Annales Academiæ Scientiarum Fennicæ Mathematica Volumen 27, 2002, 97–108

BILIPSCHITZ APPROXIMATIONS OF QUASICONFORMAL MAPS

Christopher J. Bishop

SUNY at Stony Brook, Mathematics Department Stony Brook, NY 11794-3651, U.S.A.; bishop@math.sunysb.edu

Abstract. We show that for any K -quasiconformal map of the upper half plane to itself and any ε > 0, there is a (K + ε)-quasiconformal map of the half plane with the same boundary values which is also biLipschitz with respect to the hyperbolic metric.

1. Introduction

If f is a K -quasiconformal map of the upper half space, H, to itself, then it is well known that one can find a quasiconformal g: H → H which agrees with f on the real line and which is also biLipschitz with respect to the hyperbolic metric. Moreover, the quasiconformal constant of g can be bounded in terms of K (e.g., [1], [5]). The purpose of this note is to show that the quasiconformal constant of g may be taken as close to K as we wish.

Theorem 1.1. Given K < ∞ and ε > 0, there is a C < ∞ so that if f is a K -quasiconformal map of the upper half-plane H to itself, then there is a (K + ε)-quasiconformal map g: H → H which is C -biLipschitz with respect to the hyperbolic metric on H and which agrees with f on R = ∂H.

One cannot take ε = 0 in this result. A K -quasiconformal self-map of the upper half-plane is called uniquely extremal if it is the only K -quasiconformal extension of its boundary values. If the result above held with ε = 0 then every uniquely extremal map would be hyperbolically biLipschitz with constant depending only on K. But Theorem 11 of [4] implies that any K -quasiconformal map can be uniformly approximated by K -quasiconformal uniquely extremal maps. Since not every quasiconformal map is hyperbolically biLipschitz, neither can every uniquely extremal map, and hence the result above cannot hold for ε = 0. I thank V. Marković for showing this argument to me.

Theorem 1.1 arose from [3]. That paper considered finding the best constant in a theorem of Dennis Sullivan involving quasiconformal maps from a plane domain Ω to the disk and the question arises of whether the best constant is the same if we also require these maps to be biLipschitz with respect to the hyperbolic metrics.

2000 Mathematics Subject Classification: Primary 30C62.

The author is partially supported by NSF Grant DMS 9800924.

Theorem 1.1 shows that this is the case. I thank Al Marden for his comments on [3] which led me to formulate and prove the result.

Now, we sketch the proof of Theorem 1.1, leaving certain estimates to be proven later. We start by factoring f = f_{n} ∘ ⋯ ∘ f_{1} as a composition of n maps, each with quasiconformal constant K^{1/n} ≈ 1 + log K/n (Lemma 2.1). If μ_{j} = (f_{j})z̄/(f_{j})_{z} is the Beltrami coefficient associated to f_{j}, then we smooth μ_{j} out to get a new Beltrami coefficient μ̃_{j}, by convolving with a smooth, radial bump function with respect to the hyperbolic metric in H which is supported in a disk of hyperbolic radius δ (and similarly for the lower half-plane). We will show that ∥μ̃_{j}∥_{∞} ≤ ∥μ_{j}∥_{∞}(1 + Cδ^{2}) (Corollary 2.8) and so the corresponding map, denoted f̃_{j}, is quasiconformal with constant K^{1/n}(1+C log Kδ^{2}/n). We will also show f̃_{j} is biLipschitz for the hyperbolic metric with constant 1+C log K/δn (Lemma 2.12). Although f̃_{j} might not equal f_{j} on the boundary, we shall show that g_{j} = f_{j} ∘ f̃_{j}^{−1} is quasisymmetric with constant 1 + C(log K/n)^{2} , independent of δ, if n is large enough (Corollary 2.11). Thus g_{j} can be extended to a quasiconformal mapping (also denoted g_{j}) of the plane which is quasiconformal and hyperbolically biLipschitz both with constant 1 + C(log K/n)^{2} (Lemma 2.3). Thus the map G_{n} = (g_{n} ∘ f̃_{n}) ∘ ⋯ ∘ (g_{1} ∘ f̃_{1}) agrees with f on the boundary, has quasiconformal constant

K_{n} = [K^{1/n}(1 + C log Kδ^{2}/n)(1 + C(log K/n)^{2})]^{n} = K^{1+Cδ^{2}}(1 + O(1/n)),

and biLipschitz constant

B_{n} = [(1 + C log K/(nδ))(1 + C(log K/n)^{2})]^{n} = K^{C/δ}(1 + O(1/n)).

Taking n large enough and δ small enough, we can make K_{n} as close to K as we wish. This proves the theorem, except for proving the lemmas mentioned above. Throughout the paper C will denote a generic constant which may be different at different places. We will use subscripts when it is important to recall a particular value.

I thank the referee for a most careful reading of the paper and numerous comments, corrections and suggestions which greatly improved it.

2. Proof of the lemmas

Most of the facts we need are proven in Ahlfors’ book [1]. In several cases we need a result with a sharper estimate than is stated there, so we give the necessary argument. We begin with some notation.

We let B_{r} = B(0, r) ⊂ C and define

∥f∥_{p,r} = (∫_{B_{r}} |f|^{p} dx dx)^{1/p};

P denotes the Cauchy transform

(1)  P h(w) = \frac{−1}{π} ∫ h(z)(\frac{1}{z − w} − \frac{1}{z}) dx dy,

and T denotes the Beurling transform

T h(w) = −\frac{1}{π} lim_{ε→0} ∫_{|z−w|>ε} \frac{h(w)}{(z − w)^{2}} dx dy,

which is known to be bounded on L^{p} with norm K_{p} for each 1 < p < ∞. Moreover, K_{p} → 1 as p → 2 and for any k < 1, we define k(p) > 2 so that kK_{p} < 1 for 2 ≤ p ≤ k(p).

Given a measurable function μ on the plane with ∥μ∥_{∞} ≤ k < 1, there is a K -quasiconformal map (with K = (k + 1)/(k − 1)) of the plane to itself which satisfies the Beltrami equation fz̄ = μf_{z}. We may normalize so that 0, 1, ∞ are fixed points of the map and call this solution f^{μ}. If μ has compact support then we may also normalize so that f(0) = 0 and f_{z} − 1 ∈ L^{p} for some p > 2 (depending on k). This solution will be denoted F^{μ}.

Lemma 2.1. If f is a K -quasiconformal mapping of the plane then we can write f = f_{n} ∘ ⋯ ∘ f_{1} where each map is K^{1/n} -quasiconformal.

For a proof, see page 99 of Ahlfors’ book [1]. A map h: R → R is called k -quasisymmetric if

\frac{1}{k} ≤ \frac{h(x + t) − h(x)}{h(x) − h(x − t)} ≤ k.

The following are standard results.

Lemma 2.2. Suppose f is a K -quasiconformal map of the plane to itself which maps H to itself. Then the restriction to R is a k -quasisymmetric map for some k depending only on K. There is a K_{0} > 1 and a C_{0} < ∞, so that if K = 1 + ε ≤ K_{0} then we may take k ≤ 1 + C_{0}ε.

The first part is Theorem 1 of Chapter IV of [1]. The small constant case is proven in [2].

Lemma 2.3. Suppose f: R → R is a k -quasisymmetric map. Then f has an extension to a K -quasiconformal mapping of H to itself which is K -biLipschitz with respect to the hyperbolic metric, where K depends only on k. There is a k_{1} > 1 and C_{1} < ∞ so that if k = 1 + ε ≤ k_{1} , then we may take K ≤ 1 + C_{1}ε.

Proof. The first part is in Chapter IV of [1]. To prove the small constant case, associate to each dyadic interval I = [j2^{−n}, (j + 1)2^{−n}] = [a, b] on the line the “Whitney box” Q_{I} = I × [2^{−n−1}, 2^{−n}] in the upper half plane. Think of this a pentagon with the five vertices

(a, |I|), (b, |I|), (b, \frac{1}{2}|I|), (\frac{1}{2}(a + b), \frac{1}{2}|I|), (a, \frac{1}{2}|I|).

To each dyadic interval I = [a, b] we define the point z_{I} = (a, |I|) (the “upper left corner” of the Whitney box Q_{I}) and define the extension f at the point z_{I} by

f(z_{I}) = (f(j2^{−n}), |f(I)|).

This defines f at every vertex of every Whitney box and we may extend it into the boxes in a piecewise linear way if k is close enough to 1. It is easy to check from the definition of quasisymmetry that the image of Q_{I} is a five sided polygon with angles and side lengths differing only O(ε) from those of Q, and this proves the result.

Recall from page 99 of [1] that if μ satisfies the symmetry condition μ(z̄) = \overline{μ(z)} then the corresponding map f^{μ} maps the real line into itself. From now on, we will assume we are dealing with such maps.

Lemma 2.4. Given α < 1, R < ∞, there is a k < 1 and a constant C_{2} = C_{2}(α, R) < ∞ so that the following holds. Suppose ∥μ∥_{∞} ≤ k < 1 and μ is supported in the ball B_{R} = \{z : |z| < R\} and that μ(z̄) = \overline{μ(z)}. Then for any point z ∈ B_{R},

|F^{μ}(z) − z| ≤ C_{2}∥μ∥_{∞}|z|^{α}

and for any z ∈ B_{1} ,

\frac{1}{C_{2}} |z|^{1/α} ≤ |F^{μ}(z)| ≤ C_{2}|z|^{α}.

Proof. Equation (10) of Section V.B of [1] implies that

(2)  |F^{μ}(z_{1}) − F^{μ}(z_{2})| ≤ C(k, R)∥μ∥_{∞}|z_{1} − z_{2}|^{α} + |z_{1} − z_{2}|,

for some α < 1 which depends only on k and such that α → 1 as k → 0 (one may take α = 1 − 2/p where p is such that kK_{p} < 1 with K_{p} the norm of the Beurling transform T on L^{p}, p > 2). Applying this to z_{2} = 0 for both F^{μ} and its inverse clearly implies the second estimate, so we need only prove the first.

If we apply (2) to z_{1} = 0 and z_{2} = x ∈ [−R, R] and set ε = ∥μ∥_{∞}|z|^{α}, we see that

|F^{μ}(x)| ≤ C(k, R)ε + |x| ≤ |x| + Cε.

Since F^{μ} is an orientation preserving homeomorphism on R, and applying the same argument to the inverse map, we get |F^{μ}(x) − x| ≤ Cε, as desired.

Now consider a point z = x + iy ∈ B_{R} ∩ H. The estimate above implies |F^{μ}(z)| ≤ |z| + Cε. Considering the inverse map shows

−Cε ≤ |F^{μ}(z)| − |z| ≤ Cε.

Let t = |z|(x/|x|). As above, we can deduce

−Cε ≤ |F^{μ}(z) − F^{μ}(t)| − |z − t| ≤ Cε,

and hence by the previous paragraph,

−2Cε ≤ |F^{μ}(z) − t| − |z − t| ≤ 2Cε.

Since the circles {w : |w| = |z|} and {w : |w − t| = |z − t|} intersect at an angle which is bounded away from zero, we see that the 2Cε neighborhoods of these circles in H intersect in a set of diameter at most 20Cε and this intersection contains both z and F^{μ}(z). This proves the lemma.

Lemma 2.5. If k < 1 is small enough then there is a constant C_{3} = C_{3}(k) so that the following holds. Suppose ∥μ∥_{∞} ≤ k < 1. Then

∥f_{z}^{μ} − 1∥_{p,1} ≡ (∫_{B_{1}} |f_{z}^{μ} − 1|^{p} dx dy)^{1/p} ≤ C_{3}∥μ∥_{∞}

for all 2 ≤ p ≤ p(k).

Proof. This is the Lemma in Section V.C of [1], although the statement there only claims that ∥f_{z}^{μ} − 1∥_{p} → 0 as ∥μ∥_{∞} → 0. We sketch the proof making the necessary changes.

First assume μ is supported in B_{R} and let ε = ∥μ∥_{∞}. It is proven on page 100 of [1] that ∥F_{z}^{μ} − 1∥_{p} ≤ C∥μ∥_{p} ≤ CεR^{2/p} if p < p(k). Since f^{μ} = F^{μ}/F^{μ}(1), Lemma 2.4 implies

\begin{aligned}∥f_{z}^{μ} − 1∥_{p,1} & = ∥\frac{F_{z}^{μ}}{F^{μ}(1)} − 1∥_{p,1} = |1 − \frac{1}{F^{μ}(1)}| + \frac{1}{F^{μ}(1)}∥F_{z}^{μ} − 1∥_{p,1} \\ & ≤ 2C_{2}ε + \frac{C(R)}{1 − Cε}ε ≤ Cε. \end{aligned}

Now write f̌(z) = 1/f(1/z). We want to show

(3)  ∥f̌_{z}^{μ} − 1∥_{p,R} ≤ C(R)ε,

when μ has support in B_{R}. Just as above, it suffices to show ∥F̌_{z}^{μ} − 1∥_{p,R} ≤ Cε. Note that F̌^{μ} is analytic on {z : |z| < 3r} where r = 1/(3R). For an analytic function f on a ball B(x, r) it is easy to see by the mean value property and Hölder’s inequality that

|f(x)| ≤ \frac{1}{πr^{2}} ∫_{B(x,r)} |f| ≤ \frac{1}{(πr^{2})^{1/p}} ∥f∥_{L^{p}(B(x,r))}.

Thus by the maximum principle,

\begin{aligned}∫_{|z|<r} |F̌_{z}^{μ}(z) − 1|^{p} dx dy & ≤ C(r) sup_{|z|=2r} |F̌_{z}^{μ}(z) − 1|^{p} \\ & ≤ C(r) ∫_{r<|z|<3r} |F̌_{z}^{μ}(z) − 1|^{p} dx dy. \end{aligned}

On the other hand, changing variables from z to 1/z gives

∫_{r<|z|<R} |F̌_{z}^{μ}(z) − 1|^{p} dx dy = ∫_{1/R<|z|<1/r}|\frac{z^{2}F_{z}^{μ}(z)}{F^{μ}(z)^{2}} − 1|^{p} \frac{dx dy}{|z|^{4}}

= ∫_{1/R<|z|<1/r}|\frac{z^{2}(F_{z}^{μ}(z) − 1)}{F^{μ}(z)^{2}} + \frac{z^{2} − F^{μ}(z)^{2}}{F^{μ}(z)^{2}} |^{p} \frac{dx dy}{|z|^{4}}

≤ C ∫_{1/R<|z|<1/r}|\frac{z^{2}(F_{z}^{μ}(z) − 1)}{F^{μ}(z)^{2}} |^{p} + |\frac{z^{2} − F^{μ}(z)^{2}}{F^{μ}(z)^{2}} |^{p} \frac{dx dy}{|z|^{4}}

≤ C(R) ∫_{1/R<|z|<1/r} |F_{z}^{μ}(z) − 1|^{p} dx dy + C(R) ∫_{1/R<|z|<1/r} |z − F^{μ}(z)|^{p} dx dy

≤ C(R)ε^{p}.

Since the integral over {|z| < 3r} was dominated by a constant (depending only on R) times this estimate, we have proven (3).

The general case now follows just as in [1]. Write f = ǧ ∘ h where μ_{h} = μ_{f} inside the unit disk and μ_{h} = 0 outside the unit disk. Then

∥f_{z} − 1∥_{p,1} ≤ ∥[(ǧ_{z} − 1) ∘ h]h_{z}∥_{p,1} + ∥h_{z} − 1∥_{p,1}.

The second term is bounded by Cε by the first paragraph and the first term is bounded using

∥[(ǧ_{z} − 1) ∘ h]h_{z}∥_{p,1}^{p} = ∫_{B_{1}} |(ǧ_{z} − 1) ∘ h|^{p}|h_{z}|^{p} dx dy

≤ \frac{1}{1 − k^{2}} ∫_{h(B_{1})} |ǧ_{z} − 1|^{p}|h_{z} ∘ h^{−1}|^{p−2} dx dy

≤ \frac{1}{1 − k^{2}} (∫_{h(B_{1})} |ǧ_{z} − 1|^{2p} dx dy ∫_{B_{1}} |h_{z}|^{2p−4} dx dy)^{1/2}

Clearly h(B_{1}) ⊂ {|z| < R} for some R depending only on k. Thus using (3), the first integral is bounded by

∫_{B_{R}} |ǧ_{z} − 1|^{2p} dx dy ≤ Cε^{2p},

(assuming 2p < p(k); but since p(k) → ∞ as k → 0 this holds for some p > 2 if k is small enough). On the other hand

∫_{B_{1}} |h_{z}|^{2p−4} dx dy ≤ C(∫_{B_{1}} |h_{z}|^{2p} dx dy)^{1−2/p} ≤ ∥μ_{h}∥_{p,1} + ∥1∥_{p,1} ≤ C,

since ∥h_{z} − 1∥_{p} ≤ C∥μ_{h}∥_{p}.

Lemma 2.6. There is a 0 < k < 1 and a C_{4} < ∞ so that the following holds. Suppose that f is a quasiconformal mapping of the plane to itself which preserves H, fixing 0, 1 and ∞ and the Beltrami coefficient of f is μ with ∥μ∥_{∞} ≤ k. Then

|f(w) − [w − \frac{1}{π} ∫_{C} μ(z)R(z, w) dx dy]| ≤ C_{4}∥μ∥_{∞}^{2},

for all |w| ≤ 1, where

R(z, w) = \frac{1}{z − w} − \frac{w}{z − 1} + \frac{w − 1}{z} = \frac{w(w − 1)}{z(z − 1)(z − w)}.

Proof. Again, we follow the proof in [1, Section V.C], simply inserting more explicit estimates at a few points. It is shown there that

f(w) = w − \frac{1}{π} ∫_{B_{1}} fz̄(z)R(z, w) dx dy − \frac{1}{π} ∫_{B_{1}} \frac{f̌z̄(z)}{f̌(z)^{2}} zS(z, w) dx dy,

where

S(z, w) = \frac{w^{2}}{1 − wz} − \frac{w}{1 − z}

and as before f̌(z) = 1/f(1/z). Using fz̄ = μf_{z} = μ + μ(f_{z} − 1), the first integral equals

∫_{B_{1}} μ(z)R(z, w) dx dy + ∫_{B_{1}} μ(z)(f_{z}(z) − 1)R(z, w) dx dy
= ∫ μ(z)R(z, w) dx dy + O(∥μ∥_{∞}∥f_{z} − 1∥_{p,1}∥R∥_{q,1})
= ∫ μ(z)R(z, w) dx dy + O(∥μ∥_{∞}^{2}),

by Lemma 2.5 and the fact that R ∈ L^{q}, for every q < 2 (with a bound depending on q, but not on w for |w| ≤ 1).

The second integral is estimated by writing f̌z̄ = μ̌ + μ̌(f̌_{z} − 1) where μ̌(z) = (z/z̄)^{2}μ(1/z). Repeating the argument above shows the second integral is equal to

∫_{B_{1}} \frac{μ̌(z)}{f̌(z)^{2}} + \frac{μ̌(z)(f̌_{z}(z) − 1)}{f̌(z)^{2}}zS(z, w) dx dy

= ∫_{B_{1}} μ(\frac{1}{z})[\frac{1}{z̄^{2}} + \frac{z^{2} − f̌(z)^{2}}{z̄^{2}f̌(z)^{2}} ]zS(z, w) dx dy + ∫_{B_{1}} \frac{μ̌(f̌_{z} − 1)}{f̌(z)^{2}}zS(z, w) dx dy

= ∫_{B_{1}} μ(\frac{1}{z})\frac{1}{z̄^{2}} zS(z, w) dx dy + I + II.

Using Lemma 2.4, we see

\frac{1}{C} |z|^{1/α} ≤ |f̌(z)| ≤ C|z|^{α}, |z − f̌(z)| ≤ C∥μ∥_{∞}|z|^{α},

so we can estimate I by

\begin{aligned}I & ≤ |∫_{B_{1}} μ(1/z)\frac{z^{2} − f̌(z)^{2}}{z̄^{2}f̌(z)^{2}} zS(z, w) dx dy| \\ & ≤ C∥μ∥_{∞}^{2} ∫_{B_{1}} |z|^{2α−1−2/α}S(z, w) dx dy ≤ C∥μ∥_{∞}^{2}C(α), \end{aligned}

if 2α − 1 − 2/α > −2 (recall that we may take α as close to 1 as we wish, if k is small enough). To estimate II, note that for 1/p + 1/q = 1. Lemma 2.5 implies

\begin{aligned}II & = ∫_{B_{1}} \frac{μ̌(z)(f̌_{z}(z) − 1)}{f̌(z)^{2}}zS(z, w) dx dy ≤ C∥μ∥_{∞}∥f̌_{z} − 1∥_{p}∥\frac{zS(z, w)}{f̌(z)^{2}} ∥_{q} \\ & ≤ ∥μ∥_{∞}^{2}∥z^{1−2/α}S(z, w)∥_{q}. \end{aligned}

Fix some q < 2, and take k so small that α > 2q/(2 + q), which implies the L^{q} norm is finite (with bound depending only on α, hence only on k). Thus,

f(w) = w − \frac{1}{π} ∫_{B_{1}} μ(z)R(z, w) dx dy − \frac{1}{π} ∫_{B_{1}} μ(\frac{1}{z})\frac{1}{z̄^{2}} zS(z, w) dx dy + O(∥μ∥_{∞}^{2}).

Changing variables from z to 1/z in the second integral converts the integrand to the same form as the first (but now over {|z| > 1}). Hence,

f(w) = w − \frac{1}{π} ∫_{R^{2}} μ(z)R(z, w) dx dy + O(∥μ∥_{∞}^{2}),

as desired.

Lemma 2.7. There is a δ_{0} > 0 and a C_{5} < ∞ such that the following holds. Suppose 0 < δ ≤ δ_{0} and that φ is a decreasing continuous function of compact support on [0, δ). Then

∫_{H} φ(ρ(z, i)) \frac{dx dy}{y^{2}} ≤ ∫_{H} φ(ρ(z, i)) dx dy ≤ (1 + C_{5}δ^{2}) ∫_{H} φ(ρ(z, i)) \frac{dx dy}{y^{2}} .

Proof. If we can prove this when φ(t) = χ_{[0,s]}(t), 0 < s ≤ δ, is the characteristic function of an interval, then by linearity it holds for decreasing step functions. By passing to uniform limits it holds for all decreasing continuous functions of compact support. Thus it suffices to prove that if H(s) denotes the hyperbolic area of a hyperbolic ball B ⊂ H is a hyperbolic ball of radius s centered at i, and E(s) is its Euclidean area, then H(s) ≤ E(s) ≤ (1 + Cδ^{2})H(s).

Considering where B hits the imaginary axis, and using dρ = |dz|/y, it is easy to see the Euclidean diameter of B is (1/y) − y where y = e^{−s} and hence its Euclidean area is

E(s) = \frac{π}{4} (\frac{1}{y} − y)^{2} = \frac{π}{4} (e^{s} − e^{−s})^{2} = π(s^{2} + \frac{1}{3}s^{4} + O(s^{6})).

To compute the hyperbolic area we move to the disk and assume B is a hyperbolic ball of radius s centered at the origin. Since dρ = 2|dz|/(1 − |z|^{2}), we see that the Euclidean radius of B is r = (e^{s} − 1)/(e^{s} + 1). Thus the hyperbolic area of B is

\begin{aligned}H(s) & = 4 ∫_{0}^{2π} ∫_{0}^{r} \frac{t}{(1 − t^{2})^{2}} dt dθ = 8π ∫_{0}^{r^{2}} \frac{1}{(1 − u)^{2}} \frac{du}{2} \\ & = 4π \frac{r^{2}}{1 − r^{2}} = π(e^{s} − 2 + e^{−s}) = π(s^{2} + \frac{s^{4}}{12} + O(s^{6})), \end{aligned}

and hence

\frac{E(s)}{H(s)} = \frac{s^{2} + \frac{1}{3}s^{4} + O(s^{6})}{s^{2} + \frac{1}{12}s^{4} + O(s^{6})} = 1 + \frac{1}{4}s^{2} + O(s^{4}),

which proves the lemma.

Now suppose f is a K -quasiconformal selfmap of H which fixes 0, 1 and ∞. Let μ = μ_{f} be the associated symmetric Beltrami coefficient. Given a 0 < δ < 1, choose a φ on [0, δ] so that φ(z, i) = φ(ρ(z, i)) is a smooth, positive function and

∫_{H} φ(z, i) \frac{dx dy}{y^{2}} = 1.

This can clearly be done so that φ also satisfies

|φ| ≤ \frac{C}{δ^{2}} , |φ′| ≤ \frac{C}{δ^{3}} ,

for some C < ∞ which is independent of δ. Suppose z = x + iy and define

ψ(z, w) = \frac{1}{y^{2}} φ(ρ(z, w)).

If w = u + iv then a simple linear change of variables shows

\begin{aligned}∫_{H} ψ(z, w) dx dy & = ∫_{H} φ(z, i) \frac{dx dy}{y^{2}} = 1, \\ ∫_{H} ψ(z, w) du dv & = ∫_{H} φ(i, w) du dv ≤ 1 + Cδ^{2}. \end{aligned}

Now for y > 0, define

μ̃(z) = ∫_{H} μ(w)ψ(z, w) dw,

and symmetrically for y < 0.

Corollary 2.8. With μ and μ̃ as above, ∥μ̃∥_{∞} ≤ ∥μ∥_{∞}(1 + C_{6}δ^{2}).

This is immediate from the preceding remarks. Also note for later

Lemma 2.9. μ̃ is a smooth function off R with

|∇μ̃(x + iy)| ≤ ∥μ∥_{∞} \frac{C}{δ} \frac{1}{|y|}.

Proof. To prove this, let z_{1} = x_{1} + iy_{1} , z_{2} = x_{2} + iy_{2} , ε = |z_{1} − z_{2}| and assume, without loss of generality, that y_{1} ≤ y_{2} and ε < δy_{1} . Then (as a function of w) |ψ(z_{1}, w) − ψ(z_{2}, w)| is supported on a hyperbolic 3δ ball B around z_{1} (and hence has Euclidean area ≤ Cδ^{2}y_{1}^{2} ). Hence

|μ̃(z_{1}) − μ̃(z_{2})| ≤ ∥μ∥_{∞} ∫_{H} |ψ(z_{1}, w) − ψ(z_{2}, w)| dw
= ∥μ∥_{∞} ∫_{H} |y_{1}^{−2}φ(z_{1}, w) − y_{2}^{−2}φ(z_{2}, w)| dw
≤ ∥μ∥_{∞}[∫_{B} y_{1}^{−2}|φ(z_{1}, w) − φ(z_{2}, w)| dw + ∫_{B} |y_{1}^{−2} − y_{2}^{−2}|φ(z_{2}, w) dw]
≤ ∥μ∥_{∞}[Cεδ^{−3}y_{1}^{−1} ∫_{B} y_{1}^{−2} dw + Cδ^{−2} ∫_{B} εy_{1}^{−3} dw]
≤ ∥μ∥_{∞}[Cεδ^{−1}y_{1}^{−1} + Cεy_{1}^{−1}].

Dividing by ε = |z_{1} − z_{2}| gives the desired estimate.

Let f̃ = f μ̃ be the quasiconformal map with Beltrami coefficient μ̃ which fixes 0, 1, ∞.

Lemma 2.10. There is a 0 < k < 1 and C_{7} < ∞ so that the following holds. Suppose f is a K -quasiconformal selfmap of H with ∥μ∥_{∞} ≤ k < 1 and let f̃ be the map obtained by smoothing μ = μ_{f} as above. Then

(4)  max_{x∈[0,1]} |f(x) − f̃(x)| ≤ C_{7}∥μ∥_{∞}^{2}.

Proof. To prove (4) we use Lemma 2.6. The main point is that for a fixed w ∈ R, R(z, w) is analytic in z (off the real line) and hence harmonic. In 2 dimensions, Euclidean harmonic functions are the same as hyperbolically harmonic functions (e.g. [6]). Thus R is hyperbolically harmonic and thus satisfies the mean value property with respect to averaging over hyperbolic balls. Thus if ζ ∈ [0, 1],

∫_{H} R(z, ζ)ψ(z, w) dx dy = ∫_{H} R(z, ζ)φ(ρ(z, w)) \frac{dx dy}{y^{2}} = R(w, ζ).

Hence

\begin{aligned}_{∫H} & μ̃(z)R(z, ζ) dx dy = ∫_{H}[∫_{H} μ(w)ψ(z, w) du dv]R(z, ζ) dx dy \\ & = ∫_{H} μ(w)[∫_{H} R(z, ζ)ψ(z, w) dx dy] du dv = ∫_{H} μ(w)R(w, ζ) du dv. \end{aligned}

Thus if we apply Lemma 2.6 to both f and f̃, the integral terms are identical and hence cancel, giving

|f(ζ) − f̃(ζ)| ≤ 2C_{4}∥μ∥_{∞}^{2},

for all ζ ∈ [0, 1].

Corollary 2.11. With f and f̃ as above, g = f ∘ f̃^{−1} is (1 + C_{8}∥μ∥_{∞}^{2})- quasisymmetric on the real line.

Proof. For any two real numbers a < b, we want to estimate

\frac{g(b) − g(\frac{1}{2}(a + b))}{g(\frac{1}{2}(a + b)) − g(a)}.

After re-normalizing by linear maps, we may as well assume a = g(a) = 0 and b = g(b) = 1, so we want to show

\frac{1}{1 + C∥μ∥_{∞}^{2}} ≤ \frac{1 − g(\frac{1}{2})}{g(\frac{1}{2}) − 0} ≤ 1 + C∥μ∥_{∞}^{2}.

Thus it is enough to show that

|g(\frac{1}{2}) − \frac{1}{2}| = |f(f̃^{−1}(\frac{1}{2})) − \frac{1}{2} | = O(∥μ∥_{∞}^{2}).

This follows by taking x = f̃^{−1}(\frac{1}{2}) in (4).

Lemma 2.12. There is a 0 < k < 1 and a C_{9} < ∞ so that the following holds. Suppose f is a quasiconformal map of H to itself and its Beltrami coefficient μ satisfies ∥μ∥_{∞} ≤ ε ≤ k and

|∇μ(x + iy)| ≤ M \frac{ε}{y} .

Then f is (1 + C_{9}Mε)-biLipschitz with respect to the hyperbolic metric.

Proof. Since the inverse of f satisfies the same hypotheses, it suffices to show f is Lipschitz. Also, after rescaling by linear maps, we may just give a Lipschitz estimate at i.

Write f = g ∘ h where on H we have μ_{h} = μ_{f} on B_{1} = B(i, \frac{1}{4}), μ_{h} = 0 off B_{2} = B(i, \frac{1}{2}) and |∇μ_{h}(x + iy)| ≤ 2Mε/y (and symmetrically on the lower half-plane). Thus g is analytic on h(B_{1}), which contains a ball B_{3} of fixed radius (depending only on k) around h(i). Thus by the mean value theorem for analytic functions

|g_{z}(h(i))| ≤ \frac{1}{|B_{3}|} ∫_{B_{3}} |g_{z}| dx dy ≤ 1 + C ∫_{B_{3}} |g_{z} − 1| dx dy ≤ 1 + C∥μ∥_{∞},

by Lemma 2.5.

To estimate the Lipschitz constant for h, we follow the proof from [1, Lemma 3, Section V.B] that continuity of μ implies differentiability of h. It is shown there that h_{z} = λ = e^{σ} and hz̄ = μλ = μe^{σ}, with

(5)  σ = P (μ_{h}v + (μ_{h})_{z}) + K,

where P denotes the Cauchy transform, K is a constant chosen so that σ(w) → 0 as w → ∞ and v satisfies v = T (μ_{h}v) + T ((μ_{h})_{z}), where T is the Beurling transform. Since μ_{h} and (μ_{h})_{z} are in every L^{p}, p < ∞, this equation can be solved via the geometric series

v = T ((μ_{h})_{z}) + T (μT ((μ_{h})_{z}) + T μT (μT ((μ_{h})_{z}) + …,

and v satisfies

∥v∥_{p} ≤ \frac{C∥(μ_{h})_{z}∥_{p}}{1 − C∥μ_{h}∥_{∞}} .

Thus if 1/p + 1/q = 1, p > 2,

|σ| ≤ (∥\frac{1}{z − w} ∥_{q} + ∥\frac{1}{z}∥_{q})(∥μv∥_{p} + ∥(μ_{h})_{z}∥_{p}) ≤ C_{q}(ε∥v∥_{p} + CMε) ≤ CMε.

Thus

|h_{z}| ≤ e^{σ} ≤ 1 + CMε,

108 Christopher J. Bishop

and similarly for |hz̄| = |μ_{h}h_{z}| ≤ CMε.

Finally we have to estimate K in (5). Taking w → ∞ in (1), we see that

K = \frac{1}{π} ∫ (μ_{h}v + (μ_{h})_{z}) \frac{dx dy}{z} .

Hence, using the same estimates as above, we get

|K| ≤ ∥\frac{1}{z}∥_{q}∥μ_{h}∥_{∞}∥v∥_{p} + ∥\frac{1}{z}∥_{q}∥(μ_{h})_{z}∥_{p} ≤ Cε.

This proves the desired estimate.

We have now proven all the estimates used in the proof of Theorem 1.1, so the result is established.

References

[1] Ahlfors, L.V.: Lectures on Quasiconformal Mappings. - Math. Studies 10, Van Nostrand, 1966.

[2] Beurling, A., and L. Ahlfors: The boundary correspondence under quasiconformal mappings. - Acta Math. 96, 1956, 125–142.

[3] Bishop, C.J.: Quasiconformal Lipschitz maps, Sullivan’s convex hull theorem and Brennan’s conjecture. - Ark. Mat. 40, 2002, 1–26.

[4] Božin, V., N. Lakic, V. Marković, and M. Mateljević: Unique extremality. - J. Anal. Math. 75, 1998, 299–338.

[5] Douady, A., and C.J. Earle: Conformally natural extension of homeomorphisms of the circle. - Acta Math. 157, 1986, 23–48.

[6] Nicholls, P.J.: The Ergodic Theory of Discrete Groups. - London Math. Soc. Lecture Note Ser. 143, Cambridge University Press, 1989.

Received 4 October 2000