Skip to main content
22 events
when toggle format what by license comment
Oct 8, 2020 at 15:06 vote accept Henry
Oct 7, 2020 at 9:41 vote accept Henry
Oct 8, 2020 at 15:06
Oct 5, 2020 at 16:56 history edited YCor CC BY-SA 4.0
formatting
Oct 5, 2020 at 15:30 comment added MaoWao @PietroMajer Typically this term is used for functions that satisfy this inequality with a constant independent of the dimension.
Oct 5, 2020 at 15:04 comment added Pietro Majer As to 1 note that since all norms are equivalent in finite dimension, "Lipschitz" denotes the same class of maps (although with different constants).
Oct 5, 2020 at 14:41 comment added Yemon Choi Henry, regarding the equations in the "review paper" you keep mentioning, there is a vital extra assumption that the operators commute with each other. Without that assumption the result you want fails to hold (indeed, if it didn't fail, then "operator Lipschitz" functions would just be "Lipschitz functions")
Oct 5, 2020 at 14:40 answer added Mikael de la Salle timeline score: 9
Oct 5, 2020 at 14:38 answer added Yemon Choi timeline score: 8
Oct 5, 2020 at 14:37 comment added Henry I'm wondering if considering functions on finite closed subsets of the real line will help. It looks like in this case you're considering twice differentiable functions across the reals. In the 2016 review paper I linked, they make the point that indeed the constants are not always equal, but in the case of a closed subset of the real line they are (see the remark just above Equation (3.1.3))
Oct 5, 2020 at 14:29 comment added Mateusz Wasilewski The survey you mention is about Lipschitz estimates for the operator norm, at least most of it is. As Christian Remling mentions, for reasonably smooth functions there is some estimate, but it is usually more complicated than simply taking the Lipschitz constant of a function.
Oct 5, 2020 at 14:18 comment added Christian Remling @Henry: Actually, I'm slightly more skeptical now. I used something very similar here: math.ou.edu/~cremling/research/preprints/gen-toda.pdf . See eq. (2.1). The constant is not the Lipschitz constant of $f$.
Oct 5, 2020 at 14:01 comment added Henry Thanks for the pointer @ChristianRemling, I'll have a look
Oct 5, 2020 at 13:57 comment added Christian Remling I think the inequality $\|f(M)-f(N)\|\le k \|M-N\|$ must definitely be true when enough assumptions on $f$ are made ($f\in C^2$ should work) because then calculus techniques for the map $M\mapsto f(M)$ become available. See Pedersen, Operator differentiable functions (2000) (I don't have the paper available right now, so can't check the details).
Oct 5, 2020 at 13:46 history edited Henry CC BY-SA 4.0
Cleared up norm subscript notation
Oct 5, 2020 at 13:38 history edited Henry CC BY-SA 4.0
added 100 characters in body
Oct 5, 2020 at 13:33 comment added Henry In my question I'm using $\left\Vert A \right\Vert_F = \sqrt{\sum_{i,j}A_{i,j}^2}$ as the Frobenius norm and $\left\Vert A \right\Vert_2 = \sup_{x \not = 0} \frac{\left\Vert Ax \right\Vert_2}{\left\Vert x\right\Vert_2}$ where the norms for these vectors are the Euclidean norm. In Kittaneh they write $\left\Vert A \right\Vert_2$ as the Hilbert–Schmidt operator which I think is the Frobenius norm in the finite case (it also looks like they're using the Frobenius norm as I've defined it in the proof)
Oct 5, 2020 at 12:41 comment added Branimir Ćaćić I’m pretty sure that $\lVert \cdot \rVert_F = \lVert \cdot \rVert_2$, with F standing for Frobenius. That said, I get the impression that “Frobenius inner product” is more frequently used to refer to the finite-dimensional special case of the Hilbert–Schmidt inner product than it is as a true synonym.
Oct 5, 2020 at 12:27 history edited YCor CC BY-SA 4.0
edited title
Oct 5, 2020 at 12:18 comment added YCor what's $F$ in $\|\cdot\|_F$? in Kittaneh it appears as $\|\cdot\|_2$.
Oct 5, 2020 at 12:17 history edited YCor CC BY-SA 4.0
formatting, provided free AMS link
Oct 5, 2020 at 11:32 review First posts
Oct 5, 2020 at 11:44
Oct 5, 2020 at 11:26 history asked Henry CC BY-SA 4.0