Show that the map $T(x)=x+a \sin^2( \pi x)$ is uniquely ergodic on the unit circle

Question

The task is to prove that the map $T(x)=x+a \sin^2( \pi x)$ on the unit circle is uniquely ergodic for $a < \frac{1}{\pi}$. I tried to show that Birkhoff sums of the characters with respect to $T: S^1 \to S^1$ all converge uniformly to a constant, but I don't think that will work.

Answer 1

Succinctly, your idea does work.

For posterity, this is (part of) Exr.4.7.3 in Brin & Stuck's Introduction to Dynamical Systems (p.90).

Here is a humble graph: https://www.desmos.com/calculator/tgrhizoot6. It's also useful to have the phase portrait in mind; it looks like this:

(I've taken this image from Hasselblatt & Katok's A First Course in Dynamics, p.130. Note the resemblance to the projective action of shear matrices; indeed horocycle flows are also uniquely ergodic and unique ergodicity is a topological property (see Topological Invariance of Unique Ergodicity for the latter statement).)

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For a fixed $a\in]0,1/\pi[$, put $T=T_a: \mathbb{T}\to \mathbb{T}, x\mapsto x+a\sin^2(\pi x)$. Note that $T$ has a unique fixed point $0\in\mathbb{T}$, so that $\delta_0$ is an invariant Borel probability measure. Thus to show that $T$ is uniquely ergodic it suffices to show that for any continuous $\phi:\mathbb{T}\to \mathbb{R}$ and for any $x\in \mathbb{T}$:

$$\lim_{n\to\infty}\dfrac{1}{n}\sum_{k=0}^{n-1}\phi\circ T^k(x)=\phi(0).$$

Since $0$ is a fixed point we may assume $x\in\mathbb{T}\setminus0$. Also note that pointwise convergence is sufficient; uniform convergence will be automatic (see e.g. Einsiedler & Ward's Ergodic Theory with a view toward Number Theory, Thm.4.10 on p.105). Let us lift everything to the unit interval $[0,1]$ without changing notation.

Observe that $0< x < T(x) < T^2(x) < \cdots < T^n(x) < \cdots < 1$ and $\lim_{n\to\infty} T^n(x)=1$. Fix $\epsilon\in\mathbb{R}_{>0}$. Then since $\phi$ is uniformly continuous, for some $\delta\in\mathbb{R}_{>0}$ if $y$ and $z$ are $\delta$-close then $\phi(y)$ and $\phi(z)$ are $\epsilon$-close. There is an $N\in\mathbb{Z}_{>0}$ such that for any $n\in\mathbb{Z}_{\geq N}$ we have $|T^n(x)-1|<\delta$; thus

$$\exists N\in\mathbb{Z}_{>0},\forall n\in\mathbb{Z}_{\geq N}: |\phi\circ T^n(x)-\phi(1)|<\epsilon.$$

Then we have, for $n\in\mathbb{Z}_{\geq N}$,

\begin{align*} \left|\dfrac{1}{n}\sum_{0\leq k<n}\phi\circ T^k(x) -\phi(1)\right| &\leq \dfrac{1}{n}\sum_{0\leq k<n}\left|\phi\circ T^k(x) -\phi(1)\right|\\ &= \dfrac{1}{n}\left(\sum_{0\leq k<N}\left|\phi\circ T^k(x) -\phi(1)\right|+\sum_{N\leq k<n}\left|\phi\circ T^k(x) -\phi(1)\right|\right)\\ &\leq \dfrac{1}{n}(N2|\phi|_{C^0}+ (n-N)\epsilon) = \epsilon + \dfrac{N(2|\phi|_{C^0}-\epsilon)}{n}. \end{align*}

Taking $n\to\infty$ we are done since $\epsilon$ was arbitrary. (Note that $1\sim0$ when we project back down to the circle.)