Is it true that if the average of a continuous function $f:\mathbb{R}^2\rightarrow[0,1]$ over a unit circle centered around $(x,y)$ is $f(x,y)$ for all $(x,y)\in\mathbb{R}^2$, then $f$ is necessarily constant?



On the average of continuous functions $f:\mathbb{R}^2\rightarrow[0,1]$
Yes, any such $f$ is constant. In fact, if we relax the condition so that $f$ is only required to be bounded below, but not above, then it is still true that $f$ is constant. This can be proven by martingale theory, as can the statement that harmonic functions bounded below are constant ([Liouville's theorem](http://en.wikipedia.org/wiki/Harmonic_functions#Liouville.27s_theorem)).


Let $X\_1,X\_2,\ldots$ be a sequence of independent random random variables uniformly distributed on the unit circle, set $S\_n=\sum\_{m=1}^nX\_m$ and let $\mathcal{F}\_n$ be the sigma-algebra generated by $X\_1,X\_2,\ldots,X\_n$. Then, $S\_n$ is a random walk in the plane, and is recurrent. Your condition is equivalent to $\mathbb{E}[f(S\_{n+1})\vert\mathcal{F}\_n]=f(S\_n)$. That is, $f(S\_n)$ is a martingale. It is a standard result that a martingale which is bounded below converges to a limit, with probability one. However, as $S\_n$ is recurrent, this only happens if $f$ is constant almost everywhere. By continuity of $f$, it must be constant everywhere.


For the same argument applied to functions $f\colon\mathbb{Z}^2\to\mathbb{R}$, see Byron Schmuland's answer to [this math.SE question](https://math.stackexchange.com/q/51926/1321).




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In general, for a continuous function $f\colon\mathbb{R}^2\to\mathbb{C}$, if $f(x,y)$ is the average of $f$ on the unit circle centered at $(x,y)$ then it *does not* follow that $f$ is harmonic. So, we cannot prove the result directly by applying Liouville's theorem. As an example (based on the comments by Gerald Edgar and by me), consider $f(x,y)=\exp(ax)$. The average of $f$ on the unit circle centered at $(x,y)$ is
$$
\frac{1}{2\pi}\int\_0^{2\pi}f(x+\cos t,y+\sin t)\,dt=\frac{1}{2\pi}f(x,y)\int\_0^{2\pi}e^{a\cos t}\,dt=f(x,y)I\_0(a).
$$
Here, $I\_0(a)$ is the [modified Bessel function of the first kind](http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html). Whenever $I\_0(a)=1$ then $f$ satisfies the required property. This is true for $a=0$, in which case $f$ is constant, but there are also nonzero solutions such as [$a\approx1.88044+6.94751i$](http://www.wolframalpha.com/input/?i=solve+I_0(z)%3D1+for+7.19+<+|z|+<+7.2). In that case $f$ satisfies the required property but is not harmonic.


