For which $s\in\mathbb R$, is $H^s(\mathbb T)$ a Banach algebra?

Solution 1:

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 $s>k/2$.

Solution 2:

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.

Solution 3:

Please see this question and the accepted answer. The proof works the same in the torus case.