(**Preliminaries:**) 1.) Let $S\subset\mathbb{R}^n$ and define $\mathcal{M}(S) = \{\text{$\mu$ a Borel measure}: \text{$0 < \mu(S) < \infty$ and $\mathrm{support}(\mu)\subset S$}\}$.


2.) Define the Fourier transform of a measure as $\hat{\mu}(x):= \int\_{\mathbb{R}^n}e^{-2\pi i\xi \cdot x}d\mu(\xi)$.


3.) Define the Fourier dimension of a set $S\subset\mathbb{R}^n$ as


$$\mathrm{dim}\_FS := \sup\{s\in \left[0,n\right]: \exists \mu\in \mathcal{M}(S):\forall x\in\mathbb{R}^n:\left|\hat{\mu}(x)\right|\leq \left|x\right|^{-s/2}\}$$


(**Remark:**) I was a bit hesitant to post this question here since there is a non-zero chance for it being completely trivial, but as I (honest to God) could not find any discussion on this in books such as Mattila's *Fourier Analysis and Hausdorff Dimension* or *Geometry of Sets and Measures in Euclidean Spaces* or in exercise sections of such relevant books, here I am.


(**Question:**) Given $-\infty<a<b<\infty$, what is the Fourier dimension of the interval $\left[a, b\right]\subset\mathbb{R}$ and what measure $\mu\in \mathcal{M}\left(\left[a, b\right]\right)$ gives it?


My naïve first thought was to just use the one dimensional Lebesgue measure restricted to $\left[a, b\right]$. However, then for $x\neq 0$ we get


$$|\hat{\mu}(x)|^2 = \left|\frac{i}{2\pi x}\left(e^{-2\pi i xb} - e^{-2\pi ixa}\right)\right|^2 = \frac{1 - \cos\left(2\pi x(b - a)\right)}{2\pi^2x^2}$$


whence $|\hat{\mu}(x)|^2 \leq \left|x\right|^{-s}\Longleftrightarrow 1 - \cos\left(2\pi x(b - a)\right)\leq 2\pi^2\left|x\right|^{2-s}, s\in \left[0,1\right]$


By taking $e.g. b = 4, a = -2, x = 0.05$ we see that if $\left[a, b\right]$ were to have a Fourier dimension equal to one, then $\mu$ will not give it as the LHS is equal to $\approx 1.309$ while the RHS is equal to $\approx 0.987$.


Any ideas how the measure $\mu$ should be constructed? Also, do you happen to know a good source which works through a bit more elementary examples (like this) of the Fourier dimension?



Calculating the Fourier dimension of a real interval $\left[a, b\right]$
If $-\infty\le a<b\le\infty$, then $\dim\_F[a,b]=\infty$.


Indeed, without loss of generality $[a,b]=[-1,1]$. Take any natural $n$ and let
$$\nu\_n:=\frac1{c\_n}\,\mu\_{1/2}\*\cdots\*\mu\_{1/2^n},$$
where $\mu\_L$ is the uniform distribution on $[-L,L]$ and $c\_n:=\prod\_{k=1}^n 2^k$. Then for all real $t\ne0$
$$\hat\nu\_n(t):=\int\_{-\infty}^\infty e^{itx}\nu\_n(dx)
=\frac1{c\_n}\,\prod\_{k=1}^n\hat\mu\_{1/2^k}(t)
=\frac1{c\_n}\,\prod\_{k=1}^n\frac{\sin(t/2^k)}{t/2^k}$$
and hence
$$|\hat\nu\_n(t)|\le\frac1{c\_n}\,\prod\_{k=1}^n\frac1{|t|/2^k}=|t|^{-n},$$
for all natural $n$. Also, $\nu\_n$ is a nonzero measure with support contained in $[-1,1]$. So, $\dim\_F[-1,1]=\infty$. $\quad\Box$




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This assumes the more natural definition
$$\mathrm{dim}\_FS := \sup\{s\in \left[0,\infty\right): \exists \mu\in \mathcal{M}(S)\ \forall x\in\mathbb{R}^d\setminus\{0\}\ \left|\hat{\mu}(x)\right|\leq \left|x\right|^{-s/2}\}$$
instead of the strange definition
$$\mathrm{dim}\_FS := \sup\{s\in \left[0,d\right]: \exists \mu\in \mathcal{M}(S):\forall x\in\mathbb{R}^d:\left|\hat{\mu}(x)\right|\leq \left|x\right|^{-s/2}\}$$
in the OP.
Using the definition in the OP (with $\mathbb{R}^d$ corrected to $\mathbb{R}^d\setminus\{0\}$), we would of course get $\dim\_F[-1,1]=1$.


I also used a slightly different definition of the Fourier transform, without the extra factor $-2\pi$ in the exponent. Of course, this rescaling does not affect $\dim\_F$.


