# Derivation of a “Weighted” Mean Value Thm

I am trying to derive the following equation for an analytic function, $f$, defined on the unit disc, $D$

\begin{equation*} f(z)=\frac{1}{\pi}\int_{D} \frac{f(w)}{(1- \bar{w}z)^{2}}dA(w). \end{equation*}

Question: Anyone have any ideas on how to derive this formula?

I've been able to verify that the above is true for $z =0$. In attempting to derive for the general case, I used conformal maps to arrive at

\begin{equation*} f(z)=\frac{1}{\pi}\int_{D} f(w) \left| \frac{-1+|z|^{2}}{(-1+w\bar{z})^{2}} \right|^{2} dA(w) \end{equation*}

Unfortunately this is not the formula I was trying to derive,

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What is $A(\omega)$ –  anonymous Oct 8 '10 at 5:26
@Chandru1: I guess it's area measure; $dA(w)=dx dy$ if $w=x+iy$. –  Hans Lundmark Oct 8 '10 at 11:31

First, you're missing a factor of $1/\pi$ on the right-hand side, as can be seen by taking $f=1$ and $z=0$.
To prove the identity, you can use Green's formula, which translated into complex variables takes the form $$\int_C F(w,\bar{w}) dw = 2i \iint \frac{\partial F}{\partial \bar{w}} dA(w).$$ This transforms your area integral into a line integral around the unit circle $C$. To evaluate this line integral, use that $\bar{w}=1/w$ when $|w|=1$; this gives you an integral in $w$ alone (no $\bar{w}$) so that you can use Cauchy's formula to finish the job.