Real and Complex Analysis (Rudin)

290 REAL AND COMPLEX ANALYSIS
Suppose (a) is false. Then there is a sequence {lXn} in 0 such that IXn -+ p,
f(1X2n)-+ W1' f(1X 2n + 1)-+ W2' and W1 "# W2. Choose Y as in Definition 14.16,
and put r(t) = f(y(t», for 0 ~ t < 1. Put Kr = g(i5(O; r», for 0 < r < 1. Then
Kr is a compact subset of O. Since y(t)-+ P as t-+ 1, there exists a t* < 1
(depending on r) such that y(t) ¢ Kr if t* < t < 1. Thus 1 r(t) 1 > r if
t* < t < 1. This says that 1 r(t) 1-+ 1 as t-+ 1. Since r(t2n)-+ W1 and
r(t2n+ 1)-+ W2' we also have 1 W1 1 = 1 w21 = 1.
It now follows that one of the two open arcs J whose union is
T - ({wd u {W2}) has the property that every radius of U which ends at a
point of J intersects the range of r in a set which has a limit point on T.
Note that g(r(t» = y(t) for 0 ~ t < 1 and that 9 has radial limits a.e. on T,
since 9 E H oo • Hence
lim g(reil) = P (a.e. on J), (1)
r-1
since g(r(t»-+ pas t-+ 1. By Theorem 11.32, applied to 9 -
p, (1) shows that
9 is constant. But 9 is one-to-one in U, and we have a contradiction. Thus
W1 = W2' and (a) is proved.
Suppose (b) is false. If we multiply f by a suitable constant of absolute
value 1, we then have P1 "# P2 butf(P1) =f(P2) = 1.
Since P1 and P2 are simple boundary points of 0, there are curves Yi with
parameter interval [0, 1] such that Yi([O, 1» c 0 for i = 1 and 2 and Yi(l) =
Pi' Put ri(t) =f(Yi(t». Then ri([O, 1» c U, and r 1(1) = r 2(1) = 1. Since
g(ri(t» = Yi(t) on [0, 1), we have
lim g(ri(t» = Pi (i = 1,2). (2)
1-1
Theorem 12.10 implies therefore that the radial limit of gat 1 is P1 as well as
P2' This is impossible if P1 "# P2'
IIII
14.19 Theorem If 0 is a bounded simply connected region in the plane and if
every boundary point of 0 is simple, then every conformal mapping of 0 onto U
extends to a homeomorphism offi onto U.
PROOF Suppose f E H(O), f(O) = U, and f is one-to-one. By Theorem 14.18
we can extendfto a mapping offi into U such thatf(lXn)-+f(z) whenever
{lXn} is a sequence in 0 which converges to z. If {zn} is a sequence in fi which
converges to z, there exist points IXn E 0 such that IlXn - Zn 1 < lin and
1 f(lXn) - f(zn) 1 < lin. Thus IXn-+ Z, hence f(lXn)-+ f(z), and this shows that
f(zn)-+ f(z).
We have now proved that our extension off is continuous on fi. Also
U cf(fi) c U. The compactness of U implies that f(fi) is compact. Hence
f(fi) = U.