Suppose that $a_n$ is a sequence of real numbers such that $\sum_na_nb_n$ converges whenever $\sum_n b_n^2 \lt \infty$. Show that $\sum_{n=1}^{\infty}a_n^2 \lt \infty$.

My try: I defined $T: l^2 \to \mathbb{R}$ by sending $(b_1,b_2,\ldots,b_n,\ldots,) \to \sum_{n=1}^{\infty}a_nb_n$. Then $T$ is linear. I want to show that $T$ is bounded and then that will give us the result for $b^n=(a_1,a_2,\ldots,a_n,0,\ldots,0)$, $$T\left(\frac{b^n}{\sqrt{\sum_{j=1}^n a_j^2}}\right)=\frac{a_1^2+a_2^2+\ldots a_n^2}{\sqrt{\sum_{j=1}^n a_j^2}}=\sqrt{\sum_{j=1}^n a_j^2} \le \|T\|, \forall n \in \mathbb{N}$$ which implies that $$\sum_{j=1}^{\infty} a_j^2 =\lim_{n \to \infty} \sum_{j=1}^n a_j^2 \le \|T\|^2 \lt \infty$$

The only thing which remains to be shown now is that $T$ is bounded for which I tried to evoke the Closed Graph Theorem. Suppose that $b^n=(b^n(1),b^n(2),\ldots,b^n(j),\ldots) \in l^2 $ converge to $0$ and $T(b^n) \to y$. Let $\epsilon \gt 0$. Then there exists $n_0 \in \mathbb{N}$ such that for all $n \ge n_0$, we have $\|b^n\|_2 \lt \frac{\epsilon}{2}$ which in particular implies that $|b^n(j)| \lt \frac{\epsilon}{2}$ for all $j$ and all $n \ge n_0$. Since $T(b^n) \to y$, there exists $n_1 \in \mathbb{N}$ such that for all $n \ge n_1$, $|T(b^n)-y| \lt \frac{\epsilon}{2}$.

Then for all $n \ge \max{(n_0,n_1)}$$$|y| \le |y-T(b^n)|+|T(b^n)| \lt \epsilon+\frac{\epsilon}{2}(|a_1|+|a_2|+\ldots+|a_k|)$$ (Note: Since $T(b^n) \lt \infty$, the tail of the series goes to $0$ which means that $|\sum_{j \ge k} a_jb^n(j)| \lt \frac{\epsilon}{2}$ )

This is not what I intended to show. Can I conclude from here that $y=0$ since $\epsilon \gt 0$ is arbitrary? For me the problem is that apriori, I don't have a way to get away with the $|a_j|$'s, since they depend on the choice of $b^n$.

Note: There is an answer to this question here: If $\sum a_n b_n <\infty$ for all $(b_n)\in \ell^2$ then $(a_n) \in \ell^2$. But I wanted to know if I can go via this route and get to the answer and if not, why so.

Thanks for the help!!

  • $\begingroup$ Try the other way. Given that $a_n$ is not square summable, find a sequence $b_n$ which is square summable, but it isn't true that $a_nb_n$ is summable. $\endgroup$ – Teresa Lisbon Jul 19 '17 at 19:28
  • $\begingroup$ Your hint to doing this : Let $A_n$ be the partial sums of $a_n^2$, and let $b_n = \frac{a_n}{A_n}$. Now show the above properties. Hence, you are avoiding any kind of function space analysis this way. $\endgroup$ – Teresa Lisbon Jul 19 '17 at 19:31
  • $\begingroup$ @астонвіллаолофмэллбэрг How do I proceed with my attempt? $\endgroup$ – tattwamasi amrutam Jul 19 '17 at 19:33
  • $\begingroup$ I have literally given you what $b_n$ is, now all you have do to is go out and prove what is to be proved. It's very mechanical from here. $\endgroup$ – Teresa Lisbon Jul 19 '17 at 19:34
  • $\begingroup$ See math.stackexchange.com/questions/37647/…, or, for the "try the other way" version, math.stackexchange.com/questions/1957276/…. $\endgroup$ – Jim Ferry Jul 19 '17 at 19:36

You may avoid using the Uniform Boundedness Principle, but lack of continuity of $T$ means that you need to get another control of taking limits. In the present case, you may use monotone convergence, which ultimately relies upon the lattice structure of $\ell^2$. This is also implicit in solutions in previous posts done by hand (without Baire) although perhaps not explicitly stated.

We may assume that every $a_n\geq 0$ (this was in fact done in a previous post, and possibly with the intention of providing a hint, I leave it to you to reduce the problem to this case).

Let $K=\ell^2_+({\Bbb N}) = \{ \beta =(b_n) \in \ell^2: b_n\geq 0\}$ be the positive cone in $\ell^2$

For every $\beta\in K$ our assumption is that $0\leq T\beta= \sum_n a_n b_n < +\infty$. As you mention it suffices to show that $T$ is a bounded linear functional. So suppose it is not. Then you may find a sequence of vectors $\beta_1,\beta_2,...$ in $K$ with the properties that for all $j\geq 1$:

(0) : $\|\beta_j\| \leq 1$

(1) : $T\beta_{j} = 4^j$

Now define the ($\ell^2$-convergent) sum $x = \sum_{j\geq 1} 2^{-j} \beta_j \in K$. Every $\beta_j$ is a positive sequence, so when calculating $Tx$ we only add positive numbers. By monotonicity, we have for every $j\geq 1$:

$$ Tx \geq 2^{-j} T\beta_j = 2^j$$ So $Tx=+\infty$ contrary to hypothesis.

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EDIT: As pointed out in the comments, I was using a slightly different method. Rather, I was proving the map $(b_n)\mapsto(a_nb_n)$ is a bounded linear map from $\ell^2\to\ell^1$. The conclusion of course remains the same, but the method is not quite what the OP wanted.

It's possible to approach the question this way, although it is not the best method. We want to show if $b^n\to0$ in $\ell^2$ and $T(b^n)\to y$ in $\ell^1$ then $y=0$, so it is sufficient to show that $y_k=0$ for each $k$. Fix $\varepsilon>0$. Let $n_0$ be large enough that $\max\{\|b^n\|_2,\|T(b^n)-y\|_1\}<\varepsilon$ for all $n\ge n_0$. Observe that $$|a_kb^n(k)-y_k|\le\|T(b^n)-y\|_1<\varepsilon,$$ so if we can show that $(a_n)$ is a bounded sequence it will follow that $$|y_k|\le\varepsilon+|a_k|\|b^n\|_2<(1+M)\varepsilon$$ where $M=\sup_n|a_n|$, and so we will be done. Suppose for a contradiction that $(a_n)$ is not bounded. Then there is a subsequence $(n_k)$ such that $|a_{n_k}|\ge k$. Let $b=(b_n)$ be the sequence defined by

$$b_n=\begin{cases} \frac1k&\text{if }n=n_k,\\ 0&\text{otherwise.} \end{cases}$$

Clearly $b\in\ell^2$, but $|a_{n_k}b_{n_k}|\ge1$ for all $k$, so $a_nb_n\not\to0$ and in particular $\sum_na_nb_n$ cannot converge. This completes the proof.

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  • $\begingroup$ I thought that $y$ is a real number for the $T(b^n)$ end up in $\mathbb{R}$ $\endgroup$ – tattwamasi amrutam Jul 19 '17 at 20:18
  • $\begingroup$ Ah, I think I was looking at $T:\ell^2\to\ell^1$ given by $(b_n)\mapsto(a_nb_n)$. I guess this is not quite the same method as you had in mind. $\endgroup$ – Jason Jul 19 '17 at 20:21
  • $\begingroup$ Yes. My end space is $\mathbb{R}$ $\endgroup$ – tattwamasi amrutam Jul 19 '17 at 20:28

For each $n$, define $a^{(n)}=\{a_{1},a_{2},\ldots,a_{n},0,0,\ldots\}\in l^{2}$. Define $\theta^{n}\in(l^{2})^{\ast}$ by $\theta^{n}(x)=\langle a^{(n)},x\rangle=\sum_{k=1}^{n}a_{k}x_{k}$.

For each $x\in l^{2}$, the sequence $\{\theta^{n}(x)\}_{n}$ is bounded because the infinite series $\sum_{k=1}^{\infty}a_{k}x_{k}$ is convergent. By Uniform Boundedness Principle, $\sup_{n}||\theta^{n}||=M<\infty$. By Riese Representation Theorem, $\sum_{k=1}^{n}a_{k}^{2}=||a^{(n)}||^{2}=||\theta^{n}||^2\leq M^2$. It follows that $\sum_{k=1}^{\infty}a_{k}^{2}<\infty$.

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I believe your assertion is false: Consider the case $a_n$ = 1 and $b_n$ = $1/n^2$.

CORRECTION: Sorry, this isn't a valid counterexample - I mis-read the implied 'for all' for sequences $b_n$. Please excuse my haste.

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  • $\begingroup$ I believe that the assertion is true. They have been proved already. Look into the link in my question $\endgroup$ – tattwamasi amrutam Jul 19 '17 at 20:13
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    $\begingroup$ You can delete this answer PMar $\endgroup$ – tattwamasi amrutam Jul 19 '17 at 20:23

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