Let $f:S^1\to S^1$ be an orientation-preserving circle diffeomorphism with irrational rotation number (see [here](http://en.wikipedia.org/wiki/Rotation_number)). Then the system $(S^1,f)$ admits a unique invariant measure, say $\mu\_f$. 


Let $\displaystyle \lambda(f,x)=\lim\_{n\to\infty}\frac{1}{n}\log D\_xf^n$ be the Lyapunov exponent at $x$. Then $\lambda(f,x)$ is independent of $x$, and is denoted by $\lambda(f)$.


**Question.** Is $\lambda(f)$ always zero? 


Note that $\lambda(f)$ equals to the integral $\int \log D\_x f\; d\mu\_f(x)$. Take the Denjoy map for example (a non-transitive diffeomorphism with irrational rotation number).


**Remark.** If $\log Df$ has bounded variation, then Denjoy proved that $e^{-V}\le Df^{q\_n}(x)\le e^V$ for all $x$ (a *much stronger* property), where $q\_n, n\ge 1$ are the denominators of rational approximates. In particular this implies $\lambda(f,x)=0$ for all $x$.


**Edit again.** I thought the integral might be easier to compute. Then I realized that the estimation of $\lambda(f)$ is quite straight forward, and answered my own question. 



Lyapunov exponent for circle diffeomorphisms
Note that $\int \log D\_xf^n \, dm(x)\le \log \int D\_xf^n \, dm(x)=\log 1=0$ for all $n\ge 1$. Therefore (by uniform convergence for unique ergodic system)


$\displaystyle \lambda(f)=\int \lambda(f,x) dm(x)=\int\lim\_{n\to\infty}\frac{1}{n}\log D\_x f^n dm(x)=\lim\_{n\to\infty}\frac{1}{n}\int\log D\_x f^n dm(x)\le 0$.


Then applying to the inverse $f^{-1}$, one gets $\lambda(f^{-1},x)\le 0$, too. Therefore, $\lambda(f,x)=0$ for all $x$. Or equally, $\mu\_f(\log Df)=0$.


