Does Ahlfors–David regularity of a measure imply its Fourier asymptotic behavior?

Question

Let $\mu$ be a Borel probability measure on $R^d$. If $\mu$ satisfies $\mu(B(x,r))\le Cr^\alpha$ for any $x\in R^d$ and $r>0$, then Strichartz (Fourier asymptotics of fractal measures, J. Funct. Anal. 1990) proved that $$\limsup\limits_{R\to \infty}\frac{1}{R^{d-\alpha}}\int_{|x|\le R}|\widehat{\mu}(x)|^2dx\le C_2.$$ Here $\widehat{\mu}(x)=\int e^{-2\pi itx}d\mu(t)$ is the Fourier transformation of $\mu.$

My question is: if $\mu$ is Ahlfors–David regular, i.e., $C^{-1}r^\alpha\le\mu(B(x,r))\le Cr^\alpha$ for any $x\in R^d$ and $r>0$, can we have $$C_1\le \liminf\limits_{R\to \infty}\frac{1}{R^{d-\alpha}}\int_{|x|\le R}|\widehat{\mu}(x)|^2dx\le\limsup\limits_{R\to \infty}\frac{1}{R^{d-\alpha}}\int_{|x|\le R}|\widehat{\mu}(x)|^2dx\le C_2?$$

Here $C, C_1, C_2$ are all constants.

Does someone know related results or can prove it? Thanks!

Answer 1

(There are some details missing in this answer. Time permitting, I will try to expand it.)

It suffices to find a non-negative continuous function $\phi$ with $\phi(0) > 0$ and Fourier transform bounded and supported in $B(0, 1)$. In dimension one, $\phi(x) = (1 - \cos x)/(\pi x^2)$, $\hat\phi(z) = (1 - |z|)_+$ is a good example.

Then $\phi_R(x) = R^d \phi(R x)$ has Fourier transform $\hat\phi(R^{-1} z)$ supported in $B(0, R)$, and we have $$ \int_{|z| \le R} |\hat\mu(z)|^2 dz \ge \frac{1}{\|\hat\phi\|_\infty} \int_{|z| \le R} |\hat\phi_R(z) \hat\mu(z)|^2 dz = \frac{2\pi}{\|\hat\phi\|_\infty} \int_{\mathbb{R}^d} |\mu * \phi_R(x)|^2 dx $$ by Plancherel's theorem. Now, $\phi(x) \ge C_1 \mathbb{1}_{B(0, r)}(x)$ for some $C_1, r > 0$, and hence $$ \int_{|z| \le R} |\hat\mu(z)|^2 dz \ge \frac{2 \pi C_1 R^{2d}}{\|\hat\phi\|_\infty} \int_{\mathbb{R}^d} |\mu * \mathbb{1}_{B(0, r/R)}(x)|^2 dx . $$ Ahlfors regularity implies that $\mu * \mathbb{1}_{B(0, r/R)}(x) \ge C_2 (r/R)^\alpha$ on a set of measure at least $C_3 (r/R)^{d-\alpha}$ (see Edit below). We conclude that $$ \int_{|z| \le R} |\hat\mu(z)|^2 dz \ge \frac{C_1 R^{2d}}{\|\hat\phi\|_\infty} \, (C_2 (r/R)^{\alpha})^2 C_3 (r / R)^{d - \alpha} = C_4 R^{d - \alpha} , $$ as desired.

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Edit: If $r > 0$, then $$ \int_{\mathbb{R}^d} \mu * \mathbb{1}_{B(0, r)}(x) dx = C_1 r^d \mu(\mathbb R^d) $$ (by Fubini), and $$ \mu * \mathbb{1}_{B(0, r)}(x) = \mu(B(x, r)) \leqslant C_2 r^\alpha $$ for every $x$ (by Ahlfors regularity). (Well, we assume this for $x$ in the support of $\mu$ only, but extension to general $x$ is standard.) It follows that the Lebesgue measure of $$ \{ x : \mu(B(x, r)) > 0 \} $$ is at least $C_3 r^{d - \alpha}$. However, if $\mu(B(x, r)) > 0$, then $\mu(B(x, 2r)) \geqslant C_4 r^\alpha$ (again by Ahlfors regularity). We conclude that $$ \mu * \mathbb{1}_{B(0, 2 r)}(x) \geqslant C_4 r^\alpha \quad \text{on a set of Lebesgue measure at least } C_3 r^{d - \alpha} . $$ This is of course equivalent to the property used in the answer above.