Timeline for Existence of a Gelfand triple involving the Arens–Eells space (aka Lipschitz free space)
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 24, 2021 at 15:58 | vote | accept | Yury Korolev | ||
| Dec 22, 2021 at 19:10 | comment | added | Yury Korolev | Yes, sorry, I should have said $X^*$. I would actually be happy with a finite-dimensional compact $X$ (unit ball in $R^n$ with some metric, not necessarily euclidean). I was thinking of something like a Sobolev space $H^k$ perhaps, which embeds continuously into some Hölder spaces (which are Lipschitz w.r.t. an appopriate metric on the unit ball). But the embedding of $AE$ into a Sobolev space is not clear to me. | |
| Dec 21, 2021 at 22:08 | comment | added | Bill Johnson | As for " I would like to identify $H$ with some space of functions/distributions on $𝑋$", did you mean $X^*$ instead of $X$. It is not very natural to embed spaces of functions on $X$ back into $X$. | |
| Dec 21, 2021 at 22:04 | comment | added | Bill Johnson | Well, If $X=C[0,1]$, then you can use $L_2(0,1)$ as the Hilbert space. That is pretty "natural". Every separable Banach space embeds isometrically isomorphically into $C[0,1]$, but it is a stretch to claim that there is a "natural" isometric embedding of every separable space into $C[0,1]$. | |
| Dec 21, 2021 at 15:32 | comment | added | Yury Korolev | Thank you all very much! Is there a ‘natural’ identification of such a Hilbert space? In certain cases (as discussed in this question), $AE$ is linearly isomorphic to $\ell^1$ and $Lip_0$ to $\ell^\infty$, so we could take $\mathcal H = \ell^2$, but I would like to identify $\mathcal H$ with some space of functions/distributions on $X$. | |
| Dec 21, 2021 at 13:28 | comment | added | Nik Weaver | Also note that this works for arbitrary separable $X$, it doesn't have to be compact. | |
| Dec 21, 2021 at 12:05 | comment | added | Philip Brooker | @Ryan: Yes, because you may assume that the operator Bill describes has dense range in the Hilbert space (by replacing the Hilbert space by the norm closure of the range of the operator if necessary; this subspace is also a Hilbert space). Under the assumption that the operator described by Bill has dense range, the adjoint of this operator is a continuous, injective linear operator from the dual of the Hilbert space into the Lipschitz space. Identifying the Hilbert space with its dual space gets you the rest of the way. | |
| Dec 21, 2021 at 5:07 | comment | added | Ryan | Is it similarly “obvious” hilbert space embeds into the lipschitz space? | |
| Dec 20, 2021 at 23:13 | history | answered | Bill Johnson | CC BY-SA 4.0 |