Can $f, g \in L^{1}(\mathbb{R})$ imply $ \lim_{x \to \infty} (f * g)(x) = 0 $

Question

Can $f, g \in L^{1}(\mathbb{R})$ imply $$ \lim_{x \to \infty} (f * g)(x) = 0 $$ or not?

$*$ denotes convolution here: $$ (f*g)(x) = \int_{\mathbb{R}} f(t)g(x-t)\ dt $$

I read that when $f \in L^{p}$ and $g \in L^q$, we can prove $f * g(x)$ vanishes at infinity, using Holder's inequality to estimate $\int_{B(0,R)^C}| f(t)g(x-t) | dt$. But in the case of $L^1(\mathbb{R})$ I have difficulty in the estimation. I guess it is true but not so sure. Any help is appreciated!

Answer 1

By Wiener's theorem the map $T_f: L^1(\mathbb{R}) \to L^1(\mathbb{R})$ given by $g \mapsto f \ast g$ is invertible when its Fourier transform $\widehat{f}$ is bounded bellow. That will give you plenty of counterexamples. Indeed, prety much every function $h$ in $L^1$ is a convolution by taking $$h = f \ast \Big( \frac{1}{\widehat{f}} \, \widehat{h} \Big)^{\vee} = f \ast g$$

Answer 2

The convolution of two $L^{1} (\mathbb R)$ functions need not be continuous. It is in $L^{1} (\mathbb R)$; it is an equivalence class of functions. So the question does not make sense.