## (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.

This entry was posted in math.CA and tagged , . Bookmark the permalink.