Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.)

share|improve this question

1 Answer 1

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.

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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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