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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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1 Answer 1

up vote 2 down vote accepted

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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Thanks a lot! I knew there was something wrong with the definition I had. –  Your Ad Here Nov 4 '12 at 16:28

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