Posted by: matheuscmss | March 16, 2009

R. Kaufmann’s proof of J. M. Marstrand’s theorem

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.

–Statement of Marstrand’s theorem–

Define {v_\theta=(\cos\theta,\sin\theta)\in\mathbb{R}^2}, {\theta\in\mathbb{R}} and denote by {\pi_\theta} the orthogonal projection of the plane {\mathbb{R}^2} onto the line {L_\theta=\mathbb{R}\cdot v_\theta}. In other words, if we identify {L_\theta} with the real line {\mathbb{R}} (via {x\in\mathbb{R}\mapsto x\cdot v_\theta\in L_\theta}, then, given {u=(u_1,u_2)\in\mathbb{R}^2}, we have {\pi_\theta(u)=u\cdot v_\theta = u_1\cos\theta + u_2\sin\theta}.

Theorem 1 (J. Marstrand’s theorem) Let {K\subset\mathbb{R}^2} be a subset with Hausdorff dimension {HD(K)>1}. Then, its projection {\pi_\theta(K)} has positive Lebesgue measure for (Lebesgue) almost every {\theta}.

An interesting consequence of this theorem (in the context of stable intersections of Cantor sets of the real line) is the following fact:

Corollary 2 Let {K_1} and {K_2} be two Cantor sets of the real line {\mathbb{R}} such that

\displaystyle HD(K_1)+HD(K_2)>1.

Then, {K_1-\lambda K_2} has positive Lebesgue measure for almost every {\lambda\in\mathbb{R}}.

Proof: Firstly, we recall the following well-known property of the Hausdorff dimension: {HD(K_1\times K_2)\geq HD(K_1)+HD(K_2)>1} (see K. Falconer’s book for more details). Secondly, we notice that the projection {\pi_\theta(K_1\times K_2)} of {K_1\times K_2} is {\pi_\theta(K_1\times K_2)=\cos\theta\cdot (K_1-\lambda K_2)} where {\lambda=-\tan\theta}. Since {\cos\theta\neq 0} for {-\pi/2<\theta<\pi/2} and we can parametrize {\lambda\in\mathbb{R}} via {\lambda=-\tan\theta} with {-\pi/2<\theta<\pi/2}, we get the desired corollary from Marstrand’s theorem. \Box

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).

–Kaufmann’s proof of Marstrand theorem

We begin with a preliminary reduction.

Lemma 3 Let {K\subset\mathbb{R}^2} be a subset of the plane with Hausdorff dimension {d=HD(K)}. Given {d'<d}, we can find {K'\subset K} such that {0<m_{d'}(K')<\infty} and

\displaystyle m_{d'}(K'\cap B_r(x))\leq C r^{d'}

for all {x\in\mathbb{R}^2} and {0<r\leq 1}. Here, {m_s} denotes the Hausdorff {s}-measure.

Proof: See theorem 5.6 of K. Falconer’s book. \Box

In particular, during the proof of Marstrand’s theorem, we can assume that {K\subset\mathbb{R}^2} verifies {0<m_d(K)<\infty} and, for some constant C>0, m_d(K\cap B_r(x))\leq C r^d (for any 0<r\leq 1 and x\in\mathbb{R}^2). In fact, given K\subset\mathbb{R}^2 with d=HD(K)>1, we can fix {1<d'<d} and apply the previous lemma in order to obtain {K'\subset K} verifying {0<m_{d'}(K')<\infty} and {m_{d'}(K'\cap B_r(x))\leq C r^{d'}}. Hence, if we can prove Marstrand’s theorem for {K'} (i.e., {\pi_\theta(K')} has positive Lebesgue measure for almost every {\theta}), the same result holds for {K} because {\pi_\theta(K')\subset\pi_\theta(K)}.

So, in the sequel, we’ll make the following hypothesis:

Main Assumption. {K\subset\mathbb{R}^2} satisfies {1<d=HD(K)}, {0<m_d(K)<\infty} and {m_d(K\cap B_r(x))\leq C r^d} for any {x\in\mathbb{R}^2} and { 0 < r\leq 1 }.

In this situation, we can define a finite measure on {\mathbb{R}^2} by the formula {\mu(A):=m_d(A\cap K)}. For each {\theta\in\mathbb{R}}, we denote by {\mu_\theta:=(\pi_\theta)_*\mu} the unique measure on {\mathbb{R}} satisfying {\int_{\mathbb{R}} f d\mu_\theta = \int_{\mathbb{R}^2} (f\circ\pi_\theta) d\mu} for every continuous function {f}.

Note that the support of {\mu_\theta} is contained in {\pi_\theta(K)}. Hence, our task can be reduced to show that the support of {\mu_\theta} has positive Lebesgue measure for almost every {\theta}. In this direction, we have the following simple criterion:

Lemma 4 Let {\eta} be a finite measure with compact support on {\mathbb{R}} and denote by {\widehat{\eta}(p)=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}e^{-ixp}d\eta(x)} the Fourier transform of {\eta}. Suppose that

\displaystyle 0<\int_{\mathbb{R}}|\eta(p)|^2 dp<\infty.

Then, the support of {\eta} has positive Lebesgue measure.

Proof: Plancherel’s theorem guarantees that {\phi(x) := \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{ixp} \widehat{\eta}(p) dp} is a well-defined square-integrable function on {\mathbb{R}} such that {d\eta = \phi dx} and {\int_{\mathbb{R}}|\phi(x)|^2 dx = \int_{\mathbb{R}}|\widehat{\eta}(p)|^2 dp>0}. Thus, the support of {\eta} (which is the support of {\phi}) can’t have zero Lebesgue measure. \Box

In view of this lemma, the proof of Marstrand’s theorem will be complete once we show that {0<\int_{\mathbb{R}}|\widehat{\mu_\theta}(p)|^2 dp<\infty} for almost every {\theta}.

Observe that {0<\int_{\mathbb{R}}|\widehat{\mu_\theta}(p)|^2 dp} for all {\theta}. Indeed, if this integral vanishes (for some {\theta}), then {\int_{\mathbb{R}}|\phi(x)|^2 dx =0} (where {\phi(x) := \frac{1}{\sqrt{2\pi}} \int_{\mathbb{R}} e^{ixp} \widehat{\mu_\theta}(p) dp}). In particular, {\phi(x)=0} for almost every {x}. Since {d\mu_\theta=\phi dx}, we have {\mu_\theta(\mathbb{R})=\int_{\mathbb{R}} \phi(x) dx = 0}, i.e., {0<m_d(K)=\mu(\mathbb{R}^2)=\mu_\theta(\mathbb{R})=0}, a contradiction.

Now, we’ll end the proof of theorem 1 by showing that {\widehat{\mu_\theta}} is square-integrable for almost every {\theta}. Note that

\displaystyle |\widehat{\mu_\theta}(p)|^2 = \frac{1}{2\pi}\int\int e^{i(y-x)p} d\mu_\theta(x) d\mu_\theta(y) = \frac{1}{2\pi}\int\int e^{ip(v-u)\cdot v_\theta} d\mu(u) d\mu(v).

In particular,

\displaystyle |\widehat{\mu_\theta}(p)|^2+|\widehat{\mu_{\theta+\pi}}(p)|^2 = \frac{1}{\pi}\int\int \cos(p(v-u)\cdot v_\theta) d\mu(u) d\mu(v).

Thus,

\displaystyle \begin{array}{rcl} \int_0^{2\pi}|\widehat{\mu_\theta}(p)|^2 d\theta &=& \frac{1}{2\pi}\int_0^{2\pi}\int\int\cos(p(v-u)\cdot v_\theta) d\mu(u) d\mu(v) d\theta \\ &=& \frac{1}{2\pi}\int\int\left(\int_0^{2\pi}\cos(p(v-u)\cdot v_\theta) d\theta\right)d\mu(u) d\mu(v). \end{array}

Observe that the integral {\int_0^{2\pi}\cos(p(v-u)\cdot v_\theta) d\theta} doesn’t depend on the direction of {(v-u)} (but only on {\|v-u\|}). Therefore, denoting by {J_0(z)=\frac{1}{2\pi}\int_0^{2\pi} \cos(z\cos\theta) d\theta}, we get

\displaystyle \int_0^{2\pi}|\widehat{\mu_\theta}(p)|^2 d\theta = \int\int J_0(p\|v-u\|)d\mu(u) d\mu(v).

Hence, for any {a>0},

\displaystyle \begin{array}{rcl} \int_{-a}^{a}\int_0^{2\pi}|\widehat{\mu_\theta}(p)|^2 d\theta dp &=& \int\int\int_{-a}^{a} J_0(p\|v-u\|) dp d\mu(u) d\mu(v) \\ &=& \int\int\frac{\left(\int_{-a\|v-u\|}^{a\|v-u\|} J_0(z) dz\right)}{\|v-u\|} d\mu(u) d\mu(v). \end{array}

Since we know that the integral {\int_{\infty}^{\infty}J_0(z)dz} of the Bessel function {J_0(z)} is convergent, we see that there exists a real number {A>0} (independent of {a>0}) such that

\displaystyle \int_{-a}^{a}\int_0^{2\pi}|\widehat{\mu_\theta}(p)|^2 d\theta dp\leq A\int\int\frac{1}{\|v-u\|} d\mu(u) d\mu(v).

In particular,

\displaystyle \int_{-a}^{a}\int_0^{2\pi}|\widehat{\mu_\theta}(p)|^2 d\theta dp\leq A\int\int\frac{d\mu(u) d\mu(v)}{\|v-u\|}.

Because

\displaystyle \int\int\frac{d\mu(u) d\mu(v)}{\|v-u\|}\leq \mu(\mathbb{R}^2)\cdot\sup\limits_{u\in\mathbb{R}^2}\int\frac{d\mu(v)}{\|v-u\|}

we can concatenate the these two previous estimates to obtain

\displaystyle \int_{-a}^{a}\int_0^{2\pi}|\widehat{\mu_\theta}(p)|^2 d\theta dp\leq A\mu(\mathbb{R}^2)\cdot\sup\limits_{u\in\mathbb{R}^2}\int\frac{d\mu(v)}{\|v-u\|}. \ \ \ \ \ (1)

We claim that the right-hand side of (1) is finite. Indeed, for any {u\in\mathbb{R}^2}, we have

\displaystyle \begin{array}{rcl} \int\frac{d\mu(v)}{\|v-u\|} &=& \int_{\|v-u\|\geq 1}\frac{d\mu(v)}{\|v-u\|} + \sum\limits_{n=1}^\infty \int_{\frac{1}{2^n}\leq \|v-u\|< \frac{1}{2^{n-1}}} \frac{d\mu(v)}{\|v-u\|} \\ &\leq& \mu(\mathbb{R}^2)+ \sum\limits_{n=1}^\infty 2^n\mu(B_{1/2^{n-1}}(u)). \end{array}

Using the main assumption {m_d(K\cap B_r(x))\leq Cr^d} for any { 0 <r \leq 1} and x\in\mathbb{R}^2 (where d=HD(K)>1), we obtain

\displaystyle \sup\limits_{u\in\mathbb{R}^2}\int\frac{d\mu(v)}{\|v-u\|}\leq \mu(\mathbb{R}^2) + \frac{C}{\frac{1}{2} - \frac{1}{2^d}}<\infty. \ \ \ \ \ (2)

Putting the equations (1) and (2) together, using Fubini’s theorem and letting {a\rightarrow\infty}, we conclude

\displaystyle \int_0^{2\pi}\left(\int_\mathbb{R}|\widehat{\mu_\theta}(p)|^2 dp \right) d\theta\leq A\mu(\mathbb{R}^2)\cdot\left(\mu(\mathbb{R}^2) + \frac{C}{\frac{1}{2} - \frac{1}{2^d}}\right)<\infty.

Therefore, it follows that {\widehat{\mu_\theta}} is square-integrable for almost every {\theta}, so that the proof of Marstrand’s theorem is complete.

Posted in expository, math.DS, Mathematics | Tags: , , , ,

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Responses

  1. Dear Matheus: some corrections.

    – The statement of Lemma 3 has a typographic error, as the d’-Hausdorff dimension should be smaller than \infty.

    – In the paragraph after the Main Assumption, the support of \mu_teta is a subset of \pi_\teta(K).

    By: Yuri on May 22, 2009
    at 5:52 pm

    Reply

    • Dear Yuri: thanks for the corrections!

      By: matheuscmss on May 23, 2009
      at 10:17 am

      Reply

  2. Dear Matheus:

    For the lemma $4$, if $\eta$ is not absolutely continuous with respect to the Lebesgue measure, whether it is plausible that we write $d\eta (x)=\phi(x) dx$. Whether one should understand it as in the distribution sense.

    By: yaoxiao on October 10, 2017
    at 1:08 pm

    Reply

    • Dear yaoxiao,

      You’re right: one should interpret it in the distributional sense.

      Best,

      Matheus

      By: matheuscmss on October 10, 2017
      at 1:16 pm

      Reply


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