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.

## 1 Answer

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.

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$2more comments