One way to define the maximal Hilbert transform of a function, $f$, is by $$\mathcal{H}[f](x):=\sup_{\varepsilon>0} \left| \int_{|x-t|\geq\varepsilon} \frac{f(t)}{x-t} \, dt\right|, \quad x\in\mathbb{R},$$ and $\mathcal{H}$ has been shown to be bounded from $L_p(\mathbb{R})\to L_p(\mathbb{R})$, $1<p<\infty$.
If one only considers the Hilbert transform (i.e. the principal value of the integral over $\mathbb{R}$ without absolute value on the outside), then one can show that an iterated version of the Hilbert transform in higher dimensions is $L_p(\mathbb{R}^d)\to L_p(\mathbb{R}^d)$ bounded. I can think of two ways to formulate a similar maximal version. Restricting our attention to two dimensions, consider the first:
$$\mathcal{H}_1[f](x_1,x_2):= \sup_{\varepsilon>0} \left|\int_{|x_2-t_2|\geq\varepsilon} \int_{|x_1-t_1|\geq\varepsilon} \frac{f(t_1,t_2)}{(x_1-t_1)(x_2-t_2)} \, dt_1 \, dt_2\right|.$$
Or one could consider different parameters:
$$\mathcal{H}_2[f](x_1,x_2):=\underset{\varepsilon,\delta>0}\sup\;\left|\int_{|x_2-t_2|\geq\varepsilon}\int_{|x_1-t_1|\geq\delta}\dfrac{f(t_1,t_2)}{(x_1-t_1)(x_2-t_2)}dt_1dt_2\right|.$$
The question then is: are these bounded on $L_p(\mathbb{R}^2)$? I thought there would be a fairly simple way to use the univariate result, but so far such an argument has eluded me.
