Real and Complex Analysis (Rudin)

64 REAL AND COMPLEX ANALYSIS
PROOF Let A and B be the two factors on the right of (1). If A = 0, thenf = 0
a.e. (by Theorem 1.39); hencefg = 0 a.e., so (1) holds. If A> 0 and B = 00, (1)
is again trivial. So we need consider only the case 0 < A < 00, 0 < B < 00.
Put
This gives
f
F=­ A'
1 FP dp, = 1 Gq dp, = 1.
If x E X is such that 0 < F(x) < 00 and 0 < G(x) < 00, there are real
numbers sand t such that F(x) = e'/p, G(x) = e t / q • Since lip + 1/q = 1, the
convexity of the exponential function implies that
It follows that
for every x E X. Integration of (6) yields .
lFG dp,::S; p-l + q-l = 1, (7)
by (4); inserting (3) into (7), we obtain (1).
Note that (6) could also have been obtained as a special case of the
inequality 3.3(8).
To prove (2), we write
(f + g)P =f· (f + g)P-l + g . (f + g)p-l. (8)
HOlder's inequality gives
f f· (f + g)P-l ::S; {f fP} l/ P {f (f + g)(P-l)q} l/ q . (9)
Let (9') be the inequality (9) with f and g interchanged. Since (p - l)q = p,
addition of (9) and (9') gives
(3)
(4)
(5)
(6)
Clearly, it is enough to prove (2) in the case that the left side is greater
than 0 and the right side is less than 00. The convexity of the function tP for
o < t < 00 shows that