Timeline for Non-surjective isometries of $l_p$
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 23, 2023 at 10:47 | comment | added | YCor | The whole point of a permutation is that the domain and the range are the same. If you relabel the the range, you have to relabel the domain too. This is pointless anyway, since your formulation is immediate to fix (writing "injective self-map" instead of permutations, since permutation precisely correspond to surjective isometries. | |
| Apr 21, 2023 at 20:16 | comment | added | Bill Johnson | Look up the Banach-Lamperti Theorem. | |
| Apr 21, 2023 at 15:49 | comment | added | fedja | And for $p<2$ and two unit vectors, one can use the same limit and compare it to $2$ instead of infinity... | |
| Apr 21, 2023 at 15:28 | comment | added | fedja | For $p>2$ it is trivial: just consider $\lim_{t\to 0+}\frac{\|x+ty\|^p+\|x-ty\|^p-2\|x\|^p}{t^p}$. It is finite if the supports are disjoint and infinite otherwise. | |
| Apr 21, 2023 at 15:08 | comment | added | fedja | This boils down to the two-dimensional question whether the images of two standard basis vectors must have disjoint supports. Looks like it must be true but the formal proof still escapes me (though I haven't thought of it too long yet :-) ) | |
| Apr 21, 2023 at 10:52 | comment | added | Markus | Indeed! If $(x_n)$ are disjointly supported (not necessarily finitely support even), then one also gets isometries. | |
| Apr 21, 2023 at 10:42 | comment | added | S Argyros | Concerning the obvious isometries I think you should use sequences of normalized and disjoint supported vectors instead of block sequences. | |
| Apr 21, 2023 at 10:29 | history | edited | Markus | CC BY-SA 4.0 |
added 7 characters in body
|
| Apr 21, 2023 at 10:29 | comment | added | Markus | Good point. Yes, I mean linear non-surjective isometries. | |
| Apr 21, 2023 at 10:14 | comment | added | Pietro Majer | Do you assume linearity? (It is for free in the surjective case by the Banach-Mazur Thm, but it is not guaranteed for non surjective isometries) | |
| Apr 21, 2023 at 9:52 | comment | added | Markus | You are right about that. Stil, does it make a difference in the end? I am not sure. For your example, after relableing the range we still get a permutation. If I am considering all block basis, am I missing an isometry by saying permutation instead of injective self-map? | |
| Apr 21, 2023 at 9:38 | comment | added | YCor | No, the injective self-map $n\mapsto n+1$ of $\mathbf{N}_{\ge 0}$ is not a permutation on its range $\mathbf{N}_{\ge 1}$. | |
| Apr 21, 2023 at 9:37 | comment | added | Markus | I don't think it makes a difference in this case. The injective self-map is a permutation on the range, and a subsequence of a block basis is also a block basis. | |
| Apr 21, 2023 at 9:34 | history | edited | YCor |
edited tags
|
|
| Apr 21, 2023 at 9:34 | comment | added | YCor | The second time you're writing "permutation" (which means bijective self-map), you probably mean "injective self-map". | |
| Apr 21, 2023 at 9:18 | history | asked | Markus | CC BY-SA 4.0 |