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Here is the question:

{Let $A = [a_{ij}]_{i,j = 1}^{\infty}$ be an infinite matrix of real numbers and suppose that, for any $x \in \ell^2,$ the sequence $Ax$ belongs to $\ell^2.$ Prove that the operator $T,$ defined by $T(x) = Ax,$ is a bounded operator on $\ell^2.$

I got a hint to use the closed graph theorem to show that it is continuous and hence bounded, but now I do not know how to show that this operator is closed or to show that its graph is closed. Could anyone help me in doing so please?

Definition:

A linear operator $T : X \rightarrow Y$ between normed linear spaces $X,Y$ is closed if for every $\{x_{n}\} \subset X$ we have $$x_{n} \rightarrow x_{0}, Tx_{n} \rightarrow y_{0} \implies Tx_{0} = y_{0}.$$

Definition: Let $T : X \rightarrow Y$ be a linear map between normed linear spaces $X,Y.$ The graph of $T$ is the set $\{(x, Tx) : x \in X \}.$

Still, I do not know how to prove that the given operator in question is closed.

Here is my trial:

Assume that $x = (\xi_{j}) \in \ell^2, $ then by definition of real sequences in $\ell^2$ we have $\sum_{j=1}^{\infty}|\xi_{j}|^2 < \infty$ which means that the series of the square of the elements converges (I also do not want the square of the elements to converge, I want the elements themselves. so I do not know how to overcome this problem also. Any help will be appreciated ), say to $x_{0}.$ Also, by assumption we have that the sequence $Ax$ belongs to $\ell^2,$ then $T\xi_{j}$ is convergent also say to $y_{0}.$ Now how can I show that $Tx_{0} = y_{0}$?

I tried another method here(but it did not work):

Does $\sum_{i = 1}^{\infty} |\beta_{i}|^2 < \infty$ implies that $A $ satisfies $ \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} |a_{ij}|^2 < \infty $?

  • For each $i$ show the continuity of $x\mapsto A_i(x)=\sum\limits_{j=1}^\infty a_{i,j}x_j=\lim\limits_{n\to\infty}\sum\limits_{j=1}^n a_{i,j}x_j$ by using the Banach-Steinhaus theorem. – Jochen Apr 01 '20 at 15:56
  • @Jochen can I use the uniform boundedness theorem ? –  Apr 01 '20 at 15:59
  • Yes, the uniform boundedness principle of Banach and Steinhaus. – Jochen Apr 01 '20 at 16:04
  • Could you provide some details of using it as I see that the result follows directly from it without any proof .... am I correct?@Jochen –  Apr 01 '20 at 16:11

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