Convergence criterion in the domain of an unbounded operator

Question

Cross-post from math.sx.

My question is somewhat close to this one, but the counterexamples given there do not apply here.

Setup. Given a Hilbert space $\mathcal H$, a closed operator $A$ and a convergent sequence $(x_n)_{n\in\mathbb N}\subset \mathcal D(A)$, I want to ensure that $\lim_{n\to\infty}x_n \in \mathcal D(A)$.
It is not necessary that $\lim_{n\to\infty} Ax_n = Ax$.

My intuition says that something like a uniform upper bound on $\|Ax_n\|$ should be a sufficient condition. Let me illustrate that in the following setting for normal operators.

Lemma. Let $A$ be a normal operator on $\mathcal H$, $x_n$ a convergent sequence in $\mathcal D(A)$ and assume $\liminf_{n\to\infty}\|Ax_n\|<\infty$. Then $\lim_{n\to\infty} x_n \in \mathcal D(A)$.
Proof. We use the spectral theorem and Fatou's lemma for weakly convergent measures. Let $E_A$ be the projection valued measure associated to $A$. Then $$ \int_{\mathbb C} |\lambda|^2 d\langle x,E_A(\lambda)x\rangle \le \liminf_{n\to\infty} \int_{\mathbb C} |\lambda|^2 d\langle x_n,E_A(\lambda)x_n\rangle = \liminf_{n\to\infty}\|Ax_n\|^2<\infty. $$ Hence, $x\in\mathcal D(A)$. $\square$

Question. What extensions of above lemma are known? Does anyone have a counterexample? References or own proofs both are warmly welcome.

Answer 1

If you have a uniform upper bound on $\|Ax_n\|$ then you can extract a weakly convergent subsequence $Ax_{n_k}$. Denote $\lim_n Ax_{n_k} = y_{\infty}$ to be the weak limit. Since $A$ is closed, it is also weakly closed. Since $A$ is a weakly closed operator, we have weakly $y_{\infty} = Ax_{\infty}$ and $x_{\infty} \in D(A)$ as desired. (Noting that $\lim_n x_{n_k} = \lim_n x_{n} =: x_{\infty}$ strongly and weakly).

Actually, this shows that every subsequence of $ Ax_{n}$ contains a further subsequence that converges weakly to the same $y_{\infty}$. Thus, we can actually conclude that $A x_n$ converges to $Ax_{\infty}$ weakly. If we assume the stronger condition that $\|Ax_n\| \rightarrow \|Ax_{\infty}\| < \infty$, which implies $\|Ax_n\|$ is uniformly bounded, then we have $Ax_n \rightarrow A{x_\infty}$ strongly as well.

Edit: I incorrectly stated/implied that uniform boundedness of $\|Ax_n\|$ gave a (strongly) convergent subsequence of $Ax_n$. This actually only gives a weakly convergent subsequence, which suffices for the proof.

Answer 2

The answer by Lars van der Laan gives a positive answer for the Hilbert space case (which was considered in the question), and it also works on reflexive Banach spaces.

It might be worthwhile to add that the answer is no, in general, on non-reflexive Banach spaces. A simple counterexample is:

Example. Endow the space $E = C_0((0,1])$ of continuous functions on $[0,1]$ that vanish in $0$ with the supremum norm, and consider the operator $A$ given by \begin{align*} D(A) & = \{u \in E: \, u \text{ is differentiable and } u' \in E \}, \\ Au & = -u'. \end{align*} This is a closed operator with compact resolvent (and even a semigroup generator).

Now, consider the function $f \in E$ given by $f(x) = x$. Then $f$ can easily be approximated in sup norm by functions from $D(A)$ whose derivatives are bounded in sup norm. Yet, $f \not\in D(A)$ since $f' \not \in E$.