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Timeline for Legendre-Fenchel transform

Current License: CC BY-SA 4.0

13 events
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Dec 6, 2021 at 10:34 comment added user12345678 Thank you very much. It was most helpful. What I actually need is contained in the answer to this question: mathoverflow.net/questions/341403/…
Dec 5, 2021 at 17:49 comment added Pietro Majer And here $F^*$ has never full domain, for the condition implies that $\text{dom}F^*$ does not contain lines (which btw in finite dimension even implies it is bounded, and that $F$ is Lipschitz, I think)
Dec 4, 2021 at 9:59 history edited YCor
edited tags
Dec 4, 2021 at 9:10 comment added Fedor Petrov How do you define the domain: the set of $y$ for which it is finite?
S Dec 4, 2021 at 7:50 history suggested CommunityBot CC BY-SA 4.0
typo in the title
Dec 4, 2021 at 7:42 review Suggested edits
S Dec 4, 2021 at 7:50
Dec 3, 2021 at 11:09 comment added Dirk Maybe there is some confusion regarding "continuous on its domain". If the domain is not the full space, the conjugate will never be continuous on the domain, because it has to jump to $+\infty$ on the boundary. So are you asking if the conjugate has full domain?
Dec 3, 2021 at 11:08 comment added Dirk This does not seem to be true. If $F\equiv 0$, then the conjugate is the indicator of $\{0\}$ which is not continuous in $0$.
Dec 3, 2021 at 9:20 history edited Daniele Tampieri CC BY-SA 4.0
(Very) Minor Math Jaxing (used $\langle\cdot,\cdot\rangle$ instead of $<\cdot,\cdot>$ and formatting
Dec 2, 2021 at 16:02 comment added Anthony Quas @mlk: convex functions don’t have to be continuous on the boundary of their domain.
Dec 2, 2021 at 13:51 comment added mlk I might be mistaken here, but $F^*$ should be a convex function and convex functions are themselves continuous (on their domain).
Dec 2, 2021 at 9:37 history edited user12345678 CC BY-SA 4.0
added 1 character in body
Dec 2, 2021 at 9:17 history asked user12345678 CC BY-SA 4.0