1549055259-Ubiquitous_Quasidisk__The__Gehring_

8.3. UNIFORM DOMAINS AND LINEAR LOCAL CONNECTIVITY 105

r

FIGURE 8.7

PROOF. Fix zo E R^2 , 0 < r < oo, and suppose that z 1 , z2 E D n B (z 0 , r). We

must show that z1 and z2 can be joined in DnB(zo, er). By hypothesis there exists

an arc a joining z 1 and z 2 in D with

diam(a):::; a lz1 - z2I :::; 2ar.

If z E a, then

lz - zol:::; lz - z1I + lz1 - zol < diam(a) + r:::; (2a + l)r:::; er,

and hence a joins Z1 and Z2 in D n B(zo, er) as required.

Suppose next that z 1 , .i2 E D \ B(zo, r). Again by hypothesis there exists an

arc (3 joining z 1 and z2 in D with

min iz - Zj I :::; b dist(z, 8D)

J=l,2

for each z E (3. Suppose

(3 c/. D \ B ( zo, r / c).

Then there exists a point z E (3 with

r r

lz-zol < - < e - --2b + 1 '

and for j = 1, 2 we have

2b

lz· J - z l > - iz· J - zol - lz - zol > --2b + 1 r.

Thus
1 2 r
dist(z, 8D) 2: - min iz - z1I > -b--r 2: iz - zol + -
b J=l,2 2 + 1 c
and hence B(z 0 ,r/e) CD. But this implies that D \ B(zo,r/c) is connected and
hence that z 1 , z 2 can b e j oined by an arc in this set. Thus z 1 , z 2 can always be
joined in D \ B(z 0 , r /e). D

If Dis a uniform domain in R^2 , then D satisfies the hypotheses in Theorem 8.3.l
with a = (3 and a = b. Hence uniform domains are linearly locally connected.
We shall see in the following three sections that a simply connected domain
which is linearly locally connected is a quasidisk and hence, by what has gone