Timeline for Explicit isomorphism between $L^2(\mathbb{R}^2)$ and $L^2(\mathbb{R})$?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 3, 2021 at 14:39 | vote | accept | Slava Rychkov | ||
| Aug 1, 2021 at 5:41 | comment | added | Slava Rychkov | Thanks all for your comments! Maybe the Peano curve one will be the one which will do the trick for me (since I am looking at a particular application). It's funny because I am currently 10km from Cuneo, Piedmont, Italy where Giuseppe Peano was born and where there is a monument to the Peano curve: slrobertson.com/galleries/europe/italy/piedmont/scenic/… | |
| Aug 1, 2021 at 2:48 | comment | added | Piotr Hajlasz | I used your result (Linear extension operators...) in my research years ago in the paper P. Hajłasz, P. Koskela, H. Tuominen, Sobolev embeddings, extensions and measure density condition J. Funct. Anal. 254 (2008), 1217--1234. Actually, the original version of the paper was rejected by Maz'ya, because I was not aware of your result :) | |
| Jul 31, 2021 at 22:31 | comment | added | Joseph Van Name | In the case where $1\leq p<\infty,p\neq 2$, Banach (Lamperti generalized this result) has shown that the only isometries $U:L^{p}[0,1]\rightarrow L^{p}[0,1]$ (which are not necessarily onto) are the mappings where $Uf=h\cdot(f\circ\phi)$ for some $h\in L^{p}$ and Borel mapping $\phi:[0,1]\rightarrow[0,1]$ (this holds in the real and complex case). Therefore, any isometry of $L^{2}$-spaces that is not of this form must only be an isometry where $p=2$. | |
| Jul 31, 2021 at 21:03 | comment | added | Piotr Hajlasz | @TerryTao My answer is related to yours, but has some additional features. | |
| Jul 31, 2021 at 21:02 | answer | added | Piotr Hajlasz | timeline score: 5 | |
| Jul 31, 2021 at 20:45 | comment | added | Joseph Van Name | A measure probability algebra is a pair $(A,\mu)$ where $A$ is a $\sigma$-complete Boolean algebra and $\mu(\sum_{k=1}^{\infty}a_{k})=\sum_{k=1}^{\infty}(a_{k})$ and $\mu(a)=0$ iff $a=0$ and $\mu(1)=1$. Define a metric $d$ on $A$ by letting $d(x,y)=\mu(x\oplus y)$. Caratheodory has proven that all the atomless separable measure probability algebra is isomorphic, and such an isomorphism lifts to an isomorphism between the $L^{2}$-spaces. This is an abstraction of Terry Tao's example. More results like these can be found in Royden's book Real Analysis (3rd edition) Ch. 15. | |
| Jul 31, 2021 at 19:42 | comment | added | Terry Tao | Pullback by the Peano curve gives an isomorphism between $L^2([0,1]^2)$ and $L^2([0,1])$. | |
| Jul 31, 2021 at 19:34 | comment | added | Robert Furber | You can take an explicit Schauder basis for $L^2(\mathbb{R})$ (e.g. Hermite polynomials) and use your favourite explicit enumeration of $\mathbb{N}^2$ to construct a unitary isomorphism using the tensored basis for $L^2(\mathbb{R}^2)$. | |
| Jul 31, 2021 at 19:11 | history | asked | Slava Rychkov | CC BY-SA 4.0 |