Existence Theorem on Quasiconformal Mappings Seddik Gmira

To cite this version:

Seddik Gmira. Existence Theorem on Quasiconformal Mappings. 2016. hal-01285906

HAL Id: hal-01285906 https://hal.archives-ouvertes.fr/hal-01285906 Preprint submitted on 9 Mar 2016

HAL is a multi-disciplinary open access L’archive ouverte pluridisciplinaire HAL, est archive for the deposit and dissemination of sci- destinée au dépôt et à la diffusion de documents entific research documents, whether they are pub- scientifiques de niveau recherche, publiés ou non, lished or not. The documents may come from émanant des établissements d’enseignement et de teaching and research institutions in France or recherche français ou étrangers, des laboratoires abroad, or from public or private research centers. publics ou privés. Existence Theorem on Quasiconformal Mappings Seddik Gmira

Quasiconformal mappings are, nowadays, recognized as a useful, important, and fundamental tool, applied not only in the theory of Teichmüller spaces, but also in various …elds of complex analysis of one variable such as the theories of Riemann surfaces, of Kleinian groups, of univalent functions. In this paper we prove the existence theorem of the solution of the Bel- trami di¤erential equation, and we give a fundamental variational formula for quasiconformal mappings, due to L.Ahlfors and L.Bers.

1 Quasiconformal Mapping

We consider an orientation-preserving homeomorphism f, which is at least partially di¤erentiable almost every where on a domain in C, satisfying the Beltrami equation D

fz = fz

As a natural generalization of the notion of conformal mappings we consider the following

1.1 Analytic De…nition De…nition 1 Let f be an orientation-preserving homeomorphism of a domain into C. f is quasiconformal (qc) on if D D 1. f is absolutly continuous on lines (ACL)

2. There exists a constant k, 0 k < 1 such that

fz k fz j j j j almost every where on : D Setting K = (1 + k) = (1 k), we say that f is K-quasiconformal mapping on . We call the in…mum of K > 1 such that f is qc, the maximal D dilatation of f, and denote it by Kf : Example 1 An a¢ ne mapping f (z) = az+bz+c; (a; b; c C; b < a ) 2 j j j j is qc: k = b = a and hence Kf = ( a + b ) =( a b ) j j j j j j j j j j j j 1 Example 2 Set f (z) = z , z . This f is an orientation- 1 z 2 2 preserving of the unit disk butj notj quasiconformal. In general, existence of the partial derivatives fz and fz is not enough to guarantee good properties of f applicable to further investigations. However in the case of homeomorphisms Gehring and Lheto obtained the following remrkable result. Proposition 1 If a homeomorphism f of a domain onto C has the D partial derevatives fx and fy almost.every where on , then f is totally di¤erentiable almost.every where on . D Proposition 2 Let f be a quasiconformalD mapping of a domain . Then D the partial derevatives fz and fz are totally square integrable on . Proof. Let (E) be the area of f (E) of a measurable Borel subsetD E of A , and Jf (z) the density of the set function (with respect to the Lebesgue measureD dxdy). The Lebesgue theorem impliesA

Jf (z) dxdy (E) A ZE Then, f is totally di¤erentiable at almost every z and at each such 2 2 2 D a point z we have Jf (z) = fz fz : Since f is quasiconformal, then j j j j 2 2 1 fz fz Jf (z) a.e. on j j j j 1 k2 D Proposition 3 For every quasiconformal mapping f of a domain , D the partial derivatives fz and fz are coincident with those in the sens of distribution. Namely for every element ' 01 ( ), the set of all smooth functions on with compact supports it follows2 C thatD D

fz:'dxdy = f:' dxdy z ZZD ZZD fz:'dxdy = f:' dxdy z ZZD ZZD Lemma 1 For a given p > 1, let f be a continuous function on a domain p whose distributional derivatives fz and fz are locally L on . Then for D D every compact subset F of , there is a sequence fn 1 in ( ) such D f g1 C01 D that fn converges to f uniformly on F , with

p lim (fn) fz ddxdy = 0 n j z j !1 ZZF p lim (fn) fz ddxdy = 0 n j z j !1 ZZF 2 Proof. Fix 01 ( ) with = 1 in some neighbourhood of F . Then f 2 C D p has a compact support. Further (f)z and (f)z exist and belong to L ( ). Set D C: exp 1 ; z 1 z 2 ' (z) = j j 2 ( 0; z C 2 Here, is the unit disk. Choose a constant C so that

' (z) dxdy = 1 ZZ 2 Further, for every n > 0, set 'n (z) = n ' (nz) ; z C and 2

fn (w) = 'n (f)(w) = 'n (w z)(f)(z) dxdy; w C 2 ZZC Then for every n, we have fz = ' (f) and fz = ' (f) . Morever, n z n z fn ( ) for every su¢ ciently large n, and fn converges to f uniformly 2 C01 D on F as n . Since (f)z = fz and (f)z = fz on F and since we can show equalities:! 1

p p lim (fn) (f) dxdy = 0; lim (fn) (f) dxdy = 0 n j z zj n j z zj !1 ZZD !1 ZZD Lemma 2 (Weyl) Let f be a continous function on whose distrib- D utional derivative fz is locally integrable on . If fz = 0 in the sens of distributions on , then f is holomorphic on D. D D Proof. For an arbitrarily relatively subdomain D1of , we construct an 1 D L -smoothing sequence fn 1 for f with respect to D1 as in the proof of f g1 Lemma 1, we see that (f)z = 0 in some neighbourhood of D1, we see also that (fn)z = 0 on D1 for every su¢ ciently large n. Since fn converges to f uniformly on D1 as n , f is holomorphic on this arbitrary D1. ! 1 1.2 Geometric De…nition

A quadrelateral (Q; q1; q2; q3; q4) is a pair of a Jordan closed domain Q and four points q1; q2; q3; q4 @Q, which are naturally distinct and located in this order with respect to2 the positive orientation of the boundary @Q. Proposition 4 For every quadrelateral (Q; q1; q2; q3; q4), there is a homeomorphism h of Q onto some rectangle = [0; a] [0; b](a; b > 0) which is conformal in the interior IntQ and satis…esR

h (q1) = 0; h (q2) = a

h (q3) = a + ib; h (q4) = ib

3 Morever a=b is independant of h. The value a=b is called the module of the quadrelateral (Q; q1; q2; q3; q4) and denotet by (Q). Proof. The Riemann mapping theorem impliesM the existence of a conformal mapping h1 : IntQ H (upper half-plane). By Carathéodory’sTheorem ! h1 can be extended to a homeomorphism of Q onto H R. Using a suitable Möbius transformation, we may assume that [

h1 (q1) = 1; h1 (q2) = 1 h1 (q3) = h1 (q4) > 1 Set k = 1=h1 (q3), and w dz h2 (w) = ; z H 2 2 2 0 (1 z ) (1 k z ) 2 Z Then, h2 is a conformal mappingp of H onto the interior of some rectangle [ K;K] [0;K ](K;K > 0). Hence we see that 0 0 h (z) = h2 h1 (z) + K; z Q 2 is a desired mapping. Now, let h : Q = [0; a] 0; d be another ! R mapping which satis…es the same conditions as the last proposition.h i Using the Schwarz re‡exion principale thee map e e e

1 h h can be extended to an element ofe Aut (C). Hence with suitable complex numbers c and d, we have

h (z) = ch (z) + d; (z IntQ) 2 Since e h (q1) = h (q1) = 0; h (q1) > 0 and h (q2) > 0 we conclude that c > 0 and d = 0. Hearafter, for every quadrelateral e e (Q; q1; q2; q3; q4) and a homeomorphism f of Q onto C, we consider f (Q) as a quadrelateral with vertices f (q1) ; f (q2) ; f (q3) ; f (q4). Lemma 3 Every K-qc mapping f of a domain satis…es D 1 M (Q) M (f (Q)) KM (Q) K Proof. Fix mappings h : Q R = [0; a] [0; b] and h : f (Q) R = ! 1 ! [0; a] 0; b as before. Then F = h f h is a qc mapping of the IntR onto e e h i e e e 4 the IntR which maps 0; a, ib and a + ib to 0; a, ib and a + ib, respectively. In particular, F (z) is ALC on R, and hence for almost every y [0; b], we 2 have e e e e e a @F a a F (a + iy) F (iy) = (x + iy) dx ( Fz + Fz ) dx j j @x j j j j Z0 Z0

e Since JF dxdy R = ab; intgrating both sides of th above R A inequalityRR over [0; b] e ee 2 2 (ab) ( Fz + Fz ) dxdy j j j j ZZR Fz + Fz 2 2 e j j j jdxdy Fz Fz dxdy R Fz Fz R j j j j ZZ 0 j j j j ZZ 0 Kdxdy JF dxdy K (ab) ab R R ZZ 0 ZZ 0 e where R0 = w R : Fz (w) = 0 . Thus we have M (fe(Q)) KM (Q). f 2 6 g 1 Next replacing F by ih f (ih) (or considering (Q; q2; q3; q4; q1)), the same argument gives e 1 K M (f (Q)) M (Q)

De…nition 2 A homeomorphism f of a domain into C which preserves the orientation is quasiconformal on , if there isD a constant K 1 such that, (f(Q)) K (Q) holds for everyD quadrelateral Q in M M D

Theorem 1 Let f be a Kf -quasiconformal mapping of onto , then D D 1 1. The inverse f is also Kf -quasiconformal e 2. K-quasiconformally is conformally invariant: Namely conformal mappings h and h of domains and respectively, the composed mapping D D h f h is also Kf -quasiconformal e e 3. For every Kg-qc mapping g of f ( ), the composed mapping g f is e D Kf Kg-quasiconformal.

5 Proof. 1. Lemma 3 gives

1 1 1 f (Q) (Q)) Kf f (Q ) Kf M M M 2. By the conformal invariance of the module 3. Clear from the de…nition Lemma 4 Every qc-mapping f of a domain satis…es D 2 2 fz fz dxdy = (E) E j j j j A ZZ for every subset E of , where (E) = E dxdy: Proof. Let E be a rectangleD containedA in . If f is absolutly continuous RR on the boundary @E, then in view of PropositionD 8, we …nd L2-smoothing sequence fn 1 for f with respect to E (Lemma 1). Set fn = un + ivn. Green’slemmaf g implies

(um)x (vn)y (um)y (vn)x dxdy = umdvn E @E ZZ n o Z Let m and n , we obtain ! 1 ! 1

(u)x (v)y (u)y (v)x dxdy = udv E @E ZZ n o Z Here we write f = u + iv. The right hand side is interpred as the line integral of udv along the Jordan curve @f (E). By assumption, @f (E) is recti…able. Hence we can show that

udv = dudv = (E) A Z@f(E) ZZf(E) Since f is ACL, every rectangle contained in can be approximated by such rectangles. D Proposition 5 If f is quasiconformal on a domain , then fz = 0 almost.every where on . D 6 D Proof. The set E = z : fz = 0 is measurable and fz = 0 a.e. on f 2 D g 1 E. Hence.f (E) has area zero. Since f is also quasiconformal, then E = 1 f (f (E)) has area zero Next, for every quasiconformal mapping f of a domain , we can consider D fz f = fz

6 almost every where on . This f is a bounded measurable function and satis…es D Kf 1 ess: sup f (z) < 1 z Kf + 1 22D We call f the complex dilatation of f on . Proposition 6 For every quasiconformalD mappings f and g of a domain , D fz g f g f 1 f = f 1 z f g almost every where on 1 D 1 Proof. g f is quasiconformal on f ( ) by Theorem 1. g f is di¤er- D 1 entiable on f ( ) by Proposition 1. Hence g f is di¤erentiable on f ( ) D 1 D except for a subset E of measure zero. f (E) is also of measure zero by lemma 4. Hence Theorem 1 implies that, f and g f are di¤erentiable at z and f (z), respectively a.e on . At such a point z, the chain rule is valid. Writing w = f (z), we get D

1 1 gz = g f f:fz + g f f:f w w z 1 1 gz = g f f:fz + g f f:fz w w By similar argument as before, we can show that

1 fz = 0, g = 0, and g f f = 0 6 6 w 6 for almost every z . 2 D 2 Existence Theorem on Quasiconformal Mappings

We have seen that a quasiconformal mapping f of a domain induces a bounded measurable function on , which satis…es D f D

ess: sup f (z) < 1 z 22D Next, we shall prove the converse. Namely, for every measurable function with ess: sup (z) < 1, we construct a quasiconformal mapping, whose z j j complex dilatation22D is equal to . We consider the complex Banach space L ( ) of bounded measurable functions on a domain , with the norm 1 D D = ess: sup (z) k k1 z j j 22D 7 Consider the set B ( )1 = L1 ( ); < 1 of Beltrami coe¢ - cients on . D f 2 D k k1 g PropositionD 7 Let B ( ) . If there exists a quasiconformal map- 2 D 1 ping f with the complex dilatation f = , then for every conformal mapping h of f ( ), the mapping h f has the same complex dilatation . Con- D 1 versely, for every quasiconformal mapping g with g = , the map g f is a conformal mapping of f ( ). D Proof. As before, we have = = , and 1 = 0 almost.every h f f g f 1 where on f ( ). It follows that g f is 1-quasiconformal and hence is conformal. D To solve the Beltrami di¤erential equation fz = fz; B (C) , consider 2 1 …rst, solving the @-problem. If we get a suitable representation f = G (fz), then we have a relation

fz = G (fz)z = G (fz)z

Rewriting this relation in the form fz = F (), we obtain a solution

f = G (F ())

of the Beltrami equation. On the other hand to reconstruct f from fz we use the classical Cauchy transfotmation. For every p with 1 P < , 1 we consider the complex Banach space LP (C) of all measurable functions f on C such that f p dxdy < j j 1 ZZC The following Pompeiu’sformula is essential to solve the @-problem. Proposition 8 Fix p, 2 < p < , and let f be a continuous function P 1 on C such that, fz and fz L (C). Then f satis…es 2 1 f (z) dz 1 f (z) f () = z dxdy, D 2i z z 2 Z@D ZZD for every open disk D in C. 1 1 q 1 Proof. Let q < 2 such that, p + q = 1. Since 1= (z &) L (D), the assertion follows by Hölder’s inequality. Take anj Lp smoothingj 2 sequence

fn 1 for f with respect to D and …x a point & D . For every n , Green’s fformulag gives 2

1 fn (z) dz 1 (fn)z (z) fn () = dz dz 2i z 2i z ^ Z@D ZZD

8 p Since fn f uniformly on D, and (fn) fz in L (D) respectively. ! z ! Next, we de…ne a linear operator P on LP (C) as follows 1 1 1 P h () = h (z) dxdy z z ZZC Lemma 5 For every p with 2 < p < and for every h Lp (C), P h 1 2 is a uniformly Hölder continuous function on C with exponent (1 2p) and satis…es P h (0) = 0. Morever (P h)& = h on C in the sens of distribution. Proof. Hölder equality implies 1 P h (&) h < j j k kp (z &) 1 q Further, if & = 0, by changing the variable, we have 6 q q 1 2 2q 1 dxdy = & dxdy z (z &) j j z (z 1) ZZC ZZC Hence there is a costant K depending only on p such that p 1 2=p P h (&) Kp h & , & Cr 0 j j k kp j j 2 f g Since P h (0) = 0 by de…nition, this inequality is valid even when & = 0: Now set h1 (z) = h (z + &1). Then we have

P h1 (&2 &1) = P h (&2) P h (&1) We conclude that,

1 2p P h (&2) P h (&1) Kp h &2 &1 j j k kp j j Thus P h is a uniformly Hölder continuous with exponent 1 2p: For the second assertion we take a sequence hn 1 in 01 (C) such that f g1 C h hn 0 as n . Then for every hn k kp ! ! 1 1 @ hn (z + &) (P hn) (&) = dxdy & @& z ZZC 1 (h ) (z) = n z dxdy z & ZZC Hence Green’sformula implies

hn (z) (P hn)& (&) = lim dz = hn (&) " 0 z & =" z & ! Zfj j g 9 In particular, for every ' 01 (C), we get 2 C

hn'dxdy = P hn'zdxdy, ' 01 (C) 2 C ZZC ZZC

Since h hn p 0 as n , P hn converges to P h locally on (uni- k k ! ! 1 1 2=p formaly on any compact subset) C by P h (&) Kp h p & . Hence, when we let n the above equalityj gives j k k j j ! 1

h:'dxdy = P h:'zdxdy, ' 01 (C) 2 C ZZC ZZC

Next, we need a suitable integral representation for (P h)z. For this purpose, let h 01 (C). And Green’sformula gives 2 C 1 h (z) (P h) (&) = z dz dz & 2i z & ^ ZZC 1 h (z) 1 h (z) = lim dz + 2 dz dz " 0 (2i z & =" z & 2i z & >" (z &) ^ ) ! Zfj j g ZZfj j g Since the …rst term in the right side converges to zero as " 0, the second term is essential. Let T be the linear operator de…ned by !

1 h (z) T h (&) = lim 2 dxdy , h 01 (C) " 0 ( z & >" (z &) ) 2 C ! ZZfj j g

Lemma 6 Every h 01 (C) satis…es 2 C (P h) = T h, on C and T h = h z k k2 k k2

Proof. We have already seen (P h)z = T h on C for every h 01 (C) and that 2 C 1 1 T h 2 = (P h) P h dz dz = (P h) P h dz dz k k2 2i z z ^ 2i zz ^ ZZC ZZC 1 1 = (P h) h dz dz = h (P h) dz dz 2i z ^ 2i z ^ ZZC ZZC = h 2 k k2

We see that, the operator T is extended to a bounded linear operator on L2 (C) into itself with norm 1. Since P is an operator on Lp (C) with p > 2,

10 T is also an operator on L2 (C). Then, we see by the following classical Calderon-Zygmund’s theorem that T gives a bounded linear operator on LP (C)(p > 2) into itself. Theorem 2 (Calderon-Zygmund) For every p with 2 p < 1

Cp = sup T h p h 1(C), h =1 k k 2C0 k kp is …nite. Hence the operator T is extended to a bounded operator of Lp (C). Morever, Cp is continuous with respect to p. In particular Cp satis…es

lim Cp = 1 p 2 ! In fact "Calderon-Zygmund’sTheorem" gives the following Proposition 9 For every number p > 2 and every h Lp (C) 2

(P h)z = T h on C in the sens of distribution. P Proof. Let hn 1 be a sequence in 01 (C) approximating h in L (C). Then f g C

T hn:' dxdy = P hn:'z dxdy, ' 01 (C) 2 C ZZC ZZC p Here P hn P h locally uniformally on C and T hn T h in L (C) respectively, as!n . Hence we have the assertion. ! ! 1 2.1 Existence of the Normal Solutions Theorem 3 Fix k such that 0 k < 1 arbitrarily and take p > 2 with kCp < 1. Then for every B (C)1 with a compact support; k, there exists a unique continuous2 function f such that k k1

p f (0) = 0; fz 1 L (C) 2 and satisfying fz = fz on C in the sens of distribution. This f is called the normal solution of the Beltrami equation for : Proof. First, we drive a condition which the partial derivative fz of the normal solution should satisfy. Since fz = fz has a compact support, and p p since fz 1 L (C), fz L (C) also. Thus we can consider P (fz). 2 2

11 Set F (z) = f (z) P (fz)(z) , z C 2 Then F (z) is continuous and F (0) = 0. Moreover Fz = 0 in the sens of distribution. Hence F (z) is holomorphic on C by Weyl’slemma. On the p other hand, since fz 1 and T (fz)(z) belong to L (C), so does F 0 1. Thus we conclude that F 0 (z) = 1, i.e., F (z) = z + a, a C. Since f (0) = 0, then we have 2 f (z) = P (fz)(z) + z, z C 2 Then we obtain fz = T (fz) + 1 Suppose that, there is another normal solution g. Then

gz = T (gz) + 1

By Calderon-Zygmund’sTheorem, we obtain

fz gz = T (fz) T (gz) kCp fz gz k kp k kp k kp

Since kCp < 1 by assumption, we get fz = gz a.e on C. Then the Beltrami equation also gives fz = gz a.e on C. Hence again by Weyl’s lemma f g and f g are holomorphic on C, which in turn implies that f g should be constant. Since f (0) = g (0) = 0 we conclude that f = g. Finally, the existence of the normal solution follows from fz = T (fz)+1. In fact repeat substituting the whole right hand side for fz on the tight hand side. Then we …nd

fz 1 = T (fz) = T ( (1 + T (fz))) = T + T (T (fz)) = T + T (T ) + T (T (T )) + ::::::

This series converges in Lp (C), since the operator norm of h T (h) p ! 2 L (C) is not greater than kCp < 1. Set

p h = T + T (T ) + T (T (T )) + :::::; then h L (C) 2 We shall show that, the desired solution for the Beltrami equation is

f (z) = P ( (h + 1)) (z) + z

12 In fact (h + 1) Lp (C), for has compact support. Hence Lemma 2 5 implies that f is continuous, f (0) = 0 and fz = (h + 1). Morever, by Proposition 9 we have:

fz = T ( (h + 1)) + 1 = h + 1

p Hence, f satis…es the Beltrami equation fz = fz, and fz 1 L (C) : 2

2.2 Basic Properties of Normal Solutions Corollary 1 Under the conditions of Theorem 3, the following inequalities hold: 1 fz p p , and k k 1 kCp k k Kp 1 2p f (&1) f (&2) p &1 &2 + &1 &2 j j 1 kCp k k j j j j for every &1;&2 C. Proof. Since h2= T (h) + T , then we have:

h kCp h + k kp k kp k kP

The …rst inequakity holds for fz = (h + 1). For the second we have f (z) = P (fz)(z) + 1 as before,

f (&1) f (&2) P (fz)(&1) P (fz)(&2) + &1 &2 j j j j j j From this, the second inequality holds easly. Furthermore, the normal solutions depend on the Beltrami coe¢ cients as follows. Corollary 2 For 0 k < 1 and p > 2. Let 1 be a sequence in f ng1 B (C)1with the following conditions:

1. n < k for every n k k1 2. every n has a support contained in z C : z < M with a suitable constant independent of n f 2 j j g

3. n converges to some B (C) a.e. on C as n 2 1 ! 1

13 Let fn be the normal solution for , and f be the normal solution for . Then fn f uniformaly on C as n , and ! ! 1

lim (fn) fz = 0: n z p !1 k k

Proof. First, since fz = T (fz) + 1, then we have

(fn) fz T ( (fn) fz) + T ( ) fz k z kp k n z kp k n kp kCp (fn) fz + CP ( ) fz k z kp k n kp Hence we have

Cp (fn)z fz p ( n) fz p k k 1 kCp k k

Since the support of all n are uniformaly bounded, and since n converges to a.e.on C as n , then we have ! 1

lim (fn) fz = 0 n z p !1 k k Next, we get

f (&) fn (&) = P (fz (fn) )(&) j j j z j 1 2 Kp ( ) fz + k (fn) fz & p k n kp k z kp j j n o for every & C. Thus fn f uniformaly on C as n . Since 2 ! ! 1 fn f is holomorphic in a …xed neighbourhood of for every n, we ! 1 conclude that fn converges to f uniformaly on C.

2.2.1 Existence Theorem In fact the existence of a quasiconformal mapping is also valid for a general complex dilatation B (C) 2 1 Theorem 4 For every Beltrami coe¢ cient B (C) , there exists a 2 1 homeomorphim f of C onto itself which is a quasiconformal mapping of C with complex dilatation . Morever f is uniquely determined by the following normalization conditions

f (0) = 0, f (1) = 1 and f ( ) = 1 1 We call this f the canonical -quasiconformal mapping of C.and denote it by f :

14 Proof. The uniqueness of f is given by Proposition 7 and the normalization conditions. For the existence, we suppose that has compact support. Let F be the normal solution for . Then Theorem 3 implies that F (z) =F (1) is the desired one. Now suppose that = 0 almost every where in some neighbourhood of the origin. Pulling back by the Möbus transformation 1 (z) = z If we set 1 z2 (z) = ; z C z z2 2 then B (C)1, and hase a compact support. Hence as before, there exists 2 1 the canonical -quasiconformal mapping f of C. At every such point z , the quasiconformale mapping 1e e f (z) = f (1=z)

is also totally di¤erentiable, so usinge the chain rule we have z2 1 (z) = = (z) , a.e. on f z2 z C Clearly, f sats…es the normalizatione conditions. Hence f is the desired function. Finally, suppose that is a general Beltrami coe¢ cient. Now, suppose that is a general Beltrami coe¢ cient. We set

(z) ; z C (z) = 2 1 0; z 2 where is the unit disk. Then f 1 exists as before. Finally Set

(f 1 ) 1 z 1 1 2 = (f ) 1 (f 1 ) 1 z ! Then, f 2 exists, because 2 has a compact support. Mreover g = f 2 f 1 is quasiconformal and we can see that g = a.e. Clearly g is the desired function. Hereafter, we state several applications of the existenc theorem Proposition 10 Every quasiconformal mapping of the disk onto a Jordan domain D is extended to a homeomorphism on onto D Proof. Fix such a quasiconformal mapping f : D, and set = . ! f By setting = 0 on C , we can consider B (C)1. Hence by Theorem 2 1 3, there exists the canonical -quasiconformal f of C. Set g = f f . Then g is a 1-quasiconformal mapping of D. Hence g is a conformal mapping of D. Since f () is a Jordan domain, Carathéodory’s theorem gives the

15 1 extension g to a homeomorphism of D onto f (). Since f = g f , we obtain the assertion. Proposition 11 There exists no quasiconformal mapping of the disk onto C Proof. Suppose that f : C is a quasiconformal conformal mapping 1 ! Then f is also quasiconformal. Set = 1 ; then there exists the canon- f 1 ical -quasiconformal mapping of . On the other hand, since g (C) = , 1 Liouville’sTheorem implies that g should be a constant. Proposition 12 Let be an element of B (H)1. Then there exists a quasiconformal mapping w : H H ! with complex dilatation : Moreover, such a mapping w(which can be extended to a homeomorphism of H = H R onto itself)is uniquely determined by the following normalization conditions:[

w (0) = 0, w (1) = 1, and w ( ) = 1 1 We call this unique w, the canonical -quasiconformal mapping of H, and denote it by w. Proof. The uniqueness is given by the normalization conditions as before. For the existence, set

(z) ; z H 2 (z) = 0; R 8 2 (z); z H = C R < 2 e By the uniqueness theorem,: the canonical -qc mapping f of C satis…es e f (z) = f (z) e e e In particular we see that f R = R. Since f preserves orientation, f (H) = H. Hence the restriction of f onto H is the desiret qc mapping. e e e e

3 Dependence on Beltrami coe¢ cients

Some of the most important useful facts on the canonical quasiconformal mapping of C concern dependence of the canonical quasiconformal mapping on the Beltrami coe¢ cient.

16 Theorem 5 Let (t) be a family of Beltrami coe¢ cients depending, f g on t R or C. Suppose that (t) 0 as t 0, and that (t) is di¤erentiable2 at t = 0 : k k1 ! !

(t)(z) = t (z) + t (t)(z) ; z C 2 with suitable L1 (C) and (t) L1 (C) such that (t) 0 as t 0. Then 2 2 k k1 ! ! f (t) & f [](&) = lim & 0 t ! exists for every & C, and the convergence is locally uniform on C. Moreover : 2 f [] has the integral representation

1 & (& 1) f [](&) = (z) dxdy z (z 1) (z &) ZZC Proof. [IT ] for example. Corollary 3 Let (t) be a family of Beltrami coe¢ cients depending f g on t R or C. Suppose that (t) is di¤erentiable at t = 0 : 2 (t)(z) = (z) + t (z) + t (t)(z) ; z C 2 with suitable B (C)1, L1 (C) such that (t) 0 as t 0: Then 2 2 k k1 ! ! (t) f (&) = f (&) + tf [](&) + ( t ) ;& C j j 2 locally uniformaly on C as t 0, where ! 1 f (&)(f (&) 1) ((f ) (z))2 f [](&) = (z) z dxdy f (z)(f (z) 1) (f (z) f (&)) ZZC (t) 1 Proof. Set ft = f (f ) . Then the complex dilatation (t) of ft is given by: (t) (f )z 1 (t) = f = (f ) t 1 : (t) (f ) z ! Hence, (t) is written as (t) = t + ( t ) in L1 (C), where j j (f )z 1 = 2 (f ) 1 (f ) ! j j z

17 Apply Theorem 5 to this family ft . Then we can conclude that f g ft (&) & t converges to

1 & (& 1) f (&) = (z) dxdy z (z 1) (z &) ZZC locally uniformally on C. Hence, changing the variable z in this integrale 1 to (f ) (z) and noting that

(t) f f (ft f0) = f t t we get the assertion. References [L.Ahlfors] Lectures on Quasiconformal Mappings, D.Van Nostrand, Princeton, New Jersey, 1966 [F.W.Ghering] Charaterstic Proporties of Quasidisks, Les presses de l’université de Montreal, 1982 [Y.Imayoshi and M.Taniguchi] An introduction to Teichmüller Spaces, Springer-Verlag 1992 [O.Lehto] Quasiconformal Mappings in the plane, 2nd Ed.,Springer- Verlag, Berlin and New York, 1972