ommend the lassi monograph Le tures on

n-dimensional quasi onformal mappings

by Väisälä [30℄ and the monograph [4℄. These notes are based on ourses given at the University of Jyväskylä in 1997, 2004 and 2008 and at the University of Mi higan in 2002.

The

urrent notes are the out ome of several iterations. We wish to thank all the people who have provided us with lists of typos. In our experien e, most of the material an be overed in a one semester, graduate level topi s ourse. Regarding the sour es for the presented material, we wish to highlight [8℄, [14℄ and [30℄. There are rather few histori al omments in what follows, and the in lusions or omissions of referen es are essentially random.

1 The metri denition We begin by introdu ing the so- alled metri denition of quasi onformality. To this end, let

(X, | · |), (Y, | · |) be metri x ∈ X and r > 0. Dene

spa es and

homeomorphism. Let

Lf (x, r) := sup{|f (x) − f (y)| : |x − y| ≤ r}, lf (x, r) := inf{|f (x) − f (y)| : |x − y| ≥ r}, 3

f : X → Y

and

Hf (x, r) :=

Lf (x, r) . lf (x, r) l f (x,r)

f f(x) r

x

f(B)

B

L f (x,r)

Figure 1: The denition of A homeomorphism

f

Lf (x, r)

and lf (x, r)

is quasi onformal if there exists

H 0,

is quasi onformal in all

dimensions. 4) There is quasi onformal mapping the von Ko h snowake urve. 5) Ea h dieomorphism

G ⊂⊂ Ω.

f : Ω → Ω′

f : R2 → R2

su h that

f (S 1 (0, 1))

is

is quasi onformal in every subdomain

Let us begin by onsidering 1) in the plane. Write z = x + iy and f (z) = u(x, y) + iv(x, y) for a onformal mapping f, where u, v are real fun tions. Then f is analyti and the Ja obian determinant Jf of f is stri tly positive. By the Cau hy-Riemann equations we have that

ux = vy , uy = −vx . 4

Thus

   ux uy ux uy . = Df (x, y) = −uy ux vx vy 

Jf (x, y) = (ux )2 +(uy )2 = |∇u|2 = |∇v|2 and that ∇u·∇v = 0. Moreover, also the two olumns of Df (x, y) are perpendi ular and both of length |∇u|. Thus, given a ve tor h, we have that We on lude that

|Df (x, y)h| = |∇u||h|. By the ( omplex) dierentiability of

f

we on lude that

lim sup Hf (x, r) = 1 r→0

everywhere. Noti e also that

|Df (x, y)|2 = Jf (x, y) everywhere, where |A| f ′ (x + iy) = ux (x + iy) have that

= sup|h|≤1 |Ah| is the usual operator norm. Sin e − iuy (x + iy) for the omplex derivative f ′ , we also |Df (x, y)| = |f ′(x + iy)|, where the latter term is the modulus of

the omplex derivative and the former again the operator norm. For 2) one easily he ks that

f

is indeed quasi onformal, with

lim sup Hf (x, r) = 2 r→0

everywhere. The radial mapping des ribed in 3) requires already some eort, see Chapter 10 below. We will also dis uss the mapping referred to in 4) in more detail in Chapter 10.

Jf (x) of f is lo ally bounded away |Df (x)| = sup|h|≤1 |Df (x)h| is lo ally bounded. Thus,

Regarding 5), noti e that the Ja obian from zero and that given

G ⊂⊂ Ω,

we have that

for some onstant

|Df (x)|n ≤ KJf (x) K

and all

x ∈ G.

This implies that

|Df (x)| ≤ K ′ min |Df (x)h| |h|=1

′ n−1 (in fa t, we may take K = |K| ). The ′ quasi onformality then follows with H = K using the dierentiability of f.

with some onstant

K′

in

G

The metri denition is easy to state but it is hard to dedu e properties of quasi onformal mappings dire tly from it.

For example, it is not lear

from the denition if quasi onformal mappings form a group. The problem is that the denition is an innitesimal one.

In the next hapter we show

that it implies a global estimate whi h is easier to work with. 5

2 From lo al to global In this hapter we prove the following global estimate and introdu e the ma hinery needed for its proof.

f : Ω → Ω′ be H -quasi onformal, where Ω, Ω′ ⊂ Rn n ≥ 2. Then Hf (x, r) ≤ H ′(H, n) whenever B(x, 7r) ⊂ Ω.

2.1 Theorem. Let domains,

are

To help to understand the fundamental ideas of the proof, let us begin with a simpler setting.

2.1 Spe ial ase Suppose that

Ω = Ω′ = R2

and assume that

f is orientation preserving. Jf (x) > 0. It follows that

Assume that at

x

with

f

Let

be a dieomorphism.

x ∈ Ω.

Then

f

is dierentiable

max |Df (x)e| ≤ H min |Df (x)e| |e|=1

|e|=1

and

|Df (x)|2 ≤ HJf (x),

Hf (x0 , r) ≤ H ′ . We L := Lf (0, r) and l := lf (0, r).

see Subse tion 11.2 in the appendix. Let us show that may assume that Dene

x0 = 0 = f (x0 ). Denote   if |y| ≤ l 1   if |y| ≥ L v(y) = 0 . L  log  |y|  L if l ≤ |y| ≤ L log

Then

  0 |∇v(y)| = 0  

l

1 |y| log

if if

L l

6

if

|y| < l |y| > L . l < |y| < L

Now

Z

R2

1 log Ll

|∇v(y)|2 dy =

1 log Ll

=

1 log Ll 2π . = log Ll

=

Let

!2 Z

l≤|y|≤L

!2 Z

L

l

!2

Z

0

dy |y|2

2π

1 r dϕ dr r2

(log L − log l)2π

u(x) = v(f (x)). Then (see Subse tion 11.2 in the appendix) Z Z 2 |∇u(x)| dx ≤ |∇v(f (x))|2|Df (x)|2 dx R2 R2Z ≤H |∇v(f (x))|2 |Jf (x)| dx 2 ZR =H |∇v(y)|2 dy R2

2πH . log Ll

= Now, su h

u = 1 on f −1 (B(0, l)) and u = 0 on f −1 (R2 \ B(0, L)). Let w0 , z0 be −1 that |w0 | = |z0 | = r, w0 ∈ f (R2 \ B(0, L)) and z0 ∈ f −1 (B(0, l)). Set ( √ w0 if |w0 − z0 | ≥ 2r √ . w = w2 0 +z0 r if |w0 − z0 | < 2r |w0 +z0 |

πr < t < r, S 1 (w, t) interse ts both f −1 (R2 \B(0, L)) and 4 1 Now, sin e u os illates from 0 to 1 on S (w, t), we have Then, for

1≤ for ea h

πr 4

Z

Hölder

S 1 (w,t)

|∇u| ≤ (2πt)

< t < r. Thus Z Z 2 |∇u| ≥ B(0,2r)

where

C

is independent of

r πr 4

r.

Z

S 1 (w,t)

1/2

Z

|∇u|

S 1 (w,t)

2



Hen e

L ≤ exp(CH). l This gives us the desired global ontrol. 7

dt ≥

|∇u| Z

r πr 4

2

f −1 (B(0, l)).

1/2

1 =C 2πt

(1)

2.2 Relaxing the regularity assumption We ontinue with the planar setting. We begin by disposing of the use of the

hain rule. Let us dene

ρ(x) =

( |Df (x)|

1 |f (x)| log L l

on

0

Z

R2

γ

ρ2 ≤

2πH . log Ll

S 1 (w, t) whi h onne ts f −1 (R2 \ B(0, L)) f ◦ γ onne ts R2 \ B(0, L) to B(0, l), and so Z Z ds ≥ 1, ρ ds ≥ L S 1 (w,t) f ◦γ |y| log l

is a subar of

then

πr < t < r are as above. 4

oordinates, we on lude that

where

w

.

elsewhere

Then

If

f −1 (B(0, L) \ B(0, l)) =: A

and

Z

B(0,2r)

to

f −1 (B(0, l)),

Reasoning as in (1), using polar

ρ2 ≥ C > 0.

We no longer require the hain rule, but it still looks like we need

f

to be

dierentiable. To relax this assumption, let us try to dis retize the denition

ρ. Re all that we wish to bound L/l from above. We may L ≥ 2l. S A = f −1 (B(0, L) \ B(0, l)) ⊂ Bj , where Bj 's are balls. Set

of our fun tion

thus assume that Suppose

 −1 X L diam(f (Bj )) 1 ρ(x) = log χ2Bj (x). l diam(Bj ) dist(0, f (Bj )) Then

Z

−1 X  Z diam(f (Bj )) 1 L χ2Bj ds. ρ ds = log l diam(Bj ) dist(0, f (Bj )) S 1 (w,t) S 1 (w,t)

If the

Bj 's

are small, then

Z

S 1 (w,t)

χ2Bj ds ≥

8

diam(Bj ) 2

whenever

Bj ∩ S 1 (w, t) 6= ∅.

Hen e

−1  1 L ρ ds ≥ log l 2 S 1 (w,t)

Z

X

Bj ∩S 1 (w,t)6=∅

diam(f (Bj )) . dist(0, f (Bj ))

f (Bj ) are so small that ea h f (Bj ) interse ts at most i i−1 two annuli Ai = B(0, 2 l) \ B(0, 2 l). Write ⌊t⌋ for the integer part of a real number t. Then Assume that the sets

L

X

Bj ∩S 1 (w,t)6=∅

⌊log2 ⌋ 1 Xl diam(f (Bj )) ≥ dist(0, f (Bj )) 4 i=1

diam(f (Bj )) dist(0, f (Bj ))

X

diam(f (Bj )) 2i l

Bj ∩S 1 (w,t)6=∅,f (Bj )∩Ai 6=∅

L

⌊log2 ⌋ 1 Xl ≥ 4 i=1

X

Bj ∩S 1 (w,t)6=∅,f (Bj )∩Ai 6=∅

L

⌊log2 ⌋ 1 Xl 2i−1 l ≥ 4 i=1 2i l

≥ and so

Z

L 1 log2 , 8 l

S 1 (w,t)

whenever

πr 4

< t < r.

As before, this gives

Z When we try to estimate integral



L log l

−2 Z

ρ ds ≥ C > 0

B(0,2r)

B(0,2r)

ρ2 ≥ C > 0.

kρkL2 (B(0,2r))

from above, we are fa ed with the

k X diam(f (Bj )) 1

(2)

1 χ2Bj (x) diam(Bj ) dist(f (Bj ), 0)

!2

dx.

(3)

Problems: 1) How to sele t balls

Bj

so that we an nd an ee tive estimate on our

integral? This requires ontrol on the overlap of the balls 2) How to get rid of the annoying

2 9

in

χ2Bj ?

Bj .

3) Even if we an handle 1) and 2), how an we handle dimensions

n ≥ 3?

Noti e here that the proof of (2) strongly used the fa t that we are in the plane. We next introdu e the te hnology that will allow us to handle the above problems.

2.3 Covering theorems We will later use overing theorems to sele t the above balls

Bj .

We begin

with a overing lemma that holds in all metri spa es whose losed balls are

ompa t.

2.2 Theorem. (Vitali) Let

B be a olle tion of losed balls in Rn su h that

sup{diam B : B ∈ B} < ∞. B1 , B2 , . . . (possibly a nite sequen e) Bi ∩ Bj = ∅ for i 6= j and [ [ B⊂ 5Bj .

Then there are su h that

from this olle tion

B∈B

For a proof we refer the reader to [20℄. Let us anyhow briey explain the

B onsists of balls B(x, rx ), x ∈ A and A is bounded. Let M = supx∈A rx . Choose a ball B1 = B(x, rx ) so that rx > 3M/4. Continue by onsidering points in A \ 3B1 , and repeating the rst step (now letting M1 = supy∈A\3B ry ) and after that 1

idea in a simple ase. Suppose that the family where

ontinue by indu tion. In the Eu lidean setting, a sub olle tion often an be hosen so that we only have uniformly bounded overlap for the over.

2.3 Theorem. (Besi ovit h) Let su h that the set

A

B

be a olle tion of losed balls in

onsisting of the enters is bounded.

ountable (possibly nite) sub olle tion

χA (x) ≤ for all

x.

The sele tion of the balls

overing theorem.

Bj

X

B1 , B2 , . . .

Rn

Then there is a

su h that

χBj (x) ≤ C(n)

eventually will be made using the Besi ovit h

In more general settings, say, in the Heisenberg group,

Besi ovit h fails. The reason it holds in the Eu lidean setting, is basi ally the following fa t: 10

B(x1 , r1 ) and B(x2 , r2 ) so that 0 ∈ B(x1 , r1 ) ∩ B(x2 , r2 ), x1 ∈ / B(x2 , r2 ) and x2 ∈ / B(x1 , r1 ). Then the angle between the ve tors x1 and x2 is at least 60 degrees. Suppose that we are given

For a proof of the Besi ovit h overing theorem, we again refer to [20℄.

2.4 The maximal fun tion We will need maximal fun tions to dispose of the onstant

2 in the term χ2Bj

in (3). Maximal fun tions turn out to be important for other things as well. 1 n Let u ∈ Llo (R ). The non- entered maximal fun tion of u is

Z sup − Mu(x) =

x∈B(y,r) B(y,r)

Here and in what follows,

when

A

Z Z 1 − v= v |A| A A

is measurable with

measure of

|u|.

A.

0 < |A| < ∞,

and

|A|

refers to the Lebesgue

2.4 Remarks. 1) A

ording to the Lebesgue dierentiation theorem ( f. Remarks 4.3), Mu(x)

≥ |u(x)|

almost everywhere. This fa t is not be needed in this se tion. 2) There are many other maximal fun tions. For example the restri ted,

entered maximal fun tion

C Mδ u(x)

C 3) We always have M∞ u(x) 4) Noti e that

{Mu > t}

Z = sup −

0 t}, for some B(y, r) ontaining x.

measurable. Indeed, if the denition that

|u|.

2.5 Theorem. 1) If

u ∈ L1

and

t > 0,

then

|{Mu > t}| ≤ 11

5n t

R

{Mu>t}

|u| ≤

5n kuk1 . t

R (Mu)p ≤ C(p, n) |u|p . R Proof. 1) We may assume that M := {Mu>t} |u| < ∞. For ea h x ∈ {Mu > t} there is a ball B su h that x ∈ B and Z − |u| > t. 2) If

u ∈ Lp , p > 1,

then

R

B

Then

−1

|B| < t and thus

Z

B

|u|

diam(B) < C(n)t−1 ||u||1. If

y ∈ B,

then Mu(y)

B ⊂ {Mu > t}. So Z Z 1 1 |u| ≤ |B| < |u|. t B t {Mu>t}∩B >t

and thus

By the Vitali overing theorem we nd pairwise disjoint balls above so that

{Mu > t} ⊂

|{Mu > t}| ≤

X

S

5Bj .

|5Bj | = 5

n

B1 , B2 , . . .

as

Then

X

Z Z 5n 5n X |u| ≤ |u|. |Bj | ≤ t t {Mu>t} Bj

2) Re all the Cavalieri prin iple:

Z

p

|v| = p =p =p

Z Z

|v(x)|

tp−1 dt dx

Z Z0 ∞

Z

0

∞

tp−1 χ{|v|>t} dt dx

0

tp−1 |{|v| > t}| dt.

t > 0. Dene g(x) = |u(x)| χ{|u(x)|> 2t } (x). Then |u(x)| ≤ g(x) + 2t and so t t Mu(x) ≤ Mg(x) + . Thus {x : Mu(x) > t} ⊂ {x : Mg(x) > }. Now, the 2 2

Fix

Cavalieri prin iple, part 1) of our theorem and the Fubini theorem yield the

12

estimate

Z

p

(Mu(x)) = p

Z

∞

Z0 ∞

tp−1 |{Mu(x) > t}| dt

t tp−1 |{Mg(x) > }| dt 2 Z0 ∞ n 2·5 ≤p tp−1 ||g||1 t 0 Z ∞ n Z p−1 2 · 5 ≤p t |u| dx dt t 0 {|u(x)|> 2t }| Z ∞ Z n p−2 ≤2·5 p χ{|u(x)|> t } |u| dx dt t 2 Rn 0 Z Z 2|u(x)| = 2 · 5n p |u(x)| tp−2 dt dx n 0 ZR 2p 5n p = |u|p . p−1 ≤p

2

2.6 Remark. 1) Let us single out, for future referen e, the estimate

2 · 5n |{Mu(x) > t}| ≤ t

Z

{|u(x)|> 2t }

|u| dx

from the above proof.

u ∈ Lp (Ω), p > 1.R Applying TheoremR 2.5 to the zero (Mu)p ≤ C(p, n) Ω |u|p. Similarly, extension of u we on lude that Ω the inequality in part 1) of this remark an be restri ted to Ω when u ∈ L1 (Ω).

2) Suppose that

The ase

p=1

2.7 Example. If 1

n

L (R )

unless

u

was not left out by a

ident from the previous theorem.

u(x) = χB(0,1) (x),

then Mu

is the zero fun tion.

6∈ L1 (Rn ).

In fa t,

Mu ∈ /

The following lemma from [7℄ will allow us to handle problem 2) stated after formula 3.

13

2.8 Lemma. (Bojarski) Fix

aj ≥ 0

and

λ > 1.

Then

k Proof.

The ase

X

1 ≤ p < ∞.

Let

aj χλBj kp ≤ C(λ, p, n)k

p = 1 is lear.

Let

X

B1 , B2 , . . .

be balls in

Rn ,

aj χBj kp .

p > 1. Then, by the Lp −Lp/(p−1) -duality

(see Subse tion 11.3 in the appendix),

k

X

aj χλBj kp =

sup kϕk

p ≤1 p−1

Z X . a χ ϕ j λB j

Now, using monotone onvergen e and Theorem 2.5 we estimate

Z X X Z ≤ aj |ϕ| a χ ϕ j λB j λBj Z X ≤ aj |λBj | − |ϕ| λBj Z X ≤ aj λn Mϕ Bj Z X n =λ aj χBj Mϕ X p ≤ λn k aj χBj kp kMϕk p−1 X p . ≤ λn C(p, n)k aj χBj kp kϕk p−1

The laim follows.

2

2.5 Upper gradients and Poin aré inequalities In this se tion we give a substitute for (1). We will later show that it allows us to prove an analog of (2) in all dimensions. A Borel fun tion

g≥0

is an upper gradient of

|u(x) − u(y)| ≤ whenever

x 6= y ∈ U

and

γx,y

Z

u

in

U,

if

g ds

(4)

γx,y

is a re tiable urve that joins

x

to

y

in

U.

Here we agree that inequality (4) holds, whatever an expression we have on the left hand side, if the given line integral is innite, and that both and

u(y)

are nite if the integral in question onverges. 14

u(x)

The re tiability of

γ : [a, b] → Rn k−1 X j=1

above means that, for some

M < ∞,

|γ(tj+1 ) − γ(tj )| ≤ M

a = t1 < t2 < · · · < tk = b and k ≥ 2. The supremum of su h sums over all k ≥ 2 and all partitions is then the length of γ. Re all that n ea h re tiable urve γ : [a, b] → R of length l an be parametrized by γ0 : [0, l] → Rn so that |γ0′ (t)| = 1 for a.e. t and γ0 is 1-Lips hitz, i.e. |γ0 (t) − γ0 (s)| ≤ |t − s| for all t, s ∈ [0, l]. Then Z Z g ds := g(γ0(t)) dt. whenever

γ

[0,l]

For all this see [30℄.

2.9 Examples. 1)

u ∈ C 1 , g = |∇u|.

This is simply the fundamental theorem of al ulus

u ◦ γ0

of a single

< ∇u(γ0(t)), γ0′ (t) > dt.

(5)

for the absolutely (even Lips hitz) ontinuous fun tion

variable:

u(γ(l)) − u(γ(0)) = 2)

u

Lips hitz,

g

Z

[0,l]

the pointwise Lips hitz  onstant Lipu(x)

r→0

3)

Noti e that

(u ◦ γ0 )′ (t) ≤ Lipu(t)

u

g ≡ ∞.

anything,

innite.

|u(x) − u(y)| . r |x−y|≤r

= lim sup sup

for almost every

t.

In this ase, the right hand side of (4) is always

Integration of (4), the Fubini theorem and spheri al oordinates give us the important Poin aré inequality.

u ∈ L1 (B(x0 , r)) ⊂ Rn , n ≥ 2, and let g ∈ L (B(x0 , r)), 1 ≤ p < ∞, be an upper gradient of u in B(x0 , r). 2.10 Theorem. (Poin aré inequality) Let p

Then

Z −

B(x0 ,r)

Here

R uB := −B(x0 ,r) u.

Z |u − uB | ≤ C(n)r −

B(x0 ,r)

15

g

p

1/p

.

Proof.

x ∈ B = B(x0 , r). Then Z Z 1 |u(x) − u(y)| dy ≤ g(x + t(y − x))|y − x| dt dy B B 0 Z 1Z = g(x + t(y − x))|y − x| dy dt 0 B   Z 1Z |z − x| −n t dz dt ≤ g(z) t 0 B∩B(x,2tr) Z 1Z ≤ 2r g(z)t−n dz dt

Let

Z

0

≤ 2r

Z

Z Z B

B

x

g(z)

B

≤ Cn r Integrating with respe t to

B∩B(x,2tr) Z 1

n

Z

B

|z−x| 2r

t−n dt dz

g(z) dz. |z − x|n−1

we obtain the estimate

Z Z

g(y) dy dx |y − x|n−1 ZB B Z 1 dx dy g(y) = Cn r n n−1 B B |y − x| Z ′ n+1 g. ≤ Cn r

|u(x) − u(y)| dy dx ≤ Cn r

n

B

Now

Z Z − |u(x) − uB | dx = − B

B

Z Z Z − u(x) − u(y) dy dx ≤ − − |u(x) − u(y)| dy dx. B

B

B

Combining the above estimates, we obtain the desired inequality for The general ase follows by Hölder's inequality.

p = 1. 2

2.11 Remarks. 1) The Poin aré inequality also holds when when

x 0. Then Z g n ≥ δ(δ0 , n) > 0. 2.12 Theorem. Let u be ontinuous in B(x0 , 3r), g

B(x0 ,3r)

R Proof. Let a = −B(x0 ,r) u. Assume that a ≤ 1/2. ri = 2−i r, i ≥ −1, Bi = B(x, ri ). Then Z u(x) = lim uBi = lim − u. i→∞

Now

Let

x ∈ F

and write

i→∞ B i

X 1 ≤ |u(x) − uB(x0 ,r) | ≤ |uBi − uBi+1 | + |uB0 − uB(x0 ,r) |. 2 i≥0

17

Also,

B(x0 , r) ⊂ B(x, 2r)

and thus a simple estimate and the Poin aré in-

equality yield

|uB0 − uB(x0 ,r) | ≤ |uB(x,r) − uB(x,2r) | + |uB(x0 ,r) − uB(x,2r) | Z n ≤2·2 − |u − uB(x,2r) | B(x,2r)

Z ≤ C(n)2r −

g

n

B(x,2r)

≤ C(n)(2r)

1/n

 Z −1 (2r)

Similarly

|uBi − uBi+1 | ≤

1/n

1/n C(n)ri

g

n

B(x,2r)

1/n

.

1/n  Z −1 n ri g . Bi

Thus

1/n  Z ∞ X 1 1/n −1 n ≤ C(n)ri ri g 2 i=−1 Bi 1/n  Z 1/n n −1 ≤ C(n)r sup t g . 0 1/2,

X

diam(5Bk ) ≤ C(n)r

then we use

E

instead of

F

may be repla ed with

XZ

18

Bk

above.

Bi B(x0 , r).

2.13 Remark. By hoosing the balls

B(x0 , 3r)

B1 , B2 , . . .

n

as above su h that

g ≤ C(n)r

Z

F ⊂

g n.

B(x0 ,3r)

2

more leverly, one an show that

2.6 Proof of Theorem 2.1. B(x0 , r). We may assume that x0 = 0 = f (x0 ). Lf (x0 ,r) Re alling that we wish to bound Hf (x0 , r) = from above, we may lf (x0 ,r) further assume that L ≥ 2l, where L := Lf (x0 , r) and l := lf (x0 , r). Let We prove the estimate for

A := f −1 ((B(0, L) \ B(0, l)) ∩ Ω′ ) ∩ B(0, 6r). x ∈ A,

For ea h

pi k

0 < rx < r/30

H(x, rx ) < 2H

su h that

and

diam(f (B(x, rx ))) < l/4.

By the Besi ovit h overing theorem we nd a sub olle tion of

{B(x, rx )}x

so that

χA (x) ≤ for all

Pi k

x.

Be ause

f

rj < rxj < 2rj

X

{Bj }j = {B(xj , rj )}j

χBj (x) ≤ C(n)

is a homeomorphism, also

so that

X

χf (Bj ) (x) ≤ C(n).

diam(f (B(xj , rxj ))) ≤ 2 diam(f (Bj )). Be ause

over

A,

A

is ompa t, already a nite number of the balls

ˆj = B(xj , rx ) B j

ˆ1 , . . . , B ˆk . Dene B −1 X  k 1 diam(f (Bˆj )) L χ (x). ρ(x) = log ˆj ) dist(0, f (B ˆj )) 2Bˆj l diam(B 1

say

Then

−1 X  k 1 diam(f (Bj )) L χ4Bj (x). ρ(x) ≤ 8 log l diam(B ) dist(0, f (B )) j j 1 By Lemma 2.8

Z

!n −n Z X  k 1 diam(f (B )) L j χBj (x) dx ρn dx ≤ C(n) log l diam(B ) dist(0, f (B )) j j 1  −n X n k  L diam(f (Bj )) ≤ C(n) log l dist(0, f (Bj )) 1 −n X  k L |f (Bj )| . ≤ C(n, H) log n l dist(0, f (B )) j 1 19

Denote

k X 1

Ai = {x : 2i−1 l ≤ |x| ≤ 2i l} i

0 X |f (Bj )| ≤ dist(0, f (Bj ))n 1

and so

Z

and

i0 = ⌊log Ll / log 2⌋ + 1.

Then

i

X

f (Bj )∩Ai 6=∅

0 X |f (Bj )| |B(0, 2i+1l))| ≤ C(n) , dist(0, f (Bj ))n (2(i−2) l)n 1

1−n  L . ρ dx ≤ C(n, H) log l n

(6)

−1 Noti e that f (Rn \ B(0, L)) ontains a ontinuum F that joins S n−1 (0, r) n−1 to S (0, 2r) in B(0, 2r). Be ause f (B(0, r)) is open, it is easy to he k −1 that B(0, l) ⊂ f (B(0, r)). Dene E = f (B(0, l)). Then E is a ontinuum,

diam(E) ≥ r, diam(F ) ≥ r, and E, F ⊂ B(0, 2r). If γ is a re tiable urve E to F , then f ◦ γ joins B(0, l) to Rn \ B(0, L). Reasoning as in 2.2 we see that Z ρ ds ≥ ε0 > 0

that joins

γ

where

ε0

does not depend on

f, r

or

γ.

Dene

1 u(x) = inf ε0

Z

ρ ds, γx

x to F. Then u = 0 ρ that ρ is bounded.

where inmum is taken over all re tiable urves that join

F and u ≥ 1 in E. Remember Let u(y) > u(x). Then in

from the denition of

|u(y) − u(x)| ≤ for all re tiable urves of

u.

Note that

u

γx,y

onne ting

Z

x

γx,y

to

ρ ds ε0

y.

Thus

ρ is an upper gradient ε0

is Lips hitz be ause

|u(x) − u(y)| ≤

ρ(z) |x − y|. z∈B(x,2|x−y|) ε0 sup

By Theorem 2.12 we on lude that

Z

B(0,6r)

A bound on

L/l

ρn dx ≥ εn0 δ > 0.

follows ombining (6) and (7), as desired.

2.14 Remarks. 20

(7)

1) The assumption that

n≥2

was needed to ensure that the exponent

1 − n in (6) is negative. Thus the proof does not extend to the ase n = 1. This is no a

ident. The simple quasi onformal mapping f (x) = x+ exp(x) of a single variable shows that the laim of Theorem 2.1 fails for n = 1. 2) We only needed that

lim inf Hf (x, r) ≤ H r→0

x ∈ Ω for our homeomorphism in the proof of Theorem 2.1.

for all

Thus

the quasi onformality assumption an be relaxed to this ondition.

lim sup or lim inf of Hf (x, r) is really ne essary. To this end, let E ⊂ [0, 1] be 1 the -Cantor set. Then the Cantor fun tion ξ : [0, 1] → [0, 1] maps 3 E to a set of positive length. Let Ω =]0, 1[×R and dene f (x, y) = (x + ξ(x), y). Then lim sup Hf (x, r) = 1

3) It is now natural to inquire if the uniform boundedness of the

r→0

outside

E ×R

and

f : Ω → f (Ω)

is a homeomorphism that takes a set

of zero area to a set of positive area. We will soon prove that a quasi onformal mapping annot do this (one an also show dire tly using the properties of the Cantor fun tion that f annot be quasi onformal). 1 We an repla e the -Cantor set in this example with any Cantor set, 3 even of Hausdor dimension zero. Consequenty, uniform boundedness of

Hf (x)

outside a set of dimension one when

n=2

does not su e

for the uniform boundedness of Hf (x, r). In higher dimensions, one reR above by Rn−1 to see that the analog of dimension one is then

pla es

n − 1. On the other hand, if

lim inf Hf (x, r) ≤ H r→0

outside a set of si onformal. when

n = 2:

σ -nite (n − 1)-measure,

one an prove that

f

is qua-

This is rather easily seen from our previous arguments Let

E˜

σ -nite length. Instead of ˜ . Dene ρ x ∈ A, do this for A \ E

be the ex eptional set of

pi king small balls entered at ea h as before. Then still

Z

−1  L . ρ ≤ C(H) log l 2

21

What about the lower bound? Let us refer to our previous argument in 2.2. It ould well happen that our balls do not over the subar of

S 1 (w, t). However, one an prove that, for almost every t > 0, the set E˜ ∩ S 1 (w, t) is ountable. Then the balls we sele ted over the subar 1 of S (w, t) up to a ountable set for almost every t > 0. Hen e the 1 images of the balls over f (S (w, t)) up to a ountable set. Thus Z ρ ds ≥ ε0 > 0 S 1 (w,t)

for almost every

t > 0.

The general setting is similar in spirit to that

in the plane.

H ′ depends on H, n. It is not known if the laim ′

ould hold with some H that does not depend on the dimension n. This ′ is an interesting open problem. One annot in general take H = H even when f is a onformal mapping of the unit disk onto a simply

4) In the above proof,

onne ted planar domain. 5) Theorem 2.1 extends to a rather abstra t setting. Let

1

Q-regular

,

Q > 1,

Poin aré inequality with exponent

f :X →Y

X, Y

be Ahlfors

suppose that losed balls are ompa t, and the

p=Q

holds for both

X

and

Y.

If

is quasi onformal, then

Hf (x, r) ≤ H ′ for all

x ∈ X , r > 0.

In fa t, even

lim inf Hf (x, r) ≤ H r→0

for all

x

su es.

These results an be proved by suitably modifying

the argument that we used above, see [5℄. The real di ulty is in ir umventing the Besi ovit h overing theorem. The size of ex eptional sets is not yet entirely understood in this general setting, see [18℄. 6) A metri spa e a onstant

C

X

is alled linearly lo ally onne ted (LLC), if there is

so that

i) ea h pair of points in any ball

CB , 1A

B

an be joined by a ontinuum in

and

metri measure spa e

X

is Ahlfors

Q-regular,

if there is a onstant

C −1 rQ ≤ µ(B(x, r)) ≤ CrQ for all

x, r

for some Borel measure

µ. 22

C

so that

ii) ea h pair of points outside any ball −1 in X \ C B.

B an be joined by a ontinuum

The spa es in 5) are LLC. This onne tivity ondition is used to nd substitutes for the sets

E

and

F

in the proof of Theorem 2.1.

3 Quasisymmetri mappings By Theorem 2.1 we know that quasi onformality implies the uniform lo al boundedness of

Hf (x, r).

We introdu e the equivalent on ept of quasisym-

metry that turns out to be very useful. Let

X

morphism. A

for all

η : [0, ∞) → [0, ∞) be a homeohomeomorphism f : X → Y is η -quasisymmetri (η -qs), if   |f (a) − f (x)| |a − x| ≤η |f (b) − f (x)| |b − x|

and

Y

be metri spa es and let

a 6= x 6= b.

3.1 Remark. If

f

is

η -quasisymmetri , Hf (x, r) =

then

Lf (x, r) ≤ η(1). lf (x, r)

So, quasisymmetri mappings are quasi onformal. We next prove that quasi onformal mappings are lo ally quasisymmetri .

f : B(x0 , 3r0 ) → Ω′ ⊂ Rn be a homeomorphism su h that Hf (x, r) ≤ H for all x ∈ B(x0 , r0 ) and 0 < r < 2r0 . Then f|B(x ,r ) is 0 0 η -quasisymmetri , where η depends only on n and H . 3.2 Theorem. Let

Proof. Case 1:

Let a 6= x 6= b t > 1. Write

be points in

B(x0 , r0 )

aj = x + j|b − x| for

j = 0, 1, . . . , k ,

where

k = ⌊t⌋.

and let

t = |a − x|/|b − x|.

a−x |a − x|

Then

|f (aj ) − f (aj−1)| ≤ H|f (aj−1) − f (aj−2 )|, for

j ≥ 2,

and so

|f (aj ) − f (aj−1)| ≤ H j−1|f (a1 ) − f (x)| ≤ H j |f (b) − f (x)|. 23

Sin e

|f (a) − f (ak )| ≤ H|f (ak ) − f (ak−1)|,

we obtain

|f (a) − f (x)| ≤ |f (a) − f (ak )| + k+1

k X j=1

|f (aj ) − f (aj−1 )|

≤ (k + 1)H |f (b) − f (x)| ≤ (t + 1)H t+1 |f (b) − f (x)|. b

x = a0 a1 ak

a

Figure 2: Case 1 Case 2:

t < 1/9.

Denote

bj = x + 3−j (b − x),

for

j ≥ 0,

and

Bj = B((bj + bj−1 )/2, 3−j |b − x|), for

j ≥ 1.

Let

j ≤ k = ⌊log3 (1/t)⌋.

Then

|a − x| ≤ |bj − x|

and so

|f (a) − f (x)| ≤ H|f (bj ) − f (x)| ≤ H 2 |f (bj ) − f (bj−1 )| ≤ H 2 diam(f (Bj )). This implies that

|f (a) − f (x)|n ≤ C(H, n)|f (Bj )|.

a x

Bk bk b1

b

B1

Figure 3: Case 2 Sin e the balls

Bj

are pairwise disjoint and

f (Bj ) ⊂ f (B(x, |b − x|)) ⊂ B(f (x), H|f (b) − f (x)|), 24

we obtain

n

k|f (a) − f (x)| ≤ C(H, n)

k X j=1

|f (Bj )|

≤ C(H, n)|B(f (x), H|f (b) − f (x)|)| ≤ C ′ (H, n)|f (b) − f (x)|n . Thus

|f (a) − f (x)| ≤ C ′′ (H, n)(log(1/t))−1/n . |f (b) − f (x)|

Case 3:

1/9 ≤ t ≤ 1.

Clearly

|f (a) − f (x)| ≤ H. |f (b) − f (x)| Sele t a homeomorphism the above bounds.

η : [0, ∞[→ [0, ∞[

that is greater than or equal to

2

3.3 Remarks. 1) The proof goes through if are

Q-regular.

and

s

is LLC and both

depending on

n

X

and

Y

H so that the restri tion of f to B(x0 , r) is η ˜-quasisymmetri with η˜(t) = C max{ts , t1/s }.

2) In fa t, one an hoose

C

f : X → Y, X

and

This requires a bit more work.

Ω, Ω′ ⊂ Rn , where n ≥ 2. Suppose that f : Ω → Ω′ is quasi onformal and let 0 < λ < 1. Then there is an η = η(n, H, λ) so that the restri tion of f to B(x, λd(x, ∂Ω)) is η -quasisymmetri whenever x ∈ Ω.

3.4 Corollary. Let

Proof. By Theorem 2.1, the assumptions of Theorem 3.2 are satised for balls B(x, d(x, ∂Ω)/15). If 1/15 < λ < 1, one then iterates the quasisymmetry estimate for the ase λ = 1/15 so as to obtain quasisymmetry in B(x, λd(x, ∂Ω)) (with a new ontrol fun tion η that also depends on λ). 2

It is easy to he k, from the denition, that quasisymmetri mappings form a group. The following proposition follows dire tly from the denition.

25

f : A1 → A2 be η1 -quasisymmetri and let g : A2 → A3 be η2 -quasisymmetri . Then f −1 : A2 → A1 is ηˆ-quasisymmetri , where ηˆ(0) = 0 and 1 ηˆ(t) = −1 1 , η1 ( t ) 3.5 Proposition. Let

for

t > 0,

and

g ◦ f : A1 → A3

is

η˜-quasisymmetri ,

where

η˜(t) = η2 (η1 (t)).

As a onsequen e of Corollary 3.4 and Proposition 3.5 we now on lude that quasi onformal mappings also form a group.

This annot be easily

proven from the denition.

f : Ω1 → Ω2 be H1 -quasi onformal and let g : Ω2 → Ω3 −1 be H2 -quasi onformal. Then f is H(H1 , n)-quasi onformal and g ◦ f is H(H1 , H2 , n)-quasi onformal. 3.6 Theorem. Let

η = η(n, H) so that the restri tion of f to −1 any ball B = B(x, d(x, ∂Ω1 )/2) is η -quasisymmetri . Then f : f (B) → B is η ˆ-quasisymmetri by Proposition 3.5. Given y ∈ Ω2 , hoose x = f −1 (y), let B = B(x, d(x, ∂Ω)/2), noti e that B(y, r) ⊂ f (B) for r < lf (x, d(x, ∂Ω1 )/2), −1 and apply Remark 3.1 to f . Proof.

By Corollary 3.4 there is

The quasi onformality of the omposition follows by a similar argument.

2

3.7 Remark. Let 2

B (0, 1) → Ω

Ω ⊂ R2

be bounded and simply onne ted.

Let

f :

be quasi onformal. Then the following are equivalent:

1)

f

is quasisymmetri .

2)

Ω

is LLC.

3) There is a quasi onformal mapping

g : R2 → R2

so that

g|B2 (0,1) = f .

The fa t that 1) implies 2) is easy to prove. By Corollary 3.4, 3) yields 1). The remaining impli ations are harder. To see that 2) implies 1), one reasons as in the proof of Theorem 2.1 using Remark 2.13 and a suitable ase study. The fa t that 1) implies 3) an be shown relying on te hniques from Chapter 10 below.

26

4 Gehring's lemma and regularity of quasi onformal mappings We will prove that quasi onformal mappings are dierentiable almost everywhere, preserve the null sets for Lebesgue measure, and belong to the 1,p Sobolev lass Wlo for some p = p(n, H) > n. This amounts to absolute

ontinuity of the omponent fun tions of

f

on almost all lines parallel to the

oordinate axes (in the domain in question) and lo al

p-integrability

of the

lassi al partial derivatives.

4.1 The volume derivative It will be important for us to pull ba k the Lebesgue measure under our quasi onformal mapping.

4.1 Proposition. Let

f : Ω → Ω′

be a homeomorphism. Then

|f (B(x, r))| r→0 |B(x, r)|

µ′f (x) = lim exists almost everywhere in

Ω, Z E

for ea h Borel set

belongs to

L1lo (Ω)

and

µ′f (x) dx ≤ |f (E)|

E ⊂ Ω, with equality whenever |A| = 0 implies |f (A)| = 0.

This is a dire t onsequen e of the following Radon-Nikodym theorem when one hooses

µ(A) = |f (A)|

and

λ(A) = |A|.

4.2 Theorem. (Radon-Nikodym) Let

Ω ⊂ Rn . exists

µ

and

λ

be Radon measures on

Then

λ-a.e.,

µ(B(x, r)) r→0 λ(B(x, r))

D(µ, λ, x) := lim

is lo ally integrable with respe t to

Z

E

for ea h Borel set with respe t to

E

λ,

and

D(µ, λ, x) dλ(x) ≤ µ(E)

with equality if an only if

λ.

27

µ

is absolutely ontinuous

Re all that a measure

µ

is Radon, if

sets are measurable,

µ(K) < ∞

µ(U) = sup{µ(K) : K ⊂ U for open

U,

for ompa t sets, Borel

ompa t}

and

µ(A) = inf{µ(U) : A ⊂ U for arbitrary

open}

A.

We refer the reader to [20℄ for a proof of the Radon-Nikodym theorem. It is a rather dire t appli ation of a overing theorem that we have not dis ussed. Let us however briey explain how a weaker version of Proposition 4.1 ′

an be justied using the overing theorems from 2.3. Instead of µf , let us

onsider

u(x) = lim sup r→∞

|f (B(x, r))| , |B(x, r)|

u. Let E ⊂ Ω be a Borel set. We may assume that E ⊂⊂ Ω. Given k ∈ Z, write Ek = {x ∈ E : 2k−1 < u(x) ≤ 2k }, and set E 0 = {x ∈ E : u(x) = 0}, E∞ = {x ∈ E : u(x) = ∞}. Consider rst E∞ . Let 0 < r < d(E, ∂Ω) and x M ≥ 1. For ea h x ∈ E∞ , we nd 0 < rx < r so that and let us assume that we already know the Borel measurability of

|B(x, r)| ≤ M|f (B(x, rx ))|. By the Vitali overing theorem, we nd pairwise disjoint balls above and so that

E∞ ⊂ ∪j 5B j . Thus X |E∞ | ≤ 5n |B j | ≤ 5n M −1 | ∪j f (B j )|.

B 1, B 2, · · ·

as

F ⊂ Ω, independent of M, so that ∪j B j ⊂ F. | ∪j f (B j )| ≤ |f (F )| < ∞. By letting M tend to innity, we on lude that |E∞ | = 0. Fix then ε > 0 and let k ∈ Z. Pi k an open set Uk so that f (Ek ) ⊂ Uk and |Uk | < |f (Ek )| + ε. For ea h x ∈ Ek , pi k 0 < rx < d(E, ∂Ω) so that There exists a ompa t set Thus

2k−1 |B(x, rx )| ≤ |f (B(x, rx )| and

f (B(x, rx )) ⊂ Uk .

Using the Vitali overing theorem as above, we on-

lude that

2k−1 5−n |Ek | ≤ |Uk | < |f (Ek )| + ε, 28

and letting

ε → 0,

we infer that

2k−15−n |Ek | ≤ |f (Ek )|.

(8)

Regarding the opposite inequality, we hoose an open set so that

|Uk | < |Ek | + ε.

Given

x ∈ Ek

pi k then

rx

Uk

ontaining

Ek

so that

2k |B(x, rx )| ≥ |f (B(x, rx ))|

B(x, rx ) ⊂ Uk . By the Besi ovit h overing B 1 , B 2 , · · · as above and so that X χEk (x) ≤ χB j ≤ Cn χUk . and

theorem, we nd balls

j

Summing over

j

and letting

ε → 0,

we on lude that

|f (Ek )| ≤ 2k Cn |Ek | Summing over

k

(9)

R

in (8) and (9), and noti ing that

Cn−1 |f (E

\ E∞ )| ≤

Z

E

E0

u = 0,

we arrive at

u(x) ≤ Cn |f (E)|,

(10)

Cn depends only on n. Re alling that |E∞ | = 0, we may repla e E \E∞ E, provided f maps sets of measure zero to sets of measure zero. One an establish (10) with Cn = 1 by substituting a suitable more

where with

rened overing theorem [20℄ for the Besi ovit h and Vitali overing theorems ′ above. The almost everywhere existen e of the limit in the denition of µf ′ also follows from suitable versions of (8) and (9). The measurability of µf is rutine.

4.3 Remarks. 1) (Lebesgue's dierentiation theorem) Let

Z lim −

r→0 B(x,r) for almost every

1 u ∈ Llo .

Then

u(y) dy = u(x)

x.

Proof. By onsidering the positive and negative parts of we may assume that n able E ⊂ R . Then

u ≥ 0. µ is a

Dene

µ(E) =

Z lim −

u E

u

separately,

for Lebesgue measur-

Radon measure and the Radon-Nikodym

theorem gives

Z

R

E r→0 B(x,r)

u(y) dy dx =

Z

E

Thus the laim follows. 29

u dx.

2) The Lebesgue dierentiation theorem an be improved to: If

p ≥ 1,

then

Z lim −

r→0 B(x,r) for almost every

x.

p u ∈ Llo ,

|u(y) − u(x)|p dy = 0

This follows by applying the Lebesgue dierentation uq (y) = |u(y) − q|p , q ∈ Q.

theorem to the fun tions 3) Let that

E ⊂ Rn

be Lebesgue measurable. From 1), with

lim

r→0 for almost every

x ∈ E.

u = χE ,

we see

|E ∩ B(x, r)| =1 |B(x, r)|

x in the Lebesgue dierentation theorem is not essential. Indeed, onsider the olle tion Q onsisting of all ubes 1 Q ⊂ Rn . If u ∈ Llo , then, for almost every x, Z lim − u(y) dy = u(x)

4) The use of balls entered at

j→∞ Q j

Qj ∈ Q satisfy ∩j Qj = {x}. This an be proved, for example, by rst noti ing that the laim is trivial if u is ontinuous, approximatwhenever

ing a general lo ally integrable fun tion by ontinuous ones, and by

ontrolling the error terms via the weak boundedness (as in part 1) of Theorem 2.5) of the maximal operator [26℄.

4.2 The maximal stre hing Set

Lf (x) = lim sup r→0

4.4 Lemma. Let

f : Ω → Ω′

Lf (x, r) . r

be a homeomorphism.

The fun tion

Lf

is

Borel measurable and

µ′f (x) ≤ Lf (x)n ≤ Hf (x)n µ′f (x) for almost every

Proof.

x ∈ Ω.

In parti ular,

The Borel measurability of

ompa t subset

E

of

Lf ∈ Lnlo (Ω) when f Lf

follows from the fa t that, given a

Ω, {x ∈ E : Lf (x) < t} = 30

is quasi onformal.

[

Ai ,

where the sets

Ai =



1 |f (x + h) − f (x)| ≤t− x∈E: |h| i

for all

 0 < |h| < d(E, ∂Ω)/i

x ∈ Ω, 0 < r < d(x, ∂Ω). n  Lf (x, r) |f (B(x, r))| ≤ . |B(x, r)| r f.

are losed by ontinuity of

Let

Then

Now



Lf (x, r) r

n

≤



Lf (x, r) lf (x, r)

n 

lf (x, r) r

Hen e the laim follows by letting

Noti e that, at a point terms of

Lf (x).

However,

absolute ontinuity of Indeed,

Lf (x) = 1

f

≤



Lf (x, r) lf (x, r)

n

|f (B(x, r))| . |B(x, r)|

tend to zero.

2

Df (x) exists, |Df (x)| is ontrolled in integrability of Lf does not a priori guarantee x,

where

on almost all lines parallel to the oordinate axes.

almost everywhere for the homeomorphism

3) of Remarks 2.14, but to the

r

n

f

f

from part

is not absolutely ontinuous on any line parallel

x-axis.

We are thus lead to modify the denition of ′ For a homeomorphism f : Ω → Ω and ε > 0 dene

Lεf (x) = sup r≤ε

Lf .

Lf (x, r) . r

ε ε Then Lf is Borel measurable. Note that ε 7→ Lf (x) is in reasing and that Lεf (x) → Lf (x), as ε → 0. ε It is easy to he k that, in dimension one, lo al integrability of Lf guarantees the absolute ontinuity of f. The following result is a generalization of this fa t.

4.5 Lemma. Let

f : Ω → Ω′

be a homeomorphism, and let

|f (x) − f (y)| ≤

Z

γ

x to y in Ω. In parti ular, 2Lεf is an upper fun tions fi of f in Ω, and of the fun tion

u(x) = |f (x) − f (x0 )|, whenever

x0 ∈ Ω

Then

2Lεf ds

for all re tiable urves onne ting gradient of the omponent

ε > 0.

is xed. 31

Proof. Fix x, y ∈ Ω and let γ = γ0 : [0, l] → Ω, be a re tiable urve joining x to y . Assume rst that d := diam(γ([0, l])) < ε. Let z ∈ γ([0, l]). Then γ([0, l]) ⊂ B(z, d) and so |f (x) − f (y)| ≤ diam(f (γ([0, l]))) ≤ 2Lf (z, d). Hen e

|f (x) − f (y)| ≤

Z

[0,l]

2Lf (γ(s), d) ds ≤ l

Z

[0,l]

d ≥ ε, hoose 0 = t1 < · · · < tk = l su h 1 ≤ i < k , and use the triangle inequality. If

2Lf (γ(s), d) ds ≤ d

that

Z

γ

2Lεf ds.

diam(γ([ti , ti+1 ])) < ε,

for

The rest of the laim follows from the fa ts that

|fi (x) − fi (y)| ≤ |f (x) − f (y)|, for

1 ≤ i ≤ n,

and

|u(x) − u(y)| = |f (x) − f (x0 )| − |f (y) − f (x0 )| ≤ |f (x) − f (y)|.

2

f is a homeomorphism. In k fa t, the on lusion holds for ea h ontinuous f : Ω → R , k ≥ 1. ε We next show that Lf is lo ally p-integrable for all p < n, provided f is The above proof did not employ the fa t that

quasisymmetri .

f be η -quasisymmetri 0 < ε < diam(B)/100. Then

4.6 Lemma. Let and let

in

2B ,

where

B = B(x0 , r0 ) ⊂ Rn ,

|{x ∈ B : Lεf (x) > t}| ≤ [5η(1)η(2)/t]n |f (B)| for

t > 0.

Proof.

If

Lεf (x) > t,

then there exists

0 < rx ≤ ε

su h that

Lf (x, rx ) > t. rx Et = {x ∈ B : Lεf (x) > t}. By the Vitali overing theorem we nd pairwise disjoint losed balls B1 = B(x1 , r1 ), B2 = B(x2 , r2 ), . . . as above so

Write

32

that

S

Et ⊂

5B j .

Thus

X |Bj | ≤ 5n |B(0, 1)|t−n Lf (xj , rj )n X ≤ |B(0, 1)| [5η(1)/t]n lf (xj , rj )n X ≤ [5η(1)/t]n |f (B(xj , rj ))|

|Et | ≤ 5n

X

≤ [5η(1)/t]n |f (2B)|. By quasisymmetry,

|f (2B)| ≤ |B(0, 1)|Lf (x0 , 2r0 )n ≤ |B(0, 1)|lf (x0 , r0 )n η(2)n ≤ |f (B)|η(2)n.

2

f be η -quasisymmetri in 2B , where B = B(x0 , r0 ) ⊂ Rn , 0 < ε < diam(B)/100. Then  p/n Z |f (B)| ε p − (Lf ) ≤ C(n, η, p) |B| B

4.7 Lemma. Let and let

for

1 ≤ p < n.

Proof.

Applying the Cavalieri formula and the previous lemma we see that

Z

B

(Lεf )p

Z

∞

tp−1 |{x ∈ B : Lεf (x) > t}| dt 0 Z Z ∞ t0 =p + 0 t0 Z Z t0 p−1 t |B| dt + C(n, η, p)|f (B)| ≤p

=p

0

=

Solve for

t0

|B|tp0

tp−n−1 dt

t0

+

. C(n, η, p)|f (B)|tp−n 0

so that the two terms are equal.

f : Ω → Ω′ be Ω, Ω′ ⊂ Rn are domains, n ≥ 2.

4.8 Corollary. Let where

∞

2

quasisymmetri (or quasi onformal), 1,n n n Then f ∈ Wlo (Ω, R ): |f | ∈ Llo (Ω),

the omponent fun tions are absolutely ontinuous on almost all lines par-

Ω, and the n to Llo (Ω).

allel to the oordinate axes in

oordinate fun tions belong

33

lassi al partial derivatives of the

u : Ω → R on almost all lines that, for (n − 1)− almost every

Re all that absolute ontinuity of a fun tion

Ω requires (x2 , · · · , xn ), u(t, x2 , · · · , xn ) is absolutely ontinuous on ea h ompa t line segment in the x1 -dire tion in Ω, as a fun tion of t, and analogously when x1 above is repla ed by xj , j = 2, · · · , n.

parallel to the oordinate axes in

0 < ε < 1 so that Lεf ∈ L1 (Q), see Lemma 4.7. Fix a oordinate dire tion, say x1 . Fiber Q by line segments parallel to the x1 -axis. Denote J(x2 , . . . , xn ) = {y ∈ Q : y2 = x2 , . . . , yn = xn }. By the Fubini theorem Lεf ∈ L1 (J(x2 , . . . , xn )) for (n − 1)-almost every (x2 , . . . , xn ). Let J = J(x2 , . . . , xn ) be su h a line segment. By Lemma 4.5 we have, for 1 ≤ j ≤ n, Z |fj (t1 , x2 , . . . , xn ) − fj (t2 , x2 , . . . , xn )| ≤ 2Lεf ds,

Proof.

Fix a ube

Q

with

Q ⊂ Ω,

and pi k

J(t1 ,t2 )

where

J(t1 , t2 ) = {x ∈ J : t1 ≤ x1 ≤ t2 }.

Sin e Lebesgue integral is ab-

fj is x ∈ J.

solutely ontinuous with respe t to Lebesgue measure, it follows that absolutely ontinuous on

J

and that

∂1 fj (x)

exists at almost every

Furthermore,

|∂1 fj (x)| ≤ Lf (x), for su h points.

The above learly shows that

fj

is absolutely ontinuous

on almost all lines parallel to the oordinate axes in Ω. Next, from Lemma n 4.4 we know that Lf ∈ Llo (Ω). Be ause a quasi onformal mapping is lo ally quasisymmetri by Corollary 3.4, the laim follows.

2

4.9 Remark. The previous results do not allow us to on lude that a quasisymmetri mapping of the real line onto itself is absolutely ontinuous. p-integrability of Lεf for p < 1 and thus Lemma 4.5 gives no estimate on the os illation of f. This Indeed, in the proof of Lemma 4.7 we only obtain the

does not mean any weakness in our te hnique be ause one an give examples of quasisymmetri mappings Next we will show that

f : R → R that fail to be absolutely ontinuous.

p Lf ∈ Llo (Ω)

for some

p = p(n, H) > n.

4.3 Gehring's lemma The following result is the starting point for the higher integrability of

34

Lf .

4.10 Lemma. (Reverse Hölder Inequality) Let on

2B ⊂ R

n

. Then

f

be

η -quasisymmetri

Z 1/n Z n − Lf ≤ C(n, η)− Lf . B

B

Proof. There is nothing to be proved when n = 1. Thus assume that n ≥ 2. Let ε > 0 be small. Suppose B = B(x0 , r0 ). Dene u(x) = |f (x) − f (x0 )|. ε Then, by Lemma 4.5, 2Lf is an upper gradient of u and thus, by the Poin aré inequality, Z Z − |u − uB | ≤ C(n)r0 − Lεf . B

Sin e

Lεf

B

is lo ally integrable, the monotone onvergen e theorem implies

that

Z Z − |u − uB | ≤ C(n)r0 − Lf . B

Now

(11)

B

Z Z 1 2−n uB = − |f (x) − f (x0 )| ≥ |f (x) − f (x0 )| ≥ Lf (x0 , r0 ), |B| B\ 12 B η(2) B and there is a

δ = δ(n, η) > 0

su h that

u(x) = |f (x) − f (x0 )| ≤ whenever

2−n Lf (x0 , r0 ), 2η(2)

x ∈ δB . Thus Z Z 1 (uB − u) ≥ C(n, η)Lf (x0 , r0 ). − |u − uB | ≥ |B| δB B

This, ombined with (11), gives

Z Lf (x0 , r0 ) ≤ C(n, η)r0 − Lf .

(12)

B

So, by Lemma 4.4 and Proposition 4.1,

Z 1/n Z 1/n |f (B)|1/n n ′ − Lf ≤ C(n, η) − µf ≤ C(n, η) r0 B B Z Lf (x0 , r0 ) ≤ C(n, η) ≤ C(n, η)− Lf . r0 B 2

35

4.11 Remarks. 1) Applying Hölder's inequality to the right hand side of (12) we obtain the estimate

1/p Z p Lf (x0 , r0 ) ≤ C(n, η)r0 − Lf B

for

p ≥ 1.

In parti ular, with

p = n,

we have

Lf (x0 , r0 ) ≤ C(n, η) 2) If

X

is

Z

B

Lnf

1/n

.

Q-regular and if we have a p-Poin aré inequality for some p < Q,

then the proof of Lemma 4.10 gives

1/p 1/Q Z Z p Q ≤ C(data) − Lf − Lf B

B

when

f

is

η -quasisymmetri .

3) The reverse Hölder inequality also holds with balls repla ed by ubes (assuming that

Q,

f

√

nQ). Indeed, let √ B = B(x0 , r0 ) ⊂ diam(B). Then Q ⊂ nB. By 1), Z Lf (x0 , r0 ) ≤ Cr0 − Lf ,

is quasisymmetri on

where the edge length of

Q

is

B

and by quasisymmetry

√ diam(f (Q)) ≤ 2η( n)Lf (x0 , r0 ). Following the proof of Lemma 4.10 we see that

Z 1/n diam(f (Q)) n − Lf ≤C , r0 Q and we on lude that

1/n Z Z Z n − Lf ≤ C− Lf ≤ C− Lf . Q

B

Q

One an also verify the reverse Hölder inequality dire tly for ubes, without using the Poin aré inequality. Let us sket h this in dimension

36

f is η -quasisymmetri on Q. Assume for notational 2 simpli ity that Q = [−1, 1] . By quasisymmetry, √ √ diam(f (Q)) ≤ 2η( 2)|f (1, t)−f (0, 0)| ≤ 2η(1)η( 2)|f (1, t)−f (−1, t)| two. Suppose that

for ea h

−1 ≤ t ≤ 1.

have that

As at the end of the proof of Lemma 4.10, we

Z 1/2 2 − Lf ≤ C(n, η) diam(f (Q)). Q

The laim follows by noti ing (see the proof of Corollary 4.8)

|f (1, t) − f (−1, t)| ≤ for almost every points

(1, t)

and

Z

2Lf ds

Jt

−1 ≤ t ≤ 1, where Jt is the line segment between (−1, t), and then integrating with respe t to t.

the

As the rst onsequen e of the reverse Hölder inequality we show that quasi onformal mappings preserve the lass of sets of measure zero.

f : Ω → Ω′ be quasi onformal, where Ω, Ω′ ⊂ Rn , |f (E)| = 0, if and only if |E| = 0. In parti ular, Z |f (E)| = µ′f dx,

4.12 Corollary. Let

n ≥ 2.

Then

E

for Borel (and all Lebesgue measurable) sets

E,

and

f

maps Lebesgue mea′ surable sets to Lebesgue measurable sets. Moreover, µf (x) > 0 almost everywhere.

Proof. Let |E| = 0. We may assume that E is bounded and E ⊂ Ω. Pi k n open U ⊃ E so that U ⊂⊂ Ω. Then Lf ∈ L (U) by Lemma 4.4. Given ε > 0, we further nd an open set V with E ⊂ V ⊂ U and |V | < ε. For ea h x ∈ E , pi k a ball B(x, rx ) so that B(x, 15rx ) ⊂ V . By the Vitali overing theorem, we nd su h balls B 1 , B 2 , . . . so that B i ∩ B j = ∅ when i 6= j and E ⊂ ∪5B j . Then f (E) ⊂ f (∪5B j ) and X X |f (∪5B j )| ≤ |f (5B j )| ≤ η(5)C(n) Lf (xj , rj )n and so, by part 1) of Remark 4.11,

|f (∪5B j )| ≤ C(n, η)

XZ

Bj

Lnf

= C(n, η)

Z

∪Bj

37

Lnf

≤ C(n, η)

Z

V

Lnf .

ε → 0, we on lude that |f (E)| = 0. The only if  part follows from −1 fa t that f is also quasi onformal. By the Radon-Nikodym theorem, Z |f (E)| = µ′f dx (13)

Letting the

E

E. Let E ⊂ Ω be Lebesgue measurable. Pi k a Borel set F ⊃ E so that |F \ E| = 0. Then f (F ) is a Borel set, f (E) ⊂ f (F ) and |f (F ) \ f (E)| = |f (F \ E)| = 0. It follows that f (E) is Lebesgue measurable ′ and that (13) holds also for E. Suppose nally that µf (x) = 0 in E with |E| > 0. Then Z for all Borel sets

|f (E)| =

whi h ontradi ts the fa t that

E

µ′f dx = 0,

|f (E)| = 0

if and only if

|E| = 0.

2

We ontinue with a powerful tool from harmoni analysis, the CalderónZygmund de omposition, and some onsequen es of this de omposition. The dyadi de omposition of a ube with fa es parallel to the fa es of where

i = 1, 2, . . .

Q0

Q0

Q ⊂ Q0 l(Q) = 2−i l(Q0 ),

onsists of open ubes

and of edge length

refer to the generation in the onstru tion.

The ubes

Q0 up to a set of measure zero and the losures in of the ubes in a xed generation over Q0 ; there are 2

ubes of edge −i length 2 l(Q0 ) in the ith generation and the ubes orresponding to the same

in ea h generation over

generation are pairwise disjoint. For almost every

Q0 ⊃ Q1 ⊃ . . . of ubes T {x} = Qi . In what follows, Q, Q0 , Qx

x ∈ Q0 , there is a (unique)

de reasing sequen e

in the dyadi de omposition so

that

et . are ubes.

4.13 Theorem. (Calderón-Zygmund de omposition) Let Q0

L1 (Q0 ),

and suppose that

⊂ Rn , u ∈

Z t ≥ − u ≥ 0. Q0

Then there is a sub olle tion that

Qi ∩ Qj = ∅

when

i 6= j ,

{Qj }

from the dyadi de omposition of

Q0

so

Z t < − u ≤ 2n t Qj

for ea h

Proof.

j,

and

u(x) ≤ t

for almost every

S

Qj .

xT∈ Q0 there is a de reasing sequen e {Qj } of {x} = Qj . By the Lebesgue dierentiation theorem

For almost every

dyadi ubes so that

x ∈ Q0 \

38

(see part 3) of Remarks 4.3)

Z lim − u = u(x)

j→∞ Q j

x. Let u(x) > t and assume that the above holds for x {Qj }. Then there must be maximal Qx := Qj(x) so that Z − u > t.

for almost every su h with the sequen e

Qx

For this ube we have

Z Z n t t.

ubes Qx .

The dyadi maximal fun tion of a measurable fun tion to a ube

Q0 )

It is then easy

2

u

(with respe t

is dened by

Z MQ0 u(x) = sup − |u|, x∈Q⊂Q0 Q

where the supremum is taken over all ubes de omposition of

Q0

Q

and whose losures ontain

that belong to the dyadi

x.

4.14 Remark. As for the usual maximal fun tion, we have the weak type estimate

2 · 5n |{x ∈ Q0 : MQ0 u(x) > t}| ≤ t

Z

{x∈Q0 :|u(x)|> 2t }

|u|

for the dyadi maximal fun tion. Moreover,

Z

Q0

for

p > 1.

p

(MQ0 u) ≤ C(p, n)

Z

Q0

|u|p

The proof of the weak type estimate is a tually easier than for the

usual maximal operator be ause no overing theorem is needed. The following simple onsequen e of the Calderón-Zygmund de omposition is essentially the onverse of the weak type estimate for the dyadi maximal fun tion. 39

4.15 Lemma. Let

u ∈ L1 (Q0 )

Z

{x∈Q0 :|u(x)|>t}

and suppose

R t ≥ −Q0 |u|.

Then

|u| ≤ 2n t|{x ∈ Q0 : MQ0 u(x) > t}|.

Proof. By the Calderón-Zygmund

ubes Q1 , Q2 , . . . so that Z

de omposition we nd pairwise disjoint

t < − |u| ≤ 2n t Qj

for all

j,

and

|u(x)| ≤ t Z

almost everywhere in

{x∈Q0 :|u(x)|>t}

|u| ≤ ≤

XZ

Qj

X

Q0 \

|u|

S

Qj .

Then

2n t|Qj |

≤ 2n t|{x ∈ Q0 : MQ0 u(x) > t}|, be ause

Z MQ0 u(x) ≥ − |u| > t Qj

for ea h

x ∈ Qj .

2

We are now ready to prove an important result. For histori al reasons, it is only alled a lemma (Gehring's lemma). I learned the trun ation tri k employed in the proof below from Xiao Zhong.

4.16 Lemma. (Gehring's lemma, 1973) Let and suppose that

1/q Z Z q − |u| ≤ C− |u| Q

for all dyadi sub ubes

u ∈ Lq (Q0 ), 1 < q < ∞

Q ⊂ Q0 .

Then there is

s = s(q, n, C) > q

1/s Z Z s 1/s − |u| ≤ 2 C− |u|. Q0

In parti ular,

Proof.

(14)

Q

so that

(15)

Q0

u ∈ Ls (Q0 ).

We begin by noti ing that

MQ0 (|u|q )(x) ≤ C q MQ0 u(x) 40

(16)

R

− |u|q . Combining Lemma 4.15, (16) Q0 and the weak type estimate from Remark 4.14, we on lude that

x ∈ Q.

for ea h

t ≥ t0 :=

Let then

Z

{x∈Q0 :|u(x)|q >t}

|u|q ≤ 2n t|{x ∈ Q0 : MQ0 (|u|q )(x) > t}| ≤ 2n t|{x ∈ Q0 : MQ0 u(x) > C −q t1/q }| Z n+1 n 1− 1q ≤2 5 t |u|, {x∈Q0 :|u(x)|> 21 C −q t1/q }

provided

t ≥ t0 . Z

Consequently, for these values of 1

{x∈Q0 :|u(x)|q >t}

|u|q ≤ Cn t1− q

t,

Z

{x∈Q0 :|u(x)|>δt1/q }

|u|,

(17)

−1 −q . Multiplying both sides of (17) where Cn depends only on n and δ = 2 C p−2 by t and integrating over the interval [t0 , j], where j > t0 is xed, results in

Z

Z

j p−2

t

{x∈Q0 :|u(x)|q >t}

t0

Write

Z

q

|u| dxdt ≤

j p−1−1/q

t

t0

q

b(j, s, u(x)) = min{j, s|u(x)| }

Z

s > 0.

when

{x∈Q0 :|u(x)|>δt1/q }

|u| dxdt. (18)

Noti e that

b(j, s, u(x)) ≤ sb(j, 1, u(x)) when

s ≥ 1.

By the Fubini theorem,

Z

j p−1−1/q

t

{x∈Q0 :|u(x)|>δt1/q }

t0

=

Z

Z

Q0

|u|

Z

|u| dxdt

b(j,δ−q ,u(x))

tp−1−1/q dtdx

t0

Z

b(j, δ −q , u(x))p−1/q |u| dx Q0 Z −1 1−pq b(j, 1, u(x))p−1/q |u| dx ≤ q(pq − 1) δ ZQ0 b(j, 1, u(x))p−1 |u|q . ≤ q(pq − 1)−1 δ 1−pq ≤ q(pq − 1)

−1

Q0

Similarly,

Z

j p−2

t t0

Z

q

{x∈Q0 :|u(x)|q >t} −1

= (p − 1)

Z

Q0

|u| dxdt =

Z

Q0

|u|

q

Z

b(j,1,u(x))

t0


q b(j, 1, u(x))p−1 − tp−1 |u| dx. 0 41

tp−2 dtdx

Combining the above estimates for the left and right hand sides of (18) we

on lude that

Z Z Z q p ′ pq q p−1 q ′ − min{j, |u(x)| } |u(x)| ≤ C (− |u| ) ≤ C C (− |u|)pq Q0

Q0

Q0

−1

C ′ = ((p − 1)−1 − q(pq − 1)−1 δ 1−pq ) , provided C ′ > 0. We used (14) ′ at the last step. Choosing p > 1 so that C = 2 allows us to on lude the

laim via the monotone onvergen e theorem. 2 where

Ω ⊂ Rn and 1 ≤ p ≤ ∞, we let W 1,p (Ω) denote the olp le tion of all fun tions u ∈ L (Ω) that are absolutely ontinuous on almost all lines parallel to the oordinate axes in Ω and whose lassi al partial derivap 1,p tives belong to L (Ω). Then W (Ω, Rn ) refers to mappings f : Ω → Rn 1,p whose ea h omponent fun tion fj , j = 1, · · · , n, belongs to W (Ω). The 1,p 1,p n denitions of Wlo (Ω) and Wlo (Ω, R ) should then be obvious. Given a domain

f : Ω → Ω′ be quasi onformal, where Ω, Ω′ ⊂ Rn , p = p(n, H) > n and a onstant C = C(n, p, H) so that

4.17 Corollary. Let

n ≥ 2. 1)

There is

1,p f ∈ Wlo (Ω, Rn )

and

1/p Z Z 1/n p n − Lf ≤ C − Lf Q

whenever 2) If

2Q ⊂ Ω

Q

2Q ⊂ Ω. and

E⊂Q

is measurable, then

|f (E)| ≤C |f (Q)| Proof.



|E| |Q|

1−n/p

.

1) By Remark 4.11 we have

Z 1/n Z n − Lf ≤ C− Lf Q

whever

2Q ⊂ Ω.

Q

The Sobolev regularity and the asserted inequality follow

from Gehring's lemma be ause

f

is absolutely ontinuous on almost all lines

parallel to the oordinate axes and

|∂j fi (x)| ≤ Lf (x) 42

for almost every

x,

see Corollary 4.8 and its proof.

2) By Corollary 4.12, Lemma 4.4, Hölder's inequality, Proposition 4.1, and part 1) we see that

|f (E)| =

Z

E

≤C

µ′f

Z

≤C E

Lpf

Z

E

Lnf

n/p

|E|1−n/p

n/p Z p |E|1−n/p |Q|n/p ≤ C − Lf Z Q ≤ C− Lnf |E|1−n/p |Q|n/p Q |{z} ≤Cµ′f

≤ C|f (Q)||E|1−n/p |Q|n/p−1.

2

4.4

Ap -weights

We will briey point out the onne tion between

Ap -weights

and reverse

Hölder inequalities. The results of this se tion will not be needed later on. We refer the reader to [26℄ for proofs of the fa ts presented in this se tion. 1 Let w ∈ Llo , w > 0 almost everywhere. If −∞ < s < t < ∞ and |E| > 0,

then

1/s Z 1/t Z s t − w ≤ − w . E

So, when

p > 1,

E

we have that

1−p Z Z 1/p 1/(1−p) p − w ≤ − w B

for ea h ball

Ap - lass),

B.

B

We say that

w

is an

Ap -weight (belongs to the Mu kenhoupt

if for all balls

1/p Z 1−p Z p 1/(1−p) − w ≤ Cp,w − w , B

when

1 < p < ∞,

and

B

Z − w ≤ C1,w essinfB w, B

43

pS = 1. Clearly A1 ⊂ Ap ⊂ Aq when 1 ≤ p ≤ q. We nally set A∞ = p>1 Ap . One of the onne tions between Ap - lasses and reverse Hölder inequalities when

is given by the following result.

1 w ∈ Llo , w > 0 almost everywhere. exist q > 1 and C su h that

4.18 Fa t. Let only if there

Then

w ∈ A∞

if and

1/q Z Z q − w ≤ C− w B

for all balls

B

B.

4.19 Corollary. Let

A∞ .

n ≥ 2.

If

f : Rn → Rn

is quasi onformal, then

w ∈ Ap , p > 1, one an use the above that w ∈ Aq for some q < p that depends

Given prove

Ap -weights

µf ∈

reverse Hölder inequality to on

n, p, Cp,w .

are of their own interest. One of their important properties is

that they work well with maximal fun tions.

4.20 Fa t. Let

1 < p < ∞. Z

The inequality

p

(Mu) w ≤ C

holds for ea h Lebesgue measurable One an further hara terize

onstants

C

and

δ

u

A∞

Z

|u|p w

if and only if

w ∈ Ap .

by the following ondition. There are

so that

R δ  w |E| E R ≤C |Q| w Q

(19)

Q and ea h measurable E ⊂ Q. Given a domain G, let us write A∞ (G) for the olle tion of all w for whi h (19) holds with uniform onstants for ea h ube Q ⊂ G with diam(Q) ≤ d(Q, ∂G). Then, in dimensions n ≥ 2, ′ a homeomorphism f : Ω → Ω is quasi onformal if and only if, for ea h −1 ∈ A∞ (f (G)) for ea h w ∈ A∞ (G) and w ◦ f ∈ subdomain G ⊂ Ω, w ◦ f A∞ (G) for ea h w ∈ A∞ (f (G)) with uniform bounds in both ases. For this for ea h ube

see [25℄.

44

4.5 Dierentiability almost everywhere We begin with an almost everywhere dierentiability result that goes ba k 1,p u ∈ Wlo (Ω) means that u is lo ally p-

to Cesari and Calderón. Re all that

integrable, absolutely ontinuous on almost all lines parallel to the oordinate axes in

Ω

and that the lassi al partial derivatives are lo ally

4.21 Theorem. Let

p>n

and let

dierentiable almost everywhere.

1,p u ∈ Wlo (Ω)

p-integrable.

be ontinuous. Then

u

is

This result is optimal in the sense that there exist ontinuous fun tions 1,n in Wlo that are nowhere dierentiable. We need a few te hni al results for the proof of this theorem.

4.22 Lemma. Let set

1,1 u ∈ Wlo (Ω)

and

Z ur (x) = −

Ω0 ⊂⊂ Ω.

Given

0 < r < d(Ω0 , ∂Ω),

u(y) dy

B(x,r)

for

x ∈ Ω0 .

Then

ur ∈ C 1 (Ω0 )

and

Z ∇ur (x) = −

B(x,r)

∇u(y) dy.

Proof. Fix 0 < r < d(Ω0 , ∂Ω), x ∈ Ω0 and 1 ≤ j ≤ n. Let 0 < |t| < d(Ω0 , ∂Ω) − r . By the absolute ontinuity of u on almost all lines parallel to the xj -axis in Ω, Z u(y + tej ) − u(y) = ∂j u(y + sej ) ds [0,t]

for almost all

y ∈ B(x, r).

Integrating this estimate and invoking the Fubini

theorem we infer that

Z ur (x + tej ) − ur (x) u(y + tej ) − u(y) =− dy t t B(x,r) Z Z =− − ∂j u(y + sej ) ds dy B(x,r) [0,t] Z Z =− − ∂j u(y + sej ) dy ds. [0,t] B(x,r) Z Z =− − ∂j u(y) dy ds. [0,t] B(x+sej ,r) {z } | =:f (s)

45

Sin e

∂j u ∈ L1 (Ω),

f

it follows that

Z ∂j ur (x) = lim −

t→0 [0,t]

is ontinuous. Hen e

Z f (s) ds = f (0) = −

∂j u(y) dy.

B(x,r)

2

v ∈ Lp (λB), 1 ≤ p < ∞,

4.23 Lemma. Suppose that c

0 < ε < d(B, λB ),

set

Z vε (x) = −

where

λ > 1.

Given

v(y) dy

B(x,ε)

for

x ∈ B.

Then

vε → v

in

Lp (B).

Proof. Let w ∈ Lp (λB). Let 0 ≤ ψε ∈ L∞ be su h n sptψε ⊂ B(0, ε). Extend w as zero to R \ λB . Then Z wψε := ψε (y)w(x − y) dy

that

R

ψε = 1

and

Rn

is bounded on

B: |wψε (x)| ≤ kψε kL∞

Choose now

ψε (y) = wε = wψe .

and write

|wε | =

Z

ψε (y)

1/p

Rn

Z

λB

|w|.

1 χB(0,ε) (y) |B(0, ε)|

By the Hölder inequality,

|w(x − y)|ψε (y)

(p−1)/p

≤

Z

Rn

ψε (y)|w(x − y)| dy

and so

Z

p

B

Z

Z

ψε (y)|w(x − y)|p dy dx ZλB Z = ψε (y) |w(x − y)|p dx dy n λB ZR p ≤ |w| .

|wε | ≤

Rn

λB

If

w

is ontinuous on

λB ,

then

kw − wε kLp (B) → 0, 46

p

1/p

as

ε → 0.

Let

δ > 0.

Re all that ontinuous fun tions are dense in

see Subse tion 11.3 in the appendix. Choose a ontinuous

w

Lp (λB),

su h that

kv − wkLp (λB) < δ, and take

ε>0

so small that

kw − wε kLp (B) < δ. Then

kv − vε kLp (B) ≤ kv − wkLp (B) + kw − wε kLp (B) + k wε − vε kLp (B) < 3δ. | {z } =(w−v)ε

Thus

vε → v

in

Lp (B).

4.24 Corollary. If

Z

u ∈ W 1,1 (B),

B

2

then

|u − uB | dx ≤ C diam(B)

Z

B

|∇u| dx.

Proof. Let 0 < δ < 1. Then, for 0 < r < δ/2, ur is well (1 − δ)B . Thus, by the usual Poin aré inequality, Z Z |ur (x) − (ur )(1−δ)B | dx ≤ C(1 − δ) diam(B) (1−δ)B

dened and

(1−δ)B

C1

in

|∇ur (x)| dx.

r → 0 we see that this inequality holds for u (vr tends to v in ∈ L1 and r → 0 by Lemma 4.23). The laim follows by letting δ → 0; noti e that u(1−δ)B → uB . 2 By letting L1 when v

4.25 Corollary. Let

u ∈ W 1,p (5B)

and let

|u(x) − u(y)| ≤ C(n, p)|x − y| for all Lebesgue points

x, y ∈ B

of

u.

47

1−n/p

p > n. Z

Then

B(x,2|x−y|)

|∇u|

p

1/p

Proof. Let x, y ∈ B be Lebesgue points of u. Dene Bi = B(x, 2−i |x − y|) for i ≥ 0. Then, by Corollary 4.24 and the Hölder inequality, |u(x) − uB0 | ≤

∞ X i=1

|uBi−1 − uBi |

∞ Z X ≤2 − |u − uBi | n

i=0

≤ C(n, p) ≤ C(n, p)

Bi

∞ X

i=0 ∞ X i=0

1/p Z p 2 |x − y| − |∇u| −i

Bi

−i

(2 |x − y|)

≤ C(n, p)|x − y|

1−n/p

1−n/p

Z

Bi

Z

B(x,|x−y|)

|∇u|

|∇u|

p

p

1/p

1/p

.

Similarly,

|u(y) − uB(y,|x−y|) | ≤ C(n, p)|x − y| Moreover, denoting we have

1−n/p

Z

B(y,|x−y|)

|∇u|

p

1/p

.

Bx = B(x, |x − y|), By = B(y, |x − y|) and ∆ = Bx ∩ By ,

|uBx − uBy | ≤ |uBx − u∆ | + |u∆ − uBy | Z Z ≤ − |u − uBx | + − |u − uBy | ∆

∆

! Z Z ≤ C(n) − |u − uBx | + − |u − uBy | By

Bx

Z 1−n/p − ≤ C(n, p)|x − y|

B(x,2|x−y|)

The laim follows by the triangle inequality.

1,p u ∈ Wlo (Ω), p > n,

then

Z u˜(x) = lim sup − r→0

48

1/p

. 2

4.26 Remarks. 1) If

|∇u|

p

B(x,r)

u

is ontinuous and satises the modulus of ontinuity given in the orollary. This easily follows from the previous orollary. Noti e that, by

u˜ = u almost everywhere. We 1,n

all u ˜ the ontinuous representative of u. A fun tion u ∈ Wlo (Ω) does not need to have a ontinuous representative when n > 1. An example −1 −1 of this is u(x) = log log |x| , |x| < e . the Lebesgue dierentiation theorem,

2) The ontinuous representative of

u˜

and the fa t that

u˜ = u

1,p u˜ belongs to Wlo (Ω):

By the ontinuity

almost everywhere, we have that

u˜(x) = lim(˜ u)r (x) = lim ur (x) r→0

r→0

∂j (ur )(x) = (∂j u)r (x) for all x and 1 ≤ j ≤ n. QR ⊂⊂ Ω and R1 ≤ j ≤ n. Sin e (∂j u)r → ∂j u in L1 (Q), it follows that (∂ u)r → J ∂j u for almost every line segment J ⊂ Q J j parallel to the xj -axis. Let J be su h a line segment with endpoints x and y . Then Z Z Z u˜(x)−˜ u(y) = lim(ur (x)−ur (y)) = lim ∂j (ur ) = lim (∂j u)r = ∂j u. for all

x.

By Lemma 4.22,

Fix a ube

r→0

u˜

It follows that that

∂j u˜ = ∂j u

r→0

r→0

J

J

J

is absolutely ontinuous on almost all lines in

Ω

and

almost everywhere, as desired.

We are now ready to prove Theorem 4.21.

Proof. almost

1,p u ∈ Wlo (Ω) be ontinuous. Then, by part 2) every x0 , ∇u(x0 ) exists and Z lim − |∇u(x) − ∇u(x0 )|p dx = 0. Let

of Remark 4.3, at

r→0 B(x ,r) 0

Fix su h an

x0

and dene

w(x) = u(x) − u(x0 ) − ∇u(x0 ) · (x − x0 ). 1,p w ∈ Wlo (Ω)

Then

By Corollary 4.25,

and

∇w(x) = ∇u(x) − ∇u(x0 )

Z |w(x) − w(x0 )| ≤ C(n, p)|x − x0 | −

B(x0 ,5|x−x0 |)

whenever

∇u(x) p

|∇u(y) − ∇u(x0 )| dy

exists.

1/p

Thus

lim

x→x0

|u(x) − u(x0 ) − ∇u(x0 ) · (x − x0 )| |w(x) − w(x0 )| = lim = 0. x→x0 |x − x0 | |x − x0 | 49

.

2

4.27 Remark. By Theorem 4.21, Lips hitz fun tions are dierentiable almost everywhere. This immediately implies that ea h Lips hitz mapping f : Rn → Rn is almost everywhere dierentiable.

1,p Ω ⊂ Rn , re all that Wlo (Ω, Rn ) denotes the olle tion n of mappings f : Ω → R whose ea h omponent fun tion fj , j = 1, · · · , n, 1,p belongs to Wlo (Ω). Given a domain

f : Ω → Ω′ be quasi onformal, where Ω, Ω′ ⊂ Rn , 1,p n ≥ 2, are domains. Then f belongs to Wlo (Ω, Rn ) for some p > n and, for almost every x ∈ Ω, f is dierentiable at x with Jf (x) 6= 0 and satises 4.28 Corollary. Let

|Df (x)|n ≤ Hf (x)n−1 |Jf (x)|.

Proof.

By Corollary 4.17 and Theorem 4.21 applied to the oordinate fun 1,p n tions of f , f belongs to Wlo (Ω, R ), for some p > n, and is dierentiable almost everywhere.

f

Suppose that Then

where

is dierentiable at

and that

Jf (x0 ) = det Df (x0 ) = 0.

|f (B(x0 , r))| ≤ (|Df (x0 )| + ε(r))n−1 r n−1 ε(r)r,

ε(r) → 0,

as

r → 0.

Thus

µ′f (x0 ) = lim

r→0

Be ause

x0

µ′f > 0

|f (B(x0 , r))| = 0. |B(x0 , r)|

almost everywhere by Corollary 4.12 and

almost everywhere, Suppose that

f

Jf 6= 0

f

is dierentiable

almost everywhere.

is dierentiable at

x0

with

Then

Jf (x0 ) = det Df (x0 ) 6= 0.

|Df (x0 )| ≤ Hf (x0 ) min |Df (x0 )h|. |h|=1

Be ause

|Jf (x0 )| ≥



min |Df (x0 )h|

|h|=1

see Subse tion 11.2 we on lude that

n−1

|Df (x0 )|,

|Df (x0 )|n ≤ Hf (x0 )n−1 |Jf (x0 )|. 2

50

4.29 Remarks. 1) The exponent

n−1

for

H

in Corollary 4.28 is optimal. This is seen

by onsidering the quasi onformal mapping

f (x) = Ax,

where

A

is a

diagonal matrix whose diagonal entrees are all 1 expe t for a single entry whi h is, say, 2.

f : Ω → Ω′ , both domains in Rn , is a homeomorphism and dierentiable at x, y ∈ Ω, then either Jf (x) ≥ 0 and Jf (y) ≥ 0 or Jf (x) ≤ 0 and Jf (y) ≤ 0. This an be proved using the so- alled topologi al de-

2) If

gree, whi h we have not introdu ed.

Combining this with Corollary

4.28 allows us to on lude that, given a quasi onformal mapping f, den ned in a domain Ω ⊂ R , n ≥ 2, either Jf (x) > 0 almost everywhere in

Ω

or

Jf (x) < 0

almost everywhere in

Ω.

1,p f ∈ Wlo (Ω, Rn ), where Ω ⊂ Rn , n ≥ 2 is a domain, is a homeomorphism and p > n − 1 (p ≥ 1 in the plane), then f is dierentiable almost everywhere, see [23℄. If p = n − 1 and n ≥ 3, then f need not be

3) If

dierentiable anywhere. The positive results are non-trivial. For the 1,n−1

ounterexample, one pi ks a ontinuous fun tion u ∈ Wlo (Rn−1 ) of

n−1

variables that fails to be dierentiable anywhere and denes

f (x1 , · · · , xn ) = (x1 , · · · , xn−1 , xn + u(x1 , · · · , xn−1 )). p < n − 1, it 1,p f ∈ Wlo (Ω, Rn )

4) If

is not known if the Ja obian of a homeomorphism

an hange its sign.

For

p > n − 1,

the Ja obian

determinant annot hange its sign by 1) and 2) and this is expe ted to also hold when

Added:

p = n − 1.

Hen l and Malý, Ja obians of Sobolev homeomorphisms, to

appear in Cal . Var. have very re ently shown that one an relax the

p > n − 1 to p > pn , where pn is the integer part of n/2, p3 = 1. The ase 1 < p ≤ pn remains open when n > 3.

assumption espe ially

5 The analyti denition In this hapter we give an analyti denition for quasi onformality by establishing the following hara terization of quasi onformality.

5.1 Theorem. Suppose that Ω, Ω′

⊂ Rn are domains, n ≥ 2.

be a homeomorphism. Then the following are equivalent: 1)

f

is quasi onformal. 51

Let

f : Ω → Ω′

2) There exists

η

f |B

su h that

2B ⊂ Ω. 3)

1,1 f ∈ Wlo (Ω, Rn )

η -quasisymmetri

is

and there is

K

for ea h ball

B

with

su h that

|Df (x)|n ≤ K|Jf (x)| almost everywhere in

Ω.

5.2 Remark. It follows that either

Jf < 0

almost everywhere in

Ω,

Jf > 0

almost everywhere in

Ω

or that

see Corollary 4.28 and Remarks 4.29.

We already saw in Chapter 3 that 1) and 2) are equivalent and Corollary 4.28 shows that

1)

implies

3).

In order to dedu e

1)

from

3)

we introdu e

some preliminary results. Re all the notation

|f (B(x, r))| r→0 |B(x, r)|

µ′f (x) = lim

that we used for homeomorphisms. One of our aims is to show that

Jf

is

lo ally integrable for a homeomorphism that is lo ally in the Sobolev lass 1,1 ′ . This will be done by relating Jf to µf . It is rather easy to do this at Wlo the points of dierentiability of our homeomorphism. The problem is that, in 1,1 n dimensions n ≥ 3, our regularity assumption f ∈ Wlo (Ω, R ) (see Remarks 4.29) does not by itself guarantee dierentiability even at a single point. In order to over ome this, we will use Lips hitz approximations to

f,

but the

prize we have to pay is that these Lips hitz mappings need not be inje tive. n Given a ontinuous mapping f : Ω → R , we write

µ′f (x) = lim sup r→0

Be ause

f (B(x, r))

|f (B(x, r))| . |B(x, r)|

is ompa t and so measurable,

µ′f (x)

is indeed dened.

We annot however apply the Radon-Nikodym theorem as we did in the

onne tion with Proposition 4.1: a measure when

f

µ(A) = |f (A)|

does not ne essarily dene

fails to be inje tive. We will be able to get around this

problem.

5.3 Lemma. Let f Then

1,1 : Ω → Rn be ontinuous and assume that f ∈ Wlo (Ω, Rn ).

|Jf (x)| ≤ µ′f (x) almost everywhere in

Ω. 52

The proof of this result will be based on a sequen e of lemmas.

5.4 Lemma. Let tiable at

x0 ∈ Ω.

f : Ω → Rn

be ontinuous and assume that

f

is dieren-

Then

|Jf (x0 )| = µ′f (x0 ).

We already saw in the proof of Corollary 4.28 that if Jf (x0 ) = 0 ′ dierentiable at x0 , then µf (x0 ) = 0. Suppose that Jf (x0 ) 6= 0. We may assume that x0 = 0 = f (x0 ). Be ause Jf (0) 6= 0, the inverse matrix (Df (0))−1 exists. Dene g(x) = (Df (0))−1f (x). Then g is dierentiable at

Proof. and f is

0, Dg(0) = I ,

and moreover,

|f (B(0, r))| = |Df (0) g(B(0, r))| = |Jf (0)||g(B(0, r))|. Thus it su es to show that

|g(B(0, r))| = 1. r→0 |B(0, r)| lim

Be ause

g

is dierentiable at

0

and

Dg(0) = I ,

|g(x) − x| ≤ ε(|x|)|x|, where

ε(|x|) → 0

as

|x| → 0.

(20)

It follows that

|B(0, r + ε(r)r)| |g(B(0, r))| ≤ = (1 + ε(r))n −→ 1, |B(0, r)| |B(0, r)|

as

r → 0,

so espe ially

lim sup r→0

|g(B(0, r))| ≤ 1. |B(0, r)|

For the opposite inequality we use the fa t that (21)

0 < r < rε . This follows from Lemma 11.10 in the appendix, sin e now |g(x) − x| ≤ ε for |x| < rε by inequality (20). Thus by (21) we obtain for r < rε that for given

so

ε>0



B 0, (1 − ε)r ⊂ g B(0, r)

whenever

|g(B(0, r))| |B(0, (1 − ε)r)| ≥ = (1 − ε)n −→ 1, |B(0, r)| |B(0, r)| lim inf r→0

This proves the lemma.

|g(B(0, r))| ≥ 1. |B(0, r)|

53

as

ε → 0,

2

A ⊂ Rn

5.5 Lemma. (M Shane extension) Let

L-Lips hitz,

that is

x, y ∈ A. ˜ A = f. f|

for all that

Proof.

Let

|f (x) − f (y)| ≤ L|x − y| √ n m there exists a ( mL)-Lips hitz f˜ : R → R

Then

m = 1.

f : A → Rm

and

be

su h

Dene

˜ = inf {f (a) + L|x − a|}. f(x) a∈A

Then

f˜(x) = f (x)

when

x ∈ A:

Sin e

f

is

f (x) ≤ f (a) + L|x − a| ˜ ≥ f (x). f(x) n Given x, y ∈ R ,

and so

Also, learly we have that

L-Lips hitz when

on

A,

x, a ∈ A,

f˜(x) ≤ f (x).

f˜(x) = inf {f (a) + a∈A

L|x − a| } | {z }

≤L(|y−a|+|y−x|)

≤L|y − x| + f˜(y). Be ause this also holds with

x

y,

repla ed by

we on lude that

f˜

is

L-

Lips hitz. Let us then onsider the ase

f˜ = (f˜1 , . . . , f˜m )

m ≥ 2.

For given

f = (f1 , . . . , fm )

dene

as in the previous ase. Now

|f˜(x) − f˜(y)|2 =

m X 1

|f˜i(x) − f˜i (y)|2 ≤ mL2 |x − y|2,

and the laim follows.

2

5.6 Remark. By hoosing a suitable extension dierent from the M Shane extension, one ould require above

f˜ to

be

L-Lips hitz.

This an be done

using the so- alled Kirszbaum extension.

u ∈ W 1,1 (3B) and ε > 0. |B \ Aε | < ε and u|Aε is Lips hitz.

5.7 Lemma. Let that

54

Then there is a set

Aε ⊂ B

so

Proof. Write B = B(x0 , r0 ). Let x, y ∈ B be Lebesgue points of u. Choose Bj = B(x, 2−j |x − y|) for j ≥ 0 and Bj = B(y, 2j+1|x − y|) for j < 0. Then by the Poin aré inequality (as in the proof of Theorem 2.12),

|u(x) − u(y)| ≤

∞ X −∞

|uBj − uBj+1 | ≤

Z ∞ X ≤ Cn rj − |∇u| −∞

∞ X −∞

Z Cn − |u − uBj | Bj

Bj

≤ Cn |x − y| M3r0 |∇u(x)| + M3r0 |∇u(y)| ≤ 2Cn |x − y|λ when both have

x

and

y

Cn λ-Lips hitz

Badλ

belong to the set

{z ∈ B :

ontinuity outside the set

M3r0 |∇u(z)|

= {z ∈ B : M3r0 |∇u(z)| > λ} ∪ {z ∈ B : z




≤ λ}.

Thus we

non-Lebesgue point of

u}.

By Remark 2.6,

5n 2 |Badλ | ≤ λ

Z

|

and the laim follows.

{|∇u(z)|> λ }∩3B 2

|∇u| = o

{z

−→ 0

λ→∞

}

1 λ


2

u is Cn λ-Lips hitz in B \ Badλ , 1 where |Badλ | = o . Use the M Shane extension theorem to extend the λ restri tion of u to this set as Cn λ-Lips hitz fun tion uλ to all of B. Then 5.8 Remark. The above proof shows that


Z

B

Z

|∇u − ∇uλ | ≤

Badλ

|∇u| + |∇uλ| ≤

Z

Badλ

|∇u| + Cn λo

be ause

∇uλ (x) = ∇u(x)

1 λ




−→ 0

λ→∞

(22)

Gλ = B \ Badλ . Reason: If E ⊂ Ω is measurable, ∂i v and ∂i w exist almost everywhere in E and v = w on E , then ∂i v = ∂i w almost everywhere in E : Simply noti e that almost every point x of E is of linear density one in the xi -dire tion. at almost every point

x

of

One an do even better. Consider the set

′ Badλ

= {x ∈ B : 55

M3r0 u(x)

≥ λ}.




′ 1 Then |Badλ | = o . So, when λ is large, the distan e from any point in λ ′ ′ Badλ to B \ Badλ is at most one. Thus the M Shane extension uλ of u from B \ (Badλ ∪ Bad′λ ) is Cn λ-Lips hitz and bounded in absolute value by 2Cn λ on

B.

It follows that

Z

|u − uλ | + |∇u − ∇uλ | −→ 0. λ→∞

B

The nal estimate of the pre eding remark yields the following orollary:

5.9 Corollary. If

u ∈ W 1,1 (3B), then there is a sequen e (ϕj )∞ 1

of Lips hitz

fun tions su h that

|{x ∈ B : ϕj (x) 6= u(x)}| → 0 and

Z

B

as

j → ∞.

|u − ϕj | + |∇u − ∇ϕj | → 0

5.10 Remarks. 1) One an get rid of the onstant 3 above (see Figure 4)

y

x

Figure 4: Remark 5.10 (1).

2) The same argument as above gives the orollary for

Z

B

|u − ϕj |p + |∇u − ∇ϕj |p → 0

as

n

Wlo (Ω, R ).

Let

B⊂Ω

be a ball

and with

j → ∞.

f : Ω → Rn is ontinuous and f ∈ with 3B ⊂ Ω. It su es to prove that

Proof of Lemma 5.3. Assume that 1,1

W 1,p

|Jf (x)| ≤ µ′f (x)

for a.e.

56

x ∈ B.

ε > 0. Pi k a Lips hitz mapping f˜ : Rn → Rn su h that for the set B = {x ∈ B : f˜(x) 6= f (x)} we have |B| < ε, see Corollary 5.9. Be ause f˜ is Lips hitz, it is dierentiable almoste everywhere in B \ B; see Remark ′ 4.27. By Lemma 5.4, |Jf˜(x)| = µ ˜(x) at the points of dierentiability. By the f reasoning in Remark 5.8, see (22), Jf˜(x) = Jf (x) almost everywhere in B \ B . ′ ′ So it su es to prove that µ ˜(x) ≤ µf (x) almost everywhere in G = B \ B . f Let x ∈ G. Then Let

˜ |f(B(x, r) ∩ G)| + |f˜(B(x, r) ∩ B)| |f˜(B(x, r))| ≤ |B(x, r)| |B(x, r)| |f (B(x, r))| Ln |B(x, r) ∩ B| ≤ + , |B(x, r)| |B(x, r)| and the laim follows be ause the last term tends to zero for almost every

x∈G

by Remarks 4.3 3).

5.11 Corollary. Let f If

2

1,1 : Ω → Ω′ be a homeomorphism with f ∈ Wlo (Ω, Rn ).

|Df (x)|n ≤ K|Jf (x)| almost everywhere in

Ω

for some

1 ≤ K < ∞,

then

1,n f ∈ Wlo (Ω, Rn ).

′ By Lemma 5.3, |Jf (x)| ≤ µf (x) almost everywhere in ′ 1 follows be ause µf ∈ Llo (Ω), by Proposition 4.1.

Proof.

5.12 Lemma. Let let

′

u : Ω → [0, ∞)

f : Ω → Ω′

be a homeomorphism,

be Borel measurable. Then

Z

Ω

u(f (x))|Jf (x)| ≤

Z

Ω.

The laim

1,1 f ∈ Wlo (Ω, Rn ),

2

and

u.

Ω′

′ j j+1 Proof. Let S a > 1 and ′set Gj = {y ∈ Ω : a < u(y) ≤ a } for j ∈ Z. ′ Then Ω \ Gj = {y ∈ Ω : u(y) = 0}. Thus, by Proposition 4.1 and Lemma

57

5.3,

Z

u(f (x))|Jf (x)| =

Z

u(f (x))|Jf (x)|

f −1 (∪Gj )

Ω

=

XZ

u(f (x))|Jf (x)| f −1 (Gj )

XZ

aj+1 µ′f (x) dx Z X XZ j+1 u dy = a u. ≤ a |Gj | ≤ a

≤

f −1 (Gj )

Gj

Let

a→1

Ω′

to omplete the proof.

2

1,n f : Ω → Ω′ be a homeomorphism, f ∈ Wlo (Ω, Rn ) n 1 ′ and |Df (x)| ≤ K|Jf (x)| almost everywhere in Ω. If u is C on Ω , then 1,n u ◦ f ∈ Wlo (Ω) and Z Z n |∇(u ◦ f )| ≤ K |∇u|n .

5.13 Lemma. Let

Ω′

Ω

Proof.

Clearly

u◦f

is absolutely ontinuous on almost all lines parallel to

Ω be ause f is and u is lo ally Lips hitz. Be ause u ◦ f is lo ally bounded, it thus su es to show the lo al n-integrability of |∇(u ◦ f )| and the asserted inequality. Let f˜ be as in the proof of Lemma ˜ almost 5.3. Then f˜ is dierentiable almost everywhere and Df (x) = D f(x) everywhere in G (see Remark 5.8). Thus, using the usual hain rule and

the oordinate axes in

Proposition 11.1, we see that

|∇(u ◦ f )(x)|n = |∇u(f (x))Df (x)|n ≤ |Df (x)|n |∇u(f (x))|n ≤ K|Jf (x)||∇u(f (x))|n almost everywhere in where in

Ω.

Proof of dene

G.

It follows that this inequality holds almost every-

Use Lemma 5.12 to omplete the proof.

3) ⇒ 1)

in Theorem 5.1. Let

B = B n (x0 , r0 ) ⊂ 2B ⊂ Ω,

l, L as in the proof of Theorem 2.1, see Figure 5.

58

2

and

We may again assume

that

L ≥ 2l.

By Corollary 5.11,

1,n f ∈ Wlo (Ω, Rn ).

  1      0

and set

if if

Dene

|y − f (x0 )| ≤ l |y − f (x0 )| ≥ L

1 1 u(y) = log − log   |y − f (x0 )| L   if l ≤ |y − f (x0 )| ≤ L ,  L   log l R uε (y) = −B(y,ε) u(z)dz for ε > 0. Then uε is C 1 by Lemma 4.22 Ω

and

l

f(x) f

f(B)

x r B

L

Figure 5: thus, by Lemma 5.13,

f B(x, r)

1,n uε ◦ f ∈ Wlo (Ω) and Z Z n |∇(uε ◦ f )| ≤ K |∇uε |n .

Z

Ω′

when

ε → 0

(23)

Ω′

Ω

Next




n

|∇uε | →

Z

Ω′

|∇u|n = ωn−1 log Ll


1−n

(24)

by Lemma 4.22 and dominated onvergen e (also by Lemma

4.22 and Lemma 4.23). Here

ωn−1

is the

unit sphere.

(n − 1)-dimensional measure

of the

f −1 (B n (f (x0 ), l)) is a onne ted set ontaining x0 and its n−1 (x0 , r). Furthermore, f −1 (Rn \B(f (x0 ), L)) has an open

losure interse ts S n−1

omponent G whose losure interse ts both S (x0 , r) and S n−1 (x0 , 3r ). We 2 −1 n (B (f (x0 ), l)) and F ⊂ G, both of diameter may then sele t ontinua E ⊂ f at least r0 /4 so that uε = 1 on E and uε = 0 on F for all su iently small ε. Thus Z |∇(uε ◦ f )|n ≥ δn > 0 (25) Noti e that

Ω1

59

ε > 0 be ause of the size of u ◦ f ∈ W 1,1 (2B); noti e that the

0-

and

1-sets

of

uε

for all su iently small

the

and the fa t that

proof of Theorem 2.12

only assumed a Poin aré inequality, whi h holds in our setting by Corollary 4.24. A bound on

L/l

and so also quasi onformality of

f

follow by ombining

(23), (24) and (25).

5.14 Remarks.

2

H K≤

1) Regarding the relationship between the onstants

and K in parts 1) and 3) of Theorem 5.1, we have the estimates H n−1 and H ≤ exp(Cn K 1/(n−1) ).

The rst of these is sharp and ontained in Corollary 4.28 and the se ond follows from the proof of Theorem 5.1 above. The se ond estimate an be improved to

Hf (x) ≤ K

almost everywhere, but, for

example, for the simple planar quasi onformal mapping dened by

f (x, y) = (x, 2y) in the upper losed half plane and by f (x, y) = (x, y/2) in the lower half plane, one has K = 2 and H = 4. On the other hand, one an onstru t examples (in the plane [19℄) that show the sharpness of the given global bound on

H.

2) Noti e that the analyti denition requires the pointwise inequality at almost every point. One ould then expe t that the metri denition

ould also be slightly relaxed. This is indeed the ase in the sense that ′ a homeomorphism f : Ω → Ω satises

1,n f ∈ Wlo (Ω, Rn )

and

|Df (x)| ≤ K min |Df (x)h| |h|=1

lim inf r→0 Hf (x, r) < ∞ outside a set of σ -nite (n − 1)-measure and lim inf r→0 Hf (x, r) ≤ K almost everywhere. Above, lim sup instead of lim inf naturally works as well. almost everywhere if and only if

6 K -quasi onformal mappings Let us all from now on a homeomorphism and

|Df (x)|n ≤ K|Jf (x)|

K -quasi onformal (K -q )

f : Ω → Ω′ a.e. in

with

1,1 f ∈ Wlo (Ω, Rn )

Ω

a

ording to the analyti denition. We will typ-

i ally abuse the notation and only talk about K -quasi onformal mappings ′ n below. Above, Ω, Ω ⊂ R are domains and we assume that n ≥ 2. Noti e

that ea h onformal

f

is

1-q .

60

f

6.1 Remark. If (i) (ii)

f

1,p f ∈ Wlo (Ω, Rn )

(v)

(vi)

f

then

is dierentiable almost everywhere,

(iii) either (iv)

K -q ,

is

for some

Jf (x) > 0

a.e. in

Ω

p = p(n, K) > n, or

Jf (x) < 0

a.e. in

Ω,

is lo ally Hölder ontinuous,

|f (E)| =

R

|Jf | whenever E ⊂ Ω is measurable, α  |f (E)| |E| ≤C whenever E ⊂ Q ⊂ 2Q ⊂ Ω, |f (Q)| |Q| E

where

C = C(n, K), 0 < α = α(n, K). All this follows by ombining our previous results. By Theorem 3.6 we know that quasi onformal mappings form a group. It turns out that the analyti denition allows us to give sharp estimates on the asso iated onstants of quasi onformality.

f1 : Ω1 → Ω2 be K1 -q f2 ◦ f1 : Ω1 → Ω3 is K1 K2 -q .

6.2 Theorem. Let Then

Proof.

We already know that

f2 ◦ f1

and

f2 : Ω2 → Ω3

be

K2 -q .

is quasi onformal be ause the three

dierent denitions (in Theorem 5.1) give the same lass of mappings. Thus 1,1 1,p f2 ◦ f1 ∈ Wlo (Ω1 , Rn ) (and even Wlo (Ω1 , Rn ) for some p > n). Now f2 is dierentiable almost everywhere in in

Ω2 , f1

is dierentiable almost everywhere

Ω1 , and be ause f1 annot map a set of positive measure to a set of measure

zero, we on lude that

D(f2 ◦ f1 )(x) = Df2 (f1 (x))Df1 (x) for almost every

x ∈ Ω1 .

In parti ular, for su h a point

x,

|D(f2 ◦ f1 )(x)|n = |Df2 (f1 (x))Df1 (x)|n . For almost every

x ∈ Ω1 , |Df1 (x)|n ≤ K1 |Jf1 (x)|,

and for almost every

y = f1 (x) ∈ Ω2 , |Df2 (f1 (x))|n ≤ K2 |Jf2 (f1 (x))|. 61

Be ause

f1

an not map a set of positive measure to a set of measure zero,

both inequalities hold for almost every

x ∈ Ω1 .

Thus

|D(f2 ◦ f1 )(x)|n ≤ K2 K1 |Jf2 (f1 (x))||Jf1 (x)| = K2 K1 |Jf2 ◦f1 (x)| for almost every

x ∈ Ω1 .

6.3 Theorem. Let

2

f : Ω → Ω′

6.4 Remark. The onstants

be

K -q .

K1 K2

and

Then

K n−1

f −1 : Ω′ → Ω

is

K n−1 -q .

in Theorem 6.2 and Theorem

6.3 are sharp. To see this, simply onsider the linear quasi onformal mapasso iated to the diagonal matri es A1 , A2 and A where the 1/(n−1) 1/(n−1) , of A2 is K2 and all the rest are 1, rst diagonal entry of A1 is K1 pings

f1 , f2 , f

and the

n−1

rst diagonal entries of

A

are all

K

and the last one is 1.

For the proof of Theorem 6.3 we need some elementary linear algebra:

6.5 Proposition. If

det A 6= 0

and

|A|n ≤ K| det A|,

then

|A−1 |n ≤ K n−1 | det A−1 | . Proof.

By Proposition 11.2 in the appendix, we nd two orthonormal bases

so that the matrix of

A

with respe t to these bases is diagonal. Noti e that

the asso iated hanges of bases preserve lenghts. Thus the operator norms −1 −1 of A and A and the determinants of A, A

an be readily read of from this diagonal representation may assume that

with

D

of

A

(see Lemma 11.4 in the appendix). We

  λ1 . . . 0   D =  ... . . . ...  0 . . . λn

|λ1 | ≥ |λ2 | ≥ . . . ≥ |λn | > 0. Then   1/λ1 . . . 0  .  .. ..  . D −1 =  ... . 0 . . . 1/λn

Be ause

|λ1 |n ≤ K|λ1 . . . λn |,

1 1 = |D | = n |λn | |λn | −1 n

we have that



1 |λn |

n−1 62

≤

|λj | ≤ K|λn |

for ea h

j.

Thus

K n−1 = K n−1 | det A−1 |. |λ1 . . . λn |

2 Proof of Theorem 6.3. and so

f

−1

1,n

′

n

∈ Wlo (Ω , R ).

Also

f

x, f

measure and, at almost every ular, for almost every

f −1

We already know that

is quasi onformal

preserves the null sets for the Lebesgue

is dierentiable with

x∈Ω

Jf (x) 6= 0.

In parti -

I = D(f −1 ◦ f )(x) = Df −1(f (x))Df (x). So

 −1 Df −1 (f (x)) = Df (x)

x ∈ Ω and so also for almost every y = f (x) ∈ Ω′ . Be ause f is K -q , we have |Df (x)|n ≤ K| det Df (x)|, and onsequently Proposition 6.5 gives the laim. 2 for almost every

6.6 Remark. Combining Corollary 4.25, almost everywhere dierentiability of q mappings and Corollary 4.17 we see that ea h

K -q

mapping is lo ally

Hölder- ontinuous:

p>n

|f (x) − f (y)| ≤ C|x − y|

1−n/p

Z

B

|Df |

p

1/p

Z 1/p p − |Df | = C |x − y| r ZB Gehring ˜ − y|1−n/p r n/p − |Df |, ≤ C|x 1−n/p

n/p

B

p = p(n, K) > n. depends on K, n. It

f

where

Thus

is Hölder ontinuous with some exponent

that

is then natural to ask for the best possible Hölder

exponent.

6.7 Theorem. Let

f : Ω → Ω′

be

K -q .

|f (x) − f (y)| ≤ C(n, K) diam f (B) where

C1 = C1 (n),

whenever

x, y ∈ B .

63

If

7B ⊂ Ω,



|x − y| diam B

then

C1 /K

,

Proof. Let g : Ω1 → Ω2 be K -q , let y0 ∈ Ω2 and let y ∈ Ω2 |y − y0 | < d(y0, ∂Ω2 )/3. Write r = d(y0, ∂Ω2 )/2. We dene   1, if |z − y0 | ≤ |y − y0 |      if |z − y0 | ≥ r 0, 1 1 v(z) = log − log   |z − y0 | r   , if |y − y0 | ≤ |z − y0 | ≤ r . r    log |y − y0 | Write

u = v◦g

and extend

u

as zero to the exterior of

Ω1 .

satisfy

Then, as in the

proof of Theorem 5.1,

Z

Ω1

where

ωn−1

|∇u|n ≤ K

is the

Z

Ω2

|∇v|n ≤ K ωn−1 log

(n − 1)-dimensional



r |y − y0 |

 !1−n

,

(26)

measure of the unit sphere. Suppose

that we ould show that

Z

Ω1

|∇u|n ≥ ωn−1

log



C(n)Lg−1 (y0 , r) |g −1(y) − g −1(y0 )|

 !1−n

Then, ombining (26) and (27) and the fa t that the support of

ontained in

Ω,

.

(27)

u is ompa tly

we would on lude that

−1

−1

|g (y) − g (y0 )| ≤ C(n)Lg−1 (y0 , r)



|y − y0 | r

K −1/(n−1)

.

K n−1 -q (see Theorem 6.3 ) mapping g = f −1 : Ω′ → Ω, the laim would easily follow with C1 = 1. It is not easy to establish (27), but it is not hard to prove the lower bound with some onstant Cn , whi h is Applying this to the

su ient for the laim of our theorem: −1 Write L = Lg −1 (y0 , r), x0 = g (y0 ) and

Z −

B(x0 ,3s)

then

Z

B(x0 ,3s)

s = |g −1(y) − g −1(y0 )|.

If

u ≤ 23 ,

|∇u|n ≥ δ(n) > 0

by the proof of the orresponding earlier estimate (Theorem 2.12); noti e −1 that u = 1 on the ompa t, onne ted set g (B(y0 , |y − y0 |)) of diameter at 64

s. Noti e further that u(x) = 0 on Rn \ B(y0, L). Pi k w ∈ S n−1 (y0 , 2L). Then Z − u = 0. least

B(w,L)

Now, we may assume that

Z −

B(x0 ,3s)

u≥

2 3

Z −

and

u = 0,

B(w,L)

and thus (see Figure 6)

3L 2L

L B0 x0

w

g−1( y) L Bk−1 Bk

Figure 6: Choi e of

1 3

≤ ≤

k X j=1

k X

Bj 's

|uBj − uBj−1 | ≤ C˜

j=0

Z

Bj

|∇u|n

˜ + 1)(n−1)/n ≤ C(k | {z } ≤c log

cL s

in the proof of 6.7

k X

Crj

!1/n Z − |∇u|n Bj

j=0

!1/n Z

S

This gives the desired lower bound for

Bj

R

65

|∇u|n

Ω1

!1/n

|∇u|n .

.

2

6.8 Remarks. 1) Given a domain

Ω

with

E ∩ F = ∅,

set

Ω ⊂ Rn , n ≥ 2, and ompa t sets E, F ⊂

capn (E, F ; Ω) =

inf

u∈A(E,F ;Ω)

Z

Ω

|∇u|n ,

where

 1,n A(E, F ; Ω) = u ∈ C(Ω ∪ E ∪ F ) ∩ Wlo (Ω) : u ≥ 1

in

E

and

u≤0

in

F .

the onformal apa ity (varionational n- apa ity, n- apa ity) of E and F with respe t to Ω. As a part of the proof of Theorem 5.1 we

This is alled

essentially showed the fa t that

whenever


capn f −1 (E), f −1(F ); Ω ≤ K capn (E, F ; Ω′ )

E, F ⊂ Ω′

f : Ω → Ω′

are ompa t and

The basi estimates are: (i) If

E ⊂ E ′, F ⊂ F ′

and

Ω ⊂ Ω′ ,

is

K -q .

then

capn (E, F ; Ω) ≤ capn (E ′ , F ′ ; Ω′ ) . (ii) If

B(x, r) ⊂ B(x, R) ⊂ Ω ,

then



capn B(x, r), S n−1(x, R); Ω = capn B(x, r), S n−1(x, R); B(x, R) ωn−1 ≤
n−1 . log Rr

In fa t, the inequality an also be reversed: If

u ∈ A(B(x, r), S n−1(x, R); B(x, R)) is C 1 , then the fundamental the-

orem of al ulus and Hölder's inequality give

1≤ ≤ ≤

Z

R

|∇u(tw)| dt

r

Z

R

|∇u(tw)|t

r

Z

R r

dt t

n−1 n−1 − n n

 n−1 Z n

r

dt

R n n−1

|∇u(tw)| t

1/n dt

w ∈ S n−1 (0, 1). The desired inequality follows by raising both n−1 sides of this inequality to power n and integrating over S (0, 1) with respe t to w. Approximation then gives the same for general test fun for every

tions. 66

(iii) If

E, F ⊂ B(x, r)

are ontinua with

min{diam E, diam F } ≥ δ1 > 0 , r

then

capn (E, F ; B(x, r)) ≥ δ(δ1 , n) > 0. (iv) If

Ω

is bounded and

E⊂Ω

is a ontinuum, then

capn (E, ∂Ω; Ω) ≥ 

ωn−1 log

C(n) diam Ω diam E

This is not trivial; one uses symmetrization [11℄.

n−1

One in fa t also has the estimates [11℄

given ontinua

capn (E, F ; Rn ) ≥

ωn−1 (log(C(n)(1 + t)))n−1

E, F ⊂ Rn ,

t=

where

d(E,F ) , and min{diam(E),diam(F )}

capn (E, F ; Rn+ ) ≥ capn (E, F ; Rn )/2 E, F ⊂ Rn+ .

when

2) If we use (iv) in the proof of the Hölder- ontinuity estimate, we see that one an take

C1 = 1,

so that the Hölder exponent is

f (x) = x|x|−(1−1/K)

is

α = 1/K. This

is sharp:

K -q .

1/K (in terms of  H , 1/H in 1,p dimension two). By Corollary 4.25, f ∈ Wlo is lo ally Hölder- ontinuous n with exponent 1 − . To obtain the Hölder exponent 1/K via Corollary 4.25, p 1,pK with one would need f to be in the Sobolev lass Wlo 3) The Hölder exponent we found is thus

pK = The radial mapping

nK . K −1

f (x) = x|x|−(1−1/K) belongs to

1,p Wlo

exa tly when

p

is stri tly less than this

Ω, Ω′ ⊂ Rn 1,p then f ∈ Wlo

6.9 Conje ture. Let

be domains, where

K -quasi onformal,

for all

This holds when

n=2

p < pK .

pK .

n ≥ 2.

If

f : Ω → Ω′

is

by results by Astala [2℄. In higher dimensions, the

onje ture would follow if a ertain onje ture in al ulus of variations gets proved [16℄. 67

7 Sobolev spa es and onvergen e of quasi onformal mappings We will show that quasi onformality is stable under lo ally uniform onvergen e in the following sense.

7.1 Theorem. Let ′

fj → f : Ω → Ω

fj : Ω → Ωj

K -q for ea h j ≥ 1, and suppose that If f is a homeomorphism, then f is K -q .

be

lo ally uniformly.

7.2 Remarks. 1) In the plane, one obtains the following on lusion in terms

fj : Ω → Ωj are quasi onformal in terms of the metri denition with Hfj (x) = lim supr→0 Hfj (x, r) ≤ H almost everywhere in Ω for ea h j. If the sequen e (fj )j onverges lo ally uniformly ′ to a homeomorphism f : Ω → Ω , then f is quasi onformal with Hf (x) ≤ H almost everywhere in Ω. To see this noti e rst that ea h fj is H -q by Corollary 4.28. Thus Theorem 7.1 shows that f is H -q . By Corollary 4.28 we know that f is dierentiable at almost every x with Jf (x) 6= 0. Fix su h an x. As in the proof of Proposition 6.5, we may assume that Df (x) is diagonal with diagonal entries λ1 , λ2 satisfying |λ1 | ≥ |λ2 | > 0. Then

of the metri denition.

Suppose that

λ21 ≤ H|λ1λ2 |

Hf (x) ≤ H, as desired. n n 2) Let n ≥ 3. There is a sequen e of q mappings fj : R → R so that fj → f lo ally uniformly, the (metri ) H -dilatations of fj are all almost everywhere bounded by some H0 > 1 and the H -dilatation of f is not essentially bounded by H0 . Su h examples have been found by Iwanie [15℄. and it follows that

|λ1 | ≤ H|λ2|.

This implies that

3) The assumption that the limit fun tion be a homeomorphism is not superuous. Indeed, the sequen e by setting

fj (x) = x/j

(fj )j

of 1-quasi onformal mappings dened

onverges lo ally uniformly to the onstant fun tion

f (x) ≡ 0.

4) One an hara terize the lass of

K -quasi onformal

ompleteness property related to Theorem 7.1.

mappings by a

We will return to this in

Chapter 8. In order to prove Theorem 7.1 we need a better understanding of the Sobolev spa es than what immediately follows from the denition that we have used this far. We begin by stating a hara terization for the membership in the Sobolev lass and by sket hing its proof.

7.3 Theorem. (Denitions of Sobolev spa es.) Let

p < ∞ , Ω ⊂ Rn .

Then the following are equivalent: 68

u ∈ Lp (Ω), 1 ≤

u˜ ∈ W 1,p (Ω)

1)

(ACL)

2)

(H) There is a sequen e (ϕj )j ⊂ C 1 (Ω) (∇ϕj )j is Cau hy in Lp (Ω).

3)

(W)

There is

For ea h

1≤j≤n

with

u˜ = u

4) There is

so that

vj ∈ Lp (Ω) so Z Z u ∂j ϕ = − vj ϕ

there is

Ω

for ea h

almost everywhere.

ϕj → u

in

Lp (Ω)

and

that

Ω

ϕ ∈ C0∞ (Ω). u˜

g ∈ Lp (Ω) so that u˜ = u gradient of u ˜ in Ω.

and

is an upper

Proof. (sket h) 2) ⇒ 1): Passing to a subsequen e, for almost every x. We dene

almost everywhere in

we may assume that

(ϕj (x))j

Ω

and

g

onverges

u˜(x) = lim ϕj (x) j→∞

u˜(x) = 0 for the remaining x ∈ Ω. in Ω. By the fundamental theorem of

whenever the limit exists, and set, say, Then

u˜(x) = u(x)

almost everywhere

al ulus applied to the fun tions absolute ontinuity in

Ω

ϕj

and the Hölder inequality, one obtains

on the lines for whi h both

Z

I

for ea h ompa t subinterval

|∇u − ∇ϕj |p −→ 0 j→∞

I

in

Ω

and

limj→∞ ϕj (x)

exists for some

x ∈ I.

By the Fubini theorem, this holds for almost all lines parallel to the oordinate axes. It also easily follows that the lassi al partial derivatives of almost everywhere in

Ω

u˜ exist

and that they are obtained as limits of the partial

derivatives of the approximating fun tions.

1) ⇒ 2): We already proved in Chapter 5 that

this manner by Lips hitz fun tions, provided

u an be approximated in Ω = Rn . In this ase, the laim

follows by taking averages, see Lemma 4.22. For the general ase, one uses a P∞ ∞ partition of unity: 0 ≤ ψi ≤ 1, ψi ∈ C0 (Ω) su h that 1 ψi = 1 in Ω and the supports have bounded overlap. Considering uψi , the statement easily

follows.

1) ⇒ 3): Integrate by parts,

vj

is the lassi al partial derivative.

69

3) ⇒ 2): We use the (smooth) onvolution approximation: Let

ψ1 (x) = where

If

C

is hosen so that

p v ∈ Llo ,

set

R

( 0,

C exp

Rn



1 |x|2 −1

ψ1 dx = 1.



|x| ≥ 1

, |x| < 1 ,

Dene

x 1 ψε (x) = n ψ1 . ε ε

ε

v (x) = (ψε ∗ v)(x) =

Z

ψε (x − y)v(y) dy ,

B(x, ε) ⊂⊂ Ω. If v ∈ Lp (Rn ), then v ε → v in Lp (Rn ), see the ε Lemma 4.23. Also v (x) → v(x) when x is a Lebesgue point of u. Fix x ∈ Ω and ε > 0 small ompared to d(x, ∂Ω). Now

when

proof of

uε (x + hei ) − uε (x) h      Z 1 x + hei − y x−y 1 u(y) dy = n ψ1 − ψ1 ε Ωh ε ε {z } |  ∂ψ 1 ∂ψ1 x − y ε = εn (x − y) −→ h→0 ε ∂xi ε ∂xi Z ∂ψε −→ (x − y) u(y) dy h→0 Ω ∂xi by the dominated onvergen e theorem:

Z Z h i 1 1 · · · u(y) dy ≤ k∇ψ1 k∞ |u| dy . ε G G h

Thus

and be ause

∂uε (x) = ∃ ∂xi

ψε

Z

Ω

∂ψε (x − y) u(y) dy ∂xi

is smooth, we see that

∂uε (x) = ∂xi

uε

Z

is

C 1.

Moreover, when

∂ψε (x − y) u(y) dy ∂xi Z ∂ψε (x − y) u(y) dy =− ∂yi Z = ψε (x − y) vi(y) dy . 70

u ∈ W 1,p ,

vi ∈ Lp (Rn ), then this onvolution sequen e onverges to vi When u is given, use a partition of unity to redu e the setting to

If

Lp (Rn ). n that of R . in

2) ⇒ 4): Re all that we have already shown that 2) implies 1).

Pi k a

1

(ϕj )j of C -fun tions in the norm kϕkLp (Ω) + k∇ϕkLp (Ω) so p that ϕj → u and ∇ϕj → ∇u in L (Ω). Then a subsequen e of (ϕj ) onverges to u almost everywhere and we dene u ˜ as the pointwise limit of su h a xed subsequen e. Write E for the set where this subsequen e does not onverge. We set u ˜(x) = 0 when x ∈ E. We may assume that Cau hy sequen e

k∇u − ∇ϕj kLp (Ω) ≤ 2−j . Let

γ

be a re tiable urve. If

lim

j→∞

Z

p

γ

|∇u − ∇ϕj | = 0,

and if further the sequen e

(ϕj (x))j

of

γ

(see (5); in fa t

γ

(ϕj (y))j

Then there is

|∇u − ∇ϕj | = 0,

u˜, |∇u| along γ

then onverges for all

γ

j → ∞.

γ

so that

Z when

j→∞

onverges for some

gradient inequality holds for the pair re tiable urve

lim

then

Z

x ∈ γ, then the upper

and along any sub urve

y ∈ γ ).

Consider then a

|∇u − ∇ϕj | 9 0.

δ > 0 so that Z |∇u − ∇ϕj | ≥ δ

(28)

γ

for innitely many

j.

Now

Z X γ

and

Z

Rn

Dene

j

|∇u − ∇ϕj | ds = ∞

X

|∇u − ∇ϕj |

g = h(x) + |∇u(x)| +

X

p

≤ 1.

|∇u − ∇ϕj | ,

h(x) to be innite if x ∈ E and h(x) = 0 when x ∈ / E. We may assume that g is a Borel fun tion. It now easily follows that g is an upper gradient of u ˜. where we set

71

4) ⇒ 1): This is immediate from the denitions.

p

2

1,p We now easily obtain the important weak ompa tness property of W (Ω), p > 1. Re all that vjk ⇀ v in L (Ω) refers to weak onvergen e, see Subse -

tion 11.3 in the appendix.

(uj )j be bounded in W 1,p (Ω), 1 < p < ∞. Then there p p that ujk ⇀ u in L (Ω) and ∇ujk ⇀ ∇u in L (Ω) for a

7.4 Corollary. Let

u ∈ W

1,p

(Ω) so subsequen e (ujk )k . is

Proof. Both (uj )j and v = (v1 , . . . , vn )

(∇uj )j L (Ω) so

and in

p

ujk ⇀ u

are bounded in

Lp (Ω).

Thus there exist

u

that

∇ujk ⇀ v

and

in

Lp (Ω),

see Subse tion 11.3 in the appendix. Now

Z

Ω

Z

∂i ϕujk = − ↓ ∂i ϕ u

=

Ω

ϕ ∂i ujk Ω

Z ↓ − ϕvi Ω

by the weak onvergen e, when

(v1 , . . . , vn ).

Z

ϕ ∈ C01 (Ω).

Thus

u ∈ W 1,p (Ω)

and

∇u = 2

p = 1. For example, Ω = B 2 (0, 1) and uj (x) = min{1, max{0, jx2 }}, we have that uj ⇀ 2 u = χB+2 (0,1) , where B+ (0, 1) = B 2 (0, 1) ∩ {(x1 , x2 ) : x2 > 0}. Thus the only potential weak limit of a subsequen e of (uj )j is u. Moreover, our sequen e (uj )j is bounded in W 1,1 (Ω) and u ∈ / W 1,1 (Ω). 7.5 Remark. Corollary 7.4 does not extend to the ase

when

Proof of Theorem 7.1. with

η

j

B ⊂ 2B ⊂⊂ Ω.

Then

fj | 3 B is η -quasisymmetri 2

(Corollary 3.4 and Theorem 5.1). It follows from the 3 uniform onvergen e of the mappings fj that f is η -quasisymmetri on B . 2 Be ause B was arbitrary, we on lude from Theorem 5.1 that f is K1 -q in

Ω

independent of

Fix

K1 . It remains to be proven that we B = B(x, r) be as above. For ea h ε > 0

for some Let

may hoose there is

jε

f (B(x, r − ε)) ⊂ fj (B) ⊂ f (B(x, r + ε)) 72

K1 = K.

su h that (29)

j ≥ jε .

for

Indeed, it su es to he k that

B(x, r − ε) ⊂ f −1 (fj (B)) ⊂ B(x, r + ε). The se ond in lusion follows using the uniform onvergen e of our sequen e 3 −1 on f ( B). Regarding the rst in lusion, and the uniform ontinuity of f 2 noti e that, given ε ˜ there is jε˜ so that

|f −1 ◦ fj (y) − y| ≤ ε˜ for all

y

with

|x − y| = r

when

applying Lemma 11.10 to

j ≥ jε˜.

Thus the desired in lusion follows by


1 −1 f ◦ fj (rz + x) − f −1 ◦ fj (x) . r Be ause f is quasi onformal and |∂B| = 0, we on lude from Corollary 4.12 that |f (∂B)| = 0, Thus, it follows from Remark 6.1 and (29) that Z Z j→∞ |Jfj | = |fj (B)| −→ |f (B)| = |Jf | . h(z) =

B

Now

B

Z

B

in

|Dfj | ≤ K

Z

B

|Jfj | ≤ M

M be ause |fj (B)| → |f (B)| < ∞. Moreover, there is M ′ < ∞ ′ that |fj (x)| ≤ M for x ∈ B for all j . Thus the sequen e (fj ) is bounded 1,n n W (B, R ) and so a subsequen e onverges to some g ∈ W 1,n (B, Rn )

for some nite so

n

weakly, i.e.

fjk ⇀ g, Dfjk ⇀ Dg Be ause

Ln (Ω) .

fj → f uniformly on B we on lude that g = f . Thus Z Z Z n n |Dfjk |n |Df | = |Dg| ≤ lim inf k→∞ BZ B B Z |Jfk | = K |Jf | . ≤ K lim inf k→∞

Let

in

x∈Ω

B

B

|Df (x)|n and |Jf (x)|. Then Z n |Df | ≤ K lim − |Jf | = K|Jf (x)| ,

be a Lebesgue point both for

Z |Df (x)| = lim − n

r→0 B(x,r)

r→0 B(x,r)

and the proof is omplete.

2

73

fj : B → fj (B) ⊂ Rn be K -q . Assume that the 1,1 sequen e (fj )j is bounded in W (B; Rn ). Then a subsequen e onverges 1,n lo ally uniformly to a ontinuous mapping f ∈ W (Ω; Rn ). If f is a homemorphism, then f is K -q . 7.6 Corollary. Let

Proof.

Fix

˜ ⊂ 2B ˜ ⊂ B. B

By Remark 6.6

|fj (x) − fj (y)| ≤ C|x − y|

1− n p

|B|

1 −1 p

˜ ⊂ 2B ˜ ⊂ B , and so our sequen e x, y ∈ B ˜ with some η ea h fj is η -quasisymmetri in B

Z

B

|Dfj |

whenever

is equi ontinuous on

Also,

independent of

j,

˜ B.

whi h

together with the estimate

Z

B

implies that

|fj | ≤ M ′

on

˜. B

|fj | ≤ M < ∞

Invoking the Arzela-As oli theorem we may

apply Theorem 7.1 to on lude the laim.

2

8 On 1-quasi onformal mappings As mentioned earlier, ea h onformal mapping is 1-q . Thus there are plenty of 1-q mappings in the plane. However, the stru ture of global 1-q mappings is simple in all dimensions

n ≥ 2.

This also holds for 1-quasi onformal

mappings a

ording to the metri denition, see part 1) of Remarks 5.14.

8.1 Theorem. Let

M >0 for all

so that

x, y ∈ Rn .

f : Rn → Rn

be

1-q , n ≥ 2.

Then there is a onstant

|f (x) − f (y)| = M|x − y|

Theorem 8.1 does not extend to the ase

n = 1 (for the metri denition)

as is seen by onsidering the 1-quasi onformal mapping f : R → R dened by f (x) = x3 . We postpone the proof of Theorem 8.1 for a while and ontinue with a version of the Liouville theorem a

ording to whi h there are very few 1-q mappings in dimensions

8.2 Theorem. Let Then

f

n ≥ 3.

This result is due to Gehring.

Ω, Ω′ ⊂ Rn , n ≥ 3,

be domains and

is the restri tion of a Möbius transformation to

74

f : Ω → Ω′ Ω.

be

1-q .

Re all that a Möbius transformation is a nite omposition of ree tions with respe t to spheres and hyperplanes. The proof of Theorem 8.2 will be based on the usual Liouville theorem whi h assumes

a priori regularity of the mappings in question.

Ω, Ω′ ⊂ Rn be domains, n ≥ 3, and f : Ω → Ω be 1-q , f ∈ C (Ω) and Jf > 0 in Ω. Then f is the restri tion of a Möbius transformation to Ω.

8.3 Theorem. (Liouville) Let ′

3

We omit the proof and refer the reader to [17℄ for a proof.

Proof of Theorem 8.2. We may assume that in

Ω,

see Remark 5.2. Noti e that

f

is lo ally

are Hölder ontinuous with exponent

1

Jf (x) ≥ 0 almost everywhere −1 Lips hitz and so is f (both

by part 2) of Remarks 6.8. Thus

|x − y| ≤ |f (x) − f (y)| ≤ C|x − y| C when

x

y

is xed and

is su iently lose to

it is lo ally bounded.

Consequently

(almost everywhere).

Be ause

almost everywhere with

x; C

may depend on

x

but

Jf

is bounded away from zero lo ally n is 1-q , we have that |Df (x)| = Jf (x)

f Jf (x) > 0.

Fix su h an

x.

We on lude from basi

linear algebra (see Proposition 11.3 and Proposition 11.4 in the appendix) that

for ea h

h ∈ Rn .

|Df (x)h| = Jf (x)1/n |h| Thus

ad Df (x) = Jf (x)1−2/n Df (x)t by Proposition 11.5.

Let

ej

be one of the oordinate ve tors.

Then the

previous equation shows that

ad Df (x)ej = Jf (x)1−2/n ∇fj (x). |∇fj (x)| = |Df (x)t ej | = Jf (x)1/n . We thus on lude from 1 Proposition 11.8 that fj is n-harmoni in Ω, and thus C by Proposition 11.6. Be ause |∇fj (x)| is (lo ally) bounded away from zero, it follows from ∞ Proposition 11.7 that f is C -smooth. The laim thus follows from Theorem Noti e further that

8.3.

8.4 Lemma. Let

f : Rn → Rn

be a homeomorphism so that



f S n−1 (x, r) = S n−1 f (x), Rx,r 75

for all

x ∈ Rn , r > 0 .

Then there is

M >0

so that

|f (x) − f (y)| = M|x − y| for all

x, y ∈ Rn .

Proof.

Let us rst observe that lines get mapped to lines: If

x

z

y

z is the midpoint

f(x) f(z) f(y)

Figure 7: Line segment is mapped to a line segment of

[x, y],

then

f (z)

lies on

[f (x), f (y)]

and furthermore

|f (x) − f (z)| = |f (z) − f (y)|, as we an see from Figure 7. By iterating, we see that for a given line is

ML

so that

|f (x) − f (y)| = ML |x − y|

whenever

L there

x, y ∈ L.

L

ML r

r x

f(x) ML’r

r L’

Figure 8: when Let then

ML = ML′ .

L and L′

L ∩ L′ 6= ∅

L = L′ , then the setting looks like in Figure 8 and thus ML = ML′ .

be lines. Suppose rst that

Otherwise

76

L ∩ L′ 6= ∅.

If

If

L ∩ L′ = ∅,

pi k

L′′

so that

Proof of Theorem 8.1.

for all

x ∈ Rn

and

L ∩ L′′ 6= ∅

and

L′ ∩ L′′ 6= ∅.

It su es to show that



f S n−1 (x, r) = S n−1 f (x), Rx,r

r>0

2

(30)

(Lemma 8.4 will then give the laim). Fix

x

By using translations, rotations and dilations, we may assume that

r = 1, f (e1 ) = e1

and

Set

and r . x = 0,

B(0, 1) ⊂ f (B(0, 1)).

 W = f : Rn → Rn : f

1-q , f (0) = 0, f (e1 ) = e1

is

and

B(0, 1) ⊂ f (B(0, 1))

a = supf ∈W |f (B(0, 1))|. Then a < ∞ be ause ea h su h f is η -qs with a xed η and so f (B(0, 1)) ⊂ B(0, η(1)). We will show that a = |B(0, 1)|. Clearly a ≥ |B(0, 1)|. Suppose a > |B(0, 1)| and pi k a sequen e (fj )j of mappings in W so that |fj (B(0, 1))| → a. Then (fj ) is bounded in W 1,n (2B). Indeed Dene

fj (B(0, 2)) ⊂ B(0, η(1)η(2)), and so

Z

2B

n

|Df | ≤

Z

2B

|Jf | ≤ c0 .

fjk → g uniformly in B(0, 3/2) for some mapping subsequen e (fjk )k . Be ause fjk (0) = 0 and fjk (e1 ) = e1 and

Thus, by Corollary 7.6,

g

and some

ea h

g

fjk

is

η -quasisymmetri ,

it follows from the uniform onvergen e that

is a homeomorphism. Invoking Corollary 7.6 again, we on lude that

g

is

B(0, 1) ⊂ g(B(0, 1)) (and g(B(0, 1)) \ B(0, 1) ontains some non-trivial open set U be ause |g(B(0, 1))| = a > 1. Clearly |g(U)| > 0. Consider h = g ◦ g . Now h ∈ W and
|h(B(0, 1))| = |g g(B(0, 1)) | ≥ |g(B(0, 1)) ∪ g(U)| ≥ a + |g(U)| > a, 1-q . As in the proof of Theorem 7.1, we see that

that

|g(B(0, 1))| = a).

Thus

g ∈ W.

Noti e that

a. We have proven that a = |B(0, 1)|. Returning to our xed mapping f, this shows that |f (B(0, 1))| = |B(0, 1)|. By assumption, B(0, 1) ⊂ f (B(0, 1)), and we on lude that f (B(0, 1)) = B(0, 1). It follows that f (B(x, r)) = B(f (x), Rx,r ) for all x, r. This implies (30). 2 whi h ontradi ts the denition of

77

8.5 Remark. The proof of Theorem 8.1 was based on a ompa tness argument. In fa t, ompa tness an be used to hara terize quasi onformality in the following sense.

T : Rn → Rn similarity if there is a onstant λ > 0 so n that |T (x)−T (y)| = λ|x−y| for all x, y ∈ R . Next, we say that a family F of n n homeomorphisms f : R → R is omplete with respe t to similarities if, for ea h f ∈ F and all similarities T, S, the omposite mapping g = T ◦ f ◦ S also n n belongs to F . We all a homemorphism f : R → R normalized if f (0) = 0 and f (e1 ) = e1 , where e1 is the unit ve tor in the x1 -dire tion. Then the family F is said to satisfy the ompa tness ondition if every innite set of normalized mappings in F ontains a subsequen e whi h onverges lo ally We all a mapping

uniformly to a homeomorphism.

We have the following result: Let a family

Rn → Rn , n ≥ 2,

F

the ompa tness ondition if and only if there is

f ∈F

is

of homeomorphisms

be omplete with respe t to similarities. Then

K -q .

1≤K 0 and then applies the equivalen e of the metri and analyti is

proves that there is and every

denitions.

Hf (x, r). By the ompa tness property it easily follows that there is H < ∞ so that |f (x)| ≤ H for ea h n−1 normalized f ∈ F and all x ∈ S (0, 1). Given f ∈ F , x, and r > 0, pi k n−1 y ∈ S (x, r) that realizes lf (x, r). Map e1 to y and 0 to x using a similarity S, f (x) to 0 and f (y) to e1 using a similarity T, and apply the above bound to g = T ◦ f ◦ S. Here is a sket h of a proof of the estimate on

9 Mapping theorems We begin by dis ussing the planar setting. It is onvenient to use omplex R2 with C and write a point z ∈ C as z = x + iy, 1,1 where x, y are real. Let f ∈ Wlo (Ω; C) be ontinuous, where Ω ⊂ C is a

notation: we identify

f (z) = u(z) + iv(z) with u, v real-valued, we noti e that v have, at almost every z, partial derivatives ux , uy , vx , vy with x, y. Then ∂x f (z) = ux (z) + ivx (z),

domain. Writing both

u

and

respe t to

∂y f (z) = uy (z) + ivy (z). 78

We will employ the derivatives

∂f, ∂f

dened by

1 ∂f (z) = (∂x f (z) − i∂y (f )), 2 1 ∂f (z) = (∂x f (z) + i∂y (f )). 2 Re alling the Cau hy-Riemann equations

ux = vy , uy = −vx , ∂f (z) = 0 if f is analyti . In fa t, for a ontinuous f ∈ Wlo (Ω; C), ∂f (z) = 0 almost everywhere only when f is analyti . iα Let us further denote by ∂α f (z) the derivative of f in the dire tion e (if it happens to exist). In the real notation, this is simply Df (x, y)(cos α, sin α) if f is dierentiable at the point (x, y) and it is easy to he k that, in our we noti e that 1,1

omplex notation,

∂α f (z) = ∂f (z)eiα + ∂f (z)e−iα . In fa t, one has for ea h

(31)

h∈C

Df (z)h = ∂f (z)h + ∂f (z)h, where

∂α f (z)

h

is the omplex onjugate of

h

(for

h = x + iy, h = x − iy ).

Now

has maximal length when the two ve tors in the sum (31) point to

the same dire tion, i.e. when

α + arg ∂f (z) = −α + arg ∂f (z) (modulo

2π ),

and minimal length when these two ve tors point to opposite

dire tions. Here

arg w

denotes the argument of a omplex number

the maximal dire tional derivative has the value

|∂f (z)| + |∂f (z)| and orresponds to the hoi e

1 α = (arg ∂f (z) − arg ∂f (z)) 2 and one has the minimal value

||∂f (z)| − |∂f (z)|| 79

w.

Thus

orresponding to

α=

π 1 + (arg ∂f (z) − arg ∂f (z)). 2 2

Moreover,

|Jf (z)| = |(|∂f (z)| + |∂f (z)|)(|∂f (z)| − |∂f (z)|)| = ||∂f (z)|2 − |∂f (z)|2 |. µ : C → C satisfy ||µ||L∞ < 1. f : C → C so that

9.1 Theorem. Let formal mapping

Then there is a quasi on-

∂f (z) = µ(z)∂f (z) almost everywhere. This is a very strong existen e theorem. Noti e that everywhere be ause 9.1 shows that

almost everywhere.

f

almost

|Df (z)|2 1 + |µ(z)| = |Jf (z)| 1 − |µ(z)| Moreover, for almost every z, Hf (z) =

and the dierential

Jf (z) 6= 0

is quasi onformal. Thus the dis ussion before Theorem

Df (z)

1 + |µ(z)| 1 − |µ(z)|

maps disks

B(z, r)

entered at

z

to ellipses with

major axes of the length

2|Df (z)|r = 2|∂f (z)|r(1 + |µ(z)|) and minor axes of the length

2|∂f (z)|r(1 − |µ(z)|). The orientation of these ellipses is not determined by the olle tion of all ellipses and the minor axis is

Hf (z)

E

with enter

α= Df (z)

so that the ratio of the major

and the angle determined by the minor axis and

the real line is

Then the dierential

x

µ(z). However, onsider

1 arg µ(z). 2

maps these ellipses to dis s entered at

f (z).

We will omit the proof of Theorem 9.1 and refer the reader to [4℄ for the proof and further extensions of this existen e theorem. Let us re all the Riemann mapping theorem, see [23℄ for a proof. 80

9.2 Theorem. (Riemann Mapping Theorem) Ea h simply onne ted domain

Ω(C

is onformally equivalent to the unit disk.

′ It follows that, given simply onne t, proper subdomains Ω, Ω of the ′ plane, there is a onformal mapping f : Ω → Ω . We ontinue with a quasi-

onformal version of this statement.

9.3 Theorem. (Measurable Riemann Mapping Theorem) Let Ω, Ω′

(

C be simply onne ted subdomains and suppose that µ : Ω → C satises kµkL∞ < 1. Then there is a quasi onformal mapping f : Ω → Ω′ so that ∂f (z) = µ(z)∂f (z) In fa t,

Proof.

f

is

1 + kµk∞ 1 − kµk∞

Given

Ω, Ω′

a.e. in

Ω.

-q . and

µ,

we extend

µ

as zero to the rest of

C.

Then

Theorem 9.1 gives us a quasi onformal mapping as asserted, ex ept for the ′ requirement that f (Ω) = Ω . In any ase, f (Ω) is a simply onne ted proper subdomain of

C,

and thus the usual Riemann mapping theorem provides us ′ with a onformal mapping g : f (Ω) → Ω . Setting f˜ = g ◦ f, it is easy to

he k using the  hain rules

∂(g ◦ h) = ∂g(h)∂h + ∂g(h)∂h, that

f˜ has

∂(g ◦ h) = ∂g(h)∂h + ∂g(h)∂h, all the required properties.

2

We dedu e from Theorem 8.2 that there is no Riemann mapping theorem in higher dimensions.

9.4 Corollary. Let ball or a half spa e.

f : B n → f (B n ) ⊂ Rn

be

1-q , n ≥ 3.

Then

f (B n )

is a

One ould still hope for a quasi onformal Riemann mapping theorem for

n ≥ 3.

Unfortunately, this hope is futile:

Ω ⊂ R3 be as f : B (0, 1) → Ω.

9.5 Example. Let mal mapping

in Figure 9. Then there is no quasi onfor-

3

81

g(t)=t2

Ft E −1

x1

0

Ω

Figure 9: Domain

Ω

of the example 9.5

Reason : Suppose there is a quasi onformal mapping f : B 3 (0, 1) → Ω. Pi k 2 a ir le Ft of radius 2t around the usp at the level x1 = t and let E = [−1, 0] on x1 -axis. Then (see Figure 10) 
2
 2 capn (E, Ft ; Ω) ≤ capn B (t, 0, 0), 2t , S (t, 0, 0) , t ; Ω ω2 = 2 −→ 0 when t → 0 . t log 2t2

Ft 0

Figure 10:

is

K -q

B (t, 0, 0) , 2t2

2t 2




and

x1


S 2 (t, 0, 0) , t .

K , it follows that   cap3 f −1 (E), f −1 (Ft ); B 3 (0, 1) ≤ K cap3 (E, Ft ; Ω) −→ 0 Be ause

f

t

t

for some

82

when

t → 0.

But, on the other hand,

min

n

for all t, and thus

diam f −1 (E), diam f −1 (Ft )
d f −1 (E), f −1 (Ft )

o

≥ 10−6

  cap3 f −1 (E), f −1 (Ft ); B 3 (0, 1) ≥ δ(3, 10−6) > 0.

To be pre ise, we have heated a bit above. Indeed, E interse ts the boundary −1 of Ω and thus it is not lear if f (E) is ompa t (nor even if f −1 has an extension to the points

Ej ⊂ E

−1, 0).

It is easy to x this by repla ing

E

with

whi h is the segment [−1 + 1/j, −1/j] with j su iently large. −1 Noti e that f (y) ne essarily tends to the boundary of B 3 (0, 1) when y

∂Ω.

tends to

By the above example, not every topologi ally ni e

Ω ⊂ Rn , n ≥ 3,

is

quasi onformally equivalent to the unit ball. One does not in fa t know any general geometri riteria for this equivalen e. The following result due to Gehring gives a su ient ondition for quasi onformal equivalen e.

For a

proof see [30℄.

∂Ω is dieomorphi n mapping f : B (0, 1) → Ω.

9.6 Theorem. If

onformal

to

S n−1 (0, 1),

then there is a quasi-

Based on Corollary 9.4 it is natural to ask if domains in are

K -q

equivalent to the unit ball for a suitably small

K

Rn , n ≥ 3,

that

are more regular

than one a priori expe ts. This turns out to be true in the sense that they are even quasisymmetri ally equivalent to the unit ball.

n ≥ 3. There exists K0 = K0 (n) > 1 su h that if f : B → f (B ) ⊂ Rn is K -q , 1 ≤ K < K0 and bounded, then f is n quasisymmetri . In parti ular, f extends to a homeomorphism f˜ : B → f (B n ).

9.7 Theorem. Let n

n

This theorem is from [3℄, [27℄. The proof heavily relies on results due to Reshetnyak [23℄ that essentially give an asymptoti version of Theorem 8.2 when the distortion

K

tends to 1.

9.8 Remark. There are still plenty of quasi onformal mappings. For example, there is a quasi onformal mapping

|∂Ω| = ∞.

See [31℄ for this.

83

f : B n → f (B n ) ⊂ Rn

so that

10 Examples of quasi onformal mappings 10.1 Example. (Basi mappings)

Linear transformations:

1)

If

quasi onformal.

Radial stret hings:

2)

f

is

K -q ,

Let

where

f : Rn → Rn

is linear and invertible, then

f (x) = x|x|a−1 =

( an−1 K= a−1

if if

x |x|a , where |x|

0 < a < ∞.

f

is

Then

a≥1 0 < a < 1.

In the planar setting, it is easy to establish this estimate on K by using f : C → C, f (z) = z|z|a−1 = z (1+a)/2 z (a−1)/2 .

omplex notation. Indeed, let Then

1

1

1

1

∂f (z) = 21 (a − 1)z 2 (1+a) z 2 (a−3)

∂f (z) = 12 (a + 1)z 2 (a−1) z 2 (a−1) , so

µ(z) = Thus

∂f a−1 z = . ∂f a+1 z

|µ(z)| = |a − 1|/(a + 1), and the

desired estimate follows by the dis us-

sion in the beginning of Chapter 9.

The higher dimensional setting requires a bit more thinking. We leave

f maps balls entered x 6= 0. The matrix Df (x) is

this to the reader with the following hints. First of all, at the origin to balls entered at the origin. Let

x lies on the x1 -axis and the required estimate then easily follows. Also, the image of B(x, r) in this ase is approximatively determined by the image ellipsoid of B(x, r) under the linear transformation orresponding to Df (x). Next, given x 6= 0, the image of B(x, r) under f is, modulo a rotation, the image of B(z, r), where z lies on the x1 -axis and satises |z| = |x|, and f is dierentiable at x. Combining this with the approximation from diagonal when

above gives the laim.

Folding maps:

3) n

R

,

n ≥ 2;

this

(r, ϕ, z) be the ylindri al oordinates of x = (x1 , . . . , xn ) ∈ n−2 means that r > 0, 0 ≤ ϕ < 2π , z ∈ R , and Let

x1 = r cos ϕ , x2 = r sin ϕ 0 < α, β ≤ 2π , 0 < ϕ < β}. Then K -q , where Let

and

z = (x3 , . . . , xn ) .

Ωα = {(r, ϕ, z) : 0 < ϕ < α}, Ωβ = {(r, ϕ, z) : mapping f : Ωα → Ωβ , (r, ϕ, z) → (r, (β/α)ϕ, z) is

and let the

( (β/α)n−1 K= α/β 84

for for

α≤β α>β.

The estimate on

K

is obtained using the diagonal representation of

Df (x)

obtained using suitable orthonormal oordinates.

Ωα

Ωβ

f β

α Figure 11: Folding map

4)

Cone map:

f : Ωα → Ωβ

(R, ϕ, θ) be the spheri al oordinates R > 0, 0 ≤ ϕ < 2π , 0 ≤ θ ≤ π , and

Let

this means that

x1 = R sin θ cos ϕ x2 = R sin θ sin ϕ

and

of

(x1 , x2 , x3 ) ∈ R3 ;

x3 = R cos θ .

0 < α ≤ π the domain Cα = {(R, ϕ, θ) : 0 ≤ θ < α} is alled a one of angle α. The mapping f : Cα → Cβ , (R, ϕ, θ) → (R, ϕ, βθ/α) (see Fig. 12 for the spe ial ase where β = π/2), is K -q for 0 < α ≤ β < π , where

For

K= For

β =π

β 2 sin α . α2 sin β

the quasi onformality fails. Use similar oordinates as for 3) to

verify the laim.

f

g r= π/2

α

β=π/2 Cβ = H

Figure 12: Maps

5)

f : Cα → Cπ/2

Cone to an innite ylinder:

H = Cπ/2 .

Let

C∞

C

Let

H

g : H → C∞ .

be the half-spa e determined by

be the innite ylinder 85

and

8

Cα

C∞ = {(r, ϕ, x3 ) : r ≤ π/2}

g : H → C∞ , whi h maps the point (R, ϕ, θ) ∈ H (spheri al) to (r = θ, ϕ, x3 = log R) ∈ C∞ ( ylindri al), is π 2 /4-q ; see Figure 12. Espe ially, for ea h one Cα of angle 0 < α < π, there is a quasi onformal mapping h : Cα → C∞ . (in ylindri al oordinates).

Then

n ≥ 2 and 0 < λ < n, 0 < λ′ < n, there is a K -q map f : R → R and Cantor sets E, E ′ of Hausdor ′ ′ ′ dimensions λ, λ , respe tively, so that f (E) = E . Here K depends on n, λ, λ . 10.2 Example. (Dust to dust) Given n

n

Reason : Let I = [0, 1]n ⊂ Rn and Ii , i = 1, . . . , 2n be the dyadi sub ubes 1 1 n of I with side length . Fix 0 < s < and for ea h i = 1, . . . 2 pi k a 2 2 similarity mapping gi : I 7→ Ii : x 7→ sx + ai , where ai ∈ Ii is hosen so that the enters of Ii and Qi = gi (I) oin ide. Let [ Fj = (32) gi1 ◦ gi2 ◦ . . . ◦ gij (I) . 1≤i1 ,i2 ...,ij ≤2n

F1 ⊃ F2 ⊃ . . . . Moreover, the ubes gi1 ◦ . . . ◦ gij (I) ′ and gi′ ◦ . . . ◦ gi′ (I) are disjoint if ik 6= ik for some 1 ≤ k ≤ j . We dene a 1 j n Cantor set Cs by setting ∞ \ Csn = Fj . (33) It is easy to see that

j=1

log 12 n Then the Hausdor dimension of Cs is n , see [20℄. log s 1 1 ′ and 0 < s < and the orresponding Cantor onstru tions Fix 0 < s < 2 2 n n as above. It is easy to see that there exists a K -quasi onformal f1 : R → R −1 ′ so that f1 (x) = x outside I , and f1 (x) = gi ◦ gi (x) if x ∈ Qi , where K 1 −s′ 1 1 n depends basi ally only on the ratio 21 . For example, dene ψ : [− , ] → 4 4 −s 2 [− 14 , 14 ]n by

ψ(x) =

( s′

and nally set for

s

x

x,

when

1 −s′ 2 1 −s 2

x ∈ Ii

+

s and 2 s ≤ when 2

0 ≤ q||x||max ≤
s′ −s

4||x||max( 12 −s)

||x||max ≤ 14 ,

that

f1 (x) = ψ(x − bi ) + bi , where dene

(34)

bi denotes the enter of Ii (whi h also is the enter of Qi ). On I c , we f1 to be identity. It is an easy exer ise to he k that f1 satises the

desired properties (see Figure 13 in the two-dimensional ase).

86

Figure 13: The initial map

fj

We dene a sequen e of fun tions dened, we dene the mapping

fj+1

f1

indu tively:

by setting

assuming that

fj+1 (x) = fj (x)

fj

outside

is

Fj

and

fj+1 (x) = gi′1 ◦ fj ◦ gi−1 , 1

if

x ∈ gi1 ◦ · · · ◦ gij (I)

(35)

x ∈ Fj . It is easy to he k that fj is a homemorphism that maps Fj Fj′ . Moreover, be ause ea h gi and gi−1 is 1-q , ea h fj is K -q with the

onstant K orresponding to the onstru tion of f1 above. It is immediate from the onstru tion that the sequen e (fj )j of K -q s s′ maps onverges uniformly to a homemorphism f that maps Cn onto Cn . From Theorem 7.1 we dedu e that f is K -q . when

onto

f : Rn+ → Rn+ be a K -q map that maps bounded sets to bounded sets. Then f is quasisymmetri and thus f extends n n to a (quasisymmetri ) homeomorphism f˜ : R+ → R+ . Dene   f (x) if xn > 0 ˆ f(x) = f˜(x) if xn = 0   f (x) if xn < 0 , 10.3 Example. (Ree tion) Let

where

x = (x1 , x2 , . . . , −xn ).

Reason :

Then

ˆ : Rn → Rn f(x)

is

K -q .

Repeat the argument we used to prove that the analyti denition

implies the metri denition (Theorem 5.1) to see that (see Figure 14). For the K -quasi onformality of 1,n n fˆ ∈ Wlo . For ea h bounded G ⊂ R we have

Z

G+

n

|Df | ≤ K

Z

87

G+

fˆ it

|Jf | < ∞,

f

is quasisymmetri

su es to he k that

8

8 z y

f(z) f(y) r

x

f(x) xn

xn Figure 14:

be ause

f

f : Rn+ → Rn+

maps bounded sets to bounded sets. Similarly,

Thus we only need to he k that

Z

∂i fj ϕ = −

This is trivial when

Z

fj ∂i ϕ

for all

i = 1, . . . , n − 1;

R

G−

ˆ n < ∞. |D f|

ϕ ∈ C0∞ (Rn ).

almost every line parallel to the rst

n − 1 oordinate axes lies either in the upper half spa e or in the lower one. For i = n, integrate by parts along lines up to boundary in both sides; the boundary term showing up gets an elled be ause f is ontinuous. 10.4 Example. (Lifting) Let Then there is a quasi onformal

Reason :

f : Rn → Rn be quasisymmetri , n ≥ 1. n+1 → Rn+1 so that fˆ|Rn = f . mapping fˆ : R

n = 1 dene  Z 1  Z 1 1 fˆ(x, y) = 2 f (x + ty) + f (x − ty) dt, f (x + ty) − f (x − ty) dt For

0

for

y > 0

0

and use ree tion (Example 10.3).

This is the Beurling-Ahlfors

extension. The high dimensional ase is hard essentially be ause of topologi al dif ulties. The setting

n ≥ 4

n=2

is due to Ahlfors [1℄,

to Tukia and Väisälä [29℄.

ould simply assume that

f

n=3

to Carleson [9℄ and

Noti e that, in dimensions

be quasi onformal. For

n=1

n ≥ 2,

we

one really needs

to assume quasisymmetry be ause there exist quasi onformal mappings of the real line that fail to be quasisymmetri .

f : R → f (R) ⊂ R2 be 2 2 mapping fˆ : R → R so

10.5 Example. (Generalized lifting) Let

qua-

sisymmetri . Then there is quasi onformal

that

fˆ|R = f .

88

h

h(z) z

Ω1

y f

h(x)

x Ω Figure 15: Conformal

Reason :

h(y)

2

h : R2+ → Ω1

(See Figure 15.) One an show using the fa t that

quasisymmetri that

Ω1

f : R → ∂Ω1

is

is LLC. Then the Riemann mapping theorem gives 2 us a onformal mapping h : R+ → Ω1 and g is quasisymmetri by the usual arguments (we may assume that h maps bounded sets to bounded sets; see the proof of Theorem 5.1). We an then extend h to a quasisymmetri ˜ : R2+ → Ω1 . Now h ˜ −1 ◦ f : R → R is also quasisymmetri . By mapping h 2 2 ˜ −1 ◦ f . lifting, there is a quasi onformal mapping g : R → R so that g|R = h ˜ ◦ g : R2+ → Ω1 is quasisymmetri and f1 |R = f . Repeat the Then f1 = h same pro edure to obtain a quasisymmetri mapping

f2 |R = f ,

and dene

fˆ in

pie es.

f2 : R2− → Ω2

so that

f : Rn → f (Rn ) ⊂ f˜ : Rn+1 → Rn+1 , when

10.6 Remark. There are quasisymmetri mappings

Rn+1 that n ≥ 2.

do not extend to a homeomorphism

ˆ is a quasi ir le if there is a γ⊂C 1 that γ = f (S ) or γ \ {∞} = f (R).

10.7 Denition. A Jordan urve

onformal mapping Above, of

C.)

ˆ C

f :C→C

so

refers to the Riemann sphere (the one-point ompa ti ation

One an he k that ea h quasi onformal mapping

to a homeomorphism the Riemann sphere.

ˆ → C; ˆ fˆ : C

γ

is a quasi ir le

(2) one of the omponents of

f :C→C

extends

this extension is also quasi onformal on

10.8 Remark. The following are equivalent: (1)

quasi-

C\γ

is LLC

(3) both omponents are LLC

89

(4) If

z, w, y ∈ γ

and

y

is between

z

and

w,

then

|z − y| + |w − y| ≤ C|z − w| with

C>0

independent of

z, w

and

y. γ

y

w

z 10.9 Example. (The snowake mapping) Take pie ewise linear mappings

fk : [0, 1] → C

as in Figure 16. Then extend the onstru tion to entire

R

as

f1

0

1

0

1

f2

0

1

0

1

Figure 16: First iterations of the snowake map in Figure 17 to obtain pie ewise linear mappings

fˆk : R → C.

The mappings

f

2

0

1

3

0

Figure 17: Extension of

fˆk

1

f2

3 to

R

are uniformly quasisymmetri . By Arzela-As oli, we obtain a quasisym-

metri mapping snowake urve.

f : R → γ ⊂ C, where γ The mapping f satises the

is a version of the von Ko h estimate

log 3 log 3 1 |x − y| log 4 ≤ |f (x) − f (y)| ≤ C |x − y| log 4 C

for

x, y ∈ [0, 1].

Eventually, take a quasi onformal extension 90

f˜ : C → C

of

f.

10.10 Remarks. 1) One an hange the onstru tion so that, for a given 1 2

< α ≤ 1,

there is

fα

so that

1 |x − y|α ≤ |f (x) − f (y)| ≤ C |x − y|α C for

x, y ∈ R.

2) In higher dimensions, similar onstru tions have been made by David and Toro for

α

lose to

1

[10℄.

Referen es [1℄ Ahlfors, L.:

Extension of quasi onformal mappings from two to three

[2℄ Astala, K.:

Area distortion of quasi onformal mappings. A ta Math.

dimensions. Pro . Nat. A ad. S i. U.S.A. 51 (1964) 768771. 173 (1994), no. 1, 3760.

[3℄ Astala, K. and Heinonen J.:

On quasi onformal rigidity in spa e and

plane. Ann. A ad. S i. Fenn. Ser. A I Math. 13 (1988), no. 1, 8192.

Ellipti Partial Dierential Equations and Quasi onformal Mappings in Plane. Prin eton Univer-

[4℄ Astala, K., Iwanie , T. and Martin, G.: sity Press, 2009.

[5℄ Balogh, Z.M., Koskela, P. and Rogovin, S.:

quasi onformal mappings on urves.

Absolute ontinuity of

Geom. Fun t. Anal. 17 (2007),

no. 1, 645664.

The boundary orresponden e under quasi onformal mappings. A ta Math. 96 (1956), 125142.

[6℄ Beurling, A. and Ahlfors, L.:

[7℄ Bojarski,

B.:

Remarks on Sobolev imbedding inequalities.

Com-

plex Analysis, Joensuu 1987, 5268. Le ture Notes in Math., 1351, Springer, Berlin, 1988.

Analyti al foundations of the theory of quasi onformal mappings in Rn . Ann. A ad. S i. Fenn. Ser. A I Math.

[8℄ Bojarski, B. and Iwanie , T.: 8 (1983), no. 2, 257324. [9℄ Carleson, L.:

The extension problem for quasi onformal mappings.

Contributions to analysis (a olle tion of papers dedi ated to Lipman Bers), pp. 3947. A ademi Press, New York, 1974.

91

[10℄ David, G. and Toro, T.:

Reifenberg at metri spa es, snowballs, and

embeddings. Math. Ann. 315 (1999), no. 4, 641710.

[11℄ Gehring, F.W.:

Symmetrization of rings in spa e. Trans. Amer. Math.

So . 101 (1961) 499519.

The Lp -integrability of the partial derivatives of a quasi onformal mapping. A ta Math. 130 (1973), 265277.

[12℄ Gehring, F.W.:

Approximation in Sobolev spa es of nonlinear expressions involving the gradient. Ark. Mat. 40 (2002), no. 2,

[13℄ Hajªasz, P. and Malý, J.: 245274.

Quasi onformal maps in metri spa es with ontrolled geometry. A ta Math. 181 (1998), no. 1, 161.

[14℄ Heinonen, J. and Koskela, P.:

[15℄ Iwanie , T.:

The failure of lower semi ontinuity for the linear dilata-

tion. Bull. London Math. So . 30 (1998), no. 1, 5561.

[16℄ Iwanie , T.:

Nonlinear Cau hy-Riemann operators in Rn .

Trans.

Amer. Math. So . 354 (2002), no.5, (2002), 19611995. [17℄ Iwanie , T. and Martin, G.:

analysis.

Geometri fun tion theory and non-linear

Oxford Mathemati al Monographs. The Clarendon Press,

Oxford University Press, New York, 2001.

Ex eptional sets for the denition of quasi onformal mappings in metri spa es. Int. Math. Res. Noti es 16

[18℄ Koskela, P. and Wildri k, K.: (2008).

[19℄ Lehto, O., Virtanen, K.I. and Väisälä, J.:

Contributions to the distor-

tion theory of quasi onformal mappings. Ann. A ad. S i. Fenn. Ser. A I No. 273 (1959), 14 pp.

Geometry of sets and measures in Eu lidean spa es. Fra tals and re tiability. Cambridge Studies in Advan ed Mathemati s,

[20℄ Mattila, P.:

44. Cambridge University Press, Cambridge, 1995. [21℄ Mu kenhoupt, B.:

Weighted norm inequalities for the Hardy maximal

fun tion. Trans. Amer. Math. So . 165 (1972), 207226.

[22℄ Reshetnyak, Yu. G.:

Stability theorems in geometry and analysis.

Mathemati s and its Appli ations, 304. Kluwer A ademi Publishers Group, Dordre ht, 1994.

92

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Quasiregular mappings. Ergebnisse der Mathematik und

ihrer Grenzgebiete (3) [Results in Mathemati s and Related Areas (3)℄, 26. Springer-Verlag, Berlin, 1993. [24℄ Rudin, W.:

Real and omplex analysis.

Third edition. M Graw-Hill

Book Co., New York, 1987. [25℄ Staples, S.:

Maximal fun tions, A∞ -measures and quasi onformal

maps. Pro . Amer. Math. So . 113 (1991), no. 3, 689700.

Harmoni analysis: real-variable methods, orthogonality, and os illatory integrals. With the assistan e of Timothy S. Murphy.

[26℄ Stein, E.M.:

Prin eton Mathemati al Series, 43. Monographs in Harmoni Analysis, III. Prin eton University Press, Prin eton, NJ, 1993.

Continuation of spatial quasi onformal mappings that are lose to onformal. Sibirsk. Mat. Zh. 28 (1987), no. 6, 126

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Quasisymmetri embeddings of metri

spa es. Ann. A ad. S i. Fenn. Ser. A I Math. 5 (1980), no. 1, 97114.

[29℄ Tukia, S. and Väisälä, J.:

n

Quasi onformal extension from dimension

to n + 1. Ann. of Math. (2) 115 (1982), no. 2, 331348.

[30℄ Väisälä, J.:

Le tures on n-dimensional quasi onformal mappings. Le -

ture Notes in Mathemati s, Springer-Verlag, Berlin, 1971. [31℄ Väisälä, J.:

Quasi onformal maps and positive boundary measure.

Analysis 9 (1989), no. 1-2, 205216.

11 Appendix 11.1 Conformal mappings of a square onto a re tangle Let us explain why one annot map a square onformally to a re tangle whi h is not a square, so that the verti es get mapped to the verti es. Noti e that this statement is a bit ambiguous. Indeed, the onformal mapping is a priori only dened in the open square and thus the meaning of verti es being mapped to verti es is not lear.

Be ause of the simple geometry of

both of these domains, one an easily verify that any onformal mapping ne essarily extends to a homeomorphism of the losed square onto the losed re tangle. Thus, we are laiming that there is no homeomorphism between

93

a losed square and and a losed, non-square re tangle whi h is onformal in the open square and maps the sides of the square to the sides of the re tangle.

Q

Let us all the square

R

and the re tangle

f. By that Q =

and the mapping

translating and s aling, we may assume without loss of generality

]0, 1[×]0, 1[ and that R =]0, 1[×]0, L[ for some L > 0. We may further assume that the verti al sides of Q get mapped to the verti al sides of R. Consider the line segment Iy = {(t, y) : 0 ≤ t ≤ 1} for 0 < y < 1. Be ause f (Iy ) joins the verti al sides of R, we on lude that Z |Df (x, y)| dx ≥ L. Iy

By Hölder's inequality we dedu e that

2

L ≤ Integrating with respe t to

y

Z

Iy

|Df (x, y)|2 dx.

and using the inequality

|Df (x, y)|2 ≤ Jf (x, y) that follows from the Cau hy-Riemann equations ( f. Se tion 1) we arrive at

2

L ≤ where

|R|

Z

Q

2

|Df (x, y)| dx dy ≤

is the area of

R.

Sin e

Z

Q

Jf (x, y) dx dy ≤ |R|,

|R| = L,

we on lude that

opposite inequality follows by reversing the roles of

Q

and

R

L ≤ 1.

The

in the above

argument.

11.2 Some linear algebra Let

A

be a

n × n-matrix.

The operator norm of

A

is dened by

|A| = sup |Ah|, |h|=1

|h|, |Ah| are the eu lidean lengths of the given ve tors. uses the norm sX a2ij ||A||HS =

where also

Sometimes one

i,j

alled the Hilbert-S hmidt norm. These two norms are learly equivalent.

94

11.1 Proposition. For ea h

h ∈ Rn ,

we have that

|hA| ≤ |A||h|. To see that this estimate holds, noti e rst that

|hA| = |At ht |.

t It thus su es to show that |A | h ∈ Rn so that |At h| = |A|. Now

≤ |A|.

To this end, hoose a unit ve tor

|At |2 =< At h, At h >=< AAt h, h >≤ |A||At h||h| ≤ |A||At |,

and the laim follows. Noti e that we a ually proved above that t t (A ) = A, we arrive at the equality

|At | ≤ |A|.

Re alling that

|At | = |A|.

(36)

We ontinue with a result a

ording to whi h linear mappings an always be represented by diagonal matri es.

L : Rn → Rn be a linear mapping. Then there are orthonormal bases {v1 , . . . , vn } and {w1 , . . . , wn } so that the matrix of L with 11.2 Proposition. Let

respe t to these bases is diagonal.

Proof.

Pi k

v1 ,

with

|v1 | = 1,

so that

|Lv1 | = sup|h|=1 |Lh|.

then the zero matrix will do (use then the standard bases). Lv1 w1 = |Lv . Assume for simpli ity that |Lv1 | = 1. 1| If v⊥v1 , then Lv⊥Lv1 .

If

|Lv1 | = 0,

We will take

Claim:

|v| = 1 w⊥Lv1 .

Suppose not. We an assume that

Lv = cLv1 + w ε > 0. Now

when

ε > 0

for some

c > 0,

where

hLv, Lv1 i > 0. Then |L(v1 + εv)| ≥ 1 + cε for

and that Thus

|L(v1 + εv)| 1 + cε 1 + 2cε + c2 ε2 ≥√ > 1, = |v1 + εv| 1 + ε2 1 + ε2 is small, whi h ontradi ts the fa t

This proves the above laim. Then pi k

v2

with

|v2 | = 1

1 = |Lv1 | = sup|h|=1 |Lh|.

and so that

|Lv2 | =

sup |h|=1,h⊥v1

|Lh|.

Repeat the argument as in the rst step. After

n

steps, we have found the

required basis.

2

We dedu e the following familiar property of linear transformations. 95

11.3 Proposition. If balls to ellipsoids.

Proof.

L : Rn → Rn

By the linearity of

L,

L

maps

L(B (0, 1)) is an

ellip-

is linear and one-to-one, then

it su es to show that

n

soid. Relying on the pre eding proposition, we may assume that the matrix

orresponding to

L is diagonal.

Sin e

L is one-to-one, the

diagonal entries of

this matrix are non-zero. The laim follows.

2

Let us re all the standard fa t that, under a linear transformation, the measure of the image of a set

E

E

is obtained by multiplying the measure of

by the absolute value of the determinant of the matrix representing the

linear transformation.

11.4 Proposition. We have

|AE| = | det A||E| for ea h measurable set

E.

Noti e that our laim is trivial when

A

is diagonal. Thus the previous

proposition essentially gives our laim. The only problem is that one would need the fa t that the determinant does not depend on the hoi e of the orthonormal bases involved. A rigorous elementary proof of our laim an be found in [24℄. To ea h by setting

n × n-

matrix

A

we asso iate the adjun t matrix

ad A,

dened

(ad A)ji = det A′ij , where

(ad A)ji

refers to the entry of

the matrix obtained from

A

ad A

i

at row

and olumn

by repla ing the entry at row

i

j

and

A′ij

and olumn

is

j

with 1 and all the other entries in the orresponding row and olumn by 0. −1 If A is invertible, then ad A = A det A, and, more generally,

A ad A = I det A, where

I

(37)

is the identity matrix.

11.5 Proposition. Let

λ>0

and suppose that

A

|Ah| = λ|h| for all

h ∈ Rn .

Then

ad A = (det A)1−2/n At . 96

satises

Proof.

Clearly

11.4 shows that

|AE| = λn |E|

for ea h measurable set.

Thus Proposition

λ = (det A)1/n . we

B = λ1 A.

|Bh| = |h| t

onlude that also |B | = 1. Fix h with |h| = 1. Then

Write

Then

for all

(38)

h ∈ Rn ,

and so

|B| = 1.

From (36)

1 =< Bh, Bh >=< B t Bh, h >, and be ause

|B t Bh| ≤ |B t ||B||h| = 1,

we on lude that

B t Bh = h.

It follows that

B −1 = B t .

Now

At = λB t = λB −1 = λ2 A−1 .

(39)

Combining (39) with (37) and (38) we on lude that

ad A = A−1 det A = (det A)1−2/n At , as desired.

11.3

2

Lp-spa es Lp (Ω), 1 ≤ p < ∞, onsists of (equivalen e lasses) of measurable u with Z |u|p < ∞.

Re all that fun tions

Ω

We write

||u||Lp = ||u||p := Furthermore,

L∞ (Ω)

Z

Ω

|u|

p

1/p

Ω that are ||u||L∞ = ||u||∞ is the essential supremum of |u| ′ ′ set p = p/(p − 1), and we dene 1 = ∞. With this

onsists of those measurable fun tions on

essentially bounded. Then over

Ω.

If

.

1 < p < ∞, we

notation, we have the Minkowski

||u + v||p ≤ ||u||p + ||v||p and Hölder

||uv||1 ≤ ||u||p||v||p′ 97

inequalities. One often needs the following spheri al oordinates. Given a Borel fun u ∈ L1 (B n (0, 1)) we have that

tion

Z

u=

B(0,1)

Z

S n−1 (0,1)

Z

1

u(tw)tn−1 dtdw.

0

p We say that a sequen e (ui )i onverges to u in L (Ω) if all these fun tions p belong to L (Ω) and if ||u − ui ||p → 0 when i → ∞. We then write ui → u in p L (Ω). If ui → u in Lp (Ω), then there is a subsequen e (uii )k of (ui )i whi h pointwise almost everywhere. For 1 ≤ p < ∞, ontinuous p p fun tions are dense in L (Ω) : given u ∈ L (Ω) one an nd ontinuous ui p with ui → u both in L (Ω) and almost everywhere. This an easily seen by

onverges to

u

rst approximating

u by simple fun tions, then approximating the asso iated

measurable sets by ompa t sets and nally approximating the hara teristi fun tions of the ompa t sets by ontinuous fun tions. p p/(p−1) (Ω) when 1 < p < ∞. Then The dual of L (Ω) is L

||u||p =

sup ||ϕ||

p =1 p−1

||uϕ||1.

One of the inequalities easily follows by Hölder's inequality and the other by ϕ to be a suitable onstant multiple of |u|p−1.

hoosing

We also need the following weak ompa tness property: if (uj )j is a p bounded sequen e in L (Ω), 1 < p < ∞, then there is a subsequen e (ujk )k p and a fun tion u ∈ L (Ω) so that

lim

k→∞ for ea h

ϕ ∈ Lp/(p−1) (Ω).

Z

ujk ϕ =

Ω

Z

uϕ Ω

We then write

uuk ⇀ u. This notation should in prin iple in lude the exponent

p,

but the exponent

in question is typi ally only indi ated when its value is not obvious. fun tion

u,

This

alled the weak limit, is unique and satises

||u||p ≤ lim inf ||ujk ||p . k→∞

The existen e of the weak limit

p < ∞,

u

follows from the fa t that

is reexive. Furthermore, the norm estimate on

u

Lp (Ω), 1
dx = 0

ϕ ∈ C0∞ (Ω).

11.6 Proposition. Ea h fun tion

C 1,α -smooth,

where

u, p-harmoni

fun tion in

Ω,

is (lo ally)

α = α(p, n).

Noti e that when p = 2, our p-harmoni fun tion is in fa t harmoni and C ∞ -smooth. In general, for p 6= 2, this is not true. The di ulty lies in

then

the fa t that our equation is the degenerate. This is in fa t the only obsta le as the the next proposition asserts.

11.7 Proposition. Let (lo ally). Then

u

is

C

∞

u

be

p-harmoni

-smooth. 99

in

Ω

and

C1

with

|∇u(x)| > 0

Proofs for the regularity results above an be found for example in the paper Regularity of the derivatives of solutions to ertain degenerate ellipti equations by J.L.Lewis in Indiana Math. J. 32 (1983), pp. 849858. In the planar setting, the oordinate fun tions of a onformal mapping are harmoni . In higher dimensions, they turn out to be

n-harmoni .

This

is based on the following result.

11.8 Proposition. Let is a domain. Let

ej

1,n−1 f ∈ Wloc (Ω, Rn )

and

ϕ ∈ C0∞ (Ω),

where

be one of the oordinate ve tors. Then

Z

Ω

Ω ⊂ Rn

< ad Df (x)ej , ∇ϕ(x) > dx = 0.

C 2 -smooth, the laim follows from a dire t omputation using 2 the denition of ad Df (x) and the fa t that, for a C -smooth fun tion u, 1,n−1 ∂j ∂j u(x) = ∂i ∂j u(x). To relax the regularity assumption to f ∈ Wloc (Ω, Rn ), approximate f by smooth mappings and observe that the entries of ad Df (x) ane (n−1)-fold produ ts of the partial derivatives of the oordinate mappings of f. If

f

is

11.5 Fixed point theorem and related results The following result is the Brouwer xed point theorem.

G : B(0, 1) → B(0, 1) is ontinuous, then there is at least x ∈ B(0, 1) (i.e. G(x) = x).

11.9 Theorem. If one xed point

Noti e that, in dimension one, the laim easily follows from the mean value theorem. The higher dimensional version an be rather easily redu ed to the Hairly Ball Theorem a

ording to to whi h an even dimensional sphere does not admit any ontinuous eld of non-zero tangent ve tors. This redu tion and a surprisingly simple analyti proof of this lassi al topologi al result an be found in the paper Analyti proofs of the `hairy ball theorem' and the Brouwer xed point theorem, by John Milnor in the Ameri an Mathemati al Montly, Vol. 85, No. 7, pp. 521524. We employ the xed point theorem to prove the following result that

ould also be established using degree theory.

11.10 Lemma. Let

Then

n

h : B (0, 1) → Rn

|h(x) − x| ≤ ε
B(0, 1 − ε) ⊂ h B(0, 1) .

100

be a ontinuous mapping satisfying when

|x| = 1.

Proof.

Assume that there is

F (x) =

( h(2x)

2|x| − 1


x

|x|

x0 ∈ B(0, 1 − ε) \ h(B(0, 1))


x + 2 1 − |x| h |x|

when when

and dene

|x| < 1 2

1 2

≤ |x| ≤ 1 .



F B(0, 1/2) = h B(0, 1) i and F (x) = x if |x| = 1. h h(x) x 1 ≤ |x| ≤ 1 we see that F (x) ∈ |x| , |h(x)| , and so |F (x)| > 1−ε (see Also, for 2
Figure 18). Thus x0 ∈ / F B(0, 1) . Let g : Rn → Rn be a homeomorphism

Then

F

is ontinuous,

x h x | |

F(x) x |x | x 1− ε 1

Figure 18: so that

g(x) = x

if

|x| = 1

and

F (x)

when

g(x0 ) = 0

1 2

≤ |x| ≤ 1.

and dene for


g F (x)
. G(x) = − g F (x)

x ∈ B(0, 1)

G : B(0, 1) → {x : |x| = 1} is ontinuous, and if |x| = 1, then G(x) = −x. This means that G does not have a xed point, whi h ontradi ts the previous Brouwer's xed point theorem. 2 Then

The te hniques from algebrai topology that are usually used to prove the Brouwer xed point theorem also yield related results. One of them is invarian e of domain whi h is a stronger version of the previous lemma.

11.11 Theorem. Let and one-to-one. Then phism.

Ω ⊂ Rn be a domain and f : Ω → Rn be ontinuous f (Ω) is a domain and f : Ω → f (Ω) is a homeomor-

Noti e that the laim is trivial when stri tly in reasing or stri ly de reasing.

101

n = 1.

Indeed, then

f

is either