Define Hilbert Transform (HT) as the convolution with the function $1/x$. E. Stein proves in his book Singular Integrals and Differentiability Properties of Functions that HT, when understood as a singular integral operator, is a bounded operator on $L^p(\mathbb{R})$ for $p\in (1, \infty)$.
I am wondering if HT has compact commutator with multiplication by $C_0(\mathbb{R})$ on $L^p(\mathbb{R})$?
More precisely, if $T \in \mathscr{L}(L^p(\mathbb{R}))$ denotes the Hilbert transform, and $f \in C_0(\mathbb{R})$, is it true that $Tf - fT \in \mathbb{K}(L^p(\mathbb{R}))$? If it is true, would you please give me a reference? Thank you!
The commutator is compact if and only if $f$ is in VMO. Source: On the compactness of operators of Hankel type by A. Uchiyama (free access; note that $VMO$ is called $CMO$ there). Since $C_0(\mathbb R)\subset VMO$, the answer to your question is positive.