Doubly infinite matrices $A=(a_{i,j})_{i,j=\infty}^{\infty}$ Let $A=(a_{i,j})_{i,j=\infty}^{\infty}$, where $$ \|A\|:=\sum_{r=-\infty}^{\infty}\sup_{j}|a_{j,j+r}|<\infty. $$ I want to show that for all matrices $\|AB\|\leq\|A\|\|B\|$. I obverse that $$ (AB)_{i,j}=\sum_{k=-\infty}^{\infty}a_{i,k}b_{k,j}=c_{i,j} $$ Hence, \begin{align} \|AB\|&=\sum_{r=-\infty}^{\infty}\sup_{j}|c_{j,j+r}|\\\ &=\sum_{r=-\infty}^{\infty}\sup_{j}|\sum_{k=-\infty}^{\infty}a_{j,k}b_{k,j+r}|\\\ &\leq \sum_{r=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\sup_{j}|a_{j,k}b_{k,j+r}|\\\ &\leq \sum_{r=-\infty}^{\infty}\sum_{k=-\infty}^{\infty}\sup_{j}|a_{j,k}|\cdot|b_{k,j+r}|\\\ &\leq ? \end{align} How can I finish this proof? Any hints?

$ \| A B \| = \sum_r \sup_j |\sum_k a_{j,k} b_{k,j+r}| $

$ \leq \sum_r \sum_k |a_{j(r),k}| |b_{k,j(r)+r}| $ \--- $j(r)$ is where the $\sup$ is attained (possibly up to $\epsilon$...)

$ = \sum_r \sum_k |a_{j(r),k+j(r)}| |b_{k+j(r),j(r)+r}| $

$ = \sum_k \sum_r |a_{j(r),k+j(r)}| |b_{k+j(r),j(r)+r}| $

$ \leq \sum_k \sup_j |a_{j,k+j}| \sum_r |b_{k+j(r), j(r) + r}| $

$ \stackrel{r = r' + k}{=} \sum_k \sup_j |a_{j,k+j}| \sum_{r'} |b_{k+j(r'+k), k+j(r'+k) + r'}| $

$ \leq \sum_k \sup_j |a_{j,k+j}| \sum_{r'} \sup_\ell |b_{\ell, \ell + r'}| $