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

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!

• I suppose by the limit you mean that $(f\star g)(x)\stackrel{x\to\infty}\longrightarrow0$ almost everywhere? – Math1000 May 10 '18 at 4:12
• @Math1000 In the $L_p,L_q$ case, we can even prove $f *g$ is uniformly continuous (I read else where ). Does it hold in this case? – Edward Wang May 10 '18 at 4:48
• @Math1000 I'm kind of confused what $(f * g)(x)\stackrel{x\to\infty}\longrightarrow0$ almost everywhere exactly mean? – Edward Wang May 10 '18 at 6:17

## 2 Answers

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 $$h = f \ast \Big( \frac{1}{\widehat{f}} \, \widehat{h} \Big)^{\vee} = f \ast g$$

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.

• So is there any example that the convolution is not continuous? – Edward Wang May 10 '18 at 6:15
• Let $f(x)=\frac 1 {\sqrt |x|}$ on $(-1,1)$ and 0 elsewhere. Then $f$ is integrable but $(f*f)(0)=\int_{-1}^{1} \frac 1 {|x|} \, dx=\infty$. – Kavi Rama Murthy May 11 '18 at 7:31
• It may not make sense literally, but if you replace limit by essential limit, it is a perfectly valid question. – tomasz May 11 '18 at 12:22