Sobolev space $H^s(\mathbb{R}^n)$ is an algebra with $2s>n$ How do you prove that the Sobolev space $H^s(\mathbb{R}^n)$ is an algebra if $s>\frac{n}{2}$, i.e. if $u,v$ are in $H^s(\mathbb{R}^n)$, then so is $uv$? Actually I think we should also have $\lVert uv\rVert_s \leq C \lVert u\rVert_s \lVert v\rVert_s$. Recall that $\lVert f\rVert_s=\lVert(1+|\eta|^2)^{s/2}\,\hat{f}(\eta)\rVert$, the norm on $H^s(\mathbb{R}^n)$. This is an exercise from Taylor's book, Partial differential equations I. 
 A: One way to see this is by an argument similar to proof of a "trace theorem": first, for $f,g\in H^s(\mathbb R^n)$ with $s\ge 0$, $f\otimes g\in H^s(\mathbb R^{2n})$ because $1+|x|^2+|y|^2\le (1+|x|^2)(1+|y|^2)$. Next, prove an easy form of a "trace theorem", namely, that restriction from $\mathbb R^N$ to $\mathbb R^{N-n}$ maps $H^s$ to $H^{s-{n\over 2}}$ for $s> {n\over 2}$.
Edit: in response to the comments of @ElmarZander, ... The question, as originally posed, cannot be quite right, no. The argument sketched here shows that for $s>n/2$ the product of two elements in $H^s(\mathbb R^n)$ is in $H^{s-n/2-\varepsilon}$ for every $\epsilon>0$.
I do not know whether higher-dimensional results can be sharpened, but for $n=1$ it is easy to do explicit examples showing the limitation: take $\hat{f}=\hat{g}$ to be $|x|^{-3/4-\varepsilon}$ for $x\ge 1$ and $0$ otherwise. These are in $H^{1/2+\varepsilon'}(\mathbb R)$. Then the convolution has a lower bound $x^{-1/2-2\varepsilon}$, I believe, so $fg$ is not in $H^{1/2+\varepsilon'}$.
A: Note that
$$
\begin{split}
(1+|\xi|^2)^p
&\leq (1+2|\xi-\eta|^2+2|\eta|^2)^p\\
&\leq 2^p(1+|\xi-\eta|^2+1+|\eta|^2)^p\\
&\leq c(1+|\xi-\eta|^2)^p + c(1+|\eta|^2)^p,
\end{split}
$$
for $p>0$, where $c=\max\{2^{p},2^{2p-1}\}$. 
Put $\langle\xi\rangle=\sqrt{1+|\xi|^2}$.
Then we have
$$
\begin{split}
\langle\xi\rangle^s |\widehat{uv}(\xi)|
&\leq \int \langle\xi\rangle^s |\hat{u}(\xi-\eta)\hat{v}(\eta)|\,\mathrm{d}\eta\\
&\leq c\int \langle\xi-\eta\rangle^s |\hat{u}(\xi-\eta)\hat{v}(\eta)|\,\mathrm{d}\eta
+ c\int \langle\eta\rangle^s |\hat{u}(\xi-\eta)\hat{v}(\eta)|\,\mathrm{d}\eta\\
&\leq c|\langle\cdot\rangle^s\hat u|*|\hat v| + c|\hat u|*|\langle\cdot\rangle^s\hat v|,
\end{split}
$$
which, in light of Young's inequality, implies
$$
\|uv\|_{H^s} \leq c\|u\|_{H^s} \|\hat v\|_{L^1} + c\|\hat u\|_{L^1}\|v\|_{H^s}.
$$
Finally, we note that $\|\hat u\|_{L^1}\leq C\,\|u\|_{H^s}$ when $s>\frac{n}2$.
