# Showing $f(\zeta)=\frac{1}{\pi}\int_{|z|<1}\frac{f(z)dxdy}{(1-\bar{z}\zeta)^2}$

I'm doing practice physics qualifying exam problems and came across this one I didn't know how to solve:

Show that if $f(x)$ is bounded and analytic for $|z|=|x+iy|<1$, then $$f(\zeta)=\frac{1}{\pi}\int_{|z|<1}\frac{f(z)dxdy}{(1-\bar{z}\zeta)^2}$$ Hint: First express the area integral in polar coordinates, then transform one of the integrals to a suitable line integral of a rational function that can be evaluated using the calculus of residues.

I tried using $z=re^{i\theta}$ and messing around with the integral, but after a long writeout I am left with a puddle of muddled thoughts. Could someone explain the next steps?

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You can prove this by Green formula: $$\int_{|z|=1}F(z)dz=2i\int_{|z|<1}\frac {\partial F}{\partial \bar{z}}dxdy.$$ By the Cauchy Int Formula $$f(\zeta)=\frac{1}{2\pi i }\int_{|z|=1} \frac{f(z)}{z-\zeta}dz.$$

Let $F(z)=\frac{f(z)}{z-\zeta}$, on the circle $|z|=1$, we have $z=\frac{1}{\bar z}$, so $F(z)=\frac{f(z)}{z-\zeta}=\frac{\bar zf(z)}{1-\bar z\zeta}$, by easy computation , we can get $\frac {\partial F}{\partial \bar{z}}=\frac{f(z)}{(1-\bar z\zeta)^2}$. Finally, using the Green Formula, we get $$f(\zeta)=\frac{1}{\pi }\int_{|z|<1}\frac{f(z)dxdy}{(1-\bar z\zeta)^2}.$$

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Great, thanks a lot for your help. –  Peter May 4 '12 at 17:38
@Peter My pleasure. –  Riemann May 5 '12 at 1:53

The function $$K(\zeta,z)=\frac{1}{\pi}\frac{1}{(1-\bar{z}\zeta)^2}$$ is known as the Bergman reproducing kernel.

Hint:

1. Compute the series expansion of $g(x)=\frac{1}{(1-x)^2}$, e.g. by using that $$\frac{1}{(1-x)^2} =\frac{d}{dx}\frac{1}{1-x}$$

2. First prove the statement for $f_n(z)=z^n$, e.g. by using the fact $$\int_0^{2\pi}e^{ikt}dt=0,\text{ for all integers k\ne0}$$

Hopefully leaving the fun parts to you...

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Thanks for the tip. I'll try working it out, and hopefully won't run into any more problems. –  Peter May 4 '12 at 17:38