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edited Jun 10, 2014 at 3:48

This follows from two facts:

$(H^s \cap L^\infty) (\mathbb{T})$ is a Banach algebra (see for example in the framework of fractional spaces $W^{s, p}$ ($W^{s, 2} = H^s$) Bourgain, Brezis, Mironescu, Lifting in Sobolev spaces, 2000),
if $s > \frac{1}{2}$, then $H^{s} (\mathbb{T}) \subset L^\infty (\mathbb{T})$: if $u \in H^s (\mathbb{T})$ and $t \in \mathbb{T}$, then $$\vert u (t) \vert \le \sum_{k \in \mathbb{Z}} \vert c_k (u)\vert \le \Big(\sum_{k \in \mathbb{Z}} \frac{1}{(k^2 + 1)^s}\Big)^\frac{1}{2}\Big(\sum_{k \in \mathbb{Z}} \lvert c_k (u) \vert^2 (k^2 + 1)^s\Big)^\frac{1}{2} = C \Vert u \Vert_{H^s},$$ where $c_k (u)$ is the $k$-th Fourier coefficient of the function $u$. In general $W^{s, p} (M)$ is continuously embedded into $L^\infty (M)$ if $sp > \dim M$ (see for example Adams, Sobolev spaces, Academic Press, 1975, theorem 7.57).

More generally $W^{s, p} (M)$ is a Banach algebra if $sp > \dim M$.

This follows from two facts:

$(H^s \cap L^\infty) (\mathbb{T})$ is a Banach algebra (see for example in the framework of fractional spaces $W^{s, p}$ ($W^{s, 2} = H^s$) Bourgain, Brezis, Mironescu, Lifting in Sobolev spaces, 2000),
if $s > \frac{1}{2}$, then $H^{s} (\mathbb{T}) \subset L^\infty (\mathbb{T})$: if $u \in H^s (\mathbb{T})$ and $t \in \mathbb{T}$, then $$\vert u (t) \vert \le \sum_{k \in \mathbb{Z}} \vert c_k (u)\vert \le \Big(\sum_{k \in \mathbb{Z}} \frac{1}{(k^2 + 1)^s}\Big)^\frac{1}{2}\Big(\sum_{k \in \mathbb{Z}} \lvert c_k (u) \vert^2 (k^2 + 1)^s\Big)^\frac{1}{2} = C \Vert u \Vert_{H^s},$$ where $c_k (u)$ is the $k$-th Fourier coefficient of the function $u$.

More generally $W^{s, p} (M)$ is a Banach algebra if $sp > \dim M$.

This follows from two facts:

$(H^s \cap L^\infty) (\mathbb{T})$ is a Banach algebra (see for example in the framework of fractional spaces $W^{s, p}$ ($W^{s, 2} = H^s$) Bourgain, Brezis, Mironescu, Lifting in Sobolev spaces, 2000),
if $s > \frac{1}{2}$, then $H^{s} (\mathbb{T}) \subset L^\infty (\mathbb{T})$: if $u \in H^s (\mathbb{T})$ and $t \in \mathbb{T}$, then $$\vert u (t) \vert \le \sum_{k \in \mathbb{Z}} \vert c_k (u)\vert \le \Big(\sum_{k \in \mathbb{Z}} \frac{1}{(k^2 + 1)^s}\Big)^\frac{1}{2}\Big(\sum_{k \in \mathbb{Z}} \lvert c_k (u) \vert^2 (k^2 + 1)^s\Big)^\frac{1}{2} = C \Vert u \Vert_{H^s},$$ where $c_k (u)$ is the $k$-th Fourier coefficient of the function $u$. In general $W^{s, p} (M)$ is continuously embedded into $L^\infty (M)$ if $sp > \dim M$ (see for example Adams, Sobolev spaces, Academic Press, 1975, theorem 7.57).

More generally $W^{s, p} (M)$ is a Banach algebra if $sp > \dim M$.

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created Jun 6, 2014 at 13:12

This follows from two facts:

$(H^s \cap L^\infty) (\mathbb{T})$ is a Banach algebra (see for example in the framework of fractional spaces $W^{s, p}$ ($W^{s, 2} = H^s$) Bourgain, Brezis, Mironescu, Lifting in Sobolev spaces, 2000),
if $s > \frac{1}{2}$, then $H^{s} (\mathbb{T}) \subset L^\infty (\mathbb{T})$: if $u \in H^s (\mathbb{T})$ and $t \in \mathbb{T}$, then $$\vert u (t) \vert \le \sum_{k \in \mathbb{Z}} \vert c_k (u)\vert \le \Big(\sum_{k \in \mathbb{Z}} \frac{1}{(k^2 + 1)^s}\Big)^\frac{1}{2}\Big(\sum_{k \in \mathbb{Z}} \lvert c_k (u) \vert^2 (k^2 + 1)^s\Big)^\frac{1}{2} = C \Vert u \Vert_{H^s},$$ where $c_k (u)$ is the $k$-th Fourier coefficient of the function $u$.

More generally $W^{s, p} (M)$ is a Banach algebra if $sp > \dim M$.