If $f,g\in L^1\cap L^\infty$, then $(f*g)(x)\to0$ as $\abs{x}\to\infty$.Suggest a slogan
Let \(f,g\in L^1(\RR^n)\) be essentially bounded. Then the convolution \(f*g\) is defined at every \(x\in\RR^n\) and \[ \lim_{\abs{x}\to\infty}(f*g)(x)=0 . \]
Convolutions of bounded integrable functions vanish at infinity
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