Two very different types of dynamical behaviour are illustrated by a pair of very well-known examples on the circle: the doubling map and an irrational rotation. On the unit circle in , the doubling map is given by , while an irrational rotation is given by for some irrational .

Lebesgue measure (arc length) is invariant for both transformations. For the doubling map, it is just one of many invariant measures; for an irrational rotation, it turns out to be the only invariant measure. We say that the doubling map exhibits *hyperbolic* behaviour, while the irrational rotation exhibits *elliptic* behaviour.

Systems with hyperbolicity have many invariant measures, as we saw in a series of previous posts. The goal of this post is to recall a proof that the opposite situation is true for an irrational rotation, and that in particular every orbit equidistributes with respect to Lebesgue measure; then we consider orbits generated by random rotations, where instead of rotating by a fixed angle , we rotate by either or , with the choice of which to use being made at each time step by flipping a coin.

**1. Invariant measures via **

First we recall some basic facts from ergodic theory for topological dynamical systems. Given a compact metric space and a continuous map , let denote the space of Borel probability measures on . Writing for the space of all continuous functions , recall that is the space of all continuous linear functionals . Then is (isomorphic to) the space of finite complex Borel measures on . *(This last assertion uses the fact that is a compact metric space and combines various results from “Linear Operators” by Dunford and Schwartz, but a more precise reference will have to wait until I have the book available to look at.)*

Using this fact together with the polar decomposition for finite complex Borel measures, we have the following: for every , there is and a measurable function such that

Note that although is endowed with the operator norm, we will usually think of it as a topological vector space with the weak* topology. Thus embeds naturally into , and (1) shows that every element of can be described in a canonical way in terms of .

Let be a countable set whose span is dense in . Then to every we can associate the sequence , where depending on the context we may index using either or . If is bounded then for every , and so such a defines a linear map .

Because is determined by the values (by linearity and continuity of , and density of the span of ), the map is 1-1. In particular, it is an isomorphism onto its image, which we denote by

Note that because is separable and is not.

It is straightforward to see that is continuous, and its inverse is also continuous on . Thus we can translate questions about , and in particular about , into questions about .

Remark 1It is a nontrivial problem to determine which elements of correspond to elements of , and also to determine which of those sequences correspond to actual measures (elements of ). We will not need to address either of these problems here.

The action induces an action on by , and hence on by duality. This action is given by

In particular, also induces an action by

A measure is -invariant iff it is a fixed point of . Let be the action induced by on ; that is,

If can be chosen so that takes a particularly nice form, then this can be used to understand what invariant measures has, and how empirical measures converge.

Let us say more clearly what is meant by convergence of empirical measures. Given and , let be the empirical measure along the orbit segment . Let be the set of weak* accumulation points of the sequence . By compactness of , the set is non-empty, and it is a standard exercise to show that every measure in is -invariant.

In particular, if is uniquely ergodic, then it only has one invariant measure , and so for every . In this case we have for every , and it is reasonable to ask how quickly this convergence occurs.

**2. Irrational rotations **

Now we specialise to the case of an irrational rotation. Let be the unit circle, fix irrational, and let be given by . We will show that is -invariant iff it is Lebesgue measure, and then examine what happens in a broader setting.

Given , let , and let . Then the span of contains all functions that are polynomials in and , thus it is a subalgebra of that contains the constant functions, separates points, and is closed under complex conjugation. By the Stone–Weierstrass theorem, this span is dense in , and since is bounded the discussion from above gives an isomorphism , where we suppress in the notation. This isomorphism is given by

for a general , and for we write

The sequence is the sequence of *Fourier coefficients* associated to the measure . The choice of means that the action induces on takes a simple form: the Fourier coefficients of and are related by

Thus if is invariant, we have for all . Because is irrational, we have for all , and so . Thus the only non-zero Fourier coefficient is . Because is an isomorphism between and , this shows that the only with is Lebesgue measure. In particular, is uniquely ergodic, with Lebesgue as the only invariant measure, and thus for any , the empirical measures converge to Lebesgue.

**3. Random rotations **

Consider a sequence of points , and let be the average of the point masses on the first points of the sequence:

We say that the sequence *equidistributes* if converges to Lebesgue on in the weak* topology.

The previous sections showed that if the points of the sequence are related by , where is irrational, then the sequence equidistributes. A natural generalisation is to ask what happens when the points are related not by a fixed rotation, but by a randomly chosen rotation.

Here is one way of making this precise. Let be the set of infinite sequences of 1s and 2s, and let be the -Bernoulli measure on , so that all sequences of length are equally likely. Fix real numbers and , and fix . Given , consider the sequence given by

Then one may ask whether or not equidistributes almost surely (that is, with probability 1 w.r.t. ). The remainder of this post will be dedicated to proving the following result.

Theorem 1If either of or is irrational, then equidistributes almost surely.

Remark 2The proof given here follows a paper by Lagarias and Soundararajan, to which I was referred by Lucia on MathOverflow.

Using Fourier coefficients as in the previous section, we have that equidistributes iff all the non-constant Fourier coefficients of converge to zero — that is, iff as for all . This is Weyl’s criterion for equidistribution.

Fix a value of , which will be suppressed in the notation from now on. Write for the absolute value of the th Fourier coefficient of , and note that

The outline of the proof is as follows.

- Show that there is a constant such that the expected value of is at most .
- Given , show that there is a constant such that the probability that exceeds is at most .
- Find an exponentially increasing sequence such that if , then for every .
- Use the Borel–Cantelli lemma to deduce that with probability 1, exceeds only finitely often, hence exceeds only finitely often. Since was arbitrary this shows that .

** Step 1 **

Given , let denote the distance between and the nearest integer.

*Proof:* Let be such that , and define recursively by , so that .

Using the fact that , we have

If , then

where the sums are over all for . Since there are values of with , we have

and since we have

Together with (11), this gives

Using the fact that and that , we have

and so

Finally, for we have , which proves the bound in (10).

Because one of is irrational, the denominator in (10) is positive, and so , which completes Step 1.

** Step 2 **

Given , we have

and so by Lemma 2 we have

Putting completes step 2.

** Step 3 **

Given , we have

and so

In particular, if for some , we have

for all . If , then (12) implies that for all .

** Step 4 **

Let be the event that . By Part 2, we have , and because increases exponentially in by (13), we have . By the Borel–Cantelli lemma, this implies that with probability 1, there are only finitely many values of for which .

By the previous part, this in turn implies that there are only finitely many values of for which . In particular, , and since was arbitrary, we have . Thus the sequence satisfies Weyl’s criterion almost surely, which completes the proof of Theorem 1.