Week 5 Worksheet Solutions

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1. 1. The periodic sequence \[ x = \mathtt{0101010101010101010101010101\cdots} \] has a finite orbit because $T^2(x) = x$.
   2. The periodic sequence \[ x = \mathtt{0101010101010101010101010101\cdots} \] has an orbit that is not dense, because if \[ y = \mathtt{1111111111111111111111111111\cdots} \] then $\met(T^n(x),y) \ge \tfrac{1}{4}$ for all $n \in \N$.
   3. Let $n \mapsto w_n$ be an enumeration of all finite strings of zeroes and ones. Let $x$ be the sequence obtained by contatenating all the $w_n$. For any $y \in \{0,1\}^\N$ and any $\epsilon > 0$ there is $r \in \N$ so that $Tr(x)$ begins with $y(1) \cdots y(m)$ so that $\met(T^r(x),y) \l \epsilon$ if $m$ is chosen large enough.
2. 1. One calculates \[ \begin{aligned} T^{n+1}(x,y) &{} = T(T^n(x,y)) \\ &{} = T(x + n \alpha, y + nx + \tbinom{n}{2} \alpha ) \\ &{} = (x + n \alpha + \alpha, y + nx + \tbinom{n}{2} + x + n\alpha) \\ &{} = (x + (n+1)\alpha, y + (n+1)x + \tbinom{n+1}{2} \alpha) \end{aligned} \] by induction.
   2. When $\alpha$ is irrational.
3. Let $R \g 0$ be a bound on the sequence $n \mapsto x_n$. Fix $k \in \N$. We have \[ \left| \sum_{n=1}^N x_n - \sum_{n=k}^{N+k} x_n \right| \le |x_1| + \cdots + |x_k| + |x_{N+1}| + \cdots + |x_{N+k}| \le 2kR \] which converges to zero upon division by $N$, giving the desired result.
4. No, it is too lethargic. Indeed \[ \log(2^{N+1}) - \log(2^N) = \log 2 \approx 0.30103 \] so if we choose (as we may) a sequence $k(N)$ such that $\log(2^{k(N)}) \bmod 1$ converges, say to $\rho$, then the frequency \[ \dfrac{|\{ 1 \le n \le 2^{k(N+1)} : \log(n) \notin [\rho,\rho + \log(2) )\}|}{2^{k(N+1)}} \] will not be larger than $\tfrac{1}{2}$ yet the set \[ [0,1) \setminus [\rho,\rho + \log 2) \] has length strictly larger than $\tfrac{1}{2}$.
5. 1. Say that $x_n$ is uniformly distributed in $X$ if \[ \lim_{N \to \infty} \dfrac{|\{ 1 \le n \le N : x_n \in [a,b) \times [c,d) \}|}{N} = (d-c)(b-a) \] for all $0 \le a \l b \le 1$ and all $0 \le c \l d \le 1$.
   2. Never, because the sequence never belongs to the set $[\tfrac{1}{8},\tfrac{2}{8}) \times [\tfrac{6}{8},\tfrac{7}{8})$.
6. 1. By passing to a subsequence one can arrange for \[ N \mapsto \dfrac{1}{N} \sum_{n=1}^N f(T^n x) \] to converge for every continuous function $f : X \to \R$ by an application of the Stone-Weierstrass theorem. The map $\phi : \mathsf{C}(X) \to \R$ defined by \[ \phi(f) = \lim_{N \to \infty} \dfrac{1}{r(N)} \sum_{n=1}^{r(N)} f(T^n(x)) \] is then a bounded linear functional on $\mathsf{C}(X)$ and there is therefore, by the Riesz representation theorem, a measure $\mu$ on $X$ with \[ \lim_{N \to \infty} \dfrac{1}{r(N)} \sum_{n=1}^{r(N)} f(x_n) = \int f \intd \mu \] \for all $f \in \mathsf{C}(X)$.
   2. Yes, because both the sequences $r$, $s$ and the measure $\mu$,$\nu$ can be different. Explicitly, one can define a sequence $n \mapsto x_n$ by $x_n = 0$ whenever $2^{2k-1} \l n \le 2^{2k}$ and $x_n = \tfrac{1}{2}$ whenever $2^{2k} \l n \le 2^{2k+1}$.
7. Since the orbit $\{n \alpha \bmod 1 : n \in \N \}$ is dense in $[0,1)$ we can find $n_1,\dots,n_6$ with \[ \bigcup_{i=1}^6 (\tfrac{1}{4},\tfrac{3}{4}) - n_i \alpha = [0,1) \]Say that $E \subset \N$ is syndetic if there is $r(1),\dots,r(k)$ with \[ \N = (E - r(1)) \cup \cdots \cup (E - r(k)) \] and taking \[ E = \{ n \in \N : n \alpha \bmod 1 \in (\tfrac{1}{4},\tfrac{3}{4}) \} \] gives the desired result.