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Let $X$ be a Banach space and $\{T_{\alpha}\}_{\alpha\in I}\subset B(X)$ be a (possible uncountable) family of bounded linear operators on $X$ such that $\sum T_{\alpha}$ converges to the identity operator in the strong operator topology. This means that $\sum T_{\alpha} x=x \ \forall x\in X$. We know that for any summable series, all but countably many of the terms must be $0$. Hence for each $x\in X$, $T_{\alpha}x=0$ for all but countably many terms.

Can we then conclude that all but countably many of the $T_{\alpha}$ themselves are $0$? This should happen, for example if $X$ is a separable Hilbert space. What about a separable Banach space? Is there an example of a space where this need not happen?

Thanks in advance.

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Let $X=\ell_2(\omega_1)$, where $\omega_1$ is the first uncountable ordinal. For each $\gamma<\omega_1$ let $T_\gamma$ be the orthogonal projection onto the line generated by $e_\gamma$. Then $\sum_{\gamma}T_\gamma$ converges to the identity in the strong operator topology.

What is your argument that it is impossible on a separable Hilbert space?

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  • $\begingroup$ For the argument for Hilbert spaces you can read my answer which is for seperable Banach spaces. $\endgroup$ – Tim B. Nov 30 '16 at 10:01
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It is true for separable Banach spaces as well. Let $(x_i)_{i\in\mathbb N}$ be a dense subset of $X$. Let $J_i\subset I$ be countable such that $T_\alpha x_i = 0$ if $\alpha \notin J_i$. Then $J := \bigcup_{i\in \mathbb N}J_i$ is countable. If $\alpha\in I\setminus J$ then $T_\alpha x_i = 0$ for all $i\in\mathbb N$, hence, by countinuity, for all $x\in\overline{\{x_i~\vert~ i\in\mathbb N\}} = X$. We conclude $T_\alpha = 0$ for each $\alpha\in I\setminus J$.

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