Average size of the Fourier--Stieltjes transform of the fractal measures

Question

For $0<\theta<1/2$ define $\mu_\theta$ to be the uniform (self-similar) measure on the Cantor set obtained from the dissection pattern $(1-2\theta,\theta)$. For example, when $\theta=1/3$ the $\mu_\theta$ is the uniform Cantor measure on the classical middle-third Cantor set. It was obtained by Hille--Tamarkin that the Fourier--Stieltjes transform of $\mu_\theta$ is given by $$\widehat{\mu}_\theta(\xi) = \prod_{k=1}^\infty \cos(\theta^k\xi).$$ It is usually a very hard question to find the actual decay rate of $\widehat{\mu}_\theta(\xi)$ as $\xi\to\infty$. It is a classical theorem by Wiener--Wintner that $$\frac{1}{T}\int_{-T}^T|\widehat{\mu}_\theta(\xi)|^2\,\mathrm{d}\xi = O\left(T^{-s}\right)$$ where $s=-\frac{\log 2}{\log\theta}$ is the Hausdorff dimension of the corresponding Fractal.

Is the last estimate essentially sharp for all $\theta$? By essentially sharp I mean a power-saving improvement is not possible.

Is there some theta where any improvement is possible?

Answer 1

No, we cannot have $\int_{-T}^T |\widehat{\mu}|^2 \lesssim T^{1-s-\epsilon}$. This would imply that $$ I_{s+\epsilon/2}(\mu) = \int d\mu(x)\int d\mu(y) |x-y|^{-s-\epsilon/2} = c \int |t|^{s+\epsilon/2-1}|\widehat{\mu}(t)|^2 \, dt < \infty ; $$ to see that this last integral is finite, cut it into dyadic pieces $|t|\simeq 2^n$. This in turn gives the corresponding Cantor set $C$ positive $s+\epsilon/2$ Riesz capacity. This is impossible for a set of Hausdorff dimension $< s+\epsilon/2$.

(The Riesz capacities behave just like the Hausdorff measures, they switch from being positive to zero at a number that one can call the capacitary dimension, and this equals the Hausdorff dimension. A good source for all this is Mattila, Geometry of sets and measures in Euclidean space; see Theorem 8.9 there for this last statement.)