# Calculating the Fourier dimension of a real interval $\left[a, b\right]$

(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 $$-\infty, what is the Fourier dimension of the interval $$\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?

• Any $f\in C_0^{\infty}$ will have $\widehat{f}=O(|x|^{-N})$ for any $N$. Mar 1, 2023 at 18:20

If $$-\infty\le a, then $$\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, $$\dim_F[-1,1]=\infty$$. $$\quad\Box$$

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 $$\dim_F$$.

• @ChristopheLeuridan : Thank you for your edits. Mar 2, 2023 at 17:02

Your definition of the Fourier dimension is wrong. In the sup, it should be "there exists a constant C such that the Fourier transform of $$\mu$$ is bounded by $$C |\xi|^{-s/2}$$".

• We may replace any $\mu$ with $C^{-1}\mu$, so the definition supplied is equivalent. May 30, 2023 at 0:21
• Ah yes thank you, I read the definition too fast. I'm more used to work with probability measures. May 30, 2023 at 8:15