Continuous functions on $\mathbb{R}^2$ with special property

Question

The following problem is from Miklos Schweitzer competition (Year 1983, Problem 7):

Prove that if the function $f: \mathbb{R}^{2}\to [0, 1]$ is continuous, and its average on every circle of radius one equals the function value at the center, then $f$ is constant.

There is some discussion here but no elementary solution (from my point of view). I would like to see a solution that avoids Fourier transforms. But perhaps someone could elaborate on Kent Merryfield's observations and give a complex-analytic solution?

From the number of upvotes this meta question has received, I will assume that its okay to ask such questions here.

Answer 1

A proof from the martingale convergence theorem. (I got this from Persi Diaconis back when we were in graduate school.)

Consider i.i.d. random variables $X_n$, distributed according to normalized arc length on the unit circle in $\mathbb R^2$. These are defined on some sample space $(\Omega,\mathcal F, \mathbb P)$. Define a stochastic basis $(\mathcal F_n)$, defining $\mathcal F_n \subset \mathcal F$ to be the sigma-field generated by $X_1,\cdots,X_n$. Define $S_n = X_1+\cdots +X_n$, a random walk in $\mathbb R^2$ suggested by this problem.

Almost surely, the sequence $(S_n)$ is dense in the plane. Such random walks in $2$ dimensions are recurrent: another fact needed here. (The probability of returning to a small neighborhood of the origin after $n$ steps is asymptotically proportional to $1/n$ and the series $\sum 1/n$ diverges, so by Borel-Cantelli, we return infinitely often.)

Now suppose we have our continuous function $f : \mathbb R^2 \to [0,1]$ with the averaging property. Consider the real stochastic process $$ Y_n = f(S_n) $$ The conditional expectation obeys $$ \mathbb E[Y_{n+1} | \mathcal F_n] = Y_n . $$ This is exactly the averaging property, and the reason $X_n$ was defined as it was. Thus: $(Y_n)$ is a martingale. The values are in $[0,1]$ so it is a bounded martingale. Therefore by the martingale convergence theorem, $Y_n$ converges a.s. But, with probability one, $S_n$ is dense in the plane. So (recall $f$ is continuous) the only way $f(S_n)$ can converge is for $f$ to be constant in the plane.

Answer 2

An almost-proof from the central limit theorem. This write up took far more words than I expected. The point is that $f(x_0, y_0)$ is the average of $f$ on a circle centered around $(x_0, y_0)$. The value of $f$ at each point on that circle is, in turn, the average of $f$ at circles centered round the points of the unit circle. This means that $f(x_0,y_0)$ can be written as some sort of two dimensional average of the values of $f$ on a disc of radius $2$ around $(x_0, y_0)$. Repeating this argument, $f$ can be written as a weighted average of the values of $f$ on a disc of radius $n$ and, by the central limit theorem, as $n \to \infty$, the weighting function will start to look like a bell curve of width $\approx \sqrt{n}$. So $f(x_1,y_1) - f(x_2,y_2)$ is the integral of $f$ against a difference of two bell curves, whose centers are displaced from each other by $(x_1-x_2, y_1-y_2)$. As $n \to \infty$, the two bell curves will get very wide and flat, so the difference between their centers becomes insignificant and the integral goes to zero. Okay, now for the proof.

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The hypothesis is that $$f(x,y) = \frac{1}{2 \pi} \int_{\theta} f(x+\cos \theta, y+\sin \theta) d \theta \quad (\ast)$$ Plugging formula $(\ast)$ into itself, $$f(x,y) = \frac{1}{4 \pi^2} \int_{\theta} \int_{\phi} f(x+\cos \theta+\cos \phi, y+\sin \theta + \sin \phi) d \theta d \phi$$ which, if I got the Jacobian factors right, turns into $$f(x,y) = \frac{1}{2 \pi^2} \int_{u,v} f(x+u,y+v) \sqrt{(u^2+v^2)(2-u^2-v^2)} du dv$$ I don't care about the details of this formula, what I care is that it looks like $$f(x,y) = \int_{u,v} f(x+u, y+v) \mu_2(u,v) du dv$$ for some function $\mu_2$.

If I plug $(\ast)$ into itself to get a triple integral, I will get some different formula $$f(x,y) = \int_{u,v} f(x+u, y+v) \mu_3(u,v) du dv$$ and, in general, we have $$f(x,y) = \int_{u,v} f(x+u, y+v) \mu_n(u,v) du dv$$ where $\mu_n$ is the $n$-fold convolution of the measure of the unit circle with itself. If you are uncomfortable convolving measures and want to convolve functions, take $n$ even and define $\mu_n = \mu_2 \ast \mu_2 \ast \cdots \ast \mu_2$, where I have convolved $n/2$ times.

By the central limit theorem in $\mathbb{R}^2$, $$\mu_n(x, y) \approx \frac{C}{n} \exp\left(-\frac{x^2+y^2}{n \sigma} \right)$$ Here $C$ and $\sigma$ are positive real numbers I didn't feel like computing. This proof will be nonrigorous to the extent that I don't replace that $\approx$ by some precise limiting notion; I was too lazy to scan through the different notions of convergence for which the central limit theorem has been proved and figure out which one I want.

Now, fix $(x_1, y_1)$ and $(x_2, y_2) \in \mathbb{R}^2$. So $$f(x_1, y_1) - f(x_2, y_2) =$$ $$ \int_{(u,v) \in \mathbb{R}^2} f(x_1+u, y_1+v) \mu_n(u,v) dx dy - \int_{(u,v) \in \mathbb{R}^2} f(x_2+u, y_2+v) \mu_n(u,v) dx dy=$$ $$\int_{(a,b) \in \mathbb{R}^2} f(a,b) \left( \mu_n(a-x_1, b-y_1) - \mu_n(a-x_2, b-y_2) \right) da db \approx$$ $$ \int_{(a,b) \in \mathbb{R}^2} f(a,b) \frac{C}{n} \left( \exp\left( - \frac{(a-x_1)^2+(b-y_1)^2}{\sigma n} \right) - \exp\left( - \frac{(a-x_2)^2+(b-y_2)^2}{\sigma n} \right) \right) da db.$$

Acting as if the $\approx$ were an $=$, we have $$|f(x_1,y_1) - f(x_2, y_2)| \leq$$ $$\frac{C}{n} \int_{(a,b) \in \mathbb{R}^2} \left| \exp\left( - \frac{(a-x_1)^2+(b-y_1)^2}{\sigma n} \right) - \exp\left( - \frac{(a-x_2)^2+(b-y_2)^2}{\sigma n} \right) \right| da db.$$

We will show that $$\lim_{n \to \infty} \frac{C}{n} \int_{(a,b) \in \mathbb{R}^2} \left| \exp\left( - \frac{(a-x_1)^2+(b-y_1)^2}{\sigma n} \right) - \exp\left( - \frac{(a-x_2)^2+(b-y_2)^2}{\sigma n} \right) \right| da db = 0$$ and, thus, once we get the right form of the central limit theorem to cite, $f$ is constant.

So it remains to compute $$\lim_{n \to \infty} \frac{C}{n} \int_{(a,b) \in \mathbb{R}^2} \left| \exp\left( - \frac{(a-x_1)^2+(b-y_1)^2}{\sigma n} \right) - \exp\left( - \frac{(a-x_2)^2+(b-y_2)^2}{\sigma n} \right) \right| da db $$ By translational and rotational symmetry, we may assume that $(x_1, y_1)$ and $(x_2, y_2)$ are of the form $(r,0)$ and $(-r,0)$ for some positive $r$.

$$\frac{C}{n} \int_{(a,b) \in \mathbb{R}^2} \left| \exp\left( - \frac{(a-r)^2+b^2}{\sigma n} \right) - \exp\left( - \frac{(a+r)^2+b^2}{\sigma n} \right) \right| da db =$$ $$\frac{C}{n} 2 \int_{b=-\infty}^{\infty} \int_{a=0}^{\infty} \exp\left( - \frac{(a-r)^2+b^2}{\sigma n} \right) - \exp\left( - \frac{(a+r)^2+b^2}{\sigma n} \right) da db =$$ $$\frac{2C}{n} \int_{b=-\infty}^{\infty} \int_{a=0}^{2r} \exp\left( - \frac{(a-r)^2+b^2}{\sigma n} \right) da db = \frac{2C}{n} \int_{c=-r}^r \exp \frac{-c^2}{\sigma n} dc \int_{b=-\infty}^{\infty} \exp\left( \frac{-b^2}{\sigma n} \right) db$$ $$=\frac{2 C \sqrt{\pi \sigma}}{\sqrt{n}} \int_{c=-r}^r \exp \frac{-c^2}{\sigma n} dc = 2 C \sqrt{\pi \sigma} \int_{x=-r/\sqrt{n}}^{r/\sqrt{n}} \exp\left(- \frac{x^2}{\sigma} \right) dx.$$ (We have made the subsititutions $c=a-r$ and $x=c/\sqrt{n}$.) Clearly, the last integral goes to $0$.