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!
P.S: cross-posted from MSE here: https://math.stackexchange.com/questions/676833/does-hilbert-transform-commute-with-function-multiplication-modulo-compact-on-l