For which $s\in\mathbb R$, is $H^s(\mathbb T)$ a Banach algebra? According to Theorem 4.39 in Adams & Fournier Sobolev Spaces: 
If $mp>n$, $m\in\mathbb N$, then $W^{m,p}(\Omega)$ is a Banach algebra $($i.e., if $u,v\in W^{m,p}(\Omega)$, then $uv\in W^{m,p}(\Omega)$ as well$)$, provided that $\Omega\subset\mathbb R^n$ satisfies the cone condition.
Clearly, $(0,2\pi)$ does satisfy the cone condition. However, does the above Theorem 4.39 apply for $m$ real? 
In particular, is it true that $H^s(\mathbb T)$ is a Banach algebra, for $s>1/2$ and there is a $c>0$, such that
$$
\|uv\|_{H^s} \le c\,\|u\|_{H^s}\|v\|_{H^s},
$$
for all $u,v \in H^s(\mathbb T)$?
 A: I provide a detailed answer to my question:
Proposition.  If $s>1/2$, then there exists a $c_s>0$, such that for every 
$u,\upsilon\in H^s(\mathbb T)$
\begin{equation*}
\| u\upsilon\|_{H^s} \,\le\, c_s\, \|u\|_{H^s} \|\upsilon\|_{H^s},
\end{equation*} 
Proof. If $s>1/2$, then 
$H^s(\mathbb T)\subset L^\infty(\mathbb T)$ and for every 
$u\in H^s(\mathbb T)$,
$$
\|u\|_{L^\infty}\le c\|u\|_{H^s},
$$
where $c>0$ does not depend on $u$. The
proposition is now a corollary of the following result: 
Lemma.  Let $s\ge 0$, then there exists
a constant $c_s>0$, such that 
\begin{equation*}
\| u\upsilon\|_{H^s} \,\le\, 
c_s\, \big( \|u\|_{H^s}\|\upsilon\|_{L^\infty}+\|u\|_{L^\infty}\|\upsilon\|_{H^s}\big)
\end{equation*}
for every $u,\upsilon \in H^s(\mathbb T)\cap L^\infty(\mathbb T)$.
Proof. First note that the $H^s-$norm is defined as
$\|w\|_{H^s}^2=\sum_{k\in\mathbb Z}(1+k^2)^s|\widehat{w}_k|^2$, for
$w(x)=\sum_{k\in\mathbb Z}\widehat{w}_k\mathrm{e}^{\mathrm{i}kx}$; in particular,
for $s=0$, the $H^s-$norm coincides with the $L^2-$norm. 
Using the inequality
$$
\big(1+(x+y)^2\big)^{s/2} \le c\,\big( (1+x^2)^{s/2}+(1+y^2)^{s/2}\big),
$$
where $x,y\in\mathbb R$ and $c=\max\{2^{s/2},2^{s-1}\}$, we obtain
\begin{align*}
(1+k^2)^{s/2} \big|\widehat{u\upsilon}(k)\big| \,=\,& (1+k^2)^{s/2}
\Big|\sum_{\ell\in\mathbb Z} \widehat{u}(k-\ell)\widehat{\upsilon}(\ell)\,\Big| \\
\,\le\, & c\sum_{\ell\in\mathbb Z} \big( (1+(k-\ell)^2)^{s/2}+(1+\ell^2)^{s/2}\big)
\lvert\widehat{u}(k-\ell)\rvert\lvert\widehat{\upsilon}(\ell)\rvert \\
\,=\, &  c\sum_{\ell\in\mathbb Z}  (1+(k-\ell)^2)^{s/2}
\lvert\widehat{u}(k-\ell)\rvert\lvert\widehat{\upsilon}(\ell)\rvert
 +c\sum_{\ell\in\mathbb Z}  (1+\ell^2)^{s/2}
\lvert\widehat{u}(k-\ell)\rvert\lvert\widehat{\upsilon}(\ell)\rvert.
\end{align*}
Therefore
\begin{align*}
(1\!+\!k^2)^{s} \big|\widehat{u\upsilon}(k)\big|^2 \le&   2c^2
\Big(\!
\sum_{\ell\in\mathbb Z} 
(1\!+\!(k\!-\!\ell)^2)^{s/2}
\lvert\widehat{u}(k\!-\!\ell)\rvert \lvert\widehat{\upsilon}(\ell)\rvert 
\Big)^{\!2} \\ &+2c^2\Big(\!
\sum_{\ell\in\mathbb Z} 
(1\!+\!\ell^2)^{s/2}\lvert\widehat{u}(k\!-\!\ell)\rvert
\lvert\widehat{\upsilon}(\ell)\rvert 
\Big)^{\!2}. \qquad\qquad\qquad\qquad\qquad (\star)
\end{align*}
If we set
$$
f(x)=\!\sum_{k\in\mathbb Z} \lvert \widehat{u}_k\rvert\mathrm{e}^{\mathrm{i}kx},\,
g(x)=\!\sum_{k\in\mathbb Z} \lvert \widehat{\upsilon}_k\rvert\mathrm{e}^{\mathrm{i}kx},\,
F(x)=\!\sum_{k\in\mathbb Z}  (1\!+\!k^2)^{s/2}\lvert\widehat{u}_k\rvert\mathrm{e}^{\mathrm{i}kx},\,
G(x)=\!\sum_{k\in\mathbb Z}  (1\!+\!k^2)^{s/2}\lvert\widehat{\upsilon}_k\rvert\mathrm{e}^{\mathrm{i}kx},
$$
then $(\star)$ provides that
\begin{align*}
(1+k^2)^{s} \big|\widehat{u\upsilon}(k)\big|^2 \le   2c^2
\Big(   \big\lvert\widehat{fG}(k)\big\rvert^2+\big\lvert\widehat{Fg}(k)\big\rvert^2\Big),
\end{align*}
and hence
\begin{align*}
\|u\upsilon\|_{H^s}^2&=\sum_{k\in\mathbb Z}(1+k^2)^{s} \big|\widehat{u\upsilon}(k)\big|^2 
\le   2c^2
\sum_{k\in\mathbb Z}\Big(   \big\lvert\widehat{fG}(k)\big\rvert^2
+\big\lvert\widehat{Fg}(k)\big\rvert^2\Big)
=2c^2\Big(\|fG\|_{L^2}^2+\|Fg\|_{L^2}^2\Big) \\
&\le 2c^2\big(\|f\|_{L^\infty}^2\|G\|_{L^2}^2+\|F\|_{L^2}^2\|g\|_{L^\infty}^2\big)
=2c^2\big(\|u\|_{L^\infty}^2\|\upsilon\|_{H^s}^2+\|u\|_{H^s}^2\|\upsilon\|_{L^\infty}^2\big),
\end{align*}
which concludes the proof. $\qquad$ Ὅ.Ἔ.Δ.
Note. An almost identical proof provides that $\,H^s(\mathbb T^k)\,$ is a Banach algebra for $s>k/2$. 
A: The answer is yes. As long as a Sobolev space embeds into $C^0$, it forms an algebra. This boils down to  $$|f^2(x)-f^2(y)| \le 2\sup|f|\, |f(x)-f(y)|\tag1 $$ Indeed, the norm of  $H^s(\mathbb T)$, $0<s<1$, has an equivalent expression
$$
\|f\|_{H^s}^2 \approx \iint_{\mathbb T\times\mathbb T}\frac{|f(x)-f(y)|^2}{|x-y|^{1+2s}}\,dx\,dy \tag2
$$
From (1) and (2) it follows that $f^2\in H^s$ whenever  $f\in H^s$. Hence $H^s$ is closed under multiplication in this case.  
When $k<s<k+1$ for some integer $k\ge 1$, reason inductively using the derivatives of $f$. For example, suppose $1<s<2$. Since $(f^2)' = 2ff'$ and   $f,f'\in H^{s-1}$, it follows from the above that $(f^2)' \in H^{s-1}$. Hence $f^2\in H^s$.
When $s=k$ is an integer, expand $(f^2)^{k}$ in terms of the derivatives of $k$; all terms are in $L^2$.

The fact that $H^s(\mathbb T)$ embeds into $C^0$ is a form of Morrey-Sobolev embedding. It can be proved most easily from the Fourier series definition of $H^s$: 
$$\sum_n (1+ n^2 )^{s} |\hat f(n)|^2<\infty$$
which by Cauchy-Schwarz implies $\sum_n   |\hat f(n)|<\infty $, hence the Fourier series converges uniformly, and $f$ is continuous.
A: Please see this question and the accepted answer. The proof works the same in the torus case.
