2
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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 $R_T-R_S$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.