\[
\newcommand{\C}{\mathbb{C}}
\newcommand{\haar}{\mathsf{m}}
\newcommand{\P}{\mathcal{P}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Q}{\mathbb{Q}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\d}{\mathsf{d}}
\newcommand{\g}{>}
\newcommand{\l}{<}
\newcommand{\intd}{\,\mathsf{d}}
\newcommand{\Re}{\mathsf{Re}}
\newcommand{\area}{\mathop{\mathsf{Area}}}
\newcommand{\met}{\mathop{\mathsf{d}}}
\newcommand{\emptyset}{\varnothing}
\DeclareMathOperator{\borel}{\mathsf{Bor}}
\]
Week 5 Tutorial Solutions
- Yes, because for every irrational $\alpha$ the orbit
\[
\{ T^n(0) : n \in \N \}
\]
is dense in $[0,1)$ by a result from class and therefore in particular intersects $(0,\tfrac{1}{4})$.
- Yes, because for $S(x) = x + 2\alpha \bmod 1$ the orbit
\[
\{ S^n(0) : n \in \N \} = \{ T^{2n}(0) : n \in \N \}
\]
is dense in $[0,1)$.
- Yes, because for $S(x) = x + 2\alpha \bmod 1$ the orbit
\[
\{ S^n(\alpha) : n \in \N \} = \{ T^{2n+1}(0) : n \in \N \}
\]
is dense in $[0,1)$.
-
Not necessarily: take
\[
\alpha = \tfrac{1}{2} - \tfrac{\sqrt{2}}{10}
\]
and calculate
\[
\begin{gathered}
T^{0}(0) = 0
\qquad\qquad
T^1(0) = \alpha = \tfrac{1}{2} - \tfrac{\sqrt{2}}{10}
\\
T^2(0) = 2\alpha = - \tfrac{\sqrt{2}}{5}
\qquad
T^3(0) = \tfrac{1}{2} - \tfrac{3 \sqrt{2}}{10}
\end{gathered}
\]
which is always outside $(0,\tfrac{1}{4})$.
- Not necessarily: take
\[
\alpha = \tfrac{1}{2} - \tfrac{\sqrt{2}}{100^{10000000000}}
\]
and verify for $0 \le n \le 7^{1000}$ we have $n \alpha \bmod 1$ in $(\tfrac{1}{8},\tfrac{7}{8})$.
- Not necessarily: take
\[
\alpha = \tfrac{1}{8} + \tfrac{\sqrt{2}}{100}
\]
and verify for $0 \le n \le 3$ we have $n \alpha \bmod 1$ in $(\tfrac{1}{8},\tfrac{7}{8})$.
- Yes. Writing $I = (-\tfrac{1}{16},\tfrac{1}{16})$ the intervals
\[
I, I - \alpha, I - 2 \alpha, I - 3\alpha,\dots, I - 7, I - 8
\]
cannot be pairwise disjoint, as each has measure $\tfrac{1}{4}$. Thus there are $r \l s$ in $\{0,1,\dots,8\}$ with
\[
I-r \alpha \cap I-s \alpha
\]
non-empty. It must then be the case that $(s-r)\alpha \bmod 1$ belongs to $(-\tfrac{1}{8},\tfrac{1}{8})$.
- Take $E = [0,1) \setminus \{ T^n(x) : n \in \N \}$. As the orbit is countable it has zero measure and therefore $\lambda(E) = 1$. But $1_E(T^n(x)) = 0$ for all $n \in \N$ and therefore the limit is zero.
- Take
\[
E = [0,1) \setminus \bigcup_{j \in \N} \{ T^n(x) : 2^{2j} \le n \l 2^{2j +2} \}
\]
and note that
\[
N \mapsto \dfrac{1}{N} \sum_{n=0}^{N-1} 1_E(T^n(x))
\]
has different accumulation points for $N$ of the form $2^{2m}$ and $N$ of the form $2^{2m+1}$.
- We know, because $\alpha$ is irrational, that
\[
\int f \intd \lambda = \lim_{N \to \infty} \dfrac{1}{N} \sum_{n=0}^{N-1} f(T^n(x)) = 0
\]
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.