How can I find the limit?

Question

Let $f\in$$L^1(\mathbb{R})$and $g\in$$L^1(\mathbb{R})$$\cap$$L^\infty(\mathbb{R})$. My question is how can I find the value $\lim_{x \to \infty}f\ast g (x)$ ? I know that $f\ast g $ is continuous and $f\ast g\in$$L^1(\mathbb{R})$. So I guess $\lim_{x \to \infty}f\ast g (x)=0$. Is true or false ?

Answer 1

Let $\epsilon > 0.$ Then there exists a continuous $f_0$ with compact support such that $\|f-f_0\|_1 < \epsilon.$ Let the support of $f_0$ be $[-M,M].$ Then

$$(1)\,\,\,\,|\int_{\mathbb {R}}f(x-t)g(t)\,dt\,| \le \int_{\mathbb {R}}|f(x-t)-f_0(x-t)|\,|g(t)|\,dt + \int_{\mathbb {R}}|f_0(x-t)|\,|g(t)|\,dt$$ $$ \le \epsilon\|g\|_\infty + \|f_0\|_\infty\int_{x-M}^{x+M}|g(t)|\,dt.$$

Because $g\in L^1,$ the last integral $\to 0$ as $x\to \infty.$ Thus the $\limsup$ of (1) is $\le \epsilon\|g\|_\infty .$ Since $\epsilon$ is arbitrary, we have $\lim_{x\to \infty}f*g(x)=0$ as desired.