Using Furstenberg's skew-product for $\alpha n^2$ equidistributed.

Question

I am having trouble proving that if $\alpha_1,\alpha_2$ are irrational, then the sequence $(\alpha_1n,\alpha_2n^2)$ is equidistributed in $\mathbb{T}^2$. It is straightforward to use Furstenberg's skrew product when $\alpha_1=\alpha_2$ but I can't see a way of doing the general case. Any hints?

Answer 1

Hints:

The usual approach is to first find a measurable map of the $2$-torus such that under iteration you get precisely $(\alpha_1 n,\alpha_2 n^2)\bmod1$.

Then you should show that Lebesgue measure is ergodic, most probably by showing that all but one Fourier coefficient vanish.

After that you should apply Birkhoff's ergodic theorem to an arbitrary continuous function on the $2$-torus. For example, by considering a continuous function depending only on the second component you are actually looking at the uniform distribution of $\alpha_2 n^2\bmod1$.