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
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thank you. I thought in this direction, but for unknown reason i could not get it. Now it is clear. $\endgroup$ – demolishka Jan 19 '17 at 8:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.