I am looking for the solution to the problem:
Let $A$ be a self-adjoint operator on a Hilbert space $\mathcal{H}$. Show that $A$ is bounded if and only if $$\lim_{h\to 0}\|e^{ihA}-{\rm id}_\mathcal{H}\|=0.$$ Here $h \in\mathbb R$ and $i^2=-1$.
The if part is easy by Taylor expansion of $e^{ihA}$ and some estimation. But I do not know how to prove the only if part. I tried to prove the equivalent: $A$ is unbounded then $$\lim_{h\to 0}\|e^{ihA}-{\rm id}_\mathcal{H}\|\ne 0.$$
I stared with definition of unbounded operator $A$, i.e. there exists a sequence $x_n\in D(A)$ such that $$\dfrac{\|Ax_n\|}{\|x_n\|}\to \infty.$$ But how to derive from this that there exists a sequence $y_n\in D(A)$ such that $$\lim_{h\to 0}\frac{\|e^{ihA}y_n-y_n\|}{\|y_n\|}\to \infty$$
By the Spectral Theorem we can assume that $\mu$ is a measure on (some $\sigma$-algebra of subsets of ) $X$, $D\subset L^2(\mu)$, and $$A:D\to L^2(\mu)$$is given by $$Af(x)=m(x)f(x)$$for some measurable function $m:X\to\Bbb R$.
Now we have to show that $m\in L^\infty$ if and only if $$||e^{ihm}-1||_\infty\to0\quad(h\to0),$$which is not hard.