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!



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


