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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
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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
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Aug 2, 2018 at 7:10 history edited Idonknow CC BY-SA 4.0
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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
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Aug 1, 2018 at 11:59 history edited Idonknow CC BY-SA 4.0
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Aug 1, 2018 at 0:47 history edited Idonknow CC BY-SA 4.0
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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
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Jul 31, 2018 at 4:49 history asked Idonknow CC BY-SA 4.0