84 6. TWO-SIDED CRITERIA
where f = 'ljJ-^1 o ¢ : S^1 --+ S^1 denotes the sewing homeomorphism associated with
D. We have proved that
(6.4.4) length(li) = length(J2) implies length(f(li)) = length(f(h)).
By performing two rotations, if necessary, we may assume that f(l) = l. Then by
continuity ( 6.4.4) implies tha t f ( -1) = -1. Continuing as in the proof of Theorem
3.8.1, we see that f fixes every point ei^8 on S^1 withe= m7r/2n, n = 1,2, ... and
m = 1, ... , 2 · 2n. By continuity f is the identity m apping on S^1. Thus D must be
a disk as in the proof of Theorem 3.8.l. D
A harmonic doubling condition in a Jordan domain is not enough to guarantee
that domain to be a quasidisk. In fact the doubling Jordan domains are exactly
the John disks.
THEOREM 6.4.5 (Kim [101], Kim and Langmeyer [102]). A bounded Jordan
domain D in R^2 is doubling if and only if it is a John di sk.
The next result, proved by Jerison and Kenig [89], characterizes quasidisks in
terms of the doubling condition for harmonic measure. We will follow the proof
given in Broch-Hag-Junge [29]. See also Garnett-Marshall [45].
THEOREM 6.4.6 (Jerison and Kenig [89]). A bounded Jordan domain D in R^2
is a quasidisk if and only if D and D* both are doubling.PROOF. For sufficiency, suppose that D and D satisfy the harmonic doubling
condition with respect to z 0 E D and z 0 E D, with the same constant b. We will
show that then D is a quasisymmetric domain and hence a quasidisk by Theorem
3 .8.8. To this end let 11 and 12 be adjacent arcs of 8D with
(6.4.7) w(zo, 1 1; D) = w(zo, 12; D).
Without loss of generality we m ay assume that
(6.4.8) diam (/1) ::; diam (/2).
In the following we will use the notation w(r) = w(z 0 , 1; D) and w = w(z 0 , 1; D).
Denote by z 1 and z2 the endpoints of 11 and by z 2 and z 3 the endpoints of 12.
For z, w E 11 U 12 we let 1(z, w) denote the arc of 8D from z to w contained in
/1 U /2 and order points a long 11 U 12 in the direction from z 1 to z3.
We choose wo, W1, ... , Wn on 12 as follows. Let w 0 = z 2. Suppose that points
wo, w1, ... , Wk have been chosen along 12 = 1(z 2 , z 3 ) so that for all 0 ::; i::; k - 1,
wi+ 1 is the fir st point with
diam(r(wk, Wk+1)) = diam(r(z1, wk)).If diam(r(wk, z3)) < diam(r(wki z 1 )) , then we stop and set n = k. Otherwise,
continue. This process must stop with n ::; b, for by using the doubling condition
in D we h ave that
This implies tha t
n-l
nw(r1) S:: b L w(r(wk, Wk+1)) S:: bw(r2),
k=O
and by (6.4.7) we must conclude that n::; b.