Control of the bilinear indicator cube testing property

Abstract

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

\(\mathcal{BICT}_{T^{\alpha }}\left( \sigma ,\omega \right) \equiv \sup_{Q\in \mathcal{P}^{n}}\sup_{E,F\subset Q}\frac{1}{\sqrt{\left\vert Q\right\vert_{\sigma }\left\vert Q\right\vert _{\omega }}}\left\vert \int_{F}T_{\sigma}^{\alpha }\left( \mathbf{1}_{E}\right) \omega \right\vert ,\)

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 \(A_{2}^{\alpha }\left( \sigma ,\omega\right)\), provided the product measure \(\sigma \times \omega\) is diagonally reverse doubling (in particular if it is reverse doubling) with exponent exceeding \(2\left(n-\alpha \right)\).

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 \(2\left( n-\alpha \right)\) and satisfies \(A_{2}^{\alpha }\left( \mu ,\mu \right)<\infty\) and \(\mathcal{BICT}_{I^{\alpha }}\left( \mu ,\mu \right)=\infty\), 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 \(\mathcal{BICT}_{H}\left(\sigma,\omega \right)\) holds for the Hilbert transform \(H\) in case \(\alpha=0\), but we show that \(\mathcal{BICT}_{H^{\operatorname{dy}}}\left(\sigma ,\omega\right)\) is not controlled by the Muckenhoupt condition \(\mathcal{A}_{2}^{\alpha }\left( \omega ,\sigma \right)\) for the dyadic Hilbert transform \(H^{\operatorname{dy}}\) and doubling weights \(\sigma ,\omega $\).