A compactness argument for small high frequencies

Question

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*}

Answer 1

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.