= Solution
Write $U=\widehat u$, $V=\widehat v$, and $\langle\xi\rangle=(1+\xi^2)^{1/2}$. By the <Sobolev space> hypotheses, $A=\langle\xi\rangle^sU$ and $B=\langle\xi\rangle^tV$ belong to the <L2 space>. Their product is an <integrable function> by <Cauchy-Schwarz inequality>. Thus $UV=\langle\xi\rangle^{-s-t}AB$ is a <locally integrable function> defining a <tempered distribution>, with estimate
$$
\left|\int UV\phi\,d\xi\right|
\leq\|u\|_{H^s}\|v\|_{H^t}
\sup_\xi\bigl(\langle\xi\rangle^{-s-t}|\phi(\xi)|\bigr).
$$
The supremum is controlled by finitely many <Schwartz seminorms>. Define the <convolution of Sobolev distributions> by
$$
\boxed{u*v:=\mathcal F^{-1}(UV)\in\mathcal S'(\mathbb R).}
$$
It is independent of the representatives chosen for $U,V$, and the <convolution theorem> shows that it agrees with ordinary <convolution> for <Schwartz functions>. More generally, approximate $u,v$ by <Schwartz functions> in their respective <Sobolev norms>; the displayed estimate shows convergence of the products in <tempered distributions>, so their ordinary <convolutions> converge to this definition. For $u,v\in L^1$ that also satisfy the stated <Sobolev space> hypotheses, the ordinary <convolution theorem> gives the same result.
There need not be any real $r$ with $u*v\in H^r(\mathbb R)$. Choose
$$
U(\xi)=V(\xi)=\xi^{-1/3}\mathbf1_{(0,1)}(\xi),\qquad
u=\mathcal F^{-1}U,\quad v=\mathcal F^{-1}V.
$$
Here $\mathbf1_{(0,1)}$ is an <indicator function>. Since $\int_0^1\xi^{-2/3}\,d\xi=3$ and every weight $(1+\xi^2)^q$ is bounded above and below by positive constants on $[0,1]$, both $u$ and $v$ belong to every <Sobolev space>. But their <convolution> has <Fourier transform> $UV=\xi^{-2/3}\mathbf1_{(0,1)}$, and for every real $r$,
$$
\int(1+\xi^2)^r|UV|^2\,d\xi
\geq\min(1,2^r)\int_0^1\xi^{-4/3}\,d\xi=\infty.
$$
Thus $\boxed{u,v\in\bigcap_{q\in\mathbb R}H^q(\mathbb R)\quad\text{but}\quad u*v\notin\bigcup_{r\in\mathbb R}H^r(\mathbb R)}$. The obstruction is failure of local square integrability at zero; a weight controlling large <frequencies> cannot fix it.
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