Analysis Problem of the Day 8

Today’s problem appeared as Problem 12 on UCLA’s Spring 2018 Analysis Qual:

Problem 12: Let $f$ be a bounded holomorphic function on the unit disc $\mathbb{D} :=\{|z|<1\}.$ Prove that for any $w \in \mathbb{D}$ we have

$f(w) = \frac{1}{\pi} \int_{\mathbb{D}} \frac{f(z)}{(1-\overline{z}w)^2} d\mu(z).$

Solution: The first thing that might come to one’s mind when looking at this formula is Cauchy’s integral formula

$f(w) = \int_{\partial \mathbb{D}} \frac{f(z)}{z-w} dz = \int_{\partial \mathbb{D}} \frac{f(z) \overline{z}}{1-\overline{z}w} dz,$

where the simplification in the second formula is due to $z \overline{z} = |z|^2 =1$ on $\partial \mathbb{D}.$ Since we want to switch this to an area integral, the trick turns out to be to use Green’s theorem:

$f(w)= \int_{\partial \mathbb{D}} \frac{f(z) \overline{z}}{1-\overline{z}w} dz = \frac{1}{\pi} \int_{\mathbb{D}} \frac{\partial \frac{f(z) \overline{z}}{1-\overline{z}w}}{\partial \overline{z}} d\mu(z) = \frac{1}{\pi} \int_{\mathbb{D}} \frac{f(z)}{(1-\overline{z}w)^2} d\mu(z),$

and thus one is finished.

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