1549055259-Ubiquitous_Quasidisk__The__Gehring_

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74 5. CRITERIA FOR EXTENSION


is connected, and hence w 1 is also not a cut point of fJD. 0


We turn now to the proof of the result in Example 5.3.2. For sufficiency we may
assume that Dis bounded since Dis a Jordan domain by the lemma above. Next
choose a disk B C D with two boundary points z 1 , z 2 E fJD and choose w 1 , w2 E fJD
such that
lw1 - w2I = diam(D).


Because D is a Jordan domain, there exists a conformal self-mapping f of D such
that J (zj) = Wj for j = 1, 2 where J is the homeomorphic extension of f to D.
By hypothesis, f* is the restriction of a Mobius transformation to D and hence
f (B) is a disk in D with


diam(j(B)) = diam(D).


Hence f (B) = D.


The following quasiconformal counterpart of Example 5.3.2 yields another char-
acterization for quasidisks.


THEOREM 5.3.4 (Gehring-Hag [55], Rickman [149]). A simply connected do-
main D is a K -quasidisk if and only if it is a quasiconformal extension domain
with constant c, where K and c depend only on each other.

5.4. Bilipschitz mappings
Suppose that D is a disk, that C = fJD, and that f : C-tC' is a homeomor-
phism. Then C' is the boundary of a Jordan domain D'. A well-known theorem of
Schoenflies (Newman [140]) asserts that f has a homeomorphic extension g to D
which maps D onto D'. What is the analogue of this result for bilipschitz mappings?


DEFINITION 5.4.1. A Jordan domain D with C = fJD has the bilipschitz exten-
sion property if there exists a constant c ~ 1 such that each L-bilipschitz mapping
f: C-tC' has a cL-bilipschitz extension g: D-tD'.


EXAMPLE 5.4.2. For each L > 1 there exists a bounded Jordan domain D with
C = fJD and an L-bilipschitz mapping f of C which has no M-bilipschitz extension
to D for any constant M < oo.


To see this, fix L > 1 and let
D = {z = x + iy: lxl < 1, a(lxl^1 /^2 -1) < y < 1}

where a= (L - 1)/2L. Then D has an outward directed cusp with tip at z = -ia
and


f(x+iy) =x+ilYI


is L-bilipschitz on C = fJD. Next if g is an M-bilipschitz extension off to D and
if


I = It = { z = x + i y E D : y = t}

for -a< t < 0, then g(I) c D and


M > length(g(I))



  • length(!)


as t-7 - a, a contradiction.


a2
> ---700


  • a+t

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