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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 $f(\zeta)=\frac{1}{\pi}\int_{|z|<1}\frac{f(z)\,dx\,dy}{(1-\bar{z}\zeta)^2}$ 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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    @Peter My pleasure.2012-05-05
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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.2012-05-04