Real and Complex Analysis (Rudin)

274 REAL AND COMPLEX ANALYSIS
is hoI om orphic in the interior of K N • Thus f has precisely the prescribed
principal parts in the interior of K N , and hence in 0, since N was arbitrary.
IIII
Simply Connected Regions
We shall now summarize some properties of simply connected regions (see
Sec. 10.38) which illustrate the important role that they play in the theory of
holomorphic functions. Of these properties, (a) and (b) are what one might call
internal topological properties of 0; (c) and (d) refer to the way in which 0 is
embedded in S2; properties (e) to (h) are analytic in character; U) is an algebraic
statement about the ring H(O). The Riemann mapping theorem 14.8 is another
very important property of simply connected regions. In fact, we shall use it to
prove the implication U)-+ (a).
13.11 Theorem For a plane region 0, each of the following nine conditions
implies all the others.
(a) 0 is homeomorphic to the open unit disc U.
(b) 0 is simply connected.
(c) Indy (IX) = Of or every closed path y in 0 andfor every IX E S2 - O.
(d) S2 - 0 is connected.
(e) Every f E H(O) can be approximated by polynomials, uniformly on compact
subsets ofO.
(f) For every f E H(O) and every closed path y in 0,
if(Z) dz = O.
(g) To every f E H(O) corresponds an F E H(O) such that F' = f.
(h) Iff E H(O) and Ilf E H(O), there exists agE H(O) such that f = exp (g).
U) Iff E H(O) and Ilf E H(O), there exists a ({J E H(O) such thatf = ({J2.
The assertion of (h) is that f has a "holomorphic logarithm" 9 in 0; U)
asserts that f has a "holomorphic square root" ({J in 0; and (f) says that the
Cauchy theorem holds for every closed path in a simply connected region.
In Chapter 16 we shall see that the monodromy theorem describes yet
another characteristic property of simply connected regions.
PROOF (a) implies (b). To say that 0 is homeomorphic to U means that there
is a continuous one-to-one mapping r/! of 0 onto U whose inverse r/! -1 is also
continuous. If y is a closed curve in 0, with parameter interval [0, 1], put
H(s, t) = r/! -1(tr/!(y(s»).
Then H: 12-+ 0 is continuous; H(s, 0) = r/!-1(0), a constant; H(s, 1) = y(s);
and H(O, t) = H(I, t) because y(O) = y(I). Thus 0 is simply connected.