Let $U\subset\mathbb{R}^n$ be open and $u\in C(U)$. Show the following statements are equivalent:
$$ \frac{1}{|B(x,r)|}\int_{B(x,r)}u(y)\,dy=u(x) $$
$$\frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}u(y)dS(y)=u(x)$$
I know that when the second equation holds, $u$ is harmonic by the converse of the mean-value property and so the first equation holds as well.
But for the other implication, I have no clue. I've tried to set $$\phi(r)=\frac{1}{|\partial B(x,r)|}\int_{\partial B(x,r)}u(y)dS(y)$$ Then $$\phi'(r)=\frac{r}{n}\cdot\frac{1}{|B(x,r)|}\int_{B(x,r)}\Delta u(y)\,dy$$ Taking $\Delta u\neq0$, say $\Delta u>0$, we see that $\phi'(r)>0$, thus $\phi(r)$ is strictly increasing. Now $$\int_{B(x,r)}u(y)\,dy=\int_{0}^r\int_{\partial B(x,r)}u(y)\,dS(y)=\int_0^r|\partial B(x,r)|\cdot\phi(s)ds$$ but then I don't see how to continue
The equivalence of these two conditions is very basic. It's not necessary to pass through harmonicity, the only relevant fact is that $$ \int_{B(x,r)} u(y) \, dy = \int_0^r \int_{\partial B(x,s)} u(y) \, dS(y) ds. $$ To see this, denote the function $f(r) := \int_{B(x,r)} u(y) \, dy$, noting that now $f'(r) = \int_{\partial B(x,r)} u(y) \, dS(y)$. The equivalence of two mean value properties translates to the following: $$ f(r) = \omega_n r^n \cdot u(x) \text{ for all } r>0 \quad \Longleftrightarrow \quad f'(r) = n \omega_n r^{n-1} \cdot u(x) \text{ for all } r>0. $$ The implication $\Rightarrow$ is evident. Proving $\Leftarrow$, at first we only infer that $f(r) = \omega_n r^n \cdot u(x) + C$. But $f(r) \to 0$ as $r \to 0$ (we're integrating a bounded function on smaller and smaller balls), so this constant $C$ has to be zero.