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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?

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Sorry, I mistyped my question. I meant f takes values in [0,1]! –  Nimr Jul 21 '11 at 19:56
Note there exists function with this poperty. Holomorphic function are suitable if you allow complex valued functions and real harmonic functions in the case before. There are no real valued holomorphic function except constant functions, there are no bounded harmonic functions, so you're finished if you can show that f is indeed harmonic. –  plusepsilon.de Jul 21 '11 at 20:06
With the restriction to [0,1], I can't see the link with harmonic functions anymore. –  Nimr Jul 21 '11 at 20:09
No, i just suggest. If the integral formula holds, then it is sufficient to show that this implies that $f$ is harmonic. Then you can argue that since $f$ is bounded and harmonic, it must be constant. –  plusepsilon.de Jul 21 '11 at 20:13
@George: although not solving the current problem, it is still quite interesting that $f(x,y)=e^{1.88044 x} \cos(6.947506 x)$ has the unit-circle averaging property, even though it is not harmonic! –  Gerald Edgar Jul 22 '11 at 23:38
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2 Answers

up vote 13 down vote accepted

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).

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.

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. 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 $a\approx1.88044+6.94751i$. In that case $f$ satisfies the required property but is not harmonic.

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These functions are called harmonic functions. One the simplest examples is $f(x,y)=xy$.

More generally, the real part of any holomorphic function $\mathbb C\to \mathbb C$ is a harmonic function $\mathbb C\to \mathbb R$.

Added later:
As mentioned by Gerald, harmonic functions are characterized by the property that $$\int_0^1f(z+re^{2\pi\theta})d\theta=f(z),\qquad \forall r\ge 0,\quad \forall z\in \mathbb C.$$ I don't know whether that property for $r=1$ implies that property for all $r\ge 0$.
Partial answer to the edited question:

If you require the function to be bounded, then I think that yes, that should force it to be constant. Liouville's theorem states that any bounded holomorphic function $\mathbb C\to \mathbb C$ is constant. There is also a version of Liouville's theorem for harmonic functions, so yes: the function is constant.

Gap in the argument:
▹ why is the function harmonic?

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He said unit circle. Harmonic functions have this property for all circles. Why do you say $xy$ is simplest? Shy not just $x$? –  Gerald Edgar Jul 21 '11 at 19:26
I think the question was, if $f: \mathbb{C} \rightarrow \mathbb{R}$ with $\int_0^{1} f( z + e^{2 \pi i \theta}) d \theta = f(z)$ implies that $f$ is constant. I am not seeing this being answered here. Can you please elaborate your answer? –  plusepsilon.de Jul 21 '11 at 19:49
@Gerald: You're right I missed the simplest one. @pm: It is not true that $\int_0^1f(z+e^{2\pi\theta})d\theta=f(z)$ $\forall z$ implies $f$ is constant. I have provided a counterexample. –  André Henriques Jul 21 '11 at 19:58
Sorry, I mistyped my question. I meant f takes values in [0,1]! –  Nimr Jul 21 '11 at 19:58
Dear André Henriques, I misread that you had implied that the formula implies that $f$ is indeed harmonic, but you only stated that the formula holds for harmonic functions and answered the question. Sorry for this. But I guess now the question is more interesting, since there are no bounded nonconstant harmonic functions=) –  plusepsilon.de Jul 21 '11 at 20:10
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