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 $\ell^2.$
**Here is my trial: **
We will use the uniform bounded theorem Principle. Using that $T_{N}x = \sum_{j=1}^{N} a_{ij}x_{j}$
I have proved that $T_N$ is bounded with $\|T_N \| \leq (\sum_{j=1}^{N} |a_{ij}|^2)^{1/2}.$.... am I correct?
Now I am stuck in proving that $(\|T_{N}x\|)_{N \in \mathbb{N}}$ is a bounded sequence for each fixed $x.$ I need a condition on $x$ like it has finitely many nonzero terms or any other condition or here is what the given that for each x the sequence $Ax \in l^2$ comes into play?Could anyone help me in this, please?
Hint: You are told that $\ Ax\in l^2\ $. Instead of $\ T_N\ $, consider the operators $\ S_{MN}\ $ defined by $$ (S_{MN}x)_i=\cases{\sum_\limits{j=1}^N a_{ij}x_j& if $\ i\le M\ $\\ 0 & if $\ i>M\ $,} $$ and note that $\ \left\|S_{MN}x\right\|_{l^2}\le \|Ax|_M\| _{l^2}\ $ for all $\ x\ $, where $\ x|_M\ $ is the sequence obtained from $\ x\ $ by replacing all its entries beyond the $\ M$-th to zero.