# Mean Value Property on Ball

$$\textbf{Problem}$$ Let $$\Omega$$ be open and connected in $$\mathbb{R}^n$$ for $$n\geq 2$$, and suppose that $$u \in C^2(\Omega)$$. Prove that the following statements are all equivalent. \begin{align*} (\textrm{i}) \quad u \textrm{ is harmonic in } \Omega\\ (\textrm{ii}) \quad \textrm{If } \overline{B_r(x)} \subset \Omega, \textrm{ then}\\ &u(x)=\frac{1}{\textrm{Vol} \partial B_r(x)}\int _{\partial B_r(x)} u(y)dS(y).\\ (\textrm{iii}) \quad \textrm{If } \overline{B_r(x)} \subset \Omega, \textrm{ then}\\ &u(x)=\frac{1}{\textrm{vol}B_r(x)}\int_{B_r(x)} u(y) dy. \end{align*}

I proved (i) and (ii) are equivalent. I want to know how to prove (i) and (iii) are equivalent.

I knew that this problem was already uproaded. However, I don't know the Poisson's Integral fromula on $$\mathbb{R}^n$$....

Any help is appreciated..

Let's see that $$1)$$ implies $$3)$$: \begin{align}\frac{1}{\textrm{vol}B_r(x)}\int_{B_r(x)} u(y) dy&=\frac{1}{\textrm{vol}B_r(x)} \int_0^r\int_{\partial B_s(x)}u(y)dS(y)ds\\ &=\frac{1}{\textrm{vol}B_r(x)} \int_0^r\textrm{vol}\partial B_s(x)u(x)ds\\ &=u(x)\frac{1}{\textrm{vol}B_r(x)}\textrm{vol}B_r(x)=u(x) .\end{align} The second equality is given by the already proved equivalence between $$1)$$ and $$2)$$.
Let's see that $$3)$$ implies $$1)$$:
Define $$\phi(r)=\frac{1}{\textrm{vol}\partial B_r(x)}\int_{\partial B_r(x)}u(y)dS(y).$$ You can check (and probably you already did proving the first equivalence) that $$\phi'(r)=\frac{r}{n\textrm{vol}B_r(x)}\int_{B_r(x)}\Delta u(y)dy.$$ Assume that $$\Delta u$$ is not identically $$0$$, we can find a point $$x$$ and an open ball $$B_\rho(x)$$ such that $$\Delta u(y)>0$$ in $$B_\rho(x)$$ (you can use the same argument if $$\Delta u(y)<0$$). This implies that $$\phi'(s)>0$$ for $$0, thus $$\phi$$ is strictly increasing in this interval. Finally, you can conclude \begin{align}\frac{1}{\textrm{vol}B_\rho(x)}\int_{B_\rho(x)} u(y) dy&=\frac{1}{\textrm{vol}B_\rho(x)} \int_0^\rho\int_{\partial B_s(x)}u(y)dS(y)ds\\ &=\frac{1}{\textrm{vol}B_\rho(x)} \int_0^\rho\textrm{vol}\partial B_s(x)\phi(s)ds\\ &>\phi(0)=u(x),\end{align} which is a contradiction.