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? 



Lipschitz bound on semigroups
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 $R\_T-R\_S$.


