# Local mean value property?

$\textbf{Meanvalue property}:$ Let $G\subseteq \mathbb{C}$ be open and assume that $\overline{K(a,\rho)}\subseteq G$. We say that a harmonic function $h:G\to \mathbb{R}$ posseses the mean value property if $$h(a)=\int_{0}^{2\pi}h(a+\rho e^{i\theta})d\theta$$

$\textbf{Converse to mean value property}$:

If $h:G\to \mathbb{R}$ is a continuous function that satisfies the local mean value property i.e. given $a\in G$ there exists $\rho>0$ such that $$h(a)=\int_{0}^{2\pi}h(a+r e^{i\theta})d\theta$$ then $h$ is harmonic on $G$.

I cannot see why the mean value property is local only in the second case. Since in the first case the choice of $\rho$ does indeed depend on a?

Also is there any way to combine the two theorems it into an if and only statement?

• The MVP is not properly stated.
– zhw.
Dec 12, 2017 at 1:24
• @zhw how come? `? Dec 12, 2017 at 14:11
• The quantifiers are out of order. I would state it this way: $\textbf{Mean value property}:$ Let $G\subset \mathbb{C}$ be open and assume $h:G\to \mathbb{R}$ is continuous. We say $h$ has the mean value property in $G$ if $$h(a)=\int_{0}^{2\pi}h(a+\rho e^{i\theta})d\theta$$ whenever $\overline{K(a,\rho)}\subseteq G$. Thm: If $h$ is harmonic in $G,$ then $h$ has the mean value property in $G.$ @s
– zhw.
Dec 12, 2017 at 18:07

Knowing that the first theorem holds for every $\rho>0$ (such that $\overline{K(a,\rho)}\subset G$) gives a much stronger theorem than if it were true just for one $\rho$.
An if and only if follows. Stated informally, if $h$ is continuous the following are equivalent: (i) harmonic (ii) mean-value property (iii) local mean-value property.
• @DavidCUllrich I still have trouble understanding what is ment by "the mean value value theorem holds globally," as the value of $\rho$ will always depends on $a$? Dec 11, 2017 at 19:54
• What it means is that it works whenever the closed disk is contained in $G$. Dec 11, 2017 at 21:02