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Let $T:{\rm Dom}(T)\subset X\to Y$ be a closed operator between Hilbert spaces, i.e. ${\rm Dom}(T)$ is dense and $G_T$ is a closed subset.

Q For $\{x_n\}\to x \in {\rm Dom}(T)$, can we say $T(x_n)\to T(x)$?

Idea to show is that:

  1. For any closed subset $S\subset {\rm Dom}(T)$, we want to show $T\vert_S$ is continuous (by closed graph theorem and closable of the restriction map).
  2. Choose $S=\{x_n\}\cup\{x\}$, the continuity of $T|_S$ implies that $T(x_n)\to T(x)$.

Is this right? Or is there a counterexample?

Dean Miller
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DLIN
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1 Answers1

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No, your condition would mean $T$ is actually bounded. As a counterexample, consider, say, $T$ is multiplication by $n$ along the diagonal in $l^2(\mathbb{N})$. Then $x_n = e_n/n$ are in the domain and converges to $x = 0$, which is also in the domain. But $T(x_n) = e_n$ does not converge to $T(0) = 0$. (Your argument does not work because, among other things, closed graph theorem requires your map to be defined on a closed subspace, not just a closed subset.)

David Gao
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  • Though the question in your title is correct. Restricting closed operators to closed subspaces does give you continuous maps. If we assume further in your question that $T(x_n)$ actually converges, then the definition of closed operators will tell you that the sequence converges to $T(x)$, but as my example demonstrates, it can simply happen that $T(x_n)$ does not converge at all. – David Gao Oct 17 '23 at 04:35
  • Thanks a lot. I see the mistake(s) I made . – DLIN Oct 18 '23 at 00:33