According to Mattila, Geometry of sets and measures in Euclidean spaces, p. 168, the Fourier dimension $\text{dim}_F(A)$ of $A\subseteq \mathbb R^n$ is the unique number in $[0,n]$ such that for any $0<s<\text{dim}_F A$ there exists a non-zero Radon measure $\mu$ with spt $\mu\subset A$ and $|\hat\mu(x)|\le |x|^{-s/2}$ for $x\in\mathbb R^n$, and that for $\text{dim}_F(A)<s\le n$ no such measure exists.

Does this condition on $\mu$ for $\text{dim}_F(A)>0$ imply a nontrivial lower bound on the $\mu$-measure of a ball of radius $r$, $\mu(B(x,r))$ in terms of $r$, for $\mu$-almost all $x$? For instance, a lower bound like $r^s$ for a constant $s$?

(The reason for asking being that positive Hausdorff dimension is enough to get such an upper bound and I'm wondering if one can get lower bounds by strengthening the assumption "positive Hausdorff dimension" to "positive Fourier dimension" or something similar.)


Fourier dimension doesn't directly say anything about lower bounds for the mass of balls. Fourier dimension is smaller or equal than Hausdorff dimension, and in order to give a bound of the form $\mu(B(x,r))\ge r^s$, the exponent $s$ needs to be large.

This suggests that one should seek a notion of dimension that gives larger values than Hausdorff dimension. In fact, what you need here is exactly packing dimension.

If $\mu$ is a finite Borel measure on $\mathbb{R}^n$, let us define its (upper) packing dimension as $$ \dim_P\mu = \inf\{ \dim_P A: \mu(\mathbb{R}^n\setminus A)=0\}, $$ where $\dim_P(A)$ is the packing dimension of $A$: for the definition, see for example Chapter 2 in

Falconer, Kenneth. Techniques in fractal geometry. John Wiley & Sons, Ltd., Chichester, 1997

There is an alternative characterization of packing dimensions of measures in terms of mass decay (see Proposition 10.3 in Falconer's book): $$ \dim_P\mu = \inf \left\{s: x:\limsup_{r\to 0} \frac{\log\mu(B(x,r))}{\log r}\le s \text{ for $\mu$-a.e. } x \right\} $$

In particular, if $s>\dim_P\mu$, then for $\mu$ almost all $x$ it holds that $\mu(B(x,r)) \ge r^s$ for all small enough $r$ (depending on $x$).

Hausdorff and packing dimensions (of sets and measures) coincide often for "natural" fractals, but not always.

Note, finally, that if $s>n$, then for any Radon measure $\mu$ on $\mathbb{R}^n$ one has $\mu(B(x,r)) \ge r^s$ for all sufficiently small $r$ (depending on $x$). This follows from the above comments (since packing dimension of a measure can't exceed $n$: take $A=\mathbb{R}^n$ in the definition), but it can also be proved directly with the help of standard covering lemmas.

  • $\begingroup$ I guess what I really want for the Fourier dimension question is more like $\mu(B(x,r))\ge r^s$ for all $x$, not just $\mu$-almost every $x$, but then that doesn't exactly make sense either... $\endgroup$ Nov 10 '15 at 23:34
  • $\begingroup$ You can ask for $\mu(B(x,r))\ge r^s$ for all $x$ in the support of $\mu$, there are certainly many measures satisfying this for some $s$, but I still don't see any relation to Fourier dimension. $\endgroup$ Nov 11 '15 at 0:29
  • $\begingroup$ It is known that $\limsup \mu(B_r)/r^s$ is relatively well behaved and one has analogs of the Lebesgue differentiation theorem, but I thought the OP was asking the much more difficult question about the behavior of $\liminf\ldots$. $\endgroup$ Nov 11 '15 at 1:13

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