## (Discrete) Polyá-Knopp Inequality

A while ago I uploaded a note to the Classic Analysis section of the website on the proof of Hardy’s inequality. Hardy’s inequality says that if a function $f \in L^{p}(0,\infty)$, for $1, then setting $F(x):=\frac{1}{x}\int_{0}^{x}\left|f(t)\right|dt$, we have the inequality

$\displaystyle\left\|F\right\|_{L^{p}}\leq\dfrac{p}{p-1}\left\|f\right\|_{L^{p}}$

The discrete version of Hardy’s inequality is that if $(a_{n})_{n=1}^{\infty}$ is a nonnegative sequence of reals in $\ell^{p}$, for $1. Then

$\displaystyle\sum_{n=1}^{\infty}\left|\dfrac{1}{n}\sum_{k=1}^{n}a_{k}\right|^{p}\leq\left(\dfrac{p}{p-1}\right)^{p}\sum_{k=1}^{\infty}\left|a_{k}\right|^{p}$

The proof for the discrete case in my note on Hardy’s inequality is incorrect as it stands–my apologies!. I will correct it later.

What I want to discuss today, though, isn’t Hardy’s inequality per se, but rather the inequality that results when you let $p\rightarrow\infty$ in Hardy’s inequality. This limiting case is alternatively called the discrete Polyá-Knopp inequality or Carleman’s inequality:

If $(a_{n})_{n=1}^{\infty}$ is a sequence of reals in $\ell^{1}$, then

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

I will present two proofs of this result, which are featured in the survey “CARLEMAN’S INEQUALITY – HISTORY, PROOFS AND SOME NEW GENERALIZATIONS” by Johansson, Persson, and Wedestig. Interestingly, neither of these proofs were originally presented by Carleman. The first proof uses Hardy’s inequality and a clever limiting argument, which illustrates the claim that the Polyá-Knopp inequality is a limiting case of Hardy’s. The downside to this proof is that it doesn’t show that the inequality is strict if the sequence $a_{n}$ is not identically zero. The second is less advanced in that it just uses the strict convexity of the arithmetic mean-geometric mean (AM-GM) inequality as well as a crude estimate for $e^{k}$, $k\in\mathbb{Z}^{\geq 0}$. The upside to this proof is that we get the strictness part.

Proof 1. Let $p>1$. Then

$\begin{array}{lcl}\displaystyle\left(\dfrac{1}{n}\sum_{k=1}^{n}a_{k}^{\frac{1}{p}}\right)^{p}&=&\displaystyle\exp\left(p\log\left(\dfrac{1}{n}\sum_{k=1}^{n}a_{k}^{\frac{1}{p}}\right)\right)\\[.9 em]&=&\displaystyle\exp\left(\dfrac{\log\left(\sum_{k=1}^{n}a_{k}^{\frac{1}{p}}\right)-\log\left(\sum_{k=1}^{n}a_{k}^{0}\right)}{p^{-1}}\right)\end{array}$

Using the definition of the derivative and the continuity of the $\exp$ function, we see that

$\begin{array}{lcl}\displaystyle\lim_{p\rightarrow\infty}\left(\dfrac{1}{n}\sum_{k=1}^{n}a_{k}^{\frac{1}{p}}\right)^{p}=\exp\left(\dfrac{d}{dx}\left[\log\left(\sum_{k=1}^{n}a_{k}^{x}\right)\right]_{x=0}\right)&=&\displaystyle\exp\left(\left[\dfrac{\sum_{k=1}^{n}\log(a_{k})a_{k}^{x}}{\sum_{k=1}^{n}a_{k}^{x}}\right]_{x=0}\right)\\[.9 em]&=&\displaystyle\exp\left(\dfrac{1}{n}\sum_{k=1}^{n}\log(a_{k})\right)\\[.9 em]&=&\displaystyle\left(a_{1}\cdots a_{n}\right)^{\frac{1}{n}}\end{array}$

By the dominated convergence theorem and Hardy’s inequality,

$\begin{array}{lcl}\displaystyle\sum_{n=1}^{\infty}\left(a_{1}\cdots a_{n}\right)^{\frac{1}{n}}=\lim_{p\rightarrow\infty}\sum_{n=1}^{\infty}\left(\dfrac{1}{n}\sum_{k=1}^{n}a_{k}^{\frac{1}{p}}\right)^{p}&\leq&\displaystyle\lim_{p\rightarrow\infty}\left(\dfrac{p}{p-1}\right)^{p}\sum_{k=1}^{\infty}\left(a_{k}^{\frac{1}{p}}\right)^{p}\\&=&\displaystyle e\sum_{k=1}^{\infty}a_{k}\end{array}$,

where we use the result

$\displaystyle\lim_{p\rightarrow\infty}\left(\dfrac{p}{p-1}\right)^{p}=\lim_{p\rightarrow\infty}\left(1+\dfrac{1}{p-1}\right)^{p-1}\left(1+\dfrac{1}{p-1}\right)=1\cdot e=e$

$\Box$

Proof 2. First, an elementary lemma.

Lemma. For all $k\in\mathbb{Z}^{\geq 0}$, $\dfrac{(k+1)^{k}}{k!}\leq e^{k}$, where the inequality is strict for $k\in\mathbb{Z}^{\geq 1}$.

Proof. We have that $(1+\frac{1}{n})^{n} for all $n \in \mathbb{Z}^{\geq 1}$ (see my note on compound interest). Hence,

$\begin{array}{lcl}\displaystyle e^{k}=\underbrace{e\cdots e}_{\text{k times}}&>&\displaystyle\left(1+\dfrac{1}{1}\right)^{1}\left(1+\dfrac{1}{2}\right)^{2}\cdots\left(1+\dfrac{1}{k}\right)^{k}\\[.9 em]&=&\displaystyle\left(\dfrac{1+1}{1}\right)\left(\dfrac{2+1}{2}\right)^{2}\cdots\left(\dfrac{k+1}{k}\right)^{k}\\[.9 em]&=&\displaystyle\dfrac{(k+1)^{k}}{k!}\end{array}$

$\Box$

Suppose $(a_{n})_{n=1}^{\infty}$ is a nonnegative sequence of reals such that $\sum_{n=1}^{\infty}a_{n}$ converges. Then

$\begin{array}{lcl}\displaystyle\sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}na_{n}\sum_{k=n}^{\infty}\left[\dfrac{1}{k}-\dfrac{1}{k+1}\right]&=&\displaystyle\sum_{n=1}^{\infty}na_{n}\sum_{k=n}^{\infty}\dfrac{1}{k(k+1)}\\[.9 em]&=&\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)}\sum_{k=1}^{n}ka_{k}\\[.9 em]&=&\displaystyle\sum_{n=1}^{\infty}\dfrac{1}{n+1}\left(\dfrac{a_{1}+\cdots+na_{n}}{n}\right)\end{array}$

where we use the fact that the order of summation of an absolutely convergent series is irrelevant. We can apply the AM-GM inequality to $\frac{a_{1}+\cdots+na_{n}}{n}$ to obtain

$\displaystyle\sum_{n=1}^{\infty}a_{n}\geq\sum_{n=1}^{\infty}\dfrac{\left(n!a_{1}\cdots a_{n}\right)^{\frac{1}{n}}}{n+1}=\sum_{n=1}^{\infty}\left(\dfrac{n!}{(n+1)^{n}}\right)^{\frac{1}{n}}\left(a_{1}\cdots a_{n}\right)^{\frac{1}{n}}\geq e^{-1}\sum_{n=1}^{\infty}\left(a_{1}\cdots a_{n}\right)^{\frac{1}{n}}$,

where the last inequality follows from an application of the lemma. I claim that the inequality is in fact strict if not all the terms $a_{n}$ are zero. Indeed, suppose not. Then equality must hold in each application of the AM-GM inequality. Set $c=a_{1}$. Suppose $a_{k}=\frac{c}{k}$. Then

$\displaystyle\dfrac{a_{1}+\cdots+ka_{k}+(k+1)a_{k+1}}{k+1}=\dfrac{c+\cdots+k\frac{c}{k}+(k+1)a_{k+1}}{k+1}=\dfrac{kc+(k+1)a_{k+1}}{k+1}$

and

$\begin{array}{lcl}\displaystyle\left(a_{1}\cdots ka_{k}(k+1)a_{k+1}\right)^{\frac{1}{k+1}}=\left(c\cdots k\dfrac{c}{k}(k+1)a_{k+1}\right)^{\frac{1}{k+1}}&=&\displaystyle\left(c^{k}(k+1)a_{k+1}\right)^{\frac{1}{k+1}}\\&=&\displaystyle c^{\frac{k}{k+1}}\left((k+1)a_{k+1}\right)^{\frac{1}{k+1}}\end{array}$

But by strict convexity, equality holds if and only if $c=(k+1)a_{k+1}\Longleftrightarrow a_{k+1}=\frac{c}{k+1}$. By induction, we see that $a_{n}=\frac{c}{n}$, for all $n\in\mathbb{Z}^{\geq 1}$. But since the harmonic series diverges, this contradicts the convergence of $\sum_{n=1}a_{n}$. $\Box$