Convergence of sums using Hilbert space techniques

Question

Let $a_n$ be a sequence of real numbers such that $\sum_{n=1}^{\infty} a_nb_n < \infty$ for any sequence $b_n$ satisfying $\sum_{n=1}^{\infty}b_n^2 < \infty$. Prove that $\sum_{n=1}^{\infty}a_n^2 < \infty$.

This question is in the chapter on Hilbert spaces. What sort of technique is required here? Seems like a baby analysis question.

Answer 1

First note that, for any $(b_1, b_2, \ldots) \in \ell^2$, the hypothesis implies the slightly stronger condition $$ \sum_{n \ge 1} |a_i b_i| < \infty, $$ because we can pick the sign of $b_i$ so that $a_i b_i > 0$.

Now for $n \ge 1$, let $f_n : \ell^2 \to \mathbb{R}$ be the map defined by $$ \newcommand{\norm}[1]{\left\| #1 \right\|} f_n(b_1, b_2, \ldots) = \sum_{i=1}^n a_ib_i $$ For any $\vec{b} = (b_1, b_2, \ldots)$, $f_n(b)$ is equal to the inner product of $\vec{b}$ with the vector $(a_1, a_2, \ldots, a_n, 0, 0, \ldots) \in \ell^2$. Hence $f_n$ is a bounded linear functional on $\ell^2$. Moreover, for a fixed $\vec{b} \in \ell^2$, $$ \left| \sup_{n \ge 1} f_n \vec{b} \right| < \sum_{i=1}^\infty |a_ib_i| < \infty $$ Therefore, by the Uniform Boundedness Principle, we have that the $f_n$ are uniformly bounded, i.e. there exists $C$ such that $\norm{f_n} \le C$ for all $n$. Thus, $$ \sum_{i=1}^\infty |a_i|^2 = \lim_{n \to \infty} \sum_{i=0}^n |a_i|^2 = \lim_{n \to \infty} \norm{f_n}^2 \le C^2 < \infty. \square $$

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Direct proof that $\boldsymbol{\norm{f_n} = \sqrt{\sum_{i=0}^n |a_i|^2}}$

For any $v = (v_1, v_2, \ldots)$, by Holder's inequality (in fact, the Cauchy-Schwarz inequality in this case), we have $$ \left|f_n v\right| = \left| \sum_{i=1}^n a_i v_i \right| \le \sum_{i=1}^n \left| a_i v_i \right| \le \left( \sum_{i=1}^n \left| a_i \right|^2 \right)^{1/2} \left( \sum_{i=1}^n \left| v_i \right|^2 \right)^{1/2} \le \left( \sum_{i=1}^n \left| a_i \right|^2 \right)^{1/2} \norm{v}_2 $$ Thus $f$ is bounded, with norm at most $\left( \sum_{i=1}^n \left| a_i \right|^2 \right)^{1/2}$. In fact the norm is exactly this value, because setting $v = (a_1, a_2, a_3, \ldots, a_n, 0, 0, \ldots)$ we have $$ |f_n v| = \sum_{i=1}^n |a_i|^2 = \sqrt{\sum_{i=1}^n |a_i|^2} \sqrt{\sum_{i=1}^n |a_i|^2} = \left( {\sum_{i=1}^n |a_i|^2}\right)^{1/2} \norm{v}_2. $$

Answer 2

Suppose $\sum_n a_n^2 = \infty$ and all is positive

define $S_n = \sum_{k=1}^n a_k^2$ and $p_n = \min\{q > p_{n-1}: \sum_{k=1}^q a_k^2 > 2^n\}$ with $p_0 = 0$. Then $p_n$ increases to infinity as $n \to \infty$

Then if we take $$b_k^2 = \frac{a^2_k}{S_{p_n}^{\frac{3}{2}}}$$ for $ p_{n-1}<k \leq p_n$, we have $\sum_{n} b_n^2 < \sum_n \frac{1}{\sqrt{S_{p_n}}} < \sum_n \frac{1}{(\sqrt{2})^n} < \infty$.

So we have $\sum_{n} a_nb_n < \infty$ while at the same time remark that $\sum_n a_nb_n > (S_{p_m})^{\frac{1}{4}}, \forall m$, which means that $S_{p_m}$ is bounded and thus converge to a finite limit.

Since $S_n$ is increasing, the convergence of one of its subsequence implies the convergence of $S_n$, which is contradictory with $\sum_n a_n^2 = \infty$