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A function $f$ on $\mathbb R^n$ is called harmonic if $\Delta f=0$ where $\Delta$ is the Laplacian. A function $f$ is harmonic if and only if $f$ satisfies mean value property (MVP), i.e. averaging of $f$ on any sphere is the value of $f$ at the centre. Can there be a non-harmonic function $f$ which satisfies MVP for spheres of some radii? In other words, to conclude that a function is harmonic do I have to test MVP for spheres of all positive radii (or at least on a dense set of radii)?

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Here's a related MO question:… –  Andrey Rekalo Apr 3 '12 at 8:48
You probably need some extra condition on $f$ to say that MVP implies Harmonic; say, $f$ continuous. (At the very least, you'd need Lebesgue measurability etc. even to define the Mean Value as an integral!) Assuming $f$ continuous, local MVP is enough: for each point $x$ there exists $\delta(x) > 0$ such that MVP holds on all spheres centred at $x$ of radius $< \delta(x)$. But for example, $f(x,y) = \sgn(x)$ has the local MVP on $R^2$, but is not continuous. –  Zen Harper Apr 3 '12 at 16:46
Also, assuming $f$ is continuous, then if we fix the centre $x$ and let $MV(\delta)$ be the Mean Value of $f$ over the ball of radius $\delta > 0$ centred at $x$, then $MV$ is a continuous function of $\delta$ by Dominated Convergence (or just uniform continuity of $f$), so you can assume the set of radii is closed. But in fact, the case $f$ continuous is apparently already partially solved, according to the MO question and answers referred to in Andrey Rekalo's above comment. –  Zen Harper Apr 3 '12 at 16:57

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