Lecture 7

Equidistribution theorem¶

Let γ be an irrational number. The equidistribution theorem says that the sequence $\gamma, 2\gamma,\cdots$ mod 1 is uniformly distributed on $\mathbb{T}$.

Lemma¶

For every $f\in \mathscr{C}^0(\mathbb{T})$, $\lim_{n\to \infty}\frac{\sum_{k = 1}^{n-1} f(k\gamma)}{n} = \int_{0}^1 f(x)dx$.

Proof. Let $A$ be the set of functions on $\mathbb{T}$ such that the equality holds, then it sufficies to show that $A$ contains all continuous functions. It’s easy to observe that $A$ is closed under linear combination. Then it sufficies to show the following:

• A contains $e^{2\pi i m \theta}$ for $m\in \mathbb{Z}$.
• A is closed under taking uniform limits, in other words, if $F_N \in A$, $F_N\to F$ uniformly, then $F\in A$.

We first show that $A$ is closed under uniform limits. Let $F_N,F$ be chosen as above. We must show that $|\int_{0}^1 F(x)dx - \frac{\sum_{k = 1}^{n-1}F(k\gamma)}{n}| \to 0$ when $n\to \infty$. Let $\epsilon > 0$, choose $N$ such that $\sup_{x\in [0,1]} |F_N(x) - F(x)|<\epsilon$. Then

then choose $n$ large enough so that the rest term $<\epsilon$.

To show that $A$ contains $e^{2\pi i m \theta}$, observe that in this case ${\sum_{k = 1}^{n-1} f(k\gamma)} =\sum_{k = 1}^{n-1} \omega^k$, where $\omega = e^{2\pi i m \gamma}$. Since γ is not rational, $\omega\neq 1$ for every $m\in \mathbb{Z}$ so that $\sum_{k = 0}^{n-1}\omega^k = \frac{1-\omega^n}{1-\omega}$. Then $|\frac{\sum_{k = 1}^{n-1} f(k\gamma)}{n}|\leq \frac{1}{n}\frac{2}{1-\omega}\to 0$. This shows $e^{2\pi i m \theta}\in A$ for every $m\in \mathbb{Z}$.

Now, since every continuous function on $\mathbb{T}$ is a uniform limit of trignomic polynomials (Fejer theorem), we conclude that $\mathscr{C}^0(\mathbb{T})\subset A$, the lemma is proved.

Now it reduces to show that $\chi_{(a,b)}\in A$. Let $f_{\epsilon}^+,f_{\epsilon}^-$ be continuous piecewise linear functions defined by

and similarly

(see page 110, figure 3 of Stein-Shakarchi for graph) then by definition they satisfy

Let $S_N:=\frac{1}{N}\sum_{n = 1}^N \chi_{(a,b)}(n\gamma)$. By (4), $\frac{1}{N}\sum_{n = 1}^N f_{\epsilon}^-(n\gamma)\leq S_N\leq\frac{1}{N}\sum_{n = 1}^N f_\epsilon^+(n\gamma)$.

Apply the lemma, for $N$ large enough we have $\frac{1}{N}\sum_{n = 1}^N f_\epsilon^+(n\gamma) \leq \int_{0}^1 f_\epsilon^+ + \epsilon$ and $\frac{1}{N}\sum_{n = 1}^N f_\epsilon^-(n\gamma) > \int_{0}^1 f_\epsilon^- - \epsilon$, so

By (6),

This implies $\lim_{N\to \infty}S_N = b-a$.

Weyl criterion¶

A natural question to ask is how about a general sequence $\gamma_n$ on $[0,1]$ instead of $n \gamma$. From the argument above we see that the key was to show that the associate set $A$ contains $e^{2\pi i n x}$ for all $n \in \mathbb{Z}$. Then we obtain

Theorem (Weyl criterion)¶

A sequence $\gamma_n$ is equidistributed on $\mathbb{T}$ if and only if for all non-zero $k\in \mathbb{Z}$,