Does a continuous function have a continuous integral function in a discrete dynamical system?

Question

Let $X$ be a compact manifold (or the closure of a Euclidean domain if that helps significantly) and $T\colon X\to X$ a homeomorphism. Let us say that a function $v\colon X\to\mathbb R$ is the integral function of $w\colon X\to\mathbb R$ if $w(x)=v(T(x))-v(x)$ for all $x\in X$. It is clear that a function has an integral function if and only if it sums to zero over all periodic orbits of $T$. By this summing to zero I mean that if $T^n(x)=x$, then $\sum_{k=1}^nw(T^k(x))=0$. The integral function is unique up to constants that can be chosen independently for each trajectory of $T$.

If a continuous function $w$ has an integral function, does it also have a continuous integral function? That is, can the constants for different trajectories chosen in such a consistent way as to make the integral function continuous? I assume the answer depends on the properties of the dynamical system given by $T$, but I have not found a proof (under any additional assumptions) or a counterexample.

Answer 1

There's a lot of literature on this question, the keywords are "cohomological equation" and "Livsic theorem". I don't have the references handy at the moment, but the main idea is that if the system is Anosov and periodic obstructions vanish, then you get extra regularity in the solution to the cohomological equation. Namely, if your data (the function w) is Holder then so is the solution v.

On the other hand, without any hyperbolic properties there's no reason to expect a continuous solution, or even an integrable one. Let T be an irrational rotation and w be constant and nonzero. There are no periodic points, so that condition is vacuously satisfied, but clearly there's no integrable solution to the cohomological equation, otherwise w would have zero integral, since Lebesgue measure is invariant.