Set $\mathbb{T}$ the unit circle, $dm$ the Lebesgue measure on $\mathbb{T}$ and $\mathbb{C}^+=\left\{z\in \mathbb{C},s.t.\,\Im(z)>0 \right\}$ the upper half plane.

It is well-known that the Cauchy transform $K$ defined by $$(Kf)(z)= \int_{\mathbb{T}} \frac{f(\xi)}{1-z\bar{\xi}} dm(\xi), \quad z\in \mathbb{D}$$ (or equivalently the harmonic conjugate or the Hilbert transform) is bounded from the Zygmund space $$ L\log L(\mathbb{T})=\left\{f\in L^1(\mathbb{T})\,s.t.\,\sup_{0\leq r<1}\int_{\mathbb{T}} \left|f(r\xi)\right| \log^+(\left|f(r\xi)\right|)dm(\xi)<\infty\right\}$$ to the classical (complex) Hardy space $H^1(\mathbb{D})$ (see the definition).

Now, on the upper plane $\mathbb{C}^+$, the Cauchy transform $K$ is defined by $$(Kf)(z)= \frac1{2i\pi}\int_{\mathbb{R}} \frac{f(x)}{x-z} dx, \quad z\in \mathbb{\mathbb{C}^+}.$$ Is it bounded from $$ L\log L(\mathbb{R})=\left\{f\in L^1(\mathbb{R})\,s.t.\,\int_{\mathbb{R}} \left|f(x)\right|\log^+(\left|f(x)\right|)dx<\infty\right\}$$ to $H^1(\mathbb{C}^+) $ (see the definition) ?

I am very grateful for any idea/comment on this matter. References are very welcome.


1 Answer 1


I am not sure but I would think that your $L\log L$ space should be replaced by the space of functions $f$ such that $\int_R f(x)\,dx=0$ (this is the necessary condition) and $$ \int_R |f(x)|\left[\log^+ |f(x)| + \log^+ |x|\right]\, dx <\infty. $$

Also you should see the paper

D. Cruz-Uribe, SFO and A. Fiorenza, $L\log L$ results for the maximal operator in variable $ L^p$ spaces, Trans. Amer. Math. Soc. 361 (2009), 2631-2647, doi:10.1090/S0002-9947-08-04608-4.

  • $\begingroup$ Remarks like this would be comments rather than answers. Once you got enough reputation (50 points) by writing answers and questions, you can comment on every question and answer. Oh, and welcome to this site! $\endgroup$
    – Dirk
    Aug 7, 2019 at 7:38
  • $\begingroup$ I've moved over the reference from your other answer, as it should have been added to this one. $\endgroup$
    – David Roberts
    Aug 7, 2019 at 8:02

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