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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.

Caleb Stanford
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Toeplitz
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  • What do you mean "whenever" $\sum b_n^2 < \infty$? So you are saying $a_n$ is a fixed sequence, but $b_n$ is not a fixed sequence? These things need to be specified, in stead of just saying they are both sequences. Regards – Caleb Stanford Apr 28 '14 at 20:57
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    I have the same book as Justin. The problem should give you a fixed a_n, not a fixed b_n. – Johnny Apple Apr 28 '14 at 21:55
  • @JohnnyApple Thanks! I'll go ahead and edit that in. – Caleb Stanford Apr 28 '14 at 22:14
  • And if you want the "baby analysis" proof, see http://math.stackexchange.com/questions/58565/is-there-a-constructive-proof-of-this-characterization-of-ell2 – Nate Eldredge Apr 29 '14 at 01:42

2 Answers2

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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, $$ \sum_{i=1}^\infty |a_i|^2 = \lim_{n \to \infty} \sum_{i=0}^n |a_i|^2 = \lim_{n \to \infty} \norm{f_n}^2 \le C^2 < \infty. \square $$

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 $$ |f_n v| = \sum_{i=1}^n |a_i|^2 = \sqrt{\sum_{i=1}^n |a_i|^2} \sqrt{\sum_{i=1}^n |a_i|^2} = \left( {\sum_{i=1}^n |a_i|^2}\right)^{1/2} \norm{v}_2. $$

Caleb Stanford
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  • Hmm, I wonder why the un-upvote. – Caleb Stanford Apr 29 '14 at 00:28
  • I don't follow the last equality, with the limits. Why is the norm of $f_n$ that sum? How do you know it occurs at $a_i$? – Johnny Apple Apr 29 '14 at 01:02
  • @JohnnyApple If $H$ is a Hilbert space and $x$ is an element of $H$, then $f_x$ defined by $f_x(y) = \langle x, y \rangle$ is a bounded linear operator on $H$ with norm $\left| x \right|$. In this case $x = (a_1, a_2, \ldots, a_n, 0, 0, \ldots)$, and $\left|{f_n}\right| = \left|x\right|$. – Caleb Stanford Apr 29 '14 at 01:07
  • @JohnnyApple What do you mean "it occurs at $a_i$"? – Caleb Stanford Apr 29 '14 at 01:09
  • I just mean I don't see why the norm of $f_x(y)$ is $||x||$. – Toeplitz Apr 29 '14 at 04:26
  • @Justin Well, it is a standard fact about Hilbert spaces. If $H$ is a Hilbert space then $H$ is fully isomorphic to its dual space. In particular, the isomorphism is given by $x \mapsto f_x$ defined in my previous comment, and since this is an isomorphism, $| x| = | f_x |$. – Caleb Stanford Apr 29 '14 at 04:31
  • @Justin You can also prove it directly. If you want I can add a proof to my answer. – Caleb Stanford Apr 29 '14 at 04:33
  • only if it isn't an inconvenience. I hadn't actually heard that fact. – Toeplitz Apr 29 '14 at 04:39
  • @Justin I added a proof. Also there was a typo at the end, $|f_n| < C < \infty$ should have been $|f_n|^2 < C^2 < \infty$. – Caleb Stanford Apr 29 '14 at 04:55
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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 $\sum_n a_n^2 = \infty$

Petite Etincelle
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