In Problems and Solution in Mathematics by Ta-Tsien, exercise 5123, the mean value theorem is used as:

\begin{equation} \text{log} |F(0)| = \frac{1}{2 \pi} \int_0^{2\pi} \text{log}|F(re^{i\theta})| d\theta \end{equation}

Considering the expression provided by http://en.wikipedia.org/wiki/Harmonic_function#The_mean_value_property:

\begin{equation} u(x) = \frac{1}{n\omega_n r^{n-1}}\int_{\partial B(x,r)} u\, d\sigma \end{equation}

I expected the expression

\begin{equation} \text{log} |F(0)| = \frac{1}{2 \pi r} \int_0^{2\pi} \text{log}|F(re^{i\theta})| d\theta \end{equation}

As $n=2$ we can deduce $r^{n-1} = r$. What is wrong in my reasoning ?


The point is that $d\sigma$ denotes the $n-1$-dimensional (Hausdorff) measure on the sphere that gives you

$$\int_{\partial B(x,r)} 1\,d\sigma = n\omega_n r^{n-1}.$$

For the one-dimensional sphere in $\mathbb{R}^2$ or $\mathbb{C}$, that means

$$d\sigma = r\cdot d\theta.$$

The two factors $r$ then cancel,

$$\frac{1}{2\pi r} \int_{\lvert z\rvert= r} \log \lvert F(z)\rvert\,d\sigma = \frac{1}{2\pi r} \int_0^{2\pi} \log \lvert F(re^{i\theta})\rvert\,rd\theta = \frac{1}{2\pi} \int_0^{2\pi} \log \lvert F(re^{i\theta})\rvert\,d\theta.$$

  • $\begingroup$ This answer is very clear, thank you for naming the measure. $\endgroup$ – vkubicki Jul 26 '14 at 16:47

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