I study convolution, and there is a theorem show that if $f, g$ are functions such that $f \in L^p$ for $1 < p < \infty$ and $g \in L^q$ where $q$ is the conjugate exponent of $p$. Then $$\lim_{|x| \rightarrow \infty} (f*g)(x) = 0.$$ Then it reamrks that $p$ does not include the extreme case when $p=1$ and $q = \infty$ or $p= \infty$ 0r $q=1$. That is, there is a function $f \in L^1$ and $g \in L^\infty$ such that $\lim_{|x| \rightarrow \infty}f*g(x) \neq 0.$
Since $g$ is in $L^\infty$, it is bounded almost every where. Since $f \in L^1$, I usually have a picture of it as $\frac{1}{\sqrt{x}}$ or $e^{-x}$ which is decay at infinity.
However, $e^{-x}$ decays faster than any polynomial at $\infty$ and polynomial does not belong to $L^\infty$ on $\mathbb{R}$.
So I am not quite sure what function I am looking for.
Maybe the following works: Put $f(x)=e^{-(x^2)}$ and $g(x)=1$, then $$f*g(x)=\int_\mathbb{R} e^{-\tau^2}g(\tau-x)\, d\tau=\int_\mathbb{R} e^{-\tau^2}1\, d\tau=const\neq 0$$