Timeline for An equivalence relation for norms
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24 events
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| Apr 8, 2014 at 11:01 | history | edited | alvarezpaiva | CC BY-SA 3.0 |
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| Apr 8, 2014 at 10:50 | history | edited | alvarezpaiva | CC BY-SA 3.0 |
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| Apr 7, 2014 at 19:19 | history | edited | alvarezpaiva | CC BY-SA 3.0 |
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| Apr 7, 2014 at 14:04 | comment | added | alvarezpaiva | @BenoîtKloeckner, I think I have it cornered now (see my answer). However, the relation to the modulus of convexity is still a bit slippery for me. In two dimensions, what is the relation between that and the (distributional) Laplacian of the norm? | |
| Apr 7, 2014 at 13:59 | answer | added | alvarezpaiva | timeline score: 4 | |
| Apr 7, 2014 at 13:04 | comment | added | Benoît Kloeckner | Also, thinking about $\ell_p$'s one is tempted to say that if the group of linear mappings that preserves the strong equivalence class of a norm acts transitively on directions, then the norm should be strongly equivalent to $\ell_2$. | |
| Apr 7, 2014 at 13:02 | comment | added | Benoît Kloeckner | @alvarezpaiva: indeed I did not look at Willie's answer enough. I do not have any very precise idea; but you should probably kill the symmetry between $x$ and $y$ by writing them $z+v$ and $z-v$, thinking about $z$ as the direction you are looking at, and about $v$ as a perturbation of this direction. Most interesting things should happen when $v$ is relatively small. | |
| Apr 7, 2014 at 9:52 | comment | added | alvarezpaiva | @BenoîtKloeckner: yes, that's in the Willie's answer below. The modulus of convexity seems related, but as far as i've been able to see, it is not a precise fit. Do you have something explicit in mind? | |
| Apr 7, 2014 at 9:49 | comment | added | Benoît Kloeckner | As said by Suvrit, you are more or less (probably more than less) asking that the two norms have the same modulus of convexity "in each direction". In particular, no two $\ell_p$ norms are strongly equivalent. | |
| Apr 7, 2014 at 8:45 | answer | added | Willie Wong | timeline score: 2 | |
| Apr 7, 2014 at 8:12 | comment | added | Willie Wong | Sorry, I meant that for each vertex $v$ of the unit sphere of norm 1, there exists (exactly one) vertex $w$ of the unit sphere of norm 2, such that $v = k w$ for some $k > 0$. (Projective over $\mathbb{R}_+$...) | |
| Apr 7, 2014 at 8:10 | comment | added | alvarezpaiva | However, note that $\ell_\infty$ and $\ell_1$ on the plane are not strongly equivalent. What do you mean by "in the projective sense"? | |
| Apr 7, 2014 at 7:40 | comment | added | Willie Wong | And I think this is the sharpest general criterion you can have in the polygonal case: consider the standard $\ell_1$ and the norm $\| \vec{x}\| = |x_1| + 2|x_2|$ on the plane, They are strongly equivalent with $\lambda = 2$. | |
| Apr 7, 2014 at 7:40 | comment | added | alvarezpaiva | @WillieWong: thanks! That's a simple, elegant argument. Having exactly the same geodesics for normed spaces with polytopal norms is quite restrictive. | |
| Apr 7, 2014 at 7:36 | comment | added | Willie Wong | When the unit sphere is a polygon (or a polyhedron), in the convex cone attached to each of the faces the norm is a linear function. In particular, for two points inside the same convex cone attached to the faces, we have that $$ \|x\| + \|y\| - \|x + y\| = 0 $$ Thus if norm 1 has polygonal unit sphere and norm 2 is strongly equivalent, then the restriction of norm 2 to the convex cones of the faces of norm 1 are also linear. This immediately implies that the unit sphere of norm 2 is also a polygon (or a polyhedron) with the same (in the projective sense) vertices. | |
| Apr 7, 2014 at 6:07 | comment | added | alvarezpaiva | @AliTaghavi: I suspect that the normed plane whose unit circle is a regular haxagon is not strongly equivalent to any $\ell_p$ norm. | |
| Apr 6, 2014 at 8:57 | comment | added | Ali Taghavi | Is it true to say that every norm on the plane is stongly equivalent to some $\parallel.\; \parallel_{p}$? | |
| Apr 4, 2014 at 7:32 | history | edited | alvarezpaiva | CC BY-SA 3.0 |
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| Apr 4, 2014 at 7:27 | comment | added | alvarezpaiva | @Suvrit: Thanks. I'll modify the question to reflect this last comment. | |
| Apr 3, 2014 at 21:20 | comment | added | Suvrit | btw the second condition implies the first. | |
| Apr 3, 2014 at 18:33 | history | edited | alvarezpaiva | CC BY-SA 3.0 |
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| Apr 3, 2014 at 17:03 | comment | added | Suvrit | sounds like modulus of convexity might be a good keyword... | |
| Apr 3, 2014 at 16:17 | history | edited | alvarezpaiva | CC BY-SA 3.0 |
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| Apr 3, 2014 at 16:12 | history | asked | alvarezpaiva | CC BY-SA 3.0 |