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 $\Omega$.