After my previous post on a Fourier-analysis approach to Furstenberg’s problem, I would like to present another application of the Fourier transform of measures: a simple proof of Marstrand theorem about projections of Cantor sets. Before entering the details, let me tell you what are my motivations on this topic: besides the beauty of the Fourier-analysis approach to a relevant geometrical problem, this particular proof of Marstrand theorem was generalized by C. Moreira and J. C. Yoccoz in their work on stable intersections of regular Cantor sets so that it could be applied to the study of homoclinic tangencies (Dynamical Systems) and the Markov/Lagrange spectrum (Number Theory) (as we briefly described in this post). In particular, since I’m planning to make a series of posts around these works of C. Moreira and J. C. Yoccoz, I think that a discussion of Kaufmann’s proof of Marstrand theorem is the best place to start this project.
Define , and denote by the orthogonal projection of the plane onto the line . In other words, if we identify with the real line (via , then, given , we have .
An interesting consequence of this theorem (in the context of stable intersections of Cantor sets of the real line) is the following fact:
Then, has positive Lebesgue measure for almost every .
Proof: Firstly, we recall the following well-known property of the Hausdorff dimension: (see K. Falconer’s book for more details). Secondly, we notice that the projection of is where . Since for and we can parametrize via with , we get the desired corollary from Marstrand’s theorem.
Now we turn to the proof of Marstrand’s theorem following the exposition of the book of Palis and Takens (pages 64, 65, 66 and 67).
We begin with a preliminary reduction.
for all and . Here, denotes the Hausdorff -measure.
Proof: See theorem 5.6 of K. Falconer’s book.
In particular, during the proof of Marstrand’s theorem, we can assume that verifies and, for some constant , (for any and ). In fact, given with , we can fix and apply the previous lemma in order to obtain verifying and . Hence, if we can prove Marstrand’s theorem for (i.e., has positive Lebesgue measure for almost every ), the same result holds for because .
So, in the sequel, we’ll make the following hypothesis:
Main Assumption. satisfies , and for any and .
In this situation, we can define a finite measure on by the formula . For each , we denote by the unique measure on satisfying for every continuous function .
Note that the support of is contained in . Hence, our task can be reduced to show that the support of has positive Lebesgue measure for almost every . In this direction, we have the following simple criterion:
Then, the support of has positive Lebesgue measure.
Proof: Plancherel’s theorem guarantees that is a well-defined square-integrable function on such that and . Thus, the support of (which is the support of ) can’t have zero Lebesgue measure.
In view of this lemma, the proof of Marstrand’s theorem will be complete once we show that for almost every .
Observe that for all . Indeed, if this integral vanishes (for some ), then (where ). In particular, for almost every . Since , we have , i.e., , a contradiction.
Now, we’ll end the proof of theorem 1 by showing that is square-integrable for almost every . Note that
Observe that the integral doesn’t depend on the direction of (but only on ). Therefore, denoting by , we get
Hence, for any ,
Since we know that the integral of the Bessel function is convergent, we see that there exists a real number (independent of ) such that
We claim that the right-hand side of (1) is finite. Indeed, for any , we have
Therefore, it follows that is square-integrable for almost every , so that the proof of Marstrand’s theorem is complete.