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 counter example? References or own proofs both are warmly welcome.
This criterion holds for all closed operators. If $(x_n)$ is convergent and $(Ax_n)$ is bounded, then $(x_n,Ax_n)$ is bounded. Hence there exists a weakly convergent subnet $(x_{n_j},A x_{n_j})$; let's call its limit $(x,y)$. Clearly, $x=\lim_{n\to\infty} x_n$. Since the graph of $A$ is a closed subspace, it is weakly closed by the Hahn-Banach theorem. Thus $(x,y)$ is contained in the graph and therefore $x\in \mathcal D(A)$.