For each $f\in L^1(\RR^n)$, $\norm{\tau_hf-f}_1\to0$ as $h\to0$.Suggest a slogan
For \(h\in\RR^n\) and \(f\in L^1(\RR^n)\) let \((\tau_hf)(x)\coloneqq f(x-h)\). For every \(f\in L^1(\RR^n)\), \[ \lim_{h\to 0}\norm{\tau_h f - f}_1 = 0 . \]
Each \(\tau_h\) is a linear isometry of \(L^1(\RR^n)\), so the family \(\theset{\tau_h\suchthat h\in\RR^n}\) is equicontinuous as a family of maps \(L^1(\RR^n)\to L^1(\RR^n)\). Together with the pointwise convergence \(\tau_hf\to f\) of the theorem, this makes the convergence uniform on compact subsets \(K\subseteq L^1(\RR^n)\): \[ \lim_{h\to0}\sup_{f\in K}\norm{\tau_hf-f}_1=0 . \]