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An analyst’s traveling salesman theorem for sets of dimension larger than one

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Abstract

In his 1990 Inventiones paper, P. Jones characterized subsets of rectifiable curves in the plane via a multiscale sum of \(\beta \)-numbers. These \(\beta \)-numbers are geometric quantities measuring how far a given set deviates from a best fitting line at each scale and location. Jones’ result is a quantitative way of saying that a curve is rectifiable if and only if it has a tangent at almost every point. Moreover, computing this square sum for a curve returns the length of the curve up to multiplicative constant. K. Okikiolu extended his result from subsets of the plane to subsets of Euclidean space. G. David and S. Semmes extended the discussion to include sets of (integer) dimension larger than one, under the assumption of Ahlfors regularity and using a variant of Jones’ \(\beta \)-numbers. This variant has since been used by others to give structure theorems for rectifiable sets and to give upper bounds for the measure of a set. In this paper we give a version of P. Jones’ theorem for sets of arbitrary (integer) dimension lying in Euclidean space. Our main result is a lower bound for the d-dimensional Hausdorff measure of a set in terms of an analogous sum of \(\beta \)-type numbers. We also show an upper bound of this type. The combination of these results gives a Jones theorem for higher dimensional sets. While there is no assumption of Ahlfors regularity, or of a measure on the underlying set, there is an assumption of a lower bound on the Hausdorff content. We adapt David and Semmes’ version of Jones’ \(\beta \)-numbers by redefining them using a Choquet integral, allowing them to be defined for arbitrary sets (and not just sets of locally finite measure). A key tool in the proof is G. David and T. Toro’s parametrization of Reifenberg flat sets (with holes).

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Acknowledgements

The authors would like to thank Peter Jones, Xavier Tolsa, and Tatiana Toro for their helpful discussions. In fact, some of the core ideas arose from ongoing work between the first author and Xavier Tolsa as well as the second author and Peter Jones. We also thank Silvia Ghinassi and Michele Villa for their careful proofreading of the manuscript. Finally, we would like to thank the anonymous referee who had many useful comments and suggestions that greatly improved the paper.

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Correspondence to Raanan Schul.

Additional information

Communicated by Loukas Grafakos.

J. Azzam was supported by the ERC Grant 320501 of the European Research Council (FP7/2007-2013). R.  Schul was partially supported by NSF DMS 1361473.

Appendix

Appendix

Proof of Lemma 2.1

Let

$$\begin{aligned} S_{j}=\{x:f_{j}(x)=\max \{f_{i}(x)\}>0\}. \end{aligned}$$

Then \(\bigcup \mathrm {supp}\,f_{i}=\bigcup S_{j}\). If \(x\in S_{j}\), then there are at most C many indices i for which \(f_{i}>0\), and so \(\sum f_{i}(x)\le Cf_{j}(x)\). Therefore,

$$\begin{aligned} \int \left( \sum f_{i} \right) ^{p} d{\mathscr {H}}^{d}_{\infty }&=\int {\mathscr {H}}^{d}_{\infty }\left( \left\{ \sum f_{i}>\lambda \right\} \right) \lambda ^{p-1}d\lambda \\&\le \sum _{j} \int {\mathscr {H}}^{d}_{\infty }\left( S_{j}\cap \left\{ \sum f_{i}>\lambda \right\} \right) \lambda ^{p-1}d\lambda \\&\le \sum _{j}\int {\mathscr {H}}^{d}_{\infty } \left( \{Cf_{j} >\lambda \} \right) \lambda ^{p-1}d\lambda \\&= C^{p} \sum _{j} \int fd{\mathscr {H}}^{d}_{\infty }. \end{aligned}$$

\(\square \)

Proof of Lemma 2.2

Extend f to be continuous on all of \({\mathbb {R}}^{n}\). Since E is compact and f is continuous, the set

$$\begin{aligned} E^{t} \,{:=}\, \{x\in E: f(x)\ge t\} \end{aligned}$$

is also compact for each \(t>0\). It is not hard to show (using compactness) that if

$$\begin{aligned} E_{j}^{t} \,{:=}\, \{x\in E_{j}: f(x)\ge t\} \end{aligned}$$

then, since \(\bigcap E_{j}^{t}= E^{t}\), we have

$$\begin{aligned} \lim _{j\rightarrow \infty } {\mathscr {H}}_{\infty }^{d}(E_{j}^{t})= {\mathscr {H}}^{d}_{\infty }(E^{t}) \quad \text{ for } t>0. \end{aligned}$$

Hence, by the monotone convergence theorem,

$$\begin{aligned} \int _{0}^{\infty } {\mathscr {H}}^{d}(E^{t})dt =\lim _{j\rightarrow \infty }\int _{0}^{\infty } {\mathscr {H}}^{d}(E_{j}^{t})dt. \end{aligned}$$
(13.1)

Now we observe that, for any function g and F any set,

$$\begin{aligned} \int _{F} g d{\mathscr {H}}^{d}_{\infty }&=\int _{F} {\mathscr {H}}^{d}_{\infty } (\{x\in F: g(x)>t\})dt \nonumber \\&\le \int _{F} {\mathscr {H}}^{d}_{\infty } (\{x\in F: g(x)\ge t\})dt \end{aligned}$$
(13.2)

and

$$\begin{aligned} \int _{0}^{\infty }&{\mathscr {H}}^{d}_{\infty } (\{x\in F: g(x)\ge t\})dt \nonumber \\&\le \inf _{\alpha \in (0,1)} \int _{0}^{\infty }{\mathscr {H}}^{d}_{\infty } (\{x\in F: g(x)> \alpha t\})dt \nonumber \\&=\inf _{\alpha \in (0,1)} \alpha ^{-1} \int _{0}^{\infty } {\mathscr {H}}^{d}_{\infty } (\{x\in F: g(x)> t\})dt =\int g {\mathscr {H}}^{d}_{\infty } \end{aligned}$$
(13.3)

Combining (13.2) and (13.3) gives

$$\begin{aligned} \int _{F} g d{\mathscr {H}}^{d}_{\infty } = \int _{0}^{\infty } {\mathscr {H}}^{d}_{\infty } (\{x\in F: g(x)\ge t\})dt \end{aligned}$$
(13.4)

Thus, applying this to \(g=f\) and F equal to either E or \(E_{j}\),

\(\square \)

Proof of Lemma 2.3

First assume that E is open. Without loss of generality, we may assume \({\mathscr {H}}^{d}_\infty (E)=1\). Note that \(f\mathbb {1}_{E}\) is still lower semicontinuous since E is open. By the corollary on page 118 of [1], for \(f\ge 0\) lower semicontonuous,

$$\begin{aligned} \int fd{\mathscr {H}}^{d}_{\infty } \sim _{n} \sup \left\{ \int fd\mu : \mu \in L^{1,d}({\mathbb {R}}^{n}), \quad ||\mu ||=1 \right\} . \end{aligned}$$
(13.5)

Where \(L^{1,d}({\mathbb {R}}^{n})\) is the Morrey space of Radon measures with the norm

$$\begin{aligned} ||\mu ||=\mathop {\mathop {\sup }\limits _{x\in {\mathbb {R}}^{n}}}\limits _{r>0} |\mu |(B(x,r))r^{-d}. \end{aligned}$$

Note that if \(A_{i}\) is a cover of E, then each \(A_{i}\) is contained in a ball of radius \({{\mathrm{\,diam\,}}}A_{i}\), and so

$$\begin{aligned} \mu (E)\le \sum \mu (A_{i})\le \sum ||\mu ||({{\mathrm{\,diam\,}}}A_{i})^{d} \end{aligned}$$

and infimizing over all such covers gives

$$\begin{aligned} \mu (E)\le ||\mu ||{\mathscr {H}}^{d}_{\infty }(E)=||\mu ||=1. \end{aligned}$$

Thus, if \(\frac{1}{p}+\frac{1}{q}=1\),

$$\begin{aligned} \int fd\mu&\le \left( \int f^{p}d\mu \right) ^{\frac{1}{p}}\left( \int _{E} d\mu \right) ^{\frac{1}{q}} \le \left( ||\mu ||\int f^{p}d{\mathscr {H}}^{d}_{\infty } \right) ^{\frac{1}{p}} \mu (E)^{\frac{1}{q}}\\&= \left( \int f^{p}d{\mathscr {H}}^{d}_{\infty } \right) ^{\frac{1}{p}}. \end{aligned}$$

Supremizing over all \(\mu \) and using (13.5) once more gives (2.4).

Now assume E be a compact set and f a continuous function on E. Again, we may assume \({\mathscr {H}}^{d}_{\infty }(E)=1\). Extend f to a continuous function on all of \({\mathbb {R}}^{n}\) and let

$$\begin{aligned} E_{j}=\{x\in {\mathbb {R}}^{n}:\,\mathrm {dist}(x,E)<j^{-1}\}. \end{aligned}$$

Since we know (2.4) for open sets, we may apply it to the sets \(E_{j}\). Since f and \(f^{p}\) are continuous, and since the \(E_{j}\) are open, contain E, and converge to E in the Hausdorff metric, we may use Lemma 2.2 and get

$$\begin{aligned} \int _{E} fd{\mathscr {H}}^{d}_{\infty }&{\mathop {=}\limits ^{(2.3)}} \lim _{j\rightarrow \infty } \int _{E_{j}} f d{\mathscr {H}}^{d}_{\infty } {\mathop {\lesssim }\limits ^{(2.4)}} \lim _{j\rightarrow \infty } \left( \int _{E_{j}}f^{p} d{\mathscr {H}}^{d}_{\infty } \right) ^{\frac{1}{p}}\\&{\mathop {=}\limits ^{(2.3)}} \left( \int _{E}f^{p} d{\mathscr {H}}^{d}_{\infty } \right) ^{\frac{1}{p}}. \end{aligned}$$

\(\square \)

Proof of Lemma 2.22

Let \(\alpha '=1-\alpha \). We will first prove that for all \(\lambda >0\)

$$\begin{aligned}&{\mathscr {H}}^{d}_{\infty } \left( \left\{ x\in F_{1}:\sum _{j\in {\mathscr {X}}}\mathbb {1}_{B_{j}'}f(z_{j})>\lambda \right\} \right) \nonumber \\&\quad \lesssim {\mathscr {H}}^{d}_{\infty } \left( \left\{ x\in F_{1}:\sum _{j\in {\mathscr {X}}}\mathbb {1}_{\alpha 'B_{j}'}f(z_{j})>\lambda \right\} \right) . \end{aligned}$$
(13.6)

Let

$$\begin{aligned} A=\left\{ x\in F_{2}:\sum _{j\in {\mathscr {X}}}\mathbb {1}_{\alpha 'B_{j}'}f(z_{j})>\lambda \right\} . \end{aligned}$$

Let \({\mathscr {I}}\) be a collection of balls covering A so that

$$\begin{aligned} {\mathscr {H}}^{d}_{\infty }(A)\sim _{d} \sum _{B\in {\mathscr {I}}} (2r_{B})^{d}. \end{aligned}$$
(13.7)

Let

$$\begin{aligned} {\mathscr {X}}^{\lambda }=\{j\in {\mathscr {X}}: f(z_{j})>\lambda \}. \end{aligned}$$

Note that as \(|z_{j}-z_{j}'|<\alpha r_{B_{j}}\),

$$\begin{aligned} \alpha ' B_{j}'= B(z_{j}',\alpha ' r_{B_{j}})\subseteq B(z_{j},(\alpha ' +\alpha )r_{B_{j}})=B_{j}. \end{aligned}$$

Thus, since the balls \(B_{j}\) are disjoint for \(j\in {\mathscr {X}}\), so are the balls \(B_{j}'\), and hence we have

$$\begin{aligned} A=F_{2}\cap \bigcup _{j\in {\mathscr {X}}^{\lambda }} \alpha ' B_{j}'. \end{aligned}$$

We will define a new collection of balls as the limit of a sequence of collections \({\mathscr {I}}(j)\) which we define inductively as follows. Assume \({\mathscr {X}}^{\lambda }={\mathbb {N}}\) and set \({\mathscr {I}}(0)=\emptyset \). Now assume that for some \(j>0\), \({\mathscr {I}}(j-1)\) has already been defined. Let

$$\begin{aligned} {\mathscr {I}}_{j}=\left\{ B\in {\mathscr {I}}: B\cap \alpha 'B_{j}'\ne \emptyset \right\} . \end{aligned}$$
  1. 1.

    If there is \(B\in {\mathscr {I}}_{j}\) for which \(r_{B}\ge \frac{\alpha }{2} r_{B_{j}}\), we let

    $$\begin{aligned} {\mathscr {I}}(j)={\mathscr {I}}(j-1)\cup \left\{ \frac{4}{\alpha }B \right\} \end{aligned}$$

    and note that \(\frac{4}{\alpha } B\supseteq B_{j}\) since \(B\cap B_{j}\supseteq B\cap \alpha 'B_{j}'\ne \emptyset \) and \(r_{B_{J}}\le \frac{2}{\alpha } r_{B}\) by assumption.

  2. 2.

    If \(r_{B}<\frac{\alpha }{2} r_{B_{j}}\) for all \(B\in {\mathscr {I}}_{j}\), we let

    $$\begin{aligned} {\mathscr {I}}(j)={\mathscr {I}}(j-1)\cup \{B_{j}\}. \end{aligned}$$

    Note that in this case, \(B\subseteq B_{j}\) for all \(B\in {\mathscr {I}}_{j}\).

We let \({\mathscr {I}}'=\bigcup {\mathscr {I}}(j)\). In this way, every \(B_{j}\) is contained in a ball from \({\mathscr {I}}'\), that is,

$$\begin{aligned} \bigcup _{j\in {\mathscr {X}}^{\lambda }}B_{j} \subseteq \bigcup _{B\in {\mathscr {I}}'}B. \end{aligned}$$

For \(i=1,2\), let \({\mathscr {I}}_{i}'\) be those balls in \({\mathscr {I}}_{i}\) added in case i and let \({\mathscr {X}}^{\lambda ,2}\) be those j for which case 2 happened. Since the \(B_{j}\) are disjoint, and \(B_{j}\supseteq \alpha 'B_{j}'\cap F_{2}\),

$$\begin{aligned} \sum _{B\in {\mathscr {I}}} r_{B}^{d}\ge & {} \sum _{j\in {\mathscr {X}}^{\lambda ,2} }\sum _{B\in {\mathscr {I}}(j)} r_{B}^{d} \gtrsim \sum _{j\in {\mathscr {X}}^{\lambda ,2} } {\mathscr {H}}^{d}_{\infty } \left( \alpha 'B_{j}'\cap F_{2} \right) {\gtrsim }_{\alpha } \sum _{j\in {\mathscr {X}}^{\lambda ,2} } r_{B_{j}}^{d}\nonumber \\\gtrsim & {} \sum _{B\in {\mathscr {I}}_{2}'} r_{B}^{d}. \end{aligned}$$
(13.8)

Also,

$$\begin{aligned} \sum _{B\in {\mathscr {I}}} r_{B}^{d} \ge \left( \frac{\alpha }{4} \right) ^{d} \sum _{B\in {\mathscr {I}}_{1}'} r_{B}^{d} \end{aligned}$$
(13.9)

and hence

$$\begin{aligned} {\mathscr {H}}^{d}_{\infty } (A) {\mathop {\gtrsim }\limits ^{(13.7)}} \sum _{B\in {\mathscr {I}}} r_{B}^{d} \mathop {\mathop {\gtrsim }\limits ^{(13.8)}}\limits ^{(13.9)} \sum _{B\in {\mathscr {I}}'} r_{B}^{d} \ge {\mathscr {H}}^{d}_{\infty }\left( \bigcup _{j\in {\mathscr {X}}^{\lambda }} B_{j} \right) . \end{aligned}$$
(13.10)

Let

$$\begin{aligned} A'=\left\{ x\in F_{1}:\sum _{j\in {\mathscr {X}}}\mathbb {1}_{B_{j}}f(z_{j})>\lambda \right\} . \end{aligned}$$

If \(x \in A'\), since the \(B_{j}\) are disjoint for \(j\in {\mathscr {X}}\), there is \(x\in B_{j}\) for some \(j\in {\mathscr {X}}^{\lambda }\), and so \(x\in \bigcup _{j\in {\mathscr {X}}^{\lambda }}B_{j} \), hence

$$\begin{aligned} {\mathscr {H}}^{d}_{\infty }(A') \le {\mathscr {H}}^{d}_{\infty }\left( \bigcup _{j\in {\mathscr {X}}^{\lambda }} B_{j} \right) {\mathop {\lesssim }\limits ^{(13.10)}} {\mathscr {H}}^{d}_{\infty } (A) \end{aligned}$$

and this finishes the proof of (13.6). Hence (2.48) follows by integrating (13.6). \(\square \)

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Azzam, J., Schul, R. An analyst’s traveling salesman theorem for sets of dimension larger than one. Math. Ann. 370, 1389–1476 (2018). https://doi.org/10.1007/s00208-017-1609-0

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