Proof using the closed graph theorem

[A_B]

Hi,

I'm stuck on a problem in functional analysis. Let x be a sequence on the Natural nummers such that for any square summable sequence y, the product sequence xy is absolutely summable. Then x is square summable.

Hint : Use the Closed graph theorem.

If I can prove the map Tx : y -> xy had a closed graph then It Follows from the Closed graph theorem that Tx is bounded and therefore that x is square summable, but I can't seem to show that the graph is Closed. Am I following the right path? Any hints?Thanks

 

[Hawkeye18]

Yes, you are on the right path. You need to show that your operator ##T_x:\ell^2\to\ell^1##, ##T_x y = xy##, where the product means entrywise product is closed. This fact is very simple, almost a triviality: it follows from the fact that convergence in any ##\ell^p## implies the entry wise convergence.

Let me explain; to avoid mess with double indices, I'll use the "functional notation" for sequences, i.e. a sequence ##x## will be ##x=(x(1), x(2), \ldots)##. Since ##|y(k)| \le \|y\|_{\ell^p}## (for all ##k## and ##p##) we conclude that if ##\|y_n-y\|_{\ell^2} \to 0## then $$\lim_{n\to\infty} y_n(k) = y(k)$$ for all ##k##. Similarly, if additionally ##\|x y_n -z\|_{\ell^1}\to 0## then for all ##k## $$\lim_{n\to\infty} x (k)y_n(k) =z(k).$$ But on the other hand we already know that ##\lim_{n\to\infty} y_n(k) = y(k)##, so $$\lim_{n\to\infty} x (k)y_n(k) =x(k)y(k),$$ which means that ##z=xy##.

So the operator ##T_x## is closed.

 

[A_B]

As is so often the case with mathematics,the solution to the problem seems so obvious once it is known. This concludes for me a few days of staring at a blank page.

Thank You, Hawkeye. You saved me a lot of time.