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CROSS-RATIO DISTORTION AND DOUADY-EARLE
EXTENSION: II. QUASICONFORMALITY AND ASYMPTOTIC
CONFORMALITY ARE LOCAL
By
JUN HU∗AND OLEG MUZICIAN
Abstract. Let f be an orientation-preserving circle homeomorphism and 8the Douady-Earle extension of f . In this paper, we show that the quasiconfor-
mality and asymptotic conformality of 8 are local properties; i.e., if f is quasi-
symmetric or symmetric on an arc of the unit circle, then 8 is quasiconformal or
asymptotically conformal nearby. Furthermore, our methods enable us to conclude
the global quasiconformality and asymptotic conformality from local properties.In the quasiconformal case, our methods also enable us to provide an upper bound
for the maximal dilatation of 8 on a neighborhood of the arc in the open unit disk
in terms of the cross-ratio distortion norm of f on the arc.
1 Introduction
Let f be an orientation-preserving homeomorphism of the unit circle S1 and 8 the
Douady-Earle extension of f to the closed unit disk D. In [6], Douady and Earle
show that 8 is quasiconformal if f admits a quasiconformal extension to D. It
is proved in [9] that 8 is asymptotically conformal if f admits an asymptotically
conformal extension to D. Using the Beurling-Ahlfors extension [5], we know that
if f is quasisymmetric (respectively, symmetric), then f admits a quasiconformal
(respectively, asymptotically conformal) extension to D. Therefore, the Douady-
Earle extension 8 of f is quasiconformal (respectively, asymptotically conformal)
if f is quasisymmetric (respectively, symmetric). Recently, we developed a new
method to derive the quasiconformality of 8 directly from the quasisymmetry of
f on S1. Our method also provides an upper bound for the maximal dilatation
K (8) of 8 on D in terms of the cross-ratio distortion norm || f ||cr of f on S1
[16]. In this paper, we generalize that method to show that the quasiconformality
of 8 and the upper bound for K (8) in [16] are all local properties; see Theorem
1. This generalized method also enables us to conclude global quasiconfromality
∗The research is partially supported by PSC-CUNY research awards.
JOURNAL D’ANALYSE MATHEMATIQUE, Vol. 117 (2012)
DOI 10.1007/s11854-012-0021-7
249
250 JUN HU AND OLEG MUZICIAN
of 8 from the local property; see Corollary 1. We then show that asymptotic
conformality of 8 also depends locally on the symmetry of f ; see Theorem 2.
Given a quadruple Q = {a, b, c, d} consisting of four points a, b, c, d on the
unit circle S1 arranged in counterclockwise order, we denote the following cross-
ratio of Q by
(1) cr(Q) =(b − a)(d − c)
(c − b)(d − a).
It is easy to check that a quadruple Q has cr(Q) = 1 if and only if the geodesic ac
from a to c is perpendicular to the geodesic bd from b to d . Given an orientation-
preserving homeomorphism f of S1, the cross-ratio distortion norm of f is
defined as
(2) || f ||cr = supcr(Q)=1
| ln cr( f (Q))|,
where
(3) cr( f (Q)) =( f (b) − f (a))( f (d ) − f (c))
( f (c) − f (b))( f (d ) − f (a)).
If || f ||cr is finite, then f is said to be quasisymmetric. (This definition is equiva-
lent to the one requiring bounded ratio distortions under f on all symmetric triples
on the unit circle S1.) Similarly, f is quasisymmetric on an open arc I on S1
if || f |I ||cr is finite. Furthermore, f is symmetric if ln cr( f (Q)) converges to 0
uniformly as s(Q) approaches to 0 for all Q with cr(Q) = 1, where s(Q) is defined
to be the minimum of |b − a|, |c − b|, |d − c| and |a − d |. Similarly, we say f
is symmetric on an open arc I on S1 if the previous convergence holds for all
quadruples Q ⊂ I . For applications of the cross-ratio distortion norm to Thurston
earthquake maps, see [10], [13] and [14].
Now we can give the precise statements of our results.
Theorem 1. Let f be an orientation-preserving homeomorphism of the unit
circle S1 and 8 the Douady-Earle extension of f to the closed unit disk D. Let
p ∈ S1 and Ip be an open arc on S1 containing p and symmetric with respect to
p. If || f |Ip||cr < ∞, then there exists an open hyperbolic half plane Up with p at
the middle of its boundary on S1 such that ln K (8|Up) ≤ C1|| f |Ip
||cr + C2 for two
universal positive constants C1 and C2, where K (8|Up) is the maximal dilatation
of 8 on Up.
Corollary 1 (Douady-Earle [6]). If f is a quasisymmetric homeomorphism
of S1, then the Douady-Earle extension 8 of f is quasiconformal.
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 251
Remark 1. Corollary 1 is called the Douady-Earle Extension Theorem.
Using this theorem and a functional property of the extension developed in [6]
(see Proposition 1 in Section 4), one can construct a short proof (by contradiction)
for a weaker version of Theorem 1; that is, if f is quasisymmetric on Ip, then 8 is
quasiconformal on a neighborhood Up of p. Theorem 1 provides three additional
features.
(1) The proof does not use the Douady-Earle Extension Theorem at any step,
and therefore one can deduce that theorem as a consequence.
(2) It provides an upper bound for the maximal dilatation of 8 on Up in terms
of the cross-ratio distortion norm of f on Ip.
(3) The Euclidean size of Up only depends on the size of Ip, not the position of
Ip.
Remark 2. Note that the proof of the Douady-Earle Extension Theorem pre-
sented in [6] can only be applied to derive the global quasiconformality of 8 when
the global quasisymmetry of f is assumed.
Remark 3. Given an orientation-preserving circle homeomorphism f , the
Douady-Earle extension 8 of f is quasiconformal on any closed disk contained in
the unit open disk D; but the maximal dilatation of 8 may explode as the closed
disk exhausts D. A condition was introduced in Lemma 3.6 of [20] on a lifting
h of f to the real line so that there is an upper bound depending only on the two
constants used in the condition for the quasiconformality of 8 in a neighborhood
of 0. Note that this is a different type of local property. Furthermore, the quan-
titative condition used in that lemma is not invariant under pre-composition and
post-composition by a conformal isometry on D. By contrast, the cross-ratio dis-
tortion norm is invariant under both compositions, which is significantly useful in
our work.
Theorem 2. Under the same notation as introduced in the statement of Theo-
rem 1, if f is an orientation-preserving homeomorphism of S1 and is symmetric on
a neighborhood Ip of p in S1, where p ∈ S1, then 8( f ) is asymptotically conformal
on a neighborhood Up of p in D, with Up ∩ S1 ⊂ int(Ip).
Theorem 2 implies the following results.
Corollary 2 (Wang [22]). If f is a quasisymmetric homeomorphism of S1
and is symmetric on an open arc Ip on S1, then there exists an open hyperbolic
half plane Up such that 8 is asymptotically conformal on Up.
Corollary 3 (Earle-Markovic-Saric [9]). If f is a symmetric homeomorphism
of S1, then the Douady-Earle extension 8 of f is asymptotically conformal.
252 JUN HU AND OLEG MUZICIAN
Remark 4. The study of the space T0 of all symmetric circle homeomor-
phisms was initiated by Gardiner and Sullivan in [12]. The Teichmuller structure
and other metric structures on T0 and the quotient space T/T0 of the universal Te-
ichmuller space T by T0 were extensively investigated in [7], [8], and [11]. For
example, the coincidence of the Teichmuller and Kobayashi metrics on T0 is de-
duced in [8] as an immediate consequence of an approximation theorem developed
in [11]. Recently, a short and direct proof of this result was presented in [15].
The strategy for proving Theorem 2 is essentially similar to that for the short
proof mentioned in Remark 1, but it applies our Theorem 1 instead of the Douady-
Earle Extension Theorem.
The paper is arranged as follows. Some background is provided in Section
2. Theorem 1 is proved in Section 3. Finally, in Section 4, we give a proof of
Corollary 2, then show Theorem 2, and at the end sketch the short proof mentioned
in Remark 1 for a weaker version of Theorem 1.
2 Some background on the Douady-Earle extension
Let f be an orientation-preserving homeomorphism of the unit circle S1 and 8 the
Douady-Earle extension of f . Let us first briefly recall how 8 is defined. Given
a point z on the open unit disk D centered at the origin, let ηz be the harmonic
measure on S1 viewed from z (normalized to have the measure of S1 equal to 1),
and let f∗(ηz) be the push-forward of the measure ηz by f . It is shown in [6] that
there exists a unique point w ∈ D such that
(4)
∫
S1
ζ − w
1 − wζdf∗(ηz)(ζ ) = 0,
which is called the conformal barycenter of the measure f∗(ηz) and denoted by
B( f∗(ηz)). The Douady-Earle extension 8 of f is defined as
8(z) =
B( f∗(ηz)) for z ∈ D,
f (z) for z ∈ S1.
The extension 8 has the following two important features.
(1) 8 is an orientation-preserving homeomorphism on the closed disk D and a
real analytic diffeomorphism on D.
(2) 8 is conformally natural in the sense that for any two conformal isometries
A and B on D, 8(A ◦ f ◦ B) = A ◦ 8( f ) ◦ B; see [6].
Using the conformal naturality of 8, we can estimate the maximal dilatation of 8
at an arbitrary point in D by doing the same work with a normalized extension at
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 253
the origin. By a normalized extension 8 we mean 8(0) = 0. Now assume that
the Douady-Earle extension 8 of f is normalized. The Beltrami coefficient of
8 at the origin is explicitly expressed in [6], and we use this explicit expression
in this paper. Given any point z ∈ D, 8(z) is the unique point w ∈ D such that
F (z, w) = 0, where
(5) F (z, w) =1
2π
∫
S1
f (ξ ) − w
1 − w f (ξ )·
1 − |z|2
|z − ξ |2|dξ |.
If we let
(6) c1 =∂F
∂z(0, 0) =
1
2π
∫
S1
ξ f (ξ )|dξ |, c−1 =∂F
∂z(0, 0) =
1
2π
∫
S1
ξ f (ξ )|dξ |
and
(7) d1 =∂F
∂w(0, 0) = −1, d−1 =
∂F
∂w(0, 0) =
1
2π
∫
S1
f (ξ )2|dξ |,
then
(8)∂w
∂z(0) = −
(∂F/∂w)(0, 0)(∂F/∂z)(0, 0) − (∂F/∂w)(0, 0)(∂F/∂z)(0, 0)
|(∂F/∂w)(0, 0)|2 − |(∂F/∂w)(0, 0)|2
and
(9)∂w
∂z(0) = −
(∂F/∂w)(0, 0)(∂F/∂z)(0, 0) − (∂F/∂w)(0, 0)(∂F/∂z)(0, 0)
|(∂F/∂w)(0, 0)|2 − |(∂F/∂w)(0, 0)|2.
Furthermore, the Jacobian of w = 8( f ) at 0 equals
(10)
∣∣∣∣∂w
∂z(0)
∣∣∣∣2
−
∣∣∣∣∂w
∂z(0)
∣∣∣∣2
=|c1|
2 − |c−1|2
|d1|2 − |d−1|2.
Now let h : R → R be a lifting of f to the real line R satisfying h(u + 2π) =
h(u) + 2π. Then one easily obtains
(11) |d1|2 − |d−1|
2 = 2
(1
2π
)2 ∫ 2π
0
∫ 2π
0
sin2(h(s) − h(t))dsdt.
It is given in [6] that
(12) |c1|2 − |c−1|
2 =
(1
2π
)2 ∫ π
u=0
sin u
∫ 2π
t =0
H (t, u)dtdu
with
H (t, u) = sin(h(t + u) − h(t)) + sin(h(t + 2π) − h(t + u + π))
+ sin(h(t + π + u) − h(t + π)) + sin(h(t + π) − h(t + u)).
254 JUN HU AND OLEG MUZICIAN
Let α j , j = 1, 2, 3, 4, be the differences input in the sine function in H (t, u).
Clearly, all α j ’s are nonnegative and∑4
j =1 α j = 2π. Applying the trigonometric
summation formula, we can express H [2] as
(13) H (t, u) =
4∑
j =1
sin α j = 4 sin
(α1 + α2
2
)sin
(α1 + α3
2
)sin
(α2 + α3
2
).
From (11), it is easy to see that |d1|2 − |d−1|
2 > 0, and from (12) and (13) one can
see that |c1|2 − |c−1|
2 > 0, since H (t, u) ≥ 0 for all t and u and is not identically
equal to 0. It follows that the Jacobian of 8 at the origin is positive. By the
conformal naturality, the Jacobian of 8 is positive at every point z ∈ D. This is
how Douady and Earle prove 8 to be a diffeomorphism of D [6].
Remark 5. Growing out of ideas in [3] and unpublished ideas of Milnor, an
effective algorithm for finding the images of points under 8, called the MAY it-
erator, was formally introduced in [1]. Conformally natural extensions were then
introduced in [2] for monotone degree ±1 continuous circle maps by replacing
conformal barycenters with attractors of the MAY iterators corresponding to dif-
ferent points on D. On the other hand, again by using conformal barycenters,
such extensions are defined and studied in [17] for continuous circle maps under
which the push-forward measures of the Lebesgue measure have no atom. Finally,
by using a piecewise gluing method, conformally natural extensions are given to
arbitrary continuous circle maps in [21].
3 Local quasiconformality
In this section, we generalize a method developed in [16] to prove Theorem 1 and
Corollary 1. We first introduce some terminology and notation.
Given two points X and Y on the unit circle S1, we denote by XY the closed
circular arc on S1 from X to Y in counterclockwise direction and by (X, Y ) the open
circular arc defined in the same way. We denote by |XY | the length of XY and by
|XY | the Euclidean distance between X and Y on the complex plane. Given n
points X1, X2, . . ., Xn on S1 arranged in the counterclockwise direction, we denote
by X1X2 · · · Xn the circular arc on S1 from X1 to Xn through the other points in the
counterclockwise direction.
Let A, B , C and D be the endpoints of two perpendicular diameters of S1,
arranged in the counterclockwise direction. The following concept is intended to
be defined for a semicircle on S1 in an arbitrary position. In order to introduce
numerical values relatively easily, we assume that A is located at the point −1; see
Figure 1.
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 255
Figure 1. An illustration for the scale of I c.
Referring to Figure 1, let E be the point on S1 such that the hyperbolic geodesic
on D connecting E to B is perpendicular to the one connecting A to D . Using the
cross-ratio condition for perpendicularity, we find that E = (−4 + 3i)/5. The
length of the shorter circular arc between E and A is then strictly between π/6
and π/4. Let τ be the length of the shorter circular arc between D and E . Then
π/4 < τ < π/3. Now let H = eiπ/4 and G = ei5π/4, and let F be the point on the
unit circle such that the arc FG, in the counterclockwise direction from F to G,
equals τ. Then the arc EF is of 3π/4 − 2τ, which is between π/12 and π/4.
Let J (denoted by I c) be a closed circular arc contained in the open arc (E, F)
and of length less than 3π/4 − 2τ, and let Jc (that is, I ,) be the complement of J
in S1. Then the semicircle BCD has the following two properties:
(i) BCD is contained in I , and
(ii) the length of one component of I \ BCD is greater than τ and the length of
the other is greater than π/4 + τ.
For future convenience, we introduce two definitions. Given a semicircle BCD
contained in a circular arc I , we say that BCD is τ-contained in I if the above
two conditions are satisfied. (Figure 1 gives an example.) Secondly, given three
points A, B and C on S1, we call the other endpoint of the geodesic passing through
C and perpendicular to the geodesic connecting A to B the hyperbolic middle
point between A and B with respect to C.
Note first that since BCD is a semicircle τ-contained in I , the other endpoint A
of the diameter passing through C and perpendicular to the diameter connecting B
to D belongs to I . (Refer to Figure 1 as illustration.)
256 JUN HU AND OLEG MUZICIAN
Lemma 1. Suppose that f is an orientation-preserving homeomorphism of S1
and the Douady-Earle extension 8( f ) of f fixes the origin. Let I be a circular arc
on S1 containing a semicircle τ-contained in I . Assume that || f |I ||cr is finite. Then
for any semicircle BCD (with C at the middle) which is τ-contained in I , f (BCD)
has length greater than or equal to ǫ3 = 2 arcsin(√
3/(4e|| f |I ||cr + 2√
3)).
Furthermore, if A is the other endpoint of the diameter passing through C and
perpendicular to the diameter BD, then the semicircle ABC is contained in I (but
not τ-contained in I), and f (ABC) also has length greater than or equal to ǫ3.
Finally, both f (DAB) and f (CDA) also have length greater than or equal to ǫ3.
This lemma is the first main estimate of the paper. It is a local version of
[16, Lemma 1]. While their proofs follow similar ideas, the proof for the local
version requires considerably more effort. In fact, the intuition for the proof is
quite simple; that is, if f is the restriction of a Mobius transformation on S1 and
8( f ) fixes the origin, then f has to be a rotation around the origin. Therefore, all
assertions given in the lemma hold in an obvious way.
Our lemma states that if the cross-ratio distortion norm of f on a large enough
circular arc I ⊂ S1 is bounded by a positive constant M , and if 8( f ) satisfies
the normalization condition 8( f )(0) = 0, then the images of certain semicircles
contained in I cannot be too short. Two properties are used in the proof. One
is that the image of any circular arc I ′ on S1 of Lebesgue measure (normalized
so that the measure of S1 equals 1) strictly greater than 1/2 cannot be too short;
otherwise the µ f -measure (the push-forward of the Lebesgue measure under f ) of
that image set (strictly greater than 1/2) is almost concentrated at a point and then
fails the normalization condition 8( f )(0) = 0. The other property used is that if a
circular arc I ′ is contained in I with spaces on both sides in I , then a bound under
f on cross-ratio distortions on I implies a bound on ratio distortions under f on
I ′. This property implies that if a subinterval I ′′ of I ′ is proportional in length to
I ′, then the images of I ′′ and I ′ under f are also proportional in length.
Before we present the details of the proof, let us notice that the normalization
condition 8( f )(0) = 0 is preserved under pre-composition or post-composition of
rotations around the origin. Moreover, the cross-ratio distortion norm of f on a
circular arc I ⊂ S1 is also preserved under pre-composition or post-composition
with a Mobius transformation (in the case of pre-composition, the composed map
has the same cross-ratio distortion norm on the pre-image of I under the pre-
composed Mobius transformation). For these reasons, in the proof, we may
pre-compose and/or post-compose f with rotations around the origin to arrange
considered configurations into convenient positions. In these configurations, qual-
itative or quantitative assertions can be justified relatively easily. For brevity, in
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 257
Figure 2. A reference figure for Step 1 in the proof of Lemma 1.
what follows, whenever we apply such an action to have a configuration in a de-
sired position, we simply say that by rotational invariance, we may assume that
the considered configuration is in the desired position.
Proof. Let A be the other endpoint of the diameter perpendicular to the di-
ameter connecting B and D . Denote the images of A, B, C, and D under f by
A′, B ′, C ′ and D ′, respectively. By rotational invariance, we may first assume that
AC is horizontal, and hence BD is vertical; see the left part in Figure 2.
Step 1. In this step, we show that either A′B ′C ′ or B ′C ′D ′ has arc length at
least π/3. The proof only requires the normalization condition 8( f )(0) = 0. By
rotational invariance, we further assume in this step that two images A′ and D ′ are
complex conjugates.
Let x′ be the x-coordinate of A′. The normalization condition 8( f )(0) = 0
means 0 =∫S1 zdµ f (z) =
∫S1 xdµ f (z) + i
∫S1 ydµ f (z). Then
0 =
∫
S1
xdµ f (z) ≥ (−1)µ f (D ′A′) + x′µ f ( A′B ′C ′D ′) = (−1)
(1
4
)+ x′
(3
4
).
Thus x′ · 3/4 + (−1) · 1/4 ≤ 0. Then x′ ≤ 1/3 < 1/2. Hence | A′B ′C ′D ′| ≥ 2π/3.
Therefore either |A′B ′C ′| ≥ π/3 or |B ′C ′D ′| ≥ π/3. In order to obtain a lower
bound for |B ′C ′D ′|, we only need to handle the situation that |A′B ′C ′| ≥ π/3 and
|B ′C ′D ′| < π/3. We now proceed to Step 2.
Step 2. We show that there exists a constant δ3 > 0 such that |B ′C ′D ′| ≥ δ3.
The proof makes use of the assumptions that 8( f )(0) = 0, || f |I ||cr is finite, and the
semicircle BCD is τ-contained in I . We divide the proof into two cases according
to whether |C ′D ′A′| ≤ π or |C ′D ′A′| > π. The two proofs are similar.
258 JUN HU AND OLEG MUZICIAN
Figure 3. A reference figure for Case 1 in Step 2 of the proof of Lemma 1.
Case 1: |C ′D ′A′| ≤ π. Let E be the other endpoint of the geodesic passing
through B and perpendicular to the geodesic connecting A and D , and let E ′ be
the image of E under f . Since the semicircle BCD is τ-contained in I , the point
E belongs to I . For convenience in this case, by rotational invariance, we assume
that B ′ and E ′ are complex conjugates.
Since |DE | > π/4, one can see that µ f ( B ′C ′D ′E ′) > 5/8 (Refer to the left
part of Figure 3). Let x be the x-coordinate of B ′. The normalization condition
8( f )(0) = 0 implies that 0 =∫S1 xdµ f (z) ≥ (−1)µ f (E ′B ′)+ xµ f ( B ′C ′D ′E ′). Then
x ≤µ f (E ′B ′)
µ f ( B ′C ′D ′E ′)<
3/8
5/8=
3
5<
√
3
2.
Thus | B ′C ′D ′E ′| > π/3. On the other hand, using the assumptions that |B ′C ′D ′| <
π/3 (just before going into Step 2) and |C ′D ′A′| ≤ π (in this case), we obtain
| B ′C ′D ′E ′| < | B ′C ′D ′A′| < |B ′C ′D ′| + |C ′D ′A′| ≤π
3+ π =
4π
3.
Thus the Euclidean distance |B ′E ′| between B ′ and E ′ is greater than or equal to
1; that is, |B ′E ′| ≥ 1.
The method used in Step 1 also shows | B ′C ′D ′A′| ≥ 2π/3. Notice that we have
just proved that | B ′C ′D ′A′| ≤ 4π/3. Thus |A′B ′| ≥√
3.
Let M = || f |I ||cr . Considering the distortion of f on the quadruple {E,A,B,D},
we obtain e−M ≤ |E ′A′||B ′D ′|/(|A′B ′||D ′E ′|) ≤ 2|B ′D ′|/(√
3|D ′E ′|) and then
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 259
Figure 4. A reference figure for Case 2 in Step 2 of the proof of Lemma 1.
√
3e−M/2 ≤ |B ′D ′|/|D ′E ′|. Thus√
3e−M /2
1 +√
3e−M/2≤
|B ′D ′|
|B ′D ′| + |D ′E ′|≤
|B ′D ′|
|B ′E ′|.
Since |B ′E ′| ≥ 1, e−M/((2/
√
3) + e−M)
≤ |B ′D ′|/|B ′E ′| ≤ |B ′D ′|. Letting δ3 =
e−M/((2/
√
3) + e−M), we obtain |B ′D ′| ≥ δ3.
Now set ǫ3 = 2 arcsin(δ3/2). Then | f (BCD)| ≥ ǫ3.
Case 2: |C ′D ′A′| > π. In this case, |A′B ′C ′| < π. Let K be the other endpoint
of the geodesic passing through D and perpendicular to the geodesic connecting
A to B and K ′ = f (K ). By rotational invariance, we assume here that D ′ and K ′
are complex conjugates. In similar fashion to Case 1 (refer to Figure 4), we obtain
| f (BCD)| ≥ ǫ3.
We now show that the length of f (ABC) is also greater than or equal to ǫ3.
Since the hyperbolic middle point between C and D with respect to A and the one
between D and A with respect to C belong to I , one can show that | f (ABC)| ≥ ǫ3
in the same way as for f (BCD).
Finally, the proof for | f (DAB)| ≥ ǫ3 follows the same method. We need not
worry that the hyperbolic middle point between C and D with respect to B or the
one between B and C with respect to D falls into the region I c. For the same
reason, | f (CDA)| ≥ ǫ3. �
Corollary 4. Under the same assumptions as in Lemma 1, set
M = || f ||cr , δ4 =e−5M
64(2/√
3 + e−M )3, ǫ4 = 2 arcsin(δ4/2).
260 JUN HU AND OLEG MUZICIAN
Figure 5. A reference figure for the proof of Corollary 4.
Suppose that β is a circular arc on S1 having length between π/4 and π/2 and
which can be extended in one direction (either clockwise or counterclockwise) to
a semicircle BCD that is τ-contained in I . Then the length of f (β) is greater than
or equal to ǫ4.
Proof. It suffices to show that the conclusion holds when β has length π/4.
We divide the proof into two steps. In Step 1, we show that the conclusion holds
for β with length π/2; in Step 2, we show the case for β with length π/4.
Step 1. Let A, B, C, D, A′, B ′, C ′ and D ′ be the same given points as in Lemma
1 or its proof, and assume that β is the quarter circle between C and D . There is
no need to arrange these points in special positions as done in the proof of the
previous lemma. Let ǫ3 and δ3 be the constants obtained in Lemma 1. Note that
ǫ3 = 2 arcsin(δ3/2). By the previous lemma, | f (BCD)| ≥ ǫ3 and | f (DAB)| ≥ ǫ3.
Thus |B ′D ′| ≥ δ3. Similarly, we obtain |A′C ′| ≥ δ3.
Let M = || f |I ||cr . We show that if |C ′D ′| is less than or equal to δ3/2, then it
is greater than or equal to e−Mδ32/8. By the triangle inequality,
|B ′C ′| ≥ |B ′D ′| − |C ′D ′| ≥ δ3 −δ3
2=
δ3
2.
Similarly, |D ′A′| ≥ |C ′A′| − |C ′D ′| ≥ δ3/2. Considering the cross-ratio distortion
under f on the quadruple {A, B, C, D} and using the definition of || f |I ||cr , we
obtain
e−M ≤|A′B ′||C ′D ′|
|B ′C ′||D ′A′|≤
2|C ′D ′|
δ32/4
≤8|C ′D ′|
δ32
.
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 261
Figure 6. A reference figure for the proof of Lemma 2.
Thus |C ′D ′| ≥ e−M δ32/8.
Now we have shown that if |C ′D ′| ≤ δ3/2, then |C ′D ′| ≥ e−M δ32/8. This
implies that |C ′D ′| ≥ min{e−M δ32/8, δ3/2}. Since δ3 = e−M/
((2/
√
3)+e−M)
< 1,
e−Mδ32/8 < δ3/2. Thus |C ′D ′| ≥ e−Mδ3
2/8 in general. Set δ = e−M δ32/8 and
keep in mind that δ < δ3 < 1, a fact used in the next step.
Step 2. We continue to use the notation of Step 1. Let H and G be the middle
points on S1 between C and D and assume that H is on the short arc between C
and D; see Figure 5. Without loss of generality, it suffices to show in this step that
|H ′D ′| cannot be too small.
Suppose that |H ′D ′| ≤ δ/2. From the previous step, |C ′D ′| ≥ δ . By the
triangle inequality, |C ′H ′| ≥ |C ′D ′|−|H ′D ′| ≥ δ−δ/2 = δ/2. Since the semicircle
GBH is also τ-contained in I , Lemma 1 implies that |G′H ′| ≥ δ3. Also using
δ < δ3, we obtain |D ′G′| ≥ |G′H ′| − |H ′D ′| ≥ δ3 − δ/2 ≥ δ3/2. From the
definition of M = || f ||cr , we obtain
e−M ≤|G′C ′||H ′D ′|
|C ′H ′||D ′G′|≤
2|H ′D ′|
δδ3/4≤
8|H ′D ′|
δδ3
.
Thus |H ′D ′| ≥ e−Mδδ3/8.
In fashion similar to the discussion at the end of the previous step, we conclude
that |H ′D ′| ≥ δ4, where δ4 = min{e−Mδδ3/8, δ/2} = e−Mδδ3/8 (since δ < δ3 <
1). Then δ4 = e−2Mδ 33 /64 = e−5M/
(4(2/
√
3)+e−M)3
. Now let ǫ4 = 2 arcsin(δ4/2).
Then the length of the arc H ′D ′ is greater than or equal to ǫ4. �
262 JUN HU AND OLEG MUZICIAN
Lemma 2. Suppose that f is an orientation-preserving homeomorphism of
S1 and 8( f )(0) = 0. Let I be a circular arc on S1 such that the length of I c is less
than π/24. Assume that || f |I ||cr ≤ M. Then there exists a positive constant δ1,
depending only on M, such that |d1|2 − |d−1|
2 ≥ δ1, where d1 and d−1 are given
by (7).
Proof. By rotational invariance, we may assume that the arc I is in the same
position as introduced at the very beginning of this section. Let Ŵ be the union of
two such circular arcs of length τ on S1, one starting at i and going counterclock-
wise and the other starting at ei5π/4 and going clockwise; that is, Ŵ = DE ∪ FG in
Figure 6. Since τ is a particular value chosen at the beginning of this section, the
shorter circular arc between the two components of Ŵ has length 3π/4−2τ, which
is greater than π/12. Let α = (3π/4−2τ)/4. Then α ≥ π/48. In counterclockwise
order, we add to each component of Ŵ a small circular arc, namely MD and NF ,
of length α and 2α, respectively; see Figure 6. Since the length of I c is less than
π/24, we assume that I c is contained in the shorter arc between DE and NF . By
(11), we have
|d1|2 − |d−1|
2 = 2
(1
2π
)2 ∫ 2π
0
∫ 2π
0
sin2(h(s) − h(t))dsdt
= 2
(1
2π
)2 ∫ 2π
0
∫ t
t−2π
sin2(h(s) − h(t))dsdt
≥ 2
(1
2π
)2 ∫ 2π
0
∫ t−π/4
t−π/4−α
sin2(h(s) − h(t))dsdt.
Furthermore, the double integral in the previous line is greater than or equal to
∫ π/2
π/2−α
∫ t−π/4
t−π/4−α
sin2(h(s) − h(t))dsdt +
∫ 3π/2
3π/2−α
∫ t−π/4
t−π/4−α
sin2(h(s) − h(t))dsdt,
which can be combined as∫ π/2
π/2−α
∫ t−π/4
t−π/4−α
[sin2(h(s) − h(t)) + sin2(h(s + π) − h(t + π))]dsdt.
Note that if π/2 − α ≤ t ≤ π/2 and t − π/4 − α ≤ s ≤ t − π/4, then in
counterclockwise order, the arc between eis and eit, or ei(s+π) and ei(t+π), or B =
ei3π/2 and C = ei2π (refer to Figure 6), can be extended in one direction to a
semicircle τ-contained in I and has length between π/4 and π/2. By Corollary 4,
we conclude that the values |h(s)−h(t)|, |h(s+π)−h(t+π)|, and |h(3π/2)−h(2π)|
are greater than or equal to ǫ4. Thus
|h(s) − h(t)| + |h(s + π) − h(t + π)| ≤ 2π − ǫ4.
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 263
Figure 7. An illustration for the proof of Lemma 3.
Therefore one of these two summands is less than or equal to π − ǫ4/2, and hence
is between ǫ4/2 and π − ǫ4/2. It follows that the value of either sin2(h(s) − h(t))
or sin2(h(s + π) − h(t + π)) is greater than or equal to sin2(ǫ4/2) for any π/2 − α ≤
t ≤ π/2 and t − π/4 − α ≤ s ≤ t − π/4. Thus
|d1|2 − |d−1|
2 ≥ 2
(1
2π
)2
α2 sin2
(ǫ4
2
)=
α2δ 24
8π2.
Letting δ1 = α2δ 24 /(8π2) with δ4 = e−5M/
(64(2/
√
3 + e−M )3)
completes the
proof. �
Lemma 3. Under the same assumptions of Lemma 2, there exists a positive
constant δ2, depending only on M, such that |c1|2 − |c−1|
2 ≥ δ2, where c1 and c−1
are given by (6).
Proof. Let τ and α be the constants introduced in the proof of the previous
lemma. By rotational invariance, we may assume that A, B , C and D are four
points equally distributed on S1 in counterclockwise order with D = ei(π/4+2α). Let
DE be the arc from D to E in the counterclockwise direction with length τ, and
let H denote the point eiπ and GH be the arc from G to H in the counterclockwise
direction with length τ. Then the long arc from G to E in the counterclockwise
direction has length π + π/4 + 2τ + 2α. Hence the short arc EG from E to G
in the counterclockwise direction has length 3π/4 − 2τ − 2α = 2α, since α =
(3π/4 − 2τ)/4. Therefore, the length of EG is greater than π/24. Now we assume
that I c is contained in the arc EG.
264 JUN HU AND OLEG MUZICIAN
Assume that t ∈ (0, α) and u ∈ (π/4, π/4 + α). Define β1 to be the arc on S1
from eit to ei(t+u) in counterclockwise direction, β2 from ei(t+u) to ei(t+π), β3 from
ei(t+π) to ei(t+u+π), and finally β4 from ei(t+u+π) to eit. For brevity, we set αk = | f (βk)|
for k = 1, 2, 3, 4.
Since the semicircle β3 ∪ β4 is τ-contained in I , we obtain, using Lemma 1,
ǫ3 ≤ α1 + α2 ≤ 2π − ǫ3. Similarly, using the semicircle β4 ∪ β1 and the same
lemma, we obtain ǫ3 ≤ α2 + α3 ≤ 2π − ǫ3.
Clearly, π/4 ≤ |β1|, |β3|, |β4| ≤ π/2. More importantly, each of β1, β3 and β4
can be extended in one direction to a semicircle that is τ-contained in I . Therefore,
by Corollary 4, αk ≥ ǫ4 for k = 1, 3, 4. Thus ǫ4 < 2ǫ4 ≤ α1 + α3 ≤ 2π − α4 ≤
2π − ǫ4. Then by (13),
H (t, u) = 4 sin
(α1 + α2
2
)sin
(α2 + α3
2
)sin
(α1 + α3
2
)
≥ 4 sin2
(ǫ3
2
)sin
(ǫ4
2
)= 4
(δ3
2
)2 δ4
2.
By (12),
|c1|2 − |c−1|
2 =
(1
2π
)2 ∫ π
0
∫ 2π
0
sin(u)H (t, u)dtdu
≥
(1
2π
)2 ∫ π/4+α
π/4
sin(u)
∫ α
0
H (t, u)dtdu.
All together, we obtain
|c1|2 − |c−1|
2 ≥
(1
π
)2 ∫ π/4+α
π/4
sin(u)
∫ α
0
(δ3
2
)2 δ4
2dtdu ≥
√
2δ 23 δ4α
2
16π2.
Letting δ2 =√
2δ 23 δ4α
2/(16π2) with δ4 = e−5M/(64(2/
√
3 + e−M )3), we obtain a
lower bound for |c1|2 − |c−1|
2. �
With the use of Lemmas 2 and 3, the proof of our Theorem 1 is almost identical
to that of Theorem 1 in [16].
Proof of Theorem 1. Using the notation in the statement of the theorem, we
can choose an open hyperbolic half plane Up with p in the middle of its boundary
on S1 such that for any point z in Up, there exists a conformal isometry A of D such
that A(z) = 0 and A(Ip) has length greater than 2π − 2α, where A is a hyperbolic
isometry with axis the geodesic passing through z and the origin. Let I = A(Ip).
Then I c has length less than 2α. Now let B be another isometry of D such that B
maps 8( f )(z) to 0. Then B ◦ f ◦ A−1 satisfies the conditions in Lemmas 2 and 3,
and it suffices to show that the maximal dilatation of B ◦ f ◦ A−1 at 0 is bounded
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 265
from above by a constant. Note that the cross-ratio distortion norm of B ◦ f ◦ A−1
on I equals the cross-ratio distortion norm of f on Ip. For brevity, let us continue
to use f for B ◦ f ◦ A−1; we are assuming that 8( f )(0) = 0.
Since 0 < |d1|2 − |d−1|
2 ≤ |d1|2 = 1 and |c1|
2 − |c−1|2 ≥ δ2 > 0 (Lemma 3),
(10) implies |(∂w/∂z)(0)|2 − |(∂w/∂z)(0)|2 ≥ δ2. Thus
∣∣∣∣(∂w/∂z)(0)
(∂w/∂z)(0)
∣∣∣∣2
≤ 1 −δ2
|(∂w/∂z)(0)|2.
It is easy to see that the numerator of the expression (9) for (∂w/∂z)(0) has absolute
value less than or equal to 2. Then by (9) and Lemma 2, we obtain |(∂w/∂z)(0)|2 ≤
2/(|d1|
2 − |d−1|2)
≤ 2/δ1. Therefore, the complex dilatation
κ(8)(0) =(∂w/∂z)(0)
(∂w/∂z)(0)
of 8 at 0 satisfies |κ(8)(0)|2 = |(∂w/∂z)(0)/(∂w/∂z)(0)|2 ≤ 1−δ1 ·δ2/2. Then the
maximal dilatation K (8)(0) = (1 + |κ(8)(0)|)/(1− |κ(8)(0)|) satisfies K (8)(0) ≤
(1 + k)/(1 − k), where k =√
1 − δ1 · δ2/2.
Since 1 + k ≤ 2 and α ≥ π/48, by the expressions of δ1 and δ2 in the previous
two lemmas, we obtain
K (8)(0) ≤(1 + k)2
1 − k2≤
8
δ1δ2
=16(8)(8)π4
√
2α4δ 23 δ 3
4
≤
√
2(83)(484)
δ 23 δ 3
4
,
where δ3 = e−M/(2/√
3+e−M ), δ4 = e−5M/(64(2/
√
3+e−M )3), and M = || f |Ip
||cr .
Therefore, ln K (8( f ))(z) ≤ 17|| f |Ip||cr + C for a positive constant C, where
z ∈ Up. �
Using Theorem 1 and the compactness of S1, we easily obtain a proof of Corol-
lary 1.
4 Locally asymptotic conformality
In this section, we show that local symmetry implies locally asymptotic confor-
mality, that is, Theorem 2. First we present the proof of Corollary 2. This provides
us an opportunity to lay out the strategy used to obtain local properties. The main
ingredients of this proof include the Douady-Earle Extension Theorem (Corol-
lary 1), a normal family of quasiconformal extensions to the disk with bounded
maximal dilatations, and Proposition 1, below. The proof of Theorem 2 employs
Theorem 1 and normal families of Douady-Earle extensions when restricted to
subdomains of the unit disk. A first obstacle to our proof is that the extensions are
266 JUN HU AND OLEG MUZICIAN
not globally quasiconformal on the disk, and hence the family of the extensions
is not normal on the whole disk. To overcome this difficulty, we use an exhaus-
tion by quasiconformal limiting maps defined on subdomains and show that the
exhaustion map is quasiconformal on the whole disk. Another important step is
to show that the sequence of extensions, when restricted on the unit circle, still
converges uniformly to the restriction of the exhaustion map on S1; this leads to a
contradiction by applying Proposition 1. This is done in Lemma 5.
Proposition 1 (Douady-Earle [6]). Let H(S1) (respectively, H(D)) be the
space of all orientation-preserving homeomorphisms of S1 (respectively, D),
equipped with C0-topology, and Diff(D) the space (with the C∞-topology) of dif-
feomorphisms of the open unit disk. The map 8 : H(S1) → Diff(D) ∩ H(D) is
continuous.
Lemma 4. Let f be an orientation-preserving homeomorphism of S1. If
cr( f (Q)) = 1 for any quadruple Q = {a, b, c, d} ⊂ S1\{z0} with cr(Q) = 1,
then f is a Mobius transformation.
Proof. Let Q be a quadruple consisting of four points a, b, c, and d on S1
arranged in counterclockwise order with cr(Q) = 1. Let A be the Mobius transfor-
mation A such that A ◦ f fixes a, b, and c. Using hyperbolic midpoints, one can
see geometrically that cr(Q) = cr(A ◦ f (Q)) = 1 implies A ◦ f (d ) = d . Similarly,
A◦ f has to fix all hyperbolic binary cuts between a and c with respect to d and all
hyperbolic binary cuts between c and a with respect to b except z0. By continuity,
A ◦ f fixes every point on S1. Hence f = A−1. �
Proof of Corollary 2. Suppose that F is not asymptotically conformal on
Up. Then there exist ǫ > 0 and a sequence of points {zn}∞n=1 in D converging
to z0 ∈ Up ∩ S1 such that the complex dilatation κ(F )(zn) of F at zn satisfies
|κ(F )(zn)| ≥ ǫ for each n.
For each n, let wn = F (zn), and let An and Bn be two Mobius transformations
such that An(zn) = 0, Bn(wn) = 0 and An(0) = −zn. Let Fn = Bn ◦ F ◦ A−1n and
fn = Bn ◦ f ◦A−1n . Then Fn(0) = 0 and K (Fn) = K (F ). By the conformal naturality
of Douady-Earle extensions, Fn = 8( fn).
Since Fn maps S1 to S1, we can use the reflection principle to extend Fn to
the Riemann sphere C. Denote this extension by F ′n. Then {F ′
n}∞n=1 is a family of
quasiconformal mappings on C with a constant maximal dilatation and fixes three
common points. Thus it is a normal family. Passing to a subsequence, we may
assume that {F ′n}
∞n=1 converges uniformly to a quasiconformal map F ′
∗ which has
the same maximal dilatation as each F ′n, and maps the unit circle onto itself. Let
F∗ = F ′∗|D. Then Fn → F∗ in the C0-topology on H(D).
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 267
Let f∗ = F∗|S1 . Then fn → f∗ in the C0-topology on H(S1). By Proposition 1,
Fn converges to F∗ in the C1-topology on Diff(D). This implies that
κ(Fn)(0) → κ(F∗)(0) as n → ∞.
Hence |κ(F∗)(0)| ≥ ǫ.
On the other hand, we can show that κ(F∗)(0) = 0, a contradiction. It suffices
to show that f∗ is a Mobius transformation. By Lemma 4, it is enough to show that
for any quadruple Q = {a, b, c, d} ⊂ S1\{−z0} with cr(Q) = 1, cr( f∗(Q)) = 1.
When n is large enough, Q ⊂ An(Ip), and hence A−1n (Q) ⊂ Ip. Therefore,
A−1n (Q) → z0. Thus
Cr( f∗(Q)) = limn→∞
cr( fn(Q)) = limn→∞
cr(Bn ◦ f ◦ A−1n (Q))
= limn→∞
cr( f (A−1n (Q))) = 1.
�
Lemma 5. Let { fn}∞n=1 be a sequence of homeomorphisms of S1 and {Im}∞m =1
an increasing sequence of closed circular arcs contained in S1\{−p}, with⋃∞m =1 Im = S1\{−p}. Assume fn(−p) = −p for each n. If fn converges in the
C0-topology to a homeomorphism f∗ of S1 on any arc Im and f∗(−p) = −p, then
fn converges to f∗ on S1 in the C0-topology.
Proof. Let ǫ be a positive real number much smaller than π. Using the con-
ditions that f∗ is a homeomorphism, f∗(−p) = −p, and {Im}∞m =1 is an increasing
sequence of closed circular arcs with⋃∞
m =1 Im = S1\{−p}, we obtain a natural
number m1 such that |S1\Im| < ǫ and |S1\ f∗(Im)| < ǫ for all m ≥ m1.
Since fn converges to f∗ uniformly on Im1, there exists a natural number n1
such that ||( fn − f∗)|Im1||C0 < ǫ for any n ≥ n1. We take n1 so large that for any
point z ∈ Im1, the length of the shorter circular arc on S1 between fn(z) and f∗(z) is
less than ǫ for any n ≥ n1. In particular, let a1 and a2 be the end points of Im1and
fn(ai), f∗(ai) the shorter circular arc on S1 between fn(ai) and f∗(ai) for i = 1, 2,
then | fn(ai), f∗(ai)| < ǫ for any i = 1, 2 and any n > n1. It follows that for any
n ≥ n1,
|S1\ fn(Im1))| < |S1\ f∗(Im1
)| + | fn(a1), f∗(a1)| + | fn(a2), f∗(a2)| < 3ǫ.
Now, using the conditions that −p ∈ S1 \ Im1and fn(−p) = f∗(−p) = −p, we
268 JUN HU AND OLEG MUZICIAN
obtain for any point z ∈ S1 \ Im1and any n ≥ n1,
| fn(z) − f∗(z)| = | fn(z) − fn(−p) + f∗(−p) − f∗(z)|
≤ | fn(z) − fn(−p)| + | f∗(−p) − f∗(z)|
≤ |S1\ fn(Im1))| + |S1\ f∗(Im1
)|
< 3ǫ + ǫ = 4ǫ.
Overall, we conclude that for any n ≥ n1, ||( fn − f∗)|S1 ||C0 < 4ǫ. �
Proof of Theorem 2. We apply the same strategy used in the proof of
Corollary 2. Since f is assumed to be only a homeomorphism (not necessarily
quasisymmetric), it requires more effort to show there exists a sequence of exten-
sions Fn fixing the origin and converging uniformly to a homeomorphism f∗ on
S1. The previous lemma was developed for this purpose.
Suppose that the conclusion of the theorem fails. Then for any neighborhood
Up of p in D, F is not asymptotically conformal on Up. Take such a neighborhood
Up with Up ∩ S1 ⊂ int(Ip). Then there exist ǫ > 0 and a sequence {zn}∞n=1 of
points zn ∈ Up converging to a point z0 ∈ Up ∩ S1 such that the complex dilatation
κ(Fn)(0) of Fn at the origin satisfies |κ(Fn)(0)| ≥ ǫ for each n. To simplify notation,
assume (as we may) that z0 = p.
Note that the above neighborhood Up can be chosen as the intersection of an
open disk centered at p in the complex plane and the open unit disk D. By Theorem
1, we may assume that F is quasiconformal on Up. We consider Up as such a
neighborhood of p. Here Up denotes the closure of Up in the closed disk D, as
well as in the complex plane.
For each n, we define two Mobius transformations, An and Bn, from D onto D
such that An(zn) = 0, An(0) = −zn, Bn(F (zn)) = 0 and Bn(F (A−1n (−p))) = −p. Set
Gn = Bn ◦ F ◦ A−1n . Clearly, Gn(0) = 0 and Gn(−p) = −p.
Let Dn = An(Up). Then the closure Dn equals An(Up). Clearly, 0 ∈ Dn for
each n. We also observe that each map Gn is quasiconformal on Dn with the same
maximal dilatation as F on Up.
Each Mobius transformation An maps the diameter passing through zn to itself,
zn to 0, and 0 to −zn. Since zn → p as n → ∞, the axis Ln of An converges to
the diameter L from p to −p, and the angle between the boundary of Up and Ln
converges to the angle between the boundary of Up and L, which is a right angle.
Then the image of the boundary of Up in D under An shrinks to the point −p in the
Euclidean metric. Hence the image Dn = An(Up) exhausts D as n → ∞, that is,⋃∞n=1 Dn = D and
⋃∞n=1 Dn = D \ {−p}. Furthermore, one can inductively choose
a subsequence {nk}∞k =1 such that Dnk
⊂ Dnk+1for each k. To simplify the notation,
CROSS-RATIO DISTORTION AND DOUADY-EARLE EXTENSION 269
we assume that Dn ⊂ Dn+1 for each n.
Reflecting through the unit circle S1, we can obtain, for each n, an extension
of Gn to the Riemann sphere C. Denote this reflection by G′n; it fixes 0, −p and
∞. Let D ′m = Dm ∪ Dm, where Dm is the reflection of Dm through the unit circle.
It follows from the previous paragraph that D ′n ⊂ D ′
n+1 and⋃∞
m =1 D ′m = C\{−p}.
Again by passing to a subsequence, one may further assume that D ′n ⊂ int(D ′
n+1).
In summary, we now know that each G′n fixes 0 and ∞, and for each D ′
m , the
images of D ′m under all G′
n’s omit a common point −p. Moreover, {G′n|D ′
m}∞n=m is
a sequence of quasiconformal maps with maximal dilatations bounded above by
a constant K , equal to the maximal dilatation of F |Up. It follows that {G′
n|D ′m}∞n=m
is a normal family; hence there exists a subsequence {F (1)n }∞n=1 ⊂ {G′
n}∞n=1 such
that {F (1)n }∞n=1 converges to a quasiconformal map F (1)
∗ defined on D ′1 with maxi-
mal dilatation bounded by K . Similarly, there exists a subsequence {F (2)n }∞n=1 ⊂
{F (1)n }∞n=1 such that {F (2)
n }∞n=1 converges to a quasiconformal map F (2)∗ defined on
D ′2 with maximal dilatation bounded by K . Inductively, we have subsequences
{F (m)n }∞n=1 ⊂ {F (m−1)
n }∞n=1 such that {F (m)n }∞n=1 converges to a quasiconformal map
F (m)∗ defined on D ′
m with maximal dilatation bounded by K .
Note that for each m, F (m+1)∗ = F (m)
∗ on D ′m. Let F ′
∗ =⋃∞
m =1 F (m)∗ . Then F ′
∗ is a
quasiconformal map defined on C\{−p} with maximal dilatation bounded by K .
Let Rn = D ′n\D ′
1. Then Rn is an annulus with modulus M (Rn) approaching ∞
as n → ∞. Since M (F ′∗(Rn)) ≥ M (Rn)/K , M (F ′
∗(Rn)) → ∞ as n → ∞. This
implies that F ′∗ maps C\{−p} onto itself, which then, with F ′
∗(−p) = −p, becomes
a homeomorphism of C mapping S1 onto S1.
For each n, let Fn = F (n)n |
Dand F∗ = F ′
∗|D. Then {Fn}∞n=0 converges to a
homeomorphism F∗ uniformly on every Dm. Let fn = Fn|S1 and f∗ = F∗|S1 . Then
fn converges to f∗ on each Im = Dm ∩ S1 in the C0-topology.
Since fn, f∗ and Im satisfy the hypotheses of Lemma 5, we conclude that fn →
f∗ on S1 in the C0-topology. It then follows from Proposition 1 that Fn converges
to F∗ on D in the C1-topology.
The remainder of the proof follows exactly the last part of the proof of Corol-
lary 2. �
One sees easily that Corollary 2 is a special case of Theorem 2, and Corollary
3 follows from Theorem 2 and the compactness of S1.
Finally, we sketch the short proof mentioned in Remark 1 for a weaker version
of Theorem 1; if f is quasisymmetric on an open arc Ip of S1 then the Douady-
Earle extension 8 of f is quasiconformal on a hyperbolic half plane near p. Some
details are different from the proofs of Corollary 2 and Theorem 2.
270 JUN HU AND OLEG MUZICIAN
Suppose that the extension 8 is not quasiconformal on any hyperbolic half
plane near p. Then there exists a sequence of points zn approaching Ip such that
the maximal dilatation K (8)(zn) of 8 at zn approaches ∞ as n → ∞. Passing to
a subsequence, we may assume that the sequence {zn}∞n=1 converges to a point z0
in the interior of Ip. Let wn = 8(zn) and let An be a Mobius transformation fixing
z0 and mapping zn to 0. Choose a sequence {Bn}∞n=1 of Mobius transformations
such that the image of Ip under fn = Bn ◦ f ◦ A−1n converges to the unit circle with
a point omitted. Using the Beurling-Ahlfors extension ([5], or refer to [18], [4]
or [19]) of f which is quasiconformal on a hyperbolic half plane near p, we can
show that { fn}∞n=m is a normal family on Am(Ip) for each m ≥ 1. Again passing
to subsequences, we may further assume that fn converges uniformly on Am(Ip)
for each m ≥ 1. Finally, using a similar method of exhaustion, we can show
that fn converges uniformly to a homeomorphism g which can be proved to be
quasisymmetric.
Now let 8n = Bn ◦ 8 ◦ A−1n . To reach a contradiction, we need to show that
the maximal dilatation of 8n at 0 is bounded. To do so, it suffices to show that
8n(0) stays in a compact subset of the open unit disk. Indeed, by Proposition 1,
8n(0) converges to 8(g)(0) as n → ∞. Again by Proposition 1, the maximal
dilatation of 8n at 0 is close to that of 8(g) at 0 when n is large enough, and thus
the maximal dilatations are bounded, which is the desired contradiction.
Acknowledgements. The authors thank Professors Frederick Gardiner,
Linda Keen, and Dragomir Saric for discussions and suggestions, especially Saric
for explaining his idea for the short proof mentioned in Remark 1. The authors are
also grateful to the referee for providing many comments and suggestions and to
the editor for helping us to improve the paper.
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Jun Hu
DEPARTMENT OF MATHEMATICS
BROOKLYN COLLEGE OF CUNY
BROOKLYN, NY 11210 USA
AND
GRADUATE CENTER OF CUNY
365 FIFTH AVENUE, NEW YORK, NY 10016 USA
email: junhu@brooklyn.cuny.edu, JHu1@gc.cuny.edu
Oleg Muzician
DEPARTMENT OF MATHEMATICS
CUNY GRADUATE CENTER
365 FIFTH AVENUE, NEW YORK, NY 10016 USA
email: olegmoyz@hotmail.com
(Received January 20, 2011 and in revised form January 14, 2012)