Let \(f,g:\mathbb R\to\mathbb R\) be continuous and suppose \[ \lim_{|x|\to\infty}f(x)=0, \qquad \int_{-\infty}^{\infty}|g(x)|\,dx<\infty. \] Define \[ h(x)=\int_{-\infty}^{\infty}f(x-y)g(y)\,dy. \] Prove that \[ \lim_{|x|\to\infty}h(x)=0. \]
Solution
1. The function \(f\) is bounded on \(\RR\).
Since \(f(x)\to0\) as \(\abs{x}\to\infty\), there is \(R>0\) such that \(\abs{f(x)}\leq1\) for \(\abs{x}\geq R\). On the compact interval \([-R,R]\), continuity makes \(f\) bounded. Hence there is \(M<\infty\) with \[ \abs{f(x)}\leq M \] for all \(x\in\RR\).
2. Fix \(\varepsilon>0\). There is \(R>0\) such that\[ M\int_{\abs y>R}\abs{g(y)}\,dy<\frac{\varepsilon}{2}. \]
Since \(g\in L^1(\RR)\), its tail integrals tend to zero. If \(M=0\), the claim is immediate; otherwise choose \(R\) so that \[ \int_{\abs y>R}\abs{g(y)}\,dy<\frac{\varepsilon}{2M}. \]
3. There is \(S>0\) such that, whenever \(\abs{x}>S\),\[ \int_{\abs y\leq R}\abs{f(x-y)g(y)}\,dy < \frac{\varepsilon}{2}. \]
Set \[ G_R=\int_{\abs y\leq R}\abs{g(y)}\,dy. \] If \(G_R=0\), the integral is zero. Otherwise, because \(f(t)\to0\) as \(\abs t\to\infty\), choose \(T>0\) such that \[ \abs{f(t)}<\frac{\varepsilon}{2G_R} \] whenever \(\abs t>T\). Put \(S=T+R\). If \(\abs{x}>S\) and \(\abs y\leq R\), then \[ \abs{x-y}\geq\abs{x}-\abs y>T. \] Therefore \[ \int_{\abs y\leq R}\abs{f(x-y)g(y)}\,dy < \frac{\varepsilon}{2G_R} \int_{\abs y\leq R}\abs{g(y)}\,dy = \frac{\varepsilon}{2}. \]
4. If \(\abs{x}>S\), then\[ \abs{h(x)}<\varepsilon. \]
Split the defining integral at \(\abs y=R\). By the triangle inequality, steps 2 and 3, and the bound \(\abs f\leq M\), \[ \begin{aligned} \abs{h(x)} &\leq \int_{\abs y\leq R}\abs{f(x-y)g(y)}\,dy + \int_{\abs y>R}\abs{f(x-y)g(y)}\,dy\\ &< \frac{\varepsilon}{2} + M\int_{\abs y>R}\abs{g(y)}\,dy\\ &< \varepsilon. \end{aligned} \]
5.\[ \lim_{\abs{x}\to\infty}h(x)=0. \]
Step 4 is exactly the \(\varepsilon\) definition of this limit.
6. Q.E.D.
Step 5 is the required conclusion.