2
$\begingroup$

Consider an almost periodic trigonometric polynomial $f(t)=e^{i2\pi t} + e^{i 2\pi \lambda t}$ for some irrational $\lambda$. I'm interested in distribution of such polynomial. In other words, is there a Borel measure $\mu$ defined on $X = Cl(f(\mathbb{R}))=2\mathbb{D}$ ($\mathbb{D}$ is the unit disk) such that for every Borel subset $E \subset X$

$$\lim_{T \to +\infty}\frac{1}{T}\int\limits_{0}^{T}\chi_{E}(f(t))dt = \mu(E).$$

The main problem is that we can't define a flow over trajectory due to its self-intersections. So we can't analyze this trajectory via standard dynamical system methods. It is well-known that the line $(t, \lambda t)$ is uniformly distributed on torus $\mathbb{T}^2=\mathbb{R}^2 / \mathbb{Z}^2$ w.r.t. Lebesgue measure. So we may expect the same distribution for $f$.

$\endgroup$

1 Answer 1

1
$\begingroup$

Map the torus $\mathbb T^2$ to $2 \mathbb D$ by the projection $\pi: (\theta, \phi) \mapsto e^{i\theta} + e^{i\phi}$. The trajectory is uniformly distributed on $\mathbb T^2$, and what you have is the image of that trajectory under the projection. Thus $\mu(E) = m(\pi^{-1}(E))$ where $m$ is normalized Lebesgue measure.

$\endgroup$
1
  • $\begingroup$ Thank you. I thought in this direction, but for unknown reason i could not get it. Now it is clear. $\endgroup$
    – demolishka
    Commented Jan 19, 2017 at 8:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .