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 $$ LlogL(\mathbb{T})=\left\{f\in L^1(\mathbb{T})\,s.t.\,\sup_{0\leq r<1}\int_{\mathbb{T}} \left|f(r\xi)\right| \log^+(\left|f(r\xi)\right|)dm(\xi)<\infty\right\}$$ 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 $$ LlogL(\mathbb{R})=\left\{f\in L^1(\mathbb{R})\,s.t.\,\int_{\mathbb{R}} \left|f(x)\right|\log^+(\left|f(x)\right|)dx<\infty\right\}$$ to $H^1(\mathbb{C}^+) $ (see the definition) ?

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


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