Timeline for If $E\oplus_\phi E \cong E\oplus_\psi E,$ does it imply that $\phi= \psi$?
Current License: CC BY-SA 4.0
21 events
when toggle format | what | by | license | comment | |
---|---|---|---|---|---|
Jan 4, 2019 at 2:19 | vote | accept | Idonknow | ||
Aug 18, 2018 at 19:36 | answer | added | Dirk Werner | timeline score: 5 | |
S Aug 9, 2018 at 2:52 | history | bounty ended | Idonknow | ||
S Aug 9, 2018 at 2:52 | history | notice removed | Idonknow | ||
Aug 7, 2018 at 21:30 | answer | added | Mikhail Ostrovskii | timeline score: 5 | |
Aug 5, 2018 at 8:58 | comment | added | Duchamp Gérard H. E. | @Idonknow No, my goal was to help you for generality through polynomiality; not extremal points ($(||t_1 u\oplus t_2v||_p)^p$ is (generalized) polynomial in $t_i\geq 0$). As $E\not=\{0\}$, take $u$ with norm $1$. Suppose now $E\oplus_p E\simeq E\oplus_q E$ and let $f$ be an isometry between them. I think that a careful examination of $||u\oplus tu||_p^p$ and that of the image of $u\oplus tu$ could help. | |
Aug 3, 2018 at 13:59 | history | edited | Idonknow | CC BY-SA 4.0 |
edited title
|
Aug 3, 2018 at 12:46 | comment | added | Tobias Fritz | Since $E\cong F$ implies $E\oplus_\psi E \cong F\oplus_\psi F$, you might as well phrase the question like this: does $E\oplus_\phi E \cong E\oplus_\psi E$ imply $\phi = \psi$? (Nitpick: and you may want to assume $E\neq 0$, since otherwise the answer is trivially no.) | |
Aug 2, 2018 at 7:15 | history | edited | Idonknow | CC BY-SA 4.0 |
added 154 characters in body
|
Aug 2, 2018 at 7:10 | history | edited | Idonknow | CC BY-SA 4.0 |
added 154 characters in body
|
S Aug 2, 2018 at 7:09 | history | bounty started | Idonknow | ||
S Aug 2, 2018 at 7:09 | history | notice added | Idonknow | Draw attention | |
Aug 2, 2018 at 7:05 | comment | added | Idonknow | @DuchampGérardH.E. If $E = \mathbb{R},$ then I guess we can look at their set of extreme points and deduce that $p=q?$ Because for two norms to be the same, they must have the same set of extreme points.I think this is a characterization, that is, two norms are the same iff they have the same set of extreme points. | |
Aug 2, 2018 at 0:31 | history | edited | Idonknow | CC BY-SA 4.0 |
added 90 characters in body
|
Aug 1, 2018 at 11:59 | history | edited | Idonknow | CC BY-SA 4.0 |
added 81 characters in body
|
Aug 1, 2018 at 0:47 | history | edited | Idonknow | CC BY-SA 4.0 |
edited body
|
Jul 31, 2018 at 17:24 | comment | added | user114263 | In the infinite dimensional case, the unit balls need not have any extreme points. | |
Jul 31, 2018 at 13:49 | comment | added | Idonknow | As you might have noticed, but just in case, I do not assume that $p$ and $q$ are holder conjugate. | |
Jul 31, 2018 at 13:25 | comment | added | cevasix | Objection to your final statement since in the case where $E$ is the real line. then the cases $1$ and $\infty$ are isometric. Rather too silly to be worth mentioning but for the fact that it suggests looking at the extreme points of the unit balls. If they are what I suspect (i.e. situated on the axes and the diagonals---except for $p=2$), then this would prove your conjecture. | |
Jul 31, 2018 at 5:28 | history | edited | Idonknow | CC BY-SA 4.0 |
edited body
|
Jul 31, 2018 at 4:49 | history | asked | Idonknow | CC BY-SA 4.0 |