(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<p<\infty, 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<p<\infty. 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}<e 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

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One Response to (Discrete) Polyá-Knopp Inequality

  1. Pingback: (Integral) Polyá-Knopp Inequality | Math by Matt

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