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Let $f:S^1\to S^1$ be an orientation-preserving circle diffeomorphism with irrational rotation number (see here). 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 Herman 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.

Let $f:S^1\to S^1$ be an orientation-preserving circle diffeomorphism with irrational rotation number (see here). 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.

Let $f:S^1\to S^1$ be an orientation-preserving circle diffeomorphism with irrational rotation number (see here). 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?

Take the Denjoy map for example (a non-transitive diffeomorphism with irrational rotation number).

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

Edit: Note that $\lambda(f)$ equals to the integral $\int \log D_x f\; d\mu_f(x)$. (I thought this integral might be easier to compute. Then I realized that the estimation of $\lambda(f)$ is quite straight forward, and answered my own question.)

Let $f:S^1\to S^1$ be an orientation-preserving circle diffeomorphism with irrational rotation number (see here). 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 Herman 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.

Let $f:S^1\to S^1$ be an orientation-preserving circle diffeomorphism with irrational rotation number (see here). 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 equal to the integral $\int \log D_x f\; d\mu_f(x)$.

If $\log Df$ has bounded variation, then $\lambda(f,x)=0$ (I think it is proved by Herman) and hence $\int \log D_x f\; d\mu_f(x)=0$.

What about the general case? Is $\int \log D_x f\; d\mu_f(x)$ always zero?

Take the Denjoy map for example (a non-transitive diffeomorphism with irrational rotation number).

Let $f:S^1\to S^1$ be an orientation-preserving circle diffeomorphism with irrational rotation number (see here). 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?

Take the Denjoy map for example (a non-transitive diffeomorphism with irrational rotation number).

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

Edit: Note that $\lambda(f)$ equals to the integral $\int \log D_x f\; d\mu_f(x)$. (I thought this integral might be easier to compute. Then I realized that the estimation of $\lambda(f)$ is quite straight forward, and answered my own question.)