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 $\ell^2.$
My question is:
I got a hint to use the uniform boundedness principle here but I do not know why, could anyone explain this to me, please? what makes me when I look at a problem to decide that it should be solved by UBP?
EDIT:
1-I have taken this proposition: "The series $\sum_{n =1}^{\infty} a_{n} b_{n}$ converges absolutely for every convergent sequence $\{b_{n}\}$ iff $\sum_{n =1}^{\infty} |a_{n}|$ converges." will it be helpful here in our case? the problem is that here in our case we are in $l^2.$
2-Also, Should it be better to use the uniform boundedness principle or the following theorem to solve the problem given above?
Theorem:
Let $X,Y$ be Banach spaces and let $\{T_{n}\}_{n=1}^{\infty}$ and $T$ be operators in $\mathcal{L}(X,Y).$ then $\lim_{n} T_{n}x = Tx,$ for all $x \in X,$ iff
(a)the sequence $\{T_{n}\}$ is bounded;
(b)lim_{n} T_{n}x exists on a dense subset of $X.$
Best Answer
First let's show a simpler version (1-dimensional): If $\sum_i a_i x_i < \infty$ all for $x\in\ell^2$, then $a\in \ell^2$.
You can prove this claim using uniform boundedness principle, or you can just use Riesz Representation Theorem. See this post.
Now, let's go back to your problem. It follows from the claim above that each row of $A$ is in $\ell_2$. Define $T_N$ to be the restriction of $A$ onto the first $N$ rows, that is, $$ T_N x = \left(\sum_j a_{1j}x_j,\sum_j a_{2j}x_j,\dots,\sum_j a_{Nj}x_j,0,0,\dots,\right). $$ We claim that $\|T_N\| < \infty$. Note that $$ \|T_Nx\|_2^2 = \sum_{i=1}^N \left|\sum_j a_{ij}x_j\right|^2 \leq \sum_{i=1}^N\left(\sum_j |a_{ij}|^2 \right)\left(\sum_j |x_j|^2 \right) \leq \|x\|_2^2\cdot \sum_{i=1}^N\sum_{j=1}^\infty |a_{ij}|^2, $$ thus $$ \|T_N\| \leq \left(\sum_{i=1}^N\sum_{j=1}^\infty |a_{ij}|^2\right)^{1/2}. $$ (Note that the infinite sum over $j$ is finite because of the claim at the beginning.)
Now, for each fixed $x$, observe that $\|T_Nx\|_2$ is uniformly bounded by $\|Ax\|_2$ (since $\|T_Nx\|_2$ is just part of the sum for $\|Ax\|_2<\infty$). It follows from the uniform boundedness principle that $\sup_N \|T_N\|<\infty$. Note that $\|Ax\|_2 = \lim_{N\to\infty} \|T_Nx\|_2 \leq (\sup_N \|T_N\|)\|x\|$, which implies that $A$ is bounded and $\|A\| \leq \sup_N \|T_N\|$.