Rectifiable-Reifenberg and the Regularity of Stationary ... · arXiv:1504.02043v5 [math.DG] 14 Apr 2016 RECTIFIABLE-REIFENBERG AND THE REGULARITY OF STATIONARY AND MINIMIZING HARMONIC

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016 RECTIFIABLE-REIFENBERG AND THE REGULARITY OF STATIONARY A ND

MINIMIZING HARMONIC MAPS

AARON NABER AND DANIELE VALTORTA

ABSTRACT. In this paper we study the regularity of stationary and minimizing harmonic mapsf : B2(p) ⊆

M → N between Riemannian manifolds. IfSk( f ) ≡ x ∈ M : no tangent map atx is k + 1-symmetric is the

kth-stratum of the singular set off , then it is well known that dimSk ≤ k, however little else about the structure

of Sk( f ) is understood in any generality. Our first result is for a general stationary harmonic map, where we

prove thatSk( f ) is k-rectifiable. In fact, we prove fork-a.e. pointx ∈ Sk( f ) that there exists a uniquek-plane

Vk ⊆ TxM such thateverytangent map atx is k-symmetric with respect toV.

In the case of minimizing harmonic maps we go further, and prove that the singular setS( f ), which is well

known to satisfy dimS( f ) ≤ n− 3, is in factn− 3-rectifiable with uniformlyfinite n− 3-measure. An effective

version of this allows us to prove that|∇ f | has estimates inL3weak, an estimate which is sharp as|∇ f | may not

live in L3. More generally, we show that the regularity scaler f also hasL3weak estimates.

The above results are in fact just applications of a new classof estimates we prove on thequantitative

stratificationsSkǫ,r ( f ) andSk

ǫ ( f ) ≡ Skǫ,0( f ). Roughly,Sk

ǫ ⊆ M is the collection of pointsx ∈ M for which no ball

Br(x) is ǫ-close to beingk+ 1-symmetric. We show thatSkǫ is k-rectifiable and satisfies the Minkowski estimate

Vol(Br Skǫ ) ≤ Crn−k.

The proofs require a newL2-subspace approximation theorem for stationary harmonic maps, as well as new

W1,p-Reifenberg and rectifiable-Reifenberg type theorems. These results are generalizations of the classical

Reifenberg, and give checkable criteria to determine when aset isk-rectifiable with uniform measure estimates.

The new Reifenberg type theorems may be of some independent interest. TheL2-subspace approximation

theorem we prove is then used to help break down the quantitative stratifications into pieces which satisfy these

criteria.

CONTENTS

1. Introduction 31.1. Results for Stationary Harmonic Maps 61.2. Results for Minimizing Harmonic Maps 81.3. Results under Additional Hypothesis 91.4. Outline of Proofs and Techniques 102. Preliminaries 132.1. Stationary Harmonic Maps and Monotonicity 132.2. Quantitative 0-Symmetry and Cone Splitting 142.3. Defect Measure 152.4. ǫ-regularity for Minimizers 162.5. Hausdorff, Minkowski, and packing Content 17

Date: April 15, 2016.

The first author has been supported by NSF grant DMS-1406259,the second author has been supported by SNSF grant 149539.1

http://arxiv.org/abs/1504.02043v5

2 AARON NABER AND DANIELE VALTORTA

2.6. The Classical Reifenberg Theorem 192.7. W1,p maps and rectifiability 203. TheW1,p-Reifenberg and rectifiable-Reifenberg Theorems 203.1. Statement of MainW1,p-Reifenberg and rectifiable-Reifenberg Results 214. Technical Constructions toward New Reifenberg Results 244.1. Hausdorff distance and subspaces 244.2. Distance betweenL2 best planes 264.3. Comparison betweenL2 andL∞ planes 304.4. bi-Lipschitz equivalences 304.5. Pointwise Estimates onD 355. Proof of Theorem3.3: The Rectifiable-Reifenberg 355.1. Weak Upper boundλk(S ∩ Br(x)) ≤ C(n)rk 355.2. First induction: upwards 375.3. Second induction: downwards: outline 375.4. Second induction: downwards: detailed proof 395.5. Map and manifold structure. 425.6. Volume estimates on the manifold part 445.7. Estimates on the excess set 465.8. Completion of the weak upper bounds 465.9. Rectifiability,W1,p and improved measure bounds 476. Proof of Theorems3.2and3.4 536.1. Proof of Theorem3.2: TheW1,p-Reifenberg 536.2. Proof of Theorem3.4: The Discrete Rectifiable-Reifenberg 577. L2-Best Approximation Theorems 577.1. Symmetry and Gradient Bounds 587.2. BestL2-Subspace Equations 597.3. Proof of Theorem7.1 628. The Inductive Covering Lemma 638.1. Technical Constructions for DecomposingU′r ∪U′

+65

8.2. EstimatingU> in Lemma8.1 698.3. EstimatingU0 in Lemma8.1 719. Proof of Main Theorems for Stationary Harmonic Maps 739.1. Proof of Theorem1.3 739.2. Proof of Theorem1.4 759.3. Proof of Theorem1.5 7810. Proof of Main Theorems for Minimizing Harmonic Maps 7810.1. Proof of Theorem1.6 7810.2. Proof of Theorem1.8 7911. Sharpness of the results 7911.1. Sharpness ofLp Estimates for Minimizers 7911.2. Rectifiable-Reifenberg Example I 79

RECTIFIABLE-REIFENBERG AND THE REGULARITY OF STATIONARY AND MINIMIZING HARMONIC MAPS 3

11.3. Rectifiable-Reifenberg Example II 80Acknowledgments 80References 80

1. INTRODUCTION

Let f : B2(p) ⊆ M → N be a stationary harmonic map between two Riemannian manifolds with Ncompact without boundary. We recall that the singular set off is defined as the complement of the regularset, i.e.,

S(f ) =

x ∈ M s.t. ∃r > 0 s.t. u|Br (x) is continuousC=

x ∈ M s.t. ∃r > 0 s.t. u|Br (x) is smoothC.

(1.1)

One can usef to give a stratification ofM by the singular sets

S0( f ) ⊆ · · · ⊆ Sk( f ) ⊆ · · ·S(f ) ⊆ M , (1.2)

defined by

Sk( f ) ≡ x ∈ M : no tangent map atx is k+ 1-symmetric , (1.3)

see Definition1.2 for a precise definition and more detailed discussion. A key result of [SU82] is that wehave the Hausdorff dimension estimate

dimSk ≤ k , (1.4)

However, little else is known about the structure of the singular setsSk( f ). In the stationary case, nothingis known in general. In the case of a minimizing harmonic map,[SU82] proves thatS( f ) = Sn−3( f ), and,under the additional assumption thatN is analytic, it follows from the wonderful work of [Sim96] that thetop stratumSn−3 is rectifiable. In low dimensions and when the targets are spheres, these results have beenfurther refined in [LW06] and [HL90].

The goal of this paper is to study additional properties of the stratification and the associated quantitativestratification for stationary and minimizing harmonic maps. That is, a first goal is to see for a stationaryharmonic map with a general compact target space thatSk( f ) is k-rectifiablefor all k. In fact, the proofallows us to see the following stronger result. Fork-a.e.x ∈ Sk there will exist a uniquek-planeVk ⊆ TxMsuch thateverytangent map atx will be k-symmetric with respect toV. Let us observe a subtlety to thisstatement. We are not claiming that the tangent maps at suchx are unique. We are simply claiming thatfor k-a.e. point there is a maximal dimensionalk-plane of symmetry which exists, and that this plane ofsymmetry is unique independent of which tangent map is chosen. Theorem1.5gives the precise results forthe classical stratification of a stationary harmonic map.


For minimizing harmonic maps we can improve on this result intwo ways. First, we can show that thetop stratum is not onlyn− 3 rectifiable, but has ana priori bound on itsn− 3 measure. That is, if

?B2(p)|∇ f |2 ≤ Λ , (1.5)

is theL2 bound on the energy, then we have then− 3-Hausdorff measure estimate

λn−3(Sn−3( f ) ∩ B1) ≤ C , (1.6)

whereC depends onM, N andΛ (see Sections1.1and following for precise statements).Indeed, we show a much stronger Minkowski version of this estimate, see Theorem1.6 for the complete

statement. In fact, we can prove what turns out to be a much more effective analytic version of the above.Namely, in Theorem1.8we show that|∇ f |, and in fact the regularity scaler f , havea priori weakL3 bounds.That is,

Vol(

|∇ f | > r−1

∩ B1

)

≤ Vol(

Br

|∇ f | > r−1

∩ B1

)

≤ Cr3 , (1.7)

whereBr (E) stands for the tubular neighborhood of radiusr around the setE, i.e.,

Br (E) = x s.t. d(x,E) < r =⋃

x∈E

Br (x) . (1.8)

Notice that this is indeed sharp, in that there are counterexamples showing that|∇ f | < L3, see Example11.1. Let us also point out that this sharpens results from [CN13b], where it was proven for minimizingharmonic maps that|∇ f | ∈ Lp for all p < 3. We refer the reader to Section1.2 for the precise and mostgeneral statements.

Now the techniques of this paper all center around estimating thequantitativestratification, not the stan-dard stratification itself. In fact, it is for the quantitative stratification that the most important results of thepaper hold, everything else can be seen to be corollaries of these statements.

A first version of the quantitative stratification can be found in [Alm00, section 2.25]. This conceptwas later developed in [CN13a] with the goal of proving new estimates on noncollapsed manifolds withRicci curvature bounded below and in particular Einstein manifolds, and then extended in [CN13b] to giveeffective andLp estimates on minimal and stationary harmonic maps and minimal submanifolds. It hassince been used in [CHN13b], [CHN13a], [CNV15], [FMS15], [BL15] to prove similar results in the areasof mean curvature flow, critical sets of elliptic equations,harmonic map flow, and biharmonic maps.

Before describing the results in this paper on the quantitative stratification, let us give more precisedefinitions of everything. To begin with, to describe the stratification and quantitative stratification we needto discuss the notion of symmetry associated to a solution. Specifically:

Definition 1.1. We define the following:

(1) A mapping f : Rn → N is calledk-symmetric if f (λx) = f (x) ∀λ > 0 and if there exists ak-planeVk ⊆ Rn such that for eachx ∈ Rn andy ∈ Vk we have thatf (x+y) = f (x). Note that a 0-symmetricfunction is simply a homogeneous function of degree 0.


(2) Given a mappingf : M → N andǫ > 0, we say a ballBr(x) ⊆ M is (k, ǫ)-symmetric ifr <inj(x)and there exists ak-symmetric mappingf : TxM → N such that

>Br (x) | f − f |2dvg < ǫ, where we

have used the exponential map to identifyf as a function onM.

Thus, a function isk-symmetric if it only depends onn − k variables and is radially invariant. Ak-symmetric function can therefore be naturally identified with a function on an− k− 1 sphere. A function is(k, ǫ)-symmetric on a ballBr(x) if it is L2 close to ak-symmetric function on this ball.

With the notion of symmetry in hand we can define precisely thequantitative stratification associated to asolution. The idea is to group points together based on the amount of symmetry that balls centered at thosepoints contain. In fact, there are several variants which will play a role for us, all of which are important forthe applications to the standard singular set. Let us introduce them and briefly discuss them:

Definition 1.2. For a stationary harmonic mapf : B2(p)→ N we make the following definitions:

(1) Forǫ, r > 0 we define thekth (ǫ, r)-stratificationSkǫ,r( f ) by

Skǫ,r( f ) ≡ x ∈ B1(p) : for no r ≤ s< 1 is Bs(x) a (k + 1, ǫ)-symmetric ball. (1.9)

(2) Forǫ > 0 we define thekth ǫ-stratificationSkǫ ( f ) by

Skǫ ( f ) =

⋂

r>0

Skǫ,r( f ) ≡ x ∈ B1(p) : for no 0< r < 1 is Br(x) a (k+ 1, ǫ)-symmetric ball. (1.10)

(3) We define thekth-stratificationSk( f ) by

Sk( f ) =⋃

ǫ>0

Skǫ = x ∈ B1(p) : no tangent cone atx is k+ 1-symmetric. (1.11)

Remark1.1. It is a small but important exercise to check that the standard stratificationSk( f ) as defined in(1.2) agrees with the set

⋃

ǫ>0 Skǫ . We do this carefully in Section9.3.

Let us discuss in words the meaning of the quantitative stratification, and how it relates to the standardstratification. As discussed at the beginning of the section, the stratificationSk( f ) of M is built by separatingpoints ofM based on the infinitesimal symmetries off at those points. The quantitative stratificationsSk

ǫ ( f )andSk

ǫ,r( f ) are, on the other hand, instead built by separating points of M based on how many symmetriesexist on balls of definite size around the points. In practice, the quantitative stratification has two advan-tages to the standard stratification. First, for applications to minimizing harmonic maps the quantitativestratification allows one to prove effective estimates. In particular, in [CN13b] the weakerLp estimates?

B1(p)|∇ f |3−δ,

?B1(p)|∇2 f |

3−δ2 < Cδ ∀ δ > 0 , (1.12)

on solutions were obtained by exploiting this fact. The second advantage is that the estimates on the quan-titative stratification are much stronger than those on the standard stratification. Namely, in [CN13b] theHausdorff dimension estimate (1.4) on Sk( f ) was improved to the Minkowski content estimate

Vol(Br Skǫ,r) ≤ Cδr

n−k−δ ∀δ > 0 . (1.13)


One of the key technical estimates of this paper is that in Theorem1.3we drop theδ from the above estimateand obtain an estimate of the form

Vol(Br Skǫ,r) ≤ Crn−k . (1.14)

From this we are able conclude in Theorem1.4an estimate onSkǫ of the form

Vol(Br Skǫ ) ≤ Crn−k . (1.15)

In particular, this estimate allows us to conclude thatSkǫ has uniformly finitek-dimensional measure. In fact,

the techniques will prove much more for us. They will show us that Skǫ is k-rectifiable, and that fork-a.e.

point x ∈ Skǫ there is a uniquek-planeVk ⊆ TxM such thateverytangent map atx is k-symmetric with

respect toV. By observing thatSk( f ) =⋃

Skǫ ( f ), this allows us to prove in Theorem1.5 our main results

on the classical stratification. This decomposition ofSk into the piecesSkǫ is crucial for the proof.

On the other hand, (1.14), combined with theǫ-regularity theorems of [SU82],[CN13b], allow us to con-clude in the minimizing case both the weakL3 estimate on|∇ f |, and then− 3-finiteness of the singular setof f . Thus we will see that Theorems1.6and1.8are fairly quick consequences of (1.15).

Thus we have seen that (1.14) and (1.15), and more generally Theorem1.3 and Theorem1.4, are themain challenges of the paper. We will give a more complete outline of the proof in Section1.4, how-ever let us mention for the moment that two of the new ingredients to the proof are a newL2-subspaceapproximation theorem for stationary harmonic maps, proved in Section7, and newW1,p-Reifenberg andrectifiable-Reifenberg theorems, proved in Section3. TheL2-approximation result roughly tells us that theL2-distance of a measure from being contained in ak-dimensional subspace may be estimated by integratingthe energy drop of a stationary harmonic map over the measure. To exploit the estimate we prove a newW1,p-Reifenberg type theorem. The classical Reifenberg theorem states that if we have a setS which isL∞-approximated by a subspace at every point and scale, thenS is bi-Holder to a manifold, see theorem2.18 for a precise statement. It is important for us to improve on this bi-Holder estimate, at least enoughthat we are able to control thekth-dimensional measure of the set and prove rectifiability. Inparticular, wewant to improve theCα-maps toW1,p-maps forp > k, and we will want to do it using a condition which isintegral in nature. More precisely, we will only require a form of summableL2-closeness of the subsetSto the approximating subspaces. We will see in Theorem7.1 that by using theL2-subspace approximationtheorem that the conditions of this new rectifiable-Reifenberg are in fact controllable for the quantitativestratificationsSk

ǫ , as least after we break it up into appropriate pieces.

1.1. Results for Stationary Harmonic Maps. We now turn our attention to giving precise statements ofthe main results of this paper. In this subsection we focus onthose concerning the singular structure ofstationary harmonic maps. That is, we will be considering stationary harmonic mapsf : B2(p) ⊆ M → N.In order to write the constants involved in some explicit form, let us chooseKM,KN, n > 0 to be the smallest


number such that

| secB2(p) | ≤ KM , inj(B2(p)) ≥ K−1M ,

| secN | ≤ KN, inj(N) ≥ K−1N , diam(N) ≤ KN ,

dim(M), dim(N) ≤ n . (1.16)

Now let us begin by discussing our main theorem for the quantitative stratificationsSkǫ,r( f ):

Theorem 1.3 ((ǫ, r)-Stratification of Stationary Harmonic Maps). Let f : B2(p) ⊆ M → N be a sta-tionary harmonic mapping satisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Then for eachǫ > 0 there exists

Cǫ(n,KM ,KN,Λ, ǫ) such that

Vol(

Br

(

Skǫ,r( f )

))

≤ Cǫrn−k . (1.17)

When we study the stratumSkǫ ( f ), we can refine the above to prove structure theorems on the set itself.

For the definition ofk-rectifiability, we refer the reader to the standard reference [Fed69].

Theorem 1.4(ǫ-Stratification of Stationary Harmonic Maps). Let f : B2(p) ⊆ M → N be a stationary har-monic mapping satisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Then for eachǫ > 0 there exists Cǫ(n,KM ,KN,Λ, ǫ)

such that

Vol(

Br(

Skǫ ( f )

)

)

≤ Cǫrn−k . (1.18)

In particular, we have the k-dimensional Hausdorff measureestimateλk(Skǫ ( f )) ≤ Cǫ . Further, Sk

ǫ ( f ) isk-rectifiable, and for k-a.e. x∈ Sk

ǫ there exists a unique k-plane Vk ⊆ TxM such that every tangent cone ofx is k-symmetric with respect to Vk.

Remark1.2. In fact, the techniques will prove an estimate even strongerthan the Minkowski estimate of(1.17) and (1.18). That is, one can prove a uniformk-dimensional packing content estimate. More precisely,let Br i(xi) be any collection of disjoint balls such thatxi ∈ Sk

ǫ , then we have the content estimate∑

rki ≤ Cǫ.

Finally, we end this subsection by stating our main results when it comes to the classical stratificationSk( f ). The following may be proved from the previous Theorem in only a few lines given the formulaSk( f ) =

⋃

Skǫ ( f ):

Theorem 1.5 (Stratification of Stationary Harmonic Maps). Let f : B2(p) ⊆ M → N be a stationaryharmonic mapping satisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Then for each k we have that Sk( f ) is k-rectifiable.

Further, for k-a.e. x∈ Sk( f ) there exists a unique k-plane Vk ⊆ TxM such that every tangent cone of x isk-symmetric with respect to Vk.


1.2. Results for Minimizing Harmonic Maps. In this section we record our main results for minimizingharmonic maps. Most of the results of this section follow quickly by combining the quantitative stratificationresults of Section1.1with theǫ-regularity of [SU82, CN13b], see Section2.4 for a review of these points.

Our first estimate is on the singular set S(f ) of a minimizing harmonic map. Recall that S(f ) is the set ofpoints wheref is not smooth. Our first estimate on the singular structure ofa minimizing harmonic map isthe following:

Theorem 1.6(Structure of Singular Set). Let f : B2(p) ⊆ M → N be a minimizing harmonic mappingsatisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Then S( f ) is n− 3-rectifiable and there exists C(n,KM,KN,Λ) such

that

Vol(

Br(

S( f ))

∩ B1(p))

≤ Cr3 . (1.19)

In particular, λn−3(S( f )) ≤ C.

Remark1.3. As in Theorem1.4, the same techniques prove a packing content estimate on S(f ). That is,if Br i (xi) is any collection of disjoint balls such thatxi ∈ S(f ) ∩ B1 , then we have the content estimate∑

rki ≤ C.

The above can be extended to effective Schauder estimates onf . To state the results in full strength let usrecall the notion of the regularity scale associated to a function. Namely:

Definition 1.7. Let f : B2(p) → N be a mapping between Riemannian manifolds. Forx ∈ B1(p) we definethe regularity scaler f (x) by

r f (x) ≡ max0 ≤ r ≤ 1 : supBr (x)|∇ f | ≤ r−1 . (1.20)

By definition,r f (x) ≡ 0 if f is not Lipschitz in a neighborhood ofx.

Remark1.4. The regularity scale has nice scaling property. Indeed, if we defineTX,ρ : TxM → N by

Tx,ρ(y) = f (expx(ρy)) , (1.21)

thenr f (x) = ρrTx,ρ (0). In other words, ifr ≡ r f (x) and we rescaleBr(x) → B1(0), then on the rescaled ballwe will have that|∇Tx,r | ≤ 1 onB1(0).

Remark1.5. We have the easy estimate|∇ f |(x) ≤ r f (x)−1. However, a lower bound onr f (x) is in principlemuchstronger than an upper bound on|∇ f |(x).

Remark1.6. Notice that the regularity scale is a Lipschitz function with |∇r f (x)| ≤ 1.

Remark1.7. If f satisfies an elliptic equation, e.g. is weakly harmonic, then we have the estimate

supBr f /2(x)

|∇k f | ≤ Ck r f (x)−k . (1.22)

In particular, control onr f gives control on all higher order derivatives.


Now let us state our main estimates for minimizing harmonic maps:

Theorem 1.8 (Estimates on Minimizing Harmonic Maps). Let f : B2(p) ⊆ M → N be a minimizingharmonic mapping satisfying(1.16) with

>B2(p)|∇ f |2 ≤ Λ. Then there exists C(n,KM ,KN,Λ) such that

Vol(

x ∈ B1(p) : |∇ f | > r−1)

≤ Vol(

x ∈ B1(p) : r f (x) < r)

≤ Cr3 . (1.23)

In particular, both|∇ f | and r−1f have uniform bounds in L3weak

(

B1(p))

, the space of weakly L3 functions. Infact, we also have the Hessian estimate

Vol(

x ∈ B1(p) : |∇2 f | > r−2)

≤ Cr3 , (1.24)

which in particular gives us uniform bounds for|∇2 f | in L3/2weak(B1(p)), the space of weakly L3/2 functions.

Remark1.8. The L3weak estimates are sharp in that there exist examples for which|∇ f | does not live inL3,

see Section11.1.

1.3. Results under Additional Hypothesis. The previous two subsections have focused on results for com-pletely general stationary or minimizing harmonic maps. Inthis subsection we would like to see how theseresults may be improved under further assumptions. Specifically, for stationary harmonic mapsf : M → Nwe would like to see that the regularity results may be improved to match those of minimizing harmonicmaps if we assume there are no smooth harmonic maps fromS2 into N. The idea behind this follows that of[Lin99],[CHN13a]. Additionally, though the regularity results of Sections1.1and1.2are sharp in completegenerality, they may be improved if we assume thatN has no other stationary or minimizing harmonic mapsfrom Sk into N. Precisely, the main result of this subsection for stationary harmonic maps is the following:

Theorem 1.9(Improved Estimates for Stationary Maps). Let f : B2(p) ⊆ M → N be a stationary harmonicmapping satisfying(1.16) with

>B2(p)|∇ f |2 ≤ Λ. Assume further that for some k≥ 2 there exists no smooth

nonconstant stationary harmonic maps Sℓ → N for all ℓ ≤ k. Then there exists C(n,KM ,N,Λ) such that

Vol(

x ∈ B1(p) : |∇ f | > r−1)

≤ Vol(

x ∈ B1(p) : r f (x) < r)

≤ Cr2+k . (1.25)

In particular, both|∇ f | and r−1f have uniform bounds in L2+k

weak

(

B1(p))

, the space of weakly L2+k functions.

Remark1.9. The proof of Theorem1.9follows verbatim the proof of Theorem1.8, except one replaces theǫ-regularity of theorem2.8with theǫ-regularity of theorem2.11.

In the context wheref is minimizing we have a similar improvement, though in this case we only needto assume there exists no minimizing harmonic maps fromSk→ N:


Theorem 1.10(Improved Estimates for Minimizing Maps). Let f : B2(p) ⊆ M → N be a minimizingharmonic mapping satisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Assume that for some k≥ 2 there exists no

smooth nonconstant minimizing harmonic maps Sℓ → N for all ℓ ≤ k. Then there exists C(n,KM ,N,Λ)such that

Vol(

x ∈ B1(p) : |∇ f | > r−1)

≤ Vol(

x ∈ B1(p) : r f (x) < r)

≤ Cr2+k . (1.26)

In particular, both|∇ f | and r−1f have uniform bounds in L2+k

weak

(

B1(p))

, the space of weakly L2+k functions.

Remark1.10. The proof of Theorem1.10follows verbatim the proof of Theorem1.8, except one replacestheǫ-regularity of theorem2.8with theǫ-regularity of theorem2.10.

1.4. Outline of Proofs and Techniques.In this subsection we give a brief outline of the proof of the maintheorems. To describe the new ingredients involved it will be helpful to give a comparison to the proofs ofprevious results in the area, in particular the dimension estimate (1.4) of [SU82] (which are similar in spiritto the estimates for minimal surfaces in [Alm68] and [Alm00, theorem 2.26]) and the Minkowski andLp

estimates (1.12) of [CN13b].Indeed, the starting point for the study of singular sets forsolutions of geometric equations typically

looks the same, that is, one needs a monotone quantity. In thecase of harmonic mapsf : M → N betweenRiemannian manifolds we consider the normalized Dirichletenergy

θr(x) ≡ r2−n∫

Br (x)|∇ f |2 . (1.27)

For simplicity sake let us takeM ≡ Rn in this discussion, which is really of no loss except for somesmalltechnical work (it is the nonlinearity ofN that is the difficulty). Thend

drθr ≥ 0, andθr (x) is independent ofr if and only if f is 0-symmetric, see Section2.1for more on this. Interestingly, this is the only informationone requires to prove the dimension estimate (1.4). Namely, sinceθr (x) is monotone and bounded, it mustconverge asr tends to zero. In particular, if we consider the sequence of scalesrα = 2−α then we have foreachx that

limα→∞

∣

∣

∣

∣

θrα(x) − θrα+1(x)∣

∣

∣

∣

→ 0 . (1.28)

From this one can conclude thateverytangent map off is 0-symmetric. This fact combined with some verygeneral dimension reduction arguments originating with Federer from geometric measure theory [Sim83],which do not depend on the harmonic behavior off at all, yield the dimension estimate (1.4) from [SU82].

The improvement in [CN13b] of the Hausdorff dimension estimate (1.4) to the Minkowski content esti-mate (1.13), and therefore theLp estimate of (1.12), requires exploiting more about the monotone quantityθr(x) than that it limits asr tends to zero. Indeed, an effective version of (1.28) says that for eachδ > 0 thereexistsN(Λ, δ) > 0 such that

∣

∣

∣

∣


∣

∣

∣

< δ , (1.29)

holds for all except for at mostN scalesα ∈ α1, . . . , αN ⊆ N. Thesebad scales where (1.29) fails maydiffer from point to point, but the number of such scales is uniformly bound. This allows one to conclude


that for all but at mostN-scales thatBrα(x) is (0, ǫ)-symmetric, see Section2.2. To exploit this information anew technique other than dimension reduction was required in [CN13b]. Indeed, in [CN13b] the quantitative0-symmetry of (1.29) was instead combined with the notion of cone splitting and an energy decompositionin order to conclude the estimates (1.12),(1.13). Since we will use them in this paper, the quantitative 0-symmetry and cone splitting will be reviewed further in Section 2.2.

Now let us begin to discuss the results of this paper. The mostchallenging aspect of this paper is theproof of the estimates on the quantitative stratifications of Theorems1.3 and1.4, and so we will focus onthese in our outline. Let us first observe that it might be advantageous to replace (1.29) with a version thatforces an actual rate of convergence, see for instance [Sim96]. More generally, if one is in a context wherean effective version of tangent cone uniqueness can be proved then this may be exploited. In fact, in thecontext of critical sets of elliptic equations one can follow exactly this approach, see the authors work [NV]where versions of Theorems1.3 and1.4 were first proved in this context. However, in the general contextof this paper such an approach fails, as tangent cone uniqueness is not available, and potentially not correct.

Instead, we will first replace (1.29) with the following relatively simple observation. Namely, for eachxthere existsN(Λ, δ) and a finite number of scalesα1, . . . , αN ⊆ N such that

∑

α j<α<α j+1

∣

∣

∣

∣


∣

∣

∣

< δ . (1.30)

That is, not only does the energy drop by less thanδ between these scales, but the sum of all the energydrops is less thanδ between these scales.

Unfortunately, exploiting (1.30) turns out to besubstantiallyharder to use than exploiting either (1.28) oreven (1.29). In essence, this is because it is not a local assumption in terms of scale, and one needs estimateswhich can see many scales simultaneously, but which do not require any form of tangent cone uniquenessstatements. Accomplishing this requires two new ingredients, a new rectifiable-Reifenberg type theorem,and a newL2-best subspace approximation theorem for stationary harmonic maps, which will allow us toapply the rectifiable-Reifenberg. Let us discuss these two ingredients separately.

We begin by discussing the newW1,p-Reifenberg and rectifiable-Reifenberg theorems, which are intro-duced and proved in Section3. Recall that the classical Reifenberg theorem, reviewed inSection2.6, givescriteria under which a set becomesCα-Holder equivalent to a ballB1(0k) in Euclidean space. In the contextof this paper, it is important to improve on this result so that we have gradient and volume control of our set.Let us remark that there have been many generalizations of the classical Reifenberg theorem in the litera-ture, see for instance [Tor95] [DT12], however those results have hypotheses which are much too strong forthe purposes of this paper. Instead, we will focus on improving theCα-equivalence to aW1,p-equivalence.This is strictly stronger by Sobolev embedding, and ifp > k then this results in volume estimates and arectifiable structure for the set. More generally, we will require a version of the theorem which allows formore degenerate structural behavior, namely a rectifiable-Reifenberg theorem. In this case, the assumptionswill conclude that a setS is rectifiable with volume estimates. Of course, what is key about this result isthat the criteria will be checkable for our quantitative stratifications, thus let us discuss these criteria briefly.


Roughly, ifS ⊆ B1(0n) is a subset equipped with thek-dimensional Hausdorff measureλk, then let us definethek-dimensional distortion ofS by

DkS(y, s) ≡ s−2 inf

Lks−k

∫

S∩Bs(y)d2(z, Lk) dλk(z) , (1.31)

where the inf is taken over allk-dimensional affine subspaces ofRn. That is,Dk measures how farS is frombeing contained in ak-dimensional subspace.

Our rectifiable-Reifenberg then requires this be small onS in an integral sense, more precisely that

r−k∫

S∩Br (x)

∑

rα≤r

Dk(y, rα)dλk(y) < δ2 . (1.32)

For δ sufficiently small, the conclusions of the rectifiable-Reifenberg Theorem3.3 are that the setS isrectifiable with effective bounds on thek-dimensional measure. Let us remark that one cannot possiblyconclude better than rectifiable under this assumption, seefor instance the Examples of Section11.

Thus, in order to prove the quantitative stratification estimates of Theorems1.3and1.4, we will need toverify that the integral conditions (1.32) hold for the quantitative stratificationsSk

ǫ ( f ), Skǫ,r( f ) on all balls

Br (x). In actuality the proof is more complicated. We will need to apply a discrete version of the rectifiable-Reifenberg, which will allow us to build an iterative covering of the quantitative stratifications, and each ofthese will satisfy (1.32). This will allow us to keep effective track of all the estimates involved. However,let us for the moment just focus on the main estimates which allows us to turn (1.32) into information aboutour harmonic maps, without worrying about such details.

Namely, in Section7 we prove a new and very general approximation theorem for stationary harmonicmaps. As always in this outline, let us assumeM ≡ Rn, the general case is no harder and we simply workon an injectivity radius ball. Thus we consider a stationaryharmonic mapf : B16(0n) → N, as well as anarbitrary measureµ which is supported onB1(0n). We would like to study how closely the support ofµ canbe approximated by ak-dimensional affine subspaceLk ⊆ Rn, in the appropriate sense, and we would like toestimate this distance by using properties off . Indeed, if we assume thatB8(0n) is not (k+ 1, ǫ)-symmetricwith respect tof , then for an arbitraryµ we will prove in Theorem7.1that

infLk⊆Rn

∫

d2(x, Lk) dµ ≤ C∫

|θ8(x) − θ1(x)|dµ , (1.33)

whereC will depend onǫ, energy bound off , and the geometry ofM andN. That is, if f does not havek+1 degrees of symmetry, then how closely the support of an arbitrary measureµ can be approximated by ak-dimensional subspace can be estimated by looking at the energy drop of f alongµ. In applications,µ willbe the restriction toB1 of some discrete approximation of thek-dimensional Hausdorff measure onSk

ǫ , andthus the symmetry assumption onf will hold for all balls centered on the support ofµ.

In practice, applying (1.33) to (1.32) is subtle and must be done inductively on scale. Additionally, inorder to prove the effective Hausdorff estimatesλk(Sk

ǫ ∩Br) ≤ Crk we will need to use the Covering Lemma8.1 to break up the quantitative stratification into appropriate pieces, and we will apply the estimates tothese. This decomposition is based on a covering scheme firstintroduced by the authors in [NV]. Thus forthe purposes of our outline, let us assume we have already proved the Hausdorff estimateλk(Sk

ǫ ∩Br) ≤ Crk,


and use this to be able to apply the rectifiable-Reifenberg inorder to conclude the rectifiability of the singu-lar set. This may feel like a large assumption, however it turns out the proof of the Hausdorff estimate willitself be proved by a very similar tactic, though will require an inductive argument on scale, and will use thediscrete rectifiable-Reifenberg of Theorem3.4in place of the rectifiable-Reifenberg of Theorem3.3.

Thus let us choose a ballBr and letE ≡ supBrθr (y). Let us consider the subsetSk

ǫ ⊆ Skǫ ∩ Br defined by

Skǫ ≡ y ∈ Sk

ǫ ∩ Br : θ0(y) > E − η , (1.34)

whereη = η(n,KM ,KN,Λ, ǫ) will be chosen appropriately later. We will show now thatSkǫ is rectifiable.

Sinceη is fixed and the ballBr is arbitrary, the rectifiability of all ofSkǫ follows quickly from a covering

argument. Thus, let us estimate (1.32) by plugging in (1.33) and the Hausdorff estimate to conclude:

r−k∫

Skǫ

∑

rα≤r

Dk(x, rα) dλk

= r−k∫

Skǫ

∑

rα≤r

(

infLk

r−2−kα

∫

Skǫ∩Brα (x)

d2(y, Lk)dλk(y))

dλk(x)

≤ Cr−k∫

Skǫ

∑

rα≤r

(

r−kα

∫

Skǫ∩Brα (x)

|θ8rα(y) − θrα(y)|dλk(y))

dλk(x)

= Cr−k∑

rα≤r

r−kα

∫

Skǫ

λk(Skǫ ∩ Brα(y))|θ8rα (y) − θrα(y)|dλk(y) ,

≤ Cr−k∫

Skǫ

∑

rα≤r

|θ8rα(y) − θrα(y)|dλk(y) ,

≤ Cr−k∫

Skǫ

|θ8r(y) − θ0(y)|dλk(y)

≤ Cr−kλk(Skǫ ) · η

< δ2 . (1.35)

where in the last line we have chosenη = η(n,KM ,KN,Λ, ǫ) so that the estimate is less than the requiredδ from the rectifiable-Reifenberg. Thus we can apply the rectifiable-Reifenberg of Theorem3.3 in order toconclude the rectifiability of the setSk

ǫ , which in particular proves thatSkǫ is itself rectifiable, as claimed.

2. PRELIMINARIES

2.1. Stationary Harmonic Maps and Monotonicity. The key technical tool available to a stationary har-monic mapf : B2(p) ⊆ M → N, which is not available to a weakly harmonic map, is that of a monotonequantity. Namely, givenx ∈ B1(p) andr ≤ 1 we can consider the normalized Dirichlet energy defined by

θr(x) ≡ r2−n∫

Br (x)|∇ f |2 . (2.1)


If M ≡ Rn , then for eachx fixed we have thatθr (x) is exactly a monotone increasing function ofr. Moreprecisely, we have that

ddrθr (x) = 2r2−n

∫

∂Br (x)

∣

∣

∣

∣

∂ f∂r

∣

∣

∣

∣

2= 2r2−n

∫

∂Br (x)|〈∇ f , nx〉|

2 . (2.2)

Hence, we see that ifθr(x) is independent ofr then we have thatf is 0-symmetric with respect tox. Moregenerally, ifθs(x) = θr(x) then f is radially symmetric on the annulusAs,r(x) = Br (x) \Bs (x). Motivated bythis, we see that what we are really interested in is the amount the Dirichlet energy drops from one scale tothe next, and thus we define

Ws,r(x) ≡ θr(x) − θs(x) ≥ 0 . (2.3)

Oftentimes we will want to enumerate our choice of scale, so let us define the scalesrα ≡ 2−α for α ≥ 0, andthe corresponding Dirichlet energy drop:

Wα(x) ≡Wrα,rα−3(x) ≡ θrα−3(x) − θrα(x) ≥ 0 . (2.4)

In the general case, i.e., whenM , Rn, essentially the same statements may be made, howeverθr(x) isnow only almostmonotone, meaning thateCrθr(x) is monotone for some constant which depends only onthe geometry ofM. From this one can prove that at every point, every tangent map is 0-symmetric, whichis the starting point for the dimension estimate (1.4) of [SU82]. In Section2.2 we will discuss quantitativeversions of this point, first introduced in [CN13b] and used in this paper as well, and also generalizationswhich involve higher degrees of symmetry. These points werefirst used in [CN13b] to prove Minkowskiestimates on the quantitative stratification of a stationary harmonic map. They will also play a role in ourarguments, though in a different manner.

2.2. Quantitative 0-Symmetry and Cone Splitting. In this subsection we review some of the quantitativesymmetry and splitting results of [CN13b], in particular those which will play a role in this paper.

The first result we will discuss acts as aneffectiveformulation of the fact that every tangent map is 0-symmetric. Namely, the quantitative 0-symmetry of [CN13b] states that for eachǫ > 0 and pointx ∈ B1, allbut a finite number of the balls

Brα(x)

α∈N are (0, ǫ)-symmetric, whererα ≡ 2−α. Precisely:

Theorem 2.1(Quantitative 0-Symmetry [CN13b]). Let f : B2(p) ⊆ M → N be a stationary harmonic mapsatisfying(1.16) with

>B2(p)|∇ f |2 ≤ Λ and letǫ > 0 be fixed. Then the following hold:

(1) There existsδ(n,KN,Λ, ǫ) > 0 such that for each x∈ B1(p) if |θr(x) − θδr(x)| < δ and KM < δ, thenBr(x) is (0, ǫ)-symmetric.

(2) For each x∈ B1(p) there exists a finite number of scalesα1, . . . , αK ⊆ Nwith K ≤ K(n,KM ,KN,Λ, ǫ)such that for r< (rα j /2, 2rα j ) we have that Br(x) is (0, ǫ)-symmetric.

Remark2.1. In [CN13b] it is stated that the constant depends on the manifoldN, not justKN, however it iseasy to check that onlyKN is important in the proof. See the proof of Lemma7.2for a relevant argument.

Remark2.2. The assumptionKM < δ is of little consequence, since this just means focusing theestimateson balls of sufficiently small radius after rescaling.


Another technical tool that played an important role in [CN13b] was that of cone splitting. This will beused in this paper when proving the existence of unique tangent planes of symmetry for the singular set, sowe will discuss it here. In short, cone splitting is the idea that multiple 0-symmetries add to give rise to ak-symmetry. To state it precisely let us give a careful definition of the notion of independence of a collectionof points:

Definition 2.2. We say a collection of pointsx1, . . . , xℓ ∈ Rn is independent if they are linearly indepen-dent. We say the collection isτ-independent ifd(xk+1, spanx1, . . . , xk) > τ for eachk. If x1, . . . , xℓ ∈ Mthen we say the collection isτ-independent with respect tox if d(x, x j ) < inj(x) and the collection isτ-independent when written in exponential coordinates atx.

Now we are in a position to state the effective cone splittingof [CN13b]:

Theorem 2.3(Cone Splitting [CN13b]). Let f : B3(p) ⊆ M → N be a stationary harmonic map satisfying(1.16) with

>B2(p)|∇ f |2 ≤ Λ and letǫ, τ > 0 be fixed. Then there existsδ(n,KN,Λ, ǫ, τ) > 0 such that if

KM < δ and x0, . . . , xk ∈ B1(p) are such that

(1) B2(x j) are (0, δ)-symmetric,(2) x0, . . . , xk are τ-independent at p,

then B1(p) is (k, ǫ)-symmetric.

We end with the following, which one can view as a quantitative form of dimension reduction. The proofis standard and can be accomplished by a contradiction argument:

Theorem 2.4(Quantitative Dimension Reduction). Let f : B2(p) ⊆ M → N be a stationary harmonic mapsatisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Then for eachǫ > 0 there existsδ(n,KM ,KN,Λ, ǫ), r(n,KM ,KN,Λ, ǫ) >

0 such that if B2(p) is (k, δ)-symmetric with respect to some k-plane Vk, then for each x∈ B1(p) \ Bǫ(Vk) wehave that Br(x) is (k+ 1, ǫ)-symmetric.

2.3. Defect Measure. Since it will play a role for us in one technical aspect of thispaper, let us recall inthis section the defect measure of a sequence of stationary harmonic maps. We begin with a definition:

Definition 2.5. Let fi : B2(p) → N be a sequence of stationary harmonic maps satisfying (1.16) with>B2(p) |∇ fi |2 ≤ Λ. Then, after possibly passing to a subsequence,fi has a weakW1,2 limit f , and we can

consider the measure

|∇ fi |2dvg

∗ |∇ f |2 dvg + ν , (2.5)

where this last convergence is intended in the weak sense of Radon measures. The measureν is nonnegativeby Fatou’s lemma, and it is called the defect measure associated to fi ∗ f .

Remark2.3. We can also allowfi : B2(pi) → N to be defined on different manifolds. In practice, this willoccur under the assumptionKMi → 0, so thatν becomes a measure onB2(0n).


The following theorem is one of the main accomplishments of [Lin99]:

Theorem 2.6(Rectifiability of Defect Measure). [Lin99, lemma 1.7]If ν is a defect measure as in definition2.5, then it is n− 2 rectifiable.

A key tool in the proof of the above, which will be useful in this paper as well, is the following 0-symmetryresult:

Theorem 2.7. [Lin99, lemma 1.7 (ii)]Let fi : B2(pi) ⊆ Mi → N be a sequence of stationary harmonicmaps satisfying(1.16) with

>B2(p)|∇ fi |2 ≤ Λ. Assume KMi < δi → 0 and that

∣

∣

∣θi,2(pi) − θi,δi

∣

∣

∣ < δi → 0. Then

the defect measure|∇ fi |2dvg ∗ |∇ f |2 dvg+ ν is 0-symmetric, that is, both the function f andν are invariant

under dilation around the origin.

2.4. ǫ-regularity for Minimizers. In this subsection we quickly review theǫ-regularity theorems of [CN13b],which themselves build on the difficult work of [SU82]. This will be our primary technical tool in upgradingthe structural results on stationary harmonic maps to the regularity results for minimizing harmonic maps.Recall the definition of the regularity scale given in Definition 1.7, then the main theorem of this subsectionis the following:

Theorem 2.8(Minimizing ǫ-Regularity [CN13b]). Let f : B2(p) ⊆ M → N be a minimizing harmonic mapsatisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Then there existsǫ(n,KN,Λ) > 0 such that if KM < ǫ and B2(p) is

(n− 2, ǫ)-symmetric, thenr f (p) ≥ 1 .

Remark2.4. As previously remarked, the assumptionKM < ǫ is of little consequence, since this just meansfocusing the estimates on balls of sufficiently small radiusafter rescaling.

The following stationary version was proved in [CHN13a], and is essentially just a combination of thedefect measure ideas of [Lin99], theǫ-regularity of [Bet93], and a contradiction argument:

Theorem 2.9(Stationaryǫ-Regularity [CHN13a]). Let f : B2(p) ⊆ M → N be a minimizing harmonic mapsatisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Assume that there exists no nonconstant stationary harmonic map

S2→ N. Then there existsǫ(n,N,Λ) > 0 such that if KM < ǫ and B1(p) is (n− 2, ǫ)-symmetric, then

r f (p) ≥12.

Let us also discuss some improvements of the aboveǫ-regularity theorems in the case where there are nononconstant harmonic maps fromSk into N. These are the keyǫ-regularity results needed for Theorems1.9and1.10. We begin with the minimizing case:


Theorem 2.10(Improved Minimizingǫ-Regularity [CHN13a]). Let f : B2(p) ⊆ M → N be a minimizingharmonic mapping satisfying(1.16) with

>B2(p) |∇ f |2 ≤ Λ. Assume further that for some k≥ 2 there exists

no nonconstant minimizing harmonic maps Sℓ → N for all ℓ ≤ k. Then there existsǫ(n,N,Λ) such that ifKM < ǫ and B1(p) is (n− k − 1, ǫ)-symmetric, then

r f (p) ≥12.

Finally, we end with the stationary version of the above theorem:

Theorem 2.11(Improved Stationaryǫ-Regularity [CN13b]). Let f : B2(p) ⊆ M → N be a stationaryharmonic mapping satisfying(1.16) with

>B2(p)|∇ f |2 ≤ Λ. Assume further that for some k≥ 2 there exists

no nonconstant stationary harmonic maps Sℓ → N for all ℓ ≤ k. Then there existsǫ(n,N,Λ) such that ifKM < ǫ and B1(p) is (n− k − 1, ǫ)-symmetric, then

r f (p) ≥12.

Remark2.5. Note that in both theorems above, nothing would change if we only required thatN admits nocontinuous(or evensmooth) nonconstant stationary (or minimizing) harmonic mapSℓ → N for all ℓ ≤ k.Indeed, the tangent maps at singular points of non-smooth stationary (or minimizing) harmonic maps giverise to smooth stationary (or minimizing) harmonic maps from lower dimensional spheres.

2.5. Hausdorff, Minkowski, and packing Content. In this subsection we give a brief review of the notionsof Hausdorff, Minkowski, and packing content. We will also use this to recall the definition of Hausdorffmeasure. The results of this subsection are completely standard, but this gives us an opportunity to introducesome notation for the paper. For a more detailed reference, we refer the reader to [Mat95, Fed69]. Let usbegin with the notions of content:

Definition 2.12. Given a setS ⊆ Rn andr > 0 we define the following:

(1) Thek-dimensional Hausdorffr-content ofS is given by

λkr (S) ≡ inf

∑

ωkrki : S ⊆

⋃

Br i (xi) andr i ≤ r

. (2.6)

(2) Thek-dimensional Minkowskir-content ofS is given by

mkr (S) ≡ (2r)s−nVol (Br (S)) . (2.7)

(3) Thek-dimensional packingr-content ofS is given by

pkr (S) ≡ sup

∑

ωkrki : xi ∈ S andBr i (xi) are disjoint andr i ≤ r

. (2.8)

These definitions make sense for anyk ∈ [0,∞), though in this paper we will be particularly interested ininteger valuedk. Notice that ifS is a compact set thenλk

r (S),mkr (S) < ∞ for anyr > 0, and that we always

have the relations

λkr (S) . mk

r (S) . pkr (S) . (2.9)


In particular, bounding the Hausdorff content is less powerful than bounding the Minkowski content, whichis itself less powerful than bounding the packing content.

Primarily in this paper we will be mostly interested in content estimates, because these are the mosteffective estimates. However, since it is classical, let usgo ahead and use the Hausdorff content to define ameasure. To accomplish this, let us more generally observe that if r ≤ r′ thenλk

r (S) ≥ λk′r (S). In particular,

we can define the limit

λk0(S) ≡ lim

r→0λk

r (S) = supr>0

λkr (S) .

It turns out thatλk0 is a genuine measure.

Definition 2.13. Given a setS ⊆ Rn we define itsk-dimensional Hausdorff measure byλk(S) ≡ λk0(S).

Similar constructions can be carried out for the Minkowski and packing content. In particular, we candefine

mk0(S) ≡ lim sup

r→0mk

r (S) , mk0(S) ≡ lim inf

r→0mk

r (S) , (2.10)

pk0(S) = lim

r→0pk

r (S) = infr>0

pkr (S) . (2.11)

Definition 2.14. Given a setS ⊆ Rn we define its Hausdorff and Minkowski dimension (or box-dimension)by

dimH S ≡ inf

k ≥ 0 : λk0(S) = 0

,

dimM S ≡ inf

k ≥ 0 : mk0(S) = 0

. (2.12)

Remark2.6. Note that we could define an upper and lower Minkowski dimension by

dimMS ≡ inf

k ≥ 0 : mk0(S) = 0

, dimMS ≡ inf

k ≥ 0 : mk0(S) = 0

. (2.13)

In general, dimMS ≤ dimMS, where the inequality may be strict. However, for the purposes of this paperwe will only be interested in theupperMinkowski dimension.

As an easy example consider the rationalsQn ⊆ Rn. Then it is a worthwhile exercise to check thatdimH Qn

= 0, while dimM Qn= n.

A very important notion related to measures is thedensityat a point. Although this is standard, forcompleteness we briefly recall the definition of Hausdorff density, and refer the reader to [Mat95, chapter 6]for more on this subject.

Definition 2.15. Given a setS ⊂ Rn which isλk-measurable, andx ∈ Rn, we define thek-dimensionalupper and lower density ofS at x by

θ⋆k(S, x) = lim supr→0

λk(S ∩ Br (x))

ωkrk, θk

⋆(S, x) = lim infr→0

λk(S ∩ Br (x))

ωkrk. (2.14)

In the following, we will use the fact that for almost any point in a set with finiteλk-measure, the densityis bounded from above and below.


Proposition 2.16( [Mat95]). Let S⊂ Rn be a set withλk(S) < ∞. Then for k-a.e. x∈ S :

2−k ≤ θ⋆k(S, x) ≤ 1 , (2.15)

while for k-a.e. x∈ Rn \ S

θ⋆k(S, x) = 0 . (2.16)

2.6. The Classical Reifenberg Theorem.In this Section we recall the classical Reifenberg Theorem,aswell as some more recent generalizations. The Reifenberg theorem gives criteria on a closed subsetS ⊆B2 ⊆ Rn which determine whenS ∩ B1 is bi-Holder to a ballB1(0k) in a smaller dimensional Euclideanspace. The criteria itself is based on the existence ofgoodbest approximating subspaces at each scale. Westart by recalling the Hausdorff distance.

Definition 2.17. Given two setsA, B ⊆ Rn, we define the Hausdorff distance between these two by

dH(A, B) = inf r ≥ 0 s.t. A ⊂ Br (B) and B ⊂ Br (A) . (2.17)

Recall thatdH is a distance on closed sets, meaning thatdH(A, B) = 0 impliesA = B.

The classical Reifenberg theorem says the following:

Theorem 2.18(Reifenberg Theorem [Rei60, Sim]). For each0 < α < 1 andǫ > 0 there existsδ(n, α, ǫ) > 0such that the following holds. Assume0n ∈ S ⊆ B2 ⊆ Rn is a closed subset, and that for each Br(x) ⊆ B1

with x ∈ S we have

infLk

dH(

S ∩ Br(x), Lk ∩ Br(x))

< δ r , (2.18)

where theinf is taken over all k-dimensional affine subspaces Lk ⊆ Rn. Then there existsφ : B1(0k) → Swhich is a Cα bi-Holder homeomorphism onto its image with[φ]Cα , [φ−1]Cα < 1+ ǫ and S∩B1 ⊆ φ(B1(0k)).

Remark2.7. In fact, one can prove a little more. In particular, under thehypothesis of the previous theorem,there exists a closed subsetS′ ⊂ Rn such thatS′ ∩ B1 (0) = S ∩ B1 (0) and which is homeomorphic toa k-dimensional subspace 0n ∈ T0 ⊆ Rn via theCα bi-Holder homeomorphismφ : T0 → S′. Moreover,|φ(x) − x| ≤ C(n)δ for all x ∈ T0 andφ(x) = x for all x ∈ T0 \ B2 (0).

One can paraphrase the above to say that ifS can be well approximated on every ball by a subspace intheL∞-sense, thenS must be bi-Holder to a ball in Euclidean space.

Let us also mention that there are several more recent generalizations of the classic Reifenberg theo-rem. In [Tor95], the author proves a strengthened version of (2.18) that allows one to improve bi-Holder tobi-Lipschitz. Unfortunately, for the applications of thispaper the hypotheses of [Tor95] are much too re-strictive. We will require a weaker condition than in [Tor95], which is more integral in nature, see Theorem3.2. In exchange, we will improve the bi-Holder of the classical Reifenberg toW1,p.


We will also need a version of the classical Reifenberg whichonly assumes that the subsetS is containednear a subspace, not conversely that the subspace is also contained nearS. In exchange, we will only con-clude the set is rectifiable. A result in this direction was first proved in [DT12], but again the hypotheses aretoo restrictive for the applications of this paper, and additionally there is a topological assumption necessaryfor the results of [DT12], which is not reasonable in the context in this paper. We will see how to appropri-ately drop this assumption in Theorem3.3.

2.7. W1,p maps and rectifiability. In the paper we will be using the structure ofW1,p maps forp > k inorder to conclude rectifiable structures on sets. For the reader’s convenience, here we recall a standard resultabout rectifiability andW1,p maps:

Lemma 2.19. LetΩ ⊂ Rk be an open domain, and let f: Ω→ Rn be a W1,p map with p> k. Then for allK ⊂ Ω, f (K) is a k-rectifiable set.

In order to prove this we need, by definition, to show that there exists a countable sequence of Lipschitzmapsfi : Rk → Rn such that

λk

f (K) \⋃

i

fi(Rk)

= 0 . (2.19)

By a classical result, it is possible forp > k to approximate allW1,p maps with Lipschitz maps. Morespecifically, for allǫ > 0, there exists a Lipschitz mapfǫ which coincides withf up to a setEǫ of smallk-dimensional measure inRk. In particular, we have

Eǫ ≡

x ∈ Rk s.t. f (x) , fǫ(x)

, λk(Eǫ ) < ǫ . (2.20)

For a detailed reference, see for example [EG92, theorem 3, sec 6.6.3] or [Zie89, section 3.10]). The onlything left to prove is thatf (Eǫ ) has small measure:

Lemma 2.20. Let f be the continuous representative of a W1,p map as above with p> k. Then for allmeasurable subsets E⊂ Ω we have

λk( f (E)) ≤ C(k, n, p) ‖∇ f ‖kLp(Ω)

(

λk(E))1− k

p. (2.21)

This lemma follows from standard Morrey-type estimates, see for example [BMT13, proposition 2.4].

3. THE W1,p-REIFENBERG AND RECTIFIABLE-REIFENBERG THEOREMS

In this Section we state very general Reifenberg type theorems for subsets of Euclidean spaces, theirproofs are carried out in Sections5 and 6. Although this is a key ingredient of the proof of our maintheorems, the results of this Section are of an independent nature, and may be of some separate interest.Therefore, we will attempt to write them so that this Sectionmay be read independently of the rest of thepaper, with the exception of the appropriate Preliminary Sections.


As described in Section2.6, the classical Reifenberg theorem tells us that if a closed subsetS ⊆ B2 ⊆ Rn

can be well approximated on every ball by somek-dimensional affine space in theL∞ sense, thenS ∩ B1

must itself be homeomorphic, in fact bi-Holder, to a ballB1(0k) in a smaller dimensional Euclidean space.

For the applications of this paper, we will need to improve onthe classical Reifenberg theorem in severalways. At the most basic level,Cα-equivalence is not strong enough. We will require control on the gradientof the mappings, since this is the essence of rectifiability and volume control. For this purpose we willobtainW1,p-control on our mappings fork < p < ∞. We will deal with this first in Theorem3.2, where wewill also replace theL∞ closeness of the Reifenberg theorem with anL2 closeness condition, which turnsout to be more natural in the applications.

A second manner in which we will need improvement is that we will need to allow for the existence ofholes. That is, in applications our setS may be rectifiable, but it might not be homeomorphic to a ball,thisis much too strong. To deal with this we can weaken the hypothesis of the Reifenberg theorem and onlyassume that on each ballBr(x) there is ak-dimensional subspaceV such that each point of our setS is closeto V, but not conversely that each point of the subspaceV is close toS. We study this case in Theorem3.3,where we will show under natural conditions that such a setS is rectifiable with volume bounds, which iseasily seen to be a sharp result under the assumptions of Theorem3.3.

The last version of the Reifenberg theorem that we prove has amore discrete nature, and it is the one wewill use to obtain the volume bounds in the rest of the paper. In this case, our setS = xi will be a discreteset of points paired with radiir i > 0 such thatBr i (xi) are disjoint balls. We will associate to this collectionthe measureµ =

∑

ωkrki δxi , and under the appropriate assumptions show in Theorem3.4 that the volume of

µ continues to enjoy the packing upper boundµ(Br(x)) < Crk for all radii r > r i . In the applications, wewill prove the main theorems of this paper by building a series of covers of the quantitative stratificationsSkǫ , and we will apply the discreteW1,p-Reifenberg to each of this to obtain volume bounds. Only at the last

stage will we apply Theorem3.3in order to obtain the rectifiability of the sets as well.

It is worth mentioning that some interesting generalizations of Reifenberg’s theorem have been proved innumorous sources. Generalizations in the spirit of this paper are explored in [DS93], [Tor95], and [DT12].However, in each of these cases the requirements of these theorems are too stringent to be applicable in oursituation. Very recently, and using techniques independent from this work, the authors in [AT15],[Tol15]proved necessary and sufficient conditions for thek-rectifiability of a set which are closely related to theresults of this paper. However, these results lack anyapriori control over the volume of the sets in question,which is fundamental in this paper to the applications.

3.1. Statement of Main W1,p-Reifenberg and rectifiable-Reifenberg Results.There are three main re-sults we wish to discuss in this subsection, each more general than the last. In order to keep the statementsas clean and intuitive as possible we will introduce the following definitions, and discuss them briefly.


Definition 3.1. Let µ be a measure onB2 with r > 0 andk ∈ N. Then we define thek-dimensionaldisplacement by

Dkµ(x, r) ≡ inf

Lk⊆Rnr−(k+2)

∫

Br (x)d2(y, Lk) dµ(y) , (3.1)

if µ(Br(x)) ≥ ǫnrk ≡ (1000n)−7n2rk, and Dk

µ(x, r) ≡ 0 otherwise, where the inf’s are taken over allk-dimensional affine subspacesLk ⊆ Rn. If S ⊆ B2, then we can define itsk-displacementDk

S(x, r) byassociating toS thek-dimensional Hausdorff measureλk

S restricted toS.

Remark3.1. Note that in literatureD is usually referred to as the Jones’ numberβ2.

Remark3.2. One can replaceǫn by any smaller lower bound, zero included, and the proofs andstatementswill all continue to hold. Our particular choice is based on constants which will be obtained in Section4.

Remark3.3. Notice that the definitions are scale invariant. In particular, if we rescaleBr → B1 and letS bethe induced set, thenDk

S(x, r)→ DkS(x, 1).

Remark3.4. Notice the monotonicity given by the following: IfS′ ⊆ S, thenDkS′(x, r) ≤ Dk

S(x, r).

Remark3.5. It is easily seen from the definition thatDkµ(x, r) is controlled above and below byDk

µ(x, r/2)and Dk

µ(x, 2r), as long asµ is not too small onBr (x). In particular, if µ(Br(x)) ≥ 2kǫnrk, then for ally ∈ S ∩ Br (x), Dk

µ(x, r) ≤ 2k+2Dkµ(y, 2r). As a corollary we have the estimate

µ(Br(x)) ≥ 2kǫnrk=⇒ Dk

µ(x, r) ≤ 2k+2?

Br (x)Dkµ(y, 2r)dµ(y) . (3.2)

The first of the main results of this part of the paper is not strictly used elsewhere in the paper, but it isa natural generalization of the Reifenberg and gives intuition and motivation for the rest of the statements,which are essentially more complicated versions of it. Thus, we have decided it is worth discussing inde-pendently. The first theorem of this section is the following:

Theorem 3.2(W1,p-Reifenberg). For eachǫ > 0 and p ∈ [1,∞) there existsδ(n, ǫ, p) > 0 such that thefollowing holds. Let S⊆ B2 ⊆ Rn be a closed subset with0n ∈ S , and assume for each x∈ S ∩ B1 andBr(x) ⊆ B2 that

infLk

dH(

S ∩ Br(x), Lk ∩ Br(x))

< δr , (3.3)∫

S∩Br (x)

(∫ r

0Dk

S(x, s)dss

)

dλk(x) < δ2rk . (3.4)

Then the following hold:

(1) there exists a mappingφ : Rk → Rn which is a1 + ǫ bi-W1,p map onto its image and such thatS ∩ B1 (0n) = φ(B1(0k)) ∩ B1 (0n). In particular?

B1(0k)|∇φ|p ≤ 1+ ǫ ,

?S∩B1(0n)

∣

∣

∣∇φ−1∣

∣

∣

p≤ 1+ ǫ . (3.5)

(2) S ∩ B1(0n) is rectifiable.


(3) For each ball Br(x) ⊆ B1 with x∈ S we have

(1− ǫ)ωkrk ≤ λk(S ∩ Br(x)) ≤ (1+ ǫ)ωkr

k . (3.6)

Remark3.6. The result (2) follows from (1). Indeed by a standard result,if φ is aW1,p map withp > k, thenits image is ak-rectifiable set (see Lemma2.19). We get (3) by applying the result of (1) to all smaller ballsBr(x) ⊆ B1, since the assumptions of the theorem hold on these balls as well.

Note that theW1,p estimates forφ−1 are justified in Lemma6.2.

Remark3.7. Note that forp > k, by Sobolev embeddings aW1,p map is also aCα map withα = 1− kp.

Remark3.8. As it is easily seen, the requirement thatS is closed is essential for this theorem, and inparticular for the lower bound on the Hausdorff measure. As an example, consider any setS ⊆ Rk whichis dense but has zero Hausdorff measure. In the following theorems, we will not be concerned with lowerbounds on the measure, and we will be able to drop the closed assumption.

Let us now consider the case when we drop the assumption (3.3) from the result. The key distinction nowis thatS only needs to be locally near a piece of ak-dimensional subspace, but not conversely. Thus, wecannot hope to obtain topological information about the setS, see Section11.3. The precise statement is thefollowing:

Theorem 3.3(Rectifiable-Reifenberg). For everyǫ > 0, there existsδ(n, ǫ) > 0 such that the followingholds. Let S⊆ B2 ⊆ Rn be aλk-measurable subset, and assume for each Br(x) ⊆ B2 with λk(S ∩ Br(x)) ≥ǫnrk that

∫

S∩Br (x)

(∫ r

0Dk

S(x, s)dss

)

dλk(x) <δ2rk . (3.7)

Then the following hold:

(1) For each ball Br(x) ⊆ B1 with x∈ S we have

λk(S ∩ Br(x)) ≤ (1+ ǫ)ωkrk . (3.8)

(2) S ∩ B1(0n) is k-rectifiable.

Remark3.9. Notice that for the statement of the theorem we do not need control over balls which alreadyhave small measure. This will be quite convenient for the applications.

Remark3.10. Instead of (3.7) we may assume the essentially equivalent estimate∑

rα≤4r

∫

S∩Br (x)Dk

S(x, rα) dλk(x) < δ2rk . (3.9)

In the applications, this will be the more convenient phrasing.

Finally, we end by stating a version of the above theorem which is more discrete in nature. This resultwill be particularly important in the proof of the main theorems of this paper:


Theorem 3.4 (Discrete Rectifiable-Reifenberg). There existsδ(n) > 0 and D(n) such that the followingholds. LetBr j (x j)xj∈S ⊆ B2 be a collection of disjoint balls, and letµ ≡

∑

xj∈Sωkrkjδxj be the associated

measure. Assume that for each Br(x) ⊆ B2 with µ(Br (x)) ≥ ǫnrk we have∫

Br (x)

(∫ r

0Dkµ(x, s)

dss

)

dµ(x) < δ2rk . (3.10)

Then we have the estimate∑

xj∈B1

rkj < D(n) . (3.11)

Remark3.11. As in Theorem3.3, instead of (3.7) we may assume the estimate

∑

rα≤2r

∫

Br (x)Dkµ(x, rα) dµ(x) < δ2rk . (3.12)

In the applications, this will be the more convenient phrasing.

For many of the applications of this paper, it is this versionof the Reifenberg which will be most impor-tant. The reasoning is that to keep uniform control on all estimates our inductive covering will need to coverat all scales. It is only at the last scale that we begin to cover the singular setsSk on sets of positive measure.

4. TECHNICAL CONSTRUCTIONS TOWARDNEW REIFENBERGRESULTS

In this section, we prove some technical lemmas needed for dealing with the relation between bestL2

subspaces. These elementary results will be used in many of the estimates of subsequent sections.

4.1. Hausdorff distance and subspaces.We start by recalling some standard facts about affine subspacesin Rn and Hausdorff distance.

Definition 4.1. Given two linear subspacesL,V ⊆ Rn, we define the Grassmannian distance between thesetwo as

dG(L,V) = dH(L ∩ B1 (0) ,V ∩ B1 (0)) = dH

(

L ∩ B1 (0),V ∩ B1 (0))

, (4.1)

Note that if dim(L) , dim(V), thendG(L,V) = 1.

For general subsets inRn, it is evident thatA ⊆ Bδ (B) does not implyB ⊆ Bcδ (A). However, ifA andB are affine spaces with the same dimension, then it is not difficult to see that this property holds. Moreprecisely:

Lemma 4.2. Let V, W be two k-dimensional affine subspaces inRn, and suppose that V∩ B1/2 (0) , ∅.There exists a constant c(k, n) such that if V∩ B1 (0) ⊆ Bδ (W∩ B1 (0)), then W∩ B1 (0) ⊆ Bcδ (V ∩ B1 (0)).Thus in particular dH(V ∩ B1 (0) ,W∩ B1 (0)) ≤ cδ.


Proof. The proof relies on the fact thatV andW have the same dimension. Letx0 ∈ V be the point ofminimal distance from the origin. By assumption, we have that ‖x0‖ ≤ 1/2. Let x1, · · · , xk ∈ V ∩ B1 (0) bea sequence of points such that

‖xi − x0‖ ≥ 1/2 and for i , j ,⟨

xi − x0, x j − x0

⟩

= 0 . (4.2)

In other words,xi − x0ki=1 is an affine base forV. Let yi

ki=0 ⊆W∩ B1 (0) be such thatd(xi , yi) ≤ δ. Then

‖yi − y0‖ ≥ 1/2− 2δ and for i , j ,∣

∣

∣

∣

⟨

yi − y0, y j − y0

⟩

∣

∣

∣

∣

≤ 2δ + 4δ2 . (4.3)

This implies that forδ ≤ δ0(n), yi − y0ki=1 is an affine base forW and for ally ∈W

y = y0 +

k∑

i=1

αi(yi − y0) , |αi | ≤ 10‖y− y0‖ . (4.4)

Now lety ∈W∩ B1 (0) be the point of maximum distance fromV, and letπ be the projection ontoV andπ⊥

the projection ontoV⊥. Then

d(y,V) = d(y, π(y)) =∥

∥

∥π⊥(y)∥

∥

∥ ≤

k∑

i=1

|αi |∥

∥

∥π⊥(yi − y0)∥

∥

∥ ≤ c′(n, k)δ . (4.5)

Sincey ∈ B1 (0), thenπ(y) ∈ V∩B1+c′δ (0), and thusd(y,V∩B1(0)) ≤ 2c′δ ≡ cδ, which proves the claim.

Next we will see that the Grassmannian distance between two subspaces is enough to control the projec-tions with respect to these planes. In order to do so, we recall a standard estimate.

Lemma 4.3. Let V,W be linear subspaces of a Hilbert space. Then dG(V,W) = dG(

V⊥,W⊥)

.

Proof. We will prove thatdG(

V⊥,W⊥)

≤ dG (V,W). By symmetry, this is sufficient.Takex ∈ V⊥ such that‖x‖ = 1, and consider thatd(x,W⊥) = ‖πW(x)‖. Let z = πW(x) andy = πV(z). We

want to show that ifdG(V,W) ≤ ǫ < 1, then‖z‖ ≤ ǫ. We can limit our study to the space spanned byx, y, z,and assume wlog thatx = (1, 0, 0), y = (0, b, 0) andz = (a, b, c). By orthogonality betweenz andz− x, wehave

a2+ b2+ c2+ (1− a)2

+ b2+ c2= 1 =⇒ a = a2

+ b2+ c2 , (4.6)

and sincez∈W, we also have‖z− y‖ ≤ ǫ ‖z‖, which implies

a2+ c2 ≤ ǫ2

(

a2+ b2+ c2

)

=⇒ a2+ c2 ≤

ǫ2

1− ǫ2b2 . (4.7)

Since the functionf (x) = x2/(1− x2) is monotone increasing forx ≥ 0, we can defineα ≥ 0 in such a waythat

a2+ c2=

α2

1− α2b2 , a = a2

+ b2+ c2=

1

1− α2b2 . (4.8)

Note that necessarily we will haveα ≤ ǫ. Now we have

1

(1− α2)2b4= a2 ≤

α2

1− α2b2

=⇒ b2 ≤ α2(

1− α2)

=⇒ ‖z‖2 = a2+ b2+ c2 ≤ α2 ≤ ǫ2 . (4.9)

This proves thatV⊥ ∩ B1 (0) ⊂ Bǫ(

W⊥)

. In a similar way, one proves the opposite direction.


As a corollary, we prove that the Grassmannian distancedG(V,W) is equivalent to the distance given by‖πV − πW‖.

Lemma 4.4. Let V,W be linear subspaces ofRn. Then for every x∈ Rn,

‖πV(x) − πW(x)‖ ≤ 2dG(V,W) ‖x‖ . (4.10)

In particular, if x ∈W⊥, then‖πV(x)‖ ≤ 2dG(V,W) ‖x‖.Conversely, we have

dG(V,W) ≤ supx∈Rn\0

‖πV(x) − πW(x)‖‖x‖

. (4.11)

Proof. The proof is just a corollary of the previous lemma. Assume wlog that‖x‖ = 1, and letx = y + zwherey = πV(x) andz= πV⊥(x). Then

‖πV(x) − πW(x)‖ = ‖y− πW(y) − πW(z)‖ ≤ ‖y− πW(y)‖ + ‖z− πW⊥(z)‖ = d(y,W) + d(z,W⊥) . (4.12)

Since‖y‖2 + ‖z‖2 = ‖x‖2 = 1, by the previous lemma we get the first estimate.The reverse estimate is an immediate consequence of the definition.

4.2. Distance betweenL2 best planes.Here we study the distance between best approximating subspacesfor our measureµ on different balls. Let us begin by fixing our notation for this subsection, and pointingout the interdependencies of the constants chosen here. Throughout this subsection, our choice of scaleρ = ρ(n,M) > 0 is a constant which will eventually be fixed according to Lemma4.7. For applications tofuture sections, it is sufficient to know that we can takeρ(n,M) = 10−10(100n)−nM−1. We also point outthat in Section5, we will fix M = 40n, and soρ will be a constant depending only onn. In particular, wecan use the very coarse estimate

ρ = 10−10(100n)−3n . (4.13)

We will also introduce a threshold valueγk = ωk40−k. The dimensional constantγk is chosen simply to bemuch smaller than any covering errors which will appear.

We will consider a positive Radon measureµ supported onS ⊆ B1 (0), and useD(x, r) ≡ Dkµ(x, r) to bound

the distances between bestL2 planes at different points and scales. By definition let us denote byV(x, r) abestk-dimensional plane onBr (x), i.e., ak-dimensional affine subspaces minimizing

∫

Br (x) d(x,V)2dµ. Notethat, in general, this subspace may not be unique. We want to prove that, under reasonable hypothesis, thedistance betweenV(x, r) andV(y, r′) is small if d(x, y) ∼ r andr′ ∼ r.

In order to achieve this, we will need to understand some minimal properties ofµ. First, we need tounderstand how concentratedµ is on any given ball. For this reason, for someρ > 0 and allx ∈ B1 (0) wewill want to consider the upper mass bound

µ(Bρ (x)) ≤ Mρk ∀x ∈ B1(0) . (4.14)

However, an upper bound on the measure is not enough to guarantee bestL2-planes are close, as thefollowing example shows:


Example4.1. Let V,V′ bek-dimensional subspaces, 0∈ V ∩ V′, and setS =(

V ∩ B1 (0) \ B1/10 (0))

∪ S′,whereS′ ⊆ V′ ∩ B1/10 (0) andµ = λk|S. Then evidentlyD(0, 1) ≤ λk(S′) andD(0, 1/10) = 0, independentlyof V andV′. However,V(0, 1) will be close toV, while V(0, 1/10)= V′. Thus, in general, we cannot expectV(0, 1) andV(0, 1/10) to be close ifµ(B1/10 (0)) is too small.

Thus, in order to prove that the best planes are close, we needto have some definite amount of measureon the set, in such a way thatS “effectively spans” ak-dimensional subspace, where by effectively span wemean the following:

Definition 4.5. Given a sequence of pointspi ∈ Rn, we say thatpi

ki=0 α−effectively span ak-dimensional

affine subspace if for alli = 1, · · · , k

‖pi − p0‖ ≤ α−1 , pi < Bα

(

p0 + spanp1 − p0, · · · , pi−1 − p0)

. (4.15)

This implies that the vectorspi − p0 are linearly independent in a quantitative way. In particular, weobtain immediately that

Lemma 4.6. If piki=0 α-effectively span the k-dimensional affine subspace

V = p0 + spanp1 − p0, · · · , pk − p0 ,

then for all x∈ V there exists a unique setαiki=1 such that

x = p0 +

k∑

i=0

αi(pi − p0) , |αi | ≤ c(n, α) ‖x− p0‖ . (4.16)

Proof. The proof is quite straightforward. Sincepi − p0ki=1 are linearly independent, we can apply the

Gram-Schmidt orthonormalization process to obtain an orthonormal basise1, · · · , ek for the linear spacespanpi − p0

ki=1. By induction, it is easy to check that for alli

ei =

i∑

j=1

α′j(p j − p0) ,∣

∣

∣α′j

∣

∣

∣ ≤ c(n, α) . (4.17)

Now the estimate follows from the fact that for allx ∈ V

x = p0 +

k∑

i=1

〈x− p0, ei〉ei . (4.18)

With these definitions, we are ready to prove that in caseµ is not too small, then its support must effec-tively span somethingk-dimensional.

Lemma 4.7. Let γk = ωk40−k. There exists aρ0(n, γk,M) = ρ0(n,M) such that if (4.14) holds for someρ ≤ ρ0 and if µ(B1 (0)) ≥ γk, then for every affine subspace V⊆ Rn of dimension≤ k − 1, there exists anx ∈ S such that Bρ (x) ⊆ B1 (0), B10ρ (x) ∩ V = ∅ andµ

(

Bρ (x))

≥ c(n, ρ) = c(n)ρn > 0.


Proof. Let V be anyk − 1-dimensional subspace, and consider the setB11ρ (V). Let Bi = Bρ (xi) be asequence of balls that cover the setB11ρ (V) ∩ B1 (0) and such thatBi/2 ≡ Bρ/2 (xi) are disjoint. IfN is thenumber of these balls, then a standard covering argument gives

Nωnρn/2n ≤ ωk−1(1+ ρ)k−1ωn−k+1(12ρ)n−k+1 ≤ 24nωk−1ωn−k+1ρ

n−k+1

=⇒ N ≤ 48nωk−1ωn−k+1

ωnρ1−k . (4.19)

By (4.14), the measure of the setB11ρ (V) is bounded by

µ(B11ρ (V)) ≤∑

i

µ (Bi) ≤ MNωkρk= 48nωkωk−1ωn−k+1

ωnMρ ≤ 105(50n)nMρ = c(n)Mρ. (4.20)

Thus if

ρ ≤ 10−5(50n)−nγk/(4M) , (4.21)

thenµ(B11ρ (V)) ≤ γk/4. In particular, we get that there must be some point ofS not in B11ρ (V). Moreeffectively, let us consider the setB1 (0) \ B11ρ (V). This set can be covered by at mostc(n, ρ) = 4nρ−n ballsof radiusρ, and we also see that

µ(

B1 (0) \ B11ρ (V))

≥3γk

4. (4.22)

Thus, there must exist at least one ball of radiusρ which is disjoint fromB10ρ (V) and such that

µ(

Bρ (x))

≥3γk

44−nρn ≥ c(n)ρn . (4.23)

Now if at two consecutive scales there are some balls on whichthe measureµ effectively spans k-dimensional subspaces, we show that these subspaces have tobe close together.

Lemma 4.8. Letµ be a positive Radon measure and assumeµ(B1 (0)) ≥ γk and that for each y∈ B1 (0) wehaveµ(Bρ2 (y)) ≤ Mρ2k, whereρ ≤ ρ0. Additionally, let Bρ(x) ⊂ B1 (0) be a ball such thatµ(Bρ (x)) ≥ γkρ

k.Then if A = V(0, 1) ∩ Bρ (x) and B = V(x, ρ) ∩ Bρ (x) are L2-best subspace approximations ofµ withd(x,A) < ρ/2, then

dH(A, B)2 ≤ c(n, ρ,M)(

Dkµ(x, ρ) + Dk

µ(0, 1))

. (4.24)

Proof. Let us begin by observing that ifc(n, ρ,M) > 2δ−1(n, ρ,M), which will be chosen later, then we mayassume without loss of generality that

Dkµ(x, ρ) + Dk

µ(0, 1) ≤ δ = δ(n, ρ,M) , (4.25)

since otherwise (4.24) is trivially satisfied. Moreover, note thatγk >> ǫn, so equation (3.1) is valid onBρ (x)and onB1 (0).

We will estimate the distancedH(A, B) by findingk+ 1 ballsBρ2 (yi) which have enough mass and effec-tively span in the appropriate senseV(x, ρ). Given the upper bounds onDk

µ, we will then be in a position toprove our estimate.


Consider anyBρ2 (y) ⊆ B1 (0) and letp(y) ∈ Bρ2 (y) be the center of mass ofµ restricted to this ball. Letalsoπ(p) be the orthogonal projection ofp ontoV(x, ρ). By Jensen’s inequality:

d(p(y),V(x, ρ))2= d(p(y), π(p(y)))2

= d

?Bρ2(y)

z dµ(z),V(x, ρ)

2

≤1

µ(Bρ2 (y))

∫

Bρ2(y)

d(z,V(x, ρ))2dµ(z) .

(4.26)

Using this estimate and Lemma4.7, we want to prove that there exists a sequence ofk+1 ballsBρ2 (yi) ⊆Bρ (x) such that

(i) µ(

Bρ2 (yi))

≥ c(n, ρ,M) > 0

(ii) π(p(yi ))ki=0 ≡ πi

ki=0 effectively spansV(x, ρ). In other words for alli = 1, · · · , k, πi ∈ V(x, ρ) and

πi < B5ρ2(

π0 + span(π1 − π0, · · · , πi−1 − π0))

. (4.27)

We prove this statement by induction oni = 0, · · · , k. For i = 0, the statement is trivially true sinceµ(Bρ (x)) ≥ γkρ

k. In order to findyi+1, consider thei-dimensional affine subspaceV(i)= span(π0, · · · , πi−1) ≤

V(x, ρ). By Lemma4.7 applied to the ballBρ (x), there exists someBρ2 (yi+1) such thatµ(

Bρ2 (yi+1))

≥

c(n, ρ,M) > 0 and

yi+1 < B10ρ2(

span(π0, · · · , πi−1))

. (4.28)

By definition of center of mass, it is clear thatd(yi , p(yi )) ≤ ρ2. Moreover, by item (i) and equation (4.26),we get

d(p(yi+1),V(x, ρ))2 ≤ c∫

Bρ2(y)

d(z,V(x, ρ))2dµ(z) ≤ cDkµ(x, ρ) ≤ cδ . (4.29)

Thus by the triangle inequality we haved(yi , πi) ≤ 2ρ2 if δ ≤ δ0(n, ρ,M) is small enough. This implies(4.27). Using similar estimates, we also proved(p(yi ),V(0, 1))2 ≤ c′Dk

µ(0, 1) for all i = 0, · · · , k. Thus bythe triangle inequality

d(πi ,V(0, 1)) ≤ d(πi , p(yi)) + d(p(yi ),V(0, 1)) ≤ c(n, ρ,M)(

Dkµ(x, ρ) + Dk

µ(0, 1))1/2

. (4.30)

Now consider anyy ∈ V(x, ρ). By item (ii ) and Lemma4.6, there exists a unique setβiki=1 such that

y = π0 +

k∑

i=0

βi(πi − π0) , |βi | ≤ c(n, ρ) ‖y− π0‖ . (4.31)

Hence for ally ∈ V(x, ρ) ∩ Bρ (xi), we have

d(y,V(0, 1)) ≤ d(π0,V(0, 1))+∑

i

|βi |d(πi − π0,V(0, 1)) ≤ c(n, ρ,M)(

Dkµ(x, ρ) + Dk

µ(0, 1))1/2

. (4.32)

By Lemma4.2, this completes the proof of (4.24).


4.3. Comparison betweenL2 and L∞ planes. Given Br (x), we denote as before byV(x, r) one of thek-dimensional subspace minimizing

∫

Br (x) d(y,V)2dµ. Suppose that the support ofµ satisfies a uniform one-sided Reifenberg condition, i.e. suppose that there existsa k-dimensional planeL(x, r) such thatx ∈ L(x, r)and

supp(µ) ∩ Br (x) ⊆ Bδr (L(x, r)) . (4.33)

Then, by the same technique used in Lemma4.8, we can prove that

Lemma 4.9. Letµ be a positive Radon measure withµ (B1 (0)) ≥ γk and such that for all Bρ (y) ⊆ B1 (0) wehaveµ(Bρ (y)) ≤ Mρk and (4.33). Then

dH(L(0, 1)∩ B1 (0) ,V(0, 1)∩ B1 (0))2 ≤ c(n, ρ,M)(

δ2+ Dk

µ(0, 1))

. (4.34)

4.4. bi-Lipschitz equivalences. In this subsection, we study a particular class of maps with nice localproperties. These maps are a slightly modified version of themaps which are usually exploited to proveReifenberg’s theorem, see for example [Rei60, Tor95, DT12], [Mor66, section 10.5] or [Sim]. The estimatesin this section are standard in literature.

We start by defining the functionsσ. For somer > 0, let xi be anr-separated subset ofRn, i.e.,

(i) d(xi , x j) ≥ r.

Let alsopi be a points inRn with

(ii) pi ∈ B10r (xi)

and letVi be a sequence ofk-dimensional linear subspaces.By standard theory, it is easy to find a locally finite smooth partition of unity λi : Rn→ [0, 1] such that

(iii) supp(λi) ⊆ B3r (xi) for all i,(iv) for all x ∈

⋃

i B2r (xi),∑

i λi(x) = 1 ,(v) supi ‖∇λi‖∞ ≤ c(n)/r ,(vi) if we set 1− ψ(x) =

∑

i λi(x), thenψ is a smooth function with‖∇ψ‖∞ ≤ c(n)/r .

Note that by (iii ), and sincexi is r-separated, there exists a constantc(n) such that for allx, λi(x) > 0 for atmostc(n) different indexes.

For convenience of notation, setπV(v) to be the orthogonal projection onto the linear subspaceV of thefree vectorv, and set

πpi ,Vi (x) = pi + πVi (x− pi) . (4.35)

In other words,πpi ,Vi is the affine projection onto the affine subspacepi +Vi. Recall thatπVi is a linear map,and so the gradients ofπVi and ofπpi ,Vi at every point are equal toπVi .

Definition 4.10. Givenxi , pi , λi satisfying (i) to (vi), and given a family of lineark-dimensional spacesVi,we define a smooth functionσ : Rn→ Rn by

σ(x) = x+∑

i

λi(x)πV⊥i(pi − x) = ψ(x)x+

∑

i

λi(x)πpi ,Vi (x) . (4.36)


By local finiteness, it is evident thatσ is smooth. Moreover, ifψ(x) = 1, thenσ(x) = x. It is clearthat philosophicallyσ is a form of “smooth interpolation” between the identity andthe projections ontothe subspacesVi. It stands to reason that ifVi are all close together, then this mapσ is close to being anorthogonal projection in the region

⋃

i B2r (xi).

Lemma 4.11. Suppose that there exists a k-dimensional linear subspace V⊆ Rn and a point p∈ Rn suchthat for all i

dG(Vi ,V) ≤ δ , d(pi , p+ V) ≤ δ . (4.37)

Then the mapσ restricted to the set U= ψ−1(0) =(∑

i λi)−1 (1) can be written as

σ(x) = πp,V(x) + e(x) , (4.38)

and e(x) is a smooth function with

‖e‖∞ + ‖∇e‖∞ ≤ c(n)δ/r = c(n, r)δ . (4.39)

Remark4.1. Thus, onU we have thatσ is the affine projection ontoV plus an error which is small inC1.

Proof. On the setU, we can define

e(x) = σ(x) − πp,V(x) = −πp,V(x) +∑

i

λi(x) ·(

πpi ,Vi (x))

=

=

∑

i

λi(x) ·(

pi − p− πV(pi − p) + πV(pi) − πVi (pi) + πVi (x) − πV(x))

. (4.40)

By (4.37) and Lemma4.4, we have the estimates

‖pi − p− πV(pi − p)‖ < δ ,∥

∥

∥πV(x− pi) − πVi (x− pi)∥

∥

∥ ≤ c(n)δ ‖x− pi‖ ≤ 13c(n)rδ . (4.41)

This implies

‖e‖L∞(U) ≤ c(n)(1+ 13r)δ ≤ c(n)δ . (4.42)

As for ∇e, we have

∇e=∑

i

∇λi(x) ·(

pi − p− πV(pi − p) + πV(pi) − πVi (pi) + πVi (x) − πV(x))

+

∑

i

λi(x)∇(

πVi (x) − πV(x))

.

(4.43)

The first sum is easily estimated, and since〈∇(πW)|x,w〉 = πW(w), we can still apply Lemma4.4 andconclude:

‖∇e‖L∞(U) ≤c(n)

rδ . (4.44)

As we have seen,σ is in some sense close to the affine projection top + V. In the next lemma, whichis similar in spirit to [Sim, squash lemma], we prove that the image throughσ of a graph overV is again agraph overV with nice bounds.


Lemma 4.12(squash lemma). Fix ρ ≤ 1 and some Br/ρ (y) ⊆ Rn, let I = xi ∩ B5r/ρ (y) be an r-separatedset and defineσ as in Definition4.10. Suppose that there exists a k-dimensional subspace V and somep ∈ Rn such that d(y, p+ V) ≤ δr and for all i:

d(pi , p+ V) ≤ δr and dG(Vi ,V) ≤ δ . (4.45)

Suppose also that there exists a C1 function g: V → V⊥ such that G⊆ Rn is the graph

G = p+ x+ g(x) for x ∈ V ∩ Br/ρ (y) ,

and r−1 ‖g‖∞ + ‖∇g‖∞ ≤ δ′. There exists aδ0(n) > 0 sufficiently small such that ifδ ≤ δ0ρ andδ′ ≤ 1, then

(i) ∀z ∈ G, r−1 |σ(z) − z| ≤ c(n)(δ + δ′)ρ−1, andσ is a C1 diffeomorphism from G to its image,(ii) the setσ(G) is contained in a C1 graphp+ x+ g(x) , x ∈ V with

r−1 ‖g‖∞ + ‖∇g‖∞ ≤ c(n)(δ + δ′)ρ−1 . (4.46)

(iii) moreover, if U′ is such that Bc(δ+δ′)ρ−1 (U′) ⊆ ψ−1(0), then the previous bound is independent ofδ′,in the sense that

r−1 ‖g‖L∞(U′∩V) + ‖∇g‖L∞(U′∩V) ≤ c(n)δρ−1 . (4.47)

For example, ifδ′ ≤ δ0(n)ρ−1, we can take U′ =⋃

i B1.5r (xi).(iv) the mapσ is a bi-Lipschitz equivalence between G andσ(G) with bi-Lipschitz

constant≤ 1+ c(n)(δ + δ′)2ρ−2.

Proof. For convenience, we fixr = 1 andp = 0. By notation, given any mapf : Rn → Rm, p ∈ Rn andw ∈ Tp(Rn) = Rn, we will denote by∇|p f [w] the gradient off evaluated atp and applied to the vectorw.

Recall that

σ(x+ g(x)) = ψ(z)(x+ g(x)) +∑

xi∈I

λi(z)(

πpi ,Vi (x+ g(x)))

, 1− ψ(x) =∑

xi∈I

λi(x) , (4.48)

where we have set for conveniencez= z(x) = x+ g(x). Defineh(x) by

(1− ψ(z))x+ h(x) ≡∑

i

λi(z)(

πpi ,Vi (x+ g(x)))

. (4.49)

Set alsohT(x) = πV(h(x)) andh⊥(x) = πV⊥(h(x)). By projecting the functionσ(x + g(x)) onto V and itsorthogonal complement we obtain

σ(x+ g(x)) ≡ σT(x) + σ⊥(x) ,

σT(x) = x+ hT (x) , σ⊥(x) = ψ(z)g(x) + h⊥(x) . (4.50)

We claim that ifδ′ ≤ 1, then∣

∣

∣hT (x)∣

∣

∣ +

∣

∣

∣∇hT(x)∣

∣

∣ ≤cδρ, (4.51)

where this bound isindependentof δ′ as long asδ′ ≤ 1. Indeed, for allx ∈ V we have

hT(x) = πV

∑

i

λi(z)(

πpi ,Vi (x+ g(x)) − x)

=

∑

i

λi(z)πV

[(

πpi ,Vi (x) − πV(x))

+ πpi ,Vi (g(x))]

(4.52)


Given (4.45) and Lemma4.4, with computations similar to (4.41), we get∣

∣

∣hT (x)∣

∣

∣ ≤ cδ(1+ ρ−1) ≤ cδρ−1. Asfor the gradient, we get for any vectorw ∈ V

∇hT |x[w] = πV

∑

i

∇λi |z[

w+ ∇g|x[w]]

(

πpi ,Vi (x+ g(x)) − x)

+

∑

i

λi(z)(

πVi (w+ ∇g[w]) − w)

, (4.53)

In particular, we obtain∣

∣

∣∇hT |x[w]∣

∣

∣ ≤∑

i

|∇λi | (1+ |∇g|) |w|∣

∣

∣πpi ,Vi (x+ g(x)) − x∣

∣

∣ +

∑

i

λi(z)(∣

∣

∣πVi (w) − w∣

∣

∣ +

∣

∣

∣πVi (∇g[w])∣

∣

∣

)

. (4.54)

For the first term, we can estimate

|∇λi | ≤ c(n) , |∇g| ≤ δ′ ≤ 1 ,∣

∣

∣πpi ,Vi (x+ g(x)) − x∣

∣

∣ ≤∣

∣

∣πpi ,Vi (x) − x∣

∣

∣ +

∣

∣

∣πVi (g(x))∣

∣

∣ . (4.55)

Sincex ∈ V with |x| ≤ ρ−1, andg(x) ∈ V⊥, by (4.45) and Lemma4.4we obtain∣

∣

∣πpi ,Vi (x+ g(x)) − x∣

∣

∣ ≤ cδρ−1 . (4.56)

As for the second term, we have∣

∣

∣

∣

∣

∣πVi (w) − w∣

∣

∣ ≤ cδ |w| ,∣

∣

∣πVi (∇g[w])∣

∣

∣

∣

∣

∣ ≤ cδδ′ |w| ≤ cδ |w| . (4.57)

Summing all the contributions, we obtain (4.51) as wanted.Thus we can apply the inverse function theorem on the function σT(x) : V → V and obtain aC1 inverse

Q such that for allx ∈ V, |Q(x) − x| + |∇Q− id| ≤ c(n)δρ−1 , and ifψ(x+ g(x)) = 1, thenQ(x) = x . So wecan write that for allx ∈ V

σ(x+ g(x)) = σT(x) + g(σT(x)) where g(x) = σ⊥(Q(x)) = h⊥(Q(x)) + ψ (z(Q(x))) g(Q(x)) . (4.58)

Arguing as above, we see thath⊥(x) is aC1 function with

∣

∣

∣h⊥(x)∣

∣

∣ +

∣

∣

∣∇h⊥(x)∣

∣

∣ ≤cδρ, (4.59)

and this bound is independent ofδ′ (as long asδ′ ≤ 1).Thus the function ˜g : V → V⊥ satisfies for allx in its domain

|g(x)| + |∇g(x)| ≤ c(n)(δ + δ′)ρ−1 , (4.60)

Moreover, for thosex such thatψ(Q(x)+g(Q(x))) = 0, the estimates on ˜g are independent ofδ′, in the sensethat |g(x)| + |∇g(x)| ≤ c(n)δρ−1 . Note that by the previous bounds we have

|Q(x) + g(Q(x)) − x| ≤ c(δ + δ′)ρ−1 , (4.61)

and so ifBc(δ+δ′)ρ−1 (U′) ⊂ ψ−1(0), then for allx ∈ U′ ∩ V, ψ(Q(x) + g(Q(x))) = 0. This proves items (ii ),(iii ). As for item (i), it is an easy consequence of the estimates in (4.51), (4.59).

Now since bothG andσ(G) are Lipschitz graphs overV, it is clear that the bi-Lipschitz map induced byπV would have the right bi-Lipschitz estimate. Sinceσ is close toπV, it stands to reason that this propertyremains true. In order to check the estimates, we need to be a bit careful about the horizontal displacementof σ.


bi-Lipschitz estimates In order to prove the estimate in (iv), we show that for allz= x+ g(x) ∈ G and forall unit vectorsw ∈ Tz(G) ⊂ Rn, we have

∣

∣

∣|∇σ|z[w]|2 − 1∣

∣

∣ ≤ c(δ + δ′)2 . (4.62)

First of all, note that ifψ(z) = 1, thenσ is the identity, and there’s nothing to prove.In general, we have that

∇σ|z[w] =

ψ(z)w+∑

i

λi(z)πVi [w]

+

z∇ψ[w] +∑

i

πpi ,Vi (z)∇λi [w]

≡ A+ B . (4.63)

Sinceψ(z) +∑

i λi(z) = 1 everywhere by definition, we have

|B| =

∣

∣

∣

∣

∣

∣

∣

∑

i

(πpi ,Vi (z) − z)∇λi [w]

∣

∣

∣

∣

∣

∣

∣

≤ csupi

∣

∣

∣πpi ,Vi (z) − z∣

∣

∣

≤ cδ′ . (4.64)

This last estimate comes from the fact thatG is the graph ofg over V with ‖g‖∞ ≤ δ′. Moreover, we can

easily improve the estimate forB in the horizontal direction using Lemma4.4. Indeed, sinceπpi ,V(z) − z =πV⊥i

(z), we have

|πVB| =

∣

∣

∣

∣

∣

∣

∣

∑

i

πV

(

πpi ,Vi (z) − z)

∇λi [w]

∣

∣

∣

∣

∣

∣

∣

≤ csupi

∣

∣

∣

∣

πV

(

πpi ,Vi (z) − z)

∣

∣

∣

∣

≤ cδ′δ . (4.65)

As for A, by adapting the proof of Lemma4.11, we get|A− πV[w]| ≤ cδ. Moreover, also in this case we getbetter estimates forA in the horizontal direction. Indeed, we have

|πV(A) − πV[w]| =

∣

∣

∣

∣

∣

∣

∣

ψ(z)πV[w] +∑

i

(

λi(z)πV[πVi [w]])

− πV[w]

∣

∣

∣

∣

∣

∣

∣

=

∣

∣

∣

∣

∣

∣

∣

∑

i

λi(z)(

πV[πVi [w] − πV[w]])

∣

∣

∣

∣

∣

∣

∣

. (4.66)

Now let w = πV[w] + πV⊥[w] = wV + wV⊥. Then we have

|πV(A) − πV[w]| ≤∑

i

λi(z)(∣

∣

∣πV[πVi [wV] − wV]∣

∣

∣ +

∣

∣

∣πV[πVi [wV⊥]]∣

∣

∣

)

=

∑

i

λi(z)(∣

∣

∣

∣

πV[πV⊥i[wV]]

∣

∣

∣

∣

+

∣

∣

∣πV[πVi [wV⊥]]∣

∣

∣

)

.

(4.67)

SinceG is the Lipschitz graph ofg overV with ‖∇g‖ ≤ cδ′, then‖πV⊥[w]‖ ≤ cδ′. Then, by Lemma4.4, wehave

|πV(A) − πV[w]| ≤ c∑

i

λi(z)(

δ2+ δδ′

)

. (4.68)

Summing up, since|πV[w]| ≤ |w| = 1, we obtain that∣

∣

∣|∇σ|z[w]|2 − 1∣

∣

∣ =

∣

∣

∣|πV⊥∇σ|z[w]|2 + |(πV∇σ|z[w] − πV[w]) + πV[w]|2 − 1∣

∣

∣ ≤ (4.69)

≤ c(δ + δ′)2+

∣

∣

∣|πV[w]|2 − 1∣

∣

∣ = c(δ + δ′)2+ |πV⊥[w]|2 ≤ c(δ + δ′)2 . (4.70)


4.5. Pointwise Estimates onD. We wish to see in this subsection how (3.4) implies pointwise estimateson D, which will be convenient in the proof of the generalized Reifenberg results. Indeed, the following isan almost immediate consequence of Remark3.5:

Lemma 4.13. Assume Br(x) ⊆ B2(0) is such thatµ(Br(x)) ≥ 4kǫnrk and∫

B2r(x)Dkµ(y, 2r) dµ(y) < δ2(2r)k.

Then there exists c(n) such that D(x, r) < cδ2. In particular, if (3.4) holds then for every Br(x) ⊆ B2(0) suchthatµ(Br(x)) ≥ 4kǫnrk we have that D(x, r) < cδ2.

5. PROOF OFTHEOREM 3.3: THE RECTIFIABLE-REIFENBERG

Here we carry out the proof of Theorem3.3.In the proof, we will fix the constantM = C1(n) ≤ 40n and therefore a positive scaleρ(n,C1(n)) = ρ(n) <

1 according to Lemma4.7. For convenience, we will assume thatρ = 2−q, q ∈ N, so that we will be able touse the sum bounds (3.9) more easily.

The constantC1(n) will be defined by the end of the proof, however it is enough toknow that it is can betaken to beC1(n) = 40n. Thus, the value of the parameterρ will depend only on the dimensionn, as alreadypointed out in (4.13). Let us also define the scalesr j = ρ

j .

5.1. Weak Upper boundλk(S∩ Br(x)) ≤ C(n)rk. We start by proving a uniform upper bound weaker than(3.8), in particular we want to show that for allx ∈ Rn andr > 0:

λk(S ∩ Br(x)) ≤ C(n)rk . (5.1)

Once we have obtained this estimate, the stronger upper bound and the rectifiability will be almost corollariesof this proof.

Given the scale invariance of the quantities involved, we donot lose generality if we prove (5.1) only forx = 0 andr = 1. The strategy for the proof is the following: first we prove thatS hasσ-finite k-dimensionalHausdorff measure, and use this information to build a suitable covering by balls with controlledλk-measure.Then we fix anyA ∈ N andrA = ρ

A and show by induction onj = A, · · · , 0 that the measure of quasi-ballsBr j (y), defined in (5.11), is bounded above as in (5.1). The definition of a quasi-ball will be such that forr = 1 the quasi-ballB1 (x) agrees with the setS∩ B1(x), up to a set of measure zero, which will prove (5.1).In order to prove the measure statement on the quasi-balls, we will use a second downward induction oni = j, · · · ,A, which is the technical heart of the construction.

σ-finiteness of the measureAs a first step towards the proof, we remark thatS must haveσ-finite k-dimensional Hausdorff measure. We will use this to reduce the proof to the case whenλk(S) < ∞. Indeed,in order for (3.7) to be true, we need in particular

∫

S∩B1(0)Dk

S(x, 1)dλk(x) ds< cδ2 . (5.2)

Define fora > 0 the setsSa =

x ∈ S s.t. DkS(x, 1) ≥ a

andS0 =

x ∈ S s.t. DkS(x, 1) = 0

. Then evi-

dently for alla > 0, λk(Sa) < cδ2/a < ∞. Moreover, ifx ∈ S0, then eitherλk(S ∩ B1 (x)) < ǫn ≤ 1, or up to


sets ofk-measure zero,S ∩ B1 (x) is contained in ak-dimensional plane, and thereforeλk(S0) < ∞. SinceS = S0 ∪

∞i=1 Si−1, S hasσ-finite k-dimensional measure, as claimed.

Now in order to prove the uniform bounds onλk(S), we can therefore assume without loss of generalitythatλk(S) < ∞. Indeed, by the monotonicity of Remark3.4, all subsets ofS satisfy (3.7). Therefore if weshow that (5.1) holds for all subsets of finitek-dimensional measure, then the estimate will hold also onS.Thus, we will assume throughout the remainder of the proof thatλk(S) < ∞.

Covering of the setS⋆ Let S⋆ ⊆ S ∩ B1 (0) be the set of points with controlled upper density, i.e.

S⋆=

x ∈ S ∩ B1 (0) s.t. 2−k ≤ θ⋆k(S, x) ≤ 1

. (5.3)

By proposition2.16, S \ S⋆ has zerok-dimensional measure, thus it is sufficient to give bounds onS⋆.We want to cover the setS⋆ by balls which have uniform upper and lower bounds on their Hausdorff

measure and such that their bestL2 plane is not too far away from their center. In order to achieve this, forall x ∈ S⋆ let rx be such that

rx = ρnx , where nx ∈ N , nx ≥ 2 , (5.4)

λk(

Bρrx (x) ∩ S)

≥1

2k+1ωkρ

k (ρrx)k ,

λk (Br (x) ∩ S) ≤ 2ωkrk , ∀r ≤ rx (5.5)

Let px be the center of mass ofBρrx (x) ∩ S with respect toλk|Bρrx(x)∩S. In particular, we have thatpx ∈ Bρrx (x). By Jensen’s inequality, we have

λk(

Bρrx (x) ∩ S)

d(px,V(x, rx))2 ≤

∫

Bρrx(x)∩Sd(y,V(x, rx))

2 ≤ rk+2x D(x, rx) , (5.6)

where the last inequality comes from Definition3.1, (4.13) andǫn = (1000n)−7n2. Using Lemma4.13and

(3.7) we therefore get that

λk(

Bρrx (x) ∩ S)

d(px,V(x, rx))2 ≤ crk+2

x δ2 . (5.7)

In particular, using our lower bound onλk (

Brx/8 (x) ∩ S)

we obtain thatd(px,V(x, rx)) ≤ cδrx, whichimplies forδ small enough that

d(x,V(x, rx)) ≤ (ρ + cδ)rx ≤ rx/100. (5.8)

Now consider the open covering ofS⋆ given by∪x∈S⋆Brx/5 (x), and extract a countable Vitali subcovering.Thus

S⋆ ⊆⋃

x∈S

Brx (x) ∩ S⋆ , (5.9)

where ifx, y ∈ S, x , y, thenBrx/5 (x) ∩ Bry/5 (y) = ∅.

Fix anyA ∈ N andr = ρA, and define the setsSrr =

x ∈ S s.t. r ≤ rx ≤ r

, Sr = Sρ2

r and

Sr =

⋃

x∈Sr

Brx (x) ∩ S⋆ . (5.10)


It is clear thatSr ր S⋆ as r → 0, so if we have bounds onλk(Sr ) which are independent of ¯r, we aredone. Thus from here on we will consider ¯r = ρA to be positive and fixed.

5.2. First induction: upwards. We are going to prove inductively onj = A, · · · , 0 that for allx ∈ Rn andr j = ρ

j ≤ 1 the measure of

Br (x) =⋃

y∈Sρrr ∩Br (x)

Bry (y) ∩ S⋆ , (5.11)

is bounded by

λk(

Br j (x))

≤ C1(n)rkj ≤ 40nrk

j . (5.12)

Note that forj = A this bound follows from the definition ofrx and (5.5). We emphasize that by constructionBry (y) ∩ S⋆ appears in the union in (5.11) only if r ≤ ry ≤ ρr. However, given (5.4), we have the inclusionSr ∩ B1 (0) ⊆ B1 (0). Note also thatBr (x) ⊆ Br(1+ρ) (x). .

The reason why we have to introduce and estimateBr j (x) instead ofBr j (x) is that we have no a prioricontrol of what happens inside any of the ballsBry (y). The bounds (5.5) are valid only on each ball as awhole. However, since our primary goal is to estimateSr ∩ B1 (0), and we have the inclusionSr ∩ B1 (0) ⊆B1 (0), there is no loss in this restriction.

Rough estimate Fix some j, and suppose that (5.12) holds for allr ≤ rk ≤ r j+1.Let us first observe that we can easily obtain a bad upper boundon λk

(

Bχr j+1 (x))

for some fixedχ > 1.

Consider the pointsy ∈ Sr ∩ Bχr j+1 (x), and divide them into two groups: the ones withry ≤ r j+2 and theones withry ≥ r j+1.

For the first group, cover them by ballsBr j+1 (zi) such thatBr j+1/2 (zi) are disjoint. Since there can be atmostc(n, χ) balls of this form, and for all of these balls the upper bound(5.12) holds, we have an inducedupper bound on the measure of this set.

As for the points withry > r j+2, by construction there can be onlyc(n, χ) many of them, and we also havethe boundry ≤ ρχr j+1, which by (5.5) impliesλk

(

Bry (y))

≤ c(n, χ)rkj+1. Summing up the two contributions,

we get the very rough estimate

λk(Bχr j+1 (x)) ≤ C2(n, χ)rkj+1 , (5.13)

whereC2 >> C1. Note that, as long as the inductive hypothesis holds,C1 and C2 are independent ofj. However, it is clear that successive repetitions of the above estimate will not lead to (5.12). Our goaltherefore is to push down this estimate to arrive at the better constant ofC1, however it will be technicallyvery convenient when applying the tools of Section4 that we may assume the worse bound in the process.

5.3. Second induction: downwards: outline. Suppose that (5.12) is true for all x ∈ B1 (0) and i = j +1, · · · ,A. Fix x ∈ Rn, and consider the set

B = Br j (x) ⊆ B(1+ρ)r j (x) .

We are going to build by induction oni ≥ j a sequence of smooth mapsσi : Rn → Rn and smoothk-dimensional manifoldsT j,i = Ti which satisfy nine properties, which we will use to eventually conclude the


proof of (5.12) for B. Let us outline the inductive procedure now, and introduce all the relevant terminology.Everything described in the remainder of this subsection will be discussed more precisely over the comingpages. To begin with, we will have at the first step that

σ j = id,

T j, j = T j = V(x, r j) ∩ B2r j (x) ⊆ Rn , (5.14)

whereV(x, r j ) is one of thek-dimensional affine subspaces which minimizes∫

Br j (x) d2(y, V) dλk. Thus, the

first manifoldT j is ak-dimensional affine subspace which best approximatesBr j (x). At future steps we canrecoverTi from Ti−1 andσi from the simple relation

Ti = σi(Ti−1) . (5.15)

We will see thatσi is a diffeomorphism when restricted toTi−1, and thus each additional submanifoldTi isalso diffeomorphic toRk. As part of our inductive construction we will build at each stage a covering ofTi

given by

Br j (Ti) ∩ Br j (x) ∼i

⋃

s= j

⋃

y∈I sb

Brs (y)⋃

y∈I sf

(

Brs (y) ∩ Sr)

⋃

y∈I ig

Br i (y) , (5.16)

where given any two distinct ballsB1 andB2 in the covering,B1/5 andB2/5 are disjoint. HereIg, Ib, andI f represent thegood, bad, andfinal balls in the covering. A final ballBr i (y) with y ∈ I i

f is a ball such that

y ∈ S and ry = r i , and the other balls in the covering are characterized as good or bad according to howmuch measure they carry. Good balls are those with large measure, bad balls the ones with small measure.More precisely, we have

λk(Br i (y))

≥ γkrki , if y ∈ I i

g ,

λk(Br i (y))

< γkrki , if y ∈ I i

b , (5.17)

whereγk = ωk40−k. We will see that, over each ballBrs (y) in this covering,Ti can be written as a graphover the best approximating subspaceV(y, rs) with good estimates.

Our goal in these constructions is the proof of (5.12) for the ball B = Br j (x), and thus we will need torelate the submanifoldsTi , and more importantly the covering (5.16), to the setB. Indeed, this covering ofTi almost covers the setB, at least up to an excess setEi−1. That is,

B ⊆ Ei−1 ∪

i⋃

s= j

⋃

y∈I sb

Brs (y)⋃

y∈I sf

(

Brs (y) ∩ Sr)

⋃

y∈I ig

Br i (y) . (5.18)

We will see that the setEi−1 consists of those points ofB which do not satisfy a uniform Reifenberg condi-tion. Thus in order to prove (5.12) we will need to estimate the covering (5.16), as well as the excess setEi−1.

Let us now outline the main properties used in the inductive construction of the mappingσ j+1 : Rn→ Rn,and henceT j+1 = σ j+1(T j). As is suggested in (5.16), it is the good balls and not the bad and final balls whichare subdivided at further steps of the induction procedure.In order to better understand this construction


let us begin by analyzing the good ballsBr i (y) more carefully. On each such ball we may consider thebest approximatingk-dimensional subspaceV(y, r i). SinceBr i (y) is a good ball, one can check that mostof B∩ Br i (y) must satisfy a uniform Reifenberg and reside in a small neighborhood ofV(y, r i). We denotethose points which don’t byE(y, r i), see (5.33) for the precise definition. Then we can define the next stepof the excess set by

Ei = Ei−1

⋃

y∈I ig

E(y, r i ) . (5.19)

Thus our excess set represents all those points which do not lie in an appropriately small neighborhood ofthe submanifoldsTi . With this in hand we can then find a submanifoldT′i ⊆ Ti, which is roughly defined by

T′i ≈ Ti ∩⋃

y∈I ig

Br i (y) , (5.20)

see (5.47) for the precise inductive definition, such that

B ⊆ Ei ∪

i⋃

s= j

⋃

y∈I sb

Brs (y)⋃

y∈I sf

Brs (y)

⋃

Br i+1/4(

T′i)

≡ Ri

⋃

Br i+1/4(

T′i)

, (5.21)

where Ri represents our remainder term, and consists of those balls and sets which will not be furthersubdivided at the next stage of the induction. Now in order tofinish the inductive step of the construction,we can coverBr i+1/4

(

T′i)

by some Vitali set

Br i+1/4(

T′i)

⊆⋃

y∈I

Br i+1 (y) , (5.22)

wherey ∈ I ⊆ T′i . We may then decompose the ball centers

I i+1= I i+1

g ∪ I i+1b ∪ I i+1

f , (5.23)

based on (5.17). Now we will use Definition4.10and the best approximating subspacesV(y, r i+1) to buildσi+1 : Rn→ Rn such that

suppσi+1 − Id ⊆⋃

y∈I i+1g

B3r i+1 (y) . (5.24)

This completes the outline of the inductive construction.

5.4. Second induction: downwards: detailed proof.Let us now describe precisely the proof of thisinductive construction which will lead to (5.12). For j ≤ i ≤ A, we will define a sequence of approximatingmanifoldsTi for the setS and a sequence of smooth mapsσi such that

(i) σ j = id, T j, j = T j = V(x, r j) ∩ B2r j (x) ⊆ Rn,(ii) Ti = σi(Ti−1),(iii) for y ∈ Ti ,

d(σi+1(y), y) ≤ cδr i+1 , (5.25)

andσi+1|Ti is a diffeomorphism,


(iv) for every y ∈ Ti , Ti ∩ B2r i (y) is the graph over somek-dimensional affine subspace of a smoothfunction f satisfying

‖ f ‖∞r i+ ‖∇ f ‖∞ ≤ cδ , (5.26)

As outlined before, the manifoldsTi will be good approximations of the setS up to some “excess” set ofsmall measure. Moreover, we will also introduce the conceptof good, badandfinal balls (whose centerswill be in the setsI i

g, I ib and I i

f ), a remainder set Ri, and the manifoldsT′i ⊆ Ti . Before giving the precisedefinitions (which are in equations (5.44), (5.36), (5.34) and (5.47) respectively), let us group here all theproperties that we will need (and prove) for these objects, so that the reader can always come back to thispage to have a clear picture of what are the objectives of the proof.

(v) for every i ≥ j + 1 andy ∈ I ig, d(y, V(y, r i)) ≤ cδr i , the setTi ∩ B1.5r i (y) is the graph overV(y, r i )

of a smooth functionf satisfying (5.26), whereV(x, r) is one of thek-dimensional affine subspacesminimizing

∫

Br (x) d2(y, V) dλk,(vi) for all i, we have the inclusion

B ⊆ Br i/2

(

T′i−i

)⋃

Ri−1 , (5.27)

(vii) for every i ≥ j + 1 and ally ∈ I ig, the setT′i−1 ∩ B2r i (y) is a Lipschitz graph over the planeV(y, r i )

with d(y, V(y, r i)) ≤ cδr i ,

The last two properties needed are the key for the final volumeestimates:

(viii) we can estimate

λk(σ−1i (T′i )) +

[

#(

I ib

)

+ #(

I if

)]

ωk(r i/10)k ≤ λk(T′i−1) , (5.28)

(ix) we can estimate the excess set by

λk(E(y, r i))

r2i+1 ≤ C(n)Dk

S(y, 2r i )) . (5.29)

Inductive definitions First of all, note that we can assume without loss of generality that

λk(Br j (x)) ≥ γkrkj , (5.30)

otherwise there’s nothing to prove. With this hypothesis, we start our inductive construction by setting

I jg = x , I j

f = I jb = ∅ , T j

= V(x, r j) T′j = T j , σ j = id. (5.31)

Excess set.Let us begin by describing the construction of the excess set. Fix anyy andr i ≥ ρ−1r = ρA−1,

and assume thatBr i (y) satisfiesλk(Br i (y)) ≥ γkrki .

Thus defineV(y, r i ) to be (one of) thek-dimensional plane minimizing∫

Bri (y) d(y, V)2dλk, and define also

the excess set to be the set of points which are some definite amount away from the best planeV. Precisely,

E(y, r i ) = Br i (y) \ Br i+1/4

(

V)

, E(y, r i) =⋃

y∈Sri+2r ∩E(y,r i )

Bry (y)⋂

S⋆ . (5.32)


The points inE are in some sense what prevents the setS from satisfying a uniform Reifenberg conditionon Br i (y). By construction, all points inE have a uniform lower bound on the distance fromV, so that if weassumeλk(Br i (y)) ≥ γkrk

i , i.e. Br i (y) is a good ball, then we can estimate∫

Bri (y)\E(y,r i )d(y, V(y, r i ))

2 dλk(y) + λk(E(y, r i))

(r i+1/5)2 ≤∫

Bri (y)d(y, V(y, r i))

2 dλk(y)

≤

∫

Bri (y)d(y,V(y, r i ))

2 dλk(y) ≤∫

B(1+ρ)ri (y)d(y,V(y, (1+ ρ)r i ))

2

= ((1+ ρ)r i)k+2Dk

S(y, (1+ ρ)r i) ≤ C(n)rk+2i Dk

S(y, 2r i ) . (5.33)

Good, bad and final balls, and remainder setInductively, let us define the remainder set to be the unionof all the previous bad balls, final balls, and the excess sets:

Ri=

i⋃

s= j

⋃

y∈I sb

Brs (y)⋃

y∈I sf

Brs (y)⋃

y∈I sg

E(y, rs)

, Ri=

i⋃

s= j

⋃

y∈I sb

Brs (y)⋃

y∈I sf

(

Brs (y) ∩ Sr)

⋃

y∈I sg

E(y, rs)

. (5.34)

The setRi represents everything we want to throw out at the inductive stage of the proof. We will see laterin the proof how to estimate this remainder set itself.

Note that fori = j, i.e. at the first step of the induction, we have

Rj= E(x, r j ) , Rj

= E(x, r j) . (5.35)

Now consider the balls in the covering outside the remainderset, and separate the balls with radius≥ r i+1

from the others by defining fory ∈ I ig the sets

I i+1f (y) =

z∈(

Sr \ Ri)

∩ Br i (y) s.t. rz = r i+1

, (5.36)

and

Ji+1(y) =

z∈(

Sr i+2r \ Ri

)

∩ Br i (y) ∩ Br i+1/3

(

V(y, r i))

. (5.37)

From this we can construct the sets

I i+1f = ∪y∈I i

gI i+1

f (y) and Ji+1= ∪y∈I i

gJi+1(y) , (5.38)

Note that by construction we have

Sr \ Ri ⊆

(

⋃

z∈I i+1f

Br i+1 (z)⋃

z∈Ji+1

Brz (z))

⋂

Sr . (5.39)

Let us now consider a minimal covering of (5.39) given by

Sr \ Ri ⊆⋃

z∈I i+1f

Br i+1 (z)⋃

z∈I

Br i+1 (z) , (5.40)


whereI ⊆ T′i , and for anyp , q ∈ I i+1f ∪ I , Br i+1/5 (p) ∩ Br i+1/5 (q) = ∅. Note that this second property is

true by definition forp, q ∈ I i+1f , we only need to complete this partial Vitali covering with other balls of the

same size. To be precise, note that by (5.34) and (5.32)

Sr \ Ri \⋃

z∈I i+1f

B4r i+1/5 (z) ⊂

⋃

y∈I ig

Br i+1/4

(

V(y, r i))

⋂

⋃

y∈I ig

Br i (y)

⋂

⋃

z∈I i+1f

B4r i+1/5 (z)

C

. (5.41)

Take a finite covering of this last set by balls

Br i+1/3 (y)

y∈Y. Note that we can pick

Y∩

⋃

z∈I i+1f

Br i+1 (z)

= ∅ . (5.42)

SinceT′i is locally a Lipschitz graph overV(y, r i ) with (5.26), we can chooseY ⊂ T′i . Moreover, since wehave the inclusionSr \ Ri ⊆

⋃

y∈I ig

Br i (y), we can also chooseY ⊂⋃

y∈I ig

Br i (y).Consider a Vitali subcovering of this set, denoteI the set of centers in this subcovering. Such a sub-

covering will have the property that the balls

Br i+1/3 (y)

y∈I will be pairwise disjoint. These balls will alsobe disjoint from

⋃

z∈I i+1f

Brz/5 (z) by (5.42). The (finite version of) Vitali covering theorem ensures that⋃

y∈I Br i+1 (y) will cover the whole set in (5.41).Now by construction ofI f and the remainder set, all the balls

Brx (x)

s∈Srwith rx ≥ r i+1 have already

been accounted for. This means that

Sr \ Ri \⋃

z∈I i+1f

Br i+1 (z) ⊂⋃

y∈I

Br i+1 (y) , (5.43)

as desired.We split the balls with centers inI into two subsets, according to how much measure they carry. In

particular, let

I i+1g =

y ∈ I s.t. λk(

Br i+1 (y))

≥ γkrki+1

, I i+1b =

y ∈ I s.t. λk(

Br i+1 (y))

< γkrki+1

. (5.44)

5.5. Map and manifold structure. Let

λi+1s

= λs be a partition of unity such that for eachys ∈ I i+1g

• supp(λs) ⊆ B3r i+1 (ys)• for all z ∈ ∪ys∈I i+1

gB2r i+1 (ys),

∑

sλs(z) = 1• maxs ‖∇λs‖∞ ≤ C(n)/r i+1.

For everyys ∈ I i+1g , let V(ys, r i+1) to be (one of) thek-dimensional subspace that minimizes

∫

Bri+1(ys)d(z,V)2dλk.

By Remark3.4and by (5.11), we can estimate

r−k−2i+1

∫

Bys(r i+1)d(z, V(ys, r i+1))2dλk(z) ≤ (1+ ρ)k+2D(ys, (1+ ρ)r i+1) . (5.45)

Let ps ∈ B(1+ρ)r i+1 (ys) be the center of mass ofλk|Bri+1(ys). It is worth observing thatps ∈ Vys,r i+1.


Define the smooth functionσi+1 : Rn→ Rn as in Definition4.10, i.e.,

σi+1(x) = x+∑

s

λi+1s (x)πV(ys,r i+1)⊥ (ps − x) . (5.46)

With this function, we can define the sets

Ti+1 = σi+1(Ti) , T′i+1 = σi+1

T′i \⋃

y∈I i+1f

Bry/6 (y)⋃

y∈I i+1b

Br i+1/6 (y)

. (5.47)

Fix anyy ∈ I i+1g , and letz ∈ I i

g be such thaty ∈ Br i (z). By induction,Ti ∩ B10r i+1 (y) ⊆ Ti ∩ B1.5r i (z) is thegraph of aC1 function overV(z, r i ). Consider the pointsys = I i+1

g ∩ B5r i+1 (y). By construction and usingan estimate similar to (5.8), it is easy to see thatd(ys, V(z, r i)) ≤ cδr i+1, and so we can apply the estimates inLemma4.8with M = C1 by the first induction. Using condition (3.7), we obtain that for allys:

r−1i+1dH

(

V(z, r i) ∩ Br i+1 (ys) , V(ys, r i+1) ∩ Br i+1 (ys))

≤ c(

DkS(ys, (1+ ρ)r i+1) + Dk

S(z, (1+ ρ)r i))1/2≤ c(n, ρ,C1)δ .

(5.48)

This implies that, ifδ(n, ρ,C1) is small enough,Ti ∩ B10r i+1 (y) is a graph also overV(y, r i+1) satisfying thesame estimates as in (5.26), up to a worse constantc. That is, ifδ is sufficiently small, we can apply Lemma4.12and prove induction point (v).

It is important to notice that onB1.5r i+1 (y), the bound on the Lipschitz constant of the graph isindependentof the previous bound in the induction step by point (iii ) in Lemma4.12.

Points (iii ) and (iv) Points (iii ) and (iv) are proved with similar methods. We briefly sketch the proofs ofthese two points.

Let y ∈ Ti+1, and recall the functionψi+1 ≡ 1−∑

λs. If ψi+1|B2ri+1(y) is identically 1, thenσi+1|B2ri+1(y) = id,and there is nothing to prove.

Otherwise, there must exist somez′ ∈ I i+1g ∩ B5r i+1 (y), and thus there exists az ∈ I i

g such thatB3r i+1 (y) ⊆B1.5r i (z). By point (iv) in the induction,Ti∩B1.5r i (z) is a Lipschitz graph overV(z, r i ). Proceeding as before,by the estimates in Lemma4.8 and Lemma4.12, we obtain thatTi+1 ∩ B2r i+1 (y) is also a Lipschitz graphoverV(z, r i ) with small Lipschitz constant, and that|σi+1(p) − p| ≤ cδr i+1 for all p ∈ Ti .

Moreover,σi+1|Ti is locally a diffeomorphism at scaler i+1. From this we see thatσi+1 is a diffeomorphismon the wholeTi .

It is worth to remark a subtle point. In order to prove point (iii ), we cannot use inductively (iii ), we needto use point (iv). Indeed, as we have seen, given anyz ∈ I i

g, thenTi ∩ B1.5r i (z) is a Lipschitz graph of afunction f where|∇ f | ≤ cδ, and thisc is independentof the induction step we are considering by (iii ) inLemma4.12. If we tried to iterate directly the bound given by (iii ), the constantc would depend on theinduction stepi, and thus we could not conclude the estimate we want.

Now we turn our attention to the items (vi), (vii ), (viii ).Properties of the manifoldsT′i Here we want to prove the measure estimate in (5.28). The basic idea is that


bad and final balls correspond to holes in the manifoldTi , and each of these holes carries ak-dimensionalmeasure which is proportionate to the measure inside the balls. In particular, letBr i+1 (y) be a bad or a finalball. In either case, we will see thaty must be∼ r i+1-close toTi , which is a Lipschitz graph at scaler i . Thisimplies thatλk(Br i (y) ∩ Ti) ∼ rk

i , and thus we can bound the measure of a bad or final ball with themeasureof the hole we have created onTi .

In detail, point (vi) is an immediate consequence of the definition ofRi. As for point (vii ), if y ∈ I ig, then

by definition, there existsz ∈ I i−1g such thaty ∈ Br i−1 (z), andy < Ri−1. This implies thaty is far away from

the balls we discard while buildingT′i−1, in particular

B3r i (y)⋂

⋃

y∈I i−1f

Bry/6 (y)⋃

I i−1b

Br i−1/6 (y)

= ∅ (5.49)

This proves thatTi−1 ∩ B2r i (y) = T′i−1 ∩ B2r i (y), and in turn point (vii ).In order to prove the volume measure estimate, consider that

T′i \ σ−1i+1(T′i+1) ⊆

⋃

y∈I i+1f ∪I i+1

b

Br i+1/6 (y)

. (5.50)

Note that the balls in the collection

Br i+1/5 (y)

y∈I i+1f ∪I i+1

bare pairwise disjoint. Pick anyy ∈ I i+1

b , and let

z ∈ I ig be such thaty ∈ Br i (z). By definition,y ∈ T′i andλk(Br i+1 (y)) < γkrk

i+1 < 10−kωkrki+1. Sincey < Ri, by

(5.34) Br i+1/6 (y) is disjoint from the set

i⋃

s= j

⋃

y∈I sb

Brs/5 (y)⋃

y∈I sf

Brs/5 (y)

(5.51)

and thusTi ∩ Br i+1/6 (y) = T′i ∩ Br i+1/6 (y). Moreover,Ti ∩ B2r i (z) is a graph overV(z, r i ) with y ∈ T′i ⊂ Ti ,thus (forδ ≤ δ0(n) small enough)

λk(T′i ∩ Br i+1/6 (y)) ≥ ωk7−krk

i+1 . (5.52)

A similar estimate holds for the final balls. The only difference is that ify ∈ I i+1f , then it is not true in

general thaty ∈ Ti . However, by construction of the ballsBry (y), using an argument similar to the one in theproof of (5.7), we obtain thatd(y, V(z, r i)) ≤ cδr i + r i+1/10. Given (5.26), we can conclude

λk(T′i ∩ Br i+1/7 (y)) ≥ ωk10−krki+1 . (5.53)

Now it is evident from the definition ofT′i that

λk(σ−1i (T′i )) +

[

#(

I ib

)

+ #(

I if

)]

ωk(r i/10)k ≤ λk(T′i−1) . (5.54)

5.6. Volume estimates on the manifold part.Here we want to prove that for every measurableΩ ⊆ Ti

λk(σi+1(Ω)) ≤ λk(Ω) + c(n, ρ,C1)∫

S∩B(1+ρ)r j (x)D(p, 4r i+1)dλk(p) . (5.55)

The main applications will be withΩ = Ti ∩ B2r j (x) andΩ = T′i ∩ B2r j (x). In order to do that, we need toanalyze in a quantitative way the bi-Lipschitz correspondence betweenTi andTi+1 given byσi+1.


As we already know,σi+1 = id on the complement of the setG = ∪y∈I i+1g

B5r i+1 (y), so we can concentrateonly on this set.

Using the same techniques as before, and in particular by Lemmas4.8 and4.12, we prove that for eachy ∈ I i+1

g , the setTi ∩ B5r i+1 (y) is a Lipschitz graph overV(y, r i+1) with Lipschitz constant bounded by

c(n, ρ,C1)

D(y, (1+ ρ)r i+1) +∑

z∈I ig∩B5ri (y)

D(z, (1+ ρ)r i )

1/2

(5.56)

In a similar manner, we also have thatTi+1 ∩ B5r i+1 (y) is a Lipschitz graph overV(y, r i+1) with Lipschitzconstant bounded by

c(n, ρ,C1)

∑

z∈I i+1g ∩B10ri+1(y)

D(z, (1+ ρ)r i+1) +∑

z∈I ig∩B5ri (y)

D(z, (1+ ρ)r i)

1/2

. (5.57)

By the bi-Lipschitz estimates in the squash Lemma4.12, we obtain thatσi+1 restricted toTi ∩ B5r i+1 (y) is abi-Lipschitz equivalence with bi-Lipschitz constant bounded by

L(y, 5r i+1) ≤ 1+ c

∑

z∈I i+1g ∩B10ri+1(y)

D(z, (1+ ρ)r i+1) +∑

z∈I ig∩B5ri (y)

D(z, (1+ ρ)r i)

(5.58)

In order to estimate this upper bound, we use an adapted version of (3.2) and the definition of good ballsto write for allz ∈ I i+1

g

D(z, (1+ ρ)r i+1) ≤ c?

Bri+1(z)D(p, 4r i+1)dλk(p) ≤ c(n, ρ,C1)r−k

i+1

∫

Bri+1(z)D(p, 4r i+1)dλk(p) , (5.59)

and a similar statement holds forz ∈ I ig. Since by construction any pointx ∈ Rn can be covered by at most

c(n) different good balls at different scales, we can bound

L(y, 5r i+1) ≤ 1+c(n, ρ,C1)

rki+1

∫

B5ri (y)

[

D(p, 4r i+1) + D(p, 4r i )]

dλk(p) (5.60)

We can also badly estimate

D(p, 4r i+1) + D(p, 4r i) ≤ c(n, ρ)D(p, 4r i ) . (5.61)

Now let Ps be a measurable partition ofΩ ∩G such that for eachs, Ps ⊆ B5r i+1 (ys). By summing up thevolume contributions ofPs, and since evidentlyλk(Ps) ≤ crk

i+1, we get

λk(σi+1(Ω)) =∑

s

λk(σi+1(Ω ∩ Ps)) ≤∑

s

λk(Ps)

1+c

rki+1

∫

B5ri (ys)D(p, 4r i )dλ

k(p)

≤

≤ λk(Ω) + c∫

⋃

ys∈I i+1g

B5ri (ys)D(p, 4r i )dλ

k(p) ≤

≤ λk(Ω) + c(n, ρ,C1)∫

B(1+ρ)r j (x)∩SD(p, 4r i )dλ

k(p) . (5.62)


5.7. Estimates on the excess set.In this paragraph, we estimate the total measure of the excess set, whichis defined by

ET =

A⋃

i= j

⋃

y∈I ig

E(y, r i) . (5.63)

At eachy and at each scaler i such thatλk(Br i (y)) ≥ γkrki , we have by (5.33) and (3.2)

λk(E(y, r i)) ≤ c(n, ρ)rki Dk

S(y, 2r i ) ≤ c(n, ρ)rki

?B2ri (y)∩S

DkS(p, 4r i )dλ

k(p) ≤ c(n, ρ)∫

B2ri (y)∩SDk

S(p, 4r i )dλk(p)

(5.64)

Now by construction of the good balls, there exists a constant c(n) such that at each stepi, eachx ∈ Rn

belongs to at mostc(n) many balls of the form

B2r i (y)

y∈I ig. Thus for eachi ≥ j, we have

∑

y∈I ig

λk(E(y, r i )) ≤ c(n, ρ)∫

∪y∈I igB2ri (y)

DkS(p, 4r i )dλ

k(p) ≤ c(n, ρ)∫

B2r j (x)∩SDk

S(p, 4r i )dλk(p) . (5.65)

If we sum over all scales, we get

λk(

ET

)

≤ c(n, ρ)A

∑

i= j

∫

B2r j (x)∩SDk

S(p, 4r i)dλk(p) . (5.66)

Sinceρ = 2−q, it is clear that

λk(

ET

)

≤ c(n, ρ)A

∑

i= j

∫

B2r j (x)∩SDk

S

(

p, 22−qi)

dλk(p) ≤ c(n, ρ)δrkj , (5.67)

since the sum in the middle is clearly bounded by (3.9).

This estimate is exactly what we want from the excess set.

5.8. Completion of the weak upper bounds.By adding (5.55), withΩ ≡ σ−1i+1(T′i+1), and (5.28), we prove

that for all i = j, · · · ,A+ 1, ...

λk(T′i+1) +[

#(

I ib

)

+ #(

I if

)]

ωk(r i/10)k ≤ λk(T′i ) + c(n, ρ,C1)∫

B4r j (y)∩Sr

D(p, 2r i )dλk(p) . (5.68)

Adding the contributions from all scales, by (3.9) we get

λk(T′i+1) +i

∑

s= j

[

#(

I sb

)

+ #(

I sf

)]

ωk(rs/10)k ≤

≤ λk(T j ∩ B2r j (x)) + c(n, ρ,C1)i

∑

s= j

∫

B2r j (x)∩Sr

D(p, 2rs)dλk(p) ≤

≤ λk(

T j ∩ B2r j (x)) [

1+ c(n, ρ,C1)δ2]

, (5.69)

where in the last line we estimatedλk(

T j ∩ B2r j (x))

∼ rkj , sinceT j is a k-dimensional subspace, and we

bounded the sum using (3.9).


In the same way, we can also bound the measure ofTi by

λk(Ti+1) ≤ λk(

T j ∩ B2r j (x)) [

1+ c(n, ρ,C1)δ2]

≤ c(n)rkj . (5.70)

Upper estimates forλk. Given the definition ofSr , all the ballsBry (y) inside this set havery ≥ rA. Thusby (5.27)

B = Br j (x) ⊆ RA . (5.71)

In particular, this and the estimates in (5.67) and (5.69) imply

λk(B) ≤A

∑

s= j

[

γk#(

I sb

)

+ 2ωk#(

I sf

)]

(rs)k+ λk(ET) ≤

≤ C3(n)i

∑

s= j

[

#(

I sb

)

+ #(

I sf

)]

ωk(rs/10)k + λk(ET) ≤ C3(n)(1+ c(n, ρ,C1)δ)rkj . (5.72)

In this last estimate, we can fixC1(n) = 2C3(n) ≤ 40n, andρ(n,C1) according to Lemma4.7. Now, it is easyto see that ifδ(n, ρ,C1) is sufficiently small, then

λk(B) ≤ C1(n)rkj , (5.73)

which finishes the proof of the downward induction, and hencethe actual ball estimate (5.1).

5.9. Rectifiability, W1,p and improved measure bounds.We have now proved a mass bound for the setS. In this subsection, we wish to improve this mass bound, as well as prove the rectifiability conditions onS. At several stages we will have to repeat the arguments of theupward and downward inductions, whichwe will only sketch since the arguments will be almost verbatim, though in some cases technically mucheasier.

Rectifiability and W1,p estimates Now that we have proved the bound (5.1), we can use a constructionsimilar to the one just described to sharpen the upper estimate and prove rectifiability ofλk|S. The maindifference with the previous case is that we do not need to be concerned any more with any fixed coveringBry (y) of our set, and so at every step we can estimate directly the measureλk|S of a whole Euclidean ballBr (z) without the need to limit our estimate toBr (z). For the same reason, we do not need to introduce thesubspacesV(z, r) and the setsE, R, we will only deal withV(z, r),E,R. Moreover, since we don’t have tostop our construction at any positive scalery, we do not need to introduce and study the set of final ballsI i

f .As a consequence, the construction we need here is technically less involved than the one used before.

However, as opposed to the previous construction, we are concerned about what happens at an infinitesi-mal scale, and in particular we want to have uniform estimates also for the limit ¯r → 0 of various quantities.

Fix anyx ∈ Rn andr > 0 such thatBr (x) ⊆ B1 (0). We are going to prove (3.8), i.e.

λk (S ∩ Br (x)) ≤ (1+ ǫ)ωkrk , (5.74)


whereǫ > 0 is the arbitrary constant chosen at the beginning of Theorem 3.3.For convenience and wlog, we assumex = 0 andr = 1. Using the same technique as before, we build a

sequence of smooth mapsσi∞i=0 onRn and manifoldsTi such that

(i) σ0 = id, T0 = V(x, r) = V(0, 1) ⊆ Rn andφi = σi σi−1 · · ·σ0,(ii) Ti = σi(Ti−1) = φi(T0) is a smoothk-dimensional submanifold ofRn,(iii) for y ∈ Ti

d(σi+1(y), y) ≤ cδr i+1 , (5.75)

andσi+1|Ti is a diffeomorphism. In a similar way for ally ∈ T0, d(φi (y), y) ≤ cδ andφi |T0 is adiffeomorphism,

(iv) for every y ∈ Ti , Ti ∩ B2r i (y) is the graph over somek-dimensional affine subspace of a smoothfunction f satisfying

‖ f ‖∞r i+ ‖∇ f ‖∞ ≤ cδ , (5.76)

In order to do so, we define inductively oni = 0, · · · ,∞ a sequence of setsE(y, r i), I ig, I

ib,R

i and manifoldsT′i ⊆ Ti such that

(v) for everyi ≥ 1 and ally ∈ I ig, the setT′i−1 ∩ B1.5r i (y) is a Lipschitz graph over the planeV(y, r i) with

d(y,V(y, r i )) ≤ cδr i

(vi) for all i, we have the inclusion

S ∩ B1 (0) ⊆ Br i/2

(

T′i−1

)

∪ Ri−1 (5.77)

(vii) for every i ≥ j + 1 and ally ∈ I ig, the setT′i−1 ∩ B2r i (y) is a Lipschitz graph over the planeV(y, r i )

with d(y, V(y, r i)) ≤ cδr i ,(viii) we can estimate

λk(σ−1i (T′i )) + #

(

I ib

)

ωk(r i/10)k ≤ λk(T′i−1) . (5.78)

Up to minor modifications (actually simplifications), all these properties are proved in the same way asin the downward induction from the previous subsection.

The key extra-property we need is some form of control over the W1,p norm ofφi . In particular we willprove inductively that for allj ∈ N

∥

∥

∥∇φ j

∥

∥

∥

p

Lp =

?T0∩B2(0)

∣

∣

∣∇φ j

∣

∣

∣

pdλk ≤ 1+ c(n)2pδ2 , (5.79)

where hereλk is also Lebesgue measure onRk.As we will see in (5.98), in order for this estimate to work, we will need to chooseδ(n, ρ,C1, p) small

enough, and in particularδ→ 0 asp→ ∞.

Proof of the first points As mentioned before, the proof of items (i)-(viii) can be carried out in the sameway as in Section5.4, and we take from this section also the definitions of the setsI i

g, I ib, E(x, r) andRi, up


to replacingBr (y) with Br (y) andV(y, r) with V(y, r). Recall also that in this case we have no final balls, sowe can just assume thatI i

f = ∅ for all i, and that in this case we havex = 0, r = 1 and soj = 0.For convenience, we recall the definition of the excess and remainder sets:

E(y, r i ) = Br i (y) \ Br i+1/4 (V(y, r i)) , Ri=

i⋃

s=0

⋃

y∈I sb

Brs (y)⋃

y∈I sg

E(y, rs)

. (5.80)

We also introduce the notation

Ei=

i⋃

s=0

⋃

y∈I sg

E(y, rs) , Ri= Ei

i⋃

s=0

⋃

y∈I sb

Brs (y) . (5.81)

We briefly sketch again the main steps in the construction. Assuming wlog thatB1 (0) is a good ball, i.e.,thatλk(B1 (0)) ≥ γk, we first estimate the excess set on this ball. Since this set is the set of points which aresome definite amount away from the best planeV(0, 1), the definition ofD immediately gives the followingestimate, similar to (5.33):

λk(E(0, 1))

(ρ/5)2 ≤ DkS(0, 1) . (5.82)

Then we cover the non-excess part with a Vitali-type covering by ballsBρ (xi)i∈I centered on the planeV(0, 1). We classify the balls in this covering into good and bad balls, according to how much mass theycarry. Br (x) is a good ball ifλk(S ∩ Br (x)) ≥ γkrk, otherwise it’s a bad ball.

A good ball carries enough measure to apply Lemma4.8, and compare the best subspaceV(0, 1) with thenew best subspaceV(xi , ρ).

We setσ1 to be the map defined in4.10, i.e., an interpolation among all the projections ontoV(xi , ρ)i∈I1g,

and we also setT1 = σ1(T0) = σ1(V(0, 1) ∩ B1+cδ (0)). By the squash lemma4.12, σ1|T0 is a smoothdiffeomorphisms, andT1 is locally at scaler1 = ρ

1 the Lipschitz graph of a function with small Lipschitzbounds. Moreover,σ1 is a bi-Lipschitz equivalence with quantitative estimateson the Lipschitz constant. Inparticular, we get that for ally ∈ B1 (0) ∩ T0, the following version of the estimates in (5.58) holds:

L(y, 5ρ) = max

‖∇σ1‖L∞(T0∩B5ρ(y)) , ‖∇σ1‖L∞(σ1(T0∩B5ρ(y)))

≤ 1+ c

∑

z∈I1g∩B10ρ(y)

D(z, ρ) + D(0, 1)

(5.83)

In order to keep track of the measure inside the bad balls, we define the manifold “with holes”

T1 = σ1(T0) , T′1 = σ1

T0 \⋃

z∈I1b

Bρ/6 (z)

. (5.84)

SinceV(0, 1) is ak-plane, and by definition of bad balls, each “hole” in the manifold T′1 carries more measurethan the corresponding bad ball which created it, giving us the estimate

λk(σ−11 (T′1)) + λk

⋃

z∈I1b

S ∩ Bρ (z)

≤ λk(σ−11 (T′1)) + #

(

I1b

)

ωk(ρ/9)k ≤ λk(T0) . (5.85)


Now we repeat the construction inductively on the scalesr2 = ρ2, r3, · · · , and we obtain all the desiredproperties. Moreover, by summing (5.82) and its iterations, we obtain the following estimate for the totalexcess set:

λk (

E∞)

≤ c(n, ρ)δ2 , (5.86)

which is the equivalent of (5.67).

We conclude by noting the following:σ|i(z) = id for all z ∈ T0 ∩

(

∪ij=0

⋃

z∈I jbBr i/2 (z)

)

. In other words,

once a hole is created, it never changes. This implies that the iterations of (5.85) lead to

λk(

T′i ∩ B1+cδ (0))

+ 9−kγ−1k

i∑

j=0

∑

y∈I jb

λk(S ∩ Br j (y)) ≤ λk(Ti ∩ B1+cδ (0)) . (5.87)

W1,p estimates As part of the downward induction in the proof of (i)-(viii),let us define the mapsτi, j =

σ j σ j−1 · · · σi, so thatτi, j(Ti) = T j+1 andτ0, j = φ j . We will prove inductively fori = j, · · · , 0 that forall x ∈ Ti ,

r−ki

∫

Bri (x)∩Ti

∣

∣

∣∇τi, j

∣

∣

∣

pdλk ≤ 2C1(n) . (5.88)

Here the integration is simply the integration on a smoothk-dimensional subset wrt thek-dimensionalLebesgue measureλk, and the gradient of the functionsσi andτi, j is the restriction of the gradient inRn tothe corresponding manifoldTi .

Suppose that the statement is true fori + 1. Consider a covering of this ball by

Br i+1 (ys)

, then by thechain rule

∫

Bri (x)∩Ti

∣

∣

∣∇τi, j

∣

∣

∣

pdλk(y) ≤

∑

s

∫

Bri+1(ys)∩Ti

∣

∣

∣∇τi+1, j |σi (y)

∣

∣

∣

p|∇σi(y)|p dλk(y) ≤ C(n, p, ρ) . (5.89)

This gives us a first rough estimate.In order to obtain a better estimate, we will prove by induction ons= i, · · · , j that

∫

Bri (x)∩Ti

∣

∣

∣∇τi, j

∣

∣

∣

pdλk(y) ≤

∫

Bs

∣

∣

∣∇τs, j

∣

∣

∣

pdλk(y) + c(n, ρ)2p

s∑

t=i

∫

B10ri (x)∩SD(p, 10rt)dλ

k(p) , (5.90)

whereBs ≡ τi,s−1(Ti ∩ Br i (x)) ⊆ Ts ∩ B(1+cδ)r i (x). Indeed, suppose that the statement is true fors, then wehave

∫

Bs

∣

∣

∣∇τs, j

∣

∣

∣

pdλk(y) =

∫

Bs

∣

∣

∣∇τs+1, j |σs(y)

∣

∣

∣

p|∇σi(y)|p dλk(y) . (5.91)

Sinceσi is a diffeomorphism, we can change the variables and write∫

Bs

∣

∣

∣∇τs, j

∣

∣

∣

pdλk(y) =

∫

Bs+1

∣

∣

∣∇τs+1, j |y∣

∣

∣

p ∣

∣

∣∇σs|σ−1s (y)

∣

∣

∣

pdetJ(σs) dλk(y) . (5.92)

Now consider the partial covering of this set given by

B5rs (yt)

yt∈I sg. As we have seen before, outside of this

set,σs is the identity, so we don’t need to make any estimates on it.Arguing in a manner verbatim to the proof of (5.55), we can prove bi-Lipschitz estimates forσ|s. In

particular, we can use the definition of good balls and the bounds in Lemma4.8to estimate in a quantitative


way the distance between best subspaces at nearby points andscales. This allows us to use the squashLemma4.12and prove thatσs restricted toTs∩ B5rs+1 (yt) is a bi-Lipschitz map with

∥

∥

∥

∣

∣

∣∇σs|Ts

∣

∣

∣ − 1∥

∥

∥

L∞(Ts∩B5rs+1(yt))≤

c(n, ρ)

rks

∫

B5rs(yt)∩SD(p, 10rs)dλ

k(p) . (5.93)

Let Pt,Q be a measurable partition ofBs with Pt ⊆ B5rs+1 (yt) andQ ⊆ ∩B5rs+1 (yt)c, thenσs(Pt), σs(Q)

is a measurable partition ofBs+1 with σs(Q) = Q. So we get∫

Bs

∣

∣

∣∇τs, j

∣

∣

∣

pdλk(y) =

∫

Q

∣

∣

∣∇τs, j

∣

∣

∣

pdλk(y) +

∑

t

∫

σ(Pt)

∣

∣

∣∇τs+1, j |y∣

∣

∣

p ∣

∣

∣∇σs|σ−1s (y)

∣

∣

∣

pdetJ(σs) dλk(y) ≤

≤

∫

Q

∣

∣

∣∇τs+1, j

∣

∣

∣

pdλk(y) +

∑

t

∫

σ(Pt)

∣

∣

∣∇τs+1, j

∣

∣

∣

pdλk

1+c

rks+1

∫

B5rs(yt)∩SD(p, 10rs)dλ

k(p)

. (5.94)

Note that these estimates are basically the same as the ones in (5.62).By the first induction, we have the upper bound

∫

Ps

∣

∣

∣∇τs+1, j |y∣

∣

∣

p≤ c(n)rk

s+1, and thus∫

Bs

∣

∣

∣∇τs, j

∣

∣

∣

pdλk(y) ≤

∫

Bs+1

∣

∣

∣∇τs+1, j

∣

∣

∣


∑

t

∫

B5rs+1(yt)∩SD(p, 10rs)dλ

k(p) . (5.95)

Since all points inRn are covered at mostc(n, ρ) times by

B5rs+1 (yt)

t, we can simply estimate∫

Bs

∣

∣

∣∇τs, j

∣

∣

∣

pdλk(y) ≤

∫

Bs+1

∣

∣

∣∇τs+1, j

∣

∣

∣


∫

B10ri (x)∩SD(p, 10rs)dλ

k(p) . (5.96)

This proves (5.90). In order to complete the first induction, observe that

∫

Bri (x)∩Ti

∣

∣

∣∇τi, j

∣

∣

∣

pdλk(y) ≤

∫

Bs

∣

∣

∣∇τ j, j

∣

∣

∣


j∑

t=i

∫

B2ri (x)∩SD(p, 10rt)dλ

k(p) , (5.97)

Sinceτ j, j is the identity onT j , and given the measure bounds (5.70), we get the result if we chooseδ(n, ρ, p)such that the rhs is small enough, and in particular

c(n, ρ)2pA

∑

j=i

∫

S∩B2ri (x)D(p, 10r i−1)dλk(p) ≤ c(n, ρ)2pδ2 ≤ C1(n) . (5.98)

Note that, once the improved estimate (3.8) is proven (which will be done in the next subsection) it willlikewise be possible to improve thisW1,p estimate to

1

ωkrki

∫

Bri (x)∩Ti

∣

∣

∣∇τi, j

∣

∣

∣

pdλk ≤ 1+ ǫ . (5.99)

Improved measure estimatesConsider the mapsφ j : T0 → T j+1. Sinceφ j has uniformW1,p estimatesand by point (iii ), thenφ j → φ∞ uniformly, with

∫

T0∩B2(0)|∇φ∞|

p ≤ 2C1(n) . (5.100)


By (5.77), we have the inclusion for alli:

S ⊆ Ei

⋃

Br i+1/2

(

T′i)

i⋃

j=0

⋃

y∈I jb

Br j (y) , =⇒ S ∩ B1 (0) ⊆ E∞⋃

(

T′∞ ∩ B1 (0))

∞⋃

j=0

⋃

y∈I jb

Br j (y) . (5.101)

By (5.87) and (5.86), we obtain that

λk (S ∩ B1 (0)) ≤ λk(T∞) + λk(E∞) ≤ λk(T∞) + cδ2 . (5.102)

In order to give better estimates onT∞, note thatT∞ ∩ B1 (0) ⊆ φ(T0 ∩ B1+cδ (0)). Define the functionf : T0→ R by

f (x) =∞∑

i=0

supBri (φi(x))∩Ti

||∇σi | − 1| . (5.103)

By the usual double-induction argument, see for example theproofs of (5.73) and (5.88), we can estimate∫

T0∩B1+cδ(0)f (x)dλk ≤ cδ2 . (5.104)

Define the setsUa ⊆ T0 by

Ua = x ∈ T0 ∩ B1+cδ (0) s.t. f (x) > a . (5.105)

By simpleL1 estimates, we know thatλk(Ua) ≤ cδ2

a , and by Lemma2.20for p > k we get

λk(φ∞(Ua)) ≤ C(n, p) ‖∇φ∞‖kLp λ

k(Ua)p−kp ≤ C

(

cδ2

a

)1−k/p

. (5.106)

Note that, on the complement ofUa, φ∞ is a Lipschitz function with‖∇φ∞‖∞ ≤ ea+ cδ. Indeed, take any

x, y ∈ Uca, and leti be such thatr i+1/2 < |x− y| ≤ r i/2. We prove that

|φ∞(x) − φ∞(y)| ≤ (ea+ cδ) |x− y| . (5.107)

By definition,φ∞(z) = τi∞(φi(z)), and by (5.75), we easily get

|τi∞(φi(x)) − φi(x)| + |τi∞(φi(y)) − φi(y)| ≤ cδr i . (5.108)

Moreover, the definition off immediately implies a uniform Lipschitz condition onφi . Indeed,

‖∇φi‖L∞(Bri (x)) ≤

i∏

s=0

‖∇σs‖L∞(Brs(φs(x))) ≤ exp

s∑

i=0

log(

‖∇σs‖L∞(Brs(φs(x)))

)

≤ ea . (5.109)

Thus we get

|ψ(x) − ψ(y)| ≤ cδ |x− y| + ea |x− y| , (5.110)

as desired.By choosinga(ǫ) sufficiently small, andδ(n, ρ, p, ǫ) sufficiently small as well, we get

λk(T∞ ∩ B1 (0)) ≤ λk(

φ∞(Ua) ∪ φ∞(UCa )

)

≤

(

1+12ǫ

)

ωk . (5.111)

Summing up all the estimates, we prove the sharpened upper bound

λk(S ∩ B1 (0)) ≤ λk(E∞) + λk(φ∞(T0 ∩ B1+cδ (0))) ≤ (1+ ǫ)ωk . (5.112)


Rectifiability As for the rectifiability, we can restrict ourselves toS⋆ and consider a covering of this setmade by balls

Brx (x)

x∈S⋆ such thatλk(Brx (x) ∩ S) ≥ 2−k−1ωkrkx.

If we prove that for all such balls there exists a subsetG ⊆ S ∩ Brx (x) which is rectifiable and for whichλk(G) ≥ λk(Brx (x) ∩ S)/2, then an easy covering argument gives us the rectifiabilityof all of S.

By scale invariance, we assume for simplicityx = 0 andrx = 1. As we have seen before,

S ⊆ R∞ ∪ T∞ . (5.113)

SinceT∞ is the image of aW1,p map withp > k, this set is rectifiable by Lemma2.19. Moreover, by (5.87)and (5.111), we know that

i∑

j=0

∑

y∈I jb

λk(S ∩ Br j (y)) ≤ 10kγk (5.114)

Given the estimates on the excess set given in (5.86), we obtain the lower bound

λk(S ∩ T∞) ≥ λk(S) − λk(R∞) ≥ 2−k−1ωk − γk10k − cδ ≥ 2−k−2ωk , (5.115)

which therefore completes the proof.

6. PROOF OFTHEOREMS3.2 AND 3.4

As it will be clear, the proofs of these theorems are simple modifications of the proof of Theorem3.3,and actually from the technical point of view they are a lot simpler. For this reason, we will simply outlinethem, pointing out the main differences needed in these cases.

6.1. Proof of Theorem 3.2: The W1,p-Reifenberg. The proof of this theorem is almost a corollary ofTheorem3.3 and the classic Reifenberg theorem2.18. In addition to the upper volume bound proved inTheorem3.3, we also need to prove a lower volume bound on each ball.

In fact, one could just trace through the argument of Theorem3.3, and see inductively that there exists nobad balls to conclude this. Instead, we will use a standard argument (see for example [DT12, lemma 13.2])to prove the lower bound onλk(S ∩ Br (x)) given in (3.6) directly by using only the uniform Reifenbergcondition.

Lemma 6.1. Under the assumptions of Theorem2.18, for all x ∈ S such that Br (x) ⊆ B1 (0),

λk(S ∩ Br (x)) ≥ (1− cδ)ωkrk . (6.1)

Proof. By scale invariance, we assumex = 0 andr = 1. The classic Reifenberg theorem proves that thereexits a bi-Holder continuous mapφ : L→ Rn whereL is ak-dimensional plane and

(1) |φ(x) − x| ≤ cδ for all x ∈ L(2) φ(x) = x for |x| ≥ 1+ cδ(3) S ∩ B1 (0) = φ(L′), whereB1−cδ (0) ∩ L ⊆ L′ ⊆ B1+cδ (0) ∩ L.


Now let f = πL φ : L → L. This map is continuous and it is the identity outsideB1+cδ (0), and thus bytopological reasons (degree theory) it is also surjective from L to itself.

In particular, the setA = B1−3cδ (0) ∩ L is contained in the image off . By the uniform Reifenbergcondition,π−1

L (A) ∩ S ⊆ B1−2cδ (0), and by the properties ofφ, f −1(A) = φ−1(π−1L (A) ∩ S) ⊆ B1−cδ (0). Thus

φ( f −1(A)) ⊆ S.Now, sinceπL has Lipschitz constant 1, by a standard result (see [Fed69, 2.10.11])

λk(S) ≥ λk(φ( f −1(A)) ≥ λk(πL φ( f −1(A))) = λk(A) ≥ (1− 3cδ)kωkrk . (6.2)

Forδ small enough, we have the thesis.

The upper bound onλk(S) and the rectifiability are direct consequences of Theorem3.3. The only thingleft to prove is the statement aboutW1,p bounds on the map. This can be obtained by simple modifications(actually simplifications) in the proof of Theorem3.3.

As before, we denote byV(x, r) (one of) thek-dimensional subspace that minimizes∫

Br (x) d(y,V)2, and wesetL(x, r) to be one of thek-dimensional subspaces satisfying the Reifenberg condition, i.e., ak-dimensionalsubspace such thatdH(S ∩ Br (x) , L ∩ Br (x)) < δr.

First of all, note that for anyx ∈ S, r > 0, λk(Br (x)) > ωkrk/2, and so there are no bad balls in ourcovering. Thus, by Lemma4.9, we can conclude that

dH(V(x, r) ∩ Br (x) , L(x, r) ∩ Br (x)) ≤ cδ . (6.3)

As a consequence, forδ(n, ρ) small enough, all the excess sets at all scales are empty. Indeed, by (5.32),

E(x, r) = Br (x) \ Bρr/3 (V(x, r)) ∩ S ⊆ Br (x) \ B(ρ/3−cδ)r (L(x, r)) ∩ S = ∅ . (6.4)

Proceeding with the same construction as in Section5.9, we obtain a sequence of mapsφ j : T0 → Rn,φ j(T0) ≡ T j+1 converging inW1,p to someφ∞ such that for alli, S ⊆ Br i (Ti) ⊆ Br i−1 (Ti−1), thus provingthatS ⊆ T∞ ≡ φ∞(T0).

W1,p estimates for the inverseHere we want to prove that the mapφ−1∞ : S → T is also aW1,p map with

bounds. Note thatS equipped with the Euclidean distance and thek-dimensional Hausdorff measure is aAhlfors regular metric measure space, in the sense that there exists aC1(n) such that for allx ∈ S ∩ B1 (0)andr ≤ 1,

C−11 rk ≤ λk(S ∩ Br (x)) ≤ C1rk , (6.5)

whereBr (x) is the usual Euclidean ball inRn. On such spaces there are several methods of defining thespace ofW1,p maps, for instance as the closure of the lipschitz functionsunder theW1,p norm. For rectifiablespaces all such definitions are classically understood to beequivalent.

Note that sinceS is rectifiable, we can use the integration by substitution towrite∫

Sf (z)dλk(z) =

∫

φ−1∞ (S)

f (φ∞(x)) J(φ∞)|xdλk(x) , (6.6)

which will allow us to easily study integrals on our rectifiable spaces.


The next Lemma tells us that the mappingφ−1∞ is approximated by a sequence of Lipschitz maps whose

gradients form a Cauchy sequence inLp and have our desired estimates. In particular, the following Lemmawill finish our proof of theW1,p-Reifenberg result. For convenience, we introduce the notation ψ ≡ φ−1

∞ :

Lemma 6.2. There exists a sequence of functionsψt such that

(1) ψt are Lipschitz functions inRn, with Lipschitz constant bounded by c(n)t,(2) if Rt = z ∈ S s.t. ψt(z) , ψ(z), thenλk(Rt)→ 0,(3) ψt converges uniformly toψ,(4) the sequence∇ψt is a Cauchy sequence in Lp(S).

Moreover, there exists a g∈ Lp(S) such that for all t,|∇ψt | ≤ g λk-a.e. on S , and the following is valid forall t:

∫

S∩B2(0)|∇ψt |

p ≤ C(n) ,?

S∩B1(0)|∇ψt |

p ≤ 1+ ǫ . (6.7)

Proof. The proof is a standard consequence of the properties provedso far for the mapφ and the usualLusin-type approximation forW1,p functions (see for example [EG92, theorem 3, sec 6.6.3]).

First of all, we fix some notation. Given the mapsφi : T → Ti andφ∞ : T → S, we set

ψi ≡ φ−1i : Ti → T , αi ≡ φ∞ ψi : Ti → S , βi ≡ α

−1i : S→ Ti . (6.8)

Note that, by the Reifenberg construction, all these maps are Holder continuous maps.Next, we introduce a slight variant of the functionf (x) defined before. In particular, set forz ∈ S:

g(z) =∞∏

i=0

supy∈B3ri (βi (z))∩Ti

∣

∣

∣∇σ−1i

∣

∣

∣ . (6.9)

By adapting the proof of theW1,p estimates in (5.99), we prove thatg ∈ Lp(S) with∫

S∩B2(0)g(z)pdλk(z) ≤

∫

T∩B2.1(0)g(φ(x))pJ(φ∞)|xdλ

k(x) ≤ 3C1(n) . (6.10)

Moreover, we can also refine the bound to∫

S∩B1(0)g(z)pdλk(z) =

∫

T∩ψ(S∩B1(0))g(φ(x))pJ(φ∞)|xdλ

k(x) ≤ ωk(1+ ǫ) . (6.11)

Now define fort ≥ 1 the setsRt = g−1[0, t]. Sinceg ∈ Lp, then

lim supt→∞

tpλk(Rt) ≤ lim supt→∞

∫

g≥tgp→ 0 . (6.12)

Moreover,ψ|Rt is a Lipschitz function with Lipschitz constant bounded byc(n)t. Indeed, letx, y ∈ Rt, andseti to be such thatr i+1/2 < |x− y| ≤ r i/2. We prove that

|ψ(x) − ψ(y)| ≤ c(n)t |x− y| (6.13)

that is,ψ|Rt is c(n)t Lipschitz wrt the extrinsic distance onS.


By definition,ψ(z) = ψi(βi(z)). Moreover, the definition ofg immediately implies a uniform Lipschitzcondition onψi . In particular,

‖∇ψi‖L∞(Bri (βi (x))) ≤

i∏

s=0

∥

∥

∥∇σ−1s

∥

∥

∥

L∞(Brs(βs(x))) ≤ t . (6.14)

Now, by (5.25), we have that|βi(x) − x| + |βi(y) − y| ≤ c(n, ρ)δr i+1, so, forδ ≤ δ0(n, ρ) small enough, we get

|βi(x) − βi(y)| ≤ |x− y| + 2cδr i+1 ≤ 2 |x− y| , (6.15)

and thus

|ψ(x) − ψ(y)| ≤ |ψi(βi(x)) − ψi(βi(y))| ≤ t |βi(x) − βi(y)| ≤ 2t |x− y| , (6.16)

as desired.Now, defineψt to be a Lipschitz extension ofψ|Rt over allRn with the same Lipschitz constant. In partic-

ular,ψt will be a Lipschitz function onS with Lipschitz constant bounded by 2t. Note that, independentlyof t, we have the estimate

|∇ψt(z)| ≤ 4g(z) for λk − a.e.z ∈ S . (6.17)

Indeed, ifz ∈ RCt , this estimate is valid for allx for which∇ψt(x) is defined. Moreover, we prove that this

estimate is also valid for allz ∈ Rt ⊂ S which haveλk-density 1 (both with respect toRt andS) and suchthat∇ψt(x) is defined. SinceRt ⊂ S is k-rectifiable, it is well-known that the density of both thesesets is 1λk almost everywhere (see for example [AFP00, theorem 2.63]). In particular, forλk a.e.z ∈ Rt we have

limr→0

λk(Br (z) ∩ Rt)

λk(Br (z) ∩ S)= 1 . (6.18)

For y ∈ S, setyt to be an almost projection ofy ontoRt, i.e., a point such thatd(y, yt) ≤ 2d(y,Rt). Then wehave

lim supy→z

|ψt(y) − ψt(z)||y− z|

≤ lim supy→z

[

|ψt(y) − ψt(yt)||y− z|

+|ψt(yt) − ψt(z)||y− z|

]

. (6.19)

By the proof of (6.13), and since|yt − z| ≤ 2 |y− z|, we get

lim supy→z

|ψt(yt) − ψt(z)||y− z|

= lim supy→z

|ψ(yt) − ψ(z)||y− z|

≤ 4g(z) . (6.20)

Moreover, sincez is a density 1 point onRλ ⊂ S, |y− yt | cannot be too big aroundz. Indeed, if|y− yt | ≥ 2r,thenBr (y) ∩ Rt = ∅, and so

(

14

lim supy→z

|y− yt |

|y− z|

)k

= lim supy→z

λk(

B|y−yt |/2 (z) ∩ S)

λk(

B2|y−z| (z) ∩ S) ≤ lim sup

y→z

1−λk

(

B2|y−z| (z) ∩ Rt

)

λk(

B2|y−z| (z) ∩ S)

= 0 . (6.21)

By the global Lipschitz estimate onψt, we get

lim supy→z

|ψt(y) − ψt(yt)||y− z|

≤ 2t lim supy→z

|y− yt |

|y− z|= 0 . (6.22)


This proves (6.17). Now consider anyT ≥ t. SinceS is rectifiable, and sinceψt = ψT onRt, then∇ψt = ∇ψT

λk-a.e. onRt. Moreover, by (6.17), we have∫

S|∇ψT − ∇ψt |

p=

∫

RCt

|∇ψT − ∇ψt |p ≤ 2p

∫

RCt

|∇ψt |p+ 2p

∫

RCt

|∇ψT |p ≤ 2ptpλk

(

RCt

)

+ 2p∫

RCt

|g|p .

(6.23)

By (6.12), this proves point (4). Sinceψ is a uniformly continuous function, point (3) is a corollaryof point(2).

6.2. Proof of Theorem 3.4: The Discrete Rectifiable-Reifenberg.Up to minor differences, the proof ofthis theorem is essentially the same as the proof of (5.73), and is actually much simpler from the technicalpoint of view. Indeed, we do not need to define the setsE, B, V, but we can just deal with the setsE,V, B.

In particular, letrα = 2−α andr = rA for someA ∈ N. One can define the measure

µ ≡∑

xj∈S

ωkrkjδxj , µr = ωk

∑

xj∈S∩B1(0) s.t. r j≥r

rkjδxj , (6.24)

and prove inductively onα = A,A − 1, · · · , 0 that for allx ∈ B1 (0) such thatBrα/10 (x) is not contained inany of theBr j

(

x j

)

:

µr(Brα (x)) ≤ D(n)rkα . (6.25)

By proceeding with a rough covering argument as in the proof of (5.13), one obtains easily a rough upperbound onµr(Brα−1 (x)). Moreover, by mimicking the inductive constructions in Section 5.4, one can defineExcess sets, good, bad and final balls, and the mapsσi and the approximating manifoldsTi . By studying thebi-Lipschitz properties ofσi, and by keeping track of the holes created by final and bad balls in the sameway as before, one proves the inductive estimate

µr(Brα (x)) ≤ C(n)

1+∑

rβ≤2rα

∫

Brα (x)Dkµ(x, rα) dµ(x)

, (6.26)

as desired.

7. L2-BEST APPROXIMATION THEOREMS

In this Section we prove the main estimate necessary for us tobe able to apply the rectifiable-Reifenbergof Theorems3.3and3.4to the singular setsSk

ǫ ( f ) of the stratification induced for stationary harmonic maps.

Namely, we need to understand how to estimate on a ballBr(x) the L2-distance ofSkǫ from the best

approximatingk-dimensional subspace. Whenr < inj(M) this means we would like to consider subspacesLk ⊆ TxM and estimated2(x, Lk) for x ∈ Sk ∩ Br(x), where the distance is taken in the normal coordinatecharts. After rescaling this is equivalent to looking at a ball of definite size, but assumingKM << 1. The


main Theorem of this Section is stated in some generality as we will need apply it with some care whenproving Theorems1.3and1.4. We recall that by definition

Wα(x) ≡Wrα,rα−3(x) ≡ θrα−3(x) − θrα(x) ≥ 0 , (7.1)

whererα = 2−α.

Theorem 7.1 (L2-Best Approximation Theorem). Let f : B9r(p) → N be a stationary harmonic mapsatisfying(1.16) with r2−n

∫

B9(p)|∇ f |2 ≤ Λ, and letǫ > 0. Then there existsδ(n,KN,Λ, ǫ), C(n,KN,Λ, ǫ) > 0

such that if KM < δ, and B9r(p) is (0, δ)-symmetric but not(k + 1, ǫ)-symmetric, then for any finite measureµ on Br(p) we have that

Dµ(p, r) = infLk

r−2−k∫

Br (p)d2(x, Lk) dµ(x) ≤ Cr−k

∫

Br (p)W0(x) dµ(x) , (7.2)

where theinf is taken over all k-dimensional affine subspaces Lk ⊆ TpM.

Remark7.1. Throughout this section, for convenience we will fixr = 1 without essential loss of generality.

Remark7.2. The assumptionKM < δ is of little consequence, since given anyKM this just means focusingthe estimates on balls of sufficiently small radius after rescaling.

7.1. Symmetry and Gradient Bounds. In this subsection we study stationary harmonic maps which arenot (k + 1, ǫ)-symmetric on some ball. In particular, we show that this forces the gradient to be of somedefinite size when restricted to anyk+ 1-dimensional subspace. More precisely:

Lemma 7.2. Let f : B4(p) ⊆ M → N be a stationary harmonic map satisfying(1.16) with>

B4(p)|∇ f |2 ≤ Λ.

Then for eachǫ > 0 there existsδ(n,KN,Λ, ǫ) > 0 such that if KM < δ, and B4(p) is (0, δ)-symmetric but isnot (k+ 1, ǫ)-symmetric, then for every k+ 1-subspace Vk+1 ⊆ TpM we have?

A3,4(p)|〈∇ f ,V〉|2dλ ≥ δ , (7.3)

where|〈∇u,V〉|2 ≡∑k+1

1 |〈∇u, vi〉|2 for any orthonormal basisvi of V, andλ is the volume measure on M

(which is equivalent toλn).

Proof. The proof is by contradiction. So withn,KN,Λ, ǫ > 0 fixed let us assume the result fails. Thenthere exists a sequencefi : B4(pi) → Ni for which KMi < δi , with B4(pi) being (0, δi)-symmetric but not(k+ 1, ǫ)-symmetric, and such that for some subspacesVk+1

i we have that?A3,4(pi )

|〈∇ fi ,Vk+1i 〉|

2 ≤ δi → 0 . (7.4)

Then after passing to subsequences we have thatVi → Vk ⊆ Rn with

NiC1,α

−→ N , (7.5)

and

fi −→ f : B4(0n)→ N , (7.6)


where the convergence is inL2 ∩ H1weak. Now (7.4) and theH1

weakconvergence guarantees that?A3,4(0n)

|〈∇ f ,Vk+1〉|2 = 0 . (7.7)

On the other hand, the (0, δi )-symmetry of thefi, combined with theL2 convergence, tells us thatf is0-symmetric. Combining these tells us that?

B4(0n)|〈∇ f ,Vk+1〉|2 = 0 , (7.8)

and hence we have thatf is k+ 1-symmetric. Because the convergencefi → f is in L2, this contradicts thatthe fi are not (k+ 1, ǫ)-symmetric fori sufficiently large, which proves the Lemma.

7.2. BestL2-Subspace Equations.In order to prove Theorem7.1we need to identify which subspace min-imizes theL2-energy, and the properties about this subspace that allow us to estimate the distance. We beginin Section7.2.1by studying some very general properties of the second directional moments of a generalprobability measureµ ⊆ B1(p). We will then study in Section7.2.2 the Dirichlet energy of a stationaryharmonic map when restricted to these directions.

7.2.1. Second Directional Moments of a Measure.Let us consider a probability measureµ ⊆ B1(0n), andlet

xicm = xi

cm(µ) ≡∫

xi dµ(x) , (7.9)

be the center of mass. Let us inductively consider the maximum of the second directional moments ofµ.More precisely:

Definition 7.3. Let λ1 = λ1(µ) ≡ max|v|2=1

∫

|〈x− xcm, v〉|2 dµ(x) and letv1 = v1(µ) with |v1| = 1 (any of) thevector obtaining this maximum. Now let us define inductivelythe pair (λk+1, vk+1) from v1, . . . , vk by

λk+1 = λk+1(µ) ≡ max|v|2=1,〈v,vi 〉=0

∫

|〈x− xcm, v〉|2 dµ(x) , (7.10)

wherevk+1 is (any of) the vector obtaining this maximum.

Thusv1, . . . , vn defines an orthonormal basis ofRn, ordered so that they maximize the second directionalmoments ofµ. Let us define the subspaces

Vk= Vk(µ) ≡ xcm+ spanv1, . . . , vk . (7.11)

The following is a simple but important exercise:

Lemma 7.4. If µ is a probability measure in B1(0n), then for each k the functional

minLk⊆Rn

∫

d2(x, Lk) dµ(x) , (7.12)


where themin is taken over all k-dimensional affine subspaces, attains its minimum at Vk. Further, we havethat

minLk⊆Rn

∫

d2(x, Lk) dµ(x) =∫

d2(x,Vk) dµ(x) = λk+1(µ) + · · · + λn(µ) . (7.13)

Note that the best affine subspaceVk will necessarily pass through the center of massxcm by Steiner’sformula, or equivalently by Jensen’s inequality and the definition of xcm.

Now let us record the following Euler-Lagrange formula, which is also an easy computation:

Lemma 7.5. If µ is a probability measure in B1(0n), then we have that v1(µ), . . . , vn(µ) satisfy the Euler-Lagrange equations:

∫

〈x− xcm, vk〉(x− xcm)i dµ(x) = λkvik , (7.14)

where

λk =

∫

|〈x− xcm, vk〉|2 dµ(x) . (7.15)

Proof. The proof is a simple application of the Lagrange multipliers method. Inductively onk, consider thefunction f (vk, λk, λk,1, · · · , λk,k−1) : Rn ×Rk → R given by

f (vk, λk, λk,ℓ) =∫

〈x− xcm, vk〉2 dµ(x) − λk

(

|vk|2 − 1

)

− 2k−1∑

ℓ=0

λk,ℓ 〈vk, vℓ〉 . (7.16)

By the multipliers method, we obtain thatvk andλk satisfy the equations

12∇(vk) f (vk, λk, λk,ℓ) =

∫

〈x− xcm, vk〉 (x− xcm)dµ(x) − λkvk −

k−1∑

ℓ=0

λk,ℓvℓ = 0 . (7.17)

By taking the scalar product of this equation withvℓ, since〈vk, vℓ〉 = δk,ℓ, we have

λk =

∫

|〈x− xcm, vk〉|2 dµ(x) , (7.18)

λk,ℓ =

∫

〈x− xcm, vℓ〉〈x− xcm, vk〉dµ(x) =

⟨∫

〈x− xcm, vℓ〉 (x− xcm) dµ(x), vk

⟩

=

=

⟨

λℓvℓ +ℓ−1∑

s=1

λℓ,svs, vk

⟩

= 0 . (7.19)


7.2.2. Restricted Dirichlet Energies.Our goal is to now study the Dirichlet energy of a stationary harmonicmap when restricted to the directionsv1(µ), . . . , vn(µ) associated to a probability measure. The main resultof this subsection is the following, which holds for a general stationary harmonic map:

Proposition 7.6. Let f : B9 (p) ⊆ M → N be a stationary harmonic map satisfying(1.16) with KM < 2−4

and such that>

B9(p) |∇ f |2 ≤ Λ. Let µ be a probability measure on B1(p) with λk(µ), vk(µ) defined as inDefinition7.3. Then there exists C(n,KN) > 0 such that

λk

∫

A3,4(p)|〈∇ f (z), vk〉|

2 dvg(z) ≤ C∫

W0(x) dµ(x) . (7.20)

Proof. Note first that there is no harm in assuming thatxcm ≡ 0. If not we can easily translate to make thisso, in which case we still have that supp(µ) ⊆ B2. Additionally, we will simplify the technical aspect ofthe proof by assuming thatM ≡ Rn. By working in normal coordinates the proof of the general case is nodifferent except up to some mild technical work.

Now let us fix anyz ∈ A3,4. By Lemma7.5, we can inner product both sides of (7.14) by ∇ f (z) to obtainfor eachk andz∈ A3,4:

λk 〈∇ f (z), vk〉 =

∫

〈x, vk〉〈∇ f (z), x〉 dµ(x) , (7.21)

Observe that, by definition of center of mass,∫

〈x, z〉dµ(x) = 〈xcm, z〉 = 0 . (7.22)

Thus we can write

λk 〈∇ f (z), vk〉 =

∫

〈x, vk〉〈∇ f (z), x− z〉 dµ(x) . (7.23)

By Holder inequality, we can then estimate

λ2k |〈∇ f (z), vk〉|

2 ≤ λk

∫

|〈∇ f (z), x− z〉|2 dµ(x) . (7.24)

If λk , 0, integrating with respect tozon both sides we get the estimate

λk

∫

A3,4

|〈∇ f (z), vk〉|2 dvg(z) ≤

∫ ∫

A3,4

|〈∇ f (z), x− z〉|2 dvg(z) dµ(x) . (7.25)

Set for conveniencenx(z) = (z− x)/ |z− x|, i.e.,nx(z) is the radial vector fromx to z. Now for x ∈ supp(µ)we can estimate

∫

A3,4

|〈∇ f (z), x− z〉|2 dvg(z) =∫

A3,4

|〈∇ f (z), nx(z)〉|2 |x− z|2−n|x− z|n dvg(z)

≤ 6n∫

A3,4

|〈∇ f (z), nx(z)〉|2|x− z|2−n dvg(z)

≤ C(n)∫

A1,8(x)|〈∇ f (z), nx(z)〉|

2|x− z|2−n dvg(z)

= C(n)W0(x) . (7.26)


Applying this to (7.25) we get as desired

λk

∫

A3,4

|〈∇ f (z), vk〉|2 dvg(z) ≤C(n)

∫

W0(x) dµ(x) . (7.27)

7.3. Proof of Theorem 7.1. Let us now combine the results of this Section in order to prove Theorem7.1.Indeed, letµ be a measure inB1(p) ⊆ TpM. We can assume thatµ is a probability measure without any lossof generality, since both sides of our estimate scale. Let

(

λ1(µ), v1(µ))

, . . . ,(

λn(µ), vn(µ))

be the directionalsecond moments as defined in Definition7.3, with Vk the induced subspaces defined as in (7.11). UsingLemma7.4we have that

minLk⊆Rn

∫

d2(x, Lk) dµ(x) =∫

d2(x,Vk) dµ(x) = λk+1(µ) + · · · + λn(µ) ≤ (n− k)λk+1(µ) , (7.28)

where we have used thatλ j ≤ λi for j ≥ i. Therefore our goal is to estimateλk+1. To begin with, Proposition7.6tells us that for eachj

λ j

∫

A3,4(p)|〈∇ f (z), v j 〉|

2 dvg(z) ≤ C∫

W0(x) dµ(x) . (7.29)

Let us sum the above for allj ≤ k+ 1 in order to obtain

k+1∑

j=1

λ j

∫

A3,4(p)|〈∇ f (z), v j〉|

2 dvg(z) ≤ (k+ 1)C∫

W0(x) dµ(x) , (7.30)

or by using thatλk+1 ≤ λ j for k+ 1 ≥ j we get

λk+1

∫

A3,4(p)|〈∇ f (z),Vk+1〉|2 dvg(z) = λk+1

k+1∑

j=1

∫

A3,4(p)|〈∇ f (z), v j〉|

2 dvg(z) ≤ C∫

W0(x) dµ(x) . (7.31)

Now we use thatB8(p) is (0, δ)-symmetric, butnot (k+ 1, ǫ)-symmetric in order to apply Lemma7.2andconclude that

∫

A3,4(p)|〈∇ f (z),Vk+1〉|2 dvg(z) ≥ δ . (7.32)

Combining this with (7.31) we obtain

δλk+1 ≤ λk+1

∫

A3,4(p)|〈∇ f (z),Vk+1〉|2 dvg(z) ≤ C

∫

W0(x) dµ(x) , (7.33)

or that

λk+1 ≤ C(n,KN,Λ, ǫ)∫

W0(x) dµ(x) . (7.34)

Combining this with (7.28) we have therefore proved the Theorem.


8. THE INDUCTIVE COVERING LEMMA

This Section is dedicated to the basic covering lemma neededfor the proof of the main theorems of thepaper. The covering scheme is similar in nature to the one introduced by the authors in [NV] in order toprove structural theorems on critical sets. Specifically, let us consider a stationary harmonic mapf betweenRiemannian manifolds. We wish to build a covering of the quantitative stratification

Skǫ,r( f ) ∩ B1(p) ⊆ Ur ∪U+ = Ur ∪

⋃

Br i (xi) , (8.1)

which satisfies several basic properties. To begin with, thesetU+ is a union of balls satisfyingr i > r ≥ 0,and should satisfy the packing estimateωk

∑

rki ≤ C. Each ballBr i (xi) should have the additional property

that there is a definite energy drop off when compared toB1(p). To describe the setUr we should distin-guish between the caser > 0 andr ≡ 0. In the caser > 0 we will have thatUr =

⋃

Br(xri ) is a union of

r-balls and satisfies the Minkowski estimate Vol(Ur ) ≤ Crn−k. In the case whenr ≡ 0 we will have thatU0

is k-rectifiable with the Hausdorff estimateλk(U0) ≤ C. Let us be more precise:

Lemma 8.1(Covering Lemma). Let f : B16(p) ⊆ M → N be a stationary harmonic map satisfying(1.16)with

>B16(p) |∇ f |2 ≤ Λ and KM < δ(n,KN,Λ, ǫ). Let E= supx∈B1(p)∩Sk

ǫ,rθ1(x) with ǫ > 0, r ≥ 0, and k∈ N.

Then for allη ≤ η(n,KN,Λ, ǫ), there exists a covering Skǫ,r( f ) ∩ B1(p) ⊆ U = Ur ∪ U+ such that

(1) U+ =⋃

Br i (xi) with ri > r and∑

rki ≤ C(n,KN,Λ, ǫ).

(2) supy∈Bri (xi )∩Skǫ,rθr i (y) ≤ E − η.

(3) If r > 0 then Ur =⋃N

1 Br(xri ) with N ≤ C(n)r−k.

(4) If r = 0 then U0 is k-rectifiable and satisfies Vol(Bs (U0)) ≤ C(n)sn−k for each s> 0.In particular, λk(U0) ≤ C(n).

Remark8.1. The assumptionKM < δ is of little consequence, since given anyKM this just means focusingthe estimates on balls of sufficiently small radius after rescaling.

To prove the result let us begin by outlining the construction of the covering, we will then spend the restof this section proving the constructed cover has all the desired properties.

Thus let us consider someη > 0 fixed, and then define the energy scale forx ∈ B1(p) by

sx = sE,ηx ≡ inf

r ≤ s≤ 1 : supBs(x)∩Sk

ǫ,r

θηs(y) ≥ E − η

. (8.2)

Note that the energy scale implicitly depends on many constants. If r = 0 let us define the setU0 by

U0 ≡

x ∈ Skǫ,r( f ) ∩ B1(p) : sx = 0

, (8.3)

while if r > 0 let us first define the temporary covering

U′r =⋃

Br(xri ) , (8.4)

where

xri ⊆

x ∈ Skǫ,r ( f ) ∩ B1(p) : sx = r

, (8.5)


is any minimalr/5-dense set. In particular, note that the collection of balls Br/10(xri ) are disjoint.

In order to define the temporary coveringU′+= Br i(xi) let us first consider

x ∈ Skǫ,r( f ) ∩ B1(p) : sx > r

⊆⋃

sx>r

Bsx/10(x) , (8.6)

and choose from it a Vitali subcovering so that

x ∈ Skǫ,r( f ) ∩ B1(p) : sx > r

⊆⋃

i∈I

Br i/2(xi) ⊆⋃

i∈I

Br i (xi) ≡ U′+ , (8.7)

wherer i ≡ sxi . In particular, we have that the collection of ballsBr i/10(xi) are all disjoint. It is clear fromthe construction that we have built a covering

Skǫ,r( f ) ∩ B1(p) ⊆

⋃

Br/2(xri ) ∪

⋃

i∈I

Br i/2(xi) ⊆⋃

Br(xri ) ∪

⋃

i∈I

Br i(xi) ≡ U′r ∪U′+. (8.8)

Now this is not quite the covering of Lemma8.1, as the energy drop properties of (8.2) involve droppingan extraη-scales. It will be more convenient to estimate this covering, however with only minimal worklet us now build from this covering the desired covering of Lemma8.1, which is only a small modification.Indeed, consider for each ballBr i(xi) (or Br(xr

i )) a Vitali covering

Br i/2(xi) ∩ Skǫ,r ⊆

Ni⋃

1

Bηr i(xia) ≡Ni⋃

1

Br ia(xia) , (8.9)

wherexia ∈ Br i/2(xi) ∩ Skǫ,r and by a standard covering argumentNi ≤ Ni(n, η). Then we can define the

coverings

Ur ≡⋃

i

Ni⋃

a=1

Br ia(xria) ,

U+ ≡⋃

i

Ni⋃

a=1

Br ia(xia) . (8.10)

It is clear from the construction thatU+ now satisfies the energy drop condition of Lemma8.1.2 . What is leftis to show the content estimates of Lemma8.1, and from our estimates onNi it is clear withη < η(n,KN,Λ, ǫ)that it is enough to prove these estimates for the setsU′r andU′+ themselves, which is therefore the goal ofmuch of this section.

The outline of this Section is as follows. Section8.1is dedicated to proving a variety of technical lemmaswhich will be used to further decompose the setsU′

+andU′r when r > 0. The technical issue is that we

cannot directly apply the discrete Reifenberg to the setU′+, and will instead need to exchangeU′

+for a more

manageable collection of balls without losing much content. In Section8.2we will use these tools in orderto prove our content estimates onU′+ andU′r when r > 0. The proof will require an inductive argumentcombined with applications of the discrete Reifenberg of Theorem3.4 to our new covering. Finally in Sec-tion 8.3 we will prove the desired estimates onU0. The volume estimates will follow almost immediatelyfrom our previous constructions, and to prove the rectifiable statement will require a careful application of


the rectifiable Reifenberg theorem.

8.1. Technical Constructions for DecomposingU′r ∪ U′+. Let us consider the set of positive radius balls

in our covering given by

U> ≡

U′+

if r = 0 ,

U′r ∪U′+

if r > 0 .(8.11)

This section will be dedicated to proving some technical results which will be required in estimating thisset. The estimates on this set are a little delicate, the reason being that we cannot directly apply the recti-fiable Reifenberg of Theorem3.4 to this set. Instead, we will need to replaceU> with a different coveringat each stage, which will be more adaptable to Theorem3.4. Thus this subsection is dedicated to proving ahandful of technical results which are important in the construction of this new covering.

Throughout this subsection we are always working under the assumptions of Lemma8.1. Let us beginwith the following point, which is essentially a consequence of the continuity of the energy:

Lemma 8.2. For eachη′ > 0 there exists R(n,KN,Λ, η′) > 1 such that ifδ < δ(n,KN,Λ, η

′) and η ≤η(n,KN,Λ, η

′), then we have for each z∈ Br i (xi) the estimate

θRri (z) > E − η′ . (8.12)

Proof. The proof relies on a straight forward energy comparison. Namely, letx, y ∈ Bs/2 with s< 1 and letus denoted ≡ d(x, y). Then we have the estimate

θs(y) = s2−n∫

Bs(y)|∇ f |2 ≤ s2−n

∫

Bs+d(x)|∇ f |2 =

( ss+ d

)2−nθs+d(x) . (8.13)

To apply this in our context, let us note for eachxi in our covering, that by our construction ofU> there mustexistyi ∈ Br i (xi) such thatθ(R−1)r i (yi) ≥ θr i (yi ) = E − η. Let us now apply (8.13) to obtain

θRri (z) ≥

(

Rri(R− 2)r i

)2−n

θ(R−2)r i (yi) ≥( RR− 2

)2−n(

E − η)

. (8.14)

If R= R(n,KN,Λ, η′) > 0 andη ≤ η(n,KN,Λ, η

′), then we obtain from this the claimed estimate.

In words, the above Lemma is telling us that even though we have no reasonable control over the size ofθr i (xi), after we go up a controlled number of scales we can again assume that the energy density is againclose toE.

In the last Lemma the proof was based on continuity estimateson the energyθ. In the next Lemma wewish to show an improved version of this continuity under appearance of symmetry. Precisely:

Lemma 8.3(Improved Continuity ofθ). Let f : B4(p)→ N be a stationary harmonic map satisfying(1.16)with>

B4(p) |∇ f |2 ≤ Λ. Then for each0 < τ, η < 1 there existsδ(n,KN,Λ, η, τ) > 0 such that if

(1) we have KM < δ ,


(2) there exists x0, . . . , xk ∈ B1(p) which areτ-independent with|θδ(x j) − θ3(x j)| < δ,

then if Vk is the k-dimensional subspace spanned by x0, . . . , xk, then for all x, y ∈ B1(p) ∩ Bδ(Vk) andη ≤ s≤ 1 we have that|θs(x) − θs(y)| < η.

Proof. The proof is by contradiction. Thus, imagine no suchδ exists. Then there exists a sequence ofstationary harmonic mapsfi : B4(pi )→ Ni satisfying

>B4(p) |∇ f |2 ≤ Λ such that

(1) KM < δi → 0(2) there existsxi,0, . . . , xi,k ∈ B1(p) which areτ-independent with

∣

∣

∣

∣

θfiδi r

(xi, j) − θfi3 (xi, j)

∣

∣

∣

∣

< δi → 0,

however we have that there existsxi , yi ∈ B1(p) ∩ Bδi (Vki ) andη ≤ si ≤ 1 such that

∣

∣

∣

∣

θfisi(xi) − θ

fisi(yi)

∣

∣

∣

∣

≥ η. Letus now pass to a subsequence to obtain the defect measure

|∇ fi |2dvg → |∇ f |2 dvg + ν ,

xi,β → xβ ∈ B1(0n) ,

xi → x, yi → y ∈ V = spanx0, . . . , xk ,

si → s, (8.15)

whereν is a measure onB4(0n), and

s2−n

∣

∣

∣

∣

∣

∣

∫

Bs(x)

(

|∇ f |2 dvg + dν)

−

∫

Bs(y)

(

|∇ f |2 dvg + dν)

∣

∣

∣

∣

∣

∣

≥ η . (8.16)

However, we have by theorem2.7 that f andν are 0-symmetric onB2 with respect to each of the pointsx0, . . . xk. In particular, by the standard cone splitting arguments wehave thatν is k-symmetric with respectto thek-planeV on B1. In particular,ν is invariant under translation by elements ofV. However, this is acontradiction to (8.16), and therefore we have proved the result.

Now the first goal is to partitionU> into a finite collection, each of which will have a few more manage-able properties thanU> itself. More precisely:

Lemma 8.4. For each R> 1 there exists N(n,R) > 1 such that we can break up U> as a union

U> =

N⋃

a=1

Ua> =

N⋃

a=1

⋃

i∈Ia

Br i (xi) , (8.17)

such that each Ua> has the following property: if i∈ Ia, then for any other j∈ Ia we have that if xj ∈ BRri(xi),then rj < R−2r i .

Proof. Let us recall that the balls in the collectionBr i/5(xi) are pairwise disjoint. In particular, givenR > 1 if we fix a ballBr i (xi) then by the usual covering arguments there can be at mostN(n,R) ball centersx j

N1 ∈ U> ∩ BR3r i

(xi) with the property thatr j ≥ R−6r i . Indeed, ifx jN1 is such a collection of balls then

we get

ωn(6R3r i)n= Vol(B6R3r i

(xi)) ≥N

∑

1

Vol(Br j/5(x j)) ≥ Nωn(R−6r i/5)n , (8.18)


which by rearranging gives the estimateN ≤ N(n,R).

Now we wish to build our decompositionU> =⋃N

1 Ua>, whereN = N + 1 is from the first paragraph.

We shall do this inductively, with the property that at each step of the inductive construction we will havethat the setsUa

+will satisfy the desired property. In particular, for everya and i ∈ Ia, if j ∈ Ia is such that

x j ∈ BRri(xi), thenr j < R−2r i .

Begin by letting eachIa be empty. We are going to sort the pointsxii∈I into the setsIa one at a time.At each step leti ∈ I \

⋃Na=1 Ia be an index such thatr i = maxr j , where the max is taken over all indexes

in I \⋃N

a=1 Ia, i.e., over all indexes which haven’t been sorted out yet. Now let us consider the collection ofball centersy j j∈J such thatJ ⊂ I , xi ∈ BRr j(y j ) andr i ≤ r j ≤ R2r i . Note that, by construction, eithery j hasalready been sorted out in someIa, or r j = r i . Now evidentlyy j ∈ BR3r i

(xi) for all j ∈ J and so by the firstparagraph the cardinality ofJ is at mostN(n,R). In particular, there must be someIa such thatIa∩ y j = ∅.Let us assigni ∈ Ia to this piece of the decomposition. Clearly the decomposition

⋃

Ua> still satisfies the

inductive hypothesis after the addition of this point, and so this finishes the inductive step of construction.Since at each stage we have chosenxi to have the maximum radius, this process will continue indefinitelyto give the desired decomposition ofU>.

Now with a decomposition fixed, let us consider for each 1≤ a ≤ N the measures

µa ≡∑

xi∈Ua>

ωkrki δxi . (8.19)

The following is a crucial point in our construction. It tells us that each ballB10r i (xi) either has smallµa-volume, or the pointxi must have large energy at scaler i . Precisely:

Lemma 8.5. Let D, η′ > 0 be fixed. There exists R= R(n,KN,Λ, η′,D, ǫ) > 0 such that if we consider the

decomposition(8.17), and if

(1) KM < δ(n,KN,Λ, η′, ǫ) andη ≤ η(n,KN,Λ, η

′, ǫ), ri < 10−4,(2) we haveµa(B10r i (xi)) ≥ 2ωkrk

i ,(3) for all ball centers yj ∈ Ar i/10,10r i (xi) ∩ Ua and for s= 10−2nD−1r i , we have thatµa(Bs(y j)) ≤ Dsk,

then we have thatθ(xi , η′r i) ≥ E − η′.

Proof. Let us begin by choosingη′′ << η′, which will be fixed later in the proof so thatη′′ = η′′(n,KN,Λ, ǫ),and let us also defineτ ≡ 10−3nD−1. Let δ(n,KN,Λ, η

′) be from Lemma8.3 so that the conclusions holdwith 10−1η′, and letδ′(n,KN,Λ, η

′′, ǫ) be chosen so that conclusions of Lemma8.2 are satisfied withη′′ ifη ≤ η(n,KN,Λ, η

′′, ǫ). Now throughout we will assumeη < δ andKM < minδ, δ′. We will also chooseR= R(n,KN,Λ, η

′′,D, ǫ) > maxτ−1, δ−1, δ′−1 so that Lemma8.2is satisfied withη′′.

Since it will be useful later, let us first observe thatr i ≥ Rr. Indeed, if not then for each ball centery j ∈ Ar i/10,10r i (xi) ∩ Ua

> we would haver ≤ r j < R−2r i < r. This tells us that there can be no ball centers in


Ar i/10,10r i (xi) ∩Ua>. However, by our volume assumption we have that

µa(Ar i/10,10r i (xi))

= µa(B10r i (xi))

− µa(Br i/10(xi))

≥ 2ωkrki − ωkr

ki ≥ ωkr

ki , (8.20)

which contradicts this. Therefore we must have thatr i ≥ Rr.

Now our first real claim is that under the assumptions of the Lemma there exists ball centersy0, . . . , yk ∈

Ua> ∩ Ar i/10,10r i (xi) which areτr i-independent in the sense of Definition2.2. Indeed, assume this is not the

case, then we can find ak− 1-planeVk−1 such that

yii∈Ia ∩ Ar i/10,10r i (xi) ⊆ Bτr i

(

Vk−1)

. (8.21)

In particular, by coveringBτr i (V) ∩ B10r i (xi) ∩ Ia by C(n)τ1−k ≤ 10nτ1−k balls of radius 10τr i with centersin Ia, and using our assumption thatµa(Bτr i (y)) ≤ Dτkrk

i , we are then able to conclude the estimate

µa(

Ar i/10,10r i (xi))

≤ µa(

Bτr i (V) ∩ B10r i (xi))

≤ 102nDτrki < ωnrk

i . (8.22)

On the other hand, our volume assumption guarantees that

µa(Ar i/10,10r i (xi))

= µa(B10r i (xi))

− µa(Br i/10(xi))

≥ 2ωkrki − ωkr

ki ≥ ωkr

ki , (8.23)

which leads to a contradiction. Therefore there must existk+ 1 ball centersy0, . . . , yk ∈ Ar i/10,10r i (xi) whichareτr i-independent points, as claimed.

Let us now remark on the main consequences of the existence ofthesek + 1 points. Note first that foreachy j we have thatθR−1r i

(y j ) > E − η′′, since by the construction ofUa> we have thatr j ≤ R−2r i , and

therefore we can apply Lemma8.2. Thus we havek + 1 points inB10r i (xi) which areτr i-independent, andwhose energies areη′′-pinched. To exploit this, let us first apply Lemma8.3 in order to conclude that foreachx ∈ Bδ(V), whereV is the plane spanned by thek+ 1 independent points just determined, we have

θη′r i (x) ≥ θη′r i (y j ) − |θη′r i (x) − θη′r i (y j)| ≥ E − η′′ − 10−1η′ > E − η′ . (8.24)

In particular, if we assume thatxi is such thatθη′r i (xi) < E − η′, then we must have thatr−1i d(x j ,V) ≥ δ =

δ(n,KN,Λ, η′).

Therefore, let us now assumeθη′r i (xi) < E − η′, and thus to prove the Lemma we wish to find a contra-diction. To accomplish this notice that we have ourk+ 1 pointsy0, . . . , yk ∈ B10r i which areτr i-independentand for which|θ20r i (y j) − θR−1r i

(y j)| < η′′. Therefore by applying the cone splitting of Theorem2.3we havefor eachǫ′ > 0 that if η′′ ≤ η′′(n,KN,Λ, ǫ

′) thenB10r i (xi) is (k, ǫ′)-symmetric with respect to thek-planeVk. However, sinced(xi ,V) > δr i , we have by Theorem2.4 that if ǫ′ ≤ ǫ′(n,KN,Λ, ǫ) then there existssomeτ′ = τ′(n,KN,Λ, ǫ) such thatBτ′r i (xi) is (k+ 1, ǫ)-symmetric. However, we can assume after a furtherincrease thatR = R(n,KN,Λ,D, ǫ) > 4τ′−1, and thus we have thatτ′r i > 4R−1r i > 4r. This contradicts thatxi ∈ Sk

ǫ,r , and thus we have contradicted thatθη′r i (xi) < E − η′, which proves the Lemma.


8.2. Estimating U> in Lemma 8.1. Recall the setU> defined in (8.11), which consists of all positive radiiballs in our covering. We now wish to estimate this set in thissubsection. First let us pickD′ = D′(n) ≡216nD(n), whereD(n) is from Theorem3.4. Then for someη′ fixed we can chooseR as in Lemma8.5. Itis then enough to estimate each of the setsUa

>, as there are at mostN = N(n,KN,Λ, ǫ, η, η′) pieces to the

decomposition. Thus we will fix a setUa> and focus on estimating the content of this set. Let us begin by

observing that ifr > 0 then we have the lower boundr i ≥ r. Otherwise, let us fix anyr > 0 and restrictourselves to the collection of balls inUa

> with r i ≥ r. There is no loss in this so long as the estimates we willprove will be independent of our choice ofr, and thus by lettingr → 0 we will obtain estimates on all ofUa

>.

Now let us make the precise statement we will prove in this subsection. Namely, consider any of ballcentersxii∈Ia and any radius 2−4r i ≤ rα ≤ 2, whererα = 2−α. Then we will show that

µa(Brα(xi))

≤ 25nD(n)rkα , (8.25)

whereD(n) is the constant from the discrete Reifenberg theorem. Let us observe that once we have proved(8.25) then we have finished the proof of the Covering Lemma, as we will then have the estimate

∑

rki = µ

a(B2(xi))

≤ C(n) . (8.26)

We prove (8.25) inductively onα. To begin notice that for eachxi if β is the largest integer such that2−4r i ≤ rβ, then the statement clearly holds, by the definition of the measureµa. In fact, we can go furtherthan this. For eachxi , let r′i ∈ [r i/10, 10r i ] be the largest radius such that for all 10−1r i ≤ s≤ r′i we have

µa(Bs(xi))

≤ 2ωksk . (8.27)

In particular, we certainly have the much weaker estimateµa(Bs(xi))

≤ 25nD(n)sk, and hence (8.25) is alsosatisfied for all 2−4r i ≤ rβ ≤ r′i . Notice that we then also have the estimate

ωkrki ≤ µ

a(Br i/10(xi))

≤ µa(Br ′i(xi)

)

≤ 2ωk(

r′i)k. (8.28)

Now let us focus on proving the inductive step of (8.25). Namely, assumeα is such that for allxi with2−4r i ≤ rα+1 < 2 we have that (8.25) holds. Then we want to prove that the same estimate holds forrα.Let us begin by seeing that a weak version of (8.25) holds. Namely, for any indexi ∈ Ia and any radiusrα ≤ s≤ 8rα, by coveringBs(xi) by at most 28n balls Brα+1(y j) of radiusrα+1 we have the weak estimate

µa(Bs(xi))

≤∑

µa(Brα+1(y j))

≤ D′(n)sk , (8.29)

where of courseD′(n) >> 25nD(n).

To improve on this, let us fix ani ∈ Ia and the relative ball centerxi ∈ Ua> with 2−4r i ≤ rα. Now let

x j j∈J = xii∈I ∩ Brα(xi) be the collection of ball centers inBrα(xi). Notice first that ifr′j > 2rα for any ofthe ball centersx j, then we can estimate

µa(Brα(xi))

≤ µa(B2rα(x j))

≤ 2ωk(

2rα)k≤ 25nD(n)rn

α , (8.30)

so that we may fairly assumer′j ≤ 2rα for every j ∈ J. Now for each ballBr ′j(x j) let us define a new ball

Br j (y j) which is roughly equivalent, but will have some additionaluseful properties needed to apply the


discrete Reifenberg. Namely, for a given ballBr ′j(x j), let us consider the two optionsr′j < 10r j or r′j = 10r j .

If r′j < 10r j , then we lety j ≡ x j with r j ≡ r j . In this case we must have thatµa(B10r j (x j)) > 2ωn(

r j)k, and

thus we can apply Lemma8.5 in order to conclude thatθη′ r j (y j) ≥ E − η′. In the case whenr′j = 10r j is

maximal, lety j ∈ Br j (x j) ∩ Skǫ,r with y j , x j be a point such thatθηr j (y j) = E − η, such a point exists by the

definition ofr j , and let ¯r j ≡ 9r j . In either case we then have the estimates

θη′ r j (y j ) ≥ E − η′ ,

ωk10−krkj ≤ µ

(

Br j/10(y j))

≤ µa(Br j (y j))

≤ 4ωkrki ,

Br j (x j) ⊆ Br j (y j) . (8.31)

Now let us consider the coveringUa> ∩ Brα(xi) ⊆

⋃

Br j/10(y j ), and choose from it a Vitali subcovering

Ua> ∩ Brα(xi) ⊆

⋃

Br j (y j) , (8.32)

such thatBr j/10(y j) are disjoint, where we are now being lose on notation and referring to y j j∈J as theball centers from this subcovering. Let us now consider the measure

µ′ ≡∑

j∈J

ωk

( r j

10

)kδyj . (8.33)

That is, we have associated to the disjoint collectionBr j/10(y j) the natural measure. Our goal is to provethat

µ′(

Brα(xi))

≤ D(n)rkα . (8.34)

Let us observe that if we prove (8.34) then we are done. Indeed, using (8.31) we can estimate

µa(Brα(xi))

≤∑

µa(Br j (y j))

≤ 4ωk

∑

rkj = 4 · 10kµ′

(

Brα(xi))

≤ 25nD(n)rkα , (8.35)

which would finish the proof of (8.25) and therefore the Lemma.

Thus let us concentrate on proving (8.34). We will want to apply the discrete Reifenberg in this case tothe measureµ′. Let us begin by proving a weak version of (8.34). Namely, for any ball centery j from oursubcovering and radius ¯r j < s ≤ 4rα let us consider the setzk = yss∈J ∩ Bs(y j) of ball centers insideBs(y j). Since the ballsBrk/5(zk) are disjoint we have that ¯rk ≤ 10s. Using this, (8.28), and (8.31) we canestimate

µ′(

Bs(y j))

=

∑

zk∈Bs(yj )

ωk10−krkk ≤ C(n)

∑

zk∈Bs(yj )

µa(Brk/8(zk)) ≤ C(n)µa(B2s(y j )) ≤ C(n)sk , (8.36)

where of courseC(n) >> 25nD(n).

Now let us finish the proof of (8.34). Thus let us pick a ball centery j ∈ Brα(xi) and a radiuss< 4rα. Notethat by the first equation in (8.31) and theorem2.3we have thatBs(y j) can be made arbitrarily 0-symmetricby choosingη′ sufficiently small. Ifµ′(Bs(y)) ≤ ǫnsk thenDµ′(y j , s) ≡ 0 by definition, and ifs≤ r j/10 thenDµ′(y, s) ≡ 0, since the support ofµ′ in Br i/10(y j ) contains at most one point and thus is precisely contained


in a k-dimensional subspace. In the case whens> r i/10 andµ(Bs(y)) > ǫnsk then by Theorem7.1we havethat

Dµ′(y j , s) ≤ C(n,KN,Λ, ǫ)s−k

∫

Bs(y)Ws(z) dµ′(z) . (8.37)

By applying this to allr ≤ t ≤ swe can estimate

s−k∫

Bs(x)Dµ′(y, t) dµ′(y) ≤ Cs−k

∫

Bs(x)t−k

∫

Bt(y)Wt(z) dµ′(z) dµ′(y)

= Cs−kt−k∫

Bs(x)µ′(Bt(y))Wt(y) dµ′(y)

≤ Cs−k∫

Bs(x)Wt(y) dµ′(y) , (8.38)

where we have used our estimate onµ′(Bt(y)) from (8.36) in the last line. Let us now consider the case whent = rβ ≤ s≤ 2rα. Then we can sum to obtain:

∑

rβ≤s

s−k∫

Bs(x)Dµ′(y, rβ) dµ′(y) ≤ C

∑

r ′y≤rβ≤s

s−k∫

Bs(x)Wrβ(y) dµ′(y)

= Cs−k∫

Bs(x)

∑

ry≤rβ≤s

Wrβ(y) dµ′(y)

≤ C s−k∫

Bs(x)

∣

∣

∣θ4s(y) − θry(y)∣

∣

∣ dµ′(y) ≤ C(n,KN,Λ, ǫ)η′ , (8.39)

where we are using (8.31) in the last line in order to see that∣

∣

∣θs(y) − θry(y)∣

∣

∣ ≤ η′. Now let us chooseη′ ≤ η′(n,KN,Λ, ǫ) such that

∑

rβ≤s

s−k∫

Bs(x)Dµ′(y, rβ) dµ′(y) ≤ δ2 , (8.40)

whereδ is chosen from the discrete rectifiable-Reifenberg of Theorem3.4. Since the estimate (8.40) holdsfor all Bs ⊆ B2rα(x), we can therefore apply Theorem3.4to conclude the estimate

µ′(Brα(x′i )) ≤ D(n)rk

α . (8.41)

This finishes the proof of (8.31), and hence the proof of Lemma8.1for the setsUr andU+.

8.3. Estimating U0 in Lemma 8.1. We now finish the proof of Lemma8.1. Let us begin by proving theMinkowski estimates onU0. Indeed, observe for anyr > 0 thatU0 ⊆ Ur , and thus we have the estimate

Vol(Br (U0)) ≤ Vol(Br (Ur)) ≤ ωnrn · N ≤ C(n,KN,Λ, ǫ)rn−k , (8.42)

which proves the Minkowski claim. In particular, we have as aconsequence thek-dimensional Hausdorffmeasure estimate

λk(U0) ≤ C(n,KN,Λ, ǫ) . (8.43)


In fact, let us conclude a slightly stronger estimate, sinceit will be a convenient technical tool in the remain-der of the proof. IfBs(x) is any ball withx ∈ B1 ands< 1

2, then by applying the same proof to the rescaledball Bs(x)→ B1(0), we can obtain the Hausdorff measure estimate

λk(U0 ∩ Br(x)) ≤ Crk . (8.44)

To finish the construction we need to see thatU0 is rectifiable. We will in fact apply Theorem3.3in orderto conclude this. To begin with, letµ ≡ λk

∣

∣

∣

U0be thek-dimensional Hausdorff measure, restricted toU0.

Let Bs(y) be a ball withy ∈ B1 ands < 12, now we will argue in a manner similar to Section8.2. Thus, if

µ(Bs(y)) ≤ ǫnsk thenDU0(y, s) ≡ 0, and by theorem2.3and Theorem7.1we have that

Dµ(y, s) ≤ C(n,KN,Λ, ǫ)s−k

∫

Bs(y)Ws(z) dµ(z) . (8.45)

By applying this to allt ≤ swe have

s−k∫

Bs(x)Dµ(y, t) dµ(y) ≤ Cs−k

∫

Bs(x)t−k

∫

Bt(y)Wt(z) dµ(z) dµ(y)

= Cs−kt−k∫

Bs(x)µ(Bt(y))Wt(y) dµ(y)

≤ Cs−k∫

Bs(x)Wt(y) dµ(y) , (8.46)

where we have used our estimate (8.44) in the last line. Let us now consider the case whent = rβ = 2−β ≤ s.Then we can sum to obtain:

∑

rβ≤s

s−k∫

Bs(x)Dµ(y, rβ) dµ(y) ≤ C

∑

rβ≤s

s−k∫

Bs(x)Wrβ(y) dµ(y)

= Cs−k∫

Bs(x)

∑

rβ≤s

Wrβ(y) dµ(y)

≤ C s−k∫

Bs(x)

∣

∣

∣θs(y) − θ0(y)∣

∣

∣ dµ(y) ≤ C(n,KN,Λ, ǫ)η , (8.47)

where we have used two points in the last line. First, we have used our estimateµ(Bs(x)) ≤ Csk. Second,we have used that by the definition ofU0, for each point in the support ofµ we have that

∣

∣

∣θs(y) − θ0(y)∣

∣

∣ ≤ η.Now let us chooseη ≤ η(n,KN,Λ, ǫ) such that we have

∑

rβ≤s

s−k∫

Bs(x)Dµ(y, rβ) dµ(y) ≤ δ2 , (8.48)

whereδ is chosen from Theorem3.3. Thus, by applying Theorem3.3 we see thatU0 is rectifiable, whichfinishes the proof of Lemma8.1in the context ofU0.


9. PROOF OFMAIN THEOREMS FORSTATIONARY HARMONIC MAPS

In this section we prove the main theorems of the paper concerning stationary harmonic maps. With thetools of Sections3, 7, and8 developed, we will at this stage mainly be applying the covering of Lemma8.1 iteratively to arrive at the estimates. When this is done carefully, we can combine the covering lemmawith the cone splitting in order to check that fork-a.e.x ∈ Sk

ǫ there exists a uniquek-dimensional subspaceVk ⊆ TxM such that every tangent map off at x is k-symmetric with respect toV.

For the proofs of the Theorems’ of this section let us make thefollowing remark. For anyδ > 0 we cancoverB1(p) by a collection of balls

B1(p) ⊆N⋃

1

BK−1M δ(pi) , (9.1)

whereN ≤ C(n,KM)δ−n. Thus ifδ = δ(n,KM ,KN,Λ, ǫ) and we can analyze each such ball, then this givesus estimates on all ofB1(p). In particular, by rescalingBK−1

M δ(pi) → B1(pi), we see that we can assume inour analysis thatKM < δ without any loss of generality. We shall do this throughout this section.

9.1. Proof of Theorem 1.3. Let f : B2(p) ⊆ M → N be a stationary harmonic map satisfying (1.16) with>B2(p)|∇ f |2 ≤ Λ. With ǫ, r > 0 fixed, let us chooseη(n,KN,Λ, ǫ) > 0 andδ(n,KN,Λ, ǫ) > 0 as in Lemma8.1.

By the remarks around (9.1) we see that we can assume thatKM < δ, which we will do for the remainder ofthe proof.

Now let us begin by first considering an arbitrary ballBs(x) with x ∈ B1(p) andr < s ≤ 1, potentiallyquite small. We will use Lemma8.1in order to build a special covering ofSk

ǫ,r ∩ Bs(x). Let us define

Ex,s ≡ supy∈Bs(x)∩Sk

ǫ,r

θs(y) , (9.2)

and thus if we apply Lemma8.1with η(n,KN,Λ, ǫ) fixed toBs(x), then we can build a covering

Skǫ,r ∩ Bs(x) ⊆ Ur ∪ U+ =

⋃

Br(xri ) ∪

⋃

Br i(xi) , (9.3)

with r i > r. Let us recall that this covering satisfies the following:

(a) rk−nVol(Br U′r) + ωk∑

rki ≤ C(n) sk.

(b) supy∈Bri (xi ) θr i (y) ≤ Ex,s − η.

Now that we have built our required covering on an arbitrary ball Bs(x), let us use this iteratively to buildour final covering ofSk

ǫ,r( f ). First, let us apply it toB1(p) is order to construct a covering

Skǫ,r( f ) ⊆ U1

r ∪ U1+ =

⋃

Br

(

xr,1i

)⋃

Br1i

(

x1i

)

, (9.4)

such that

rk−nVol(

Br

(

U1r

))

+ ωk

∑(

r1i

)k≤ C(n,KN,Λ, ǫ) , (9.5)


and with

supy∈Br1i

(x1i )∩Sk

ǫ,r

θr1i(y) ≤ Λ − η . (9.6)

Now let us tackle the following claim, which is our main iterative step in the proof:

Claim: For eachℓ there exists a constantCℓ(ℓ, n,KM ,KN,Λ, ǫ) and a covering

Skǫ,r( f ) ⊆ Uℓ

r ∪Uℓ+=

⋃

Br

(

xr,ℓi

)⋃

Brℓi

(

xℓi)

, (9.7)

with rℓi > r, such that the following two properties hold:

rk−nVol(

Br

(

Uℓr

))

+ ωk

∑

(

rℓi)k≤ Cℓ(ℓ, n,KN,Λ, ǫ) ,

supy∈B

rℓi(xℓi )∩Sk

ǫ,r

θrℓi(y) ≤ Λ − ℓ · η . (9.8)

To prove the claim let us first observe that we have shown this holds for ℓ = 1. Thus let us assume wehave proved the claim for someℓ, and determine from this how to build the covering forℓ + 1 with someconstantCℓ+1(ℓ + 1, n,KM ,KN,Λ, ǫ), which we will estimate explicitly.

Thus with our covering determined at stageℓ, let us apply the covering of (9.3) to each ball

Brℓi

(

xℓi)

inorder to obtain a covering

Skǫ,r ∩ Brℓi

(xℓi ) ⊆ Ui,r ∪ Ui,+ =

⋃

j

Br

(

xri, j

)⋃

j

Br i, j

(

xi, j

)

, (9.9)

such that

rk−nVol(

Br(

Ui,r))

+ ωk

∑

j

(r i, j )k ≤ C(n,KN,Λ, ǫ)

(

rℓi)k,

supy∈Bri, j (xi, j )∩Sk

ǫ,r

θr i, j (y) ≤ Λ − (ℓ + 1)η . (9.10)

Let us consider the sets

Uℓ+1r ≡ Uℓ

r

⋃

i

Ui,r ,

Uℓ+1+ ≡

⋃

i, j

Br i, j (xi, j) . (9.11)

Notice that the second property of (9.8) holds forℓ + 1 by the construction, hence we are left analyzing thevolume estimate of the first property. Indeed, for this we combine our inductive hypothesis (9.8) for Uℓ and


(9.10) in order to estimate

rk−nVol(

Br

(

Uℓ+1r

))

+ ωk

∑

i, j

(r i, j )k ≤ rk−nVol

(

Br

(

Uℓr

))

+

∑

i

(

rk−nVol(

Br(

Ui,r))

+ ωk

∑

j

(r i, j)k)

≤ Cℓ +C∑

i

(rℓi )k

≤ C(n,KM ,KN,Λ, ǫ) ·Cℓ(ℓ, n,KM ,KN,Λ, ǫ)

≡ Cℓ+1 . (9.12)

Hence, we have proved that if the claim holds for someℓ then the claim holds forℓ + 1. Since we havealready shown the claim holds forℓ = 1, we have therefore proved the claim for allℓ.

Now we can finish the proof. Indeed, let us takeℓ = ⌈η−1Λ⌉ = ℓ(η,Λ). Then if we apply the Claim to

such anℓ, we must have by the second property of (9.8) that

Uℓ+ ≡ ∅ , (9.13)

and therefore we have a covering

Skǫ,r ⊆ Uℓ

r =

⋃

i

Br(xi) . (9.14)

But in this case we have by (9.8) that

Vol(

Br

(

Skǫ,r( f )

))

≤ Vol(

Br

(

Uℓr

))

≤ C(n,KN,Λ, ǫ)rn−k , (9.15)

which proves the Theorem.

9.2. Proof of Theorem 1.4. There are several pieces to Theorem1.4. To begin with, the volume estimatefollows easily now that Theorem1.3has been proved. That is, for eachr > 0 we have that

Skǫ ( f ) ⊆ Sk

ǫ,r( f ) , (9.16)

and therefore we have the volume estimate

Vol(

Br

(

Skǫ ( f )

))

≤ Vol(

Br

(

Skǫ,r( f )

))

≤ C(n,KM,KN,Λ, ǫ)rn−k . (9.17)

In particular, this implies the much weaker Hausdorff measure estimate

λk(Skǫ ( f )) ≤ C(n,KM,KN,Λ, ǫ) , (9.18)

which proves the first part of the Theorem.

Let us now focus on the rectifiability ofSkǫ . We consider the following claim, which is ther = 0 version

of the main Claim of Theorem1.3. We will be applying Lemma8.1, which requiresKM < δ. As in the proofof Theorem1.3we can just assume this without any loss, as we can coverB1(p) by a controlled number ofballs of radiusK−1/2

M δ, so that after rescaling we can analyze each of these balls with the desired curvatureassumption. Thus let us consider the following:


Claim: If KM < δ, then for eachℓ there exists a coveringSkǫ ( f ) ⊆ Uℓ

0 ∪Uℓ+ = Uℓ

0

⋃

Brℓi(xℓi ) such that

(1) λk(Uℓ0) + ωk

∑(

rℓi)k≤ Cℓ(ℓ, n,KN,Λ, ǫ).

(2) Uℓ0 is k-rectifiable.

(3) supy∈Brℓi

(xℓi )∩Skǫθrℓi

(y) ≤ Λ − ℓ · η

The proof of the Claim follows essentially the same steps as those for the main Claim of Theorem1.3.For base stepℓ = 0, we consider the decompositionSk

ǫ ⊆ U00 ∪ U0

+ whereU00 = ∅ andU0

+ = B1(p).Now let us assume we have proved the claim for someℓ, then we wish to prove the claim forℓ+ 1. Thus,

let us consider the setUℓ+

from the previous covering step given by

Uℓ+=

⋃

Brℓi(xℓi ) . (9.19)

Now let us apply Lemma8.1to each of the ballsBrℓi(xℓi ) in order to write

Skǫ ∩ Brℓi

(xℓi ) ⊆ Ui,0 ∪Ui,+ = Ui,0 ∪⋃

j

Br i, j (xi, j) , (9.20)

with the following properties:

(a) λk(Ui,0) + ωk∑

j rki, j ≤ C(n,KN,Λ, ǫ, p)(rℓi )

k,(b) supy∈Bri, j (xi, j )∩Sk

ǫθr i, j (y) ≤ Λ − (ℓ + 1)η,

(c) Ui,0 is k-rectifiable.

Now let us define the sets

Uℓ+10 =

⋃

Ui,0 ∪Uℓ0 ,

Uℓ+1+=

⋃

i, j

Br i, j (xi, j) . (9.21)

Conditions (2) and (3) from the Claim are clearly satisfied. We need only check condition (1). Using (a)and the inductive hypothesis we can estimate that

λk(Uℓ+10 ) + ωk

∑

i, j

(

r i, j)k≤ λk(Uℓ

0) +∑

i

(

λk(Ui,0) + ωk

∑

j

(

r i, j)k)

,

≤ Cℓ +C(n,KN,Λ, ǫ)∑

i

(

rℓi)k

≤ C(n,KN,Λ, ǫ) ·Cℓ

≡ Cℓ+1 . (9.22)

Thus, we have proved the inductive part of the claim, and thusthe claim itself.

Let us now finish the proof thatSkǫ ( f ) is rectifiable. So let us takeℓ = ⌈η−1

Λ⌉ = ℓ(η,Λ). Then if we applythe above Claim toℓ, then by the third property of the Claim we must have that

Uℓ+ ≡ ∅ , (9.23)


and therefore we have the covering

Skǫ ⊆ Uℓ

0 , (9.24)

whereUℓ0 is k-rectifiable with the volume estimateλk(Uℓ

0) ≤ C, which proves thatSkǫ is itself rectifiable.

Finally, we prove that fork a.e. x ∈ Skǫ there exists ak-dimensional subspaceVx ⊆ TxM such thatevery

tangent map atx is k-symmetric with respect toVx. To see this we proceed as follows. For eachη > 0 let usconsider the finite decomposition

Skǫ =

⌈η−1Λ⌉

⋃

α=0

Wk,αǫ,η , (9.25)

where by definition we have

Wk,αǫ,η ≡

x ∈ Skǫ : θ0(x) ∈

[

αη, (α + 1)η)

. (9.26)

Note then that eachWk,αǫ,η is k-rectifiable, and thus there exists a full measure subsetWk,α

ǫ,η ⊆ Wk,αǫ,η such that

for eachx ∈ Wk,αǫ,η the tangent cone ofWk,α

ǫ,η exists and is a subspaceVx ⊆ TxM.Now let us consider such anx ∈ Wk,α

ǫ,η , and letVkx be the tangent cone ofWk,α

ǫ,η at x. For all r << 1sufficiently small we of course have|θr (x) − θ0(x)| < η. Thus, by the monotonicity and continuity ofθ wehave for allr << 1 sufficiently small and ally ∈ Wk,α

ǫ,η ∩ Br(x) that |θr(y) − θ0(y)| < 2η. In particular, byTheorem2.1we have for eachy ∈Wk,α

ǫ,η ∩ Br(x) thatBr(y) is (0, δη)-symmetric, withδη → 0 asη→ 0. Nowlet us recall the cone splitting of Theorem2.3. Since the tangent cone atx is Vk

x, for all r sufficiently smallwe can findk + 1 pointsx0, . . . , xk ∈ Br(x) ∩Wk,α

ǫ,η which are 10−1r-independent, see Definition2.2, andfor which B2r(x j) are (0, δη)-symmetric. Thus, by the cone splitting of Theorem2.3 we have thatBr(x) is(k, δη)-symmetric with respect toVk

x for all r sufficiently small, whereδη → 0 asη→ 0. In particular, everytangent map atx is (k, δη)-symmetric with respect toVx, whereδη → 0 asη→ 0.

Now let us consider the sets

Wkǫ,η ≡

⋃

α

Wk,αǫ,η . (9.27)

SoWkǫ,η ⊆ Sk

ǫ is a subset of fullk-dimensional measure, and for every pointx ∈ Wkǫ,η we have seen thatevery

tangent map of is (k, δη)-symmetric with respect to someVx ⊆ TxM, whereδη → 0 asη → 0. Finally let usdefine the set

Skǫ ≡

⋂

j

Wkǫ, j−1 . (9.28)

This is a countable intersection of full measure sets, and thus Skǫ ⊆ Sk

ǫ is a full measure subset. Further, wehave for eachx ∈ Sk

ǫ thateverytangent map must be (k, δ)-symmetric with respect to someVx, for all δ > 0.In particular, every tangent map atx must be (k, 0) = k-symmetric with respect to someVx. This finishes theproof of the Theorem.


9.3. Proof of Theorem 1.5. Let us begin by observing the equality

Sk( f ) =⋃

ǫ>0

Skǫ ( f ) =

⋃

β∈N

Sk2−β( f ) . (9.29)

Indeed, if x ∈ Skǫ ( f ), then no tangent map atx can be (k + 1, ǫ/2)-symmetric, and in particulark + 1-

symmetric, and thusx ∈ Sk( f ). This shows thatSkǫ ( f ) ⊆ Sk( f ). On the other hand, ifx ∈ Sk( f ) then we

claim there is someǫ > 0 for which x ∈ Skǫ ( f ). Indeed, if this is not the case, then there existsǫi → 0 and

r i > 0 such thatBr i (x) is (k+ 1, ǫi)-symmetric. Ifr i → 0 then we can pass to a subsequence to find a tangentmap which isk + 1-symmetric, which is a contradiction. On the other hand, ifr i > r > 0 then we see thatBr(x) is itself k + 1-symmetric, and in particulareverytangent map atx is k + 1-symmetric. In either casewe obtain a contradiction, and thusx ∈ Sk

ǫ ( f ) for someǫ > 0. Therefore we have proved (9.29).

As a consequence,Sk( f ) is a countable union ofk-rectifiable sets, and therefore is itselfk-rectifiable. Onthe other hand, Theorem1.4tells us that for eachβ ∈ N there exists a setSk

2−β( f ) ⊆ Sk

2−β( f ) of full measure

such that

Sk2−β ⊆

x : ∃Vk ⊆ TxM s.t. every tangent map atx is k-symmetric wrtV

. (9.30)

Hence, let us define

Sk( f ) ≡⋃

Sk2−β( f ) . (9.31)

Then we still have thatSk( f ) hask-full measure inSk( f ), and if x ∈ Sk( f ) then for someβ we have thatx ∈ Sk

2−β, which proves that there exists a subspaceV ⊆ TxM such that every tangent map atx is k-symmetric

with respect toV. We have finished the proof of the theorem.

10. PROOF OFMAIN THEOREMS FORM INIMIZING HARMONIC MAPS

In this section we prove the main theorems of the paper concerning minimizing harmonic maps. That is,we finish the proofs of Theorem1.6 and Theorem1.8. In fact, the proofs of these two results are almostidentical, though the first relies on Theorem1.4and the latter on Theorem1.3. However, for completenesssake we will include the details of both.

10.1. Proof of Theorem1.6. We wish to understand better the size of the singular set S(f ) of a minimizingharmonic map. Let us recall that theǫ-regularity of theorem2.8 tells us that if f is minimizing, then thereexistsǫ(n,KN,Λ) > 0 with the property that ifx ∈ B1(p) and 0< r < r(n,KM ,KN,Λ) is such thatB2r(x) is(n− 2, ǫ)-symmetric, thenr f (x) ≥ r. In particular,x is a smooth point, and we have forǫ(n,KM ,KN,Λ) > 0that

Sing(f ) ∩ B1(p) ⊆ Sn−3ǫ ( f ) . (10.1)

Thus by Theorem1.4there existsC(n,KM ,KN,Λ) > 0 such that for each 0< r < 1 we have

Vol(

Br(

Sing(f ))

∩ B1(p))

≤ Vol(

Br

(

Sn−3ǫ ( f )

))

≤ Cr3 . (10.2)


This of course immediately implies, though of course is muchstronger than, the Hausdorff measure estimate

λn−3(Sing(f ) ∩ B1(p))

≤ C , (10.3)

which finishes the proof of the Theorem.

10.2. Proof of Theorem 1.8. We begin again by considering theǫ-regularity of theorem2.8. This tells usthat if f is minimizing, then there existsǫ(n,KN,Λ) > 0 with the property that ifx ∈ B1(p) and 0< r <r(n,KM ,KN,Λ) is such thatB2r(x) is (n − 2, ǫ)-symmetric, thenr f (x) ≥ r. In particular, we have for suchǫ, r that

x ∈ B1(p) : r f (x) < r ⊆ Sn−3ǫ,r ( f ) . (10.4)

Thus by Theorem1.4there existsC(n,KM ,KN,Λ) > 0 such that for each 0< r < 1 we have

Vol(

Brx ∈ B1(p) : r f (x) < r)

≤ Vol(

Br

(

Sn−3ǫ,r ( f )

))

≤ Cr3 , (10.5)

which proves the second estimate of (1.23). To prove the first we observe that|∇ f |(x) ≤ r f (x)−1, and toprove (1.24) we use the remark after Definition1.7 to conclude that|∇2 f |(x) ≤ C(n,KM ,KN)r f (x)−2. Thisconcludes the proof of the Theorem.

11. SHARPNESS OF THE RESULTS

In this section we present a few examples which motivate the sharpness of our results.

11.1. Sharpness ofLp Estimates for Minimizers. This example is completely standard, we wish to simplypoint out some of its properties. Namely, consider the mapping f : B2(03)→ S2 given by the projection

f (x) =x|x|. (11.1)

This is a minimizing harmonic map (see [Lin87, CG89]). It is easy to compute that

|∇ f |(x) ≈ r−1f (x) ≈

1|x|. (11.2)

We therefore get that|∇ f |, r−1f (x) have uniform estimates inL3

weak, however neither belong toL3. In particu-lar, this shows that the estimates of Theorem1.8are sharp.

11.2. Rectifiable-Reifenberg Example I. Let us begin with an easy example, which shows that the recti-fiable conclusions of Theorem3.3 is sharp. That is, one cannot hope for better structural results under thehypothesis. Indeed, consider anyk-dimensional subspaceVk ⊆ Rn, and letS ⊆ Vk ∩ B2(0n) be an arbitrarymeasurable subset. Then clearlyD(x, r) ≡ 0 for eachx andr > 0, and thus the hypotheses of Theorem3.3are satisfied, howeverS clearly need not be better than rectifiable. In the next example we shall see thatSneed not even come from a single rectifiable chart, as it does in this example.


11.3. Rectifiable-Reifenberg Example II. With respect to the conclusions of Theorem3.3 there are twonatural questions regarding how sharp they are. First, is itpossible to obtain more structure from the setSthan rectifiable? In particular, in Theorem3.2 there are topological conclusions about the set, is it possibleto make such conclusions in the context of Theorem3.3? In the last example we saw this is not the case.Then a second question is to ask whether we can at least find a single rectifiable chart which covers thewhole setS. This example taken from [DT12, counterexample 12.4] shows that the answer to this questionis negative as well.

To build our examples let us first consider a unit circleS1 ⊆ R3. Let M2 ⊃ S1 be a smooth Mobius striparound this circle, and letSǫ ⊆ M2∩Bǫ(S1) ≡ M2

ǫ be an arbitraryλ2-measurable subset of the Mobius strip,contained in a small neighborhood of theS1. In particular,Area(Sǫ ) ≤ Cǫ → 0 asǫ → 0. It is not hard,though potentially a little tedious, to check that assumptions of Theorem3.3hold for δ→ 0 asǫ → 0.

However, we have learned two points from these example. First, sinceSǫ was an arbitrary measurablesubset of a two dimensional manifold, we have that it is 2-rectifiable, however that is the most which maybe said ofSǫ . That is, structurally speaking we cannot hope to say betterthan 2-rectifiable about the setSǫ .More than that, sinceSǫ is a subset of the Mobius strip, we see that even thoughSǫ is rectifiable, we cannoteven coverSǫ by a single chart fromB1(02), as there are topological obstructions, see [DT12] for more onthis.

ACKNOWLEDGMENTS

We would like to thank prof.Camillo de Lellisfor his very precise comments on this paper and its earlierversions.

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