# inserting absolute value in Hilbert transform and a discrete version of Hilbert transform

It is well known that the Hilbert transform $H(f)(x)=p.v. \int\frac{f(x-y)}{y}dy$ is bounded on $L^p(\mathbb{R})$ for $p\in(1,\infty)$. I want to consider some variants of $H$.

1) What happens if we intersect absolute value? i.e. Consider $\bar{H}(f)(x)=p.v. \int|\frac{f(x-y)}{y}|dy$. Does $\bar{H}$ make sense and satisfy any boundedness properties?

2) Consider the discrete version of $H$ defined by $H_df(n)=\sum_{m\neq0}\frac{f(n-m)}{m}$. Is $H_d$ bounded on $l^p$ for some $p$? I can't find any reference (textbook or paper) on this seemingly nature operator.

As for (2), look at Proposition 1.3 of: M. J. Marsden, F. B. Richards, and S. D. Riemenschneider, Cardinal spline interpolation operators on $\ell_p$ data, Indiana Univ. Math. J. 24(1975), 677-689; Erratum, ibid., 25(1976), 919. The result of this is that it is bounded on $\ell_p$ for every $1<p<\infty$.
For (1), I can't give you a proof at the moment, but I think that the discrete version of $\overline{H}$ is not necessarily bounded on $\ell_2$. You can find this in Hardy's book "Inequalities" just before Section 8.13 (it is p.214 in the version I have). He considers the bilinear form $$\sum_{i}\sum_{j\neq i}\frac{x_iy_j}{|i-j|},$$ and shows that for $$x_i = y_i = i^{-1/2}(log\; i)^{-1},\quad i>1$$ and $x_1=y_1=x_2$, the sum in the display is divergent, and so the bilinear form is not $\ell_2\to\ell_2$ bounded. On the other hand, if you take out the absolute value, the corresponding form is bounded, though it is more difficult to show.
• I neglected to mention, if it is not clear, you can do a change of variables in the Hilbert transform to look at is as $p.v.\int\frac{f(y)}{x-y}dy$, and same for the discrete version. – Keaton Apr 10 '15 at 22:20