## Thursday, 6 September 2012

### A clarifying view of Weyl's criterion

Consider an arbitrary sequence $(x_n)$ on the circle $\mathbf{T} = \mathbf{R}/\mathbf{Z}$. If $(x_n)$ is sufficiently well behaved, then often it will have a sort of limiting distribution, which would be usefully represented by a measure $\mu$. For instance, if $x_n$ alternates between $0$ and $1/2$ somewhat regularly, then this limiting distribution $\mu$ is an atomic measure localised at $0$ and $1/2$. Other sequences, such as $\alpha n \text{ mod$1$}$ with $\alpha$ irrational, are equidistributed, by which I mean that the limiting distribution $\mu$ is just the uniform measure. Finally, some sequences do not have a well defined limiting distribution at all, e.g. the sequence which spends 1 year at home, then 10 years at its summer home, then 100 years at home, then 1000 years at the summer home, etc.

There are two ways to make this rigorous:
1. We say $(x_n)$ has a limiting distribution if the limit $$D(f) = \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N f(x_n)$$exists for every continuous function $f:\mathbf{T}\to\mathbf{C}$. In this case $D$ defines a linear functional on $C(\mathbf{T})$ such that $D(1)=1$ and $f\geq 0$ implies $D(f)\geq 0$, so Riesz's representation theorem implies that $D$ is represented by a probability measure $\mu$: $$D(f) = \int_\mathbf{T} f\, d\mu.$$
2. There's another way of saying (pretty much) the same thing, in case you don't like Riesz. Namely, for each $N$ consider the atomic measure $\mu_N$ which gives a mass $1/N$ to each of $x_1, ..., x_N$, and then take a weak limit of some subsequence of $(\mu_N)$ (recall that the space of regular Borel measures on $\mathbf{T}$ is weakly compact). We say that $(x_n)$ has a limiting distribution if this weak limit is independent of subsequence, i.e. if $(\mu_N)$ weakly converges.
Next we say that $(x_n)$ is equidistributed if it has the uniform distribution as its limiting distribution. Now a probability measure on $\mathbf{T}$ is the uniform distribution iff all its nonzero Fourier coefficients vanish. But if $\mu$ is the limiting distribution of $(x_n)$ then $$\hat{\mu}(k) = \int_0^1 e^{-i2\pi k x} d\mu(x) = \lim_{N\to\infty} \frac{1}{N}\sum_{n=1}^N e^{-i2\pi k x_n},$$so this comes down to exactly Weyl's criterion.

Taking the weak limit point of view rather than Riesz's, we can additionally conclude the existence of the limiting distribution from Weyl's criterion: if you take the weak limit $\mu$ of some subsequence, Weyl's criterion implies that this $\mu$ is uniform. Since all convergent subsequences then converge to the same limit, the sequence is actually convergent.

In fact, now that I'm thinking about it, this reasoning gives a sort of generalized Weyl's criterion for the existence of a limiting distribution. Namely, the "Fourier coefficients" $$\lim_{N\to\infty}\frac{1}{N}\sum_{n=1}^N e^{-i2\pi k x_n}$$all exist (they need not be $0$) iff $(x_n)$ has a limiting distribution.

#### 1 comment:

1. See
http://www.echolalie.org/curves for an illustration.