5
$\begingroup$

Let $0<\beta<1$ and $ f \colon [0,1] \to [0,1]$ be $\beta$ Hölder continuous with constant $C$. Let $H$ be a Hilbert space and $A,B$ be self adjoint operators on $H$, such that $\sigma(A+B),\sigma(A) \subset [0,1]$. Then we can define $f(A+B)$ and $f(B)$ by the continuous functional calculus. Do we then have the estimate $$ \left \lvert \operatorname{tr} (f(A+B)-f(A)) \right \rvert \le C \lVert B \rVert_\beta^\beta$$ EDIT: The semi-norm $\lVert B \rVert_\beta$ is the Schatten von Neumann semi-norm.

This does hold for commutating operators $A,B$ and it seems to hold for 2x2 matrices, if i calculated correctly. There is also the stronger hypothesis, that for any unitary equivalent norm $\lVert \cdot \rVert$, we have the estimate $$ \left \lVert f(A+B) - f(A) \right \rVert \le C \lVert \lvert B \rvert^\beta\rVert$$ I am aware of the question Hölder continuity for operators and its answer, but this is different, as the trivial counter example does not hold. The special case $f(t)=t^\beta$ is stated as true in an answer to that question.

Improve this question
$\endgroup$
6
  • $\begingroup$ For the first statement, it is sufficient to show, that for the ordered singular values, we have the inequality $\lvert s_i(A+B)-s_i(A) \rvert \le s_i(B)$ using the assumption $0 \le A,A+B \le1$. $\endgroup$
    – Paul Pfeiffer
    Oct 20, 2020 at 15:07
  • $\begingroup$ If you are working in infinite dimensions, are you including the assumption that B is trace-class? Your question talks of self-adjoint operators on Hilbert spaces but these might not have SVD etc $\endgroup$
    – Yemon Choi
    Oct 20, 2020 at 15:16
  • $\begingroup$ My first approach does not work. Even for commutating operators, we would need to rearrenge the singular values of $B$ to make this work. This is probably not a good approach. $\endgroup$
    – Paul Pfeiffer
    Oct 20, 2020 at 15:16
  • $\begingroup$ If $B$ is not trace class, or more general not in the $\beta$- Schatten von Neumann class, the right hand side is infinite and hence the statement is tautolgical. So you may assume $B$ to be trace class. $\endgroup$
    – Paul Pfeiffer
    Oct 20, 2020 at 15:18
  • 1
    $\begingroup$ That approach does not work. For my application, the assumption, that $A$ is trace class is satisfied, but I am also interested in the case, where $A$ is not trace class. $\endgroup$
    – Paul Pfeiffer
    Oct 20, 2020 at 15:51

1 Answer 1

Reset to default
7
$\begingroup$

Such questions have been much studied, in particular by Aleksandrov and Peller. Probably the most relevant reference is the paper Functions of operators under perturbations of class $S_p$ by Aleksandrov and Peller, J. Funct. Anal. 258 (2010). Zbmath link or mathscinet link.

In particular it is proved there (Theorem 9.14) that for every $\beta<1$ and $p \leq 1$, there is a $\beta$-Hölder-continuous function $f$ and operators $A$, $B$ such that $B \in S_{1}$ such that $f(A)-f(B)$ does not belong to $S_{1/\beta}$. In particular, $B \in S_\beta$ and $f(A) -f(B)$ does not belong to $S^1$.

Remarkably, this is optimal (Theorem 9.13): for every $p>1$, and every $\beta$-Hölder continuous function $f$, $f(A+B) - f(A)$ belongs to $S_{p/\beta}$ whenever $B$ belongs to $S_p$.

In the same paper, sufficient conditions on $f$ which imply that $\|f(A+B)-f(A)\|_1 \leq \|B\|_\beta^\beta$ are derived.

Improve this answer
$\endgroup$
3
  • $\begingroup$ That does show, that my general claim is false. There is still a very significant gap between the necessary and sufficient conditions for such an $f$ in this paper. The sufficient condition implies Lipschitz, which is not satisfied by the renji entropy functions I am studying. The necessary condition is the Hölder continuity I claimed to be sufficient. Still, it is an answer to the question, although not the one I had hoped for. $\endgroup$
    – Paul Pfeiffer
    Oct 21, 2020 at 1:13
  • $\begingroup$ @PaulPfeiffer What are these "Renji entropy functions" that you are mentionning ? $\endgroup$
    – Mikael de la Salle
    Oct 21, 2020 at 14:49
  • $\begingroup$ $h_\alpha(t)= (1-\alpha)^{-1} \ln(t^\alpha+(1-t)^\alpha$ for $\alpha \not =1$. But I think, that they can be approximated sufficiently close by $C_c^2$ functions, which are in $B^1_{\infty,1}$, if i understood the Bezov spaces correctly. Hence this does help with my application. $\endgroup$
    – Paul Pfeiffer
    Oct 21, 2020 at 16:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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