Does the this converse of the MVT hold true for harmonic functions?

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Let $\Omega$ be a bounded connected open region in $\Bbb R^n$, and $u$ is merely continuous on $\Omega$ that satisfies this MVT: $$u(x_0)=\frac1{|S_r(x_0)|}\int_{S_r(x_0)}u(y)dy$$ where $S_r(x_0)$ is the sphere of radius $r$ centred at $x_0$ and lies within $U$. Then is it true that $\Delta u=0$ in all of $U$?

The only difficulty lies in proving that $u$ is actually $C^2$ i.e. $\Delta u$ should at least exist and be continuous.


Background: this problem comes from an attempt to prove Harnack's first theorem, which asserts that if a sequence of functions $\{u_k\}$ are $C(\bar U)$ and are harmonic in $U$ and converge uniformly on $\partial U$, then they also converge uniformly to a continuous function $u$ on $\bar U$ which is harmonic in $U$.

My attempt: first use Cauchy's criteria to write something like $|u_m(x)-u_n(x)|<\epsilon$ on $\partial U$, then since $u_m-u_n$ is harmonic and thus satisfies the strong maximum principle, therefore the above inequality holds throughout $\bar U$ hence making $u_k$ converge uniformly to some $u\in C(\bar U)$. But is it possible to further show $u\in C^2(U)$?

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Take any closed ball in the domain and restrict $u$ to its boundary. Use the Poisson kernel to extend it to a harmonic function $f$ on the ball. The difference $u-f$ also has the mean value property and vanishes on the boundary of the ball. Use these properties to conclude that both the maximum and minimum of $u-f$ must be $0$ on the ball. (It must be constant on any concentric sphere around a point where it has a maximum or minimum.)