Cute Series Problem

The following problem comes from Walter’s Rudin classic text Real and Complex Analysis. It’s Exercise 7 of Chapter 4 (Elementary Hilbert Space Theory).

Suppose (a_{n})_{n=1}^{\infty} is a sequence of positive numbers such that \sum_{n=1}^{\infty}a_{n}b_{n}<\infty for any nonnegative sequence (b_{n})_{n=1}^{\infty}\in\ell^{2}. Then \sum_{n=1}^{\infty}a_{n}^{2}<\infty.

Proof. Suppose that \sum_{n=1}^{\infty}a_{n}^{2}=\infty. By induction, we can construct a countable collection of disjoint finite sets \left\{E_{k}\right\}_{k=1}^{\infty} which partition \mathbb{Z}^{\geq 1} and satisfy

\displaystyle 1<\sum_{n\in E_{k}}a_{n}^{2}<\infty

We will construct a nonnegative sequence (b_{n})_{n=1}^{\infty} such that \sum_{n=1}^{\infty}a_{n}b_{n}=\infty, but \sum_{n=1}^{\infty}b_{n}^{2}<\infty. For each k\in\mathbb{Z}^{\geq 1}, define

\displaystyle c_{k}:=\dfrac{1}{k}\left(\sum_{n\in E_{k}}a_{n}^{2}\right)^{-1}

and for n\in E_{k}, define b_{n}:=c_{k}a_{n}. Since the partial sums of the series \sum_{n=1}^{\infty}b_{n}^{2} are nonnegative, the order of summation is irrelevant, provided the series converges. Thus,

\begin{array}{lcl}\displaystyle\sum_{n=1}^{\infty}b_{n}^{2}=\sum_{k=1}^{\infty}\sum_{n\in E_{k}}c_{k}^{2}a_{n}^{2}\left(\sum_{n\in E_{k}}a_{n}^{2}\right)^{-2}&<&\displaystyle\sum_{k=1}^{\infty}\sum_{n\in E_{k}}\dfrac{a_{n}^{2}}{k^{2}}\left(\sum_{n\in E_{k}}a_{n}^{2}\right)^{-1}\\&=&\displaystyle\sum_{k=1}^{\infty}\sum_{n\in E_{k}}\dfrac{1}{k^{2}}\\&=&\displaystyle\sum_{k=1}^{\infty}\dfrac{1}{k^{2}}<\infty\end{array},

since \left(\sum_{n\in E_{k}}a_{n}^{2}\right)^{-1}<1, and

\displaystyle\sum_{n=1}^{\infty}a_{n}b_{n}=\sum_{k=1}^{\infty}\sum_{n\in E_{k}}\dfrac{a_{n}^{2}}{k}\left(\sum_{n\in E_{k}}a_{n}^{2}\right)^{-1}=\sum_{k=1}^{\infty}\sum_{n\in E_{k}}\dfrac{1}{k}=\sum_{k=1}^{\infty}\dfrac{1}{k}=\infty

We arrive at a contradiction, which completes the proof. \Box

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