We have seen that a measure-preserving transformation $T$ on a probability space $(X,\B,\mu)$ induces an isometry of the space $\Lp(X,\B,\mu)$ by composition. That is, if $f$ belongs to $\Lp(X,\B,\mu)$ then so too does \[ Tf = f \circ T \] and moreover $\| Tf \|_\mathsf{2} = \| f \|_\mathsf{2}$.
We restrict now our attention to the special case where our probability space is $[0,1)$ equipped with Lebesgue measure $\lambda$. By paying the small price of working with complex instead of real-valued functions, we will produce an orthonormal basis of $\Lp(X,\B,\mu)$ that will allow us to more easily verify dynamical properties of measure-preserving transformations on the unit interval.
Write $\Lp[2,\C]([0,1))$ for the collection of functions $f : [0,1) \to \C$ that are Borel measurable and for which \[ \int |f|^2 \intd \lambda \l \infty \] where we identify two complex-valued functions $f,g$ if \[ \int |f-g|^2 \intd \lambda \] holds. Next, define for each $k \in \Z$ the function \[ \chi_k(x) = e^{2 \pi i k x} = \cos(2 \pi k x) + i \sin(2 \pi k x) \] for each $x \in [0,1)$. Since $|\chi_k| = 1$ for each $k \in \Z$ and $\lambda([0,1)) = 1$ each of the functions $\chi_k$ belongs to $\Lp[2,\C]([0,1))$.
In dealing with complex-valued functions we must modify slightly the inner product we work with. Define \[ \langle f, g \rangle = \int f \cdot \overline{g} \intd \lambda \] whenever $f,g$ belong to $\Lp[2,\C]([0,1))$. With respect to this inner product we have \[ \langle \chi_k, \chi_j \rangle = \int \chi_k \cdot \overline{\chi_j} \intd \lambda = \int\limits_0^1 e^{2 \pi i (k-j)x} \intd x = \begin{cases} 1 & j = k \\ 0 & j \ne k \end{cases} \] which is to say that the collection $\{ \chi_k : k \in \Z \}$ of functions is an orthonormal system. In fact it is a basis of $\Lp[2,\C]([0,1))$ in the sense that \[ f = \sum_{n=1}^\infty \langle f, \chi_k \rangle \chi_k \] for every $f$ in $\Lp[2,\C]([0,1))$. Note that this is distinct from the meaning of basis in linear algebra: we are not asserting that every $f$ is a finite linear combination of functions from our orthonormal family; instead we are using the norm to say that \[ \lim_{N \to \infty} \left\| f - \sum_{n=-N}^N \langle f, \chi_n \rangle \chi_n \right\|_\mathsf{2} = 0 \] for every $f$ in $\Lp[2,\C]([0,1))$.
The $\chi_k$ form a basis of $\Lp[2,\C]([0,1))$.
We will take this for granted. The key ingredients in the proof are Bessel's inequality, the fact the continuous functions are dense in $\Lp[2,\C]([0,1))$, and the fact that every continuous function on $[0,1)$ is a uniform limit of linear combinations of our orthonormal system. ▮
Every irrational rotation is ergodic.
Write $T(x) = x + \alpha \bmod 1$ for some irrational $\alpha$. Fix $f$ in $\Lp[2,\C]([0,1))$ that is $T$ invariant. We can write \[ f = \sum_{n \in \Z} c(n) \chi_n \] and calculate that \[ Tf = \sum_{n \in \Z} c(n) T\chi_n = \sum_{n \in \Z} c(n) e^{2 \pi i n \alpha} \chi_n \] and conclude that $c(n) = c(n) e^{2 \pi i n \alpha}$ for all $n \in \Z$. As $\alpha$ is irrational we must have $c(n) = 0$ for all non-zero $n$. Thus $f = c(0)$ is constant. ▮
The map $T(x) = 2x \bmod 1$ on $[0,1)$ is ergodic.
Fix $f$ in $\Lp[2,\C]([0,1))$ that is $T$ invariant. We can write \[ f = \sum_{n \in \Z} c(n) \chi_n \] and calculate that \[ Tf = \sum_{n \in \Z} c(n) T\chi_n = \sum_{n \in \Z} c(n) \chi_{2n} \] and conclude that $c(n) = c(2n)$ for all $n \in \Z$. Calculating \[ \| f\|_\mathsf{2}^2 = \langle f, f \rangle = \sum_{n \in \Z} |c(n)|^2 \] it is not possible for $c$ to take the same non-zero value infinitely often. Thus $c(n) = 0$ for all $n \ne 0$ and $f = c(0)$ is constant. ▮
For the irrational rotation $T(x) = x + \alpha$ each of the functions $\chi_n$ is an eigenfunction. We define the class of dynamical systems with this property.
A measure-preserving transformation $T$ on a probability space $(X,\B,\mu)$ has discrete spectrum when $\Lp[2,\C](X,\B,\mu)$ has an orthonormal basis consisting of eigenfunctions.
Ergodic transformations with discrete spectrum can be modelled by systems that look like irrational rotations.
If a measure-preserving transformation $T$ is ergodic and has discrete spectrum then it is isomorphic to a measure-preserving transformation $S$ on a compact, Abelian group defined by $S(g) = g + a$ for some $a$ in $G$ with the property that $\{ na : n \in \Z \}$ is dense.
Can the doubling map $T(x) = 2x \bmod 1$ be modelled by an irrational rotation? Perhaps it is, in disguise, a system with discrete spectrum. In fact, the doubling map is mixing and systems that are ergodic and mixing cannot have non-constant eigenfunctions.
The doubling map $T(x) = 2x \bmod 1$ on $[0,1)$ is mixing.
Fix $f,g$ in $\Lp[2,\C]([0,1))$. Write \[ f = \sum_{k \in \Z} c(k) \chi_k \qquad g = \sum_{s \in \Z} d(s) \chi_s \] and fix $\epsilon \g 0$. There are $K,M \in \N$ such that \[ \left\| f - \sum_{|k| \le K} c(k) \chi_k \right\|_\mathsf{2} \l \epsilon \qquad \left\| g - \sum_{|m| \le M} d(m) \chi_m \right\|_\mathsf{2} \l \epsilon \] both hold. Write $f_K$ and $g_M$ respectively for the truncated sums above. Now \[ | \langle f, T^n g \rangle - \langle f_M, T^n g_K \rangle| \l \epsilon (\| f \|_\mathsf{2} + \| g \|_\mathsf{2} ) \] for all $n \in \N$ using the Hölder inequality. Choose $N$ with $2^N \g M$. For all $n \ge N$ we have \[ \langle f_M, T^n g_K \rangle = \sum_{|m| \le M} \sum_{|k| \le K} c(m) \overline{d(k)} \langle \psi_m, \psi_{2^n k} \rangle = c(0) \overline{d(0)} \] because $2^n k = m$ only happens when $m = 0$ and $k = 0$. Thus \[ | \langle f, T^n g \rangle - c(0) \overline{d(0)} | \l \epsilon (\| f \|_\mathsf{2} + \| g \|_\mathsf{2} ) \] for all $n \ge N$ and we are done. ▮
We can now distinguish the irrational rotation and the doubling map using the Koopman operator. Indeed, suppose that $f$ is an eigenfunction for the doubling map $T$ with eigenvalue $\eta$. Then \[ |\langle f, T^n f \rangle| = |\langle f, \eta^n f \rangle| = |\overline{\eta^n}| \| f \|_\mathsf{2}^2 = \| f \|_\mathsf{2}^2 \] whereas mixing gives \[ \langle f, T^n f \rangle \to \langle f, 1 \rangle \langle 1, f \rangle = 0 \] if $\eta \ne 1$ because eigenfunctions with distinct eigenvalues are orthogonal. Thus $\| f \|_\mathsf{2} = 0$ which is not possible and $T$ has no eigenfunctions.