Existence Theorem on Quasiconformal Mappings

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Existence Theorem on Quasiconformal Mappings

Seddik Gmira

To cite this version:

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

Existence Theorem on Quasiconformal Mappings

Seddik Gmira

Quasiconformal mappings are, nowadays, recognized as a useful, impor- tant, 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 homeomorphismf, which is at least partially di¤erentiable almost every where on a domain D inC, satisfying the Beltrami equation

fz= fz

As a natural generalization of the notion of conformal mappings we con- sider the following

1.1 Analytic De…nition

De…nition 1 Let f be an orientation-preserving homeomorphism of a do- main Dinto C. f is quasiconformal (qc) on D if

1. f is absolutly continuous on lines (ACL) 2. There exists a constantk,0 k <1such that

jf_zj kjf_zj almost every where onD:

Setting K = (1 +k)=(1 k), we say that f is K-quasiconformal map- ping on D. We call the in…mum of K >1 such thatf is qc, the maximal dilatation off, and denote it byKf:

Example 1An a¢ ne mappingf(z) =az+bz+c;(a; b; c2C; jbj<jaj) is qc: k=jbj=jajand henceK_f = (jaj+jbj)=(jaj jbj)

Example 2 Set f(z) = ^z

1 jzj², z 2 . This f is an orientation- preserving of the unit disk but not quasiconformal.

In general, existence of the partial derivativesfz andfz is not enough to guarantee good properties off 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 D onto C has the partial derevatives f_x and f_y almost.every where on D, then f is totally di¤erentiable almost.every where onD.

Proposition 2Letfbe a quasiconformal mapping of a domainD. Then the partial derevativesf_z and f_z are totally square integrable on D.

Proof. Let A(E) be the area of f(E) of a measurable Borel subset E of D, andJ_f(z)the density of the set functionA(with respect to the Lebesgue measuredxdy). The Lebesgue theorem implies

Z

E

J_f(z)dxdy A(E)

Then,f is totally di¤erentiable at almost everyz2 D and at each such a pointz we have Jf(z) =jfzj² jfzj²:Since f is quasiconformal, then

jf_zj² jf_zj² 1

1 k²J_f(z) a.e. onD

Proposition 3 For every quasiconformal mapping f of a domain D, the partial derivatives f_z and f_z are coincident with those in the sens of distribution. Namely for every element ' 2 C0¹(D), the set of all smooth functions onDwith compact supports it follows that

Z Z

D

f_z:'dxdy =

Z Z

D

f:'_zdxdy Z Z

D

f_z:'dxdy =

Z Z

D

f:'_zdxdy

Lemma 1For a givenp >1, letf be a continuous function on a domain Dwhose distributional derivatives f_z and f_z are locallyL^p on D. Then for every compact subset F of D, there is a sequence ff_ng¹1 in C0¹(D) such thatfn converges tof uniformly onF, with

nlim!1

Z Z

Fj(f_n)_z f_zj^pddxdy = 0

nlim!1

Z Z

Fj(fn)_z fzj^pddxdy = 0

Proof. Fix 2 C0¹(D) with = 1 in some neighbourhood of F. Then f has a compact support. Further( f)_zand( f)_z exist and belong toL^p(D).

Set

'(z) = (

C:exp ¹

1 jzj² ; z2

0; z2C

Here, is the unit disk. Choose a constant C so that Z Z

'(z)dxdy= 1

Further, for everyn >0, set '_n(z) =n²'(nz); z2Cand fn(w) ='_n ( f) (w) =

Z Z

C

'_n(w z) ( f) (z)dxdy; w 2C Then for everyn, we havefz ='_n ( f)_z andfz ='_n ( f)_z. Morever, f_n2 C0¹(D) for every su¢ ciently large n, andf_n converges to f uniformly on F asn ! 1. Since( f)_z =f_z and ( f)_z =f_z on F and since we can show equalities:

nlim!1

Z Z

Dj(f_n)_z ( f)_zj^pdxdy= 0; lim

n !1

Z Z

Dj(f_n)_z ( f)_zj^pdxdy= 0 Lemma 2 (Weyl) Let f be a continous function on D whose distrib- utional derivative f_z is locally integrable on D. If f_z = 0 in the sens of distributions onD, thenf is holomorphic onD.

Proof. For an arbitrarily relatively subdomain D₁of D, we construct an L¹-smoothing sequence ff_ng¹1 for f with respect to D₁ as in the proof of Lemma 1, we see that( f)_z = 0 in some neighbourhood of D1, we see also that (f_n)_z = 0 on D₁ for every su¢ ciently large n. Since f_n converges to f uniformly onD₁ asn ! 1,f is holomorphic on this arbitrary D₁. 1.2 Geometric De…nition

A quadrelateral(Q;q₁; q₂; q₃; q₄) is a pair of a Jordan closed domain Qand four points q₁; q₂; q₃; q₄ 2 @Q, which are naturally distinct and located in this order with respect to the positive orientation of the boundary@Q.

Proposition 4For every quadrelateral(Q;q₁; q₂; q₃; q₄), there is a home- omorphismh of Qonto some rectangle R= [0; a] [0; b] (a; b >0)which is conformal in the interiorIntQand satis…es

h(q₁) = 0; h(q₂) =a h(q₃) = a+ib; h(q₄) =ib

Morevera=bis independant of h. The value a=bis called the module of the quadrelateral(Q;q1; q2; q3; q4) and denotet by M(Q).

Proof. The Riemann mapping theorem implies the existence of a conformal mappingh1 :IntQ !H (upper half-plane). By Carathéodory’s Theorem h1 can be extended to a homeomorphism of QontoH[R. Using a suitable Möbius transformation, we may assume that

h₁(q₁) = 1; h₁(q₂) = 1 h₁(q₃) = h₁(q₄)>1 Setk= 1=h1(q3), and

h2(w) = Z w

0

p dz

(1 z²) (1 k²z²); z 2H

Then,h2is a conformal mapping ofHonto the interior of some rectangle [ K; K] [0; K⁰] (K; K⁰>0). Hence we see that

h(z) =h₂ h₁(z) +K; z 2Q is a desired mapping. Now, let eh:Q !Re = [0;ea] h

0;dei

be another mapping which satis…es the same conditions as the last proposition. Using the Schwarz re‡exion principale the map

eh h ¹

can be extended to an element of Aut(C). Hence with suitable complex numberscand d, we have

eh(z) =ch(z) +d; (z2IntQ) Since

eh(q₁) =h(q₁) = 0; h(q₁)>0 and eh(q₂)>0

we conclude that c > 0 and d = 0. Hearafter, for every quadrelateral (Q;q₁; q₂; q₃; q₄)and a homeomorphismf ofQontoC, we considerf(Q)as a quadrelateral with verticesf(q1); f(q2); f(q3); f(q4).

Lemma 3 Every K-qc mapping f of a domainD satis…es 1

KM(Q) M(f(Q)) KM(Q)

Proof. Fix mappings h : Q ! R = [0; a] [0; b] and eh :f(Q) ! Re = [0;ea] h

0;ebi

as before. ThenF =eh f h ¹is a qc mapping of theIntRonto

theIntRe which maps 0; a,iband a+ibto0;ea,ieb andea+ieb, respectively.

In particular, F(z) is ALC on R, and hence for almost every y 2[0; b], we have

e

a jF(a+iy) F(iy)j= Z a

0

@F

@x (x+iy)dx Z a

0

(jFzj+jFzj)dx Since R R

RJ_Fdxdy A Re = eaeb; intgrating both sides of th above inequality over [0; b]

(eab)²

Z Z

R

(jF_zj+jF_zj)dxdy

2

Z Z

R⁰

jF_zj+jF_zj jF_zj jF_zjdxdy

Z Z

R⁰ jF_zj² jF_zj² dxdy Z Z

R⁰

Kdxdy Z Z

R⁰

JFdxdy K(ab) eaeb

whereR⁰ =fw2R:F_z(w)6= 0g. Thus we haveM(f(Q)) KM(Q).

Next replacingF by ieh f (ih) ¹ (or considering(Q;q₂; q₃; q₄; q₁)), the same argument gives

1 M(f(Q))

K M(Q)

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

Theorem 1 Letf be a K_f-quasiconformal mapping ofD onto De, then 1. The inversef ¹ is alsoK_f-quasiconformal

2. K-quasiconformally is conformally invariant: Namely conformal map- pingshandehof domainsDandDerespectively, the composed mapping eh f h is alsoKf-quasiconformal

3. For everyK_g-qc mapping g of f(D), the composed mapping g f is KfKg-quasiconformal.

Proof. 1. Lemma 3 gives 1

K_fM f ¹(Q) M(Q)) K_fM f ¹(Q ) 2. By the conformal invariance of the module

3. Clear from the de…nition

Lemma 4 Every qc-mapping f of a domain Dsatis…es Z Z

E jf_zj² jf_zj² dxdy=A(E) for every subsetE ofD, where A(E) =R R

Edxdy:

Proof. Let E be a rectangle contained inD. If f is absolutly continuous on the boundary @E, then in view of Proposition 8, we …nd L²-smoothing sequence ffng¹1 for f with respect to E (Lemma 1). Set fn = un+ivn. Green’s lemma implies

Z Z

E

n

(u_m)_x(v_n)_y (u_m)_y(v_n)_xo

dxdy= Z

@E

u_mdv_n

Letm ! 1 and n ! 1, we obtain Z Z

E

n

(u)_x(v)_y (u)_y(v)_xo

dxdy= Z

@E

udv

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

Z

@f(E)

udv= Z Z

f(E)

dudv=A(E)

Since f is ACL, every rectangle contained inDcan be approximated by such rectangles.

Proposition 5 If f is quasiconformal on a domain D, then f_z 6= 0 almost.every where onD.

Proof. The set E = fz2 D:fz = 0g is measurable and fz = 0 a.e. on E. Hence.f(E) has area zero. Sincef ¹ is also quasiconformal, then E = f ¹(f(E)) has area zero

Next, for every quasiconformal mapping f of a domain D, we can con- sider

f = f_z f_z

almost every where on D. This _f is a bounded measurable function and satis…es

ess:sup

z22D f(z) Kf 1

K_f+ 1 <1 We call _f the complex dilatation of f on D.

Proposition 6 For every quasiconformal mappingsf andgof a domain D,

g f ¹ f = fz

fz

g f

1 _{f g}

almost every where onD

Proof. g f ¹ is quasiconformal on f(D)by Theorem 1. g f ¹ is di¤er- entiable onf(D)by Proposition 1. Henceg f ¹ is di¤erentiable onf(D) except for a subset E of measure zero. f ¹(E) is also of measure zero by lemma4. Hence Theorem 1implies that, f and g f are di¤erentiable atz and f(z), respectively a.e on D. At such a point z, the chain rule is valid.

Writingw=f(z), we get

g_z = g f ¹ _w f:f_z+ g f ¹ _w f:f_z g_z = g f ¹ _w f:f_z+ g f ¹ _w f:f_z By similar argument as before, we can show that

fz 6= 0,g6= 0, and g f ¹ _w f 6= 0 for almost everyz2 D.

2 Existence Theorem on Quasiconformal Mappings

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

ess:sup

z22D f(z) <1

Next, we shall prove the converse. Namely, for every measurable function with ess:sup

z22D j (z)j <1, we construct a quasiconformal mapping, whose complex dilatation is equal to . We consider the complex Banach space L¹(D) of bounded measurable functions on a domainD, with the norm

k k₁=ess:sup

z22D j (z)j

Consider the set B(D)₁ =f 2L¹(D) ;k k₁<1g of Beltrami coe¢ - cients onD.

Proposition 7 Let 2 B(D)₁. If there exists a quasiconformal map- pingf with the complex dilatation _f = , then for every conformal map- pinghoff(D), the mappingh f has the same complex dilatation . Con- versely, for every quasiconformal mapping g with _g = , the map g f ¹ is a conformal mapping of f(D).

Proof. As before, we have _{h f} = _f = , and _{g f} 1 = 0 almost.every where on f(D). It follows that g f ¹ is 1-quasiconformal and hence is conformal.

To solve the Beltrami di¤erential equationfz = fz; 2B(C)₁, consider

…rst, solving the @-problem. If we get a suitable representation f =G(f_z), then we have a relation

f_z=G(f_z)_z =G( f_z)_z

Rewriting this relation in the form f_z=F( ), we obtain a solution f =G( F( ))

of the Beltrami equation. On the other hand to reconstruct f from f_z we use the classical Cauchy transfotmation. For every p with 1 P <1, we consider the complex Banach spaceL^P(C) of all measurable functionsf

onC such that Z Z

Cjfj^pdxdy <1

The following Pompeiu’s formula is essential to solve the@-problem.

Proposition 8 Fix p, 2 < p < 1, and let f be a continuous function onC such that, fz and fz 2L^P (C). Then f satis…es

f( ) = 1 2 i

Z

@D

f(z)dz z

1Z Z

D

fz(z)

z dxdy, 2D

for every open diskDinC.

Proof. Let q < 2 such that, ¹_p + ¹_q = 1. Since j1=(z &)j^q 2 L¹(D), the assertion follows by Hölder’s inequality. Take an L^p smoothing sequence ff_ng¹1 forf with respect toDand …x a point& 2D. For everyn, Green’s formula gives

f_n( ) = 1 2 i

Z

@D

fn(z)dz z

1 2 i

Z Z

D

(fn)_z(z)

z dz^dz

Since fn !f uniformly onD, and (fn)_z !fz inL^p(D) respectively.

Next, we de…ne a linear operatorP on L^P(C)as follows P h( ) = 1 Z Z

C

h(z) 1 z

1

z dxdy

Lemma 5 For every p with 2 < p < 1 and for every h 2 L^p(C), P h is a uniformly Hölder continuous function onCwith exponent(1 2p)and satis…esP h(0) = 0. Morever(P h)_& =h onC in the sens of distribution.

Proof. Hölder equality implies jP h(&)j 1

khkp (z &) _q<1 Further, if & 6= 0, by changing the variable, we have

Z Z

C

1 z(z &)

q

dxdy =j&j^{2 2q} Z Z

C

1 z(z 1)

q

dxdy

Hence there is a costant K_p depending only onp such that jP h(&)j Kpkhkpj&j^{1 2=p},& 2Crf0g

Since P h(0) = 0by 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,

jP h(&₂) P h(&₁)j K_pkhkpj&₂ &₁j^{1 2p}

Thus P his a uniformly Hölder continuous with exponent 1 2p:

For the second assertion we take a sequencefhng¹1 inC0¹(C) such that kh hnkp !0 asn ! 1. Then for every hn

(P h_n)_&(&) = 1 @

@&

Z Z

C

h_n(z+&)

z dxdy

= 1 Z Z

C

(hn)_z(z) z & dxdy Hence Green’s formula implies

(P hn)_&(&) = lim

" !0

Z

fjz &j="g

hn(z)

z & dz =hn(&)

In particular, for every '2 C0¹(C), we get Z Z

C

h_n'dxdy= Z Z

C

P h_n'_zdxdy,'2 C0¹(C)

Sincekh h_nkp !0asn ! 1,P h_nconverges toP hlocally on (uni- formaly on any compact subset) C by jP h(&)j K_pkhkpj&j^{1 2=p}. Hence, when we letn ! 1 the above equality gives

Z Z

C

h:'dxdy= Z Z

C

P h:'_zdxdy,'2 C0¹(C)

Next, we need a suitable integral representation for (P h)_z. For this purpose, leth2 C0¹(C). And Green’s formula gives

(P h)_&(&) = 1 2 i

Z Z

C

h_z(z)

z &dz^dz

= lim

" !0

( 1 2 i

Z

fjz &j="g

h(z)

z &dz+ 1 2 i

Z Z

fjz &j>"g

h(z)

(z &)²dz^dz )

Since the …rst term in the right side converges to zero as" !0, the second term is essential. LetT be the linear operator de…ned by

T h(&) = lim

" !0

( 1 Z Z

fjz &j>"g

h(z) (z &)²dxdy

)

,h2 C0¹(C) Lemma 6 Every h2 C0¹(C) satis…es

(P h)_z =T h, on Cand kT hk2 =khk2

Proof. We have already seen (P h)_z =T h on C for every h 2 C0¹(C) and that

kT hk²2 = 1 2i

Z Z

C

(P h)_z P h _zdz^dz = 1 2i

Z Z

C

(P h) P h _zzdz^dz

= 1

2i Z Z

C

(P h) h _zdz^dz= 1 2i

Z Z

C

h(P h)_zdz^dz

= khk²2

We see that, the operatorT is extended to a bounded linear operator on L²(C)into itself with norm1. SinceP is an operator onL^p(C)withp >2,

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 L^P(C) (p >2)into itself.

Theorem 2 (Calderon-Zygmund) For everypwith 2 p <1

C_p = sup

h2C¹0 (C),khkp=1kT hkp

is …nite. Hence the operatorT is extended to a bounded operator ofL^p(C).

Morever,Cp is continuous with respect top. In particularCp satis…es

plim!2Cp= 1

In fact "Calderon-Zygmund’s Theorem" gives the following Proposition 9 For every number p >2and everyh2L^p(C)

(P h)_z =T h onC in the sens of distribution.

Proof. Let fh_ng¹1 be a sequence in C0¹(C) approximating h in L^P (C).

Then Z Z

C

T hn:' dxdy= Z Z

C

P hn:'_z dxdy,'2 C0¹(C)

Here P hn ! P h locally uniformally on C and T hn !T h inL^p(C) respectively, asn ! 1. Hence we have the assertion.

2.1 Existence of the Normal Solutions

Theorem 3 Fix k such that 0 k < 1 arbitrarily and take p > 2 with kC_p < 1. Then for every 2 B(C)₁ with a compact support; k k₁ k, there exists a unique continuous functionf such that

f(0) = 0; f_z 12L^p(C) and satisfying

f_z= f_z

onC in the sens of distribution. Thisf is called the normal solution of the Beltrami equation for :

Proof. First, we drive a condition which the partial derivative f_z of the normal solution should satisfy. Sincef_z = f_z has a compact support, and sincefz 12L^p(C),fz2L^p(C) also. Thus we can considerP(fz).

Set

F(z) =f(z) P(fz) (z),z2C

Then F(z) is continuous and F(0) = 0. Moreover F_z = 0 in the sens of distribution. HenceF(z) is holomorphic on Cby Weyl’s lemma. On the other hand, sincefz 1andT(fz) (z)belong toL^p(C), so doesF⁰ 1. Thus we conclude thatF⁰(z) = 1, i.e.,F(z) =z+a,a2C. Sincef(0) = 0, then we have

f(z) =P(fz) (z) +z,z2C Then we obtain

fz =T( fz) + 1

Suppose that, there is another normal solution g. Then gz =T( gz) + 1

By Calderon-Zygmund’s Theorem, we obtain

kf_z g_zkp =kT( f_z) T( g_z)kp kC_pkf_z g_zkp

Since kC_p < 1 by assumption, we get f_z = g_z a.e on C. Then the Beltrami equation also gives f_z = g_z 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) = 0we conclude that f =g.

Finally, the existence of the normal solution follows fromf_z=T( f_z)+1.

In fact repeat substituting the whole right hand side forfzon the tight hand side. Then we …nd

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

= T +T( T( f_z))

= T +T( T ) +T( T( T )) +::::::

This series converges inL^p(C), since the operator norm ofh !T( h)2 L^p(C) is not greater thankC_p<1. Set

h=T +T( T ) +T( T( T )) +:::::; thenh2L^p(C) We shall show that, the desired solution for the Beltrami equation is

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

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

f_z=T( (h+ 1)) + 1 =h+ 1

Hence,f satis…es the Beltrami equationfz = fz, and fz 12L^p(C):

2.2 Basic Properties of Normal Solutions

Corollary 1 Under the conditions of Theorem 3, the following inequalities hold:

kfzkp

1

1 kCp k kp, and jf(&₁) f(&₂)j Kp

1 kC_p k kpj&₁ &₂j^{1 2p}+j&₁ &₂j for every&₁; &₂2C.

Proof. Since h=T( h) +T , then we have:

khkp kC_pkhkp+k kP

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

jf(&₁) f(&₂)j jP(f_z) (&₁) P(f_z) (&₂)j+j&₁ &₂j 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 f ng¹1 be a sequence in B(C)₁with the following conditions:

1. k nk₁< k for everyn

2. every _n has a support contained infz2C:jzj< Mgwith a suitable constant independent ofn

3. _n converges to some 2B(C)₁ a.e. onCasn ! 1

Letfn be the normal solution for , andf be the normal solution for . Thenfn !f uniformaly onCasn ! 1, and

nlim!1k(fn)_z fzkp= 0:

Proof. First, sincef_z =T( f_z) + 1, then we have

k(fn)_z fzkp kT( _n(fn)_z fz)kp+kT( _n)fzkp

kCpk(fn)_z fzkp+CP k( _n)fzkp

Hence we have

k(f_n)_z f_zk_p C_p

1 kCp k( _n)f_zkp

Since the support of all _n are uniformaly bounded, and since _n con- verges to a.e.onC asn ! 1, then we have

nlim!1k(f_n)_z f_zk_p = 0 Next, we get

jf(&) fn(&)j = jP(fz (fn)_z) (&)j K_pn

k( _n)f_zkp+kk(f_n)_z f_zk_po j&j^{1 2p} for every & 2 C. Thus f_n ! f uniformaly on C as n ! 1. Since fn ! f is holomorphic in a …xed neighbourhood of 1 for every n, we conclude thatfn converges to f uniformaly onC.

2.2.1 Existence Theorem

In fact the existence of a quasiconformal mapping is also valid for a general complex dilatation 2B(C)₁

Theorem 4 For every Beltrami coe¢ cient 2 B(C)₁, there exists a 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) = 1and f(1) =1

We call this f the canonical -quasiconformal mapping of C.and denote it byf :

Proof. The uniqueness off 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 thatF (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

(z) = _z¹ If we set

e(z) = 1 z

z² z²; z2C

thene 2B(C)₁, and has a compact support. Hence as before, there exists the canonicale-quasiconformal mappingf^eofC. At every such point _z¹, the quasiconformal mapping

f(z) = 1 f^e(1=z)

is also totally di¤erentiable, so using the chain rule we have

f(z) = z² z²e 1

z = (z), a.e. on C

Clearly, f sats…es the normalization conditions. Hencef is the desired function. Finally, suppose that is a general Beltrami coe¢ cient. Now, suppose that is a general Beltrami coe¢ cient. We set

1(z) = (z); z2C

0; z2

where is the unit disk. Then f ¹ exists as before. Finally Set

2 = ¹

1 ₁

(f ¹)_z (f ¹)_z

!

(f ¹) ¹

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. Clearlyg is the desired function.

Hereafter, we state several applications of the existenc theorem

Proposition 10 Every quasiconformal mapping of the disk onto a Jordan domainDis extended to a homeomorphism on onto D

Proof. Fix such a quasiconformal mappingf : !D, and set = _f. By setting = 0 onC , we can consider 2B(C)₁. Hence by Theorem 3, there exists the canonical -quasiconformal f of C. Set g = f f ¹. Thengis a 1-quasiconformal mapping ofD. Hencegis a conformal mapping of D. Since f ( ) is a Jordan domain, Carathéodory’s theorem gives the

extensiong to a homeomorphism ofDonto f ( ). Since f =g ¹ f , we obtain the assertion.

Proposition 11 There exists no quasiconformal mapping of the disk onto C

Proof. Suppose thatf : ! C is a quasiconformal conformal mapping Thenf ¹ is also quasiconformal. Set = _f 1;then there exists the canon- ical -quasiconformal mapping of . On the other hand, sinceg ¹(C) = , Liouville’s Theorem implies thatg ¹ 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 homeomor- phism 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 uniquew, the canonical -quasiconformal mapping of H, and denote it byw .

Proof. The uniqueness is given by the normalization conditions as before.

For the existence, set e(z) =

8<

:

(z); z2H

0; 2R

(z); z2H =C R

By the uniqueness theorem, the canonicale-qc mappingf^e ofCsatis…es f^e(z) =f^e(z)

In particular we see that f^e R = R. Since f^e preserves orientation, f^e(H) =H. Hence the restriction off^e onto H is the desiret qc mapping.

3 Dependence on Beltrami coe¢ cients

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

Theorem 5 Let f (t)g be a family of Beltrami coe¢ cients depending, on t 2 R or C. Suppose that k (t)k₁ ! 0 as t ! 0, and that (t) is di¤erentiable att= 0 :

(t) (z) =t (z) +t (t) (z); z2C

with suitable 2 L¹(C) and (t) 2 L¹(C) such that k (t)k₁ ! 0 as t !0. Then

f[ ] (&) = lim

& !0

f ^(t) &

t

exists for every& 2C, and the convergence is locally uniform onC. Moreover

:

f[ ] has the integral representation f[ ] (&) = 1 Z Z

C

(z) &(& 1)

z(z 1) (z &)dxdy Proof. [IT]for example.

Corollary 3 Let f (t)g be a family of Beltrami coe¢ cients depending ont2R orC. Suppose that (t) is di¤erentiable att= 0 :

(t) (z) = (z) +t (z) +t (t) (z); z2C

with suitable 2 B(C)₁, 2L¹(C) such that k (t)k₁ ! 0 as t ! 0:

Then

f ^(t)(&) =f (&) +tf [ ] (&) + (jtj); &2C locally uniformaly onCast !0, where

f [ ] (&) = 1 Z Z

C

(z) f (&) (f (&) 1) ((f )_z(z))²

f (z) (f (z) 1) (f (z) f (&))dxdy Proof. Set ft = f ^(t) (f ) ¹. Then the complex dilatation (t) of ft is given by:

(t) = _ft = (t) 1 : (t)

(f )_z (f )_z

!

(f ) ¹

Hence, (t) is written as (t) =t + (jtj) inL¹(C), where

= 1 j j² (f )_z (f )_z

!

(f ) ¹

Apply Theorem 5 to this family fftg. Then we can conclude that ft(&) &

t converges to

f (&) = 1 Z Z

C

(z) &(& 1)

z(z 1) (z &)dxdy

locally uniformally onC. Hence, changing the variablezin this integrale to(f ) ¹(z) and noting that

f ^(t) f

t = (f_t f₀)

t f

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