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Oct 21, 2020 at 1:13 vote accept Paul Pfeiffer
Oct 20, 2020 at 22:44 history became hot network question
Oct 20, 2020 at 16:43 answer added Mikael de la Salle timeline score: 7
Oct 20, 2020 at 15:51 comment added Paul Pfeiffer 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.
Oct 20, 2020 at 15:43 comment added Yemon Choi And what about $A$? Your reference to singular values in the comment above suggests you want both A+B and B to be trace-class.
Oct 20, 2020 at 15:19 history edited Paul Pfeiffer CC BY-SA 4.0
mentioned Schatten semi norm
Oct 20, 2020 at 15:18 comment added Paul Pfeiffer 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.
Oct 20, 2020 at 15:16 comment added Paul Pfeiffer 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.
Oct 20, 2020 at 15:16 comment added Yemon Choi 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
Oct 20, 2020 at 15:07 comment added Paul Pfeiffer 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 $0 \le A,A+B \le1$.
S Oct 20, 2020 at 14:50 history suggested gmvh
Added top-level tag and "inequalities" tag (since the question is about establishing a certain inequality)
Oct 20, 2020 at 14:49 review Suggested edits
S Oct 20, 2020 at 14:50
Oct 20, 2020 at 14:45 history edited Paul Pfeiffer CC BY-SA 4.0
even stronger second hypothesis
Oct 20, 2020 at 14:38 history asked Paul Pfeiffer CC BY-SA 4.0