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!

  • 1
    $\begingroup$ 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. $\endgroup$
    – Zarrax
    Jun 21, 2022 at 22:39

1 Answer 1


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.)


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