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.


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, Graduate Texts in Mathematics. 194. Berlin: Springer. xxi, 586 p. (2000). ZBL0952.47036.

  • $\begingroup$ Hello, thank you. By the way, it is very nice that you answer me, because I actually have a question for you: In all your papers on delay equations you always assume $[-1,0]$ as the delay interval. Is this just for convenience and any $[-c,0]$ would work or is there a reason you choose $[-1,0]$? $\endgroup$ – Clement G. May 2 '18 at 11:19
  • $\begingroup$ Just convenience. It is a question of choosing the unit measurement of time. $\endgroup$ – András Bátkai May 2 '18 at 11:24

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