For $0<\theta<\frac{1}{2}$, denote by $C_\theta$ the Cantor set with dissection ratio $\theta$, i.e. the Cantor set obtained from dissection parttern $(\theta, 1-2\theta,\theta)$. It is known that $C_\theta$ carries a uniform measure $\mu_\theta$ which is usually called Cantor measure. And it is not hard to show that the Fourier–Stieltjes transform of $\mu_\theta$ is (up to scaling and constant multiple)
$$\hat{\mu}_\theta(\xi)=\prod^{\infty}_{k=1} \cos(\theta^k\xi)$$
But unlike integrable function we do not have Riemann-Lebesgue lemma for these Cantor measures. In fact a theorem of Erdős and Salem says $\hat{\mu}_\theta (\xi)=o(1)$ as $|\xi|\rightarrow\infty$ if and only if $\theta^{-1}$ is a not PV number. On the other hand, it is known that for some $\theta^{-1}$ not a PV number, $\hat{\mu}_\theta (\xi)$ does not decay at any positive rate, even though $\hat{\mu}_\theta (\xi)=o(1)$.
My question is whether there exists $\theta$ such that $\hat{\mu}_\theta (\xi)=O(|\xi|^{-\alpha})$ for some $\alpha>0$? How much is known about the precise decay rate of $\hat{\mu}_\theta (\xi)$?
Thanks in advance.
Edit: To make the second question precise, I was actually wondering if there exists Salem set (as pointed out by Pablo) among these Cantor sets. So the rate of decay I was expecting is (or arbitrarily close to) half of the Hausdorff dimension $\frac{1}{2}\dim_H(C_\theta)=\frac{1}{2}\frac{\log(1/2)}{\log(\theta)}$.
Best Answer
The first part of the question is easy: the power decay is typical. Let's look, say, at $\theta\in[1/3,1/2)$. Take any $\xi>0$ and consider the sequence $\nu_k=\pi^{-1}\xi\theta^k$ up to the moment it goes below $1$. Let $m$ be the number of terms in this sequence. Let $n_k$ be the nearest integer to $\nu_k$. The power decay will be ensured if we show that a noticeable portion of the differences $|\nu_k-n_k|$ are not too small for all sufficiently large $m$.
Fix $\alpha,\delta>0$ and assume that we have at most $\alpha m$ differences $|\nu_k-n_k|$ exceeding $\delta$. Let's trace the sequence $n_k$ backwards. Assume that $n_k$, $n_{k+1}$, and $n_{k+2}$ approximate the corresponding $\nu's$ with error less than $\delta$. Then we have $n_k\approx n_{k+1}^2n_{k+2}^{-1}$ with the relative error about $\delta(2n_{k+1}^{-1}+n_{k+2}^{-1})<\frac 13 n_k^{-1}$ if $\delta$ is chosen small enough. But then $n_k$ is completely determined by $n_{k+1}$ and $n_{k+2}$. The same argument shows that even if we only know that the approximation error is at most $\frac 12$, we can have just some fixed number $A$ of choices for $n_k$.
Now, let's count the "bad" sequences of $m$ terms. We have ${m\choose \alpha m}$ ways to select the "bad approximation positions". Then we have some constant number of starting pairs $n_{m-1}, n_m$. When going backwards, we have only $3\alpha m$ places where we have any freedom. Thus, we have just $C{m\choose \alpha m}A^{3\alpha m}\le Ce^{q(\alpha)m}$ bad sequences where $q(\alpha)\to 0$ as $\alpha\to 0$. On the other hand, $n_0$ and $n_1$ determine $\theta$ with an error of order $n_1^{-1}\le 2^{-m}$. Thus, the measure of $\theta$ that are bad for some particular $m$ is exponentially small in $m$ if $\alpha$ is small enough. The rest should be clear.
The second part of the question is harder to answer. I have no doubt that people have figured out most of what would be worth figuring out here but I doubt very much they bothered to publish any of it or, if they did, the papers made it past the referees. If you need something particular, state a precise question and we'll see if we can figure it out.
Edit in response to the edit of the question.
Now, that is far too optimistic. Take large $\lambda=\theta^{-1}$. Take any integer. Multiply by $\lambda$. Correct the product to the nearest integer. Multiply by $\lambda$. Correct the product to the nearest integer, and so on. After $m$ steps, you'll get $\xi\approx\lambda^m$ while the cumulative effect of all corrections in each position is at most $\theta+\theta^2+\theta^3+\dots\approx\theta$, so all the cosines are about $1-\theta^2$ and you get $\alpha$ not much better than $\frac{\theta^2}{\log(1/\theta)}$