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Let $\mu$ be a real measure on the circle $\mathbf{T}$. Then the function $$f(z)=\int_\mathbf{T} \mathrm{Im}\left(\frac{\zeta+z}{\zeta-z}\right) d\mu(\zeta)$$ is harmonic on the unit disc and its radial limits exist almost everywhere.

Does the Hilbert transform of $\mu$ equal the radial limit function of $f$ on $\mathbf{T}$ almost everywhere and why?

The Hilbert transform of $\mu$ is $$(H\mu)(z)=p.v.\frac{1}{\pi}\int_\mathbf{T}\frac{d\mu(\zeta)}{z-\zeta}$$ for $z\in\mathbf{T}$. (Please check, I'm not sure)

I have come so far to see, that the Hilbert transform and the radial limits of $f$ behave somewhat similarly as can be seen from noting that $$\mathrm{Im}\left(\frac{e^{it}+re^{is}}{e^{it}-re^{is}}\right)\rightarrow \cot(s/2)$$ as $r\rightarrow 1$ and $\cot(s/2)$ behaves similarly as $2/s$ when $|s|$ is small.

(I would also appreciate if somebody knew the answer just for the case that $\mu$ has continuous density w.r.t. the Lebesgue measure on $\mathbf{T}$ or similar special cases.)

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up vote 2 down vote accepted
+50

Your definition of the Hilbert transform on the unit circle is not correct, but the answer to your question is positive. Check for instance Chapter 11.5 in "Inequalities: A Journey into Linear Analysis" for a definition of the Hilbert transform on the circle and properties for some functional analytical spaces.

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