Let $A\_n$ be a family of (bounded) self-adjoint operator converging pointwise to some (unbounded) self-adjoint operator $A,$ i.e. for all $x$ in the domain of $A$


$$\left\lVert A\_n x-Ax \right\rVert \rightarrow 0.$$


Does this imply that $e^{it A\_n}$ converges pointwise to $e^{itA}$?


I know it holds, if the $A\_n$ commute with each other as in this case 


$$\left\lVert T\_n(t)x-T\_m(t)x \right\rVert \le \int\_0^t \left\lVert \frac{d}{ds} T\_m(t-s)T\_n(s) x \right\rVert \ ds \le t \left\lVert A\_n x- A\_m x \right\rVert$$


where $T\_n =e^{itA\_n}.$


Searching the literature myself I noticed that this might be related (especially the frist paragraph of the question):


[click me.](https://mathoverflow.net/questions/178904/strongly-convergent-series-of-bounded-self-adjoint-operators?rq=1)



Pointwise convergence in functional calculus
There are easyer and more direct ways to prove it, but this follows immediatelly as a special case from the Trotter-Kato approximation theorem, see Theorem III.4.8 in 


*Engel, Klaus-Jochen; Nagel, Rainer*, [**One-parameter semigroups for linear evolution equations**](http://dx.doi.org/10.1007/b97696), Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). [ZBL0952.47036](https://zbmath.org/?q=an:0952.47036).


