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.

If the discrete version is unbounded, then I would say it is good reason to suspect that the continuous version you mention is unbounded as well.

| cite | improve this answer | |
  • $\begingroup$ 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. $\endgroup$ – Keaton Apr 10 '15 at 22:20
  • $\begingroup$ In the paper you mentioned, the authors only said "the discrete Hilbert transform is bounded" without giving a proof. $\endgroup$ – Tony B Apr 12 '15 at 19:05
  • $\begingroup$ Ah okay, I just copied what I had cited in a paper, I forgot they didn't supply the proof. I don't have access to it at the moment, but do they cite a paper by Marsden and Moreka (I probably have spelled that wrong, and there should be a third author, either Richards or Riemenschneider)? The proof might be in there. If not I will look again on Monday, I know I have seen it before. $\endgroup$ – Keaton Apr 12 '15 at 19:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.