Here is the link to the solution:
Why should we use the uniform boundedness principle here?
And here is the solution:
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][1].
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 $\|A\| \leq \sup_N \|T_N\|$.
But I do not understand this step:
$\sum_{i=1}^N \left|\sum_j a_{ij}x_j\right|^2 \leq \sum_{i=1}^N\left(\sum_j |a_{ij}|^2 \right)\left(\sum_j |x_j|^2 \right)$
I know that this is by Cauchy-Schwartz, but should not $a_{ij}x_j$ be inside an absolute value to apply Cauchy Schwartz?
