Sobolev space $H^s(\mathbb{R}^n)$ is an algebra with $2s>n$

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 $s>\frac{n}2$.

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 $H^{1/2+\varepsilon'}$.