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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 $\Omega$.

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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 $$\frac1{m(B(p,r))}\int_{B(p,r)}v<M.$$

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  • $\begingroup$ What is the Poisson integral? All I can think of is integrating $\phi$ against the Poisson kernel, but isn't that the solution of Laplace's equation on the 2-dimensional unit ball? I was asking for any dimension, not just 2. This is a nice answer for 2 dimensions, but can it be used for $n$ dimensions? $\endgroup$ – MickG Jul 11 '15 at 19:05
  • $\begingroup$ There is a Poisson kernel for the unit ball in $\mathbb R^n$. Try Wikipedia... $\endgroup$ – David C. Ullrich Jul 11 '15 at 19:09
  • $\begingroup$ Here it is. Hadn't heard of it just yet. $\endgroup$ – MickG Jul 11 '15 at 19:20
  • $\begingroup$ Final note: apply the argument to any ball and voilà, harmonicity is proved. $\endgroup$ – MickG Jul 11 '15 at 19:28
  • $\begingroup$ When you say «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 surface averages», do you mean "volume" instead of the second "surface"? $\endgroup$ – MickG Jul 11 '15 at 19:30

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