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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!!!

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  • $\begingroup$ I am confused. What is your $\ell^2 $, sequences or matrices? $\endgroup$
    – Fedor Petrov
    Mar 27, 2018 at 17:56
  • $\begingroup$ @FedorPetrov the matrix $A$ acts on $\ell^{2}$. $\endgroup$
    – Raphael
    Mar 28, 2018 at 13:55

1 Answer 1

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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).

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  • $\begingroup$ Thank you for your answer! However, would you mind being more specific? I have tried this approach and I get stuck when trying to prove that $$\|A^{(N)}e_{j}-Ae_{j}\|_{2}\to 0$$ as $N \to \infty.$ It is clear to me that all the entries of the columns of $A^{(N)}$ converge to those of $A$, however convergence in $\ell^{2}$-norm doen't seem immediate. Could you give more details? I am very sorry for asking this dumb question. $\endgroup$
    – Raphael
    Mar 28, 2018 at 14:00
  • $\begingroup$ @Raphael As I wrote, this follows from the dominated convergence theorem. You want to prove that $\sum_i |A^{(N)}_{i,j}-A_{i,j}|^2$ tends to zero. You know that each term in the sum goes to zero. But also that the $i$-th term is less than $(1+\|\eta\|_\infty)^2 |A_{i,j}|^2$, which is summable as a function of $i$. $\endgroup$
    – Mikael de la Salle
    Mar 28, 2018 at 16:57

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