Let $\eta$ be a continuous bounded function on $(0, \infty)^{2}$ so that $\eta(0,0)=1$. Let $A$ be a bounded operator on $\ell^{2}(\mathbb{Z}\_{\geq 0})=\ell^{2}$ (by bounded operator I will always mean such an object) and let $A\_{j,k}$ be its "matrix entries" with respect to the standard basis of $\ell^{2}$.


Suppose now that for any integer $N\geq 1$ and any bounded operator $A$, the infinite matrices $A^{(N)}$ with entries $$A^{(N)}\_{j,k}=\eta(\frac{j}{N}, \frac{k}{N})A\_{j,k}$$ form a uniformly bounded sequence, meaning that there exists $C>0$, independent of $N$, so that $$\|A^{(N)}\|\_{op}\leq C \|A\|\_{op},$$ where $\|\cdot\|\_{op}$ denotes the usual operator norm.


It is very easy to show that under these assumptions, one can have uniform convergence of $A^{(N)}\to A$ as $N\to \infty$ (i.e. $\|A^{(N)}-A\|\_{op}\to 0$) whenever $A$ is a finite matrix (as we have entrywise convergence). By density of the finite matrices and an $\varepsilon/3$-argument it is not so difficult to show that the same happens whenever we restrict to the ideal of compact operators.


What can we say about the convergence of $A^{(N)}\to A$ for an infinite matrix? Of course, the best one can hope for, dropping the compactness of $A$, is convergence in the Strong Operator Topology, but I cannot find an argument that proves my statement. Any help would be appreciated!!! 



Convergence of sequence of images of Schur multipliers
By the uniform bound on $\|A^{(N)}\|$ and linearity, SOT convergence follows from the $\ell^2$-norm convergence, for every $i$, of the $i$-th column $A^{(N)} e\_i$ to the $i$-th column $A e\_i$. This convergence is straightforward (say by the dominated convergence theorem).


