Hölder curves and parameterizations in the Analyst's Traveling Salesman theorem
HH ¨OLDER CURVES AND PARAMETERIZATIONS IN THE ANALYST’STRAVELING SALESMAN THEOREM
MATTHEW BADGER, LISA NAPLES, AND VYRON VELLIS
Abstract.
We investigate the geometry of sets in Euclidean and infinite-dimensionalHilbert spaces. We establish sufficient conditions that ensure a set of points is containedin the image of a (1 /s )-H¨older continuous map f : [0 , → l , with s >
1. Our resultsare motivated by and generalize the “sufficient half” of the Analyst’s Traveling SalesmanTheorem, which characterizes subsets of rectifiable curves in R N or l in terms of aquadratic sum of linear approximation numbers called Jones’ beta numbers. The originalproof of the Analyst’s Traveling Salesman Theorem depends on a well-known metriccharacterization of rectifiable curves from the 1920s, which is not available for higher-dimensional curves such as H¨older curves. To overcome this obstacle, we reimagine Jones’non-parametric proof and show how to construct parameterizations of the intermediateapproximating curves f k ([0 , tube approximations that ensure the approximating curves converge to a H¨older curve. As an application tothe geometry of measures, we identify conditions that guarantee fractional rectifiabilityof pointwise doubling measures in R N . Contents
1. Introduction 2
Part I. Proof of the H¨older Traveling Salesman Theorem
72. Preliminaries 73. Traveling Salesman algorithm 114. Mass of intervals 225. H¨older parametrization 366. Lipschitz parameterization 43
Part II. Applications and Further Results
Date : April 3, 2019.2010
Mathematics Subject Classification.
Primary 28A75; Secondary 26A16, 28A80, 30L05, 65D10.
Key words and phrases.
H¨older curves, parameterization, fractional rectifiability, Wa˙zewski’s theorem,Analyst’s Traveling Salesman problem, Jones’ beta numbers, tube approximations.M. Badger was partially supported by NSF DMS grants 1500382 and 1650546. L. Naples was partiallysupported by NSF DMS grant 1650546. V. Vellis was partially supported by NSF DMS grant 1800731. a r X i v : . [ m a t h . C A ] A p r MATTHEW BADGER, LISA NAPLES, AND VYRON VELLIS
Appendix A. Tours on connected, finite simple graphs 69Appendix B. From Lipschitz to H¨older parameterizations 70References 721.
Introduction
The ubiquitous Traveling Salesman problem [LLRKS85, GP02, ABCC06] is to find atour of edges on a finite graph that returns to the initial vertex and has the shortestpossible length. The Analyst’s Traveling Salesman problem [Jon90, Sch07a] is to finda rectifiable curve that contains a finite or infinite, bounded set of points in a metricspace that has the shortest possible length. The former problem always has a solutionyet is computationally hard, while the latter problem may or may not have any solutionat all. A sophisticated example from Geometric Measure Theory of a bounded point setthat is not contained in any rectifiable curve is a
Besicovitch irregular set [Bes28] (see § R [Jon90], R N [Oki92], l [Sch07b],the first Heisenberg group [LS16a, LS16b], Carnot groups [CLZ19, Li19], Laakso-typespaces [DS17], and in general metric spaces [Hah05, Hah08, DS19]. Applications of Jones’and Okikiolu’s solution of the Analyst’s TSP in R N have been given in Complex Analysis[BJ94, Bis02, Bis11], Dynamics and Probability [BJ97, BJPP97], Geometric MeasureTheory [BS15, BS17], Harmonic Analysis [Tol03], and Metric Geometry [NP11, AS12].Let E ⊂ R N be a nonempty set and let Q ⊂ R N be a bounded set of positive diameter(such as a ball or a cube). Following [Jon90], the Jones beta number β E ( Q ) is defined by β E ( Q ) := inf (cid:96) sup x ∈ E ∩ Q dist( x, (cid:96) )diam Q ∈ [0 , , where (cid:96) ranges over all straight lines in R N , if E ∩ Q (cid:54) = ∅ , and by β E ( Q ) = 0, if E ∩ Q = ∅ .Let ∆( R N ) denote the family of dyadic cubes in R N ,∆( R N ) := { [2 k m , k ( m + 1)] × · · · × [2 k m N , k ( m N + 1)] : m , . . . , m N , k ∈ Z } . Given a cube Q and a scaling factor λ >
0, we let λQ denote the concentric dilate of Q by λ . Analyst’s Traveling Salesman Theorem ([Jon90, Oki92]) . A bounded set E ⊂ R N iscontained in a rectifiable curve Γ = f ([0 , if and only if (1.1) S E := (cid:88) Q ∈ ∆( R N ) β E (3 Q ) diam Q < ∞ . More precisely,(1) If Γ is any curve containing E , then diam E + S E (cid:46) N length(Γ) .(2) If S E < ∞ , then there exists a curve Γ ⊃ E such that length(Γ) (cid:46) N diam E + S E . ¨OLDER CURVES AND PARAMETERIZATIONS 3 We may refer to statements (1) and (2) as the necessary half and sufficient half of theAnalyst’s Traveling Salesman theorem, respectively. The theorem is valid if the lengthof a curve Γ = f ([0 , f . A curious feature of theknown proofs of the sufficient half of the Analyst’s TST (see [Jon90] or [BS17]) is thata rectifiable curve Γ containing the set E is constructed as the limit of piecewise linearcurves Γ k containing a 2 − k -net for E without constructing a parameterization of Γ k or Γ.This aspect of the proof breaks the analogy with the classical TSP, in which one is askedto find a minimal tour of a graph.In this paper, we provide a parametric proof of the sufficient half of the Analyst’s TST,which more closely parallels the classical TSP. Beyond its intrinsic interest, the methodthat we provide is important, because it allows us to establish multiscale tests to ensurethat a bounded set of points in R N is contained in a (1 /s ) -H¨older continuous curve with s ∈ (1 , N ). Rectifiable curves correspond precisely to the class of Lipschitz curves ( s = 1).Remarkably, in the H¨older Traveling Salesman theorem (see §§ approximationby thin tubes . For a self-contained statement of the “parametric” Analyst’s TST, see § § H¨older Traveling Salesman Theorem(s).
A (1 /s ) -H¨older curve Γ in R N is theimage of a continuous map f : [0 , → R N satisfying the H¨older condition, | f ( x ) − f ( y ) | ≤ H | x − y | /s for all x, y ∈ [0 , , where s ∈ [1 , ∞ ) and H is a finite constant independent of x and y . A 1-H¨older curveis also called a Lipschitz curve or a rectifiable curve . While non-trivial rectifiable curvesalways have topological dimension 1 and asymptotically resemble a unique tangent line H almost everywhere, (1 /s )-H¨older curves with s > • an m -dimensional cube in R N ( m ≤ N ) is a (1 /m )-H¨older curve; • the von Koch snowflake is a log (3)-H¨older curve; and, • the standard Sierpi´nski carpet is a log (3)-H¨older curve.In fact, Remes [Rem98] proved that any compact, connected self-similar set K ⊂ R N ofHausdorff dimension s that satisfies the open set condition is a (1 /s )-H¨older curve. Forrelated work on space-filling curves generated by graph-directed iterated function systems,see Rao and Zhang [RZ16].Towards a H¨older version of the Analyst’s Traveling Salesman theorem, the first andthird authors proved in [BV18] as a test case that if s > E ⊂ R N is bounded, and (cid:88) Q ∈ ∆( R N ) Q ∩ E (cid:54) = ∅ , side Q ≤ (diam Q ) s < ∞ , MATTHEW BADGER, LISA NAPLES, AND VYRON VELLIS then E is contained in a (1 /s )-H¨older curve. By establishing a parametric version ofJones’ proof of the sufficient half of the Analyst’s TST, we are able to obtain the followingsubstantial improvement. Theorem 1.1 (H¨older Traveling Salesman I) . For all N ≥ and s > , there exists β ∈ (0 , such that if E ⊂ R N is bounded and (1.2) S s, + E := (cid:88) Q ∈ ∆( R N ) β E (3 Q ) ≥ β (diam Q ) s < ∞ , then E is contained in a (1 /s ) -H¨older curve. More precisely, E ⊂ Γ = f ([0 , for some (1 /s ) -H¨older map f : [0 , → R N with H¨older constant H (cid:46) N,s diam E +(diam E ) − s S s, + E . Condition (1.2) implies that at H s almost every point, the set E asymptotically lies insufficiently thin tubes . Theorem 1.1 provides a sufficient test that identifies all subsets ofsome well-known H¨older curves such as snowflakes of small dimension. However, becauseof the richness of H¨older geometry, a condition using Jones beta numbers alone such as(1.2) cannot be expected to hold for all subsets of every H¨older curve. Indeed (1.2) failswhen E is a carpet or a square. For expanded discussion and related examples, see § β in Theorem 1.1, see Lemma 2.8 and Remark5.6. The following variant of Theorem 1.1 is an immediate corollary, whose hypothesisdoes not require knowledge of β . Corollary 1.2 (H¨older Traveling Salesman II) . Suppose that N ≥ , s > , and p > .If E ⊂ R N is bounded and (1.3) S s,pE := (cid:88) Q ∈ ∆( R N ) β E (3 Q ) p (diam Q ) s < ∞ , then E is contained in a (1 /s ) -H¨older curve. More precisely, E ⊂ Γ = f ([0 , for some (1 /s ) -H¨older map f : [0 , → R N with H¨older constant H (cid:46) N,s diam E + β − p (diam E ) − s S s,pE , where β is the constant appearing in Theorem 1.1. A good exercise is to prove that any bounded set E in R N satisfying condition (1.3)with s > s -dimensional Hausdorff measure. In § E (hence H s ( E ) = 0) such that E is not contained in any (1 /s )-H¨older curvewith 1 ≤ s < N . Thus, Corollary 1.2 is nonvacuous.1.2. Overview of the proof of Theorem 1.1.
In order to properly discuss the proofof Theorem 1.1, we quickly sketch the proof of the sufficient half of the Analyst’s TST.The proof splits into three steps. In the first step, one uses the Jones beta numbers β E (3 Q ) (in particular, whether they are large or small) to construct a sequence of finite, ¨OLDER CURVES AND PARAMETERIZATIONS 5 connected graphs G k in R N with straight edges that converge in the Hausdorff distanceto a compact, connected set G containing E . Each graph G k is obtained by refining G k − and resembles a flat arc near points of E that look flat at scale 2 − k . In step two, one usesthe structure of the graphs G k and the Pythagorean theorem to prove the existence of aconstant C > H ( G k +1 ) ≤ H ( G k ) + C (cid:88) Q ∈ ∆( R N )side Q (cid:39) − k β E (3 Q ) diam Q. Condition (1.1) and Go(cid:32)l¸ab’s semicontinuity theorem (e.g. see [AO17]) ensure that H ( G ) ≤ lim inf k →∞ H ( G k ) < ∞ . Thus, the first two parts of the proof yield a compact, connectedset G containing E with H ( G ) < ∞ . The final step is to invoke Wa˙zewski’s theoremto conclude existence of a Lipschitz parameterization for G : if G ⊂ R N is connected,compact, and H ( G ) < ∞ , then there exists a Lipschitz map f : [0 , → R N such that G = f ([0 , H ( G ) < ∞ promotes connectedness of G to local connectedness (because G is a curve).In the H¨older setting, there are at least two obstacles to following the approach above.First and foremost, a naive analogue of Wa˙zewski’s theorem cannot hold for H¨older maps,since the condition H s ( G ) < ∞ does not imply a continuum is locally connected when s > G is assumed to be an Ahlfors s -regular curve with finite H s measure, we cannot conclude that G is a (1 /s )-H¨oldercurve; we provide examples in § H s with s >
1. Thus, in a proof of a H¨older Traveling Salesman theorem,estimating the Hausdorff measure of approximating sets has no direct use.To overcome these obstacles, we reimagine the proof of the Analyst’s TST, and in § { I k } k ≥ of [0 ,
1] and a sequence ofpiecewise linear maps { f k : [0 , → R N } k ≥ that parameterize approximating graphs G k .Each map f k is built by carefully refining f k − to ensure that (cid:107) f k − f k − (cid:107) ∞ (cid:46) − k . Thisguarantees that the maps f k have a uniform limit f whose image contains the Hausdorfflimit of G k (and hence E ). To prove that f is H¨older continuous, one must estimategrowth of the Lipschitz constants of the maps f k (see Appendix B for the basic method).In §
4, we introduce a notion of mass of intervals I ∈ I k , defined using the s -power ofdiameters of images f l ( J ) of intervals J ⊂ I , l ≥ k . This lets us record estimates in thedomain of the map rather than its image, and in §
4, we provide a mass-centric analogueof (1.4) that is adapted to the H¨older setting. In turn, this lets us estimate the Lipschitzconstants of the maps f k and complete the proof of the H¨older Traveling Salesman theoremin §
5. For completeness, we use our method to reprove and strengthen the sufficient halfof the Analyst’s TST in § Wa ˙zewski type theorem for flat continua.
The Hahn-Mazurkiewicz Theorem(e.g. see [HY88, Theorem 3.30]) asserts that a set E ⊂ R N is a continuous image of [0 , MATTHEW BADGER, LISA NAPLES, AND VYRON VELLIS if and only if E is compact, connected, and locally connected. The Wa˙zewski Theorem(for an attribution, see [AO17]) asserts that E ⊂ R N is a Lipschitz image of [0 ,
1] if andonly if E is compact, connected, and H ( E ) < ∞ . It is an easy exercise to check thatevery (1 /s )-H¨older continuous image of [0 ,
1] is compact, connected, locally connected,and has H s ( E ) < ∞ , but the converse fails when s > § R N ?The method of proof of the H¨older Traveling Salesman theorems leads to the followingWa˙zewski type theorem for flat continua. For the proof of Proposition 1.3, see §
8. A set E ⊂ R n is called Ahlfors s -regular if there exist 0 < c ≤ C < ∞ such that(1.5) cr s ≤ H s ( E ∩ B ( x, r )) ≤ Cr s for all x ∈ E and 0 < r ≤ diam E. We say that E is lower (upper) Ahlfors s -regular if the first (second) inequality in (1.5)holds for all x ∈ E and 0 < r ≤ diam E . Proposition 1.3.
There exists a constant β ∈ (0 , such that if s > and E ⊂ R N iscompact, connected, H s ( E ) < ∞ , E is lower Ahlfors s -regular with constant c , and (1.6) β E (cid:0) B ( x, r ) (cid:1) ≤ β for all x ∈ E and < r ≤ diam E, then E = f ([0 , for some injective (1 /s ) -H¨older continuous map f : [0 , → R N withH¨older constant H (cid:46) s c − H s ( E )(diam E ) − s . Inclusion of lower Ahlfors regularity in the hypothesis of Proposition 1.3 is justifiable,because it holds automatically when s = 1, i.e. every non-trivial connected set is lowerAhlfors 1-regular. When s >
1, a non-trivial (1 /s )-H¨older curve is not necessarily lowerAhlfors s -regular, and, in fact, could have zero H s measure. Nevertheless, Mart´ın andMattila [MM93] proved that if Γ is a (1 /s )-H¨older curve in R N with H s (Γ) >
0, thenlim inf r ↓ H s (Γ ∩ B ( x, r )) r s > H s -a.e. x ∈ Γ . Even if it can be weakened, the lower regularity hypothesis in Proposition 1.3 cannot becompletely dropped: In § s > β ∈ (0 , E ⊂ R N with H s ( E ) < ∞ satisfying (1.6) such that E is not contained in a (1 /s )-H¨older curve.Sharp estimates on the Minkowski dimension of sets satisfying (1.6) were provided byMattila and Vuorinen [MV90]; for generalized Mattila-Vuorinen type sets , see [BL15].1.4.
Related Work.
As noted above, one motivation for this paper is to develop toolsto analyze the structure of Radon measures. See § , . This problem is still open, but some positive steps were recently takenby Azzam and Schul [AS18] for Hausdorff content lower regular sets . Also see [Vil18]. ¨OLDER CURVES AND PARAMETERIZATIONS 7
Acknowledgements.
We would like to thank Mika Koskenoja for providing us with acopy of M. Remes’ thesis [Rem98]. The first author would also like to thank Guy C.David for a useful conversation at an early stage of this project. We thank an anonymousreferee for their close and careful reading of the initial manuscript.
Part I. Proof of the H¨older Traveling Salesman Theorem
In the first part of the paper, §§ §
2, we introduce notationand essential concepts used in the proof, including nets, flat pairs, and variation excess.In §
3, we present a refined version of Jones’ Traveling Salesman construction, which takesa nested sequence ( V k ) ∞ k =0 of ρ k r -separated sets, approximating lines { (cid:96) k,v : v ∈ V k } ∞ k =0 ,and associated errors { α k,v : v ∈ V k } ∞ k =0 and outputs a sequence of partitions I k of [0 , f k such that f k ([0 , ⊃ V k . In §
4, we define and estimate adiscrete s -variation of the maps f k , which is adapted to the partitions I k of the domain.When s >
1, the total s -mass M s ([0 , f k fills therole that 1-dimensional Hausdorff measure H plays in Jones’ proof of the Analyst’s TST.In §
5, we use the algorithm of § § §
6, we use our method to obtain a stronger version of thesufficient half of the Analyst’s Traveling Salesman theorem. The construction presentedbelow can be carried out in any finite or infinite-dimensional Hilbert space.2.
Preliminaries
Given numbers x, y ≥ a , . . . , a n , we may write x (cid:46) a ,...,a n y if thereexists a positive and finite constant C depending on at most a , . . . , a n such that x ≤ Cy .We write x (cid:39) a ,...,a n y to denote x (cid:46) a ,...,a n y and y (cid:46) a ,...,a n x . Similarly, we write x (cid:46) y or x (cid:39) y to denote that the implicit constants are universal.2.1. Ordering flat sets.
The following lemma shows that if a discrete set is sufficientlyflat at the scale of separation, then there exists a natural linear ordering of its points.Estimates (2.1) and (2.2) are consequences of the Pythagorean theorem.
Lemma 2.1 ([BS17, Lemma 8.3]) . Suppose that V ⊂ R N is a -separated set with card( V ) ≥ and there exist lines (cid:96) and (cid:96) and a number α ∈ (0 , / such that dist( v, (cid:96) i ) ≤ α for all v ∈ V and i = 1 , . Let π i denote the orthogonal projection onto (cid:96) i . There exist compatible identifications of (cid:96) and (cid:96) with R such that π ( v ) ≤ π ( v (cid:48) ) if and only if π ( v ) ≤ π ( v (cid:48) ) for all v, v (cid:48) ∈ V .If v and v are consecutive points in V relative to the ordering of π ( V ) , then (2.1) H ([ u , u ]) ≤ (1 + 3 α ) · H ([ π ( u ) , π ( u )]) for all [ u , u ] ⊆ [ v , v ] . Moreover, (2.2) H ([ y , y ]) ≤ (1 + 12 α ) · H ([ π ( y ) , π ( y )]) for all [ y , y ] ⊆ (cid:96) . MATTHEW BADGER, LISA NAPLES, AND VYRON VELLIS
Suppose that V , (cid:96) , and π are given as in Lemma 2.1 and let v, v , v ∈ V . Givenan orientation of (cid:96) (that is, an identification of (cid:96) with R ), we say v is to the left of v and v is to the right of v if π ( v ) < π ( v ). We say v is between v and v if π ( v ) ≤ π ( v ) ≤ π ( v ) or π ( v ) ≤ π ( v ) ≤ π ( v ). Lemma 2.2.
Suppose that V ⊂ R N is a δ -separated set with card( V ) ≥ and there existsa line (cid:96) and a number α ∈ (0 , / such that dist( v, (cid:96) ) ≤ αδ for all v ∈ V .
Enumerate V = { v , . . . , v n } so that v i +1 is to the right of v i for all ≤ i ≤ n − . Then (2.3) n − (cid:88) i =1 | v i +1 − v i | s ≤ (1 + 3 α ) s | v − v n | s for all s ≥ . Moreover, if card( V ) ≥ , then (2.4) n − (cid:88) i =1 | v i +1 − v i | s ≤ ((1 + 3 α ) | v − v n | − δ ) s + δ s for all s ≥ . Proof.
Let π denote the orthogonal projection onto (cid:96) and put x i := π ( v i ). Then | x i +1 − x i | ≤ | v i +1 − v i | ≤ (1 + 3 α ) | x i +1 − x i | for all 1 ≤ i ≤ n − , where the first inequality holds since projections are 1-Lipschitz and the second inequalityholds by Lemma 2.1. Assume s ≥ V ) ≥
3. Then n − (cid:88) i =1 | v i +1 − v i | s (1 + 3 α ) s ≤ (cid:32) n − (cid:88) i =1 | x i +1 − x i | s (cid:33) + | x n − x n − | s ≤ (cid:32) n − (cid:88) i =1 | x i +1 − x i | (cid:33) s + | x n − x n − | s = ( | x n − x | − | x n − x n − | ) s + | x n − x n − | s ≤ (cid:18) | x n − x | − δ α (cid:19) s + (cid:18) δ α (cid:19) s ≤ (cid:18) | v n − v | − δ α (cid:19) s + (cid:18) δ α (cid:19) s , where the penultimate inequality holds because for any M > (cid:15) ∈ (0 , M ), and s ≥ f ( t ) = t s + ( M − t ) s defined on [ (cid:15), M − (cid:15) ] attains its maximum at t = (cid:15) . Thisestablishes (2.4). Inequality (2.3) follows from a similar (and easier) computation. (cid:3) Nets, flat pairs, and variation excess.
Let ( X, | · | ) denote the Hilbert space l ( R ) of square summable sequences or the Euclidean space R N for some N ≥ V = { ( V k , ρ k ) } k ≥ be a sequence of pairs of nonempty finite sets V k in X andnumbers ρ k >
0. Assume that there exist x ∈ X , r > C ∗ ≥
1, and 0 < ξ ≤ ξ < V satisfies the following properties. ¨OLDER CURVES AND PARAMETERIZATIONS 9 ( V0 ) When k = 0, we have ρ = 1. For all k ≥
0, we have ξ ρ k ≤ ρ k +1 ≤ ξ ρ k .( V1 ) When k = 0, we have V ⊂ B ( x , C ∗ r ).( V2 ) For all k ≥
0, we have V k ⊂ V k +1 .( V3 ) For all k ≥ v, v (cid:48) ∈ V k , we have | v − v (cid:48) | ≥ ρ k r .( V4 ) For all k ≥ v ∈ V k +1 , there exists v (cid:48) ∈ V k such that | v − v (cid:48) | < C ∗ ρ k +1 r .With C ∗ and ξ given, define the associated parameter A ∗ := C ∗ − ξ > C ∗ . In addition to (V0)–(V4), assume that for each k ≥ v ∈ V k we are given a number α k,v ≥ (cid:96) k,v in X such that( V5 ) sup x ∈ V k +1 ∩ B ( v, A ∗ ρ k r ) dist( x, (cid:96) k,v ) ≤ α k,v ρ k +1 r . We call the line (cid:96) k,v an approximating line at ( k, v ).The formulation of (V5) is motivated by [BS17, Proposition 3.6]. We remark that thenumber ρ k +1 r appearing on the right hand side of (V5) is the scale of separation of pointsin V k +1 . While we allow X = l ( R ), each V k can be identified with a subset of R N k forsome increasing sequence N k if convenient, because each V k is finite and V k ⊂ V k +1 . Lemma 2.3.
Let v ∈ V k be such that α k,v ≤ / and fix an orientation for (cid:96) k,v .(1) If v (cid:48) ∈ V k ∩ B ( v, A ∗ ρ k r ) is the first point to the right of v , then there exist fewerthan . C ∗ points of V k +1 between v and v (cid:48) (inclusive).(2) There exist fewer than . C ∗ points of V k +1 ∩ B ( v, C ∗ ρ k +1 r ) to the right of v .Proof. The points in V k +1 ∩ B ( v, A ∗ ρ k r ) are ρ k +1 r -separated and are linearly orderedby Lemma 2.1. Let v (cid:48) be the first point in V k ∩ B ( v, A ∗ ρ k r ) to the right of v andlet w , . . . , w m denote the points in V k +1 that lie between v and v (cid:48) (inclusive). By (V4),each point w i belongs to B ( v, C ∗ ρ k +1 r ) ∪ B ( v (cid:48) , C ∗ ρ k +1 r ). Let π k,v denote the orthogonalprojection onto (cid:96) k,v . By (V3) and (2.1), 1 . | π ( w i ) − π ( w j ) | > | w i − w j | ≥ ρ k +1 r forall distinct i, j , since (1 + 3(1 / ) < .
1. It follows that there are fewer than 1 . C ∗ points w i in V k +1 ∩ B ( v, C ∗ ρ k +1 r ) to the right of v and fewer than 1 . C ∗ points w i in V k +1 ∩ B ( v (cid:48) , C ∗ ρ k +1 r ) to the left of v (cid:48) . The first claim follows. A similar argument givesthe second claim. (cid:3) Definition 2.4 (flat pairs) . Fix a parameter α ∈ (0 , / k ≥
0, define
Flat ( k )to be the set of pairs ( v, v (cid:48) ) ∈ V k × V k such that(1) ρ k r ≤ | v − v (cid:48) | < A ∗ ρ k r ,(2) α k,v < α and v (cid:48) is the first point in V k ∩ B ( v, A ∗ ρ k r ) to the left or to the rightof v with respect to ordering induced by (cid:96) k,v .Define the corresponding set of geometric line segments L k = { [ v, v (cid:48) ] : ( v, v (cid:48) ) ∈ Flat ( k ) } .Note that the collection Flat ( k ) of flat pairs is not symmetric in the sense that ( v, v (cid:48) ) ∈ Flat ( k ) does imply ( v (cid:48) , v ) ∈ Flat ( k ), because α k,v does not control α k,v (cid:48) . Lemma 2.5.
Let e , e , e be distinct elements of L k for some k ≥ .(1) Edges e and e intersect at most in a common endpoint.(2) Edges e , e and e do not have a common point.Proof. Let e = [ v , v (cid:48) ], e = [ v , v (cid:48) ], and e = [ v , v (cid:48) ] represent distinct elements of L k , where ( v i , v (cid:48) i ) ∈ Flat ( k ) for all i ∈ { , , } . If two or more of the edges intersect in acommon point, say { e i : i ∈ I } for some I ⊂ { , , } with card( I ) ≥
2, then those edgesare contained in B ( v j , A ∗ ρ k r ) for each j ∈ I , since each edge has diameter at most14 A ∗ ρ k r . Note that V k is a ρ k r separated set, dist( v i , (cid:96) k,v j ) ≤ α k,v j ρ k +1 r < α k,v j ρ k r for all i, j ∈ I , and α k,v j ≤ /
16. Thus, by Lemma 2.1, the vertices { v i , v (cid:48) i : i ∈ I } are consistently linearly ordered according to their projections onto (cid:96) k,v j for each j ∈ I .Claims (1) and (2) follow immediately, since the segments in L k emanating from a vertex v ∈ V k with α k,v < α are only drawn to the first vertex in V k ∩ B ( v, A ∗ ρ k r ) to the leftor right of v with respect to the projection onto (cid:96) k,v . (cid:3) Given a pair ( v, v (cid:48) ) ∈ Flat ( k ), let V k +1 ( v, v (cid:48) ) denote the set of all points x ∈ V k +1 ∩ B ( v, A ∗ ρ k r ) such that x lies between v and v (cid:48) (including v and v (cid:48) ). Definition 2.6 (variation excess) . For all s ≥
1, for all k ≥
0, and for all ( v, v (cid:48) ) ∈ Flat ( k ),define the s -variation excess τ s ( k, v, v (cid:48) ) by τ s ( k, v, v (cid:48) ) | v − v (cid:48) | s = max (cid:40)(cid:32) n − (cid:88) i =1 | v i +1 − v i | s (cid:33) − | v − v (cid:48) | s , (cid:41) , where V k +1 ( v, v (cid:48) ) = { v , . . . , v n } with v = v , v n = v (cid:48) , and v i +1 is the first point to theright (or left) of v i for all 1 ≤ i ≤ n − Lemma 2.7.
For all k ≥ and ( v, v (cid:48) ) ∈ Flat ( k ) , we have τ ( k, v, v (cid:48) ) ≤ α k,v .Proof. Let V k +1 ( v, v (cid:48) ) = { v , . . . , v n } , where v = v , v n = v (cid:48) , and v i +1 is to the right of v i for all 1 ≤ i ≤ n −
1. By Lemma 2.2, with s = 1, n − (cid:88) i =1 | v i +1 − v i | ≤ (1 + 3 α k,v ) | v n − v | = (1 + 3 α k,v ) | v − v (cid:48) | . Rearranging the inequality gives τ ( k, v, v (cid:48) ) ≤ α k,v . (cid:3) We now demonstrate that when s >
1, the variation excess τ s ( k, v, v (cid:48) ) is zero wheneverthe set V k +1 ( v, v (cid:48) ) lies in a sufficiently thin tube. Lemma 2.8 (tube control) . For all s > , there exists (cid:15) s,C ∗ ,ξ ,ξ ∈ (0 , / such that if α k,v ≤ (cid:15) s,C ∗ ,ξ ,ξ , then τ s ( k, v, v (cid:48) ) = 0 for all ( v, v (cid:48) ) ∈ Flat ( k ) .Proof. Let ( v, v (cid:48) ) ∈ Flat ( k ) and enumerate V k +1 ( v, v (cid:48) ) = { v , . . . , v n } so that v = v , v n = v (cid:48) , and v i +1 is to the right of v i for all 1 ≤ i ≤ n −
1. If n = 2, then (cid:80) n − i =1 | v i +1 − v i | s = ¨OLDER CURVES AND PARAMETERIZATIONS 11 | v − v (cid:48) | s . Thus, suppose that n ≥
3. By Lemma 2.2, with δ = ρ k +1 r and α = α k,v , n − (cid:88) i =1 | v i +1 − v i | s ≤ | v − v (cid:48) | s (cid:20)(cid:18) (1 + 3 α k,v ) − ρ k +1 r | v − v (cid:48) | (cid:19) s + (cid:18) ρ k +1 r | v − v (cid:48) | (cid:19) s (cid:21) ≤ | v − v (cid:48) | s (cid:20)(cid:18) (1 + 3 α k,v ) − ξ A ∗ (cid:19) s + (cid:18) ξ A ∗ (cid:19) s (cid:21) =: A α k,v | v − v (cid:48) | s , because ξ − ρ k +1 r ≤ | v − v (cid:48) | ≤ A ∗ ξ − ρ k +1 r and the function f ( t ) = ((1 + 3 α k,v ) − t ) s + t s on [ ξ / A ∗ , ξ ]takes its maximum at t = ξ / A ∗ . Since s >
1, the coefficient A (cid:15) → (cid:18) − ξ A ∗ (cid:19) s + (cid:18) ξ A ∗ (cid:19) s < (cid:15) → . Thus, by continuity, there exists (cid:15) (cid:48) > A (cid:15) (cid:48) = 1. Let (cid:15) s,C ∗ ,ξ ,ξ = min { (cid:15) (cid:48) , / } . Then (cid:80) n − i =1 | v i +1 − v i | s ≤ | v − v (cid:48) | s whenever α k,v ≤ (cid:15) s,C ∗ ,ξ . (cid:3) Traveling Salesman algorithm
For the rest of §§ X denote the Hilbert space l ( R ) or R N , let ( V k ) k ≥ bea sequence of sets in X and let ρ k > § α ∈ (0 , /
16] in Definition 2.4. For eachinteger k ≥
0, we will construct(1) two collections of pairwise disjoint, open intervals in [0 ,
1] denoted by B k (called“bridge intervals”) and E k (“edge intervals”),(2) two collections of pairwise disjoint, nondegenerate closed intervals in [0 ,
1] denotedby F k (“frozen point intervals”) and N k (“non-frozen point intervals”), and(3) a continuous map f k : [0 , → X that satisfy the following properties. (P1) The four collections B k , E k , F k , N k are mutually disjoint and for any x ∈ [0 , I contained in their union such that x ∈ I . (P2) The map f k | I is affine on each I ∈ E k ∪ B k and the map f k | J is constant on each J ∈ F k ∪ N k . (P3) For all I ∈ E k , we have diam f k ( I ) < A ∗ ρ k r . (P4) The map f k | (cid:83) E k is 2-to-1; that is, for every x ∈ (cid:83) E k there exists a unique x (cid:48) ∈ (cid:83) E k \ { x } such that f k ( x ) = f k ( x (cid:48) ). (P5) If ( v, v (cid:48) ) ∈ Flat ( k ), then there exists I ∈ E k such that f k ( I ) joins v with v (cid:48) .Conversely, if a and b are endpoints of an interval I ∈ E k and α k,f k ( a ) < α , then( f k ( a ) , f k ( b )) ∈ Flat ( k ). (P6) For each I ∈ F k ∪ N k , the image f k ( I ) ∈ V k and for each v ∈ V k there exists aunique I ∈ N k such that f k ( I ) = v . (P7) If J ∈ N k is such that f k ( J ) is an endpoint of f k ( I ) for some I ∈ E k , then thereexists I (cid:48) ∈ E k (possibly I (cid:48) = I ) such that f k ( I (cid:48) ) = f k ( I ) and J ∩ I (cid:48) (cid:54) = ∅ . Lemma 3.1.
Assume (P5) holds at stage k ≥ . Let a < b < a (cid:48) < b (cid:48) be such that ( a, b ) and ( a (cid:48) , b (cid:48) ) belong to E k , f k ( b ) = f k ( a (cid:48) ) , and α k,f k ( b ) ≤ α . Then either(1) f k (( a, b )) = f k (( a (cid:48) , b (cid:48) )) or(2) f k ( b ) lies between f k ( a ) and f k ( b (cid:48) ) with respect to order induced by (cid:96) k,f k ( b ) .Proof. This is an immediate consequence of (P5) and Lemma 2.5. (cid:3) In § E , B , N , F , and f . In § E k +1 , B k +1 , N k +1 , F k +1 , and f k +1 in §§ § § Step . Fix a point v ∈ V . Let G be the (not necessarily connected) graph withvertices V and edges L . Suppose that G has components G (1)0 , . . . , G ( l )0 with v ∈ G (1)0 . Case 1.
Suppose that V = { v } . Then set E = ∅ , B = ∅ , N = { [0 , } and F = ∅ .Define also f : [0 , → X with f ( x ) = v for all x ∈ [0 , Case 2.
Suppose that card( V ) ≥ l = 1, that is, G is connected. We applyProposition A.1 for v with ∆ = [0 , G = G and we obtain a collection of intervals I and a continuous map g . By Lemma 2.5, each point v ∈ V has valence at most 2 in G and there exists a component J v of g − ( v ) such that if e is an edge of G that contains v as an endpoint, then e has a preimage I ∈ I such that I ∩ J v (cid:54) = ∅ . Let N be the collectionof all such intervals J v .Set E = I , B = ∅ , N = N , define F to be the components of [0 , \ (cid:83) ( E ∪ B ∪ N ),and let f = g . Properties (P1)–(P7) follow from Proposition A.1. Case 3.
Suppose that card( V ) ≥ l ≥
2, that is, G is disconnected.For each j = 2 , . . . , l fix a vertex u j of G ( j )0 . Let { I , . . . , I l − } be a collection of openintervals, enumerated according to the orientation of [0 , , { J , . . . , J l − } bethe components of I \ (cid:83) l − j =1 I j enumerated according to the orientation of [0 , G = G (1)0 , v and ∆ = J , we obtain a family of open intervals I , amap g : J → G (1)0 and a family N of closed intervals. Similarly, for each j = 2 , . . . , l ,applying Proposition A.1 for G = G ( j )0 , u j and ∆ = J j , we obtain a family of openintervals I j , a map g j : J j → G ( j )0 and a family N j of closed intervals. There exists acontinuous map g : [0 , → X that extends the maps g j such that(1) g ( J j +1 ) = v for each j ∈ { , . . . , l − } ;(2) g | I j is affine for each j ∈ { , . . . , l − } and g ( I j − ) = g ( I j ) = [ u j , v ] for each j ∈ { , . . . , l − } .Set E = (cid:83) lj =1 I j , B = { I , . . . , I l − } , N = (cid:83) lj =1 N j , define F to be the componentsof [0 , \ (cid:83) ( E ∪ B ∪ N ), and let f | [0 ,
1] = g . Properties (P1)–(P7) follow fromProposition A.1. ¨OLDER CURVES AND PARAMETERIZATIONS 13 Inductive hypothesis.
Suppose that for some k ≥ B k , E k of open intervals in [0 , F k , N k of nondegenerate closed intervals in[0 , f k : [0 , → X , which satisfy properties (P1)–(P7).We will define a new map f k +1 : [0 , → X and new collections B k +1 , E k +1 , F k +1 , N k +1 , B k +1 = (cid:91) I ∈ B k ∪ E k ∪ F k ∪ N k B k +1 ( I ) , E k +1 = (cid:91) I ∈ B k ∪ E k ∪ F k ∪ N k E k +1 ( I ) , F k +1 = (cid:91) I ∈ B k ∪ E k ∪ F k ∪ N k F k +1 ( I ) , N k +1 = (cid:91) I ∈ B k ∪ E k ∪ F k ∪ N k N k +1 ( I ) , where B k +1 ( I ), E k +1 ( I ), F k +1 ( I ), N k +1 ( I ) are collections of intervals in I that we definebelow. In particular: • In § f k +1 | I for I ∈ B k . • In § § f k +1 | I for I ∈ E k . • In § f k +1 | I for I ∈ F k . • In § § f k +1 | I for I ∈ N k .3.3. Step k + 1 : intervals in B k . For any I ∈ B k we set B k +1 ( I ) = { I } , E k +1 ( I ) = ∅ , F k +1 ( I ) = ∅ , N k +1 ( I ) = ∅ , and we define f k +1 | I = f k | I . In other words, bridge intervalsare frozen and we make no changes on them.3.4. Step k +1 : intervals in E k with a at least one endpoint with flat image. Herewe consider those intervals I = ( a I , b I ) ∈ E k such that one of the α k,f k ( a I ) , α k,f k ( b I ) is lessthan α . If no such interval exists, we move to § { I, I (cid:48) } where f k ( I ) = f k ( I (cid:48) )and f k ( I ) ∩ f k ( J ) = ∅ for all J ∈ E k \ { I, I (cid:48) } . Fix now such a pair { I, I (cid:48) } . We choose oneof the two intervals I, I (cid:48) to start with, say I .Without loss of generality, assume that α k,f k ( a I ) < α . Let (cid:96) be the approximating linefor ( k, f k ( a I )), oriented so that f k ( a I ) lies to the left of f k ( b I ). Let V k +1 ,I denote thepoints in V k +1 ∩ B ( f k ( a I ) , A ∗ ρ k r ) that lie between f k ( a I ) and f k ( b I ) with respect to (cid:96) ,including f k ( a I ) and f k ( b I ). Enumerate V k +1 ,I from left to right, V k +1 ,I = { v , . . . , v l } . That is, v i lies to the left of v i +1 for all i ∈ { , . . . , l − } , v = f k ( a I ), and v l = f k ( b I ). Remark 3.2.
By Lemma 2.3, we have that l < . C ∗ .Let { I , . . . , I l − } be a collection of open intervals in I with mutually disjoint closures,enumerated according to the orientation of [0 ,
1] so that the left endpoint of I coincideswith a I and the right endpoint of I l − coincides with b I . Let N k +1 ( I ) be the componentsof [0 , \ (cid:83) l − i =1 I i and F k +1 ( I ) = ∅ . Define E k +1 ( I ) = { I i : | v i − v i +1 | < A ∗ ρ k +1 r } , B k +1 ( I ) = { I i : | v i − v i +1 | ≥ A ∗ ρ k +1 r } . Then define f k +1 | I continuously so that Figure 1.
The image of f k +1 ( I ): black arrows denote images of intervalsin E k ∪ B k , green arrows denote images of intervals in E k +1 , and red arrowsdenote images of intervals in B k +1 .(1) f k +1 is affine on each J ∈ E k +1 ( I ) ∪ B k +1 ( I ) and constant on each J ∈ N k +1 ( I ) ∪ F k +1 ( I );(2) for each j = 1 , . . . , l − f k +1 ( I j ) = [ v j , v j +1 ] mapping the left endpoint of I j onto v j and the right endpoint of I j onto v j +1 .See Figure 1 for the image of f k +1 ( I ).Once we have defined the four families and f k +1 for I , we work as follows for I (cid:48) . Firstnote that V k +1 ,I = V k +1 ,I (cid:48) . Define ψ I (cid:48) ,I : I (cid:48) → I to be the unique orientation-reversinglinear map between I (cid:48) and I . Define E k +1 ( I (cid:48) ) = { ψ I (cid:48) ,I ( J ) : J ∈ E k +1 ( I ) } and B k +1 ( I (cid:48) ) = { ψ I (cid:48) ,I ( J ) : J ∈ B k +1 ( I ) } This time, however, we set F k +1 ( I (cid:48) ) to be the components of I (cid:48) \ (cid:83) l − i =1 I (cid:48) i and N k +1 ( I (cid:48) ) = ∅ .Define also f k +1 | I (cid:48) continuously so that f k +1 | I (cid:48) = ( f k +1 | I ) ◦ ψ I (cid:48) ,I . Lemma 3.3.
For i = 1 , , let I i ∈ E k be an interval with at least endpoint having flatimage and let I (cid:48) i ∈ E k +1 ( I i ) . If f k ( I ) (cid:54) = f k ( I ) , then f k +1 ( I (cid:48) ) ∩ f k +1 ( I (cid:48) ) = ∅ .Proof. Suppose that I (cid:48) ∈ E k +1 ( I ), I (cid:48) ∈ E k +1 ( I ), and f k +1 ( I (cid:48) ) ∩ f k +1 ( I (cid:48) ) (cid:54) = ∅ . Because f k +1 ( I (cid:48) ) and f k +1 ( I (cid:48) ) do not include their endpoints (since intervals in E k +1 are open), weconclude that f k +1 ( I (cid:48) ) = f k +1 ( I (cid:48) ) by Lemma 2.5. Now, the endpoints of the four intervals f k ( I ), f k ( I ), f k +1 ( I (cid:48) ), and f k +1 ( I (cid:48) ) lie in the 30 A ∗ ρ k r neighborhood of any flat endpointof f k ( I ) or f k ( I ). In particular, the endpoints of the four intervals are linearly orderedby Lemma 2.1, and the endpoints of f k +1 ( I (cid:48) i ) lie between the endpoints of f k ( I i ) by theconstruction of E k +1 ( I i ). Because f k +1 ( I (cid:48) ) = f k +1 ( I (cid:48) ), this forces f k ( I ) = f k ( I ). (cid:3) Step k + 1 : intervals in E k with no endpoints with flat image. Suppose that I = ( a I , b I ) ∈ E k is such that α k,f k ( a I ) ≥ α and α k,f k ( b I ) ≥ α . Then set E k +1 ( I ) = ∅ , B k +1 ( I ) = { I } , N k +1 ( I ) = ∅ , F k +1 ( I ) = ∅ , and f k +1 | I = f k | I . In other words, edgeintervals with no endpoints with flat image become bridge intervals and remain bridgeintervals for the rest of the construction.3.6. Step k + 1 : intervals in F k . For any I ∈ F k we set E k +1 ( I ) = ∅ , B k +1 ( I ) = ∅ , F k +1 ( I ) = { I } , N k +1 ( I ) = ∅ and we set f k +1 | I = f k | I . In other words, frozen pointintervals in F k remain frozen for the rest of the construction. ¨OLDER CURVES AND PARAMETERIZATIONS 15 Step k + 1 : intervals in N k with flat image. We now consider the intervals I ∈ N k for which α k,f k ( I ) < α . If no such interval exists we proceed to § I be such an interval and let (cid:96) be the approximating linefor ( k, f k ( I )). We consider three cases.3.7.1. Non-terminal vertices.
Suppose that there exist distinct v, v (cid:48) ∈ V k \ { f k ( I ) } suchthat ( f k ( I ) , v ) and ( f k ( I ) , v (cid:48) ) are in Flat ( k ). By (P5), Lemma 3.1 and the induction step,there exist J, J (cid:48) ∈ E k such that f k ( I ) and v are the endpoints of f k ( J ), while f k ( I ) and v (cid:48) are the endpoints of f k ( J (cid:48) ). Hence, all points of V k +1 between v and v (cid:48) are contained inthe image of f k +1 ( J ∪ J (cid:48) ) defined in § E k +1 ( I ) = ∅ , B k +1 ( I ) = ∅ , N k +1 ( I ) = { I } , F k +1 ( I ) = ∅ and f k +1 | I = f k | I .3.7.2. Suppose that there exists unique v ∈ V k \ { f k ( I ) } suchthat ( f k ( I ) , v ) ∈ Flat ( k ). Fix an orientation for (cid:96) so that v lies to the left of f k ( I ). Asin § V k +1 that lie between f k ( I ) and v are all contained in f k +1 ( J )for some J ∈ E k . Let V k +1 ,I denote the set that includes f k ( I ) and all points in V k +1 ∩ B ( f k ( I ) , C ∗ ρ k +1 r ) that lie to the right of f k ( I ). Enumerate V k +1 ,I = { v , . . . , v l } fromleft to right. That is, v = f k ( I ) and v l is the rightmost point of V k +1 ,I Remark 3.4. card V k +1 ,I ≤ . C ∗ by Lemma 2.3.If V k +1 ,I = { f k ( I ) } , then set E k +1 ( I ) = ∅ , B k +1 ( I ) = ∅ , N k +1 ( I ) = { I } , F k +1 ( I ) = ∅ and f k +1 | I = f k | I .If V k +1 ,I (cid:54) = { f k ( I ) } , then let G k +1 ,I be the graph with vertices the points in V k +1 ,I andedges the segments { [ v , v ] , . . . , [ v l − , v l ] } . That is, G k +1 ,I forms a simple polygonal arcto the right of f k ( I ) joining f k ( I ) with v l . Let I and g be the collection and map givenby Proposition A.1 for ∆ = I , G = G k +1 ,I and v = f k ( I ). For each v (cid:48) ∈ V k +1 ,I fix acomponent of I \ (cid:83) I k +1 ( I ) that is mapped onto v (cid:48) and let N be the collection of thesecomponents. Set E k +1 ( I ) = I , B k +1 ( I ) = ∅ , N k +1 ( I ) = N , and define F k +1 ( I ) to be theset of components of I \ (cid:83) ( E k +1 ( I ) ∪ B k +1 ( I ) ∪ N k +1 ( I )). Set f k +1 | I = g . See the lefthalf of Figure 2 for the image of f k +1 ( I ).3.7.3. Suppose that there exists no point in V k \ { f k ( I ) } suchthat ( f k ( I ) , v ) ∈ Flat ( k ). That is, V k ∩ B ( f k ( I ) , A ∗ ρ k r ) = { f k ( I ) } . Set V k +1 ,I = V k +1 ∩ B ( f k ( I ) , C ∗ ρ k +1 r ) . Fix an orientation for (cid:96) and enumerate V k +1 ,I = { v , . . . , v l } from left to right. Remark 3.5. card( V k +1 ,I ) ≤ . C ∗ by Lemma 2.3.If V k +1 ,I = { f k ( I ) } , then set E k +1 ( I ) = ∅ , B k +1 ( I ) = ∅ , N k +1 ( I ) = { I } , F k +1 ( I ) = ∅ and f k +1 | I = f k | I . If V k +1 ,I (cid:54) = { f k ( I ) } , then let G k +1 ,I be the graph with vertices thepoints in V k +1 ,I and edges the segments { [ v , v ] , . . . , [ v l − , v l ] } . The remainder of theconstruction proceeds in the same way as in § f k +1 ( I ). Figure 2.
The image of f k +1 ( I ): on the left, we have f k +1 ( I ), where I isas in § f k +1 ( I ), where I is as in § Lemma 3.6.
Let I , I ∈ N k be distinct intervals as in § I ∈ E k be an intervalas in § I (cid:48) i ∈ E k +1 ( I i ) for i = 1 , , , then the segments f k +1 ( I (cid:48) ) , f k +1 ( I (cid:48) ) , and f k +1 ( I (cid:48) ) are mutually disjoint.Proof. This follows from similar arguments employed in the proof of Lemma 3.3. (cid:3)
Step k + 1 : intervals in N k with non-flat image. In this final part of thealgorithm, we define E k +1 ( I ), B k +1 ( I ), N k +1 ( I ), F k +1 ( I ) and f k +1 | I for those I ∈ N k suchthat α k,f k ( I ) ≥ α . Let { I , . . . , I n } be an enumeration of such intervals. The constructionin this case resembles that in Step 0.We start with I . Let V k +1 ,I be the set of points in V k +1 ∩ B ( f k ( I ) , C ∗ ρ k +1 r ) that arenot images of some I ∈ N k +1 defined in § § L k +1 ,I be the set of edges in L k +1 that have an endpoint in V k +1 ,I . Then define ˜ V k +1 ,I to be the union of V k +1 ,I andthe set of all endpoints of edges in L k +1 ,I . By the triangle inequality, the set ˜ V k +1 ,I issubset of B ( v, A ∗ ρ k +1 r ). Finally, let G k +1 ,I denote the graph with vertices ˜ V k +1 ,I andwith edges L k +1 ,I . We note that the graph G k +1 ,I may be connected or disconnected.If ˜ V k +1 ,I = { f k ( I ) } , then we simply set E k +1 ( I ) = ∅ , B k +1 ( I ) = ∅ , N k +1 ( I ) = { I } , F k +1 ( I ) = ∅ and f k +1 | I = f k | I .For the remainder of § V k +1 ,I contains at least two points. Let G (1) k +1 ,I , . . . , G ( l ) k +1 ,I denote the connected components of G k +1 ,I , labeled so that G (1) k +1 ,I is the component containing f k ( I ). There are two cases.3.8.1. Connected graph.
Suppose that G k +1 ,I is connected. Apply Proposition A.1 for∆ = I , G = G k +1 ,I and v = f k ( I ) to obtain a collection of intervals I and a continuousmap g . If v (cid:48) ∈ V k +1 ,I , then v (cid:48) has valence at most 2 by Lemma 2.5. Hence, by PropositionA.1, there exists a component J v (cid:48) of g − ( v (cid:48) ) with the following property:If v (cid:48) is the endpoint of some e ∈ L k +1 ,I , then there exists I ∈ I such that g ( I ) = e and I ∩ J v (cid:48) (cid:54) = ∅ . ¨OLDER CURVES AND PARAMETERIZATIONS 17 Let N be the collection of the fixed intervals J v (cid:48) where v (cid:48) ∈ ˜ V k +1 ,I . Now define a set E ⊆ I with two rules:(1) If e ∈ L k +1 ,I has both its endpoints in V k +1 ,I then both components of g − ( e )are in E .(2) If e ∈ L k +1 ,I has one endpoint v (cid:48) ∈ V k +1 ,I and another in ˜ V k +1 ,I \ V k +1 ,I , thenonly one component of g − ( e ) (one that intersects J v (cid:48) ) is in E .Set E k +1 ( I ) = E , B k +1 ( I ) = I \ E , N k +1 ( I ) = N , and define F k +1 ( I ) to be the setof components of I \ (cid:83) ( E k +1 ( I ) ∪ B k +1 ( I ) ∪ N k +1 ( I )) and f k +1 | I = g .3.8.2. Several components.
Suppose that l ≥
2; that is, G k +1 ,I is disconnected. SeeFigure 3 for the image of f k +1 ( I ). We will add some edges which will make the graphconnected and the preimage of these edges will be bridge intervals. To this end, for each j ∈ { , . . . , l } fix some point v j ∈ V k +1 ,I ∩ G ( j ) k +1 ,I . Let { I , , . . . , I , l − } be a collectionof open intervals, enumerated according to the orientation of [0 , I . Let also { J , , . . . , J , l − } be the components of I \ (cid:83) l − j =1 I ,j enumerated according to the orientation of [0 , § I of open intervals in J , , a subset E ⊂ I ,a family N containing some components of J , \ (cid:83) I and a continuous map g : J , → G (1) k +1 ,I . Similarly, for each j ∈ { , . . . , l } we obtain a family I j of open intervals in J , j − , a subset E j ⊂ I j , a family N j containing some components of J , j − \ (cid:83) I j anda continuous map g j : J , j − → G ( j ) k +1 ,I . There exists a continuous map g : I → X thatextends the maps g j such that(1) g ( J , j +1 ) = f k ( I ) for each j ∈ { , . . . , l − } ;(2) g | I ,j is affine for all j ∈ { , . . . , l − } and g ( I , j − ) = g ( I , j ) = [ v j , f k ( I )] forall j ∈ { , . . . , l − } .Define edge intervals E k +1 ( I ) = (cid:83) l j =1 E j and bridge intervals B k +1 ( I ) = (cid:83) l − j =1 I ,j ∪ (cid:83) l j =1 ( I j \ E j ). Set N k +1 ( I ) = (cid:83) j ∈{ , , ,..., l } N j and define F k +1 ( I ) to be the set ofcomponents of I \ (cid:83) ( E k +1 ( I ) ∪ B k +1 ( I ) ∪ N k +1 ( I )). Also, set f k +1 | I = g .3.8.3. Inductive hypothesis.
Inductively, suppose that for i ∈ { , . . . , r − } we have de-fined E k +1 ( I i ), B k +1 ( I i ), F k +1 ( I i ), N k +1 ( I i ) and f k +1 | I i . We work now for I r . Let V k +1 ,I r be the set of points in V k +1 ∩ B ( f k ( I ) , C ∗ ρ k +1 r ) that are not images of some I ∈ N k +1 defined in § § I , . . . , I r − . Let L k +1 ,I r be the setof edges in L k +1 that have an endpoint in V k +1 ,I r and let ˜ V k +1 ,I r be the set of endpoints ofedges in L k +1 ,I r . Let now G k +1 ,I r be the (not necessarily connected) graph with verticesthe set ˜ V k +1 ,I r and with edges the set L k +1 ,I r . To continue, repeat the procedure carriedout for I mutatis mutandis . Remark 3.7.
By the choice of set N for I , it follows that if I ∈ N k +1 ( I ) and if f k +1 ( I )is the endpoint of f k +1 ( J ) for some J ∈ E k +1 ( I ), then there exists J (cid:48) ∈ E k +1 ( I ) (possibly J (cid:48) = J ) such that f k +1 ( J ) = f k +1 ( J (cid:48) ) and I ∩ J (cid:48) (cid:54) = ∅ . The same is true for all I j . Figure 3.
The image of f k +1 ( I ): Blue segments represent edges in L k +1 ,black arrows represent images of intervals in E k ∪ B k , green arrows representimages of intervals in E k +1 ( I ) and red arrows represent images of intervalsin B k +1 ( I ). Lemma 3.8.
Let J ∈ N k be as in § J ∈ E k be as in § J ∈ N k be asin § J (cid:48) i ∈ E k +1 ( J i ) for i = 1 , , , then the segments f k +1 ( J (cid:48) ) , f k +1 ( J (cid:48) ) , f k +1 ( J (cid:48) ) aremutually disjoint.Proof. By Lemma 3.6 we know that f k +1 ( J (cid:48) ) and f k +1 ( J (cid:48) ) are disjoint. Fix an interval J ∈ N k as in § J ∈ E k is as in § J ∈ N k is as in § J (cid:48) ∈ E k +1 ( J ) and J (cid:48) ∈ E k +1 ( J (cid:48) ). By Lemma 2.5, either f k +1 ( J (cid:48) ) ∩ f k +1 ( J (cid:48) ) (cid:54) = ∅ or f k +1 ( J (cid:48) ) = f k +1 ( J (cid:48) ). However, we have defined L k +1 ,J as those elements in L k +1 thatare not contained in f k +1 ( J ), where J ∈ E k is as in § J ∈ N k is as in § f k +1 ( J (cid:48) ) ∩ f k +1 ( J (cid:48) ) = ∅ . (cid:3) Properties (P1)–(P7) for Step k + 1 . We have now defined B k +1 , E k +1 , F k +1 , N k +1 and f k +1 : [0 , → X . It remains to prove that f k +1 is continuous and that properties(P1)–(P7) are satisfied by the new collections of intervals and f k +1 . Properties (P1), (P2),and (P3) follow immediately from the construction. Continuity of f k +1 . By design, the map f k +1 is continuous on every point interior toan interval in E k ∪ B k ∪ N k ∪ F k . If x is an endpoint of some interval in E k ∪ B k ∪ N k ∪ F k ,then f k +1 ( x ) = f k ( x ). Thus, continuity of f k +1 at x follows from continuity of f k at x . Property (P6).
The first claim of (P6), that f k +1 ( I ) ∈ V k +1 for all I ∈ N k +1 ∪ F k +1 ,is immediate from the construction. To check the second claim of (P6), fix v ∈ V k +1 . By(V4), there exists v (cid:48) ∈ V k such that | v − v (cid:48) | < C ∗ ρ k +1 r . By the inductive step, thereexists I ∈ N k such that f k ( I ) = v (cid:48) . There are two cases. ¨OLDER CURVES AND PARAMETERIZATIONS 19 Case 1.
Suppose that α k,v (cid:48) < α . Then, following the discussion in § v = f k +1 ( J ) for some J ∈ N k ( J (cid:48) ) and J (cid:48) ∈ E k as in § v = f k +1 ( J ) for some J ∈ N k ( I ). Case 2.
Suppose that α k,v (cid:48) ≥ α . Following the construction of the graph G k +1 ,I and the design of the map f k +1 | I , if v is not the image of some J ∈ N k +1 ( J (cid:48) ), where J (cid:48) ∈ E k ∪ B k ∪ N k \ { J (cid:48) } , then there exists J ∈ N k +1 ( I ) such that v = f k +1 ( J ). Property (P4).
Fix I ∈ E k +1 . There are three cases. Case 1.
Suppose that I ∈ E k +1 ( I ) for some I ∈ E k . By the inductive hypothesis,there exists unique I (cid:48) ∈ E k \ { I } such that f k ( I (cid:48) ) = f k ( I ) while f k ( I ) ∩ f k ( J ) = ∅ for all J ∈ E k \ { I , I (cid:48) } . By construction, there exists I (cid:48) ∈ E k +1 ( I (cid:48) ) such that f k +1 ( I ) = f k +1 ( I (cid:48) ).Again by construction, f k +1 ( I ) ∩ f k +1 ( J ) = ∅ for all J ∈ E k +1 ( I ) ∪ E k +1 ( I (cid:48) ) \ { I, I (cid:48) } . ByLemma 3.3, Lemma 3.6 and Lemma 3.8, f k +1 ( I ) does not intersect any f k +1 ( J ) for any J ∈ E k +1 ( J (cid:48) ) and J (cid:48) ∈ E k ∪ N k \ { I , I (cid:48) } . Case 2.
Suppose that I ∈ E k +1 ( I ) for some I ∈ N k as in § I (cid:48) ∈ E k +1 ( I ) \ { I } such that f k +1 ( I ) = f k +1 ( I (cid:48) ) while f k +1 ( I ) ∩ f k +1 ( J ) = ∅ forall J ∈ E k +1 ( I ) \ { I, I (cid:48) } . Moreover, by Lemma 3.6 and Lemma 3.8, f k +1 ( I ) does notintersect any f k +1 ( J ) for any J ∈ E k +1 ( J (cid:48) ) and J (cid:48) ∈ E k ∪ N k \ { I } . Case 3.
Suppose that I ∈ E k +1 ( I ) for some I ∈ N k as in § f k +1 ( I ) ∩ f k +1 ( I (cid:48) ) = ∅ for all I (cid:48) ∈ E k +1 ( J ) and all J ∈ E k as in § J ∈ N k as in § f k +1 | I , there are two possibilities. Case 3a.
Suppose that both endpoints of f k +1 ( I ) are in V k ,I . Then there exists aninterval I (cid:48) ∈ E k +1 ( I ) \ { I } such that f k +1 ( I ) = f k +1 ( I (cid:48) ). On the other hand, f k +1 ( I ) (cid:54)∈ L k +1 ,J for any J ∈ N k \ { I } . Thus, by Lemma 2.5, if J (cid:48) ∈ E k +1 ( J ) and J ∈ N k \ { I } ,then f k +1 ( J ) ∩ f k +1 ( J (cid:48) ) = ∅ . Case 3b.
Suppose that only one endpoint of f k +1 ( I ) is in V k ,I . In this case, byconstruction, f k +1 ( I ) ∩ f k +1 ( J ) = ∅ for all J ∈ E k +1 ( I ) \{ I } . Moreover, there exists unique I (cid:48) ∈ N k \ { I } as in § V k +1 ,I (cid:48) contains the other endpoint of f k +1 ( I ). As with I , there exists unique I (cid:48) ∈ E k +1 ( I (cid:48) ) such that f k +1 ( I (cid:48) ) = f k +1 ( I ) while f k +1 ( J ) ∩ f k +1 ( I ) = ∅ for all J ∈ E k +1 ( I (cid:48) ). Finally, by the construction and Lemma 2.5, f k +1 ( I ) ∩ f k +1 ( J ) = ∅ for all J ∈ E k +1 ( J (cid:48) ) and all J (cid:48) ∈ N k \ { I , I (cid:48) } as in § Property (P5).
To prove the first claim in (P5), fix ( v, v (cid:48) ) ∈ Flat ( k + 1). Let v bethe point of V k closest to v and let I ∈ N k be such that f k ( I ) = v . There are four cases. Case 1.
Suppose that α k,v < α and v is non-terminal (see § v and v (cid:48) lie to the left of v (with respect to (cid:96) − k, v ) or both lie to the right of v . Inany case, [ v, v (cid:48) ] is the preimage of some I ∈ E k +1 ( J ) under f k +1 where J ∈ E k and f k ( J )is an edge with endpoint f k ( I ). Case 2.
Suppose that α k,v < α and v is 2-sided terminal (see § v, v (cid:48) ] isthe preimage of some I ∈ E k +1 ( I ) under f k +1 . Case 3.
Suppose that α k,v < α and v is 1-sided terminal (see § v and v (cid:48) lie to the left of v (with respect to (cid:96) k,v ) or both lie to the right of v .Depending on their position, we work as in Case 1 or Case 2. Case 4.
Suppose that α k,v ≥ α . By definition of graph G k +1 ,I in § v, v (cid:48) ] is the image of some J ∈ E k +1 under f k +1 .To prove the second claim of (P5), fix ( a, b ) ∈ E k +1 such that one of its endpoints hasflat image. Without loss of generality, assume α k +1 ,f k +1 ( a ) < α . Case 1.
Suppose that ( a, b ) ∈ E k +1 ( I ) for some I ∈ N k as in § f k +1 on such intervals, ( f k +1 ( a ) , f k +1 ( b )) ∈ Flat ( k + 1). Case 2.
Suppose that ( a, b ) ∈ E k +1 ( I ) for some I ∈ N k as in § f k +1 on such intervals, no point of V k +1 ∩ B ( f k +1 ( a ) , A ∗ ρ k +1 r ), V k +1 ∩ B ( f k +1 ( a ) , A ∗ ρ k +1 r ) ⊂ V k +1 ∩ B ( f k ( I ) , A ∗ ρ k r ) , lies strictly between f k +1 ( a ) and f k +1 ( b ) with respect to (cid:96) k,f k ( I ) . The same is true withrespect to (cid:96) k +1 ,f k +1 ( a ) by Lemma 2.1. Thus, ( f k +1 ( a ) , f k +1 ( b )) ∈ Flat ( k + 1). Case 3.
Suppose that ( a, b ) ∈ E k +1 ( I ) for some I ∈ E k as in § Property (P7).
To check the final property, fix J ∈ N k +1 and choose I ∈ E k +1 suchthat f k +1 ( J ) is an endpoint of f k +1 ( I ). There are several cases. Case 1.
Suppose J ∈ N k +1 ( J ) for some J ∈ N k as in § I (cid:48) ∈ E k +1 ( J ) such that f k +1 ( I (cid:48) ) = f k +1 ( I ). By the construction of N k +1 ( J ) in § I (cid:48) ∩ J (cid:54) = ∅ . Case 2.
Suppose J ∈ N k +1 ( J ) for some J ∈ E k as in § I (cid:48) ∈ E k +1 ( J ) such that f k +1 ( I (cid:48) ) = f k +1 ( I ). The interval I (cid:48) satisfies I (cid:48) ∩ J (cid:54) = ∅ . Case 3.
Suppose J ∈ N k +1 ( J ) for some J (cid:48) ∈ N k as in § Case 3a.
Suppose that f k +1 ( J ) (cid:54) = f k ( J ). Then by the choice of N k +1 ( J ), there exists I (cid:48) ∈ E k +1 ( J ) such that f k +1 ( I (cid:48) ) = f k +1 ( I ) and I (cid:48) ∩ J (cid:54) = ∅ . Case 3b.
Suppose that f k +1 ( J ) = f k ( J ) and there exists ˜ I ∈ E k +1 ( J ) such that f k +1 ( ˜ I ) = f k +1 ( I ). As in Case 3a, the claim follows from the choice of N k +1 ( J ). Case 3c.
Suppose that f k +1 ( J ) = f k ( J ) and there exists no ˜ I ∈ E k +1 ( J ) such that f k +1 ( ˜ I ) = f k +1 ( I ). In this case, f k ( J ) is the endpoint of f k ( I ) for some I ∈ E k . By theinductive hypothesis and (P4), there exists at least one and at most two intervals I (cid:48) ∈ E k such that f k ( I ) = f k ( I (cid:48) ) and I (cid:48) ∩ J (cid:54) = ∅ . On one hand, if there is only one interval I (cid:48) ,then J is as in § J = J . Hence there exists I (cid:48) ∈ I (cid:48) such that f k +1 ( I (cid:48) ) = f k +1 ( I )and I (cid:48) ∩ J (cid:54) = ∅ . On the other hand, if there are two intervals I (cid:48) , I (cid:48)(cid:48) , then one of them has aclosure which intersects J , say I (cid:48) . Then there exists I (cid:48) ∈ I (cid:48) such that f k +1 ( I (cid:48) ) = f k +1 ( I )and I (cid:48) ∩ J (cid:54) = ∅ .3.10. Choices in the Traveling Salesman algorithm. In §§ α ∈ (0 , /
16] determines the set
Flat ( k ) of flat pairs. The constant14 in the definition of Flat ( k ) is chosen to facilitate the estimates in § α k,v is chosen tobe larger than (1 + 3(1 / ) · ·
14. For example, see the proof of Lemma 2.5. ¨OLDER CURVES AND PARAMETERIZATIONS 21 (C1) If
I, I (cid:48) ∈ E k satisfy f k ( I ) = f k ( I (cid:48) ) and are as in § I has at least oneendpoint x with α k,f k ( x ) < α ), then either N k +1 ( I ) = F k +1 ( I (cid:48) ) = ∅ or vice-versa.(C2) If I ∈ N k is as in § α k,f k ( I ) < α and V k ∩ B ( f k ( I ) , A ∗ ρ k r ) = { f k ( I ) } ),then there may exist up to two different ways to parameterize the graph G k +1 ,I therein.(C3) If I ∈ N k is as in § α k,f k ( I ) ≥ α ) and G (1) k +1 ,I , . . . , G ( l ) k +1 ,I are the graphcomponents of the graph G k +1 ,I therein, then(C3a) we choose the order in which we parameterize the graph components and(C3b) in each graph component, there exists up to two choices of parameterization.Similar choices are made in the step 0.(C4) We get to choose the enumeration of intervals I in N k such that α k,f k ( I ) ≥ α .The algorithm can be made more flexible by permitting four additional choices. Let˜ α ∈ (0 , α ) and ˜ A > A ∗ .(C5) Suppose that I ∈ N k . • If α k,f k ( I ) < ˜ α then we treat I as in § f k ( I ) as a flat vertex. • If α k,f k ( I ) ≥ α then we treat I as in § f k ( I ) as a non-flatvertex. • If α k,f k ( I ) ∈ [ ˜ α , α ) then we can either treat I as in § § v ∈ V k is chosen to be considered “flat” by (C5). Let (cid:96) be theapproximating line for ( k, v ) and let v (cid:48) ∈ V k be such that there exists no v (cid:48)(cid:48) ∈ V k ∩ B ( v, ˜ Aρ k r ) such that π (cid:96) ( v (cid:48)(cid:48) ) is between v and v (cid:48) . • If | v − v (cid:48) | < A ∗ ρ k r , then ( v, v (cid:48) ) ∈ Flat ( k ). • If | v − v (cid:48) | ≥ ˜ Aρ k r , then ( v, v (cid:48) ) (cid:54)∈ Flat ( k ). • If | v − v (cid:48) | ∈ [14 A ∗ ρ k r , ˜ Aρ k r ), then we are free to choose whether ( v, v (cid:48) ) iscontained in Flat ( k ) or not.(C7) Similarly to (C6), suppose that I ∈ E k is as in § f k ( I ) has at least oneendpoint x whose image is “flat” by (C5). Let { v , . . . , v l } and { I , . . . , I l − } beas in § • If | v i − v i +1 | < A ∗ ρ k +1 r , then we set I i ∈ E k +1 . • If | v i − v i +1 | ≥ ˜ Aρ k +1 r , then we set I i ∈ B k +1 . • If | v i − v i +1 | ∈ [14 A ∗ ρ k +1 r , ˜ Aρ k +1 r ), then we can choose in each instancewhether I i ∈ E k +1 ( I ) or I i ∈ B k +1 ( I ).(C8) Suppose that { I , . . . , I n } are the intervals in N k that have a non-flat image.Suppose also that we have defined f k +1 on I , . . . , I r − and on intervals in N k thathave an image chosen to be flat. Let v ∈ V k +1 be a point which is not the imageof some I ∈ N k +1 ( J ), where J ∈ { I , . . . , I r − } or J ∈ N k is as in § • If v ∈ B ( f k ( I r ) , A ∗ ρ k r ), then v ∈ V k +1 ,I r . • If v ∈ B ( f k ( I r ) , ˜ Aρ k r ) \ B ( f k ( I r ) , A ∗ ρ k r ), then we may choose whether v ∈ V k +1 ,I r or not.Note that the (C5) has subsequent implications on the treatment of intervals I ∈ E k .For instance, if both endpoints if I have images chosen to be non-flat, then I ∈ B k +1 ; otherwise, we treat I as in § L k and together with(C8) affects the parametrization near non-flat vertices. Remark 3.9 (coherence) . In the original, non-parametric Analyst’s Traveling Salesmanconstruction, Jones [Jon90] required the coherence property (V2), i.e. V k ⊂ V k +1 . The firstauthor and Schul [BS17] established a non-parametric Traveling Salesman construction,which replaced (V2) with the weaker property that for all v ∈ V k , the set v (cid:48) ∈ V k +1 ∩ B ( v, C ∗ ρ k r )is nonempty. This relaxation was crucial for the proof of the main result in [BS17], whichcharacterized Radon measures in R N that are carried by rectifiable curves. We would liketo emphasize that in the parametric Traveling Salesman construction described above,we heavily rely on (V2). At this time, we do not know how to build a parameterizationunder the relaxed condition of [BS17].4. Mass of intervals
In this section, we use the construction of §
3, to assign mass to intervals defined in § M s on the domain of the maps fills the role that the Hausdorff measure H of the image plays in the proof of the sufficient half of the Analyst’s TST given in[Jon90] or [BS17]. The main result of this section is Proposition 4.11, which bounds thetotal mass of [0 ,
1] by a sum involving the flatness approximation errors α k,v and variationexcess τ s ( k, v, v (cid:48) ) defined in § k ≥
0, set I k := E k ∪ B k ∪ N k ∪ F k and I := (cid:91) k ≥ I k . For each I ∈ I k , set I k +1 ( I ) := E k +1 ( I ) ∪ B k +1 ( I ) ∪ N k +1 ( I ) ∪ F k +1 ( I ). Remark 4.1. If I ∈ I k ∩ I m for some m (cid:54) = k , then f k | I = f m | I .4.1. Trees over intervals.
Given k ≥ I ∈ I k , we define a finite tree T over ( k, I )to be a finite subset of (cid:83) m ≥ ( { m } × I m ) satisfying the following three conditions.(1) The pair ( k, I ) ∈ T . If ( m, J ) ∈ T , then m ≥ k and J ⊂ I .(2) If ( m, J ) ∈ T and there exists J (cid:48) ∈ I m +1 ( J ) such that ( m + 1 , J (cid:48) ) ∩ T , then { m + 1 } × I m +1 ( J ) ⊆ T .(3) If ( m, J ) ∈ T for some m > k and J ∈ I m ( J (cid:48) ), then ( m − , J (cid:48) ) ∈ T .The first condition says that the root of the tree is ( k, I ) and its elements are descendantsof ( k, I ). The second condition says that if one child ( m + 1 , J (cid:48) ) of ( m, J ) is in T , thenevery child of ( m, J ) is in T . The third condition says that if ( m, J ) is in T , then all itsancestors up to ( k, I ) are in T .We extend this notion to the entire domain by defining a finite tree T over [0 ,
1] to bea set of the form T = { [0 , } ∪ (cid:91) I ∈ I T I , ¨OLDER CURVES AND PARAMETERIZATIONS 23 where T I is a finite tree over (0 , I ). A finite tree over [0 ,
1] may be thought to belong tostep k = − T be a finite tree over ( k, I ). The boundary ∂T of T is defined by ∂T := { ( m, J ) ∈ T : ( { m + 1 } × I m +1 ( I )) ∩ T = ∅} . The depth m ( T ) of T is the integer defined by m ( T ) := max { m ≥ m, J ) ∈ T } = max { k ≥ m, J ) ∈ ∂T } . If m ( T ) ≥
1, the parent tree p ( T ) is defined by p ( T ) := T \ ( { m ( T ) } × I m ( T ) )Note that m ( p ( T )) = m ( T ) − Remark 4.2. If T is a finite tree over ( k, I ) and ∂T = { ( k , J ) , . . . , ( k n , J n ) } , then theintervals J , . . . , J n partition I . That is, for all x ∈ I , there exists a unique i ∈ { , . . . , n } such that x ∈ J i .4.2. Mass of intervals.
For all s ≥ k ≥
0, and intervals I ∈ I k , define the s -mass M s ( k, I ) of ( k, I ) by M s ( k, I ) := sup T (cid:88) ( k (cid:48) ,I (cid:48) ) ∈ ∂T (diam f k (cid:48) ( I (cid:48) )) s ∈ [0 , ∞ ] , where the supremum is taken over all finite trees over ( k, I ). This notion extends to [0 , M s ([0 , (cid:88) I ∈ I M s (0 , I ) ∈ [0 , ∞ ] . Lemma 4.3.
Let k ≥ and I ∈ I k .(1) If I ∈ B k , then M s ( k, I ) = (diam f k ( I )) s .(2) If I ∈ F k , then M s ( k, I ) = 0 .(3) M s ( k, I ) ≥ (cid:80) I (cid:48) ∈ I k +1 ( I ) M s ( k + 1 , I (cid:48) ) .(4) If I ∈ I k ∩ I m for some m ≥ , then M s ( k, I ) = M s ( m, I ) . Before proving Lemma 4.3, we make two clarifying remarks. First, it is possible foran interval I ∈ N k to have M s ( k, I ) > f k ( I ) s = 0. This happenswhenever I ∈ N k and E k +1 ( I ) ∪ B k +1 ( I ) is non-empty. Second, Lemma 4.3(4) implies thatgiven I ∈ I k , the mass M s ( k, I ) is defined independently of the step of the constructionin which I appears. Nevertheless, we include the step k in definition of the mass toimprove exposition of the estimates in § § Proof of Lemma 4.3.
For the first claim, note that if I ∈ B k and m ≥ k + 1, then I m ( I ) = { I } . Therefore, if T is a finite tree over I of depth m , then ∂T = { ( m, I ) } .Thus, M s ( k, I ) = sup m ≥ k (diam f m ( I )) s = (diam f k ( I )) s .For the second claim, note that if I ∈ F k and m ≥ k + 1, then I m ( I ) = { I } and f m ( I )is a point. Therefore, if T is a finite tree over I of depth m , then ∂T = { ( m, I ) } . Thus, M s ( k, I ) = sup m ≥ k (diam f m ( I )) s = 0. For the third claim, let us first assume that M s ( k + 1 , J ) = ∞ for some J ∈ I k +1 ( I ).Fix M > T J over ( k + 1 , J ) such that (cid:88) ( k (cid:48) ,J (cid:48) ) ∈ T J (diam f k (cid:48) ( J (cid:48) )) s > M. The collection T = T J ∪ { ( k, I ) } ∪ ( { k + 1 } × I k +1 ( I )) is a finite tree over ( k, I ) and ∂T J ⊂ ∂T . Hence M s ( k, I ) ≥ (cid:88) ( k (cid:48) ,I (cid:48) ) ∈ ∂T (diam f k (cid:48) ( I (cid:48) )) s ≥ (cid:88) ( k (cid:48) ,J (cid:48) ) ∈ ∂T J (diam f k (cid:48) ( J (cid:48) )) s > M. We conclude that M s ( k, I ) = ∞ .Alternatively, assume that M s ( k + 1 , J ) is finite for all J ∈ I k +1 ( I ). Fix (cid:15) >
0. Foreach interval J ∈ I k +1 ( I ), let T J be a finite tree over ( k + 1 , J ) such that (cid:88) ( k (cid:48) ,J (cid:48) ) ∈ ∂T J (diam f k (cid:48) ( J (cid:48) )) s > M s ( k + 1 , J ) − (cid:15) card( I k +1 ( I )) . Then the collection T = { ( k, I ) } ∪ (cid:83) J ∈ I k +1 ( I ) T J is a finite tree over ( k, I ) with ∂T = (cid:83) J ∈ I k +1 ( I ) ∂T J . Therefore, M s ( k, I ) ≥ (cid:88) J ∈ I k +1 ( I ) (cid:88) ( k (cid:48) ,J (cid:48) ) ∈ ∂T J (diam f k (cid:48) ( J (cid:48) )) s > (cid:88) J ∈ I k +1 ( I ) M s ( k + 1 , J ) − (cid:15). The third claim follows by taking (cid:15) ↓ I ∈ I k ∩ I m for some m and k , say without loss ofgenerality that m > k . Because I ∈ I k ∩ I m , we have I n ( I ) = { I } for all k ≤ n ≤ m .Thus, iterating the third claim, M s ( k, I ) ≥ M s ( m, I ). For the opposite inequality, let T be a finite tree over ( k, I ). If ( m, I ) (cid:54)∈ T , then T = { ( k, I ) , . . . , ( l, I ) } for some k ≤ l ≤ m − T (cid:48) = { ( m, I ) } . If ( m, I ) ∈ T , then T ⊃ { ( k, I ) , . . . , ( m − , I ) } andwe set T (cid:48) = T \ { ( k, I ) , . . . , ( m − , I ) } so that T (cid:48) is a finite tree over ( m, I ) and ∂T = ∂T (cid:48) .In either case, (cid:88) ( k (cid:48) ,I (cid:48) ) ∈ ∂T (diam f k (cid:48) ( I (cid:48) )) s = (cid:88) ( k (cid:48) ,I (cid:48) ) ∈ ∂T (cid:48) (diam f k (cid:48) ( I (cid:48) )) s and it follows that M s ( k, I ) ≤ M s ( m, I ). (cid:3) When s = 1, the 1-mass is comparable to the Hausdorff measure H of the image. Lemma 4.4.
For each k ≥ and each I ∈ I k , M ( k, I ) = lim m →∞ (cid:88) J ∈ I m J ⊂ I diam f m ( J ) ≥ lim sup m →∞ H ( f m ( I )) . If there exists n ∈ N such that f m | (cid:83) B k is at most n -to-1 for all m ≥ k , then M ( k, I ) (cid:39) n lim inf m →∞ H ( f m ( I )) ¨OLDER CURVES AND PARAMETERIZATIONS 25 Proof.
Fix k ≥ I ∈ I k . By definition of the mass and (P2), M ( k, I ) ≥ lim sup m →∞ (cid:88) J ∈ I m J ⊂ I diam f m ( J ) ≥ lim sup m →∞ H ( f m ( I )) . To establish the other direction, let T be a finite tree over I of depth m ≥ k andenumerate with ∂T = { ( k , I ) , . . . , ( k n , I n ) } , where each k i ≤ m . For each i = 1 , . . . , n ,let I i be the set of all intervals J ∈ I m such that J ⊂ I i . Then (cid:88) J ∈ I m J ⊂ I diam f m ( J ) = n (cid:88) i =1 (cid:88) J ∈I i diam f m ( J ) ≥ n (cid:88) i =1 diam f k i ( I i ) = (cid:88) ( l,J ) ∈ ∂T diam f l ( J ) . Therefore,sup m ≥ k (cid:88) J ∈ I m J ⊂ I diam f m ( J ) ≤ M ( k, I ) = sup T (cid:88) ( l,J ) ∈ ∂T diam f l ( J ) ≤ lim inf m →∞ (cid:88) J ∈ I m J ⊂ I diam f m ( J ) . This shows that M ( k, I ) = lim m →∞ (cid:88) J ∈ I m J ⊂ I diam f m ( J ) . By (P4), the maps f m | (cid:83) E m are 2-to-1. In §
3, we did not examine overlaps of imagesof bridge intervals. By modifying the algorithm, the overlap of images of bridge intervalscan be made 2-to-1 (see the proof of Proposition 5.7). Nevertheless, suppose that weknow the overlaps of images of bridge intervals is at most n -to-1 for some n ≥
2. Then M ( k, I ) = lim m →∞ (cid:88) J ∈ I m J ⊂ I diam f m ( J ) ≤ n lim inf m →∞ H ( f m ( I )) . (cid:3) While Lemma 4.4 does not hold when s >
1, we always have the following comparisonbetween the s -mass and the Hausdorff measure H s of the closure of the points in (cid:83) ∞ k =0 V k . Lemma 4.5.
For all s ≥ , H s (cid:16)(cid:83) ∞ k =0 V k (cid:17) (cid:46) s,C ∗ ,ξ M s ([0 , . Proof.
Fix δ > m ∈ N sufficiently large such that 2 C ∗ ξ m +12 r / (1 − ξ ) ≤ δ .By (V0), (V2), and (V4), the collection { B ( v, C ∗ ρ m +1 r / (1 − ξ )) : v ∈ V m } is a cover of (cid:83) ∞ k =0 V k with elements of diameter at most 2 C ∗ ρ m +1 r / (1 − ξ ) ≤ C ∗ ξ m +12 r ≤ δ . Let T be the maximal finite tree over [0 ,
1] of depth m , i.e. T = (cid:83) mk =0 I k . Then H sδ (cid:16)(cid:83) ∞ k =0 V k (cid:17) ≤ (cid:88) v ∈ V m (cid:18) C ∗ ρ m +1 r − ξ (cid:19) s ≤ (cid:18) C ∗ ξ − ξ (cid:19) s (cid:88) ( m,I ) ∈ ∂T (diam f m ( I )) s (cid:46) s,C ∗ ,ξ M s ([0 , . Taking δ ↓ (cid:3) We include Lemma 4.4 and Lemma 4.5 for completeness. We will not use either lemmain any the estimates below.
Terminal vertices and phantom mass.
Let I ∈ N k be an interval such that α k,f k ( I ) < α . We classify f k ( I ) according to the arrangement of nearby points in V k +1 . • If I is as in § f k ( I ) is called a non-terminal vertex in V k . • If I is as in § f k ( I ) is called a V k . • If I is as in § f k ( I ) is called a V k .Motivated by [Jon90] and [BS17], for each k ≥ P k ⊂ { k } × N k and for each ( k, I ) ∈ P k define a number p k,I >
0, which we call the phantom mass at( k, I ). The phantom mass p k,I will let us pay for the length of edges between vertices in V k +1 nearby f k ( I ) that do not lie between vertices in V k nearby f k ( I ) (i.e. the blue edgesin Figure 2). To start, define an auxiliary parameter P depending only on s , C ∗ , and ξ by requiring that [ P + 2(1 . C ∗ ) s ] ξ s = P. That is,(4.1) P = 2(1 . C ∗ ) s − ξ s . For each k ≥
0, define P k = { ( k, I ) : I ∈ N k , α k,f k ( I ) < α and f k ( I ) is 1- or 2-sided terminal in V k } . For each k ≥ k, I ) ∈ P k , assign p k,I := (cid:40) P ρ sk r s , if f k ( I ) is 2-sided terminal P ρ sk r s , if f k ( I ) is 1-sided terminal . Lemma 4.6.
Let I ∈ N k be an interval such that α k,f k ( I ) < α . If f k ( I ) is 1-sidedterminal, then (cid:88) J ∈ I k +1 ( I ) (diam f k +1 ( J )) s < . C ∗ ) s ρ sk +1 r s . If f k ( I ) is 2-sided terminal, then (cid:88) J ∈ I k +1 ( I ) (diam f k +1 ( J )) s < . C ∗ ) s ρ sk +1 r s . Proof.
Suppose v = f k ( I ) is 1-sided terminal and let { v , . . . , v n } be an enumeration of thepoints in V k +1 ∩ B ( v, C ∗ ρ k +1 r ) starting from v = v and moving consecutively towardsthe terminal direction. Then (cid:88) J ∈ I k +1 ( I ) (diam f k +1 ( J )) s = 2 n (cid:88) i =1 | v i +1 − v i | s ≤ α k,v ) s | v − v n | s < . s ( C ∗ ρ k +1 r ) s by Lemma 2.2, since 1 + 3 α k,v ≤ / < .
1. The case that v is 2-sided terminalfollows from a similar computation. (cid:3) ¨OLDER CURVES AND PARAMETERIZATIONS 27 Special bridge intervals.
Given k ≥
1, we define B ∗ k := { ( k, I ) : I ∈ B k ( J ), J ∈ E k − is as in § } . Recall from § k, I ) ∈ B ∗ k , then 14 A ∗ ρ k r ≤ diam f k ( I ) < A ∗ ρ k − r . Lemma 4.7. If I ∈ B ∗ k , then
16 (diam f k ( I )) s ≥ P ρ sk r s . Proof.
Because diam f k ( I ) ≥ A ∗ ρ k r , it suffices to check that (14 A ∗ ) s ≥ P . Recallingthe definition of A ∗ , we find that(14 A ∗ ) s = (cid:18) C ∗ − ξ (cid:19) s ≥ . C ∗ ) s − ξ s = 6 P. Here 1 / (1 − ξ ) s ≥ / (1 − ξ s ) because 0 < ξ < s ≥ (cid:3) Let T be a finite tree over [0 , m be the depth of T and let 0 ≤ k ≤ m be aninteger. Define B ∗ k ( T ) := { ( k, I ) ∈ B ∗ k : there exists ( k, J ) ∈ T ∩ ( { k } × N k ) such that J ∩ I (cid:54) = ∅} . Although the sets B ∗ k ( T ) are not necessarily subsets of ∂T , we show in the next lemmathat each element in ∂T generates at most two elements in (cid:83) mk =1 B ∗ k ( T ). Lemma 4.8.
Let T be a finite tree over [0 , with depth m ≥ and let k ≤ m .(1) For each ( k, I ) ∈ B ∗ k ( T ) , there exists a unique ( l, J ) ∈ ∂T such that I ⊂ J . Infact, J ∈ E l ∪ B l .(2) For each ( l, J ) ∈ ∂T such that J ∈ E l ∪ B l , there exist at most two distinct ( k, I ) ∈ (cid:83) mi =0 B ∗ i ( T ) such that I ⊂ J .Proof. The first claim of (1) follows immediately from Remark 4.2. For the second claim,let ( k, I ) ∈ B ∗ k ( T ). There are two cases. Case 1.
If ( k, I ) ∈ T , then I n ( I ) = { I } and I ∈ B n for all n > k , since I is a bridgeinterval. Therefore, ( l, I ) ∈ ∂T for some l ≤ m and I ∈ B l . Case 2.
Suppose now that ( k, I ) (cid:54)∈ T . If J was in N l ∪ F l , then the closure of I wouldbe contained in the interior of J and I would intersect only intervals I (cid:48) ∈ I l (cid:48) for which( l (cid:48) , I (cid:48) ) (cid:54)∈ T , which is a contradiction.For (2), fix ( l, J ) ∈ ∂T with J ∈ E l ∪ B l and assume that { ( k i , J i ) : i = 1 , , } are distinct elements in (cid:83) mj =0 B ∗ j ( T ) such that J i ⊂ J for i = 1 , ,
3. Without loss ofgenerality, assume that J is between J and J , in the orientation of J . Then there exists l (cid:48) ≥ l + 1 and J (cid:48) ⊆ J such that J (cid:48) ∈ N l (cid:48) and ( l (cid:48) , J (cid:48) ) ∈ T . But then ( l, J ) (cid:54)∈ ∂T which is acontradiction. (cid:3) We now impose an additional restriction on α . Lemma 4.9.
For all C ∗ , ξ , and ξ , there exists α ∈ (0 , / such that if α ≤ α , thenfor all k ≥ , ( v, v (cid:48) ) ∈ Flat ( k ) , and y, y (cid:48) ∈ V k +1 ( v, v (cid:48) ) , we have | y − y (cid:48) | ≤ | v − v (cid:48) | . Proof.
Enumerate V k +1 ( v, v (cid:48) ) = { v , . . . , v n } from left to right, so that v = v and v n = v (cid:48) .Let y = v l and y (cid:48) = v m for some 1 ≤ l < m ≤ n . If y = v and y (cid:48) = v (cid:48) , the conclusionis trivial. Thus, let us suppose that there exists at least one point to the left of y or theright of y (cid:48) , say without loss of generality that l ≥
2. Let x i = π (cid:96) k,v for all 1 ≤ i ≤ n .Then, arguing as in the proof of Lemma 2.2, | v − v | + | v l − v m | α ≤ | x − x | + | x l − x m | ≤ | x − x n | ≤ | v − v (cid:48) | . Because V k +1 is ρ k +1 separated, | v l − v m | ≤ (1 + 3 α ) | v − v (cid:48) | − | v − v | ≤ (1 + 3 α ) | v − v (cid:48) | − ρ k +1 r = | v − v (cid:48) | (cid:18) α − ρ k +1 r | v − v (cid:48) | (cid:19) . Now | v − v (cid:48) | ≤ A ∗ ρ k r ≤ A ∗ ξ − ρ k +1 r . Hence | v l − v m | ≤ | v − v (cid:48) | provided that1 + 3 α − ξ A ∗ ≤ . Thus, we can take (cid:3) (4.2) α := min (cid:40) , (cid:18) ξ A ∗ (cid:19) / (cid:41) . Together Lemma 4.8 and Lemma 4.9 yield the following result.
Corollary 4.10.
Assume α ≤ α . If T is a finite tree over [0 , of depth m , then, m (cid:88) k =1 (cid:88) ( k,J ) ∈ B ∗ k ( T ) (diam f k ( J )) s ≤ (cid:88) ( l,J ) ∈ ∂T (diam f l ( J )) s . An upper bound for the total mass.
Here we prove the following propositionwhich gives an upper bound for the mass of [0 ,
1] in terms of the variation excess τ s ( k, v, v (cid:48) )of flat pairs ( v, v (cid:48) ) ∈ Flat ( k ) defined in § Proposition 4.11.
Assume that α ≤ α (see Lemma 4.9). For all s ≥ , M s ([0 , (cid:46) s,C ∗ ,ξ r s + ∞ (cid:88) k =0 (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk r s + ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ρ sk r s . The proof of Proposition 4.11 reduces to proving the following lemma (cf. [BS17, (9.4)]).Recall the definition of the parent tree from § Lemma 4.12.
Let k ≥ be an integer, let T be a finite tree over [0 , of depth k + 1 andlet p ( T ) be the parent tree. There exists a constant C > depending only on s , C ∗ , and ¨OLDER CURVES AND PARAMETERIZATIONS 29 ξ such that (cid:88) ( l,I ) ∈ ∂T (diam f l ( I )) s + (cid:88) ( k +1 ,I ) ∈ P k +1 ∩ ∂T p k +1 ,I ≤ (cid:88) ( l,I ) ∈ ∂p ( T ) (diam f l ( I )) s + (cid:88) ( k,I ) ∈ P k ∩ ∂p ( T ) p k,I + 13 (cid:88) ( k +1 ,I ) ∈ B ∗ k +1 ( T ) (diam f k +1 ( I )) s + C (cid:88) v ∈ V k α k,v ≥ α ρ sk r s + C (cid:88) w ∈ V k +1 α k +1 ,w ≥ α ρ sk +1 r s + C (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk r s . (E)We prove Lemma 4.12 in § Proof of Proposition 4.11.
Assume that α ≤ α . Let T be a finite tree over [0 ,
1] ofdepth m . Set T m = T and for each 1 ≤ k ≤ m (if any) set T k − = p ( T k ). Note that T = { } × I . By Lemma 4.12, for all 1 ≤ k ≤ m (if any), (cid:88) ( l,I ) ∈ ∂T k (diam f l ( I )) s + (cid:88) ( k,I ) ∈ P k ∩ ∂T k p k,I ≤ (cid:88) ( l,I ) ∈ ∂T k − (diam f l ( I )) s + (cid:88) ( k − ,I ) ∈ P k − ∩ ∂T k − p k − ,I + 13 (cid:88) ( k,I ) ∈ B ∗ k ( T k ) (diam f k ( I )) s + C (cid:88) v ∈ V k − α k − ,v ≥ α ρ sk − r s + C (cid:88) w ∈ V k α k,w ≥ α ρ sk r s + C (cid:88) ( v,v (cid:48) ) ∈ Flat ( k − τ s ( k − , v, v (cid:48) ) ρ sk − r s . Iterating the latter inequality, (cid:88) ( l,I ) ∈ ∂T (diam f l ( I )) s ≤ (cid:88) I ∈ E ∪ B (diam f ( I )) s + (cid:88) I ∈ P p ,I + 13 m (cid:88) k =1 (cid:88) I ∈ B ∗ k ( T ) (diam f k ( I )) s + 2 C m (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ρ sk r s + C m (cid:88) k =1 (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk r s . Since α ≤ α , we obtain (cid:80) mk =1 (cid:80) I ∈ B ∗ k ( T ) (diam f k ( I )) s ≤ (cid:80) ( l,I ) ∈ ∂T (diam f l ( I )) s byCorollary 4.10. This is the only place in the proof of the Proposition that we use therestriction α ≤ α . Therefore, (cid:88) ( l,I ) ∈ ∂T (diam f l ( I )) s ≤ (cid:88) I ∈ E ∪ B (diam f ( I )) s + 3 (cid:88) I ∈ P p ,I + 6 C m (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ρ sk r s + 3 C m − (cid:88) k =0 (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk r s . There are now two alternatives. On one hand, suppose that α ,v < α for some v ∈ V .Then V projects onto an (1 + 3(1 / ) − r separated set in (cid:96) ,v of diameter at most C ∗ r by Lemma 2.1. Hence card V (cid:46) C ∗ (cid:88) I ∈ E ∪ B (diam f ( I )) s + (cid:88) I ∈ P p ,I (cid:46) s,C ∗ ,ξ r s . On the other hand, suppose that α ,v ≥ α for all v ∈ V . Then (cid:88) I ∈ E ∪ B (diam f ( I )) s + (cid:88) I ∈ P p ,I (cid:46) s,C ∗ ,ξ (cid:88) v ∈ V α ,v ≥ α r s In either case, we arrive at (cid:88) ( l,I ) ∈ ∂T (diam f l ( I )) s (cid:46) s,C ∗ ,ξ r s + m − (cid:88) k =0 (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk r s + m (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ρ sk r s (cid:46) s,C ∗ ,ξ r s + ∞ (cid:88) k =0 (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk r s + ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ρ sk r s . Since T was an arbitrary tree over [0 , M s ([0 , (cid:3) Proof of Lemma 4.12.
The proof is divided into five estimates (E1), (E2), (E3),(E4) and (E5), whose sum gives (E). Towards this end, we split the left hand side of (E)into smaller sums by making the following four decompositions.Firstly, ∂T can be partitioned as E ∪ E ∪ E ∪ B ∪ B ∪ B ∪ B ∪ F ∪ ( ∂T ∩ ∂p ( T )),where E = { ( k + 1 , I ) ∈ ∂T : I ∈ E k +1 ( J ) and J ∈ N k is as in § } , E = { ( k + 1 , I ) ∈ ∂T : I ∈ E k +1 ( J ) and J ∈ N k is as in § } , E = { ( k + 1 , I ) ∈ ∂T : I ∈ E k +1 ( J ) and J ∈ E k is as in § } , B = { ( k + 1 , I ) ∈ ∂T : I ∈ B k +1 ( J ) and J ∈ N k is as in § } , B = { ( k + 1 , I ) ∈ ∂T : I ∈ B k +1 ( J ) and J ∈ E k is as in § } , B = { ( k + 1 , I ) ∈ ∂T : I ∈ B k +1 ( J ) and J ∈ E k is as in § } , B = { ( k + 1 , I ) ∈ ∂T : I ∈ B k +1 ( J ) and J ∈ B k is as in § } , F ⊆ { k + 1 } × ( N k +1 ∪ F k +1 ) . Secondly, P k +1 ∩ ∂T can be partitioned as P ∪ P ∪ P , where P = { ( k + 1 , I ) ∈ ∂T ∩ P k +1 : I ∈ N k +1 ( J ) and J ∈ N k is as in § } , P = { ( k + 1 , I ) ∈ ∂T ∩ P k +1 : I ∈ N k +1 ( J ) and J ∈ N k is as in § } , P = { ( k + 1 , I ) ∈ ∂T ∩ P k +1 : I ∈ N k +1 ( J ) and J ∈ E k is as in § } . ¨OLDER CURVES AND PARAMETERIZATIONS 31 Thirdly, ∂p ( T ) can be partitioned as E (cid:48) ∪ E (cid:48) ∪ F (cid:48) ∪ ( ∂T ∩ ∂p ( T )), where E (cid:48) = { ( k, I ) ∈ ∂p ( T ) \ ∂T : I ∈ E k is as in § } , E (cid:48) = { ( k, I ) ∈ ∂p ( T ) \ ∂T : I ∈ E k is as in § } , B (cid:48) = { ( k, I ) ∈ ∂p ( T ) \ ∂T : I ∈ B k is as in § } , F (cid:48) ⊆ { k } × ( N k ∪ F k ) . Fourthly, set B ∗∗ k +1 ( T ) = B ∗ k +1 ( T ) (cid:116) B ∗ k +1 ( T ). Then define collections B ∗ and B ∗ asfollows. If I ∈ N k , α k,f k ( I ) < α , I k +1 ( I ) ⊂ ∂T , and f k ( I ) is an endpoint of the image f k +1 ( J ) of ( k + 1 , J ) ∈ B ∗ k +1 , then we include a copy of ( k + 1 , J ) from B ∗∗ k +1 ( T ) in B ∗ .If K ∈ N k and f k ( K ) is the endpoint of f k +1 ( I ) for some ( k + 1 , I ) ∈ B that lies strictlybetween the endpoints of the image f k ( J ) of the associated edge interval J ∈ E k , thenwe include a copy of ( k + 1 , I ) from B ∗∗ k +1 ( T ) in B ∗ . Because each bridge has only twoendpoints, we can choose the included copies so that B ∗ ∪ B ∗ ⊂ B ∗∗ k +1 . Before proceeding to the estimates, we remark that (cid:88) ( k,I ) ∈F (cid:48) (diam f k ( I )) s = (cid:88) ( k +1 ,I ) ∈F (diam f k +1 ( I )) s = 0 , because f k is constant on each interval in N k ∪ F k by (P2).Estimate 1. Here we deal with phantom masses and new intervals coming from some I ∈ N k whose image is not flat. In particular, we will show that there exists C dependingonly on s , C ∗ , and ξ such that (cid:88) ( k +1 ,I ) ∈E ∪B (diam f k +1 ( I )) s + (cid:88) ( k +1 ,J ) ∈P p k +1 ,J ≤ C (cid:88) v ∈ V k α k,v ≥ α ρ sk r s + C (cid:88) w ∈ V k +1 α k +1 ,w ≥ α ρ sk +1 r s . (E1)Since { I ∈ N k : α k,f k ( I ) ≥ α and I k +1 ( I ) ⊂ ∂T } ⊆ { I ∈ N k ∩ ∂p ( T ) : α k,f k ( I ) ≥ α } , inequality (E1) follows from the inequality (cid:88) I ∈ N k { k +1 }× I k +1 ( I ) ⊂ ∂Tα k,fk ( I ) ≥ α (cid:88) J ∈ E k +1 ( I ) ∪ B k +1 ( I ) (diam f k +1 ( I )) s + (cid:88) J ∈ N k +1 ( I )( k +1 ,J ) ∈ P k +1 p k +1 ,J ≤ C (cid:88) I ∈ N k { k +1 }× I k +1 ( I ) ⊂ ∂Tα k,fk ( I ) ≥ α ρ sk r s + C (cid:88) I ∈ N k { k +1 }× I k +1 ( I ) ⊂ ∂Tα k,fk ( I ) ≥ α (cid:88) J ∈ N k +1 ( I ) α k +1 ,fk +1( J ) ≥ α ρ sk +1 r s . (4.3)To prove (4.3), fix any I ∈ N k such that { k + 1 } × I k +1 ( I ) ⊂ ∂T and α k,f k ( I ) ≥ α .Recall that N k +1 ( I ) is in one-to-one correspondence with V k +1 ,I defined in § are now two possibilities. On one hand, suppose that α k +1 ,w < α for some w ∈ V k +1 ,I .Then V k +1 ,I projects onto an (1 + 3(1 / ) − ρ k +1 r separated set in (cid:96) k +1 ,w of diameterat most 2 C ∗ ρ k +1 r by Lemma 2.1. Hence card I k +1 ( I ) (cid:46) C ∗ (cid:88) J ∈ E k +1 ( I ) ∪ B k +1 ( I ) (diam f k +1 ( J )) s + (cid:88) J ∈ N k +1 ( I )( k +1 ,J ) ∈ P k +1 p k +1 ,J (cid:46) s,C ∗ ,ξ ρ sk +1 r s (cid:46) s,C ∗ ,ξ ρ sk r s . On the other hand, suppose that α k +1 ,w ≥ α for all w ∈ V k +1 ,I . Then (cid:88) J ∈ E k +1 ( I ) ∪ B k +1 ( I ) (diam f k +1 ( J )) s + (cid:88) J ∈ N k +1 ( I )( k +1 ,J ) ∈ P k +1 p k +1 ,J (cid:46) s,C ∗ ,ξ (cid:88) J ∈ N k +1 ( I ) ρ sk +1 r s . Because the sets V k +1 ,I for different I ∈ N k are pairwise disjoint (see § I yields (4.3).Estimate 2. We estimate the new phantom masses and new intervals coming from some I ∈ N k such that α k,f k ( I ) < α and I k +1 ( I ) ⊂ ∂T . In particular, we show that (cid:88) ( k +1 ,I ) ∈E (diam f k +1 ( I )) s + (cid:88) ( k +1 ,I ) ∈P p k +1 ,I ≤ (cid:88) ( k,I ) ∈ P k ∩ ∂p ( T ) p k,I + 16 (cid:88) ( k +1 ,I ) ∈B ∗ (diam f k +1 ( I )) s . (E2)This estimate is responsible for the choice of the constant 14 A ∗ appearing in the definitionof Flat ( k ), and thus, for the constant 30 A ∗ appearing in the definition of α k,v . Inequality(E2) is equivalent to (cid:88) I ∈ N k { k +1 }× N k +1 ( I ) ⊂ ∂Tα k,fk ( I ) <α (cid:88) J ∈ E k +1 ( I ) (diam f k +1 ( I )) s + (cid:88) J ∈ N k +1 ( I )( k +1 ,J ) ∈ P k +1 p k +1 ,J ≤ (cid:88) I ∈ N k { k +1 }× N k +1 ( I ) ⊂ ∂Tα k,fk ( I ) <α p k,I + 16 (cid:88) ( k +1 ,I (cid:48) ) ∈B ∗ (diam f k +1 ( I )) s . (4.4)To prove (4.4) fix I ∈ N k such that { k + 1 } × N k +1 ( I ) ⊂ ∂T and α k,f k ( I ) < α . There arenine cases (1a, 1b, 2a, 2b, 2c, 2d, 3a, 3b, 3c). We sincerely apologize to the reader.For the first four cases (1a, 1b, 2a, 2b), we show that(4.5) (cid:88) J ∈ E k +1 ( I ) (diam f k +1 ( J )) s + (cid:88) J ∈ N k +1 ( I )( k +1 ,J ) ∈ P k +1 p k +1 ,J ≤ p k,I . Case 1.
Suppose f k ( I ) is 2-sided terminal in V k . ¨OLDER CURVES AND PARAMETERIZATIONS 33 Case 1a. If N k +1 ( I ) = { I } , then f k +1 ( I ) is 2-sided terminal in V k +1 , E k +1 ( I ) = ∅ and the new phantom mass p k +1 ,I = 2 P ρ sk +1 r s is dominated by the old phantom mass p k,I = 2 P ρ sk r s . Hence (4.5) holds. Case 1b.
Assume that N k +1 ( I ) contains at least two elements (see Figure 2 above). Inthis case, at most two elements of N k +1 ( I ) map to 1-sided terminal vertices in V k +1 . ByLemma 4.6, p k +1 ,J + p k +1 ,J + (cid:88) J ∈ I k +1 ( I ) (diam f k +1 ( J )) s ≤ P ρ sk +1 r s + 4(1 . C ∗ ) s ρ sk +1 r s ≤ P ρ sk r s = p k,I , because P was chosen to be sufficiently large such that[ P + 2(1 . C ∗ ) s ] ξ s ≤ P. Thus, (4.5) holds in this case, as well.
Case 2.
Suppose that f k ( I ) is 1-sided terminal in V k . Let ( v , v (cid:48) ) be the unique elementin Flat ( k ) with v = f k ( I ) and let v be the first vertex in V k +1 between v and v (cid:48) in thedirection going from v to v (cid:48) . By property (P7), we can find an interval L in E k such that f k ( L ) = ( v , v (cid:48) ) and I ∩ L (cid:54) = ∅ . Let K be an interval in I k +1 ( L ) such that f k ( K ) = ( v , v ).There will be four cases, depending on whether N k +1 ( I ) contains one or more elementsand whether K belongs to E k +1 ( L ) or B k +1 ( L ). Case 2a.
Suppose that N k +1 ( I ) = { I } and K ∈ E k +1 ( L ). Then f k +1 ( I ) is 1-sidedterminal in V k +1 , E k +1 ( I ) = ∅ , and the new phantom mass p k +1 ,I = P ρ sk +1 r s is dominatedby the old phantom mass p k,I = P ρ sk r s . Hence (4.5) holds. Case 2b.
Suppose that N k +1 ( I ) contains at least two elements (see Figure 2 above) and K ∈ E k +1 ( L ). Then at most one element of N k +1 ( K ) maps to a 1-sided terminal vertexin V k +1 . By Lemma 4.6, p k +1 ,J + (cid:88) J ∈ I k +1 ( I ) (diam f k +1 ( J )) s ≤ P ρ sk +1 r s + 2(1 . C ∗ ) s ρ sk +1 r s ≤ P ρ sk r s = p k,I , because P was chosen to be sufficiently large such that[ P + 2(1 . C ∗ ) s ] ξ s ≤ P. Thus, (4.5) holds, once again.
Case 2c.
Suppose that N k +1 ( I ) = { I } and K ∈ B k +1 ( L ). Then f k +1 ( I ) is 2-sidedterminal in V k +1 and E k +1 ( I ) = ∅ . The new phantom mass that must be paid for is p k +1 ,I = 2 P ρ sk +1 r s . In this case, we pay for one half of p k +1 ,I with p k,I = P ρ sk r s and useLemma 4.7 to pay for the other half of p k +1 ,I with (diam f k +1 ( K )) s , where K ∈ B ∗ k +1 ( T ).That is, p k +1 ,I ≤ p k,I + 16 (diam f k +1 ( K )) s . Case 2d.
Suppose that N k +1 ( I ) contains at least two points and K ∈ B k +1 ( L ). Then f k +1 ( I ) is 1-sided terminal in V k +1 , E k +1 ( I ) is nonempty, and up to one of the new verticesdrawn could be 1-sided terminal in V k +1 , as well. In this case, p k +1 ,I + p k +1 ,J + (cid:88) J ∈ E k +1 ( I ) (diam f k +1 ( J )) s (cid:124) (cid:123)(cid:122) (cid:125) ≤
16 (diam f k +1 ( K )) s + p k,I . by Lemma 4.7, Lemma 4.6, and the choice of P . Case 3.
Suppose f k ( I ) is not terminal in V k . Then E k +1 ( I ) = ∅ . It remains to pay for p k +1 ,I as needed. Let L , L , K , and K be defined by analogy with L and K from Case2, but corresponding to the two distinct flat pairs ( f k ( I ) , v (cid:48) ) and ( f k ( I ) , v (cid:48)(cid:48) ). There arethree cases, depending on whether K and K both edge intervals, one of K or K is anedge interval and the other is a bridge interval, or both K and K are bridge intervals. Case 3a.
Suppose that K belongs to E k +1 ( L ) and K belongs to E k +1 ( L ). Then f k +1 ( I ) is non-terminal in V k +1 . Hence ( k + 1 , I ) (cid:54)∈ P k +1 and both sides of (4.5) are zero.In other words, there is nothing to pay for in Case 3a. Case 3b.
Suppose that one of K or K is an edge interval and the other is a bridgeinterval, say without loss of generality that K ∈ E k +1 ( L ) and K ∈ B k +1 ( L ). Then f k +1 ( I ) is 1-sided terminal in V k +1 and p k +1 ,I = P ρ sk +1 r s ≤ (diam f k +1 ( K )) s by Lemma4.7. Case 3c.
Suppose that K belongs to B k +1 ( L ) and K belongs to B k +1 ( L ). Then f k +1 ( I ) is 2-sided terminal in V k +1 and p k +1 ,I = 2 P ρ sk +1 r s ≤
16 (diam f k +1 ( K )) s + 16 (diam f k +1 ( K )) s . Adding up the estimates in the nine cases, we obtain (E2).Estimate 3. On one hand, ( k, I ) ∈ B (cid:48) if and only if ( k + 1 , I ) ∈ B (see § k, I ) ∈ B (cid:48) , we have diam f k ( I ) = diam f k +1 ( I ). Thus, (cid:88) ( k +1 ,I ) ∈B (diam f k +1 ( I )) s = (cid:88) ( k,I ) ∈B (cid:48) (diam f k ( I )) s . On the other hand, when both endpoints of the image of an edge interval are non-flat,the edge interval becomes a bridge interval and pays for itself (see § (cid:88) ( k +1 ,I ) ∈B (diam f k +1 ( I )) s = (cid:88) ( k,I ) ∈E (cid:48) (diam f k ( I )) s . All together, we have(E3) (cid:88) ( k +1 ,I ) ∈B ∪B (diam f k +1 ( I )) s = (cid:88) ( k,I ) ∈E (cid:48) ∪B (cid:48) (diam f k ( I )) s . Estimate 4. Next, we control the new phantom masses at endpoints of images f k +1 ( J )of bridge intervals J ∈ B k +1 ( I ) coming from some edge interval I ∈ E k as in § ¨OLDER CURVES AND PARAMETERIZATIONS 35 that the endpoint lies between the endpoints of f k ( I ). Specifically, we show that (cid:88) ( k +1 ,I ) ∈P p k +1 ,I ≤ (cid:88) ( k +1 ,I ) ∈B ∗ (diam f k +1 ( I )) s . (E4)Inequality (E4) is equivalent to (cid:88) ( k,I ) ∈E (cid:48) (cid:88) J ∈ N k +1 ( I )( k +1 ,J ) ∈ P k +1 p k +1 ,J ≤ (cid:88) ( k +1 ,I ) ∈B ∗ (diam f k +1 ( I )) s . (4.6)To prove (4.6), fix ( k, I ) ∈ E (cid:48) and J ∈ N k +1 ( I ) be such that ( k + 1 , J ) ∈ P k +1 . Then f k +1 ( J ) lies strictly between the endpoints of f k ( I ) and f k +1 ( J ) is either 1- or 2-sidedterminal in V k +1 . On one hand, if f k +1 ( J ) is 1-sided terminal, then there exists preciselyone element ( k + 1 , K ) ∈ B ∗ such that f k +1 ( J ) is an endpoint of f k +1 ( K ). In this case, p k +1 ,J = P ρ sk +1 r s ≤
16 (diam f k +1 ( K )) s by Lemma 4.7. On the other hand, if f k +1 ( J ) is 2-sided terminal, then there exist twoelements ( k + 1 , K ) and ( k + 1 , K ) in B ∗ such that f k +1 ( J ) is the common endpoint of f k +1 ( K ) and f k +1 ( K ). In this case, p k +1 ,J = 2 P ρ sk +1 r s ≤
16 (diam f k +1 ( K )) s + 16 (diam f k +1 ( K )) s by Lemma 4.7. Remark 4.13.
In Estimates 2 and Estimate 4, each endpoint of the image f k +1 ( I ) of( k + 1 , I ) ∈ B ∗ ∪ B ∗ is used once and each f k +1 ( I ) has only two endpoints. Hence16 (cid:88) ( k +1 ,I ) ∈B ∗ ∪B ∗ (diam f k +1 ( I )) s ≤ (cid:88) ( k +1 ,I ) ∈ B ∗∗ k +1 ( T ) = 13 (cid:88) ( k +1 ,I ) ∈ B ∗ k +1 ( T ) . Estimate 5. In this final estimate, we deal with new intervals in ∂T coming from anedge interval in ∂p ( T ) which has an endpoint with flat image. We will show that (cid:88) ( k +1 ,I ) ∈E ∪B (diam f k +1 ( I )) s ≤ (cid:88) ( k,I ) ∈E (cid:48) (diam f k +1 ( I )) s + (14 A ∗ ) s (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk r s . (E5)For each I ∈ E (cid:48) , pick an endpoint x I of I such that α k,f k ( x I ) < α and let y I be the otherendpoint of I . Estimate (E5) follows immediately from (cid:88) ( k,I ) ∈E (cid:48) (cid:88) J ∈ E k +1 ( I ) ∪ B k +1 ( I ) (diam f k +1 ( J )) s ≤ (cid:88) ( k,I ) ∈E (cid:48) ((diam f k ( I )) s + (14 A ∗ ) s τ s ( k, f k ( x I ) , f k ( y I ))) ρ sk r s . (4.7) To show (4.7), fix ( k, I ) ∈ E (cid:48) and enumerate E k +1 ( I ) ∪ B k +1 ( I ) = { I , . . . , I n } . Let v = f k ( x I ) and let v (cid:48) = f k ( y I ). By definition of τ s ( k, v, v (cid:48) ) and (P3), we have n − (cid:88) i =1 (diam f k +1 ( I i )) s ≤ | v − v (cid:48) | s + τ s ( k, v, v (cid:48) ) | v − v (cid:48) | s ≤ (diam f k ( I )) s + (14 A ∗ ) τ s ( k, v, v (cid:48) ) ρ sk r s . Summing over all pairs ( k, I ) ∈ E (cid:48) , we obtain (4.7).Adding (E1), (E2), (E3), (E4), and (E5), we arrive at (E). This completes the proof ofLemma 4.12. 5. H¨older parametrization In § § § § C ∗ , ξ , and ξ , let α be defined by (4.2). That is, α = min (cid:40) , (cid:18) ξ (1 − ξ )42 C ∗ (cid:19) / (cid:41) . Theorem 5.1 (H¨older Traveling Salesman with Nets) . Assume that X = l ( R ) or X = R N for some N ≥ . Let s ≥ , let V = ( V k , ρ k ) k ≥ be a sequence of finite sets V k in X and numbers ρ k > that satisfy properties (V0)–(V5) defined in § α ∈ (0 , α ] and (5.1) S s V := ∞ (cid:88) k =0 (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk + ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ρ sk < ∞ , then there exists a (1 /s ) -H¨older map f : [0 , → X such that f ([0 , ⊃ (cid:83) k ≥ V k and theH¨older constant of f satisfies H (cid:46) s,C ∗ ,ξ ,ξ r (1 + S s V ) . Proof of Theorem 5.1.
In this subsection, w will always denote a finite word inthe alphabet N = { , , . . . } , including the empty word ∅ . We denote the length of a word w by | w | .The conclusion holds trivially if (cid:83) k ≥ V k is a singleton. Thus, in addition to S s V < ∞ ,we may assume that (cid:83) k ≥ V k contains at least two points. Because α ≤ α , Proposition4.11 gives(5.2) 0 < M s ([0 , (cid:46) s,C ∗ ,ξ r s (1 + S s V ) < ∞ . To proceed, we start by renaming the intervals in { [0 , } ∪ I . Denote ∆ ∅ = [0 ,
1] andwrite I = { ∆ , . . . , ∆ n ∅ } , enumerated according to the orientation of [0 , w with | w | = k , we have defined ∆ w ∈ I k . Suppose also that I k +1 (∆ w ) = { J , . . . , J n w } , ¨OLDER CURVES AND PARAMETERIZATIONS 37 enumerated according to the orientation of [0 , i ∈ { , . . . , n w } , denote∆ wi = J i . Denote by W the set of all finite words with letters from N for which an interval∆ w has been defined.Next, we use the masses of intervals defined in § w .Define ∆ (cid:48)∅ = [0 , { ∆ (cid:48) , . . . , ∆ (cid:48) n ∅ } be a partition of ∆ (cid:48)∅ , enumerated according to theorientation of [0 , (cid:48) i is open (resp. closed) if and only if ∆ i is open (resp. closed), and(2) diam ∆ (cid:48) i = M s (0 , ∆ i ) / M s ([0 , M s ([0 , (cid:80) n ∅ i =1 M s (0 , ∆ i ). Inductively, suppose that aninterval ∆ (cid:48) w ⊂ [0 ,
1] has been defined for some w ∈ W such thatdiam ∆ (cid:48) w ≥ M s ( | w | − , ∆ w ) M s ([0 , (cid:48) w is open (resp. closed) if and only if ∆ w is open (resp. closed). Let { ∆ (cid:48) w , . . . , ∆ (cid:48) wn w } be a partition of ∆ (cid:48) w , enumerated according to the orientation of [0 , (cid:48) wi is open (resp. closed) if and only if ∆ wi is open (resp. closed), and(2) diam ∆ (cid:48) wi ≥ M s ( | w | , ∆ wi ) / M s ([0 , E (cid:48) k = { ∆ (cid:48) w : ∆ w ∈ E k } and similarly define the families B (cid:48) k , N (cid:48) k , F (cid:48) k , and I (cid:48) k . For each k ≥
0, define a continuousmap F k : [0 , → X by F k | ∆ (cid:48) w = ( f k | ∆ w ) ◦ φ w for all w ∈ W , where φ w is the unique increasing affine homeomorphism mapping ∆ (cid:48) w onto ∆ w when ∆ (cid:48) w is nondegenerate and φ w maps to any point in ∆ w when ∆ (cid:48) w is a singleton. (The latterpossibility occurs only when ∆ (cid:48) w belongs to F (cid:48) k or N (cid:48) k .)We now prove two auxiliary results for the sequence ( F k ) k ≥ . Lemma 5.2.
For all k ≥ and x ∈ [0 , , we have | F k +1 ( x ) − F k ( x ) | ≤ A ∗ ξ r ρ k . Proof.
Fix x ∈ [0 ,
1] and w ∈ W such that x ∈ ∆ w and | w | = k . Let also i ∈ { , . . . , n w } be such that x ∈ ∆ (cid:48) wi .If ∆ (cid:48) w ∈ B (cid:48) k , then ∆ (cid:48) w = ∆ (cid:48) wi , ∆ w = ∆ wi , and F k | ∆ (cid:48) w = F k +1 | ∆ (cid:48) wi . We conclude that | F k +1 ( x ) − F k ( x ) | = 0.If ∆ (cid:48) w ∈ F (cid:48) k , then ∆ (cid:48) w = ∆ (cid:48) wi = { x } , and F k ( x ) = F k +1 ( x ). Hence | F k +1 ( x ) − F k ( x ) | = 0.If ∆ (cid:48) w ∈ E (cid:48) k , then | F k +1 ( x ) − F k ( x ) | ≤ F k (∆ (cid:48) w ) = 2 diam f k (∆ w ) ≤ A ∗ ρ k +1 r .If ∆ (cid:48) w ∈ N (cid:48) k , then | F k +1 ( x ) − F k ( x ) | ≤ f k (∆ w ) ≤ diam ˜ V k +1 ,I ≤ A ∗ ρ k +1 r ,where ˜ V k +1 ,I is a set defined in § (cid:3) Lemma 5.3.
For all k ≥ and x, y ∈ [0 , , | F k ( x ) − F k ( y ) | ≤ M s ([0 , r − s ρ − sk | x − y | (cid:46) s,C ∗ ,ξ r (1 + S s V ) ρ − sk | x − y | . Proof.
Fix k ≥ x, y ∈ [0 , x < y .We consider three cases. In the first two cases, the points x and y belong to the sameinterval ∆ (cid:48) w , | w | = k , while in the third case they belong to different intervals. Case 1. If x, y ∈ ∆ (cid:48) w ∈ N (cid:48) k ∪ F (cid:48) k , then | F k ( x ) − F k ( y ) | = 0 | x − y | , because the map F k | ∆ (cid:48) w is constant. Case 2.
Suppose that x, y ∈ ∆ (cid:48) w ∈ E (cid:48) k ∪ B (cid:48) k . Since F k | ∆ (cid:48) w is affine, | F k ( x ) − F k ( y ) | ≤ M s ([0 , f k (∆ w ) M s ( | w | − , ∆ w ) | x − y |≤ M s ([0 , f k (∆ w ) − s | x − y |≤ M s ([0 , r − s ρ − sk | x − y | , by (V3) and the assumption s ≥
1. Thus, by (5.2), | F k ( x ) − F k ( y ) | (cid:46) s,C ∗ ,ξ r (1 + S s V ) ρ − sk | x − y | . Case 3.
Suppose that x ∈ ∆ (cid:48) w and y ∈ ∆ (cid:48) u for some ∆ (cid:48) w , ∆ (cid:48) u ∈ I (cid:48) k with ∆ (cid:48) w ∩ ∆ (cid:48) u = ∅ .By the preceding cases and the Fundamental Theorem of Calculus, | F k ( x ) − F k ( y ) | ≤ (cid:90) yx |∇ F k ( t ) | dt ≤ M s ([0 , r − s ρ − sk | x − y | (cid:46) s,C ∗ ,ξ r (1+ S s V ) ρ − sk | x − y | . (cid:3) We are now ready to prove Theorem 5.1.
Proof of Theorem 5.1.
Define F : [0 , → X pointwise by F ( x ) := F ( x ) + ∞ (cid:88) k =0 ( F k +1 ( x ) − F k ( x )) . By Lemma 5.2, F is well defined and continuous in all [0 , F is (1 /s )-H¨older continuous with H¨older constant H ≤ ξ (cid:18) M s ([0 , r − s + 60 A ∗ r ξ − ξ (cid:19) (cid:46) s,C ∗ ,ξ ,ξ r (1 + S s V ) . Finally, for any integer k ≥ m ≥ k , we have V k ⊂ F m ([0 , V k ⊂ F ([0 , k ≥ (cid:3) Corollaries to Theorem 5.1 and Proof of Theorem 1.1.Corollary 5.4 (tube approximation) . For all s > , C ∗ ≥ , and < ξ < ξ < ,there exists α ∗ > with the following property. Assume that X = l ( R ) or X = R N forsome N ≥ . Let V = ( V k , ρ k ) k ≥ be a sequence of finite sets in X and numbers ρ k > satisfying properties (V0)–(V5) of § C ∗ , ξ , and ξ . If S s, + V := ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ∗ ρ sk < ∞ , ¨OLDER CURVES AND PARAMETERIZATIONS 39 then there exists a (1 /s ) -H¨older map f : [0 , → X such that (cid:83) k ≥ V k ⊂ f ([0 , and theH¨older constant of f satisfies H (cid:46) s,C ∗ ,ξ ,ξ r (1 + S s, + V ) .Proof. By Lemma 2.8, there exists (cid:15) s,C ∗ ,ξ ,ξ ∈ (0 , /
16] such that if k ≥ v ∈ V k , and α k,v ≤ (cid:15) s,C ∗ ,ξ,ξ , then τ s ( k, v, v (cid:48) ) = 0 for all ( v, v (cid:48) ) ∈ Flat ( k ). Set α ∗ = min { (cid:15) s,C ∗ ,ξ ,ξ , α } (a careful inspection shows (cid:15) s,C ∗ ,ξ ,ξ is strictly smaller than α ). Thus, with α = α ∗ , ∞ (cid:88) k =0 (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) ρ sk + ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ∗ ρ sk = ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ∗ ρ sk < ∞ . The conclusion follows immediately by Theorem 5.1. (cid:3)
Corollary 5.5.
Alternatively, if S s,p V := ∞ (cid:88) k =0 (cid:88) v ∈ V k α pk,v ρ sk < ∞ for some p > , then there exists a (1 /s ) -H¨older map f : [0 , → X such that (cid:83) k ≥ V k ⊂ f ([0 , and theH¨older constant of f satisfies H (cid:46) s,C ∗ ,ξ ,ξ r (1 + ( α ∗ ) − p S s,p V ) .Proof. Inspecting the definitions of the two sums, S s, + V ≤ ( α ∗ ) − p S s,p V . (cid:3) We now turn to the proof of Theorem 1.1.
Remark 5.6.
For all x ∈ R N and r >
0, a minimal dyadic cube Q in R N such that x ∈ Q and 3 Q contains B ( x, r ) satisfies diam 3 Q ≤ Cr for some C = C ( N ) > Proof of Theorem 1.1.
Let N ≥ s > β > E ⊂ R N is a bounded set such that S s, + E = (cid:88) Q ∈ ∆( R N ) β E (3 Q ) ≥ β (diam Q ) s < ∞ . Pick any x ∈ E and set r = diam E . Define V = { x } . Assume that V k has beendefined for some k . Choose a maximal 2 − ( k +1) -separated set in E such that V k +1 ⊃ V k .Then the sequence V = ( V k , − k ) k ≥ satisfies conditions (V0)–(V4) in § C ∗ = 2and ξ = ξ = 1 /
2. Note that A ∗ = C ∗ − ξ = 4 , A ∗ = 120 . For all k ≥ v ∈ V k , let Q k,v be a minimal dyadic cube such that v ∈ Q k,v and 3 Q k,v contains B ( v, · − k r ) and choose (cid:96) k,v be a line such thatsup x ∈ E ∩ Q dist( x, (cid:96) k,v ) = β E (3 Q ) diam 3 Q. Then, by Remark 5.6, there exists C = C ( N ) > α k,v := 12 − k +1 r sup x ∈ V k +1 ∩ B ( v, · − k r ) dist( x, (cid:96) k,v ) ≤ diam 3 Q − k +1 r β E (3 Q ) ≤ Cβ E (3 Q ) . We now specify that β = α ∗ / C , where α ∗ is the constant from Corollary 5.4 and C the constant from Remark 5.6. Because each dyadic cube Q in R N is associated to some( k, v ) at most C ( n ) times, it follows that S s, + V (cid:46) N r − s S s, + E < ∞ . Therefore, by Corollary 5.4, there exists a (1 /s )-H¨older m fap f : [0 , → R N such that (cid:83) k ≥ V k ⊂ f ([0 , f satisfies H (cid:46) N,s r (1 + S s, + V ) (cid:46) N,s diam E + S s, + E (diam E ) s − . Because (cid:83) k ≥ V k is dense in E , the curve f ([0 , E . (cid:3) A refinement of Theorem 5.1.
The parameterization in Theorem 5.1 can be madein such a way so that the sequence of maps F k obtained are essentially 2-to-1 in the senseof the following proposition. Proposition 5.7.
Let V = ( V k , ρ k ) k ≥ and α satisfy the hypothesis of Theorem 5.1 andlet x ∈ V . There exists a sequence of piecewise linear maps F k : [0 , → X with thefollowing properties.(1) For all k ≥ , F k (0) = x = F k (1) .(2) For all k ≥ , there exists G k ⊂ [0 , such that F k | G k is 2-1 and F k ([0 , \ G k ) isa finite set.(3) For all k ≥ , F k ([0 , ⊃ V k ; for all x ∈ V k +1 , dist( x, F k ([0 , ≤ C ∗ ρ k +1 r .(4) For all k ≥ , (cid:107) F k − F k +1 (cid:107) ∞ (cid:46) C ∗ ,ξ ρ k +1 r .(5) For all k ≥ , the map F k is Lipschitz with Lip( F k ) (cid:46) s,C ∗ ,ξ ,ξ r (1 + S s V ) ρ − sk .The maps F k converge uniformly to a (1 /s ) -H¨older map F : [0 , → X whose imagecontains (cid:83) k ≥ V k , the parameterization F starts and ends at x in the sense of (1), andthe H¨older constant of F satisfies H (cid:46) s,C ∗ ,ξ ,ξ r (1 + S s V ) .Proof. Following the algorithm of §
3, we construct for each k ≥
0, four families E k , B k , F k , N k of intervals in [0 ,
1] and a continuous piecewise linear map f k : [0 , → X that satisfy(P1)–(P7). In Step 0, we may assume that f (0) = f (1) = x . Thus, f k (0) = f k (1) = x for all k ≥
0. Moreover, for all x ∈ V k +1 ,dist( x, F k ([0 , ≤ dist( x, V k ) < C ∗ r ρ k +1 by (V4). From the construction, (cid:107) f k − f k +1 (cid:107) ∞ (cid:46) A ∗ ρ k +1 r . Hence (cid:107) f k − f k +1 (cid:107) ∞ (cid:46) C ∗ ,ξ ρ k +1 r . We have shown that the maps f k satisfy properties (1), (3) and (4).As for property (2), we already know from (P4) that f k | (cid:83) E k is 2-to-1. We proceed tomodify f k on each I ∈ B k . From the algorithm in §
3, recall that for each I ∈ B k , thereexists unique I (cid:48) ∈ B k , I (cid:48) (cid:54) = I such that f k ( I ) = f k ( I (cid:48) ). Enumerate B k = { I , I (cid:48) , . . . , I l , I (cid:48) l } , where f k ( I j ) = f k ( I (cid:48) j ). Starting with I = ( a , b ), define ˜ f k | I so that ¨OLDER CURVES AND PARAMETERIZATIONS 41 (a) ˜ f k | I is piecewise linear and continuous, ˜ f k ( a ) = f k ( a ) and ˜ f k ( b ) = f k ( b );(b) H ( ˜ f k ( I )) ≤ | f k ( a ) − f k ( b ) | + ρ k r ;(c) ˜ f k ( I ) ∩ (cid:83) I ∈ E k f k ( I ) is a finite set.Let ψ : I (cid:48) → I be the unique orientation-reversing linear map from I (cid:48) onto I . Thendefine ˜ f k | I (cid:48) = ( ˜ f k | I ) ◦ ψ . For induction, assume that we have defined ˜ f k on I , I (cid:48) , . . . , I r − , I (cid:48) r − . Define ˜ f k | I r as with I , only this time we require that the set˜ f k ( I r ) ∩ (cid:32) r − (cid:91) i =1 ˜ f k ( I i ) ∪ r − (cid:91) i =1 ˜ f k ( I (cid:48) i ) ∪ (cid:91) I ∈ E k f k ( I ) (cid:33) be finite. Let ψ r : I (cid:48) r → I r be the unique orientation-reversing linear map from I (cid:48) r onto I r and define ˜ f k | I (cid:48) r = ( ˜ f k | I r ) ◦ ψ r . Extending ˜ f k | I = f k | I for all I ∈ E k ∪ N k ∪ F k , we obtaina sequence ˜ f k of maps that satisfy properties (1)–(4).The rest of the proof is similar to that of Theorem 5.1 and we only sketch the steps.Define the M s for each I ∈ I k and define the collections of intervals { ∆ w } and { ∆ (cid:48) w } .For each w , let φ w : ∆ (cid:48) w → ∆ w be the unique affine homeomorphism from ∆ (cid:48) w onto ∆ w and let F k | ∆ (cid:48) w = ( ˜ f k | ∆ w ) ◦ φ w . Although the maps f k are different from ˜ f k , we have by(b) that diam f k ( I ) (cid:39) ξ diam ˜ f k ( I ) for all k ≥ I ∈ I k . Thus, Lemma 5.2 andLemma 5.3 still hold with constants depending at most on s , C ∗ , ξ and ξ . Therefore,the maps F k satisfy properties (1)–(5). (cid:3) A Carleson condition for an upper Ahlfors s -regular curve. Replacing (5.1)in the main theorem with a Carleson-type condition ensures that the H¨older curve isupper Ahlfors regular. This answers a question posed to us by T. Orponen.
Theorem 5.8.
Assume that X = l ( R ) or X = R N for some N ≥ . Let s ≥ , let V = (( V k , ρ k )) k ≥ be a sequence of finite sets V k in X and numbers ρ k > that satisfyproperties (V0)–(V4) defined in § Λ ≥ C ∗ and Λ ∗ := Λ / (1 − ξ ) . Suppose for all k ≥ and v ∈ V k +1 , we are given a line (cid:96) k,v and α k,v ≥ such that ( (cid:102) V5) sup x ∈ V k +1 ∩ B ( v, ∗ ρ k r ) dist( x, (cid:96) ) ≤ α k,v ρ k +1 . If Λ (cid:29) ξ ,ξ C ∗ , α ∈ (0 , α ] , and there exists M < ∞ such that for all j ≥ and w ∈ V j , (5.3) S s V ( j, w ) := ∞ (cid:88) k = j (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) v,v (cid:48) ∈ B ( w, Λ ρ j r ) τ s ( k, v, v (cid:48) ) ρ sk + ∞ (cid:88) k = j (cid:88) v ∈ V k α k,v ≥ α v ∈ B ( w, Λ ρ j r ) ρ sk ≤ M ρ sj , then there exists a (1 /s ) -H¨older map f : [0 , → X such that f ([0 , ⊃ (cid:83) k ≥ V k and thecurve f ([0 , is upper Ahlfors s -regular with constant depending on at most s , C ∗ , ξ , ξ , and M . Proof.
By (V0) and (V4), excess ( (cid:83) ∞ k =1 V k , V ) ≤ ρ r + ρ r + · · · = r − ξ ≤ A ∗ r , whereexcess( A, B ) = sup x ∈ A inf y ∈ B | x − y | whenever A and B are nonempty sets in X . Hence (cid:83) ∞ k =0 V k ⊂ B ( x , A ∗ r ) by (V1).Thus, there exists a (1 /s )-H¨older map f : [0 , → X such that Γ := f ([0 , ⊃ (cid:83) ∞ k =0 V k and the H¨older constant of f satisfies H f (cid:46) s,C ∗ ,ξ ,ξ r (1 + M ) by Theorem 5.1, since S s V = S s V (0 , w ) ≤ M . In particular, H s (Γ) ≤ H sf H ([0 , (cid:46) s,C ∗ ,ξ ,ξ ,M r s . Let x ∈ Γ and let 0 < r ≤ diam Γ. Because (cid:83) ∞ k =0 V k ⊂ B ( x , A ∗ r ), we have diam Γ ≤ A ∗ r . If r ≥ r , then H s (Γ ∩ B ( x, r )) ≤ H s (Γ) (cid:46) s,C ∗ ,ξ ,ξ ,M r s (cid:46) s,C ∗ ,ξ ,ξ ,M r s . Otherwise, 0 < r < r , say ρ j +1 r ≤ r < ρ j r for some integer j ≥ w ∈ V j such that | w − x | = dist( x, V j ). By Lemma 5.2, (V0), and (V4),dist( x, f j ([0 , ≤ ξ − ξ A ∗ ρ j r . Because α ≤ α , the longest line segment drawn between vertices in V j has length atmost 14 A ∗ ρ k − r . By (V0), it follows thatexcess( f j ([0 , , V j ) ≤ ξ A ∗ ρ j r . Thus, dist( x, V j ) (cid:46) ξ ,ξ A ∗ ρ j r (cid:28) ξ ,ξ Λ ρ j r . For all k ≥
0, define (cid:101) ρ k := ρ j + k ρ j . Define (cid:101) V := V j ∩ B ( w, Λ ρ j r ). Then, for each k ≥
1, recursively define (cid:101) V k to be set of all x ∈ V j + k ∩ B ( w, Λ ρ j r ) such that dist( x, (cid:101) V k − ) ≤ Λ ρ j + k r . Then (cid:101) V = ( (cid:101) V k , (cid:101) ρ k ) k ≥ satisfy(V0)–(V4) with respect to (cid:101) x = w and (cid:101) r = ρ j r , (cid:102) C ∗ = Λ, (cid:101) ξ = ξ , and (cid:101) ξ = ξ . For all k ≥ v ∈ (cid:101) V k , assign (cid:101) (cid:96) k,v := (cid:96) j + k,v and (cid:101) α k,v = α j + k,v . Then (cid:101) V satisfies (V5) withrespect to (cid:101) (cid:96) k,v and (cid:101) α k,v by ( (cid:102) V5). Moreover, by (5.3), S s (cid:101) V = ∞ (cid:88) k =0 (cid:88) ( v,v (cid:48) ) ∈ (cid:94) Flat ( k ) (cid:101) τ ( k, v, v (cid:48) ) (cid:101) ρ ks + ∞ (cid:88) k =0 (cid:88) v ∈ (cid:102) V k (cid:103) α k,v ≥ α (cid:101) ρ ks = S V ( j, w ) ρ sj ≤ M. Thus, by Theorem 5.1, there is a (1 /s )-H¨older map g with H¨older constant H g (cid:46) s, Λ ,ξ ,ξ (cid:101) r (1 + M ) such that g ([0 , (cid:83) k ≥ (cid:101) V k . Because the algorithm in § x, V j ) (cid:46) ξ ,ξ A ∗ ρ j r (cid:28) ξ ,ξ Λ ρ j r , and r < ρ j r , we can guarantee that g ([0 , f ([0 , ∩ B ( x, r ) provided that Λ is sufficiently large. Therefore, H s (Γ ∩ B ( x, r )) ≤ H sg H ([0 , (cid:46) s, Λ ,ξ ,ξ , (cid:101) r (1 + M ) (cid:46) s,C ∗ ,ξ ,ξ ,M ( ρ j r ) s (cid:46) s,C ∗ ,ξ ,ξ ,M r s , where the final inequality holds because ρ j +1 r ≤ r . (cid:3) ¨OLDER CURVES AND PARAMETERIZATIONS 43 Lipschitz parameterization
Using the method above, we obtain the following refinement of the sufficient half of theAnalyst’s TST in Hilbert space, which is originally due to Jones [Jon90] in the Euclideancase and due to Schul [Sch07b] in the infinite-dimensional case.
Proposition 6.1 (Sufficient half of the Analyst’s Traveling Salesman with Nets) . Assumethat X = l ( R ) or X = R N for some N ≥ . Let V = ( V k , ρ k ) k ≥ be a sequence of finitesets V k in X and numbers ρ k > that satisfy properties (V0)–(V5) defined in § (6.1) S V := ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ρ k < ∞ , then for every x ∈ V , we can find a sequence of piecewise linear maps F k : [0 , → X with the following properties.(1) For all k ≥ , F k (0) = x = F k (1) .(2) For all k ≥ , there exists G k ⊂ [0 , such that F k | G k is 2-1 and F k ([0 , \ G k ) isa finite set.(3) For all k ≥ , F k ([0 , ⊃ V k ; for all x ∈ V k +1 , dist( x, F k ([0 , ≤ C ∗ ρ k +1 r .(4) For all k ≥ , (cid:107) F k − F k +1 (cid:107) ∞ (cid:46) C ∗ ,ξ ρ k +1 r .(5) For all k ≥ , the map F k is Lipschitz with Lip( F k ) (cid:46) C ∗ ,ξ ,ξ r (1 + S V ) .The maps F k converge uniformly to a Lipschitz map F : [0 , → X whose image contains (cid:83) k ≥ V k , the parameterization F starts and ends at x in the sense of (1), and the Lipschitzconstant of F satisfies L (cid:46) C ∗ ,ξ ,ξ r (1 + S V ) .Proof. Let α = α (see (4.2)), which depends only on C ∗ , ξ , and ξ . If ( v, v (cid:48) ) ∈ Flat ( k ),then τ ( k, v, v (cid:48) ) ≤ α k,v by Lemma 2.7. Thus, by definition of S V (see Theorem 5.1), S V = ∞ (cid:88) k =0 (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ ( k, v, v (cid:48) ) ρ k + ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α ρ k ≤ ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v <α α k,v ρ k + 1 α ∞ (cid:88) k =0 (cid:88) v ∈ V k α k,v ≥ α α k,v ρ k ≤ α S V . The conclusion now follows from Proposition 5.7. (cid:3)
Part
II.
Applications and Further Results In §
7, we give an application of the H¨older Traveling Salesman theorem to the geometryof measures. In particular, we obtain sufficient conditions for a pointwise doubling measurein R N to be carried by (1 /s )-H¨older curves, s >
1. This extends the work [BS15, BS17]by the first author and Schul, which characterizes 1-rectifiable Radon measures in R N in terms of geometric square functions. In §
8, we use the method of Part I to obtaina Wa˙zewski type theorem for flat continua, which we described above in § in §
9, we present examples of H¨older curves and of sets that are not contained in anyH¨older curve to highlight the rich geometry of sets in R N and illustrate the strengths andlimitations of our principal results.7. Fractional rectifiability of measures
One goal of geometric measure theory is to understand the structure of a measurein R N through its interaction with families of lower dimensional sets. For an extendedintroduction, see the survey [Bad19] by the first author. In this section, we use the H¨olderTraveling Salesman theorem to establish criteria for fractional rectifiability of pointwisedoubling measures in terms of L p Jones beta numbers. In particular, we extend part ofthe recent work of the first author and Schul [BS17] on measures carried by rectifiablecurves to measures carried by H¨older curves (see Theorem 7.5). The study of fractional(that is, non-integer dimensional) rectifiability of measures was first proposed by Mart´ınand Mattila [MM93, MM00] and examined further by the first and third author [BV18].7.1.
Generalized rectifiability.
Let A be a nonempty family of Borel sets in R N andlet µ be a Borel measure on R N . We say that µ is carried by A if there exists a sequence( A i ) ∞ i =1 of sets in A such that µ ( R N \ (cid:83) i A i ) = 0. At the other extreme, we say that µ is singular to A if µ ( A ) = 0 for all A ∈ A . If µ is σ -finite, then µ can be uniquelywritten as the sum of a Borel measure µ A carried by A and a Borel measure µ ⊥A singularto A (e.g. see the appendix of [BV18]). These definitions encode several commonly usednotions of rectifiability of measures (see [Bad19]).Let 1 ≤ m ≤ N −
1. Let A denote the family of Lipschitz images of [0 , m in R N .We say that a Borel measure µ is m -rectifiable if µ is carried by A ; we say that µ is purely m -unrectifiable if µ is singular to A . A Borel set E ⊂ R n with 0 < H m ( E ) < ∞ is called m -rectifiable or purely m -unrectifiable if H m E , the m -dimensional Hausdorffmeasure restricted to E , has that property. The classes of 1-rectifiable sets and purely1-unrectifiable sets are also called Besicovitch regular sets and
Besicovitch irregular sets,respectively, in reference to the pioneering investigations by Besicovitch [Bes28, Bes38]into the geometry of 1-sets in the plane.
Example 7.1.
Let Γ , Γ , . . . be a sequence of rectifiable curves in R N and let a , a , . . . be a sequence of positive weights. Then the measure µ = (cid:80) i a i H Γ i is 1-rectifiable.Note that if the closure of (cid:83) i Γ i is R N and the weights are chosen so that (cid:80) i a i H (Γ i ) = 1,we get a 1-rectifiable Borel probability measure µ whose support is R N . Example 7.2.
Let C ⊂ R be the middle halves Cantor set (formed by replacing [0 , , ] ∪ [ ,
1] at iterating). Then E = C × C ⊂ R is a Cantor set of Hausdorffdimension one, 0 < H ( E ) < ∞ , E is Ahlfors 1-regular in the sense that H ( E ∩ B ( x, r )) (cid:39) r for all x ∈ E and 0 < r ≤ diam E, and E is Besicovitch irregular (e.g. see [MM93]). In particular, the set E is compact andmeasure-theoretically one-dimensional, but E is not contained in any rectifiable curve. ¨OLDER CURVES AND PARAMETERIZATIONS 45 L p Jones beta numbers and rectifiability.
Let µ be a Radon measure on R N ,that is, a locally finite Borel regular measure, let 1 ≤ m ≤ N −
1, let p >
0, let x ∈ R N ,let r >
0, and let L be an m -dimensional affine subspace of R N . We define(7.1) β ( m ) p ( µ, x, r, L ) := (cid:18)(cid:90) B ( x,r ) (cid:18) dist( z, L ) r (cid:19) p dµ ( z ) µ ( B ( x, r )) (cid:19) /p , (7.2) β ( m ) p ( µ, x, r ) := inf L β ( m ) p ( µ, x, r, L ) , where the infimum is taken over all m -planes L in R N . The quantity β ( m ) p ( µ, x, r ) is calledthe m -dimensional L p Jones beta number of µ in B ( x, r ). The L p Jones beta numberswere introduced by David and Semmes [DS91, DS93] to study quantitative rectifiabilityof Ahlfors regular sets and boundedness of singular integral operators. The normalizationof the measure in (7.1) that we have chosen (i.e. dividing by µ ( B ( x, r ))) ensures that β ( m ) p ( µ, x, r ) ∈ [0 ,
1] and β ( m ) p is invariant under dilations T λ ( z ) = λz in the sense that(7.3) β ( m ) p ( µ, x, r ) = β ( m ) p ( T λ [ µ ] , λx, λr ) , T λ [ µ ]( E ) = µ ( λ − E )for all µ , x ∈ R N , r >
0, and λ >
0. By monotonicity of the integral,(7.4) sµ ( B ( y, s )) /p β ( m ) p ( µ, y, s ) ≤ rµ ( B ( x, r )) /p β ( m ) p ( µ, x, r ) when B ( y, s ) ⊂ B ( x, r ) . In a pair of papers, Tolsa [Tol15] and Azzam and Tolsa [AT15] characterize m -rectifiableRadon measures µ on R N with µ (cid:28) H m in terms of L Jones beta numbers. The restriction µ (cid:28) H m is equivalent to the upper density bound lim sup r ↓ r − m µ ( B ( x, r )) < ∞ µ -a.e.(e.g. see [Mat95, Chapter 6]) and implies that the Hausdorff dimension of the measureis at least m (see [MMR00]). The proof that (7.6) implies the measure µ is m -rectifiableuses an intricate stopping time argument in conjunction with David and Toro’s Reifenbergalgorithm for sets with holes [DT12] to construct bi-Lipschitz images of R m inside R N thatcarry µ . For related developments, see [ENV17, Ghi18]. Theorem 7.3 (see [Tol15], [AT15]) . Let µ be a Radon measure on R N . Assume that (7.5) 0 < lim sup r ↓ µ ( B ( x, r )) r m < ∞ for µ -a.e. x ∈ R N . Then µ is m -rectifiable if and only if (7.6) (cid:90) β ( m )2 ( µ, x, r ) µ ( B ( x, r )) r m drr < ∞ for µ -a.e. x ∈ R N . In [BS17], the first author and Schul characterize 1-rectifiable Radon measure µ on R N in terms of L p Jones beta numbers and the lower density lim inf r ↓ r − µ ( B ( x, r )).In contrast with Theorem 7.3, the main theorem in [BS17] does not require an a priori relationship between the null sets of µ and H , nor a bound on the Hausdorff dimension of µ . To lighten the notation, we present Badger and Schul’s theorem for pointwise doublingmeasures and refer the reader to [BS17, Theorem A] for the full result. Although theclasses of measures satisfying (7.5) and (7.7) have no direct relationship with each other, a posteriori an m -rectifiable measure satisfying (7.5) also satisfies (7.7). The proof that(7.8) implies the measure µ is 1-rectifiable uses a technical extension of the sufficient halfof the Analyst’s Traveling Salesman theorem. See [BS17, Proposition 3.6]. Theorem 7.4 (see [BS17, Theorem E]) . Let µ be a Radon measure on R n and let p ≥ .Assume that µ is pointwise doubling in the sense that (7.7) lim sup r ↓ µ ( B ( x, r )) µ ( B ( x, r )) < ∞ for µ -a.e. x ∈ R N .Then µ is 1-rectifiable if and only if (7.8) (cid:90) β (1) p ( µ, x, r ) rµ ( B ( x, r )) drr < ∞ for µ -a.e. x ∈ R N . Sufficient conditions for fractional rectifiability.
The following theorem is anapplication of the H¨older Traveling Salesman theorem and generalizes the “sufficient half”of Theorem 7.4 (also see [BV18, Theorem A]). The exponents p and q in the H¨older case( s >
1) are less restrictive than in the Lipschitz case ( s = 1). Theorem 7.5.
Let µ be a Radon measure on R N , let s > , and let p, q > . Then µ (cid:26) x ∈ R n : (cid:90) β (1) p ( µ, x, r ) q r s µ ( B ( x, r )) drr < ∞ and lim sup r ↓ µ ( B ( x, r )) µ ( B ( x, r )) < ∞ (cid:27) is carried by (1 /s ) -H¨older curves. At the core of the proof of Theorem 7.5 is the following lemma.
Lemma 7.6.
Let µ be a Radon measure in R N , and let s > and p, q > be fixed.Given x ∈ R N and parameters M > , θ > , and P > , let A denote the set of points x ∈ B ( x , / such that (7.9) (cid:90) β (1) p ( µ, x, r ) q r s µ ( B ( x, r )) drr ≤ M, (7.10) µ ( B ( x, r )) ≤ P µ ( B ( x, r )) for all r ∈ (0 , , and let A (cid:48) denote the set of points in A such that (7.11) µ ( A ∩ B ( x, r )) ≥ θµ ( B ( x, r )) for all r ∈ (0 , . Then A (cid:48) is contained in a (1 /s ) -H¨older curve Γ = f ([0 , with H¨older constant dependingon at most N , s , p , q , M , P , θ , and µ ( A ) .Proof. Let { A (cid:48) k } k ≥ be a nested sequence of 2 − k -nets in A (cid:48) , so that the sets V k ≡ A (cid:48) k and scales ρ k = 2 − k satisfy conditions (V0)–(V4) of § r = 1, C ∗ = 2, ξ = ξ = 1 /
2. Note that A ∗ = C ∗ − ξ = 4 and 30 A ∗ = 120 . ¨OLDER CURVES AND PARAMETERIZATIONS 47 By (7.9),
M µ ( A ) ≥ (cid:90) A (cid:90) β (1) p ( µ, x, r ) q r s µ ( B ( x, r )) drr dµ ( x )= ∞ (cid:88) k =9 (cid:90) − k − ( k +1) (512 r ) s (cid:90) A β (1) p ( µ, x, r ) q µ ( B ( x, r )) dµ ( x ) drr (7.12)where in the second line we used the change of variables r (cid:55)→ r (note 512 = 2 ) andTonelli’s theorem. Now, the open balls { B ( y, − ( k +1) ) : y ∈ A (cid:48) k } are pairwise disjoint,because the points in A (cid:48) k are separated by distance at least 2 − k . Thus,(7.13) M µ ( A ) ≥ ∞ (cid:88) k =9 (cid:90) − k − ( k +1) r s (cid:88) y ∈ A (cid:48) k (cid:90) A ∩ B ( y, − ( k +1) ) β (1) p ( µ, x, r ) q µ ( B ( x, r )) dµ ( x ) (cid:124) (cid:123)(cid:122) (cid:125) I ( k,y,r ) drr . Next, we bound I ( k, y, r ) from below. Fix k ≥ y ∈ A (cid:48) k , and r ∈ [2 − ( k +1) , − k ].Suppose that x ∈ A ∩ B ( y, − ( k +1) ). Then(7.14) µ ( B ( x, r )) ≤ P µ ( B ( x, r )) ≤ P µ ( B ( y, r )) ≤ P µ ( B ( y, · − k ))by (7.10). Since B ( y, · − k ) ⊂ B ( x, · − k ) ⊂ B ( x, r ), it follows that β (1) p ( y, · − k ) ≤ (cid:18) r · − k (cid:19) (cid:18) µ ( B ( x, r )) µ ( B ( y, · − k )) (cid:19) /p β (1) p ( µ, x, r ) ≤ P /p β (1) p ( µ, x, r )(7.15)by (7.4). Hence(7.16) I ( k, y, r ) ≥ − q P − − q/p β (1) p ( µ, y, · − k ) q µ ( B ( y, · − k )) (cid:90) A ∩ B ( y, − ( k +1) ) dµ ( x ) . Invoking doubling again, µ ( B ( y, · − k )) ≤ P µ ( B ( y, − ( k +1) )). Thus, by (7.11),(7.17) 1 µ ( B ( y, · − k )) (cid:90) A ∩ B ( y, − ( k +1) ) dµ ( x ) ≥ P − µ ( A ∩ B ( y, − ( k +1) )) µ ( B ( y, − ( k +1) )) ≥ P − θ. Therefore,(7.18) I ( k, y, r ) ≥ − q P − − q/p θ β (1) p ( µ, y, · − k ) q Combining (7.13) and (7.18), we obtain(7.19) 3 q P q/p θ − M µ ( A ) ≥ ∞ (cid:88) k =9 (cid:32)(cid:90) − k − ( k +1) r s drr (cid:33) (cid:88) y ∈ A (cid:48) k β (1) p ( µ, y, · − k ) q . In particular, we conclude that(7.20) ∞ (cid:88) k =9 (cid:88) y ∈ A (cid:48) k β (1) p ( µ, y, · − k ) q − ks ≤ s − − s q P q/p θ − M µ ( A ) < ∞ For each k ≥ y ∈ A (cid:48) k , let (cid:96) k,v be any line such that(7.21) β (1) p ( µ, y, · − k , (cid:96) k,v ) ≤ β (1) p ( µ, y, · − k ) . We will now bound the distance of points in A (cid:48) k +1 ∩ B ( y, · − k ) to (cid:96) k,v . Fix any point z ∈ A (cid:48) k +1 ∩ B ( y, · − k ) and let t − k +1 = dist( z, (cid:96) k,v ). Then β p ( µ, y, · − k , (cid:96) k,v ) q ≥ (cid:32) t − ( k +1) · − k (cid:33) q (cid:32) µ ( B ( z, t − ( k +1) )) µ ( B ( y, · − k )) (cid:33) q/p ≥ (cid:18) t (cid:19) q P − ( q/p ) log (1920 /t ) ≥ (cid:18) t (cid:19) q +( q/p ) log ( P ) , (7.22)where in the second line we used doubling of µ at z . It follows that(7.23) α k,v := 2 k +1 sup z ∈ A (cid:48) k +1 ∩ B ( y, · − k ) dist( z, (cid:96) k,v ) ≤ C ( p, q, P ) β p ( µ, y, · − k , (cid:96) k,v ) η , where η [ q + ( q/p ) log ( P )] = q . Therefore, all together,(7.24) ∞ (cid:88) k =9 (cid:88) y ∈ A (cid:48) k α q +( q/p ) log ( P ) k,v − ks ≤ C ( s, p, q, M, P, θ, µ ( A )) < ∞ . Finally, by Corollary 5.5, the set A (cid:48) is contained in the Hausdorff limit of A (cid:48) k and thisis contained in a (1 /s )-H¨older curve Γ = f ([0 , s , p , q , M , P , θ , and µ ( A ). (cid:3) Theorem 7.5 follows from countably many applications of Lemma 7.6 and a standarddensity theorem for Radon measures in R N . See the proof of [BV18, Theorem 6.7], wherea similar argument is employed. We leave the details to the reader.8. H¨older parameterization of flat continua
The goal of this section is to prove Proposition 1.3, which we now restate.
Proposition 8.1.
There exists a constant β ∈ (0 , such that if s > and E ⊂ R N iscompact, connected, H s ( E ) < ∞ , E is lower Ahlfors s -regular with constant c , and (8.1) β E (cid:0) B ( x, r ) (cid:1) ≤ β for all x ∈ E and < r ≤ diam E, then E = f ([0 , for some injective (1 /s ) -H¨older continuous map f : [0 , → R N withH¨older constant H (cid:46) s c − H s ( E )(diam E ) − s . The proposition is trivial if E is a singleton (in which case (8.1) is vacuous). Thus, wemay assume that E ⊂ R N is a continuum ; that is, E is compact, connected, and containsat least two points. Furthermore, as the hypothesis and the conclusion are scale-invariant,we may assume without loss of generality that diam E = 1. To complete the proof of theproposition, we mimic the proof of Theorem 5.1, but with a few modifications. In § §
3. Then, in § M s ([0 , c − H s ( E ), which fills the role that Proposition ¨OLDER CURVES AND PARAMETERIZATIONS 49 mutatis mutandis .At the core of the proof of Proposition 8.1 is the following property, which is satisfied bycontinua that are sufficiently flat at all locations and scales. We defer a proof of Lemma8.2 to § Lemma 8.2.
Suppose that E ⊂ R N is a continuum satisfying β E (cid:0) B ( x, r ) (cid:1) ≤ − for all x ∈ E and < r ≤ diam E. Then for all distinct x, y ∈ E and for all z ∈ [ x, y ] , there exists z (cid:48) ∈ E such that π (cid:96) x,y ( z (cid:48) ) = z and | z − z (cid:48) | ≤ − | x − y | , where (cid:96) x,y denotes the line containing x and y . Traveling Salesman algorithm for flat continua.
Fix a constant β > E ⊂ R N be a continuum satisfying the hypothesis of Proposition8.1. Without loss of generality, we assume that diam E = 1. Pick x and y such that | x − y | = 1, set r = 1, C ∗ = 2, and ξ = ξ = 1 /
2. Then A ∗ = C ∗ − ξ = 4 , A ∗ = 56 , A ∗ = 120 . Furthermore, the scales ρ k = 2 − k satisfy (V0).Set V = { x , y } , α ,x = 4 β E (cid:0) B ( x , (cid:1) and α ,y = 4 β E (cid:0) B ( y , (cid:1) , and let (cid:96) ,x and (cid:96) ,y be best fitting lines corresponding to β E on the closed balls B ( x ,
1) and B ( y , k ≥ ≤ j ≤ k , we have defined sets V j , numbers α j,v ≥ v ∈ V j , and lines (cid:96) j,v for all v ∈ V j satisfying (V5). Choose V k +1 to be anymaximal 2 − ( k +1) -separated subset of E such that V k +1 ⊃ V k . For each v ∈ V k +1 , set α k +1 ,v := 2 r k +1 − ( k +2) β E (cid:0) B ( v, r k +1 ) (cid:1) ≤ β E (cid:0) B ( v, r k +1 ) (cid:1) , where r k +1 := min { · − ( k +1) , } , and let (cid:96) k +1 ,v be a best fitting line corresponding to β E (cid:0) B ( v, r k ) (cid:1) . The reader may check that the sequence of sets ( V k ) ∞ k =0 satisfy properties(V1)–(V5) in 2.2.We now specify α = 512 β ≤ /
16. This ensures that α k,v < α for all k ≥ v ∈ V k . Moreover, β is sufficiently small that we may invoke Lemma 8.2 for E . With α fixed, carry out a modified version of the algorithm in §
3, in which (P4), (P6), and (P7)are replaced by:(P4’) f k | (cid:83) E k is one-to-one.(P6’) For each I ∈ N k ∪ F k , the image f k ( I ) ∈ V k . For every v ∈ V k , there exists aunique interval I ∈ N k ∪ F k such that v = f k ( I ).(P7’) If I k = E k ∪ B k ∪ N k ∪ F k is enumerated according to the natural order in [0 , I k = { I , . . . , I l +1 } , then the intervals alternate between elements of N k ∪ F k and E k . (Thus, the family B k = ∅ .) Moreover, card N k = 2 and I , I l +1 ∈ N k .The vertices f k ( I ) and f k ( I l +1 ) are 1-sided terminal in V k , while each other vertex f k ( I j +1 ) is non-terminal in V k for all 1 ≤ j ≤ l − We now sketch some steps in the modified algorithm.8.1.1.
Step 0.
Partition [0 ,
1] = [0 , / ∪ (1 / , / ∪ [2 / ,
1] and assign E = { (1 / , / } , B = ∅ , N = { [0 , / , [2 / , } , F = ∅ . Also set f ([0 , / x and f ([2 / , y , and define f | (1 / , /
3) to be the restrictionof the affine map which interpolates between x and y . Verifying properties (P1), (P2),(P3), (P4’), (P5), (P6’), and (P7’) is straightforward. We omit the details.8.1.2. Induction Step.
Suppose that E k , B k , N k , F k , and f k have been defined and satisfyproperties (P1), (P2), (P3), (P4’), (P5), (P6’), and (P7’).By (P7’) and the induction assumption, the procedure in § § I ∈ E k as written, except assign all closed subintervals N k +1 ( I ) ∪ F k +1 ( I ) generated in I to F k +1 ( I ) instead of N k +1 ( I ). Also set N k +1 ( I ) = ∅ .Below, we check that B k +1 ( I ) = ∅ ; see Lemma 8.3.Because α k,v < α for all v ∈ V k , the procedure in § § § I k containing 0 and 1 belong to N k and f k maps each of them onto a 1-sided terminal vertex in V k . Let I be the interval in N k containing 0 and let v = f k ( I ). Let ( v, v (cid:48) ) ∈ Flat ( k )be the unique flat pair with first coordinate v . Choose an orientation for (cid:96) k,v so that [ v, v (cid:48) ] lies on the right side of v . Enumerate the points in V k +1 ∩ B ( f k ( I ) , C ∗ ρ k +1 r ) on the left side of v (including v ) as v l , v l − , . . . , v = v ,starting at the leftmost vertex and working right. The construction splitsinto three cases. Case 1a. If l = 1, then no new points appeared to the left of v and weset N k +1 ( I ) = I and f k +1 | I = f k | I . Set E k +1 ( I ) = B k +1 ( I ) = F k +1 ( I ) = ∅ . Case 1b. If l = 2, then one new point appeared to the left of v , the newpoint is 1-sided terminal in V k +1 and v is non-terminal in V k +1 . Subdivide I = [0 , a ] = [0 , a ] ∪ ( a, a ) ∪ [ a, a ], set N k +1 ( I ) = [0 , a ], E k +1 ( I ) =( a, a ), F k +1 ( I ) = [ a, a ], and B k +1 = ∅ . Define f k +1 | I by assigning f k +1 ([0 , a ]) = v , f k +1 | ( a, a ) to be the restriction of the affine map thatinterpolates between v and v , and f k +1 ([ a, a ]) = v . Case 1c. If l ≥
3, then subdivide I into 2 l − N k +1 ( I )and the subsequent closed intervals to F k +1 ( I ). Assign the open intervalsin E k +1 ( I ) and set B k +1 ( I ) = ∅ . The map f k +1 | I is the piecewise linearmap starting at v l , connecting v l to v l − , . . . , connecting v to v , which isconstant on the intervals in N k +1 ( I ) ∪ F k +1 ( I ) and constant speed on theintervals in E k +1 ( I ). ¨OLDER CURVES AND PARAMETERIZATIONS 51 Carry out a similar construction for the interval J in N k containing 1,but modified so that N k +1 ( J ) contains only one interval and that intervalcontains 1.Because α k,v < α for all v ∈ V k , the procedure in § Lemma 8.3.
For all k ≥ , B k = ∅ . Moreover, for all I ∈ E k , diam f k ( I ) < · − k .Proof. We need to check that, for flat continua, the procedure in § I ∈ E k , let v and v (cid:48) denote the endpoints of f k ( I ). Choose anorientation of (cid:96) k,v so that v (cid:48) lies to the right of v . Enumerate V k +1 ( v, v (cid:48) ) = { v , . . . , v n } ,where v = v , v n = v (cid:48) , and v i +1 is the first point to the right of v i for all 1 ≤ i ≤ n − B k +1 ( I ) (cid:54) = ∅ . Then | v j +1 − v j | ≥ A ∗ ρ k +1 = 56 · − ( k +1) for some 1 ≤ j ≤ n − x = ( v j + v j +1 ) / v j and v j +1 . By Lemma 8.2, thereexists y ∈ E such that | y − x | ≤ (1 / | v j +1 − v j | . Thus,dist( y, V k +1 ) ≥ dist( x, V k +1 ) − | y − x | ≥ | v j − v j − | > · − ( k +1) . This contradicts our assumption that V k +1 is a maximal 2 − ( k +1) -separated set for E .Therefore, B k +1 ( I ) = ∅ for all I ∈ E k . The only other instances in the algorithm wherebridge intervals could be generated are in the procedures in §§ α k,v < α for all k ≥ v ∈ V k , these procedures were never used.Similarly, suppose to get a contradiction that there exists I ∈ E k such that diam f k ( I ) ≥ · − k . Let v and v (cid:48) denote the endpoints of f k ( I ), and let x = ( v + v (cid:48) ) / y ∈ E such that | y − x | < (1 / | v − v (cid:48) | . Thendist( y, V k ) ≥ dist( x, V k ) − | x − y | ≥ | v − v (cid:48) | ≥ − k . This contradicts our assumption that V k +1 is a maximal 2 − ( k +1) -separated set for E . (cid:3) Verifying properties (P1), (P2), (P3), (P4’), (P5), (P6’), and (P7’) for E k +1 , B k +1 , N k +1 , F k +1 , and f k +1 is again routine. We leave the details to the reader.8.2. Mass estimate.
Let M s ([0 , § Lemma 8.4. M s ([0 , ≤ s c − H s ( E ) .Proof. Fix a finite tree T over [0 ,
1] of depth m (see §
4) and suppose that ∂T = { ( k (cid:48) , J ) , ( k , I ) , . . . , ( k (cid:48) l , J l ) , ( k l , I l ) , ( k (cid:48) l +1 , J l +1 ) } , enumerated according to the orientation of [0 ,
1] so that { I , . . . , I l } ⊆ (cid:91) k ≥ E k and { J , . . . , J l +1 } ⊆ (cid:91) k ≥ ( N k ∪ F k ) . The first interval J ∈ N k (cid:48) and the last interval J l +1 ∈ N k (cid:48) l +1 , since they contain 0 and 1,respectively. The remaining intervals J i ∈ F k (cid:48) i , because they do not contain 0 or 1. For each 1 ≤ i ≤ l , let x i denote the midpoint of f k i ( I i ). Claim 8.5.
One one hand, the set E ∩ B ( x i , (3 / − k i ) is nonempty for all ≤ i ≤ l .On the other hand, the family (cid:8) E ∩ B ( x i , (1 / − k i ) : 1 ≤ i ≤ l (cid:9) is pairwise disjoint.Proof. Given 1 ≤ i ≤ l , let v i and v (cid:48) i denote the endpoints of f k i ( I i ). By Lemma 8.3, | v i − v (cid:48) i | < · − k i . Hence there exists z i ∈ E ∩ B ( x i , (1 / | v i − v (cid:48) i | ) ⊂ E ∩ B ( x i , (3 / · − k i ) by Lemma 8.2.Suppose in order to reach a contradiction that there exists z ∈ E ∩ B ( x i , (1 / − k i ) ∩ B ( x j , (1 / − k j )for some i (cid:54) = j with 0 ≤ k i ≤ k j ≤ m . Case 1 . Suppose that k j ≥ k i + 3. Then | v j − x i | ≤ | v j − x j | + | x j − z | + | z − x i | < · − k j + 14 2 − k j + 14 · − k i < · − k i and similarly for v (cid:48) j . Because { v j , v (cid:48) j } ⊂ B ( x i , − k i ), we conclude that v j and v (cid:48) j liebetween v i and v (cid:48) i with respect to the linear ordering of V i ∩ B ( v i , · − k i ). It followsthat I j is contained in I i , but I j (cid:54) = I i . This contradicts the assertion that I i ∈ ∂T . Case 2.
Suppose that k j ≤ k i + 2. Then | v i − v j | ≤ | v i − x i | + | x i − z | + | z − x j | + | x j − v j | <
32 2 − k i + 14 · − k i + 14 · − k j + 32 2 − k j < · − k j , | v (cid:48) i − v j | ≤ | v (cid:48) i − v i | + | v i − v j | < · − k i + 8 · − k j ≤ · − k j . In particular, v i , v (cid:48) i , v j , v (cid:48) j belong to V j ∩ B ( v j , · − k j ), which is linearly ordered byLemma 2.2, where v j and v j +1 are consecutive points. Assume that v i and v (cid:48) i both lie onthe left or the right side of [ v j , v (cid:48) j ], say without loss of generality that the appear from leftto right as v j , v (cid:48) j , v i , v (cid:48) i . Then | x i − v (cid:48) j | ≥ | v (cid:48) i − v i | ≥ · − k i > · − k i . It follows that B ( v j , · − k j ) ∩ B ( v i , · − k i ) is empty, which contradicts our assumption.Thus, one of v i or v (cid:48) i lies on the left side of [ v j , v (cid:48) j ] and the other lies on the right side.Then I j ⊂ I i . If k j ≥ k i + 1, then we reach the same contradiction as in Case 1. If k j = k i ,then it follows that I j = I i , which contradicts our assumption that i (cid:54) = j . (cid:3) We now continue with the proof of Lemma 8.4. By Claim 8.5, we can find balls B ( z i , (1 / − k i ) centered in E for all 1 ≤ i ≤ l , which are pairwise disjoint. Moreover,because E is lower Ahlfors regular, we have c [(1 / − k i ] s ≤ H s ( E ∩ B ( z i , (1 / − k i ))for all 1 ≤ i ≤ l . Therefore, by Lemma 8.3 and additivity of measures on disjoint sets, (cid:88) ( k,I ) ∈ ∂T (diam f k ( I )) s = l (cid:88) i =1 (diam f k i ( I i )) s ≤ l (cid:88) i =1 (3 · − k i ) s ≤ s c − H s ( E ) . ¨OLDER CURVES AND PARAMETERIZATIONS 53 Because T was an arbitrary finite tree over [0 , M s ([0 , (cid:3) Corollary 8.6. If k ≥ , I ∈ I k , and a is an endpoint of I , then M s ( k, I ) ≤ s c − H s ( E ∩ B ( f k ( a ) , · − k )) . Proof.
Let T be a finite tree over ( k, I ). By Lemma 8.3, the image of any edge intervalin ∂T is contained in a ball centered at f k ( a ) of radius at most3 · − k + 3 · − ( k +1) + · · · = 6 · − k . The conclusion follows by repeating the proof of Lemma 8.4. (cid:3)
Proof of Proposition 8.1.
With Lemma 8.4 in hand, follow the proof of Theorem5.1 in § mutatis mutandis . Construct families of intervals E (cid:48) k , N (cid:48) k , and F (cid:48) k as in § • If I ∈ E k , say I ∈ E k ( I ) for some I ∈ E k − ∪ N k − , then the correspondinginterval I (cid:48) ∈ E (cid:48) k satisfies(8.2) diam I (cid:48) = M s ( k, I ) M s ([0 , E k ( I )) diam I − (cid:88) J ∈ I k ( I ) M s ( k, J ) M s ([0 , . • If I ∈ F k , then the corresponding interval I (cid:48) ∈ F (cid:48) k satisfies diam I (cid:48) = 0. That is, I (cid:48) is a singleton. • If I ∈ N k , then the corresponding interval I (cid:48) ∈ N (cid:48) k satisfies diam I (cid:48) = M s ( k, I ) M s ([0 , Lemma 8.7.
For any (cid:15) > , there exists k ≥ such that diam I ≤ (cid:15) for all k ≥ k and I ∈ I (cid:48) k .Proof. Fix (cid:15) >
0. Note that if J ∈ E l , then card( E l +1 ( I )) ≥ f l ( J ) ≥ − l and V l +1 is a maximal 2 − ( l +1) -separated set in E .Suppose that I ∈ I k . Set J k = I . Inductively, given J l ∈ I l , let J l − denote the uniqueinterval in I l − with J l ⊂ J l − . That is, I = J k ⊂ J k − ⊂ · · · ⊂ J ⊂ J = [0 , . For each i ∈ { , . . . , k } , let J (cid:48) i ∈ I (cid:48) i be the interval associated to J i . We claim that foreach i ∈ { , . . . , k } (8.3) diam J (cid:48) i ≤ i (cid:88) l =0 l − i M s ( l, J l ) M s ([0 , . We prove (8.3) by induction. For i = 0 the claim is clear. Assume that (8.3) is true for0 ≤ i < k . If J i + i ∈ F i +1 ( J i ), then diam J (cid:48) i +1 = 0 and (8.3) is clear. If J i + i ∈ F i +1 ( J i ), then diam J (cid:48) i +1 = M s ( i + 1 , J i +1 ) / M s ([0 , J i + i ∈ E i +1 ( J i ),then by (8.2) and the induction hypothesis,diam J (cid:48) i +1 ≤ M s ( i + 1 , J i +1 ) M s ([0 , J (cid:48) i ≤ i +1 (cid:88) l =0 l − ( i +1) M s ( l, J l ) M s ([0 , . This establishes (8.3).Choose an integer l sufficiently large so that 2 − l l ≤ (cid:15)/
2. If k ≥ l , then by (8.3) andthe fact that M s ([0 , ≥ (diam E ) s = 1,diam I (cid:48) ≤ l − (cid:88) l =0 l − k M s ( l, J l ) M s ([0 , k (cid:88) l = l l − k M s ( l, J l ) M s ([0 , ≤ (cid:15) M s ( l , J l ) . Thus, by Corollary 8.6,diam I (cid:48) ≤ (cid:15) s c sup x ∈ E H s ( E ∩ B ( x, · − l ))whenever k ≥ l . Now, because E is compact, H s ( E ) < ∞ , and H s has no atoms,lim n →∞ sup x ∈ E H s ( E ∩ B ( x, · − n )) = 0 . Hence, by choosing l even larger if necessary, we can ensure thatsup x ∈ E s c H s ( E ∩ B ( x, · − l )) < (cid:15) . Therefore, diam I (cid:48) < (cid:15) for all k ≥ l provided that l is sufficiently large depending on (cid:15) and E . (cid:3) Following the proof of Theorem 5.1 in § F k : [0 , → R N and a (1 /s )-H¨older continuous map F : [0 , → R N satisfying the following properties.(1) For each k ≥
0, there exist (possibly degenerate) closed intervals [0 , a k ] and [ b k , F k | [0 , a k ] and F | [0 , b k ] are constant maps and F (0) , F (1) ∈ V k .(2) For each k ≥
0, the map F k | ( a k , b k ) is an injective piecewise linear map connecting F k (0) to F k (1) along line segments between points in V k of length at most 3 · − k .(3) If x ∈ V k \ { F k (0) , F k (1) } for some k ≥
0, then F − k ( x ) = F − j ( x ) for all k ≥ j (because intervals in F j are frozen).(4) The maps F k converge uniformly to F and F ([0 , (cid:83) ∞ k =0 V k .(5) The H¨older constant of F satisfies H ≤ ξ (cid:18) M s ([0 , r − s + 60 A ∗ r ξ − ξ (cid:19) (cid:46) s c − H s ( E )(diam E ) − s . It remains to show that F ([0 , E and F is injective.On one hand, since each V k is a maximal 2 − k -separated set in E , F ([0 , ⊃ (cid:83) ∞ k =0 V k = E ¨OLDER CURVES AND PARAMETERIZATIONS 55 by (3). On the other hand, if x ∈ F ([0 , x = F ( t ), thendist( x, E ) ≤ lim inf k →∞ dist( F k ( t ) , E ) = 0by (2). Thus, F ([0 , E .To check injectivity, we first establish a lemma. We say that an interval I separates two numbers x < y if x < z < y for all z ∈ I . Lemma 8.8.
Let ≤ x < y ≤ . If k is the least integer k ≥ such that there exists aninterval in E (cid:48) k separating x and y , then | F k ( x ) − F k ( y ) | (cid:38) − k for all k ≥ k .Proof. Fix 0 ≤ x < y ≤
1, let k be as in the statement of the lemma and let k ≥ k . Let I ∈ E k be such that I separates x and y . The proof is divided into two cases. Case 1.
Assume that k ≤ k + 3. By Remark 3.5 and minimality of k , there existat most 13 intervals I ∈ E (cid:48) k separating x from y . Therefore, there exist consecutiveintervals J , . . . , J l ∈ I k such that x, y ∈ (cid:83) li =1 J i , I ⊂ (cid:83) li =1 J i and l ≤
15. Let a be anendpoint of I . Since diam F k ( J i ) ≤ · − k for all i ∈ { , . . . , l } , the points F k ( x ) and F k ( y ) are in B := B ( F k ( a ) , · − k ). Let (cid:96) be a best fitting line for B and let π be theorthogonal projection on (cid:96) . Since β E ( B ) ≤ − , the points of F k ( (cid:83) li =1 J i ) ∩ B are linearlyordered according to their projection on (cid:96) . In particular, | z − w | ≤ | π ( z ) − π ( w ) | for all z, w ∈ F k ( (cid:83) li =1 J i ) ∩ B . Thus, | F k ( x ) − F k ( y ) | ≥ | π (cid:96) ( F k ( x )) − π (cid:96) ( F k ( y )) | ≥ diam π (cid:96) ( F k ( I )) ≥
12 diam F k ( I ) ≥
12 2 − k . Case 2.
Assume that k ≥ k + 4. For each integer i ≥
0, let P i ( x, y ) be the end-points of intervals in E (cid:48) k +4+ i lying between x and y . Let x i (resp. y i ) be the leftmost(resp. rightmost) element of P i ( x, y ). For all i ≥ x ≤ x i +1 ≤ x i < y i ≤ y i +1 ≤ y, and I separates x from y . As in Case 1, if (cid:96) is a best fitting line for F k ( a ), then | F k +3 ( x ) − F k +3 ( y ) | ≥ | π (cid:96) ( F k +3 ( x )) − π (cid:96) ( F k +3 ( y )) | ≥ diam π (cid:96) ( F k ( I )) ≥
910 2 − k . Now, each x i +1 (resp. y i +1 ) is contained in the closure of an interval in E (cid:48) k +4+ i which has x i (resp. y i ) as an endpoint. This fact along with Lemma 8.3 yields | F k +4+ i ( x i ) − F k +5+ i ( x i +1 ) | ≤ · − k − − i , | F k +4+ i ( y i ) − F k +5+ i ( y i +1 ) | ≤ · − k − − i . Therefore, by the triangle inequality, | F k ( x ) − F k ( y ) | ≥ | F k +4 ( x ) − F k +4 ( y ) | − ∞ (cid:88) i =0 | F k +4+ i ( x i ) − F k +5+ i ( x i +1 ) |− ∞ (cid:88) i =0 | F k +4+ i ( y i ) − F k +5+ i ( y i +1 ) |≥
910 2 − k − · − k − − · − k − . Hence | F k ( x ) − F k ( y ) | ≥ (3 / − k and the proof is complete. (cid:3) Suppose that x, y ∈ [0 ,
1] with x < y . By Lemma 8.7, there exist intervals I ∈ E (cid:48) k thatseparate x and y provided that k is sufficiently large. If k is the least such integer, then | F ( x ) − F ( y ) | (cid:38) − k > F is injective and completesthe proof of Proposition 8.1.8.4. Proof of Lemma 8.2.
We first give an auxiliary estimate.
Lemma 8.9.
Let E ⊂ R N , x ∈ E , and r > . If y ∈ E ∩ B ( x, r ) , | y − x | ≥ β E ( B ( x, r )) r ,and (cid:96) x,y is the line passing through x and y , then dist( z, (cid:96) x,y ) ≤ β E ( B ( x, r )) (cid:18) . r | y − x | (cid:19) r for all z ∈ E ∩ B ( x, r ) . Proof.
Let z ∈ E ∩ B ( x, r ), z (cid:54) = x . Let (cid:96) be a best fitting line for E in B ( x, r ). Thendist( x, (cid:96) ), dist( y, (cid:96) ), and dist( z, (cid:96) ) are bounded above by β E ( B ( x, r ))2 r . Let (cid:96) x = (cid:96) − x .Then x ∈ (cid:96) x and dist( y, (cid:96) x ) and dist( z, (cid:96) x ) are bounded above by 2 β E ( B ( x, r ))2 r . If y ∈ (cid:96) x ,then we have dist( z, (cid:96) x,y ) = dist( z, (cid:96) x ) ≤ β E ( B ( x, r ))2 r and we are done.To continue, suppose that y (cid:54)∈ (cid:96) x and let y (cid:48) = π (cid:96) x ( y ) and let z (cid:48) = π (cid:96) x ( z ). Define w = x + | z (cid:48) − x || y (cid:48) − x | ( y − x ) ∈ (cid:96) x,y . Since z (cid:48) ∈ (cid:96) x between x and y (cid:48) , we have that z (cid:48) = x + | z (cid:48) − x || y (cid:48) − x | − ( y (cid:48) − x ). Therefore,dist( z (cid:48) , (cid:96) x,y ) ≤ | z (cid:48) − w | = | y (cid:48) − y | | z (cid:48) − x || y (cid:48) − x | ≤ | y (cid:48) − y | r | y (cid:48) − x | . Thus, by the triangle inequality,dist( z, (cid:96) x,y ) ≤ | z (cid:48) − z | + | y (cid:48) − y | r | y (cid:48) − x | ≤ β E ( B ( x, r )) (cid:18) r | y (cid:48) − x | (cid:19) r. Since dist( y, (cid:96) x ) ≤ β E ( B ( x, r ))2 r ≤ (1 / | x − y | , we have1 . | y (cid:48) − x | ≥ (1 + 3(1 / ) | y (cid:48) − x | ≥ | y − x | by Lemma 2.1, applied with V = { x, y } . Therefore,dist( z, (cid:96) x,y ) ≤ β E ( B ( x, r )) (cid:18) . r | y − x | (cid:19) r. (cid:3) ¨OLDER CURVES AND PARAMETERIZATIONS 57 We now give a proof of the key lemma.
Proof of Lemma 8.2.
Without loss of generality, we may assume that diam E = 1. Fix x, y ∈ E and let n ≥ − ( n +1) < | x − y | ≤ − n . Case 1.
Suppose that n ∈ { , } . Let (cid:96) be the line containing x and y . Since β E ( B ( x, ≤ − and since | x − y | > − , by Lemma 8.9 we have thatsup w ∈ E dist( w, (cid:96) ) = sup w ∈ E ∩ B ( x, dist( w, (cid:96) ) ≤ (cid:18) . | x − y | (cid:19) β E ( B ( x, ≤ . ≤ < | x − y | . Therefore, E is contained inside the tube T := B ( x, ∩ B ( (cid:96), − | x − y | ). Let D be aclosed ( N − z , perpendicular to (cid:96) and of radius 2 − | x − y | . In otherwords, D is the set of all points in B ( z, − ) whose projection on (cid:96) is z . Then D cuts T into two pieces, one containing x and another containing y . By connectedness of E , wemust have D ∩ E (cid:54) = ∅ . Case 2.
Suppose that n ≥
2. The procedure here is roughly the same as that inCase 1, with the difference that the tube T is replaced by a more complicated set. Byconnectedness of E , for each k ∈ { , . . . , n − } , there exists a point y k ∈ B ( x, − k ) ∩ E .For each k ∈ { , . . . , n − } let (cid:96) k be the line containing x and y k . Let also (cid:96) n be the linecontaining x and y .Working as in Case 1, we can show that for each k ∈ { , . . . , n } , E ∩ B ( x, − ( k − ) ⊂ T k := B ( x, − ( k − ) ∩ B ( (cid:96) k , − − ( k − ) . Since diam E = 1, we also have E ⊂ T . For each k ∈ { , . . . , n − } , let T k, , T k, be thetwo components of T k \ B ( x, − k ). Set T = T , ∪ · · · ∪ T n − , ∪ T n ∪ T n − , ∪ · · · ∪ T , . The sets T , , . . . , T n − , , T n , T n − , , . . . , T , intersect at most in pairs. In particular,(1) if i ∈ { , } , then T ,i ∩ T m,j = ∅ unless m ∈ { , } and j = i ;(2) if k ∈ { , . . . , n − } (if any) and i ∈ { , } , then T k,i ∩ T m,j = ∅ unless m ∈{ k − , k, k + 1 } and j = i ;(3) if i ∈ { , } , then T n − ,i ∩ T m,j = ∅ unless m ∈ { n − , n − } and j = i ;(4) T n ∩ T m,j = ∅ unless m = n − D is an ( N − z , perpendicular to (cid:96) n and of radius2 − − n , then D cuts T n into two pieces, one containing x and another containing y .Consequently, D cuts T into two pieces, one containing x and another containing y . Byconnectedness of E and the fact that E ⊂ T , we must have D ∩ E (cid:54) = ∅ . (cid:3) Examples
In this section, we give examples of H¨older curves and of sets that are not contained inH¨older curves to illuminate Theorem 1.1, Proposition 1.3, and Theorem 5.1.9.1.
H¨older curves that are non-flat in all scales.
First up, we show that condition(1.2) in Theorem 1.1 is not necessary for a bounded set to be contained in a (1 /s )-H¨oldercurve when s >
1. In contrast, when s = 1, condition (1.1) in the Analyst’s TravelingSalesman theorem is necessary and sufficient for a bounded set to be contained in arectifiable curve.Let N ≥ ≤ m ≤ N − E ⊂ R N and an N -cube Q ⊂ R N with E ∩ Q (cid:54) = ∅ , define the m -dimensional beta number β ( m ) E ( Q ) := inf P sup x ∈ E ∩ Q dist( x, P )diam Q where the infimum is taken over all m -planes P in R N . If E ∩ Q = ∅ , set β ( m ) E ( Q ) = 0.Note that β (1) E ( Q ) = β E ( Q ) as defined in § β ( m ) E ( Q ) ≤ β ( n ) E ( Q ) whenever m ≥ n . Proposition 9.1.
For any N ≥ and any s ∈ (1 , N ] , there exists a (1 /s ) -H¨older curve E ⊂ R N such that (9.1) (cid:88) Q ∈ ∆( R N ) β ( N − E (3 Q ) ≥ (6 √ N ) − (diam Q ) s = ∞ . The construction splits into three cases. Before proceeding, we introduce some notation.Given a cube Q ⊂ R N , denote by ∆( Q ) the set of dyadic cubes in ∆( R N ) that arecontained in Q . Moreover, given positive integers m ≤ N , there exists a polynomial P N,m of degree m with the following property: If n ∈ N and { Q , . . . , Q N n } is a partition of[0 , N into N -cubes of side-length 1 /n , thencard { Q i : Q i intersects the m -skeleton of ∂ [0 , N } = P N,m ( n ) . Recall that if I , . . . , I N are nondegenerate compact intervals, and Q = I × · · · × I N isan N -cube, then the m -skeleton of Q is the union of sets I (cid:48) × · · · × I (cid:48) N where I (cid:48) j = I j for m indices j and I (cid:48) j = ∂I j for the remaining N − m indices j . Finally, we note that if K is the set of vertices of a cube Q in R N and P is an ( N − x, P ) ≥ (2 √ N ) − diam K for all x ∈ K. Case 1: s = N . We simply take E = [0 , N . It is well known that there exists a(1 /N )-H¨older parametrization f : [0 , → E . On the other hand, by (9.2), (cid:88) Q ∈ ∆( R N ) β ( N − E (3 Q ) ≥ (2 √ N ) − (diam Q ) N ≥ ∞ (cid:88) k =0 (cid:88) Q ∈ ∆([0 , N )diam Q = √ N − k (diam Q ) N = ∞ (cid:88) k =0 Nk ( √ N − k ) N = ∞ . ¨OLDER CURVES AND PARAMETERIZATIONS 59 Case 2: s ∈ (1 , N ) \ N . Let m be the integer part of s . Since the degree of P N,m isstrictly less than s and strictly lager than s −
1, we can fix n ∈ N such that(9.3) n s − ( n − s < P N,m ( n ) < n s − . By (9.3) and the Intermediate Value Theorem, there exists λ ∈ ( n , − n ) such that(9.4) P N,m ( n ) n − s + λ s = 1 . Partition [0 , N into N -cubes of side-lengths 1 /n and let { Q i } li =1 ( l = P N,m ( n )) be thosecubes that intersect the m -skeleton of [0 , N . Let also Q = [1 /n, /n + λ ] N . For each i = 0 , . . . , l , let φ i be a similarity of R N such that φ i ([0 , N ) = Q i . Finally, define E ⊂ R N , E := ∞ (cid:92) k =1 (cid:91) i ··· i k ∈{ ,...,l } k φ i ◦ · · · ◦ φ i k ([0 , N ) . Since the maps { φ , . . . , φ l } satisfy the open set condition, E is Ahlfors regular [Hut81].By (9.4) the Hausdorff dimension of E is equal to s , so E is s -regular. Lemma 9.2.
The set E is connected.Proof. Set W = (cid:83) k ≥ { , . . . , l } k with the convention that { , . . . , l } is the empty word ∅ and φ ∅ is the identity map of R N For each w ∈ W , let K w denote the 1-skeleton of φ w ([0 , N ). The proof is based now on two observations. First, by the choice of cubes Q , . . . , Q l , it follows that K w ⊂ E for all w ∈ W . Second, K w ∩ K wi (cid:54) = ∅ for all w ∈ W and i ∈ { , . . . , l } .Now fix x ∈ E . There exists a sequence of words ( w n ) n ≥ in W such that w is theempty word, w n +1 = w n i n with i n ∈ { , . . . , l } , and x ∈ (cid:84) n ≥ φ w n ( x ). The set (cid:83) n ≥ K w n is a path that joins x with the origin. Hence E is connected. (cid:3) By Lemma 9.2, the fact that the Hausdorff dimension of E is s , and Theorem 4.12 in[Rem98], there exists a (1 /s )-H¨older map f : [0 , → R N such that f ([0 , E . Itremains to show (9.1). We first prove a lemma. Lemma 9.3. If Q ∈ ∆([0 , N ) is a dyadic cube that intersects E , then there exists adyadic cube Q (cid:48) ⊂ Q such that diam Q (cid:48) ≥ (3 n ) − diam Q and β ( N − E (3 Q (cid:48) ) ≥ (6 √ N ) − .Proof. Fix x ∈ Q ∩ E and let i , i , . . . be a sequence of numbers in { , . . . , l } such that x ∈ ∞ (cid:92) k =1 φ i ◦ · · · ◦ φ i k ([0 , N ) . Let k be the smallest positive integer such that φ i ◦ · · · ◦ φ i k ([0 , N ) ⊂ Q and define K to be the set of vertices of φ i ◦ · · · ◦ φ i k ([0 , N ). Since each φ i has a scaling factor atleast 1 /n , by minimality of k we have that diam K ≥ (1 /n ) diam Q .Let Q (cid:48) be a dyadic cube in ∆(3 Q ) (possibly Q (cid:48) = Q ) of minimal diameter such that K ⊂ Q (cid:48) . We claim that(9.5) 13 diam K ≤ diam Q (cid:48) ≤ diam K. The lower inequality is clear. If diam
K < diam Q , then, since K has edges parallel to theaxes, K is contained in 3 Q for some dyadic cube Q ⊂ Q with diam Q = diam Q (cid:48) ,which is a contradiction. That establishes the upper inequality of (9.5).By (9.2), (9.5), and the fact that K ⊂ E , β ( N − E (3 Q (cid:48) ) ≥ β ( N − K (3 Q (cid:48) ) ≥ (2 √ N ) − diam K diam 3 Q = (6 √ N ) − diam K diam Q ≥ (6 √ N ) − . This proves the lemma. (cid:3)
By Ahlfors s -regularity of E , there exists a constant C > { Q ∈ ∆([0 , N ) : diam Q = √ N − k and Q ∩ E (cid:54) = ∅} ≥ C − sk . Fix a positive integer k such that 2 k > n . For k ∈ N , set Q k = { Q ∈ ∆([0 , N ) : diam Q ∈ [ √ N − k , √ N − k − k ] and β ( N − E (3 Q ) ≥ (6 √ N ) − } . By Lemma 9.3,card Q k ≥ − N card { Q ∈ ∆([0 , N ) : diam Q = √ N − k and Q ∩ E (cid:54) = ∅} ≥ C − − N sk . Therefore, (cid:88) Q ∈ ∆( R N ) β ( N − E (3 Q ) ≥ (6 √ N ) − (diam Q ) N ≥ ∞ (cid:88) k =0 (cid:88) Q ∈Q kk (diam Q ) N ≥ ∞ (cid:88) k =0 C − − N skk ( √ N ) s − s ( k +1) k = ∞ . Case 3: s ∈ { , . . . , N − } . Fix n ∈ N large enough so that P N,s − ( n ) < n s . Partition[0 , N into N -cubes with disjoint interiors and side-lengths 1 /n and let { Q , . . . , Q l } ( l = n s ) be a collection of such cubes so that the set (cid:83) lk =1 Q i is connected and containsthe ( s − , N . The rest of the construction is similar to Case 2 and is leftto the reader.9.2. Ahlfors regular curves without H¨older parametrizations.
Next, for all s > s -regular curves that are not contained in any (1 /s )-H¨older curve.The basic strategy is take a disconnected set, which is not contained in a H¨older curve,and then extend the set to transform it into an s -regular curve. We call the curves thatwe construct “Cantor ladders”. Proposition 9.4.
Let N ∈ N with N ≥ , let s ∈ (1 , N ) , and let m ∈ N with m ≤ s .There exists an Ahlfors s -regular curve E ⊂ R N , which is not contained in a ( m/s ) -H¨olderimage of [0 , m . We treat the cases s ∈ N and s (cid:54)∈ N separately. Given m ∈ N , let W m be the set offinite words formed by the letters { , . . . , m } including the empty word ∅ . We denote by | w | the number of letters a word has with the convention |∅| = 0. ¨OLDER CURVES AND PARAMETERIZATIONS 61 Case 1.
Suppose that s ∈ { , , . . . , N − } . Let D ∅ = [0 , . Given a square D w ⊂ R for some w ∈ W , let D w , D w , D w , D w be the four corner squares in D w withdiam D wi = (1 /
4) diam D w . Let C be the Cantor set in R defined by C = ∞ (cid:92) k =0 (cid:91) w ∈W | w | = k D w . For each i = 1 , . . . , | w | , define D w,i = D w × { (2 i − −| w |− } , K = ( C × [0 , ∪ (cid:91) w ∈W | w | − (cid:91) i =0 D w,i and E = K × [0 , s − × { } N − s − . Here and for the rest of § A × { } = A . Case 2.
Suppose that s ∈ (1 , N ) \ N . Let p = s − (cid:98) s (cid:99) be the fractional part of s . Let I ∅ = [0 , I w = [ a w , b w ] for some w ∈ W , let I w = [ a w , a w + 2 − p ( b w − a w )] and I w = [ b w − − p ( b w − a w ) , b w ] . Let C p denote the Cantor set in R defined by C p = ∞ (cid:92) k =0 (cid:91) w ∈W | w | = k I w . Let S be the bi-Lipschitz embedded image of ([0 , , | · | p +1 ) into R . For each w ∈ W ,let S w be a rescaled copy of S whose endpoints are the right endpoint of I w and the leftendpoint of I w . For each w ∈ W and i = 1 , . . . , | w | −
1, define S w,i = S w + (0 , (2 i − −| w |− )and define K p +1 = ( C p × [0 , ∪ (cid:91) w ∈W | w | (cid:91) i =0 S w,i and E = K p +1 × [0 , s − p − × { } N + p − s − . Verification of the desired properties of E is the same for the two cases, so we only treatCase 1. By Theorem 2.1 in [MM00], there exists no ( m/s )-H¨older map f : [0 , m → R N whose image contains C × [0 , s − × { } N − s − . We show that E is a curve in § s -regularity of E in § E is a curve. By the Hahn-Mazurkiewicz theorem [HY88, Theorem 3.30], to showthat E is a curve it is enough to show that E is compact, connected, and locally connected.For compactness, it is easy to see that K ⊂ [0 , , hence E ⊂ [0 , N . Moreover, as | w | → ∞ , the squares D w,i accumulate on C × [0 , K is closed. Consequently, E is compact.To settle both connectedness and local connectedness, we prove that there exists C > x, y ∈ E there exists a path joining x with y of diameter atmost C | x − y | / . Clearly, it suffices to show the claim for K instead of E . Fix x, y ∈ K and let w be the word in W of maximum word-length such that the projections of x and y on R × { } are contained in D w . This means that | x − y | ≥ −| w | . Choose i ∈ N such that dist( x, D w ,i ) ≤ · −| w | . If x and y are the projections of x and y onto D w ,i , respectively, thenmax {| x − x | , | y − y |} (cid:46) −| w | + 4 −| w | + | x − y | (cid:39) −| w | . There exist sequences ( w n ) n ∈ N , ( u n ) n ∈ N of words in W and sequences ( i n ) n ∈ N , ( j n ) n ∈ N ofpositive integers such that(1) | w n | = | u n | = | w | + n ;(2) the orthogonal projection of x (resp. y ) on R is contained in D w n (resp. D u n );(3) there exists x n ∈ D w n ,i n such thatmax {| x − x n | , | y − y n |} ≤ −| w |− n √ −| w |− n . Properties (1) and (2) imply that D w (cid:41) D w (cid:41) D w (cid:41) · · · and D w (cid:41) D u (cid:41) D u (cid:41) · · · ,while property (3) implies that the Hausdorff distancesdist H ( D w n ,i n , D w n +1 ,i n +1 ) (cid:46) −| w |− n and dist H ( D u n ,j n , D u n +1 ,j n +1 ) (cid:46) −| w |− n . Let γ ⊂ K be the line segment joining x with y . For each n ≥
0, let z n ∈ D w n ,i n be acorner point and let z (cid:48) n be its projection on D w n +1 ,i n +1 . Also, let p n ∈ D u n ,j n be a cornerpoint and let p (cid:48) n be its projection on D u n +1 ,j n +1 . Consider the curve γ = γ ∪ (cid:91) n ∈ N ([ x n , z n ] ∪ [ z n , z (cid:48) n ] ∪ [ z (cid:48) n , x n +1 ]) ∪ (cid:91) n ∈ N ([ y n , p n ] ∪ [ p n , p (cid:48) n ] ∪ [ p (cid:48) n , y n +1 ]) , which is a subset of K and joins x with y . Thendiam γ (cid:46) diam γ + (cid:88) n ≥ diam γ n + (cid:88) n ≥ diam σ n ≤ | x − y | + (cid:88) n ≥ ( | x n − z n | + | z n − z (cid:48) n | + | z (cid:48) n − x n +1 | )+ (cid:88) n ≥ ( | y n − p n | + | p n − p (cid:48) n | + | p (cid:48) n − y n +1 | ) (cid:46) −| w | + (cid:88) n ≥ (2 −| w |− n + 4 −| w |− n + 2 −| w |− n ) (cid:46) −| w | (cid:39) | x − y | / . E is s -regular. We show s -regularity for E . Because the product of regular compactspaces of dimension s and s is ( s + s )-regular, to show that E is s -regular, it sufficesto show that K is 2-regular. Fix x ∈ K and r ∈ (0 , diam K ).We first show that(9.6) H ( B ( x, r ) ∩ K ) (cid:38) r . If x ∈ C × [0 , C × [ − , x ∈ D w,i and r ≤
10 diam D w , then (9.6) follows from the 2-regularity of D w,i . If x ∈ D w,i and r ≥
10 diam D w , then there exists z ∈ ( C × [0 , ∩ B ( x, r ) such that B ( z, r/ ⊂ B ( x, r )and (9.6) follows from the 2-regularity of B ( z, r/ ∩ K . ¨OLDER CURVES AND PARAMETERIZATIONS 63 For the upper regularity of K , instead of working with balls B ( x, r ), it is more conve-nient to use cubes Q ( x, r ) = x + [ − r/ , r/ x ∈ K , r > . Without loss of generality, we may assume that r = 4 − k for some k ∈ N . For each k ≥ D k ( x, r ) = { D w,i : Q ( x, r ) ∩ D w,i (cid:54) = ∅ and | w | = k } . Then by the 2-regularity of C × [ − , (cid:88) k ≥ (cid:88) D w,i ∈D k ( x,r ) H ( Q ( x, r ) ∩ D w,i ) (cid:46) r . The following lemma will let us estimate the above sum. In the sequel, we denote by m ≥ D w,i ∈ D ( x, r ) with | w | = m . Lemma 9.5.
Let m ≥ be the smallest integer for which D m ( x, r ) (cid:54) = ∅ .(1) If k > m and D k ( x, r ) (cid:54) = ∅ , then k ≥ k .(2) If Q (cid:48) is the projection of Q ( x, r ) on R × { } , then for all k ≥ , card { D w : D w ∩ Q (cid:48) (cid:54) = ∅ and | w | = k } ≤ k − k . (3) For each w ∈ W , card { i : D w,i ∩ Q ( x, r ) (cid:54) = ∅} ≤ | w | +1 − k . (4) For each k ≥ , card D k ( x, r ) ≤ (1 + 4 k − k )(1 + 2 k +1 − k ) .(5) We have (cid:88) D w,i ∈D m ( x,r ) H ( D w,i ∩ Q ( x, r )) (cid:46) r . Proof.
For (1), recall that if | w | > m , then the vertical distance between D w,i and D w ,i is at least 2 −| w | . Since r = 4 − k , the cube Q ( x, r ) can not intersect any D w,i , unless4 − k ≥ −| w | . Thus, | w | ≥ k .For (2), we first note that if k ≤ k , then Q (cid:48) can intersect at most one square D w with | w | = k . We now use induction to show that for all k ≥ k ,card { D w : D w ∩ Q (cid:48) (cid:54) = ∅ and | w | = k } ≤ k − k . For k = k , it is true. Suppose that the claim is also true for some k ≥ k . Then Q (cid:48) intersects D w with | w | = k + 1 if and only if there exists w (cid:48) with | w (cid:48) | = k such that Q ∩ D w (cid:48) (cid:54) = ∅ and D w ⊂ D w (cid:48) . Since each square of generation k contains 4 squares ofgeneration k + 1,card { D w : D w ∩ Q (cid:48) (cid:54) = ∅ and | w | = k + 1 } ≤ { D w : D w ∩ Q (cid:48) (cid:54) = ∅ and | w | = k }≤ k +1 − k . For (3), fix w ∈ W . Recall that the vertical height of Q ( x, r ) is 2 r = 2 · − k and thatthe vertical distance between D w,i and D w,j with i (cid:54) = j is at least 2 −| w | . Therefore,card { i : D w,i ∩ Q ( x, r ) (cid:54) = ∅} ≥ r ) / −| w | = 1 + 2 | w | +1 − k . Claim (4) is immediate from (2) and (3).It remains to show (5). On one hand, if m > k , then by (4), card( D m ( x, r )) = 1.Hence (5) follows from the 2-regularity of squares D w,i . On the other hand, if m ≤ k ,then by (4), (cid:88) D w,i ∈D m ( x,r ) H ( D w,i ∩ Q ( x, r )) ≤ card( D m ( x, r ))(4 − m ) (cid:46) − m − k ≤ r . (cid:3) By Lemma 9.5, we have (cid:88) D w,i ∈D ( x,r ) H ( B ( x, r ) ∩ D w,i ) ≤ (cid:88) D w,i ∈D m ( x,r ) H ( D w,i ) + ∞ (cid:88) k =2 k (cid:88) D w,i ∈D k ( x,r ) H ( D w,i ) (cid:46) r + ∞ (cid:88) k =2 k k − k k − k − k . Finally, ∞ (cid:88) k ≥ k k − k k − k − k = 4 − k (cid:88) k ≥ k − k (cid:46) − k (cid:46) r . Therefore, K is 2-regular.9.3. A compact countable set that is not contained in any H¨older cube.Proposition 9.6.
For each N ∈ N , N ≥ , there exists a compact and countable set E ⊂ R N with one accumulation point such that for any m ∈ { , . . . , N − } and any s ∈ [1 , N/m ) , the set E is not contained in a (1 /s ) -H¨older image of [0 , m . Corollary 9.7.
For each N ∈ N , N ≥ , there exists a compact and countable set E ⊂ R N with one accumulation point such that E is not contained in a rectifiable curve. For each integer k ≥
0, define G k to be the union of all vertices of all dyadic cubesin R N that are contained in [0 , N and have side length 2 − k . By a simple combinatorialargument, card( G k ) = (2 k + 1) N for all k ≥ φ be the identity map, and for each k ≥
1, define a map φ k : R N → R N by φ k ( x ) = ( k + 1) − x + (cid:32) , . . . , , k (cid:88) i =1 i − (cid:33) . Set A := (cid:80) ∞ i =1 i − = π /
6, and define the set E := { (0 , . . . , , A ) } ∪ ∞ (cid:91) k =0 φ k ( G k ) . ¨OLDER CURVES AND PARAMETERIZATIONS 65 The set E is clearly countable. If ( x , . . . , x N ) ∈ E , then | x i | ≤ i = 1 , . . . , N − | x N | ≤ A . Therefore, E is bounded. Moreover, the only accumulation point of E is the point (0 , . . . , , A ) which is contained in E . Thus, E is closed.Next, we claim that(9.7) | x − y | ≥ − k ( k + 1) − for all x ∈ φ k ( G k ) and all y ∈ E \ { x } . Indeed, if x, y ∈ G k , then inequality (9.7) is clear. Otherwise, dist( G k , E \ G k ) ≥ ( k + 1) − ,and thus, (9.7) holds again.Suppose in order to get a contradiction that there exists a (1 /s )-H¨older continuous map f : [0 , m → R N such that E ⊂ f ([0 , H be the H¨older constant of f . For each k ≥ x ∈ G k , fix a point w k,x such that f ( w k,x ) = x and set B k,x = B ( w k,x , H − s − ks ( k + 1) − s ) . Inequality (9.7) implies that the balls B k,x are mutually disjoint. Moreover, it is easy tosee that each B k,x is contained in [ − , m . Therefore,1 (cid:38) m H m ([ − , m ) ≥ ∞ (cid:88) k =0 (cid:88) x ∈G k H m ( B k,x ) (cid:38) H,s ∞ (cid:88) k =0 (2 k + 1) N − skm ( k + 1) sm (cid:39) N ∞ (cid:88) k =0 k ( N − ms ) ( k + 1) s . Since
N > ms , the sum on the right hand side diverges and we reach a contradiction.9.4.
Flat curves with finite H s measure and no (1 /s ) -H¨older parametrizations. The following example shows that the assumption of lower s -regularity can not be droppedfrom Proposition 1.3. Proposition 9.8.
For any β ∈ (0 , , there exists s ∈ (1 , with the following property.For any s ∈ (1 , s ) there exists a curve E ⊂ R such that(1) H s ( E ) < ∞ and(2) β E ( Q ) < β for all Q ∈ ∆( R N ) ,but E is not contained in any (1 /s ) -H¨older image of [0 , . Before proceeding, we recall a well-known construction method for snowflakes in R . Let p = ( p , p , . . . ) be sequence of numbers in [1 / , / be the segment [0 , × { } ,oriented from (0 ,
0) to (1 , k with 4 k edges. Define Γ k +1 to be the polygonal arc constructed by replacing eachedge e of Γ k by a rescaled and rotated copy of the oriented polygonal arc in Figure 4 with p = p k , so that the new oriented arc lies to the left of e . A snowflake arc S p is obtainedby taking the limit of Γ k , just as in the construction of the usual von Koch snowflake. Remark 9.9.
For any (cid:15) >
0, there exists p ∗ > / / ≤ p k ≤ p ∗ for all k ≥
0, then β Γ k ( B ( x, r )) ≤ (cid:15)r for all k ≥ x ∈ Γ k , and r > Figure 4.
Fix β ∈ (0 , p ∗ ∈ (1 / , /
2) such that β S p ( Q ) < β for all Q ∈ ∆( R N ). Set p = ( p ∗ , p ∗ , . . . ), set s = − log 4 / log p , andfix s ∈ [1 , s ). It is well-known that there exists a (1 /s )-bi-H¨older homeomorphismΦ : [0 , → S p ; e.g., see [BH04, RV17].We now construct a self-similar Cantor set in [0 ,
1] in the following way. Let I ∅ = [0 , I w = [ a w , b w ] for some w ∈ { , } n , let I w = [ a w , a w + ( b w − a w )2 − s /s ] and I w = [ b w − ( b w − a w )2 − s /s , b w ] . Define E (cid:48) = (cid:84) ∞ n =0 (cid:83) w ∈{ , } n I w . For each component J of [0 , \ E (cid:48) , let γ J be the linesegment joining the endpoints of Φ( J ). Then define E = Φ( E (cid:48) ) ∪ (cid:91) J γ J , where the union is taken over all components J of [0 , \ E (cid:48) . Since E (cid:48) is s/s -regular andΦ is (1 /s )-bi-H¨older, H s ( E ) = H s (Φ( E (cid:48) )) + (cid:88) J H s ( γ J ) ≤ C H s/s ( E (cid:48) ) < ∞ . Since Φ( E (cid:48) ) ⊂ S p and γ J are line segments, we have β E ( Q ) < β for all Q ∈ ∆( R N ).Finally, by Theorem 2.1 in [MM00], there does not exist a (1 /s )-H¨older map f : [0 , → R whose image contains Φ( E (cid:48) ) (and consequently E ).9.5. Sharpness of exponent 1 in Theorem 5.1.
To wrap up, we show that Theorem5.1 does not hold if numbers τ s ( k, v, v (cid:48) ) are replaced by τ s ( k, v, v (cid:48) ) p with p >
1. When s = 1, this follows from the necessary half of the Analyst’s Traveling Salesman theorem.Thus, we may focus on the case s > Proposition 9.10.
Let p > , let s > be sufficiently close to , and let α > besufficiently close to . There exists a sequence of finite sets { ( V k , ρ k ) } ≥ of numbers andfinite sets in R satisfying (V0)–(V5) such that (9.8) (cid:88) v ∈ V k α k,v ≥ α ρ sk r s + (cid:88) ( v,v (cid:48) ) ∈ Flat ( k ) τ s ( k, v, v (cid:48) ) p ρ sk r s < ∞ but there does not exist a (1 /s ) -H¨older map f : [0 , → R such that (cid:83) k ≥ V k ⊂ f ([0 , . ¨OLDER CURVES AND PARAMETERIZATIONS 67 Let s > n ∈ N be constants to be specified below. Fix a number0 < q < min { /s, ( p − /s } . For each n ∈ N , let t k = (cid:115) /s (cid:18) k + n (cid:19) q − . Construct a sequence of polygonal arcs Γ k as in § p k = 1 / t k . We may assume that numbers p k are in [1 / , /
2) by taking n to be sufficiently large.For each k ≥
0, we define a finite set V k ⊂ Γ k as follows. Define V := { v , , v , } , where v , = (0 ,
0) and v , = (1 , k ≥ V k = { v k, , . . . , v k,N k } , N k = 2 k + 1 , where points v k,i are enumerated according to the orientation of Γ k . For each i =1 , . . . , k + 1, set v k +1 , i − = v k,i , and assign v k +1 , i to the point of Γ k +1 that lies be-tween v k +1 , i − and v k +1 , i +1 and is equal distance to v k +1 , i − and v k +1 , i +1 (the peak ofthe triangle in Figure 4). Define the quantities(9.9) r = 1 , C ∗ = 2 , ξ = 2 − /s , ξ = 1 + 2 − /s , ρ = 1 , ρ k = 2 − k/s ( k + 1 + n ) q (2 + n ) q . For each k ≥ v ∈ V k , define α k,v := inf (cid:96) sup x ∈ V k +1 ∩ B ( v, A ∗ ρ k r ) dist( x, (cid:96) ) ρ k +1 r , where the infimum is taken over all lines (cid:96) in R and A ∗ is as in § (cid:96) k,v be a line (cid:96) ,which realizes the number α k,v . Lemma 9.11.
There exist choices of s and n so that the following properties hold.(1) For all k ≥ and i ∈ { , . . . N k } , we have | v k,i − v k,i +1 | = ρ k .(2) The sequence { ( V k , ρ k ) } k ≥ satisfies (V0)–(V5) with the parameters given in (9.9).(3) For all k ≥ and v ∈ V k , we have α k,v ≤ α , where α is as in Definition 2.4.(4) For all k ≥ and i ∈ { , . . . , N k } , Flat ( k ) = { ( v k,i , v k,i +1 ) : i = 1 , . . . , k } and V k +1 ,i ( v k,i , v k,i +1 ) = { v k,i , v k +1 , i , v k,i } . Proof.
For (1), we work by induction. The claim is true for k = 0 by the choice of points v , and v , . Assume the claim is true for some k ≥
0. By the Pythagorean theorem, | v k +1 , i − − v k +1 , i | = | v k,i − v k +1 , i − | = (4 − + t ) / | v k,i − v k,i +1 | = (4 − + t ) / ρ k = ρ k +1 . In similar fashion, one can compute | v k +1 , i − v k +1 , i +1 | and the proof of (1) is complete.Claim (3) is immediate from Remark 9.9 by taking s sufficiently close to 1 and n sufficiently large. For (V0), we have ρ k +1 ρ k = 2 − /s (cid:18) k + 2 + n k + 1 + n (cid:19) q . Clearly, ρ k +1 > ξ ρ k . On the other hand, since 2 − /s < ξ <
1, if n is sufficientlylarge, then ρ k +1 ≤ ξ ρ k . Properties (V1), (V2), and (V5) are immediately satisfied by ourconstruction. For (V4), fix a point v k +1 , i ∈ V k +1 \ V k . By (1), we have | v k +1 , i − v k +1 , i +1 | = ρ k +1 and (V4) is satisfied.For (V3), claim (3), and claim (4), we apply induction on k . For k = 0 (V3) is immediateby the choice of parameters. For claim (3), we note that α ,v = 0 for all v ∈ V , since V contains only 2 points. For the same reason, claim (4) is satisfied when k = 0.To show (V3), we note by (3) that the closest point of V k +1 to v k +1 , i are the points v k,i and v k,i +1 . Therefore, min v ∈ V k +1 \{ v k +1 , i } | v − v k +1 , i | = | v k +1 , i − v k,i | = ρ k +1 . Similarly, by (3), the closest point of V k +1 to v k +1 , i +1 = v k,i +1 are the points v k +1 , i and v k +1 , i +1) (or only one point of these two if i = 0 or i = 2 k } ) and the above inequalityalso applies.Finally, to show (4), we apply (3) and the arguments in the proof of (V3). Namely, if v k,i ∈ V k with k ∈ { , k − } , then α v k,i < α and v k,i lies between points v k,i − and v k,i +1 .Therefore, Flat ( k ) = { ( v k,i , v k,i +1 ) : i = 1 , . . . , k } . Furthermore, the only point of V k +1 lying between v k,i and v k,i +1 is v k +1 , i . Thus, V k +1 ,i ( v k,i , v k,i +1 ) = { v k,i , v k +1 , i , v k,i } . (cid:3) We now show that there does not exist a (1 /s )-H¨older map f : [0 , → R whose imagecontains (cid:83) k ≥ V k . Contrary to the claim, assume that such a map f exists and let H beits H¨older constant. For each v k,i ∈ V k , fix w k,i ∈ [0 ,
1] such that f ( w k,i ) = v k,i . Then | w k,i − w k,j | (cid:38) H,s | v k,i − v k,j | s (cid:38) n ,q,s − k ( k + 1) sq . Therefore, 1 ≥ k min i =1 ,..., k +1 | w k,i − w k,j | (cid:38) H,s,n ,q ( k + 1) sq , which diverges as k → ∞ and we reach a contradiction.It remains to check (9.8). By Lemma 9.11, it suffices to show that ∞ (cid:88) k =0 2 k (cid:88) i =1 τ s ( k, v k,i , v k,i +1 ) p | v k,i − v k,i +1 | s < ∞ . ¨OLDER CURVES AND PARAMETERIZATIONS 69 By the Mean Value Theorem, τ s ( k, v k,i , v k,i +1 ) = | v k,i − v k +1 , i | s + | v k +1 , i − v k,i +1 | s − | v k,i − v k,i +1 | s | v k,i − v k,i +1 | s = 2 ( k + 2 + n ) sq − ( k + 1 + n ) sq ( k + 2 + n ) sq (cid:46) n ,s,q k + 1 . Finally, since sq − p < − ∞ (cid:88) k =0 2 k (cid:88) i =1 τ s ( k, v k,i , v k,i +1 ) p | v k,i − v k,i +1 | s (cid:46) n ,s,q ∞ (cid:88) k =0 k ( k + 1) p (2 − k/s ( k + 1) q ) s = ∞ (cid:88) k =0 ( k + 1) sq − p < ∞ . Appendix A. Tours on connected, finite simple graphs A finite simple graph G = ( V, E ) in a Banach space X is a finite set of points V ⊂ X (called vertices of G ) along with a set E ⊆ {{ v, v (cid:48) } : distinct v, v (cid:48) ∈ V } (called edges of G ). We may identify edges { v, v (cid:48) } in the graph with the (unoriented) line segments[ v, v (cid:48) ] in X . A graph is connected if every pair of vertices in the graph can be joined by asequence of edges in the graph. The valence of a vertex v in G is the number of edges in G that contain v . Proposition A.1.
Let G be a connected, finite simple graph in X . Assume that everyvertex in G has valence at most 2. For any vertex v in G and any nondegenerate compactinterval ∆ , there exists a collection I of open intervals, whose closures are mutuallydisjoint and contained in the interior of ∆ , and there exists a continuous map g : ∆ → G with the following properties.(1) The endpoints of ∆ are mapped onto v .(2) For every vertex v of G , there exists at least one component J of ∆ \ (cid:83) I such that g ( J ) = v . Conversely, for every component J of ∆ \ (cid:83) I , there exists a vertex v of G such that g ( J ) = v .(3) Each interval in I is mapped linearly onto some edge e of G . Conversely, for eachedge e of G , there exist exactly two intervals I ∈ I such that g ( I ) = e .(4) For any vertex v in G , there exists a component J of g − ( v ) ∩ ∆ \ (cid:83) I such thatfor any edge e containing v as an endpoint, there exists I ∈ I such that I ∩ J (cid:54) = ∅ and g ( I ) = e .Proof. If we only desired properties (1)–(3), then we could prove the proposition withoutany restriction on the valency of the vertices by simple induction on the number of edges.The restriction on the valency of the vertices ensures the graph has one of two simpleforms that make it easy to describe maps g satisfying properties (1)–(4). Thus, let G be a connected, finite simple graph in X , and assume that every vertex in G has valence atmost 2. The conclusion being trivial otherwise, we may assume that G contains at leasttwo vertices. There are two possibilities. In each case, we will construct the family I andthe map g , but leave verification of properties (1) through (4) to the reader. Case 1: Suppose that every vertex of G has valence 2 (i.e. G is a “cycle”). Then we canfind an enumeration { u , . . . , u k } of the vertices of G so that the edges of G are precisely { [ u i , u i +1 ] : i = 1 , . . . , k } , where we set u k +1 = u . Without loss of generality, assume that u = v . Let I = { I , . . . , I k } be open intervals, enumerated according to the orientationof ∆, whose closures are mutually disjoint and contained in the interior of ∆. Then thereexists a continuous, surjective map g : ∆ → G such that(1) g is linear on each I i and constant on each component of ∆ \ (cid:83) ki =1 I i ;(2) for each i ∈ { , . . . , k } , g maps I i linearly onto [ u i , u i +1 ] and maps the left endpointof I i onto u i ;(3) for each i ∈ { k + 1 , . . . , k } , g maps I i linearly onto [ u i − k , u i − k +1 ] and maps theleft endpoint of I i onto u i − k .That is, g winds twice around the graph, starting and ending at v = u . Case 2: Suppose that least one vertex of G has valence 1 (i.e. G is an “arc”) . Thenwe can find an enumeration { u , . . . , u k } of the vertices of G so that the edges of G are precisely { [ u i , u i +1 ] : i = 1 , . . . , k − } . In this case, u and u k have valence 1and all other vertices have valence 2. Assume that v = u l for some 1 ≤ l ≤ k . Let I = { I , . . . , I k − } be open intervals, enumerated according to the orientation of ∆,whose closures are mutually disjoint and contained in the interior of ∆. Then there existsa continuous, surjective map g : ∆ → G such that(1) g is linear on each I i and constant on each component of ∆ \ (cid:83) k − i =1 I i ;(2) for each 1 ≤ i ≤ l − g maps I i linearly onto [ u l − i , u l − i +1 ] and maps theleft endpoint of I i onto u l − i +1 ;(3) for each i ∈ { l, . . . , l + k − } , g maps I i linearly onto [ u i − l +1 , u i − l +2 ] and maps theleft endpoint of I i onto u i − l +1 ;(4) for each l + k − ≤ i ≤ k −
1) (if any), g maps I i linearly onto [ u k − l − i , u k − l − i ]and maps the left endpoint of I i onto u k − l − i .That is, g walks along the graph from v = u l towards u , walks from u to u k , and walksfrom u k back to u l . (cid:3) Appendix B. From Lipschitz to H¨older parameterizations
The following method of obtaining H¨older parameterizations from Lipschitz ones is wellknown, see e.g. [SS05, Lemma VII.2.8]. We include Lemma B.1 and its proof here to havea clear statement about the dependence of the H¨older constant of the map f . Lemma B.1.
Let s > , M > , < ξ ≤ ξ < , α > , β > , and j ∈ Z . Let ( X, | · | ) be a Banach space. Suppose that ρ j ( j ≥ j ) is a sequence of scales and f j : [0 , M ] → X ( j ≥ j ) is a sequence of Lipschitz maps satisfying ¨OLDER CURVES AND PARAMETERIZATIONS 71 (1) ρ j = 1 and ξ ρ j ≤ ρ j +1 ≤ ξ ρ j for all j ≥ j ,(2) | f j ( x ) − f j ( y ) | ≤ A j | x − y | for all j ≥ j , where A j ≤ αρ − sj , and(3) | f j ( x ) − f j +1 ( x ) | ≤ B j for all j ≥ j , where B j ≤ βρ j .Then f j converges uniformly to a map f : [0 , M ] → X such that | f ( x ) − f ( y ) | ≤ H | x − y | /s for all x, y ∈ [0 , M ] , where H is a finite constant depending only on max( M, /M ) , ξ , ξ , α , and β ; see (B.7) .Proof. Define f : [0 , M ] → X pointwise by f ( x ) = f j ( x ) + (cid:80) ∞ k = j ( f k +1 ( x ) − f k ( x )) . Then f exists and is the uniform limit of the maps f j by (3), because (cid:80) ∞ k = j B k < ∞ . In fact,for all j ≥ j and x ∈ [0 , M ],(B.1) | f ( x ) − f j ( x ) | ≤ ∞ (cid:88) k = j | f j +1 ( x ) − f j ( x ) | ≤ ∞ (cid:88) k = j βρ k ≤ β − ξ ρ j . Suppose that x, y ∈ [0 , M ] with x (cid:54) = y . Then there is a unique integer j ≥ j such that(B.2) M ρ sj +1 < | x − y | ≤ M ρ sj . By the triangle inequality, (2), and (B.1), | f ( x ) − f ( y ) | ≤ | f j ( x ) − f j ( y ) | + | f ( x ) − f j ( x ) | + | f ( y ) − f j ( y ) |≤ αρ − sj | x − y | + 2 β − ξ ρ j . (B.3)By the first inequality in (B.2) and (1), we have(B.4) ρ j < M /s ξ | x − y | /s . Hence, by the second inequality in (B.2), we have(B.5) ρ − sj | x − y | ≤ M ρ j < MM /s ξ | x − y | /s . Combining (B.3)–(B.5) yields | f ( x ) − f ( y ) | ≤ h | x − y | /s for all x, y ∈ [0 , M ], where(B.6) h = 1 M /s ξ (cid:18) αM + 2 β − ξ (cid:19) . If M ≥
1, then 1 /M /s ≤
1, while if M ≤
1, then 1 /M /s ≤ /M . Thus, it follows that | f ( x ) − f ( y ) | ≤ H | x − y | /s for all x, y ∈ [0 , M ], where(B.7) H = max(1 , /M ) ξ (cid:18) αM + 2 β − ξ (cid:19) = αξ max(1 , M ) + 2 βξ (1 − ξ ) max(1 , /M )depends only on max( M, /M ), ξ , ξ , α , and β . (cid:3) Remark B.2.
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