Timeline for For which norms does closest projection never increase norm?
Current License: CC BY-SA 4.0
21 events
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| May 24 at 3:43 | history | edited | Whatsumitzu | CC BY-SA 4.0 |
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| May 24 at 2:29 | comment | added | Whatsumitzu | @fedja I think that in $\mathbb R^4$ the vertices of the regular $24$-cell are the analogues of the hexagon in $\mathbb R^2$ (vertices are $(\pm1, \pm1,0,0)$ and their permutations). It's harder to check but I think that the same proof that works for the regular hexagon in $\mathbb R^2$ might work also for this polytope. | |
| May 24 at 1:52 | comment | added | fedja | Correct. In 2D there are other examples too like $\ell^p$ norm in the first and the third quadrant and $\ell^q$ norm in the second and the third one ($\frac 1p+\frac 1q=1$), the hexagon corresponding to $p=1,q=\infty$. However I don't know anything like that in $\mathbb R^n$ for $n\ge 3$. | |
| May 23 at 22:41 | comment | added | Whatsumitzu | @fedja I think that in 2D if I take the norm whose unit ball is a regular hexagon, then the property is valid | |
| May 23 at 22:39 | history | edited | Whatsumitzu | CC BY-SA 4.0 |
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| May 23 at 22:34 | history | edited | Whatsumitzu | CC BY-SA 4.0 |
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| May 23 at 22:28 | history | edited | Whatsumitzu | CC BY-SA 4.0 |
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| May 21 at 23:39 | comment | added | Whatsumitzu | Ok my previous comment about perturbations looks fishy, but the following one is more robust: a norm coming from an inner product will satisfy the property, because the property is valid for the Euclidean norm, and it is invariant under changing the norm unit ball by (invertible) linear deformations. Currently I tend to conjecture that this covers all $C^2$-regular norms in $\mathbb R^2$ (so, if property holds in 2D and the norm is $C^2$, then the unit ball is an ellipse) | |
| May 21 at 21:06 | comment | added | Whatsumitzu | @fedja I think that if Euclidean unit ball has as boundary a perfect sphere, then a small perturbation of this ball within the class of convex centrally symmetric bodies, will still have the desired norm-decreasing property from the question.. I did not try to prove it in detail though | |
| May 21 at 15:52 | comment | added | fedja | I'm curious if you (or anyone) know an example of a non-Euclidean norm with this property in dimension 3 or higher. | |
| May 21 at 15:16 | comment | added | Whatsumitzu | I want that at least one of them works, so in that case it would be accepted. But if you take a line of slope 1/2 you get a counterexample for $\ell_1$.. see edited post. | |
| May 21 at 15:14 | history | edited | Whatsumitzu | CC BY-SA 4.0 |
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| May 21 at 15:11 | history | edited | Whatsumitzu | CC BY-SA 4.0 |
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| May 21 at 14:49 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor Math Jaxing
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| May 21 at 14:28 | history | edited | Whatsumitzu | CC BY-SA 4.0 |
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| May 21 at 14:09 | history | edited | Whatsumitzu | CC BY-SA 4.0 |
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| May 21 at 14:06 | comment | added | Whatsumitzu | taking v=0 shows that $||v^\star - x||\leq ||x||$, which is not what I asked, but I donkt know MikhailKatz' comment so this may be off track | |
| May 21 at 14:02 | comment | added | Whatsumitzu | Sorry, this MikhailKatz must have removed their comment, what did the comment say?.. the max norm is not such norm, euclidean one is such norm, so what's a good property that separates the two? | |
| May 21 at 12:36 | comment | added | fedja | @MikhailKatz In the Euclidean case, yes, but consider projecting $(1,1)$ to the line spanned by $(2,1)$ under the norm $\max(|x|,|y|)$ and you will see that the life is not as simple as you suggest. | |
| May 21 at 6:45 | review | Close votes | |||
| May 25 at 18:03 | |||||
| May 21 at 5:02 | history | asked | Whatsumitzu | CC BY-SA 4.0 |