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Let $T$ be a closed unbounded (in my case also symmetric) operator on a Hilbert space $\mathcal{H}$ with dense domain $\mathcal{D}(T)$, and let $f\in \mathcal{D}(T)$. Suppose there is a dense subspace $Y$ of $\mathcal{H}$ contained in $\mathcal{D}(T)$. If I choose a sequence $f_n$ in $Y$ converging to $f$, under what conditions, if any, can I ensure that $Tf_n\to Tf$? The convergence of $Tf_n$ may be taken to be strong or weak. I know that $Tf$ is defined, but is it the limit of $Tf_n$?

Indeed, supposing my Hilbert space is $L^2(\mathbb{R}^n)$, I'm even interested in the $w^*$ convergence of the functionals $(Tf_n,\dot\:)$, where $(\dot\:,\dot\:)$ is the $L^2$-inner product. (These functionals are of course bounded on $\mathcal{D}(T)$ and so extend by continuity to $L^2(\mathbb{R}^n)$, which, by Riesz's lemma means they still take the form $(Tf_n,\dot\:)$.)

A related (and equivalent, by Banach-Steinhaus) question is, whether the sequence $Tf_n$ is uniformly bounded.

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  • $\begingroup$ Taking strong convergence, I think the condition is that $T$ is the closure of the restriction of $T$ to $Y$. See these notes p.1 . EDIT : That's if you know $\lim Tf_n$ exists. $\endgroup$ – Keith McClary Oct 31 '15 at 2:57
  • $\begingroup$ I mean, are you asking "If I choose a sequence $f_n$ in $Y$ converging to $f$, under what conditions, if any, can I ensure that $Tf_n$ converges ..." or "If I choose a sequence $f_n$ in $Y$ converging to $f$, such that $Tf_n$ converges, under what conditions, if any, can I ensure ..." ? $\endgroup$ – Keith McClary Oct 31 '15 at 20:24
  • $\begingroup$ Thanks Keith, I'm asking about the former. I know that $Tf$ and $Tf_n$ exist, and $f_n\to f$, but I don't know whether or when $Tf_n\to Tf$, or even if $Tf_n\to g$ for some $g$. It's very opaque to me. If I knew that $Tf_n\to g$, then things would be quite easy, I could use the closeness of $T$. $\endgroup$ – AlexN Nov 2 '15 at 0:03
  • $\begingroup$ There are examples in $L^2(\mathbb{R})$ where $\parallel f_n\parallel _2 \to 0$ but $\parallel \frac{d}{dx}f_n \parallel _2$ does not converge (eg., increasingly oscillatory functions). Are you asking for conditions on $T$ so that this won't happen? $\endgroup$ – Keith McClary Nov 2 '15 at 4:35
  • $\begingroup$ Yes, I'm interested in when this won't happen, given that $Tf$ is defined. The idea is I can choose the sequence $f_n$. I'm free to choose it from a dense set $Y$ contained in $\mathcal{D}(T)$, so it seems plausible that there might be ways to choose this sequence so that $Tf_n$ converges to $Tf$. For example, if $T=\Delta$, the Laplacian on $\mathbb{R}^n$, and $Y=\mathcal{S}(\mathbb{R}^n)$, is there a way to choose $f_n\in\mathcal{S}(\mathbb{R}^n)$ converging to $f\in \mathcal{D}(\Delta)=H^2(\mathbb{R}^n)$ in such a way that $\Delta f_n\to \Delta f$? $\endgroup$ – AlexN Nov 3 '15 at 1:46

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