In my last post, I went over the statement and proof of the discrete Polyá-Knopp inequality: if is a sequence in , then
In particular, if the are not all zero, then the inequality is sharp. We gave two proofs of this result: one derived the Polyá-Knopp result from Hardy’s inequality by a limiting argument, and the other one relied on the AM-GM inequality.
I now want to prove the integral version of the Polyá-Knopp inequality by a limiting argument similar to the one used in the proof of discrete version. So the reader should be familiar with the integral form of Hardy’s inequality.
Theorem 1. If , and a.e., then
Proof. Suppose . Replacing by , we may assume that is nonnegative. Then for all , by Hölder’s inequality. Then for ,
We first compute the derivative of the function defined by .
Since is dominated by , for sufficiently small. By the dominated convergence theorem,
Hence by the chain rule,
Noting that the argument of is a difference quotient, we see from the continuity of the $\exp$ function that
Let . Then by Jensen’s inequality
By Hardy’s inequality and the dominated convergence theorem,
for all . Letting , the monotone convergence theorem implies the inequality
I claim that the inequality is strict for a.e. Indeed, by Jensen’s inequality,
where we use Jensen’s inequality in the third line and Fubini’s theorem in the fourth line. If equality holds, then Jensen’s inequality implies that
But since , we see that , which contradicts that a.e.