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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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up vote 4 down vote accepted

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 $t\neq 0$. Now since $C\times C$ has dimension $2\log 2/\log 3>1$, Marstrand's theorem on projections tells us that $C+ tC$ has positive Lebesgue measure for almost every $t$. Hence, for almost every $t$ we have $\dim_F(C)=\dim_F(tC)=0$ but $\dim_F(C+ tC)=1$.

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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Regarding the Hausdorff dimension of sumsets, you might want to have a look at an important paper by Peres and Shmerkin:

There are plenty of useful references there, too.

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