Why does $|\hat{\mu}(x)|\leq|x|^{-s/2} \implies \int_{\mathbb{R}^n}|\hat{\mu}(x)|^2|x|^{s-n}dx<\infty$ for the Fourier transform of a measure $\mu$

Question

(See edit) I have seen the claim $|\hat{\mu}(x)|\leq|x|^{-s/2} \implies \int_{\mathbb{R}^n}|\hat{\mu}(x)|^2|x|^{s-n}dx<\infty$ for the Fourier transform $\hat{\mu}(\xi)\equiv \int_{\mathbb{R}^n}e^{-2\pi i x\cdot \xi}d\mu(x)$ of a finite Borel measure $\mu$ on $\mathbb{R}^n$ in quite a few places without a proof and I am honestly quite confused by it. Namely, what if the origin belongs to the support of $\hat{\mu}$? In that case there is no way that the integral $\int_{\mathbb{R}^n}|\hat{\mu}(x)|^2|x|^{s-n}dx$ can be finite, at least with the crude approximation

$$\int_{\mathbb{R}^n}|\hat{\mu}(x)|^2|x|^{s-n}dx \leq \int_{\mathbb{R}^n}|x|^{-n}dx$$

Is there some good-to-know trick that one can use with the assumed inequality to conclude that $\int_{\mathbb{R}^n}|\hat{\mu}(x)|^2|x|^{s-n}dx < \infty$? Thanks!

Edit: My current understanding is that $s$ can be negative. Moreover $\mu$ is assumed to have a compact support, although I am not sure how that is helpful in this case.

Answer 1

I don't think the implication in the OP holds, although I don't have a specific counter-example at the moment. In a related posting it is also proven that if $\int_{\mathbb{R}^n}|\widehat{\mu}(x)|^2|x|^{s-n}\,dx<\infty$, then $$\lim_{R\to\infty} R^{-n} \mathcal{L}^n(\{x\in B(0,R): |\hat\mu(x)| > |x|^{-s/2}\}) = 0$$ where $\mathcal{L}^n$ is Lebesgue's measure on $\mathbb{R}^n$. This is the closed to the OP condition in one direction.

Now, as the type of integrals in the OP are related to Hausdorff dimension, here are some observations for the OP. For simplicity, we restrict ourselves to nonnegative finite measures $\mu$ on $\mathscr{B}(\mathbb{R}^n)$ (if $\mu$ is a complex Borel meaure, then $|\mu|$, the variation measure, is nonnegative and finite).

Definition: For any $s\geq0$, the Riesz $s$-energy of $\mu$ is defined as \begin{align} I_s(\mu)=\int_{\mathbb{R}^n\times\mathbb{R}^n}\frac{1}{|x-y|^s}\,(\mu\otimes\mu)(dx,dy)=\int_{\mathbb{R}^n}\Big(\int_{\mathbb{R}^n}\frac{\mu(dy)}{|x-y|^s}\Big)\,\mu(dx)\tag{0}\label{s-energy} \end{align}

Formally, the $s$-energy integral in \eqref{s-energy} can be seen as the integral of the convolution of $k_s(y)=|y|^{-s}$ with $\mu(dy)$. Considering $k_s$ as a tempered distribution, its Fourier transform is given by $$\widehat{k_s}(\xi)=\gamma_{n,s}k_{n-s}(\xi)$$ for some constant $\gamma_{n,s}$ depending only on $n$ and $s$. Approximating $\mu$ by absolutely continuous measures with smooth Radon-Nikodym derivatives, it can be shown that

Theorem 1: For any $0<s<n$ \begin{align}\mathcal{I}_s(\mu)=\gamma_{n,s}\int_{\mathbb{R}^n}|\widehat{\mu}(\xi)|^2|\xi|^{s-n}\,d\xi\tag{1}\label{R-F} \end{align}

The following result makes the connection between the $s$-energy integrals and the dimension of sets.

Theorem 2: Let $\mathcal{H}^s$ be the $s$-dimensional Hausdorff measure on $\mathscr{B}(\mathbb{R}^n)$. For $F\in\mathscr{B}(\mathbb{R}^n)$, $\operatorname{dim}_H(F)$ denotes its Hausdorff dimension.

1. If there is a probability measure $\mu$ such that $\mu(F^c)=0$ such that $I_s(\mu)<\infty$, then $\mathcal{H}^s(F)=\infty$ and $\operatorname{dim}_{H}(F)\geq s$,
2. If $\mathcal{H}^s(F)>0$, then there exists a probability measure $\mu$ with $\mu(F^c)=0$ such that $I_u(\mu)<\infty$ for all $0<u<s$.

Now, if the condition \begin{align} |\widehat{\mu}(\xi)|\leq C |\xi|^{-s/2}\tag{2}\label{s-hausdorff} \end{align} holds for some $0<s<n$, then from \eqref{R-F} and the fact that $\mu(\mathbb{R}^n)<\infty$, we obtain $$I_t(\mu)\leq C_1\int_{|x|\leq1}|x|^{t-n}\,dx + C_2\int_{|x|>1}|x|^{t-n}|x|^{-s}\,dx<\infty$$ for all $0<t<s$. Thus, if $\mu(F^c)=0$, then $\operatorname{dim}_H(F)\geq t$.

The largest $s$ for which there is a probability measure $\mu$ with $\mu(F^c)=0$ that satisfies \eqref{s-hausdorff} is called the Hausdorff dimension of $F$.

In a related posting it is also proven that $$\lim_{R\to\infty} R^{-n} \mathcal{H}^n(\{x\in B(0,R): |\hat\mu(x)| > |x|^{-s/2}\}) = 0$$ This is the closed to the OP condition in one direction.

The book Matilla, P., Fourier Analysis and Hausdorff Dimension, Cambridge University Press, 2015 has detail explanation of the conceps and results mentioned in this posting.