It is well known and easy to prove that any Denjoy counterexample in the circle is approximated by homeomorphisms of the circle which have the same rotation number and are transitive (in particular, they are conjugated to the rotation).

My question is the following. Given a $C^1$ denjoy counterexample $f:S^1\to S^1$, does there exists a sequence $f_n$ of $C^1$ diffeomorphisms of the circle such that:

1-the rotation number of $f_n$ is the same as the one of $f$.

2-the diffeomorphisms $f_n$ are conjugated to the rotation.

3-$f_n \rightarrow f$ in the $C^1$ topology.

If the diffeomorphisms $f_n$ are of class $C^2$, there is no need to ask for the second hypothesis by the Theorem of Denjoy.

I would guess that for diophantine rotation number, one could even get that the $f_n$ are $C^1$ conjugated to the rotation. I believe that this must be known, but I could not find any reference.


1 Answer 1


Dear Raphael,

perhaps I misunderstood the question, but in order to get your sequence $f_n$, one can $C^1$-approximate $f$ by $C^2$-diffeomorphisms $g_n$ and then compose each $g_n$ with an adequate small rotations $R_{\epsilon_n}$ to adjust the rotation number (so that $f_n=R_{\epsilon_n}\circ g_n$ do the job), right?



  • $\begingroup$ This answers the question. Do you know how the conjugations are? If they can be made explicit? I know how to do them in the continuous case by constructing conjugations that map an interval into smaller ones, but for the C1 case is not as clear. And do you know if what I've said about being C1 conjugated (which your observation proves is right for diophantine numbers) holds also for Liouvilleans? $\endgroup$
    – rpotrie
    Commented May 14, 2010 at 15:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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