9.1. UNIFORM DOMAINS AND SCHWARZIAN DERIVATIVES 119
for z E 'Y between z1 and zo. This inequality then implies that f is bounded near
z1 and hence analytic at z 1. Thus we can let z---+ z 1 along 'Y to get (9.1.3). O
The following result allows us to replace the pre-Schwarzian derivative Tt =
f" / f' in Lemma 9.1.1 with the Schwarzian derivative St·
LEMMA 9.1.4. If u and v are absolutely continuous in each closed subinterval
of [a, b), if u(a) = 0, and if
U^1 <UV
a.e. in [a, b), then u:::; 0 in [a, b).
PROOF. Let
w(t) = u(t) exp (-1t v(s) ds).
Then w is absolutely continuous in each closed subinterval of [a , b),
w'(t) =exp (-1t v(s) ds) (u'(t) - u(t)v(t)):::; 0
a.e. in [a , b), and hence
w(t) :::; w(a) = 0, u(t) = w(t) exp (1t v(s) ds) :::; 0
in [a, b). 0
We then have the following analogue of Lemma 9.1.1 for the Schwarzian deriv-
ative St·
LEMMA 9.1.5. Suppose that z 1 , z2 E D, that 'Y is a rectifiable open arc joining
z 1 , z2 in D with midpoint zo, and that 0 < c < ~. If f is meromorphic and locally
injective in D with f"(zo) = 0 and if
c
(^9 .1.^6 ) ISt(z)I:::; min(s,length('Y) - s) 2 '
for z E 'Y, where s is the arclength of 'Y from z1 to z, then
I
f (z 1 ) - f (z2) I 2 c
f'(zo) - (z1 - z2) :::; 1 _ 2 c length('Y).
PROOF. By Lemma 9.1.l it is sufficient to show that
l
f"(z)I 2c
(9.1.^7 ) f'(z) :::; min(s, length('Y) - s)
for each z E 'Y· Again by symmetry we need only prove (9.1.7) for the case where z
lies between z 0 and z 2 , i.e., for s E [a, length('Y)) where again we let a= length('Y)/2.
Since f"(z 0 ) = 0, f is finite at z 0 and there exists t E (a,length('Y)) such that
f is finite at z( s) for s E [a, t), where z( s) is the arclength representation of 'Y- Let
b denote the supremum of all such numbers t, and for s E [a, b) let
2c
¢(s) -.
- length('Y) - s ' I
'lfJ(s) = f" f'(z(s)) I +¢(a).
Then ¢ and 'ljJ are absolutely continuous in each closed subinterval of [a, b) with
<P'(s) _ ~ ¢(s)^2 = 2c(l - c) > c
2 (length('Y) - s)2 - (length('Y) - s)^2