I'm trying to prove the following:

If $(a_n)$ is a sequence of positive numbers such that $\sum_{n=1}^\infty a_n b_n<\infty$ for all sequences of positive numbers $(b_n)$ such that $\sum_{n=1}^\infty b_n^2<\infty$, then $\sum_{n=1}^\infty a_n^2 <\infty$.

The context here is functional analysis homework, in the subject of Hilbert spaces.

Here's what I've thought:

Let $f=(a_n)>0$. Then the problem reads: if $\int f\overline{g}<\infty$ for all $g>0,g\in \ell^2$, then $f\in \ell^2$. This brings the problem into the realm of $\ell^p$ spaces.

I know the inner product is defined only in $\ell^2$, but it's sort of like saying: if $\langle f,g\rangle <\infty$ for all $g>0,g\in \ell^2$ then $f\in \ell^2$.

I read this as: "to check a positive sequence is in $\ell^2$, just check its inner product with any positive sequence in $\ell^2$ is finite, then you're done", which I find nice, but I can't prove it :P

From there, I don't know what else to do. I thought of Hölder's inequality which in this context states: $$\sum_{n=1}^\infty a_nb_n \leq \left( \sum_{n=1}^\infty a_n^2 \right)^{1/2} \left( \sum_{n=1}^\infty b_n^2 \right)^{1/2}$$

but it's not useful here.


3 Answers 3


Put $l_n(b) =\sum_{k=1}^n a_kb_k$. $l_n$ is linear, continuous and $\lVert l_n\rVert = \left(\sum_{k=1}^na_k^2\right)^{\frac 12}$. For all $b\in \ell^2$ the sequence $\{l_n(b)\}$ is bounded. By the principle of uniform boundedness we get that the sequence $\left\{\lVert l_n\rVert\right\}$ is bounded.

  • $\begingroup$ ... and it converges to $\lVert a\rVert$. $\endgroup$
    – Rasmus
    May 7, 2011 at 17:54
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    $\begingroup$ But I think the Principle of Uniform Boundedness is far too advanced for this homework problem. $\endgroup$
    – GEdgar
    May 7, 2011 at 21:38
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    $\begingroup$ @GEdgar: We have seen the principle of uniform boundedness in class. What strikes me as odd is that this solution uses no machinery proper to Hilbert spaces, and the problem belongs to a Hilbert space handout. $\endgroup$ May 7, 2011 at 21:44

Following the idea left by david as an answer, I'll post a detailed solution.

Let $T_n:\ell^2 \to \mathbb{C}, T_n(c)=\sum_{j=1}^n a_j c_j$. Then clearly $T_n \in (\ell^2)^*$ for every $n$.

I claim that for every $c\in \ell^2$, the limit $\lim_n T_n(c)=\sum_{j=1}^\infty a_jc_j$ exists.

Indeed, we know it does for $c\in \ell^2$ such that $c(n)\geq 0$ for all $n$. But then, for an arbitrary $c\in \ell^2$, we can decompose $c$ as:

$c=(\mbox{Re } c)^+ - (\mbox{Re } c)^- + i\left( (\mbox{Im }c)^+ - (\mbox{Im } c)^-\right)$

which proves the claim.

As a consequence we have that $\sup_n \lVert T_n(c) \rVert <\infty$ for all $c\in \ell^2$, but then by the uniform boundedness principle, $\sup_n \lVert T_n \rVert <\infty$.

Now, $T_n(c) = \langle c, a^{(n)}\rangle_{\ell^2}$ where $a^{(n)}(j)=\begin{cases} a_j & \mbox{if }j\leq n \\ 0 & \mbox{if } j>n \end{cases}$.

By Riesz' representation theorem on Hilbert spaces, we know that $\lVert T_n \rVert = \lVert a^{(n)}\rVert_2= \left( \sum_{j=1}^n a_j^2 \right)^{\frac{1}{2}}$.

To conclude, since $a_n\geq 0$ for all $n$, we have that $\left( \sum_{n=1}^\infty a_n^2 \right)^{\frac{1}{2}} = \sup_n \lVert a^{(n)} \rVert_2 = \sup_n \lVert T_n \rVert < \infty$, and then $\sum_{n=1}^\infty a_n^2 <\infty$.

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    $\begingroup$ +1: This looks correct to me. Thanks for elaborating on david's somewhat cryptic answer! $\endgroup$
    – t.b.
    May 14, 2011 at 18:49

It can be proven using the Hölder inequality — see lemma 1.8 (page 8) of these notes. (There may be mistakes; I typeset these from lecture notes.) There's no need to invoke heavy machinery like the uniform boundedness principle or even the Riesz representation theorem; it's a special case of a general result for $\ell^p$ spaces.

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    $\begingroup$ Perhaps I am missing something, but I don't see that it follows from the lemma you cite. We do not know that $(b_n) \mapsto \sum a_n b_n$ is a bounded linear functional on $\ell^2$; we only know that it is finite for each $b_n$. If you're sure about this, perhaps you could elaborate? $\endgroup$ May 14, 2011 at 20:31
  • $\begingroup$ @Nate: I think you're right. Actually, I'm beginning to suspect the proof of the lemma has a gap as well, but it may just be that it's late and my brain is not working. $\endgroup$
    – Zhen Lin
    May 15, 2011 at 0:21
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    $\begingroup$ Zhen: no, the proof of lemma 1.8 in your notes looks quite fine at a quick glance. By the way: I don't see the Riesz representation theorem as a difficult result (as you seem to imply in your answer), it's just a bit of linear algebra. Moreover, there is a very recent article by Sokal that gives a cute proof of the uniform boundedness principle without invoking Baire. I hope you like that one. $\endgroup$
    – t.b.
    May 15, 2011 at 2:07
  • $\begingroup$ Yeah, in the lemma, I don't see why $x_i = y_i^{q-1} z_i$ is in $\ell^p$. $\endgroup$ May 15, 2011 at 15:36
  • $\begingroup$ Oh, I figured it out. $|z_i|=1$ so $|x_i|^p = |y_i|^{(q-1)p} = |y_i|^q$ since $p,q$ are conjugate exponents, and $y \in \ell^q$ by assumption. Lemma 1.8 looks good, but I still don't think that it implies the desired statement. $\endgroup$ Aug 19, 2011 at 20:37

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