For $1\le p<\infty$, $\tau_h f\to f$ in $L^p$ as $h\to0$; a function is uniformly continuous exactly when $\tau_h f\to f$ in the sup norm.Suggest a slogan
For \(h\in\RR^n\) and \(f\colon\RR^n\to\CC\), let \((\tau_h f)(x)\coloneqq f(x-h)\).
If \(1\leq p<\infty\) and \(f\in L^p(\RR^n)\), then \(\norm{\tau_h f-f}_{L^p}\to0\) as \(h\to0\) [@Fol13].
A function \(f\colon\RR^n\to\CC\) is uniformly continuous if and only if \(\sup_{x\in\RR^n}\abs{f(x-h)-f(x)}\to0\) as \(h\to0\).
Part (a) fails for \(p=\infty\): for \(f\coloneqq\chi_{[0,\infty)}\in L^\infty(\RR)\) and \(h\neq0\), \(\tau_h f-f=\pm\chi_I\) for the half-open interval \(I\) between \(0\) and \(h\), so \(\norm{\tau_h f-f}_{L^\infty}=1\).