Circle average

Remarks

Example: let $f(z) = z$, with $c \in \C$ and $r > 0$. Then $M(f; c, r) = c + \frac{r}{2\pi} \int_0^{2\pi} e^{i\theta} \, d\theta = c$, since the integral of $e^{i\theta}$ over a full period is $0$. Non-example: let $f(z) = 1/\abs{z - 1}$ on $\C \setminus \set{1}$, with $c = 0$ and $r = 1$. Then $f(e^{i\theta}) = 1/(2\abs{\sin(\theta/2)})$ for $0 < \theta < 2\pi$, which is not integrable near $\theta = 0$, so the circle average is not defined. For $r = 0$ the circle average is $f(c)$. The mean value property says that $M(u; c, r) = u(c)$ when $u$ is harmonic on a neighbourhood of the closed disc of centre $c$ and radius $r$. For $r > 0$, $M(f; c, r)$ is the integral of $f$ against arc length on the circle, divided by the length $2\pi r$.