HHAUSDORFF DIMENSION OF WIGGLY METRIC SPACES
JONAS AZZAMA
BSTRACT . For a compact connected set X ⊆ (cid:96) ∞ , we define a quantity β (cid:48) ( x, r ) thatmeasures how close X may be approximated in a ball B ( x, r ) by a geodesic curve. Wethen show there is c > so that if β (cid:48) ( x, r ) > β > for all x ∈ X and r < r , thendim X > cβ . This generalizes a theorem of Bishop and Jones and answers a questionposed by Bishop and Tyson. Mathematics Subject Classification (2010):
Keywords:
Wiggly sets, traveling salesman, geodesic deviation, Hausdorff dimension,conformal dimension.
1. I
NTRODUCTION
Background and Main Results.
Our starting point is a theorem of Bishop and Jones,stated below, which roughly says that a connected subset of R that is uniformly non-flat in every ball centered upon it (or in other words, is very “wiggly”), must have largedimension. We measure flatness with Jones’ β -numbers: if K is a subset of a Hilbert space H , x ∈ K and r > , we define(1.1) β ( x, r ) = β K ( x, r ) = 1 r inf L sup { dist ( y, L ) : y ∈ K ∩ B ( x, r ) } where the infimum is taken over all lines L ⊆ H . Theorem 1. ( [1, Theorem 1.1] ) There is a constant c > such that the following holds.Let K ⊆ R be a compact connected set and suppose that there is r > such that for all r ∈ (0 , r ) and all x ∈ K , β K ( x, r ) > β . Then the Hausdorff dimension of K satisfiesdim K ≥ cβ . There are also analogues of Theorem 1 for surfaces of higher topological dimension,see for example [5].Our main theorem extends this result to the metric space setting using an alternate def-inition of β . Before stating our results, however, we discuss the techniques and stepsinvolved in proving Theorem 1 to elucidate why the original methods don’t immediatelycarry over, and to discuss how they must be altered for the metric space setting.The main tool in proving Theorem 1 is the Analyst’s Traveling Salesman Theorem ,which we state below. First recall that for a metric space ( X, d ) , a maximal ε -net is amaximal collection of points X (cid:48) ⊆ X such that d ( x, y ) ≥ ε for all x, y ∈ X (cid:48) . Mathematics Subject Classification.
Primary 28A75, Secondary 28A78, 30L10.
Key words and phrases.
Wiggly sets, β -number, traveling salesman, geodesic deviation, Hausdorff dimen-sion, conformal dimension.The author was supported by the NSF grants RTG DMS 08-38212 and DMS-0856687. See Section 2 for the definition of Hausdorff dimension and other definitions and notation. a r X i v : . [ m a t h . M G ] J a n JONAS AZZAM
Theorem 2. ( [16, Theorem 1.1] ) Let A > , K be a compact subset of a Hilbert space H , and X n ⊇ X n +1 be a nested sequence of maximal − n -nets in K . For A > , define (1.2) β A ( K ) := diam K + (cid:88) n ∈ Z (cid:88) x ∈ X n β K ( x, A − n )2 − n . There is A such that for A > A there is C A > (depending only on A ) so that for any K , β A ( K ) < ∞ implies there is a connected set Γ such that K ⊆ Γ and H (Γ) ≤ C A β A ( K ) . Conversely, if Γ is connected and H (Γ) < ∞ , then for any A > , (1.3) β A (Γ) ≤ C A H (Γ) . At the time of [1], this was only known for the case H = R , due to Jones [9]. Thiswas subsequently generalized to R n by Okikiolu [13] and then to Hilbert space by Schul[16].The proof of Theorem 1 goes roughly as follows: one constructs a Frostmann measure µ supported on K satisfying(1.4) µ ( B ( x, r )) ≤ Cr s for some C > , s = 1 + cβ and for all x ∈ K and r > . This easily impliesthat the Hausdorff dimension of K is at least s (see [12, Theorem 8.8] and that sectionfor a discussion on Frostmann measures). One builds such a measure on K inductivelyby deciding the values µ ( Q n ) µ ( Q ) for each dyadic cube Q intersecting K and for each n -thgeneration descendant Q n intersecting K , where n is some large number that will dependon β . If the number of such n -th generation descendants is large enough, we can choosethe ratios and hence disseminate the mass µ ( Q ) amongst the descendants Q n in such a waythat the ratios will be very small and (1.4) will be satisfied. To show that there are enoughdescendants, one looks at the skeletons of the n -th generation descendants of Q and usesthe second half of Theorem 2 coupled with the non-flatness condition in the satement ofTheorem 1 to guarantee that the total length of this skeleton (and hence the number ofcubes) will be large.In the metric space setting, however, no such complete analogue of Theorem 2 exists,and it is not even clear what the appropriate analogue of a β -number should be. Note, forexample, that it does not make sense to estimate the length of a metric curve Γ using theoriginal β -number, even if we consider Γ as lying in some Banach space. A simple counterexample is if Γ ⊆ L ([0 , is the image of s : [0 , → L ([0 , defined by t (cid:55)→ [0 ,t ] .This a geodesic, so in particular, it is a rectifiable curve of finite length. However, β Γ ( x, r ) (i.e. the width of the smallest tube containing Γ ∩ B ( x, r ) in L , rescaled by a factor r ) isuniformly bounded away from zero, and in particular, β A (Γ) = ∞ .In [6], Hahlomaa gives a good candidate for a β -number for a general metric space X using Menger curvature and uses it to show that if the sum in (1.2) is finite for K = X (using his definition of β X ), then it can be contained in the Lipschitz image of a subset ofthe real line (analogous to the first half of Theorem 2). An example of Schul [15], however,shows that the converse of Theorem 2 is false in general: (1.3) with Hahlomaa’s β X doesnot hold with the same constant for all curves in (cid:96) . We refer to [15] for a good summaryon the Analyst’s Traveling Salesman Problem.To generalize Theorem 1, we use a β -type quantity that differs from both Jones’ andHahlomaa’s definitions. It is inspired by one defined by Bishop and Tyson in [2] that AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 3 measures the deviation of a set from a geodesic in a metric space: if X is a metric space, B X ( x, r ) = { y ∈ X : d ( x, y ) < r ) } , and y , ..., y n ∈ B X ( x, r ) an ordered sequence,define(1.5) (. y , ..., y n ) = n − (cid:88) i =0 d ( y i , y i +1 ) − d ( y , y n ) + sup z ∈ B X ( x,r ) min i =1 ,...,n d ( z, y i ) and define(1.6) ˆ β X ( x, r ) = inf { y i }⊆ B X ( x,r ) (. y , ..., y n ) d ( y , y n ) where the infimum is over all finite ordered sequences in B X ( x, r ) of any length n .In [2], Bishop and Tyson ask whether, for a compact connected metric space X , (1.6)being uniformly larger than zero is enough to guarantee that dim X > . We answer this inthe affirmative. Theorem 3.
There is κ > such that the following holds. If X is a compact connectedmetric space and ˆ β X ( x, r ) > β > for all x ∈ X and r ∈ (0 , r ) for some r > , thendim X ≥ κβ . Instead of ˆ β , however, we work with a different quantity, which we define here fora general compact metric space X . First, by Kuratowski embedding theorem, we mayassume X is a subset of (cid:96) ∞ , whose norm we denote by | · | . Let B ( x, r ) = B (cid:96) ∞ ( x, r ) anddefine(1.7) β (cid:48) X ( x, r ) = inf s (cid:96) ( s ) − | s (0) − s (1) | + sup z ∈ X ∩ B ( x,r ) dist ( z, s ([0 , | s (0) − s (1) | where the infimum is over all curves s : [0 , → B ( x, r ) ⊆ (cid:96) ∞ and (cid:96) ( s ) = sup { t i } ni =0 n − (cid:88) i =0 | s ( t i ) − s ( t i +1 ) | is the length of s , where the supremum is over all partitions t < t < · · · < t n = 1 .In general, if s is defined on a union of disjoint open intervals { I j } ∞ j =1 , we set (cid:96) ( s | (cid:83) I j ) = (cid:88) j (cid:96) ( s | I j ) . The case in which s is just a straight line segment through the center of the ball with length r gives the estimate β (cid:48) X ( x, r ) ≤ .The quantity β (cid:48) ( x, r ) measures how well X ∩ B ( x, r ) may be approximated by a ge-odesic. To see this, note that if, for some s : [0 , → (cid:96) ∞ , the β (cid:48) ( x,r )2 | s (0) − s (1) | -neighborhood of s ([0 , contains X ∩ B ( x, r ) , then the length of s must be at least (1 + β (cid:48) ( x,r )2 ) | s (0) − s (1) | , which is β (cid:48) ( x,r )2 | s (0) − s (1) | more than the length of any ge-odesic connecting s (0) and s (1) . The quantity ˆ β similarly measures how well the portionof X ∩ B ( x, r ) may be approximated by a geodesic polygonal path with vertices in X . InFigure 1, we compare the meanings of β, ˆ β, and β (cid:48) .We will refer to the quantities (cid:96) ( s ) and (. y , ..., y n ) as the geodesic deviation of s and { y , ..., y n } respectively. We will also say ˆ β X ( x, r ) and β (cid:48) X ( x, r ) measure the geodesicdeviation of X inside the ball B ( x, r ) .Note that for the image of t (cid:55)→ [0 ,t ] ∈ L ([0 , described earlier, it is easy to checkthat ˆ β ( x, r ) = β (cid:48) ( x, r ) = 0 for all x ∈ X and r > , even though β X ( x, r ) is bounded JONAS AZZAM { β ( x, r )2 rB ( y i , β | y − y n | ) | y − y n | | s (0) − s (1) | < β | s (0) − s (1) | s ([0 , B = B ( x, r ) X F IGURE
1. In each of the three figures above is a ball B = B ( x, r ) containing a portion of a curve X . In the first picture, β ( x, r )2 r is thewidth of the smallest tube containing X ∩ B ( x, r ) . In the second, we seethat ˆ β ( x, r ) is such that for β > ˆ β ( x, r ) , there are y , ..., y n ∈ X withvertices in X ∩ B so that balls centered on the y i of radius β | y − y n | cover X ∩ B , and so that the geodesic deviation (that is, its length minus | y − y n | is at most β | y − y n | . In the last, we show that if β (cid:48) ( x, r ) < β ,there is s : [0 , → (cid:96) ∞ whose geodesic deviation and whose distancefrom any point in X ∩ B are both at most β | s (0) − s (1) | .away from zero. This, of course, makes the terminology “wiggly” rather misleading inmetric spaces, since there are certainly non-flat or highly “wiggly” geodesics in L ; weuse this terminology only to be consistent with the literature. Later on in Proposition 18,however, we will show that in a Hilbert space we have for some C > ,(1.8) β (cid:48) ( x, r ) ≤ β ( x, r ) ≤ Cβ (cid:48) ( x, r ) . That the two should be correlated in this setting seems natural as β ( x, r ) is measuring howfar X is deviating from a straight line, which are the only geodesics in Hilbert space.In Lemma 17 below, we will also show that for some C > , β (cid:48) ( x, r ) ≤ ˆ β ( x, r ) ≤ Cβ (cid:48) ( x, r ) so that Theorem 3 follows from the following theorem, which is our main result. AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 5
Theorem 4.
There is c > such that the following holds. If X is a compact connectedmetric space and β (cid:48) X ( x, r ) > β > for all x ∈ X and r ∈ (0 , r ) for some r > , thendim X ≥ c β . We warn the reader, however, that the quadratic dependence on β appears in Theorem4 and Theorem 1 for completely different reasons. In Theorem 1, it comes from usingTheorem 2, or ultimately from the Pythagorean theorem, which of course does no hold ingeneral metric spaces; in Theorem 4, it seems to be an artifact of the construction and canperhaps be improved.Our approach to proving Theorem 4 follows the original proof of Theorem 1 describedearlier: to show that a metric curve X has large dimension, we approximate it by a polyg-onal curve, estimate its length from below and use this estimate to construct a Frostmannmeasure, but in lieu of a traveling salesman theorem. (In fact, taking β (cid:48) ( x, A − n ) insteadof β ( x, A − n ) in Theorem 2 does not lead to a metric version of Theorem 2 for a similarreason that Hahlomaa’s β -number doesn’t work; one need only consider Schul’s example[15, Section 3.3.1].)1.2. An Application to Conformal Dimension.
The original context of Bishop and Tyson’sconjecture, and the motivation for Theorem 4, concerned conformal dimension. Recall thata quasisymmetric map f : X → Y between two metric spaces is a map for which there isan increasing homeomorphism η : (0 , ∞ ) → (0 , ∞ ) such that for any distinct x, y, z ∈ X , | f ( x ) − f ( y ) || f ( z ) − f ( y ) | ≤ η (cid:18) | x − y || z − y | (cid:19) . The conformal dimension of a metric space X isC-dim X = inf f dim f ( X ) where the infimum ranges over all quasisymmetric maps f : X → f ( X ) . For moreinformation, references, and recent work on conformal dimension, see for example [11].In [2], it is shown that the antenna set has conformal dimension one yet every qua-sisymmetric image of it into any metric space has dimension strictly larger than one. The antenna set is a self similar fractal lying in C whose similarities are the following: f ( z ) = z , f ( z ) = z + 12 , f ( z ) = iαz + 12 , f ( z ) = − iαz + 12 + iα where α ∈ (0 , ) is some fixed angle (see Figure 2).F IGURE
2. The antenna set with α = .To show the conformal dimension is never attained under any quasisymmetric imageof the antenna set, the authors show by hand that any quasisymmetic map of the antenna JONAS AZZAM set naturally induces a Frostmann measure of dimension larger than one. At the end of thepaper, however, the authors suggested another way of showing the same result by provingan analogue of Theorem 1 for a β -number which is uniformly large for the antenna set aswell as any quasisymmetric image of it.Theorem 4 doesn’t just give a much longer proof of Bishop and Tyson’s result, but itlends itself to more general sets lacking any self-similar structure. Definition 5.
Let c > , Y = [0 , e ] ∪ [0 , e ] ∪ [0 , e ] ⊆ R , where e j is the j th standardbasis vector in R , and let X be a compact connected metric space. For x ∈ X , r > , we say B X ( x, r ) has a c -antenna if there is a homeomorphism h : Y → h ( Y ) ⊆ B X ( x, r ) such that the distance between h ( e i ) and h ([0 , e j ] ∪ [0 , e k ])) is at least cr for allpermutations ( i, j, k ) of (1 , , . We say X is c -antenna-like if B X ( x, r ) has a c -antennafor every x ∈ X and r < diam X ,Clearly, the classical antenna set in R is antenna-like. Theorem 6.
Let X be a compact connected metric space in (cid:96) ∞ .(1) If B X ( x, r ) has a c -antenna, then β (cid:48) ( x, r ) > c . Hence, if X is c -antenna-like, wehave dim X ≥ c c .(2) Any quasisymmetric image of an antenna-like set into any metric space is alsoantenna-like and hence has dimension strictly larger than one. Note that this result doesn’t say the conformal dimension of an antenna-like set is largerthan one, only that no quasisymmetric image of it has dimension equal to one. However,see [10], where the author bounds the conformal dimension of a set from below using adifferent quantity.1.3.
Outline.
In Section 2, we go over some necessary notation and tools before proceed-ing to the proof of Theorem 4 in Section 3. In Section 4, we prove Theorem 6, and inSection 5 we compare β (cid:48) , ˆ β, and β .1.4. Acknowledgements.
The author would like to thank Steffen Rohde, Tatiana Toro,and Jeremy Tyson for their helpful discussions, and to Matthew Badger, John Garnett,Raanan Schul, and the anonymous referee for their helpful comments on the manuscript.Part of this manuscript was written while the author was at the IPAM long program Inter-actions Between Analysis and Geometry, Spring 2013.2. P
RELIMINARIES
Basic notation.
Since we are only dealing with compact metric spaces, by the Ku-ratowski embedding theorem, we will implicitly assume that all our metric spaces arecontained in (cid:96) ∞ , whose norm we will denote | · | .For x ∈ (cid:96) ∞ and r > , we will write B ( x, r ) = { y ∈ (cid:96) ∞ : | x − y | < r } ⊆ (cid:96) ∞ . If B = B ( x, r ) and λ > , we write λB for B ( x, λr ) . For a set A ⊆ (cid:96) ∞ and δ > , define A δ = { x ∈ (cid:96) ∞ : dist ( x, A ) < δ } and diam A = sup {| x − y | : x, y ∈ A } where dist ( A, B ) = inf {| x − y | : x ∈ A, y ∈ B } , dist ( x, A ) = dist ( { x } , A ) . AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 7
For a set E ⊆ R , let | E | denote its Lebesgue measure. For an interval I ⊆ R , we will write a I and b I for its left and right endpoints respectively. For s > , δ ∈ (0 , ∞ ] and A ⊆ (cid:96) ∞ ,define H sδ ( A ) = inf (cid:110)(cid:88) diam A j : A ⊆ (cid:91) A j , diam A j < δ (cid:111) , H s ( A ) = lim δ → H δ ( A ) . The
Hausdorff dimension of a set A isdim A := inf { s : H s ( A ) = 0 } . Cubes.
In this section, we construct a family of subsets of (cid:96) ∞ , tailored to a metricspace X , that have properties similar to dyadic cubes in Euclidean space. These cubesappeared in [16] (where they were alternatively called “cores”) and are similar to the so-called Christ-David Cubes ([4, 3]) in some respects, although they are not derived fromthem.Fix M > and c ∈ (0 , ) . Let X n ⊆ X be a nested sequence of maximal M − n -netsin X . Let B n = { B ( x, M − n ) : x ∈ X n } , B = (cid:91) n B n . For B = B ( x, M − n ) ∈ B n , define Q B = cB, Q jB = Q j − B ∪ (cid:91) { cB : B ∈ (cid:91) m ≥ n B m , cB ∩ Q j − B (cid:54) = ∅} , Q B = ∞ (cid:91) j =0 Q jB . Basically, Q B is the union of all balls B (cid:48) that may be connected to B by a chain { cB j } with B j ∈ B , diam B j ≤ diam B , and cB j ∩ cB j +1 for all j .For such a cube Q constructed from B ( x, M − n ) , we let x Q = x and B Q = B ( x, cM − n ) .Let ∆ n = { Q B : B ∈ B n } , ∆ = (cid:91) ∆ n . Note that, for Q ∈ ∆ n , x Q ∈ X n . Lemma 7. If c < , then for X and ∆ as above, the family of cubes ∆ satisfy the followingproperties.(1) If Q, R ∈ ∆ and Q ∩ R (cid:54) = ∅ , then Q ⊆ R or R ⊆ Q .(2) For Q ∈ ∆ , (2.1) B Q ⊆ Q ⊆ (1 + 8 M − ) B Q . The proof is essentially in [14], but with slightly different parameters. So that the readerneed not perform the needed modifications, we provide a proof here.
Proof.
Part 1 follows from the definition of the cubes Q . To prove Part 2, we first claimthat if { B j } nj =0 is a chain of balls with centers x j for which cB j ∩ cB j +1 (cid:54) = ∅ , then for C = − M − ,(2.2) n (cid:88) j =0 diam cB j ≤ C max j =0 ,...,n diam cB j . JONAS AZZAM
We prove (2.2) by induction. Let x j denote the center of B j If n = 1 , diam B ≤ diam B ,and x and x are the centers of B and B respectively, then diam B ≤ M − diam B since otherwise B , B ∈ B N for some N and M − n ≤ | x − x | ≤ diam cB diam cB cM − n < M − n since c < , which is a contradiction. Hence,diam cB + diam cB ≤ (1 + 2 M − ) diam cB ≤ C diam cB . Now suppose n > . Let j ∈ { , ..., n } and N be an integer so that(2.3) diam B j = max j =1 ,...,n diam B j = 2 M − N . Recall that all balls in B have radii that are powers of M − , so there exists an N so thatthe above happens.Note that B j − and B j cannot have the same diameter (which follows from the n = 1 case we proved earlier). Since B j has the maximum diameter of all the B j , we in factknow that diam B j − ≤ M − B j (again, recall that all balls have radii that are powers of M − ).Let i ≤ j be the minimal integer for which diam B i ≤ M − diam B j (which existsby the previous discussion) and let k ≥ j be the maximal integer such that B k ≤ M − diam B j . By the induction hypothesis, k (cid:88) j = j +1 diam cB j ≤ C max j
There are a few different constructions of families of metric subsets with prop-erties similar to dyadic cubes, see [4], [3], and [8] for example, and the references therein.Readers familiar with any of these references will see that Schul’s “cores” we have just con-structed are very different from the cubes constructed in the aforementioned references. Inparticular, each ∆ n does not partition any metric space in the same way that dyadic cubes(half-open or otherwise) would partition Euclidean space, not even up to set of measurezero). However, for each n we do have(2.7) X ⊆ (cid:91) { c − Q : Q ∈ ∆ n } , and we still have the familiar intersection properties in Lemma 7. The reason for the adhoc construction is the crucial “roundness” property (2.6). Lemma 9.
Let γ : [0 , → (cid:96) ∞ be a piecewise linear function and set Γ = γ ([0 , , whoseimage is a finite union of line segments, and let ∆ be the cubes from Lemma 7 tailored to X . Then for any Q ∈ ∆ , H ( Q. ) = 0 and | γ − ( Q. ) | = 0 .Proof. Note that since Γ is a finite polynomial curve, µ = H | Γ is doubling on Γ , meaningthere is a constant C so that µ ( B ( x, M r )) ≤ Cµ ( B ( x, r )) for all x ∈ Γ and r > . If x ∈ Q. for some Q ∈ ∆ , then there is a sequence x n ∈ X n such that | x n − x | < M − n sincethe X n are maximal M − n -nets. To each x n corresponds a ball B n = B ( x n , M − n ) ∈ B n .Let N be such that Q ∈ ∆ N . Since cB n ⊆ Q B n ∈ ∆ n , we have by Lemma 7 thateither cB n ⊆ Q (if Q B n ∩ Q (cid:54) = ∅ ) or cB n ⊆ R for some R ∈ ∆ N with Q ∩ R = ∅ .In either case, since cubes don’t contain their boundaries (since they are open), we havethat cB n ∩ Q. = ∅ . This implies that Q is porous, and it is well known that such setshave doubling measure zero. More precisely, the doubling condition on µ guarantees that lim n →∞ µ ( Q. ∩ B ( x,M − n )) µ ( B ( x,M − n )) = 1 µ -a.e. x ∈ Γ (see [7, Theorem 1.8]), but if x ∈ Q. and B n isas above, then one can show using the doubling property of µ that lim sup n →∞ µ ( Q. ∩ B ( x, M − n )) µ ( B ( x, M − n )) ≤ lim sup n →∞ µ ( B ( x, M − n ) \ B n ) µ ( B ( x, M − n )) < , and thus µ ( Q. ) = 0 .The last part of the theorem follows easily since γ is piecewise affine. (cid:3) The following lemma will be used frequently.
Lemma 10.
Let I ⊆ R be an interval, s : I → (cid:96) ∞ be continuous and I (cid:48) ⊆ I a subinterval.Then (2.8) (cid:96) ( s | I (cid:48) ) − | s ( a I (cid:48) ) − s ( b I (cid:48) ) | ≤ (cid:96) ( s | I ) − | s ( a I ) − s ( b I ) | . Proof.
We may assume (cid:96) ( s I ) < ∞ , otherwise (2.8) is trivial. We estimate (cid:96) ( s | I (cid:48) ) − | s ( a I (cid:48) ) − s ( b I (cid:48) ) | = (cid:96) ( s | I ) − (cid:96) ( s | I \ I (cid:48) ) − | s ( a I (cid:48) ) − s ( b I (cid:48) ) |≤ (cid:96) ( s | I ) − ( | s ( a I ) − s ( a I (cid:48) ) | + | s ( b I ) − s ( b I (cid:48) ) | ) − | s ( a I (cid:48) ) − s ( b I (cid:48) ) |≤ (cid:96) ( s | I ) − | s ( a I ) − s ( b I ) | . (cid:3)
3. P
ROOF OF T HEOREM
Setup.
For this section, we fix a compact connected set X satisfying the conditionsof Theorem 4. The main tool is the following Lemma, which can be seen as a very weaksubstitute for Theorem 2. Lemma 11.
Let c (cid:48) < . We can pick M large enough (by picking ε > small enough)and pick β , κ > such that, for any X satisfying the conditions of Theorem 4 for some β ∈ (0 , β ) , the following holds. If X n is any nested sequence of M − n -nets in X , there is n = n ( β ) such that for x ∈ X n with M − n < min (cid:8) r , diam X (cid:9) , (3.1) X n + n ∩ B ( x , c (cid:48) M − n ) ≥ M (1+ κβ ) n . We will prove this in Section 3.2, but first, we’ll explain why this proves Theorem 4.
Proof of Theorem 4.
Without loss of generality, we may assume r > by scaling X ifnecessary. We first consider the case that β < β . Let ∆ be the cubes from Lemma 7tailored to the metric space X with c = c (cid:48) and define inductively, ∆ (cid:48) = ∆ , ∆ (cid:48) n +1 = { R ∈ ∆ ( n +1) n : R ⊆ Q for some Q ∈ ∆ n } . By Lemma 11, for any Q ∈ ∆ (cid:48) n , if B Q = B ( x Q , cM − N ) , then(3.2) { R ∈ ∆ (cid:48) n +1 , R ⊆ Q } ≥ X N + n ∩ Q ≥ X n ∩ c (cid:48) B Q ≥ M (1+ κβ ) n and moreover, since c (cid:48) < ,(3.3) B Q ∩ B R = ∅ for Q, R ∈ ∆ n . Define a probability measure µ inductively by picking Q ∈ ∆ (cid:48) , setting µ ( Q ) = 1 and for Q ∈ ∆ (cid:48) n and R ∈ ∆ (cid:48) n +1 , R ⊆ Q (3.4) µ ( R ) µ ( Q ) = 1 { S ∈ ∆ (cid:48) n +1 : S ⊆ Q } (3.2) ≤ M − (1+ κβ ) n . Let x ∈ X , r ∈ (0 , r M ) . Pick n so that(3.5) M − n ( n +1) ≤ r < M − n n . Claim:
There is at most one y ∈ X ( n − n such that(3.6) B ( y, c (cid:48) M − ( n − n ) ∩ B ( x, r ) (cid:54) = ∅ and Q = Q B ( y,c (cid:48) M − ( n − n ) ∈ ∆ (cid:48) n − . AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 11
Indeed, if there were another such y (cid:48) ∈ X ( n − n with B ( y (cid:48) , c (cid:48) M − ( n − n ) ∩ B ( x, r ) (cid:54) = ∅ ,then M − ( n − n ≤ | y (cid:48) − y |≤ c (cid:48) M − ( n − n + dist (cid:16) B ( y, c (cid:48) M − ( n − n ) , B ( y (cid:48) , c (cid:48) M − ( n − n ) (cid:17) + c (cid:48) M − ( n − n ≤ c (cid:48) M − ( n − n + diam B ( x, r ) ≤ c (cid:48) M − ( n − n + 2 r (3.5) ≤ M − ( n − n ( c (cid:48) + M − n ) < c (cid:48) M − ( n − n < M − ( n − n since c (cid:48) < and we can pick ε < c (cid:48) so that M − n ≤ M − < c (cid:48) , which gives a contra-diction and proves the claim.Now, assuming we have y ∈ X ( n − n satisfying (3.6), B ( x, r ) ⊆ B ( y, c (cid:48) M − ( n − n + 2 r ) (3.5) ⊆ B ( y, c (cid:48) M − ( n − n + 2 M − nn ) ⊆ B ( y, c (cid:48) M − ( n − n ) = 2 B Q for M large enough (that is, for M − < c (cid:48) , which is possible by picking ε < c (cid:48) ). If Q (cid:54)∈ ∆ (cid:48) n − , then (3.3) implies B Q ∩ B R = ∅ for all R ∈ ∆ (cid:48) n − , and so µ ( B ( x, r )) ≤ µ (2 B Q ) = 0 . Otherwise, if Q ∈ ∆ (cid:48) n − , then Q ⊆ Q , so that µ ( B ( x, r )) ≤ µ (2 B Q ) (3.3) = µ ( Q ) (3.4) = M − (1+ κβ ) n ( n − µ ( Q ) (3.5) ≤ M κβ ) r − (1+ κβ ) thus µ is a (1+ κβ ) -Frostmann measure supported on X , which implies dim X ≥ κβ (c.f. [12, Theorem 8.8]).Now we consider the case when β ≥ β . Trivially, β (cid:48) ( x, r ) ≥ β ≥ β for all x ∈ X and r < r , and our previous work gives dim X ≥ κt for all t < β , hence dim X ≥ κβ . Since β (cid:48) ≤ , we must have β, β ≤ , and sodim X ≥ κβ ≥ κβ β and the theorem follows with c = 4 κβ . (cid:3) To show Lemma 11, we will approximate X by a tree containing a sufficiently densenet in X and estimate its length from below. The following lemma relates the length ofthis tree to the number of net points in X . Lemma 12.
Let X n be a maximal M − n -net for a connected metric space X where n is so that M − n < diam X . Then we may embed X into (cid:96) ∞ so that there is a connectedunion of finitely many line segments Γ n ⊆ (cid:96) ∞ containing X n such that for any x ∈ X n and r ∈ (4 M − n , diam X ) , (3.7) H (cid:16) Γ n ∩ B (cid:16) x, r (cid:17)(cid:17) ≤ M − n X n ∩ B ( x, r )) . Proof.
Embed X isometrically into (cid:96) ∞ ( N ) so that for any x ∈ X , the first X n coordi-nates are all zero. Construct a sequence of trees T j as follows. Enumerate the elements of X n = { x , ..., x X n } . For two points x and y , let A xy,i = { tx + (1 − t ) y + max { t, − t }| x − y | e i : t ∈ [0 , } where e i is the standard basis vector in (cid:96) ∞ ( N ) (i.e. it is equal to in the i th coordinate andzero in every other coordinate).Now construct a sequence of trees T j in (cid:96) ∞ ( N ) inductively by setting T = { x } and T j +1 equal to T j united with S j +1 := A x j +1 x (cid:48) j +1 ,j +1 , where x (cid:48) j +1 ∈ { x , ..., x j } and x j +1 ∈ X n \{ x , ..., x j } are such that | x j +1 − x (cid:48) j +1 | = dist ( X n \{ x , ..., x j } , { x , ..., x j } ) . Since X is connected, | x j +1 − x (cid:48) j +1 | ≤ M − n , so that H ( S j ) = H ( A x j ,x (cid:48) j ,j ) ≤ | x j − x (cid:48) j | ≤ · M − n = 8 M − n . Then Γ n := T X n is a tree contained in (cid:96) ∞ ( N ) containing X n (the reason we made thearcs S j reach into an alternate dimension is to guarantee that the branches of the tree don’tintersect except at the points X n ).To prove (3.7), note that since r > M − n and x j ∈ S j ⊆ B ( x j , M − n ) , we have H (cid:16) Γ n ∩ B (cid:16) x, r (cid:17)(cid:17) ≤ (cid:88) S j ∩ B ( x, r ) (cid:54) = ∅ H ( S j ) ≤ (cid:88) x j ∈ B ( x, r +2 M − n ) M − n ≤ X n ∩ B ( x, r )) . (cid:3) Proof of Lemma 11.
We now dedicate ourselves to the proof of Lemma 11. Again,let X be a connected metric space satisfying the conditions of Theorem 4. Without loss ofgenerality, n = 0 , so that diam X > . Embed X into (cid:96) ∞ as in Lemma 12. Fix n ∈ N .Let Γ n be the tree from Lemma 12 containing the M − n -net X n ⊆ X .Since Γ n is a tree of finite length that is a union of finitely many line segments,it is not hard to show that there is a piecewise linear arc length parametrized path γ :[0 , H (Γ n )] → Γ n that traverses almost every point in Γ n at most twice (except atthe discrete set of points X n ). The proof is similar to that of its graph theoretic analogue.Let ∆ be the cubes from Lemma 7 tailored to Γ n and fix Q ∈ ∆ . We will adjust thevalues of c > in Lemma 7 and the value ε > in the definition of M as we go alongthe proof. Note that diam X > implies diam Γ n > > (1 + εβ ) c if c < , and so Γ n (cid:54)⊆ Q . For Q, R ∈ ∆ , write R = Q if R is a maximal cube in ∆ properly containedin Q . For n ≥ and Q ∈ ∆ , define L ( Q ) = { R ∈ ∆ : R = Q } , L n ( Q ) = (cid:91) R ∈ L n − ( Q ) L ( R ) , ˜ L n ( Q ) = L n ( Q ) ∩ n − (cid:91) j =0 ∆ j , ˜ L ( Q ) = (cid:91) ˜ L n ( Q )˜ L n = ˜ L n ( Q ) , ˜ L = ˜ L ( Q ) . For Q ∈ ∆ , let λ ( Q ) = { [ a, b ] : ( a, b ) is a connected component of γ − ( Q ) } and for n ≤ n , define γ n to be the continuous function such that for all Q ∈ L n ( Q ) and [ a, b ] ∈ λ ( Q ) , γ n | [ a,b ] ( at + (1 − t ) b ) = tγ ( a ) + (1 − t ) γ ( b ) for t ∈ [0 , , AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 13 that is, γ n is linear in all cubes in ∆ n and agrees with γ on the boundaries of the cubes (seeFigure 3). (a)(b) (c) Q R ∈ L ( Q ) γ n +1 | I γ | I γ n | I F IGURE
3. In (a), we have a typical cube Q ∈ ∆ n , and some of itschildren in L ( Q ) . Note that their sizes can be radically different. In (b)are the components γ | γ − ( Q ) , where in this case γ − ( Q ) consists of twointervals, and we’ve pointed at a particular component γ | I for some I ∈ λ ( Q ) . In (c), the dotted lines represent the components of γ n | γ − ( Q ) ,which is affine in cubes in ∆ n , and hence is affine in Q , and the solidpiecewise-affine curves represent the components of γ n +1 | γ − ( Q ) , whichare affine in the children of Q (since they are in ∆ n +1 ).Lemma 11 will follow from the following two lemmas: Lemma 13.
There is K ∈ (0 , and β > (independent of n above) such that if β ∈ (0 , β ) , n < n , and Q ∈ ˜ L n , either (3.8) (cid:88) I ∈ λ ( Q ) ( (cid:96) ( γ n +1 | I ) − (cid:96) ( γ n | I )) ≥ εβ diam Q or Q ∈ ∆ Bad , where (3.9) ∆ Bad = { R ∈ ˜ L : H ∞ (Γ n ∩ R ) ≥ (1 + Kβ ) diam R } Lemma 14.
With ∆ Bad defined as above, we have (3.10) (cid:88) Q ∈ ∆ Bad β diam Q ≤ K H (Γ n ) . We’ll prove these in sections 3.3 and 3.4 respectively, but first let us finish the proof ofLemma 11.
For Q ∈ ˜ L , let n ( Q ) be such that Q ∈ L n and define d ( Q ) = (cid:88) I ∈ λ ( Q ) (cid:0) (cid:96) ( γ n ( Q )+1 | I ) − (cid:96) ( γ n ( Q ) | I ) (cid:1) . By telescoping sums and Lemma 9, we have (cid:88) Q ∈ ˜ L d ( Q ) = n − (cid:88) n =0 (cid:88) Q ∈ ˜ L n (cid:88) I ∈ λ ( Q ) ( (cid:96) ( γ n +1 | I ) − (cid:96) ( γ n | I )))= n − (cid:88) n =0 (cid:0) (cid:96) ( γ n +1 | γ − ( Q )) − (cid:96) ( γ n | γ − ( Q )) (cid:1) ≤ (cid:96) ( γ | γ − ( Q ) ) = 2 H (Γ n ∩ Q ) . (3.11)Note that diam (Γ n ∩ Q ) ≥ since Q ∈ ∆ , diam Γ n > , and Γ n is connected. This,Lemma 13, and Lemma 14 imply Kε H (Γ n ∩ Q ) ≥ Kε H (Γ n ∩ Q ) + 8 ε H (Γ n ∩ Q ) (3.10)(3.11) ≥ (cid:88) Q ∈ ∆ Bad β diam Q + 4 ε (cid:88) Q ∈ ˜ L \ ∆ Bad d ( Q ) (3.8) ≥ (cid:88) Q ∈ ∆ Bad β diam Q + (cid:88) Q ∈ ˜ L \ ∆ Bad β diam Q = (cid:88) Q ∈ ˜ L β diam Q = n − (cid:88) n =0 (cid:88) Q ∈ ∆ n β diam Q ≥ n − (cid:88) n =0 (cid:88) Q ∈ ∆ n β diam B Q = n − (cid:88) n =0 c (cid:88) Q ∈ ∆ n β diam c B Q (2.7) ≥ cn β diam (Γ n ∩ Q ) ≥ cn β so that Kcn βε ≤ H (Γ n ∩ Q ) . By Lemma 12, and since B Q has radius c , H (Γ n ∩ Q ) ≤ H (Γ n ∩ (1 + εβ ) B Q ) ≤ H (Γ n ∩ B ( x, c )) ≤ X n ∩ B ( x, c )) M − n Combining these two estimates we have, for c < c (cid:48) that δn M n β ≤ X n ∩ B ( x , c (cid:48) )) , δ = Kcε Pick n = (cid:108) δβ ε (cid:109) . Since εβ = M , we get X n ∩ B ( x , c (cid:48) )) ≥ δn M n β = n (cid:18) δεβ (cid:19) M n εβ ≥ M n +1 = M n (1+ n ) ≥ M n (1+ δβ − ) ≥ M n (1+ δ β ) since δβ ≥ , and this proves Lemma 11 with κ = δ . AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 15
Remark 15.
By inspecting the proof of Lemma 13 below, one can solve for explicit valuesof ε, c, β , and K . In particular, one can choose ε < , K < , c < , and β = , so that the supremum of permissible values of κ is at least − , and is by nomeans tight.In the next two subsections, we prove Lemma 13 and Lemma 14.3.3. Proof of Lemma 13.
Fix Q as in the statement of the lemma. For any I ∈ λ ( Q ) , (cid:96) ( γ n +1 | I ) − (cid:96) ( γ n | I ) ≥ (cid:96) ( γ n +1 | I ) − | γ n ( a I ) − γ n ( b I ) | = (cid:96) ( γ n +1 | I ) − | γ n +1 ( a I ) − γ n +1 ( b I ) | ≥ . Hence, to prove the lemma, it suffices to show that either Q ∈ ∆ Bad or there is an interval I ∈ λ ( Q ) for which (cid:96) ( γ n +1 | I ) − (cid:96) ( γ n | I ) ≥ εβ diam Q. Fix N so that Q ∈ ∆ N . Let ˜ Q ∈ ∆ N +1 be such that x Q ∈ ˜ Q ⊂ ˜ Q = Q and pick I ∈ λ ( Q ) such that γ n +1 ( I ) ∩ ˜ Q (cid:54) = ∅ . Note that γ n | I ⊆ Q is a segment withendpoints the same as γ n +1 | I , hence (cid:96) ( γ n | I ) = H ( γ n ( I )) = diam γ n ( I ) = | γ n ( a I ) − γ n ( b I ) | = | γ n +1 ( a I ) − γ n +1 ( b I ) | ≤ diam Q (3.12)Before proceeding, we’ll give a rough idea of how the proof will go. We will considera few cases, which are illustrated in Figure 4 below. γ n ( I ) γ n +1 ( I )˜ QQ zX ρ
Case 1 Case 2aCase 2b z Case 2b cont.F
IGURE
4. Illustrations of cases 1,2a, and 2b.
In the first case, we assume the diameter of γ n ( I ) is small with respect to Q ; since γ n +1 | I has the same endpoints as γ n | I and intersects the center cube ˜ Q , there must be alarge difference in length between γ n +1 ( I ) and γ n ( I ) since the former must enter Q , hit ˜ Q ,and then exit Q , and so (3.8) will hold. For the next two cases, we assume γ n ( I ) has largediameter. The second case (2a) assumes that γ n +1 ( I ) contributes more length than γ n ( I ) ,again implying (3.8) trivially. (It is possible to combine this case with (1), but we foundthis split to be somewhat convenient.) In the final case (2b) we assume the difference inlength between γ n +1 ( I ) and γ n ( I ) is small. Since β X ( B Q ) > β , we can show this impliesthe existence of z ∈ X far away from γ n +1 ( I ) (since γ n +1 | I has small geodesic deviation,so it can’t approximate all of X in B Q ). Since Γ n approximates X , we can find a largecurve ρ ⊆ Γ n entering B Q , approaching z , and then leaving B Q . The presence of both γ ( I ) and ρ inside Q implies that the total length of Γ n ∩ Q must be large, which means Q ∈ ∆ Bad .Now we proceed with the actual proof.
Case 1:
Suppose (cid:96) ( γ n ( I )) < diam Q . Since γ n +1 | I is a path entering Q , hitting ˜ Q , andthen leaving Q , we can estimate (cid:96) ( γ n +1 | I ) ≥ dist ( ˜ Q, Q c ) (2.6) ≥ dist ((1 + εβ ) B ˜ Q , B Q )= 2( cM − N − (1 + εβ ) cM − N − ) = 2 cM − N (1 − (1 + εβ ) M − ) ≥ diam B Q (cid:18) − εβ − ε β (cid:19) > (1 − εβ ) diam B Q (2.6) ≥ − εβ εβ diam Q = (cid:18) εβ εβ − εβ εβ (cid:19) diam Q ≥ (1 − εβ ) diam Q. (3.13)Thus, (cid:96) ( γ n +1 | I ) − (cid:96) ( γ n | I ) (3.13) ≥ (1 − εβ ) diam Q − diam Q ≥ diam Q if ε < , which implies the lemma in this case. Case 2:
Suppose(3.14) (cid:96) ( γ n | I ) ≥ diam Q We again split into two cases.
Case 2a:
Suppose (cid:96) ( γ n +1 | I ) ≥ (1 + εβ ) (cid:96) ( γ n | I ) . Then (cid:96) ( γ n +1 | I ) − (cid:96) ( γ n | I ) ≥ εβ(cid:96) ( γ n | I ) (3.14) ≥ εβ diam Q. Case 2b:
Now suppose(3.15) (cid:96) ( γ n +1 | I ) < (1 + εβ ) (cid:96) ( γ n ( I )) . Note that in this case, we have a better lower bound on (cid:96) ( γ n | I ) , namely,(3.16) (cid:96) ( γ n | I ) (3.15) ≥ (cid:96) ( γ n +1 | I )1 + εβ (3.13) ≥ − εβ εβ diam Q ≥ (1 − εβ ) diam Q. Let C ∈ (0 , (we will pick its value later). AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 17
Sublemma 16.
Assuming the conditions in case 2b, let I (cid:48) ⊆ I be the smallest intervalwith γ n +1 ( a I (cid:48) ) , γ n +1 ( b I (cid:48) ) ∈ (. (1 − Cβ ) B Q ) and γ n +1 ( I (cid:48) ) ∩ ˜ Q (cid:54) = ∅ . Then (3.17) (cid:96) ( γ n +1 | I (cid:48) ) − | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | ≤ εβ | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | Proof.
Since γ n +1 enters (1 − Cβ ) B Q , hits ˜ Q , and then leaves (1 + Cβ ) B Q , we have (cid:96) ( γ n +1 | I (cid:48) ) ≥ dist ( ˜ Q, (1 − Cβ ) B cQ ) (2.6) ≥ dist ((1 + εβ ) B ˜ Q , (1 − Cβ ) B cQ )= 2((1 − Cβ ) cM − N − (1 + εβ ) cM − N − )= 2 cM − N (1 − Cβ − (1 + εβ ) M − ) > diam B Q (1 − Cβ − M − )= (1 − Cβ − εβ diam B Q (2.6) ≥ − Cβ − εβ εβ diam Q = (cid:32) εβ εβ − Cβ + εβ εβ (cid:33) diam Q > (1 − Cβ − εβ ) diam Q (3.18)Hence, | γ n +1 ( a I ) − γ n +1 ( b I ) | − | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) |≤ | γ n +1 ( a I ) − γ n +1 ( a I (cid:48) ) | + | γ n +1 ( b I ) − γ n +1 ( b I (cid:48) ) |≤ (cid:96) ( γ n +1 | I \ I (cid:48) ) = (cid:96) ( γ n +1 | I ) − (cid:96) ( γ n +1 | I (cid:48) ) (3.15)(3.18) ≤ (1 + εβ ) (cid:96) ( γ n ( I )) − (1 − Cβ − εβ ) diam Q (3.13) ≤ (1 + εβ ) diam Q − (1 − Cβ − εβ ) diam Q = (3 εβ + Cβ ) diam Q (3.12)(3.14) ≤ εβ + Cβ ) | γ n +1 ( a I ) − γ n +1 ( b I ) | (3.19)Thus, | γ n +1 ( a I ) − γ n +1 ( b I ) | ≤ | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | − εβ + Cβ ) ≤ | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | (3.20)if we pick ε < and β < (recall C ∈ (0 , ). By Lemma 10, (cid:96) ( γ n +1 | I (cid:48) ) − | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | (2.8) ≤ (cid:96) ( γ n +1 | I ) − | γ n +1 ( a I ) − γ n +1 ( b I ) | (3.15) < εβ | γ n +1 ( a I ) − γ n +1 ( b I ) | (3.20) ≤ εβ | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | which proves (3.17). (cid:3) By the main assumption in Theorem 4, and because we’re assuming n = 0 so that M − n = 1 < r , β < β (cid:48) X ( x Q , (1 − Cβ ) cM − N ) ≤ (cid:96) ( γ n +1 | I (cid:48) ) − | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | + sup z ∈ (1 − Cβ ) B Q ∩ X dist ( z, γ n +1 ( I (cid:48) )) | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | (3.17) ≤ εβ | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | + sup z ∈ (1 − Cβ ) B Q ∩ X dist ( z, γ n +1 ( I (cid:48) )) | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | = 2 εβ + sup z ∈ (1 − Cβ ) B Q ∩ X dist ( z, γ n +1 ( I (cid:48) )) | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | so there is z ∈ X ∩ (1 − Cβ ) B Q withdist ( z, γ n +1 ( I (cid:48) )) ≥ ( β − εβ ) | γ n +1 ( a I (cid:48) ) − γ n +1 ( b I (cid:48) ) | (3.20) ≥ β − εβ | γ n +1 ( a I ) − γ n +1 ( b I ) | (3.14) ≥ β − εβ diam Q ≥ β diam Q (3.21)if ε < .Since γ n +1 ([0 , hits every cube in L ( Q ) , which all have diameter at most εβ ) cM − N − by (2.6) (recall N was chosen so that Q ∈ ∆ N ), Γ n ∩ Q ⊆ ( γ n +1 ([0 , εβ ) cM − N − ⊆ ( γ n +1 ([0 , cM − N − Note that since Q ∈ ˜ L n , we have N < n . Since X n ⊆ Γ n ∩ X and N < n , X ∩ (1 − Cβ ) B Q ⊆ X ∩ Q ⊆ (Γ n ∩ Q ) M − n ⊆ ( γ n +1 ([0 , cM − N − +2 M − n ⊆ ( γ n +1 ([0 , (4 cM − N − +2 M − N − ) = ( γ n +1 ([0 , (2+ c ) M − cM − N = ( γ n +1 ([0 , (2+ c ) M − diam B Q ⊆ ( γ n +1 ([0 , c M − diam B Q since c < . Since z ∈ X ∩ (1 − Cβ ) B Q , there is t ∈ [0 , such that(3.22) | γ n +1 ( t ) − z | < c M − diam B Q = εβ c diam Q and so(3.23) dist ( γ n +1 ( t ) , γ n +1 ( I (cid:48) )) ≥ dist ( z, γ n +1 ( I (cid:48) )) − | γ n +1 ( t ) − z | (3.21)(3.22) ≥ (cid:18) β − εβ c (cid:19) diam Q ≥ β diam Q for ε < c . Also, since z ∈ (1 − Cβ ) B Q , we know that B Q ⊇ B (cid:18) z, Cβ diam B Q (cid:19) (2.6) ⊇ B (cid:18) z, Cβ εβ ) diam Q (cid:19) ⊇ B (cid:18) z, Cβ diam Q (cid:19) (3.22) ⊇ B (cid:18) γ n +1 ( t ) , (cid:18) Cβ − εβ c (cid:19) diam Q (cid:19) ⊇ B (cid:18) γ n +1 ( t ) , Cβ diam Q (cid:19) (3.24)for ε < Cc . In particular, t ∈ γ − n +1 ( B Q ) . Note AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 19 dist ( γ n +1 ( t ) , γ n +1 ( I )) ≥ dist ( γ n +1 ( t ) , γ n +1 ( I (cid:48) )) − max { diam γ ([ a I , a (cid:48) I ]) , diam γ ([ b (cid:48) I , b I ]) }≥ dist ( γ n +1 ( t ) , γ n +1 ( I (cid:48) )) − (cid:96) ( γ | I/I (cid:48) ) (3.19)(3.23) ≥ β diam Q − (3 εβ + Cβ ) diam Q ≥ β diam Q for ε < and C < . Thus, since of course C < , we have B (cid:18) γ n +1 ( t ) , Cβ diam Q (cid:19) ⊆ Q \ ( γ n +1 ( I )) β diam Q In particular, γ n +1 ( t ) ∈ Q , and so by construction, t ∈ [ a, b ] for some [ a, b ] ∈ λ ( Q ) ,where γ n +1 ( a ) and γ n +1 ( b ) are both in Γ n . In particular, γ n +1 (( a, b )) is a line segmentin a cube R ∈ ˜ L ( Q ) . If ζ := γ n +1 ( a ) ∈ Γ n , then | ζ − γ n +1 ( t ) | ≤ diam R (2.6) ≤ (1 + εβ ) diam B R = 2(1 + εβ ) cM − N − ≤ (1 + εβ ) M − diam Q = (1 + εβ ) εβ diam Q ≤ εβ diam Q ≤ Cβ diam Q (3.25)for ε < C , and so(3.26) B (cid:18) ζ, Cβ diam Q (cid:19) ⊆ B (cid:18) γ n +1 ( t ) , Cβ diam Q (cid:19) ⊆ Q \ ( γ n +1 ( I )) β diam Q . Thus, since Γ n is connected and diam Γ n > diam Q > Cβ diam Q , we know there isa curve ρ ⊆ Γ n ∩ B ( ζ, Cβ diam Q ) connecting ζ to B ( ζ, Cβ diam Q ) c , and hence hasdiameter at least Cβ diam Q . Hence, H ∞ ( ρ ) ≥ diam ρ ≥ Cβ diam Q. Moreover, H ∞ ( γ ( I )) ≥ diam γ ( I ) ≥ | γ ( a I ) − γ ( b I ) | (3.12) = | γ n ( a I ) − γ n ( b I ) | (3.16) ≥ (1 − εβ ) diam Q. Hence, since any cube in L ( Q ) intersecting ρ has diameter at most εβ diam Q < β by(3.25), they are disjoint from those intersecting γ ( I ) by (3.26) if we choose ε < (sinceif they intersect γ ( I ) , they also intersect γ n +1 ( I ) by the definition of γ n +1 ). Thus, we have H ∞ ( Q ) ≥ Cβ diam Q + (1 − εβ ) diam Q ≥ (cid:18) Cβ (cid:19) diam Q for ε < C . Hence, by picking K = C , we see that Q ∈ ∆ Bad , which finishes the proofof Lemma 133.4.
Geometric martingales and the proof of Lemma 14.
For Q ∈ ∆ , define k ( Q ) tobe the number of cubes in ∆ Bad that properly contain Q , and set ∆ Bad,j = { Q ∈ ∆ Bad : k ( Q ) = j } ,Bad j ( Q ) = { R ⊆ Q : k ( R ) = k ( Q ) + j } ,G ( Q ) = (Γ n ∩ Q ) \ (cid:91) R ∈ Bad ( Q ) R. We will soon define, for each Q ∈ ∆ bad , a nonnegative weight function w Q : Γ n → [0 , ∞ ) H | Γ n -a.e. in a martingale fashion by defining it as a limit of a sequence w jQ .Each w jQ will be constant on various subsets of Γ n that partition Γ . We will actuallydecide the value of w jQ on an element A of the partition, say, by declaring the value of w jQ ( A ) := (cid:90) Γ n ∩ A w jQ d H . Then we will define w j +1 Q to be constant on sets in a partition subordinate to the previouspartition so that, on sets A in the j th partition, w j +1 Q ( A ) = w jQ ( A ) , and so forth. Wedo this in such a way that we disseminate the mass of the weight function w Q so that w Q is supported in Q , has integral diam Q , and so that w Q ( x ) ≤ Kβ ) k ( x ) − k ( Q ) , where k ( x ) is the total number of bad cubes containing x . By geometric series, this will meanthat (cid:80) Q ∈ ∆ Bad w Q Q is a bounded function, so that its total integral is at most a constanttimes H (Γ ) . However, the integral of each of these functions w Q is diam Q , and so theintegral is also equal to (cid:80) Q ∈ ∆ Bad diam Q , which gives us (3.10). This method appears in[16]. Now we proceed with the proof.First set(3.27) w Q ( Q ) = diam Q, w Q | Q c ≡ and construct w j +1 Q from w jQ as follows:(1) If R ∈ Bad j ( Q ) for some j , and S ∈ Bad ( R ) , set w j +1 Q to be constant in S sothat(3.28) w j +1 Q ( S ) = w jQ ( R ) diam S (cid:80) T ∈ Bad ( R ) diam T + H ( G ( R )) . (2) Set w j +1 Q to be constant in G ( R ) so that(3.29) w j +1 Q ( G ( R )) = w jQ ( R ) − (cid:88) S ∈ Bad ( R ) w j +1 Q ( S ) . (3) For points x not in in any R ∈ Bad j ( Q ) , set w j +1 Q ( x ) = w jQ ( x ) .Like a martingale, we have by our construction that, if R ∈ Bad j ( Q ) , then w iQ ( R ) = w jQ ( R ) for all i ≥ j , and in particular, w jQ ( Q ) = diam Q for all j ≥ .We will need the following inequality:(3.30) (cid:88) T ∈ Bad ( R ) diam T + H ( G ( R )) ≥ H ∞ ( R ∩ Γ n ) ≥ (1 + Kβ ) diam R. The first inequality comes from the fact that if δ > and A i is a cover of G ( R ) by sets sothat (cid:80) diam A i < H ( G ( R )) + δ , then { A i } ∪ Bad ( R ) is a cover of R (up to a set of H -measure zero by Lemma 9), and so (cid:88) T ∈ Bad ( R ) diam T + H ( G ( R )) + δ > (cid:88) diam A i + (cid:88) T ∈ Bad ( R ) diam T ≥ H ∞ ( R ∩ Γ n ) which gives the first inequality in (3.30) by taking δ → . The last inequality in (3.30) isfrom the definition of ∆ Bad . AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 21
For S ∈ Bad ( R ) and R ∈ Bad j ( Q ) , by induction we have w j +1 Q ( S ) diam S (3.28) = w jQ ( R ) (cid:80) T ∈ Bad ( R ) diam T + H ( G ( R )) (3.30) ≤ w jQ ( R ) diam R
11 + Kβ ≤ w Q ( Q ) diam Q Kβ ) j +1 (3.27) = 1(1 + Kβ ) j +1 (3.31)Hence, since w j +1 Q is constant in S , for x ∈ S ∩ Γ n , w j +1 Q ( x ) (3.28) = w jQ ( R ) diam S (cid:80) T ∈ Bad ( R ) diam T + H ( G ( R )) 1 H ( S ∩ Γ n ) (3.30) ≤ w jQ ( R ) (cid:80) T ∈ Bad ( R ) diam T + H ( G ( R )) 11 + Kβ (3.30) ≤ w jQ ( R ) diam R Kβ ) (3.31) ≤ w Q ( Q ) diam Q Kβ ) j +2 = 1(1 + Kβ ) j +2 . (3.32)Moreover, if x ∈ G ( R ) , w j +1 Q ( x ) = w j +1 Q ( G ( R )) H ( G ( R )) (3.29) = w jQ ( R ) − (cid:80) S ∈ Bad ( R ) w j +1 Q ( S ) H ( G ( R )) (3.28) = w jQ ( R ) H ( G ( R )) − (cid:88) S ∈ Bad ( R ) diam S (cid:80) T ∈ Bad ( R ) diam T + H ( G ( R )) = w jQ ( R ) H ( G ( R )) H ( G ( R )) (cid:80) T ∈ Bad ( R ) diam T + H ( G ( R ))= w jQ ( R ) (cid:80) T ∈ Bad ( R ) diam T + H ( G ( R )) (3.30) < w jQ ( R ) diam R
11 + Kβ (3.33) (3.31) ≤ Kβ ) j +1 (3.34)Since ∆ Bad ⊆ (cid:83) n j =0 ∆ j , and H ( (cid:83) Q ∈ ∆ Q. ) = 0 , almost every point x ∈ Q ∩ Γ n is contained in at most finitely many cubes in ∆ Bad , and hence the value of w j +1 Q ( x ) changes only finitely many times in j , thus the limit w Q = lim j w jQ is well defined almosteverywhere. For x ∈ Q ∩ Γ n , set k ( x ) = k ( R ) where R ⊆ Q is the smallest cube in ∆ Bad containing x . Then (3.32) and (3.34) imply w Q ( x ) ≤ Kβ ) k ( x ) − k ( Q ) and so (cid:88) x ∈ Q ∈ ∆ Bad w Q ( x ) ≤ k ( x ) (cid:88) j =0 Kβ ) j ≤ ∞ (cid:88) j =0 Kβ ) j = 1 + KβKβ ≤ Kβ since Kβ < . Hence, (cid:88) Q ∈ ∆ Bad diam Q = (cid:88) Q ∈ ∆ Bad (cid:90) Q w Q ( x ) d H ( x ) = (cid:90) Γ n (cid:88) x ∈ Q ∈ ∆ Bad w Q ( x ) d H ( x ) ≤ Kβ H (Γ n ) which finishes the proof of Lemma 14.4. A NTENNA - LIKE SETS
This section is devoted to the proof of Theorem 6.It is easy to verify using the definitions that being antenna-like is a quasisymmetricinvariant quantitatively, so by Theorem 4, it suffices to verify that, if X is c -antenna-like,then any ball B ( x, r ) with x ∈ X and < r < diam X has β (cid:48) ( x, r ) > c .Fix such a ball, so there is a homeomorphism h : (cid:83) i =1 [0 , e i ] → X ∩ B ( x, r ) so that(4.1) dist ( h ( e i ) , h ([0 , e j ] ∪ [0 , e k ])) ≥ cr for all permutations ( i, j, k ) of (1 , , (see Figure 5).Let s : [0 , → B ( x, r ) satisfy (cid:96) ( s | [0 , ) − | s (0) − s (1) | + sup z ∈ X ∩ B ( x,r ) dist ( z, s ([0 , < β (cid:48) ( x, r ) | s − s | =: β. Set x i = h ( e i ) for i = 1 , , and let t = inf s − (cid:32) (cid:91) i =1 B ( x i , β ) (cid:33) . This always exists since X ∩ B ( x, r ) ⊆ ( s ([0 , β . Without loss of generality, assume s ( t ) ∈ B ( x , β ) . Similarly, let(4.2) t = inf s − (cid:32) (cid:91) i =2 B ( x i , β ) (cid:33) and again, without loss of generality, assume s ( t ) ∈ B ( x , β ) .Note that h ([0 , e ] ∪ [0 , e ]) is a path connecting x to x , where the latter point is notcontained in ( s ([ t , t ])) β by our choices of t and t , although the latter point is; other-wise, there would be t ∈ [ t , t ] such that s ( t ) ∈ B ( x , β ) , contradicting the minimalityof t . Since h ([0 , e ] ∪ [0 , e ]) is connected and ( s ([ t , t ])) β contains x but not x , wecan pick a point z ∈ h ([0 , e ] ∪ [0 , e ]) so that dist ( z, s ([ t , t ])) = β . Pick ζ ∈ [ t , t ] and ζ ∈ ( t , so that(4.3) | s ( ζ ) − z | = dist ( z, s ([ t , t ])) = β and | s ( ζ ) − z | < β. AUSDORFF DIMENSION OF WIGGLY METRIC SPACES 23 s ( ζ ) zx s ( t ) s ( t ) s ( ζ ) x x h ( Y ) ⊆ X ∩ B ( x, r ) F IGURE β (cid:48) ( x, r ) | s − s | > (cid:96) ( s | [0 , ) − | s (0) − s (1) | ≥ (cid:96) ( s | [ ζ ,ζ ] ) − | s ( ζ ) − s ( ζ ) |≥ (cid:96) ( s | [ ζ ,t ] ) + (cid:96) ( s | [ t ,ζ ] ) − | s ( ζ ) − z | − | z − s ( ζ ) | (4.3) > | s ( ζ ) − s ( t ) | + | s ( t ) − s ( ζ ) | − β − β ≥ | z − x | − | s ( ζ ) − z | − | x − s ( t ) | + | x − z | − | s ( t ) − x | − | s ( ζ ) − z | − β (4.1) , (4.3) ≥ cr − β − β + cr − β − β − β = 2 cr − β ≥ c | s (0) − s (1) | − β ( x, r ) | s (0) − s (1) | which yields β (cid:48) ( x, r ) ≥ c and completes the proof of Theorem 65. C OMPARISON OF THE β - NUMBERS
For quantities A and B , we will write A (cid:46) B if there is a universal constant C so that A ≤ CB , and A ∼ B if A (cid:46) B (cid:46) A . Lemma 17.
Let X ⊆ (cid:96) ∞ be a compact connected set, x ∈ X , and < r < diam X . Then (5.1) β (cid:48) ( x, r ) ≤ ˆ β ( x, r ) (cid:46) β (cid:48) ( x, r ) . Proof.
The first inequality follows trivially from the definitions, since each sequence y , ..., y n ∈ X induces a finite polygonal Lipschitz path s in (cid:96) ∞ for which (cid:96) ( s ) − | s (0) − s (1) | = n − (cid:88) i =0 | y i − y i +1 | − | y − y n | . For the opposite inequality, let s : [0 , → (cid:96) ∞ be such that(5.2) (cid:96) ( s ) − | s (0) − s (1) | + sup z ∈ B ( x,r ) ∩ X dist ( z, s ([0 , | s (0) − s (1) | ≤ β (cid:48) ( x, r ) =: β. Let A = s − (cid:0) ( s ([0 , β | s − s | (cid:1) which is a relatively open subset of [0 , . Let a = inf A and define a = t < t < · · ·
Moreover,diam ∩ B ( x, r ) ⊆ ( s ([0 , β | s (0) − s (1) | ⊆ (cid:91) i B ( s ( t i ) , (2 β + β ) | s (0) − s (1) | ) ⊆ (cid:91) i B ( y i , (2 β + β + 2 β ) | s (0) − s (1) | ) ⊆ (cid:91) i B ( y i , β | s (0) − s (1) | ) ⊆ (cid:91) i B ( y i , β | y − y n | ) (5.5)Thus (5.4) and (5.5) imply ˆ β ( x, r ) ≤ β = 20 √ β (cid:48) ( x, r ) . (cid:3) Proposition 18. If X is a compact connected subset of some Hilbert space, then β (cid:48)(cid:48) ( x, r ) ≤ β ( x, r ) (cid:46) β (cid:48)(cid:48) ( x, r ) for x ∈ Γ and r < diam X where β (cid:48)(cid:48) ( x, r ) = inf s (cid:32)(cid:18) (cid:96) ( s ) − | s (0) − s (1) || s (0) − s (1) | (cid:19) + sup z ∈ B ( x,r ) ∩ X ] dist ( z, s ([0 , | s (0) − s (1) | (cid:33) . In particular, (5.6) β (cid:48) ( x, r ) ≤ β ( x, r ) (cid:46) β (cid:48) ( x, r ) . Note that (5.6) is tight in the sense that if X ⊆ C , ∈ X , and B (0 , ∩ Γ = [ − , ∪ [0 , iε ] , then by Theorem 6 and (5.6), for all ε > , β (0 , ≤ ε ≤ β (cid:48) (0 , ≤ β (0 , ≤ ε. However, if X ∩ B ( x, r ) = [ − , ∪ [0 , e iε ] , then for all ε > , again by (5.6) (andestimating β (cid:48)(cid:48) (0 , by letting s be the path traversing the segments [ − , ∪ [0 , e iε ] ), β (0 , ∼ ε (cid:38) β (cid:48) (0 , (cid:38) β (0 , . Proof.
For the first inequality, simply let s : [0 , → H be the line segment spanning L ∩ B ( x, r ) where L is some line passing through B ( x, r ) . Then (cid:96) ( s ) = H ( L ∩ B ( x, r )) ≥ r and hence β (cid:48)(cid:48) ( x, r ) ≤ sup z ∈ B ( x,r ) ∩ X dist ( z, s ([0 , | s (0) − s (1) | ≤ sup z ∈ B ( x,r ) ∩ X dist ( z, L ) r . Since x ∈ X , the range of admissible lines in the infimum in (1.1) can be taken to be linesintersecting B ( x, r ) . Using this fact and infimizing the above inequality over all such linesproves the first inequality in (5.6).For the opposite inequality, let s satisfy (cid:18) (cid:96) ( s ) − | s (0) − s (1) || s (0) − s (1) | (cid:19) + sup z ∈ B ( x,r ) ∩ X dist ( z, s ([0 , | s (0) − s (1) | ≤ β (cid:48)(cid:48) ( B ( x, r )) =: β. Let β ( s ) := sup t ∈ [0 , dist ( s ( t ) , [ s (0) , s (1)]) . Then by the Pythagorean theorem, there is c > so that (1 + cβ ( s ) ) | s (0) − s (1) | ≤ (cid:96) ( s ) ≤ (1 + β ) | s (0) − s (1) | so that β ( s ) ≤ c − β, . Hence, if L is the line passing through s (0) and s (1) , β ( x, r ) ≤ sup z ∈ B ( x,r ) ∩ X dist ( z, L ) ≤ sup z ∈ B ( x,r ) ∩ X dist ( z, [ s (0) , s (1)]) ≤ β ( s ) + sup z ∈ B ( x,r ) ∩ X dist ( z, s ([0 , ≤ c − β + β (cid:46) β (cid:3) R EFERENCES1. C.J. Bishop and P.W. Jones,
Wiggly sets and limit sets , Ark. Mat. (1997), no. 2, 201–224. MR 1478778(99f:30066)2. C.J. Bishop and J.T. Tyson, Conformal dimension of the antenna set , Proc. Amer. Math. Soc. (2001),no. 12, 3631–3636 (electronic). MR 1860497 (2002j:30021)3. M. Christ, A T ( b ) theorem with remarks on analytic capacity and the Cauchy integral , Colloq. Math. (1990), no. 2, 601–628. MR 1096400 (92k:42020)4. G. David, Morceaux de graphes lipschitziens et integrales singuli`eres sur une surface. , Revista matem´aticaiberoamericana (1988), no. 1, 73.5. , Hausdorff dimension of uniformly non flat sets with topology , Publ. Mat. (2004), no. 1, 187–225.MR 2044644 (2005b:49073)6. I. Hahlomaa, Menger curvature and Lipschitz parametrizations in metric spaces , Fund. Math. (2005),no. 2, 143–169. MR 2163108 (2006i:30053)7. J. Heinonen,
Lectures on analysis on metric spaces , Universitext, Springer-Verlag, New York, 2001.MR 1800917 (2002c:30028)8. T. Hyt¨onen and A. Kairema,
Systems of dyadic cubes in a doubling metric space , Colloq. Math. (2012),no. 1, 1–33. MR 29011999. P. W. Jones,
Rectifiable sets and the traveling salesman problem , Invent. Math. (1990), no. 1, 1–15.MR 1069238 (91i:26016)10. J. M Mackay,
Spaces and groups with conformal dimension greater than one , Duke Mathematical Journal (2010), no. 2, 211–227.11. J. M. Mackay and J. T. Tyson,
Conformal dimension , University Lecture Series, vol. 54, American Mathe-matical Society, Providence, RI, 2010, Theory and application. MR 2662522 (2011d:30128)12. P. Mattila,
Geometry of sets and measures in Euclidean spaces , Cambridge Studies in Advanced Mathe-matics, vol. 44, Cambridge University Press, Cambridge, 1995, Fractals and rectifiability. MR 1333890(96h:28006)13. K. Okikiolu,
Characterization of subsets of rectifiable curves in R n , J. London Math. Soc. (2) (1992),no. 2, 336–348. MR 1182488 (93m:28008)14. R. Schul, Ahlfors-regular curves in metric spaces , Ann. Acad. Sci. Fenn. Math. (2007), no. 2, 437–460.MR 2337487 (2008j:28005)15. , Analyst’s traveling salesman theorems. A survey , In the tradition of Ahlfors-Bers. IV, Contemp.Math., vol. 432, Amer. Math. Soc., Providence, RI, 2007, pp. 209–220. MR 2342818 (2009b:49099)16. ,
Subsets of rectifiable curves in Hilbert space—the analyst’s TSP , J. Anal. Math. (2007), 331–375. MR 2373273 (2008m:49205)J. A
ZZAM , U
NIVERSITY OF W ASHINGTON D EPARTMENT OF M ATHEMATICS , C–138 P
ADELFORD H ALL ,S EATTLE , WA, 98195-4350
E-mail address ::