TITLE: For holomorphic functions, does the mean-value property for $0<r<R$ imply it for $r=R$?
QUESTION [8 upvotes]: The mean-value property for holomorphic functions states that if $f$ is holomorphic in an open disc centered at $z_{0}$ of radius $R$, then
$$f(z_{0}) = \frac{1}{2\pi}\int_{0}^{2\pi}f(z_{0} + re^{i\theta})\, d\theta$$
for any $0 < r < R$. When can I say this is true for $r = R$?

REPLY [5 votes]: As Henning mentions in his comment, the condition is that $f$ is continuous on the closed disk $\{z:|z-z_0|\le R\}$. When that is the case, $f$ is uniformly continuous (because it is continuous on a compact set) and so for any $\epsilon>0$, there is a $\delta>0$ so that if $0<R-r<\delta$, then
$$
|f(z_0+Re^{i\theta})-f(z_0+re^{i\theta})|<\epsilon
$$
Therefore,
$$
\left|\frac{1}{2\pi}\int_0^{2\pi}\left(f(z_0+Re^{i\theta})-f(z_0+re^{i\theta})\right)\;\mathrm{d}\theta\right|<\epsilon
$$