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>



Does Hilbert Transform commute with Function Multiplication modulo Compact on $L^p(R)$?
The problem reduces to the case of smooth functions with compact support, since they are norm dense in $C\_0$.


Now let $f$ be a smooth function with compact support. Then $[T,f]$ is an integral operator with smooth kernel $k(x,y) := (f(x)-f(y))/(x-y)$.


There is an [easy to check sufficient condition](http://www2.math.umd.edu/~lvrmr/2008-2009-S/Classes/AMSC674/IntOps.pdf) for compactness of integral operators from $L^p$ to $L^q$ in terms of iterated norms: namely, if an integral operator with kernel $k$ has finite norm $\left[ \intop \left( \intop |k(x,y)|^{p^\ast} dy \right)^{q/p^\ast} dx \right]^{1/q}$, $1/p+1/p^\ast = 1$, then the operator is compact from $L^p$ to $L^q$. This condition is obviously satisfied by our kernel.


