1549055259-Ubiquitous_Quasidisk__The__Gehring_

(jair2018) #1
9.5. RIGID DOMAINS ARE LINEARLY LOCALLY CONNECTED 133

LEMMA 9.5.l. Suppose that ¢(t) is a real-valued function definedforO < t < oo,
that
(9.5.2)
for 0 <ti, t2 < oo, and that
_ { zei¢(1zl) if 0 < lz l < oo,
(9.5.3) f(z) - 0 if z = 0.

Then f is (1 + b )-bilipschitz in R^2.


PROOF. Choose distinct points z1, z2 E R^2 with lz 1 1::; lz2I- If z 1 '/= 0, then


by (9.5.2) while

lf(zi) - f(z2)I::; lz1 - z2I + lz1llei¢(lzil) - ei¢(lz^2 lll
::; lz1 - z2I + lz1ll¢(lz11) - ¢(lz2l)I
::; I z1 - z2 I + b I z1 I I log I z1 I - log I z2 I I
::; (1 + b)lz1 - z2I

lf(zi) - f(z2)I = lz2I ::; (1 + b)lz1 - z2I
if z 1 = 0. Next since 1-^1 is given by (9.5.3) with -¢ in place of ¢, the above
argument can be applied to f-^1 to complete the proof. 0

THEOREM 9.5.4 (Gehring [50]). Suppose that Dis a domain in R^2 with L(D) >
1 and that c is a constant where
7r
(9.5.5) log c > L(D) _ l

Then for each z 0 E R^2 and 0 < r < oo, D n 8B(z 0 , r) is empty or lies in a
component of
G = D n (B ( z 0 , er) \ B ( zo, r / c)).
PROOF. Suppose there exist points z 1 ,z 2 E Dn8B(z,r) which lie in different
components G 1 , G 2 of G. By making a change of variable we may assume that
zo = 0. Choose e E [-7r' 7r] so that Z2 = z1 ei 1:1 and let f be as in (9.5.3) where
0 if t (j. [r/c, er],


ct e
¢(t) = log--- if t E [r/c,r],
r loge
er e
log--- if t E [r, er].
t loge

Then¢ satisfies (9.5.2) with b = 7r / log c and f is L-bilipschitz in R^2 by Lemma 9.5.1,
where


(9.5.6)

Set

7r
L = 1 + -
1


  • < L(D).
    ogc


(z) = { z if z ED\ G1,
g f ( z) if z E G 1.

If U is any open disk in D, then either UC D \ G 1 , in which case g(z) = z in U, or


whence
Free download pdf