Given that the Soloblev space norm $$\|f\|_{H^s}^2=|\widehat{f}(0)|^2+\sum_{n\in\mathbb{Z}}|n|^{2s}|\widehat{f}(n)|^2.$$ and the inequality $$\|f(\cdot +\theta)-f\|_{L^2}\leq 2\pi \|f\|_{H^s}|\theta|^s. $$ How do you show that $\|S_Nf-f\|_{L^2}$ converges at rate of $N$ alone, in fact, $|S_Nf-f|_2=O(N^{-s})$. So here is what I have $$\left\|\sum_{n=-N}^N\widehat{f}(n)e^{2\pi xni}-f(x)\right\|_2$$ From here, how do you modify the inequality to obtain something like $$\|f(x+N^{-2s})-f(x)\|_{L^2}\leq2\pi |N|^{-s}$$
Thanks
By Parseval's identity
$$\left\|\sum_{n=-N}^N\widehat{f}(n)e^{2\pi xni}-f(x)\right\|_2^2 = 2\pi \sum_{|n| > N}|\widehat{f}(n)|^2.$$ $$= 2\pi N^{-2s} \sum_{|n| > N}|N^{s}\widehat{f}(n)|^2$$ $$\leq 2\pi N^{-2s} \sum_{|n| > N}|n^s\widehat{f}(n)|^2$$ $$\leq 2\pi N^{-2s}||f||_s^2$$ So $\|S_Nf-f\|_{L^2} \leq \sqrt{2\pi} N^{-s}||f||_s$ which gives you what you need. Note that actually $\|S_Nf-f\|_{L^2} = o(N^{-s})$ as $N \rightarrow \infty$.