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Suppose that the measure $\mu$ on $\mathbb{R}^n$ has Fourier dimension $\beta$, which is to say that \begin{equation*} \beta= \sup\left\{\gamma \leq n : |\hat{\mu}(x)| \leq C(1+|x|)^{-\gamma/2}\right\}. \end{equation*} The Fourier dimension is always majorized by the Hausdorff dimension (for sets, this is a consequence of Frostman's lemma) and various results in Geometric Measure Theory reveal that Fourier dimension is generally a much stronger and more stable property (for instance, when intersecting two sets, the Hausdorff "codimensions" need not add, but will if one of the sets has positive Fourier dimension, and the same holds in the context of dimensions of measures).

Because Fourier dimensions are generally much better behaved than Hausdorff dimensions, and it is immediate that the Hausdorff dimension does not drop when restricting to an open set, it is natural to ask whether this continues to hold true of Fourier dimension.

Is it true that for any measure $\mu$ of Fourier dimension $\beta$, and open set $U$, the Fourier dimension of $\mu_U$, the measure defined by \begin{equation*} \int g d\mu_U = \int_U g d\mu, \end{equation*} is also at least $\beta$?

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Take the sum of two measures with different Fourier dimensions and disjoint supports. Then pick $U$ so that it contains only the measure with the smaller dimension. Am I missing something? – user31373 Aug 12 '12 at 5:23
In this case the Fourier dimension of the restricted measure increases, because Fourier dimension is not additive. To see this, take $\mu=\mu_1+\mu_2$ with $|\widehat{\mu_i}(x)|\leq C_i(1+|x|)^{-\frac{\beta_i}{2}}$ and $\beta_1<\beta_2$. Then we have that $|\widehat{\mu}(x)|=|\widehat{\mu_1}(x)+\widehat{\mu_2}(x)|\leq \sum_{i=1}^2C_i(1+|x|)^{-\frac{\beta_i}{2}}$ which certainly does not decay at the rate of $\beta_1$ for $\beta_1<\beta_2$, but as $C'(1+|x|)^{-\frac{\beta_2}{2}}$. For $supp(\mu_i)\subset O_i$ disjoint open sets as you suggested, restricting will get us a decay of $\beta_i$. – question Aug 12 '12 at 21:19
I don't follow the conclusion of your comment. If $\beta_1<\beta_2$, then the term with exponent $-\beta_1/2$ is the bigger one, so it determines the behavior of the sum. – user31373 Aug 12 '12 at 21:22
If $\beta_1<\beta_2$, then yes it does. The sum has dimension $\beta_1$. And when we restrict, we either get $\mu_1$ which has dimension $\beta_1$, or we get $\mu_2$ which has dimension $\beta_2$. In either case, the dimension has NOT shrunk - it was $\beta_1$, and it remains at least $\beta_1$. – question Aug 12 '12 at 22:16
I see: the dimension of disjoint union is the smaller of the dimensions of pieces. I would not say that this dimension is "better behaved than Hausdorff", but that's subjective. (By the way, I am puzzled by the statement that Hausdorff dimension "does not drop": the disjoint union of a disk and a line segment has Hausdorff dimension 2, but restricting to the segment drops it to 1.) – user31373 Aug 12 '12 at 22:26

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