Let \(f,g:\mathbb R\to\mathbb R\) be continuous, suppose \[ \lim_{|x|\to\infty}f(x)=0, \qquad \int_{-\infty}^\infty |g(y)|\,dy<\infty, \] and define \[ h(x)=\int_{-\infty}^\infty f(x-y)g(y)\,dy. \] Prove that \[ \lim_{|x|\to\infty}h(x)=0. \]
Solution
Because \(f\) is continuous and tends to \(0\) at both ends, it is bounded; write \[ M=\|f\|_\infty<\infty. \] Fix \(\varepsilon>0\). Choose \(R>0\) so large that \[ \int_{|y|>R}|g(y)|\,dy<\frac{\varepsilon}{2M} \] when \(M>0\); if \(M=0\), then \(f\equiv0\) and there is nothing to prove.
Since \(f(x)\to0\) as \(|x|\to\infty\), there exists \(A>0\) such that \[ |f(z)|<\frac{\varepsilon}{2\|g\|_1} \qquad(|z|>A) \] when \(\|g\|_1>0\); if \(g=0\) a.e., the conclusion is immediate.
Now take \(|x|>A+R\). For every \(|y|\le R\) one has \[ |x-y|\ge |x|-|y|>A, \] so \[ \begin{aligned} |h(x)| &\le \int_{|y|\le R}|f(x-y)|\,|g(y)|\,dy +\int_{|y|>R}|f(x-y)|\,|g(y)|\,dy\\ &\le \frac{\varepsilon}{2\|g\|_1}\int_{|y|\le R}|g(y)|\,dy +M\int_{|y|>R}|g(y)|\,dy\\ &<\frac\varepsilon2+\frac\varepsilon2 =\varepsilon. \end{aligned} \] Hence \(h(x)\to0\) as \(|x|\to\infty\).