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I would like to revisit this question, which can be equivalently stated as:

Proposition. Let $(a_n)$ be a sequence of real (or complex) numbers such that $\sum a_n b_n$ converges for every $(b_n) \in \ell^2$. Then $(a_n) \in \ell^2$.

The proof given here by Bruno Stonek is the only one I know, which applies the uniform boundedness principle to the linear functionals $(b_n) \mapsto \sum_{n=1}^m a_n b_n$ in $(\ell^2)^*$. But I am inclined to agree with GEdgar's comment on Davide Giraudo's answer that this approach, though slick, is "far too advanced".

One could write the contrapositive as:

Let $(a_n)$ be a sequence of real numbers which is not $\ell^2$. Then there exists $(b_n) \in \ell^2$ such that $\sum a_n b_n$ diverges.

A natural way to try to prove that statement would be to explicitly construct such a $(b_n)$, by somehow manipulating $(a_n)$. Our current proof is far from constructive in that sense, since the uniform boundedness principle essentially just says that the set of such $(b_n)$ is comeager in the Hilbert space $\ell^2$, and then uses Baire to assert it is nonempty.

So my question:

Can anyone think of a way to explicitly construct $(b_n)$? Or alternatively, is there some reason to think that such a construction might be impossible?

For instance, as a wild guess, perhaps one could show that any map sending each $(a_n) \notin \ell^2$ to an appropriate $(b_n)$ would have to be nonmeasurable in some sense.

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Actually, I just noticed Theo Buehler's reference to Sokal's simple proof of the uniform boundedness principle (arxiv.org/abs/1005.1585). I think following Sokal might yield a construction; I'll look at it later. –  Nate Eldredge Aug 19 '11 at 21:33
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1 Answer 1

up vote 15 down vote accepted

One can construct explicitly such a sequence $(b_n)$ and this is probably explained in several books of examples and counterexamples in Analysis. Here we go.

Assume without loss of generality that $a_1\ne0$ and, for every $n\ge1$, let $$A_n=\sum\limits_{k=1}^na_k^2,\qquad b_n=\dfrac{a_n}{A_n}. $$ The result the OP is interested in is a consequence of the two claims below.

Claim 1: For every $(a_n)$, the sequence $(b_n)$ is in $\ell^2$.

Claim 2: For every $(a_n)$ not in $\ell^2$, the sequence $(a_nb_n)$ is not in $\ell^1$.

To prove claim 1, note that, for every $n\ge2$, $$ b_n^2=\dfrac{A_n-A_{n-1}}{A_n^2}\le\dfrac{A_n-A_{n-1}}{A_nA_{n-1}}=\dfrac1{A_{n-1}}-\dfrac1{A_n}, $$ hence $(b_n^2)$ is summable (and its sum is at most $2/a_1^2$). (This step does not use the hypothesis that $(a_n^2)$ is not summable hence that $A_n\to+\infty$).

To prove claim 2, note that, for every $N$ and every $2\le k\le N$, $$ a_kb_k=\dfrac{a_k^2}{A_k}\ge\dfrac{a_k^2}{A_N}=\dfrac{A_k-A_{k-1}}{A_N}. $$ Since $(a_n^2)$ is not summable, $A_N\to+\infty$ when $N\to\infty$. Hence, for every given $n\ge1$, there exists $N\ge n$ such that $A_N\ge2A_n$. For every such $N$, $$ \sum\limits_{k=n+1}^Na_kb_k\ge\dfrac{A_N-A_{n}}{A_N}\ge\frac12. $$ This proves that $\sum\limits_{k=n+1}^Na_kb_k$ does not converge to zero when $n$ and $N\to\infty$. Hence, $(a_kb_k)$ is not summable.

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Great! That's just what I was looking for. –  Nate Eldredge Aug 20 '11 at 13:25
    
@Theo: Thanks for the correction. (Damn'd, this one will be difficult to get rid of, mentally...) –  Did Aug 21 '11 at 11:24
    
No problem. I can imagine the difficulty. I can understand how this happens to French people, but Germans succumb to this slip, too, which I find harder to understand. By the way, you're swearing a lot, lately... :) –  t.b. Aug 21 '11 at 11:42
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