Continuous compactly supported functions are translation-continuous in $L^p$

If $n=0$, the only translation vector is zero, so the difference is identically zero. Assume $n\ge 1$. Let $K=\mathrm{supp}(f)$ and choose $R>0$ with $K\subseteq B(0,R)$. For $\mathrm{\mid }h\mathrm{\mid }\le 1$, the support of ${\tau }_{h}f-f$ lies in $\overline{B(0,R+1)}$. Apply [L1] to $f$ on the larger compact ball $\overline{B(0,R+2)}$; this includes both $x$ and $x-h$ whenever $x\in \overline{B(0,R+1)}$ and $\mathrm{\mid }h\mathrm{\mid }\le 1$.

Let $\epsilon >0$. Uniform continuity in step 1.1 gives $0<\delta \le 1$ such that $\mathrm{\mid }f(x-h)-f(x)\mathrm{\mid }<\epsilon$ for $\mathrm{\mid }h\mathrm{\mid }<\delta$ and every $x\in \overline{B(0,R+1)}$. Outside that ball both terms vanish by the support bound. Translation invariance preserves measurability; the compact ball has finite measure $V$ by [L2]. Thus $\mathrm{\parallel }{\tau }_{h}f-f{\mathrm{\parallel }}_p^p\le {\epsilon }^{p}V\mathrm{.}$

Given any target $\eta >0$, choose $\epsilon >0$ with ${\epsilon }^{p}V<{\eta }^p$ in step 2.1. Then $\mathrm{\mid }h\mathrm{\mid }<\delta$ implies $\mathrm{\parallel }{\tau }_{h}f-f{\mathrm{\parallel }}_p<\eta$. Hence $\mathrm{\parallel }{\tau }_{h}f-f{\mathrm{\parallel }}_p\to 0$ as $h\to 0$.