Real and Complex Analysis (Rudin)

394 REAL AND COMPLEX ANALYSIS
and
Hence (22) yields
100 4,00002
2n 3 p dp = 2,OOOno.
4.1 p
(24)
I F(z) - CIl(Z) I < 6,OOOw(0) (z En). (25)
Since F E H(n), Ken, and S2 - K is connected, Runge's theorem
shows that F can be uniformly approximated on K by polynomials. Hence
(3) and (25) show that (2) can be satisfied.
This completes the proof.
One unusual feature of this proof should be pointed out. We had to prove
that the given function f is in the closed subspace P(K) of C(K). (We use the
terminology of Sec. 20.1.) Our first step consisted in approximating f by . But
this step took us outside P(K), since was so constructed that in general will
not be holomorphic in the whole interior of K. Hence is at some positive
distance from P(K). However, (25) shows that this distance is less than a constant
multiple of w(o). [In fact, having proved the theorem, we know that this distance
is at most w(o), by (3), rather than 6,000 W(o).J The proof of (25) depends on the
inequality (4) and on the fact that a = 0 in G. Since holomorphic functions cp
are characterized by acp = 0, (4) may be regarded as saying that is not too far
from being holomorphic, and this interpretation is confirmed by (25).
IIII
Exercises
I Extend Mergelyan's theorem to the case in which 8 2 - K has finitely many components: Prove
that then every f e C(K) which is holomorphic in the interior of K can be uniformly approximated on
K by rational functions.
2 Show that the result of Exercise 1 does not extend to arbitrary compact sets K in the plane, by
verifying the details of the following example. For n = 1,2,3, ... , let Dn = D(lXn; rn) be disjoint open
discs in U whose union V is dense in U, such that :Ern < 00. Put K = -0 - V. Let rand Yn be the
paths
and define
os; t s; 21l,
L(f) = If(Z) dz - t 1.f(Z) dz
(fe C(K».
Prove that L is a bounded linear functional on C(K), prove that L(R) = 0 for every rational function
R whose poles are outside K, and prove that there exists anf e C(K) for which L(f) -# O.
3 Show that the function g constructed in the proof of Lemma 20.2 has the smallest supremum norm
among a1lf e H(n) such that zf(z)--+ 1 as z--+ 00. (This motivates the proof of the lemma.)
Show also that b = Co in that proof and that the inequality I b I < 4r can therefore be replaced by
I b I < r. In fact, b lies in the convex hull of the set E.