Control of the bilinear indicator cube testing property

Abstract

We show that the $\alpha$-fractional bilinear indicator/cube testing constant

${\mathcal{BICT}}_{T^{\alpha }}(\sigma ,\omega )\equiv \underset{Q\in {\mathcal{P}}^n}{sup}\underset{E,F\subset Q}{sup}\frac{1}{\sqrt{{\left|Q\right|}_{\sigma }{\left|Q\right|}_{\omega }}}\left|{\int }_F{T}_{\sigma }^{\alpha }\left({\mathbf{1}}_E\right)\omega \right|,$

defined for any $\alpha$-fractional singular integral $T^{\alpha }$ on ${\mathbf{R}}^n$ with $0<\alpha <n$, is controlled by the classical $\alpha$-fractional Muckenhoupt constant $A_2^{\alpha }(\sigma ,\omega )$, provided the product measure $\sigma \times \omega$ is diagonally reverse doubling (in particular if it is reverse doubling) with exponent exceeding $2(n-\alpha )$.

Moreover, this control is sharp within the class of diagonally reverse doubling product measures. In fact, every product measure $\mu \times \mu$, where $\mu$ is an Ahlfors-David regular measure $\mu$ with exponent $n-\alpha$, has diagonal exponent $2(n-\alpha )$ and satisfies $A_2^{\alpha }(\mu ,\mu )<\mathrm{\infty }$ and ${\mathcal{BICT}}_{I^{\alpha }}(\mu ,\mu )=\mathrm{\infty }$, which has implications for the $L^2$ trace inequality of the fractional integral $I^{\alpha }$ on domains with fractional boundary.
When combined with the main results in arXiv:1906.05602, 1907.07571 and 1907.10734, the above control of ${\mathcal{BICT}}_{T^{\alpha }}$ for $\alpha >0$ yields a $T1$ theorem for doubling weights with appropriate diagonal reverse doubling, i.e. the norm inequality for $T^{\alpha }$ is controlled by cube testing constants and the $\alpha$-fractional one-tailed Muckenhoupt constants ${\mathcal{A}}_2^{\alpha }$ (without any energy assumptions), and also yields a corresponding cancellation condition theorem for the kernel of $T^{\alpha }$, both of which hold for arbitrary $\alpha$-fractional Calderón-Zygmund operators $T^{\alpha }$.
We do not know if the analogous result for ${\mathcal{BICT}}_H(\sigma ,\omega )$ holds for the Hilbert transform $H$ in case $\alpha =0$, but we show that ${\mathcal{BICT}}_{H^{\mathrm{dy}}}(\sigma ,\omega )$ is not controlled by the Muckenhoupt condition ${\mathcal{A}}_2^{\alpha }(\omega ,\sigma )$ for the dyadic Hilbert transform $H^{\mathrm{dy}}$ and doubling weights $\sigma ,\omega \$$