# Explicit isomorphism between $L^2(\mathbb{R}^2)$ and $L^2(\mathbb{R})$?

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.

• 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 19:34
• Pullback by the Peano curve gives an isomorphism between $L^2([0,1]^2)$ and $L^2([0,1])$. Jul 31, 2021 at 19:42
• 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 20:45
• @TerryTao My answer is related to yours, but has some additional features. Jul 31, 2021 at 21:03
• 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 5:41

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.