I have ran across the following theorem but the given proof does not convince me.
Theorem Let $u, v \in H^s(\mathbb{T})$ with $s>1/2$. Then the pointwise product $uv$ is in $H^s(\mathbb{T})$ and $\lVert uv\rVert_s \le C\lVert u\rVert_s\lVert v \rVert_s$ for a constant $C$ not depending on $u$ and $v$.
In functional terms, this theorem is telling us that $H^s(\mathbb{T})$ space is a Banach algebra when $s>1/2$.
Question Can you provide a reference for the proof of this theorem?
The proof which I was given begins writing
\begin{equation}\begin{split} \lVert uv\rVert_{H^s}&=\left\lVert \sum_{k_1+k_2=k}\langle k \rangle^s \hat{u}(k_1)\hat{v}(k_2)\right\rVert_{\ell^2}\\ &=\left\lVert \sum_{k_1+k_2=k}\frac{\langle k \rangle^s}{\langle k_1\rangle^s\langle k_2\rangle^s} \hat{u}(k_1)\langle k_1\rangle^s\hat{v}(k_2)\langle k_2\rangle^s\right\rVert_{\ell^2}\\ &=\left\lVert \sum_{k_1} A(k_1, k-k_1) \hat{f}(k_1)\hat{g}(k-k_1)\right\rVert_{\ell^2}, \end{split} \end{equation} where $A(k_1, k_2)=\frac{\langle k_1+k_2\rangle^s}{\langle k_1\rangle^s\langle k_2\rangle^s}$ and $\hat{f}(k_1)=\hat{u}(k_1)\langle k_1\rangle^s, \hat{g}(k_2)=\hat{v}(k_2)\langle k_2\rangle^s$. Then using triangle and Cauchy-Schwarz inequalites one proves that $$ \lVert uv\rVert_{s}\le \lVert A(k_1, k_2)\rVert_{\ell^\infty_{k_2}\ell^2_{k_1}}\lVert u\rVert_s\lVert v \rVert_s.$$ However, it seems to me that this proves nothing since $$\lVert A(k_1,k_2)\rVert_{\ell^\infty_{k_2}\ell^2_{k_1}}=\sup_{k_2}\lVert A(\cdot, k_2)\rVert_{\ell^2} \ge \lVert A(\cdot, 0)\rVert_{\ell^2}=\infty.$$
How can one repair this?
