Converse to mean value property: ball mean value property implies harmonicity It is well-known that harmonic functions satisfy the mean value property. That is, if we set $\alpha(n)$ to be the volume of the unit $n$-ball, we have the following theorem.

Let $u$ be an harmonic function on $\Omega\subseteq\mathbb{R}^n$ an open set. Let $x\in\Omega$ and $r$ be a radius such that $B(x,r)\subseteq\Omega$. Then:
  $$\frac{1}{n\alpha(n)r^{n-1}}\int_{\partial B(x,r)}u(y)\mathrm{d}S(y)\overset{\ast}{=}u(x)\overset{\star}{=}\frac{1}{\alpha(n)f^n}\int_{B(x,r)}u(y)\mathrm{d}y.$$

I also know that $\ast$ alone implies harmonicity. That is, if $\ast$ holds for any $r$ such that $B(x,r)\subseteq\Omega$ and $x\in\Omega$, then $u$ is harmonic on $\Omega$. I was thus wondering if $\star$ also implies harmonicity in a similar way. I came across this question, but the question asks for a proof that $\ast$ implies harmonicity and the answer proves harmonicity while assuming both $\ast$ and $\star$. The Wikipedian proof in the above link seems to be much the same result, in both directions. Besides, it uses convolutions and seems, well, convoluted, and I don't know how much I am convinced that moving a laplacian around in a convolution causes no harm, i.e. $u\ast\Delta v=\Delta u\ast v$, as Wikipedia assumes. So is there a way to prove the following?

Let $\Omega\subseteq\mathbb{R}^n$ be open and $u:\Omega\to\mathbb{R}$ satisfy the "volume mean value property", that is for all $x\in\Omega$ and $r$ such that $B(x,r)\subseteq\Omega$ we have:
  $$u(x)=\frac{1}{\alpha(n)r^n}\int_{B(x,r)}u(y)\mathrm{d}y.$$
  Then $u$ is harmonic in $\Omega$.

 A: Well first we should note that the result you state is not true without some smoothness assumptions on $u$. Let's assume $u$ is continuous. We also assume $u$ is real-valued.
There's a standard proof of the result for surface averages that doesn't involve the things you're concerned about and that works just as well for volume
 averages:
Suppose $\overline{B(0,1)}\subset\Omega$. We will show that $u$ is harmonic in $B=B(0,1)$. Let $\phi$ be the restriction of $u$ to the boundary of the unit ball, and let $$f=P[\phi],$$the Poisson integral of $\phi$. Then $f$ is harmonic in $B$ and extends continuously to $\overline B$, with values on the boundary given by $\phi=u$. We will let $f$ denote the extension of the original $f$ to $\overline B$. Now define $v:\overline B\to\mathbb R$ by $$v(x)=u(x)-f(x)\quad(x\in\overline B).$$If we can show $v=0$ in $B$ we're done, since $f$ is harmonic in $B$.
Suppose that $v$ does not vanish identically in $B$. Wlog assume that $v$ is strictly positive at some point of $B$. Define $$M=\sup_{x\in B}v(x)$$ and $$K=\{x\in B\,:\,v(x)=M\}.$$
Since $v$ is continuous in $\overline B$ and vanishes on the boundary it follows that $K$ is a nonempty compact subset of $B$. Let $p$ be a point of $K$ at minimal distance to the boundary of $B$.
Now $v$ satisfies the volume mean-value property in $B$ since both $u$ and $f$ do. So for small $r>0$ we have $$M=v(p)=\frac1{m(B(p,r))}\int_{B(p,r)}v.$$But this is impossible: $v\le M$ everywhere in $B(p,r)$ and $v<M$ in a nonempty open subset of $B(p,r)$, so $$\frac1{m(B(p,r))}\int_{B(p,r)}v<M.$$
