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 $\mathsf{d}(T^n(x),y)\ge \frac{1}{4}$ for all $n\in \mathbb{N}$.
   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 $y\in \{0,1{\}}^{\mathbb{N}}$ and any $ϵ>0$ there is $r\in \mathbb{N}$ so that $Tr(x)$ begins with $y(1)\cdots y(m)$ so that $\mathsf{d}(T^r(x),y)<ϵ$ if $m$ is chosen large enough.
2. 1. One calculates $$\begin{array}{rl}{T}^{n+1}(x,y)& =T(T^n(x,y))\\ & =T(x+n\alpha ,y+nx+(\genfrac{}{}{0ex}{}{n}{2})\alpha )\\ & =(x+n\alpha +\alpha ,y+nx+(\genfrac{}{}{0ex}{}{n}{2})+x+n\alpha )\\ & =(x+(n+1)\alpha ,y+(n+1)x+(\genfrac{}{}{0ex}{}{n+1}{2})\alpha )\end{array}$$ by induction.
   2. When $\alpha$ is irrational.
3. Let $R>0$ be a bound on the sequence $n\mapsto x_n$. Fix $k\in \mathbb{N}$. We have $$|\sum _{n=1}^N{x}_n-\sum _{n=k}^{N+k}{x}_n|\le |x_1|+\cdots +|x_k|+|x_{N+1}|+\cdots +|x_{N+k}|\le 2kR$$ 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 $\log (2^{k(N)})mod1$ converges, say to $\rho$, then the frequency $${\displaystyle \frac{|\{1\le n\le 2^{k(N+1)}:\log (n)\notin [\rho ,\rho +\log (2))\}|}{2^{k(N+1)}}}$$ will not be larger than $\frac{1}{2}$ yet the set $$[0,1)\setminus [\rho ,\rho +\log 2)$$ has length strictly larger than $\frac{1}{2}$.
5. 1. Say that $x_n$ is uniformly distributed in $X$ if $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{|\{1\le n\le N:x_n\in [a,b)\times [c,d)\}|}{N}}=(d-c)(b-a)$$ for all $0\le a<b\le 1$ and all $0\le c<d\le 1$.
   2. Never, because the sequence never belongs to the set $[\frac{1}{8},\frac{2}{8})\times [\frac{6}{8},\frac{7}{8})$.
6. 1. By passing to a subsequence one can arrange for $$N\mapsto {\displaystyle \frac{1}{N}}\sum _{n=1}^{N}f(T^{n}x)$$ to converge for every continuous function $f:X\to \mathbb{R}$ by an application of the Stone-Weierstrass theorem. The map $\varphi :\mathsf{C}(X)\to \mathbb{R}$ defined by $$\varphi (f)=\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{r(N)}}\sum _{n=1}^{r(N)}f(T^n(x))$$ 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 $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{r(N)}}\sum _{n=1}^{r(N)}f(x_n)=\int f\mathsf{d}\mu$$ \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 $2^{2k-1}<n\le 2^{2k}$ and $x_n=\frac{1}{2}$ whenever $2^{2k}<n\le 2^{2k+1}$.
7. Since the orbit $\{n\alpha mod1:n\in \mathbb{N}\}$ is dense in $[0,1)$ we can find $n_1,\dots ,n_6$ with $$\bigcup _{i=1}^6(\frac{1}{4},\frac{3}{4})-n_i\alpha =[0,1)$$Say that $E\subset \mathbb{N}$ is syndetic if there is $r(1),\dots ,r(k)$ with $$\mathbb{N}=(E-r(1))\cup \cdots \cup (E-r(k))$$ and taking $$E=\{n\in \mathbb{N}:n\alpha mod1\in (\frac{1}{4},\frac{3}{4})\}$$ gives the desired result.