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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 $r = R$?

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By continuity, it ought to be enough that $f$ is continuous on the closed disc of radius $R$. (This might be relevant, for example, if $f(z)=\sqrt{z}$ and $R=z_0\in\mathbb R$). – Henning Makholm Oct 15 '11 at 20:05
up vote 3 down vote accepted

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, $$ \left|\frac{1}{2\pi}\int_0^{2\pi}\left(f(z_0+Re^{i\theta})-f(z_0+re^{i\theta})\right)\;\mathrm{d}\theta\right|<\epsilon $$

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