(Integral) Polyá-Knopp Inequality

In my last post, I went over the statement and proof of the discrete Polyá-Knopp inequality: if (a_{n})_{n=1}^{\infty} is a sequence in \ell^{1}, then

\displaystyle\sum_{n=1}^{\infty}\left|a_{1}\cdots a_{n}\right|^{\frac{1}{n}}\leq e\sum_{n=1}^{\infty}\left|a_{n}\right|,

In particular, if the a_{n} 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 f \in L^{1}(0,\infty), and f\neq 0 a.e., then

\displaystyle\int_{0}^{\infty}\exp\left(\dfrac{1}{x}\int_{0}^{x}\log\left|f(t)\right|dt\right)dx\leq e\int_{0}^{\infty}\left|f(t)\right|dt

Proof. Suppose f \in L^{1}(0,\infty). Replacing f by \left|f\right|, we may assume that f is nonnegative. Then for all p\geq 1, \left|f\right|^{\frac{1}{p}}\in L^{1}(0,\infty) by Hölder’s inequality. Then for x>0,

\begin{array}{lcl}\displaystyle\left(\dfrac{1}{x}\int_{0}^{x}\left|f(t)\right|^{\frac{1}{p}}dt\right)^{p}&=&\displaystyle\exp\left[p\log\left(x^{-1}\int_{0}^{x}\left|f(t)\right|^{\frac{1}{p}}dt\right)\right]\\[.9 em]&=&\displaystyle\exp\left[\dfrac{\log\left(\int_{0}^{x}\left|f(t)\right|^{\frac{1}{p}}dt\right)-\log\left(\int_{0}^{x}\left|f(t)\right|^{0}dt\right)}{p^{-1}}\right]\end{array}

We first compute the derivative of the function g_{x}: (0,\infty)\rightarrow\mathbb{R} defined by g_{x}(r):=\int_{0}^{x}\left|f(t)\right|^{r}dt.

\displaystyle\dfrac{g_{x}(r+h)-g_{x}(r)}{h}=\dfrac{1}{h}\int_{0}^{x}\left(\left|f(t)\right|^{r+h}-\left|f(t)\right|^{r}\right)dt=\int_{0}^{x}\left(\sum_{k=1}^{\infty}\dfrac{(h\left|f(t)\right|)^{k}}{k!}\right)dt

Since \sum_{k=1}^{\infty}\dfrac{(h\left|f(t)\right|)^{k}}{k!} is dominated by \sum_{k=1}^{\infty}\frac{\left|f(t)\right|^{k}}{k!}, for h\neq 0 sufficiently small. By the dominated convergence theorem,

\displaystyle g_{x}'(r)=\int_{0}^{x}\left|f(t)\right|^{r}\log\left|f(t)\right|dt

Hence by the chain rule,

\displaystyle\dfrac{d}{dr}\left[\log g_{x}(r)\right]=\dfrac{g_{x}'(r)}{g_{x}(r)}=\dfrac{\int_{0}^{x}\left|f(t)\right|^{r}\log\left|f(t)\right|dt}{\int_{0}^{x}\left|f(t)\right|^{r}dt}

Noting that the argument of \exp is a difference quotient, we see from the continuity of the $\exp$ function that

\displaystyle\lim_{p\rightarrow\infty}\left(\dfrac{1}{x}\int_{0}^{x}\left|f(t)\right|^{\frac{1}{p}}dt\right)^{p}=\exp\left(\dfrac{g_{x}'(0)}{g_{x}(0)}\right)=\exp\left[\dfrac{1}{x}\int_{0}^{x}\log\left|f(t)\right|dt\right]

Let R\gg 0. Then by Jensen’s inequality

\displaystyle\left(\dfrac{1}{x}\int_{0}^{x}\left|f(t)\right|^{\frac{1}{p}}dt\right)^{p}\leq\dfrac{1}{x}\int_{0}^{x}\left|f(t)\right|dt\in L^{1}(R^{-1},R)

By Hardy’s inequality and the dominated convergence theorem,

\begin{array}{lcl}\displaystyle\int_{R^{-1}}^{R}\exp\left(\dfrac{1}{x}\int_{0}^{x}\log\left|f(t)\right|dt\right)dx=\lim_{p\rightarrow\infty}\int_{R^{-1}}^{R}\left(\dfrac{1}{x}\int_{0}^{x}\left|f(t)\right|^{\frac{1}{p}}dt\right)^{p}dx&\leq&\displaystyle\lim_{p\rightarrow\infty}\left(\dfrac{p}{p-1}\right)^{p}\int_{0}^{\infty}\left|f(t)\right|dt\\&=&\displaystyle e\int_{0}^{\infty}\left|f(t)\right|dt\end{array}

for all R>0. Letting R\uparrow\infty, the monotone convergence theorem implies the inequality

\displaystyle\int_{0}^{\infty}\exp\left(\dfrac{1}{x}\int_{0}^{x}\log\left|f(t)\right|dt\right)dx\leq e\int_{0}^{\infty}\left|f(t)\right|dt

\Box

I claim that the inequality is strict for f \neq 0 a.e. Indeed, by Jensen’s inequality,

\begin{array}{lcl}\displaystyle\int_{0}^{\infty}\exp\left(\dfrac{1}{x}\int_{0}^{x}\left[\log t + \log\left|f(t)\right|-\log t\right]dt\right)dx&=&\displaystyle\int_{0}^{\infty}\exp\left(\dfrac{1}{x}\int_{0}^{x}\log\left|tf(t)\right|dt\right)\exp\left(-\dfrac{x\log x}{x}+1\right)dx\\[.9 em]&=&\displaystyle\int_{0}^{\infty}\dfrac{e}{x}\exp\left(\dfrac{1}{x}\int_{0}^{x}\log\left|tf(t)\right|dt\right)dx\\[.9 em]&\leq&\displaystyle\int_{0}^{\infty}\dfrac{e}{x^{2}}\int_{0}^{x}\left|tf(t)\right|dtdx\\[.9 em]&=&\displaystyle\int_{0}^{\infty}\left|tf(t)\right|\left(\int_{t}^{\infty}\dfrac{e}{x^{2}}dx\right)dt\\[.9 em]&=&\displaystyle e\int_{0}^{\infty}\left|f(t)\right|dt\end{array}

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

\displaystyle\log\left|tf(t)\right|=C \Longrightarrow tf(t)=C\text{ a.e.}

But since f\in L^{1}(0,\infty), we see that C=0, which contradicts that f\neq 0 a.e.

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