Let \(f\in L^1(\mathbb R)\). Prove that \[ \lim_{h\to0}\int_{\mathbb R}|f(x+h)-f(x)|\,dx=0. \]
Solution
Fix \(\varepsilon>0\). Choose \(g\in C_c(\mathbb R)\) such that \[ \|f-g\|_1<\frac{\varepsilon}{3}, \] using density of \(C_c(\mathbb R)\) in \(L^1(\mathbb R)\).
Because \(g\) is continuous with compact support, it is uniformly continuous. If \(K\) is a compact interval containing the support of \(g\) and all its sufficiently small translates, then for small \(h\) one has \[ \|g(\cdot+h)-g\|_1 \le m(K)\,\sup_{x\in K}|g(x+h)-g(x)| \longrightarrow0. \] Hence there exists \(\delta>0\) such that \(|h|<\delta\) implies \[ \|g(\cdot+h)-g\|_1<\frac{\varepsilon}{3}. \]
Translations preserve Lebesgue measure, so \[ \|f(\cdot+h)-g(\cdot+h)\|_1=\|f-g\|_1. \] Therefore, for \(|h|<\delta\), \[ \begin{aligned} \|f(\cdot+h)-f\|_1 &\le \|f(\cdot+h)-g(\cdot+h)\|_1 +\|g(\cdot+h)-g\|_1 +\|g-f\|_1\\ &<\varepsilon. \end{aligned} \] Thus \[ \|f(\cdot+h)-f\|_1\longrightarrow0 \qquad(h\to0). \]