Singular Integrals and $L^1$

Question

Let us consider in one dimension the Fourier multiplier $\vert D\vert$ and the derivative $iD$. Both are well-defined on the Schwartz space $\mathscr S(\mathbb R)$ with the derivative sending $\mathscr S(\mathbb R)$ onto $\mathscr S_0(\mathbb R)$ (the Schwartz functions with mean 0). On the other hand $\vert D\vert$ sends $\mathscr S(\mathbb R)$ into $H^\infty$.

Question. I guess that the $L^1$ norm of $\vert D\vert u$ and $Du$ are not equivalent and even that $\vert D\vert u$ could be outside of $L^1$. Are there explicit examples?

Answer 1

We do have $v=|D|u\in L^1$ always, basically because $\widehat{|\xi|}$ decays like $1/x^2$, so the lack of smoothness of $\widehat{v}(\xi)$ near $\xi=0$ is not so serious after all.

More precisely, the Fourier transform of $|\xi|$ is a regularized (near $x=0$) version of $1/x^2$; another way of saying it is $\widehat{|\xi|}=c(PV(1/x))'$, or see the last part of my lecture notes here for more details. So $u*\widehat{|\xi|}$ decays like $1/x^2$ and in particular is integrable.

On the other hand, $|D|u$ and $Du$ differ by a Hilbert transform, which is indeed not bounded on $L^1$.