High frequencies destroy uniform translation control

By [F1] and [F4], $\mathrm{\parallel }{f}_j{\mathrm{\parallel }}_{L^p}^p=c_p^p{\int }_0^{2\pi }\mathrm{\mid }\sin (jx){\mathrm{\mid }}^{p}dx=1$, and all supports lie in the fixed bounded set $[0,2\pi ]$, so the family is bounded and tight.

Put $h_j:=\pi \mathrm{/}j$. For $x\in (h_j,2\pi )$ both $x$ and $x-h_j$ lie in $(0,2\pi )$, so [F2] gives $f_j(x-h_j)-f_j(x)=-2c_p\sin (jx)$ and hence $\mathrm{\mid }{f}_j(x-h_j)-f_j(x){\mathrm{\mid }}^p=2^p{c}_p^p\mathrm{\mid }\sin (jx){\mathrm{\mid }}^p$; integrating over $(h_j,2\pi )$ and using [F1] with the trivial bound ${\int }_0^{h_j}\mathrm{\mid }\sin (jx){\mathrm{\mid }}^{p}dx\le h_j$ gives $\mathrm{\parallel }{\tau }_{h_j}{f}_j-f_j{\mathrm{\parallel }}_{L^p}^p\ge 2^p{c}_p^p(c_p^{-p}-h_j)=2^p(1-c_p^p{h}_j)\to 2^p$.

Hence ${\mathrm{lim\hspace{0.17em}inf}}_j\mathrm{\parallel }{\tau }_{h_j}{f}_j-f_j{\mathrm{\parallel }}_{L^p}\ge 2>0$ although $h_j=\pi \mathrm{/}j\to 0$, so the family is not uniformly translation continuous; by [F3] it is not relatively compact in $L^p(\mathbb{R})$, and the refuted claim is false. Countable Choice is used by the scaling interface [F1] and the necessity lemma [F3].