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](http://en.wikipedia.org/wiki/Fourier_transform#Fourier.E2.80.93Stieltjes_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](http://en.wikipedia.org/wiki/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 $\frac{1}{2}\dim\_H(C\_\theta)=\frac{1}{2}\frac{\log(1/2)}{\log(\theta)}$.



Fourier decay rate of Cantor measures
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.



> 
> So the rate of decay I was expecting is (or arbitrarily close to) half of the Hausdorff dimension $\frac12 dim\_H(C\_\theta)=\frac 12\frac{\log(1/2)}{\log(\theta)}$.
> 
> 
> 


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 $\frac{\theta^2}{\log(1/\theta)}$


