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I am trying to show that if $D$ is the open unit disk, $f$ is holomorphic in a neighborhood of the closure of $D$, and $w$ is an arbitrary point in $D$, then $f(w)=\frac{1}{\pi}\iint_{D}\frac{f(z)}{(1-\bar{z}w)^{2}}dxdy$.

Previously I already had $f(w)=\frac{1}{2\pi}\int_{\partial D}\frac{f(z)}{1-\bar{z}w}|dz|$ so I tried applying Green's theorem but I wasn't successful.

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up vote 2 down vote accepted

This is the reproducing property of the Bergman kernel in the unit disk, and it is derived in several books. It is valid for the Bergman space of square-integrable holomorphic functions in the unit disk, a weaker condition than the one you are given.

There is an ad-hoc way to check the formula as follows. Expand $f$ into a power series $f(z)=\sum\limits_{n=0}^\infty a_n z^n$ and integrate in polar coordinates, using the identity $\sum\limits_{k=0}^\infty (k+1) \zeta^k = \frac{1}{(1-\zeta)^2}$ for $|\zeta|<1$, to get $$ \begin{align*} \frac1\pi \iint_D & \frac{f(z)}{(1-\bar{z}w)^2} \, dx \, dy = \frac1\pi \int_0^1 \int_0^{2\pi} \sum_{n=0}^\infty a_n r^n e^{itn} \sum_{k=0}^\infty (k+1) r^k e^{-itk} w^k \, r \, dr \, dt \\ &=\frac{1}{\pi}\sum_{n,k=0}^\infty a_n (k+1) w^k \int_0^1 r^{n+k+1} \, dr \int_0^{2\pi} e^{it(n-k)} \, dt \end{align*} $$ Now observe that the $t$-integral is $0$ whenever $n\ne k$, and is $2\pi$ otherwise, so you end up with $$ 2 \sum_{n=0}^\infty a_n (n+1) w^n \int_0^1 r^{2n+1} \, dr = \sum_{n=0}^\infty a_n w^n = f(w) $$

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I guess in this setting, another way of going about it would be $f(w)=\int_{\partial D}\frac{f(z)}{z-w}dz=\int_{\partial D}\frac{\bar{z}f(z)}{1-\bar{z}w}dz$ and then apply Green's theorem. – Yong Oct 9 '12 at 20:34

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