A Quantification of a Besicovitch Non-linear Projection Theorem via Multiscale Analysis

Abstract

The Besicovitch projection theorem states that if a subset E of the plane has finite length in the sense of Hausdorff measure and is purely unrectifiable (so its intersection with any Lipschitz graph has zero length), then almost every orthogonal projection of E to a line will have zero measure. In other words, the Favard length of a purely unrectifiable 1-set vanishes. In this article, we show that when linear projections are replaced by certain non-linear projections called curve projections, this result remains true. In fact, we go further and use multiscale analysis to prove a quantitative version of this Besicovitch non-linear projection theorem. Roughly speaking, we show that if a subset of the plane has finite length in the sense of Hausdorff and is nearly purely unrectifiable, then its Favard curve length is very small. Our techniques build on those of Tao, who in (Proc Lond Math Soc 98:559–584, 2009) proves a quantification of the original Besicovitch projection theorem.

Appendix A: Technical Results

In this section, we prove that the sets of high-multiplicity pairs and high-density pairs are closed. We also show that $\Delta$ is parametrically closed.

Lemma A.1

($H_n$ is closed) Let $H_n$ be as in Definition 3.4. For any $1 \le n \le N$, $H_n$ closed.

Proof

Fix n and let $\left\{ \left( e_m, \alpha _m \right) \right\} _{m=1}^\infty \subset H_n$ be a sequence of points such that $\left( e_m, \alpha _m \right) \rightarrow \left( e, \alpha \right)$. We need to show that $C_{e,\alpha }$ is of high multiplicity at a scale index at most n. Since $\left( e_m, \alpha _m \right) \in H_n$, then the curve $C_m := C_{e_m, \alpha _m}$ contains $N^{1/100}$ points that are $r_{n}^-$-separated. Let $e_{m,1}, e_{m, 2}, \ldots , e_{m, N^{1/100}}$ denote the points on the curve $C_m$, where they are ordered so that $e_{m, i}$ is to the left of $e_{m, j}$ whenever $i < j$. Moreover, for any $i = 1, \ldots , N^{1/100} - 1$, $\left| e_{m,i+1} - e_{m,i}\right| \ge r_{n}^-$. Since $\left\{ e_{m,1}\right\} _{m=1}^\infty \subset E$ is bounded (since E is compact), then it contains a convergent subsequence, $\left\{ e_{m(1,k),1}\right\} _{k=1}^\infty$. Since $\left\{ e_{m(1,k), 2}\right\} _{k=1}^\infty$ is also bounded, then it too contains a convergent subsequence, $\left\{ e_{m(2,k), 2}\right\} _{k=1}^\infty$. Continuing on with this diagonalization process, we extract a subsequence $\left\{ m_k\right\} _{k=1}^\infty \subset {\mathbb {N}}$, where $m_k = m(N^{1/100}, k)$, so that for every $j = 1, \ldots , N^{1/100}$, $\displaystyle \lim _{k \rightarrow \infty }e_{m_k, j} = e_j$. Since E is compact, then $e_j \in E$ for each j. We first show that $\left\{ e_j\right\} _{j=1}^{N^{1/100}}$ is $r_{n}^-$-separated. Let $\varepsilon > 0$. There exists $K \in {\mathbb {N}}$ so that whenever $k \ge K$, we have $\left| e_{m_k, j} - e_j\right| < \frac{\varepsilon }{2}$ for any $j \in \left\{ 1, \ldots , N^{1/100}\right\}$. It follows that for any $j \in \left\{ 1, \ldots , N^{1/100}-1\right\}$,

$$\begin{aligned} r_{n}^-&\le \left| e_{m_K, j+1} - e_{m_K, j}\right| = \left| e_{m_K,j+1} - e_{j+1} + e_{j+1} - e_{m_K, j} + e_j - e_{j} \right| \\&\le \left| e_{m_K,j+1} - e_{j+1}\right| + \left| e_{j+1} - e_{j} \right| + \left| e_{m_K, j} - e_j\right| < \left| e_{j+1} - e_{j} \right| + \varepsilon . \end{aligned}$$

Since $\varepsilon > 0$ was arbitrary, we conclude that $\left| e_{j+1} - e_{j} \right| \ge r_{n}^-$, showing that $\left\{ e_j\right\} _{j=1}^{N^{1/100}}$ is $r_{n}^-$-separated. Finally, since $\left( e_m, \alpha _m \right) \rightarrow \left( e, \alpha \right)$, then $C_m \rightarrow C_{e, \alpha }$. In particular, $C_{m_k} \rightarrow C_{e,\alpha }$. Since each $e_{m_k, j} \in C_{m_k}$, we deduce that $e_j \in C_{e, \alpha }$. It follows that $\left\{ e_j\right\} _{j=1}^{N^{1/100}} \subset C_{e,\alpha }$, completing the proof. $\square$

Lemma A.2

($D_n$ is closed) Let $D_n$ be as in Definition 3.6. For any $1 \le n \le N$, $D_n$ is closed.

Proof

Fix n and let $\left\{ \left( e_m, \alpha _m \right) \right\} _{m=1}^\infty \subset D_n$ so that $\left( e_m, \alpha _m \right) \rightarrow \left( e, \alpha \right)$. Since $\left( e_m, \alpha _m \right) \in D_n$, then there exists $J_m$ with $\left| J_m\right| \ge r_{n}^-$ and $\mu \left( \Phi _{\alpha _m,+}^{-1}\left( J_m \right) \right) \ge N^{1/100} \left| J_m\right|$. Write $J_m = \left[ a_m, b_m\right]$. Since $\left\{ a_m\right\} _{m=1}^\infty \subset {\mathbb {R}}$ is bounded (since E is compact), then there exists a convergence subsequence $\left\{ a_{m(1,k)}\right\} _{k=1}^\infty$. Similarly, $\left\{ b_{m(1,k)}\right\} _{k=1}^\infty \subset {\mathbb {R}}$ is bounded, so there is a convergent subsequence $\left\{ b_{m(2, k)}\right\} _{k=1}^\infty$. With $m_k = m(2, k)$, both $\left\{ a_{m_k}\right\} _{k=1}^\infty$ and $\left\{ b_{m_k}\right\} _{k=1}^\infty$ are convergent sequences in ${\mathbb {R}}$, with limits a and b, respectively. Define $J = \left[ a, b\right]$. Since $b_m - a_m = \left| J_m\right| \ge r_{n}^-$ for all $m \in {\mathbb {N}}$, then taking limits shows that $b - a = \left| J\right| \ge r_{n}^-$ as well. Taking limits and appealing to continuity also shows that $\mu \left( \Phi _{\alpha ,+}^{-1}\left( J \right) \right) \ge N^{1/100} \left| J\right|$. In particular, $\Phi _{\alpha ,+}^{-1}\left( J \right)$ has high density at scale index n. Since $C_{e_m, \alpha _m} \subset \Phi _{\alpha _m,+}^{-1}(J_m)$ for each $m \in {\mathbb {N}}$, then another limiting argument shows that $C_{e, \alpha } \subset \Phi _{\alpha ,+}^{-1}(J)$, completing the proof. $\square$

Lemma A.3

($\Delta$ is parametrically closed) For $\Delta$ be as defined in (3.17), $\Delta$ is parametrically closed.

Proof

Since H is parametrically open by definition and $H_{n_0 + N^{-3/100} N}$ is closed (by Lemma A.1), and therefore parametrically closed, then $\Delta H := H_{n_0 + N^{-3/100} N} \setminus H$ is parametrically closed. Similarly, since D is open by definition and $D_{n_2 + N^{-10/100} N}$ is closed (by Lemma A.2), then $\Delta D := D_{n_2 + N^{-10/100} N} \setminus D$ is closed, and consequently parametrically closed. It can be shown (following arguments similar to those used for each $H_n$) that $P_{n_1}$ is closed. Since H is parametrically open, then $P_{n_1} \setminus H$ is also parametrically closed. It follows that $\Delta$, the union of three parametrically closed sets, is itself parametrically closed.

$\square$