Real and Complex Analysis (Rudin)

344 REAL AND COMPLEX ANALYSIS
17.17 Theorem Suppose 0 < p ~ 00, f E HP, and f is not identically O. Then
log I f* I E L 1 (n, the outer function
Qf(Z) = exp -2
n -" e
{If" e it + Z .}
~ log I f*(e'~ I dt
is in HP, and there is an inner function M f such that
Z
(1)
(2)
Furthermore,
log I f(O) I ~ 2n 1 f" _}Og I f*(ei~ I dt. (3)
Equality holds in (3) if and only if M f is constant.
The functions M f and Qf are called the inner and outer factors off, respectively;
Qf depends only on the boundary values of I f I.
PROOF We assume first that f E HI. If B is the Blaschke product formed with
the zeros of f and if g = fiB, Theorem 17.9 shows that g E HI; and since
I g* I = I f* I a.e. on T, it suffices to prove the theorem with g in place off
So let us assume that f has no zero in U and that f(O) = 1. Then log I f I
is harmonic in U, log I f(O)l = 0, and since log = log+ - log-, the mean
value property of harmonic functions implies that
1 f" . 1 f" .
2n _}og- If(re'~1 dO = 2n _}Og+ If(re'~ dO ~ IIfllo ~ IIflll (4)
for 0 < r < 1. It now follows from Fatou's lemma that both log+ I f* I and
log- I f* I are in L 1 (n, hence so is log I f* I.
This shows that the definition (1) makes sense. By Theorem 17.16, Qf E
HI. Also, I QJI = If* I =F 0 a.e., since log If* IE Ll(n. If we can prove that
(z E U), (5)
thenflQf will be an inner function, and we obtain the factorization (2).
Since log I Qf I is the Poisson integral of log I f* I, (5) is equivalent to the
inequality
log If I ~ P[log If*I], (6)
which we shall now prove. Our notation is as in Chap. 11: P[h] is the
Poisson integral of the function h E L 1 (T).
For I Z I ~ 1 and 0 < R < 1, putfR(Z) = f(Rz). Fix Z E U. Then
log I fR(Z) I = P[log+ I fR I ](z) - P[log- I fR I ](z). (7)