# Multivariate Maximal Hilbert Transform

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.

• Sorry I wrote that kind of off the cuff, I edited the formula to relay what I intended. This does look close to the Riesz transform, but slightly different. – Keaton Hamm Feb 26 '15 at 5:28
• I'm a little confused--I think in the one-variable definition you want the region of integration to be $|x-t|\geq \epsilon$. But then the corresponding regularization of the multivariate version doesn't work, since the region $\|x-t\|_2\geq \epsilon$ still contains points where the denominator of the integrand vanishes. Indeed, it seems one would have to remove a neighborhood of the union of the coordinate hyperplanes (translated by $x$), and of course any such region is unbounded. – Mike Jury Apr 10 '15 at 20:22
• You are right, that is something I missed. If one modifies the definition to look at something like an iterated version of $\mathcal{H}$ then this I think is well defined, and possibly sufficient to consider. I will try to formulate it correctly and edit the question later. – Keaton Hamm Apr 10 '15 at 21:39