# Deducing identities from the mean value property of a real-valued harmonic function

Let $u$ be a real-valued harmonic function in the plane. We know that $u$ satisfies the mean value property in the following form: for all $r$ > $0$

(i) $$u(z_o) = \frac{1}{2\pi}\int_{0}^{2pi}u(z_o + re^{i\theta})d\theta$$

The question asks us to use the above formula to deduce the following identities for all $r$ > $0$

(ii)

$$u(z_o) = \frac{1}{2\pi r}\int_{_dD(z_o, r)} u(x,y)ds$$

(iii)

$$u(z_o) = \frac{1}{\pi r^2}\int\int_{_D(z_o, r)} u(x,y)dxdy$$

where $D(z_o, r)$ = {$z\in\mathbb{C}: |z-z_o| < r$}

Thoughts so far: The identities seem relatively straight forward, but I'm having a hard time deducing them from (i). I know $z = z_o + re^{i\theta}$ is a parametrization of the circle of radius $r$ centered at $z_o$. In class we showed that if we use this parametrization in Cauchy's Integral Formula, then we obtain (i), so it doesn't make sense to me to substitute $z$ back into (1). I'm not sure where to start really.

Sorry if the solution turns out to be very simple. It seems to me it should be and i'm probably missing something obvious.

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sorry, i know it's not much to work with – matt Oct 21 '12 at 1:10
here's another shot at it: if we substitute $z = z_o + re^{i\theta}$ into (i), then $dz = rie^{i\theta}d\theta = rzd\theta$. we know $s = r\theta$, so $ds = rd\theta$.So, we have $u(z_o) = \frac{1}{2\pi r}\int_{_dD(z_o,r)}u(z)ds$. And substituting $u(z) = u(x + iy) = u(x,y)$ gives us (ii). – matt Oct 21 '12 at 2:23
for (iii) I have been trying to use Green's theorem, but so far I haven't worked it out correctly. – matt Oct 21 '12 at 2:26

For part ii) substiute $z = z_o + re^{i\theta}$ and rewrite $d\theta = \frac{ds}{r}$ (since $s = r\theta$).

Then,

$$u(z_o) = \frac{1}{2\pi r}\int_{_dD(z_o, r)} u(x,y)ds$$

Since $u(z) = u(x + iy) = u(x,y)$.

For part iii) rewrite:

$$\int\int_{_D(z_o, r)} u(x,y) = \int_0^r (\int_{_dD(z_o, \rho)} u(x,y)ds)d\rho$$

By part i) the inner on the right hand side evaluates to $2\pi \rho u(z_o)$, so we have:

$$2\pi u(z_o) \int_0^r \rho d\rho = \pi r^2 u(z_o)$$

Therefore,

$$u(z_o) = \frac{1}{\pi r^2}\int\int_{_D(z_o, r)} u(x,y)dxdy$$

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