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Suppose $f(x,y)$ is a bounded harmonic function in the unit disk $D = \{z = x + iy : |z| < 1 \} $ and $f(0,0) = 1$. Show that $$\iint_{D} f(x,y)(1 - x^2 - y^2) ~dx ~dy = \frac{\pi}{2}.$$

I'm studying for a prelim this August and I haven't taken Complex in a long time (two years ago). I don't know how to solve this problem or even where to look unless it's just a game with Green's theorem-any help? I don't need a complete solution, just a helpful hint and I can work the rest out on my own.

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2 Answers 2

Harmonic functions have the mean value property, that is $$ \frac1{2\pi}\int_0^{2\pi}f(z+re^{i\phi})\,\mathrm{d}\phi=f(z) $$ If we write the integral in polar coordinates $$ \begin{align} \iint_Df(x,y)(1-x^2-y^2)\,\mathrm{d}x\,\mathrm{d}y &=\int_0^1\int_0^{2\pi}f(re^{i\phi})(1-r^2)\,r\,\mathrm{d}\phi\,\mathrm{d}r\\ &=2\pi\int_0^1f(0)(1-r^2)\,r\,\mathrm{d}r\\ &=2\pi f(0)\cdot\frac14\\ &=\frac\pi2 \end{align} $$

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Perfect, thank you very much. –  C.Cheshire Jul 30 '12 at 5:14
    
+1 Very nice indeed ! –  DonAntonio Jul 30 '12 at 5:18

Begin with the mean value property of harmonic functions: $\int_0^{2 \pi} f(r \cos \theta, r \sin \theta) d\theta =f(0,0) 2 \pi $.

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