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][1]). 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][2]. ---------- 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][3]. 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 [$a\approx1.88044+6.94751i$][4]. In that case $f$ satisfies the required property but is not harmonic. [1]: http://en.wikipedia.org/wiki/Harmonic_functions#Liouville.27s_theorem [2]: http://math.stackexchange.com/q/51926/1321 [3]: http://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html [4]: http://www.wolframalpha.com/input/?i=solve+I_0(z)%3D1+for+7.19+<+|z|+<+7.2