1549055259-Ubiquitous_Quasidisk__The__Gehring_

28 2. GEOMETRIC PROPERTIES

PROOF. Choose z 1 , z 2 EE\ B(z 0 , r) and let

2 z - zo

f(z ) = r I

12

+ zo.

z - zo

Then f (z 1 ) , f (z 2 ) E f (E) n B(z 0 , r ). Next condition 1° implies that these points

can be joined by a connected set I in f(E) n B(z 0 , er). Hence 1-^1 (1) joins z 1 and

z 2 in E \ B(z 0 , r / c) and condition 2° holds. D

The connection cited above between linear local connectivity and the three-

point condition yields the following reformulation of Theorem 2.2.5.

COROLLARY 2.4.3 (Walker [165]). A Jordan domain D is a K-quasidisk if and

only if 8D is linearly locally connected with constant c, where K and c depend only

on each other.

In contrast to the three-point condit ion, the property of linear local connectivity

can b e applied to a sdoma in as well as to its boundary. For example, a sector D of

angle a is linearly locally connected with c = csc(a/ 2).

The converse of the Jordan curve theorem implies that a simply connected

domain D is a Jordan domain if and only if D is locally connected at ea ch point of

• 2
  R (Newman [140]).
  We have the following counterpart of this r esult for quasidisks.
  THEOREM 2.4.4 (Gehring [49]). A simply connected domain D is a K -quasidisk
  if and only if it is linearly locally connected with constant c, where K and c depend
  only on each other.
  Moreover D is a disk or ha lf-plane if and only if it is linearly locally connected
  with constant c = l. See Langmeyer [ 110 ].
  Finally we show that in general, t he notion of linear local connectivity is invari-
  ant with respect to Mi:ibius transformations. Our proof depends on the following
  result.
  
  
  • 2
    LEMMA 2.4.5. Suppose that 1 < c < oo and that E 1 , E 2 c R are sets which
    are separated by an annulus
    { z : a < I z - zo I < ac}.
    If f is a Mobius transformation, then f(E 1 ) and f(E2) are separated by an annulus
    {w: b < lw - wol < bg(c)}
    where g(c) = c^112 + c^112 - l. This bound is sharp.
  
  
  
  

PROOF. By means of a preliminary similarity mapping we may assume that

a = 1 and z 0 = 0. Let

C1 = { z : lz l = 1}, C2 = { z : lz l = c}.

Since the inversion i(z) = c/z interchanges C 1 and C 2 , we m ay also assume that

(2.4.6) diamf(C1) ~ diamf(C2).

This implies that f(Ci) is a circle and j(C 2 ) is a circle or line. Hence there exists
a line L' through the center of f(Ci) which contains the center of f(C 2 ) if f(C 2 )
is a circle or is orthogonal to f(C 2 ) if f(C 2 ) is a line. Now L = f-^1 (L') is a line
through the origin which is orthogonal to the conce ntric ci rcles C 1 and C 2.