Cross-post from [math.sx](https://math.stackexchange.com/questions/4198077/convergence-criterion-in-the-domain-of-an-unbounded-operator).


My question is somewhat close to [this](https://math.stackexchange.com/questions/1653867/is-a-self-adjoint-operator-continuous-on-its-domain) 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](https://en.wikipedia.org/wiki/Fatou%27s_lemma#Extensions_and_variations_of_Fatou%27s_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.



Convergence criterion in the domain of an unbounded operator
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.


