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This question came up when my supervisors, my colleague, and I were considering arithmetic progressions in sets of fractional dimension. In particular, we were interested in "extracting" Salem sets from other sets, and we came across the following question:

We define the Fourier dimension of $E \subseteq \mathbb{R}$, $\mathrm{dim}_F(E)$, as the supremum of $\beta \in [0,1]$ such that for some probability measure $\mu$ supported on $E$,

$ |\widehat{\mu}(\xi)| \leq C|\xi|^{-\beta/2}. $

Suppose $E_1,E_2$ are two subsets of $\mathbb{R}$. What is the relationship, if any, between $\mathrm{dim}_F(E_1 + E_2)$ and $\mathrm{dim}_F(E_1)$, $\mathrm{dim}_F(E_2)$?

A quick calculation shows

$ \mathrm{dim}_F(E_1 + E_2) \geq \min(\mathrm{dim}_F(E_1) + \mathrm{dim}_F(E_2),1), $

but can we do any better than this? Or can strict inequality hold?

For some motivation, we may look to Hausdorff dimension instead (it is known that $\mathrm{dim}_F(E) \leq \mathrm{dim}_H(E)$; a set $E$ such that $\mathrm{dim}_F(E) = \mathrm{dim}_H(E)$ is called a Salem set). I believe Falconer gave an example of sets $E_1$ and $E_2$ such that $\mathrm{dim}_H(E_1) = 0 = \mathrm{dim}_H(E_2)$, yet $\mathrm{dim}_H(E_1+E_2) = 1$.

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It is possible that $\dim_F(E_1)=\dim_F(E_2)=0$ yet $\dim_F(E_1+E_2)=1$, so there is no inequality in the opposite direction.

In fact, Falconer's example of sets $E_1, E_2$ such that $\dim_H(E_1)=\dim_H(E_2)=0$ but $\dim_H(E_1+E_2)=1$ already works. In Falconer's example, not only $E_1+E_2$ has dimension $1$, but in fact $E_1+E_2$ is an interval. Hence $\dim_F(E_1)=\dim_F(E_2)=0$ (since $\dim_F$ is bounded above by $\dim_H$) but $\dim_F(E_1+E_2)=1$.

There are also examples which are not forced by the Hausdorff dimensions of the sets. Indeed, let $C$ be the ternary Cantor set. It is a classical result of Kahane and Salem that if $\mu$ is any measure supported on $C$, then $\widehat{\mu}(\xi)\nrightarrow 0$ as $|\xi|\to\infty$; in particular, $\dim_F(C)=0$. Clearly, the same is true for any dilate $t C$ with $t\neq 0$.

On the other hand, it is well known that $C+C$ equals the interval $[0,2]$. Indeed, since $C$ and also $tC$ have Newhouse thickness equal to $1$, Newhouse's gap lemma (the endpoint version), guarantees that $C+tC$ has nonempty interior for all $t\neq 0$, and therefore also Fourier dimension $1$ (note that I'm not claiming that the indicator function of $C+tC$ has Fourier decay, just that there is a measure supported on (an interval in) $C+tC$ that does).

Hence, for all $t$ we have $\dim_F(C)=\dim_F(tC)=0$ but $\dim_F(C+ tC)=1$. This shows that the opposite inequality fails in a rather dramatic function: it does not even hold typically in the sense of Marstrand's Theorem.

Morally speaking, there is no reason why $$\dim_F(E_1+E_2)=\min(1,\dim_F(E_1)+\dim_F(E_2))$$ should hold in general. Leaving Hausdorff dimension considerations aside, if $E_1$ or $E_2$ are not Salem, this tells us that there are some resonances in the construction of the sets at a set of frequencies (possibly very sparse). These special frequencies will in general be lost in the sum $E_1+E_2$ (unless $E_1$ and $E_2$ also resonate to each other in some strong form), so one would expect that $\dim_F(E_1+E_2) > \dim_F(E_1)+\dim_F(E_2)$. However, I suspect it is not trivial at all to give specific examples where $\dim_H(E_1+E_2)<1$ (because proving Salemness or even some good decay of Fourier coefficients is usually hard).

On the other hand, equality $\dim_F(E_1+E_2)=\min(1,\dim_F(E_1)+\dim_F(E_2))$ can certainly hold. Indeed, it is easy to see this is always the case when $E_1$ and $E_2$ are Salem and additionally one of them has coinciding Hausdorff and upper box-counting dimension (which is the case for all known constructions of Salem sets). Indeed, denoting upper box-counting dimension by $\dim_B$, it is well known that $\dim_H(A+B)\le \dim_H(A)+\dim_B(B)$, so in this case $\dim_H(E_1+E_2)\le \dim_H(E_1)+\dim_H(E_2)$, and it follows from salem-ness and $\dim_F\le \dim_H$ that $$ \dim_F(E_1+E_2)\le \dim_F(E_1)+\dim_F(E_2). $$

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  • $\begingroup$ You wrote Marstrand's theorem implies C+tC has positive Lebesgue measure (for almost every $t$), hence C+tC has Fourier dimension 1 (for almost every $t$). I don't think positive Lebesgue measure alone implies full Fourier dimension. See math.stackexchange.com/questions/149660/… $\endgroup$ – MichaelGaudreau Sep 3 '18 at 15:50
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    $\begingroup$ @MichaelGaudreau, you're right, it's unclear to me now whether a set of positive Lebesgue measure always has Fourier dimension 1. The fact that the indicator function of the set has no fast decay is not enough to conclude the opposite, though. In this case, however, $C+tC$ has nonempty interior by Newhouse's gap Lemma so the Fourier dimension is $1$. I will edit my answer accordingly. $\endgroup$ – Pablo Shmerkin Sep 8 '18 at 12:47
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Regarding the Hausdorff dimension of sumsets, you might want to have a look at an important paper by Peres and Shmerkin:

http://arxiv.org/abs/0705.2628

There are plenty of useful references there, too.

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