Let us consider the function $I(x):=\frac{1}{2\epsilon}1_{[-\epsilon, \epsilon]}(x)$. Which are the functions $g: \mathbb R \to [0,1]$ such that
$$g(x) = f*I(x)$$
for some $f: \mathbb R \to [0,1]$?
It can be proved, by the fundamental theorem of integral calculus that, any $g$ of this form must be $\frac{1}{\epsilon}-$lipschitz, but I do not think this condition is sufficient.