Relation between $H^s$ for $0 \leq s \leq 1$ and Fourier decay

Question

My convention for the Fourier transform is $\hat f(n) = \int_0^1 e^{-2 \pi i n x} f(x) dx$. For any $s \in \mathbb{R}$, define $$ \lVert f \rVert_{H^s}^2 = \lvert \hat f(0) \rvert^2 + \sum \lvert n \rvert^{2s}\lvert \hat f(n) \rvert^2.$$ I want to show that for $0 \leq s \leq 1$, $$\lVert f(x+h) - f(x) \rVert_2^2 \leq 4\pi^2 \lVert f \rVert_{H^s}^2 \lvert h \rvert^{2s}.$$ I used Parseval on the left-hand side to obtain \begin{align*} \lVert f(x+h) - f(x) \rVert_2^2 &= \sum \lvert \hat f(n)e^{2\pi i n h} - \hat f(n)\rvert^2 \\ &= \sum \lvert \hat f(n) \rvert^2 \lvert e^{2\pi i n h} - 1 \rvert^2 \\ &\leq \sum \lvert \hat f(n) \rvert^2 \lvert 2\pi i n h \rvert^2 \\ &= 4\pi^2 \sum \lvert \hat f(n) \rvert^2 \lvert nh \rvert^2 \end{align*} This is almost what I want, but I can't figure out the right Holder conjugates to conclude.

Answer 1

The inequality $\lvert e^{2\pi i n h} - 1 \rvert \le \lvert 2\pi i n h \rvert$ is too wasteful when $|nh|$ is large. The left hand side is never greater than $2$. So, for $|nh|>1$ or so, you are better off using $2$ as an upper bound. This results in the sum split in two: $$\sum_{|nh|\le 1} |\hat f(n)|^2 |2\pi i n h|^2 + 4 \sum_{|nh|> 1} |\hat f(n)|^2$$ In the first sum $S_1$, reduce the exponent of $nh$ to $2s$ by using $|nh|\le 1$: $$S_1\le 4\pi^2 \sum_{|nh|\le 1} |\hat f(n)|^2 |nh|^{2s} = 4\pi^2 h^{2s}\sum_{|nh|\le 1} |\hat f(n)|^2 |n|^{2s}$$ In the second, do the opposite: bring $|nh|^{2s}$ in: $$S_2 \le 4 \sum_{|nh|> 1} |\hat f(n)|^2|nh|^{2s} = 4h^{2s}\sum_{|nh|> 1} |\hat f(n)|^2|n|^{2s}$$ Finally, combine the sums: $$S_1+S_2\le 4\pi^2 h^{2s}\sum_{n\in\mathbb{Z}} |\hat f(n)|^2|n|^{2s}$$