I'm doing exercises from book about vector calculus. And there is problem which I'm not sure what to do. Let's see the hypothesis. Let, $D \subseteq \mathbb{R}^2$ an open set. Let, $u:\overline{D} \to \mathbb{R} \in C^2$. Suposse that, $\vec{p} \in D$, and, $0<\rho \leq R$. Consider, a ball $B_R:= B(\vec{p},R)$. If $\nabla^2u = 0 $ at $D$ then : $$\frac{1}{2\pi R}\int_{\partial B_R} u \, ds = u(\vec{p})$$
So far, I know two facts, which the author suggested to readers : $$ \lim_{\rho \to 0} \frac{1}{\rho}\int_{\partial B_\rho}u \,ds = 2\pi u(\vec{p}) $$ $$ \frac{d}{d\rho} \left ( \frac{1}{\rho}\int_{\partial B_\rho}u \,ds \right) = \frac{1}{\rho}\iint_{B_\rho}\nabla^2 u \,ds = 0 $$ That's means : $$\frac{1}{\rho}\int_{\partial B_\rho}u \,ds = constant$$
Therefore, the expression above, should be equal to $2\pi u(\vec{p})$. No matter what values $\rho$ takes in particular if I take $\rho = R$. Then we are done. But I'm not really sure if the argument above works or I'm ignoring something.
Any help is appreciated