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82 6. TWO-SIDED CRITERIA

Next, assume that D is a Lip 0 -extension domain, i.e. that (6.3.2) is satisfied.

Consider f analytic in D satisfying (6.2.7). Then

Jf(z1) - f(z2)J:::; Oo(z1,z2):::; do(z1,z2):::; MJz1 - z2J°'

and Jlfllo:::; M = M(D, a). D

Now suppose that D is a Lip 0 -extension domain for the class of analytic func-

tions only. It follows from Theorem 6.2.3 that if a n analytic function satisfies (6.2.2)

in D, then it lies in locLip 0 (D) and hence in Lip 0 (D). Thus D has the Hardy-

Littlewood property of order a, and we obtain the following corollary to Theorem

6.3.5.

COROLLARY 6.3.6. If a simply connected domain is a Lip 0 -extension domain

for the class of analytic functions, then it is a Lip 0 -extension domain.

Next we prove that the Lip 0 -extension domains possess property 1° in Defini-

tion 2.4.1.

THEOREM 6.3. 7 (Gehring-Martio [66]). Suppose that a domain D is a Lip 0 -

extension domain. Then there is a constant c 2 1 depending only on a and M in

(6.2.2) such that for each zo E R^2 and each r > 0 points in D n B(z 0 , r) can be

joined in D n B(zo, er).

PROOF. Choose z1,z2 E DnB(zo,r) where we may assume that 8DnB(z 0 ,r)

contains some point w; otherwise c = 1 will work. By Theorem 6.3.l there exists a

curve 'Y joining z 1 and z 2 in D so that

(6.3.8) !, dist(z , 8D)°'-^1 ds:::; 2MJz1 - z 2 j°' :::; 2M(2r)°'.

Suppose that 'Y is not contained in D n B(z 0 , er), where c > 1. Then

!, dist(z, 8D)°'-^1 ds 2 /,Jz - wj°'-^1 ds 2 !, (jz - zol + r)°'-^1 ds

l

cr 2r°'

2 2 (t + r)°'-^1 dt = - ((c + 1)°' - 2°').

r a

This, together with (6.3.8), yields a bound for c depending only on a and M. D

COROLLARY 6.3.9 (Lappalainen [111]). If D is a Lip 0 -extension domain and

if 0 < a :::; (3 :::; 1, then D is also a Lip 13 -extension domain.

PROOF. Fix z1,z2 ED and choose zo = (z1 + z2)/ 2, r = jz 1 -z2J/2. Then z 1

and z2 can be joined by a curve 'Yin B(z 0 , er), c = 2(M + 1)^1 1°, so that (6.3.8) is

satisfied, where Mis as in (6.3.8).

If B(zo, er) n 8D = 0, then

{ dist(z,8D)/3-l ds:::; r/3-^1 2r = 2r/3:::; 2jz 1 - z 2 j/3.

}[z1,z2]

If B(zo, er) n 8D =/. 0, then dist(z, 8D) :::; 2cr for every z E 'Y C B(z 0 , er) and

2M(2r)°' 2 !, dist(z , &D)°'-^1 ds 2 (2cr)°'-/3 !, dist(z, &D)/3-l ds,

i.e. !, dist(z, 8D)/3-l ds :::; 2c/3-<> Mjz 1 - z2 Jf3. D