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Let $u$ be a continuous function on an open set $U$ of the complex plane. We say that $u$ satisfies the circle mean value property at a point $z_0\in U$ if $$ u(z_0)=\frac{1}{2\pi}\int_0^{2\pi}u(z_0+re^{i\theta})d\theta$$

for all $r$ sufficiently small such that the disc centered at $z_0$ with radius $r$> is contained in $U$. We say that $u$ satisfies the disc mean value property at a point $z_0$ if $$u(z_0)=\frac{1}{\pi r^2}\iint_{D(z_0,r)}u dxdy$$

I think the two properties are related. In particular i'd like to show that the first implies the second. Is this an application of Green's Thm maybe?

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I think that the key fact here is the expression of the area element in polar coordinates: $$dxdy=\rho d\rho d\theta.$$ – Giuseppe Negro Dec 30 '12 at 16:20
up vote 1 down vote accepted

The circle mean value property indeed implies the disc mean value property. To see this suppose without loss of generality that $z_0=0$ and introduce polar coordinates $$\begin{cases} x=\rho \cos \theta \\ y=\rho \sin \theta \end{cases} $$ We have for the area element the formula $$dxdy=\rho d\rho d\theta ,$$ meaning that we can rewrite an integral on the disc as a superposition of integrals over circles: \begin{equation} \begin{split} \frac{1}{\pi r^2} \iint_{D(r)}u(x+iy)\, dxdy&= \frac{1}{\pi r^2}\int_0^r\rho\, d\rho \int_0^{2\pi} u(\rho e^{i\theta})\, d\theta\\ &=\frac{2}{ r^2}\int_0^r \rho\,d\rho u(0)\\&=u(0). \end{split} \end{equation}

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