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Let $\mathcal{B}(\mathcal{H})$ be the Banach space of bounded linear operators on a complex, separable, infinite-dimensional Hilbert space $\mathcal{H}$. It is well known that $\mathcal{B}(\mathcal{H})$ is a Banach space, hence, it makes sense to consider a sequence $\{\mathbf{A}_{n}\}$ of bounded linear operators in $\mathcal{H}$ and ask for the convergence of the series:

$$ \sum_{n=1}^{+\infty}\mathbf{A}_{n} $$ with respect to the operator norm on $\mathcal{B}(\mathcal{H})$. Actually, the completeness of $\mathcal{B}(\mathcal{H})$ allows us to say that if the series of complex numbers:

$$ \sum_{n=1}^{+\infty}||\mathbf{A}_{n}|| $$ converges, then $\sum_{n=1}^{+\infty}\mathbf{A}_{n}$ is absoultely summable and thus summable. I would like to know if similar results hold when we consider the convergence of the series with respect to the strong operator topology or with respect to the weak operator topology on $\mathcal{B}(\mathcal{H})$. I know that if the sequence $\{\mathbf{S}_{N}\}$ defined as:

$$ \mathbf{S}_{N}=\sum_{n=1}^{N}\mathbf{A}_{n} $$ converges in the strong (weak) operator topology, then $\sum_{n=1}^{+\infty}\mathbf{A}_{n}$ is summable in the strong (weak) operator topology. Nevertheless, this criterion uses the sequence of partial sums, while I would like to know if there exist a criterion of summability which does not involve the sequence of partial sums, just as is the case for the summability with respect to the operator norm topology.

Thanks You.

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    $\begingroup$ For the strong operator topology, we have the analogous sufficient criterion, if $$\sum_{n = 1}^\infty \lVert \mathbf{A}_n x\rVert < +\infty$$ for all $x$, then the series converges in the strong operator topology. I doubt whether it is very useful, however. $\endgroup$ – Daniel Fischer Jul 3 '15 at 17:54
  • $\begingroup$ @DanielFischer Thank You. Where can I find a proof for this statement? Furthermore, why do you doubt it to be useful? $\endgroup$ – SepulzioNori Jul 3 '15 at 18:00
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    $\begingroup$ The proof is easy, $y_n=\mathbf{A}_n x$ is an absolutely summable family in a complete normed space, hence it is summable, i.e. $\sum \mathbf{A}_n x$ is norm-convergent (and the limit is stable under arbitrary reorderings of the series). That holds for all $x$, and that means the series is convergent in the SOT (the limit operator is trivially linear, and continuous by Banach-Steinhaus). I doubt the usefulness, since I can't readily imagine a situation where one knows that $\sum \lVert\mathbf{A}_n x\rVert<+\infty$ for all $x$ without having stronger information like absolute summability. $\endgroup$ – Daniel Fischer Jul 3 '15 at 18:06

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