Representation Operators of Metric and Euclidian Charges Philippe Bouafia

Representation operators of metric and Euclidian charges Philippe Bouafia

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Philippe Bouafia. Representation operators of metric and Euclidian charges. General Mathematics [math.GM]. Université Paris Sud - Paris XI, 2014. English. NNT : 2014PA112004. tel-01015943

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THÈSE

Présentée pour obtenir

LE GRADE DE DOCTEUR EN SCIENCES DE L’UNIVERSITÉ PARIS-SUD XI

Spécialité: Mathématiques par Philippe Bouafia École doctorale de Mathématiques de Paris Sud Laboratoire d’Analyse Harmonique

Blow-up analysis of multiple valued stationary functions

Soutenue le 7 janvier 2014 devant la Commission d’examen: M. Petru Mironescu (Rapporteur, Université Lyon I) M. Camillo De Lellis (Rapporteur, Universität Zürich) M. Guy David (Directeur de thèse, Université Paris Sud) M. Thierry De Pauw (Promoteur de thèse, IMJ) M. Gilles Godefroy (Professeur d’université, IMJ) M. Stefan Wenger (Full professor, University of Fribourg) Thèse préparée au Département de Mathématiques d’Orsay Laboratoire de Mathématiques (UMR 8628), Bât. 425 Université Paris-Sud 11 91 405 Orsay CEDEX Abstract

On étudie les fonctions multivaluées vers un espace de Hilbert. Après avoir introduit une bonne notion de p énergie, on donne une définition possible d’espace de Sobolev et on prouve un théorème d’existence des p minimiseurs. Puis on considère les fonctions bivaluées de deux variables, stationnaires pour les déformations au départ et à l’arrivée. On démontre qu’elles sont localement lipschitziennes et on utilise cette régularité pour montrer la convergence forte dans W 1,2 vers leur unique éclatement en un point. L’ensemble de branchement d’une telle fonction est la réunion localement finie de courbes analytiques qui se rencontrent en faisant des angles égaux. Nous donnons aussi un exemple de fonction discontinue et stationnaire seulement pour les déformations au départ. Dans un deuxième temps, on prouve qu’il n’existe pas de rétraction uniformément continue de l’espace des champs vectoriels continus vers le sous-espace de ceux dont la divergence est nulle en un sens distributionnel. On généralise ce résultat en toute codimension en utilisant la notion de m charge et à tout ensemble X Rn vérifiant une hypothèse géométrique mineure. ⊂

Keywords : Stationary multiple valued function, branch set, normal currents, charges, metric calibrations.

Abstract

We study multiple valued functions with values in a Hilbert space. We introduce a possible definition of Sobolev spaces and the rightful notion of p energy. We prove the existence of p minimizers. Then we consider two-valued real functions of two variables which are stationary with respect to both domain and range transformations. We prove their local Lipschitz continuity and use it to establish strong convergence in W 1,2 to their unique blow-up at any point. We claim that the branch set of any such function consists of finitely many real analytic curves meeting at nod points with equal angles. We also provide an example showing that stationarity with respect to domain transformations only does not imply continuity. In a second part, we prove that there does not exist a uniformly continuous retraction from the space of continuous vector fields onto the subspace of vector fields whose divergence vanishes in the distributional sense. We then generalise this result using the concept of m charges on any subset X Rn satisfying a mild geometric condition, there is no uniformly ⊂ continuous representation operator for mcharges in X.

Keywords : Stationary multiple valued function, branch set, normal currents, charges, metric calibrations.

Contents

Introduction 8

I Almgren’s multiple valued functions 15

1 Preliminaries 16 1.1 Symmetric powers ...... 16 1.2 Concatenation and splitting ...... 18 1.3 Measurability ...... 19 1.4 Lipschitz extensions ...... 21 1.5 Diﬀerentiability ...... 28

2 Embeddings 37

n 2.1 Whitney bi-Hölder embedding — The case Y = ℓ2 (K) ...... 37 2.2 Splitting in case Y = R ...... 38

n 2.3 Almgren-White locally isometric embedding — The case Y = ℓ2 (R) . . . 39 ∗ 2.4 Lipeomorphic embedding into Lipy0 (Y ) ...... 45

3 Sobolev classes 48

3.1 Deﬁnition of Lp(X, QQ(Y )) ...... 48 3.2 Analog of the Fréchet-Kolmogorov compactness Theorem ...... 49

1 3.3 Deﬁnition of Wp (U; QQ(Y )) ...... 52 3.4 The p energy ...... 56 3.5 Extension ...... 62 3.6 Poincaré inequality and approximate diﬀerentiability almost everywhere . 64 3.7 Trace ...... 70 3.8 Analog of the Rellich compactness Theorem ...... 72 3.9 Existence Theorem ...... 74

4 Blowup analysis of stationary multiple valued functions 75 4.1 Squeeze and squash variations ...... 75 4.2 Regularity results ...... 79 4.3 Blowing up ...... 84 4.4 Branch set ...... 90 4.5 Stationary surfaces in R3 ...... 96

II Representation of charges 98

5 Charges and cohomology 99 5.1 Preleminaries on Federer-Fleming currents ...... 99 5.2 Topologies on the space of normal currents ...... 101 5.3 Charges and duality with normal currents ...... 103 5.4 A representation theorem for charges ...... 104 5.5 Charges vanishing at inﬁnity ...... 106

6 Functional analytical tools 108 6.1 Duality of vector lattices ...... 108 6.2 Representation of abstract M spaces ...... 110 6.3 Representation of abstract L spaces ...... 112 6.4 p absolutely summing operators ...... 114

6.5 Lp spaces ...... 116 6.6 Grothendieck theorem ...... 118 6.7 A splitting theorem ...... 122

7 Continuous representation operators 124 7.1 Main theorem ...... 124 7.2 Adaptation to charges of positive codimension ...... 126

8 Appendix to Part II 128

References 132 8 Introduction

The Plateau problem

When we observe a soap film spanning a wire frame, we can be interested in the principles that enable it to exist in certain geometric configurations and not others. The Plateau problem amounts to asking what are the possible shapes it can assume. The liquid surface acts as an elastic membrane. In a first time, we can say that the total surface energy of such a soap film is proportionnal to its area, though such a statement leaves behind situations with multiple sheets. A shape is a physical solution of the Plateau problem only if it cannot change to a configuration with less energy. Therefore, we will be interested in certain generalized surfaces of R3 that are stable, i.e stable critical points of the area functional.

The first physicist to have studied the geometry of soap films appears to be the Belgian Joseph Plateau. In his honor, various mathematical questions dealing with the geometry of soap-film-like surfaces are referred to as the Plateau problem. The first difficulty is that there is no universal agreement on what is meant by a “surface” and its “area”. Here is a list of different formulations, to see how this problem led to a bushy mathematical literature 9

• immersed surfaces: the ﬁrst progress in this direction was independently achieved in the thirties by Tibor Radó and Jesse Douglas. Both relied on setting up minimization problems and complex analysis. However, their work does not extend in higher dimensions.

• sets with ﬁnite perimeter, introduced by De Giorgi.

• rectiﬁable currents: introduced by Federer and Flemming. They generalize sets with ﬁnite perimeter to codimensions greater than 1.

• varifolds: they can useful for the study of stationary surfaces which are not necessarily minimizing. However, they lack a boundary operator.

• Almgren’s (M, 0, δ) minimal sets: they are accurate models of physical solutions, but the lack of a compactness theorem makes it diﬃcult to prove the existence of an area minimizer.

Establishing the existence and regularity of solutions to the Plateau problem in general dimensions and codimensions is one of the most challenging problem in geometric measure theory. In the eighties, Frederick J. Almgren proved the following theorem:

Theorem 1. Let T be an m dimensional mass minimizing current in Rn. Consider the set reg(T ) of all points x supp T supp ∂T for which there exists an open neighborhood U of x such that supp T ∈ supp ∂T\ U is an C∞ manifold. Then the set sing(T ) := supp T (supp ∂T reg(T\)) has Hausdorﬀ∩ dimension less than m 2. \ ∪ − This theorem is optimal, as shown by the case of complex varieties. Its proof has required development of several new geometric and analytic techniques, central among which is the utilization of multiple valued functions to study branching phenomena. We believe stationary varifolds can be well approximated by stationary multiple valued functions, and that one can deduce from this approximation some regularity results. The following question is open, even in the case Q = 2.

Conjecture 1. Suppose

(A) V is a 2 dimensional stationary varifold in R3 with integer multiplicities,

(B) ε [0, 1), Q N∗, ∈ ∈ (C) V B3(0, 1) π(Q + ε) k k ≤ Then there exists γ (0, 1) such that H 2(sing V B3(0, γ)) = 0. ∈ ∩

Multiple valued functions

Let Y be a metric space and Q be an integer. We deﬁne QQ(Y ) to be the subset of measures in Y : Q Q (Y ) := y : y , . . . , y Y . Q  i 1 Q ∈  Xi=1  J K   10

We will adopt the compact notation

Q y1, . . . , yQ = yi . Xi=1 J K J K Elements of QQ(Y ) are just unordered Q tuples of points in Y . What we mean by a multiple valued function in the sense of Almgren is just a QQ(Y ) valued function.

A natural metric on QQ(Y ) is given by

Q 2 ( y1, . . . , yQ , z1,...,zQ ) := min v d(yi,zσ(i)) , G σ∈SQ u uXi=1 t where d is the metricJ of Y . K J K Typical examples of multiple valued functions include inverses of non injective functions, such as the map z C √z Q2(C). Another important example is given by slicing a (mass minimizing∈ or7→ stationary)∈ integral current in vertical directions in a neighborhood of a point with multiplicity Q, or slicing in a cylinder in which a current has no boundary. In the ﬁrst part of his monumental work, Almgren establishes the foundations of multiple valued calculus in case Y is Euclidean. His theory is based on the existence of a Euclidean embedding ξ : Q (Rn) RN(n,Q) Q → n such that ξ is Lipschitzian, ξ(QQ(R )) is a polyhedral cone and there is a Lipschitz retraction map ρ : RN(n,Q) ξ(Q (Rn)). → Q A Rademacher type theorem follows easily. Also, we can naturally deﬁne a generalized m n Dirichlet integral and Sobolev functions from R to QQ(R ) as one would between two Riemannian manifolds. Such an “isometric” embedding is unknown when Y is a Banach space, even when Y = ℓ2. Thus we are led to develop further the intrinsic approach pioneered in [Gob06] and [DLS11] (yet we cannot dispense completely with the locally isometric embedding, in particular when proving the lower semicontinuity of the energy).

m Though a Lipschitz function f : R QQ(Y ) (when Y is a Banach space with the Radon-Nikodým property) does not generally→ decompose into Q Lipschitz “branches”, we nevertheless establish their differentiability almost everywhere, for an appropriate notion of a derivative Df that controls the variations of f. In case Y is finite dimensional, this had been obtained by F.J. Almgren [AST00], the third author [Gob06], and C. De Lellis and E. Spadaro [DLS11]. Our proof in the infinite dimensional setting follows essentially that given in the last two references.

Sobolev Spaces

m Letting U = U(0, 1) be the unit ball of ℓ2 , we consider the Borel measurable maps f : U Q (ℓ ) with ﬁnite L “norm” → Q 2 p p m |f|Lp := (f, Q 0 ) dL < . ZU G ∞ J K 11

Their Lp semidistance is defined as 1/p p m dp(f1, f2) := (f1, f2) dL . ZU G 1 The Sobolev maps f Wp (U; QQ(ℓ2)) are defined to be the limits in this Lp semidistance of sequences of Lipschitz∈ maps f : U Q (ℓ ) such that j → Q 2 p m sup |Dfj| dL < . j ZU ∞ This sort of “weak density” of Lipschitz Q-valued maps among Sobolev ones is justified, n in case Y = ℓ2 is finite dimensional, by the fact that U is an extension domain and that im ξ is a Lipschitz retract of RN . That Sobolev Q-valued maps extend from U to the m whole ℓ2 , with the appropriate control, is a matter of routine verification. p We define the p energy Dirp(f; U) of a Sobolev Q-valued map f by relaxation

p p m Dirp(f; U) := inf |Dfj| dL : fj are Lipschitz and converge weakly to f , ZU where the weak convergence means that (fj) is a sequence of Sobolev maps such that limj dp(f, fj) = 0 and p m sup |Dfj| dL < . j ZU ∞ Thus Dirp is automatically lower semicontinuous with respect to convergence in the Lp semidistance, and we then embark on showing that f is diﬀerentiable almost everywhere and that the following intuitive formula holds:

p p m Dirp(f; U) = |Df| dL . ZU For this purpose we need to know the corresponding statement for finite dimensional ap- n proximating Sobolev maps U QQ(ℓ2 ), a convergence result for the finite dimensional approximations, a Poincaré inequality→ from which a stronger (Luzin type) approximation by Lipschitz Q-valued maps follows. The differentiability almost everywhere of a Sobolev Q-valued map now becomes a consequence of our aforementioned Rademacher type result. At that point we also obtain that

p p m Dirp(f; U) = |Df| dL , ZU thus the lower semicontinuity sought for. We prove the existence of a useful trace “operator” T , verifying the following continuity property: If (fj) is a sequence of Sobolev converging weakly to f, then limj dp(T (f), T (fj)) = 0. Finally, the multiple valued Rellich compactness Theorem relies on a Fréchet-Kolmogorov compactness Theorem and a new embedding Theorem.

Given a Lipschitz g : ∂U QQ(ℓ2) and 1 < p < , our main result states that the minimization problem → ∞

p m minimize U |Df| dL among f R W 1(U; Q (ℓ )) such that T (f) = g  ∈ p Q 2 admits a solution. 12

Stationary multiple valued functions

The first step in Almgren’s proof of Theorem 1 establishes the same kind of dimensional bound for some multiple valued functions, namely, those minimizing the Dirichlet energy. Not much is known when we replace Dirichlet minimizing functions with critical points of the Dirichlet functional. In particular, the following question was discussed in [AA86]: What can we say about the branch set of such stationary multiple valued functions ? This question is motivated by the study of the singular set of stationary integral currents. An example showing that its dimension can exceed m 2 is provided by a multiple valued function whose graph consists of two “opposite” hyperplanes.− Let us consider two-valued functions of two variables, we assume that they are stationary with respect to transformations of the domain – usually such functions are called stationary harmonic, yet we will follow Almgren’s terminology and use the term squeeze stationary. Our first result is that squeeze stationary functions are locally Lipschitz continuous. To prove this fact, we can rely on complex-analytic arguments since we work in two dimensions. Indeed, squeeze stationarity of a function f is equivalent to the Hopf form of f being holomorphic ; the Hopf form carries information on the derivatives, which implies regularity of f. Thus branching is the only type of singularity which may occur. Our main theorem states that, under the conditions m = 2, Q = 2 and n = 1, a stationary function has a singular set of dimension 1, which consists of finitely many real analytic curves in the interior and C1 up to the boundary. The extremities of the curves are precisely those branch points around which one does not decompose f as a sum of two harmonic functions. We follow a usual blow-up method. Our proof is divided into three steps.

(1) Existence of blow-ups: rescalings of a stationary multiple valued function converge to a tangent object, namely, a homogeneous stationary function. To prove this fact, we use Almgren’s frequency function; this function was shown to be nondecreasing as a consequence of stationarity – up to some minor technicalities. This fact plays a role analogous to the monotonicity of the density ratio in Allard’s regularity theorem. We also need a compactness argument to allow sequences of rescalings to converge ; one delicate point here is that unlike for Dirichlet minimizing functions, we cannot use comparison arguments. Moreover, the convergence must be strong since the equations of stationarity are nonlinear. Note that our compactness result strongly relies on the Lipschitz continuity of stationary functions.

(2) Classiﬁcation of blow-ups. As we work in low dimension, it is not diﬃcult to completely classify homogeneous stationary functions by hand. Several facts are worth noting. First, the homogeneity degree of a blow-up corresponds to its frequency. Second, since convergence of rescalings is strong in W 1,2, the Hopf forms converge during the blow-up process. This implies that there is no loss of energy. Finally, as we are able to list a posteriori all homogeneous stationary functions, we can deduce the uniqueness of blow-ups. The blow-up at one point depends only on the value of the Hopf form and its successive derivatives.

(3) Regularity of the branch set. Consider a stationary function f. We need some 13

topological results to compare the local behavior of the branch set of f with that of its blow-ups. The blow-ups vary continuously because of the continuity of the Hopf form ; we then deduce that the branch set is C∞ and then improve the regularity to real analyticity.

We hope that this work can be extended to Q-valued stationary functions, for Q 3. ≥

Charges

To understand the notion of charge, one needs some acquaintance with currents, which have been developed by Federer and Fleming (see [Fed96] for an extensive presentation). We know that differential forms can be seen as acting on oriented submanifolds by integration N ω. 7→ ZN Inspired by this duality, one may define charges as continuous linear functionals on a peculiar class of currents, known as normal currents. Unlike submanifolds, normal currents fortunately have a linear space structure, but the suitable topology is not obvious. For example, if we endow the space of normal currents with the flat norm topology, we obtain flat cochains instead of charges. This topology does not behave correctly. Besides, we do not control the normal mass of a converging sequence of normal currents. The localized topology on normal currents goes over this difficulty; it is therefore perfectly suited to define charges. Generalizing Stokes’ formula, it is possible to define a weak coboundary operator d for charges; charge cohomology is the associated cohomology. We give the proof of a significant representation theorem that helps understand the link between charges and continuous differential forms. Namely, for each m charge α, there exists a continuous m-form ω and a continuous (m 1) form ζ such that − α = ω + dζ.

Hence, charge cohomology is the natural notion to deal with continuous differential forms. Still let us mention that there exists an analogue representation theorem for flat cochains, due to Wolfe which asserts that flat cochains are L∞ differential forms whose weak derivatives are L∞. Unlike Wolfe’s theorem, the representation theorem for charges is rather abstract; it does not yield any particular continuous representatives to a given charge. A natural question is whether there is a canonical choice for ω and ζ in the representation formula. Section 7 is the original part on this subject: it states that there is no uniformly continuous choice α (ω,ζ). 7→ Sections 1, 2 and 3 of Part I are taken from the article [BGP12] written in collab- oration with Thierry De Pauw and Jordan Goblet. Section 4 is taken from [Bou13], whereas the last part on charges is taken from [Bou11]

General notations

The set of nonnegative integers is N and the one of positive integers is denoted by N∗. 14

n n We will denote by R or more accurately ℓ2 the Euclidean space of dimension n, and by ℓ2 the Hilbert space of square summable sequences. The symmetric diﬀerence of sets is denoted by . ⊖ If (Y, d) is a metric space and m is a nonnegative integer, we deﬁne, for any subset A Y and δ > 0, ⊂

m m Hδ := α(m) (diam Ai) : A i∈I Ai,I countable , ( ⊂ ∪ ) Xi∈I m where α(m) is the m volume of the unit ball in ℓ2 . The Hausdorﬀ measure of A is

m m H (A) = sup Hδ (A). δ>0

m m m We write L instead of H whenever Y = ℓ2 . A function f : X Y between two metric spaces is called Lipschitz continuous whenever → d (f(x), f(y)) Lip f := sup Y : x = y X < . ( dX (x, y) 6 ∈ ) ∞

The closed unit ball of a Banach space X is denoted by BX . The operator norm of a continuous linear map T : X Y between two normed spaces is deﬁned by → T := sup T (x) : x B . k k {k kY ∈ X }

When µ is a measure on a metric space and a X, we let the m density of µ at a be the quantity (whenever it exists) ∈

µ(B(a, r)) Θm(µ, a) := lim . r→0 α(m)rm 15 Part I Almgren’s multiple valued functions

Summary 1 Preliminaries 16 1.1 Symmetric powers ...... 16 1.2 Concatenation and splitting ...... 18 1.3 Measurability ...... 19 1.4 Lipschitz extensions ...... 21 1.5 Diﬀerentiability ...... 28

2 Embeddings 37 n 2.1 Whitney bi-Hölder embedding — The case Y = ℓ2 (K) ...... 37 2.2 Splitting in case Y = R ...... 38 n 2.3 Almgren-White locally isometric embedding — The case Y = ℓ2 (R) 39 ∗ 2.4 Lipeomorphic embedding into Lipy0 (Y ) ...... 45

3 Sobolev classes 48

3.1 Definition of Lp(X, QQ(Y )) ...... 48 3.2 Analog of the Fréchet-Kolmogorov compactness Theorem ...... 49 1 3.3 Definition of Wp (U; QQ(Y )) ...... 52 3.4 The p energy ...... 56 3.5 Extension ...... 62 3.6 Poincaré inequality and approximate differentiability almost everywhere 64 3.7 Trace ...... 70 3.8 Analog of the Rellich compactness Theorem ...... 72 3.9 Existence Theorem ...... 74

1.1 Symmetric powers

Let Q be a positive integer and let Y be a metric space. Our aim is to consider unordered Q-tuples of elements of Y . For instance, letting Y = C and letting P be a polynomial of degree Q with coeﬃcients in C, the roots of P form such an unordered Q-tuple of complex numbers. Thus the elements under consideration need not be distinct; if some agree they should be counted with their multiplicity.

Formally the collection QQ(Y ) of unordered Q-tuples in Y may be deﬁned as the Q quotient of the Cartesian product Y under the action of the symmetric group SQ. An element σ S is a permutation of 1,...,Q . It acts on Y Q in the obvious way : ∈ Q { } Y Q Y Q :(y , . . . , y ) (y , . . . , y ) . → 1 Q 7→ σ(1) σ(Q)

We will denote by y1, . . . , yQ the equivalence class of (y1, . . . , yQ) in QQ(Y ), so that in particular y1, . . . , yQ = yσ(1), . . . , yσ(Q) for every σ SQ. On occasions we shall J K ∈ also denote by v a generic element of QQ(Y ). Another way of thinking of a member v = y , . . . , yJ Q (YK) isJ to identify it withK the finite measure µ = Q δ where 1 Q ∈ Q v i=1 yi δyi is the Dirac mass with atom yi . The support of v QQ(Y ) is, by definition, the J K { } ∈ P support of the corresponding measure, supp v = supp µv = y1, . . . , yQ where y1, . . . , yQ is a numbering of v, i.e. a map y : 1,...,Q Y such{ that v = }y , . . . , y . The { } → 1 Q multiplicity of y supp v is defined as µv y . ∈ { } J K We now define a metric on QQ(Y ) associated with the given metric d of Y . Let

Q ′ ′ ′ 2 ( y1, . . . , yQ , y1, . . . , yQ ) = min v d(yi, yσ(i)) . G σ∈SQ u uXi=1 J K J K t We will sometimes use the notation 2 for in order to avoid confusion with two other useful metrics: G G

Q ′ ′ ′ 1( y1, . . . , yQ , y1, . . . , yQ ) = min d(yi, yσ(i)) , G σ∈SQ Xi=1 J K J K and ′ ′ ′ ∞( y1, . . . , yQ , y1, . . . , yQ ) = min max d(yi, yσ(i)) . G σ∈SQ i=1,...,Q Thus , and J are equivalentK J metricsK on Q (Y ). G1 G2 G∞ Q We begin with the following easy proposition.

Proposition 1. The metric space (Y, d) is complete (resp. compact, separable) if and only if (Q (Y ), ) is complete (resp. compact, separable) for every Q N∗. Q G ∈

Proof. As (Q1(Y ), ) is isometric to (Y, d) we need only to check that (QQ(Y ), ), Q 2, inherits theG properties of (Y, d). Let v be a Cauchy sequence in Q (YG). ≥ { j} Q It suﬃces to establish the convergence of some subsequence, still denoted vj , and we choose it so that (v , v ) < 2−j. We let v = y , . . . , y and we choose{ inductively} G j j+1 0 0,1 0,Q J K 1.1 - Symmetric powers 17

2 Q 2 a numbering vj = yj,1, . . . , yj,Q so that (vj, vj+1) = i=1 d(yj,i, yj+1,i) . It follows that each y , i = 1,...,Q, is a CauchyG sequence in Y . Letting y denote its limit, { j,i}j P i and v = y , . . . , y J , it remainsK to observe that (v, v )2 Q d(y , y )2 0. This 1 Q G j ≤ i=1 i j,i → proves that if Y is complete then so is QQ(Y ). Regarding totalP boundedness we notice that if a ﬁniteJ subsetK A Y and r > 0 are so that ⊂

Y = Bd(y,r) y[∈A then

QQ(Y ) = BG∞ ( y1, . . . , yQ ,r) . Q (y1,...,y[Q)∈A J K As for separability, if D is dense countable subset of (Y, d), then so is QQ(Y ) y , . . . , y :(y , . . . , y ) DQ in (Q (Y ), ). ∩ { 1 Q 1 Q ∈ } Q G∞ J K A Q-valued function from a set X to Y is a mapping f : X QQ(Y ). A multiple- valued function from X to Y is a Q-valued function for some positive→ integer Q. In case X is a metric space, the notion of continuity (in particular Lipschitz continuity) of such f now makes sense. If A is a σ-algebra of subsets of X we say that f is A-measurable (or simply measurable when A is clear from the context) whenever f −1(B) A for every Borel subset B Q (Y ). ∈ ⊂ Q Our coming observation will reveal ubiquitous. We deﬁne the splitting distance of v = y , . . . , y Q (Y ) as follows: 1 Q ∈ Q J K min d(y , y ): i, j = 1,...,Q and y = y if card supp v > 1 split v = { i j i 6 j} + if card supp v = 1 .  ∞ 

n ′ Lemma 1 (Splitting Lemma). Let v = y1, . . . , yQ QQ(R ) and v QQ(Y ) be such ′ ∈′ ′ ′ ∈ that (v, v ) split v/2. Choose a numbering of v = y1, . . . , yQ QQ(Y ) so that G′ ≤ J K ∈ d(yi, yi) split v/2, i = 1,...,Q. It follows that ≤ J K Q ′ ′ 2 (v, v ) = v d(yi, yi) G u uXi=1 t (and the analogous statement for and ). G1 G∞

Proof. We ﬁrst observe that in case split v = the conclusion indeed holds true. Thus ∞ we assume that split v < . Let σ SQ and i = 1,...,Q. We aim to show that ′ ′ ∞ ∈ d(yi, yi) d(yσ(i), yi). In case yσ(i) = yi this is obvious. Otherwise, assuming if possible ≤ ′ ′ that d(yσ(i), yi) < d(yi, yi) we would infer from the triangle inequality

split v d(yσ(i), yi) ≤ ′ ′ d(yσ(i), yi) + d(yi, yi) ≤ ′ < 2d(yi, yi) split v , ≤ 18 Preliminaries a contradiction. Since i = 1,...,Q is arbitrary we obtain

Q Q d(y , y′)2 d(y , y′)2 . i i ≤ σ(i) i Xi=1 Xi=1 Since σ S is arbitrary, the proof is complete. ∈ Q Proposition 2. The function σ : QQ(Y ) N : v card supp v is lower semicontinuous. → 7→

′ Proof. It follows easily from the deﬁnition of split v that if v, v QQ(Y ) and if (v′, v) < split v/2 then card supp v′ card supp v. ∈ G∞ ≥

1.2 Concatenation and splitting

Let Q1 and Q2 be positive integers. We deﬁne the concatenation operation

: Q (Y ) Q (Y ) Q (Y ):(v , v ) v v ⊕ Q1 × Q2 → Q1+Q2 1 2 7→ 1 ⊕ 2 as follows. Write v = y , . . . , y and v = y , . . . , y , and put v v = 1 1,1 1,Q1 2 2,1 2,Q2 1 ⊕ 2 y1,1, . . . , y1,Q1 , y2,1, . . . , y2,Q2 . We observe that this operation is commutative, i.e. v1 J K J K ⊕ v2 = v2 v1. We notice the following associativity property. If Q1,Q2 and Q3 are J ⊕ K positive integers and vj QQj (Y ), j = 1, 2, 3, then (v1 v2) v3 = v1 (v2 v3) so that v v v is well∈ defined. It is thus possible to⊕ iterate⊕ the definition⊕ to⊕ the 1 ⊕ 2 ⊕ 3 concatenation of any finite number of members of some QQj (Y ). In this new notation we readily have the identity

y , . . . , y = y ... y = Q y . 1 Q 1 ⊕ ⊕ Q ⊕i=1 i J K J K J K J K We leave the obvious proof of the next result to the reader. Proposition 3. Let Q ,...,Q N∗. The concatenation operation 1 k ∈ Q (Y ) Q (Y ) Q (Y ):(v , . . . , v ) v ... v Q1 × · · · × Qk → Q1+···+Qk 1 k 7→ 1 ⊕ ⊕ k is Lipschitz continuous.

In fact if each Q (Y ) appearing in the statement is equipped with the metric , and Q G1 if the Cartesian product is considered as an ℓ1 “product”, then the Lipschitz constant of the above mapping equals 1.

Given Q maps f1, . . . , fQ : X Y we deﬁne their concatenation f : X QQ(Y ) by the formula → → f(x) = f (x), . . . , f (x) = Q f (x) , x X. 1 Q ⊕i=1 i ∈ Abusing notation in the obviousJ way we shallK also writeJ K

f = f1, . . . , fQ .

In writing f as above we will call f1, . . .J , fQ branchesK of f. It is most obvious that such splitting of f into branches is always possible, and equally evident that branches are 1.3 - Measurability 19 very much not unique unless X is a singleton. It ensues from the above proposition that if fi : X Y , i = 1,...,Q, are measurable (resp. continuous, Lipschitz continuous) then so is→ their concatenation f = Q f . Now, if f has some of these properties, can ⊕i=1 i it be split into branches f1, . . . , fQ having the same property? The answer is positive for measurability, as we shall see momentarily,J K but not for continuity. Consider f : C Q2(C) deﬁned by f(z) = √z, √z . Thus f is (Hölder) continuous (for a recent account→ of such continuity, consult− e.g. [Bri10]). We claim however that f does not decompose into two continuousJ branches.K In fact we shall argue that the restriction of f to the unit circle, still denoted f,

f : S1 Q (S1): z √z, √z → 2 7→ − does not admit a continuous selection. SupposeJ if possibleK that there are continuous maps f , f : S1 S1 such that f = f , f . Let g : S1 S1 : z z2. From the 1 2 → 1 2 → 7→ identity idS1 = g f1 we infer that 1 = deg(g f1) = deg(g) deg(f1) = 2 deg(f1), contradicting deg(◦f ) Z. J K ◦ ◦ 1 ∈

1.3 Measurability

This section is also contained in [DLS11]. The process of splitting v Q (Y ) (such ∈ Q that split v < ) into v1 QQ1 (Y ) and v2 QQ2 (Y ), Q = Q1 + Q2 and Q1 = 0 = Q2, is locally well-deﬁned∞ and∈ continuous. ∈ 6 6

Proposition 4. Let v Q (Y ) be such that s = split v < . There then exist ∈ Q ∞ Q1,Q2 > 0 with Q = Q1 + Q2 and continuous mappings

ψ : Q (Y ) v′ : (v, v′) < s/2 Q (Y ) , k = 1, 2 , k Q ∩ { G∞ } → Qk such that v′ = ψ (v′) ψ (v′) . 1 ⊕ 2

Proof. Choose 1 Q1 < Q and a numbering v = y1, . . . , yQ so that y1 = = yQ1 and ≤ ′ ··· ′ d(y1, yi) s for every i = 1 + Q1,...,Q. Let v QQ(Y ) be such that ∞(v, v ) < s/2 ≥ ′ ′ ′ ∈J K ′ G ′ and choose a numbering v = y1, . . . , yQ such that d(yi, yi) ∞(v, v ) for every i = 1,...,Q. We claim that the mappings ≤ G J K ψ (v′) = y′ , . . . , y′ Q (Y ) 1 1 Q1 ∈ Q1 and J K ψ (v′) = y′ , . . . , y′ Q (Y ) 2 1+Q1 Q ∈ Q−Q1 ′ ′′ ′′ are well-deﬁned. Suppose indeed thatJ v = y1 , .K . . , yQ is another numbering such that ′′ ′ ′ d(yi, yi ) ∞(v, v ), i = 1,...,Q. If some yi with i = 1,...,Q1 were equal to some ′′ ≤ G J K yj with j = 1 + Q1,...,Q then the triangle inequality would imply that d(yi, yj) 2 (v, v′), a contradiction since 2 (v, v′) < s. Thus y′ , . . . , y′ = y′′, . . . , y′′ . ≤ G∞ G∞ 1 Q1 1 Q1 ′ ′ ′ Since readily v = ψ1(v ) ψ2(v ), it only remainsJ to checkK thatJ ψ1 and ψK2 are ′ ⊕ ′ continuous. Let vj be a sequence in the domain of ψ1 converging to v . Assume { } ′ ′ ′ ′ j is large enough for ∞(v , vj) < δ := min (split v )/2, s/2 ∞(v, v ) . Choose a ′ ′ G ′ ′ { ′ − G }′ numbering vj = yj,1, . . . , yj,Q such that d(yi, yj,i) < δ. Notice that d(yi, yj,i) < s/2 for J K 20 Preliminaries

′ ′ ′ each i = 1,...,Q, and consequently ψ1(vj) = yj,1, . . . , yj,Q1 . Finally we infer from the Splitting Lemma that J K Q1 (ψ (v′), ψ (v′ )) d(y′, y′ ) G1 1 1 j ≤ i j,i Xi=1 Q d(y′, y′ ) ≤ i j,i Xi=1 = (v′, v′ ) . G1 j The continuity of ψ2 is proved in the same way.

When Y is a metric space we let BY denote the σ-algebra of Borel subsets of Y . Proposition 5. Let (X, A) be a measurable space and let Y be a separable metric space.

(A) If f , . . . , f : X Y are (A, B )-measurable then f = f , . . . , f is 1 Q → Y 1 Q (A, BQQ(Y ))-measurable. J K

(B) If f : X QQ(Y ) is (A, BQQ(Y ))-measurable then there exist (A, BY )-measurable maps f ,→ . . . , f : X Y such that f = f , . . . , f . 1 Q → 1 Q J K Proof. (A) Since (QQ(Y ), ∞) is separable (Proposition 1), each open subset of QQ(Y ) is G −1 a ﬁnite or countable union of open balls. Thus it suﬃces to show that f (BG∞ (v,r)) A whenever v Q (Y ) and r > 0. Writing v = y , . . . , y we simply notice that ∈ ∈ Q 1 Q −1 f (B ∞ (v,r)) = X x : (f(x), v)

= X x : min max d(fi(x), yσ(i))

(B) The proof is by induction on Q. The case Q = 1 being trivial, we henceforth assume that Q 2. We start by letting F = QQ(Y ) v : card supp v = 1 . Notice F is ≥ −1 ∩{ } closed, according to Proposition 2, thus A0 = f (F ) A. There readily exist identical (A, B )-measurable maps f 0, . . . , f 0 : A Y such∈ that f ↾ A = f 0, . . . , f 0 . We Y 1 Q 0 → 0 1 Q next infer from Proposition 4 that to each v QQ(Y ) F there correspond a neighbor- v v ∈ ∗ \ v vJ K hood Uv of v in QQ(Y ) F , integers Q1,Q2 N such that Q = Q1 +Q2, and continuous v \ ∈ v v maps ψ : Uv QQv (Y ), k = 1, 2, such that ψ ψ = id . Since QQ(Y ) F is sep- k → k 1 ⊕ 2 Uv \ arable we find a sequence v such that Q (Y ) F = N∗ U . Thus we find a { j} Q \ ∪j∈ vj disjointed sequence Bj of Borel subsets of QQ(Y ) such that QQ(Y ) F = j∈N∗ Bj { } −1 ∗ \ ∪ ∗ and Bj Uvj for every j. Define Aj = f (Bj) A, j N . For each j N the induction⊂ hypothesis applies to the two multiple-valued∈ functions∈ ∈

vj v ψk (f ↾ Aj): Aj QQ j (Y ), k = 1, 2 ◦ → k j j j j to yield (A, BY )-measurable decompositions f1 , . . . , f vj and f vj , . . . , fQ (the Q1 1+Q1 numberings are chosen arbitrarily). We deﬁne f : X Y , i = 1,...,Q, by letting J i K J K f ↾ A = f j, j N∗. It is now plain that each f is→ (A, B )-measurable and that i j i ∈ i Y f = f1, . . . , fQ .

J K 1.4 - Lipschitz extensions 21

1.4 Lipschitz extensions

The Lipschitz extension Theorem 2 is due to F.J. Almgren in case Y is finite dimensional (see [AST00, 1.5]), a former version is found in [AA86] for a different notion of multiple- valued function). Here we merely observe that it extends to the case when Y is an arbitrary Banach space (in case Q = 1 this observation had already been recorded in [JLS84], the method being due to H. Whitney [Whi34]). This exposition is very much inspired by that of [DLS11] (see also [LS05] for a comprehensive study of the extension techniques used here). This extension Theorem in case Y is finite dimensional n is equivalent to the fact that QQ(R ) is an absolute Lispschitz retract (see Theorem 5). The latter is proved “by hand” in [AST00, 1.3]. Given a map f : X Y between two metric spaces, and r > 0, we recall that the oscillation of f at r is defined→ as

osc(f; r) = sup d (f(x ), f(x )) : x , x X and d (x , x ) r [0, + ] . { Y 1 2 1 2 ∈ X 1 2 ≤ } ∈ ∞ Proposition 6. Let Q 2. Assume that ≥ (1) X and Y are metric spaces, x X, and δ := diam X < ; 0 ∈ ∞ (2) f : X (Q (Y ), ) and f(x ) = y (x ), . . . , y (x ) ; → Q G∞ 0 1 0 Q 0

(3) There are i1, i2 1,...,Q such thatJ K ∈ { } d (y (x ), y (x )) > 3(Q 1) osc(f; δ) . Y i1 0 i2 0 −

∗ It follows that there are Q1,Q2 N such that Q1 + Q2 = Q, and f1, f2 : X QQ(Y ) such that f = f f and osc(f∈; ) osc(f; ), j = 1, 2. → 1 ⊕ 2 j · ≤ · Proof. We let J denote the family of all those J 1,...,Q such that i J and for ⊂ { } 1 ∈ every j1, j2 J, ∈ d (y (x ), y (x )) 3(card J 1) osc(f; δ) . (1) Y j1 0 j2 0 ≤ − Notice that J = (because i J ), and let J J be maximal with respect to 6 ∅ { 1} ∈ 1 ∈ inclusion. Also deﬁne J2 = 1,...,Q J1, so that J2 = : according to hypothesis (3), J contains at least i . We notice{ that}\ for every j J6 and∅ every j J one has 2 2 1 ∈ 1 2 ∈ 2

dY (yj1 (x0), yj2 (x0)) > 3 osc(f; δ), (2) otherwise J j J which contradicts the maximality of J . 1 ∪ { 2} ∈ 1 For each x X we choose a numbering f(x) = y (x), . . . , y (x) such that ∈ 1 Q

∞(f(x0), f(x)) = max dY (Jyi(x0), yi(x)) . K G i=1,...,Q

We let Q1 = card J1, Q2 = card Q2, and we deﬁne fj : X QQ(Y ), j = 1, 2, by the formula f (x) = y (x): i J , so that f = f f . → j i ∈ j 1 ⊕ 2 For each pair x, x′ X we choose σ ′ S such that J ∈ K x,x ∈ Q ′ ′ ∞(f(x), f(x )) = max dY (yi(x), yσ ′ (i)(x )) . G i=1,...,Q x,x 22 Preliminaries

We now claim that σx,x′ (J1) = J1 and σx,x′ (J2) = J2, and this will readily ﬁnish the proof. Assume if possible that there exist j1 J1 and j2 J2 such that σx,x′ (j1) = j2. Thus d (y (x), y (x′)) (f(x), f(x′)), and∈ it would follow∈ from Equation (2) that Y j1 j2 ≤ G∞

3 osc(f; δ) < dY (yj1 (x0), yj2 (x0)) d (y (x ), y (x)) + d (y (x), y (x′)) + d (y (x′), y (x )) ≤ Y j1 0 j1 Y j1 j2 Y j2 j2 0 (f(x ), f(x)) + (f(x), f(x′)) + (f(x′), f(x )) ≤ G∞ 0 G∞ G∞ 0 3 osc(f; δ) , ≤ a contradiction.

∗ Proposition 7. For each Q N there is a constant c7(Q) 1 with the following property. Assume that ∈ ≥

(1) X and Y are Banach spaces; (2) C X is a closed ball; ⊂ (3) f :(∂C, ) (Q (Y ), ) is Lipschitz. k · k → Q G∞ It follows that f admits an extension fˆ :(C, ) (Q (Y ), ) such that k · k → Q G∞ Lip fˆ c (Q) Lip f , ≤ 7 and max (fˆ(x), v): x C (6Q + 2) max (f(x), v): x ∂C {G∞ ∈ } ≤ {G∞ ∈ } for every v Q (Y ). ∈ Q Proof. There is no restriction to assume that C = B(0,R), R > 0, is a ball centered at the origin. Note that it is enough to construct a Lipschitz extension fˆ of f on a dense subset of C, for example on C 0 . The proof is by induction on Q, and we start with the case Q = 1. Choose x ∂C\{.} We deﬁne 0 ∈ x x Rx fˆ(x) = 1 k k f(x0) + k kf , x C 0 . − R ! R x ! ∈ \{ } k k This is readily an extension of f to B(0,R) 0 . In order to estimate its Lipschitz constant, we let x, x′ B(0,R) 0 , we put r\{= }x , r′ = x′ , and we assume r r′. ′ ′′ rx ∈ \{ } ′′ k k k k ≤ We deﬁne x = ′ , such as x = x and we observe that r k k k k x Rx x′′ Rx′′ fˆ(x) fˆ(x′′) = k kf k kf k − k R x ! − R x′′ !

k k ′ k k r Rx Rx = f f R x ! − x′ !

k k k′ k x x r(Lip f) ≤ x − x′

k k′′ k k = (Lip f) x x k − k (Lip f)( x x′ + x′ x′′ ) ≤ k − k k − k = (Lip f)( x x′ + x′ x ) k − k k k − k k 2(Lip f) x x′ , ≤ k − k 1.4 - Lipschitz extensions 23 and

′′ ′ ′ ′′ ′ x x Rx fˆ(x ) fˆ(x ) = k k − k k f f(x0) k − k R x′ ! − !

′ k k x x k k − k k 2R Lip f R · 2(Lip f) x x′ ≤ k − k Therefore, Lip fˆ 4 Lip f . ≤ Moreover, if v Y and x C, we compute ∈ ∈ x x Rx fˆ(x) v 1 k k f(x0) v + k k f v k − k ≤ − R ! k − k R x ! −

k k max f(ξ) v . ≤ ξ∈∂C k − k We are now ready to treat the case when Q > 1. First case. Assume there are i , i 1,...,Q and x ∂C such that 1 2 ∈ { } 0 ∈ y (x ) y (x ) > 3Q osc(f; 2R) k i1 0 − i2 0 k 3(Q 1) osc(f; 2R) . ≥ − where f(x0) = y1(x0), . . . , yQ(x0) . We infer from Proposition 6 (applied with X = ∂C) that f decomposes into f = f f with f : ∂C Q (Y ) and Lip f Lip f, 1 ⊕ 2 j → Qj j ≤ j = 1, 2. The inductionJ hypothesisK implies the existence of extensions fˆ : C Q (Y ) j → Qj of f , such that Lip fˆ c (Q ) Lip f , j = 1, 2. We put fˆ = fˆ fˆ and we notice that j j ≤ 7 j j 1 ⊕ 2 Lip fˆ max c (Q ), c (Q ) Lip f ≤ { 7 1 7 2 } and osc(fˆ; 2R) osc(f; 2R) . ≤ Let K > 0 a constant to be determined later.

• First subcase. Suppose that

K osc(fˆ; 2R) (v, f(x )). ≤ G∞ 0 Then, for any x C, one has ∈ (v, fˆ(x)) (v, f(x )) + (f(x ), fˆ(x)) G∞ ≤ G∞ 0 G∞ 0 ∞(v, f(x0)) + osc(fˆ; 2R) ≤ G −1 (1 + K ) ∞(v, f(x0)) ≤ −1 G (1 + K ) max ∞(v, f(ξ)). ≤ ξ∈∂C G

• Second subcase. Suppose that

K osc(f,ˆ 2R) > (v, f(x )). G∞ 0 24 Preliminaries

We will use the same notations as in the proof of Proposition 6. Recall that f (x) = y (x) for x ∂C and l 1, 2 . l ⊕i∈Jl i ∈ ∈ { } We choose a numbering v = v , . . . , v such that (v, f(x )) = max v J K 1 Q G∞ 0 1≤i≤Q k i − yi(x0) . We set vJ1 = i∈J1 vi and vJ2 = i∈J2 vi . We claim that for any x C, k ⊕ J K ⊕ ∈ (v, fˆ(x))J =K max( (v , fˆ (xJ)),K (v , fˆ (x))). (3) G∞ G∞ J1 1 G∞ J2 2 This together with the inductive hypothesis will complete the proof. Suppose if possible that (3) is not valid. Switching J1 and J2 if necessary, it follows that there ˆ are j1 J1 and j2 J2 with ∞(v, f(x)) = vj1 yˆj2 (x) . Therefore, using (1) and (2),∈ ∈ G k − k

K osc(fˆ; 2R) > (v, f(x )) G∞ 0 (v, fˆ(x)) (fˆ(x), f(x )) ≥ G∞ − G∞ 0 v yˆ (x) osc(fˆ; 2R) ≥ k j1 − j2 k − y (x ) y (x ) ≥ k i1 0 − i2 0 k y (x ) y (x ) y (x ) v − k i1 0 − j1 0 k − k j1 0 − j1 k y (x ) y (x ) y (x ) yˆ (x) − k i2 0 − j2 0 k − k j2 0 − j2 k osc(fˆ; 2R) − y (x ) y (x ) ≥ k i1 0 − i2 0 k 3(card J 1) osc(fˆ; 2R) (v, f(x )) − 1 − − G∞ 0 3(card J 1) osc(fˆ; 2R) (f(x ), fˆ(x)) − 2 − − G∞ 0 osc(fˆ; 2R) − (3Q 3(Q 2) K 2) osc(fˆ; 2R) ≥ − − − − If K = 2, one gets a contradiction.

Second case. Assume that for every i , i 1,...,Q and for every x ∂C one has 1 2 ∈ { } ∈ y (x) y (x) 3Q osc(f; 2R) 6QR Lip f . k i1 − i2 k ≤ ≤ where f(x) = y1(x), . . . , yQ(x) is an arbitrary numbering. We pick some x0 ∂C 1 ∈ and we definey î : C 0 Y by J \{ } → K x Rx R x yî(x) = k kyi + − k ky1(x0) , R x ! R k k x C 0 and i = 1,...,Q. We define fˆ : C 0 Q (Y ) by fˆ(x) = ∈ \{ } \{ } → Q yˆ1(x),..., yˆQ(x) , x C 0 . We first show that Lip(fˆ ↾ ∂B(0,r)) Lip f, 0

1 Note we don’t claim any regularity about the yi nor they ˆi, not even measurability 1.4 - Lipschitz extensions 25

We notice that r yˆ (x) yˆ (x′) = y (˜x) y (˜x′) k i − σ(i) k Rk i − σ(i) k r (Lip f) x˜ x˜′ ≤ R k − k = (Lip f) x x′ . k − k Therefore

ˆ ˆ ′ ′ ′ ∞(f(x), f(x )) max yˆi(x) yˆσ(i)(x ) (Lip f) x x . G ≤ i=1,...,Q k − k ≤ k − k

Next, given x ∂C, we choose j 1,...,Q such that yj(x) y1(x0) (f(x), f(x )) ∈ (Lip f)2R. For each 0∈< { t < t }1 and i = 1,...,Qk one− has k ≤ G∞ 0 ≤ 1 2 ≤ yˆ (t x) yˆ (t x) = (t t ) y (x) y (x ) k i 2 − i 1 k 2 − 1 k i − 1 0 k (t t )( y (x) y (x) + y (x) y (x ) ) ≤ 2 − 1 k i − j k k j − 1 0 k (t t ) (3Q + 1) 2R Lip f ≤ 2 − 1 = t x t x (6Q + 2) (Lip f) , k 2 − 1 k thus (fˆ(t x), fˆ(t x)) (6Q + 2) (Lip f) t x t x . G∞ 2 1 ≤ k 2 − 1 k We conclude from the triangle inequality that for any x, x′ C 0 , r = x ,r′ = x′ ∈ \{ } k k k k ′ ′ ′ rx rx ′ ∞(fˆ(x), fˆ(x )) ∞ fˆ(x), fˆ + fˆ , fˆ(x ) G ≤ G r′ !! G r′ ! ! ′ ′ x rx ′ (Lip f) x x + 6Q max c7(k) + 2 x ≤ − k k r′ 1≤k

′ 6Q max c7(k) + 4 (Lip f) x x ≤ 1≤k

Regarding the second part of the Proposition, we choose some v = v1, . . . , vQ , ordered such that ∞(f(x0), v) = max1≤i≤Q yi(x0) vi . Let x C 0 and σ such that G k − k ∈ \ J K Rx Rx ∞ f , v = max yσ(i) vi . G x ! ! 1≤i≤Q x ! −

k k k k

One has

ˆ x Rx R x ∞(f(x), v) max k kyσ(i) + − k ky1(x0) vi G ≤ 1≤i≤Q R x ! R −

k k x Rx R x = max k k yσ(i) vi + − k k(y1 (x0) vi) 1≤i≤Q R x ! − ! R −

k k Rx ∞ f , v + max y1(x0) vi ≤ G x ! ! 1≤i≤Q k − k k k Rx ∞ f , v + max ( yi(x0) vi + yi(x0) y1(x0) ) ≤ G x ! ! 1≤i≤Q k − k k − k k k 2 max ∞(f(ξ), v) + 3Q osc(f; 2R) . ≤ ξ∈∂C G 26 Preliminaries

Note that

osc(f; 2R) = max ∞(f(ξ1), f(ξ2)) ξ1,ξ2∈∂C G

max ( ∞(f(ξ1), v) + ∞(v, f(ξ2))) ≤ ξ1,ξ2∈∂C G G

2 max ∞(f(ξ), v) . ≤ ξ∈∂C G Thus the proof is complete.

∗ ∗ Theorem 2. For every Q N and every m N there exists a constant c2(m, Q) 1 with the following property.∈ Assume that ∈ ≥

(1) X is a ﬁnite dimensional Banach space with m = dim X, and A X is closed; ⊂ (2) Y is a Banach space;

(3) f : A (Q (Y ), ) is Lipschitz. → Q G∞

It follows that f admits an extension fˆ : X (Q (Y ), ) with → Q G∞ Lip fˆ c (m, Q) Lip f , ≤ 2 and sup (fˆ(x), v): x X c (m, Q) sup (f(x), v): x A {G∞ ∈ } ≤ 2 {G∞ ∈ } for every v Q (Y ). ∈ Q

m Proof. Because they are lipeomorphic, there is no restriction to assume that X and ℓ∞ m coincide: an isomorphism T : X ℓ∞ will multiply the constant c2(m, Q) by a factor T T −1 (where denotes the→ operator norm), yet one can always ﬁnd a T such that kT k·k T −1k is smallerk·k than a constant depending only on m, since the Banach-Mazur kcompactumk · k k of m dimensional spaces is bounded. We consider a partition of X A into dyadic semicubes C with the following property \ { j}j dist(C ,A) j diam C < dist(C ,A) 2 ≤ j j for every j N. With each C and k = 0, . . . , m we associate it k-skeleton S (C ), i.e. ∈ j k j S (C ) = Clos C and S (C ) is the collection of those maximal k dimensional convex m j { j} k j subsets of the (relative) boundary of each F Sk+1(Cj). We also set Sk = j∈NSk(Cj). We now deﬁne, by upwards induction on k,∈ mappings ∪

fˆ : A S Q (Y ) k ∪ k → Q [ which coincide with f on A and such that

Lip fˆ ↾ F S (C ) C(k, Q) Lip f (4) k ∩ k j ≤ [ for each F S (C ), j N, (where C(k, Q) is a constant depending only on k and ∈ k+1 j ∈ Q). Furthermore, if k 1 then fˆ is an extension of fˆ . ≥ k k−1 1.4 - Lipschitz extensions 27

Deﬁnition of fˆ . With each x A S we associate ξ A such that x ξ = 0 ∈ ∪ 0 x ∈ k − xk dist(x, A), and we put fˆ0(x) = f(ξx). For x A we obviously have fˆ0(x) = f(x). If x C then ∈ ∈ j x ξ = dist(x, A) diam C + dist(C ,A) 3 diam C . k − xk ≤ j j ≤ j Consequently, if x, x′ C then ∈ j ′ ′ ′ ξ ξ ′ ξ x + x x + x ξ ′ 7 diam C = 7 x x . k x − x k ≤ k x − k k − k k − x k ≤ j k − k Thus ′ ′ (fˆ (x), fˆ (x )) = (f(ξ ), f(ξ ′ )) 7(Lip f) x x . G∞ 0 0 G∞ x x ≤ k − k This indeed proves (4) in case k = 0.

Definition of fˆk by induction on k 1. We say a k-face F Sk is minimal if there is ′ ′ ≥ ′ ∈ no k-face F Sk such that F F and F = F . We observe that each k-face contains a minimal one,∈ and that two distinct⊂ minimal6 k-faces have disjoint (relative) interiors. If F S is a minimal k-face then its “ boundary” ∂F (relative to the k dimensional ∈ k affine subspace containing it) equals F S (C ) where C is so that F S (C ), ∩ ∪ k−1 j j ∈ k j hence Lip fˆ ↾ ∂F C(k 1,Q) Lip f according to the induction hypothesis (4). Thus k−1 ≤ − Proposition 7 guarantees the existence of an extension fˆk of fˆk−1 from ∂F to F so that Lip(fˆ ↾ F ) c (Q)C(k 1,Q) Lip f. This completes the definition of fˆ to S . By k ≤ 7 − k ∪ k construction fˆk verifies (4) for every minimal k-face F Sk. Since each k-face is the union of (finitely many) minimal k-faces all contained in∈ the same k dimensional affine subspace of X, it is an easy matter to check that (4) is also verifies for arbitrary F S . ∈ k According to Proposition 7, for k 1 and v Q (Y ), one has ≥ ∈ Q ˆ −1 ˆ sup ∞(v, fk(x)) 2Q Lip φk Lip φk sup ∞(v, fk−1(x)) x∈A∪(∪Sk) G ≤ x∈A∪(∪Sk−1) G where φk denote a lipeomorphism from a k ball to a k cube. Moreover, one has easily

sup ∞(v, fˆ0(x)) = sup ∞(v, f(x)) . x∈A∪(∪S0) G x∈A G Those two facts implies that

sup (fˆ (x), v): x X c (m, Q) sup (f(x), v): x A {G∞ m ∈ } ≤ 2 {G∞ ∈ } if c (m, Q) (2Q)m m (Lip φ Lip φ−1). 2 ≥ k=1 k k Q ′ We now check that fˆm is Lipschitz. Let x, x X and we deﬁne the line segment [x, x′] = X x + t(x′ x) : 0 t 1 . We distinguish∈ between several cases according ∩ { − ≤ ≤ } ′ ′ to the positions of these points. First case : if x, x A, the clearly ∞(fˆm(x), fˆm(x )) = ′ ′ ∈ ′ G ∞(f(x), f(x )) (Lip f) x x . Second case : x, x Clos Cj for some j N. It G ≤ k − k′ ∈ ′ ∈ then follows that ∞(fˆm(x), fˆm(x )) C(m, Q)(Lip f) x x according to (4). Third ′ G ≤ k − k ′ case : [x, x ] A = . One then checks that J = N j :[x, x ] Clos Cj = ∩ ∅ ∩ { ∩ ′6 ∅} is ﬁnite and we apply the previous case to conclude that also ∞(fˆm(x), fˆm(x )) C(m, Q)(Lip f) x x′ . Fourth case : x A and x′ A. We chooseG j N such that≤ k − k 6∈ ∈ ∈ x C and we choose arbitrarily x′′ S (C ). It follows that ∈ j ∈ 0 j x x′′ diam C dist(C ,A) x x′ k − k ≤ j ≤ j ≤ k − k 28 Preliminaries and

′′ ′′ ′ x ξ ′′ x x + x ξ ′′ diam C + 3 diam C 4 x x . k − x k ≤ k − k k − x k ≤ j j ≤ k − k Thus

(fˆ (x), fˆ (x′)) (fˆ (x), fˆ (x′′)) + (fˆ (x′′), fˆ (x′)) G∞ m m ≤ G∞ m m G∞ 0 0 ′′ ′ (Lip fˆ ↾ Clos C ) x x + (Lip f) ξ ′′ x ≤ m j k − k k x − k 6(C(m, Q) + 1)(Lip f) x x′ . ≤ k − k Fifth case : [x, x′] A = and either x or x′ does not belong to A. We let a (resp. a′) denote the point [x,∩ x′]6 ∅A closest to x (resp. x′) and we observe that ∩ (fˆ (x), fˆ (x′)) G∞ m m (fˆ (x), fˆ (a)) + (f(a), f(a′)) + (fˆ (a′), fˆ (x′)) ≤ G∞ m m G∞ G∞ m m 6(C(m, Q) + 1)(Lip f) x a ≤ k − k + (Lip f) a a′ + 6(C(m, Q) + 1)(Lip f) a′ x′ k − k k − k 6(C(m, Q) + 1)(Lip f) x x′ . ≤ k − k

Given a pair of Banach spaces X and Y we here denote by c(X,Y,Q) the best constant occurring in Theorem 2 corresponding to these Banach spaces. Thus c(X,Y,Q) c2(dim X,Q) < in case X is ﬁnite dimensional. Kirszbraun’s Theorem ≤ m N ∞ ∗ says that c(ℓ2 , ℓ2 , 1) = 1 for every n, N N , thus it follows from Theorem 5 that m n ∈ m c(ℓ2 , ℓ2 ,Q) Lip ρn,Q is bounded independently of m. Is it true that c(ℓ2 , ℓ2,Q) < for every Q >≤ 1? That would be an analog of Kirszbraun’s Theorem for multiple-valued∞ functions. On the other hand, it is well-known that c(X, ℓ∞, 1) = 1 for every X. Is it true that c(X, ℓ∞,Q) < for every Q > 1 and every ﬁnite dimensional Banach space X? See also the discussion∞ at the end of section 2.3.

1.5 Diﬀerentiability

n The results contained in this section are standard in case Y = ℓ2 is Euclidean. The notion of (approximate) differentiability was introduced (under the name (approximate) affine approximatability) by F.J. Almgren in [AST00]. We call unambiguously differentiable what Almgren calls strongly affinely approximatable. “Intrinsic” proofs (i.e. avoiding the embedding defined in section 2.3) of the analog of Rademacher’s Theorem have been given in [Gob06] and [DLS11]. In this section X is a finite dimensional Banach space, m = dim X, λ is a Haar measure on X, and Y is a separable Banach space.

We say that g : X QQ(Y ) is affine (resp. linear) if there are affine maps A1,...,AQ → Q from X to Y (resp. linear maps L1,...,LQ from X to Y ) such that g = i=1 Ai (resp. Q ⊕ g = i=1 Li ). Our first task is to observe that the Ai’s are uniquely determined by g. ⊕ J K J K 1.5 - Differentiability 29

′ ′ Q Lemma 2. Let A1,...,AQ,A1,...,AQ be affine maps from X to Y , g = i=1 Ai , ′ Q ′ ′ ⊕ g = i=1 Ai , and S X. If g(x) = g (x) for every x S and λ(S) > 0, then there ⊕ ⊂ ′ ∈ J K exists σ SQ such that Ai = Aσ(i), i = 1,...,Q. ∈J K Proof. For each σ S we define ∈ Q W = X x : A (x) = A′ (x), i = 1,...,Q , σ ∩ { i σ(i) } and we notice that W is an affine subspace of X. If x S then x W for some σ ∈ ∈ σ σ S . Thus S S W . Therefore there exists σ such that λ(W ) > 0, hence ∈ Q ⊂ ∪σ∈ Q σ σ Wσ = X.

Let f, g : X QQ(Y ) be Borel measurable, and a X. We say that f and g are approximately tangent→ at a if for every ε > 0, ∈ Θm(λ x : (f(x), g(x)) > ε x a , a) = 0 . { G k − k} It is plain that the distance can be replaced by 1 or ∞ without changing the scope of the deﬁnition. G G G

′ Proposition 8. Let g, g : X QQ(Y ) be affine and approximately tangent at some a X. It follows that g = g′. → ∈ Q ′ Q ′ ′ ′ Proof. Write g = i=1 Ai , g = i=1 Ai , where A1,...,AQ,A1,...,AQ are affine from ⊕ ′ ⊕′ ′ ′ ′ X to Y , and Ai = Li + bi, Ai = Li + bi, bi, bi Y and the Li’s and Li’s are linear. With 0 < ε 1 we associateJ K J K ∈ ≤ G = X x : (g(x), g′(x)) ε x a ε ∩ { G1 ≤ k − k} m so that Θ (λ Gε, a) = 1 by assumption, because Gε is Borel measurable. Define η = inf A (a) A′ (a) : i, j = 1,...,Q and A (a) = A′ (a) (0, ]. {k i − j k i 6 j } ∈ ∞ Suppose η < , the case η = being easier to prove. Choose δ > 0 small enough for ∞ ∞ δ(1 + 2Q max L ,..., L , L′ ,..., L′ ) < η, {k 1k k Qk k 1k k Qk} where denotes the operator norm k · k L = sup L(x) : x 1 . k k {k k k k ≤ } s Let x G B(a, δ) and write x = a + h. There exists σ S such that ∈ ε ∩ ∈ Q ε h (g(a + h), g′(a + h)) k k ≥ G1 Q = A (a + h) A′ (a + h) k i − σ(i) k Xi=1 Q = L (a) + b L′ (a) b′ + L (h) L′ (h) k i i − σ(i) − σ(i) i − σ(i) k (5) Xi=1 Q = A (a) A′ (a) + L (h) L′ (h) k i − σ(i) i − σ(i) k Xi=1 Q Q A (a) A′ (a) L (h) L′ (h) , ≥ k i − σ(i) k − k i − σ(i) k Xi=1 Xi=1 30 Preliminaries whence

Q A (a) A′ (a) k i − σ(i) k Xi=1 ε h + 2Q h max L ,..., L , L′ ,..., L′ < η ≤ k k k k {k 1k k Qk k 1k k Qk} ′ since h δ. The deﬁnition of η then implies that Ai(a) = Aσ(i)(a) for each i = 1,...,Qk .k Multiplying ≤ (5) by t > 0 we obtain

Q ε th L (th) L′ (th) k k ≥ k i − σ(i) k Xi=1 Q = A (a + th) A′ (a + th) k i − σ(i) k Xi=1 (g(a + th), g′(a + th)) . ≥ G1 In other words, we have established that for every 0 < ε 1 and every t > 0, if a + h G B(a, δ) then a + th G . ≤ ∈ ε ∩ ∈ ε Letting ε = k−1, k N∗, we choose 0

λ Gεk > 0  ∗  k∈\N   and the conclusion now follows from Lemma 2.

′ Corollary 1. If f : X QQ(Y ) is Borel measurable, g, g : X QQ(Y ) are both aﬃne and both approximately→ tangent to f at a, then g = g′. →

Proof. Observe that g and g′ are approximately tangent (to each other) at a and apply Proposition 8.

Let f : X Q (Y ) and a X. We say that f is approximately differentiable at a → Q ∈ if there exists an affine Q-valued g : X QQ(Y ) which is approximately tangent to f at a. According to the above corollary,→ the existence of such g implies its uniqueness. Q It will be subsequently denoted as Af(a). Writing Af(a) = i=1 Ai we shall see Q ⊕ later that Af(a)(a) = i=1 Ai(a) equals f(a) in case f is approximately continuous ⊕ J K Q at a. Concatenation of the linear parts Li = Ai Ai(0) yields Df(a) = i=1 Li J K − ⊕ J K 1.5 - Differentiability 31 which is uniquely determined by Af(a). It may occur (but not too often, as we shall later see) that for some pair of distinct indexes i and j one has Ai(a) = Aj(a), yet Li = Lj. We now state a definition to rule this out. We say that f is unambiguously approximately6 differentiable at a if Af(a) fulfils the following additional requirement: for every i, j = 1,...,Q, if Ai(a) = Aj(a) then Li = Lj. Example 1. The affine 2-valued map

g = R Q (R): x x x → 2 7→ ⊕ − is everywhere (approximately) diﬀerentiable, but notJ K unambiguouslyJ K so at 0.

The need for unambiguous diﬀerentiation appears when stating the Euler-Lagrange equation for minimizing multiple-valued maps with respect to range deformation (so- called “squash deformation” by F.J. Almgren), see e.g. [AST00, Theorem 2.6(4)] or section 4.1. Recall the function σ deﬁned in Proposition 2.

∗ Lemma 3. Assume v QQ(Y ), put k = σ(v), and let Q1,...,Qk N and y1, . . . , yk ∈k k ∈ 1 ∈ Y be such that v = j=1Qj yj and Q = j=1 Qj. For every 0 < r < 2 split v the ⊕ ′ ′ ′ following holds. Whenever v QQ(Y ) is suchP that ∞(v, v )

We now give a criterion implying unambiguous approximate diﬀerentiability. Since σ f takes values in Z+, it is approximately continuous at a point if and only if it is approximately◦ constant at that point.

Proposition 9. Let f : X Q (Y ) be Borel measurable, and a X. Assume that → Q ∈ (A) f is approximately continuous at a;

(B) σ f is approximately constant at a; ◦ (C) f is approximately diﬀerentiable at a. 32 Preliminaries

It follows that f is unambiguously approximately diﬀerentiable at a, and that Af(a) = f(a).

Q Proof. Write Af(a) = i=1 Ai , and deﬁne α = max A1 ,..., AQ , where A denotes the operator norm⊕ of the linear part of A. Put{k k k= σ(kf(a)).k} There existk k f (a), . . . , f (a) Y and positiveJ K integers Q ,...,Q with k Q = Q such that 1 k ∈ 1 k j=1 j P f(a) = k Q f (a) . ⊕j=1 j j 1 Let 0 0 deﬁne ≤

G = X x : (f(x), f(a)) ε and σ(f(x)) = k ε ∩ { G1 ≤ and (f(x), Af(a)(x)) ε x a . G1 ≤ k − k}

Given η > 0 there exists r1(ε, η) > 0 such that λ(G B(a, r)) (1 η)λ(B(a, r)) ε ∩ ≥ − whenever 0

k f (a) A (x) ε(1 + x a ) 2ε . k j − i k ≤ k − k ≤ jX=1 i∈XIx,j This already implies that f(a) = Af(a)(a). Now let x, x′ G B(a, r) and j, j′ ∈ ε ∩ ∈ 1,...,k . If i I I ′ ′ then { } ∈ x,j ∩ x ,j ′ ′ f (a) f ′ (a) f (a) A (x) + A (x) A (x ) + A (x ) f ′ (a) k j − j k ≤ k j − i k k i − i k k i − j k 4ε + α x x′ ≤ k − k < split f(a) 1.5 - Diﬀerentiability 33 according to our choice of ε and r, thus j = j′. This in turn readily implies that Ix,j = Ix′,j =: Ij, j = 1,...,k. It follows from (7) above that if x Gε B(a, r) and i, i′ I , j = 1,...,k, then ∈ ∩ ∈ j A (x) A ′ (x) A (x) f (x) + f (x) A ′ (x) k i − i k ≤ k i − j k k j − i k 2ε x a . ≤ k − k Since η > 0 and ε > 0 are arbitrarily small we see that Ai and Ai′ are approximately tangent at a. Thus Ai = Ai′ according to Proposition 8 applied with Q = 1. Finally ′ ′ if i Ij, i Ij′ , and j = j then Ai(a) = fj(a) = fj′ (a) = Ai′ (a). The proof is complete.∈ ∈ 6 6 Example 2. Consider

2 0 if x < 0 f : R Q2(R): x → 7→  x2 x2 if x 0 .  J K ⊕ − ≥ One readily checks that f is (approximately)J K continuousJ K at 0 and unambiguously (approximately) differentiable at 0, yet σ f is not approximately constant at 0. ◦ We are ready to state and prove a useful generalization of Rademacher’s Theorem. We recall that X is a finite dimensional Banach space, and Y a Banach space. In case f : X Q (Y ) is approximately differentiable at a X we let Af(a) = Q A and → Q ∈ ⊕i=1 i we define Li = Ai Ai(0), i = 1,...,Q, the linear part of the affine approximation. We introduce the new− notation J K Df(a) = Q L Q (Hom(X,Y )) ⊕i=1 i ∈ Q where Hom(X,Y ) denotes the space of linearJ K operators X Y (these are automatically continuous). Letting the latter be equipped with some norm→ we let |||· ||| Q 2 |Df(a)| = 2(Df(a),Q 0 ) = v Li . G u ||| ||| uXi=1 J K t Theorem 3. Let f : X QQ(Y ) be Lipschitz continuous and assume that Y has the Radon-Nikodým property.→ It follows that

(A) For λ almost every a X, f is unambiguously approximately diﬀerentiable at a, and Af(a)(a) = f(a); ∈

(B) The map X Q (Hom(X,Y )) : x Df(x) is (B , BQ ) measurable; → Q 7→ X Q(Hom(X,Y )) (C) If f is approximately differentiable at a X then it is differentiable at a in the sense that ∈ (f(x), Af(a)(x)) lim G = 0 , x→a x a k − k and |Df(a)| √Q Lip f; ≤ (D) For every injective Lipschitzian curve γ : [0, 1] X such that γ′(t) = 1 and f → k k is approximately differentiable at γ(t) for L 1 almost every 0 t 1, one has ≤ ≤ 1 2(γ(1), γ(0)) |Df(x)|dH (x) . G ≤ Zim γ 34 Preliminaries

Proof. For each k = 1,...,Q deﬁne

B = X x : σ(f(x)) = k . k ∩ { }

Notice Bk is Borel since σ is Borel measurable according to Proposition 2. Fix k and 1 let a Bk. Put ηa = split f(a). Choose 0 < r < 2 ηa small enough for 1(f(x), f(a)) < 1 η whenever∈ x B(a, r). It follows from Lemma 3 that there existG positive integers 8 a ∈ Q ,...,Q such that k Q = Q and each f(x), x B(a, r) B , can be decomposed 1 k j=1 j ∈ ∩ k as P f(x) = k Q f (x) ⊕j=1 j j in such a way that f (a) f (x)

In order to prove conclusion (B) we use the same notation Bk, a Bk and r > 0 as above. It follows that the restriction ∈

Df : B B(a, r) Q (Hom(X,Y )) : x Q Df k ∩ → Q 7→ ⊕i=1 j is Borel measurable according to Proposition 5(A), because eachJ x K Df (x) is itself 7→ j Borel measurable. Since Bk is Lindelöf the restriction Df ↾ Bk is Borel measurable for each k = 1,...,Q, and the Borel measurability of Df follows immediately. The proof of the first part of conclusion (C) is inspired by [Fed96, Lemma 3.1.5] and exactly similar to [Gob06]. In order to prove the second part of conclusion (C) we 1.5 - Differentiability 35 assume that f is differentiable at a and we write Af(a) = Q A and L = A A (0), ⊕i=1 i i i − i i = 1,...,Q. Observe that Ai(h) = Ai(0)+Li(h) = fi(a)+Li(h), i = 1,...,Q, according to (A). Observe that for each x X we have J K ∈

Q (Af(a)(x), f(a))2 x a 2 L 2 (8) G2 ≤ k − k  k ik  Xi=1   and, given x, let σ S be a permutation such that ∈ Q

Q (Af(a)(x), f(a))2 = f (a) + L (x a) f (a) 2 . (9) G2 k i i − − σ(i) k Xi=1 Assuming that f (a) = f (a), for some i = 1,...,Q, and that i 6 σ(i) 1 x a max L ,..., L split f(a) k − k {k 1k k Qk} ≤ 2

1 2 we infer that the right member of (9) is bounded below by 4 (split f(a)) , in contradiction Q 2 1 with (8) provided x a i=1 Li < 2 split f(a). Thus, if x a is small enough k − kq k k k − k then (9) becomes P

Q 2 v Li(x a) = 2(Af(a)(x), f(a)) u k − k G uXi=1 t (Af(a)(x), f(x)) + (f(x), f(a)) . ≤ G2 G2 Upon letting x a we obtain →

Q 2 sup v Li(h) : h X and h 1 Lip f . (10) u k k ∈ k k ≤ ≤ uXi=1  t   Let j = 1,...,Q be such that Lj = max L1 ,..., LQ . The above inequality k k {k k k 1 k} implies that L Lip f. Finally |Df(a)| = Q L 2 2 √Q Lip f. k jk ≤ i=1 k ik ≤ It remains to establish conclusion (D). WeP deﬁne g : [0, 1] R by the formula → ′ g(t) = 2(γ(t), γ(0)), 0 t 0. We will show that g is Lipschitzian and that g (t) |Df(γ(Gt))| at each t such≤ that≤ f is diﬀerentiable at γ(t), so that our conclusion| will| ≤ become a consequence of a Theorem of Lebesgue applied to g:

1 ′ 1 2(γ(1), γ(0)) = g(1) g(0) = g (t)dL (t) G − Z0 1 |Df(γ(t))|dL 1(t) = |Df(x)|dH 1(x) ≤ Z0 Zim γ according to the area formula applied to γ. Write Df(γ(t)) = Q L (γ(t)) . For each ⊕i=1 i J K 36 Preliminaries t, t + h [0, 1] one has ∈ g(t + h) g(t) = (f(γ(t + h)), γ(0)) (f(γ(t)), γ(0)) − G2 − G2 (f(γ(t + h)), f(γ(t)) ≤ G2 which shows that Lip g Lip(f γ); and assuming further that f is diﬀerentiable at γ(t), we obtain: ≤ ◦

(Af(γ(t))(γ(t + h)), f(γ(t))) ≤ G2 + 2(Af(γ(t))(γ(t + h)), f(γ(t + h))) G 1 Q 2 L (γ(t))(γ(t + h) γ(t)) 2 ≤  k i − k  Xi=1 + ε γ(t + h) γ(t)  k − k where the last inequality holds provided h is small enough according to ε, split f(γ(t)) and L1(γ(t) ,..., LQ(γ(t)) (recall the proof of (C)). Dividing by h , letting h 0, andk recallingk thatk Lip γ k 1 we infer that g′(t) |Df(γ(t))| provided| | that g→is diﬀerentiable at t. ≤ | | ≤

Given f : X Q (Y ) and a X we now deﬁne → Q ∈ (f(x), f(a)) lipa f := lim sup sup G . r→0 x a x∈B(a,r) k − k

If f is Lipschitz then clearly lipa f Lip f < for every a X. We leave it to the reader to check the following partial≤ “product rule”:∞ if f and λ∈ : X R are Lipschitz then → lip (λf) (lip λ)|f(a)| + λ(a) (lip f) . (11) a ≤ a | | a Proposition 10. If f : X Q (Y ) is Lipschitz then → Q lip f |Df(a)| Q(lip f) a ≤ ≤ a q for L m almost every a X. ∈ Proof. The second inequality is proved in exactly the same as Theorem 3(C) by noticing that in (18) Lip f can be replaced by lipa f. In order to prove the first inequality we assume that f is differentiable at a and that a is a Lebesgue point of x |Df(x)|. Given 0 < ε < 1 we define 7→

G = X x : |Df(x)| ε + |Df(a)| . ε ∩ { ≤ } There exists r > 0 such that for every 0

With each h H we associate the line segment Sj joining a+h and x+h, and we deﬁne the “cylinder”∈ C = Sh . h[∈H We observe that C B(a, 2ρ) and that ⊂ L m(C) = ρα(m 1)εm−1ρm−1 = εm−1α(m 1)ρm . − − Therefore,

L m(C G ) = L m(C) L m(C Gc) ∩ ε − ∩ ε εm−1α(m 1)ρm 2−mεmα(m 1)(2ρ)m ≥ − − − = εm−1α(m 1)ρm−1(1 ε)ρ . − − According to Fubini’s Theorem, Chebyshev’s inequality and Theorem 3, there exists h H such that ∈ H 1(S G ) (1 ε)ρ h ∩ ε ≥ − 1 and f is diﬀerentiable H almost everywhere on Sh. For such h, recalling Theorem 3(D), we infer that

(f(x + h), f(a + h)) |Df(z)|dH 1(z) G ≤ ZSh = |Df(z)|dH 1(z) + |Df(z)|dH 1(z) c ZSh∩Gε ZSh∩Gε (ε + |Df(a)|)ρ + Q(Lip f)ερ . ≤ q Since h ερ, the triangle inequality implies that k k ≤ (f(x), f(a)) (ε + |Df(a)|)ρ + (2 + Q)(Lip f)ερ , G ≤≤ q thus (f(x), f(a)) G ε + |Df(a)| + (2 + Q)(Lip f)ε . x a ≤ k − k q

2 Embeddings

n 2.1 Whitney bi-Hölder embedding — The case Y = ℓ2 (K)

Here we report on [Whi72, Appendix V]. We let K = R or K = C. We start by recalling the usual embedding

η : Q (K) KQ : v (η (v), . . . , η (v)) . Q → 7→ 1 Q

Given v = x1, . . . , xQ we let ηi(v) K, i = 1,...,Q, be the coeﬃcients of the Weier- strass polynomial of v: ∈ J K Q Q P (x) = (x x ) = xQ + η (v)xQ−i K[x] . v − i i ∈ iY=1 Xi=1 38 Embeddings

Readily the ηi(v) are the Q symmetric functions of Q variables, and their (Lipschitz) continuity follows. In case K = C, η is a bijection and η−1 is Hölder continuous (see e.g. [Mar66, Theorem (1,4)]). We now treat the case of Kn. We will deﬁne a mapping

η : Q (Kn) KN Q → n n where N = N(n, Q). Given u C and v = x1, . . . , xQ QQ(K ) we deﬁne a polynomial ∈ ∈ Q J K P (u, x) = (x u, x ) K[u , . . . , u , x] v − h ii ∈ 1 n iY=1 whose coeﬃcients ηα(v) form the components of η:

Q Q α1 αn Q−i Pv(u, x) = x + ηα(v)u1 . . . un x . i=1 α∈Nn X |αX|=i One computes that Q + n N(n, Q) = 1 . n ! − One shows ([Whi72, Appendix V Theorem 6A]) that η is injective, continuous, that n N −1 η(QQ(K )) is closed in K , and that η is continuous as well. In case K = C, it n follows from the Proper Mapping Theorem that η(QQ(C )) is an irreducible analytic N n variety in C , see [Whi72, Chapter 5 Theorem 5A]. In fact η(QQ(C )) is a Hölder continuous retract of CN .

2.2 Splitting in case Y = R

We now state an easy and important observation on how to compute the distance G2 of two members of QQ(R). The order of R plays the essential role. This is taken from [AST00, 1.1(4)].

′ ′ Proposition 11. Let v, v QQ(R) and choose numbering v = y1, . . . , yQ and v = ′ ′ ∈ ′ ′ ′ y1, . . . , yQ such that y1 y2 ... yQ and y1 y2 ... yQ. It follows that ≤ ≤ ≤ ≤ ≤ ≤ J K J K Q ′ ′ 2 2(v, v ) = v yi yi . G u | − | uXi=1 t Proof. We must show that for each σ S one has ∈ Q Q Q y y′ 2 y y 2 . (12) | i − i| ≤ | i − σ(i)| Xi=1 Xi=1 Let σ S be nontrivial and let i < j be such that σ(i) > σ(j). Observe that ∈ Q 0 2(y y )(y′ y′ ) ≤ j − i σ(i) − σ(j) = 2y y′ 2y y′ + 2y y′ + 2y y′ − i σ(i) − j σ(j) j σ(i) i σ(j) = (y y′ )2 + (y y′ )2 (y y′ )2 (y y′ )2 . i − σ(i) j − σ(j) − j − σ(i) − i − σ(j) n 2.3 - Almgren-White locally isometric embedding — The case Y = ℓ2 (R) 39

In other words, (y y′ )2 + (y y′ )2 (y y′ )2 + (y y′ )2 . (13) i − σ(j) j − σ(i) ≤ i − σ(i) j − σ(j) In case Q = 2 the proof is complete. We next assume that (12) holds for Q and we shall show it holds for Q + 1. Let σ SQ+1. If σ(Q + 1) = Q + 1 then we define σ¯ = σ ↾ 1,...,Q S and we simply∈ notice that { } ∈ Q Q+1 Q Q+1 y y′ 2 y y′ 2 + y y′ 2 = y y′ 2 . | i − i| ≤ | i − σ¯(i)| | Q+1 − Q+1| | i − σ(i)| Xi=1 Xi=1 Xi=1 In case σ(Q + 1) = Q + 1 we choose k 1,...,Q such that σ(k) = Q + 1. We apply inequality (13) above6 with i = k and j ∈= {Q + 1 to} obtain (y y′ )2 + (y y′ )2 (y y′ )2 + (y y′ )2 . (14) k − σ(Q+1) Q+1 − σ(k) ≤ k − σ(k) Q+1 − σ(Q+1) We define σ¯ S as follows: ∈ Q+1 σ¯(i) = σ(i) if i = k and i = Q + 1 6 6 σ¯(k) = σ(Q + 1) σ¯(Q + 1) = σ(k) = Q + 1 . Since σ¯(Q + 1) = Q + 1 the first case considered above together with (14) yield

Q+1 Q+1 Q+1 y y′ 2 y y′ 2 y y′ 2 . | i − i| ≤ | i − σ¯(i)| ≤ | i − σ(i)| Xi=1 Xi=1 Xi=1

2.3 Almgren-White locally isometric embedding — The case n Y = ℓ2 (R)

n n This section is devoted to the case Y = ℓ2 , i.e. R equipped with its Euclidean norm and inner product , . Proposition 12 and Theorem 4 are due to F.J. Almgren [AST00,k · k 1.2]. The presentationh· ·i we give here is (inspired by) that of C. De Lellis and E.N. Spadaro [DLS11]. Part (B) of Theorem 4 is due to B. White. Let e Rn be such that e = 1. We deﬁne a map ∈ k k π : Q (Rn) RQ e Q → by the requirement that πe( y1, . . . , yQ ) be the list of inner products

J y1, eK ,..., yQ, e . h i h i numbered in increasing order. Notice that we need indeed to explain how we choose to Q order these real numbers if we want the values of πe to belong to R , for otherwise they would merely belong to QQ(R). n Proposition 12. Let e1, . . . , en be an orthonormal basis of R . The mapping ξ : Q (Rn) RQn : v (π (v),..., π (v)) 0 Q → 7→ e1 en has the following properties: 40 Embeddings

(A) Lip ξ0 = 1;

n ′ n (B) For every v QQ(R ) there exists r > 0 such that for each v QQ(R ), if (v, v′)

n Q ′ 2 ′ 2 ′ ξ0(v) ξ0(v ) = yτj (i), ej yτ (i), ej k − k |h i − h j i| jX=1 Xi=1 which, by Proposition 11, is bounded by

n Q y , e y , e 2 ≤ |h i ji − h σ(i) ji| jX=1 Xi=1 Q = y y 2 k i − σ(i)k Xi=1 = (v, v′)2 . G2

n (B) Let v QQ(R ) and write v = y1, . . . , yQ . For each j = 1, . . . , n choose τj SQ ∈ 1 ∈ such that yτj (1), ej ... yτj (Q), ej . Deﬁne r = 2 min split πej (v): j = 1, . . . , n ′ h ni ≤ ≤ h Ji ′ K { ′ ′ ′ } and let v QQ(R ) be such that 2(v, v ) < r. Choose a numbering v = y1, . . . , yQ so that ∈ G Q J K (v, v′)2 = y y′ 2 . G2 k i − ik Xi=1 Notice that for every j = 1, . . . , n one has

′ max yτj (i), ej yτ (i), ej i=1,...,Q |h i − h j i| ′ ′ 1 max yτj (i) yτ (i) 2(v, v ) < split πej (v) ≤ i=1,...,Q k − j k ≤ G 2 which implies, according to the Splitting Lemma, Proposition 11 and the deﬁnition of

πej , that

Q Q ′ 2 ′ 2 ′ 2 ′ yτj (i), ej yτ (i), ej = 2(πej (v), πej (v )) = yτj (i), ej yτ (i), ej |h i − h j i| G |h i − h j i| Xi=1 Xi=1 n 2.3 - Almgren-White locally isometric embedding — The case Y = ℓ2 (R) 41

′ ′ ′ where τj SQ is such that yτ ′ (1), ej ... yτ ′ (Q), ej . Therefore, ∈ h j i ≤ ≤ h j i

J n KQ Q ξ (v) 2 = y , e 2 = y 2 = (v, Q 0 )2 . k 0 k |h i ji| k ik G2 jX=1 Xi=1 Xi=1 J K

Remark 1. The Lipschitz mapping ξ0 deﬁned above is usually not injective. Consider for instance the case when Q = 2, n = 2, and let e1, e2 be an orthonormal basis of 2 ′ R . We deﬁne v = e1 + e2, e1 . It follows that ξ0(v) = ( 1, 1, 0, 1) = ξ0(v ) where ′ − ′ − v = e1, e1 + e2 . Clearly v = v . − J 6 K J K The lack of injectivity of ξ0 is overcome by considering a lot of orthonormal bases instead of just one, i.e. we shall replace ξ0 by many copies of ξ0 corresponding to various bases. The main observation to obtain injectivity is the following.

Proposition 13. Given integers n and L there are ε > 0 and unit vectors e1, . . . , eK n−1 n ∈ S with the following property. For every v1, . . . , vL R there exists k = 1,...,K such that ∈ e , v ε v |h k li| ≥ k lk for each l = 1,...,L.

Proof. We ﬁrst notice that the measure H n−1 Sn−1 is doubling, i.e. there exists C 1 ≥ such that H n−1(Sn−1 B(e, 2r)) CH n−1(Sn−1 B(e, r)) whenever e Sn−1 and r > 0. Given e Sn−1 and ε∩ > 0 we deﬁne≤ the slab ∩ ∈ ∈ S = Sn−1 w : e, w < ε . e,ε ∩ { |h i| } Now we choose ε > 0 small enough for

H n−1(Sn−1) H n−1(S ) e,ε ≤ 3CL whenever e Sn−1. We choose a maximal collection of points e , . . . , e Sn−1 such that ∈ 1 k ∈ the (open) balls B(ek, ε), k = 1,...,K, are pairwise disjoint. Such a collection exists 42 Embeddings because H n−1(Sn−1) is ﬁnite and H n−1(Sn−1 B(e, ε)) does not depend on e Sn−1. n−1 K ∩ ∈ By maximality, we have that S = k=1 U(ek, 2ε). n Let now v1, . . . , vL R be arbitrary.S We deﬁne L = 1,...,L l : vl = 0 and ∈ −1 { } ∩ { 6 } for l L we set wl = vl vl . Our claim is that for some k, ek does not belong to any of the∈ slabs S , l L.| Suppose| if possible that for each k = 1,...,K, e S where wl,ε ∈ k ∈

S = Swl,ε . L l[∈

If l L corresponds to k so that ek Swl,ε then in fact at least “half” the ball B(ek, ε) ∈ ∈n−1 1 n−1 Sn−1 must be contained in Swl,ε, thus H (S B(ek, ε)) 2 H ( B(ek, ε)). We would then obtain ∩ ≥ ∩

K H n−1(Sn−1) H n−1(Sn−1 B(e , 2ε)) ≤ ∩ k kX=1 K C H n−1(Sn−1 B(e , ε)) ≤ ∩ k kX=1 K 2C H n−1(S B(e , ε)) ≤ ∩ k kX=1 2CH n−1(S) ≤ n−1 2C H (Swl,ε) ≤ L Xl∈ 2 H n−1(Sn−1) , ≤ 3 a contradiction. Theorem 4. There exist an integer N = N(n, Q), a real number α = α(n, Q) 1 and a mapping ≤ ξ : Q (Rn) RN Q → with the following properties.

(A) For every v, v′ Q (Rn), α (v, v′) ξ(v) ξ(v′) (v, v′); ∈ Q G2 ≤ k − k ≤ G2 n ′ n (B) For every v QQ(R ) there exists r > 0 such that for each v QQ(R ), if (v, v′)

′ ′ ′ On the other hand, letting v = y1, . . . , yQ and v = y1, . . . , yQ , we infer from Propo- sition 13 that there exists k = 1,...,K such that J K J K e , y y′ ε y y′ |h 1,k i − ji| ≥ k i − jk for every i, j = 1,...,Q. Let σ SQ be such that yσ(1), e1,k ... yσ(Q), e1,k and let τ S be such that y′ , e∈ ... y′ , eh . Observei ≤ that≤ h i ∈ Q h τ(1) 1,ki ≤ ≤ h τ(Q) 1,ki

Q (v, v′)2 y y′ 2 G2 ≤ k σ(i) − τ(i)k Xi=1 Q ε−2 y , e y′ , e 2 ≤ |h σ(i) 1,ki − h τ(i) 1,ki| Xi=1 = ε−2 π (v) π (v′) 2 k e1,k − e1,k k ε−2 ξ(v) ξ(v) 2 . ≤ k − k

We now turn to proving conclusions (B) and (C). Given v Q (Rn) and k = ∈ Q 1,...,K we choose rk > 0 according to Proposition 12(B). Let r = min r1,...,rK . If v Q (Rn) and (v, v′)

K ξ(v) ξ(v′) 2 = ξ (v) ξ (v′) 2 = K (v, v′)2 . k − k k k − k k G2 kX=1 Also, regarding conclusion (C), we observe that for every v Q (Rn), ∈ Q ξ(v) 2 = K (v, Q 0 ) , k k G2 J K according to Proposition 12(C). This means that the mapping K−1/2ξ veriﬁes the conclusions of the present proposition.

B. White’s addition (B) to F.J. Almgren’s embedding Theorem 4 has the following rather useful consequence. Here the linear spaces Hom(Rm, Rν)(ν = n or ν = N) are equipped with the norm m ν L = L(e ), e 2 v j k ||| ||| uj=1 k=1h i uX X t corresponding to the canonical bases of Rm and Rν.

m n n Proposition 14. Assume that f : R QQ(R ), a R , and that both f and ξ f are diﬀerentiable at a.2 It follows that → ∈ ◦

|Df(a)| = D(ξ f)(a) . ||| ◦ ||| 2For f this is in the sense of Theorem 3(C) 44 Embeddings

Proof. For each j = 1, . . . , m we have

2 2 (ξ f)(a + tej) (ξ f)(a) ∂j(ξ f)(a) = lim k ◦ − ◦ k k ◦ k t→0 t2 (f(a + te ), f(a))2 = lim G2 j t→0 t2

(according to Theorem 4(B))

(Af(a)(a + te ), f(a))2 = lim G2 j t→0 t2

(because f is diﬀerentiable at a)

Q 2 i=1 fi(a) Aσt(i)(a) Lσt(i)(tej) = lim k − 2 − k , t→0 P t where, as usual, Df(a) = Q A , L = A A (0), i = 1,...,Q and σ is a permutation ⊕i=1 i i i− i t σ SQ for which the quantity ∈ J K Q 2 i=1 fi(a) Aσ(i)(a) Lσ(i)(tej) k − 2 − k P t is minimal. Since the above limit exists and is ﬁnite, we infer that σ S must be t ∈ Q such that fi(a) = Aσt(i)(a) when t is small enough, i = 1,...,Q. Thus,

Q ∂ (ξ f)(a) 2 = L (e ) 2 , k j ◦ k k i j k Xi=1 and in turn,

m D(ξ f)(a) 2 = ∂ (ξ f)(a) 2 ||| ◦ ||| k j ◦ k jX=1 m Q = L (e ) 2 k i j k jX=1 Xi=1 Q = L 2 = |Df(a)|2 . ||| i||| Xi=1

Theorem 5. Let N = N(n, Q) and ξ be as in Theorem 4. There exists a Lipschitz retraction ρ : RN ξ(Q (Rn)) . → Q

N n n −1 Proof. Apply Theorem 2 with X = ℓ2 , A = ξ(QQ(R )), Y = ℓ2 and f = ξ . Letting fˆ be a Lipschitz extension of f, the mapping ρ = ξ fˆ veriﬁes the conclusion. ◦ The exact same proof shows that there exists a Hölder continuous retraction

ρ˜ : CN η(Q (Cn)) → Q ∗ 2.4 - Lipeomorphic embedding into Lipy0 (Y ) 45 where N = N(n, Q) and η are as in Section 2.1. This follows indeed from the fact that η−1 is Hölder continuous (reference [Mar66, Theorem (1,4)]). In the same vein one can prove the following, based on [BL00, Theorem 1.12] and Theorem 5: If ω : R+ R+ is m n → concave then for every A ℓ2 and every f : A QQ(ℓ2 ) such that osc(f; ) ω, there ˆ m⊂ n → ˆ · ≤ exists an extension f : ℓ2 QQ(ℓ2 ) of f such that osc(f; ) (Lip ρn,Q)ω. Here ρn,Q is the Lipschitz retraction of→ Theorem 5, and Q (ℓn) is equipped· ≤ with its metric . Q 2 G2 We recall that a metric space Z is an absolute Lipschitz retract if and only if each isometric embedding Z Z′ into another metric space Z′ has a Lipschitz right inverse ρ : Z′ Z. In other words,→ Z is a Lipschitz retract of any of its metric superspaces. This is→ equivalent to asking that any partially deﬁned Lipschitz map into Z extends N to a Lipschitz map into Z, see [BL00, Proposition 1.2]. For instance ℓ∞ is an absolute Lipschitz retract, and hence the following holds.

n Corollary 2. QQ(R ) is an absolute Lipschitz retract.

It is unknown whether QQ(Y ) is an absolute Lipschitz retract if Y is also one. Are QQ(ℓ∞) and QQ(C[0, 1]) absolute Lipschitz retracts? Are they absolute uniform retracts?

2.4 Lipeomorphic embedding into Lipy0 (Y )∗

Let (Y, y0) be a pointed metric space, i.e. a metric space Y together with a distinguished point y Y . We denote by Lip (Y ) the collection of those Lipschitz continuous 0 ∈ y0 functions u : Y R vanishing at y0. This is a Banach space equipped with the norm u = Lip u. With→ each v = y , . . . , y Q (Y ) we associate a linear functional k kLip 1 Q ∈ Q

J K Q ζ (v) : Lip (Y ) R : u u(y ) . (15) 0 y0 → 7→ i Xi=1

One readily checks that ζ0(v) is continuous and

Q ζ (v) ∗ d(y , y ) . k 0 k(Lip Y ) ≤ i 0 Xi=1

In particular ζ (v) ∗ Q(diam Y ) so that ζ is bounded when Y is. Notice also k 0 k(Lip Y ) ≤ 0 that ζ0(Q y0 ) = 0. We shall now show that

J K ζ : Q (Y ) Lip (Y )∗ 0 Q → y0 is a lipeomorphic embedding.

Theorem 6. There exists cQ > 0 such that for every pointed metric space (Y, y0) and every v, v′ Q (Y ) one has ∈ Q ′ ′ ′ c (v, v ) ζ (v) ζ (v ) ∗ (v, v ) . QG1 ≤ k 0 − 0 k(Lip Y ) ≤ G1 46 Embeddings

Proof. We start with the second inequality. Let v, v′ Q (Y ) and choose numberings ∈ Q v = y , . . . , y and v′ = y′ , . . . , y′ so that (v, v′) = Q d(y , y′). It is clear that 1 Q 1 Q G1 i=1 i i P J K J K Q Q ′ ′ ζ (v) ζ (v ) ∗ = sup u(y ) u(y ) : u Lip (Y ) 0 0 (Lip Y )  i i y0 k − k i=1 − i=1 ∈  X X

 and Lip u 1 ≤   Q d(y , y′)  ≤ i i Xi=1 = (v, v′) . G1

We now turn to proving the ﬁrst inequality, by induction on Q. If Q = 1 then ′ ′ the inequality is veriﬁed with c1 = 1. Indeed, given v = y1 and v = y1 we let u(y) = d(y1, y) d(y1, y0) so that u Lipy (Y ), Lip u 1, and − ∈ 0 ≤ J K J K ′ ′ ′ ′ ζ (v) ζ (v ) ∗ u(y ) u(y ) = d(y , y ) = (v, v ) . k 0 − 0 k(Lip Y ) ≥ | 1 − 1 | 1 1 G1 We now assume the conclusion holds for Q and we establish it for Q + 1. Let v, v′ Q+1 ′ Q+1 ′ ∈ QQ+1(Y ) and write v = i=1 yi and v = i=1 yi . We let α > 0 to be determined later, and we distinguish⊕ between two cases. ⊕ J K J K First case. We assume that

′ ′ dist(supp µ , supp µ ′ ) = min d(y , y ): i, j = 1,...,Q + 1 > α (v, v ) . v v { i j } G1

′ ′ ′ We deﬁne u0 : (supp µv) (supp µv ) R by letting u0(yi) = 0 and u0(yi) = α 1(v, v ), i = 1,...,Q + 1. It is∪ most obvious→ that Lip u 1 and we letu ˆ be an extensionG 0 ≤ 0 of u0 to Y such that Lipu ˆ0 1, whose existence follows from the McShane-Whitney Theorem. Finally we let u =≤u ˆ uˆ (y )✶ and we observe that 0 − 0 0 Y

Q+1 Q+1 ′ ′ ′ ζ (v) ζ (v ) ∗ u(y ) u(y ) = α(Q + 1) (v, v ) . 0 0 (Lip Y ) i i 1 k − k ≥ i=1 − i=1 G X X

Second case. We assume that

′ dist(supp µ , supp µ ′ ) α (v, v ) . v v ≤ G1

′ ′ Choose i0, j0 1,...,Q + 1 such that d(yi0 , yj0 ) = dist(supp µv, supp µv ). Deﬁne v,˜ v˜′ Q (Y )∈ by { } ∈ Q v˜ = y andv ˜′ = y′ . ⊕i6=i0 i ⊕j6=j0 j According to the induction hypothesisJ K there exists u LipJ K (Y ) with Lip u 1 and ∈ y0 ≤

′ 1 ′ cQ ′ u(y ) u(y ) ζ (˜v) ζ (˜v ) ∗ (˜v, v˜ ) . i − j ≥ 2k 0 − 0 k(Lip Y ) ≥ 2 G1 i6=i0 j6=j0 X X

∗ 2.4 - Lipeomorphic embedding into Lipy0 (Y ) 47

Since readily (˜v, v˜′) + d(y , y′ ) (v, v′) we infer that G1 i0 j0 ≥ G1 Q+1 Q+1 ′ ′ ζ (v) ζ (v ) ∗ u(y ) u(y ) 0 0 (Lip Y ) i j k − k ≥ i=1 − j=1 X X c Q (˜v, v˜′) u(y ) u(y′ ) ≥ 2 G1 − | i0 − j0 | c c Q (v, v′) Q d(y , y′ ) d(y , y′ ) ≥ 2 G1 − 2 i0 j0 − i0 j0 cQ cQ ′ α 1 + 1(v, v ) . ≥ 2 − 2 G We now choose α > 0 small enough for cQ α 1 + cQ > 0 and we set 2 − 2 cQ cQ cQ+1 = min α(Q + 1), α 1 + 2 − 2 so that, in both cases,

′ ′ ζ (v) ζ (v ) ∗ c (v, v ) . k 0 − 0 k(Lip Y ) ≥ Q+1G1

For the remaining part of this section we assume that Y is compact. We notice that the formula (15) deﬁning u, ζ0(v) makes sense also when u C(Y ). For convenience we consider an embeddingh in the Banachi subspace ∈

C (Y ) = C(Y ) u : u(y ) = 0 . y0 ∩ { 0 }

In order to formally distinguish between the cases when u Lipy0 (Y ) and u Cy0 (Y ) we introduce a diﬀerent notation ∈ ∈

Q ζ(v): C (Y ) R : u u(y ) . y0 → 7→ i Xi=1

The Banach space C(Y ) is equipped with its usual maximum norm C(Y ) and we notice that ζ(v) is continuous and k · k

ζ(v) ∗ Q. k kC(Y ) ≤

In the sequel we shall use the lipeomorphic embedding ζ0 in conjunction with the well-known metrization property of weakly* compact subsets of C(Y )∗ in the following form. We recollect the standard proof for completeness.

Proposition 15. Let (Y, y0) be a pointed metric space and assume that Y is compact. The ball M = C (Y )∗ T : T Q Q y0 ∩ { k k ≤ } is weakly* compact, and the restriction to MQ of the weak* topology is metrized by

d(T,T ′) = sup T (u) T ′(u) : u Lip (Y ) and Lip u 1 . {| − | ∈ y0 ≤ } 48 Sobolev classes

Proof. The weak* compactness of MQ is a consequence of the Banach-Alaoglu Theorem. We let Td denote the topology on MQ associated with the metric d above, and Tσ∗ the ∗ restriction to MQ of the weak* topology of C(Y ) . For convenience we denote by −1 φ : MQ[Td] MQ[Tσ∗ ] the identity map, and by φ its inverse, so we must recall why both are continuous.→

Let Tj be a sequence in MQ, T MQ, and assume that d(T,Tj) 0. Let u C(Y{) and} ε > 0. Since u is uniformly∈ continuous there exists a Lipschitz continuous→ ∈ uˆ : Y R such that u uˆ C(Y ) < ε (see e.g. [BL00, Proposition 2.1(i)]). It follows that → k − k

lim sup u, T Tj lim sup u,ˆ T Tj + lim sup u u,ˆ T Tj j |h − i| ≤ j |h − i| j |h − − i|

= lim sup uˆ uˆ(y0)✶Y ,T Tj + lim sup u u,ˆ T Tj j |h − − i| j |h − − i| −1 lim sup(1 + Lipu ˆ) d(T,Tj) + 2εQ ≤ j = 2εQ . The arbitrariness of ε shows that φ is continuous.

Assume that Tλ λ∈Λ is a directed family in MQ converging weakly* to some T . As a family of functionals{ } deﬁned on C(Y ), T : λ Λ is equicontinuous. Furthermore { λ ∈ } S = Lipy0 (Y ) u : Lip u 1 is compact in C(Y ) according to Ascoli’s Theorem. Therefore the pointwise∩ { (i.e.≤ weak*)} convergence of the directed family on C(Y ) implies its uniform convergence on S (see e.g. [Edw95, Proposition 0.4.9]). This is exactly saying that d(T,T ) 0. λ →

3 Sobolev classes

3.1 Deﬁnition of Lp(X, QQ(Y ))

Let (Y, y0) be a pointed metric space as usual, let (X, A, µ) be a measure space, and let 1 p < . We denote by Lp(X, QQ(Y )) the collection of mappings f : X QQ(Y ) verifying≤ ∞ the following requirements: →

(A) f is (A, BQQ(Y )) measurable; (B) The function X R : x (f(x),Q y )p is µ summable. → 7→ G2 0 In the remaining part of this paper we shallJ abbreviateK

|f(x)| = (f(x),Q y ) , G2 0 x X, and we keep in mind that no ambiguity shouldJ K occur from the lack of mention of∈y in the abbreviation3. If f L (X, Q (Y )) we also set the notation 0 ∈ p Q 1 p p |f|Lp = |f| dµ . ZX 3 In case Y is a Banach space it will be implicitly assumed that y0 = 0 3.2 - Analog of the Fréchet-Kolmogorov compactness Theorem 49

Of course Lp(X, QQ(Y )) is not a linear space. It is most obvious that the formula

1 p p dp(f, g) = 2(f, g) dµ ZX G deﬁnes a semimetric on Lp(X, QQ(Y )). As in the scalar case, we have: Proposition 16. Assume that Y is a complete metric space. It follows that Lp(X, QQ(Y ))[dp] is a complete semimetric space, and each Cauchy sequence contains a subsequence converging pointwise almost everywhere.

Proof. We merely recall the usual argument. Let f be a d Cauchy sequence in { j} p Lp(X, QQ(Y )). Deﬁne inductively an increasing sequence of integers kj such that d (f , f )p 4−j, and let X = X x : (f (x), f (x))p 2−j ,{j }N∗. Thus p kj+1 kj ≤ j ∩ { G2 kj+1 kj ≥ } ∈

1 p p 1 j dp(fkj+1 , fkj ) 2(fkj+1 , fkj ) dµ j µ(Xj) . 4 ≥ ≥ ZXj G ≥ 2 −j ∗ ′ ′ −i+1 Therefore µ(Xj) 2 , j N . Letting Xi = j≥iXj it follows that µ(Xi) 2 . ≤∗ ′ ∈ ∪ ≤ Deﬁning N = i∈N Xi it immediately follows that N is negligible. Observe that if x X N then f ∩(x) is a Cauchy sequence, hence convergent, owing to the completeness∈ \ { kj }j of QQ(Y ) (Proposition 1). The limiting f : X QQ(Y ) is (A, BQQ(Y )) measurable. For every j N∗, → ∈ p p dp(f, fkj ) = 2(f, fkj ) dµ ZX G p = lim 2(fk ′ ,kk ) dµ j′ j j ZX G p lim inf 2(fk ′ ,kk ) dµ j′ j j ≤ ZX G p = lim inf dp(fk ′ , fk ) j′ j j according to Fatou’s Lemma, and

|f| dp(f, fk ) + |fk | < Lp ≤ j j Lp ∞ according to the triangular inequality. The proof is complete.

3.2 Analog of the Fréchet-Kolmogorov compactness Theorem

Theorem 7. Assume that 1 p < and that: ≤ ∞

(A) (X, BX , λ) is a ﬁnite dimensional Banach space with a Haar measure λ deﬁned on the σ algebra BX of Borel subsets of X; (B) Y is a compact metric space, and y Y ; 0 ∈ (C) F L (X, Q (Y )) is a family subjected to the following requirements: ⊂ p Q (i) sup |f| : f F < ; { Lp ∈ } ∞ 50 Sobolev classes

(ii) For every ε > 0 there exists a neighbourhood U of 0 in X such that

sup d (τ f, f): f F < ε { p h ∈ } whenever h U, where (τ f)(x) := f(x + h); ∈ h (iii) For every ε > 0 there exists a compact K X such that ⊂ sup d (f, f ): f F < ε , { p K ∈ } where f(x) if x K fK (x) = ∈ Q y if x K.  0 6∈  J K It follows that F is relatively compact in Lp(X, QQ(Y ))[dp].

Proof. In this proof we will abbreviate = ∗ . In view of the completeness of k · k k · k(Lip Y ) Lp(X, QQ(Y )) (Proposition 16) we need only to show that F is totally bounded. Let ε > 0 and choose U and K according to hypotheses (C)(ii) and (C)(iii). There is no restriction to assume that Clos U is compact. We next consider a continuous function ϕ : X R+ such that supp ϕ U and ϕdλ = 1. Given f F we consider the map → ⊂ X ∈ R ζ f : X Lip (Y )∗ ◦ K → y0 and we observe that it is (BX , BLip (Y )∗ )-measurable, separably valued (in fact im ζ y0 f im ζ and the latter is compact according to the continuity of ζ, Theorem 6, and the◦ K ⊂ compactness of QQ(Y ), Proposition 1). It therefore ensues from the Pettis measurability Theorem, [DU77, Chap. II §1 Theorem 2], that ζ f is strongly measurable, i.e. ◦ K the pointwise λ almost everywhere limit of a sequence of (BX , BLip (Y )∗ )-measurable y0 functions with ﬁnite range. Furthermore ζ fK is bounded (because Y is) and compactly supported (because ζ(Q y ) = 0), so that◦ the Lebesgue integral ζ f dλ < . 0 X k ◦ K k ∞ Thus ζ fK is Bochner integrable. We deﬁne the convolution productR of ϕ and ζ fK by means◦ of the BochnerJ integral:K ◦

(ϕ ζ fK )(x) = (B) ϕ(h)(ζ fK )(x + h)dλ(h) , x X. ∗ ◦ ZX ◦ ∈

We now claim that each ϕ (ζ fK ) is continuous and, in fact, that the family ∗ ∗ ◦ ′ C(X, Lipy0 (Y ) ) ϕ (ζ fK ): f F is equicontinuous. Given x, x X we simply observe that ∩ { ∗ ◦ ∈ } ∈ (ϕ ζ f )(x) (ϕ ζ f )(x′) k ∗ ◦ K − ∗ ◦ K k ′ = (B) (ϕ(h) ϕ(h + x x ))(ζ fK )(x + h)dλ(h) ZX − − ◦ 1− 1 1 p p p ′ −1 p ϕ(h) ϕ(h + x x ) p dλ(h) (ζ f K )(x + h) dλ(h) ≤ ZX | − − | ZX k ◦ k ′ ′ 1− 1 osc(ϕ, x x X )λ(U + BX (0, x x X )) p |f| , ≤ k − k k − k Lp according to [DU77, Chap. II §2 Theorem 4(ii)], Hölder’s inequality, and Theorem 6. The equicontinuity follows from the uniform continuity of ϕ and hypothesis (C)(i). 3.2 - Analog of the Fréchet-Kolmogorov compactness Theorem 51

∗ We denote by C the closed convex hull of im ζ in the Banach space Lipy0 (Y ) . As im ζ is compact it ensues from Mazur’s Theorem that C is compact as well. Furthermore, the deﬁnition of the convolution product guarantees that (ϕ ζ fK )(x) C for every x X. It therefore follows from Ascoli’s Theorem, [Edw95,∗ 0.4.11],◦ that∈ the family ∈ ∗ ∗ C(X, Lipy0 (Y ) ) ϕ (ζ fK ): f F is relatively compact in Cc(X, Lipy0 (Y ) ) with respect to uniform∩ { convergence∗ ◦ (note∈ that} supp(ϕ ζ f ) K + Clos U, a compact set ∗ ◦ K ⊂ independent of f). Consequently there are f 1, . . . , f κ F such that for every f F there exists k 1,...,κ with ∈ ∈ ∈ { }

− 1 (ϕ ζ f )(x) (ϕ ζ f k )(x) < ελ(K + Clos U) p (16) k ∗ ◦ K − ∗ ◦ K k for every x X. ∈ Now given f F we choose k so that (16) holds and we aim at showing that k ∈ dp(f, f ) < Dε where D is a suitable constant; this will complete the proof. We start with the observation that

d (f, f k) d (f, f ) + d (f , f k ) + d (f k , f k) 2ε + d (f , f k ) p ≤ p K p K K p K ≤ p K K according to hypothesis (C)(iii). Next we infer from Theorem 6 and (16) that

1 k k p p cQdp(fK , fK ) (ζ fK ) (ζ fK ) dλ ≤ X k ◦ − ◦ k Z 1 p p (ζ fK ) (ϕ ζ fK ) dλ ≤ X k ◦ − ∗ ◦ k Z 1 k p p + (ϕ ζ fK ) (ϕ ζ fK ) dλ K+Clos U k ∗ ◦ − ∗ ◦ k Z 1 k k p p + (ϕ ζ fK ) (ζ fK ) dλ X k ∗ ◦ − ◦ k Z 1 p p (ζ fK ) (ϕ ζ fK ) dλ + ε ≤ X k ◦ − ∗ ◦ k Z 1 k k p p + (ϕ ζ fK ) (ζ fK ) dλ ZX k ∗ ◦ − ◦ k Thus it remains only to ﬁnd a uniform small upper bound of

p (ζ fK ) (ϕ ζ fK ) dλ ZX k ◦ − ∗ ◦ k whenever f F . Let x X, abbreviate µ = λ ϕ, and observe that ∈ ∈ (ζ f )(x) (ϕ ζ f )(x) k ◦ K − ∗ ◦ K k = (B) ϕ(h) (ζ fK )(x) (ζ fK )(x + h) dλ(h) ZX ◦ − ◦

1(fK (x), τhfK (x))dµ(h) . ≤ ZX G It then follows from Jensen’s inequality applied to the probability measure µ, and from 52 Sobolev classes

Fubini’s Theorem that

p (ζ fK )(x) (ϕ ζ fK )(x) dλ(x) X k ◦ − ∗ ◦ k Z p dλ(x) 1(fK (x), τhfK (x))dµ(h) ≤ ZX ZX G p dλ(x) 1(fK (x), τhfK (x)) dµ(h) ≤ ZX ZX G p/2 p = Q dµ(h)dp(fK , τhfK ) ZU p/2 p Q sup dp(fK , τhfK ) . ≤ h∈U Consequently, 1 p p (ζ fK ) (ϕ ζ fK ) dλ 3 Qε X k ◦ − ∗ ◦ k ≤ Z q according to hypotheses (C)(ii) and (iii). Therefore,

d (f , f k ) c−1(1 + 6 Q)ε , p K K ≤ Q q and ﬁnally, d (f, f k) 2 + c−1(1 + 6 Q) ε . p ≤ Q q

1 3.3 Deﬁnition of Wp (U; QQ(Y ))

In this section X is a finite dimensional Banach space with Haar measure λ, U X is either X itself or a bounded open subset having the extension property4, Y is a Banach⊂ space having the Radon-Nikodým property, and 1 < p < . The space Hom(X,Y ) is given a norm . We recall that each Lipschitz map f ∞: U Q (Y ) extends to a ||| · ||| → Q Lipschitz map fˆ : X QQ(Y ) according to Theorem 2, and that fˆ is differentiable at → Q λ almost every x U, according to Theorem 3. For such x, writing Df(x) = i=1 Li , we recall that we∈ have defined ⊕ J K Q 2 |Df(x)| = v Li . u ||| ||| uXi=1 t 1 We define the Sobolev class Wp (U; QQ(Y )) to be the subset of Lp(U; QQ(Y )) consisting of those f : U QQ(Y ) for which there exists a sequence fj of Lipschitz mappings X Q (Y ) with→ the following properties { } → Q (1) f L (U; Q (Y )) and |Df |pdλ < for every j = 1, 2,...; j ∈ p Q U j ∞ (2) sup |Df |pdλ < ; R j U j ∞ (3) d (f,R f ) 0 as j . p j → → ∞ 4 1 1 i.e. for every 1 < p < there exists an extension operator Wp(U) Wp(X) for classical Sobolev spaces; for instance U has∞ Lipschitz boundary → 1 3.3 - Definition of Wp (U; QQ(Y )) 53

In case U is bounded, (1) is redundant.

1 1 We define Wp(U; QQ(Y )) to be the quotient of Wp (U; QQ(Y )) relative to the equivalence relation f1 f2 iff λ f1 = f2 = 0. We now recall the definition of F.J. Almgren’s ∼ n { 6 } m n Sobolev class Yp(U; QQ(ℓ2 )). Here X = ℓ2 and Y = ℓ2 . This is simply the collection of Borel functions f : U RN (where N = N(n, Q) is as in 4) such that f is a member of → 1 N n m the classical Sobolev space Wp (U; R ), and f(x) ξ(ℓ2 ) for L almost every x U. This is reminiscent of the definition of Sobolev mappings∈ between Riemannian mani-∈ n folds, except for QQ(ℓ2 ) is not a Riemannian manifold, but merely a stratified space. n m We also let Yp(U; QQ(ℓ2 )) denote the corresponding quotient relative to equality L m ν almost everywhere. We finally recall that Hom(ℓ2 , ℓ2) is equipped with the following norm m ν L = L(e ), e 2 v j k ||| ||| uj=1 k=1h i uX X t that appears in the following result.

m n Theorem 8. Assuming that X = ℓ2 and Y = ℓ2 , the mapping

Υ: W 1(U; Q (ℓn)) Y (U; Q (ℓn)) : f ξ f p Q 2 → p Q 2 7→ ◦ yields a bijection Υ : W1(U; Q (ℓn)) Y (U; Q (ℓn)) , p Q 2 → p Q 2 and

(1) (f(x),Q 0 )pdL m(x) = Υ(f)(x) pdL m(x); U G2 U k k R R (2) If f is LipschitzJ K then

|Df(x)|pdL m(x) = DΥ(f)(x) pdL m(x) . ZU ZU ||| |||

n 1 n Proof. We first show that ξ f Yp(U; QQ(ℓ2 )) whenever f Wp (U; QQ(ℓ2 )). It is clear that f : U RN is Borel◦ measurable∈ and also that ∈ → p m p m (ξ f)(x) dL (x) = 2(f(x),Q 0 ) dL (x) < , ZU k ◦ k ZU G ∞ according to Theorem 4(C), thus ξ f is a memberJ ofK the classical Lebesgue space N ◦ Lp(U; R ) and conclusion (1) is proved. Assuming that f be also Lipschitz then so is ξ f, thus conclusion (2) holds according to Proposition 14 (in conjunction with Theorem 3◦ and the classical Rademacher Theorem), whence ξ f belongs to the classical Sobolev 1 N ◦ 1 n space Wp (U; R ). If we now return to merely assuming that f Wp (U; QQ(ℓ2 )) in ∈ n our definition, then there exists a sequence fj of Lipschitz maps X QQ(ℓ2 ) such p m { } p m → that supj U |Dfj| dL < and limj U 2(f, fj) dL = 0. We infer from conclusions (1) and (2) that ξ f is∞ a bounded sequenceG in W 1(U; RN ). Since W1(U; RN ) is a R { ◦ j} R p p reflexive Banach space, there exists a subsequence ξ fk(j) converging weakly to some g W 1(U; RN ). Since U has the extension property,{ ◦ the weak} convergence corresponds ∈ p to convergence in Lp : p m lim (ξ fk(j)) g dL = 0 , j ZU k ◦ − k 54 Sobolev classes and therefore ξ f = g L m almost everywhere, which readily implies that ξ f n ◦ ◦ ∈ Yp(U; QQ(ℓ2 )). We next observe that the equivalence class of Υ(f) depends only upon the equivalence class of f, because ξ maps null sets to null sets. Since the same is true about ξ−1, we infer n that Υ is injective. It remains to show that Υ is surjective. Let g Yp(U; QQ(ℓ2 )). n ∈ There is no restriction to assume that g(x) ξ(ℓ2 ) for all x U, and we define −1 ∈ 1 ∈ N f = ξ g; it is obviously Borel measurable. Since g Wp (U; R ) and U has the extension◦ property, there existsg ˆ W 1(Rn; RN ) such that∈g ˆ is compactly supported in ∈ p a neighborhood of U andg ˆ ↾ U = g. Choosing ϕεj a smooth compactly supported approximation to the identity, we define { } f = ξ−1 ρ (ϕ gˆ) . j ◦ ◦ εj ∗ We observe that the f : X Q (ℓn) are Lipschitz and j → Q 2 p m −1 −1 p m 2(fj, f) dL = 1(ξ ρ (ϕεj gˆ), ξ g) dL ZU G ZU G ◦ ◦ ∗ ◦ −p p m α(n, Q) ρ (ϕεj gˆ) ρ g dL ≤ ZU k ◦ ∗ − ◦ k −p p p m α(n, Q) (Lip ρ) ϕεj gˆ g dL ≤ ZU k ∗ − k 0 as j . → → ∞

Finally, if fj and ρ (ϕεj gˆ) are both diﬀerentiable at a U, then Proposition 14 implies that ◦ ∗ ∈ |Df (a)|p = |D(ξ−1 ρ (ϕ gˆ))(a)|p j ◦ ◦ εj ∗ = D(ρ (ϕ gˆ))(a) p ||| ◦ εj ∗ ||| (Lip ρ)p D(ϕ gˆ)(a) p . ≤ ||| εj ∗ ||| Since this occurs that L m almost every a U, according to Theorem 3 and the classical Rademacher Theorem, we infer that ∈

p m p p m sup |Dfj(a)| dL (Lip ρ) sup D(ϕεj gˆ) dL j ZU ≤ j ZU ||| ∗ ||| (Lip ρ)p Dgˆ pdL m . ≤ ZU ||| ||| Thus f W 1(U; Q (ℓn)). ∈ p Q 2 It is worth observing that in case p = 1 the above Theorem would not be valid, as n our definition would yield a space of mappings U QQ(ℓ2 ) of bounded variation rather than Sobolev. → We recall that U is assumed to have the extension property. This means that there exists a continuous linear operator E : W1(U; RN ) W1(Rm; RN ) . p → p 1 m N 1 m n Given f Wp(R ; R ) and f f, one easily checks that ρ f Yp (R ; QQ(ℓ2 )) and that the∈ equivalence class of ρ ∈f depends only upon that of◦f.∈ Thus the formula ◦ E˜(f) = Υ−1 (ρ E (Υ (f))) ◦ 1 3.3 - Definition of Wp (U; QQ(Y )) 55 defines an “extension mapping” W1(U; Q (ℓn)) W1(Rm; Q (ℓn)) . p Q 2 → p Q 2 Proposition 17. Let f W 1(U; Q (ℓn)) and t 0. Define ∈ p Q 2 ≥ A = U x : (f(x),Q 0 )p + (M DE(Υ(f)) )p (x) tp , t ∩ G2 ||| ||| ≤ n o where M denotes the maximal functionJ K operator and E(Υ(f)) is a representant of the class E(Υ(f)). Then there exists a Lipschitzian map h : U Q (ℓn) such that → Q 2 (1) h(x) = f(x) for L m almost every x A ; ∈ t (2) Lip h 4m+1α(n, Q)−1c (m, Q)t where α(n, Q) is as in Theorem 4; ≤ 2 (3) (h(x),Q 0 ) c (m, Q)t for every x U; G2 ≤ 2 ∈ m (4) For L almostJ K every x At, f is approximately differentiable at x and |Df(x)| = |Dh(x)|. ∈

Proof. Write u = E(Υ(f)) W 1(Rm; RN ). We let A˜ denote the Borel subset of A ∈ p t t consisting of those x such that u(x) = limε→0+ (ϕε u)(x) where ϕε ε>0 is a given ′ ∗ { } ′ approximate identity. Given distinct x, x A˜t we let Ω = U(x, 2r) U(x , 2r), where r = x x′ > 0, and we infer from [MZ97,∈ Lemma 1.50] (adapted in∩ the obvious way to thek case− k of vectorvalued maps) that

m m 4 Du(y) m u(x) udL ||| m|||−1 dL (y) −Z −Ω ≤ mα(m) ZΩ x y k − k and m ′ m 4 Du(y) m u(x ) udL |||′ m|||−1 dL (y) −Z −Ω ≤ mα(m) ZΩ x y k − k

It follows from the potential estimate [MZ97, Theorem 1.32(i)] that Du(y) Du(y) ||| ||| dL m(y) ||| ||| dL m(y) Ω x y m−1 ≤ U(x,2r) x y m−1 Z k − k Z k − k mα(m)2 x x′ M ( Du )(x) . ≤ k − k ||| ||| Since the same holds with x replaced by x′, we obtain u(x) u(x′) 4m+1t x x′ k − k ≤ k − k ′ whenever x, x A˜t. The first three conclusions now follow from Theorems 4 and 2. Conclusion (4)∈ follows from the fact that h and f are approximately tangent at each Lebesgue density point of At, together with the differentiability Theorem 3. Remark 2. We shall see in Proposition 24 that the constant in (2) does not in fact depend upon n. Corollary 3. Let f W 1(U; Q (ℓn)). It follows that f and Υ(f) are approximately ∈ p Q 2 differentiable L m almost everywhere, and that |Df(x)| = DΥ(f)(x) ||| ||| at each point x U where both are approximately differentiable. ∈ 56 Sobolev classes

Proof. That Υ(f) be approximately diﬀerentiable (in the usual sense) L m almost everywhere follows from standard Sobolev theory (see e.g. [MZ97, Theorem 1.72]). The analogous property of f follows from Proposition 17(4) and the arbitrariness of t 0. The last conclusion is a consequence of Proposition 14. ≥

3.4 The p energy

In this section, X, Y , U and p are subject to the same requirements as in the last 1 section, and sometimes more. Given f Wp (U; QQ(Y )) and an open subset V U, we deﬁne the p energy of f in V by the∈ formula ⊂

p p Dir (f; V ) = inf lim inf |Dfj| dλ : fj is a sequence p j ( ZV { }

of Lipschitz mappings U QQ(Y ) such that dp(f, fj) 0 as j . → → → ∞) We notice that Dirp(f; V ) Dirp(f; U) < . Clearly, p ≤ p ∞ 1 Proposition 18. Given f Wp (U; QQ(Y )) and an open subset V U, there exists a sequence of Lipschitz mappings∈ U Q (Y ) such that lim d (f, f )⊂ = 0 and → Q j p j p p Dir (f; V ) = lim |Dfj| dλ . p j ZV

k Proof. For each k N 0 choose a sequence fj j of Lipschitz mappings U QQ(Y ) such that d (f k, f)∈ \{0 as}j , and { } → p j → → ∞ 1 Dirp(f; V ) lim inf |Df k|pdλ < + Dirp(f; V ) . p j j p ≤ ZV k k −1 There exists an integer j(k) such that dp(f, fj(k))

p 1 k p 1 p Dirp(f; V ) |Dfj(k)| dλ < + Dirp(f; V ) . − k ≤ ZV k The sequence f k suits our needs. { j(k)}k As the p-energy is deﬁned by relaxation, we easily prove its lower semicontinuity with respect to weak convergence.

1 Proposition 19. Let f, f1, f2,... be members of Wp (U; QQ(Y )) and assume that d (f, f ) 0 as j . It follows that p j → → ∞ p p Dir (f; V ) lim inf Dir (fj; V ) p ≤ j p for every open subset V U. ⊂

Proof. For each j = 1, 2,... there exists a Lipschitz mapping gj : U Q(Y ) such that d (f , g ) j−1 and → p j j ≤ p 1 p |Dgj| dλ + Dirp(fj; V ) , ZV ≤ j 3.4 - The p energy 57 by deﬁnition of Dirp(f ; V ). Since d (f, g ) 0 as j , we have p j p j → → ∞

p p p Dir (f; V ) lim inf |Dgj| dλ lim inf Dir (fj; V ) . p j j p ≤ ZV ≤

If W Y is a linear subspace and P : Y W is a continuous linear retract we deﬁne ⊂ → Q (P ): Q (Y ) Q (W ) Q Q → Q by the formula QQ(P )( y1, . . . , yQ ) = P (y1),...,P (yQ) . It is a trivial matter to check that J(Q (P )(v)K, Q J(P )(v′)) (Lip PK)Q (v, v′) G2 Q Q ≤ 2 whenever v, v′ Q (Y ). ∈ Q Proposition 20. Assume that

(1) W Y is a linear subspace and P : Y W is a continuous linear retract; ⊂ → (2) g : U Q (Y ) is approximately differentiable at a U; → Q ∈ (3) Hom(X,Y ) and Hom(X,W ) are equipped with norms such that P L (Lip P ) L whenever L Hom(X,Y ). ||| ◦ ||| ≤ ||| ||| ∈ It follows that Q (P ) g is approximately differentiable at a and Q ◦ |D(Q (P ) g)(a)| (Lip P )|Dg(a)| . Q ◦ ≤ Proof. Write Ag(a) = Q A , with A : X Y affine maps. Observe that ⊕i=1 i i → J K Q 2 (QQ(P ) g)(x), QQ(P ) i=1 Ai (x) ap lim G ◦ ◦ ⊕ x→a x a J K k − k Q 2 g(x), i=1 Ai (x) (Lip P )ap lim G ⊕ = 0 . ≤ x→a x a k − Jk K Q Q Since QQ(P ) i=1 Ai = i=1 P Ai , and the P Ai are affine as well, we infer that ◦ ⊕ ⊕ ◦ ◦ Q QQ(P ) g is differentiable at a and A(QQ(P ) g)(a) = i=1 P Ai . Next note that if ◦ J K J K ◦ ⊕ ◦ Li is the linear part of Ai, then P Li is the linear part of P Ai. Consequently, ◦ J◦ K Q Q |D(Q (P ) g)(a)|2 = P L 2 (Lip P )2 L 2 = (Lip P )2|Dg(a)|2 . Q ◦ ||| ◦ i||| ≤ ||| i||| Xi=1 Xi=1

Proposition 21. Assume that

(1) W Y is a linear subspace and ι : W Y is the canonical injection; ⊂ → (2) g : U Q (W ) is diﬀerentiable at a U; → Q ∈ 58 Sobolev classes

(3) Hom(X,Y ) and Hom(X,W ) are equipped with norms such that ι L = L whenever L Hom(X,W ). ||| ◦ ||| ||| ||| ∈

It follows that Q (ι) g is diﬀerentiable at a and Q ◦ |D(Q (ι) g)(a)| = |Dg(a)| . Q ◦

Proof. The proof is similar to that of Proposition 20.

Hypotheses (3) of Proposition 20 and 21 are veriﬁed in two cases of interest. First when is the operator norm. Second when |||· ||| m L = ν L(u ) e ||| |||  k j kY j jX=1   m m where ν is a norm on R , m = dim X, e1, . . . , em is the canonical basis of R , and u1, . . . , um is a basis of X.

Proposition 22. Assume that

(1) W Y is a linear subspace, P : Y W is a continuous linear retraction, and ι : W⊂ Y is the canonical injection;→ → (2) g W 1(U; Q (W )); ∈ p Q (3) Hom(X,Y ) and Hom(X,W ) are equipped with norms such that P L (Lip P ) L whenever L Hom(X,Y ), and ι L = L whenever||| ◦ L||| ≤ Hom(X,W||| |||). ∈ ||| ◦ ||| ||| ||| ∈

It follows that Q (ι) g W 1(U; Q (Y ) and Q ◦ ∈ p Q (Lip P )−pDirp(g; V ) Dirp(Q (ι) g; V ) Dirp(g; V ) , p ≤ p Q ◦ ≤ p for every V U open. ⊂

Proof. Choose a sequence gj of Lipschitz mappings U QQ(W ) such that dp(g, gj) p { } p → → 0 and Dirp(g; V ) = limj V |Dgj| dλ, according to Proposition 18. Notice that QQ(ι) g are Lipschitz mappings U Q (Y ) and that ◦ R → Q

lim sup dp(QQ(ι) g, QQ(ι) gj) lim dp(g, gj) = 0 . j ◦ ◦ ≤ j

Therefore,

p p Dir (QQ(ι) g; V ) lim inf |D(QQ(ι) gj)| dλ p j ◦ ≤ ZV ◦ p lim sup |Dgj| dλ ≤ j ZV p = Dirp(g; V ) , 3.4 - The p energy 59 according to Proposition 21. The case V = U of this computation implies that QQ(ι) 1 ◦ g Wp (U; QQ(Y )), by deﬁnition of this Sobolev class, and the general case yields the second∈ inequality of our conclusion.

The other way round choose a sequence fj of Lipschitz mappings U QQ(Y ) such that d (Q (ι) g, f ) 0 and Dirp(Q ({ι) }g; V ) = lim |Df |pdλ. Notice→ that p Q ◦ j → p Q ◦ j V j the mappings QQ(P ) fj : U QQ(W ) are Lipschitz and, sinceR g = QQ(P ) QQ(ι) g, one has ◦ → ◦ ◦

d (g, Q (P ) f ) = d (Q (P ) Q (ι) g, Q (P ) f ) p Q ◦ j p Q ◦ Q ◦ Q ◦ j d (Q (ι) g, f ) 0 as j . ≤ p Q ◦ j → → ∞ Thus,

p p Dir (g; V ) lim inf |D(QQ(P ) fj)| dλ p j ≤ ZV ◦ p p (Lip P ) lim inf |Dfj| dλ j ≤ ZV = (Lip P )pDirp(Q (ι) g; V ) . p Q ◦

n For the remaining part we will only consider the cases when either Y = ℓ2 for some n N 0 or Y = ℓ2, and X is a finite dimensional Banach space as usual. The norm ∈ on\{ Hom(} X,Y ) is associated with a basis u , . . . , u of X as follows: |||· ||| 1 m m L = L(u ) 2 v j ||| ||| uj=1 k k uX t where is the Hilbert norm on Y . According to last remark, Propositions 20 and 21 apply.k · k When Y = ℓ and n N 0 we also define an n dimensional subspace of 2 ∈ \{ } Y , Wn = span e1, . . . , en , and we let Pn : Y Wn be the orthogonal projection and ι : W Y be{ the canonical} injection. → n n → n The following guarantees that the p energy is the expected quantity in case Y = ℓ2 . Notice the statement makes sense since g is almost everywhere approximately differentiable (recall Corollary 3).

Proposition 23. If g W 1(U; Q (ℓn)) for some n N 0 then ∈ p Q 2 ∈ \{ }

p p Dirp(g; V ) = |Dg| dλ ZV for every open set V U. ⊂

Proof. If g is a sequence of Lipschitz mappings U Q (ℓn) such that d (g, g ) 0 { j} → Q 2 p j → as j , then ξ g ξ gn Lp 0 as j where ξ is the Almgren embedding described→ ∞ in Theoremk ◦ 4.− Thus◦ k → → ∞

p p D(ξ g) dλ lim inf D(ξ gj) dλ j ZV ||| ◦ ||| ≤ ZV ||| ◦ ||| 60 Sobolev classes according to classical ﬁnite dimensional Sobolev theory: the above functional is weakly lower semicontinuous because it satisﬁes the hypotheses of [Dac08, Section 3.3, Theorem 3.4]. It then follows from Corollary 3 that

p p |Dg| dλ lim inf |Dgj| dλ . j ZV ≤ ZV Choosing the sequence g according to Proposition 18 we infer that { j} p p |Dg| dλ Dirp(g; V ) . ZV ≤

We turn to proving the reverse inequality. We let u = E(Υ(g)) W 1(Rm; RN ) so ∈ p that the maximal function M( Du ) Lp(U) (see e.g. [MZ97, Theorem 1.22]). For each j N 0 we deﬁne ||| ||| ∈ ∈ \{ } A = U x : (g(x),Q 0 )p + M( Du )p(x) jp j ∩ { G2 ||| ||| ≤ } and we infer that J K

p p p lim j λ(U Aj) lim ( 2(g(x),Q 0 ) + M( Du ) (x)) dλ(x) = 0 . j j \ ≤ ZU\Aj G ||| ||| n J K We let gj : U QQ(ℓ2 ) be the Lipschitz mapping associated with f = g and t = j in Proposition 17.→ We see that

1 p p lim dp(gj, g) = lim 2(gj, g) dλ(x) j j ZU\Aj G ! 1 p p lim 2(gj,Q 0 ) dλ(x) j ≤ ZU\Aj G ! 1 J K p p + lim 2(g, Q 0 ) dλ(x) j ZU\Aj G ! 1 1 J K p p p p lim (c2(m, Q)j λ(U Aj)) + lim 2(g, Q 0 ) dλ(x) j j ≤ \ ZU\Aj G ! = 0 , J K thus Dirp(g; V ) lim inf |Dg |pdλ. Furthermore, p ≤ j V j R p p p lim inf |Dgj| dλ lim inf |Dgj| dλ + lim sup |Dgj| dλ j j ZV ≤ ZV ∩Aj j ZU\Aj p p p lim inf |Dg| dλ + Q 2 lim sup (Lip gj) dλ j ≤ ZV ∩Aj j ZU\Aj |Dg|pdλ ≤ ZV p p(m+1) −p p p + lim sup Q 2 4 α(n, Q) c2(m, Q) j λ(U Aj) j \ = |Dg|pdλ . ZV This completes the proof. 3.4 - The p energy 61

Theorem 9. Let f W 1(U; Q (ℓ )). The following hold. ∈ p Q 2 (A) Q (P ) f W 1(U; Q (ℓn)) for each n N 0 ; Q n ◦ ∈ p Q 2 ∈ \{ } (B) For every open set V U one has ⊂ p p Dir (f; V ) = lim |D(QQ(Pn) f)| dλ ; p n ZV ◦ (C) The sequence |D(Q (P ) f)|p is nondecreasing λ almost everywhere and { Q n ◦ }n bounded in L1(U).

Proof. (A) Choose a sequence fj of Lipschitz mappings U QQ(ℓ2) such that d (f , f) 0 and sup |Df |pdλ{ <} . Notice that the Q (P )→f : U Q (ℓn) are p j → j U j ∞ Q n ◦ → Q 2 Lipschitz, R lim dp(QQ(Pn) fj, QQ(Pn) f) lim dp(fj, f) = 0 j ◦ ◦ ≤ j and p p sup |D(QQ(Pn) fj)| dλ sup |Dfj| dλ < j ZU ◦ ≤ j ZU ∞ according to Proposition 20. Thus f W 1(U; Q (ℓn)). ∈ p Q 2 (B) We note that for every x U one has ∈ lim 2 f(x), (QQ(ιn) QQ(Pn) f)(x) = 0 , n G ◦ ◦ and also (f, Q (ι ) Q (P ) f) (f, Q 0 ) + (Q (ι ) Q (P ) f, Q 0 ) G2 Q n ◦ Q n ◦ ≤ G2 G2 Q n ◦ Q n ◦ 2 (f, Q 0 ) . ≤ G2 J K J K Thus J K lim dp(f, QQ(ιn) QQ(Pn) f) = 0 n ◦ ◦ according to the Dominated Convergence Theorem. Therefore, p p Dir (f; V ) lim inf Dir (QQ(ιn) QQ(Pn) f; V ) p ≤ n p ◦ ◦ p = lim inf Dir (QQ(Pn) f; V ) n p ◦ p = lim inf |D(QQ(Pn) f)| dλ , n ZV ◦ according respectively to Propositions 19, 22 and 23. The other way round, we choose a sequence fj of Lipschitz mappings U QQ(ℓ2) such that dp(fj, f) 0 and p { } p → → Dirp(f; V ) = limj V |Dfj| dλ, according to Proposition 18. For each ﬁxed n we have

R p p |D(QQ(Pn) f)| dλ lim inf |D(QQ(Pn) fj)| dλ j ZV ◦ ≤ ZV ◦ (according to the proof of (A) and Proposition 19 and 23)

p lim inf |Dfj| dλ j ≤ ZV (according to Proposition 20)

p = Dirp(f; V ) . 62 Sobolev classes

Therefore lim sup |D(Q (P ) f)|pdλ Dirp(f; V ). n V Q n ◦ ≤ p (C) That the sequenceR |D(Q (P ) f)|p be nondecreasing follows as in the proof { Q n ◦ }n of Proposition 20; its boundedness in L1(U) is a consequence of (B).

|δf|(x) = lim |D(QQ(Pn) f)(x)| . (17) n ◦ It follows therefore from Theorem 9(B) that

p p Dirp(f; V ) = |δf| dλ . (18) ZV

3.5 Extension

The following is the obvious analog of [MZ97, Theorem 1.63]. Theorem 10. Let U = U(0, 1) be the unit ball in Rm. There exists a mapping

E : W 1(U; Q (ℓ )) W 1(Rm; Q (ℓ )) p Q 2 → p Q 2 with the following properties.

(A) For every f W 1(U; Q (ℓ )) one has E(f)(x) = f(x) for every x U; ∈ p Q 2 ∈ (B) For every f , f W 1(U; Q (ℓ ) one has 1 2 ∈ p Q 2 1 d (E(f ),E(f )) 2 p d (f , f ); p 1 2 ≤ p 1 2 (C) For every f W 1(U; Q (ℓ )) one has ∈ p Q 2 p m p p p p Dir (E(f); R ) 1 + Q 2 2 Dir (f; U) + |f| ; p ≤ p p (D) For every x Rm U(0, 2) one has E(f)(x) = Q 0 ; ∈ \ (E) If 0 C ℓ2 is convex and f(x) QQ(C) for everyJ Kx U, then E(f)(x) QQ(C) for every∈ ⊂x Rm. ∈ ∈ ∈ ∈

Proof. We start the proof by associating with each Lipschitz map f : U QQ(ℓ2) a Lipschitz map E (f): Rm Q (ℓ ) verifying (A), (B), (D) and (E) above (for→ Lipschitz 0 → Q 2 maps f, f1, f2) and (C) replaced with

(C’) For every Lipschitz f : U Q (ℓ ) one has → Q 2 p m p m p |DE0(f)(x)| dL (x) C(m, p) |Df(x)| dL (x) + |f| . m p ZR ≤ ZU 3.5 - Extension 63

Given f we write f(x) = Q f (x) , x U, and we deﬁne ⊕i=1 i ∈ J Q (2K x 1)f (x) if x 1 g(x) = ⊕i=1 k k − i k k ≥ 2 Q 0 if x < 1 .  J K k k 2 The conscientious reader will checkJ K that g is Lipschitz on U. In fact, it follows from Proposition 10 and the paragraph preceding it (in particular equation (17)) that

|Dg(x)| Q |Df(x)| + 2|f(x)| ≤ ! q for almost every x U(0, 1) B(0, 1/2). Therefore, ∈ \ p m p p p m p m |Dg| dL Q 2 2 |Df| dL + |f| dL . (19) ZU ≤ ZU ZU

We now define a Lipschitz mapping ϕ : U(0, 3/2) U(0, 1) B(0, 1) B(0, 1/2) by the formula \ → \ 2 ϕ(x) = 1 x , x − ! k k and E0(f) by 3 Q 0 if x 2 ≥ 3 E0(f)(x) = g(ϕ(x)) if 1 x <  J K ≤ k k 2 f(x) if x 1 . k k ≤  We notice that conclusions (A), (D) and (E) are verified by E0(f). Regarding conclusions (B) and (C’) we first observe that the differential of x/ x at a point x = 0 is the orthogonal projection onto the plane orthogonal to x. Thereforek k Jϕ = 1 and6 we apply the change of variable formula:

p p m p m dp(E0(f1),E0(f2)) (f1, f2) dL + (g1 ϕ, g2 ϕ) dL ≤ ZU G ZB(0,3/2)\B(0,1) G ◦ ◦ p m p m (f1, f2) dL + (g1 ϕ, g2 ϕ) JϕdL ≤ ZU G ZB(0,3/2)\B(0,1) G ◦ ◦ p m 2 (f1, f2) dL ≤ ZU G (because (g , g ) (f , f )), and similarly, G 1 2 ≤ G 1 2 p m p m p m |DE0(f)| dL |Df| dL + |D(g ϕ)| dL m ZR ≤ ZU ZB(0,3/2)\B(0,1) ◦ |Df|pdL m + (|Dg|p ϕ) dL m ≤ ZU ZB(0,3/2)\B(0,1) ◦ (because Lip ϕ 1) ≤

|Df|pdL m + (|Dg|p ϕ) JϕdL m ≤ ZU ZB(0,3/2)\B(0,1) ◦ p p p m p m 1 + Q 2 2 |Df| dL + |f| dL ≤ U U Z Z according to (19). 64 Sobolev classes

We now define E(f), f W 1(U; Q (ℓ )), as follows. We choose a sequence f of ∈ p Q 2 { j} Lipschitz mappings U QQ(ℓ2) associated with f as in Proposition 18 and we observe → m that E0(fj) is Cauchy in Lp(R ) : for E(f) we choose a limit of this sequence (that verifies{ conclusion} (A)). That conclusions (B), (C), (D) and (E) are valid is now a matter of routine verification.

3.6 Poincaré inequality and approximate diﬀerentiability almost everywhere

We start with a modiﬁcation of Theorem 9.

1 Theorem 11. Let f Wp (U; QQ(ℓ2)). There then exist a sequence fn of Lipschitz mappings U Q (ℓ )∈and a sequence A of Borel subsets of U such{ that} → Q 2 { n}

(A) limn dp(fn, f) = 0;

(B) For every open set V U, ⊂

p p Dir (f; V ) = lim |Dfn| dλ . p n ZV

m m (C) limn L (U An) = 0 and, for each n, |Dfn(x)| |δf|(x) for L almost every x A ; \ ≤ ∈ n (D) lim |Df (x)| = |δf|(x) for L m almost every x U. n n ∈

1 n Proof. With each n N 0 we associate gn = QQ(Pn) f Wp (U; QQ(ℓ2 )) as well as ∈ \{ } ◦ ∈ u = (g ,Q 0 )p + M( DΥ(g ) )p L (U) . n G2 n ||| n ||| ∈ 1 Letting A = U x : u (x) tp J weK can choose t > 0 large enough for n ∩ { n ≤ n} n

m p 1 max L (U An), c17(n, m, Q) undλ < , (20) ( \ ZU\An ) n where we have put c (n, m, Q) = 4m+1α(n, Q)−1c (m, Q). We then let h : U Q (ℓn) 17 2 n → Q 2 be a Lipschitz mapping associated with gn and tn as in Proposition 17, and we deﬁne f = Q (ι ) h : U Q (ℓ ) which is Lipschitz as well. We observe that n Q n ◦ n → Q 2 d (f , f) d (Q (ι ) h , Q (ι ) Q (P ) f) + d (Q (ι ) Q (P ) f, f) p n ≤ p Q n ◦ n Q n ◦ Q n ◦ p Q n ◦ Q n ◦ d (h , g ) + d (Q (ι ) Q (P ) f, f) . ≤ p n n p Q n ◦ Q n ◦ Notice that

lim dp(QQ(ιn) QQ(Pn) f, f) = 0 n ◦ ◦ 3.6 - Poincaré inequality and approximate diﬀerentiability almost everywhere 65 according to the Dominated Convergence Theorem, whereas 1 p p dp(hn, gn) = 2(hn, gn) dλ U G Z 1 p p = 2(hn, gn) dλ ZU\An G ! 1 1 p p p p 2(hn,Q 0 ) dλ + 2(gn,Q 0 ) dλ ≤ ZU\An G ! ZU\An G ! 1 1 J K p J K p p p p c2(m, Q) tndλ + 2(gn,Q 0 ) dλ ≤ ZU\An ! ZU\An G ! 1 p J K (1 + c2(m, Q)) undλ ≤ ZU\An ! 0 as n , → → ∞ from what conclusion (A) follows. Consequently,

p p Dir (f; V ) lim inf |Dfn| dλ . p n ≤ ZV Furthermore, for each n we have

p p |Dfn| dλ = |Dhn| dλ ZV ZV (according to Proposition 22)

p p 2 p p |Dgn| dλ + Q c17(n, m, Q) tndλ ≤ ZV ∩An ZV \An (according to Proposition 17)

p p p |D(QQ(Pn) f)| dλ + Q 2 c17(n, m, Q) undλ . ≤ ZV ◦ ZU\An It now follows from (20) and Theorem 9(B) that

p lim sup |Dfn| dλ n ZV p p p lim |D(QQ(Pn) f)| dλ + lim sup Q 2 c17(n, m, Q) undλ n ≤ ZV ◦ n ZU\An p = Dirp(f; V ) . This proves conclusion (B). The ﬁrst part of conclusion (C) is a consequence of (20) and the second part follows m from the fact that hn = gn on An, therefore Dhn(x) = Dgn(x) at L almost every x An, and for those x it follows from Proposition 21, the deﬁnition of |δf| and Theorem∈ 9(C) that |Df (x)| = |D(Q (ι ) h )(x)| |Dh (x)| n Q n ◦ n ≤ n = |Dg (x)| = |D(Q (P ) f)(x)| |δf|(x) . n Q n ◦ ≤ Conclusion (D) is an easy consequence of (B) and (C). 66 Sobolev classes

We are now ready to prove the analog of the Poincaré inequality.

Theorem 12. There exists a constant c12(m) 1 with the following property. Let f W 1(U; Q (ℓ )), 1 q p, and let V U be≥ a bounded open convex subset of U. ∈ p Q 2 ≤ ≤ ⊂ It follows that for L m almost every x V , ∈ q q m q+m−1 |δf| (z) m 2(f(x), f(y)) dL (y) (diam V ) dL (z) . V G ≤ V z x m−1 Z Z k − k Furthermore there exists v Q (ℓ ) such that ∈ Q 2 1− 1 (diam V )m m q m c q q m 2(f(x), v) dL (x) 12(m)(diam V ) m |δf| dL . ZV G ≤ L (V ) ! ZV

Proof. We start with the case when f is Lipschitz. Given x V it follows from Theorem 3(D) that ∈

1 1 1 2(f(x), f(y)) |Df(z)|dH (z) = x y |Df(x + t(y x))|dL (t) G ≤ ZSx,y k − k Z0 − m for L almost every y V , where Sx,y denotes the line segment joining x and y. Now, given s > 0, we observe∈ that

q m 2(f(x), f(y)) dL (y) ZV G diam V 1 q m−1 = dL (s) 2(f(x), f(y)) dH (y) Z0 ZV ∩∂B(x,s) G diam V |Df(z)|q (21) sq+m−2dL 1(s) dL m(z) ≤ 0 V ∩B(x,s) z x m−1 Z Z k − k |Df(z)|q (diam V )q+m−1 dL m(z) . ≤ V z x m−1 Z k − k 1 We now merely assume that f Wp (U; QQ(ℓ2)) and we choose a sequence of Lipschitz mappings f as in Theorem∈ 11. Thus (21) applies to each f . Let x V be such { n} n ∈ 3.6 - Poincaré inequality and approximate diﬀerentiability almost everywhere 67 that limn 2(f(x), fn(x)) = 0. In order to establish our ﬁrst conclusion we can readily assume thatG |δf|q(x) V R : z → 7→ z x m−1 k − k is summable. In that case, it follows from Theorem 11(A) and (D), from (21) and from the Dominated Convergence Theorem that

q m q m 2(f(x), f(y)) dL (y) = lim 2(fn(x), fn(y)) dL (y) n ZV G ZV G |Df (z)|q (diam V )q+m−1 lim n dL m(z) ≤ n V z x m−1 Z k − k |δf|q(z) = (diam V )q+m−1 dL m(z) . V z x m−1 Z k − k

We now turn to proving the second conclusion. Integrating the inequality above with respect to x, and applying standard potential estimates (see e.g. [MZ97, Lemma 1.31] applied with p = 1) we obtain

m q m dL (x) 2(f(x), f(y)) dL (y) ZV ZV G |δf|q(z) (diam V )q+m−1 dL m(x) dL m(z) ≤ V V z x m−1 Z Z k − k q+m−1 m 1 q m C(m)(diam V ) L (V ) m |δf| dL . ≤ ZV Thus there exists x V such that ∈ 1 q+m−1 m m q m (diam V ) L (V ) q m 2(f(x), f(y)) dL (y) C(m) m |δf| dL . ZV G ≤ L (V ) ZV Letting v = f(x) completes the proof.

m 1 Proposition 24. Let U = U(0, 1) be the unit ball in R , let f Wp (U; QQ(ℓ2)) and t 0. Deﬁne ∈ ≥ A = U x : (f(x),Q 0 )p + (M|δE(f)|)p (x) tp , t ∩ G2 ≤ n o where M denotes the maximal function operatorJ K and E denotes the extension operator deﬁned in Theorem 10. There then exists a Lipschitzian map h : U Q (ℓ ) such that → Q 2

(1) h(x) = f(x) for L m almost every x A ; ∈ t

(2) Lip h 6mα(m)c??(m, Q)t; ≤ (3) (h(x),Q 0 ) c (m, Q)t for every x U; G2 ≤ 2 ∈ m (4) For L almostJ K every x At, f is approximately diﬀerentiable at x and |Df(x)| = |Dh(x)|. ∈ 68 Sobolev classes

Proof. The proof is similar to that of Proposition 17. We abbreviate fˆ = E(f). We choose a countable dense set D Rm and we consider the collection V of subsets V of ⊂ Rm of the type V = U(x, r) U(x′,r) where x, x′ D and r Q+. Thus V is countable ∩ ∈ m ∈ and for each V V there exists NV V such that L (V NV ) = 0 and for every x V N one∈ has ⊂ \ ∈ \ V |δfˆ|(z) (fˆ(x), fˆ(y))dL m(y) (diam V )m dL m(z) , (22) V G ≤ V z x m−1 Z Z k − k ′ m according to Theorem 12 applied with q = 1. Let N = V ∈V NV . Given x, x R N we choose r Q+ such that ∪ ∈ \ ∈ x x′ 0

3(diam V )m) |δfˆ|(z) G = V y : (fˆ(x), fˆ(y)) < dL m(z) , ∩  G L m(V ) V z x m−1   Z k − k  as well as  

3(diam V )m) |δfˆ|(z) G′ = V y : (fˆ(x′), fˆ(y)) < dL m(z) . ∩  G L m(V ) V z x′ m−1   Z k − k  One readily infer from (22) that 

L m(V ) max L m(V G), L m(V G′) < , \ \ 3 n o and hence G G′ = . We choose y G G′ and we set v = fˆ(y). Thus ∩ 6 ∅ ∈ ∩ 3(2r)m |δfˆ|(z) (fˆ(x), v) dL m(z) G ≤ α(m)rm V z x m−1 Z k − k and 3(2r)m |δfˆ|(z) (fˆ(x′), v) dL m(z) . G ≤ α(m)rm V z x′ m−1 Z k − k It follows from the potential estimate [MZ97, Lemma 1.32(i)] that

|δfˆ|(z) |δfˆ|(z) dL m(z) dL m(z) V z x m−1 ≤ U(x,2r) z x m−1 Z k − k Z k − k mα(m)(2r)M |δfˆ| (x) 3mα(m) x x′ M |δfˆ| (x) , ≤ ≤ k − k 3.6 - Poincaré inequality and approximate diﬀerentiability almost everywhere 69 and similarly

|δfˆ|(z) dL m(z) 3mα(m) x x′ M |δfˆ| (x′) . V z x′ m−1 ≤ k − k Z k − k ′ ′ If furthermore x, x At then max M(|fˆ|)(x),M(|fˆ|)(x ) t and it ensues from the above inequalities that∈ { } ≤

(fˆ(x), fˆ(x′)) 6mα(m) x x′ t . G ≤ k − k One now concludes like in Proposition 17.

The following is the analog of Proposition 23 for an inﬁnite dimensional target.

Corollary 4. Let U = U(0, 1) be the unit ball in Rm and let f W 1(U; Q (ℓ )). It ∈ p Q 2 follows that f is approximately diﬀerentiable L m almost everywhere and that

p p m Dirp(f; V ) = |Df(x)| dL (x) ZV for every open set V U. ⊂

+ m Proof. Letting tj be an increasing unbounded sequence in R we observe that L (U A ) 0 as j { } (where A is defined as in the statement of Proposition 24) because\ tj → → ∞ tj both (fˆ( ),Q 0 )) and M |δfˆ| belong to L (Rm). Letting h be a Lipschitz mapping G · p j U QQ(ℓ2) which coincides with f almost everywhere on At , we easily infer that → J K j f is approximately differentiable at each Lebesgue point x Atj of Atj at which hj ∈m is approximately differentiable. Since this is the case of L almost every a Atj according to Theorem 3, our first conclusion follows. ∈ In order to prove our second conclusion, consider a point x U of approximate differentiability of f. Reasoning as in the proof of Proposition 20∈ we write Af(x) = Q A and we infer that for each integer n, A(Q (P ) f)(a) = Q P A . Since ⊕i=1 i Q n ◦ ⊕i=1 n ◦ i the linear part of Pn Ai is Pn Li, where Li is the linear part of Ai, we see that J K ◦ ◦ J K Q Q m |D(Q (P ) f)(x)|2 = P L 2 = P (L (e )) 2 . Q n ◦ ||| n ◦ i||| k n i j k Xi=1 Xi=1 jX=1 Thus

Q m 2 2 lim |D(QQ(Pn) f)(x)| = lim Pn(Li(ej)) n ◦ n k k Xi=1 jX=1 Q m = L (e )) 2 k i j k Xi=1 jX=1 = |Df(x)|2 .

Therefore |Df(x)| = |δf|(x), according to (17), and the conclusion follows from (18). 70 Sobolev classes

3.7 Trace

Proposition 25. Let U = U(0, 1) be the unit ball in Rm. For every ε > 0 there exists θ > 0 such that

u pdH m−1 θ u pdL m + ε u pdL m Z∂U | | ≤ ZU | | ZU k∇ k whenever u : Clos U R is Lipschitz. →

Proof. Givenε ˆ > 0 we choose a smooth function ϕ : [0, 1] [0, 1] such that ϕ(0) = ϕ(1) = 1 and → p 1 q εˆ = ϕq Z0 where q is the exponent conjugate to p, and we put

p 1 q θˆ = ϕ′ q . Z0 | | For every x ∂U and y U we observe that ∈ ∈ 1 d u(x) u(y) = ϕ(t)u(y + t(x y)) dL 1(t) | − | 0 dt − ! Z

1 = ϕ′(t)u(y + t(x y)) + ϕ(t) u (y + t(x y)), x y dL 1(t) 0 − h∇ − − i! Z 1 1 1 q 1 p ϕ′ q u(y + t(x y)) pdL 1(t) ≤ 0 | | 0 | − | Z Z 1 1 1 q 1 p + ϕq u(y + t(x y)) p x y pdL 1(t) , Z0 Z0 k∇ − k k − k Therefore,

1 u(x) u(y) p 2p−1θˆ u(y + t(x y)) pdL 1(t) | − | ≤ Z0 | − | 1 + 2p−1 x y pεˆ u(y + t(x y)) pdL 1(t) . (23) k − k Z0 k∇ − k In order to integrate with respect to x ∂U, we ﬁrst note that the jacobian of the map [0, 1] ∂U U :(t, x) y + t(x y)∈ at (t, x) equals x y tm−1. Since x y 2, the area× formula→ therefore7→ implies− that k − k k − k ≤

Since the similar inequality holds for the gradient term, we infer from (23) that u(z) p u(x) u(y) pdH m−1(x) 2p+m−3θˆ | | dL m(z) ∂U | − | ≤ U y z m−1 Z Z k − k u(z) p + 22p+m−3εˆ k∇ k dL m(z) . U y z m−1 Z k − k Thus,

u(x) pdH m−1(x) 2p−1mα(m) u(y) p Z∂U | | ≤ | | u(z) p u(z) p + 22p+m−4θˆ | | dL m(z) + 23p+m−4εˆ k∇ k dL m(z) , U y z m−1 U y z m−1 Z k − k Z k − k according to the triangle inequality. We now integrate with respect to y U and, referring to the potential estimate [MZ97, Lemma 1.31], we obtain ∈

u pdH m−1 2p−1m u(y) pdL m(y) Z∂U | | ≤ ZU | | u(z) p + 22p+m−4α(m)−1θˆ dL m(y) | | dL m(z) U U y z m−1 Z Z k − k u(z) p + 23p+m−4α(m)−1εˆ dL m(y) k∇ k dL m(z) U U y z m−1 Z Z k − k m(2p−1 + 22p+m−4θˆ) u pdL m ≤ ZU | | + m23p+m−4εˆ u pdL m . ZU k∇ k

Remark 3. It follows in particular from Proposition 25 that

u pdH m−1 C u pdL m + u pdL m Z∂U | | ≤ ZU | | ZU k∇ k for some C > 0. Thus there exists a unique continuous trace operator T : W1(U) L (∂U; H m−1) p → p deﬁned by T (u) = u whenever u is Lipschitz. Of course, being continuous, T is also weakly continuous. The inequality in Proposition 25 shows that T is completely con- 1 tinuous, i.e. if uk converges weakly in Wp(U) then T (uk) converges strongly in m−1 { } { N } Lp(∂U; H ). Using Proposition 25 (more precisely, an R valued version) in conjunction with the embedding Theorem 4 we obtain that for every ε > 0 there exists θn > 0 such that

p m−1 p m (f1, f2) dH θn (f1, f2) dL Z∂U G ≤ ZU G p m p m + ε |Df1| dL + |Df2| dL ZU n whenever f1, f2 : U QQ(ℓ2 ) are Lipschitz. The dependence of θ upon n is caused by a constant α(n, Q)−1 →(the biLipschitz constant of the Almgren embedding). This leads to 1 n a proper deﬁnition of a trace “operator” for maps f Wp(U; QQ(ℓ2 )) but not for maps 1 ∈ f Wp(U; QQ(ℓ2)). We use a diﬀerent approach in our next result, avoiding altogether the∈ embedding of Theorem 4. 72 Sobolev classes

Theorem 13. There exists a map

T : W 1(U; Q (ℓ )) L (∂U; Q (ℓ )) p Q 2 → p Q 2 verifying the following properties.

(A) If f : Clos U Q (ℓ ) is Lipschitz then T (f)(x) = f(x) for every x ∂U; → Q 2 ∈ (B) For every ε > 0 there exists θ > 0 such that

p m−1 p m (T (f1), T (f2)) dH θ (f1, f2) dL Z∂U G ≤ ZU G p m p m + ε |Df1| dL + |Df2| dL ZU whenever f , f W 1(U; Q (ℓ ); 1 2 ∈ p Q 2 (C) There exists C > 0 such that for every f W 1(U; Q (ℓ )), ∈ p Q 2

|T (f)|pdH m−1 C |f|pdL m + |Df|pdL m . Z∂U ≤ ZU ZU

1 Proof. Owing to definition of Wp (U; QQ(ℓ2)) (the weak density of Lipschitz maps), and to Propositions 16, 18 and 4, it suffices to show that the map T defined for Lipschitz f by (A), verifies conclusions (B) and (C) for Lipschitz f1, f2, f.

Given f1, f2 : Clos U QQ(ℓ2) we deﬁne u : Clos U R by the formula u(x) = → m → (f1(x), f2(x)), x U. Given x U and h R such that x + h U we infer from the Gtriangle inequality∈ that ∈ ∈ ∈

u(x + h) u(x) (f (x + h), f (x + h)) (f (x), f (x + h)) | − | ≤ |G 1 2 − G 1 2 | + (f (x), f (x + h)) (f (x), f (x)) |G 1 2 − G 1 2 | (f (x + h), f (x)) + (f (x + h), f (x)) . ≤ G 1 1 G 2 2 This shows at once that u is Lipschitz. Furthermore Proposition 10 implies that

u(x) |Df (x)| + |Df (x)| k∇ k ≤ 1 2 at each x U where u, f and f are diﬀerentiable. Conclusion (B) now follows from ∈ 1 2 Proposition 25, and conclusion (C) is a consequence of (B) with f1 = f, f2 = Q 0 and ε = 1. J K

3.8 Analog of the Rellich compactness Theorem

Lemma 4. Let f W 1(Rm; Q (ℓ )) and h Rm. It follows that ∈ p Q 2 ∈

(f(x + h), f(x))pdL m(x) h p |Df|pdL m . m m ZR G ≤ k k ZR 3.8 - Analog of the Rellich compactness Theorem 73

Proof. According to Propositions 16 and 18, it suﬃces to prove it when f is Lipschitz as well. In that case it follows from Theorem 3(D) and Jensen’s inequality that

1 p (f(x + h), f(x))p h |Df(x + th)|dL 1(t) G ≤ k k Z0 1 h p |Df(x + th)|pdL 1(t) . ≤ k k Z0 The conclusion follows upon integrating with respect to x Rm. ∈ m Theorem 14. Let U = U(0, 1) be the unit ball in R and let fj be a sequence in 1 { } Wp (U; QQ(ℓ2)) such that

(1) There exists a compact set C ℓ2 such that fj(x) QQ(C) for every x U and every j = 1, 2,...; ⊂ ∈ ∈

(2) sup |Df |pdL m < . j U j ∞ R It follows that there exists a subsequence f and f W 1(U; Q (ℓ )) such that { k(j)} ∈ p Q 2 limj dp(f, fk(j)) = 0.

Proof. We show that the compactness Theorem 7 applies to the sequence E(fj) in m { } Lp(R ; QQ(C)). Our hypothesis (1) and Theorem 10(D) guarantee that the extension E(fj) take their value in QQ(C). We now check that the hypotheses of Theorem 7 are veriﬁed:

(i) follows from the fact that C is bounded, thus

p m m p p |E(fj)| dL 2 α(m)Q 2 (diam C) m ZR ≤ for every j = 1, 2,...; (ii) follows from Lemma 4 and Theorem 10(C):

p m p p m sup (E(fj)(x + h),E(fj)(x)) dL (x) h |DE(fj)| dL m m j ZR G ≤ k k ZR p p m p m h C(m, p, Q) sup |fj| dL + |Dfj| dL ≤ k k j ZU ZU

(iii) follows from the fact that E(fj) = E(fj)K for each j = 1, 2,..., where K = B(0, 2), according to Theorem 10(D).

ˆ m ˆ Thus there exists f Lp(R ; QQ(ℓ2)) such that limj dp(E(fk(j)), f) = 0. It remains to ∈ ˆ 1 notice that the restriction f ↾ U belongs to Wp (U; QQ(ℓ2)). This is because for each m j = 1, 2,... we can choose a Lipschitz map gj : R QQ(ℓ2) such that dp(E(fk(j), gj) < −1 p m −1 p →m ˆ j and Rm |Dgj| dL j + Rm |E(fk(j))| dL . Thus limj dp(f ↾ U, gj ↾ U) = 0 p ≤ and sup |D(g ↾ U)| dL m < . jR U j ∞R RemarkR 4. It would be interesting to know whether or not all the results proved so far in this paper hold when the range ℓ2 is replaced by an inﬁnite dimensional Banach space Y which is separable, a dual space, and admits a monotone Schauder basis. 74 Sobolev classes

3.9 Existence Theorem

Lemma 5. Assume that

(A) X is a compact metric space;

(B) Y is a metric space;

(C) g : X Q (Y ) is continuous. → Q It follows there exists a compact set C Y such that g(x) Q (C) for every x Y . ⊂ ∈ Q ∈ Proof. We let C = Y y : y supp g(x) for some x X . One easily checks that C is closed, thus it suﬃces∩ { to show∈ it is totally bounded.∈ Since} im g itself is compact, given ε > 0 there are x1, . . . , xκ X such that for each x X there exists k = 1,...,κ ∈ Q k ∈ with (f(x), f(xk)) < ε. We write f(xk) = i=1 yi . It it now obvious that C κ GQ k ⊕ ⊂ k=1 i=1 BY (yi , ε). ∪ ∪ J K m Theorem 15. Let U = U(0, 1) be the unit ball in R and let g : ∂U QQ(ℓ2) be Lipschitz. It follows that the minimization problem →

p m minimize U |Df| dL among f R W 1(U; Q (ℓ )) such that T (f) = g  ∈ p Q 2 admits a solution.

Proof. The class of competitors is not empty according to the extension Theorem 2. We let C ℓ be a compact set associated with g in Lemma 5 and we let C be the 0 ⊂ 2 convex hull of C0 0 (so that C is compact as well). We denote by P : ℓ2 C the nearest point projection.∪ { } Given a minimizing sequence f we consider the→ sequence { j} QQ(P ) fj which, we claim, is minimizing as well. That these be Sobolev maps, and {form a minimizing◦ } sequence, follows from the inequalities

′ p m ′ p m (QQ(P ) f, QQ(P ) f ) dL (f, f ) dL ZU G ◦ ◦ ≤ ZU G (recall the paragraph preceding Proposition 20) and

p m p m |D(QQ(P ) f)| dL |Df| dL ZU ◦ ≤ ZU ′ (because Lip P 1) valid for every Lipschitz f, f : U QQ(ℓ2), and hence for every ′ 1 ≤ → f, f Wp (U; QQ(ℓ2)) as well. It also follows from these inequalities and Theorem 10(B) and (C)∈ that T (Q (P ) f) = T (f) Q ◦ whenever f W 1(U; Q (ℓ )). Thus T (Q (P ) f ) = g, j = 1, 2,.... Since all ∈ p Q 2 Q ◦ j these QQ(P ) fj take their values in QQ(C), it follows from Theorem 14 that there are integers k◦(1) < k(2) < . . . and f W 1(U; Q (ℓ )) such that lim d (f, f ) = 0. ∈ p Q 2 j p k(j) Theorem 13(B) implies that T (f) = g. Proposition 18 and Corollary 4 guarantee the required lower semicontinuity. 4.1 - Squeeze and squash variations 75 4 Blowup analysis of stationary multiple valued functions

In this last section, we will study the regularity of stationary multiple valued functions. Subsection 4.1 contains the basic definitions of stationarity. We will focus on one codimensional multiple valued functions, i.e with values in QQ(R). We recall that because of the order structure of R, QQ(R) is considerably simpler to describe than any cone QQ(Y ), see subsection 2.2. Indeed, the embeddings ξ and ξ0 conincide, and ξ Q x = (x , . . . , x ) whenever x x . Therefore, we can define Q (R) i=1 i 1 Q 1 ≤ · · · ≤ Q Q to be the set P J K Q (R) := RQ (x , . . . , x ) RQ : x x . Q ∩ { 1 Q ∈ 1 ≤ · · · ≤ Q} It follows from subsections 1.5, 2.2 and 2.3 that if f : U QQ(R) is a Lipschitz continuous multiple valued map defined on an open set in Rm→(or more generally lying in a Sobolev class), f = (f1, . . . , fQ) then Df = Q Df ⊕i=1 i almost everywhere on U. Therefore, the differentialJ K Df can be seen as the usual (Df , . . . , Df ) Hom(Rn, RQ). 1 Q ∈ All spaces will be given their canonical Euclidean structure. Since no confusion can arise, we will denote by such a Euclidean norm, except for the normed spaces (R, ) and (C, ), for examplek·k |·| | · | 1/2 Q n ∂f 2 Df := i . k k  ∂x  i=1 j=1 j X X  

4.1 Squeeze and squash variations

When dealing with stationarity, one must precise with respect to which class of small pertubations. We expect not only stationary multiple valued functions to derive from Lipschitz approximations of stationary varifolds, but also to be stable under vertical and horizontal dilatations. This stability is very important when blowing up. There- fore, there are only two possible types of deformation, deformation of the domain, or of the codomain, respectively called squeeze and squash transformations. This terminology is due to F.J. Almgren. The deﬁnitions given below are very similar to those of stationarity between Riemannian manifolds. In this framework, stationary maps with respect to domain transformations are usually called stationary-harmonic, whereas those stationary with respect to codomain transformations are called weakly harmonic. A multiple valued function f W 1,2(Ω, R) is called squash stationary whenever for all ψ = ψ(x, u) C∞(Ω R, R),∈ one has ∈ c × Q d 2 DΨt = 0, where Ψt(x) := fi(x) + tψ(x, fi(x)) (24) dt |t=0 Ω k k Z Xi=1 J K and f(x) := (f1(x), . . . , fQ(x)). The following is well-known, see for example [?, Proposition 6] 76 Blowup analysis of stationary multiple valued functions

Proposition 26. f is squash stationary if and only if for all ψ C∞(Ω R, R), ∈ c × Q Dfi,Dxψ(x, fi(x)) dx + Dfi,Duψ(x, fi(x)) Dfi(x) dx = 0. (25) Ωh i Ωh ◦ i Xi=1 Z Z Proof. The following holds for all u v and x Ω: ≤ ∈ ∂ψ ∂ψ −1 tψ(x, v) tψ(x, u) t(v u) v u whenever t − ≤ − ∂u ≤ − ≤ ∂u ∞ ∞

Thus, for small values of t,

f (x) + tψ(x, f (x)) f (x) + tψ(x, f (x)). 1 1 ≤ · · · ≤ Q Q Consequently,

Q DΨ 2(x) = Df + t(D ψ(x, f (x)) + D ψ(x, f (x)) Df (x)) 2. k tk k i x i u i ◦ i k Xi=1 Integrating over Ω and then derivating at t = 0 yields (25).

∞ m Let ϕ Cc (Ω, R ). The map Φt(x) := x + tϕ(x) deﬁnes a diﬀeomorphism of Ω ∈ 1,2 onto itself for small values of t. A multiple valued map f W (Ω, QQ(R)) is called squeeze stationary whenever ∈

d 2 D(f Φt) = 0. (26) dt |t=0 ZΩ k ◦ k ∞ m for all choices of ϕ Cc (Ω, R ). One can similarly derive the Euler-Lagrange equation for squeeze stationary∈ maps (see also [?, Proposition 6]): Proposition 27. f is squeeze stationary if and only if for all ϕ C∞(Ω, Rm) ∈ c 2 Df, Df Dϕ Df 2div ϕ = 0. (27) ZΩh ◦ i − ZΩ k k (where Df Dϕ stands for (Df Dϕ, . . . , Df Dϕ)). ◦ 1 ◦ Q ◦ Proof. As for any x Ω and t small enough, ∈ f (x + tϕ(x)) f (x + tϕ(x)), 1 ≤ · · · ≤ Q the squared gradient of f Φ equals ◦ t Q D(f Φ )(x) 2 = Df (Φ (x)) + tDf (Φ (x)) Dϕ(x) 2 k ◦ t k k i t i t ◦ k Xi=1 Q = Df (Φ (x)) 2 + 2t Df (Φ (x)), Df (Φ (x)) Dϕ(x) k i t k h i t i t ◦ i Xi=1 Q + t2 Df (Φ (x)) Dϕ(x) 2 k i t ◦ k Xi=1 = Df(Φ (x)) 2 + 2t Df(Φ (x)), Df(Φ (x)) Dϕ(x) + O(t2) k t k h t t ◦ i 4.1 - Squeeze and squash variations 77

We make the substitution y := Φt(x) in the following integral

2 2 −1 −1 D(f Φt) = Df(y) + 2t Df(y), Df(y) Dϕ(Φt (y)) Jac Φt (y)dy Ω k ◦ k Ω k k h ◦ Z Z + O(t2)

−1 We estimate the Jacobian Jac Φt (y):

−1 1 2 2 Jac Φt (y) = = 1 ttr Dϕ + O(t ) = 1 tdiv ϕ + O(t ). det(id + tDϕ) − −

When integrating and derivating at 0, we just keep the ﬁrst-order term in the preceding equations. This proves (27).

When Q = 1, Equation (25) simply asserts that f is a harmonic function. However, a squeeze stationary function is not necessarily harmonic, as shown by the function x x . 7→ | | A multiple valued function is called stationary when it is stationary with respect to both squeeze and squash variations. A basic example of stationary multiple valued function is the sum of harmonic functions f = 1≤i≤Q hi . The next proposition proves that those notions of stationarity are local. ThisP follows from the linearity of Equation (27) (resp. (25)) with respect to ϕ (resp. ψ). J K

Proposition 28. Let (Ωι)ι∈I on open cover of Ω. If f is squeeze ( resp. squash) stationary on each Ωι, then it also is on Ω.

Proof. Let us choose a locally ﬁnite open cover (Wι)ι∈I subordinate to (Ωι)ι∈I and a ∞ ∞ m partition of unity ϕι Cc (Ω, R), with supp ϕι Wι. If ϕ Cc (U, R ), ϕ can be written as a ﬁnite sum∈ ⊆ ∈ ϕ = ϕιϕ. Xι∈I Equation (27) is true with ϕ replaced by ϕιϕ. Summing over ι, we prove that f is squeeze stationary. We proceed in a similar fashion for squash stationarity.

We end this section by two useful results, the combination of which implies: if f is stationary, then so is f ( η f). When considering problems on stationary multiple valued functions, one can⊕ − often◦ replace f with its translation f ( η f), that is, require that f should be centered (η f = 0). This enables simpliﬁcations,⊕ − ◦ in particular when Q = 2. A two valued centered◦ multiple valued function f is just the data of a nonnegative function f + (the nonnegative part of f) where f = f + + f + . − 1,2 Proposition 29. Let f W (Ω, QQ(R)) be squash stationary.J It followsK J thatK η f is harmonic. ∈ ◦

∞ + Proof. Let φ Cc (Ω) be arbitrary. Let h : R R be a smooth function such that h = 1 on [0, 1]∈ and h = 0 on [2, ). Let a > 0. We→ now plug ψ(x, u) := φ(x)h(u/a) in Equation (25). One has ∞

Q 1 ′ 2 h(fi/a) Dfi, Dφ + φh (fi/a) Dfi = 0. Ω h i a Ω k k Xi=1 Z Z 78 Blowup analysis of stationary multiple valued functions

By Lebesgue dominated convergence, we control the ﬁrst term:

Q lim h(fi/a) Dfi, Dφ = Q D(η f), Dφ , a→∞ Ω h i Ωh ◦ i Z Xi=1 Z whereas the second term has limit zero. Indeed,

1 Q φ h′ φh′(f /a) Df 2 k k∞k k∞ Df 2 0. i i a ZΩ i=1 k k ≤ a ZΩ k k → X

Therefore D( η f), Dφ = 0. Since φ is arbitrary, η f is harmonic. Ωh ◦ i ◦ R 1,2 Proposition 30. Let f W (Ω, QQ(R)) be stationary, and h :Ω R be a harmonic function. It follows that ∈f h is stationary. → ⊕ Proof. Let us ﬁrst check that f h is squeeze stationary. The ﬁrst term in the left-hand side of Equation (27) applied to⊕f h is ⊕ Q 2 D(f h),D(f h) Dϕ = 2 D(fi + h),D(fi + h) Dϕ Ωh ⊕ ⊕ ◦ i Ωh ◦ i Z Xi=1 Z Q = 2 Dfi, Dfi Dϕ + Dh, Dfi Dϕ Ωh ◦ i Ωh ◦ i Xi=1 Z Z

+ Dfi,Dh Dϕ + Dh,Dh Dϕ (28) ZΩh ◦ i ZΩh ◦ i = 2 Df, Df Dϕ + 2Q Dh, D(η f) Dϕ ZΩh ◦ i ZΩh ◦ ◦ i + 2Q D(η f),Dh Dϕ + 2Q Dh,Dh Dϕ ZΩh ◦ ◦ i ZΩh ◦ i The second term of Equation (27) is equal to

D(f h) 2div ϕ = Df 2div ϕ + Q Dh 2div ϕ (29) ZΩ k ⊕ k ZΩ k k ZΩ k k + 2Q D(η f),Dh div ϕ. ZΩh ◦ i f is squash stationary, thus by Proposition 29, η f is harmonic. So is η f + h. Since harmonicity implies squeeze stationarity, ◦ ◦

2 D(η f + h),D(η f + h) Dϕ D(η f + h) 2div ϕ = 0 (30) ZU h ◦ ◦ ◦ i − ZU | ◦ | 2 D(η f),D(η f) Dϕ D(η f) 2div ϕ = 0 (31) ZU h ◦ ◦ ◦ i − ZU | ◦ | After calculations, one sees that (28) (29) = (27) + Q(30) Q(31) = 0, i.e f h is squeeze stationary. − − ⊕ Now we prove that f h is squash stationary. Let ψ(x, u) C∞(Ω R), we set ⊕ ∈ c × Λ(x) := i ψ(x, fi(x)+h(x)) and Ψt(x) := i fi(x)+h(x)+tψ(x, fi(x)+h(x)) where f(x) := i fi(x) . According to [AST00, 2.3.(4)], one has P J K P J K P J Dir(ΨK t, Ω) = QDir(h, Ω) + Dir(Ψt ( h), Ω) ⊕ − + 2Q Dh, D(η f) + 2Qt Dh, D(η Λ) . ZΩh ◦ i ZΩh ◦ i 4.2 - Regularity results 79

The derivative of the second term in the right-hand side of the above equation is zero. This is easily seen when testing the squash stationarity of f with ψ(x, u) := ψ(x, u + h(x)). Thus d Dir(Ψt, Ω) = 2Q Dh, D(η Λ) . dt |t=0 ZΩh ◦ i Since η Λ W 1,2(Ω), the squash variation of f h is zero by harmonicity of h. ◦ ∈ 0 ⊕

4.2 Regularity results

When m = 2, we have the following criterion for squeeze stationarity:

1,2 2 Proposition 31. Suppose f W (Ω, QQ(R)), where Ω R . f is squeeze stationary if and only if the Hopf form ∈ ⊂

Q 2 2 1 ∂fi ∂fi ∂fi ∂fi Hf (z)dx dy := (z) (z) 2i (z) (z) dx dy ∧ Q  ∂x ! − ∂y ! − ∂x ∂y  ∧ Xi=1   is holomorphic on Ω.

Proof. Let ϕ C∞(Ω). We apply Equation (27) to (ϕ, 0) and (0, ϕ): ∈ c Q ∂f 2 ∂f 2 ∂ϕ Q ∂f ∂f ∂ϕ i i + 2 i i = 0, Ω  ∂x ! − ∂y !  ∂x Ω ∂x ∂y ∂y Xi=1 Z Xi=1 Z   Q ∂f 2 ∂f 2 ∂ϕ Q ∂f ∂f ∂ϕ i i + 2 i i = 0. Ω  ∂y ! − ∂x !  ∂y Ω ∂x ∂y ∂x Xi=1 Z Xi=1 Z   This amounts to saying that QHf is weakly holomorphic, hence holomorphic by Weyl’s lemma.

This important property is used in [Sch84] to prove that a squeeze stationary map (actually with values in an arbitrary Riemannian manifold) is continuous. We refer to [Lin01, Chapter 3] and [Lin12] for a proof more in the spirit of multiple valued functions, which gives some information on the modulus of continuity. More regularity is unknown. In the special case Q = 2, a centered squeeze stationary multiple valued function (η f = 0) can be written f := f + + f + , where f + W 1,2(Ω) is nonnegative.◦ We compute its Hopf form: − ∈ J K J K ∂f + 2 ∂f + 2 ∂f + ∂f + ∂f + ∂f + 2 Hf = Hf + = 2i = i . ∂x ! − ∂y ! − ∂x ∂y ∂x − ∂y !

+ 2 A straightforward computation gives Hf = Df . Since Hf is continuous, we deduce that f +, hence f is locally Lipschitz.| | k k | |

Corollary 5. A centered squeeze stationary function f :Ω Q2(R) is locally Lipschitz on Ω. → 80 Blowup analysis of stationary multiple valued functions

The branch set of a stationary map f :Ω Q (R) (Ω R2) is the set → 2 ⊂ Branch(f) := z Ω: a R, f(z) = 2 a . { ∈ ∃ ∈ } Since f and f ( η f) have the same branch set, we shallJ K usually suppose f is centered, in which⊕ − case◦ Branch(f) = z : f +(z) = 0 . This deﬁnition makes sense because of corollary 5 { } Note that a stationary may not be Lipschitz continuous as soon as Q = 3. For example, let f : B2 Q (R) be deﬁned by → 3 2θ 2θ 2π 2θ 4π f(z) := r2/3 sin + r2/3 sin + + r2/3 sin + . s 3 !{ s 3 3 !{ s 3 3 !{ f is stationary and 2/3-Hölderian. Indeed, a straightforward computation gives 2 −2/3 1,2 2 Df = 4r /3, therefore f W (B , Q3(R)). Hf = 0 is holomorphic, hence f is squeezek k stationary by Proposition∈ 31. Moreover, f is locally the sum of three harmonic sheets on B2 0 , thus by Proposition 28, f is squash stationary on B2 0 . We continue this section\{ } with a simple and useful removability criterion for squash\{ } stationary functions, which will prove that f is squash stationary on all B2. This proposition will also be needed to construct an example of a non continuous squash stationary multiple valued function.

2 1,2 Proposition 32. If z0 Ω R and f W (Ω, QQ(R)) is squash stationary on Ω z , then it is on all∈Ω. ⊂ ∈ \{ 0} Proof. Let φ : R2 R be a smooth function such that φ 1 on B2(0, 1/2), 0 φ 1 2 → 2 2 ≡ ≤ ≤ on R , and φ 0 on R B (0, 1). We set for all r > 0, φr(z) := φ(z/r). Let ∞ ≡ \ 1 2 2 ψ Cc (U R, R). We decompose ψ into ψ = ψ +ψ with ψ (z, u) = φr(z z0)ψ(z, u), ∈ × 1 − where supp ψ U z0 R. By squash stationarity, Equation (25) is satisﬁed when plugging in ψ1.⊆ On\{ the other} × hand,

2 Dfi,Dzψ (z, fi(z)) dz = φr(z z0) Dfi,Dzψ(z, fi(z)) dz ZΩh i ZΩ − h i + ψ(z, fi(z)) Dfi, Dφr(z z0) dz ZΩ h − i ψ ∞ Dφ ∞ Dzψ ∞ + k k k k Dfi 2 ≤ k k r ! ZB (z0,r) k k √π (r D ψ + ψ Dφ ) Dir(f ,B2(z ,r))1/2 ≤ k z k∞ k k∞k k∞ i 0 and this tends to 0 as r 0. One easily proves that → 2 lim Dfi(z),Duψ (z, fi(z)) Dfi(z) dz = 0 r→0 ZΩh ◦ i as well, which proves that f is squash stationary.

Squash stationary functions are not necessarily continuous, although they satisfy a Caccioppoli inequality (see next Proposition 33). A usual regularity proof combining Poincaré and Caccioppoli inequalities does not work: due to the lack of algebraic operations on QQ(R), it is not possible to translate a multiple valued function and preserve its squash stationarity at branch points. 4.2 - Regularity results 81

m Proposition 33. Let f :Ω R QQ(R) a squash stationary multiple valued function and φ C∞(Ω), ⊂ → ∈ c φ2 Df 2 4 Dφ 2 (f, Q 0 )2 (32) ZΩ k k ≤ ZΩ k k G J K Proof. Let a > 0 and χ : R R be a compactly supported smooth function such that → ′ 2 χ(u) = u on [ a, a], χ(u) u on R and χ ∞ 1+1/a. We set ψ(z, u) := φ (z)χ(u). Equation (25)− yields:| | ≤ | | k k ≤

Q Q 2 2 ′ 2 φχ(fi) Dfi, Dφ + Dfi φ χ (fi) = 0. Ω h i Ω k k Xi=1 Z Xi=1 Z Letting a , we end up with → ∞ Q 2 2 φ Df = 2 φfi Dfi, Dφ (33) Ω k k − Ω h i Z Xi=1 Z Q 2 φfi Dfi Dφ ≤ Ω | |k kk k Xi=1 Z 2 φ (f, Q 0 ) Df Dφ ≤ ZΩ | |G k kk k 1/2 1/2 2 Dφ 2 J(f,K Q 0 )2 φ2 Df 2 , ≤ ZΩ k k G ZΩ k k which ﬁnally proves (32). J K

2 2 Now we construct a squash stationary Q2(R) valued function on B := B (0, 1) which is non continuous at the origin. It is not surprising that the structure of range variations is not enough to ensure continuity: T. Rivière [Riv95] already proved the existence of everywhere non continuous weakly harmonic maps, taking values in the 2 n 2 sphere S . Note that QQ(R ) in general is, as S , non negatively curved in the sense of Alexandrov, see [Zhu07].

Consider the sequence of segments (Kn)n≥n0 deﬁned by

3 3 K := [2−n 2−n , 2−n + 2−n ] 0 . n − × { }

We will choose n0 big enough, for the moment suppose n0 2. Thus the Kn are mutually disjoint and lie in B2. We will consider the open set ≥

G := B2 0 K . \ { } ∪ n n[≥n0   Consider the nonnegative map φ : B2 R deﬁned by n → 1 + 2−n φn(z) := Cn ln , z 2−n ! | − | (thus φ (2−n) = ) where C is chosen such that φ 1 on K , that is: n ∞ n n ≥ n −n −1 1 + 2 Cn := ln 3 . 2−n ! 82 Blowup analysis of stationary multiple valued functions

The functions φn/Cn are harmonic and locally uniformly bounded in G. Since Cn n 1/(n3 ln 2), the series ∼ ∞ φn n=n X0 converges locally uniformly on G to a harmonic function. Deﬁne

∞ v(z) := 1 φn(z). − n=n X0 We have that v 1 and ≤

z Kn, lim sup v(ξ) 0. ∀ ∈ ξ→z ≤ n[≥n0 −(p+1) −p We set zp := (2 + 2 )/2 for any p n0. We readily check that zp G and n n , z 2−n 2−n/4. And ≥ ∈ ∀ ≥ 0 | p − | ≥ −n 1 + 2 n 1 v(zp) = Cn ln −n Cn ln 4(1 + 2 ). − zp 2 ! ≤ nX≥n0 | − | nX≥n0 n −2 The last series converges since Cn ln 4(1+2 ) n n . If n0 is big enough, 1 v(zp) 1/2, thus v(z ) 1/2. Thus lim sup v(z) > 0. ∼ − ≤ p ≥ z→0 2 2 2 Now, we denote by B+ the upper half disk B+ := B z : z > 0 ( z stands for the imaginary part of z). Let u W 1,2(B2 ) solving the∩ { problemℑ u =} 1ℑ on ∂B2 ∈ + + \ ([ 1, 1] 0 ), u|Kn = 0 for each n n0, and u minimizes its Dirichlet energy. We can show− that× {u }is continuous up to all boundary≥ points but the origin (see [Ran95, Section 4.2]). That is lim u(ξ) = 0, z Kn ξ→z ∀ ∈ n[≥n0 and lim u(ξ) = 1 whenever z ∂B2 ([ 1, 1] 0 ). ξ→z ∈ + \ − × { } The functionu ˜ W 1,2(B2) defined by reflection ∈ u(z) x 0 u˜(z) = u(z) ℑx ≥ 0 ( ℑ ≤ is harmonic on G. Otherwise, one finds a competitor w ofu ˜:

2 2 2 2 Dir(w,B+) + Dir(w,B−) < Dir(˜u, B ) = 2Dir(u, B+).

2 2 Either w 2 or z B w(z) contradicts the minimality of Dir(u, B ). |B+ ∈ + 7→ +

Sinceu ˜ is nonnegative, lim supξ→z,ξ∈G(v u˜)(ξ) 0 for all z ∂G 0 . We conclude that v u˜. To justify this, let us mention this− general≤ version of∈ the maximum\{ } principle. To this≤ end, recall that a set E Rm is polar when there is a superharmonic function ϑ on Rm such that E θ = ⊆. In particular, countable subsets of Rm are polar. ⊆ { ∞} Theorem 16. Let h be a subharmonic function Ω R which is bounded above. If ∂Ω is non polar and → lim sup h(ξ) 0 ξ→z,ξ∈G ≤ for all z ∂Ω E, where E is a polar set. It follows that h 0. ∈ \ ≤ 4.2 - Regularity results 83

Therefore lim supξ→0 u˜(ξ) lim supξ→0 v(ξ) > 0.u ˜ is non continuous at the origin. We now deﬁne f W 1,2(B2, Q≥ (R)) ∈ 2 f := u˜ + u˜ . − We see that f decomposes into two harmonicJ K J sheetsK on G, thus f is stationary there. f is not continuous at 0, otherwise u would be. Let z be an inner point in Kn,(z 3 3 2−n 2−n , 2−n + 2−n ). Let r be small such as B2(z,r) does not contain the end6∈ points{ − of K . Letu ˆ : B2}(z,r) R deﬁned by n → u˜(ξ) ξ 0 uˆ(ξ) = u˜(ξ) ℑξ ≥ 0 ( − ℑ ≤

2 Thusu ˆ is continuous becauseu ˜ = 0 on Kn. Obviously,u ˆ is harmonic on B+(z,r) 2 2 ∪ B−(z,r) and the mean value ofu ˆ in an open ball contained in B (z,r) and centered at a point in Kn is zero. Thus, by the following Proposition 34,u ˆ is harmonic.

Proposition 34. Let Ω be an open set in Rm, and u :Ω R a continuous function, such that for each x Ω, there is r > 0 satisfying B(x, r )→ Ω and ∈ x x ⊆ 1 m u = u(x), r rx. L (B(x, r)) ZB(x,r) ∀ ≤ It follows that u is harmonic on Ω.

Proof. We can suppose Ω to be an open ball, and f to be uniformly continuous. Let h be the harmonic function solving the Dirichlet problem h = f . Then h f has |∂Ω |∂Ω − zero boundary values. Let A := x Ω:(h f)(x) = supΩ(h f) . A is obviously closed in Ω. A is also open. Indeed,{ if∈ x A then− − } ∈

m (h f)(y)dy = L (B(x, rx))(h f)(x). ZB(x,rx) − −

This implies that h f is constant in B(x, rx). Since Ω is connected, A = Ω or A = . In both cases, h f−attains its maximum in ∂Ω, which means that h f. Similarly,∅ one has h f. − ≤ ≥

We come back to our example. Since

f 2 = uˆ + uˆ |B (z,r) − f is locally the sum of two harmonic sheetsJ aroundK J zK. Thus f is squash stationary on 2 B ( 0 n≥n0 ∂Kn). Now we can remove all these singularities. To be rigorous, we ﬁrst\ assert{ } ∪f ∪is squash stationary on B2 0 . Indeed, we test squeeze stationary with a test function ψ with compact support\{ in (}B2 0 ) R and apply the removability proposition 32 ﬁnitely many times. Again, removing\{ } × the origin, f must be squash stationary on B2. 84 Blowup analysis of stationary multiple valued functions

4.3 Blowing up

1,2 2 From now on, we set m = Q = 2. Let f W (Ω, Q2(R)). If B (z,r) ⋐ Ω, we deﬁne (unless the denominator is zero) ∈

2 r 2 Df B (z,r) k k N(z,r) := 2 . 2R (f, 2 0 ) ∂B (z,r) G R Theorem 17. Let f :Ω Q2(R) a nonzero stationaryJ K multiple valued function on a domain Ω. →

(1) For all z Ω, the function r N(z,r), r (0, dist(z, ∂Ω)) is well-deﬁned and nondecreasing.∈ 7→ ∈

(2) If N(z,r) ν for small values of r, then f is ν homogeneous on a neighborhood ≡ 1,2 2 ν of z, i.e there is g W (∂B , Q2(R)) such that f(ξ) = ξ z g((ξ z)/ ξ z ) on a neighborhood∈ of x. | − | − | − |

Proof. We assume, without loss of generality, that z = 0. Let us introduce the following notations D(r) := Df 2,H(r) := (f, 2 0 )2 2 2 ZB (0,r) k k Z∂B (z,r) G for any r < dist(0, ∂Ω). We simply write N(r) instead of N(0,rJ ).K We recall from elementary analysis that r D(r) is an absolutely continuous map and that 7→ D′(r) = Df 2dH 1 2 Z∂B (0,r) k k for L 1 almost every r (0, dist(0, ∂Ω)). We also observe that H is a well deﬁned function because of the continuity∈ of f, see Corollary 5. Note that if for some r, H(r) = 0, i.e f vanishes on ∂B2(0,r), then f 0 on all Ω. Indeed, let ε > 0 be such that B2(0,r + ε) ⋐ Ω and φ :Ω R be a smooth≡ function 2 2 → such that φ 1 on B (0,r), φ 0 on Ω B (0,r + ε) and Dφ ∞ 2/ε. According to (32), one has≡ ≡ \ k k ≤ 16 Df 2 (f, 2 0 )2 16πLip(f)2 (r + ε)2 r2 . B2(0,r) k k ≤ ε2 B2(0,r+ε)\B2(0,r) G ≤ − Z Z J K Letting ε 0, one sees that Dir(f,B2(0,r)) = 0. Hence f is constant on B2(0,r) and H 0 on→B2(0,r). Since H is holomorphic, H = 0 everywhere, thus f 2 0 on Ω. f ≡ f f ≡ A change of variable in H yields J K

H(r) = r (f(rz), 2 0 )2dz. 2 Z∂B G One then deduces that: J K ∂ H′(r) = (f(rz), 2 0 )2dz + (f, 2 0 )2 dz. (34) ∂B2 G ∂B2 ∂ν G |rz Z Z Actually, this is immediate in case z J K (f(z), 2 0 )2 were smooth.J K In the general case, this follows from a density argument.7→ G J K 4.3 - Blowing up 85

Apply formula (33) with the following map φ :Ω R h → 1 if z r | | ≤ φ (z) = (r + h z )/h if r z r + h h  − | | ≤ | | ≤  0 otherwise  Letting h tend to 0, we prove that (note that the average value of φh on the annulus B2(0,r + h) B2(0,r) is 1/2): \ 2 z 1 D(r) = fi(z)Dfi(z) dH (z). (35) ∂B2(0,r) z ! Xi=1 Z | | Combining (34) and (35), one gets H(r) H′(r) = + 2D(r). (36) r One needs a last estimate using squeeze stationarity. We apply Proposition 27 with ϕ(z) = φh(z)z. As

2 ∂φh z ζ R , Dϕ(z)(ζ) = φh(z)ζ + (z) ζ, z, ∀ ∈ ∂ν * z + | | one deduces div ϕ(z) = 2φ (z) + z ∂ φ (z). Then (27) yields h | | ν h 2 2 ∂φh z 2 2 2 φh Df + 2 (z) Df(z) = 2 φh Df + z ∂νφh(z) Df(z) Ω k k Ω ∂ν z ! Ω k k Ω | | k k Z Z Z Z | | We let h 0 and ﬁnally end up with → z 2 D′(r) = Df 2 = 2 Df(z) dz. (37) ∂B2(0,r) k k ∂B2(0,r) z ! Z Z | |

Now, use (36) to infer N ′(r) 1 D′(r) H′(r) = + N(r) r D(r) − H(r) H(r)D′(r) 2D(r)2 = − D(r)H(r) Using (35), (37) and Cauchy-Schwarz inequality, one has

z 2 D(r)2 (f, 2 0 )2dH 1 Df(z) dH 1(z) ≤ ∂B2(0,r) G ∂B2(0,r) z ! Z Z ′ | | D (r) J K = H(r) 2 Thus the logarithmic derivative N ′/N is nonnegative and N is monotonous. In case N ν is constant for small values r r dist(0, ∂Ω), in particular N ′ = 0 ≡ ≤ 0 ≤ for r r , the equality case of Cauchy-Schwarz inequality tells us that for L 1 almost ≤ 0 r r , there is λ(r) R such that for i 1, 2 and H 1 almost all z ∂B2(0,r) ≤ 0 ∈ ∈ { } ∈ z fi(z) = λ(r)Dfi(z) . z ! | | 86 Blowup analysis of stationary multiple valued functions

Recalling (35), one computes

D(r) = λ(r) (f, 2 0 )2dH 1 = λ(r)H(r). 2 Z∂B (0,r) G Thus N(r) = rD(r)/H(r) = rλ(r) = ν for r J rK . To sum up, ≤ 0 ν z fi(z) = Dfi(z) . (38) z z ! | | | | An application of the area formula ﬁnally yields that (38) holds H 1 almost every on 1 2 [0, y] = ty : 0 t 1 for H almost every y ∂B (0,r0). For almost such y, the function{ ≤ ≤ } ∈ g : t (0, 1] f (ty) y ∈ 7→ i is in W 1,2[0, 1], therefore absolutely continuous, and is a solution of the ordinary diﬀer- ential equation ν g = g′ . y t y Hence t ν gy(t) = f(y). r0 This proves conclusion (2).

The limit (z) := limr→0 N(z,r) is called the frequency of f at z. We need the following proposition,N which roughly states that a limit of stationary maps is stationary. 2 Proposition 35. Let (fk) be a sequence of stationary functions fk : B Q2(R). Suppose →

(1) (fk(0))k is a bounded sequence. (2) sup Dir(f ,B2) < . k k ∞

Up to passing to a subsequence, (fk) converges locally uniformly to a stationary map f. Moreover, Df Df almost everywhere on B2. k → 2 2 Proof. We have that Dir(η fk,B ) Dir(fk,B )/2. The η fk are harmonic, therefore by the mean value property◦ for D(η ≤f ) and the boundedness◦ of the sequence (η f (0)) , ◦ k ◦ k k they are locally uniformly bounded. Up to passing to a subsequence, the η fk converge locally uniformly to a harmonic function h, and D(η f ) Dh. Since the f◦ ( η f ) ◦ k → k ⊕ − ◦ k satisfy hypotheses (1) and (2), we suppose suppose the fk to be centered, i.e fk =: + + + fk + fk where fk 0 is squeeze stationary. Let s (0, 1). By Cauchy’s integral formula,− ≥ ∈ iθ iθ J K J K 2π H + (re )e r fk + Hf (z) = i dθ k 2π 0 re θ z Z − for z B2(0, s) and r ((1 + s)/2, 1). Consequently, ∈ ∈ 2 1 2π + 2 1 2 iθ Dfk (z) Hf + (re ) rdrdθ k k ≤ 2π 1 s (1+s)/2 0 | k | − Z Z 2 + 2 2 sup Dir(fk ,B ) ≤ (1 s) π k − 1 2 = 2 sup Dir(fk,B ) (39) (1 s) π k − 4.3 - Blowing up 87

+ 2 Thus the fk are locally uniformly Lipschitz continuous on B . This and Hypothesis + (1) imply by Arzelà-Ascoli theorem the existence of a subsequence (still denoted fk ) converging locally uniformly to some f + 0. We set f := f + + f + . Now, if 2 + ≥ − + z B (0, s) and f (z) > 0, then there is a neighborhood of z on which all fk are ∈ + J K J K + positive. By squash stationarity, fk is harmonic in this neighborhood, and so is f as a + + uniform limit of harmonic functions and Dfk converges uniformly to Df in a smaller 2 2 neighborhood. We let Λs be the uniform Lipschitz constant on B (0, 2), that is, Λs is the right-hand side of (39). Let A be the set of points z B2(0, s) such that f + = 0 has Lebesgue density 1 at z. Let ε (0, 1/9). If z A and∈ r < (s z )/3 is{ small enough,} then ∈ ∈ z − | | 2 2 + L ξ B (z, 3rz): f (ξ) > 0 { ∈ 2 } < ε. (40) 9πrz 2 2 Let φ : B R be a nonnegative smooth function such that φ 1 on B (z,rz), φ 0 oﬀ B2(z, 2r→) and φ 2/r . Caccioppoli inequality (32) gives≡ ≡ z k k∞ ≤ z 16 Df + 2 f +2. (41) 2 k 2 2 k ZB (z,rz) k k ≤ rz ZB (z,2rz) 2 A is covered by the balls B (z,rz)(z A). By Besicovitch theorem, there are 1,..., β subfamilies of disjoints balls such that∈ A . We deduce F F ⊆ ∪i≤β ∪ Fi β Df + 2 Df + 2 k 2 k A k k ≤ 2 B (z,rz) k k Z Xi=1 B (z,rXz)∈Fi Z β 16 f +2. 2 2 k ≤ 2 rz B (z,2rz) Xi=1 B (z,rXz)∈Fi Z As k goes to inﬁnity, we get

β 16 lim sup Df + 2 f +2. (42) k 2 2 k→∞ A k k ≤ 2 rz B (z,2rz) Z Xi=1 B (z,rXz)∈Fi Z

2 2 2 2 Set ρz := 3√εrz. If ξ B (z, 2rz) then B (ξ, ρz) B (z, 3rz) B (0, s) and 2 2 2 ∈ ′ 2⊂ ⊂ + ′ L (B (ξ, ρz)) = 9επrz . Thus, by (40), there is ξ B (ξ, ρz) such that f (ξ ) = 0. Consequently, f +(ξ) Λ ρ . We therefore infer from∈ (40) that | | ≤ s z 16 +2 16 2 2 + 2 2 2 2 2 f L ξ B (z, 2rz): f (ξ) > 0 Λ ρ 1296πε Λ r . 2 2 2 s z s z rz ZB (z,2rz) ≤ rz { ∈ } ≤ (42) gives

β + 2 2 2 2 2 2 2 2 lim sup Dfk 1296ε ΛsL i 1296βε ΛsL (B ). k→∞ A k k ≤ F ≤ Z Xi=1 [ + Since ε is arbitrarily small, we deduce that limk→∞ Dir(fk ,A) = 0. In particular, up to passing to a subsequence, Df + 0 almost everywhere on A. k → Now we conclude: we write the Euler-Lagrange equations for stationary (27) and (25) for fk. We let k tend to . By the Lebesgue dominated convergence theorem, (27) and (25) hold also for f. ∞ 88 Blowup analysis of stationary multiple valued functions

Let r > 0, suppose f is stationary, centered and non zero in a neighborhood B2(z ,r) Ω of a branch point z , i.e f(z ) = 2 0 . The blow up of f at z is 0 ⊂ 0 0 0

f(z0 + rz)J K 2 fz0,r(z) := , z B 2 Dir(f,B (z0,r)) ∈ q Proposition 36. Let (rn) be a sequence converging to 0. Up to passing to a subsequence, fz0,rn converges locally uniformly to a nonzero (z0) homogeneous stationary multiple N (k) valued function h. Moreover, if k 0, 1,... is the ﬁrst integer such that Hf (z0) = 0, then ∈ { } 6 (k) (k) (k + 2)k! Hf (z0) Hh (0) = . (43) 4π H(k)(z ) | f 0 |

Proof. Let us suppose for convenience that z0 = 0. We write frn instead of f0,rn . Note 2 that frn (0) = 2 0 for all n and Dir(frn ,B ) = 1. Therefore we can apply Proposition 35 to (frn ). There is a subsequence, still denoted (frn ), which converges locally uniformly J K 2 to a stationary multiple valued function h : B Q2(R). Moreover, Dfrn Dh almost everywhere on B2. → → In case h = 0, then

+ + rnDf (rnz) Dfr (z) = 0 n 2 Dir(f,B (0,rn)) → q 2 2 on B (0, 1/2). Recall that the frn are uniformly Lipschitz continuous on B (0, 1/2), therefore we can apply Lebesgue dominated convergence theorem:

+ 2 1 2 + 2 Dir(f ,B (0,rn/2)) r Df (rnz) dz = 0 (44) 2 2 n + 2 Dir(f,B (0,rn)) ZB (0,1/2) k k 2Dir(f ,B (0,rn)) →

+ as n . On the other hand, Hf + is nonzero since f is not. There is an integer → ∞ k (k) k 0 and a holomorphic map g such that Hf + (z) = z g(z)/k! and g(0) = Hf + (0) = 0. Hence≥ 6 + 2 2πg(0) k+2 Dir(f ,B (0,r)) = Hf + r , 2 r→0 ZB (0,r) | | ∼ k + 2 which contradicts (44). Hence h is nonzero. Now, let r (0, 1). ∈ + 2 + 2 rDir(h ,B (0,r)) rDir(frn ,B (0,r)) Nh(0,r) = +2 = lim +2 2 n→∞ 2 ∂B (0,r) h ∂B (0,r) frn + 2 R rrnDir(R f ,B (0,rrn)) = lim +2 n→∞ 2 ∂B (0,rrn) f

= lim Nf (0R,rrn) = f (0). n→∞ N

Consequently, h is f (0) homogeneous by Theorem 17. Recall that (Hfrn ) are locally N 2 uniformly bounded and converge almost everywhere to Hh on B . Using the holomor- phicity of the Hopf forms, one easily deduces that (H(k) ) converges locally uniformly to frn 4.3 - Blowing up 89

(k) Hh . Now, to prove the last part, we use the mean value property in the ﬁrst and last equalities of: 1 H(k)(0) = H(k) h 2 2 2 h L (B (0, 1/2)) ZB (0,1/2) 1 = lim H(k) n→∞ 2 2 2 frn L (B (0, 1/2)) ZB (0,1/2) (k) k+2 1 H (rnz)r = lim f n dz n→∞ 2 2 2 2 L (B (0, 1/2)) ZB (0,1/2) Dir(f,B (0,rn)) rk = lim n H(k) n→∞ 2 2 2 2 f L (B (0, 1/2)) Dir(f,B (0,rn)) ZB (0,rn/2) rk+2 = lim n H(k)(0). n→∞ 2 f Dir(f,B (0,rn))

k (k) On the other hand, writing again Hf (z) = z g(z)/k! with g(0) = Hf (0), one has

We end this section with the classification of all homogeneous centered stationary functions h : B2 Q (R). Let us write h(reiθ) = rνg(θ), where ν is the frequency of h → 2 at 0, g : R/2πZ Q2(R) is a Lipschitz continuous map. Suppose θ0 is a non branching →+ point of h, i.e g (θ0) > 0. By continuity, there is a maximal interval Iθ0 containing θ0 + iθ on which g = 2 0 . Therefore, h is harmonic on (re : r (0, 1), θ Iθ0 , thus, there is a R 06 and ϕ R/2πZ such that h+(reiθ){ = arν sin(∈νθ + ϕ) whenever∈ } r (0, 1) ∈ \{ } J K+ ∈ + ∈ and θ Iθ0 , i.e g is trigonometric with g (θ) = a sin(νθ + ϕ). If θ1, θ2 are the end ∈ + + points of Iθ0 (in case Iθ0 = R/2πZ), then by maximality g (θ1) = g (θ2) = 0. This implies that I has a length6 π/ν. θ0 ≥ Let J be the collection of those maximal intervals Iθ : θ R/2πZ . They are mutually disjoint, and have a length bounded below. Therefore{ J∈is finite.}R/2πZ J is also a finite union of closed intervals, on which g+ is zero (thus trivially trigonometric).\ ∪ + Let us study the behaviour of g around some point in I∈J ∂I. Up to a rotation, we suppose this point is θ = 0. There is ε > 0, and a, b R such∪ that ∈ a sin νθ whenever θ [0, ε) g+(θ) = . b sin νθ whenever θ ∈ ( ε, 0] ( ∈ − We set U := B2 reiθ : r (0, 1), θ (0, ε) 1 . U := B2 ∩ {reiθ : r ∈ (0, 1), θ ∈ ( ε, 0)} ( 2 ∩ { ∈ ∈ − } Let us remark that the following holds on U1 ∂h+ 1 ∂h+ Dh+ = eiθ + i = aνrν−1ieiθ(1−ν), ∂r r ∂θ ! 90 Blowup analysis of stationary multiple valued functions

+2 2 2 2(ν−1) 2 2 2(ν−1) and Hh+ (z) = Dh = a ν z . Similarly, Hh+ (z) = b ν z on U2. Since 2 − 2 − H + is holomorphic, a = b and ν = 1 + k/2 for some k 0, 1 ... . h ∈ { } Conversely, the ν homogeneous map

iθ ν ν 3 hν,ϕ(re ) := r sin (ν(θ ϕ)) + r sin (ν(θ ϕ)) , where ν 1, , 2,... − − − ∈ 2 2 −2iνϕ 2(ν−1) is stationary. Indeed,J H + = Kν eJ z is holomorphicK thus h is squeeze hν,ϕ ν,ϕ − 2 stationary. hν,ϕ is locally the sum of two harmonic sheets on B 0 hence it is squash 1,2 \{2 } stationary there. H + is integrable, hence h is in W (B , Q (R)) and we can hν,ϕ ν,ϕ 2 conclude with Proposition| | 32. We summarize Proposition 37. The homogeneous centered stationary multiple valued functions are the ahν,ϕ, with a R, ν 1, 3/2, 2,... , ϕ R/(π/ν)Z. Moreover, H + = ∈ ∈ { } ∈ hν,ϕ ν2e−2iνϕz2(ν−1). − We deduce an important fact from Proposition 37 and (43). The Hopf form of a blow up h at z0 is necessarily of the form (2ν−2) ν Hf (z0) 2ν−2 Hh(z) = z (45) 2π H(2ν−2)(z ) | f 0 | since ν = (z ) is imposed and the (2ν 2) derivative of h at 0 is given in (43). Thus Nf 0 − (2ν−2) 1 H (z0) h = h , where 2νϕ + π = arg f . (46) √ ν,ϕ  (2ν−2)  2πν Hf (z0) | | ϕ is unambiguously deﬁned in R/(π/ν)Z. Therefore the blow up is unique. Moreover, there is no loss of energy in the blowing up, indeed

2 + 2 Dir(h,B ) = 2Dir(h ,B ) = 2 Hh = 1. 2 ZB | | This fact was not so obvious since no comparison argument can be made unlike for maps minimizing their Dirichlet energy.

4.4 Branch set

We need to distinguish between two types of branch points:

Definition 1. Let f :Ω Q2(R) be a nonzero stationary map on a domain Ω and z Branch(f). z is a branch→ point of the first type if H (z) = 0. z is a branch point of ∈ f 6 the second type if Hf (z) = 0. The set of first type branch points is denoted Branch1(f) and the set of second type branch points is denoted Branch2(f).

Equivalently, a point z Branch(f) belongs to Branch2(f) when f (z) 3/2, 2,... , it belongs to Branch∈ (f) if and only if (z) = 1. N ∈ { } 1 Nf 2 If h : B Q2(R) is one of the homogeneous stationary functions listed in the previous section,→ then Branch(h) is the restriction in B2 of a cone. Our next step is to prove that the branch set of a stationary function is locally well approximated by the branch set of its blow up. To this end, we recall the following deﬁnition: 4.4 - Branch set 91

Deﬁnition 2. Let A be any subset of Rm, and a Rm. The tangent cone Tan(A, a) at a the set of vectors v Rm such that for all ε > 0∈, there is x A Bm(a, ε) and t > 0 satisfying t(x a) ∈v < ε. Equivalently, if v = 0, v Tan(∈A, a∩) if and only if there is a sequencek (x−) of− pointsk in A a , lim x =6 a and ∈ n \{ } n x a v lim n − = . (47) n→∞ x a v k n − k k k Tan(A, a) is a cone, nonempty whenever a A. ∈ Proposition 38. Let f :Ω Q (R) be a centered stationary map on Ω R2, z → 2 ⊂ 0 ∈ Branch(f), and h be the blow up of f at z0. Then

Tan(Branch(f),z ) B2 = Branch(h). 0 ∩

Proof. Let us suppose for convenience that z0 = 0. Let v Tan(Branch(f), 0) of norm v = 1/2. There are points z Branch(f) 0 converging∈ to 0, such that k k n ∈ \{ } z lim n = v. n→∞ 2 z k nk

We set rn := 2 zn , and consider the sequence of blow ups (frn ). Up to passing to a k k 2 subsequence, we can suppose it converges to h, uniformly on B (0, 1/2). Consequently,

zn h(v) = lim frn = lim f(zn) = 2 0 , n→∞ 2 z ! n→∞ k nk J K i.e v Branch(h). ∈

Conversely, let v Branch(h) 0 . h is the limit of blow ups (frn ). Let ε ∈ \{ } 2 2 ∈ (0, 1 v ). (frn ) converges to h uniformly on B (v, ε) ⋐ B . Suppose Branch(frn ) 2 − k k + ∩ B (v, ε) = for inﬁnitely many n. Then for those values of n, frn is a positive harmonic ∅ 2 + + + function on B (v, ε). As n , frn tends to h . Thus, h is a nonnegative harmonic function on B2(v, ε). Since→ it ∞ vanishes at v, it must be zero everywhere on B2(v, ε). This is contradictory with our classiﬁcation of blow ups, see Proposition 37. Therefore, if n is big enough, then r < ε and Branch(f ) B2(v, ε) contains at n rn ∩ least some point ξ. Set z := rnξ. Then f(z) = 2 0 and z rn < ε. Moreover, −1 k k ≤ tz v < ε, where t = rn . As ε is arbitrary, v Tan(Branch(f), 0). k − k ∈ J K

In particular, when z0 is a ﬁrst type branch point, the blow up at z0 is the sum of two linear maps, therefore the tangent cone of Branch(f) at z0 is a line. Let us call ϕ(z0) the angle between the x axis and Tan(Branch(f),z0). By (46), we have

1 Hf (z0) ϕ(z0) = arg π . (48) 2 H (z ) ! − ! | f 0 |

Note that if f is a nonzero stationary Q2(R) valued function on a domain, there are locally ﬁnitely many second type branch points, as the Hopf form vanishes on each of ∞ them. We wish to study the structure of Branch1(f). We will prove that it is a C manifold. We will need the following purely topological fact. Recall a locally closed set in a topological space is just the intersection of a closed set and an open set. 92 Blowup analysis of stationary multiple valued functions

Lemma 6. Let C R be a locally closed set such that for all x C, Tan(C, x) = R. Then C is open. ⊂ ∈

Proof. Let x C, by local closedness, there is a closed interval I such that x Int(I) and C I is closed.∈ Suppose there is a closed interval [a, b] I such that C [a,∈ b] = . ∩ ⊆ ∩ 6 ∅ Either m1 := max C I ( , a] or m2 := min C I [b, + ) exists. In case m1 is ∩ ∩ −∞ ∩ ∩ ∞ − well defined, note that Tan(C, m1) is contained in the half line R , which contradicts the hypothesis. A similar contradiction arises with m2. Consequently, C I is dense in I, thus I C. This proves that C is open. ∩ ⊆ Theorem 18. Suppose f :Ω Q2(R) is stationary and not constant on a domain Ω R2. Then Branch (f) is a→ union of disjoint real analytic curves. ⊂ 1 Proof. Let z Branch (f). We will show that Branch (f) is a C∞ curve in a neigh- 0 ∈ 1 1 borhood of z0. The analyticity of Branch1(f) will be a consequence of Proposition 39. We can suppose, up to rotating and translating, that z0 = 0 and Tan(Branch(f), 0) = R 0 , i.e the function ϕ, defined by (48), verifies ϕ(0) = 0. We set × { } K := (x, y) R2 : y x . { ∈ | | ≤ | |} −1 Recall that Hf ( 0 ) consists of isolated points and ϕ is continuous, therefore we can choose r small enough{ } such that B2(0, 2r) Ω 2 ⊂ −1  B (0, 2r) Hf ( 0 ) = 2 ∩ { } ∅ .  B (0, 2r) Branch(f) K  z B2(0∩, 2r) Branch(⊂ f), Tan(Branch(f),z) = 0 R  ∀ ∈ ∩ 6 { } ×  Define π, π⊥: R2 R the projections (x, y) x,(x, y) y and → 7→ 7→ A := Branch(f) B2(0, 2r) π−1( r, r),C := π(A). ∩ ∩ − 2 Since Branch(f) B2(0, 2r) K and K π−1( r, r) B (0, √2r) B2(0, 2r), one has ∩ ⊂ ∩ − ⊂ ⊂ 2 A = Branch(f) B (0, √2r) π−1( r, r), ∩ ∩ − therefore A is closed in π−1( r, r). Consequently, C is locally closed. Moreover, for every z A, − ∈ Tan(C, π(z)) π(Tan(A, z)) = R. ⊃ C satisfies the hypothesis of Lemma 6, thus C is open. As 0 C, there is ε

Let us prove that M is continuous. Let a ( ε, ε) and (xn) a sequence in ( ε, ε) such ∈ − 2 − that xn a and (M(xn)) converges to some ℓ. B (0, √2r) is closed, therefore contains (a, ℓ). Hence→ ℓ M(a) and lim sup M(x) M(a). ≤ x→a ≤ If ρ is small enough to have B2((a, M(a)), 2ρ) B2(0, 2r) π−1( ε, ε), set ⊂ ∩ − A′ := Branch(f) B2((a, M(a)), 2ρ) π−1(a ρ, a + ρ),C′ := π(A′). ∩ ∩ − 4.4 - Branch set 93

As previously, one can apply Lemma 6 to C′. Furthermore, a C′. Consequently, there is an open interval J such that a J C′. For all x∈ J, there is some z Branch(f) B2((a, M(a)), 2ρ) π−1(∈x ).⊂ We infer that M∈(x) M(a) 2ρ. ∈ ∩ ∩ { } ≥ − Thus lim infx→a M(x) M(a) 2ρ. As ρ is arbitrarily small, this proves the lower semi-continuity. ≥ −

Let as before (xn) be a sequence in ( ε, ε) converging to a. As (xn,M(xn)) (a, M(a)), by (47), the limit points of − →

(x a, M(x ) M(a)) n − n − (x a)2 + (M(x ) M(a))2 n − n − q are (cos ϕ(a, M(a)), sin ϕ(a, M(a)). Therefore, ± M(x ) M(a) lim n − = tan ϕ(a, M(a)). n→∞ x a n − Thus M is a solution of the diﬀerential equation M ′(x) = tan ϕ(x, M(x)). Similarly we prove the same for m. As M(0) = m(0) = 0, we conclude M = m, which means that Branch(f) B2(0, 2r) π−1( ε, ε) is the graph of a C∞ function. ∩ ∩ −

The following proposition explains that we should think of points in Branch1(f) as false branch points:

Proposition 39. Suppose z Branch (f). Then f is the sum of two harmonic func- 0 ∈ 1 tions in a neighborhood of z0.

Proof. Suppose f is centered and z = 0. As H (0) = 0, there is a neighborhood V of 0 0 f 6 such that V Branch2(f) = , and a antiholomorphic function g : V C such that for ∩ 2 ∅ → +2 all z V , Hf (z) = g(z) = 0. Recall that Hf = Df almost everywhere. The equality is satisﬁed∈ everywhere on6 f + > 0 , since on that set f + is harmonic. Let us deﬁne u : V R by { } → 0 if f(z) = 2 0 u(z) := f +(z) if f(z) = 2 0 and Df +(z) = g(z)  6  f +(z) if f(z) = 2J0K and Df +(z) = g(z) − 6 J K −  We will prove that u is harmonic. As f = u J+K u , that will end the proof. − J K J K • u is continuous: this is induced by the continuity of f,

• u is a C1 map on V Branch(f), and Du = g is the restriction on V Branch(f) of a continuous map,\ \

• u 0 on V Branch(f), ≡ ∩ • for all z V Branch(f), ∈ ∩ lim Du(ξ) = g(z) ξ→z ξ∈V \Branch(f) 94 Blowup analysis of stationary multiple valued functions

is normal to V Branch(f) at z. Indeed, let us call h the blow up of f at z. By Proposition 37∩ and Equation (45), one has

2 e2iϕ(z) 1 H (z) 1 g(z) H (0) = − = f = , h 2π 2π H (z) 2π H (z) | f | | f | therefore g(z) = i H (z) eiϕ(z) is orthogonal to Tan(V Branch(f),z) (which ± | f | ∩ makes an angle ϕ(zq) with the x-axis).

These four points ensure that u is C1 on V with Du = g. Since g is anti-holomorphic, u is C2 and ∂g 1 ∂ ∂ ∂u ∂u 1 0 = = i + i = ∆u. ∂z 2 ∂x − ∂y ! ∂x ∂y ! 2 Note in particular that V Branch(f) = u−1( 0 ), thus by the implicit function theorem, ∩ { } Branch1(f) is analytic.

It remains to study the local behaviour of Branch(f) around branch points of the second type. If v R2 0 and α [0, π], we let ∈ \{ } ∈ K+(v, α) := z R2 : z, v z v cos α { ∈ h i ≥ k kk k } be the cone of direction v and aperture 2α.

Theorem 19. Let z Branch (f) and ν := (z ). In a neighborhood of z , Branch(f) 0 ∈ 2 Nf 0 0 is a union of 2ν non intersecting real analytic curves in the interior, all starting at z0 1 and C up to z0.

Proof. For convenience, let us suppose z0 = 0. According to Propositions 37 and 38, the tangent cone of Branch(f) at 0 consists of the 2ν half-lines θ = ϕ + kπ/ν , k { } ∈ 0,..., 2ν 1 (where ϕ is such that hν,ϕ/√2πν is the blow up of f at 0). Up to making {a rotation,− we} can suppose ϕ = 0. Besides, there is a holomorphic map g on Ω such that 1 H (z) = g(z)z2(ν−1), with g(0) = H(2ν−2)(0) = H(2ν−2)(0) f (2ν 2)! f −| f | − by Equations (45) and (46). Thus, by Equation (48),

1 ϕ(z) = (arg(g(z)) + 2(ν 1) arg z π) (49) 2 − −

+ for all z such that Hf (z) = 0. If z lies in K (1, π/5ν) 0 , then 2(ν 1) arg(z) 2π/5. If z is small, arg(g(z)) is6 closed to π and therefore ϕ\{(z)}= π/|2. We− can choose| ≤ r > 0 such| | that B2(0, 2r) Ω contains no second type branch points6 ± but 0, ⊂ 2ν−1 π B2(0, 2r) Branch(f) K+ ei(kπ/ν), , ∩ ⊂ 5ν k[=0 and ϕ(z) = π/2 for all z (Branch(f) B2(0, 2r) K+(1, π/5ν)) 0 . 6 ± ∈ ∩ ∩ \{ } 4.4 - Branch set 95

As in Theorem 18, deﬁne π A := Branch(f) K+ 1, π−1[0,r) B2(0, 2r) ∩ 5ν ∩ ⊂ and C := π(A). A is closed in π−1[0,r), thus C is a locally closed subset of [0,r). One proves as before that 0 C, Tan(C, 0) = R+ and Tan(C, x) = R for all x C 0 .A slight adaptation of Lemma∈ 6 implies that there is some ε (0,r] such that∈ [0\{, ε) } C. Again, let us introduce, for all x [0, ε), ∈ ⊂ ∈ M(x) := sup y :(x, y) Branch(f) K+(1, π/5ν) . m(x) := inf {y :(x, y) ∈Branch(f) ∩K+(1, π/5ν) } ( { ∈ ∩ } As in Theorem 18, we prove that M(0) = 0, M is continuous on [0, ε) and

x (0, ε),M ′(x) = tan ϕ(x, M(x)). (50) ∀ ∈ For all x [0, ε), let α(x) be ∈ π π α(x) := min α 0, : Branch(f) K+ 1, π−1[0, x] K+(1, α) . ∈ 5ν ∩ 5ν ∩ ⊂

α is nondecreasing, and limx→0 α(x) = 0 because Tan(Branch(f), 0) K+(1, π/5ν) = R+ 0 . ∩ × { }

Thus arg(x + iM(x)) α(x) 0 as x 0. Since limx→0 arg(g(x + iM(x))) = π, we infer from| Equation (49)| ≤ that ϕ→(x, M(x))→ 0 as x 0. This and (50) implies that M is diﬀerentiable at 0 and M ′(0) = 0. We deﬁne→ M˜ :(→ , ε) R −∞ → 0 if x 0 M˜ (x) := . M(x) if x ≤ 0 ( ≥ M˜ is C1, and is solution of M˜ ′(x) = γ(x, M(x)) where we set

0 if x 0 tan ϕ(x, y) if x ≤ 0 and y x tan α(x) γ(x, y) :=  ≥ | | ≤  tan ϕ(x, x tan α(x)) if x 0 and y x tan α(x)  tan ϕ(x, x tan α(x)) if x ≥ 0 and y ≥ x tan α(x) − ≥ ≤ −  γ is obviously locally Lipschitz continuous with respect to y. If we prove that α is continuous, this will imply that γ is continuous on ( , ε) R. By the Picard-Lindelöf theorem, the solution of the Cauchy problem y′ = γ(−∞x, y), y(0)× = 0 is unique. Let us call m˜ the extension by zero of m on ( , ε). We prove as well thatm ˜ is a C1 solution of this problem. Thus M = m, i.e Branch(−∞ f) is a C1 curve starting at 0 in a neighborhood of 0 in K+(1, π/5ν). We treat as well the 2ν 1 other sectors. − It remains to prove that α is continuous. We already proved that α is continuous at ′ ′ + −1 ′ 0. Let x0 (0, ε). There is (x , y ) Branch(f) ∂K (1, α(x0)) π [0, x0] with x = 0. ′ ∈ ∈ ′ ∩ ∩ 6 If x < x0, then α(x) α(x0) for all x [x , ε). Since α is nondecreasing, α α(x0) on ′ ≥ ∈ ≡ [x , x0], thus α is left continuous at x0. ′ ′ In case x = x0, then y = M(x0) and α(x0) = arg(x0 + iM(x0)). Moreover, x (0, ε), α(x) arg(x + iM(x)). ∀ ∈ ≥ 96 Blowup analysis of stationary multiple valued functions

Since M is continuous, one sees that α is lower semi-continuous at x0, as α is also nondecreasing, α is left continuous at x0. Now suppose that

ℓ := lim α(x) > α(x0). x→x0 x>x0

′ ′ + Then for all x [x0, ε), there is (x (x), y (x)) Branch(f) K (1, π/5ν) with x0 x′(x) x such that∈ ∈ ∩ ≤ ≤ arg(x′(x) + iy′(x)) ℓ. ≥ ′ There is a sequence xn x0 such that y (xn) converges to some y0.(x0, y0) Branch(f) K+(1, π/5ν)→ and arg(x + iy ) ℓ. Consequently, α(x ) ℓ. This∈ is ∩ 0 0 ≥ 0 ≥ contradictory, hence α is right continuous at x0.

4.5 Stationary surfaces in R3

In this last subsection, we describe how stationary surfaces in R3 could be studied by means of stationary multiple valued functions. We emphasize that this area is relatively untouched in comparison with the study of area minimizing currents. The major regularity result is namely Allard’s regularity theorem, see [All72]. More recently, some work has been done for stable stationary hypersurfaces, see [Wic08]. This is an attempt to remove the stability condition. First let us introduce what we mean by a stationary surface.

Deﬁnition 3. A set M Rn is called m rectiﬁable whenever countably many C1 man- ⊂ ifolds M1,M2,... such that

∞ m H M Mi = 0. \ ! i[=1

m Note that if i = j, then for H almost every x Mi Mj, the tangent spaces Tan(M , x) and Tan(6 M , x) coincide. There we can send∈ almost∩ every x M to a “tan- i j ∈ gent space” Tan(M, x). If A is H m measurable, the map x Tan(M, x) is moreover 7→ H m measurable. Let U be an open set in Rn.A m rectiﬁable varifold in U is the data of

(1) an H m measurable m rectiﬁable subset M of U, called the carrying set,

(2) a density function θ : M (0, ), H m measurable and H m locally integrable. → ∞

m Two varifolds (M1, θ1) and (M2, θ2) are meant to be equivalent whenever H (M1 m n ⊖1 M2) = 0 and θ1 = θ2 H almost everywhere on M1 M2. When X : U R is a C vector ﬁeld X = (X ,...,X ) and A G(m, n), we call∩ div X(x) the quantity→ 1 n ∈ A n div X(x) := tr(A DX(x)) = A X (x), e , A ♮ ◦ h ♮∇ i ii Xi=1 4.5 - Stationary surfaces in R3 97

n where (e1, . . . , en) is the canonical basis of R . One easily checks that if (τ1, . . . , τm) is an orthonormal basis of A, then

m div X(x) = DX(x)(τ ), τ . A h i ii Xi=1 Deﬁnition 4. A varifold (M, θ) in U is called stationary whenever

m divTan(M,x) X(x)θ(x)dH (x) = 0 (51) ZM for all continuously differentiable vector fields X : U Rn. → The notion of excess measures how much a varifold is far from being “flat”.

Deﬁnition 5. Let (M, θ) be a varifold in U n(x, z). The excess of (M, θ) in U n(x, r) is

m m 2 m E(r, t) := Tan (M, y)♮ R♮ θ(y)dH (y). Z k − k

In order to prove Conjecture 1, one needs a Lispchitz approximation theorem. The following is well known.

Theorem 20. Suppose

(A) M is 2 stationary varifold in B3(0, 1).

(B) 0 supp M, Θ2( M , 0) = 2. ∈ k k (C) α (0, 1). ∈ (D) M B3(0, 1) π(2 + α). k k ≤ then there is γ = γ(α) and an application f : B2(0, γ) R such that → (1) Lip f 1. ≤ (2) H 2((Graph supp M) B3(0, γ)) cE(0, 1). f ⊖ ∩ ≤ However, this approximation is not powerful enough for our purposes. One needs to sharpen the estimate (2) and bound the non graphical part of M by the excess to a power 1 + ε. This is mainly because the equations (26) and (27) involve the gradient to the square. 98 Part II Representation of charges

Summary 5 Charges and cohomology 99 5.1 Preleminaries on Federer-Fleming currents ...... 99 5.2 Topologies on the space of normal currents ...... 101 5.3 Charges and duality with normal currents ...... 103 5.4 A representation theorem for charges ...... 104 5.5 Charges vanishing at inﬁnity ...... 106

6.5 Lp spaces...... 116 6.6 Grothendieck theorem ...... 118 6.7 A splitting theorem ...... 122

7 Continuous representation operators 124 7.1 Main theorem ...... 124 7.2 Adaptation to charges of positive codimension ...... 126

8 Appendix to Part II 128

References 132 5.1 - Preleminaries on Federer-Fleming currents 99 5 Charges and cohomology

5.1 Preleminaries on Federer-Fleming currents

We assume n >1 and 0 m n. We let E m(Rn) be the linear space of all smooth diﬀerential m forms on R≤n. We≤ recall that each diﬀerential form ω has a unique decom- position ω = ω dx dx k1,...,km k1 ∧ · · · ∧ km 1≤k1≤···≤Xkm≤n where the ω are smooth functions. For each x Rn, we call k1,...,km ∈

ω(x) := sup ω (x)det (dx (ξ )) : ξ ,..., ξ 1 . k k  k1,...,km ki j 1≤i,j≤m k 1k k mk ≤  1≤k1≤···≤Xkm≤n  The reason why we choose to norm covectors (i.e elements of m(Rn)) this way – instead of choosing the usual Euclidean norm – is quite technical.∧ It allows to define general finite mass currents by relaxation of rectifiable currents, this will become clear in the following pages. For the moment, the choice of a norm on m(Rn) does not matter. ∧ m For each multi-index α = (α1, . . . , αm) 1, . . . , n and each compact subset K Rn we define the semi-norm ∈ { } ⊂ ω := sup ∂ ω(x) : x K , k kα,K {k α k ∈ } where α1+···+αm ∂ ωk1,...,km ∂αω(x) := dxk dxk . α1 αm 1 m ∂x1 ∂xm ∧ · · · ∧ 1≤k1≤···≤Xkm≤n ··· The family of all semi-norms induces a Fréchet topology on E m(Rn). For each k · kα,K compact subset K Rn we define D m(Rn) := ω E m(Rn) : supp ω K , which is a ⊂ K { ∈ ⊂ } closed subspace of E m(Rn). We also define the subspace D m(Rn) of smooth differential forms with compact support, and give it the localized topology (see section 8) with respect to the family := D m(Rn): K compact . C { K } m n m n n Whenever m < 0 or m > n, we set E (R ) := 0, D (R ) = 0. We let Dm(R ) be the m n n dual of D (R ), and we endow it with the weak* topology. Elements of Dm(R ) are called m currents on Rn. For any current T , we define its support to be the smallest closed subset A for which T, ω = 0 whenever supp ω Rn A. Functional restriction m n m n h i m n ∗ ⊂ \ n from E (R ) to D (R ) yields a linear map E (R ) Dm(R ), whose range consists of all currents with compact support. The boundary of→ an m dimensional current T is the (m 1) dimensional current ∂T defined by ∂T, ω := T, dω if 1 m n and ∂T = 0− otherwise. h i h i ≤ ≤ The mass of a current is the extended real number M(T ) := sup T, ω : ω D m(Rn), sup ω(x) 1 . {h i ∈ x k k ≤ } If M(T ) < , by Riesz representation theorem, there exists a unique Radon measure ∞ T on Rn, and a Borel m vector field T~, unique up to a T negligible set, such that k k k k T, ω = ω, T~ d T . n h i ZR h i k k 100 Charges and cohomology

Actually, the application of Riesz representation theorem is not so straightforward, because mRn was not given a Euclidean norm. To overcome this slightly technical diﬃ- culty,∧ we refer to [Fed96, 4.1.5]. If X is any subset of Rn, we introduce the space M (X) of currents T D (Rn) m ∈ m with ﬁnite mass, with support contained in X, normed by M. Mm(X) is a Banach space whenever X is closed.

n A current T Dm(R ) with compact support is called normal if both T and ∂T ∈ n have finite mass. For any X R , Nm(X) will denote the space of normal currents with support contained in X,⊂ normed by the normal mass N(T ) := M(T ) + M(∂T ). Nm(X) is a Banach space whenever X is closed. When f : Rn Rp is of class , T D (Rn) and f ↾ supp T is proper, we define a → ∞ ∈ m current f T D (Rp) so as to satisfy # ∈ m f T, ω = T, γf ω h # i h # i m p ∞ n whenever ω D (R ) and γ Cc (R ), γ 1 on a neighborhood of supp T f −1(supp ω).∈ Under these conditions,∈ it follows≡ that supp f T f(supp T ) and∩ # ⊂ ∂f#T = f#∂T . The flat norm of a current T N (Rn) is the number ∈ m F(T ) := inf M(T ∂S) + M(S): S N (Rn) . { − ∈ m+1 } It may not be obvious at first that F is a norm. However, we have the following proposition. Proposition 40. Let T N (Rn). Then ∈ m F(T ) := sup T, ω : max( ω , dω ) 1 . {h i k k∞ k k∞ ≤ } Proof. Let S N (Rn). Then for any ω D (Rn), we have ∈ m+1 ∈ m T, ω = T ∂S, ω + S, dω . h i h − i h i Hence T, ω M(T ∂S) + M(S) whenever max( ω , dω ) 1. Thus h i ≤ − k k∞ k k∞ ≤ sup T, ω : max( ω , dω ) 1 F(T ). {h i k k∞ k k∞ ≤ } ≤ n n ′ On the other hand, endow Dm(R ) Dm+1(R ) with the norm (ω, ω ) := ′ × k k max( ω ∞, ω ∞). We define a continuous linear form ϕ on the linear subspace k k k k n n n (ω, dω): ω Dm(R ) Dm(R ) Dm+1(R ) by setting ϕ(ω, dω) := T, ω . By { ∈ } ⊂ × n h i n Hahn-Banach theorem, we extend ϕ to a continuous linear form on Dm(R ) Dm+1(R ), which we still denote ϕ, with the same norm ×

ϕ = sup T, ω : ω D m(Rn) and max( ω , dω ) 1 . k k {h i ∈ k k∞ k k∞ ≤ } n m n ′ m+1 n Deﬁne S Dm+1(R ) by S, ω := ϕ(0, ω). For all ω D (R ), ω D (R ) such that ω ∈ 1, ω′ 1,h wei have ∈ ∈ k k∞ ≤ k k∞ ≤ T ∂S, ω + S, ω′ = ϕ(ω, ω′) ϕ . h − i h i ≤ k k Thus F(T ) M(T ∂S) + M(S) ϕ . ≤ − ≤ k k 5.2 - Topologies on the space of normal currents 101

n2 n2 m1+m2 n1+n2 Let T Dm1 (R ) and S Dm2 (R ). Then each ω D (R ) has an expansion ∈ ∈ ∈

′ ′ ′ ′ ω(x, y) = ωα1,...,αm ,β1,...,βm (x, y)dyβ1 dyβm dxα1 dxαm (52) 1 2 ∧ · · · ∧ 2 ∧ ∧ · · · ∧ 1 X ′ ′ where the sum is taken over all (α, β) Nm1 Nm2 such that m′ + m′ = m + m , ∈ × 1 2 1 2 and 1 α1 < < α ′ n1, 1 β1 < < β ′ n2. We deﬁne a current ≤ ··· m1 ≤ ≤ ··· m2 ≤ T S D (Rn1+n2 ) by × ∈ m1+m2

′ ′ T S, ω := Tx Sy ωα,β(x, y)dyβ1 dyβm dxα1 dxαm , h × i ∧ · · · ∧ 2 ∧ · · · ∧ 1 X m1 where the ﬁrst sum is taken over all α N with 1 α1 < < αm1 n1 and the m2 ∈ ≤ ··· ≤ second sum is taken over all β N with 1 β1 < < βm2 n2. Note that all ′ ∈ ≤ ··· ≤ terms in (52) for which m1 = m1 are not taken into account. From the characterizing conditions, one infers that supp(6 T S) = supp T supp S and × × ∂(T S) = ∂T S + ( 1)m1 T ∂S. × × − × One easily proves this following fact.

Proposition 41. Let T D (Rn1 ) and S D (Rn2 ). Let p : Rn1+n2 Rn2 be the ∈ m1 ∈ m2 → projection onto the last n2 coordinates. Suppose that supp T is compact. Then

T (1)S whenever m1 = 0 p#(T S) = × ( 0 otherwise

5.2 Topologies on the space of normal currents

n In this section, we study other topologies on Nm(R ) than the one inherited from the ∗ n normal mass. Let W be the weak* topology on Dm(R ), i.e the coarsest one for which the maps T T, ω , ω D m(Rn) are continuous. F will denote the topology on n 7→ h i ∈ n Nm(R ) inherited from the ﬂat norm. For any subset X of R , and c > 0, we deﬁne

N (X) := N (X) T : N(T ) c and supp T X . m,c m ∩ { ≤ ⊂ } Let us introduce the increasing system of convex sets

(X) := N (X Bn(0,k)) : k 1 C { m,k ∩ ≥ }

It is a linearly stable family in Nm(X) (see Appendix 8) consisting of sets that are closed in the F topology. Note that (X) = (Rn) ↾ N (X). We deﬁne F to be the C C m X,m localized topology of F ↾ Nm(X) with respect to (X). Using the notation of Appendix 8, one writes C

FX,m := [F ↾ Nm(X)]C(X).

n Proposition 42. If X R then FX,m = FRn,m ↾ Nm(X), and consequently FY,m = F ↾ N (Y ) for each Y⊂ X. X,m m ⊂ 102 Charges and cohomology

n Proof. As the space (Nm(R ), F) is metrizable, it is hereditarily sequential. Proposi- tion 60 assure us that inducing and localizing topologies are commutative operations, precisely

n FX,m = [F ↾ Nm(X)]C(X) = [F ↾ Nm(X)]C(R )↾Nm(X)

= FC(Rn) ↾ Nm(X) = FRn,m ↾ Nm(X)

Now if Y X, ⊂

FY,m = FRn,m ↾ Nm(Y )

= F n ↾ (N (X) N (Y )) R ,m m ∩ m = FX,m ↾ Nm(Y ).

∗ Proposition 43. If X is compact and c > 0, then F ↾ Nm,c(X) = W ↾ Nm,c(X), and these topologies turn Nm,c(X) into a compact space.

n ∗ Proof. F is a lower semicontinuous norm of Nm(R ) under the weak* topology W ↾ n ∗ n Nm(R ), hence F is larger than W ↾ Nm(R ). In particular, the identity map

id : (N (X), F ↾ N (X)) (N (X), W∗ ↾ N (X)) m,c m,c → m,c m,c ∗ n is continuous. N is also lower semicontinuous in the topology W ↾ Nm(R ), there- ∗ n fore Nm,c(X) is a W ↾ Nm(R )-closed set. By Banach-Alaoglu theorem (in general topological vector spaces, see [Rud06, Theorem 3.15]),

T D (Rn): T, ω c, T, dω c if sup ω 1, supp T X { ∈ m |h i| ≤ |h i| ≤ k k ≤ ⊂ } ∗ ∗ is compact in the weak* topology W . As a closed subset, Nm,c(X) is W -compact.

Now we show that (Nm,c(X), F ↾ Nm,c(X)) is compact; we follow the argument of [Fed96, Theorem 4.2.17, (1)] and prove that the distance associated to F makes Nm,c(X) into a totally bounded complete metric space. Choose ε > 0. As X is compact, the deformation theorem [Fed96, 4.2.9] implies that one can find a finite dimensional linear space N (Rn) and a constant γ > 0 such that the following condition YX,ε ⊂ m,c is satisfied: for each T Nm,c(X), there is P X,ε such that N(P ) γc and F(T P ) < ε. As P ∈ : N(P ) γc is a∈ bounded Y set of a finite dimensional≤ − { ∈ YX,ε ≤ } space, it is easy to infer that Nm,c(X) is covered by finitely many F-balls of radius 2ε. n By the arbitrariness of ε, Nm,c(X) is a totally bounded subset of (Nm(R ), F).

Select a Cauchy sequence (Ti) in (Nm,c(X), F) and observe (Ti) weak* converges to a continuous linear functional T : D m(Rn) R by the Banach-Steinhaus theorem → [Rud06, Theorem 2.9]. A standard argument shows that limi F(Ti T ) = 0, and the lower semicontinuity of normal mass implies N(T ) c. As X is closed,− supp T X. Hence T N (X), and (N (X), F) is complete. ≤ ⊂ ∈ m,c m,c It follows that the space (Nm,c(X), F ↾ Nm,c(X)) is compact and id is a homeomorphism.

Now we extend the deﬁnition of a pushforward f T when f : Rn Rp is locally # → Lipschitz continuous and T D (Rn) is normal. Since f is Lipschitz continuous on a ∈ m 5.3 - Charges and duality with normal currents 103 compact neighborhood U of supp T , we can suppose f to be globally Lipschitz continuous; otherwise replace f with a Lipschitz continuous extension of f ↾ U (which exists by Kirszbraun’s theorem, see [Fed96, 2.10.43]). Let (gi)i∈N be a sequence of smooth func- n tions converging uniformly to f on R and such that supi gi ∞ < . To obtain such a sequence, one can for instance regularize f. We deﬁne thek∇ mapk h ∞: [0, 1] Rn Rp ij × → by setting h (t, x) := (1 t)g (x) + tg (x). We let 0, 1 D (R) be the current ij − i j ∈ 1 1 0, 1 , ω := ω(x)dJL 1(Kx). h i Z0 Then g T g T = ∂h ( 0J, 1 K TK) + h ( 0, 1 ∂T ). Thus, j# − i# ij# × ij# × F(gj#T gi#T ) M(Jhij#K( 0, 1 T )) +J M(Khij#( 0, 1 ∂T )) − ≤ 1 × × m gj gi ((1 t) gj + t gi ) d T dt n J K J K ≤ Z0 ZR k − k − k∇ k k∇ k k k 1 m−1 + gj gi ((1 t) gj + t gi ) d ∂T dt n Z0 ZR k − k − k∇ k k∇ k k k m m−1 gj gi ∞ max(sup gi , sup gi )N(T ), ≤ k − k i k∇ k i k∇ k hence (gi#T ) is a F Cauchy sequence. Since all gi#T lie in Nm,c(X) for some c > 0 and some compact X Rp, it follows from Proposition 43 that (g T ) converges to a ⊂ i# current f#T . f#T does not depend on the choice of the gj. Proposition 44. Let T N (Rn), and let f, g : Rn Rp be locally Lipschitz continu- ∈ m → ous maps. If f ↾ supp T = g ↾ supp T , then f#T = g#T .

As a consequence of Proposition 44, if X,Y are subsets of Euclidean spaces and f : X Y is a locally Lipschitz continuous map, there exists a continuous linear map f :(N→ (X), F ) (N (Y ), F ). # m X,m → m Y,m

5.3 Charges and duality with normal currents

Now that a suitable topology has been deﬁned on Nm(X), we can introduce charges as linear functionals.

n Deﬁnition 6. An m charge in X R is an FX,m continuous linear functional on ⊂ m Nm(X). The linear space of all m charges in X is denoted CH (X).

Using Proposition 61, one easily sees that a linear functional f : N (X) R is a m → charge if and only if given ε > 0, there is a θ > 0 such that for each T Nm(X Bn(0, ε−1)), we have α, T θF(T ) + εN(T ). We next define a topology on∈CHm(X∩). For all k 1, we introduceh i ≤the seminorm ≥ α := sup α, T : α N (X Bn(0,k)) and N(T ) 1 . k kk {h i ∈ m ∩ ≤ } Proposition 45. The seminorms define a Fréchet topology in CHm(X). k · kk m Proof. First note that given α CH (X) and k 1, α k is finite. Indeed, one can find θ > 0 so that ∈ ≥ k k 1 α, T θF(T ) + N(T ) (θ + 1)N(T ) θ + 1 h i ≤ k ≤ ≤ 104 Charges and cohomology for every T N (X Bn(0,k)) with N(T ) 1. Hence α θ + 1 < . ∈ m ∩ ≤ k kk ≤ ∞ Clearly if α k = 0 for all k 1 then α = 0, so we need only to show that if (αi) is a k k m ≥ Cauchy sequence in CH (X) then it converges. The sequence (αi) converges uniformly n on each set Nm,1(X B (0,k)) to a linear functional α. Choose ε > 0 and an integer −1 ∩ n k ε . One can find αi such that α αi,T ε/2 for all T Nm,1(X B (0,k)). There≥ is θ > 0 such that for all nonzeroh −T Ni ≤(X Bn(0, 2ε−1∈)), ∩ ∈ m ∩ ε α ,T θF(T ) + N(T ). h i i ≤ 2 Now for all T N (X Bn(0, ε−1)), we have ∈ m ∩ T ε α, T = αi,T + N(T ) α αi, θF(T ) + N(T ). h i h i * − N(T )+ ≤ 2 hence we conclude α CHm(X). ∈ In the sequel, we will only consider CHm(X) with its Fréchet topology. In particular, if X is bounded, then CHm(X) is a Banach space normed by

α CHm := sup α, T : T N (X), N(T ) 1 . k k (X) {h i ∈ m ≤ } When X is closed, it follows from Proposition 43 and Proposition 59 that the Fréchet topology on CHm(X) is that of uniform convergence on bounded subsets of the space ∗ (Nm(X), FX,m) , in other words, the strong topology. By Proposition 62, we conclude that the evaluation linear map Υ : N (X) CHm(X)∗ is bijective. m → When X is compact, it is straightforward to see that Υ is a continuous linear map m ∗ from (Nm(X), N) to CH (X) normed by the dual norm of CHm(X). Since both spaces are Banach spaces, Υ is a Banach space isomorphismk ·by k the open mapping theorem. Let X Rn, Y Rp and let f : X Y be a Lipschitz continuous map. f induces a pullback⊂ map f # ⊂: CHm(Y ) CHm(→X) such that f #α, T := α, f #T for all T N (X). f # is continuous, since→ for all k N, f #α h max(Lip(i fh)m−1, Lip(i f)m) α ∈. m ∈ k kk ≤ k kk Proposition 46. Let X Y be subsets of Rn and let ι : X Y be the inclusion map. Then each charge on X can⊂ be extended to a charge on Y , i.e→ι# : CHm(Y ) CHm(X) is a surjection. →

# Proof. Since ι# : Nm(X) Nm(Y ) is the inclusion map, we see immediately that ι α = → m α ↾ Nm(X) for any α CH (Y ). The proposition hence follows from Proposition 42 and Hahn-Banach theorem.∈

5.4 A representation theorem for charges

n m n m n Let X R . We denote by Cu(X, R ) the linear space of all maps from X to R that are⊂ uniformly continuous on∧ all bounded subsets of X. Throughout, we assume∧ C (X, mRn) is given the Fréchet topology induced by the seminorms u ∧ ω := sup ω(x) : x X Bn(0,k) . | |k {k k ∈ ∩ } 5.4 - A representation theorem for charges 105

n m n m Proposition 47. For X R , the linear map Λ: Cu(X, R ) CH (X) deﬁned by Λ(ω),T := T, ω is continuous.⊂ ∧ → h i h i

Proof. Λ(ω) is a linear functional on Nm(X). We show that it is a m charge in X. Choose ε > 0. Since ω is uniformly continuous on X Bn(0, ε−1), there is c R and ∩ ∈ φ D m(Rn) such that for all x X Bn(0, ε−1), ω(x) c and ω(x) φ(x) ε. For∈ all T N (X Bn(0, ε−1))∈ and S∩ N (Rnk), k ≤ k − k ≤ ∈ m ∩ ∈ m+1 Λ(ω),T = T, ω φ + T ∂S, φ + S, dφ h i h − i h − i i εM(T ) + φ M(T ∂S) + dφ M(S) ≤ k k∞ − k k∞ εN(T ) + max( φ , dφ )(M(T ∂S) + M(S)) ≤ k k∞ k k∞ − Since S is arbitrary, Λ(ω),T εN(T ) + max( φ , dφ )F(T ). h i ≤ k k∞ k k∞ For any k 1, Λ(ω) ω , hence Λ is continuous. ≥ | |k ≤ | |k

It follows from the continuity of ∂ :(Nm+1(X), FX,m+1) (Nm(X), FX,m) that we can assign to each m charge its coboundary dα deﬁned by →

dα, T := α, ∂T . h i h i m m+1 Since for k 1, we have dα k α k, d : CH (X) CH (X) is continuous. Proposition 47≥ show that charges| | ≤ can | | be viewed as generalized→ diﬀerential forms with d as a coboundary operator. Now we introduce the Fréchet space E := C(Rn, mRn) ∧ × C(Rn, m−1Rn) and the continuous operator Θ : E CHm(Rn) deﬁned by ∧ → Θ(ω,ζ) = Λ(ω) + dΛ(ζ).

Our representation theorem 21 will prove that Θ is onto.

Lemma 7. The subspace Λ(D m(Rn)) CHm(Rn) is dense. ⊂ Proof. Choose γ CHm(Rn)∗ such that γ(Λ(ω)) = 0 whenever α D m(Rn). Recall ∈ n m n ∗ ∈ m n that the evaluation map Υ : Nm(R ) CH (R ) is a bijection. For all ω D (R ), Υ−1(α), ω = α, Λ(ω) = 0. Hence→ Υ−1(α) = 0 and α = 0. Thus, by Hahn-Banach∈ h i h i theorem [Rud06, Theorem 3.5], Λ(D m(Rn)) is dense in CHm(Rn).

The next proposition is a variation on the closed range theorem in Fréchet spaces; for a demonstration, see [Edw95, section 6]. We will only use the only if implication.

Lemma 8. Let X and Y be Fréchet spaces, and let f : X Y be a continuous linear map. f ∗(Y ∗) is strongly sequentially closed in X∗ if and only→ if f(X) is closed in Y .

n m m n Theorem 21. Let X R . For each α CH (X), there exists ω Cu(X, R ) and ζ C (X, m−1Rn) such⊂ that α = Λ(ω) +∈ dΛ(ζ). ∈ ∧ ∈ u ∧ Proof. We prove the theorem for X = Rn, for the general case is an easy consequence of the extension theorem 46 for charges. We claim that Θ∗(CHm(Rn)∗) is sequentially ∗ m n ∗ ∗ strongly closed in E . Indeed, let (Fi) be a sequence in CH (R ) such that (Θ (Fi)) converges strongly to G E ∗. Proposition 60 shows that Υ : N (Rn) CHm(Rn)∗ is ∈ m → 106 Charges and cohomology a continuous bijection when CHm(Rn)∗is given the bounded weak* topology. Since the strong topology is coarser than the bounded weak* topology, Υ is still continuous when m n ∗ −1 CH (R ) is given the strong topology. We deﬁne Ti := Υ (Fi), we have Θ∗(F ), (ω,ζ) = Θ∗ Υ(T ), (ω,ζ) h i h ◦ i i = Υ, Θ(ω,ζ) h i = Θ(ω,ζ),T h ii = Ti(ω) + ∂Ti(ζ)

∗ Since the sequence (Θ (Fi)) is uniformly bounded on each bounded subset of E , it follows from the boundedness of (ω, 0) E : ω ∞ 1 and (0,ζ) E : ζ ∞ 1 that there exists c > 0 such that{ for all i∈ 1, k k ≤ } { ∈ k k ≤ } ≥ N(Ti) = M(Ti) + M(∂Ti) ∗ ∗ = sup Θ (Fi), (ω, 0) + sup Θ (Fi), (0,ζ) kωk∞≤1h i kζk∞≤1h i

n Suppose there exists no k for which i supp Tii B (0,k). Construct recursively a ∪ ⊂ n m n subsequence (Tϕ(i)) of (Ti) such that supp Tϕ(i) B (0, i) = . There is ωi D (R ) n n ∩ 6 ∅ ∈ such that supp ωi U (0, i + 1) B (0, i), ai := Tϕ(i), ωi > 0 and ω ∞ 1. Let b := max(a−1, ⊂, a−1), and observe\ that |h i| k k ≤ j 1 ··· j B := ω C(Rn, mRn): j 1, ω jb { ∈ ∧ ∀ ≥ | |j ≤ j} is a bounded subset of C(Rn, mRn). Hence B 0 is bounded in E . As (ib ω , 0) ∧ × { } i i ∈ B 0 and Tϕ(i), ωi i, a contradiction appears. Hence, there exists k 1 such that ×{ } |h i| ≥ n ≥ the sequence (Ti) is in Nm,k(B (0,k)). By Proposition 43, there exists a subsequence n n of (Ti) that F ↾ Nm,k(B (0,k))-converges to T Nm,k(B (0,k)). Hence (Ti) converges n ∈ to T in (Nm(R ), FRn,m). Thus lim Fi = lim Υ(Ti) = Υ(T ) by the continuity of Υ. Therefore, G = Θ∗(Υ(T )), and Θ∗(CHm(Rn)) is sequentially strongly closed in Rn. By Lemma 8, the range of Θ is closed. Since Lemma 7 implies that it is dense in CHm(Rn), we therefore obtain Θ(E ) = CHm(Rn).

We brieﬂy mention that charges together with the diﬀerential operator d give birth to a new cohomology theory. Indeed, since d2 = 0, we have a linear complex

0 CH0(X) d d CHn(X) 0. → → · · · → → whose cohomology spaces are referred to as charge cohomology groups. Namely, for any integer m, we denote Bm(X) := d(CHmâ1(X)), Zm(X) := α CHm(X): dα = 0 and Hm(X) = Zm(X)/Bm(X). Hm is a contravariant functor on{ the∈ category of subsets} of a Euclidean space with Lipschitz maps. Indeed, if f : X Y is Lipschitz continuous, the map f # : CHm(Y ) CHm(X) gives rise to a map →Hm(f): Hm(Y ) Hm(X). We hope that charge cohomology→ is suitable to distinguish two Lipschitz non→ equivalent metrics on an inﬁnite torus.

5.5 Charges vanishing at inﬁnity

The paper [PT11] introduces the notion of charges vanishing at infinity. This framework is particularly well suited for the study of charges in Rn in top dimension, because the 5.5 - Charges vanishing at infinity 107 spaces under consideration are endowed with canonical norms. The space of charges n n n vanishing at infinity will be denoted by CH0(R ). It is a proper subspace of CH (R ), n n which contains the space of continuous vector fields, vanishing at infinity C0(R , R ). ∗ ∗ n Let 1 be the Sobolev conjugate exponent 1 := n/(n 1) and BV1∗ (R ) be the space ∗ of functions ϕ L1 (Rn) whose gradient ϕ is a finite vector− valued measure, i.e. ∈ ∇ 1 n n sup ϕdivv : v C (R ; R ), v ∞ 1 < . n c ZR ∈ k k ≤ ∞ The following holds [PT11, Theorem 6.1 and Section 7]

n Theorem 22. Let F be a linear functional on BV1∗ (R ). There is a continuous vector field v : Rn Rn vanishing at infinity such that for all ϕ BV ∗ (Rn) → ∈ 1 F (ϕ) = v d( ϕ) n − ZR · ∇ 1∗ n if and only if F (ϕi) 0 whenever (ϕi) is a sequence weakly converging to 0 in L (R ) satisfying sup ϕ →< — where ϕ denotes the total variation of a function i k∇ ik ∞ k∇ k ϕ BV ∗ (Rn). ∈ 1 A functional F satisfying the hypothesis of Theorem 22 is called a charge vanishing at infinity. A function f Ln(Rn) is a charge vanishing at infinity, and so is the distributional divergence of∈ a continuous vector field vanishing at infinity (see [PT11, Propositions 3.4 and 3.5]). Let us rephrase Theorem 22: there is continuous surjective linear operator div : C (Rn, Rn) CH (Rn), 0 → 0 n where CH0(R ) stands for the space of charges vanishing at infinity. Contrary to the Poisson equation, the nonhomogeneous equation div v = F lacks of uniqueness properties. In the sequel n will be a fixed integer larger than 2. I is the unit segment [0, 1]. n n We denote by C0(R , R ) the space of continuous vector fields vanishing at infinity, i.e continuous maps v : Rn Rn such that for any ε > 0 there exists a compact K Rn → n n ⊆ with v < ε outside K. We endow C0(R , R ) with the supremum norm ∞.A locally| integrable| map ϕ : Rn R is said to have bounded variation if k · k → ∞ n n ϕ := sup ϕdiv g : g C (R , R ), g ∞ 1 < . n c k∇ k ZR ∈ k k ≤ ∞ If X is a subset of Rn we denote by BV(X) the space of maps ϕ L (Rn) of bounded ∈ 1 variation and compact support in X. We endow BV(X) with the norm ϕ BV := k k (X) ϕ Rn + ϕ . k kL1( ) k∇ k ∗ We let BV ∗ (Rn) be the space of maps ϕ L1 (Rn) of bounded variation. We norm it 1 ∈ n n by ϕ BV1∗ (R ) := ϕ . Indeed BV1∗ (R ) is a norm by Gagliardo-Nirenberg-Sobolev || || ||∇n || ||·|| inequality in BV1∗ (R ) (see [PT11, Proposition 2.5]). n Note that BV(X) is a subset of BV ∗ (R ) and ∗ Rn induces a norm equivalent 1 k·kBV1 ( ) to BV in BV(X) whenever X is a Lebesgue measurable set with finite measure k · k (X) and the characteristic function ✶X has bounded variation. However BV(X) has the geometrical meaning of a normal mass. k · k 108 Functional analytical tools

n n We let CH0(R ) be the space of charges vanishing at inﬁnity. CH0(R ) can be made into a Banach space by norming it with

n F Rn := sup F (ϕ): ϕ BV ∗ (R ), ϕ 1 . k kCH0( ) { ∈ 1 k∇ k ≤ } n n n The operator div : C (R , R ) CH (R ) is deﬁned by (div v)(ϕ) := n v d( ϕ). 0 → 0 − R · ∇ R 6 Functional analytical tools

6.1 Duality of vector lattices

A linear space X over R, endowed with a (partial) order relation is called an ordered linear space if the order structure is compatible with the linear structure, that is

(1) for all x, y, z X, x y implies x + z y + z, ∈ (2) for all x, y X, λ 0, x y implies λx λy. ∈ ≥ Generally speaking, a cone in a linear space is a subset C invariant under homothetic maps x λx with λ > 0. If C is closed under addition and C ( C) = 0 , C is called a proper7→ cone . In an ordered linear space X, the set X+ :=∩ x− X :{x} 0 , called the positive cone , is a proper cone. Conversely, for any proper{ cone∈ C in a linear} space X, there exists a unique order structure on X, compatible with the linear structure, for which C is the positive cone. A vector lattice is an ordered linear space such that x y := sup x, y and x y := inf x, y exist for all x, y X. In a vector lattice, we deﬁne∨ for each{ x } X, x+∧:= x 0, x{− :=}x 0, and x := x∈ ( x), respectively called the positive part,∈ the negative part∨ , and the ∧absolute| value| of∨x.− A norm on a vector lattice X is called a lattice norm if for all x, y X, x y impliesk · k x y . A vector lattice, together with a lattice norm, is called∈ |a| normed | | vector lattice.k k ≤ k Ak normed vector lattice is called a Banach lattice when it is norm complete. Now we ﬁx a vector lattice X. An order interval is a subset of the form [x, y] := z X : x z y , for some x, y X. A linear form f : X R is called order bounded{ ∈ whenever f([x,} y]) is bounded in∈ R for any x, y X. The→order dual X∗ is the linear space of all order bounded linear forms. We endow∈ X∗ with the order relation

f g : x X+, f(x) g(x), ⇔∀ ∈ ≤ which turns X∗ into an ordered linear space. One veriﬁes that X∗ is a vector lattice with the following and operations ∨ ∧ f g(x) = sup f(y) + g(z): y,z X+, y + z = x , ∨ { ∈ } f g(x) = inf f(y) + g(z): y,z X+, y + z = x , ∧ { ∈ } whenenver x X+, and f g(x) = f g(x+) f g(x−) for any x X, and a similar formula∈ for f g. When∨ X is a normed∨ vector− lattice,∨ we need to investigate∈ the relationship between∧ the order and topological structures. First note that X+ is closed 6.1 - Duality of vector lattices 109 in X. Indeed, for all x, y X, one has x− y− x y , thus x− y− x y . The map x x− is therefore∈ Lipschitz| continuous,− | | and− we| inferk that−X+kis ≤ closed k − ask the reversed7→ image of 0 under this map. { } Proposition 48. Let X be a Banach lattice. Then a linear form is continuous if and only if it is order bounded. X∗ with its dual order structure described above and its dual norm, is a Banach lattice.

Proof. First we claim that every positive linear form on T : X R is continuous. Indeed, if T is not continuous, then T must be unbounded on the closed→ unit ball B of X. Since B (B X+) (B X+), T is unbounded on B X+. This implies the ⊂ ∩ − ∩ + 3 ∩ existence of a sequence (xn) in B X with T (xn) n for all n N. On the other ∩ −2 k k ≥ −∈2 + hand, X being complete, z := n∈N n xn exists in X, and z n xn for all n, X −2 −2 being closed. Then T (z) n PT (xn) 0 which implies T (z) n T (xn) n. Hence a contradiction. ≥ ≥ k k ≥ k k ≥ Now we claim that every order bounded linear form T is continuous. Indeed T = T + T − and T +, T − are positive, hence continuous. − We proved X∗ X∗. The reverse inclusion is easier to prove; since every order interval in X is bounded,⊂ a bounded linear form is automatically order bounded. Thus X∗ = X∗. Now X∗ is a Banach space as a dual space, and the fact that its dual norm is a lattice norm is only routine.

We now introduce two important classes of Banach lattices and examine their duality. Deﬁnition 7. An abstract M space (brieﬂy, AM-space) is a Banach lattice X such that for all x, y X+, x y = max( x , y ). If the closed unit ball of an AM space contains a largest∈ elementk ∨e, ke is calledk ak unit.k k

For instance, if K is compact Hausdorff topological space and C(K) is the space of continuous real-valued functions defined on K normed by f ∞ := sup f(x) : x K , ordered by f g if and only if f(x) g(x) for all x K,k thenk C(K){| is an AM| -space.∈ } ≥ ∈ Definition 8. An abstract Lebesgue space (briefly, AL-space) is a Banach lattice X such that for all x, y X+, x + y = x + y . ∈ k k k k k k It is straightforward to see that for any measure space (X, S, µ), the Banach space L1(X, S, µ) is an abstract Lebesgue space, hence the name. Theorem 25 will prove that conversely, any abstract Lebesgue space is a concrete Lebesgue space. Proposition 49. The dual of an AM-space is an AL-space, and the dual of an AL-space is an AM-space with unit.

Proof. Suppose X is an AM space. Note that for any positive form h in X∗, one has h = sup h(x): x 1, x X+ . k k { k k ≤ ∈ } Let f, g X∗ be positive forms, and let ε > 0. From the above statement, there exist elements∈x, y X+, such that x = y 1 and f(x) > f ε, g(y) > g ε. We have x ∈y 1, and this impliesk k k k ≤ k k − k k k k − k ∨ k ≤ f + g (f + g)(x y) f(x) + g(y) f + g 2ε. k k ≥ ∨ ≥ ≥ k k k k − 110 Functional analytical tools

By the arbitrariness of ε, f + g f + g . The reverse inequality is the triangle inequality, hence f + g =k f +k ≥g kandk Xk∗ kis an AL-space. k k k k k k On the other hand, let X be an AL-space. We define e : X R by e(x) := x+ x− for all x X. It is obvious that e is continuous positive→ linear form, and kthatk for − k all kx X, e(∈x ) = x . Consequently, e is the largest element of the closed unit ball of X∗∈. Note that| | fork allk f (X∗)+, ∈ f = inf λ 0 : f λe . k k { ≥ } Fix f, g (X∗)+. Let λ 0 such that f g λe. Then λ = λe max( f , g ). Thus, f∈ g max( f ≥, g ). On the opposite,∨ one has f g k max(k ≥ f , kg k)e,k andk so f k g ∨ kmax( ≥ f k, gk k).k Hence X∗ is an AM space. ∨ k k k k k ∨ k ≤ k k k k Proposition 49 yields further examples of abstract L spaces. For any compact Haus- dorff topological space K, the space of signed Radon measures on K, C(K)∗ (following Bourbaki’s definition, see [Bou52, III.1]) is an AL space.

6.2 Representation of abstract M spaces

This subsection is devoted to several theorems on Banach lattices of type C(K), where K is a compact Hausdorff topological space, in particular, the order theoretic Stone- Weierstrass density theorem and a representation theorem for AM-spaces with unit. Theorem 23 (Order theoretic Stone-Weierstrass theorem). Let K be a compact Haus- dorff topological space. Let e the constant 1 function on K. Let X be a vector subspace of C(K), which verifies the following properties:

(1) X is a sublattice of C(K), i.e closed under and . ∧ ∨ (2) X contains e. (3) X separates the points in K, i.e, for any x = y K, there exists f X such that f(x) = f(y). 6 ∈ ∈ 6 Then X is dense in C(K).

Proof. Let x, y be any points in K and α, β real numbers. We make the further assumption that α = β if x = y. There exists f X such that f(x) = α, f(y) = β. This is clear if x = y. If x = y there exists g X∈such that g(x) = g(y) and a suitable linear combination of e and6 g will satisfy the∈ requirement. 6

Now fix h C(K), ε > 0 and x K. For any y K, there exists fy X such that f (x) = h(x) and∈ f (y) = h(y). The∈ set U := z ∈K : f (z) > h(z) ε∈ is open and y y y { ∈ y − } contains y. Hence, the open sets Uy (y K) cover K. The compactness of K implies the existence of a finite subset I K such∈ that X = U . Using the lattice property ⊂ ∪y∈I y of X, we can define a function gx := supy∈I fy X, such that gx(y) > h(y) ε for all y K and g (x) = h(x). ∈ − ∈ x Now consider this procedure applied to each x K; we obtain a family gx : x K in X such that g (x) = h(x) for all x K and∈g (y) > h(y) ε for all{ x, y ∈ K}. x ∈ x − ∈ 6.2 - Representation of abstract M spaces 111

The set Vx := z K : gx(z) < h(z) + ε is open and contains x. Therefore, the compactness of K{ implies∈ the existence of a ﬁnite} subset J K such that K = V . ⊂ ∪x∈J x We deﬁne g := infx∈J gx X. One has g(z) ε

Let X be an AM space with unit e; we let H0 be the intersection of the hyperplane ∗ ∗ H := x X : x, e = 1 with the positive cone of X . H0 is a convex, weak* closed subset{ of∈ the dualh uniti ball} [ e, e]. It follows that H , which is called the positive face − 0 of [ e, e], is weak* compact. It is immediately veriﬁed that f H0 is an extreme point if and− only if for all h X∗, 0 h f implies that h is a scalar∈ multiple of f. Now we can prove Kakutani’s representation∈ theorem for AM spaces with unit.

Theorem 24. Let X be an AM space with unit e, and let K be the set of extreme points of the positive face H0 of the dual unit ball. Then K is non-empty and weak* compact, and the evaluation map ψ : x , x is a Banach lattice isomorphism from X to C(K) in the following sense: ψ is a7→ surjective h· i isometry which preserves the lattice operations.

Proof. Since H0 is convex and weak* compact, it follows from Krein-Milman theorem [Rud06][Theorem 3.21] that H0 is the weak* closure of the convex hull of K. We claim that f H0 is in K if and only if f is a lattice morphism. Indeed, let f K. Let x X and suppose∈ f(x+) > 0. We deﬁne a linear form h by ∈ ∈

h(y) := sup f(z) : 0 z y and z ρx+ for some ρ 0 { ≥ } whenever y X+ and h(y) := h(y+) h(y−) in the general case. It is only routine to prove that∈h is well deﬁned and bounded− (it is enough to see that it is additive and positive homogeneous on X+). It follows that 0 h f and therefore there exists ρ 0 such that h = ρf. Since h(x−) = 0, it follows that f(x−) = 0, thus for all x ≥X, inf(f(x+), f(x−)) = 0, and this statement is readily seen to be equivalent to the requirement∈ that f is a lattice morphism.

Conversely, if f H0 is a lattice morphism, then f is positive. Now let h be a positive linear form in∈ X∗ such that f h is positive, or, equivalently, 0 h f. For any x ker f, one has h(x) h( x ) − f( x ) = f(x) = 0. Thus ker h ker f and there exists∈ ρ R such| that |h ≤= ρf| .| Since≤ h| |is positive,| | ρ 0, hence f is⊂ an extreme ∈ ≥ point of H0. K is consequently the set of lattice morphisms with norm 1. From this, K is weak* closed, hence weak* compact by Banach-Alaoglu theorem. The mapping ψ is clearly a linear map from X to C(K) that preserves the lattice operations, since each f K is a lattice morphism. To show that ψ is an isometry, it is enough to show that ∈x = ψ(x) when x X+, since X and C(K) are Banach lattices. For x X+, wek havek k k ∈ ∈ x = sup f, x : f X∗, f 1 k k {h i ∈ k k ≤ } = sup f, x : f H . {h i ∈ 0} 112 Functional analytical tools

Since H0 is the closed convex hull of K and for each x X, , x deﬁnes a weak* continuous linear form on X∗, it follows that ∈ h· i

x = sup f, x : f H k k {h i ∈ 0} = sup f, x : f K h i ∈ = ψ(x) . k k Thus ψ deﬁnes an isomorphism of X onto a complete vector sublattice ψ(X) C(K), which contains the unit ψ(e) of C(K). It is straightforward that ψ(X) separates⊂ points in K, it follows from Theorem 23 that ψ(X) is dense in C(K). Hence ψ is onto.

6.3 Representation of abstract L spaces

We are looking forward to proving Kakutani’s representation theorem 25 for abstract L spaces. Given a measure space (X, S, µ), there is not necessarily a constant 1 function e in L1(X, S, µ), unless the measure µ is ﬁnite. For this reason, we need to develop some technical machinery. Let X be a vector lattice. A subset S X+ is called an orthogonal system if 0 = S and x y = 0 for each pair of distinct⊂ elements of S.A weak order unit is an element6 ∈ e X∧such that e is a maximal orthogonal system. This means that for all x X, x ∈e = 0 implies x{=} 0. ∈ ∧ Deﬁnition 9. A subset A of a vector lattice X is called solid if x A, y X and y x implies y A. An ideal is a solid vector subspace of X. ∈ ∈ | | | | ∈

Each ideal is a sublattice of X (i.e, closed under and ). For all e X, we ∨ ∧ ∈ deﬁne Xe to be the ideal generated by e , or equivalently, the intersection of all ideals containing e. Obviously, one can assume{ }e X+ by replacing it with e . In that case, ∈ | | one readily checks that X = Nn[ e, e]. e ∪n∈ − Deﬁnition 10. An element e 0 of a normed vector lattice X is called a quasi-interior + point of X if the principal ideal Xe is dense in X.

Proposition 50. Let X be a Banach lattice. For each e 0, the principal ideal X is e an AM space under the norm pe whose closed unit ball is the order interval [ e, e], with unit e, and the canonical imbedding X X is continuous. − e →

Proof. pe is the norm given by

p (x) = inf λ 0 : x λ[ e, e] . e { ≥ ∈ − } One routinely veriﬁes that p is a lattice norm and that for all x, y X X+, x y = e ∈ e ∩ k ∨ k max( x , y ). Now we prove that (Xe, pe) is a Banach space. Let (xn) be a pe Cauchy k k k k −n sequence. There exists a subsequence (xϕ(n)) such that xϕ(n+1) xϕ(n) 2 e for all + | − | − n N. Let us deﬁne un := (xϕ(n+1) xϕ(n)) and vn := (xϕ(n+1) xϕ(n)) . We have 0 ∈ u v 2−ne. Since (E, ) is− a Banach space, the series − u converges to n ∨ n k · k n∈N n P 6.3 - Representation of abstract L spaces 113 a point u X and ∈ m ∞ u u = u − n n nX=1 n=Xm+1 ∞ 2−ne = 2−me. n=Xm+1 Hence we infer m −m pe u un 2 , − ! ≤ nX=1 thus un pe converges. One obtains similar results for the series vn, therefore (xϕ(n)) pe convergesP to u v. Finally, the continuity of the inclusion mapP Xe X can be derived from the inequality− e p . → k · k ≤ k k e

We will prove that weak order units in an AL space are quasi-interior.

Lemma 9. Let X be a Banach lattice and let S denote any maximal orthogonal system + of X. For each x X , the net (xn,H ) where n N, H is a ﬁnite subset of S and x := x nu∈ order converges to x, i.e x = sup∈ x . n,H u∈H ∧ n,H n,H In particular,P if e is a weak order unit in X, then for any x X+, x = sup (x ne). ∈ n ∧

Proof. It is clear that x x for each n N and each ﬁnite H S. Suppose n,H ∈ ⊂ that z xn,H for all n and H; we have to show that z x. Fix u S. Then 0 z (x nu) = (z x) (z nu) which implies ∈ − ∧ − ∨ − 0 = ((z x) (z nu)) 0 = (z x)− (z nu)−, − ∨ − ∧ − ∧ − and so (z x)− (u n−1z)+ = 0 − ∧ − −1 −1 + for all n N. Now supn(u n z) = u. Since u is positive, sup(u n z) = u. Thus, we have ∈ − − sup (z x)− (u n−1z)+ = (z x)− u = 0. n − ∧ − − ∧ By the arbitariness of u S, it follows that (z x)− = 0 since S is a maximal orthogonal system. Hence z x. ∈ − Lemma 10. Let X be an AL space. Every increasing bounded net in X converges.

Proof. Let (xα)α∈A be a increasing net in X, bounded by a constant M. If (xα)α∈A is not a Cauchy net, there would exist a number ε > 0 and an inﬁnite subsequence

(xαn )n∈N (with αn+1 αn for n N) such that xαn+1 xαn > ε for all n N. We have ≥ ∈ k − k ∈ n+1 nε x x = x x 2M ≤ k αk − αk−1 k k αn+1 − α1 k ≤ kX=2 for all n N, which is contradictory. Hence (xα)α∈A is a Cauchy net, thus it converges for X is complete.∈ 114 Functional analytical tools

Proposition 51. Let X be an AL space, and let S be a maximal orthonormal system in X. The ideal generated by I is dense in X. In particular, every weak order unit in X is quasi-interior.

Proof. It is an easy consequence of Lemmas 9 and 10.

Now we can state and prove the most important theorem of this section.

Theorem 25. Let X be an AL space, there exists a locally compact Hausdorﬀ space Y and a positive Radon measure µ on Y such that X is isomorphic to L1(Y, µ). Y can be chosen to be compact if and only if X possesses a weak order unit.

Proof. First we consider the case where X has a weak order unit e. Then Xe is a dense ideal of X by 51. By Proposition 50 and Theorem 24, (Xe, pe) can be identified with C(K), for some compact Hausdorff topological space K. The map µ(x) := x+ x− defines a positive linear form µ on X for which x = µ( x ) for all xk Xk −. k Thek k k | | ∈ restriction of µ to Xe defines a positive Radon measure on K (which we again denote by µ). Since C(K) is dense in L1(K, µ), it is clear that the isomorphism (Xe, ) (C(K), ) extends uniquely to an isomorphism of Banach lattices X Lk( ·K, k µ→). k·kL1(K,µ) → 1 Now suppose E possesses no weak order unit and denote by S := eα : α A a maximal orthogonal system. Such a system exists by Zorn’s lemma. We let{ I be the∈ ideal} generated by S. By 51, I is dense in E. Moreover, I is the algebraic direct sum X . ⊕α∈A eα By Proposition 50 and Theorem 24, (Xeα , peα ) can be identified with C(Kα) for some compact Hausdorff topological space Kα. Thus, I can be identified with Cc(Y ), the space of compactly supported continuous real-valued map on the locally compact Hausdorff + − direct topological sum Y := α∈A Kα. Once again, we define µ(x) := x x for all x I. µ defines a positive Radon measure on Y for which x =k kx −dµ k fork all ∈ ` k k Y | | x I. The isomorphism (I, ) (Cc(Y ), L1(Y,µ)) extends to an isomorphismR of Banach∈ lattices E L (Y, µk). · k → k · k → 1

6.4 p absolutely summing operators

We introduce the class of p operators in Banach spaces. They are similar to Hilbert- Schmidt operators in a Hilbert space. In fact, if H is a Hilbert space, an operator T : H H is Hilbert-Schmidt if and only if it is p absolutely summing for some 1 p <→ . A weaker result is proved in Proposition 54. ≤ ∞ Deﬁnition 11. Let X, Y be Banach spaces, and 1 p < . A linear operator T : X Y is called p absolutely summing if there exists≤ a constant∞ C 0 such that for → n ≥ all choices of (xk)k=1 in X we have

p 1/p n 1/p p p ∗ T (xk) C sup ξ, xk : ξ X , xi 1 . k k ! ≤  |h i| ! ∈ k k ≤  kX=1  kX=1    The least such constant C is denoted πp(T ) and is called the p absolutely summing norm of T . 6.4 - p absolutely summing operators 115

In the limit case p = , an absolutely operator is just a bounded operator, and its absolutely summing norm∞ is∞ nothing but its operator norm. The next proposition shows that the class of p absolutely summing operators is an ideal class. The proof is only routine.

Proposition 52. Let U : W X, T : X Y , V : Y Z be bounded operators between Banach spaces. If T is p→absolutely summing,→ then VTU→ is p absolutely summing and π (VTU) V π (T ) U . p ≤ k k p k k We recall that a Hilbert Schmidt operator on a Hilbert space H is a bounded operator T : H H such that there exists a Hilbert basis (e ) for which → i i∈I T 2 := T (e ) 2 < . k kHS k i k ∞ Xi∈I

T HS is called the Hilbert-Schmidt norm and does not depend on the choice of the Hilbertk k basis.

Proposition 53. Suppose 1 r < p < . Let T be an r absolutely summing operator between Banach spaces X and≤Y . Then ∞T is p absolutely summing and π (T ) π (T ). p ≤ r

p/r−1 p Proof. Choose x1, , xn X and observe that if λk := T (xk) , then T (xk) = T (λ x ) r. Since···T is r absolutely∈ summing, we have k k k k k k k k n 1/r n 1/r p r T (xk) = T (λkxk) k k ! k k ! kX=1 kX=1 n 1/r r r ∗ πr(T ) sup λk ξ, xk : ξ X , ξ 1 . ≤  |h i| ! ∈ k k ≤   kX=1  By Hölder inequality with the conjugate indices p/(p r) and p/r, we have  − n 1/r n (p−r)/pr n 1/p r r pr/(p−r) p λk ξ, xk λk ξ, xk . |h i| ! ≤ ! |h i| ! kX=1 kX=1 kX=1 Thus, one easily infers

n 1/p n 1/p p p ∗ T (xk) πr(T ) sup ξ, xk : ξ X , ξ 1 . k k ! ≤  |h i| ! ∈ k k ≤  kX=1  kX=1   

Proposition 54. Let H be a Hilbert space, and let T : H H be a bounded operator. T is a Hilbert-Schmidt operator if and only if T is 2-absolutely→ summing.

Proof. Suppose T is 2-absolutely summing. Let (ei)i∈I be a Hilbert basis in H. There exists C 0 such that for each ﬁnite subset J I, ≥ ⊂

T (e ) 2 C2 sup ξ, e 2 : ξ H, ξ 1 C2. k j k ≤  |h ji| ∈ k k ≤  ≤ jX∈J jX∈J    116 Functional analytical tools

Hence T (e ) 2 C2 < . i∈I k i k ≤ ∞ LetP us prove the opposite implication. If T is a Hilbert-Schmidt operator, it is compact. Hence T ∗T is compact and self-adjoint. We can therefore exhibit a Hilbert ∗ ∗ basis (ei)i∈I consisting of T T eigenvectors; there exist si 0 such that T T (ei) = n ≥ siei. Let (xk)k=1 be a H valued ﬁnite sequence; we write xk := i∈I ak,iei for any k 1, , n . ∈ { ··· } P n n T (x ) 2 = T ∗T (x ), x k k k h k ki kX=1 kX=1 n = ak,isiei, ak,iei * + kX=1 Xi∈I Xi∈I n 2 = ak,isi kX=1 Xi∈I n 2 = si ak,i . ! Xi∈I kX=1 However, since we have n n a 2 = e , x 2, | k,i| h i ki kX=1 kX=1 we deduce that n n 2 2 T (xk) si sup ξ, xk : ξ H, ξ 1 . k k ≤ ! ( h i ∈ k k ≤ ) kX=1 Xi∈I kX=1

As i∈I si is the square of the Hilbert-Schmidt norm of T , and so is ﬁnite. Hence we canP conclude.

6.5 Lp spaces

Let 1 p . We denote ℓn the space Rn normed by ≤ ≤ ∞ p n 1/p p x p := xi k k | | ! Xi=1 if p < and x := max x . ∞ k k∞ i | i| Deﬁnition 12. Let X and Y be Banach spaces. The (multiplicative) Banach-Mazur distance between X and Y is deﬁned to be

d(X,Y ) := inf T T −1 , k kk k where the infimum is taken among all isomorphisms T : X Y . In particular, if X and Y are not isomorphic, d(X,Y ) = . → ∞ Definition 13. Let λ > 1. A Banach space X is said to be an Lp,λ space if for all finite dimensional subspace E X, there exists a finite dimensional subspace F X which ⊂ ⊂ contains E and d(F, ℓdim F ) λ. X is called an L space if there exists λ > 1 such that p ≤ p X is an Lp,λ space. 6.5 - Lp spaces 117

Proposition 55. Let (X, S, µ) be any measure space and 1 p . L (X, S, µ) is ≤ ≤ ∞ p an Lp,λ space for all λ > 1.

Proof. Choose E Lp(X, S, µ) a ﬁnite dimensional subspace and let (f1, , fn) be a basis of E, such that⊂ f = 1 for all k 1, , n . Let α > 0. There··· exist simple k kkp ∈ { ··· } functions g1, , gn such that fk gk α for all k 1, , n . There exists a partition of X···in S measurablek sets− Xk ≤, ,X such∈ that { ··· each }g is constant on { 1 ··· m} k Xj, for 1 j m. In other words, if we denote by F1 the subspace generated by the ≤ ≤ ✶ characteristic maps Xj , 1 j m and by E1 the subspace generated by the maps gk, 1 k n, then E F . F≤ is≤ clearly isometric to ℓm. ≤ ≤ 1 ⊂ 1 1 p Since all n dimensional spaces are isomorphic, there exists a constant C > 0 such that for all f E, f := n a f , we have ∈ k=1 k k P n n akgk f = ak(gk fk) − − k=1 p k=1 p X X n α a ≤ | k| kX=1 αC f ≤ k kp Hence if α is small enough, that is α < C−1, since n (1 αC) f p akgk (1 + αC) f p, − k k ≤ ≤ k k k=1 p X fk gk deﬁnes a isomorphism between E and E1. 7→ For all a , . . . , a R, for all l 1, , n , we have 1 n ∈ ∈ { ··· } n a a | l| ≤ | k| kX=1 C x ≤ k kp C n akgk ≤ 1 αC k=1 p − X

Hence the linear map E1 R, gk 0 if k = l and gl 1 is continuous with norm → 7→ 6 7→ ∗ C/(1 αC). By Hahn-Banach theorem, there exists an extension φl F1 such that ≤φ C/− (1 αC). We deﬁne T : F L (X, µ) by ∈ k lk ≤ − 1 → p n T (f) := f + φ (f)(f g ). k k − k kX=1 It is straightforward to see that T (gk) = fk, thus F := T (F1) contains E. Now for any f F , T (f) f nαC(1 αC)−1 f , thus ∈ 1 k − kp ≤ − k kp nαC nαC 1 f T (f) 1 + f − 1 αC k kp ≤ k kp ≤ 1 αC k kp − − If α is small enough, then T deﬁnes an isomorphism from F1 to F , and computing T , T −1 , we see k k k k nαC nαC −1 d(F, F ) 1 + 1 1 ≤ 1 αC − 1 αC − − For any λ > 1, choosing α small enough, the right-hand side can be λ. ≤ 118 Functional analytical tools

Theorem 26. Let K be a compact Hausdorﬀ topological space. For any λ > 1, C(K)∗ is an L1,λ space.

Proof. It is a straightforward consequence of the results of section 6.3 : C(K)∗ is an abstract L space and by Proposition 55 we infer that it is an L1,λ space for all λ > 1. Here, we give an alternative and somewhat more elementary proof which does not use any Banach lattice argument. Recall that C(K)∗ is the space of signed Radon measure on K normed by the total variation. Fix λ > 1. Let E C(K)∗ be a ⊂ ﬁnite dimensional subspace generated by a family µ1, . . . , µn. Each Radon measure µi (1 i n) is absolutely continuous with respect to µ := µ1 + + µn . The embedding≤ ≤ι : L (K, µ) C(K)∗, f fµ is an isometry. We denote| | by···f | L|(K, µ) 1 → 7→ i ∈ 1 the Radon-Nikodým derivative of µi with respect to µ. By Proposition 55, there exists a ﬁnite dimensional subspace F L1(K, µ) which contains fi (1 i n), such that d(F, ℓm) λ, where m := dim F .⊂ Then d(ι(F ), ℓm) λ and E ι(≤F ).≤ Hence C(K)∗ is 1 ≤ 1 ≤ ⊂ an L1,λ space.

6.6 Grothendieck theorem

Theorem 27 (Grothendieck inequality). There exists a universal constant KG, called p,m Grothendieck constant so that for any m, p > 0, any real matrix (ajk)j,k=1, having the p m property that for any ﬁnite real-valued sequences (sj)j=1, (tk)k=1 one has

p m ajksjtk max sj max tk , j k j=1 k=1 ≤ | | | | X X

p m then for any Hilbert space H and any H valued ﬁnite sequences (uj)j=1, (vk)k=1, we have

p m ajk uj, vk KG max uj max vk . j k j=1 k=1 h i ≤ k k k k X X

Proof. The proof presented here is taken from [Ble87]. It is very elementary compared to other proofs, but we eventually obtain a bad upper bound for the Grothendieck constant. Let us denote R(N) the space of real-valued sequences with ﬁnitely many nonzero terms. We equip R(N) with the inner product

x, y := x(n)y(n). h i nX∈N

Since the vectors u1, . . . , up, v1, . . . , vm live in a ﬁnite dimensional subspace of H isomet- N N (N) ric to some ℓ2 and there is an isometry ℓ2 R , we can restrict ourselves to the case H = R(N) (though this is not a Hilbert space).→ Note that we can easily assume m = p, completing the matrix (ajk) and the vectors (uj), (vk) with zeros if necessary. We deﬁne A : R(N) R(N) R by × → A(x, y) := (1 + x(n)y(n)) . nY∈N 6.6 - Grothendieck theorem 119

We have the following estimate

Let (Zn)n∈N be a sequence of independent real-valued random variables on some prob- 2 ability space such that E(Zn) = 0, E(Zn) = 1 and Zn = 1 almost surely. (Such a sequence of independent variables is provided by the| Rademacher| functions deﬁned on the unit interval [0, 1] with the Lebesgue measure, rn(x) := (1 + sign(sin 2nπx))/2). Given x R(N), we deﬁne a complex valued random variable ∈

F (x) := (1 + iZnx(n)) . nY∈N Almost surely, we have

1/2 2 F (x) (1 + x(n)2) = A(x, x)1/2 ekxk /2. | | ≤   ≤ nY∈N   For any x, y R(N), by independence of the random variables Z , ∈ n E(F (x)F (y)) = E ((1 + iZ x(n))(1 iZ y(n))) n − n nY∈N = E 1 + iZ (x(n) y(n)) + Z2x(n)y(n) n − n nY∈N = A(x, y)

(N) m m Thus, for any R valued ﬁnite sequences (xj)j=1 and (yk)k=1 which lie in the unit ball of R(N), we have

m m m m a A(x , y ) = a E(F (x )F (y )) jk j k jk j k j=1 k=1 j=1 k=1 X X X X

m m E a E(F (x )F (y )) jk j k ≤ j=1 k=1 X X

max F (xj) max F (yk) ≤ j | | k | | e ≤

Let (EJ )J≥2 be an inﬁnite partition of N in inﬁnite subsets. Let WJ be the J dimensional wedge of NJ given by

W := (n , . . . , n ) NJ : n > > n , J { 1 J ∈ 1 ··· J } and set up a one-to-one correspondance between EJ and WJ ,(J 2): αJ : EJ WJ . (N) ≥ → (N) Given an arbitrary x R , x 1, we deﬁne a vector φ(x) := (φ(x)(n)) N in R ∈ k k ≤ n∈ 120 Functional analytical tools setting φ(x)(n) := x(n1) x(nJ ) whenever n EJ and (n1, . . . , nJ ) := αJ (n). Once again, we estimate ··· ∈

1/2 ∞ φ(x) = (x(n ) x(n ))2 k k  1 ··· J  JX=2 nX∈WJ  J 1/2  ∞ 1 x(n)2 ≤  J!    JX=2 nX∈N      2 1/2  ekxk 1 x 2 ≤ − − k k (e 2)1/2 ≤ − < 1.

1/2 −1 (N) Let us deﬁne δ := (e 2) and φδ(x) = δ φ(x), for all x R , x 1. Then φ (x) 1. Fix x, y −in the unit ball of R(N) and expand A(x,∈ y): k k ≥ k δ k ≤

A(x, y) := 1 + x, y + + x(n1) x(nJ )y(n1) y(nJ ) + h i ··· n >···>n ··· ··· ··· 1 X J 2 Therefore x, y = A(x, y) 1 δ φδ(x), φδ(y) . Applying this formula recursively, we obtain forh eachiJ > 0 − − h i

J 2 l l l 2 J+1 J+1 J+1 x, y = ( δ ) (A(φδ(x), φδ(y)) 1) + ( δ ) φδ (x), φδ (y) . h i − − ! − h i Xl=0 Finally, letting J , we deduce → ∞ ∞ x, y = ( δ2)l(A(φl (x), φl (y)) 1) h i − δ δ − Xl=0

Now, we can establish Grothendieck inequality. We are given u1, . . . , um, v1, . . . , vm R(N) and by a homogeneity argument, we can assume all these terms to lie in the unit∈ ball of R(N). Then

m m ∞ m m a u , v δ2l a (A(φl (u ), φl (v )) 1) jk j k jk δ j δ k j=1 k=1 h i ≤ l=0 j=1 k=1 − X X X X X ∞

δ2l( e + 1) ≤ Xl=0 e + 1 = 3 e − The proof is ﬁnished and K (e + 1)/(3 e). G ≤ −

Theorem 28 (Grothendieck theorem). Let X be an L1,λ space, with λ > 1, H a Hilbert space. Every bounded operator T : X H is 1-absolutely summing and π1(T ) K λ T . → ≤ G k k

m Proof. First of all, we claim that for any m 1 and any bounded operator u : ℓ1 H, π (u) K u . Indeed, let u be such an operator.≥ Choose x , , x ℓm such→ that 1 ≤ Gk k 1 ··· p ∈ 1 6.6 - Grothendieck theorem 121

m p m m for all ξ ℓ∞, j=1 ξ, xj ξ ∞; write xj := k=1 ajkek, where (ek)k=1 denotes the ∈ m|h i| ≤ k k p m standard basisP in ℓ1 . For all s := (s1, , sp) PR and t := (t1,...,t m) R , note that ··· ∈ ∈

p m p m a s t s a t jk j k j jk k j=1 k=1 ≤ j=1 | | k=1 X X X X p

s t, x ≤ k k∞|h ji| jX=1 s t ≤ k k∞k k∞ Hence by Theorem 27, for any y , . . . , y H, we have 1 p ∈ p m a y , u(e ) K sup y sup u(e ) jk j k G j k j=1 k=1 h i ≤ j k k k k k X X

KG u sup yj . ≤ k k j k k In particular, if we choose y := u(x ) −1u(x ) if u(x ) = 0 (and y is any vector in the j k j k j j 6 j unit ball of H if u(xj) = 0) then

p p u(x ) = y , u(x ) k j k h j j i jX=1 jX=1 p m = a y , u(e ) jkh j k i jX=1 kX=1 K u ≤ Gk k Thus π (u) K u . 1 ≤ Gk k Now we turn back to the main result. Choose a finite family x1, . . . , xn in X and denote E the finite dimensional subspace it generates. There exists a finite dimensional m m subspace F X which contains E and such that d(F, ℓ1 ) λ. Let α : ℓ1 H be an isomorphism⊂ such that α α−1 λ. Up to normalizing,≤ we can suppose→ −1 k kk k ≤ m α = 1 and α λ. By the above claim, the π1 norm of T α : ℓ1 H verifies kπ (kT α) K kT α k ≤ T . Thus, → 1 ≤ Gk k ≤ k k n n T (x ) = T α(α−1(x )) k k k k k k kX=1 kX=1 n −1 ∞ KG T sup ξ, α (xk) : ξ ℓm , ξ ∞ 1 ≤ k k ( |h i| ∈ k k ≤ ) kX=1 n −1∗ ∞ = KG T sup α (ξ), xk : ξ ℓm , ξ ∞ 1 k k ( |h i| ∈ k k ≤ ) kX=1 n ∗ λKG T sup ζ, xk : ζ F , ξ ∞ 1 ≤ k k ( |h i| ∈ k k ≤ ) kX=1 n ∗ = λKG T sup ζ, xk : ζ X , ξ ∞ 1 k k ( |h i| ∈ k k ≤ ) kX=1 where the last equality follows from Hahn-Banach theorem. We conclude that T is 1-absolutely summing and π (T ) λK T . 1 ≤ Gk k 122 Functional analytical tools

6.7 A splitting theorem

In this part, we will present a method to linearize some maps between Banach space. A first attempt would be to differentiate them. However, even differentiation (at one point) of Lipschitz maps between most non reflexive Banach spaces is difficult to obtain, and false in general. Instead we will use the method of invariant means, which applies to general uniformly continuous functions. An invariant mean is a sort of generalized integral functional “with respect to a translation invariant probability measure”. The material here is fully exposed in [BL00][Chapter 7]. Definition 14. Let X be a Banach space. An invariant mean on X is a linear map M : ℓ (X) R such that ∞ → (1) M(1) = 1,

(2) M = 1, k k (3) M(x f(y + x)) = M(f) for every y X. 7→ ∈ We will prove that invariant means exist in all Banach space in Proposition 56. Actually, we can prove the existence of (left) invariant on every (possibly noncommuta- tive) semigroup (replacing the topological condition (2) by the equivalent requirement M(f) 0 whenever f 0). Though, we shall need invariant means only on Ba- nach spaces.≥ If Z∗ is a dual≥ space, one can construct a vector-valued invariant mean ∗ ∗ Mˆ : ℓ∞(X; Z ) Z in the same way the Pettis integral is defined from the Lebesgue integral. Indeed,→ define Mˆ (f),z := M (x f(x),z ) h i 7→ h i for every z Z. Mˆ trivially satisfies the requirements (1), (2) and (3) of Definition 14. ∈ Proposition 56. Let X be a Banach space. Then X admits an invariant mean.

∗ ∗ Proof. Let W be the subset of ℓ∞(X) consisting of the linear forms ϕ ℓ∞(X) such that ∈

(1) ϕ, 1 = 1, h i (2) ϕ, f 0 whenever f 0. h i ≥ ≥ The set W ∗ is convex and weak* compact. For any x X and ϕ W , we deﬁne ∈ ∈ T (ϕ), f = ϕ, f(x + ) . h x i h · i

Trivially, Tx(W ) W and Tx is weak* continuous. We claim that the set Fx := ϕ : T (ϕ) = ϕ is empty.⊂ To see this, fix any ϕ W and define { x } ∈ 1 n ϕ = T k(ϕ). n n x kX=0 It follows from 1 T (ϕ ) ϕ = (T n(ϕ) ϕ) x n − n n x − 6.7 - A splitting theorem 123 that any accumulation point of (ϕ ) is a fixed point of T . Hence F = . n n x x 6 ∅ By an easy induction argument and the fact that the maps Tx commute with each other, x∈F Fx = for any finite subset X. And since W is weak* compact, F ∩= . Now6 ∅ choose M F . F ⊂ ∩x∈X x 6 ∅ ∈ ∩x∈X x Theorem 29. Let X and Y be Banach spaces, let X0 be a closed subspace of X. Let S : X0 Y be a bounded linear operator, and assume that f : X Y is a uniformly → → ∗∗ continuous function such that f ↾ X0 = S. Then there is a linear operator T : X Y such that →

(1) T ↾ X0 = S, (2) T osc(f; 1). k k ≤ ∗∗ ∗∗ Proof. We will consider Y as a subspace of Y . Let M0 and M be Y valued invariant means on respectively X0 and X. Since f is uniformly continuous, the function h : y f(z + y) f(y) z 7→ − is bounded for each z X. More precisely, supy hz(y) osc(f; z ). Thus g(z) := M (h ) is well deﬁned.∈ Furthermore, for every z,zk′ X,k ≤ k k 0 z ∈ ′ g(z) g(z ) = M (h h ′ ) k − k k 0 z − z k sup hz(y) hz′ (y) ≤ y∈X0 k − k = sup f(z + y) f(z′ + y) y∈X0 k − k osc(f; z z′ ). ≤ k − k Now somez ˜ X . By invariance of translation in the X direction and the fact that ∈ 0 0 f ↾ X0 is linear, we get

g(z +z ˜) = M0(hz+˜z) = M (x f(z +z ˜ + x) f(˜z + x)) + M (x f(˜z + x) f(x)) 0 7→ − 0 7→ − = M0(hz) + M0(f(˜z)) = g(z) + f(˜z) = g(z) + S(˜z). (53) In particular, g(˜z) = S(˜z) by taking z = 0. The map T (z) := M (x g(z + x) g(x)) 7→ − is well deﬁned because we noticed g inherited the uniform continuity of f. By invariance of M, T (z + z′) = M (x g(z + z′ + x) g(z′ + x)) + M (x g(z′ + x) g(x)) 7→ − 7→ − = T (z) + T (z′) thus T is additive. Moreover, we easily have that sup T (z) : z 1 osc(g; 1) osc(f; 1). {k k k k ≤ } ≤ ≤ From this and the additivity of T , we infer that T is linear. Finally, by (53), g(x +z ˜) g(x) = S(˜z) for eachz ˜ X and x X. Therefore T (˜z) = S(˜z). − ∈ 0 ∈ 124 Continuous representation operators 7 Continuous representation operators

7.1 Main theorem

We will need this following theorem, which is of independent interest. Let us recall that n 2. ≥ n n Theorem 30. Neither BV(I ) nor BV1∗ (R ) is complemented in an L1 space. In particular none of these is itself an L1 space.

Proof. The following proof is inspired from [BB03, second proof of Proposition 2]. Let n L be an L1 space containing BV(I ). Suppose there exists a continuous left inverse π n n 1,1 n to the inclusion map ι : BV(I ) L. Let f : L2((0, 1) ) W ((0, 1) ) be the linear map such that for any multi-index→α Zn, → ∈ cos(2πα x) sin(2πα x) f : cos(2πα x)) · , sin(2πα x) · . · 7→ 1 + α 2 · 7→ 1 + α 2 | | | | q q f is easily seen to be continuous since it factors through W 1,2((0, 1)n):

f : L ((0, 1)n) W 1,2((0, 1)n) incl. W 1,1((0, 1)n) 2 −→ −→ n n Let g : L1∗ ((0, 1) ) L2((0, 1) ) be the bounded (by Sobolev embedding theorem) multiplier operator → cos(2πα x) sin(2πα x) g : cos(2πα x) · , sin(2πα x) · , · 7→ (1 + α )n/2−1 · 7→ (1 + α )n/2−1 | | | | for every α Zn. We have the following commutative diagram : ∈ Ψ

( n n L2((0, 1) ) L L2((0, 1) ) O O f ι π g 1,1 n ι1 n ι2 n W ((0, 1) ) / BV(I ) / L1∗ ((0, 1) )

1,1 n n ι1 is the inclusion map W ((0, 1) ) BV(I ). ι2 is the Sobolev inclusion map n n ⊂ BV(I ) L1∗ ((0, 1) ), and Ψ := gι2ι1f. The boundedness of the inclusion map ι1 is guaranteed⊂ by the boundedness of the extension by zero operator for BV maps (see [EG92, Section 5.4, Theorem 1]). The linear map gι2π maps an L1 space into a Hilbert space; we infer from Grothendieck theorem that it is 1-absolutely summing. n n So is Ψ : L2((0, 1) ) L2((0, 1) ) by virtue of Proposition 52. Thus it is 2-absolutely summing according to→ Proposition 53. Then by Proposition 54, it is a Hilbert-Schmidt operator. However we have 1 2 2 πα x 2 πα x 2 Ψ HS = Ψ(1) + n Ψ(cos(2 )) L2 + Ψ(sin(2 )) L2 k k k k 2 n k · k k · k α∈ZX\{0} 1 1 = 1 + n−1 2 n−2 = + , 2 n (1 + α )(1 + α ) ∞ α∈ZX\{0} | | | | 7.1 - Main theorem 125 and this is contradictory.

Now let us remark that BV ∗ (Rn) contains BV(In) and the norm induced by 1 k · n n BV1∗ (R ) is equivalent to BV(I ) (by Gagliardo-Nirenberg-Sobolev inequality). Let k k · k n n ✶In be the characteristic map of I . f ✶In f is a linear retraction from BV1∗ (R ) to BV(In), and it is bounded (see [EG92,7→ Section 5.4, Theorem 1] to justify this step). n n n Thus BV(I ) is complemented in BV1∗ (R ) and by the above part, BV1∗ (R ) cannot be complemented in an L1 space.

This result is false if n = 1. In fact, the map f (f(0), f ′) deﬁnes an isomorphism 7→ between BV [0, 1] and R C[0, 1]∗. R C[0, 1]∗ is a product of L spaces, and thus is an × × 1 L1 space. We can now prove the main theorem of this paper. If there exists a uniformly continuous right inverse s to div, then v v s(div v) is a uniformly continuous n n 7→ − retraction of C0(R , R ) onto its subspace ker div. Such a retraction is forbidden by the following result: n n Theorem 31. There is no uniformly continuous retraction from C0(R , R ) onto ker div.

Note that if we replace “uniformly continuous” with “bounded linear”, this theorem is an easy consequence of the preceding result. Indeed, a bounded linear retraction r : C (Rn, Rn) ker div would split the exact sequence 0 → 0 ker div C (Rn, Rn) div CH (Rn) 0. −→ −→ 0 −→ 0 −→ Therefore the dual sequence

0 CH (Rn)∗ C (Rn, Rn)∗ div (ker div)∗ 0 −→ 0 −→ 0 −→ −→ n ∗ n would also split. The key observation is that CH0(R ) is in fact isomorphic to BV1∗ (R ) n n n ∗ (see [PT11, Remark 5.2]). BV1∗ (R ) is complemented in the L1 space C0(R , R ) (see the following proof), and this is contradictory. Now, if we suppose there is a uniformly continuous retraction r, we wish to linearize r to obtain a contradiction. We want to apply the results of subsection 6.7. However ker div is not a dual space (otherwise, since it is separable, it would be a Radon-Nikodým space by Dunford-Pettis theorem, but the space c0 of convergent sequences is embeddable in ker div). This leads to a small complication in the proof below: we will embed ker div into its bidual (ker div)∗∗.

Proof. For sake of brevity, we denote ker div by Z. One argues by contradiction, assuming the existence of a uniformly continuous retraction r. By theorem 29, there exists a n n ∗∗ bounded linear map S : C0(R , R ) Z whose restriction to Z is the evaluation map ∗∗ → evZ : Z Z → ∗ ∗ evZ (z)(z ) := z (z). ∗ ∗∗∗ n n We let evZ∗ be the evaluation map Z Z and ι : Z C0(R , R ) be the inclusion map, and consider the following commutative→ diagram →

∗ ∗ n ∗ div n n ∗ ι ∗ 0 / CH0(R ) / C0(R , R ) // Z / 0 f O ∗ ∗ evZ evZ S∗ Z∗∗∗ 126 Continuous representation operators

∗ ∗ ∗ ∗ ∗ The map S evZ is a right inverse to ι , because one easily checks that evZ evZ = ◦ n ∗ ◦ idZ∗ . Therefore, the exact sequence splits linearly, and CH0(R ) is complemented in n n ∗ n ∗ n C0(R , R ) . Since CH0(R ) is isomorphic to BV1∗ (R ), it remains to be proven that n n C0(R , R ) is an L1 space. To do so, remark that the n sphere Sn is the Alexandroﬀ compactiﬁcation of Rn, so n n n n C0(R , R ) is isomorphic to a (closed) hyperplane of C(S , R ). This space is isomorphic to its hyperplanes (it is an immediate consequence of [AK06, Proposition 4.4.1]), thus

C (Rn, Rn)∗ C(Sn, Rn)∗ (C(Sn)n)∗ (C(Sn)∗)n, 0 ≃ ≃ ≃ n ∗ n and (C(S ) ) is an L1 space as a ﬁnite product of L1 spaces.

Let us mention what follows: a corollary of Michael’s selection theorem (see [BL00, Chapter 1, Section 3]) asserts that each surjective linear map between Banach spaces has a continuous right inverse, or equivalently, each closed subspace of a Banach space is a continuous positively homogeneous retract. But the proof of Michael’s theorem does not provide us with a concrete continuous retraction. One also obtains the existence of a non continuous linear retract C (Rn, Rn) ker div by elementary linear algebra. 0 →

7.2 Adaptation to charges of positive codimension

Let us introduce the following notations

E := C (X, mRn) C (X, m−1Rn) m,X u ∧ × u ∧ and Θ : E CHm(X) be the map (ω,ζ) Λ(ω) + dΛ(ζ). m,X m,X → 7→ The representation theorem 21 shows that the map Θm,X is onto. Representing charges deals with picking one right inverse to Θm,X . Michael’s selection theorem is also true for Fréchet spaces (the proof of [BL00, Theorem 1.1] remains valid in the case of Fréchet spaces), therefore the existence of a continuous right inverse to Θm,X is guaranteed. The following theorem proves that one cannot however require such an inverse to be uniformly continuous.

Theorem 32. Assume n 2 and 2 m n. Let X a subset of Rn which contains a bilipschitz copy of the m-cube.≥ The linear≤ map≤

Θ : E CHm(X) m,X m,X −→ has no uniformly continuous right inverse.

Proof. Suppose that Θm,X has a uniformly continuous left inverse S. The map

(ω, η) (ω, η) S(Λ(ω) + dΛ(η)) 7→ − is a uniformly continuous retraction of Em,X onto ker Θm,X . Proceeding as in Theorem 31, one proves that the dual sequence

Θ∗ m ∗ m,X ∗ ∗ 0 / CH (X) / Em,X / (ker Θm,X ) / 0 7.2 - Adaptation to charges of positive codimension 127 splits linearly. We now prove Theorem 32 in three steps. Step 1. We prove the theorem in top dimension m = n, for X = In.

n n Suppose there exists a continuous right inverse for Θn,In , from CH (I ) to En,In . We dualize it and obtain a continuous left inverse σ : M (In) M (In) BV(In) n × n−1 → ∗ n for the isometry Θn,In . Note that Mn−1(I ) is an L1 space. Indeed, M (In) = C(In, n−1Rn)∗ = C(In, Rn)∗ = (C(In)∗)n , n−1 ∼ ∧ ∼ ∼ n ∗ Proposition 26 proves that C(I ) is an L1 space and a ﬁnite product of L1 spaces is n ∗ n n still an L1 space. Similarly Mn(I ) is an L1 space and so is En,In = Mn(I ) Mn−1(I ). n × Thus BV(I ) is a complemented subspace of an L1 space and this is contradictory with Theorem 30. Step 2. We prove the theorem under the hypothesis 2 m < n and X = Im 0 Rn. ≤ ×{ } ⊂

m m Suppose Θm,Im×{0} : Em,Im×{0} CH (I 0 ) has a continuous linear right inverse τ, we dualize it and obtain a→ continuous linear× { } map τ ∗ : M (Im 0 ) M (Im 0 ) N (Im 0 ) m × { } × m−1 × { } → m × { } ∗ m n−m such that τ (T, ∂T ) = T for all T Nm(I 0 ). We let δ0 M0(I ) be the Dirac ∈ 0 ×n {−m} n∈ m mass at 0, i.e δ0(ω) := ω(0) for all ω D (R ). Let p : R R be the projection onto the last m coordinates. We deﬁne∈σ : M (Im) M (Im→) N (Im) by setting m × m−1 → m σ(T,S) = p τ ∗(δ T, δ S) # 0 × 0 × m m for all T Mm(I ) and S Mm−1(I ). σ is linear continuous, with norm bounded by τ ∗ . Now∈ let T N (Im),∈ the following holds k k ∈ m σ(T, ∂T ) = p τ ∗(δ T, δ ∂T ) = p τ ∗(δ T, ∂(δ T )) = p (δ T ). # 0 × 0 × # 0 × 0 × # 0 × δ0 is compactly supported, Lemma 41 hence implies p#(δ0 T ) = δ0(1)T = T . This is contradictory by step 1. × Step 3. We prove the theorem under the general assumption. Let X Rn such that X contains a bilischiptz copy of the m cube. Up to a bilipschitz transformation,⊂ we suppose that X contains Im 0 . Suppose there exists a continuous m ×{ } m right inverse ϕ : CH (X) Em,X for Θm,X . We denote by ι : I 0 X the inclusion map and by π : X→ Im 0 the projection onto the compact× { } convex → set Im 0 . Note that ι and π are→ Lipschitz× { } continuous. We deﬁne ψ := ι#ϕπ#. × { } m ϕ CH (X) / Em,X O π# ι#

m m ψ CH (I 0 ) / E m × { } m,I ×{0} m m ψ is a continuous left inverse for Θ : E m CH (I 0 ) since m,X m,I ×{0} → × { } # # # # # # # Θm,X ψ = Θm,X ι ϕπ = ι Θm,X ϕπ = ι π = (πι) = id, and this is impossible by step 2. 128 Appendix to Part II 8 Appendix to Part II

We now state unproved results on inductive limits in the category of locally convex spaces. Proofs can be found in [PMP09][Section 1].

Deﬁnition 15. A nonempty family of convex subsets of a linear space X is called linearly stable if every C containsC 0, and for any t R, x X and C , there exists D so that x + tC∈ C D. ∈ ∈ ∈ C ∈ C ⊂ Proposition 57. Let (X, T) be a locally convex space, and let be a linearly stable C family. There is a unique locally convex topology TC for X that satiﬁes the following conditions:

(i) T ↾ T ↾ for every C , C C ⊂ C ∈ C

(ii) for a locally convex linear space Y , a linear map f :(X, TC) Y is continuous whenever the restriction f ↾ C :(C, T ↾ C) Y is continuous→ for each C . → ∈ C

Moreover, the topology TC has the following properties:

(1) T T and T ↾ C = T ↾ C for every C , ⊂ C C ∈ C

(2) TC has a neighborhood base C TC at 0 consisting of all absorbing balanced convex sets U X such that BU ⊂C belongs to T ↾ C for each C , ⊂ ∩ ∈ C (3) E X is T bounded whenever it is a T bounded subset of some C . ⊂ C ∈ C

The locally convex topology TC is referred to as the localized topology. As an example, if X is a locally convex space, and is a neighborhoodC basis at 0 consisting of balanced convex sets, then for any U , byB the Banach-Alaoğlu theorem, the polar set ∈ B U ◦ := y X∗ : x U, y, x 1 { ∈ ∀ ∈ h i ≤ } is a compact balanced convex set for the weak* topology. The collection ◦ := U ◦ : U is a linearly stable family in X∗. The bounded weak* topology in XB∗ is deﬁned{ to∈ be B} the localization of the weak* topology with respect to ◦. It does not depend on the choice of the local basis . B B

Proposition 58. Let (X, T) be a locally convex space, and let := C1 C2 be a countable linear stable family consisting of closed convex sets.C We{ have⊂ ⊂ · · · }

(1) A sequence (xi) TC converges to 0 if and only if (xi) is a sequence in some Ck and T converges to 0.

(2) A set E X is T bounded if and only if E is a T bounded subset of some C . ⊂ C k

(3) If (X, T) is sequential, then a linear map f from (X, TC) to a locally convex space Y is continuous whenever f(xi) converges to 0 for each sequence (xi) that TC converges to 0. 7.2 - Adaptation to charges of positive codimension 129

Proposition 59. Let (X, T) be a locally convex space, and let = C1 C2 be a countable linearly stable family in X consisting of convex compactC { sets.⊂ The⊂ following · · · } statements hold:

(1) T = E X : C ,E C T ↾ C . C { ⊂ ∀ ∈ C ∩ ∈ }

(2) For any topological space Y , a map f :(X, TC) Y is continuous if and only if the restriction f ↾ C :(C , T ↾ C ) Y is continuous→ for all k. k k k → (3) A set E X is T bounded if and only if it is contained in some C . ⊂ C k

(4) The topology TC is sequentially complete.

(5) If (X, T) is sequential, then so is (X, TC).

Proposition 58 easily implies the following result, which will be useful for extensions of charge.

Proposition 60. Let (X, T) be a locally convex space, let = C1 C2 be a countable linearly stable family of closed convex sets. If Y Cis a linear{ ⊂ space⊂ of · ·X · }, and (Y, T ↾ Y ) is sequential, then TC ↾ Y = [T ↾ Y ]C↾Y . Proposition 61. Let T be the topology in a linear space X associated a norm α, and let β be a lower semicontinuous seminorm in X. Let (Xk)k≥1 be an increasing sequence of closed linear spaces, whose union is X. The subsets Ck := x X : β(x) k form a linearly stable family . For a linear form f : X R, the{ following∈ conditions≤ } are equivalent : C →

(1) f is TC continuous,

(2) ε > 0, θ > 0, x −1 X , f(x) θα(x) + εβ(x). ∀ ∃ ∀ ∈ ∪k≤ε k ≤

Proposition 62. Let (X, T) be a locally convex space, and let := C1 C2 be a linearly stable family of compact convex subsets of X. Let SCbe the{ strong⊂ topology⊂ · · · } in Y := (X, T )∗. Then the evaluation map Υ: X (Y, S)∗ is bijective. In other words, C → (X, TC) is semireﬂexive. Moreover, if (Y, S)∗ is given the bounded weak* topology, Υ is a homeomorphism.

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