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?

  • $\begingroup$ Note that if $f$ is bounded, then $x^{-\epsilon}f(x) \in L^1$ for all $\epsilon\in (0,1).$ This means if you are to find such an $f$ it must be unbounded. $\endgroup$ – zhw. Jun 24 '18 at 2:12

$f(x)=\frac 1 {x (\ln (x))^{2}}$ 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.

  • $\begingroup$ Very elegant indeed! $\endgroup$ – Ignatius Jun 24 '18 at 10:50

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