Is a self-adjoint operator continuous on its domain?

Let $H$ be a Hilbert space, and $A : D(A) \subset H \rightarrow H$ be an unbounded linear operator, with a domain $D(A)$ being dense in H. We assume that $A$ is self-adjoint, that is $A^*=A$.

Since $A$ is unbounded, we can find a sequence $x_n$ in the domain such that $x_n \rightarrow x$ with $x \notin D(A)$, meaning that $Ax_n$ does not converge.

My question is the following: Is it true that $A$ is continuous relatively to its domain? More precisely: given a converging sequence $x_n \rightarrow x$ for which both the sequence $x_n$ and the limit point $x$ lie in $D(A)$, can we conclude that $Ax_n \rightarrow Ax$? Or is there any counter-example?

I know that self-adjoint operators are closed, in the sense that their graph $D(A)\times A(D(A))$ is closed in $H \times H$, but continuity on the domain is something else.

• No it is not true. Take H to be little l^2, and A to be pointwise multiplication by the index n. – cauchyproblem Feb 14 '16 at 0:36
• The closed graph theorem implies that it is true if the domain is $H$. – Tsemo Aristide Feb 14 '16 at 0:38
• @TsemoAristide actually refers to the Hellinger–Toeplitz theorem. – Svetoslav Feb 14 '16 at 0:40
• @TsemoAristide Of course :) The point is indeed to know whether we still have some relative continuity when the domain is not the whole space. – Guillaume Garrigos Feb 14 '16 at 1:31
• Let e_n denote the sequence with 1 in the nth slot, and 0 otherwise. Define x_n = e_n/n. Then this sequence converges to 0 in little l^2, but Ax_n does not converge to 0 in little l^2 where A was described in my first comment. – cauchyproblem Feb 14 '16 at 1:45

If $A$ is unbounded, that means that for any $n$ there exists a $v\in D(A)$ such that $\|Av\|> n\|v\|$. Dividing such a $v$ by $\|v\|$, we may assume $\|v\|=1$, so we can find a sequence $(v_n)$ of unit vectors in $D(A)$ such that $\|Av_n\|>n$ for each $n$. The sequence $(v_n/n)$ then converges to $0$ but $Av_n/n$ does not converge to $A(0)=0$ since $\|Av_n/n\|>1$ for all $n$.
(More generally, the same argument shows that if $X$ and $Y$ are normed vector spaces, then a linear map $A:X\to Y$ is continuous iff it is bounded.)
• Thanks for your answer! In view to go further, I'd like to add here that, while $A$ is not norm-norm continuous on its domain, it is in fact weak-weak continuous there. Indeed, if $x_n \overset{w}{\rightarrow} x$, then for all $y \in D(A)$ we have $\langle Ax_n , y \rangle = \langle x_n, Ay \rangle \rightarrow \langle x,Ay \rangle = \langle Ax,y\rangle$. Since $D(A)$ is dense in $H$, the conclusion follows. – Guillaume Garrigos Feb 14 '16 at 10:50
Let $H$ be a Hilbert space, and let $D$ be a dense subset in $H$ under the metric topology induced by $H$'s norm. Let $A: D\rightarrow D$ be an unbounded self adjoint operator such that $$\langle Ah,g\rangle=\langle h,Ag\rangle, \forall h,g\in D$$ Thus there exists a sequence of elements such that $$d_{i}\in D, |d_{i}|=1, |A(d_{i})|\ge i^2, \forall i$$ Now if $h_{i}\rightarrow h \in H$, we can construct a new sequence $s_{i}=h_{i}+\frac{d_{i}}{i}$. So we have for sufficiently large $i$: $$s_{i}\rightarrow s, |A(s_{i})|\ge |A(h_{i})-A(\frac{d_{i}}{i})|\ge i-2|A(h)|$$ which cannot happen. Thus $A$ is unbounded on $H$ as well.