Week 5 Tutorial Solutions

Irrational Rotations

1. Yes, because for every irrational $\alpha$ the orbit $$\{T^n(0):n\in \mathbb{N}\}$$ is dense in $[0,1)$ by a result from class and therefore in particular intersects $(0,\frac{1}{4})$.
2. Yes, because for $S(x)=x+2\alpha mod1$ the orbit $$\{S^n(0):n\in \mathbb{N}\}=\{T^{2n}(0):n\in \mathbb{N}\}$$ is dense in $[0,1)$.
3. Yes, because for $S(x)=x+2\alpha mod1$ the orbit $$\{S^n(\alpha ):n\in \mathbb{N}\}=\{T^{2n+1}(0):n\in \mathbb{N}\}$$ is dense in $[0,1)$.
4. Not necessarily: take $$\alpha =\frac{1}{2}-\frac{\sqrt{2}}{10}$$ and calculate $$\begin{array}{c}{T}^0(0)=0T^1(0)=\alpha =\frac{1}{2}-\frac{\sqrt{2}}{10}\\ T^2(0)=2\alpha =-\frac{\sqrt{2}}{5}{T}^3(0)=\frac{1}{2}-\frac{3\sqrt{2}}{10}\end{array}$$ which is always outside $(0,\frac{1}{4})$.
5. Not necessarily: take $$\alpha =\frac{1}{2}-\frac{\sqrt{2}}{{100}^{10000000000}}$$ and verify for $0\le n\le 7^{1000}$ we have $n\alpha mod1$ in $(\frac{1}{8},\frac{7}{8})$.
6. Not necessarily: take $$\alpha =\frac{1}{8}+\frac{\sqrt{2}}{100}$$ and verify for $0\le n\le 3$ we have $n\alpha mod1$ in $(\frac{1}{8},\frac{7}{8})$.
7. Yes. Writing $I=(-\frac{1}{16},\frac{1}{16})$ the intervals $$I,I-\alpha ,I-2\alpha ,I-3\alpha ,\dots ,I-7,I-8$$ cannot be pairwise disjoint, as each has measure $\frac{1}{4}$. Thus there are $r<s$ in $\{0,1,\dots ,8\}$ with $$I-r\alpha \cap I-s\alpha$$ non-empty. It must then be the case that $(s-r)\alpha mod1$ belongs to $(-\frac{1}{8},\frac{1}{8})$.
8. Take $E=[0,1)\setminus \{T^n(x):n\in \mathbb{N}\}$. As the orbit is countable it has zero measure and therefore $\lambda (E)=1$. But $1_E(T^n(x))=0$ for all $n\in \mathbb{N}$ and therefore the limit is zero.
9. Take $$E=[0,1)\setminus \bigcup _{j\in \mathbb{N}}\{T^n(x):2^{2j}\le n<2^{2j+2}\}$$ and note that $$N\mapsto {\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{1}_E(T^n(x))$$ has different accumulation points for $N$ of the form $2^{2m}$ and $N$ of the form $2^{2m+1}$.
10. We know, because $\alpha$ is irrational, that $$\int f\mathsf{d}\lambda =\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))=0$$ by our uniform distribution theorem. Since $f$ is continuous, if it were not the zero function there would be $d>0$ and $(a,b)\subset [0,1)$ with $$d\cdot 1_{(a,b)}\le f$$ but then the integral of $f$ would have to be positive giving a contradiction.