As Hilbert spaces, $L^2(\mathbb{R}^2)$ and $L^2(\mathbb{R})$ are isomorphic. Of course the isomoprhism is vastly not unique. I wonder if there are any particularly nice explicit isomorphisms. E.g. I wonder if there is an integral transform $$ f(x,y) \mapsto (K f)(z)=\int dx\, dy K(x,y,z) f(x,y) $$ with a nice explicit kernel $K(x,y,z)$ which maps $L^2(\mathbb{R}^2)$ isometrically onto $L^2(\mathbb{R})$? Any example would be appreciated.
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7$\begingroup$ 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 $L^2(\mathbb{R}^2)$. $\endgroup$– Robert FurberCommented Jul 31, 2021 at 19:34
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11$\begingroup$ Pullback by the Peano curve gives an isomorphism between $L^2([0,1]^2)$ and $L^2([0,1])$. $\endgroup$– Terry TaoCommented Jul 31, 2021 at 19:42
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$\begingroup$ 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. $\endgroup$– Joseph Van NameCommented Jul 31, 2021 at 20:45
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$\begingroup$ @TerryTao My answer is related to yours, but has some additional features. $\endgroup$– Piotr HajlaszCommented Jul 31, 2021 at 21:03
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3$\begingroup$ 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/… $\endgroup$– Slava RychkovCommented Aug 1, 2021 at 5:41
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The following result was obtained by an "explicit" construction in [1]. It is related to the comment of Terry Tao. A modification of the argument allows one to replace the cube by the whole space.
Theorem. If $k\geq n$ and $1\leq p\leq \infty$, then there is an isometric isomorphism $\Phi: L^p([0,1]^k)\to L^p([0,1]^n)$ such that $\Phi(u)$ is continuous on $(0,1)^n$ for each $u\in L^p([0,1]^k)$ that is continuous on $(0,1)^k$.
I do not know if the result is true for $k<n$.
During the editorial corrections one of the results in the paper (the Homeomorphic Measures Theorem) has been stated incorrectly; the erratum is available at https://sites.google.com/view/piotr-hajasz/research/publications?authuser=0
[1] P. Hajłasz, P. Strzelecki, How to measure volume with a thread. Amer. Math. Monthly 112 (2005), no. 2, 176–179.
answered Jul 31, 2021 at 21:02