Proving ergodicity of map on circle

161 Views Asked by At

This dynamical system:

$x_t=x_0 + \omega t $ mod 1

Should be periodic if $\omega$ is rational and ergodic if irrational.

I know that if the time average of generic smooth function $A(x)$ coincides with phase average, ergodicity is proven.

So, following the proof on my textbook, i write $A(x)$ in terms of Fourier series and the time average is:

$\frac{1}{T}\sum^{T-1}_0 A(x)=A_0 + \frac{1}{T}\sum^{T-1}_0 \sum^{\infty}_{n\neq0}A_ne^{2\pi i n (x_0 + \omega t) } $

with $A_0$ the phase average.

Using the truncated geometric series formula:

$ \frac{1}{T}\sum^{T-1}_0 A(x)= A_0 +\frac{1}{T} \sum^{\infty}_{n\neq0}A_ne^{2\pi i n x_0 }\left( \frac{1- e^{2\pi i n \omega T}}{1-e^{2\pi i n \omega} }\right)$

Now, the proof given says ''if $\omega$ is irrational $n\omega$ cannot be integer, thus $|e^{2\pi in\omega}|<1 $the limit for T to $\infty$ implies that the sum vanishes and the time average coincides with $A_0$."

Now, for me is not clear at all this statement. Not mentioning that modulus of complex exponential alone is equal to 1.

Any suggest?

1

There are 1 best solutions below

0
On BEST ANSWER

As you noted, the statement $|e^{2\pi in\omega}| < 1$ is false for any $\omega$. What is needed is $e^{2\pi in\omega} \neq 1$, and is true since $\omega$ is irrational, otherwise the series you provided is not well defined.

Consider the series you wrote, $\frac{1}{T}\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\frac{1-e^{2\pi in\omega T}}{1 - e^{2\pi i n \omega}}$. Breaking the term $\frac{1-e^{2\pi in\omega T}}{1 - e^{2\pi i n \omega}}$ up using identities will help in understanding the problem.

We know that $\frac{1}{T}\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\frac{1-e^{2\pi in\omega T}}{1 - e^{2\pi i n \omega}} = \frac{1}{T}\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\frac{1}{1 - e^{2\pi i n \omega}} - \frac{1}{T}\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T}\frac{1}{1 - e^{2\pi i n \omega}}$. The term $\frac{1}{1 - e^{2\pi i n \omega}} = \frac{1}{2}\Big(1 + i\cot(\pi n \omega)\Big)$. The original sum is then broken up into four terms $$\frac{1}{2T}\Big(\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0} + i\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\cot(\pi n\omega) - \sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T} + i\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T}\cot(\pi n\omega)\Big).$$

The first term may be rewritten as $(A(x_0) - A_0)$ and the third term as $\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T} = \sum_{n\neq 0}^{\infty}A_ne^{2\pi in(x_0+\omega T)} = A(x_T) - A_0$. The second and fourth terms require a little more work.

As you can see, $$\lim_{T\to\infty}\frac{1}{2T}\Big(\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0} + i\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\cot(\pi n\omega) - \sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T} + i\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T}\cot(\pi n\omega)\Big) =$$ $$\lim_{T\to\infty}\frac{1}{2T}\Big((A(x_0) - A_0) + i\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\cot(\pi n\omega) - A(x_T) + A_0 + i\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T}\cot(\pi n\omega)\Big) = $$ $$ 0 + \lim_{T\to\infty}\frac{i}{2T}\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\cot(\pi n\omega) + 0 + \lim_{T\to\infty}\frac{i}{2T}\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T}\cot(\pi n\omega).$$

Knowing that $\cot(\pi n \omega)$ is uniformly bounded in $\mathbb{C}$, once again thanks to $\omega$ being irrational, there exists an $M>0 \in \mathbb{R}$ such that $|\cot(\pi n \omega)| < M$, for all $n \in \mathbb{Z}$. See Proving the cotangent function is uniformly bounded on the complex plane for understanding this statement.

Since $A(x_t)$ is smooth and periodic on $[0,1]$, we know that $A_n = O\Big(\frac{1}{|n|^2}\Big)$, as $|n| \to \infty$. The series $i\sum A_ne^{2\pi inx_0}\cot(\pi n \omega)$, converges absolutely since $$\sum |iA_ne^{2\pi inx_0}\cot(\pi n \omega)| = \sum O\Big(\frac{1}{|n|^2}\Big)M$$ converges to some limit $ L \in \mathbb{R}$ as $n \to \infty$. Consequently $$i\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\cot(\pi n\omega) \text{ and } i\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T}\cot(\pi n\omega)$$ converge and so $\lim_{T\to\infty}\frac{i}{2T}\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}\cot(\pi n\omega) = \lim_{T\to\infty}\frac{i}{2T}\sum_{n\neq 0}^{\infty}A_ne^{2\pi inx_0}e^{2\pi in\omega T}\cot(\pi n\omega) = 0$.

See Stein, Elias M.; Shakarchi, Rami (2003). Fourier Analysis: An Introduction. Princeton University Press for more background.