I introduced equidistributed sequences in an earlier post on the sequence formed by evaluating the sine function at primes in ascending order, and one theorem I mentioned the following theorem:

Theorem1.The following are equivalent:

1. is equidistributed distributed modulo ;

2. For any integer ,

;

3. For any Riemann integrable function ,

I omitted the proof in the aforementioned post, an omission I will now correct. The plan is to work through the proof and then use the theorem, along with some estimates for exponential sums, to present some examples of an equidistributed sequence. I first want to begin with some comments on the statement of Theorem 1.

The reader familiar with be Lebesgue integration may be wondering why ‘Riemann integrable’ appears in statement (3) instead of ‘Lebesgue integrable.’ Well, the theorem is false for a simple reason: we can modify a Lebesgue integrable function on a countable set (more generally, a null set) without changing the value . So if is equidistributed and , then we can define a new function by

such that

*Proof. *To establish equivalence, we will show that (1)(3) and (2)(1). Suppose that (1) holds and that is Riemann integrable. By considering real and imaginay parts separately, we may assume that is real-valued. First, suppose that is a constant function. Then

Now suppose that , where . Then since is equidistributed by hypothesis,

By the linearity of the integral, we see that (3) holds for all step functions . By definition of Riemann integrability, for any , there exist step functions and such that and

Hence,

and

We conclude that

(3)(1) is trivial, since we just take , for any subinterval .

Now suppose that (2) holds. As argued above, we may assume that , where . By linearly interpolating between neighborhoods of the subinterval endpoints, we can find continuous functions and such that and

Since we can repeat the argument used in (1)(2), we only need to show (3) holds for the step functions . Continuous functions can be uniformly approximated by trigonometric polynomials (see Fejer’s theorem or Stone-Weierstrass), so there exist trigonometric polynomials

such that , for . Hence, it suffices to show that

Since the argument is independent of , we take . Then using the hypothesis of (2) and the fact that , for all integers ,

$latex \begin{array}{lcl}\displaystyle\lim_{n\rightarrow\infty}\dfrac{1}{n}\sum_{j=1}^{n}g_{1}(x_{j})-\int_{0}^{1}g_{1}(t)dt&=&\displaystyle\lim_{n\rightarrow\infty}\left(\dfrac{1}{n}\sum_{j=1}^{n}-\int_{0}^{1}\right)\left[\alpha_{0}+\cdots+\alpha_{m}e^{2\pi i m x_{j}}\right]\\[.9 em]&=&\displaystyle\alpha_{0}-\alpha_{0}\\[.9 em]&=&\displaystyle0\end{array}$

(3)(2) is immediate since trigonometric polynomials are Riemann integrable.

A quick consequence of this theorem first proven by Hermann Weyl is that, for any irrational number , the sequence is equidistributed modulo . Indeed, applying the geometric summation formula, we obtain

,

for any integer nonzero integer . Since the last expression is bounded in modulus by , we see that

and equidistribution follows from condition (2) above.

Moreover, the sequence is equidistributed modulo , for any irrational number . To prove this result, we will need a few estimates for exponential sums. To simplify the writing below, we introduce the notation

Note that is not a norm; it is not positively homogeneous.

For positive integers ,

*Proof. *By the triangle inequality,

and applying the geometric summation formula,

$latex \begin{array}{lcl}\displaystyle\left|\sum_{n=a}^{b}e^{2\pi i n\gamma}\right|=\left|\dfrac{e^{2\pi i a\gamma}\left[1-e^{2\pi i(b-a+1)\gamma}\right]}{1-e^{2\pi i\gamma}}\right|&\leq&\displaystyle\dfrac{2}{\left|1-e^{2\pi i \gamma}\right|}\\[.9 em]&\leq&\displaystyle\dfrac{2}{\left|\sin 2\pi\gamma\right|}\end{array}$

The function is nonpositive on . To see this, observe that the derivative is nonpositive on this interval on a subinterval and . The symmetry of about yields the second bound.

Lemma 3.If is an irrational number, and are coprime integers, with , such that , then

for any integer .

*Proof. *Let . Then

Set , so that ranges from to . For a given , ranges from to ; for given , ranges from to . We can rewrite the expression for as

The triangle inequality together with Lemma 2 yields the estimate

We need to now count the number of for which the is . Partition the interval into intervals of length at most (e.g. intervals of the form ). Note that there are at most such intervals. For some , set . By hypothesis, we can write , where . Since we can write as , for some , we see that , which implies that the residue classes are distinct, where , and therefore the residues are distinct, since . If is not equivalent to , then

The contribution in from the residues not equal to , is at most , so

Since the residues classes are distinct and the function is symmetric about the midpoint, we see that , for as above, takes a value in at most twice. We get the estimate

,

where we use that the harmonic series grows logarithmically.

Putting our results above toegether, we obtain the estimate

Taking the square root of both sides and using the elementary inequality , we obtain the estimate

The existence of such integers in the statement of Lemma 3 follows from Dirichlet’s approximation theorem. So, for any irrational ,

Since can be taken arbitrarily large by Dirichlet’s theorem, statement (2) of Theorem 1 implies that is equidistributed.