Commutator with Hilbert transform

Question

Let $H$ be Hilbert transform on $L^p({R})$. It is well known that commutator of $H$ and function $b\in BMO$ is a bounded operator in $L^p({R})$: $$\|[H,b] u\|_p\leq c_p\|b\|_{BMO}\|u\|_p, 1<p<\infty. $$ My question is there any information about constant $c_p$ ? I am especially interested in the case $b\in L^\infty$.

Answer 1

One can get some upper bounds based on the classical results and some new results. However, I do not know whether these upper bounds are sharp. Therefore my answer should be considered as comment and not as the actual answer to your question.

I will just repeat the well-known argument and I will try to extract the information regarding the bound. Let $T$ be a Calderon--Zigmund operator and $b \in BMO$. Consider the analytic function $T_{z}(f)=e^{zb}T(e^{-zb}f)$. Note that: $$ [T,b]f=\frac{d}{dz} T_{z}(f)|_{z=0}=\frac{1}{2\pi i}\int_{|z|=r}\frac{T_{z}(f)}{z^{2}}dz, \quad r>0 $$ Then Minkowski's inequality together with weighted norm estimates justifies the following chain of inequalities \begin{align} \|[T,b]f\|_{L^{p}}\leq \frac{1}{2\pi r^{2}} \int_{|z|=r}\|T_{z}(f)\|_{L^{p}}|dz|\leq \frac{1}{r}\sup_{|z|=r}\|T_{z}(f)\|_{L^{p}}=\frac{1}{r}\sup_{|z|=r}\|T(e^{-zb}f)\|_{L^{p}(e^{pb\Re z})}\\ \leq \frac{1}{r}\sup_{|z|=r}\|T\|_{L^{p}(e^{pb\Re z}) \to L^{p}(e^{pb\Re z})}\|f\|_{L^{p}} \end{align}

Now we would like to minimize the constant $c(r)=\frac{1}{r}\sup_{|z|=r}\|T\|_{L^{p}(e^{pb\Re z}) \to L^{p}(e^{pb\Re z})}$ over $r\in (0,\infty)$.

For the large class of Calderon--Zigmund integral operators $T$ we have the following bound $\|T\|_{L^{p}(w)\to L^{p}(w)}\leq C(T,p)[w]_{A_{p}}^{\max\{\frac{1}{p-1},1\}}$

See for example Theorem 17.1 http://www.math.kent.edu/~zvavitch/Lerner_Nazarov_Book.pdf

(you can also find in the book that the inequality holds for the operators when the modulus of continuity of the kernel satisfies some integrability assumption).

Hilbert transform does satisfy this condition, and as I am aware of, the bound $[w]_{A_{p}}^{\max\{\frac{1}{p-1},1\}}$ is sharp. In the case of Hilbert transform one can extract the sharp (maybe?) bound for the constant $C(H,p)$ (from other papers). I would suggest to ask F. Nazarov regarding this question.

This implies that $c(r) \leq C(H,p) \frac{1}{r} \sup_{|z|=r} [e^{pb\Re z}]_{A_{p}}^{\max\{\frac{1}{p-1},1\}}$.

In order to get more explicit estimate one has to solve the following extremal problem. Let $\|b\|_{BMO}=\varepsilon$ and let $t>0$. Then what is the sharp upper bound of the quantity $[e^{pb t}]_{A_{p}}$ in terms of $\varepsilon, t, p$. The last extremal problem can be solved by Bellman function technique, and if somebody will be interested I can show how it works.

Remark: surely instead of estimating integral by its maximal value one can leave the integral and get the bound $C(H,p) \inf_{r>0} \frac{1}{2\pi r^{2}} \int_{|z|=r} [e^{pb\Re z}]_{A_{p}}^{\max\{\frac{1}{p-1},1\}}|dz|$