Local mean value property?

Question

$\textbf{Meanvalue property}:$ 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?

Answer 1

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 $\rho$.

And knowing that the second theorem holds even if we just assume the local meanvalue property makes that theorem stronger.

The difference is hypothesis versus conclusion: A stronger conclusion makes a stronger theorem, while a weaker hypothesis also makes a stronger theorem.

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.

((i) implies (ii) is the first theorem. (ii) implies (iii) is trivial. (iii) implies (i) is the second theorem.)