Convolution of Sobolev distributions ($u\ast v={\mathcal{F}}^{-1}(\hat{u}\hat{v})$)

For arbitrary real $s,t$, the convolution of $u\in H^s(\mathbb{R})$ and $v\in H^t(\mathbb{R})$ has a canonical definition as a tempered distribution. With $⟨\xi ⟩=(1+{\xi }^2{)}^{1/2}$, write $A=⟨\xi {⟩}^s\hat{u}$ and $B=⟨\xi {⟩}^t\hat{v}$, both in the L2 space. Then $AB\in L^1$ by Cauchy-Schwarz inequality, andThus the product defines a tempered distribution, and the displayed inverse Fourier transform defines $u\ast v$. This extends ordinary convolution of Schwartz functions by the convolution theorem and density in the Sobolev norm.

The result need not belong to any Sobolev space, even when both inputs belong to every Sobolev space. For $\hat{u}=\hat{v}={\xi }^{-1/3}{1}_{(0,1)}(\xi )$, each weighted squared input is integrable, but the squared product is ${\xi }^{-4/3}{1}_{(0,1)}(\xi )$, which is not locally integrable at zero. Every weight $(1+{\xi }^2{)}^r$ is bounded above and below there, proving the failure for all real $r$.