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This is really a simple question involving swapping the order of integration. The step I'm confused about comes from a proof in Pommerenke's "Univalent Functions," and for those of you with a reference, it is Theorem 1.2.

Let $C$ be a (closed) piecewise smooth curve homologous to zero with respect to the domain $H\subset \mathbb{C}$. We want to show that

\int_C \overline{h(w)} h'(w) dw = \dfrac{1}{\pi} \int_{H} |h'(w)|^2 \left( \int_C \dfrac{dz}{z-w} \right) d\Omega_w

where $d\Omega_w$ is the Lebesgue measure.

The first step in the proof is to swap the order of integration, but I'm not really sure why one could do that. I'm guessing it has something to do with a remark in the text about changing $H$ slightly to assume that $h$ is analytic on $\overline{H}$ rather than just on $H$.

Several things...

  1. I think I can assume that $H$ is a bounded domain.
  2. Assuming that $H$ is a bounded domain, with the additional assumption that $h$ is analytic on $\overline{H}$, I can assume that $h$ is bounded on $H$.

I'm guessing I have to use something like Fubini, but if I want to use Fubini, I'll have to show that

$\int_C \int_H \dfrac{1}{|z-w|} d\Omega_w |dz|$

is convergent.

I can see that the inner integral is convergent for all $z$ in $H$. But then I'm not sure why the iterated integral is convergent.

Is it true that

$\int_H \dfrac{1}{|z-w|} d\Omega_w$

is a continuous function in $z$? I think it's true, but I'm not sure how to show this...

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    Yes, I got that. Sorry if what I told was something you already knew; it was not entirely clear from the question whether this was something you were unsure about or just scene-setting for the real question.2012-01-02

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