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 $\frac{1}{2}\dim_H(C_\theta)=\frac{1}{2}\frac{\log(1/2)}{\log(\theta)}$.

  • $\begingroup$ What is $\xi$? What is $\hat{\mu}_\theta$? $\endgroup$
    – Will Sawin
    May 30, 2012 at 18:53
  • $\begingroup$ @Will: $\hat{\mu}_\theta$ is the Fourier transform of $\mu_\theta$, reference added. $\endgroup$
    – Syang Chen
    May 30, 2012 at 21:20
  • $\begingroup$ there's a huge literature (with which I'm not faniliar), but you must know erdos, On a family of symmetric Bernoulli convolutions (1939), and the papers that refer to it. $\endgroup$
    – mike
    May 31, 2012 at 0:10
  • $\begingroup$ @mike, Yes I have read that paper. I think that is where the name of the above theorem comes from. $\endgroup$
    – Syang Chen
    May 31, 2012 at 3:19

4 Answers 4


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

  • $\begingroup$ Sorry, I couldn't quite follow the edit. For example, why is the cumulative correction at most $\theta+\theta^2+\cdots$? On the other hand, is there still hope to obtain the desired decay rate for $\theta$ not close to $0$? $\endgroup$
    – Syang Chen
    Jun 1, 2012 at 6:21
  • $\begingroup$ 1) When you correct $\nu_k$ by $1$, the number $\nu_m$ is corrected by $\theta^{m-k}$. 2) I don't think there is any hope to get what you want for any $\theta$ (if what you want is the best rate of decay that doesn't directly contradict the dimension bound): some resonances are always there; they are just harder to catch. I'll try to comment more on that when I have more time... $\endgroup$
    – fedja
    Jun 1, 2012 at 11:47

Edit in response to the editing of the second question:

The central Cantor sets are never Salem (for any $\theta$). Here is a way to see this. If $A\subset \mathbb{R}$ is a Salem set, then $$ \dim(A+B) \ge \min(\dim(A)+\dim(B),1) $$ for any Borel set $B\subset\mathbb{R}$, where $\dim$ is Hausdorff dimension and $A+B=\{a+b:a\in A, b\in B\}$ is the arithmetic sum. This follows from the expression of the $t$-energy $\int\int |x-y|^{-t}d\mu(x)\mu(y)$ in terms of the Fourier transform, and Frostman's Lemma (let me know if you want more details).

On the other hand, it is easy to see the above fails for the central Cantor sets $C_\theta$. If $\theta<1/3$, then $C_\theta+C_\theta$ is easily seen to be self-similar set with three pieces of relative size $\theta$, so $$ \dim(C_\theta+C_\theta) = \frac{\log 3}{|\log\theta|} < \frac{2\log 2}{|\log\theta|} = 2\dim(C_\theta). $$ If $\theta\ge 1/3$, you instead consider the sum $C_\theta+C_{\theta^k}$ where $k$ is small enough that $\dim(C_\theta)+\dim(C_{\theta^k})<1$. Again it is not hard to see that $$ \dim(C_\theta+C_{\theta^k}) < \dim(C_\theta)+\dim(C_{\theta^k}), $$ from the fact that there is an "exact overlap" in $C_\theta+C_{\theta^k}$.

This argument can be turned into an effective upper bound for the decay of the Fourier transform of an arbitrary measure supported on $C_\theta$ (using the specific structure of $\mu_\theta$ one can get better bounds). But one cannot expect sharp bounds from this approach.

Original answer:

Interesting question. Some remarks that were too long to fit into a comment:

  • If $\theta^{-1}$ is a Salem number, then there is no polynomial rate of decay for $\widehat{\mu}_\theta$. I don't have a reference at hand but believe this is not too hard.

  • For a given Borel probability measure $\mu$, the supremum of all $\beta$ such that $|\widehat{\mu}(\xi)|=O(|\xi|^{-\beta/2})$ is called the Fourier dimension of $\mu$. This is always bounded above by the Hausdorff dimension of the measure (the infimum of the Hausdorff dimensions of sets of full measure). This gives an upper bound for the possible $\alpha$ in your question, since the Hausdorff dimension of $\mu_\theta$ is well known to be $\log 2/|\log \theta|$.

  • A measure is a Salem measure if its Fourier dimension is equal to its Hausdorff dimension, or in other words if its Fourier transform decays as fast as its dimension allows. Many random measures are known to be Salem measures, but there are very few deterministic examples. For example, if you randomize the construction of $\mu_\theta$ by translating each interval by some small random number (with all the choices independent), you end up with a Salem measure, see the paper "Fourier asymptotics of statistically self-similar measures" by C. Bluhm. Of course the random translations would kill the resonances that occur along sequences of frequencies of the form $c \theta^{-n}$ so this doesn't give any hint on the original question.

  • Using the specific structure of $\mu_\theta$ it is possible to give better upper bounds for $\alpha$. For example, if $\theta<1/3$ then $\mu_\theta * \mu_\theta$ is easily seen to be supported on a Cantor set of dimension $<1$, in particular it is singular. This implies that $\alpha\le 1/4$, for otherwise the self-convolution would be in $L^2$. For $\theta\in (1/4,1/3)$ this is better than the general bound coming from Fourier dimension. It is possible to improve this bound in various ways (for example, using more precise information about the self-convolution), in particular $\mu_\theta$ is not a Salem measure for any $\theta$.

  • $\begingroup$ Are you familiar with the paper you mentioned above? I have a question about (3.3) in that paper. $\endgroup$
    – Syang Chen
    Jun 1, 2012 at 6:37
  • $\begingroup$ I haven't read it in detail but can have a look, what is the question? $\endgroup$ Jun 1, 2012 at 8:33
  • $\begingroup$ The paper proves that the randomized Cantor set (keeping the dissection number constant 2) is almost surely Salem. But I couldn't quite go through the proof. Maybe I was just confused... $\endgroup$
    – Syang Chen
    Jun 2, 2012 at 3:39
  • $\begingroup$ Nice. So generally the "neat" Cantor sets (with neat positions, ratios) can not be Salem? $\endgroup$
    – Syang Chen
    Jun 2, 2012 at 3:55
  • 1
    $\begingroup$ @JasonJones . Just read your comment. I don't have a reference at hand for (1), will let you know if I find one. (2) is not known. Although it is known that power Fourier decay is the typical behaviour, there are essentially no explicit parameters for which power Fourier decay is known to hold. $\endgroup$ Feb 11, 2020 at 21:10

I would start looking here:

MR1785620 (2001m:42020)
Peres, Yuval; Schlag, Wilhelm; Solomyak, Boris,
Sixty years of Bernoulli convolutions.
Fractal geometry and stochastics, II (Greifswald/Koserow, 1998), 39–65, Progr. Probab., 46, Birkhäuser, Basel, 2000.


There seems to be a negative answer to this in this paper:

MR1900879 (2003d:42011) Hu, Tian-You; Lau, Ka-Sing Fourier asymptotics of Cantor type measures at infinity. Proc. Amer. Math. Soc. 130 (2002), no. 9, 2711–2717.

  • $\begingroup$ @Igor, Thanks. But the paper gives only the decay rate in the (ball) average sense. $\endgroup$
    – Syang Chen
    May 31, 2012 at 4:02

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