386 Chapter6
Theorem 6.33 Let A be any open set, and let a E A. If f is continuous
on A and open on A - {a}, then f is open on A.
Proof Let G be any open subset of A containing a. We wish to prove that
f(a) E Intf(G).
Let C: lz -al = r be a circle with center at a and such that CU Int C C
G. If f(a) E f(C), then f(a) belongs to the interior of f(G). Suppose that
f(a) f/. f(C), and let D =Int C -{a}, E =interior of a circle containing
f(a) and such that Enf(C) = 0. We have (E-{f(a)})nf(D) =j; 0. If B
is the component of the complement of f(D - D) containing E - {f(a)},
then, by Theorem 6.32, f(D) :J B :J E - {f(a)}. Hence f(D) :J E, and
f(a) E Int f(D) c Int f(G).
Theorem 6.34 If R is a bounded region, a E R and f is a continuous
function on fl and analytic on R - {a}, then
lf(z)I:::; ~ax lf(t)i for all z ER (6.22-1)
tER-RProof If f is constant on R - {a}, then f is constant on fl, and equality
holds in (6.22-1 ). If f is not constant on R-{a}, then f is open on R-{ a}.
By Theorem 6.33 it follows that f is open on R.
Now l.f(t)J has certainly a maximum on fl, since fl is compact. But this
maximum cannot be assumed on R, for if w 0 = f.( z 0 ), z 0 E R, for e > 0
small enough Ne(w 0 ) C f(R), since f(R) is open, qnd in Ne(w 0 ) there are
points with a modulus greater than Jwol = lf(zo)J.
Theorem 6.35 Let C: Jz - zol = r, and suppose that f is continuous on
lz - z 0 I ~ r. Then for every e > 0 there is a 8 > 0 such that
I
f(zo) - f(z) _ .f(() - f(z) I < /5
Zo - Z (-Zwhenever z E C and J( - zol < 8.
Proof For some constant K we have
lf(zo) - f(z)I < K for z EC
and. 1
lf(zo)-f(()I < 4re for Jzo -(J < 81
Choosing 8 < min (%r, r^2 e/4K, 81 ), we getJ(( -z)[f(zo)-f(z)] - (zo - z)[.f(() -J(z)]I
= J(( - zo)[f(zo)-f(z)] + (zo - z)[.f(zo) -f(()]I
< r
2
e K +r (~re)= ~r^2 e
4K 4 2