Real and Complex Analysis (Rudin)

HP-SPACES 351
Let us recall that every harmonic u that satisfies (1) is the Poisson integral of
a function u* E I.!'(T) (Theorem 11.30) if 1 < P < 00. Theorem 11.11 suggests
therefore another restatement of the problem:
If 1 < P < 00, and if we associate to each h E I.!'( T) the holomorphic function
1 f" e it + z .
(.ph)(z) = -2
11: _" e - z
-it- h(e'~ dt
do all of these functions .ph lie in HP?
Exercise 25 deals with some other aspects of this problem.
(z E U), (2)
17.25 Lemma If 1 < p ~ 2, b = 11:1(1 + p), 0( = (cos b)-1, and P = O(P(l + O(),
then
1 ~ P(cos (f))P - 0( cos P(f)
(1)
PROOF If b ~ I (f) I ~ 11:12, then the right side of (1) is not less than
-0( cos P(f) ~ -0( cos pb = 0( cos b = 1,
and it exceeds P(cos b)P - 0( = 1 if I (f) I ~ b.
IIII
17.26 Theorem If 1 < p < 00, then there is a constant Ap < 00 such that the
inequality
holds for every he I.!'(T).
More explicitly, the conclusion is that .ph (defined in Sec. 17.24) is in HP,
and that
where dO' = d()1211: is the normalized Lebesgue measure on T.
Note that h is not required to be a real function in this theorem, which
asserts that .p: I.!'-+ HP is a bounded linear operator.
PROOF Assume first that 1 < p ~ 2, that h E I.!'(T), h ~ 0, h ¢ 0, and let u be
the real part off = .ph. Formula 11.5(2) shows that u = P[h], hence u > 0 in
U. Since U is simply connected and f has no zero in U, there is age H(U)
such that g = r, g(O) > O. Also, u = I f I cos (f), where (f) is a real function
with domain U that satisfies I (f) I < 11:12.
If 0( = O(p and P = Pp are chosen as in Lemma 17.25, it follows that
II !rIP dO' ~ P I (Ur)P dO' - 0( II !rIP cos (P(f)r) dO' (3)
for 0 ~ r < 1.
(1)
(2)