Timeline for Is the Legendre transform as an operator Lipschitz?
Current License: CC BY-SA 4.0
4 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2022 at 3:23 | comment | added | Tom | Correct (provided one handles the indeterminate case mentioned). | |
| Dec 10, 2022 at 1:42 | comment | added | gdavtor | Lipschitz constant of 1 would then imply that LF is an isometry (on $C_{lsc}(\mathbb{R}^n)$), by the involution property $f^{**} = f$? This makes sense geometrically actually now that I think about it. Thank you! | |
| Dec 9, 2022 at 2:17 | vote | accept | gdavtor | ||
| Dec 9, 2022 at 0:14 | history | answered | Tom | CC BY-SA 4.0 |