Sobolev spaces and $H^s(\mathbb R^n)$

Question

Let $u \in H^s(\mathbb R^n)$ with $s>n/2$ , show that $\lim\limits _{x\to \infty } u(x) =0$

What I tried to do is to use the definition of $H^s(\mathbb R^n)$ and I proved that $\hat u \in L^2(\mathbb R^n)$

ANyone can help me to finish the proof ?

Answer 1

Show that $\sup_x \vert v(x) \vert \le C \Vert v \Vert_{H^s}$ for all Schwartz functions $v$. This proves that $H^S \hookrightarrow C_0$, the space of continuous functions vanishing at $\infty$, but you can also make an explicit argument:

For $u\in H^s$ take a sequence $v_n$of Schwartz functions with $v_n\rightarrow u$ in $H^s$. By the inequality $(v_n)$ is Cauchy in $C_0$, thus has a limit $v\in C_0$. Both modes of convergence ($H^s$ and $C_0$) are stronger than convergence in $\mathcal{S}'$ (tempered distributions), thus also $v_n\rightarrow u$ and $v_n\rightarrow v$ in $\mathcal{S}'$, hence $u=v\in C_0$.