A compactness argument for small high frequencies I would like to prove the following statement:
Let $N\geq 1$, $1\leq q<\infty$ and let be $E$ a relatively compact subset of $L^q(\mathbb{R}^N)$. Then
\begin{equation*}
\sup_{u\in E}\int_{|x|>R}|u|^q\overset{R\rightarrow\infty}{\longrightarrow}0.
\end{equation*}

If $E$ were finite, the problem would simply consist in noticing that $\lim_{R\to\infty}\int_{\{|x|\gt R}|g(x)|\mathrm dx=0$ if $g$ is an integrable function. 
As a relatively compact family behave "up to $\varepsilon$" as a finite family, we can use the following strategy.
Fix $\varepsilon\gt 0$. There is an integer $n\gt 0$ and $u_1,\dots,u_n\in E$ such that $E\subset\bigcup_{j=1}^nB_{\mathbb L^q}(u_j,\varepsilon)$. Therefore, for each $u\in E$, 
$$\int_{\{|x|\gt R\}}|u|^q\mathrm dx\leqslant \varepsilon^q+\max_{1\leqslant j\leqslant n}\int_{\{|x|\gt R\}}|u_j|^q,$$
hence 
$$\sup_{u\in E}\int_{\{|x|\gt R\}}|u|^q\mathrm dx\leqslant \varepsilon^q+\max_{1\leqslant j\leqslant n}\int_{\{|x|\gt R\}}|u_j|^q.$$
Since $n$ is fixed, using the remark in the first sentence, 
$$\limsup_{R\to\infty}\sup_{u\in E}\int_{\{|x|\gt R\}}|u|^q\mathrm dx\leqslant \varepsilon^q$$
and we can conclude as $\varepsilon$ was arbitrary.
