Let $\mathcal{F}$ be the set of continuous functions $\varphi$ from $\mathbb{C}$ to $[0,1]$ that satisfy $\begin{align}\varphi(z)=\frac{1}{2\pi}\int_{0}^{2\pi}\varphi(z+e^{i\theta})d\theta\end{align}$ for all $z\in\mathbb{C}$. For $\varphi\in\mathcal{F}$ and $\delta\gt0$, let $\varphi_\delta$ be the function defined by $\begin{align}\varphi_\delta(z)=\frac{1}{\delta^2}\int_{\delta/2}^{\delta/2}\int_{\delta/2}^{\delta/2}\varphi(z+\alpha+i\beta)d\alpha d\beta\end{align}$.
Is the set of functions $\varphi_\delta$ compact w.r.t to the sup norm?
I originally wanted to prove that $\mathcal{F}$ contains constant functions only. So far, I managed to prove that any compact subset of $\mathcal{F}$ (with some invariance under rotations and translations) contains constant functions only. If the set of $\varphi_\delta$'s is shown compact, I would achieve my original intent.
BTW. this is related to this post which has a probabilistic solution.

$\begingroup$ is the property that $\varphi$ satisfies here the same with the one in your post, just average of the unit circle? $\endgroup$– Shaoming GuoCommented Aug 22, 2011 at 23:33

$\begingroup$ I guess a weaker version of Ascoli's theorem is needed (sorry this is meant as a comment but I don't have the privilege!). $\endgroup$– user17330Commented Aug 23, 2011 at 4:49

$\begingroup$ Yes it is the same property. $\endgroup$– NTTCommented Aug 23, 2011 at 18:22
1 Answer
Your space $\mathcal F$ consists of constant functions only. The arguments referred to in the post which you quote are somewhat misleading as they use "too much": reduction to the Liouville theorem about bounded classical harmonic functions or recurrence for a compactly supported symmetric random walk in dimension 2. In fact, neither of these two properties is really needed.
One can reformulate your question in the following more general way. Let $\mu$ be a probability measure on a locally compact abelian group $G$ (particular case: $G=\mathbb R^d$). Call a function $\mu$harmonic if $f(x)=\int f(x+y)\ d\mu(y)$ for any $x\in G$. If the closure of the group generated by the support of $\mu$ is the whole group $G$, then all measurable bounded $\mu$harmonic functions are constant a.e. with respect to the Haar measure (in particular, all continuous bounded $\mu$harmonic functions are constant). This result is known as the ChoquetDeny theorem (proved in 1960).

$\begingroup$ Thanks for pointing this out! I still would like to have an answer for my original question (i.e. compactness of the set defined above). $\endgroup$– NTTCommented Aug 23, 2011 at 18:21

$\begingroup$ Since you define the functions $\phi_\delta$ as averages of functions $\phi$ from the set $\mathcal F$ which consists of constants only, all $\phi_\delta$ are also constants, aren't they? Or are you asking for an argument working without prior knowledge that $\mathcal F$ is trivial? $\endgroup$– R WCommented Aug 24, 2011 at 16:56

$\begingroup$ @RW, yes! I'm asking for an argument without prior knowledge that $\mathcal{F}$ is trivial. $\endgroup$– NTTCommented Aug 24, 2011 at 21:15

$\begingroup$ It seems that such an argument (if it exists) would still be equivalent to proving triviality of $\mathcal F$ (or a little bit weaker). The point is that if you look just at bounded harmonic functions (be it in your setup or for more general Riemannian manifolds or Markov operators), then they also form a Banach space (with respect to the sup norm), and its unit ball is compact only if the space is finite dimensional. $\endgroup$– R WCommented Aug 25, 2011 at 7:46

$\begingroup$ @RW, if I can prove that the $\varphi_\delta$s are uniformly bounded and equicontinuous, would there be a known result that can help me reach compactness? $\endgroup$– NTTCommented Aug 26, 2011 at 2:19