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This question stems from the identity listed as Equation $(6.6)$ in this paper on Pg. $11$. We want to show $$\color{blue}{\lim_{r\to\infty} \int \cos^2(rz + t)\, d\pi_{\#}\mu(z) = \frac12} \tag{6.6}$$ for all $t\in \Bbb R$.

Here, $\mu$ is a Borel probability measure on the cone $$\mathcal C^d = \{(x_1, \ldots, x_{d+1}): |(x_1, \ldots, x_d)| = |x_{d+1}|\} \subset \Bbb R^{d+1}$$ satisfying $$|\hat\mu(\xi)| \le |\xi|^{-\alpha/2} \quad\quad \forall \xi\in \Bbb R^{d+1} \tag{6.1}$$ for some $\alpha > d-1$. Further, we assume $\operatorname{supp} \mu \subset \{(\xi, |\xi|) \in \Bbb R^d\times \Bbb R: \epsilon \le |\xi| \le 1/\epsilon\}$ for some $\epsilon > 0$. Let $\pi:\Bbb R^{d+1} \to \Bbb R$ be the projection map onto the last coordinate; i.e. $\pi: (x_1, \ldots, x_{d+1}) \mapsto x_{d+1}$.

The paper suggests:

Since $d\ge 2$ and $\alpha/2 > (d-1)/2 \ge 1/2$, $(6.1)$ gives $\pi_{\#}\mu\in L^2(\Bbb R)$. Hence, $\pi_{\#}\mu \in L^1(\Bbb R)$ with $\|\pi_{\#}\mu\|_1 = 1$, and $(6.6)$ then follows by approximating $\pi_{\#}\mu$ in $L^1$ with a finite linear combination of characteristic functions of disjoint intervals.

  • I tried to follow the line of thought in the paper to first show that $\pi_{\#}\mu \in L^2(\Bbb R)$. If we are to use Theorem $3.3$ of Mattila's Fourier Analysis and Hausdorff Dimension, we shall have to show that $\pi_{\#}\mu\in \mathcal M(\Bbb R)$ and $\widehat{\pi_{\#}\mu}\in L^2(\Bbb R)$. I have \begin{align*} \widehat{\pi_{\#}\mu}(\xi) &= \int_{\Bbb R} e^{-2\pi i\xi z} \, d\pi_{\#} \mu(z) \\ &= \int_{\Bbb R^d\times\Bbb R} e^{-2\pi i\xi z}\, d\mu(x,z)\\ &= \widehat{\mu}(0, \ldots, 0, \xi) \end{align*}

So, $$\int_{\Bbb R} \left| \widehat{\pi_{\#}\mu}(\xi) \right|^2 \, d\xi = \int_{\Bbb R} |\widehat{\mu}(0, \ldots, 0, \xi)|^2\, d\xi \lesssim \int_{\Bbb R} |\xi|^{-\alpha}\, d\xi = 2\int_0^\infty \xi^{-\alpha}\, d\xi$$ Is $\int_0^\infty |\xi|^{-\alpha}\, d\xi < \infty$? Note that $\alpha \ge 1$.

  • As suggested by @J.G. in the comments, it's enough to show $$\lim_{r\to\infty} \int e^{2irz}\, d\pi_{\#}\mu(z) = \lim_{r\to\infty}\widehat{\pi_{\#}\mu}\left(-\frac{r}{\pi}\right) = 0$$

Notation.

  • $\mathcal M(\Bbb R^n)$ is the set of all Borel measures $\nu$ satisfying $0 < \nu(\Bbb R^n) < \infty$ with compact support $\operatorname{supp} \nu \subset \Bbb R^n$.
stoic-santiago
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    Hint: let $\Bbb Ef(z):=\int f(z)d\pi_#\mu(z)$ so$$\Bbb E\cos^{2}(rz+t)=\tfrac12(1+\cos2t\Bbb E\cos(2rz)-\sin2t\Bbb E\sin(2rz)),$$whence your problem reduces to showing $\Bbb E\cos(2rz)=\Bbb E\sin(2rz)=0$, i.e. $\Bbb E\exp(2irz)=0$. – J.G. Jul 03 '22 at 17:58
  • @J.G. How would you tackle the reduced problem? $$\mathbb E[e^{2irz}] = \int e^{2irz} , d\pi_{#}\mu(z) = \int e^{2irz} , d\mu(z) \stackrel{?}{=} 0$$ – stoic-santiago Jul 04 '22 at 16:49
  • To prove $e^{2irz}$ averages to $0$, I'd exploit the symmetries of the problem. – J.G. Jul 04 '22 at 16:50
  • How about showing $\pi_{#}\mu \in L^1(\Bbb R)$ and using the Riemann-Lebesgue lemma (or similar)? Regardless, could you please give some more details? @J.G. – stoic-santiago Jul 04 '22 at 18:40
  • @J.G. An answer (possibly detailed) would be great :) – stoic-santiago Jul 07 '22 at 17:00
  • I'll be honest, I don't know exactly how to make a proof rigorous given the very specific machinery in this problem. I just expect someone who did would do it that way. – J.G. Jul 07 '22 at 17:04

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