Week 10 Worksheet Solutions

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1. Fix an interval $[a,b)\subset [0,1)$. Since $$T^{-1}([a,b))=[\frac{a}{3},\frac{b}{3})\cup [\frac{a+1}{3},\frac{b+1}{3})\cup [\frac{a+2}{3},\frac{b+2}{3})$$ is a disjoint union of intervals of length $(b-a)/3$ the map $T$ is measure-preserving because intervals - together with the empty set - form a π system.
   
   Fix $$f=\sum _{k\in \mathbb{Z}}c(k){\psi }_{k}g=\sum _{r\in \mathbb{Z}}d(r){\psi }_r$$ in $L^2([0,1))$. First suppose that $Tf=f$. Since $T{\psi }_k={\psi }_{3k}$ we get $$\sum _{k\in \mathbb{Z}}c(k)\psi (k)=\sum _{k\in \mathbb{Z}}c(k){\psi }_{3k}$$ which forces $c(k)=c(3k)$ for all $k\in \mathbb{Z}$. As $c(k)\to 0$ as $|k|\to \mathrm{\infty }$ we must have $c(k)=0$ for all $k\ne 0$ and therefore $f$ is constant. This proves $T$ is ergodic.
   
   We also have $$\begin{array}{rl}⟨f,T^{n}g⟩& =\sum _{k\in \mathbb{Z}}\sum _{m\in \mathbb{Z}}c(k)\overline{d(m)}⟨{\psi }_k,{\psi }_{3^{n}m}⟩\\ & =\sum _{k\in \mathbb{Z}}c(3^{n}m)\overline{d(m)}\end{array}$$ and this goes to zero by the same reason as in the notes.
2. Fix $$f=\sum _{r,s\in \mathbb{Z}}c(r,s){\psi }_{r,s}$$ in ${\mathsf{L}}^{\mathsf{2}}([0,1{)}^2)$ and suppose that $f$ is $T$ invariant. We have $$(T{\psi }_{r,s})(x,y)={\psi }_{r,s}(x+\alpha ,y+\beta )$$ so $$T{\psi }_{r,s}=e^{2\pi i(r\alpha +s\beta )}{\psi }_{r,s}$$ and therefore $$Tf=\sum _{r,s\in \mathbb{Z}}c(r,s)e^{2\pi i(r\alpha +s\beta )}{\psi }_{r,s}$$ giving $$c(r,s)=e^{2\pi i(r\alpha +s\beta )}c(r,s)$$ for all $r,s\in \mathbb{Z}$. We conclude that $c(r,s)=0$ whenever $$r\alpha +s\beta \notin \mathbb{Z}$$ which is always the case when $\{1,\alpha ,\beta \}$ are independent over $\mathbb{Q}$.
3. Put $f={\psi }_1$. Then $T^{n}f=e^{2\pi in\alpha }f$ and $$⟨T^{n}f,f⟩=e^{2\pi in\alpha }⟨f,f⟩=e^{2\pi in\alpha }$$ for all $n\in \mathbb{N}$. Were $T$ mixing we would have $$e^{2\pi in\alpha }\to ⟨f,1⟩⟨1,f⟩=0$$ which is impossible as $|e^{2\pi in\alpha }|=1$. Hence $T$ is not mixing.
4. If $B$ is bounded and $x_n$ in ${\mathsf{L}}^{\mathsf{2}}(X,\mathcal{B},\mu )$ is a sequence converging to $y$ then $$0\le \Vert B(x_n)-B(y){\Vert }_{\mathsf{2}}=\Vert B(x_n-y){\Vert }_{\mathsf{2}}\le C\Vert x_n-y{\Vert }_{\mathsf{2}}$$ so $B(x_n)$ converges to $B(y)$. By the sequence characterization of continuity $B$ is therefore continuous.
   
   If $B$ is continuous it is continuous at zero and there is $\delta >0$ such that $\Vert x{\Vert }_{\mathsf{2}}<\delta$ implies $\Vert B(x){\Vert }_{\mathsf{2}}\le 1$. Fix $x\ne 0$. Put $$y={\displaystyle \frac{x}{\Vert x{\Vert }_{\mathsf{2}}}}{\displaystyle \frac{\delta }{2}}$$ and note that $$\Vert y{\Vert }_{\mathsf{2}}<\delta$$ so that $\Vert B(y){\Vert }_{\mathsf{2}}\le 1$. This gives $$\Vert B(x){\Vert }_{\mathsf{2}}\le {\displaystyle \frac{2}{\delta }}\Vert x{\Vert }_{\mathsf{2}}$$ for all $x$ and therefore $B$ is bounded.
5. 1. Since $T$ is measure-preserving we have $$⟨f,f⟩=\int f\cdot \bar{f}\mathsf{d}\mu =\int Tf\cdot \overline{Tf}\mathsf{d}\mu =⟨Tf,Tf⟩$$ and since $Tf=\eta f$ we get $$⟨f,f⟩=\eta \cdot \bar{\eta }⟨f,f⟩=|\eta {|}^2⟨f,f⟩$$ so $|\eta |=1$.
   2. If $Tf=\eta f$ and $Tg=\xi g$ then $$⟨f,g⟩=⟨Tf,Tg⟩=\eta \bar{\xi }⟨f,g⟩$$ so we must have $\eta \bar{\xi }=1$ or $⟨f,g⟩=0$.
   3. We calculate $T|f|=|Tf|=|\eta f|=|f|$.
   4. Since $|f|$ is invariant it must be constant by ergodicity.
   5. If $f,g$ are eigenfunctions with the same value then $f/g$ is defined almost everywhere because $|g|=1$ almost surely. Moreover $f/g$ is $T$ invariant and therefore constant, so $f=\lambda g$ for some $g\in \mathbb{C}$.