Convergence of Linear Operator in $l^2$ 
Let L be a linear operator $l^2\to l^2$, defined by the matrix $(a_{ij})_{i,j=1}^\infty$.
Assuming that: $$\sum_{j=-\infty}^\infty \sup_{i\in N}|a_{i, i+j}|$$
Converges to $K$, show that $L$ is bounded by the above sum. 

I tried the following approach:
Using the above sum, one can show that $\sum_{j=1}^\infty a_{ij}$ Converges for every $i$ (since the above sun is strictly bigger and it converges) whatelse the sequence itself converges as it is in $l^2$ , and thus:
$$||Lx||^2=\sum_{i=1}^\infty |\sum_{j=1}^\infty a_{ij}x_j|^2\leq\sum_{i=1}^\infty (\sum_{j=1}^\infty |a_{ij}x_j|)^2\leq ||x||^2\cdot\sum_{i=1}^\infty (\sum_{j=1}^\infty |a_{ij}|^2) $$
Which would mean one has to show that:
$$\sum_{i=1}^\infty (\sum_{j=1}^\infty |a_{ij}|^2)\leq(\sum_{j=-\infty}^\infty \sup_{i\in N}|a_{i, i+j})^2$$
Which makes me quite certain there is a better approach, or a simple way to do it that I am missing.
Will appreciate any tips\guidance.
Edit: clarifying after the comments, we assume that $a_{ij}=0$ if $j\leq 0$.
 A: For each $n\in\mathbb Z$, let $x_n$ denote the vector of values on the diagonal shifted over by $n$, that is,
$$
x_n=\bigl(a_{i,i+n}\colon i\in\mathbb N\text{ and }i+n\in\mathbb N\bigr).
$$
The inequality you wish to prove is equivalent to
$$
\sum_{n\in\mathbb Z}\|x_n\|_2^2\leq \left(\sum_{n\in\mathbb Z}\|x_n\|_{\infty}\right)^2.\qquad (\star)
$$
To see why this is equivalent, note that the left hand side is the square of the Frobenius norm, which is equal to the operator norm of $L$ as an operator from $\ell_2(\mathbb N)$ to $\ell_2(\mathbb N)$.
Proof of $(\star)$. Since $\|x_n\|_2\leq \|x_n\|_{\infty}$, it follows that
$$
\sum_{n\in\mathbb Z}\|x_n\|_2^2\leq \sum_{n\in\mathbb Z}\|x_n\|_{\infty}^2.
$$
On the other hand, we have that
$$
\left(\sum_{n\in\mathbb Z}\|x_n\|_{\infty}\right)^2=\sum_{n\in\mathbb Z}\|x_n\|_{\infty}^2+\sum_{n\not=m}\|x_n\|_{\infty}\ \|x_m\|_{\infty}\geq \sum_{n\in\mathbb Z}\|x_n\|_{\infty}^2,
$$
since each of the terms is non-negative. Chaining these two inequalities yields $(\star)$.
