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The Margulis-Ruelle inequality states that measure-theoretic entropy is controlled by Lyapunov exponents; more precisely, if $f$ is a $C^{1+\alpha}$ diffeomorphism on a $d$-dimensional manifold $M$ and $\mu$ is a Borel $f$-invariant ergodic probability measure with Lyapunov exponents $\lambda_1, \dots, \lambda_d$, then $h_\mu(f) \leq \sum_{\lambda_i>0} \lambda_i$.

This also holds for non-invertible $C^1$ interval maps: we have $h_\mu(f) \leq \max(0,\lambda(\mu))$, where $\lambda(\mu) = \int \log|f'(x)|\\,d\mu(x)$. (See, for example, Proposition 4.1 of [Ledrappier, Some properties of absolutely continuous invariant measures on an interval, Ergodic Theory Dynam. Systems 1 (1981), 77-93].)

Question: Has this result been proved for piecewise $C^1$ interval maps? I would be surprised if it is not true in this setting, but neither I nor anyone I've asked has been able to produce a reference to a proof in the case where $f$ is a piecewise monotonic interval map without any assumption of Markov structure.

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up vote 1 down vote accepted

You can find a complete proof of a slightly more general result in my paper "A. Barrio Blaya and V. Jimenez Lopez, On the relations between positive Lyapunov exponents, positive entropy, and sensitivity for interval maps, Discrete Contin. Dyn. Syst. 32 (2012), 433-466", see Theorem 7.1. You can download the paper from here.

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Thanks! That's exactly what I was looking for. –  Vaughn Climenhaga Dec 15 '11 at 18:49
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Do you want just a reference or a proof? I didnt know any reference but for a proof the following should work.

Let $\mu$ be an ergodic measure of a piecewise $C^1$ interval maps (with finitely many pieces!). Either the boundary of the partition of smoothness (the finite set of singularities) has zero $\mu$ measure or $\mu$ is periodic. In the last case Ruelle's inequality trivialy holds. In the first case there exists neighborhoods of the boundary with arbitrarily small $\mu$-measure and therefore by ergodicity generic $\mu$ points visits this neighborhood very rarely. Then, when you are far from the boundary, you just apply the classical proof of Ruelle's inquality (you compute the entropy at a scale small compared to the size of the neighborhood). The part where you are closed to the boundary is negligible because local multiplicity is trivial (I mean that any singularity meets the boundary of at mosts two elements of the partition). This last fact is false in higher dimension and that's why to estimate from above entropy of piecewise (even affine) map one needs to add a term of multiplicity (see for example http://www.math.uit.no/seminar/Preprints/05-04-BKMR.pdf).

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Thanks for the answer -- using the fact that the boundary carries zero entropy seems to be the way to go (I sketched out an alternate proof, which also relies on this fact). Ideally I'd like a reference to somewhere where the result is already written: a co-author and I need it in order to state that a certain example works the way one expects, but we don't think it's important enough in the context of our paper to justify adding a complete proof there. –  Vaughn Climenhaga Nov 24 '11 at 17:00
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