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.
 A: 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.)
A: 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. 
