The problem is to find $f:(0,1)\rightarrow \mathbb{C}$ such that
\begin{equation} f\in L^1(0,1) \quad \text{ and} \quad X^{-\varepsilon}f \notin L^1(0,1) \end{equation} for all $\varepsilon > 0$.
Here $X$ denotes multiplication by $x$ so $(X^{-\varepsilon}f)(x) = x^{-\varepsilon}f(x)$.
It might be that such a function does not exist, but I have been incapable of showing this fact... Any thoughts?
$f(x)=\frac 1 {x (\ln (x))^{2}}$ for $0<x<\frac 1 2 $ and $0$ for $\frac 1 2 \leq x <1$. Make the substitution $y=\ln x$ to see that this works.