Uniform distribution

Home | Assessment | Notes | Worksheets | Tutorials

In this section we will study more closely the statistical properties of irrational rotation dynamics. Fix an irrational number $\alpha$ and define $T$ on $[0,1)$ by $$T(x)=x+\alpha mod1$$ for all $x\in [0,1)$. Given $E\subset [0,1)$ we investigate for each $N\in \mathbb{N}$ the statistical quantity $${\displaystyle \frac{|\{0\le n\le N-1:T^n(x)\in E\}|}{N}}$$ and in particular wish to understand whether its limit exists as $N\to \mathrm{\infty }$.

How can we analyse this average? There is a fruitful way to rewrite it in terms of functions. We have $${\displaystyle \frac{|\{0\le n\le N-1:T^n(x)\in E\}|}{N}}={\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{1}_E(T^n(x))$$ for all $N\in \mathbb{N}$ and this suggests considering more generally averages of the form $${\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))$$ for $f:[0,1)\to \mathbb{R}$.

First we note that the limit of the above quantity, if it exists, is well-behaved under uniform convergence. Recall that $f_k$ converges to $g$ uniformly, if, for every $ϵ>0$ there is $K\in \mathbb{N}$ such that $$k\ge K\Rightarrow ||f_k-g|{|}_{\mathsf{u}}<ϵ$$ holds. Here the expression $$||h|{|}_{\mathsf{u}}=sup\{|h(x)|:x\in X\}$$ is the uniform norm of $h:X\to \mathbb{C}$. It is convenient to write uniform convergence in terms of the uniform norm, but we are not describing a new type of convergence; we have only rephrased the definition of uniform convergence we are used to from real analysis. In particular, for example, it is true that if $f_k\to g$ uniformly and each $f_k$ is continuous then $g$ will also be continuous.

Proposition

Fix a sequence $k\mapsto f_k$ of functions from $X$ to $\mathbb{C}$ that converges uniformly to a function $g:X\to \mathbb{C}$. If $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{f}_k(T^{n}x)$$ exists for every $k\in \mathbb{N}$ then $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}g(T^{n}x)=\underset{k\to \mathrm{\infty }}{lim}\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{f}_k(T^{n}x)$$ and in particular the limit on the left-hand side exists.

Proof:

Since $k\mapsto f_k$ is in particular a Cauchy sequence for the uniform norm one can show $$k\mapsto \alpha (k)=\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{f}_k(T^{n}x)$$ is a Cauchy sequence of real numbers and therefore converges to a limit, say $\beta$.

Fix $ϵ>0$. Choose $k$ so large that $||f_k-g|{|}_{\mathsf{u}}<\frac{ϵ}{2}$ and $$|\beta -\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{f}_k(T^n(x))|<\frac{ϵ}{2}$$ both hold. Choose then $M$ so large that $$N\ge M\Rightarrow |\alpha -{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{f}_k(T^n(x))|<\frac{ϵ}{2}$$ holds. For every $N\ge M$ we have $$|\alpha -{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}g(T^n(x))|<ϵ$$ by the triangle inequality. ▮

On $[0,1)$ there is a special collection of functions for which the average is straightforward to analyze and which approximate all other continuous functions with respect to the uniform norm. These are the exponential functions $${\psi }_k(x)=\exp (2\pi ikx)$$ defined for each $k\in \mathbb{Z}$.

Theorem

For every continuous function $g:X\to \mathbb{C}$ and every $ϵ>0$ there are $r(1),\dots ,r(s)$ in $\mathbb{Z}$ and $c(1),\dots ,c(s)$ in $\mathbb{C}$ such that $$||c(1){\psi }_{r(1)}+\cdots +c(s){\psi }_{r(s)}-g|{|}_{\mathsf{u}}<ϵ$$ holds.

Proof:

We will take this result for granted. It can be deduced from the Stone-Weierstrass theorem or proved directly via analysis of the Fejér kernel. ▮

Let us then analyze the average $${\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{\psi }_k(T^n(x))$$ as $N\to \mathrm{\infty }$ for each $k\in \mathbb{Z}$. Since ${\psi }_0=1$ we clearly have $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{\psi }_0(T^n(x))=1$$ for all $x\in X$ and proceed assuming $k\ne 0$. We can calculate $$\begin{array}{rl}& \sum _{n=0}^{N-1}{\psi }_k(T^n(x))\\ =& \sum _{n=0}^{N-1}\exp (2\pi ik(x+n\alpha ))\\ =& \exp (2\pi ikx){\displaystyle \frac{1-\exp (2\pi ikN\alpha )}{1-\exp (2\pi ik\alpha )}}\end{array}$$ using the geometric series formula because $\alpha$ is irrational. Now $$|\sum _{n=0}^{N-1}{\psi }_k(T^n(x))|\le {\displaystyle \frac{2}{|1-\exp (2\pi ik\alpha )|}}$$ for all $N\in \mathbb{N}$ and, since the upper-bound is independent of $N$ we have $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{\psi }_k(T^n(x))=0$$ for all non-zero $k\in \mathbb{Z}$ and all $x\in X$.

Combining all of the above, we conclude that $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))$$ exists for every $x\in X$ and every continuous function $f:X\to \mathbb{C}$. In fact, more is true. We have $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}f(T^n(x))=\underset{0}{\overset{1}{\int }}f(x)\mathsf{d}x=\int f\mathsf{d}\mu$$ for every continuous function $f:X\to \mathbb{C}$ and every $x\in X$.

To finish, let us return to $f=1_E$ where $E=[a,b)$ is a sub-interval. We cannot approximate $1_{[a,b)}$ by continuous functions with respect to the uniform norm because the function $1_{[a,b)}$ is not continuous! However, we can find for every $ϵ\ge 0$ continuous functions $g,h$ from $X$ to $[0,1]$ with $$g\le 1_{[a,b)}\le h$$ and $$0\le \int h-g\mathsf{d}\mu <ϵ$$ from which $${\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}g(T^n(x))\le {\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{1}_{[a,b)}(T^n(x))\le {\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}h(T^n(x))$$ follows for all $N\in \mathbb{N}$ and all $x\in [0,1)$. Taking the limit as $N\to \mathrm{\infty }$ the expression $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{1}{N}}\sum _{n=0}^{N-1}{1}_{[a,b)}(T^n(x))$$ is forced to exist as it is sandwiched between $$\int g\mathsf{d}\mu$$ and $$\int h\mathsf{d}\mu$$ which, as we make $ϵ$ smaller and smaller, both converge to $b-a$. Thus we have proved the following theorem.

Theorem

Fix $\alpha$ irrational. For all $0\le a<b\le 1$ we have $$\underset{N\to \mathrm{\infty }}{lim}{\displaystyle \frac{|\{0\le n\le N-1:a\le n\alpha mod1<b\}|}{N}}=b-a$$ for all $x\in [0,1)$.

We say that the sequence $n\alpha$ is uniformly distributed mod 1. It is remarkable that the result is true simultaneously for all intervals $[a,b)\subset [0,1)$.