# Denjoy’s theorem on circle diffeomorphisms

Jordan Bell
April 3, 2014

In this note I’m just presenting the proof of Denjoy’s theorem in Michael Brin and Garrett Stuck’s Introduction to dynamical systems, Cambridge University Press, 2002.

Let $S^{1}=\mathbb{R}/\mathbb{Z}$. For $\alpha\in\mathbb{R}$, define $R_{\alpha}:S^{1}\to S^{1}$ by $R_{\alpha}(x)=x+\alpha+\mathbb{Z}$.

We say that a homeomorphism $f:S^{1}\to S^{1}$ is orientation preserving if it lifts to an increasing homeomorphism $F:\mathbb{R}\to\mathbb{R}$: $\pi\circ F=f\circ\pi$.

The rotation number of an orientation preserving homeomorphism $f$ is defined by

 $\rho(f)=\lim_{n\to\infty}\frac{F^{n}(x)-x}{n}.$

One proves that this is independent both of the lift $F$ of $f$ and the point $x\in\mathbb{R}$. Some facts about the rotation number: it is an invariant of topological conjugacy, and $\rho(f)$ is rational if and only if $f$ has a periodic point. A periodic point is $x\in S^{1}$ such that $f^{n}(x)=x$ for some $n\geq 1$.

There are some lemmas in Chapter 7 that I don’t want to write out. The important theorem that we’re going to use without proof is that if $f:S^{1}\to S^{1}$ is an orientation preserving homeomorphism that is topologically transitive with irrational rotation number $\rho(f)$, then $f$ is topologically conjugate to $R_{\rho(f)}$. This reduces our problem to showing that a map is topologically transitive.

We will use the following lemma in the proof of Denjoy’s theorem.

###### Lemma 1.

Let $f:S^{1}\to S^{1}$ be a $C^{1}$ diffeomorphism and let $J$ be an interval in $S^{1}$. Let $g=\log f^{\prime}$. If the interiors of $J,f(J),\ldots,f^{n-1}(J)$ are pairwise disjoint, then for any $x,y\in J$ and any $n\in\mathbb{Z}$ we have

 $\operatorname{Var}(g)\geq|\log((f^{n})^{\prime}(x))-\log((f^{n})^{\prime}(y))|.$
###### Proof.

The intervals $[x,y],[f(x),f(y)],\ldots,[f^{n-1}(x),f^{n-1}(y)]$ are pairwise disjoint, so they are part of a partition of $[0,1]$. The total variation of $g$ is defined as a supremum over all partitions, so in particular it will be $\geq$ the sum coming from any particular partition or a subset of that partition.

 $\displaystyle\operatorname{Var}(g)$ $\displaystyle\geq$ $\displaystyle\sum_{k=0}^{n-1}|g(f^{k}(y))-g(f^{k}(x))|$ $\displaystyle\geq$ $\displaystyle\Big{|}\sum_{k=0}^{n-1}g(f^{k}(y))-g(f^{k}(x))\Big{|}$ $\displaystyle=$ $\displaystyle\Big{|}\log\prod_{k=0}^{n-1}f^{\prime}(f^{k}(y))-\log\prod_{k=0}^% {n-1}f^{\prime}(f^{k}(x))\Big{|}$ $\displaystyle=$ $\displaystyle|\log((f^{n})^{\prime}(x))-\log((f^{n})^{\prime}(y))|.$

Now we can prove Denjoy’s theorem.

###### Theorem 2.

If $f:S^{1}\to S^{1}$ is a $C^{1}$ diffeomorphism that is orientation preserving, that has irrational rotation number $\rho(f)$, and whose derivative $f^{\prime}:S^{1}\to\mathbb{R}$ has bounded variation, then $f$ is topologically conjugate to $R_{\rho(f)}$.

###### Proof.

Suppose by contradiction that $f$ is not topologically transitive. It’s a fact proved in Chapter 7 of Brin and Stuck that this implies that $\omega(x)$ is perfect and nowhere dense, and is independent of the point $x$. (Recall that $\omega(x)=\bigcap_{n\geq 1}\overline{\bigcup_{i\geq n}f^{i}(x)}$.) It follows that there is an interval $I=(a,b)$ in its complement.

The intervals $f^{n}(I)$, $n\in\mathbb{Z}$, are pairwise disjoint, for otherwise $f$ would have a periodic point. Let $\mu$ be Haar measure on $S^{1}$. Then

 $\sum_{n\in\mathbb{Z}}\mu(f^{n}(I))\leq 1.$

Let $x\in S^{1}$. Suppose for the moment that there are infinitely $n\geq 1$ such that the intervals $(x,f^{-n}(x)),(f(x),f^{1-n}(x)),\ldots,(f^{n}(x),x)$ are pairwise disjoint; we shall prove that this is true later. By applying the lemma we proved with $y=f^{-n}(x)$ we get

 $\operatorname{Var}(g)\geq\Big{|}\log\frac{(f^{n})^{\prime}(x)}{(f^{n})^{\prime% }(y)}\Big{|}=|\log((f^{n})^{\prime}(x)(f^{-n})^{\prime}(x)|.$

To see the equality in the above line it helps to write out what $(f^{-n})^{\prime}(x)$ is.

Then for infinitely many $n$ we have

 $\displaystyle\mu(f^{n}(I))+\mu(f^{-n}(I))$ $\displaystyle=$ $\displaystyle\int_{I}(f^{n})^{\prime}(x)dx+\int_{I}(f^{-n})^{\prime}(x)dx$ $\displaystyle=$ $\displaystyle\int_{I}((f^{n})^{\prime}(x)+(f^{-n})^{\prime}(x))dx$ $\displaystyle\geq$ $\displaystyle\int_{I}\sqrt{(f^{n})^{\prime}(x)(f^{-n})^{\prime}(x)}dx$ $\displaystyle=$ $\displaystyle\int_{I}\sqrt{\exp\log((f^{n})^{\prime}(x)(f^{-n})^{\prime}(x))}dx$ $\displaystyle\geq$ $\displaystyle\int_{I}\sqrt{\exp(-|\log((f^{n})^{\prime}(x)(f^{-n})^{\prime}(x)% )|)}dx$ $\displaystyle\geq$ $\displaystyle\int_{I}\sqrt{\exp(-\operatorname{Var}(g))}dx$ $\displaystyle=$ $\displaystyle\exp\Big{(}-\frac{1}{2}\operatorname{Var}(g)\Big{)}\mu(I).$

Since $\mu(I)>0$ this implies that $\sum_{n\in\mathbb{Z}}\mu(f^{n}(I))=\infty$, a contradiction. Therefore $f$ is topologically transitive, and so it is topologically conjugate to the $R_{\rho(f)}$. ∎

It is indeed necessary that $f^{\prime}$ has bounded variation. Brin and Stuck give an example on p. 161 that they attribute to Denjoy: for any irrational number $\rho\in(0,1)$, there is a nontransitive orientation preserving $C^{1}$ diffeomorphism of $S^{1}$ with rotation number $\rho$. The only condition of Denjoy’s theorem that isn’t satisfied here is that $f^{\prime}$ have bounded variation.