I recall that the Fourier transform of a function $f \in L^1 (\mathbb{R})$ is defined by $$\hat{f}(\xi) = \frac{1}{\sqrt{2 \pi}} \int_{\mathbb{R}} f(x) e^{- i x \xi} \, dx.$$ We can define that Sobolev space $H^s (\mathbb{R})$ as the set of functions for which the integral $$ \| f \|^2 _{H^s} = \int (1 + | \xi |^2 )^s | \hat{f} ( \xi)|^2 \, d \xi $$is finite. I would like to prove that for every $f \in H^s (\mathbb{R})$ and $s > 1/2$, there exists a constant $c_s$ such that $$\sup_x |f(x)| \le c_s \| f \|_{H^s}.$$
My attempt fails in two points, but maybe it is somehow "fixable": $$\begin{split} |f(x) - f(y)|^2 & = \left| \int_y^x f'(\xi) \, d\xi \right|^2 \le \int_{\mathbb{R}} |f'(\xi)|^2 \, d\xi \\ & = \int_{\mathbb{R}} |\xi|^2 |\hat{f}(\xi)|^2 \, d\xi \\ & \le \int_{\mathbb{R}} (1 + |\xi|^2)|\hat{f}(\xi)|^2 \, d\xi \end{split}$$where I used the equality between the $L^2$-norm of a function and the $L^2$-norm of its Fourier transform (corollary of Plancherel Theorem?), and the fact that $\hat{f}'(\xi) = i \xi \hat{f} ( \xi)$.
My idea here was to write $(1 + |\xi|^2) = (1 + |\xi|^2)^s (1 + |\xi|^2)^{1-s}$ in order to get the right form and the constant $c_s$, but the function $f(\xi) = (1+|\xi|^2)^{1-s}$ is bounded from above just for $s \ge 1$ and I don't know how to solve the case $1/2 < s < 1$. On the other hand, at the beginning I have $ |f(x) - f(y)|$ and not $|f(x)|$, so I am probably wrong somewhere.
Any hint to fix/idea to solve?
Thanks in advance!
Well, that way was wrong. I think I've found a solution: by the Fourier inversion formula and Cauchy-Schwarz inequality we have $$\begin{split} |f(x)| & = \left |\int_{\mathbb{R}} \hat{f} (\xi) e^{i x \xi} \, d \xi \right| = \left |\int_{\mathbb{R}} \hat{f} (\xi) e^{i x \xi} (1 + |\xi|^2)^{s/2} (1 + |\xi|^2)^{-s/2} \, d \xi \right| \\ & \le \left[ \int_{\mathbb{R}} |\hat{f}(\xi)|^2 (1 + |\xi|^2)^s \, d\xi \right]^{1/2} \left[ \int_{\mathbb{R}} (1 + |\xi|^2)^{-s} \, d \xi\right]^{1/2}\end{split}$$and if $ s > 1/2$, $$c_s=\int_{\mathbb{R}} (1 + |\xi|^2)^{-s} \, d \xi < +\infty.$$ Taking the supremum on the LHS we are done.