Let $T$ be a self-adjoint operator (possibly unbounded) and $S$ a bounded self-adjoint operator.

Then one can study the unitary groups $R_T(t):=e^{itT}$ and $R_S(t):=e^{itS}.$

Now if you think about the function $f_t(x):=e^{itx}$ then this one is Lipschitz continuous with

$$\left\lvert f_t(x)-f_t(y) \right\rvert\le t \left\lvert x-y \right\rvert$$

for real arguments $x$ and $y$ which sort of correspond to self-adjoint operators.

So it is tempting to ask whether for all $x$ in the domain of $T$

$$\left\lVert R_T(t)x-R_S(t)x \right\rVert \le t\left\lVert (T-S)x \right\rVert.$$

It turns out that it is very easy to show if $S$ and $T$ commute (because then also $R_T$ and $R_S$ commute). However, I do not know whether it still holds true in general that is if $S$ and $T$ do not commute. Perhaps some clever arguments from the functional calculus might help?


1 Answer 1


No. Consider for instance periodic $L^2$ functions and let $T=i\frac{d}{dx}$, and let $S$ be multiplication by $\sin x$. The function $\exp(i\cos x)$ is in the nullspace of $T-S$, but not in the nullspace of $R_T-R_S$.


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