Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose we have a functional question, such as the one here. Often a key step in these problems is to try to equate or chain results together (see my hint). From looking at these questions, it seems that there is often only a finite number of ways that we can attempt to join them together. Can this fact be examined more formally?

Note: I am not asking you to turn my answer from that question into something algebraic. It is that way purposely as I was only trying to give a hint, not a complete solution. What I am asking, is whether there are any results that show that the functions can only be "combined" or "equated" in a certain number of ways on some reasonably general class of problems.

share|cite|improve this question

One example is that there's an easy way to get uniqueness of functional equations in certain cases. Consider the "cohomological" equation $f(Tx) - f(x) = g(x)$ (which is important in dynamical systems, e.g. it comes up when you are trying to find an invariant $n$-form). The goal, of course, is to prove that there exists $f$ given $g$. If $T: X \to X$ is topologically transitive, then any two solutions differ by a constant. This is because if $x_0$ has a dense orbit, then the value of $f(x_0)$ determines $f(Tx_0), f(T^2x_0)$, and so on.

More generally, there is the same uniqueness for the "twisted" cohomological equation $f(Tx) - \alpha f(x) = g(x)$, where $\alpha$ is a nonzero real number.

(This is not, I think, relevant to the question, but I'll mention that one can get existence of a Hölder solution in the untwisted case for transitive Anosov diffeomorphisms of compact manifolds if the sum of $g$ over every finite orbit is zero, though it requires a clever trick. This is a result of Livsic.)

share|cite|improve this answer
This involves a lot of maths I don't know - will take me a while before I can understand this answer – Casebash Aug 10 '10 at 23:28

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.