Real and Complex Analysis (Rudin)

300 REAL AND COMPLEX ANALYSIS
Choose E, 0 < E < t. There exists an No such that
00
L I U.(s) I < E (s E 8). (3)
.=No
Let {nl' n 2 , n 3 , ••• } be a permutation of {I, 2, 3, ... }. If N ~ No, if Mis
so large that
{I, 2, ... , N} c {nh n 2 , ••• , nM}'
and if qM(S) denotes the Mth partial product of (2), then
(4)
qM - PN = PN{n (1 + u. k) -
The n k which occur in (5) are all distinct and are larger than No. Therefore
(3) and Lemma 15.3 show that
If nk = k (k = 1,2,3, ... ), then qM = PM' and (6) shows that {PN} converges
uniformly to a limit function! Also, (6) shows that
IPM - PNol ~ 21PNo iE
so that IPM I ~ (1 - 2E)lpN o l. Hence
I f(s) I ~ (1 - 2E) I PNo(S) I
which shows thatf(s) = 0 if and only if PNo(S) = O.
I}.
(M > No),
(s E S),
Finally, (6) also shows that {qM} converges to the same limit as {PN}' IIII
15.5 Theorem Suppose 0 ~ u. < 1. Then
00
if and only if L u. < 00.
.=1
PROOF If PN = (1 - u1)··· (1 - UN)' then PI ~ P2 ~"', PN > 0, hence P =
lim PN exists. If .tu. < 00, Theorem 15.4 implies P > O. On the other hand,
(5)
(6)
(7)
(8)
N
P ~ PN = n (1 - un) ~ exp {-U 1 - U2 - ... - UN}'
1
and the last expression tends to 0 as N -4 00, if .tu. = 00.
IIII
We shall frequently use the following consequence of Theorem 15.4:
15.6 Theorem Suppose f. E H(n) for n = 1, 2, 3, ... , no!. is identically 0 in
any component ofn, and
00
L 11 - f.(z) I (1)
.=1