TITLE: If $u \in H^s(\mathbb{R}^n)$ for $s > n/2$, then $u \in L^\infty(\mathbb{R}^n)$?
QUESTION [8 upvotes]: How do I use the Fourier transform to see that if $u \in H^s(\mathbb{R}^n)$ for $s > n/2$, then $u \in L^\infty(\mathbb{R}^n)$, with the bound$$\|u\|_{L^\infty(\mathbb{R}^n)} \le C\|u\|_{H^s(\mathbb{R}^n)},$$the constant $C$ depending only on $s$ and $n$?

REPLY [5 votes]: I think it is worth mentioning a proof using the Fourier analytic definition of $H^s$, if only for its succinctness.
We have
$$
\| u \|_{L^\infty} \leq \| \hat{u} \|_1 \leq \| \langle \xi \rangle^{-s} \|_2 \| \langle \xi \rangle^s \hat{u} \|_2 \leq C \| u \|_{H^s}.$$
Here $\langle \xi \rangle = \sqrt{1 + \lvert \xi \rvert^2}$. Interpolation with $L^2$ implies
$$
\| u \|_p \leq C(s) \| u \|_{H^s} \quad \forall 2 \leq p \leq \infty.
$$