CHAPTER 6Two-sided criteria
Suppose that a property of a domain D is not enough to guarantee that it is
a quasidisk. Could it still be true that if D and its exterior domain D * both have
this property, then D , and hence D*, is a quasidisk? Vve consider here five different
properties for which this question has an affirmative answer.6.1. Linear local connectivity revisited
We recall the properties 1° and 2° in Definition 2.4. l. For a set E there is to
exist a constant c 2: 1 such that
1° En B(zo, r) lies in a component of En B(zo, er) and
2° E \ B(zo, r) lies in a component of E \ B(zo, r /c)
for each zo E R^2 and each r > 0.
These properties are in a sense dual and play similar roles. Domains with one of
these properties appear quite often, and the terms linearly connected domains and
John disks have been introduced for simply connected domains with nondegenerate
boundary possessing property 1° and 2°, respectively. Linearly connected domains
and John disks have been studied extensively by Pommerenke [145], Nakki and
Vaisala [136], and others.
The following results show that these domains may be viewed as one-sided
quasi disks.
THEOREM 6.1.1 (Gehring-Martio [64]). A simply connected domain D is a
quasidisk if and only if D and D both have property 1° in Definition 2.4. l.
PROOF. If D is a quasidisk, then so is D, and it follows from Theorem 2.4.4
that both D and D have property 1 °.
To prove the converse, we note first that D is simply connected as the com-
plement of a connected set. Next &D (or &D) is a J ordan curve by the converse
of Jordan's curve theorem if oo E D (or D). If oo E &D , then &D must be locally
connected and fr ee of cut points in R^2. Since D is a domain, oo is not a cut point.
Hence &Dis a Jordan curve. For more details see Newman [140] and Pommerenke
[145].
See Figure 6.1. Assume that z 1 ,z 2 E D \ B(z 0 ,r) cannot be j oined in D \
B(zo,r/c), and let r' > r so that z 1 ,z 2 E D\B(z 0 ,r'). Then z 1 ,z 2 are separated
in D by &B(z 0 ,r/c) and hence by an open arc a of &B(zo,r/c). See Theorem
VI.7.1 in Newman [140]. Let w 1 and w 2 in &D = &D denote the endpoints of
a. Now pick w~, w~ in D n B(z 0 , r' / c) so that w 1 , w2 are accessible from w~, w~
in D n B ( zo, r' / c). These points can in turn be joined in D n B ( zo, r') and we
have an op en a rc (3 in D n B(z 0 , r') joining w 1 and w 2. The Jordan curve a U (3
in B(zo, r') must separate z 1 and z 2 in contradiction to the fact that z 1 , z2 both lie
outside B(z 0 , r'). 0
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