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 , 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 $$.
