All Questions

The Hilbert transform on the real Hilbert space $L^2(\mathbb R)$ is the singular integral operator $$ \mathcal H(f)(x) := \frac{1}{\pi} \int_{-\infty}^\infty \frac{1}{x-y} f(y) dy. $$ It satisfies $\...

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 ...