A question related to the boundedness of self-adjoint operators

Question

I am currently working on an the following exercise from Bryan C. Hall's Quantum Theory for Mathematicians:

If $A$ is a bounded self-adjoint operator (on a Hilbert space), show that $U(t) := e^{iAt}$ is continuous in the operator-norm topology.

I am kind of unsure as to how to proceed and any help/hint will be useful. Thanks

Answer 1

Note that $$ \| (e^{itA}-e^{isA})\psi\|^2=\int_{\sigma(A)}|e^{itx}-e^{isx}|^2\mathrm d\mu_\psi(x) $$ for $\mathrm d\mu_\psi$ the spectral measure corresponding to $\psi$. Then, by the mean value theorem $$ \| (e^{itA}-e^{isA})\psi\|^2\leq |t-s|\int_{\sigma(A)}|x|\mathrm d\mu_\psi(x)\\ \leq |t-s|^2\max_{x\in \sigma(A)}|x|^2\mu_\psi(\sigma(A)) $$ so $$ \|(e^{itA}-e^{isA})\|\leq |t-s|\|A\|. $$

Answer 2

You have to show that $$ \| e^{itA}-e^{isA}\| \to 0, $$ as $|t-s|\to 0$, and where $\|\cdot \|$ denotes the operator norm. Recall that one way of defining this exponential is through the power series expansion, so that $$ e^{itA}-e^{isA} = \sum_{k=0}^\infty \dfrac{(itA)^k-(isA)^k}{k!}. $$ Now write, for $k\geq 2$, \begin{equation} \begin{split} (itA)^k-(isA)^k & =(t-s)iA[(itA)^{k-1}+(itA)^{k-2}(isA)+\ldots+(itA)(isA)^{k-2}+(isA)^{k-1}]\\ & = (t-s)(iA)^k[t^{k-1}+\ldots+s^{k-1}]. \end{split} \end{equation} Can you finish from here?