Sequence of radius implies that a function is harmonic Let $u \in C(\Omega),$ where $\Omega \subset \mathbb{R}^{n}$ is open. If for all $x \in \Omega$ there exists a sequence of radius (positive) $\left\{r_{k}(x)\right\}_{k \in \mathbb{N}}$ such that $\lim _{k \rightarrow \infty} r_{k}(x)=0$ and
$$
u(x)=\dfrac{1}{n\omega_{n}r_k^{n-1}}\int_{\partial B(x,r_k(x))} u\quad\mathrm{dS} \quad \forall k \in \mathbb{N}
$$
Then $u$ is harmonic.
My attempt:
Let $y \in \Omega$ and $R > 0$ such that $\overline{B(y,R)} \subset \Omega.$ Let $v$ a harmonic function such that
\begin{array}{r}
\Delta v=0 \text{ when } x \in B(y,R) \\
v=u \text{  when } x \in \partial B(y,R)
\end{array}
Suppose by contradiction that $v \neq u$ in $\overline{B(y,R)}$ then there exists $x \in \overline{B(y,R)}$ such that
$$(v-u)(x)>0$$
Recall that $v=u$ in $\partial B(y,R)$ then $x \in B(y,R).$
So I am stuck. Any help? I was thinking about to use something like maximum principle but I am not sure, indeed I have not used the hypothesis about the radius or mean value property.
 A: We can use the following corollary
Corollary 10 (Comparison principle). Let $\Omega$ be a bounded open set, and let $u$ and $v$ be elements of $C^{2}(\Omega) \cap C(\bar{\Omega}) .$ Assume that $\Delta u \geq \Delta v$ in $\Omega$ and that $u \leq v$ on $\partial \Omega$. Then $u \leq v$ in $\Omega$.
But we need $u$ be an element of $C^{2}(\Omega).$
For that we use
Theorem: Let $\Omega \subset \mathbb{R}^N$, $u \in C(\Omega)$ be such that  $$\frac{1}{|B(x_0,R)|}\int_{B(x_0,R)}u(y)\ dy = u(x_0) = \frac{1}{|\partial B(x_0,R)|}\int_{\partial B(x_0,R)}u\ dS$$ for every ball $\overline{B(x_0,R)} \subset \Omega$. Then $u \in C^{\infty}(\Omega)$ and it is harmonic
Proof: Consider the standard mollifier: $$\rho(x) := \begin{cases}Ce^{-\frac{1}{1 - \|x\|^2}} & \text{if $\|x\|$ < 1} \\0 & \text{otherwise.}
\end{cases}$$ Here $C$ is a constant such that $\|\rho\|_{L^1} = 1.$ Let $\epsilon > 0$ and consider $$\rho_{\epsilon}(x) = \epsilon^{-N}\rho(x\epsilon^{-N}).$$
Set $\Omega_{\epsilon} = \{x \in \Omega : \text{dist}(x,\partial \Omega) > \epsilon\}$ and define for $x \in \Omega_{\epsilon}$
$$u_{\epsilon}(x) = \rho_{\epsilon} * u(x) = \int_{\Omega}\rho_{\epsilon}(x - y)u(y)\ dy.$$
The following is a well know theorem in analysis, if it is new to you you can look for a proof Analysis by Lieb and Loss or anywhere else.
**Theorem:**If $u \in C(\Omega)$, then $u_{\epsilon} \to u$ uniformly on compact subsets of $\Omega$, $u_{\epsilon} \in C^{\infty}(\Omega_{\epsilon})$ and for any multindex $\alpha$ we have $$\frac{\partial^{\alpha}u_{\epsilon}}{\partial x^{\alpha}}(x) = \int_{\Omega}\frac{\partial^{\alpha}\rho_{\epsilon}}{\partial x^{\alpha}}(x - y)u(y)\ dy.$$
Finally we can proceed with the proof!
Fix $x_0 \in \Omega_{\epsilon}$.
$$u_{\epsilon}(x_0) = \int_{B(x_0,\epsilon)}\rho_{\epsilon}(x - y)u(y)\ dy = \int_{B(0,\epsilon)}\rho_{\epsilon}(z)u(x_0 - z)\ dz = $$ $$ = \int_0^{\epsilon}r^{N - 1}\int_{\partial B(0,1)}\rho_{\epsilon}(rw)u(x_0 - rw)\ dS(w)dr = $$ $$ \int_0^{\epsilon}r^{N - 1}\rho(r)\int_{\partial B(0,1)}u(x_0 - rw)\ dS(w)dr = \int_0^{\epsilon}r^{N-1}\rho_{\epsilon}(r)\frac{\alpha_N N}{|\partial B(x_0,r)|}\int_{\partial B(x_0,r)}u(y)\ dS(y)dr $$ $$ = u(x_0)\|\rho\|_{L^1} = u(x_0).$$
This proves that $u = u_{\epsilon}$ and hence $u \in C^{\infty}(\Omega_{\epsilon})$, for every $\epsilon$.
Therefore $u$ is harmonic.
