# Hölder continuity of functional calculus

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.

• 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 $0 \le A,A+B \le1$. – Paul Pfeiffer Oct 20 at 15:07
• 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 – Yemon Choi Oct 20 at 15:16
• 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. – Paul Pfeiffer Oct 20 at 15:16
• 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. – Paul Pfeiffer Oct 20 at 15:18
• 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. – 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.
• 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. – Paul Pfeiffer Oct 21 at 1:13
• $h_\alpha(t)= (1-\alpha)^{-1} \ln(t^\alpha+(1-t)^\alpha$ 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. – Paul Pfeiffer Oct 21 at 16:02