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, since we showed that $(\ell^2,\| x\|_{2})$ is a normed space. and it is easy to show that its complete by showing that every Cauchy sequence converges using $\| .\|_{2}$, then $(\ell^2 , \| x\|_{2})$ is a Banach space and since by the given we have that for any $x \in \ell^2,$ the sequence $Ax$ belongs to $\ell^2.$ Which can be formulated mathematically as follows:
According to the givens we can define $A : \ell^2 \rightarrow \ell^2 $ by $$A x = A (\xi_{1}, \xi_{2}, ...) = (\beta_{1}, \beta_{2}, ...), $$ Where
$$ [a_{ij}]\begin{bmatrix} \xi_{1} \\ \xi_{2} \\ \vdots \end{bmatrix}=\begin{bmatrix} \beta_{1} \\ \beta_{2} \\ \vdots \end{bmatrix} $$
i.e., $\beta_{i} = \sum_{j=1}^{\infty} a_{ij} \xi_{j}$ and $ \|Ax\|^{2} = \sum_{i=1}^{\infty} |\beta_{i}|^2 < \infty $. Hence $\sup \{\|A x \|\} < \infty $ and hence $\sup \{\|T \|\} = \sup \{\|A \|\} < \infty $ as required.
Is my solution correct?
Hint: Define $T_Nx= (\sum\limits_{k=1}^{N} a_{ij}x_j)$. Verify that $T_N$ is a bounded opeartor and $(\|T_N(x)\|)$ is a bounded sequence for each fixed $x$. Apply UBP to finish the proof.