(n2αmod(n^2 \alpha \bmod 1) is equidistributed in \mathbb{T}^2\mathbb{T}^2 if \alpha \in \mathbb{R} \setminus \mathbb{Q}\alpha \in \mathbb{R} \setminus \mathbb{Q}

I did the following homework question, can you tell me if I have it right?

We want to show that the sequence (n^2 \alpha \bmod 1) is equidistributed if \alpha \in \mathbb{R} \setminus \mathbb{Q}. To that end we consider the transformation T: (x,y) \mapsto (x + \alpha, y + 2x + \alpha) on the 2-dimensional torus \mathbb{T}^2 endowed with the Lebesgue measure \lambda \times \lambda.

a) Show that the action of T on the torus is ergodic, i.e., if a measurable set A \subset \mathbb{T}^2 is invariant under T, then (\lambda \times \lambda)(A) \in \{0, 1\}. Show this by checking the following equivalent definition of ergodicity:

\forall f \in L^2( \mathbb{T}^2), we have: if f is T-invariant, then f has to be constant almost everywhere.

Hint: Use Fourier series.

My answer:

f(x,y) &= \sum_{j,k \in \mathbb{Z}} c_{jk} e^{ijx} e^{iky}
\\ &\stackrel{f = f\circ T}{=} \sum_{j,k \in \mathbb{Z}} c_{jk} e^{ij(x + \alpha)} e^{ik(y + 2x + \alpha)}
\\ &=
\sum_{j,k \in \mathbb{Z}} c_{jk} e^{ij\alpha + ik\alpha} e^{ijx + ik2x}e^{iky}
\\ &\stackrel{j \rightarrow j-2k}{=}
\sum_{j,k \in \mathbb{Z}} c_{(j-2k)k} e^{i(j-k)\alpha} e^{ijx}e^{iky} .

Now we want c_{jk} = c_{(j-2k)k} e^{i(j-k)\alpha}, and we have |c_{jk}| = |c_{(j-2k)k}| = |c_{(j-4k)k}| = \cdots, and so on.

The series only converges if |c_{(j-4k)k}| \; \xrightarrow{k \rightarrow \infty} \; 0 and so c_{jk} has to be 0, too.

b) For x \in \mathbb{T} show that \frac{1}{m} \sum_{n=1}^m T^n_\ast (\delta_x \times \lambda) \rightarrow \lambda \times \lambda using the equidistribution of (n \alpha \bmod 1).

My answer:

\frac{1}{m} \sum_{n=1}^m T_\ast^n(\delta_x \times \lambda (A \times B))
\frac{1}{m} \sum_{n=1}^m \delta_x \times \lambda ((T^{-1})^n(A \times B))
\\ &=
\frac{1}{m} \sum_{n=1}^m \delta_x \times \lambda (T^n(A \times B)) ,

where I have the last equality because it doesn’t matter which way the points are shifted. Then writing out what T^n does I get:

\frac{1}{m} \sum_{n=1}^m \delta_x \times \lambda (T^n(A \times B))
\frac{1}{m} \sum_{n=1}^m \delta_x \times \lambda ( (A + \alpha n) \times (B + \alpha n) ) \\ &=
\frac{1}{m} \sum_{n=1}^m \delta_x (A + \alpha n) \lambda(B + \alpha n)
\\ &=
\frac{1}{m} \sum_{n=1}^m \chi_{A + \alpha n}(x) \lambda(B + \alpha n) .

And then using that the Lebesgue measure \lambda is translation invariant I get:

\frac{1}{m} \sum_{n=1}^m \chi_{A + \alpha n}(x) \lambda(B + \alpha n) =
\frac{1}{m} \sum_{n=1}^m \chi_{A + \alpha n}(x) \lambda(B) .

And finally, by using the ergodic theorem:

\frac{1}{m} \sum_{n=1}^m \chi_{A + \alpha n}(x) \lambda(B)=
\lambda(B) \int_{\mathbb{T}} \chi_{A + \alpha n} (x) d \lambda(x) = \lambda(B)\lambda(A) = \lambda \times \lambda (A \times B) .

c) For \eta \in (0,1) and x,y \in \mathbb{T} define the two sequences

\mu_m &=
\frac{1}{m} \sum_{n=1}^m T^n_\ast \left(\delta_x \times \left(\frac{1}{2 \eta} \left. \lambda \right \vert_{[y-\eta, y + \eta]} \right) \right)
\nu_m &=
\frac{1}{m} \sum_{n=1}^m T^n_\ast \left(\delta_x \times \left( \frac{1}{1 – 2 \eta} \left. \lambda \right\vert_{\mathbb{T} \smallsetminus [y-\eta, y + \eta]} \right) \right)

Using exercise 3 of assignment 10 and weak^\ast-compactness of the unit ball we know that there exists a subsequence in \mathbb{N} such that both sequences converge along these subsequences. Call the limit points \mu and \nu respectively. Show that 2 \eta \mu + (1 – 2 \eta) \nu = \lambda \times \lambda.

My answer:

Exercies 3 of assignment 10 was to show that any weak^\ast limit point \mu of the sequence \mu_n = \frac{1}{n}\sum_{j=0}^{n-1} T_\ast^j \nu_n is a Borel probability measure with \mu = T_\ast \mu.

2 \eta \lim_{m \to \infty} \frac{1}{m} \sum_{n=1}^m T_\ast^n \left(\delta_x \times \left( \frac{1}{2 \eta} \left. \lambda \right\vert_{[y- \eta, y + \eta]} \right) \right) +

(1 – 2 \eta) \lim_{m \to \infty} \frac{1}{m} \sum_{n=1}^m T_\ast^n \left( \delta_x \times \left( \frac{1}{1- 2 \eta} \left. \lambda \right\vert_{\mathbb{T} \smallsetminus [y- \eta, y + \eta]} \right) \right) ,

which, by part (b), is equal to

\left. \lambda \times \lambda \right\vert_{[y – \eta , y + \eta]} + \left. \lambda \times \lambda \right \vert_{\mathbb{T} \smallsetminus [y – \eta , y + \eta]} = \lambda \times \lambda.

d) Using the following proposition, show that \mu = \lambda \times \lambda.

Proposition: A T-invariant probability measure is extremal if and only if its action is ergodic.

My answer:

Distinguish the cases \eta \geq \frac{1}{2} and \eta < \frac{1}{2}.

If \eta < \frac{1}{2} then by c) \lambda \times \lambda = 2 \eta \mu + (1 - 2 \eta) \nu and since 2 \eta < 1, by extremality, \lambda \times \lambda = \mu = \nu.

If \eta \geq \frac{1}{2} then [y - \eta , y + \eta] = \mathbb{T} and so

\mu(A \times B) &= \lim_{m \to \infty} \mu_m (A \times B)
\\ &= \lim_{m \to \infty} \frac{1}{m} \frac{1}{2 \eta} \sum_{n=1}^m \chi_A(x + \alpha n) \lambda \mid_{[y - \eta , y + \eta]} (B) \\
&\stackrel{b)}{=} \frac{1}{ 2 \eta} \lambda \times \lambda

So I think I got the sums wrong here.

e) Show that \mu_m \to \lambda \times \lambda. To that end prove and apply the following:

Lemma: Let X be a metric space and x \in X, x_n a sequence in X. Assume that each subsequence of x_n has a subsequence converging to x. Then x_n itself converges to x.

My answer:

Assume that x_n converges to y \neq x. Then there is a (sub)sequence (the sequence itself is a subsequence) not converging to x. Contradiction. Same argument for x_n diverges.

Now with c) and d), \mu_m \to \mu = \lambda \times \lambda.

I'm stuck on:

f) Show that for all f \in C(\mathbb{T}^2) and for all \varepsilon > 0, there exists \eta > 0 such that we have

\left\vert \int f d \mu_m - \int f d \omega_m \right\vert < \varepsilon ,
where \omega_m = \frac{1}{m} \sum_{n=1}^m T^n_\ast (\delta_x \times \delta_y) .

Thanks for your help!


Obviously this answer is very late, but the argument sketched in this exercise is so pretty it really deserves an answer. One thing that we will use on a couple of occasions is that T^n(x,y) = (x + n\alpha, y + 2nx + n^2\alpha), which is easily verifiable by induction.

(a) Your answer for this part is close to right. Since you are working mod 1, you should use exponentials like e^{2\pi ijx} instead of e^{ijx} when doing your Fourier analysis. Nonetheless, you correctly derive that if f\circ T = f, then c_{j,k} = c_{j-2k,k}e^{2\pi i(j-k)\alpha}. If k\neq 0, then you also correctly note that |c_{j,k}| = |c_{j-2nk,k}| for all n, and hence in order to get convergence of the Fourier series you need that c_{j,k} = 0. This argument fails when k = 0, however. If k = 0, we have derived that c_{j,0} = c_{j,0}e^{2\pi ij\alpha}. If j\neq 0, then e^{2\pi ij\alpha}\neq 1, and thus one must have c_{j,0} = 0. This proves that the only possible nonzero Fourier coefficient for f is c_{0,0}, and hence that f is constant.

(b) Your argument doesn't make use of the equidistribution of n\alpha mod 1, so I think it has some problems. A simpler argument is as follows. Let f\in C(\mathbb{T}^2), and let g\in C(\mathbb{T}) be the function g(x) = \int_{\mathbb{T}} f(x,y)\,d\lambda(y). Then \frac{1}{m}\sum_{n=1}^m\int_{\mathbb{T}^2}f\circ T^n\,d(\delta_x\times \lambda) = \frac{1}{m}\sum_{n=1}^m\int_\mathbb{T}f(x+n\alpha,y + 2nx + n^2\alpha)\,d\lambda(y) = \frac{1}{m}\sum_{n=1}^mg(x + n\alpha). Here the last equality uses the fact that \lambda is translation invariant on \mathbb{T}. Now by equidistribution of n\alpha mod 1, we get that \frac{1}{m}\sum_{n=1}^m\int_{\mathbb{T}^2} f\circ T^n\,d(\delta_x\times \lambda) = \frac{1}{m}\sum_{n=1}^mg(x + n\alpha)\to \int_\mathbb{T}g\,d\lambda = \int_{\mathbb{T}^2} f\,d(\lambda\times\lambda). This limit is precisely what it means for m^{-1}\sum_{n=1}^mT^n_*(\delta_x\times\lambda)\to \lambda\times\lambda.

(c) Pretty sure you need \eta\in (0,1/2) here, since otherwise \mu_m and \nu_m aren't probability measures. From now on I assume \eta\in (0,1/2). By weak* compactness, there is a subsequence m_k such that \mu_{m_k} converges to some \mu. Again by compactness, we can pass to a subsequence m_{k_l} of m_k to get that \nu_{m_{k_l}} converges to some \nu. From (b), we know that 2\eta\mu_{m_{k_l}} + (1 - 2\eta)\nu_{m_{k_l}}\to \lambda\times\lambda. On the other hand, 2\eta\mu_{m_{k_l}} + (1 - 2\eta)\nu_{m_{k_l}}\to 2\eta\mu + (1-2\eta)\nu, so \lambda\times\lambda = 2\eta\mu + (1-2\eta)\nu.

(d) \lambda\times \lambda = 2\eta\mu + (1-2\eta)\nu is a decomposition of \lambda\times\lambda into a convex combination of two measures \mu and \nu. Since \lambda\times\lambda is ergodic (part a), it is extremal, and hence this can only happen if \mu = \nu = \lambda\times\lambda.

(e) Note that \mu was any arbitrary limit point of the sequence \mu_m. Thus \lambda\times\lambda is the only limit point of \mu_m. The lemma then says \mu_m\to \lambda\times \lambda.

(f) Equip \mathbb{T} with the flat metric (the metric inherited from \mathbb{R}), and \mathbb{T}^2 with the product metric d. You are given f and \epsilon. Since \mathbb{T}^2 is compact, f is uniformly continuous, and hence there is an \eta>0 (with \eta<1/2) such that if (x,y),(x',y')\in \mathbb{T}^2 are such that d((x,y),(x',y'))\leq\eta, then |f(x,y) - f(x',y')|<\epsilon. The triangle inequality (and unravelling definitions) give \left|\int_{\mathbb{T}^2} f\,d\mu_m - \int_{\mathbb{T}^2}f\,d\omega_m\right|\leq \frac{1}{m}\sum_{n=1}^m\frac{1}{2\eta}\left|\int_{\mathbb{T}} f\circ T^n(x,z) - f\circ T^n(x,y)\,d\lambda|_{[y-\eta,y+\eta]}(z)\right|. Because T^n(x,z) = (x + n\alpha, z + 2nx + n^2\alpha) and T^n(x,y) = (x + n\alpha, y + 2nx + n^2\alpha), we see that if z\in [y-\eta,y + \eta], then d(T^n(x,z), T^n(x,y))\leq\eta. In particular, if z\in [y-\eta,y + \eta], then |f\circ T^n(x,z) - f\circ T^n(x,y)|<\epsilon. This shows that each of the terms in the above some is bounded above by \epsilon, and hence that the entire right hand side is bounded above by \epsilon. This finishes (f).

To complete the argument, we note that (f) shows that \omega_m\to \lambda\times\lambda for any fixed x,y\in \mathbb{T}. Applying this to x = y = 0 will give the desired equidistribution. Indeed, let g\in C(\mathbb{T}), and set f\in C(\mathbb{T}^2) to be f(x,y) = g(y). The fact that \omega_m\to \lambda\times\lambda says precisely that \frac{1}{m}\sum_{n=1}^m g(n^2\alpha) = \frac{1}{m}\sum_{n=1}^m f(n\alpha, n^2\alpha)\to \int_{\mathbb{T}^2}f\,d(\lambda\times\lambda) = \int_\mathbb{T}g\,d\lambda.

Source : Link , Question Author : Rudy the Reindeer , Answer Author : froggie

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