Let $X$ be the position operator on $L^2(\mathbb{R})$, defined by $(Xf)(x)=xf(x)$ on a suitable domain of definition.
We can then obtain an operator $\frac{1}{|X|}$ by two (equivalent, I think) ways:
- Apply the function $\frac{1}{|x|}$ on $X$ via functional calculus.
- Get $|X|$ by the polar decomposition and then take its inverse.
Now to my question: Is the operator $\frac{1}{|X|}$ compact?
Context: I am working through the proof of Theorem 11 of this paper on page 50/51, and if I understand it correctly, they make use of the fact, that $\frac{1}{|X|}$ is compact.