Given a $f \in L^1(\mathbb{R})$ is there a $g : \mathbb{R} \to \mathbb{R}$ such that $g(f) = 0$ with $g \neq 0$?

Question

Given a $f \in L^1(\mathbb{R})$ is there a $g : \mathbb{R} \to \mathbb{R}$ such that $g(f) = 0$?

I would like to solve the problem myself but don't know how to start.

Point of this question; I have a recurrence relation of the form: $z_n(x) = z_{n-1}(x) + g(z_{n-1}(x))$ and I want to know if there is a $g$ such that $\lim_{n \to \infty} z_n(x)$ can approximate any function in $L^1(\mathbb{R})$ arbitrarily-well.

Let us assume that is true and $$ \int_{\mathbb{R}} |\lim_{n\to\infty}z_n(x) - f(x)| \mathrm{d}x= 0\:; $$ this would imply that $\lim_{n\to\infty}g(z_n(x)) = g(f(x)) =0$