Is there a constructive proof of this characterization of $\ell^2$?

Solution 1:

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 $n\geqslant1$, 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 $n\geqslant2$, $a_n^2=A_n-A_{n-1}$ hence $$ b_n^2=\frac{A_n-A_{n-1}}{A_n^2}\leqslant\frac{A_n-A_{n-1}}{A_nA_{n-1}}=\frac1{A_{n-1}}-\frac1{A_n}, $$ hence the series $\sum\limits_nb_n^2$ converges (and its sum is at most $2/a_1^2$). (This step does not use the hypothesis that the sequence $(a_n^2)$ is not summable hence that $A_n\to+\infty$).

To prove claim 2, note that, for every $N$ and every $2\leqslant k\leqslant N$, $$ a_kb_k=\frac{a_k^2}{A_k}\geqslant\frac{a_k^2}{A_N}=\frac{A_k-A_{k-1}}{A_N}. $$ Since the sequence $(a_n^2)$ is not summable, $A_N\to+\infty$ when $N\to\infty$. Hence, for every given $n\geqslant1$, there exists some $N\geqslant n$ such that $A_N\geqslant2A_n$. For every such $N$, $$ \sum\limits_{k=n+1}^Na_kb_k\geqslant\frac{A_N-A_{n}}{A_N}\geqslant\frac12. $$ This proves that $\sum\limits_{k=n+1}^Na_kb_k$ does not converge to zero when $n$ and $N\to\infty$. Hence, the series $\sum\limits_ka_kb_k$ diverges.