The Fr'echet--Kolmogorov compactness criterion in $L^p({\mathbb{R}}^n)$

If $\mathcal{F}=\mathrm{\varnothing }$ it is totally bounded and its closure is empty. Otherwise fix $\epsilon >0$. By (ii) choose $R>0$ with $\mathrm{\parallel }f-f_R{\mathrm{\parallel }}_p<\epsilon$ for every $f\in \mathcal{F}$, and by (iii) choose $\delta >0$ with $\mathrm{\parallel }{\tau }_{h}f-f{\mathrm{\parallel }}_p<\epsilon$ for every $f\in \mathcal{F}$ and every $\mathrm{\mid }h\mathrm{\mid }<\delta$; let $\eta :={\eta }_{\delta \mathrm{/}2}$ be the radial mollifier at scale $\delta \mathrm{/}2$, so $\int \eta =1$, $\eta \ge 0$ and $\mathrm{supp}\eta \subseteq B(0,\delta \mathrm{/}2)$.

For every $f\in \mathcal{F}$ and every $\mathrm{\mid }y\mathrm{\mid }<\delta$, $\mathrm{\parallel }{f}_R-{\tau }_y{f}_R{\mathrm{\parallel }}_p\le \mathrm{\parallel }{f}_R-f{\mathrm{\parallel }}_p+\mathrm{\parallel }f-{\tau }_{y}f{\mathrm{\parallel }}_p+\mathrm{\parallel }{\tau }_{y}f-{\tau }_y{f}_R{\mathrm{\parallel }}_p<3\epsilon$ by step 1.1 and [F4]. Since $f_R-f_R\ast \eta =\int \eta (y)(f_R-{\tau }_y{f}_R)dy$, [F3] gives $\mathrm{\parallel }{f}_R-f_R\ast \eta {\mathrm{\parallel }}_p\le \int \eta (y)\mathrm{\parallel }{f}_R-{\tau }_y{f}_R{\mathrm{\parallel }}_{p}dy<3\epsilon$, and therefore $\mathrm{\parallel }f-f_R\ast \eta {\mathrm{\parallel }}_p<4\epsilon$.

By [F1] and [F2], every $g=f_R\ast \eta$ satisfies $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le M\mathrm{\parallel }\eta {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$ and $\mathrm{\parallel }\mathrm{\nabla }g{\mathrm{\parallel }}_{\mathrm{\infty }}\le M\mathrm{\parallel }\mathrm{\nabla }\eta {\mathrm{\parallel }}_{p^{\mathrm{\prime }}}$, and $g$ vanishes off the ball of radius $R+\delta$ because $f_R$ vanishes off the ball of radius $R$; hence the family $\mathcal{G}:=\{f_R\ast \eta :f\in \mathcal{F}\}$ is uniformly bounded and uniformly Lipschitz, and all its elements are supported in the compact ball $K:=\overline{B(0,R+\delta )}$.

The real parts $\{\mathrm{Re}g:g\in \mathcal{G}\}$ are equicontinuous and pointwise bounded on the compact metric space $K$, and likewise the imaginary parts; by [F5] both families are totally bounded in the supremum metric, and combining the finitely many real and imaginary sup-balls, $\mathcal{G}$ is totally bounded in the supremum metric over $K$. Since every $g$ is supported in $K$, the supremum over ${\mathbb{R}}^n$ equals the supremum over $K$, so for every $\vartheta >0$ the family $\mathcal{G}$ has a finite covering by $L^p$-balls of radius $\vartheta$ by [F6].

Let $r>0$ and apply steps 1.1--3.1 with $\epsilon <r\mathrm{/}10$, covering $\mathcal{G}$ by finitely many $L^p$-balls of radius $r\mathrm{/}10$. Step 2.1 then covers $\mathcal{F}$ by the same centres with radius $r\mathrm{/}2$. Discard empty intersections with $\mathcal{F}$ and choose one point of $\mathcal{F}$ in each remaining ball. Their radius-$r$ balls cover $\mathcal{F}$ by the triangle inequality, so the centres belong to $\mathcal{F}$ as required by [F6]. Thus $\mathcal{F}$ is totally bounded.

By [F6] the closure $\bar{\mathcal{F}}$ is totally bounded, and it is closed in the complete space $L^p({\mathbb{R}}^n)$, hence complete by [F7]; a complete and totally bounded metric space is compact by [F7], and compactness is equivalent to sequential compactness by [F7], so every sequence in $\mathcal{F}$ has a subsequence converging in $L^p({\mathbb{R}}^n)$. The empty case was disposed of in step 1.1, and no other choice principle is used: Countable Choice covers the completeness-to-compactness step and the finite-net selection, Dependent Choice covers the compactness-sequential equivalence.