Let $\ell^2=\ell^2(\mathbb{N})$ and let $e_1,e_2,\ldots$ be its usual basis. I have an infinite matrix $\{\alpha_{ij}\}_{i,j=1}^{\infty}$ such that $\alpha_{ij} \ge 0$ for all $i,j$ and such that there are scalars $p_i>0$ and $\beta,\gamma>0$ with
$\sum_{i=1}^{\infty}\alpha_{ij}p_{i} \le \beta p_j \quad$ , $\quad \sum_{j=1}^{\infty}\alpha_{ij}p_{j} \le \gamma p_i \quad$ for all $i,j\ge 1$
Now I want to show that there is an operator $A$ on $\ell^{2}(\mathbb{N})$ with $\langle Ae_{j},e_{i}\rangle =\alpha_{ij}$. Do you know how I must represented this operator? I must use multiplication operator?
Define $Ae_i=\sum_j a_{ij}e_j$. Boundedness of $A$ follows from Schur's test.