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.

  • $\begingroup$ 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 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 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 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 at 15:18
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    $\begingroup$ 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 at 15:51

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.

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  • $\begingroup$ 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 at 1:13
  • $\begingroup$ @PaulPfeiffer What are these "Renji entropy functions" that you are mentionning ? $\endgroup$ – Mikael de la Salle Oct 21 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 at 16:02

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