# Does Hilbert Transform commute with Function Multiplication modulo Compact on $L^p(R)$?

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 not look likely to me – Alex Gavrilov Feb 16 '14 at 11:27

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 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.
• Hi Alex, Thanks for your reply. Two questions: a) Is the operator norm of "multiplied by $f$" operator bounded by its sup (or $L^\infty$ norm)? b) For the case of $f$ smooth and cptly supported, can we use the fact that $k(x, y)$ can be approximated in sup norm by $\sum_i g_i(x)h_i(y)$? If so, is the operator norm of the "difference kernel" bounded by its sup norm? Thanks! – Clark Chong Feb 18 '14 at 21:27
• a) The operator norm of multiplication by $f$ equals the $L^\infty$ norm of $f$. b) The sup norm of the kernel on $\mathbb{R}$ doesn't bound the operator norm, since not all bounded kernels correspond to bounded operators. However, the decomposition into a sum of rank-ones is certainly useful for the "$\mathrm{supp} f \times \mathrm{supp} f$ block" of the kernel, if you view it as a "block matrix" corresponding to $L^p(\mathbb{R}) \simeq L^p(\mathrm{supp} f) \oplus L^p(\mathbb{R} \setminus \mathrm{supp} f)$. On that block this decomposition actually converges fast enough to ensure that – Alexander Shamov Feb 19 '14 at 0:48