Real and Complex Analysis (Rudin)

392 REAL AND COMPLEX ANALYSIS
A E C~(R2). Hence the last expression in (11) may be differentiated under the
integral sign, and we obtain
(a)(z) = f f (aA)(z -
R2
OJ«() de d'1
= f f J(z - ()(aA)«() de d'1
R2
= ff [f(z - () - J(z)](aA)(() de d'1. (13)
R2
The last equality depends on (9). Now (10) and (13) give (4). If we write
(13) with " and , in place of 13, we see that has continuous partial
derivatives. Hence Lemma 20.3 applies to , and (5) will follow if we can
show that 13 = 0 in G, where G is the set of all z E K whose distance from
the complement of K exceeds b. We shall do this by showing that
(z) = J(z) (z E G); (14)
note that aJ = 0 in G, since J is holomorphic there. (We r.ecall that a is the
Cauchy-Riemann operator defined in Sec. 11.1.) Now if z E G, then z - ( is in
the interior of K for all ( with I (I < b. The mean value property for harmonic
functions therefore gives, by the first equation in (11),
ft! f2"
(z) = Jo a(r)r dr Jo J(z -
rei') dO
= 2nJ(z) f a(r)r dr = J(z) f f A = J(z) (15)
for all z E G.
We have now proved (3), (4), and (5).
The definition of X shows that X is compact and that X can be covered
by finitely many open discs D 1 , ..• , D n , of radius 2b, whose centers are not in
K. Since S2 - K is connected, the center of each D j can be joined to 00 by a
polygonal path in S2 - K. It follows that each D j contains a compact connected
set E j , of diameter at least 2b, so that S2 - E J is connected and so
that K n Ej = 0.
We now apply Lemma 20.2, with r = 2b. There exist functions
R2