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

Question

For $u\in H^s(\mathbb{R}^m)$ with $s>m/2$, show that $\lim_{x\to\infty}u(x)=0$

By the Sobolev embedding theorem $H^s(\mathbb{R}^m)\hookrightarrow C_b(\mathbb{R}^m)$ and this should pretty much be it? How would I proceed rigorously?

Answer 1

We actually use the fact that $C_c^\infty(R^N)$ is dense in $H^s(R^N)$ for each $s\in R$. (Or you could use $\mathcal{S}(R^N)$ instead of $C_c^\infty$. not really matters)

Anyhow, take $(u_n)\subset C_c^\infty$ be such that $u_n\to u$ in $H^s(R^N)$. Now, by embedding result, we have $$ \|u_n-u\|_{L^\infty(R^N)}\leq C\|u_n-u\|_{H^s(R^N)}\to 0 $$

Now fix arbitrary $\epsilon>0$ and find $n'$ so large such that $$ \|u_{n'}-u\|_{L^\infty(R^N)}<\epsilon $$ for all $n\geq n'$.

Because $u_{n'}\in C_c^\infty$, there exists $r(n')>0$ so large such that $u_{n'}(x)=0$ once $|x|\geq r(n')$. Hence a.e. $x\in R^N$ with $|x|\geq r(n')$, we have $$ |u(x)|=|u(x)-u_{n'}(x)|\leq \|u-u_n\|_{L^\infty}\leq\epsilon $$ Taking into the fact that $u$ is continuous, we done.