The Fourier-dimension conjecture for cones and cylinders generated by arbitrary sets

MathDB #361368: a compact counterexample to the literal conjecture

Result

The literal statement of Conjecture 8.1 in arXiv:2401.01455v1 is false. In fact, it already fails when the generating set is a compact convex body.

Fix any integer $d\ge 1$ and take

$$A=\overline B_d=\{x\in\mathbb R^d:\lVert x\rVert\le 1\}.$$

Then

$$\dim_F(A)=d, \qquad \dim_F(C_A)=\dim_F(D_A)=d+1.$$

Thus the conjectured equality with $\dim_F(A)$ fails for both the cone and the cylinder.

Fourier dimension of a set with interior

We use the source convention that, for a Borel set $E\subset\mathbb R^N$,

$$\dim_F(E)=\sup\left\{s\in[0,N]: \begin{array}{l} \text{there is a nonzero finite measure $\mu$ supported on $E$}\\ \text{such that }\sup_{\xi\in\mathbb R^N} |\xi|^{s/2}|\widehat\mu(\xi)|<\infty \end{array}\right\}.$$

The counterexample uses probability measures, so it is unaffected by any equivalent normalization of the admissible nonzero measures.

Lemma. If a Borel set $E\subset\mathbb R^N$ has nonempty interior, then $\dim_F(E)=N$.

Proof. Choose an open ball $U$ whose closure is contained in $\operatorname{Int}(E)$, and choose $\varphi\in C_c^\infty(U)$ with $\varphi\ge0$ and $\int\varphi=1$. The probability measure

$$d\mu(x)=\varphi(x)\,dx$$

is supported on $E$. Its Fourier transform is a Schwartz function. Consequently

$$\sup_{\xi\in\mathbb R^N} |\xi|^{N/2}|\widehat\mu(\xi)|<\infty:$$

near the origin this follows from boundedness of $\widehat\mu$, and for $|\xi|\ge1$ it follows from Schwartz decay. Hence $N$ is admissible in the definition of $\dim_F(E)$. The reverse inequality is built into the ambient bound $s\in[0,N]$. Therefore $\dim_F(E)=N$. $\square$

Since $A=\overline B_d$ has interior $B_d$, the lemma gives

$$\dim_F(A)=d. \tag{1}$$

The cylinder

The cylinder generated by $A$ is

$$D_A=A\times\mathbb R =\{(y,h)\in\mathbb R^d\times\mathbb R:\lVert y\rVert\le1\}.$$

Its interior in $\mathbb R^{d+1}$ is

$$\operatorname{Int}(D_A)=B_d\times\mathbb R,$$

which is nonempty. Applying the lemma in ambient dimension $d+1$ gives

$$\dim_F(D_A)=d+1. \tag{2}$$

The cone

Write a point of $\mathbb R^{d+1}$ as $(y,h)$, with $y\in\mathbb R^d$. By definition,

$$C_A=\{(hx,h):x\in A,\ h\in\mathbb R\}.$$

This has the exact description

$$C_A=\{(y,h)\in\mathbb R^d\times\mathbb R: \lVert y\rVert\le |h|\}. \tag{3}$$

Indeed, $(y,h)=(hx,h)$ with $\lVert x\rVert\le1$ implies $\lVert y\rVert\le|h|$. Conversely, if $h\ne0$ and $\lVert y\rVert\le|h|$, then $x=y/h$ belongs to $A$; if $h=0$, the inequality forces $y=0$, which is obtained by taking $x=0$.

It follows from (3) that $C_A$ is a closed double cone and

$$\operatorname{Int}(C_A) =\{(y,h):\lVert y\rVert<|h|\}.$$

The strict-inequality set is open and nonempty. Points satisfying equality in (3) are boundary points, since arbitrarily small perturbations can make $\lVert y\rVert-|h|$ either positive or negative. The lemma therefore gives

$$\dim_F(C_A)=d+1. \tag{4}$$

Combining (1)--(4),

$$\dim_F(C_A)=\dim_F(D_A)=d+1>d=\dim_F(A),$$

which refutes the literal conjecture.

Scope

The generating set $A$ is compact, closed, Borel, and convex, and all measures used above are compactly supported probability measures. Hence the counterexample does not depend on support or measurability conventions, and adding boundedness or compactness to the hypothesis on $A$ does not repair the statement.

The example uses the explicitly allowed endpoint $s=d$. It does not settle any modified conjecture imposing $s<d$, empty interior, or another thinness hypothesis.