# $|\hat\mu(\xi)| \lesssim |\xi|^{-1/2}$ where $\mu$ is $f\mapsto \int_{\mathbb R} \psi(r) \int_{S^{1}} f(rx,r)\, d\sigma(x)\, dr$

I have questions about the proof of Theorem $$2.1$$ here. The proof is on Pg. $$10$$. I am trying to work out the $$d = 2$$ case in particular. $$\mathcal C^d = \{(x_1, \ldots, x_{d+1}): |(x_1, \ldots, x_d)| = |x_{d+1}|\} \subset \Bbb R^{d+1}$$ is the $$d$$-dimensional cone. The first part of the proof of $$\dim_F \mathcal C^2 = 1$$ on Pg. $$10$$ is to show that $$\dim_F \mathcal C^2 \ge 1$$. To do this, it is enough to consider the measure $$\mu$$ defined by $$f\mapsto \int_{\mathbb R} \psi(r) \int_{S^{1}} f(rx,r)\, d\sigma(x)\, dr$$ where $$f$$ is any non-negative Borel function, $$\sigma$$ is the rotation invariant Borel probability measure on $$S^1$$, and $$\psi$$ is a bump function on $$[1,2]$$ with $$\int \psi = 1$$. The Fourier dimension of $$A\subset\Bbb R^{d+1}$$ is: $$\dim_F A = \sup\{0\le s\le d+1: \exists\mu\in \mathcal P(A) \text{ s.t. } |\hat\mu(\xi)| \lesssim |\xi|^{-s/2},\, \forall \xi\in \Bbb R^{d+1}\}$$ where $$\mathcal P(A)$$ is the set of Borel probability measures on $$\Bbb R^{d+1}$$ satisfying $$\mu(A) = 1$$.

$$\mu$$ is the measure associated with the linear functional $$\Lambda: f\mapsto \int_{\mathbb R} \psi(r) \int_{S^{1}} f(rx,r)\, d\sigma(x)\, dr$$ in the sense that $$\Lambda f = \int f\, d\mu$$ for any non-negative Borel function $$f$$.

Question: To show $$\dim_F \mathcal C^2 \ge 1$$, we must compute the Fourier transform $$\widehat{\mu}$$ and prove that $$|\hat\mu(\xi)| \lesssim |\xi|^{-1/2},\, \forall \xi\in \mathbb R^3$$ If $$z = (rx_1,rx_2,r)$$ where $$(x_1,x_2)\in S^1$$, i.e., $$x_1^2 + x_2^2 = 1$$, then $$\widehat{\mu}(\xi) = \int e^{-2\pi i \xi\cdot z}\, d\mu(z) = \int_{\mathbb R} \psi(r) \int_{S^{1}} e^{-2\pi i (\xi_1rx_1 + \xi_2 rx_2 + \xi_3r) }\, d\sigma(x)\, dr$$ \begin{align*} |\hat\mu(\xi)| &= \left|\int_{1}^2 \psi(r) e^{-2\pi i\xi_3 r} \widehat{\sigma}(r(\xi_1,\xi_2))\, dr\right|\\ &\approx \left|\int_{1}^2 \psi(r) e^{-2\pi i\xi_3 r} J_{0}(2\pi r (\xi_1^2 + \xi_2^2)^{1/2})\, dr\right| \end{align*} since $$\widehat{\sigma^{n-1}}(x) = c(n)|x|^{(2-n)/2} J_{(n-2)/2}(2\pi|x|) \tag{\ast}$$ for all $$x\in \Bbb R^n$$ where $$J_m(r) = \frac{(r/2)^m}{\Gamma(m + 1/2) \sqrt\pi} \cdot\int_{-1}^1 e^{irt} (1-t^2)^{m-1/2} \, dt$$ for $$m > -1/2$$. In particular, $$J_{0}(r) = \frac{1}{\pi}\int_{-1}^1 e^{irt} (1-t^2)^{-1/2} \, dt$$ $$(\ast)$$ is Equation $$(3.41)$$ in Mattila's Fourier Analysis and Hausdorff Dimension. \begin{align*} |\hat\mu(\xi)|&\approx \left|\int_{1}^2 \psi(r) e^{-2\pi i\xi_3 r} \left(\int_{-1}^1 e^{2\pi ir (\xi_1^2 + \xi_2^2)^{1/2} t} (1-t^2)^{-1/2}\, dt \right)\, dr\right| \end{align*} How should I proceed?

Thanks a lot!

• This might be a difficult approach to take since you've integrated out the oscillatory integral in 2 variables but not the third one. So you have to weigh the $J_0$ factor against the $e^{-2\pi i r \xi_3}$ factor. If $|\xi_3|$ is small you can just plug in asymptotic estimates for $J_0$ and integrate, but if not, it may be difficult to balance the two factors. Jun 21, 2022 at 22:39

When the Hessian has rank $$k$$ everywhere one has an estimate $$|\hat{\mu}(\xi)| \leq C|\xi|^{-{k \over 2}}$$ by a result of Littman (see 5.8 on p 361 of Stein's Harmonic Analysis). Here $$k = 1$$ since the cone portion here has exactly one nonvanishing principal curvature at every point. The proof is basically the same as the case when the Hessian has full rank; you just show the nondegenerate estimates are uniform over $$k$$ dimensional "slices" of the surface on which one has a nondegenerate phase in $$k$$ variables, and integrate the resulting estimate. The nondegenerate case is Theorem 1 on p.348 of Stein's book.
This also follows from the decay rate estimate $$|\hat{\mu}(\xi)| \leq C|\xi|^{-{1 \over m}}$$ that holds when the surface is of type at most $$m$$ everywhere (see Theorem 2 on p. 351 of Stein's Harmonic Analysis.)