Week 10 Worksheet Solutions

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1. Fix an interval $[a,b) \subset [0,1)$. Since \[ T^{-1}([a,b)) = [ \tfrac{a}{3}, \tfrac{b}{3} ) \cup [ \tfrac{a+1}{3}, \tfrac{b+1}{3} ) \cup [ \tfrac{a+2}{3}, \tfrac{b+2}{3} ) \] 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 \[ f = \sum_{k \in \Z} c(k) \psi_k \qquad g = \sum_{r \in \Z} d(r) \psi_r \] in $L^2([0,1))$. First suppose that $Tf = f$. Since $T \psi_k = \psi_{3k}$ we get \[ \sum_{k \in \Z} c(k) \psi(k) = \sum_{k \in \Z} c(k) \psi_{3k} \] which forces $c(k) = c(3k)$ for all $k \in \Z$. As $c(k) \to 0$ as $|k| \to \infty$ we must have $c(k) = 0$ for all $k \ne 0$ and therefore $f$ is constant. This proves $T$ is ergodic.
   
   We also have \[ \begin{align*} \langle f, T^n g \rangle &{} = \sum_{k \in \Z} \sum_{m \in \Z} c(k) \overline{d(m)} \langle \psi_k, \psi_{3^n m} \rangle \\ &{} = \sum_{k \in \Z} c(3^n m) \overline{d(m)} \end{align*} \] and this goes to zero by the same reason as in the notes.
2. Fix \[ f = \sum_{r,s \in \Z} c(r,s) \psi_{r,s} \] 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 \[ Tf = \sum_{r,s \in \Z} c(r,s) e^{2 \pi i (r\alpha + s \beta)} \psi_{r,s} \] giving \[ c(r,s) = e^{2 \pi i (r \alpha + s \beta)} c(r,s) \] for all $r,s \in \Z$. We conclude that $c(r,s) = 0$ whenever \[ r \alpha + s \beta \not\in \Z \] which is always the case when $\{1,\alpha,\beta\}$ are independent over $\Q$.
3. Put $f = \psi_1$. Then $T^n f = e^{2 \pi i n \alpha} f$ and \[ \langle T^n f, f \rangle = e^{2 \pi i n \alpha} \langle f, f \rangle = e^{2 \pi i n \alpha} \] for all $n \in \N$. Were $T$ mixing we would have \[ e^{2 \pi i n \alpha} \to \langle f, 1 \rangle \langle 1, f \rangle = 0 \] which is impossible as $| e^{2 \pi i n \alpha} | = 1$. Hence $T$ is not mixing.
4. If $B$ is bounded and $x_n$ in $\Lp{2}(X,\B,\mu)$ is a sequence converging to $y$ then \[ 0 \le \| B(x_n) - B(y) \|_\mathsf{2} = \| B(x_n - y) \|_\mathsf{2} \le C \| x_n - y \|_\mathsf{2} \] 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 $\delta \g 0$ such that $\| x \|_\mathsf{2} \l \delta$ implies $\| B(x) \|_\mathsf{2} \le 1$. Fix $x \ne 0$. Put \[ y = \dfrac{x}{\| x \|_\mathsf{2}} \dfrac{\delta}{2} \] and note that \[ \| y \|_\mathsf{2} \l \delta \] so that $\| B(y) \|_\mathsf{2} \le 1$. This gives \[ \| B(x) \|_\mathsf{2} \le \dfrac{2}{\delta} \| x \|_\mathsf{2} \] for all $x$ and therefore $B$ is bounded.
5. 1. Since $T$ is measure-preserving we have \[ \langle f, f \rangle = \int f \cdot \overline{f} \intd \mu = \int Tf \cdot \overline{Tf} \intd \mu = \langle Tf, Tf \rangle \] and since $Tf = \eta f$ we get \[ \langle f, f \rangle = \eta \cdot \overline{\eta} \langle f, f \rangle = |\eta|^2 \langle f, f \rangle \] so $|\eta| = 1$.
   2. If $Tf = \eta f$ and $Tg = \xi g$ then \[ \langle f,g \rangle = \langle Tf, Tg \rangle = \eta \overline{\xi} \langle f,g \rangle \] so we must have $\eta \overline{\xi} = 1$ or $\langle f,g \rangle = 0$.
   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 $g \in \C$.