ernst kuwert, university of reiburgf course, summer term 2015...

Geometric Variational Problems

Ernst Kuwert, University of Freiburg

Course, Summer Term 2015

Draft of lectures given by Tobias Lamm, KIT

Contents

1 Introduction 5

2 Geometric variational problems 7

2.1 The Dirichlet integral . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.2 Surfaces of prescribed mean curvature in R3 . . . . . . . . . . . . . . . . . . . . 112.3 Harmonic maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 132.4 Two-dimensional geometric variational problems . . . . . . . . . . . . . . . . . . 14

3 Harmonic maps into spheres 19

3.1 Formulation as a conservation law . . . . . . . . . . . . . . . . . . . . . . . . . 193.2 Wente's inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 213.3 Regularity of harmonic maps u ∈W 1,2(D,Sn−1) . . . . . . . . . . . . . . . . . . 24

4 Hardy space 31

4.1 Higher integrability of Jacobi determinants . . . . . . . . . . . . . . . . . . . . 314.2 The Hardy space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.3 Atomic decomposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

5 Regularity of geometric variational problems 61

5.1 Gauge transformation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 615.2 Equations of the form ∆u = Ω∇u . . . . . . . . . . . . . . . . . . . . . . . . . . 63

Bibliography 71

3

Chapter 1

Introduction

These are notes of a course given in summer 2015 at the University of Freiburg. In 2007Tobias Lamm lectured on this topic at FU Berlin, and I followed his notes. The subjectof the course is the regularity theory for two-dimensional geometric variational problems, inparticular compensation methods due to Henry Wente, Frédéric Hélein and Tristan Rivière.Along the lines we introduce certain Hardy and Lorentz spaces, and present the constructionof a Coulomb gauge following Karen Uhlenbeck.

5

6 CHAPTER 1. INTRODUCTION

Chapter 2

Geometric variational problems

The purpose of this Chapter is to introduce the key examples of two-dimensional geometricvariational problems. More background information on these examples can be found in thebook of Jost [29].

2.1 The Dirichlet integral

Let G ⊂ R2 be a domain. For u ∈ C1(G,Rn) the Dirichlet energy is dened by

E(u) =1

2

∫G|Du|2 where |Du|2 = tr(DuTDu). (2.1)

This denition clearly applies in any dimension, but the following interesting feature is specicto the case of dimension two.

Theorem 2.1.1 (Conformal invariance of the Dirichlet energy). Let f : G→ G′ be a confor-mal dieomorphism of domains G,G′ ⊂ R2. Then

E(v f) = E(v) for all v ∈ C1(G′,Rn).

Proof. In dimension two, the fact that f is conformal means that

DfTDf = Df DfT = |detDf |E2.

Now D(v f) = Dv f Df , thus

|D(v f)|2 = tr(DfTDvT fDv fDf

)= tr

(DfDfTDvT fDv f

)= |detDf | |Dv|2 f.

Thus by the transformation formula

E(v f) =1

2

∫G|Dv|2 f |detDf | = 1

2

∫G′|Dv|2 = E(v).

7

8 CHAPTER 2. GEOMETRIC VARIATIONAL PROBLEMS

For u ∈ C1(G,Rn) with E(u) <∞ and ϕ ∈ C1c (G,Rn) we have the rst variation formula

d

dtE(u+ tϕ)|t=0 =

∫G〈Du,Dϕ〉. (2.2)

Assuming that u ∈ C2(G,Rn) we can integrate by parts to obtain

d

dtE(u+ tϕ)|t=0 = −

∫G〈∆u, ϕ〉.

In other words the Euler-Lagrange operator for the Dirichlet energy is −∆. We say that uis a critical point of the Dirichlet energy if the rst variation in (2.2) always vanishes. Foru ∈ C2(G,Rn) we can then conclude that u is harmonic:

∆u = uxx + uyy = 0 on G. (2.3)

The conformal invariance implies an equivariance property for the Laplace operator, moreprecisely we compute for a conformal dieomorphism f : G→ G′

−∫G〈∆(v f), ϕ〉 =

d

dtE(v f + tϕ)|t=0

=d

dtE(v + tϕ f−1)|t=0

= −∫G′〈∆v, ϕ f−1〉

= −∫G〈∆v f, ϕ〉 |detDf |.

We conclude that

∆v f =1

| detDf |∆(v f) for a conformal dieomorphism f : G→ G′. (2.4)

How comes geometry into play? For an immersion u : G→ Rn the area functional is 1

A(u) =

∫G

√det g where gαβ = 〈∂αu, ∂βu〉. (2.5)

Note that√

det g is the Jacobian Ju, in fact by denition√det g =

√|∂1u|2|∂2u|2 − 〈∂1u, ∂2u〉2 = |∂1u ∧ ∂2u| = Ju.

The rst variation of the area in direction ϕ ∈ C1c (G,Rn) is computed as follows. First

∂

∂t〈∂α(f + tϕ), ∂β(f + tϕ)〉|t=0 = 〈∂αf, ∂βϕ〉+ 〈∂βf, ∂αϕ〉.

Using this one getsd

dtA(u+ tϕ)|t=0 =

∫Ggαβ〈∂βf, ∂αϕ〉

√det g.

1greek indices α, β, . . . in 1, 2, latin indices i, j, . . . in 1, . . . , n.

2.1. THE DIRICHLET INTEGRAL 9

Here (gαβ) denotes the (symmetric) inverse matrix to (gαβ). Integrating by parts yields

d

dtA(u+ tϕ)|t=0 = −

∫G〈 ~H,ϕ〉

√det g where ~H =

1√det g

∂α(√

det ggαβ∂βu). (2.6)

We now compare the area functional to the Dirchlet energy. First we have

Ju = |∂1u ∧ ∂2u| ≤ |∂1u| |∂2u| ≤1

2(|∂1u|2 + |∂2u|2) =

1

2|Du|2.

Thus A(u) ≤ E(u) with equality if and only if u is conformally parametrized, that is

〈∂1u, ∂2u〉 = 0 and |∂1u|2 = |∂2u|2. (2.7)

Assuming that u is conformally parametrized, with induced metric gαβ = e2λδαβ and JacobianJu = e2λ, the mean curvature vector becomes

~H = e−2λ∆u. (2.8)

It follows that a conformally parametrized immersion u : G→ Rn is a minimal surface if andonly if u is Euclidean harmonic. These facts provide a close relation between the Dirichletenergy and the area functional in two dimensions.

The question whether any immersed surface admits a reparametrization which is conformalis of fundamental importance; it has local and global aspects. In 1825 Gauÿ proved thatany real-analytic surface admits locally a conformal reparametrization; this was extended tosurfaces of class C1,1 by Lichtenstein in 1911. For oriented surfaces the parameter changes arethen holomorphic, so that the immersion induces a global complex structure on the parameterdomain, which becomes naturally a Riemann surface.

The following result goes in a dierent direction: it shows that certain critical points ofgeometric variational problems satisfy automatically the conformality relations. The resultplays a crucial rôle in proving that the classical approach to the Plateau problem actuallyproduces a minimal surface. Moreover, this generalizes to other problems for surfaces withprecribed mean curvature. Historically, the result was also relevant in the regularity theory,because earlier regularity proofs needed to assume conformality [21]. In the literature oneoften nds the term stationary for a map which is critical with respect to variations of theindependent variables.

Theorem 2.1.2. Let D = (x, y) ∈ R2 : x2 + y2 < 1. Assume that u ∈ C1(D,Rn) has niteenergy and satises

d

dtE(u φt)|t=0 = 0,

for any smooth family φ : D × (−ε, ε) → D of dieomorphisms φt = φ(·, t) with φ0 = idD.Then u satises the conformally relations (2.7).

It is only asserted that u is weakly conformal, leaving open whether u is immersed or not. Theconformality may however be useful in analyzing the behavior of u at points with vanishingJacobian, showing that they are isolated and have the character of branchpoints, see [11].


Proof. We assume that u : H → Rn where H is the upper half plane. There is an explicitconformal equivalence between D and H, so that the result transfers to D via conformalinvariance. Consider a vector eld X ∈ C∞c (R2,R2) with X τ = τX, where τ(x, y) = (x,−y),in particular X2(x, 0) = 0 for all x ∈ R. The associated ow φ : R2 × R→ R2 is smooth, andφt = φ(·, t) is dieomorphic with inverse φ−t. Uniqueness for the initial value problem implies

φ(τ(z), t) = τ(φ(z, t)) for all z ∈ R2, t ∈ R.

In particular φt(R) = R and φt(H) = H. Substituting ζ = (ξ, η) = φ−t(z) we have

E(u φt) =1

2

∫H|Du(φt(ζ))Dφt(ζ)|2 dξdη =

1

2

∫H|Du(z)Dφ−t(z)

−1|2 detDφ−t(z) dxdy.

We want to dierentiate under the integral sign at t = 0. As φ(z, t) = z for z /∈ sptX, theintegrand and its derivative for t ∈ (−ε, ε) are bounded by C |Du|2 which is integrable. Wecompute

∂

∂tDφ−t(z) · v|t=0 =

∂

∂t

∂

∂sφ−t(z + sv)|s=0,t=0

=∂

∂s

∂

∂tφ−t(z + sv)|t=0,s=0

= − ∂

∂sX(z + sv)|s=0

= −DX(z) · v.

This implies

∂

∂tDφ−t(z)

−1|t=0 = DX(z),∂

∂tdetDφ−t(z) = −divX(z).

Putting X = (a, b) we nd

d

dtE(u φt)|t=0 =

∫H

(〈Du(z), Du(z)DX(z)〉 − 1

2|Du(z)|2divX(z)

)dxdy

=

∫H

(〈∂iu, ∂ju〉∂iXj − 1

2|Du|2∂iXi

)dxdy

=

∫H

(1

2(|ux|2 − |uy|2)ax + 〈ux, uy〉ay

)dxdy

+

∫H

(〈ux, uy〉bx −

1

2(|ux|2 − |uy|2)by

)dxdy.

Taking X|H ∈ C∞c (H,R2) arbitrary, we see that the integrable function

h(z) = |ux|2 − |uy|2 − 2i〈ux, uy〉

is a weak solution to the Cauchy-Riemann equations, and hence holomorphic. Next we takeX = (ϕy, ϕx) where ϕ ∈ C∞c (R2) is odd, that is ϕ(x,−y) = −ϕ(x, y). Then X is admissiblewhence ∫

H〈ux, uy〉∆ϕdxdy = 0.

Using odd reection the function 〈ux, uy〉 extends to a weakly harmonic function which isintegrable on R2. By the mean value formula, we conclude that 〈ux, uy〉 is identically zero.The Cauchy-Riemann equations now yield that h(z) is constant and in fact vanishes, againby integrability.

2.2. SURFACES OF PRESCRIBED MEAN CURVATURE IN R3 11

2.2 Surfaces of prescribed mean curvature in R3

Let D = (x, y) ∈ R2 : x2 + y2 < 1. For u ∈ C1(D,R3) we consider the functional

F(u) =1

2

∫D|Du|2 +

∫Du∗ω, (2.9)

where ω ∈ C1(R3,Λ2(R3)) is a given two-form. To interpretate the second term geometrically,let us assume for simplicity that u ∈ C2(D,R3). Consider the cone over u dened by

F : D × [0, 1]→ R3, F (z, t) = tu(z).

Writing dω = H dVR3 we get by Stokes' theorem∫D×[0,1]

F ∗(HdVR3) =

∫D×[0,1]

F ∗dω

=

∫D×[0,1]

dF ∗ω

=

∫DF (·, 1)∗ω −

∫DF (·, 0)∗ω +

∫∂D×[0,1]

F ∗ω

=

∫Du∗ω +

∫∂D×[0,1]

F ∗ω.

The C2 assumption was used when interchanging F ∗ and d. Introducing the multiplicityfunction θF (X) =

∑F (z,t)=X sign detDF (z, t), we get by the transformation formula∫

F (D×[0,1])HθF dL3 =

∫Du∗ω +

∫∂D×[0,1]

F ∗ω.

The second integral on the right depends only on u|∂D, thus it reduces to a constant whenrestricting to a class of maps with prescribed boundary values. Up to that constant, theintegral

∫D u∗ω then corresponds to the volume of the cone, weighted with the function H

and counted with multiplicities. A special case is obtained by chosing

ω =1

3XxdVR3 .

Then dω = dVR3 and F ∗ω = 0 on ∂D × [0, 1], no matter what boundary condition. Hence

1

3

∫D〈u, ux ∧ uy〉 dxdy =

∫Du∗ω =

∫F (D×[0,1])

θF dL3. (2.10)

Next we calculate the rst variation of the functional.

Lemma 2.2.1. Let F(u) be the functional in (2.9), and put dω = H dVR3 . Then for anyu ∈ C2(D,R3) and ϕ ∈ C2

c (D,R3) we have

d

dεF(u+ εϕ)|ε=0 =

∫D〈Du,Dϕ〉+

∫DH u 〈ux ∧ uy, ϕ〉. (2.11)


Proof. We consider the ane homotopy

F : D × [0, ε]→ R3, F (z, t) = u(z) + tϕ(z).

Applying Stokes' formula on D × [0, ε] we get∫D×[0,ε]

F ∗(H dVR3) =

∫D

(u+ εϕ)∗ω −∫Du∗ω +

∫∂D×[0,ε]

F ∗ω.

The last integral on the right vanishes since ∂tF (z, t) = ϕ(z) = 0 on ∂D × [0, ε]. Taking thederivative at ε = 0 yields

∂

∂ε

∫D

(u+ εϕ)∗ω|ε=0 =∂

∂ε

∫ ε

0

∫DH(F (x, y, t)) detDF (x, y, t) dxdydt|ε=0

=

∫DH(u(x, y)) detDF (x, y, 0) dxdy

=

∫DH(u(x, y)) det(ux, uy, ϕ) dxdy

=

∫DH(u(x, y))〈ux ∧ uy, ϕ〉 dxdy.

The claim follows by combining with the rst variation of the Dirichlet energy.

We see that regular critical points of F are solutions of the elliptic system

∆u = (H u) ux ∧ uy in D. (2.12)

If u is in addition a conformal immersion, with induced metric gij = e2λδij , then we canrewrite the equation in the form

e−2λ∆u = (H u) ν where ν =ux ∧ uy|ux ∧ uy|

.

We know from (2.8) that the left hand side is the mean curvature vector, hence the surface uhas prescribed mean curvature H u. A special case is when H is constant, then u is calleda constant mean curvature or CMC surface. One may then take as dierential form

ω =H

3XxdVR3 (H constant).

The partial dierential equation (2.12) is called the prescribed mean curvature equation orconstant mean curvature equation, respectively. The use of this terminology does not requiresolutions to be conformally parametrized. The geometric interpretation, however, is onlyavailable for u conformally parametrized. To obtain such solutions, one may potentially useTheorem 2.1.2. Namely for smooth dieomorphisms φt : D → D, t ∈ (−ε, ε), one has∫

D(u φt)∗ω =

∫Dφ∗tu

∗ω =

∫Du∗ω.

If u is a critical point of F with respect to variations u φt, then one concludes

d

dtE(u φt)|t=0 =

d

dtF(u φt)|t=0 = 0.

2.3. HARMONIC MAPS 13

The conformality relations now follow from Theorem 2.1.2. This applies, for example, to min-imizers of F under Plateau boundary conditions.

To derive the prescribed mean curvature equation from the vanishing of the rst variationwe have imposed strong regularity assumptions on the function u. In contrast, the existencetheory will only give us functions u ∈ W 1,2 ∩ L∞(D,R3), say. It is a key issue to show thatthese weak solutions are regular. The special case of constant mean curvature was solved byWente in 1969 [53], whereas the general case of variable mean curvature (for which H u is apriori only bounded and measurable) was proved by Riviére much later in 2007 [39].

2.3 Harmonic maps

As a second example we now introduce harmonic maps. Let M ⊂ Rn be a m-dimensionalsmooth compact submanifold where 1 ≤ m ≤ n−1. By denition, a harmonic map u : D →Mis a critical point of the Dirichlet energy under the constraint that u(D) ⊂ M ; that is onlyvariations staying in M are allowed. To derive the Euler-Lagrange equation, we need somebasic geometric facts.

Lemma 2.3.1. Let U%(M) = X ∈ Rn : dist (X,M) < %. For small % > 0, the nearestpoint projection πM : U%(M) → M is well-dened and smooth. One has for all X ∈ M theequations

DπM (X)τ = τ for all τ ∈ TXM.

DπM (X)ν = 0 for all ν ∈ TXM⊥.D2πM (X)(τ, τ) = A(X)(τ, τ) for all τ ∈ TXM.

Here A is the (normal-valued) second fundamental form of M ⊂ Rn.

Proof. We only show the formulae. For a curve γ(s) in M with γ(0) = X, γ′(0) = τ ∈ TXMwe have πM γ = γ. Dierentiating yields

(DπM γ)γ′ = γ′, i.e. DπM (X)τ = τ for s = 0.

Mow one proves πM (X + tν) = X for ν ∈ TXM⊥ and |t| small. Dierentiating at t = 0 givesthe second formula. Finally dierentiating twice along γ(s) implies

(D2πM γ)(γ′, γ′) + (DπM γ)γ′′ = γ′′.

At s = 0 one concludes D2πM (X)(τ, τ) = (γ′′)⊥ = A(X)(τ, τ).

Now assume u ∈ C1(D,Rn) has nite Dirichlet energy and maps into the submanifoldM . Forϕ ∈ C∞c (D,Rn) one computes

d

dtE(πM (u+ tϕ))|t=0 =

∫D〈Du,D

(DπM u)ϕ〉.

Assuming u ∈ C2(D) we can integrate by parts to get

d

dtE(πM (u+ tϕ))|t=0 = −

∫D〈∆u, (DπM u)ϕ〉 = −

∫D〈(DπM u)∆u, ϕ〉.


For a regular critical point, under the constraint u(D) ⊂M , it follows that

(∆u)> = (DπM u)∆u = 0. (2.13)

To get an equation which is analytically better tractable, one computes

(DπM u)∆u = ∂α((DπM u)∂αu

)− ∂α(DπM u) ∂αu

= ∂α(∂αu)− (D2πM u)(∂αu, ∂αu)

= ∆u− (A u)(∂αu, ∂αu).

We arrive at the harmonic map system

∆u = (A u)(∂αu, ∂αu). (2.14)

It should be noted that the concept of harmonic maps is naturally dened in the context ofmaps u : (M,γ) → (M, g) between Riemannian manifolds, without assuming M embedded.The energy is then locally dened by

E(u) =1

2

∫Mγαβ(g u)(∂αu, ∂βu)

√det γ.

In a geometric context this denition is preferable. However, it is dicult to avoid the em-bedding in regularity questions. There one usually needs a local coordinate representationu : U → V , u(z) = (u1(z), . . . , um(z)), to get started. But maps in the natural energy spaceW 1,2 are not necessarily continuous. Therefore a passage to local coordinates on M may beimpossible, no matter how small U is chosen. A classical example for this is as follows.

Beispiel 2.3.2. Let B be the unit ball in R3 and S2 the unit 2-sphere. By the above, a mapu : B→ S2 is harmonic if and only if

−∆u = |Du|2u. (2.15)

Namely, we have A(X)(τ, τ) = −|τ |2X, which yields (Au)(∂αu, ∂αu) = −|∂αu|2u = −|Du|2u.One checks that the following map belongs to W 1,2(B,R3) and is a weak solution:

u : B→ S2, u(x) =x

|x|.

An arbitrarily small neighborhood of the origin is mapped to the full two-sphere. We willeventually show that weak solutions are without singularities in two dimensions, however thisis not clear a priori.

2.4 Two-dimensional geometric variational problems

The goal of this section is to classify all two-dimensional variational integrals of rst orderwhich are conformally invariant. We will see that the example of the Dirichlet energy plus apullback of a 2-form already constitutes the general case, if we allow a general Riemannianmetric in the target.

2.4. TWO-DIMENSIONAL GEOMETRIC VARIATIONAL PROBLEMS 15

Let f : Rn ×R2×n → R, f = f(X,A), be an integrand. On any bounded domain G ⊂ R2, wehave the associated functional

F(u,G) =

∫Gf(u(z), Du(z)) dxdy for u : G→ Rn.

We say that F is conformally invariant if for any conformal dieomorphism φ : G → φ(G))we have the property

F(u φ−1, φ(G)) = F(u,G) for all u : G→ Rn. (2.16)

The following classication is due to Michael Grüter.

Theorem 2.4.1 ([22]). Consider a functional F(u,G) =∫G f(u,Du), such that f and D2

Afare continuous on Rn × Rn×2. Assume also that

f(X,A) > 0 whenever A 6= 0.

If F(u,G) =∫G f(u,Du) is conformally invariant, then it has the representation

F(u) =1

2

∫Gg(u)(Du,Du) +

∫Gu∗ω, (2.17)

where g is a Riemannian metric and ω is a 2-form, both continuous on Rn.

Proof. We compute using the transformation formula

F(u φ−1, φ(G)) =

∫φ(G)

f(u(φ−1(w)), Du(φ−1(w))Dφ−1(w)

)dudv

=

∫Gf(u(z), Du(z)Dφ(z)−1

)| detDφ(z)| dxdy.

Let φ : R2 → R2 be a conformal dieomorphism. Taking G = Dε(0) and u(z) = X + Az forgiven X ∈ Rn, A ∈ Rn×2, we obtain after dividing by |Dε| and letting ε 0

f(X,A) = f(X,ADφ(0)−1)|detDφ(0)|.

Chosing φ(z) = z/t for t > 0, and dierentiating twice at t = 0 shows

f(X,A) =1

2D2Af(X, 0)(A,A).

Chosing φ(z) = Sz for S ∈ SO(2) yields f(X,A) = f(X,AST), hence

D2Af(X, 0)(A,A) = D2

Af(X, 0)(AST, AST).

We now dene a symmetric and an antisymmetric form, both continuous on Rn, by

g(X)(V,W ) = D2Af(X, 0)(e1 ⊗ V, e1 ⊗W ),

ω(X)(V,W ) = D2Af(X, 0)(e1 ⊗ V, e2 ⊗W ).

The symmetry of g(X) is clear. To show that ω(X) is antisymmetric, we rst note

(eα ⊗ V )STz = 〈eα, STz〉V = 〈Seα, z〉V = (Seα ⊗ V )z.


Taking Se1 = e2, Se2 = −e1 we compute

ω(X)(V,W ) = D2Af(X, 0)(e1 ⊗ V, e2 ⊗W )

= −D2Af(X, 0)(Se2 ⊗ V, Se1 ⊗W )

= −D2Af(X, 0)((e2 ⊗ V )ST, (e1 ⊗W )ST)

= −D2Af(X, 0)(e1 ⊗W, e2 ⊗ V )

= −ω(X)(W,V ).

We conclude, using again the invariance with respect to S,

f(u,Du) =1

2D2Af(u, 0)(eα ⊗ ∂αu, eβ ⊗ ∂βu)

=1

2

(D2Af(u, 0)(e1 ⊗ ∂1u, e1 ⊗ ∂1u) +D2

Af(u, 0)(e2 ⊗ ∂2u, e2 ⊗ ∂2u))

+D2Af(u, 0)(e1 ⊗ ∂1u, e2 ⊗ ∂2u)

=1

2

(g(u)(∂1u, ∂1u) + g(u)(∂2u, ∂2u)

)+ ω(u)(∂1u, ∂2u).

Finally the assumed positivity implies that g is a Riemannian metric, namely for V 6= 0

g(X)(V, V ) = D2Af(X, 0)(e1 ⊗ V, e1 ⊗ V ) = 2f(X, e1 ⊗ V ) > 0.

This nishes the proof of the theorem.

The calculation of the Euler-Lagrange equation for the Riemannian Dirichlet energy isstraightforward using local coordinates. We compute for ϕ ∈ C∞c (G,Rn)

d

dtEg(u+ tϕ)|t=0 =

∫G

(gjk(u)∂αϕ

j∂αuk +

1

2∂igjk(u)ϕi∂αu

j∂αuk)

= −∫Gϕj(gjk(u)∆uk + ∂igjk(u)∂αu

i∂αuk − 1

2∂jgik(u)∂αu

i∂αuk)

= −∫Ggjk(u)ϕj

(∆uk + gkl(u)∂iglm(u)∂αu

i∂αum − 1

2gkl(u)∂lgim(u)∂αu

i∂αum)

= −∫Ggjk(u)ϕj

(∆uk + Γkim(u)∂αu

i∂αum).

Here the Γkij are the Christoel symbols of the metric g; we used that the term ∂αui∂αu

m issymmetric in i and m. For the pullback integral, we proceed as in the case of codimension one.Introducing the 3-Form Ω = dω, we have putting F : D× [0, ε]→ Rn, F (z, t) = u(z) + tϕ(z),

d

dε

∫G

(u+ εϕ)∗ω |ε=0 =d

dε

∫G×[0,ε]

F ∗Ω |ε=0

=

∫G

(Ω u)(ux, uy, ϕ)

=

∫Ggjk(u)ϕjgkl(u)(Ω u)(ux, uy, el).

In summary, the Euler-Lagrange operator Lf (u) of the functional (2.17) is given by

Lf (u)k = −∆uk − Γkij(u)∂αui∂αu

j + gkl(u)(Ω u)(ux, uy, el), (2.18)

2.4. TWO-DIMENSIONAL GEOMETRIC VARIATIONAL PROBLEMS 17

where Γkij are the Christoelsymbols of g and Ω = dω. The nonlinear operator

(∆Nu)k = ∆uk + Γkij(u)∂αui∂αu

j

is sometimes called the tension eld or intrinsic Laplacian of u. The system (2.18) is semilinearwith principal term given by the standard Laplacian, and a right hand side which is a quadraticform of the gradient, possibly depending nonlinearly on u. The key question is now whethera regularity theory is available for such systems. Here are the bad news.

Beispiel 2.4.2. Consider the scalar equation

−∆u = |Du|2 on G = D1/e(0).

We claim that the function u(z) = log log 1r , r = |z|, belongs toW 1,2(G) and solves the equation

in the weak sense. For this we compute

u′(r) =1

r log r,

u′′(r) = − 1

r2 log r− 1

r2 log2 r,

|Du|2 = u′(r)2 =1

r2 log2 r,

∆u = u′′(r) +1

ru′(r) = − 1

r2 log2 r.

Away from the origin the equation holds in the classical sense. Substituting r = e−t wheret ∈ [1,∞) we have ∫ 1/e

0

dr

r logs 1r

=

∫ ∞1

dt

ts=

∞ for s = 1

1s−1 for s > 1.

Thus Du ∈ L2(G,R2). Using cuto arguments one now proves that Du is the weak deriva-tive, and that the equation holds weakly on the full domain G. We also see that solutionsto the Dirichlet problem may be nonunique, since u = 0 on ∂G. Furthermore, we have acounterexample to an L1 theory for the Laplacian: ∆u is integrable while D2u is not.

The fact that u(x) = log log 1r is unbounded is important in the previous example. Namely,

if the weak solution was bounded then a regularity result by Ladyzhenskaya and Ural'tseva(1961) would imply that it is Hölder continuous; further regularity would then follow easily. Inthe case of harmonic maps the boundedness of the weak solution is for granted, by assumingthe target manifold to be compact. Nevertheless the result of Ladyzhenskaya and Ural'tsevadoes not apply, because it is limited to scalar equations. This is seen from the followingmodication of our example, due to Hildebrandt and Widman.

Beispiel 2.4.3. The map u(r) = exp(i log log 1r ) is a bounded weak solution to the system

−∆u = |Du|2Λu where Λ =

(1 −11 1

).

In summary we see that the regularity for the prescribed mean curvature equation andfor harmonic maps is subtle, and that the particular structure of the nonlinearity needs to beexploited in some way.

Chapter 3

Harmonic maps into spheres

3.1 Formulation as a conservation law

Let B = x ∈ Rd : |x| < 1 where d ≥ 2 is arbitrary. For a smooth compact submanifoldM ⊂ Rn, we consider the class of Sobolev mappings

W 1,2(B,M) = u ∈W 1,2(B,Rn) : u(x) ∈M almost everywhere. (3.1)

We say that u ∈W 1,2(B,M) is (weakly) harmonic if and only if∫B〈Du,Dφ〉 =

∫B〈A(u)(∂αu, ∂αu), φ〉 for all φ ∈W 1,2

0 ∩ L∞(B,Rn). (3.2)

Here A is the second fundamental form of M , taking values in the normal bundle of M . InChapter 1 we gave the same denition, except that we allowed only φ ∈ C∞c (B,Rn). But (3.2)is equivalent by the following approximation argument.

Given φ ∈ W 1,20 ∩ L∞(B,Rn), we rst chose ψk ∈ C∞c (B,Rn) with ψk → φ in W 1,2(B,Rn)

and almost everywhere. For R > ‖φ‖L∞(B) we dene the functions φk ∈W1,20 (B,Rn) by

φk(x) =

ψk(x) for |ψk(x)| < R,

RπSn−1(ϕk(x)) for |ϕk(x)| ≥ R.

Here πSn−1(X) = X/|X| is the projection map. For a. e. x ∈ B we have |ψk(x)| → |φ(x)| < R,which implies φk(x) = ψk(x)→ φ(x) as k →∞. This shows φk → φ almost everywhere. NowDπSn−1 has bounded norm. Thus |φk| ≤ R and |Dφk| ≤ C|Dψk|, which implies φk → φ weaklyin W 1,2(B,Rn) up to a subsequence. It follows that the left hand side of (3.2) converges. Theright hand side also passes to the limit by a. e. convergence, using the bound

|A(u)(∂αu, ∂αu)φk| ≤ CR |Du|2 ∈ L1(B).

By the standard product rule the space W 1,2∩L∞(B) is an algebra; this is used several timesbelow. Moreover if one of the functions belongs to W 1,2

0 (B,Rn), then this is also true for theproduct.

In the present section we focus on M = Sn−1. In this case there is a useful reformulationof the equation due to Yun Mei Chen and Jalal Shatah.

19

20 CHAPTER 3. HARMONIC MAPS INTO SPHERES

Theorem 3.1.1 ([8, 42]). For u ∈W 1,2(B, Sn−1) the following are equivalent:

(a) u is weakly harmonic.

(b) For any Λ ∈ Rn×n with ΛT = −Λ we have div (DuTΛu) = 0, that is∫B〈Du · gradϕ,Λu〉 = 0 for all ϕ ∈ C∞c (B). (3.3)

Proof. Let u be weakly harmonic. Along u the map ϕΛu is tangential as 〈Λu, u〉 = 0, andbelongs to W 1,2

0 ∩ L∞(B,Rn). Thus we get from the equation, using that A(u)(∂αu, ∂αu) isnormal and 〈Du,ΛDu〉 = 0,

0 =

∫B〈Du,D(ϕΛu)〉 =

∫B∂αϕ〈Du · eα,Λu〉 =

∫B〈Du · gradϕ,Λu〉.

For the reverse direction, we consider for 1 ≤ i < j ≤ n the tangent vectorelds

Λijω = (ei ⊗ ej − ej ⊗ ei)ω = ωiej − ωjei ∈ TωSn−1.

The Λijω span TωSn−1, in fact we have for any ξ ∈ TωSn−1 the representation

ξ = (ω ⊗ ξ − ξ ⊗ ω)ω =∑

1≤i<j≤n(ωiξj − ωjξi)Λijω.

For a given variation φ ∈ C∞c (B,Rn) we obtain

φ = 〈φ, u〉u+∑

1≤i<j≤nϕijΛiju, where (3.4)

ϕij = ui(φj − 〈φ, u〉uj)− uj(φi − 〈φ, u〉ui).

Now ϕij ∈W 1,20 ∩ L∞(B), and (b) holds for W 1,2

0 -functions by approximation. Thus∫B〈Du,D(ϕijΛiju)〉 =

∫B〈Du · gradϕij ,Λiju〉 = 0.

Furthermore 〈φ, u〉u ∈W 1,20 ∩ L∞(B,Rn), and∫

B〈Du,D(〈φ, u〉u)〉 =

∫B|Du|2〈u, φ〉.

The claim follows using the representation (3.4).

Equation (3.3) may also be derived by Noether's theorem. For any map u the Dirichlet energyis invariant under rotations R(t)u = exp(tΛ)u. This implies for any Ω ⊂ B

0 =d

dtE(R(t)u,Ω)|t=0 =

∫Ω〈Du,D(Λu)〉 =

∫Ω∂α〈∂αu,Λu〉 −

∫Ω〈∆u,Λu〉.

If u : B → Sn−1 is harmonic, then (∆u)> = 0 and the second integral vanishes. Since Ω wasarbitrary we get the conservation law

div (DuTu) = ∂α〈∂αu,Λu〉 = 0 in B.

3.2. WENTE'S INEQUALITY 21

The regularity of u was used in this argument: since Λu does not have compact support inΩ the weak equation does apply without further justication. On the other hand, the inter-pretation by Noether's theorem shows more clearly the key rôle of the Killing elds for theresult, that is vector elds which generate isometries. In fact, the theorem was generalized tomanifolds M with isometry group acting transitively by Frédéric Hélein in [24].

As an application of Theorem 3.1.1 we now show that the set of weakly harmonic maps isclosed under weak convergence. This is not clear, because the weak convergence of Duk → Duin L2(B,Rn×d) does not obviously imply A(uk)(Duk, Duk) → A(u)(Du,Du) in the sense ofdistributions. Still, the proof is trivial using the reformulation as a conservation law.

Corollary 3.1.2. Let uk ∈ W 1,2(B, Sn−1) be weakly harmonic, and suppose uk → u weaklyin W 1,2(B,Sn−1). Then u ∈W 1,2(B, Sn−1) is also weakly harmonic.

Proof. We have Duk → Du weakly in L2(Rn×d), and uk → u strongly in L2(B,Rn) byRellich's theorem. Then (∂αϕ)Λuk → (∂αϕ)Λu strongly in L2(B,Rn) for ϕ ∈ C∞c (B), and∫

B〈Du · gradϕ,Λu〉 = lim

k→∞

∫B〈Duk · gradϕ,Λuk〉.

The result now follows by Theorem 3.1.1.

3.2 Wente's inequality

In this section we are back to dimension d = 2. Let us start with the classical Dirichletproblem for the Poisson equation, assuming that

−∆u = f weakly in D where u ∈W 1,20 (D).

A very fundamental estimate, due to Calderon and Zygmund, asserts that if f ∈ Lp(D) where1 < p <∞, then u ∈W 2,p(D) and

‖u‖W 2,p(D) ≤ C(p) ‖f‖Lp(D).

In the case of the harmonic map or prescribed mean curvature systems, this does not applydirectly because the right hand side, being quadratic in the gradient, belongs a priori only toL1(D). The Calderon-Zygmund theory does not extend to the space L1; the diculty was al-ready observed in Example 2.4.2. Henry Wente found that this drawback can be compensatedif the right hand side has a special algebraic structure, namely when f is a Jacobi determinant.We refer to Hélein's book [26] for an in-depth discussion.

Theorem 3.2.1 ([53]). For a, b ∈W 1,2(D) there is a unique function u ∈W 1,20 (D) solving∫

D〈du, dϕ〉 =

∫Dϕa, b for all ϕ ∈W 1,2

0 ∩ L∞(D), where a, b = ayby − aybx. (3.5)

The solution belongs to C0(D), and satises the a priori estimates

||u||C0(D) + ||du||L2(D) ≤ C ‖da||L2(D)| |db||L2(D). (3.6)


Proof. The uniqueness of the solution is trivial. For existence the key step is to prove theestimates, assuming that a, b and hence u are smooth on the closed disk. The rest will follow bya simple approximation argument. We also observe before that the statement of the theoremis conformally invariant: if φ : U → D is a conformal dieomorphism with pullback metrice2λδαβ , then according to equation (2.4)

∆(u φ) = e2λ∆u φ.

On the other hand we compute

a φ, b φdx ∧ dy = d(a φ) ∧ d(b φ)

= d(φ∗a) ∧ d(φ∗b)

= φ∗(da) ∧ φ∗(db)= φ∗(da ∧ db)= e2λa, b φ dx ∧ dy.

This shows −∆(u φ) = a φ, b φ, i.e. the conformal invariance of the equation. Moreoverthe L2 norms of du, da and db and also the C0 norm of u are invariant. Now for given a ∈ Dthere is an automorphism φ of the disk with φ(0) = a. The well-known formula is

φ(z) =z + a

1 + azfor z ∈ D.

To get the C0 estimate it is then sucient to bound u at the origin, because u(a) = (uφ)(0).The value at the origin is given by Green's formula (writing r = |z| and assuming a, b smooth)

u(0) =1

2π

∫D

(log r) da ∧ db

=1

2πlimε0

∫D\Dε(0)

d((log r)a db

)− 1

2πlimε0

∫D\Dε(0)

dr

ra ∧ db

= − 1

2π

∫ 1

0

dr

r

∫ 2π

0a(r, ϕ)

∂b

∂ϕ(r, ϕ) dϕ.

Let a(r) be the mean value of a(r, ·) on [0, 2π], and estimate∣∣∣ ∫ 2π

0a(r, ϕ)

∂b

∂ϕ(r, ϕ) dϕ

∣∣∣ =∣∣∣ ∫ 2π

0(a(r, ϕ)− a(r))

∂b

∂ϕ(r, ϕ) dϕ

∣∣∣≤

∥∥a− a(r)∥∥L2(0,2π)

·∥∥ ∂b∂ϕ

∥∥L2(0,2π)

≤∥∥ ∂a∂ϕ

∥∥L2(0,2π)

·∥∥ ∂b∂ϕ

∥∥L2(0,2π)

.

In the last step we used the Poincaré inequality on (0, 2π) for functions having zero meanvalue; this follows easily by Fourier expansion. We conclude

|u(0)| ≤ 1

2π

∫ 1

0

∥∥ ∂a∂ϕ

∥∥L2(0,2π)

·∥∥ ∂b∂ϕ

∥∥L2(0,2π)

dr

r

≤ 1

2π

(∫ 1

0

∫ 2π

0

1

r2

∣∣ ∂a∂ϕ

∣∣2 rdrdϕ)1/2·(∫ 1

0

∫ 2π

0

1

r2

∣∣ ∂b∂ϕ

∣∣2 rdrdϕ)1/2

≤ 1

2π‖da‖L2(D) · ‖db‖L2(D).

3.2. WENTE'S INEQUALITY 23

The L2 bound for the gradient now follows simply by testing with u:∫D|du|2 =

∫Dua, b ≤ ‖u‖C0(D)‖da‖L2(D)‖db‖L2(D) ≤

1

2π‖da‖2L2(D)‖db‖

2L2(D).

In the smooth case the estimates are settled. Given a, b ∈ W 1,2(D) we approximate ak → a,bk → b inW 1,2(D), where ak, bk are in C

∞(D). Let uk be the solution of the Dirichlet problem

−∆uk = ak, bk in D, uk|∂D = 0.

Then uk − ul is zero on ∂D and satises

−∆(uk − ul) = ak, bk − al, bl = ak − al, bk+ al, bk − bl.

We have ‖dak‖L2(D) + ‖dbk‖L2(D) ≤ C for all k. By uniqueness and the estimates, we obtain

‖uk − ul‖C0(D) + ‖d(uk − ul)‖L2(D) ≤ C(‖d(ak − al)‖L2(D) + ‖d(bk − bl)‖L2(D)

)→ 0 as k, l→∞.

Thus uk → u in both W 1,20 (D) and C0(D), and u satises the desired estimates. To get the

weak equation for ϕ ∈W 1,20 ∩ L∞(D), we compute∫

D〈du, dϕ〉 = lim

k→∞

∫D〈duk, dϕ〉 = lim

k→∞

∫Dϕak, bk =

∫Dϕa, b.

As a consequence of Wente's theorem, we get the following regularity result.

Corollary 3.2.2. Let a, b ∈W 1,2(D) and assume that v ∈ L1(D) solves

−∫Dv∆ϕ =

∫Dϕa, b for all ϕ ∈ C∞(D), ϕ|∂D = 0. (3.7)

Then v ∈W 1,2(D) ∩ C0(D), and it satises the estimates (3.6).

Remark 3.2.3. Equation 3.7 holds for any v ∈W 1,p0 (D), p ∈ [1, 2], solving∫

D〈Dv,Dϕ〉 =

∫Dϕa, b for all ϕ ∈ C∞(D), ϕ|∂D = 0.

In fact, by Sobolev trace theory one gets using v = 0 on ∂D∫D〈Dv,Dϕ〉 =

∫∂D

v∂ϕ

∂r−∫Dv∆ϕ = −

∫Dv∆ϕ.

Proof. The remark applies to u ∈W 1,20 (D) as in Theorem 3.2.1. Subtracting yields∫

D(u− v)∆ϕ = 0 for all ϕ ∈ C∞(D), ϕ|∂D = 0.

Schwarz reection extends u−v to a (weakly, thus classically) harmonic function on R2, whichproves v ∈W 1,2(D) ∩ C0(D). In fact, v = u by the uniqueness part of Theorem 3.2.1.


3.3 Regularity of harmonic maps u ∈ W 1,2(D, Sn−1)

The regularity of two-dimensional harmonic maps into submanifolds of Rn was proved by F.Hélein around 1990, in three papers with increasing generality. The rst studied the casewhere the target manifold is a round sphere. As an auxiliary tool we will need a Hodge de-composition on the disk, which is discussed now.

For k = 0, 1, 2 let W 1,2(D,Λk) be the W 1,2-space of alternating k-linear forms. The Hodgestar, the exterior derivative d and its L2 adjoint d∗ are as in the following table1:

∗ d d∗

u ∈W 1,2(D) u dx ∧ dy ux dx+ uy dy 0

a dx+ b dy ∈W 1,2(D,Λ1) a dy − b dx (bx − ay) dx ∧ dy −(ax + by)

u dx ∧ dy ∈W 1,2(D,Λ2) u 0 uy dx− ux dy.

From the table one easily checks that d∗ = −∗d∗ and dd∗+d∗d = −∆ on forms of class W 2,2;in our notation ∆ = ∂2

x + ∂2y acting on the coecients. One further has the following rules of

partial integration, where i : ∂D → R2 is the inclusion map:∫D〈du, ω〉 −

∫Du d∗ω =

∫∂D

u i∗(∗ω) for u ∈W 1,2(D), ω ∈W 1,2(D,Λ1),∫D〈dω, θ〉 −

∫D〈ω, d∗θ〉 =

∫∂D

(∗θ) i∗ω for ω ∈W 1,2(D,Λ1), θ ∈W 1,2(D,Λ2).

The following is a special case of the decomposition theorem of W. Hodge.

Lemma 3.3.1 ([28]). Any η ∈ L2(D,Λ1) has an orthogonal decomposition

η = dα+ d∗β where α ∈W 1,20 (D), β ∈W 1,2(D,Λ2).

α, β are unique, assuming∫∂D ∗(νxβ) = 0. Moreover if d∗η = 0 then α = 0.

Proof. We start with uniqueness. For α ∈ C∞c (D), β ∈W 1,2(D,Λ2) we have∫D〈dα, d∗β〉 =

∫D〈d(dα), β〉 −

∫∂D

(∗β)i∗(dα) = 0.

This proves dW 1,20 (D) ⊥ d∗W 1,2(D,Λ2). Next assume that d∗η = 0. Multiplying the decom-

position η = dα+ d∗β with dζ and integrating yields

0 =

∫D〈η, dζ〉 =

∫D〈dα, dζ〉 for all ζ ∈W 1,2

0 (D).

Putting ζ = α shows α = 0. Now if η = 0, then d∗β = 0 which implies β constant. Thenormalization then yields β = 0, which proves the uniqueness.

1Identifying co-/vectors one may write d∗(u dx ∧ dy) = − ∗ du = ∇⊥u where ∇⊥u = (uy,−ux).

3.3. REGULARITY OF HARMONIC MAPS U ∈W 1,2(D,SN−1) 25

For existence we introduce the Sobolev space

W 1,2> (D,Λ1) = ω ∈W 1,2(D,Λ1) : i∗ω = 0,

and the bounded , symmetric operator L : W 1,2> (D,Λ1)→W 1,2

> (D,Λ1)′ given by

〈Lω,ϕ〉 =

∫D

(〈dω, dϕ〉+ 〈d∗ω, d∗ϕ〉

).

We show that L is strongly coercive. Namely we have for ω = a dx+ b dy ∈W 1,2> (D,Λ1)

〈Lω, ω〉 =

∫D

((bx − ay)2 + (ax + by)

2)

=

∫D

(|da|2 + |db|2

)+ 2

∫Dda ∧ db

=

∫D

(|da|2 + |db|2

)+

∫∂D

(abθ − aθb).

By assumption (a, b) = λ(cos θ, sin θ) along ∂D, which yields

aθ = λθ cos θ − λ sin θ,

bθ = λθ sin θ + λ cos θ.

Inserting yields abθ − aθb = λ2(cos2 θ + sin2 θ) = |ω|2, that is

〈Lω, ω〉 =

∫D|Dω|2 +

∫∂D|ω|2.

Now by a standard indirect argument using Rellich's lemma we get the estimate

‖ω‖2L2(D) ≤ C(‖Dω‖2L2(D) + ‖ω‖2L2(∂D)),

completing the proof of coercivity. The Riesz representation theorem applies to show that forany η ∈ L2(D,Λ1) there is a unique ω ∈W 1,2

> (D,Λ1) such that

〈Lω,ϕ〉 = 〈η, ϕ〉L2(D) for all ϕ ∈W 1,2> (D,Λ1).

Now ω ∈W 2,2> (D,Λ1) by regularity theory (see discussion below), thus∫

D〈dω, dϕ〉 =

∫D〈d∗dω, ϕ〉+

∫∂D

(∗dω) i∗ϕ︸︷︷︸=0

,

∫D〈d∗ω, d∗ϕ〉 =

∫D〈dd∗ω, ϕ〉 −

∫∂D

(d∗ω)i∗(∗ϕ).

Taking ϕ ∈ C∞c (D,Λ1) shows (dd∗ + d∗d)ω = η. Furthermore, any χ ∈ C∞(∂D) extends toχ ∈ C∞(D) with χ = 0 on D1/2(0). Taking ϕ = χdr and using ∗dr = dθ on ∂D yields

0 =

∫∂D

(d∗ω)i∗(∗χdr) =

∫∂D

(d∗ω)χ,

hence d∗ω = 0 on ∂D. Taking α = d∗ω and β = dω proves the desired decomposition.


We include an ad hoc argument for W 2,2 regularity. Let H be the upper halfplane andφ : H → D a conformal equivalence, with pullback metric gαβ = e2λδαβ . For 1-forms ω and2-forms θ one has the following transformation rules:

φ∗(d∗ω) = d∗g(φ∗ω) = e−2λd∗(φ∗ω),

φ∗(d∗θ) = d∗g(φ∗θ) = d∗(e−2λφ∗θ).

From this one computes putting γ = φ∗ω

φ∗(−∆ω) = e−2λ

(−∆γ − 2(dλ) d∗γ + 2(∗dλ) ∗ dγ

).

The form ω in the proof satises −∆ω = η, with boundary conditions i∗ω = 0, d∗ω = 0 on∂D. Therefore γ = φ∗ω solves the boundary value problem

−∆γ − 2(dλ) d∗γ + 2(∗dλ) ∗dγ = e2λφ∗η,

i∗∂Hγ = 0 on ∂H,

d∗γ = 0 on ∂H.

Writing γ = u dx+ v dy the rst equation on the boundary becomes u = 0 on ∂H, a Dirichletcondition. Using d∗γ = −(ux+vy) the second equation is reduced to vy = 0 on ∂H, a Neumanncondition. Now W 2,2 regularity is reduced to estimates for the scalar Laplacian under theseboundary conditions. The equations are coupled via the rst order terms, however these canbe put on the right hand side, since ω is already estimated in W 1,2 by coercivity. Recallingthe denition of α and β we arrive at the estimate

‖α‖W 1,2(D) + ‖β‖W 1,2(D) ≤ C ‖η‖L2(D). (3.8)

For η ∈ Lp(D,Λ1) where p ∈ (1,∞), one can also derive W 1,p(D) estimates for α and β usingstandard Calderon-Zygmund theory.

Theorem 3.3.2 ([23]). Any 2-dimensional harmonic map u ∈W 1,2(D,Sn−1) is continuous.

Proof. The proof is divided into two steps: rst one shows that the map is continuous, thenone shows smoothness. The second step was known previously to the work of Hélein; it is notspecic to harmonic maps but applies to general elliptic systems with quadratic nonlinearityin the gradient. Here we focus on the rst step, and deal with the second afterwards.

Using the identity ujduj = 12d|u|

2 = 0, we write the harmonic map equation in the form

−∆ui = |Du|2ui = 〈duj , uiduj − ujdui〉.

From the conservation law in Theorem 3.1.1 we know that

d∗(uiduj − ujdui) = 0 weakly for 1 ≤ i, j ≤ n.

Lemma 3.3.1, the Hodge lemma, yields forms βij = bijdx ∧ dy ∈W 1,2(D,Λ2) with

uiduj − ujdui = d∗βij = − ∗ dbij .

Inserting yields, recalling the notation a, b = axby − aybx,

−∆ui = −〈duj , ∗dbij〉 = uj , bij.


The right hand side has the desired Jacobi determinant structure. To arrange for zero bound-ary values, we let h ∈ W 1,2(D,Rn) be the harmonic extension of u|∂D. Then v = u − h ∈W 1,2(D,Rn) solves the problem

−∆vi = uj , bij in D, v = 0 on ∂D.

We have v ∈ C0(D,Rn) by Wente, see Theorem 3.2.1. But the harmonic function h is smoothin D, which proves the continuity of u = v + h.

We now take up the problem of higher regularity for systems of harmonic map type in arbitrarydimensions. This goes back to S. Hildebrandt, K.-O. Widman, and M. Wiegner

Theorem 3.3.3 ([27, 54]). Let u ∈W 1,2 ∩ L∞(B2(0),Rn) be a weak solution of the equation−∆u = A(u)(Du,Du) on B2(0) ⊂ Rm. Assume that for constants a,M <∞

|A(z)(p, p)| ≤ a |p|2 for all |z| ≤M, p ∈ Rn×m, (3.9)

‖u‖L∞(B2(0)) ≤ M. (3.10)

Then the following holds:

(1) Let α ∈ (0, 1). If aM ≤ ε0 = ε0(α) then u ∈ C0,α(B1(0),Rn).

(2) For α ∈ (23 , 1) we get further u ∈ C1,µ(B1(0),Rn), where µ = 3

2α− 1 ∈ (0, 12).

Proof. For x ∈ B1(0) and % ∈ (0, 1], let v ∈ W 1,2(B%(x),Rn) be harmonic with v − u ∈W 1,2

0 (B%(x)). We have the standard estimates

supB%(x)

|v| ≤ ‖u‖L∞(B%(x)) ≤M,

% supB%/2(x)

|Dv|+ %2 supB%/2(x)

|D2v| ≤ C

%m/2−1‖Dv‖L2(B%(x)).

For w = u− v and φ ∈W 1,20 ∩ L∞(B%(x),Rn) we infer∫B%(x)

〈Dw,Dφ〉 =

∫B%(x)

〈A(u)(Du,Du), φ〉.

We take φ = w. Using |w| ≤ |u|+ |v| ≤ 2M almost everywhere, we estimate∫B%(x)

|Dw|2 =

∫B%(x)

〈A(u)(Du,Du), w〉 ≤ 2aM

∫B%(x)

|Du|2.

Now let θ ∈ (0, 12 ]. Using |Du|2 ≤ 2(|Dv|2 + |Dw|2) we get

(θ%)2−m∫Bθ%(x)

|Du|2 ≤ 2(θ%)2−m∫Bθ%(x)

|Dv|2 + 2(θ%)2−m∫Bθ%(x)

|Dw|2

≤ Cθm(θ%)2−m∫B%(x)

|Dv|2 + 4aM(θ%)2−m∫B%(x)

|Du|2

≤ Cθ2(1 + aMθ−m

)%2−m

∫B%(x)

|Du|2.


In the last step we used that v(x) minimizes the Dirichlet energy with given boundary values.Assume for the moment that

aM ≤ θm, (3.11)

so that for any x ∈ B1(0), % ∈ (0, 1] we have the inequality

φ(x, θ%) ≤ Cθ2φ(x, %) where φ(x, %) = %2−m∫B%(x)

|Du|2.

Given % ∈ (0, 1] we choose k ∈ N0 with θk+1 < % ≤ θk, and iterate

φ(x, %) ≤ θ2−mφ(x, θk)

≤ θ2−m(Cθ2)kφ(x, 1)

≤ θ2−m−2α(Cθ2−2α)k %2α

∫B2(0)

|Du|2.

For given α ∈ [0, 1) we chose θ = θ(α) ∈ (0, 12 ] with Cθ2−2α ≤ 1, and take ε0 = θm in

assumption (1). Then (3.11) holds, and we conclude

%2−m∫B%(x)

|Du|2 ≤ C(α)%2α

∫B2(0)

|Du|2.

By Morrey's Dirichlet growth theorem u(x) is α-Hölder continuous on B1(0). To prove thatDu is also Hölder continous, go back to the inequality∫

−B%(x)

|Dw|2 =

∫−B%(x)

〈A(u)(Du,Du), w〉 ≤ a supB%(x)

|w|∫−B%(x)

|Du|2.

Now w = u − v = (u − ux,%) − (v − ux,%), where ux,% is the mean value on B%(x). We knowalready |u− ux,%| ≤ C%α, furthermore the maximum principle yields, as v = u on ∂B%(x),

supB%(x)

|v − ux,%| ≤ sup∂B%(x)

|u− ux,%| ≤ C%α.

Thus we have the estimate∫−B%(x)

|Dw|2 ≤ C%α∫−B%(x)

|Du|2 ≤ C %α%2α−2 = C%3α−2.

For convenience we put 3α− 2 =: 2µ. Clearly∫−Bθ%(x)

|Dw|2 ≤ θ−m∫−B%(x)

|Dw|2 ≤ Cθ−m%2µ,

|Dwx,θ%|2 ≤∫−Bθ%(x)

|Dw|2 ≤ Cθ−m%2µ.

Here we use the notation fx,% =∫−B%(x) f . We compute further, recalling u = v + w,∫

−Bθ%(x)

|Du− (Du)x,θ%|2 ≤ 2

∫−Bθ%(x)

|Dv − (Dv)x,θ%|2 + 2

∫−Bθ%(x)

|Dw|2 + 2|(Dw)x,θ%|2

≤ 2

∫−Bθ%(x)

|Dv − (Dv)x,θ%|2 + Cθ−m%2µ.


Using the Poincaré inequality and interior estimates for harmonic functions, we obtain∫−Bθ%(x)

|Dv − (Dv)x,θ%|2 ≤ C(θ%)2

∫−Bθ%(x)

|D2v|2

≤ C(θ%)2%−2

∫−B%(x)

|Dv − (Du)x,%|2

≤ Cθ2

∫−B%(x)

|D(v − `)|2 where `(y) = (Du)x,% y

≤ Cθ2

∫−B%(x)

|Du− (Du)x,%|2.

In the last step the minimizing property of v − ` was used. Combining we arrive at

φ(x, θ%) ≤ Cθ2φ(x, %) + Cθ−m%2µ where φ(x, %) =

∫−B%(x)

|Du− (Du)x,%|2.

Using induction, we see that

φ(x, θk) ≤ (Cθ2)kφ(x, 1) + Cθ−mθ(k−1)2µk−1∑j=0

(Cθ2−2µ)j

≤ θ2µk(

(Cθ2−2µ)k∫B2(0)

|Du|2 + Cθ−m−2µk−1∑j=0

(Cθ2−2µ)j).

Now x α ∈ (23 , 1) or equivalently µ ∈ (0, 1

2), and chose θ ∈ (0, 12 ] with Cθ2−2µ ≤ 1

2 . Then forθk+1 < % ≤ θk we infer

φ(x, %) ≤ θ−mφ(x, θk) ≤ C%2µ(∫

B2(0)|Du|2 + 1

).

Campanato's lemma implies that Du is µ-Hölder continuous on B1(0).

Corollary 3.3.4. Let u ∈W 1,2∩L∞(U,M) be a harmonic map on the open set U ⊂ Rm intothe smooth submanifold M ⊂ Rn. Assume that u(x) is continuous at x0 ∈ U , more precisely

lim%0‖u− p‖L∞(B%(x0)) = 0 for some p ∈M.

Then u(x) is smooth in a full neighborhood of x0.

Proof. By translations we may assume x0 = 0 and p = 0. The uλ : B → M , uλ(x) = u(λx),are harmonic and satisfy

‖uλ‖L∞(B2(0)) = ‖u‖L∞(B2λ(0)) → 0 as λ 0.

For xed α ∈ (0, 1), Theorem 3.3.3 yields uλ ∈ C1,µ(B1(0),Rn) for some µ > 0. ThusA(u)(Du,Du) is of class C0,µ near the origin, which means that its Newtonian potential andhence u(x) are locally C2,µ on a neighborhood of the origin. Repeated application of theSchauder estimates shows that u(x) is smooth on that neighborhood.


In subsequent work Hildebrandt-Widman and Wiegner proved Hölder continuity assumingonly aM < 1. By consequence, any harmonic map with image strictly contained in a hemi-sphere is smooth. This result is sharp in view of the example u(x) = x

|x| into Sn−1 ⊂ Sn.

The techniques of Hélein presented in this section were generalized to harmonic maps intothe sphere for m ≥ 3 by C. Evans [12]. Assuming that the map is also stationary with respectto domain variations, he obtained regularity away from a closed set of (m − 2)-dimensionalHausdor measure zero. Any hopes to prove regularity for weakly harmonic maps in dimensionm ≥ 3 without extra assumptions were dashed by T. Rivière [36]. He constructed a harmonicmap from the 3-dimensional ball B into S2 which is discontinuous on any open subset of B.

Chapter 4

Hardy space

In this chapter we study the Hardy space H1(Rn).

4.1 Higher integrability of Jacobi determinants

We start by collecting some basic results about the maximal function.

Lemma 4.1.1. Let f : Rn → [0,∞) be measurable and 0 < p <∞. Then we have∫Rnfp(x) dx = p

∫ ∞0

αp−1|x : f(x) > α| dα. (4.1)

Proof. Let χf be the characteristic function of the set (x, α) ∈ Rn × [0,∞) : f(x) > α.Then χf is Ln × L1 measurable, and

f(x)p = p

∫ f(x)

0αp−1 dα = p

∫ ∞0

αp−1χf (x, α) dα.

Integrating we get by Fubini's theorem∫Rnf(x)p dx = p

∫ ∞0

αp−1

∫Rnχf (x, α) dx dα = p

∫ ∞0

αp−1|x : f(x) > α| dα.

Denition 4.1.2. For f ∈ L1loc(Rn) we dene its maximal function Mf : Rn → [0,∞] by

Mf(x) = supr>0

1

|Br(x)|

∫Br(x)

|f(x)| dx. (4.2)

When dealing with the maximal function one needs two basic principles, namely the Vitalicovering lemma and the Calderon Zygmund decomposition. For the proof of the rst we referto [13], while the proof of the second is included for convenience.

Theorem 4.1.3 (Vitali). Let Bj, j ∈ J , be a family of closed balls in Rn with strictly positiveradius and supj∈J diam (Bj) <∞. There is a disjoint subfamily Bjkk∈N such that

⋃j∈J

Bj ⊂∞⋃k=1

5Bjk ,

31

32 CHAPTER 4. HARDY SPACE

where 5B is the concentric ball with 5 times the radius. In particular if E ⊂⋃j∈J Bj then

|E| ≤ 5n∞∑k=1

|Bjk |. (4.3)

Theorem 4.1.4 (Calderon-Zygmund [6]). Let f ∈ L1(Rn) with f ≥ 0. For any α > 0there exists a countable family G of closed cubes with pairwise disjoint interior, such that thefollowing holds:

(i) α <∫−Q f(x) dx ≤ 2nα for any Q ∈ G.

(ii) f(x) ≤ α for almost all x ∈ Rn\G, where G =⋃Q∈G Q.

(iii) |G| ≤ C

α‖f‖L1(Rn).

Proof. Chose a subdivision of Rn into congruent cubes P having volume |P | ≥ 1α‖f‖L1(Rn),

and denote this family by F0. Clearly for P ∈ F0

1

|P |

∫Pf(x) dx ≤ α. (4.4)

Using induction we now dene families Fk, Gk of cubes for k = 1, 2, . . .. For this divide eachP ∈ Fk−1 into 2n congruent subcubes using edge bisection. Then denote by Fk the subfamilyfor which (4.4) holds, and by Gk the subfamily where (4.4) fails. The union of the cubes inFk, Gk is denoted by Fk,Gk; we note Rn = Fk ∪

⋃kj=1Gj . The process is iterated as long as

Fk is nonempty. We prove the result for G =⋃Gk. Two cubes Q ∈ Gk, Q′ ∈ G` with k < `

have disjoint interior, since Q′ comes from some P ∈ Fk. If Q belongs to Gk and comes fromP ∈ Fk−1, then

α <1

|Q|

∫Qf(x) dx ≤ 2n

|P |

∫Pf(x) dx ≤ 2nα.

This proves (i). For x /∈ G, we have x ∈ Pk for a sequence Pk ∈ Fk, thus

1

|Pk|

∫Pk

f(x) dx ≤ α where |Pk| → 0.

By the Lebesgue dierentiation theorem, see Cor. 2, Sect. 1.7 of [13], the left hand sideconverges to f(x) a.e. which proves (ii). Finally (iii) follows since

|G| =∑Q∈G|Q| ≤ 1

α

∑Q∈G

∫Q|f(x)| dx =

1

α

∫G|f(x)| dx.

Our aim is to compare the function and its maximal function, in terms of integrability.The following are the key inequalities.

Lemma 4.1.5. Let f : Rn → [0,∞] be measurable and α > 0. Then for C = C(n) <∞

|x : Mf(x) > α| ≤ C

α

∫x:f(x)>α

2f(x) dx. (4.5)

4.1. HIGHER INTEGRABILITY OF JACOBI DETERMINANTS 33

Reversely, we have for constants C = C(n) <∞ and λ = λ(n) > 0

1

α

∫x:f(x)>α

f(x) dx ≤ C |x : Mf(x) > λα|. (4.6)

Proof. To prove (4.5) we chose for each x ∈ Rn with Mf(x) > α a radius rx > 0 such that∫Bxf(y) dy > α|Bx| where Bx = Brx(x).

By Vitali, Theorem 4.1.3, there are disjoint Bxk , k ∈ N, such that the set x : Mf(x) > α iscovered by the enlarged balls 5Bxk . Hence

|x : Mf(x) > α| ≤ 5n∞∑k=1

|Bxk | ≤ 5n

α

∞∑k=1

∫Bxk|f(y)| dy ≤ 5n

α

∫Rn|f(x)| dx. (4.7)

The trick to obtain the improved inequality (4.5) is to consider

f1(x) =

f(x) if f(x) ≥ α/2,0 otherwise.

Clearly f(x) ≤ f1(x) + α2 for all x ∈ Rn, which implies Mf(x) ≤Mf1(x) + α

2 and thus

x : Mf(x) > α ⊂ x : Mf1(x) >α

2.

Applying the previous estimate (4.7) to f1 yields

|x : Mf(x) > α| ≤ |x : Mf1(x) >α

2| ≤ 2 · 5n

α

∫x:f(x)>α/2

f(x) dx.

We now turn to the proof of (4.6), applying Theorem 4.1.4 by Calderon-Zygmund. For anypoint x ∈ Q with Q ∈ G we have, putting d = diamQ,

α <1

|Q|

∫Qf(y) dy ≤ |Bd(x)|

|Q|1

|Bd(x)|

∫Bd(x)

f(y) dy ≤ 2nnn/2Mf(x).

Using (ii) and (i) from Theorem 4.1.4 we infer

1

α

∫x:f(x)>α

f(x) dx ≤ 1

α

∑Q∈G

∫Qf(x) dx

≤ 2n∑Q∈G|Q|

≤ 2n|x : Mf(x) > 2−nn−n/2α|.

This proves (4.6).

As a rst application, we see that there is no dierence between f and Mf as regards Lp

integrability for 1 < p <∞.


Theorem 4.1.6 (Hardy-Littlewood). For f ∈ Lp(Rn) with 1 < p ≤ ∞ we have

‖Mf‖Lp(Rn) ≤ C‖f‖Lp(Rn) where C = C(n, p) <∞. (4.8)

Proof. The case p =∞ is trivial with C(n,∞) = 1. For 1 < p <∞ we estimate using (4.5)∫Rn|Mf(x)|p dx = p

∫ ∞0

αp−1|x : |f(x)| > α dα

≤ Cp

∫ ∞0

αp−2

∫x∈Rn:|f(x)|>α/2

|f(x)| dx dα

= Cp

∫Rn|f(x)|

∫ 2|f(x)|

0αp−2 dα dx

≤ Cp

p− 1

∫Rn|f(x)|p dx.

Remark 4.1.7. The function Mf is never in L1(Rn) unless f ≡ 0. In fact, for any R <∞we have BR(0) ⊂ B2|x|(x) for |x| ≥ R, yielding the lower bound

Mf(x) ≥ 1

|B2|x|(x)|

∫B2|x|(x)

|f(y)| dy ≥ c

|x|n

∫BR(0)

|f(y)| dy.

If ‖f‖L1(BR(0) > 0 then the right hand side is not integrable. To get an example where themaximal function is locally not integrable consider f : Rn → R given by

f(x) =1

|x|n log2 |x|χB1/e(0) ≥ 0 where e = 2.718 . . . .

We compute substituting r = e−t for 0 < % ≤ 1/e∫B%(0)

f(x) dx =

∫ %

0

dr

r log2 r=

∫ ∞− log %

dt

t2= − log %.

In particular f ∈ L1(Rn). On the other hand as B2|x|(x) ⊃ B|x|(0), we estimate for |x| ≤ 1/e

Mf(x) ≥ c(n)

|x|n

∫B2|x|(x)

f(y) dy ≥ c(n)

|x|n

∫B|x|(0)

f(y) dy ≥ − c(n)

|x|n log |x|.

The right hand side is not integrable near the origin (for the integrals see also example 2.4.2).

Contrary to the case p > 1, the L1 integrability of the maximal function Mf implies animproved integrability of f . This was discovered by E. Stein.

Theorem 4.1.8 (Stein [44]). Let f : Rn → [0,∞) be measurable having support contained ina ball B. Then one has an estimate∫

Rnf(x) log+ f(x) dx ≤ C

(B, ‖Mf‖L1(B)

). (4.9)


Proof. By (4.6) we can estimate, recalling λ = λ(n) > 0,∫Rnf(x) log+ f(x) dx =

∫Rnf(x)

∫ ∞1

χf (x, α)dα

αdx

=

∫ ∞1

1

α

∫x:f(x)>α

dx dα

≤ C

∫ ∞1|x : Mf(x) > λα| dα

= C

∫ ∞λ|x : Mf(x) > β| dβ

= C

∫x:Mf(x)>λ

Mf(x) dx.

Let us assume for simplicity that B = B1; the general case follows by scaling. For |x| > 1 wehave, using that f is supported in B,

Mf(x) ≤ C

(|x| − 1)n‖f‖L1 ,

in particular Mf(x) ≤ λ for (|x| − 1)n ≥ c ‖f‖L1 where c = c(n) > 0. Thus we have for a

radius R =(C‖f‖L1

)1/n+ 1∫

x:|x|> 32,Mf(x)>λ

Mf(x) dx ≤ C‖f‖L1Rn ≤ C(1 + ‖f‖2L1

)≤ C

(1 + ‖Mf‖2L1(B)

).

In the remaining annulus, we consider the reection

φ : B1\B 12→ B 3

2\B1, φ(x) = (2− |x|) x

|x|.

As B1 is contained in the halfplane y ∈ Rn : 〈y, x|x|〉 ≤ 1, Pythagoras implies that |y − x| ≤|y − φ(x)| for any y ∈ B, and hence Br(φ(x)) ∩ B ⊂ Br(x) ∩ B for any r > 0. We concludeusing the transformation formula∫

1<|x|< 32

Mf(y) dy ≤ C∫

12<|x|<1

Mf(φ(x)) dx ≤ C∫

12<|x|<1

Mf(x) dx.

The theorem follows by combining the estimates.

Remark 4.1.9. The set of functions for which |f | log+ |f | is integrable is called the L logLclass. As noted by Stein the above theorem is sharp, in the sense that the L logL propertyimplies the local integrability of Mf . To see this we write for any set E ⊂ Rn∫

EMf dx = 2

∫ ∞0|x ∈ E : Mf(x) > 2α| dα

≤ 2 |E|+ 2

∫ ∞1|x : Mf(x) > 2α| dα.


Now (4.5) yields, with χf (x, α) the characteristic function of |f(x)| > α,∫ ∞1|x : Mf(x) > 2α| dα ≤

∫ ∞1

(Cα

∫Rnχx:|f(x)|>α|f(x)| dx

)dα

= C

∫Rn|f(x)|

∫ ∞1

χf (x, α)dα

αdx

= C

∫Rn|f(x)| log+ |f(x)| dx.

Thus for an arbitrary set E we obtain∫EMf(x) dx ≤ C

(|E|+

∫Rn|f(x)| log+ |f(x)| dx

). (4.10)

Next we review some facts about degree theory and Jacobi determinants. Let Ω ⊂ Rn bea bounded domain of class C1. For a map u ∈ C2(Ω,Rn) the oriented multiplicity functioniu : Rn → Z is given by

iu(y) =

∑u(x)=y sign detDu(x) if y /∈ u(∂Ω) is a regular value,

0 else.(4.11)

Here y /∈ u(∂Ω) is a regular value if and only if detDu(x) 6= 0 for all x ∈ u−1y. By theinverse function theorem and compactness, each regular value has only nitely many preimagesso that the sum is dened. Our main tool in the following is the transformation formula: forany g ∈ L1(Ω), the function y 7→

∑u(x)=y g(x) is integrable on Rn and∫

Ωg(x)|detDu(x)| dLn(x) =

∫Rn

( ∑u(x)=y

g(x))dLn(y).

In particular, the set of points in Rn\u(∂Ω) which are not regular has Lebesgue measurezero. This is actually a step in the proof of the transformation formula, see [13, Section 3.3].As u(∂Ω) is also a null set, the rst alternative in the denition of iu applies almost everywhere.

Now let h ∈ C0c (Rn). By the transformation formula, we calculate∫

Rnh(y) iu(y) dLn(y) =

∫Ωh(u(x)) sign detDu(x)|detDu(x)| dLn(x)

=

∫Ωh(u(x)) detDu(x) dLn(x)

=

∫Ωu∗(h(y) dy),

where dy = dy1 ∧ . . . ∧ dyn. Inserting h = div φ where φ ∈ C1c (Rn,Rn) we infer∫

Rn(div φ) iu dLn =

∫Ωu∗(div φdy) =

∫Ωu∗d(φxdy) =

∫Ωdu∗(φxdy) =

∫∂Ωu∗(φxdy).

Taking sptφ ⊂ Rn\u(∂B) yields Diu = 0 on Rn\u(∂B), hence iu is constant on the compo-nents of that set. For general φ we compute further

u∗(φxdy)(e1, . . . , ej , . . . , en) = det(φ u, ∂1u, . . . , ∂ju, . . . , ∂nu)

= (φi u) det(ei, ∂1u, . . . , ∂ju, . . . , ∂nu)

= (−1)j−1(φi u) cof(Du)ij .


Here cofij(Du) equals (−1)i+j times the ij-minor, i.e. the subdeterminant when the i-th rowand j-th column of Du is deleted. Now assume |φ| ≤ 1, so that by Cauchy-Schwarz∣∣∣ ∫

∂Ωu∗(φxdy)

∣∣∣ ≤ ∫∂Ω|u∗(φxdy)| dHn−1 ≤

∫∂Ω|cof(Du)| dHn−1.

Recalling the denition of the variation measure |Diu| we arrive at

|Diu|(Rn) ≤∫∂Ω|cof(Du)| dHn−1. (4.12)

The following is Lemma 1.3 in [33], see also Theorem 2.10 in [43] for the case n = 3.

Lemma 4.1.10. Let Ω ⊂ Rn be open and bounded, u ∈ W 1,nloc (Ω,Rn). For any x ∈ Ω and

almost all r ∈ (0, dist(x, ∂Ω)), we have for a constant C = C(n) <∞∣∣∣ ∫Br(x)

detDudLn∣∣∣n−1n ≤ C

∫∂Br(x)

|cof(Du)| dHn−1. (4.13)

Proof. We rst assume that u ∈ C2(Ω,Rn). Using once more the transformation formula, andthe fact that iu,Br(x) is integer-valued, we have∫

Br(x)detDudLn =

∫Rniu,Br(x) dLn ≤

∫Rn|iu,Br(x)|

nn−1 dLn.

By the Sobolev embedding theorem, see [13, Sec. 5.6], and by (4.12) we can continue

(∫Rn|iu,Br(x)|

nn−1 dLn

)n−1n ≤ C |Diu,Br(x)|(Rn) ≤ C

∫∂Br(x)

|cof(Du)| dHn−1.

This proves the lemma for maps u ∈ C2(Ω,Rn). Now let u ∈ W 1,nloc (Ω,Rn), and chose uk ∈

C2(Ω,Rn) with uk → u in W 1,nloc (Ω,Rn). By Fatou we have for R < dist(x, ∂Ω)∫ R

0lim infk→∞

∫∂Br(x)

|cof(Duk)−cof(Du)| dHn−1dr ≤ limk→∞

∫BR(x)

|cof(Duk)−cof(Du)| dLn = 0.

Thus for almost all r ∈ (0,dist(x, ∂Ω)) inequality (4.13) follows by approximation.

We have now collected all ingredients to prove Müller's higher integrability theorem. Themain idea is to estimate the maximal function of the Jacobi determinant by the maximalfunction of the (n− 1)× (n− 1) minors using (4.13). The advantage is that the minors comewith a power n

n−1 , so that Theorem 4.1.6 by Hardy-Littlewood can be applied.

Theorem 4.1.11 ([33]). Let Ω ⊂ Rn, n ≥ 2, be open and bounded. If u ∈ W 1,n(Ω,Rn) hasdetDu ≥ 0 almost everywhere, then for any compact set K ⊂ Ω∫

KdetDu log+(detDu) ≤ C(K, ‖u‖W 1,n(Ω)). (4.14)


Proof. Let g = χK detDu. The result follows from Theorem 4.1.8, once we have the estimate

||Mg||L1(B) ≤ C(B, ||u||W 1,n(Ω)) for some ball B ⊃ K. (4.15)

We show an improved version where the right hand side depends only on the Lnn−1 -norm of

cof(Du). Let d = dist(K, ∂Ω). For r ≥ d/4 and all x ∈ Rn we have the trivial inequality∫−Br(x)

|g| dLn ≤ C

dn

∫Ω| detDu| dLn ≤ C

dn

∫Ω|cof(Du)|

nn−1 dLn. (4.16)

For the last step we use Du · cof(Du)T = (detDu) Id, which implies

| detDu|n = | detDu| |det cof(Du)| ≤ | detDu| |cof(Du)|n,

thus | detDu| ≤ |cof(Du)|n/(n−1). Now consider the case r ≤ d/4, in particular we mayassume dist(x, ∂Ω) > d/2. As detDu ≥ 0, Lemma 4.1.10 implies for almost all % ∈ (r, 2r)(∫

Br(x)|g| dLn

)n−1n ≤

(∫B%(x)

detDudLn)n−1

n ≤ C∫∂B%(x)

|cof(Du)| dHn−1.

Integrating on (r, 2r) and dividing by rn gives, putting cof(Du) = 0 on Rn\Ω,(∫−Br(x)

|g| dLn)n−1

n ≤ C∫−B2r(x)

|cof(Du)| dLn ≤ CM(cof(Du))(x). (4.17)

Combining (4.16) and (4.17) yields

Mg(x) ≤ CM(cof(Du))(x) +C

dn

∫Ω|cof(Du)|

nn−1 dLn.

As nn−1 > 1 we can apply Theorem 4.1.6 by Hardy-Littlewood to get

‖M(cof(Du))‖L

nn−1 (Rn)

≤ C‖cof(Du)‖L

nn−1 (Rn)

≤ C‖cof(Du)‖L

nn−1 (Ω)

.

We nally conclude, chosing any ball B ⊃ K,

‖Mg‖L1(B) ≤ C(B, ‖cof(Du)‖

Lnn−1 (Ω)

).

In [33] the estimate is stated for detDu log(2 + detDu). This follows easily from theversion above, using log(2 + s) ≤ log s+ 1 for s ≥ 2. In Section 7 of [33] a counterexample isgiven, showing that the condition detDu ≥ 0 cannot be dropped.

4.2 The Hardy space

As is well-known a bounded sequence fk in L1(Rn) may have a weak limit which is not rep-resentable by an L1(Rn) function, but only by a signed Radon measure. The Hardy spaceH1(Rn) is continuously embedded into L1(Rn) and has a norm which scales like the L1(Rn)norm. However, as opposed to L1(Rn) the unit ball in H1(Rn) is weakly sequentially compact.

To start we consider on C0c (Rn) the usual convergence φk → φ dened by

4.2. THE HARDY SPACE 39

•⋃k∈N sptφk is a bounded subset of Rn,

• φk → φ uniformly on Rn.

By the Riesz representation theorem any Λ ∈ C0c (Rn)′ has the form

Λ(φ) =

∫Rnφσdµ,

where µ is a Radon measure and σ : Rn → ±1 is µ measurable. The convolution ofΛ ∈ C0

c (Rn)′ with φ ∈ C0c (Rn) is the function

φ ∗ Λ : Rn → R, (φ ∗ Λ)(x) = Λ(φx) where φx(y) = φ(x− y).

For example (φ ∗ δ0)(x) = φ(x). We note that φ ∗ Λ ∈ C0(Rn) since the map Rn → C0c (Rn),

x 7→ φx, is continuous, as is Λ : C0c (Rn) → R. Any f ∈ L1

loc(Rn) denes canonically afunctional Λf ∈ C0

c (Rn)′ by

Λf (φ) =

∫Rnφ(y)f(y) dy,

and the notions of convolutions are consistent in the sense that

(φ ∗ Λf )(x) = Λf (φx) =

∫Rnφ(x− y)f(y) dy = (φ ∗ f)(x).

Now we introduce the following class of test functions.

T = φ ∈ C∞c (Rn) : sptφ ⊂ B1(0) and ‖Dφ‖L∞ ≤ 1. (4.18)

Clearly ‖φ‖L∞(Rn) ≤ 1 for φ ∈ T . For the rescalings we use the notation

φt(x) = t−nφ(xt

)for t > 0, (4.19)

thussptφt ⊂ Bt(0) and ‖Dφt‖L∞(Rn) ≤ C t−(n+1).

In the following denition the term grand means that one maximizes over all kernels in T ,rather than working with a specic one.

Denition 4.2.1. The grand maximal function of Λ ∈ C0c (Rn)′ is dened by

Λ∗(x) = supφ∈T supt>0 |φt ∗ Λ(x)|. (4.20)

Let φ ∈ C∞c (Rn) with sptφ ⊂ BR(0) and ‖Dφ‖L∞(Rn) = α > 0. Then the function ψ(x) =1Rαφ(Rx) belongs to T , and we calculate

(φ ∗ Λ)(x) = Λ(y 7→ φ(x− y)

)= Λ

(y 7→ Rn+1αψR(x− y)

)= Rn+1α(ψR ∗ Λ)(x).

Thus for any φ ∈ C∞c (Rn) we have the inequality

|(φ ∗ Λ)(x)| ≤ Rn+1‖Dφ‖L∞(Rn)Λ∗(x) if sptφ ⊂ BR(0). (4.21)

The Hardy space can be dened in several ways, whose equivalence is by no means obvious,see [14] or [46]. A nice introduction is due to Semmes [41].


Denition 4.2.2. H1(Rn) is the set of all Λ ∈ C0c (Rn)′ for which Λ∗ ∈ L1(Rn). We put

‖Λ‖H1(Rn) = ‖Λ∗‖L1(Rn). (4.22)

As (Λ1 + Λ2)∗ ≤ Λ∗1 + Λ∗2 and (αΛ)∗ = |α|Λ∗, the Hardy space is a normed vector space. Thefollowing lemma will allow us to consider its elements as L1 functions.

Lemma 4.2.3. The space H1(Rn) is continuously embedded into L1(Rn).

Proof. We use approximation by smoothing. Chose a xed kernel φ ∈ C∞c (Rn) with sptφ ⊂B1(0) and

∫Rn φ(x) dx = 1. Putting φ(x) = φ(−x) we have using that Λ is linear and

continuous ∫Rn

(φt ∗ Λ)(x)η(x) dx =

∫Rn

Λ(φxt ) η(x) dx

= Λ(y 7→

∫Rnφxt (y) η(x) dx

)= Λ

(φt ∗ η

)→ Λ(η) as t 0.

Using (4.21) we have the bound

|φt ∗ Λ(x)| ≤ tn+1‖Dφt‖L∞(Rn)Λ∗(x) = ‖Dφ‖L∞(Rn)Λ

∗(x) ∈ L1(Rn).

As the φt ∗ Λ are equiintegrable they converge weakly subsequentially in L1(Rn) to somef ∈ L1(Rn). Then Λf = Λ by the above, and the sublimit improves to a limit. Finally

‖f‖L1(Rn) ≤ lim inft0

‖φt ∗ Λ‖L1(Rn) ≤ ‖Dφ‖L∞(Rn)‖Λ∗‖L1(Rn).

From now on the elements of H1(Rn) are regarded as L1(Rn) functions. As pointed out atthe beginning, the following weak compactness theorem distinguishes the space H1(Rn) fromL1(Rn).

Theorem 4.2.4 (weak compactness in H1(Rn)). Let fk be a bounded sequence in H1(Rn).Then there exists an f ∈ H1(Rn), such that for a subsequence fk → f in C0

c (Rn)′, and

‖f‖H1(Rn) ≤ liminfk→∞ ‖fk‖H1(Rn). (4.23)

Proof. By Lemma 4.2.3 we have ‖fk‖L1(Rn) ≤ C, so that fk → Λ in C0c (Rn)′ after passing to

a subsequence. Now for φ ∈ T and t > 0 we have

(φt ∗ fk)(x) =

∫Rnφxt (y)fk(y) dy

k→∞−→ Λ(φxt ) = (φt ∗ Λ)(x),

which implies(φt ∗ Λ)(x) = lim

k→∞(φt ∗ fk)(x) ≤ lim inf

k→∞f∗k (x).

Take the supremum with respect to φ ∈ T and t > 0. Then by Fatou's lemma

‖Λ‖H1(Rn) = ‖Λ∗‖L1(Rn) ≤ lim infk→∞

‖f∗k‖L1(Rn) = lim infk→∞

‖fk‖H1(Rn).

Finally Λ = Λf where f ∈ L1(Rn) by Lemma 4.2.3, which nishes the proof.


Corollary 4.2.5 (Cancellation in H1(Rn)). For every f ∈ H1(Rn) we have∫Rnf(y) dy = 0. (4.24)

Proof. We rst check the scaling of the Hardy norm. For f ∈ H1(Rn) and t > 0 let ft(y) =t−nf(yt ). Compute for φ ∈ T , t > 0,

(φs ∗ ft)(x) =

∫Rnφs(x− y)t−nf

(yt

)dy

=

∫Rnφs(t (x

t− z)

)f(z) dz

= t−n(φ st∗ f)(xt

).

We estimate on the right with f∗(xt ), and then take the supremum over φ ∈ T , s > 0, to get

f∗t (x) = t−nf∗(xt

)for x ∈ Rn, t > 0.

We rst get the inequality, for equality we use (ft) 1t

= f . In particular we obtain

‖ft‖H1(Rn) = ‖f‖H1(Rn) for all t > 0.

Now for f ∈ L1(Rn) we know that ft converges to a multiple of the Dirac measure as t 0,in fact dominated convergence yields∫

Rnφ(x)ft(x) dx =

∫Rnφ(ty)f(y) dy → φ(0)

∫Rnf(y) dy.

On the other hand we must have ft → f ∈ H1(Rn) for a subsequence by Theorem 4.2.4. Asthe Dirac measure is not in H1(Rn) we conclude that

∫Rn f(y) dy = 0.

Alternative to the given argument, one can estimate more directly to show∣∣∣ ∫Rnf(y) dy

∣∣∣ ≤ C lim inf|x|→∞

f∗(x)

|x|n.

For f ∈ H1(Rn) the right hand side must be zero. In the next lemma the grand maximalfunction is compared to the maximal function of Hardy-Littlewood.

Lemma 4.2.6. Let f ∈ L1loc(Rn). The for any x ∈ Rn we have

(1) f∗(x) ≤ CMf(x),

(2) Mf(x) ≤ C f∗(x), if f ≥ 0.

Proof. To prove the rst statement we calculate for φ ∈ T , t > 0,

|(φt ∗ f)(x)| = t−n∣∣∣ ∫

Rnφ(x− yt

)f(y) dy∣∣∣

≤ ‖φ‖L∞(Rn)t−n∫Bt(x)

|f(y)| dy

≤ CMf(x).


Taking the supremum over φ ∈ T , t > 0 shows claim (1). Next we chose φ ∈ T such thatφ ≥ 1

4χB1/2(0). For f ≥ 0 we can then estimate

(φt ∗ f)(x) = t−n∫Rnφ(x− y

t

))f(y) dy

≥ t−n1

4

∫Bt/2(x)

f(y) dy

≥ c

∫−Bt/2(x)

f(y) dy.

Recalling the deniion of f∗, this implies

f∗(x) ≥ supt>0

(φt ∗ f)(x) ≥ c supt>0

∫−Bt/2(x)

f(y) dy = Mf(x).

Theorem 4.2.7. Let f ∈ L1(Rn) such that spt f ⊂ BR(0) and∫Rn f(x)dx = 0. Then

‖f‖H1(Rn) ≤ C(‖Mf‖L1(B2R(0)) + C‖f‖L1(Rn)

).

Proof. We estimate the L1 integral of f∗ by splitting into the regions B2R(0) and Rn\B2R(0).For |x| ≤ 2R the inequality f∗(x) ≤ CMf(x) from Lemma 4.2.6 yields∫

B2R(0)f∗(x) dx ≤ C ‖Mf‖L1(B2R(0)). (4.25)

For |x| ≥ 2R we have dist(x,BR(0)) = |x| −R ≥ 12 |x|, hence

φt ∗ f(x) =

∫B1(0)

φ(z)f(x− tz) dz = 0 when 0 < t <1

2|x|.

For t ≥ 12 |x| we estimate using

∫Rn f(y) dy = 0

|φt ∗ f(x)| =∣∣∣ ∫

BR(0)(φt(x− y)− φt(x))f(y) dy

∣∣∣≤ ‖Dφt‖L∞(Rn)R

∫Rn|f(y)| dy

≤ CR

|x|n+1

∫Rn|f(y)| dy.

Integrating shows ∫Rn\B2R(0)

f∗(x) dx ≤ C∫Rn|f(y)| dy. (4.26)

The theorem follows by combining (4.25) and (4.26).

Corollary 4.2.8. Let f ∈ Lp(Rn), 1 < p ≤ ∞, with spt f ⊂ BR(0) and∫Rn f(x)dx = 0.

Then f ∈ H1(Rn) and

‖f‖H1(Rn) ≤ C(p)Rn−n

p ‖f‖Lp(Rn).


Proof. The Hardy-Littlewood inequality, see Theorem 4.1.6, implies

‖Mf‖L1(B2R(0)) ≤ C Rn−n

p ‖Mf‖Lp(Rn) ≤ C(p)Rn−n

p ‖f‖Lp(Rn).

Since also ‖f‖L1(Rn) ≤ C Rn−n

p ‖f‖Lp(Rn), the claim follows from Theorem 4.2.7.

Remark 4.2.9. We give an example of a function f ∈ H1(R) with compact support, whosemaximal function is not locally integrable. Consider

f(x) =∞∑n=2

an(x)

n(log n)2where an = −nχ[− 1

n,0) + nχ(0, 1

n].

The function a1(x) belongs to H1(Rn) by Corollary 4.2.8. As an(x) = na1(nx) = (a1) 1n

(x),

we have ‖an‖H1(R) = ‖a1‖H1(R) for all n, and

‖f‖H1(R) ≤ C∞∑n=2

1

n(log n)2<∞.

On the other hand, for 12(n+1) < x ≤ 1

2n with n ≥ 1 we can estimate

f(x) ≥2n∑

k=n+1

1

(log k)2≥ n

(log 2n)2≥ 1

4x(log 1/x)2.

For 0 < x ≤ 12 we therefore have a lower bound

f(x) log f(x) ≥ 1

4x log 1/x+

1

4x(log 1/x)2log

1

4(log 1/x)2.

The second term on the right is integrable, but the rst is not. Hence f(x) does not belong tothe L logL class. By Stein's theorem 4.1.8 we conclude that Mf is not locally integrable, andthat the integrability of f∗ is due to a cancellation eect.

In 1993 Coifman, Lions, Meyer & Semmes [9] discovered that the famous div-curl lemma byMurat and Tartar is connected to Hardy space theory. We use dierential forms to state theresult in general form. We will need the following Hodge lemma.

Lemma 4.2.10. Let β ∈ Lq(Rn,Λk) where 1 < q <∞. If dβ = 0 in the sense of distributions,then there exists a form γ ∈W 1,q

loc (Rn,Λk−1) such that dγ = β, and

‖Dγ‖Lq(Rn) ≤ C ‖β‖Lq(Rn) where C = C(n, q). (4.27)

Proof. We rst show that for any β ∈ Lq(Rn,Λk) there exists a φ ∈W 2,qloc (Rn,Λk) such that

∆φ = β and ‖D2φ‖Lq(Rn) ≤ C‖β‖Lq(Rn) where C = C(n, q) <∞.

For β ∈ C∞c (Rn) we have the Newtonian potential

φ(x) =

∫Rn

Γ(x− y)β(y) dy where Γ(z) =

1

(2−n)ωn|z|2−n for n ≥ 3,

12π log |z| for n = 2.


The Lq estimate of D2φ is the classical Calderon-Zygmund inequality, see for instance [2]. Forgeneral β ∈ Lq(Rn) we approximate by βk ∈ C∞c (Rn) in Lq(Rn). Then βk = ∆φk where

lim supk→∞

‖D2φk‖Lq(Rn) ≤ C‖β‖Lq(Rn).

Subtracting a linear function, we can arrange that∫B1(0)

φk dx = 0 and

∫B1(0)

Dφk dx = 0.

Then a standard contradiction argument using Rellich's theorem implies

‖φk‖W 1,q(BR(0)) ≤ C(R, q)‖D2φk‖Lq(Rn).

After passing to a subsequence we have φk → φ in W 1,qloc (Rn), and ‖D2φ‖Lq(Rn) ≤ C‖β‖Lq(Rn).

Thus φ is the desired solution of ∆φ = β. We next show that ψ = d∗dφ solves ∆ψ = d∗dβ inthe sense of distributions. Namely for ζ ∈ C∞c (Rn,Λk) we get∫

Rn〈ψ,∆ζ〉 dx = −

∫Rn〈d∗dφ, (dd∗ + d∗d)ζ〉 dx

= −∫Rn〈dφ, dd∗dζ〉 dx−

∫Rn〈d∗φ, d∗d∗dζ〉 dx

=

∫Rn〈β, d∗dζ〉 dx.

Now if dβ = 0 then d∗dφ ∈ Lq(Rn,Λk) is harmonic. By Weyl's lemma and the mean valueinequality we conclude that d∗dφ is identically zero. This implies −dd∗φ = β, and the lemmais proved by taking γ = −d∗φ.

The following is the key observation of Coifmann, Lions, Meyer and Semmes.

Theorem 4.2.11 ([9]). Let α ∈ Lp(Rn,Λk), β ∈ Lq(Rn,Λn−k) where 1 < p, q < ∞ and1p + 1

q = 1. If dα = 0 and dβ = 0 weakly, then α ∧ β ∈ H1(Rn) and

‖α ∧ β‖H1(Rn) ≤ C ‖α‖Lp(Rn)‖β‖Lq(Rn). (4.28)

Proof. The equation dα = 0 has the alternative weak form∫Rnα ∧ dζ = 0 for all ζ ∈ C∞c (Rn,Λn−k−1). (4.29)

To see this we calculate using standard rules for the Hodge star

∗(α ∧ dζ) = (−1)k(n−k) ∗ (α ∧ ∗(∗dζ))

= (−1)k(n−k)〈α, ∗dζ〉= (−1)k(n−k)〈α, ∗d ∗ ω〉 where ζ = ∗ω= (−1)n−k−1〈α, d∗ω〉.

By Lemma 4.2.10 there exists γ ∈W 1,qloc (Rn,Λn−k−1) such that

dγ = β and ‖Dγ‖Lq(Rn) ≤ C‖β‖Lq(Rn). (4.30)


For any η ∈ C∞c (Rn) and any constant γ0 ∈ Λn−k−1, the form ζ = η(γ−γ0) ∈W 1,q(Rn,Λn−k−1)is admissible in (4.29) by approximation, thus we get∫

Rnη α ∧ β = −

∫Rnα ∧ dη ∧ (γ − γ0). (4.31)

For given φ ∈ T and t > 0, we take η(y) = φxt (y) = φt(x− y). Then |dη| ≤ t−n−1 and

|φt ∗ (α ∧ β)(x)| =∫Rnφxt α ∧ β ≤

C

t

∫−Bt(x)

|α| |γ − γ0| dy.

We now use Hölder's inequality with exponent r ∈ (1, p]. The second factor, which gets thepower s = r

r−1 ∈ [q,∞), is estimated by the Sobolev-Poincaré inequality. More precisely

|φt ∗ (α ∧ β)(x)| ≤ C

t

(∫−Bt(x)

|α|r dy) 1r(∫−Bt(x)

|γ − γ0|s dy) 1s

≤ C(∫−Bt(x)

|α|r dy) 1r(∫−Bt(x)

|Dγ|λ dy) 1λ

≤ CM(|α|r)(x)1r M(|Dγ|λ)(x)

1λ .

Here we need 1λ ≤ 1 + 1

n −1r . Take the supremum over t > 0 and integrate, then use Hölder

with exponents p, q to get∫Rn

(α ∧ β)∗ dx = C

∫RnM(|α|r)

1r M(|Dγ|λ)

1λ dx

≤ C ‖M(|α|r)‖1r

Lpr (Rn)

‖M(|Dγ|λ)‖1λ

Lqλ (Rn)

≤ C ‖|α|r‖1r

Lpr (Rn)

‖|Dγ|λ‖1λ

Lqλ (Rn)

= C ‖α‖Lp(Rn)‖Dγ‖Lq(Rn).

To apply the Hardy-Littlewood theorem 4.1.6 we needed that r < p and λ < q. We eventuallyx the parameters: rst chose r ∈ (1, p) such that

1

r<

1

p+

1

n. (4.32)

This is possible since 1p < min(1

p + 1n , 1). Then take λ ≥ 1 with

1

q<

1

λ≤ 1 +

1

n− 1

r. (4.33)

This is also possible because 1q < min(1 + 1

n −1r , 1) by (4.32). Recalling the Lq estimate from

Lemma 4.2.10, we arrive at the desired estimate

‖α ∧ β‖H1(Rn) ≤ C‖α‖Lp(Rn)‖β‖Lq(Rn).


The paper [9] states the theorem for vector elds E ∈ Lp(Rn,Rn) and B ∈ Lq(Rn,Rn)satisfying curlE = 0 and divB = 0, claiming that

‖〈E,B〉‖H1(Rn) ≤ C ‖E‖Lp(Rn)‖B‖Lq(Rn).

This result is also included in our formulation by considering the forms

α =

n∑i=1

Ei dxi and β = Bxdx1 ∧ . . . ∧ dxn =

n∑i=1

(−1)iBi dx1 ∧ . . . ∧ dxi ∧ . . . ∧ dxn.

We then have ∗(α ∧ β) = 〈E,B〉. One possible application is to Jacobi determinants of mapsu : Rn → Rn with Du ∈ Ln(Rn,Rn×n). The theorem then implies

detDu = ∗(du1 ∧ . . . ∧ dun) ∈ H1(Rn).

In the case detDu ≥ 0 this yields another proof of Müller's theorem 4.1.11: one combinesStein's theorem 4.1.8 with Lemma 4.2.6 to obtain, for any ball B ⊂ Rn,

‖ detDu log+ detDu‖L1(B) ≤ C(B, ‖M(detDu)‖L1(B)

)≤ C(B, ‖(detDu)∗‖L1(B))

= C(B, ‖ detDu‖H1(Rn)

)≤ C(B, ‖Du‖Ln(B)).

Furthermore, one obtains a generalization of Wente's theorem 3.2.1 to higher dimensions. Thisdepends on the following regularity result by Feerman and Stein [14] (see also next section).

Theorem 4.2.12 ([14]). Let f ∈ H1(Rn), and assume that u : Rn → R is a solution of

−∆u = f in Rn.

Then u = u0 + h where h : Rn → R is harmonic and u0 : Rn → R satises

‖D2u0‖H1(Rn) ≤ C‖f‖H1(Rn).

For n = 2 the space W 2,1 embeds into C0 ∩W 1,2, so that Wente's estimates are reproduced.

4.3 Atomic decomposition

In this section we prove that every element in H1 can be decomposed into so called atoms.

Denition 4.3.1. A function a ∈ L1(Rn) with the properties

spt a ⊂ B, (4.34)

||a||L∞(Rn) ≤1

|B|, (4.35)∫

Rna(x)dx = 0, (4.36)

where B ⊂ Rn is a ball, is called H1-atom.

4.3. ATOMIC DECOMPOSITION 47

An example of H1-atoms is given by the functions in (??).

Lemma 4.3.2. Let a ∈ L1(Rn) be an H1-atom. Then we have that

||a||H1(Rn) ≤ c, (4.37)

where the constant c is independent of the atom.

Proof. From the denition of an H1-atom we can assume that spt a ⊂ B = BR. Since a isadditionally in L∞(Rn) with vanishing mean value we can apply Lemma ?? (more preciselyestimate (??)) to get

||a||H1(Rn) ≤ cRn||a||L∞(Rn) ≤ c. (4.38)

This proves the Lemma.

In the next Lemma we show that a sum of H1-atoms also belongs to H1.

Lemma 4.3.3. Let ak ∈ L1(Rn), k ∈ N, be a sequence of H1-atoms and let λk, k ∈ N, besequence in R with

∑∞k=1 |λk| <∞. Then we have that f =

∑∞k=1 λkak ∈ H1(Rn) with

||f ||H1(Rn) ≤ c∞∑k=1

|λk|. (4.39)

Proof. For N ∈ N we consider the bounded function

fN (x) =

N∑k=1

λkak(x).

Using Lemma 4.3.2 and the assumption for∑|λk| we conclude that fN ∈ H1(Rn) and more-

over with the help of estimate (4.37) we get

||fN − fM ||H1 ≤N∑

k=M

|λk|||ak||H1

≤ cN∑

k=M

|λk|, (4.40)

for every N ≥M ∈ N. If we now dene

f = f1 +∞∑N=1

(fN+1 − fN ) ∈ L1(Rn),

we see that (using (4.40))

||f ||H1 ≤c|λ1|+∞∑N=1

||fN+1 − fN ||H1

≤c|λ1|+∞∑N=1

|λN |

≤c∞∑N=1

|λN |, (4.41)


which shows that f ∈ H1(Rn). Moreover we have (again using (4.40))

||f − fN ||H1 ≤∞∑k=N

||fk+1 − fk||H1

≤c∞∑k=N

|λk|. (4.42)

This shows that fN → f in H1(Rn) and proves the Lemma.

The goal of the rest of this section is to prove a statement converse to the one in Lemma4.3.3, namely that every function f ∈ H1(Rn) can be decomposed into a sum of H1-atoms.In order to do this we need a version of the Whitney decomposition which can be found forexample in [45].

Lemma 4.3.4. Let F ⊂ Rn be non-empty and closed and let Ω = Rn\F . Then there exists acollection of cubes F = Q1, . . . , Qk, . . . so that

(i) ∪kQk = Ω,

(ii) the Qk are mutually disjoint and

(iii) diamQk ≤ d(Qk, F ) ≤ 4 diamQk for every k ∈ N.

Moreover there exists 1 < b, such that if Q?k denotes the cube which has the same centeras Qk but whose side length is (1 + b) times the side length of Qk, then these cubes satisfy∪kQk = ∪kQ?k and the cubes Q?k have the bounded intersection property.

Proof. We let Q0 be the set of all cubes with side length 1 and whose corners are in Zn.Then we let Ql = 2−lQ0 be the cubes with side length 2−l and whose cornes are in (2−lZ)n.The diameter of the cubes in Ql is

√n2−l. Now we let Ω = ∪∞l=−∞Ωl where Ωl = x ∈

Ω|√n2−l+1 < d(x, F ) ≤

√n2−l+2 and we dene

F0 = ∪l∈ZQ ∈ Ql|Q ∩ Ωl 6= ∅.

For Q ∈ F0, e.g. Q ∈ Ql and x ∈ Q ∩ Ωl, we have

d(Q,F ) ≤ d(x, F ) ≤√n2−l+2 = 4 diamQ (4.43)

and

d(Q,F ) ≥ d(x, F )− diamQ ≥√n2−l+1 −

√n2−l ≥ diamQ. (4.44)

This proves (iii) of the Lemma for all cubes in Q ∈ F0 and shows in particular that Q ⊂ Ω.From the denition we therefore get

Ω = ∪Q∈F0Q. (4.45)

Next we let F be the family of the maximal cubes of F0, i.e. all cubes Q ∈ F0 such that ifQ‘ ∈ F0, Q ⊂ Q‘ ⇒ Q = Q‘. Since F 6= ∅ we see that every x ∈ Ω lies in at least one cubewith maximal side length. This implies that F still covers Ω. The fact that the cubes in Fare mutually disjoint follows directly from the denition.It remains to prove the last statement of the Lemma. We prove this result in three steps:


1) If two cubes Q1, Q2 ∈ F touch each other, then we have

1

4diamQ2 ≤ diamQ1 ≤ 4 diamQ2. (4.46)

To see this we note that since Q1 ∈ F we have d(Q1, F ) ≤ 4 diamQ1. This impliesd(Q2, F ) ≤ 4 diamQ1 + diamQ1 = 5 diamQ1, since Q1 and Q2 touch. Since Q2 ∈ Fwe conclude diamQ2 ≤ d(Q2, F ) ≤ 5 diamQ1. By construction we have diamQ2 =2k diamQ1, for some k ∈ Z, and therefore diamQ2 ≤ 4 diamQ1. The other inequalityfollows from symmetry considerations.

2) Let N = (12)n. If Q ∈ F then there are at most N cubes in F which touch Q.If Q ∈ Ql it is easy to see that there are 3n cubes (including Q) which belong to Qland touch Q. Next, each cube in Ql can contain at most 4n cubes in F which havediameter larger than or equal to 1

4 diamQ. Combining this with (4.46) nishes the proofof statement 2).

3) Let 0 < b < 14 . Then each point x ∈ Ω is contained in at most N of the cubes Q?k,

Qk ∈ F .Let Q,Qk ∈ F . We claim that Q?k intersects Q only if Qk touches Q. In fact, if weconsider the union of Qk with all cubes in F which touch Qk it is clear that this unioncontains Q?k (since the diameter of the cubes touching Qk are all larger than or equalto 1

4 diamQk). Therefore Q intersects Q?k only if Q touches Qk. Since any point x ∈ Ωis contained in a cube Q we see from step 2) that there are at most N cubes Q?k whichcontain x.

Since we also have that if Qk ∈ F then Q?k ⊂ Ω we get Ω = ∪Q∈FQ? and this nishes theproof of the Lemma.

The key step in the atomic decomposition is the following variant of the Calderon-Zygmunddecomposition.

Theorem 4.3.5. Let f ∈ H1(Rn) and let α > 0. Then there exists a decomposition f = g+b,with b =

∑∞k=1 bk, and a countable family of cubes Ck, k ∈ N, such that

(i) For a.e. x ∈ Rn we have

|g(x)| ≤ cα. (4.47)

(ii) Every function bk is supported in Ck, satises∫Ckbk(x)dx = 0 and

||bk||H1(Rn) =

∫Rnb?k(x)dx ≤ c

∫Ck

f?(x)dx. (4.48)

(iii) The countable family of cubes Ck has the bounded intersection property and if we setΩ = ∪∞k=1Ck we have

Ω = x ∈ Rn|f?(x) > α. (4.49)


Proof. For α > 0 and f ∈ H1(Rn) we set

Ω = x ∈ Rn|f?(x) > α

and note that Ω is open. (This follows from the fact that f? is obtained by taking the supover smooth functions and is therefore lower-semicontinuous. Hence the set Rn\Ω = x ∈Rn|f?(x) ≤ α is compact.) Next we apply the Whitney decomposition of Lemma 4.3.4 toΩ and F = Rn\Ω and we denote the cubes which we obtain from this decomposition by Ck‘.Moreover we can choose two constants 1 < a < b ∈ R such that, if Ck and Ck denote thecubes having the same center as Ck‘ but scaled with the factors a and b, then Ω = ∪∞k=1Ckand the family Ck has the bounded intersection property.Next we consider a positive function ξ satisfying

ξ ∈ C∞c ([−a2,a

2]n), ξ = 1 on [−1

2,1

2]n.

Then we set

ξk = ξ(x− cklk

), (4.50)

where ck denotes the center of the cube Ck‘ and lk the length of its edges. Note that ξk ∈C∞c (Ck) and ξk ≡ 1 on Ck‘. If we dene

ηk(x) =ξk(x)∑∞j=1 ξj(x)

∈ C∞c (Ck) (4.51)

it is not dicult to see that ηk forms a partition of unity subordinated to Ck (here we usethe nite intersection property of the Ck`s in order to ensure that the sum in the denominatoris nite). Moreover we observe that

lnk = |Ck‘| ≤∫Ck

ηk(x)dx ≤ |Ck| = bnlnk (4.52)

and

||∇ηk||L∞ ≤c

lk, (4.53)

for every k ∈ N, where c does not depend on k.

Now for f and ηk as above we dene

bk(x) = (f(x)− ak)ηk(x), (4.54)

where

ak =

∫Ckηk(x)f(x)dx∫Ckηk(x)dx

.

Therefore bk is supported in Ck and we see that∫Ck

bk(x)dx = 0. (4.55)


Moreover we dene

g(x) = χRn\Ω(x)f(x) + χΩ(x)

∞∑k=1

akηk(x). (4.56)

Next we claim that

|ak| ≤ c. (4.57)

In order to see this we note that

ak =

∫Ck

f(x)φk(x)dx = (f ? φ)(0),

where φ = ηk∫Ck

ηk(x)dxand therefore

||∇φ||L∞ ≤||∇ηk||L∞∫Ckηk(x)dx

≤ cl−n−1k ,

where we used (4.52) and (4.53) in the last step. Therefore we can argue as in the proof ofLemma ?? (more precisely as in the proof of (??) with z ∈ BClk(ck)∩Ωc, the constant C canbe chosen in such a way because we consider a Whitney decomposition) to get

|ak| = |(f ? φk)(0)| ≤ cf?(z) ≤ cα. (4.58)

Moreover we see that

|ak| ≤ cf?(x), (4.59)

for every x ∈ Ck. Now we want to show that the statement (i) of the Theorem holds. In thecase x 6∈ Ω we have

|g(x)| = |f(x)| ≤ f?(x) ≤ α, (4.60)

where we used the denition of Ω. If x ∈ Ω we have, using (4.58)

|g(x)| ≤∞∑k=1

|ak|ηk ≤ cα. (4.61)

Now it remains to prove statement (ii) of the Theorem.We claim that if we have

b?k(x) ≤ cf?(x), for x ∈ Ck (4.62)

b?k(x) ≤cαln+1

k

|x− ck|n+1for x 6∈ Ck, (4.63)

then we are done. To see this we estimate∫Rnb?k(x)dx =

∫Ck

b?k(x)dx+

∫Rn\Ck

b?k(x)dx

≤ c∫Ck

f?(x)dx+

∫Rn\Ck

cαln+1k

|x− ck|n+1dx.


The last integral can be estimated by (remember α < f?(x) for x ∈ Ck)∫Rn\Ck

cαln+1k

|x− ck|n+1dx ≤ cαln+1

k

∫Rn\B lk

2

(ck)

1

|x− ck|n+1dx

≤ cαlnk≤ cα|Ck|

≤ c∫Ck

f?(x)dx.

Altogether this shows that ∫Rnb?k(x)dx ≤ c

∫Ck

f?(x)dx. (4.64)

We are left with proving the estimates (4.62) and (4.63). First we write for φ ∈ T and x ∈ Ck∫Rnφt(x− y)f(y)ηk(y)dy =

∫Rnϕt,φ(y)f(y)dy,

where ϕt,φ(y) = ηk(y)φt(x− y). We estimate, using (4.53)

||∇ϕt,φ||L∞ ≤c||∇ηk||L∞

tn+ c||∇φt||L∞ (4.65)

≤c( 1

lktn+

1

tn+1). (4.66)

Now we have two cases. The rst one is that t ≤ lk. In this situation we have that||∇ϕt,φ||L∞ ≤ c

tn+1 and we can again argue as in the proof of (??) (ϕt,φ ∈ C∞c (Bt(x))) toget

supφ∈T sup0<t≤lk |∫Rnϕt,φ(y)f(y)dy| ≤ cf?(x). (4.67)

In the case t > lk we have that ||∇ϕt,φ||L∞ ≤ cln+1k

and ϕt,φ is supported in Bclk(x) (remember

ηk ∈ C∞c (Ck)), where c is chosen such that Ck ⊂ Bclk(x). Therefore we can argue as above toget

supφ∈T supt>lk |∫Rnϕt,φ(y)f(y)dy| ≤ cf?(x). (4.68)

Combining (4.67) and (4.68) we arrive at

(fηk)?(x) = supφ∈T sup0<t |

∫Rnφt(x− y)f(y)ηk(y)dy| (4.69)

≤ cf?(x). (4.70)

This implies that for x ∈ Ck we have

b?k(x) ≤ (fηk)?(x) + |ak|η?k(x) ≤ cf?(x), (4.71)


where we used (4.59) to get that |ak|η?k(x) ≤ |ak| ≤ cf?(x). This shows (4.62).We use (4.55) to get∫

Rnφt(x− y)bk(y)dy =

∫Rn

(φt(x− y)− φt(x− ck))bk(y)dy

=I1 − I2, (4.72)

where

I1(x) =

∫Rn

(φt(x− y)− φt(x− ck))f(y)ηk(y)dy

I2(x) =

∫Rn

(φt(x− y)− φt(x− ck))ak(y)ηk(y)dy.

This time we dene ψt,φ(y) = (φt(x − y) − φt(x − ck))ηk(y). Now we note that∫Rn φt(x −

y)bk(y)dy 6= 0 only if the supports of bk and φt(x−·) intersect. This is the case if t ≥ c|x−ck|.From this we get by the mean value theorem

|ψt,φ(y)| ≤|ηk(y)|||∇φ||L∞ |y − ck|

≤ clk|x− ck|n+1

. (4.73)

Another application of the mean value theorem and (4.53) yields

|∇ψt,φ(y)| ≤|∇φt(x− y)|+ |∇ηk(y)||φt(x− y)− φt(x− ck)|

≤ c

tn+1+c

lk

clktn+1

≤ c

tn+1. (4.74)

Next we argue again as in the proof of (??) (this time applied to the function ψt,φ ∈C∞c (Bclk(ck)), where c is chosen such that there exists z ∈ Bclk(ck) ∩ Ωc) to get

|ψt,φ ? f | = |I1(x)| ≤cln+1k f?(z)

|x− ck|n+1≤

cαln+1k

|x− ck|n+1, (4.75)

where we also used that by (4.74) we have |∇ψt,φ(y)| ≤ Aln+1k

with A =cln+1k

|x−ck|n+1 . On the other

hand I2 can be estimated by

|I2(x)| ≤c|ak||Ck|||ψt,φ||L∞ ≤ ccαln+1

k

|x− ck|n+1, (4.76)

where we used (4.73) and (4.58). Combining (4.72), (4.75) and (4.76) we prove (4.63) andtherefore the Theorem.

In the next Theorem we nally prove the atomic decomposition.

Theorem 4.3.6. Let f ∈ H1(Rn). Then there exists a sequence ak, k ∈ N, of H1-atoms anda sequence λk ∈ R, such that

f =∞∑k=1

λkak, (4.77)


where the convergence is in the H1-norm, and moreover

∞∑k=1

|λk| ≤ c||f ||H1(Rn). (4.78)

Proof. For every j ∈ Z we apply Theorem 4.3.5 to f ∈ H1(Rn) and α = 2j > 0. From this weget the decomposition

f(x) = gj(x) + bj(x) = gj(x) +∞∑k=1

bjk, (4.79)

and the countable family of cubes Cjkk∈N. We estimate

||f − gj ||H1 = ||bj ||H1 ≤∑k

||bjk||H1

≤ c∑k

∫Cjk

f?(x)dx

≤ c∫

Ωjf?(x)dx

= c

∫x∈Rn|f?(x)>2j

f?(x)dx, (4.80)

where we used (4.48) in the second line, the bounded intersection property of Ωj = ∪kCjk inthe third line and (4.49) in the last step. Since f? ∈ L1(Rn) we therefore conclude that

||f − gj ||H1 → 0, (4.81)

as j →∞. Since by (4.47) we also know that |gj(x)| ≤ c2j a.e. we conclude that

gj → 0, (4.82)

in the sense of distributions as j → −∞. Combining (4.81) and (4.82) we get

f = limN→∞ gN + limN→−∞ g

N

= limN→∞

N∑j=1

(gj+1 − gj) + limN→∞(−N∑j=0

(gj+1 − gj)

=∞∑

j=−∞(gj+1 − gj), (4.83)

where the convergence is in the sense of distributions. From (4.79) we have that

gj+1(x)− gj(x) = (f(x)− bj+1(x))− (f(x)− bj(x)) = bj(x)− bj+1(x). (4.84)

Therefore gj+1 − gj is supported in Ωj ⊃ Ωj+1. Hence we can write

f =∞∑

j=−∞

∞∑k=1

(gj+1 − gj)ηjk, (4.85)


where ηjk ∈ C∞c (Cjk) is the partition of unity subordinated to Cjk which was dened in the

proof of Theorem 4.3.5. From (4.47) we also get for a.e. x ∈ Rn

|gj+1(x)− gj(x)| ≤ C2j . (4.86)

Now we dene real numbers

λj,k = C2j |Bjk|, (4.87)

where C is the constant from (4.86) and Bjk is the smallest ball containing Cjk, and functions

aj,k(x) =1

λj,kAj,k, (4.88)

where Aj,k is supported in Cjk and is dened by

Aj,k = (gj+1(x)− gj(x))ηjk(x). (4.89)

This implies in particular that aj,k is supported in Bjk and because of (4.86) and (4.87) we

have

||aj,k||L∞ ≤1

λj,k||gj+1 − gj ||L∞ ≤

1

|Bjk|. (4.90)

Moreover we have that∞∑

j=−∞

∞∑k=1

|λj,k| ≤ C∞∑

j=−∞2j |Ωj |

= c

∞∑j=−∞

2j |x ∈ Rn|f?(x) > 2j|

= c

∞∑j=−∞

(

j∑k=−∞

2k)|x ∈ Rn|2j+1 ≥ f?(x) > 2j|

≤ c∞∑

j=−∞2j |x ∈ Rn|2j+1 ≥ f?(x) > 2j|

≤ c∫Rnf?(x)dx

= c||f ||H1 . (4.91)

Because of (4.85), (4.88) and (4.89) we also get that

f =

∞∑j=−∞

∞∑k=1

λj,kaj,k. (4.92)

This would yield the desired atomic decomposition of f if we would additionally have that∫Rn aj,k(x)dx = 0. Since this is not true we have to modify the decomposition.First we note that because of (4.84) and (4.89) we can write

Aj,k(x) = (bj(x)− bj+1(x))ηjk(x) = bjk(x)−∞∑l=1

bj+1l (x)ηjk(x). (4.93)


Then we dene new functions

Bj,k(x) = Aj,k(x) +

∞∑l=1

bl,kηj+1l (x), (4.94)

where

bl,k =

∫Rn b

j+1l (x)ηjk(x)dx∫

Rn ηj+1l (x)dx

. (4.95)

This implies that∫RnBj,k(x)dx =

∫RnAj,k(x)dx+

∞∑l=1

∫Rnbj+1l (x)ηjk(x)dx

=

∫Rnbjk(x)dx−

∞∑l=1

∫Rnbj+1l (x)ηjk(x)dx+

∞∑l=1

∫Rnbj+1l (x)ηjk(x)dx

=0. (4.96)

Moreover we have that

∞∑k=1

bl,k =

∑∞k=1

∫Rn b

j+1l (x)ηlk(x)dx∫

Rn ηj+1l (x)dx

=

∫Rn b

j+1l (x)dx∫

Rn ηj+1l (x)dx

= 0. (4.97)

and therefore

∞∑k=1

Bj,k =∞∑k=1

Aj,k. (4.98)

This shows that (using (4.85) and (4.89))

f =∞∑

j=−∞

∞∑k=1

Bj,k. (4.99)

Now we have to study the numbers bl,k in more detail. Since spt bj+1l ⊂ Cj+1

l we see that

bl,k 6= 0 only if Cjk ∩ Cj+1l 6= ∅. This shows that sptBj,k ⊂ Bj

k. Next we want to showthat |Bj,k(x)| ≤ c2j . To show this we need to estimate |bl,k|. First of all we note that if

Cjk ∩ Cj+1l 6= ∅ we have

c1 diamCjk ≤ diamCj+1l ≤ c2 diamCjk.

To see this we use the properties of the Whitney decomposition (Lemma 4.3.4) to conclude

diamCj+1l ≤ d(Cj+1

l , (Ωj+1)c) ≤ d(Cj+1l , (Ωj)c) ≤ diamCjk + d(Cjk, (Ω

j)c) ≤ cdiamCjk.

The other inequality follows from symmetry considerations. Hence we also have that

cljk ≤ lj+1l ≤ cljk,


where ljk denotes the side length of the cube Cjk. Therefore if we dene

φ(x) =ηjk(x)∫

Rn ηj+1l (x)dx

we have that sptφ ⊂ Bjk ⊂ B

C1lj+1l

(cj+1l ), where C1 is chosen such that B

C1lj+1l

(cj+1l ) ∩

(Ωj+1)c 6= ∅, and (∫Rn η

j+1l (x)dx ≥ (lj+1

l )n)

||∇φ||L∞ ≤c

ljk(lj+1l )n

≤ c

(lj+1l )n+1

.

Therefore we can argue as in the proof of the Calderon-Zygmund decomposition (Lemma4.3.5) to get for z ∈ B

C1lj+1l

(cj+1l ) ∩ (Ωj+1)c

|bl,k| = |∫Rnbj+1l (x)φ(x)dx| ≤ c(bj+1

l )?(z) ≤ c2j , (4.100)

where we used (4.63) (note that |z − cj+1l | ≥ clj+1

l by our choices) in the last inequality.Altogether this shows that

|Bj,k(x)| ≤ c2j +∞∑l=1

|bl,k|ηj+1l (x) ≤ c2j . (4.101)

Therefore if we dene

bj,k(x) =1

λj,kBj,k(x), (4.102)

with λj,k as in (4.87), we see that spt bj,k ⊂ Bjk,∫Rn bj,k(x)dx = 0 (here we use (4.96)) and

||bj,k||L∞ ≤C2j

λj,k≤ 1

|Bjk|. (4.103)

This shows that the bj,k`s are H1-atoms and from (4.99) we get

f =∞∑

j=−∞

∞∑k=1

λj,k bj,k. (4.104)

Together with (4.91) this proves the Theorem.

In the rest of this section we want to show how one can use the atomic decomposition toextend the Lp-theory to the situation where the right hand side is only in H1. For sake ofsimplicity we restrict ourselves to the case of the Laplace operator (for the general case see[46]).

Lemma 4.3.7. Let Γ(x) = 1n(2−n)ωn

|x|2−n, resp. Γ(x) = − 12π ln( 1

|x|), be the fundamental

solution of ∆ for n ≥ 3, resp. n = 2. Dening Kij(x) = ∂i∂jΓ(x), for every i, j ∈ 1, . . . , n,we have that


(i) ||Kij ? f ||L2 ≤ c||f ||L2, for every f ∈ L2(Rn) and

(ii)∫

2|y|≤|x| |Kij(x− y)−Kij(x)|dx ≤ c.

Proof. (i) is the classical L2 estimate for the fundamental solution of ∆ and can be found in[19]. For (ii) we note that |∂lKij | ≤ c

|x|n+1 for every i, j, l ∈ 1, . . . , n. Therefore we get for2|y| ≤ |x| with the help of the mean value theorem that

|Kij(x− y)−Kij(x)| ≤ c|y||x|n+1

. (4.105)

Using this we estimate∫2|y|≤|x|

|Kij(x− y)−Kij(x)|dx ≤ c|y|∫ ∞

2|y|r−2dr ≤ c. (4.106)

Now we can prove the regularity result for equations of the form ∆u = f ∈ H1.

Theorem 4.3.8. Let f ∈ H1(Rn) and let u ∈ W 2,1(Rn) be a solution of ∆u = f , then wehave

||∇2u||L1(Rn) ≤ c||f ||H1(Rn). (4.107)

Proof. From the above considerations it is easy to see that the result follows if we show that

||Kij ? f ||L1(Rn) ≤ c||f ||H1(Rn), (4.108)

for every i, j ∈ 1, . . . , n.To see this we rst consider an H1-atom a which is supported in a ball Br. From (i) of Lemma4.3.7 we get that

||Kij ? a||2L2 ≤ c||a||2L2(Br)

≤ c|Br|||a||2L∞

≤ c

|Br|, (4.109)

where we used the properties of an atom. From Hölder`s inequality we then get

||Kij ? a||L1(B2r) ≤ ||Kij ? a||L2

√|B2r|

≤ c. (4.110)

Now we note that

Kij ? a(x) =

∫Br

(Kij(x− y)−Kij(x))a(y)dy, (4.111)

where we used the cancellation property of an atom. From this it follows that∫Bc2r

|Kij ? a|(x)dx ≤∫Bc2r

∫Br

|Kij(x− y)−Kij(x)||a(y)|dydx

=

∫Br

|a(y)|(∫Bc2r

|Kij(x− y)−Kij(x)|dx)dy. (4.112)


Since |x| ≥ 2r and |y| ≤ r we can apply (ii) from Lemma 4.3.7 to get∫Bc2r

|Kij ? a|(x)dx ≤ c∫Br

|a(y)|dy ≤ c. (4.113)

Combining (4.110) amd (4.113) this shows that∫Rn|Kij ? a|(x)dx ≤ c. (4.114)

Now we use Theorem 4.3.6 to get f =∑

k λkak with∑

k |λk| ≤ c||f ||H1 . Then we have

||Kij ? f ||L1 ≤ c∑k

|λk|∫Rn|Kij ? ak|(x)dx

≤ c∑k

|λk|

≤ c||f ||H1 . (4.115)

This proves (4.108) and therefore the Theorem.

Remark 4.3.9. One can actually show that ∇2u ∈ H1(Rn) (see [46]).

Chapter 5

Regularity of geometric variational

problems

This chapter addresses the regularity of critical points of two-dimensional, conformally invari-ant variational integrals. The case of harmonic maps was settled by Hélein [23], whereas thegeneral result including surfaces of prescribed, variable mean curvature is due to T. Rivière[37]. It is his proof that is presented here.

5.1 Gauge transformation

In this section we prove an existence result for Coulomb gauges due to Uhlenbeck [50]. Theissue is to construct a preferred gauge for a connection on a vector bundle over a Riemannianmanifold. More precisely the theorem deals with a local situation B × Rm, where B is theunit ball in Rn and the Rm factor represents the coordinates with respect to a given frame. Aconnection is then given by a matrix-valued one-form A = Ai(x) dxi with Ai(x) ∈ Rm×m. Itinduces a notion of parallel vector elds along curves γ : [a, b]→ B along curves γ : [a, b]→ Bby the linear ordinary dierential equation

∇Avdt

= v′ +A(γ′) v = 0 where v : [a, b]→ Rm. (5.1)

We should really write (A γ)(γ′), however it is customary to omit the basepoint. Theconnection is often denoted by its local form ∇A = d + A. For simplicity we restrict to SOm

bundles; this means that the bundle is oriented and carries a Riemannian metric. The Rmfactor represents the coordinates with respect to some oriented orthonormal frame, thus thebundle metric becomes the standard scalar product 〈·, ·〉. The connections are required to becompatible in the sense of the product rule, for any vector elds φ, ψ along γ,

d

dt〈v, w〉 =

⟨∇Avdt

, ψ⟩

+ 〈v, ∇Awdt

⟩⇔

⟨A(γ′) ei, ej

⟩+⟨ei, A(γ′) ej

⟩= 0.

61

62 CHAPTER 5. REGULARITY OF GEOMETRIC VARIATIONAL PROBLEMS

Thus A is an som-valued one-form. Any oriented orthonormal frame F = v1, . . . , vm overB induces new coordinates vF , such that v = PvF for some P : B → SOm. It follows that(∇Av

dt

)F

= P−1∇Adt

(PvF )

= P−1(PvF )′ + P−1A(γ′)PvF

= v′F +(P−1dP (γ′) + P−1A(γ′)P

)vF .

The map P : B → SOm is called a gauge transformation, and the one-form P−1dP +P−1APis the transformed connection. The group of gauge transformations acts isometrically on thespace of connections with respect to the L2 distance

dist(∇A,∇B)2 =

∫B|∇A −∇B|2 dx =

∫B|A−B|2 dx.

In fact we have∫B|P−1∇AP − P−1∇BP |2 dx =

∫B|P−1(∇A −∇B)P |2 dx =

∫B|∇A −∇B|2 dx.

It is therefore natural to ask whether any gauge orbit contains an element which minimizesthe distance to the trivial connection d = ∇0. If ∇A is the desired minimizer, then we get bychosing P = exp(tχ) with χ : B → som

0 =1

2

d

dt

∫B|e−tχd(etχ) + e−tχAetχ| dx|t=0 =

∫B〈A, dχ〉.

This means that the minimizer is a weak solution to the equations

d ∗A = 0 in B, νxA = 0 on ∂B.

These are called the Coulomb or Hodge gauge conditions. The following result is due toUhlenbeck [50].

Theorem 5.1.1. Let A0 ∈ L2(B,Λ1⊗som) be a connection on B = x ∈ Rn : |x| < 1. Thenthere exists a gauge transformation P ∈ W 1,2(B, SOm), such that A = P−1dP + P−1A0P ∈L2(B,Λ1 ⊗ som) has the following properties:

(1) A solves the system d∗A = 0 in B, νxA = 0 on ∂B.

(2) ‖dP‖L2(B) + ‖A‖L2(B) ≤ C‖A0‖L2(B).

(3) There is a ξ ∈W 1,2(B,Λ2 ⊗ som) with i∗∂B(d ∗ ξ) = 0 on ∂B, such that

d∗ξ = A and ‖ξ‖W 1,2(B) ≤ C‖A‖L2(B).

Proof. As outlined we consider a minimizing sequence Pk ∈W 1,2(B, SOm) for the functional

E(P ) =

∫B|P−1dP + P−1A0P |2 dx =

∫B|dP +A0P |2 dx. (5.2)

We have the inequality

E(P ) ≥ (1− ε)∫B|dP |2 dx− Cε

∫B|A0|2. (5.3)

5.2. EQUATIONS OF THE FORM ∆U = Ω∇U 63

Moreover |Pk| = n, thus we can assume Pk → P ∈ W 1,2(B, SOm) weakly in W 1,2, stronglyin L2 and pointwise almost everywhere. As |A0Pk| = |A0| we get A0Pk → A0P in L2(B) byVitali's convergence theorem. Thus the inmum is attained by P , and

E(P ) ≤ lim infk→∞

E(Pk) ≤ E(Id) =

∫B|A0|2.

Put A = P−1dP + P−1A0P . Then ‖A‖L2(B) ≤ ‖A0‖L2(B) by the minimizing property, andfrom dP = PA−A0P we see that

‖dP‖L2(B) ≤ ‖A‖L2(B) + ‖A0‖L2(B) ≤ 2‖A0‖L2(B).

Moeoever, as explained above, A satises the weak Hodge gauge conditions∫B〈dχ,A〉 dx = 0 for all smooth χ : B → som.

To show claim (3) we employ linear Hodge theory. By Lemma 3.3.1 in Chapter 3 there existsa form ξ ∈W 1,2(B,Λ2) such that

A = d∗ξ where

∫∂B∗(νxξ) = 0 and ‖ξ‖W 1,2(B) ≤ ‖A‖L2(B).

Our proof in Chapter 3 was only in two dimensions, but its generalization is straightforward.For any smooth χ : B → som we compute by partial integration∫

B〈dχ,A〉 dx =

∫Bdχ ∧ ∗d∗ξ

= (−1)n∫Bdχ ∧ d∗ ξ

= (−1)n∫∂Bχ i∗∂B(d ∗ ξ).

The weak version of the Hodge gauge condition (1) implies i∗∂B(d ∗ ξ) = 0.

In dimension n = 2 we get ξ = 0 on the boundary. Namely for ξ = ξ0 dx1 ∧ dx2 we have

d ∗ ξ = dξ0, and the normalization becomes∫∂B ξ0 dθ = 0. The presented gauge theorem is

weak in the sense that no additional regularity of P and A is asserted. It is possible that P hassingularities which change the topological type of the bundle. In dimensions n ≤ 4 Uhlenbeckproved a stronger version where P is estimated in W 2,2, and accordingly A in W 1,2. Theseestimates depend on a smallness assumption for the L2 norm of the curvature F = dA+A∧A.For n ≤ 3 the smallness threshold can always be achieved by scaling, whereas in the criticaldimension n = 4 it is a necessary, nontrivial condition.

5.2 Equations of the form ∆u = Ω∇uLet B ⊂ R2 be the unit ball in R2. In this section we study the regularity properties ofsolutions of elliptic systems of the form

−∆u = Ω∇u, (5.4)


where u ∈ W 1,2(B,Rm) and Ω ∈ L2(B, so(m) ⊗ ∧1R2). Before coming to the detailed studylet us give some examples for systems of the type (5.4).

1) From (??) we see that harmonic maps into spheres satisfy an equation of the form (5.4)with (Ωij) = (ui∇uj − uj∇ui) ∈ L2(B, so(m)⊗ ∧1R2).

2) It is easy to see that surfaces with prescribed mean curvatureH ∈ L∞(R3) (i.e. solutionsof (??)) solve a system of the form (5.4) with

Ω = −2H(u)

0 ∇⊥u3 −∇⊥u2

−∇⊥u3 0 ∇⊥u1

∇⊥u2 −∇⊥u1 0

∈ L2(B, so(3)⊗ ∧1R2).

3) Harmonic maps into general target manifolds.Here we let u ∈ W 1,2(B,N), where N → Rm is a smooth and compact Riemannianmanifold without boundary. Then we know from the discussions in chapter 2 thatharmonic maps into N are critical points of the functional

E(u) =1

2

∫B|∇u|2dvg.

To compute the critical points of E we let ϕ ∈ C1c (B,Rm) with ϕ(x) ∈ Tu(x)N for all

x ∈ B. Then we compute

0 =d

dt|t=0E(u+ tϕ)

= −∫B

∆uϕ.

Since this is true for all such ϕ we know that

∆u ⊥ TuN.

Therefore if we let νn+1, . . . , νm be a smooth local orthonormal frame for the normalbundle near u(x) we can write

∆u(x) =m∑

i=n+1

λi(x)νi(u(x)),

where the λi are scalar functions. Using the fact that 〈∇u, νi(u)〉 = 0 for every i ∈n+ 1, . . . ,m we get

λi =〈∆u, νi(u)〉= div〈∇u, νi(u)〉 − 〈∇ju, (dkνi)(u)∇juk〉

and hence

∆u =∑i

λiνi(u)

=−m∑

i=n+1

m∑k=1

2∑j=1

〈∇ju, (dkνi)(u)∇juk〉νi(u)

=−A(u)(∇u,∇u).


Moreover, using the denition of A, we see that (using that∑

k∇ukνki (u) = 0 for everyi)

∆us =−∑i,k

〈∇u, (dkνi)(u)∇uk〉νsi (u)

=−∑i,k,l

∇uk(νsi (u)(dkνi)l(u)∇ul − νki (u)(dsνi)

l(u)∇ul),

and hence u solves an equation of the form (5.4) with

(Ωsk) = (∑i,l

(νsi (u)(dkνi)l(u)∇ul − νki (u)(dsνi)

l(u)∇ul)) ∈ L2(B, so(m)⊗ ∧1R2).

4) Conformally invariant variational problems.We consider the functional

Eω(u) =1

2

∫B

(|∇u|2 + ω(u)(∂xu, ∂yu))dx,

where ω is a C1 two-form on Rm such that the L∞-norm of dω is bounded. By Theorem2.4.1 we see that every conformally invariant energy in two-dimensions can be writtenin this way. The Euler-Lagrange equation of Eω can easily be computed to be

∆ui +Ai(u)(∇u,∇u) + λijl(u)∂xuj∂ly = 0,

where λijl(u) = dω(u)(ei, ej , el) and where eii=1,...,m is the standard basis of Rm.Using that λijl = −λjil we calculate

λijl(u)∂xuj∂ly =

1

4(λijl(u)− λjil(u))∇⊥ul∇uj .

Combining this with the result of 3) we see that the Euler-Lagrange equation of everyconformally invariant energy in two dimensions can be written in the form (5.4) with

Ωsk =∑i,l

(νsi (u)(dkνi)l(u)∇ul − νki (u)(dsνi)

l(u)∇ul)

−∑l

1

4(λskl(u)− λksl(u))∇⊥ul

∈L2(B, so(m)⊗ ∧1R2).

After having collected all these examples of systems of the type (5.4) we now state the mainTheorem of this chapter. This Theorem was only recently proved by Tristan Rivière [37] (seealso [30], [38] and [49] for related results).

Theorem 5.2.1. Let u ∈W 1,2(B,Rm) be a solution of (5.4) with Ω ∈ L2(B, so(m)⊗∧1R2).Then u is continuous and therefore by Theorem ?? as smooth as the data permits.

Proof. The Theorem will be proved in three steps.Step 1:


Lemma 5.2.2. Let m ∈ N and Ω ∈ L2(B, so(m)⊗∧1R2). Let A ∈ L∞∩W 1,2(B,M(m)) andB ∈W 1,2(B,M(m)) be solutions of

∇A−AΩ = ∇⊥B. (5.5)

Then u ∈W 1,2(B,Rm) is a solution of (5.4) with Ω i

div(A∇u+B∇⊥u) = 0. (5.6)

Proof. By a direct calculation (using that div∇⊥ = 0 and ∇u∇⊥v = −∇⊥u∇v) and using(5.5) we get

div(A∇u+B∇⊥u) =(∇A−∇⊥B)∇u+A∆u

=A(∆u+ Ω∇u).

This proves the Lemma.

Step 2:

Lemma 5.2.3. There exists ε > 0, c > 0 such that for every Ω ∈ L2(B, so(m)⊗ ∧1R2) with∫B|Ω|2dx < ε, (5.7)

there exist A ∈ L∞ ∩W 1,2(B,Gl(m)) and B ∈W 1,2(B,M(m)) satisfying∫B

(|∇A|2 + |∇B|2)dx+ ||dist(A,SO(n))||2L∞ ≤ c∫B|Ω|2 and (5.8)

∇A−AΩ−∇⊥B = 0. (5.9)

Proof. For Ω ∈ L2(B, so(m) ⊗ ∧1R2) with∫B |Ω|

2dx < ε we apply Theorem ?? to get theexistence of P ∈W 1,2(B,SO(m)) and ξ ∈W 1,2(B, so(m)) such that ξ = 0 on ∂B,

∇⊥ξ = P−1∇P + P−1ΩP. (5.10)

and

||ξ||W 1,2 + ||∇P ||L2 + ||∇P−1||L2 ≤ c||Ω||L2 . (5.11)

Claim 1: There exist A ∈W 1,2 ∩ L∞(B,M(m)) and B ∈W 1,2(B,M(m)) solving

∆A = ∇A∇⊥ξ +∇⊥B∇P in B, (5.12)

∆B = −∇⊥A∇P−1 − div(A∇ξP−1 +∇ξP−1) in B, (5.13)

∂A

∂ν= 0 and B = 0 on ∂B, (5.14)∫

BA = 0. (5.15)

To prove this claim we apply Theorem ?? (combined with remark ??) and standard L2-theoryto get

||A||W 1,2 + ||A||L∞ ≤ c||∇ξ||L2 ||∇A||L2 + c||∇P ||L2 ||∇B||L2 and (5.16)

||B||W 1,2 ≤ c||∇P−1||L2 ||∇A||L2 + c||∇ξ||L2 ||A||L∞ + c||∇ξ||L2 . (5.17)


Using (5.11) and choosing ε small enough we combine (5.16) and (5.17) to get

||A||W 1,2 + ||A||L∞ + ||B||W 1,2 ≤ c||Ω||L2 . (5.18)

The existence of the desired solution of (5.12)-(5.15) (and hence the proof of Claim 1) nowfollows from a standard xed-point argument.Next we dene A = A+ id and we see from (5.12)-(5.15) that A and B solve

∆A = ∇A∇⊥ξ +∇⊥B∇P in B, (5.19)

∆B = −∇⊥A∇P−1 − div(A∇ξP−1) in B, (5.20)

∂A

∂ν= 0 and B = 0 on ∂B, (5.21)∫

BA = |B|. (5.22)

Moreover we get from (5.18) that

||∇A||L2 + ||dist(A, SO(m))||L∞ + ||B||W 1,2 ≤ c||Ω||L2 . (5.23)

Now it is easy to see that (5.19) can be rewritten as

div(∇A− A∇⊥ξ −∇⊥BP ) = 0 (5.24)

and hence, by Lemma ??, there exists C ∈W 1,2(B,M(m)⊗ ∧1R2) such that

∇A− A∇⊥ξ −∇⊥BP = ∇⊥C. (5.25)

Since by (5.21) and the denition of ξ we have

(∇A− A∇⊥ξ −∇⊥BP ) · ν =∂A

∂ν− A∇⊥ξ · ν −∇⊥BP · ν

=0

on ∂B we can moreover assume that C = 0 on ∂B. Using a rotation by π2 (one can also view

∇⊥ as ?d and then the rotation by π2 is just another application of ?) we see that (5.25) is

equivalent to

−∇CP−1 = ∇⊥AP−1 + A∇ξP−1 +∇B,

and hence, using (5.20), we calculate

−div(∇CP−1) = ∇⊥A∇P−1 + div(A∇ξP−1) + ∆B

= 0. (5.26)

Claim 2: Every solution C of (5.26) with C = 0 on ∂B vanishes identically.To see this we apply again Lemma ?? and get the existence of D ∈ W 1,2(B,M(m) ⊗ ∧1R2)such that

∇⊥D = ∇CP−1. (5.27)


Since C = 0 on ∂B we easily see that ∂D∂ν = 0 on ∂B and we can also assume that

∫BD = 0.

Hence C and D solve

∆C = ∇⊥D∇P in B, (5.28)

∆D = ∇C∇⊥P−1 in B, (5.29)

C = 0 and∂D

∂ν= 0 on ∂B, (5.30)∫

BD = 0. (5.31)

From this we see that we can apply Theorem ?? for (5.28) and (5.29) (in this case combinedwith remark ??) to get

||∇C||L2 + ||∇D||L2 ≤ c(||∇P ||L2 ||∇D||L2 + ||∇P−1||L2 ||∇C||L2). (5.32)

By choosing ε small enough we get from (5.11) that C = D = 0 and this shows the claim.From (5.25) we now see that A and B solve

∇A− A∇⊥ξ −∇⊥BP = 0. (5.33)

Dening A = AP−1 we see that

||∇A||L2 + ||dist(A,SO(m))||L∞ ≤ c(||∇A||L2 + ||A||L∞ ||∇P−1||L2 + ||dist(A, SO(m))||L∞)

≤ c||Ω||L2 , (5.34)

where we used (5.11) and (5.23). Moreover we use (5.10) and (5.33) to calculate

0 = ∇A− A∇⊥ξ −∇⊥BP = ∇AP +A∇P −AP∇⊥ξ −∇⊥BP= A∇P −A∇P + (∇A−AΩ−∇⊥B)P

and therefore

∇A−AΩ−∇⊥B = 0. (5.35)

This nishes the proof of the Lemma.

Step 3:For every point x ∈ B we choose a radius rx > 0 such that

∫Brx (x) |Ω|

2 < ε, where ε is the

same as in Lemma 5.2.3. In the following we write Brx(x) = B. Then we can apply Lemma5.2.3 to get the existence of A and B solving (5.5). Hence we can apply Lemma 5.2.2 to seethat

div(A∇u) = ∇B∇⊥u = −∇⊥B∇u and (5.36)

∇⊥(A∇u) = ∇⊥A∇u. (5.37)

Now we apply Lemma ?? to get the existence of α ∈W 1,2(B,Rm) and β ∈W 1,2(B,Rm⊗∧1R2)such that

A∇u = ∇α+∇⊥β. (5.38)


Using (5.36) we see that α solves

∆α = div(A∇u) = −∇⊥B∇u. (5.39)

Now we denote by u the mean value of u on B 12and let u ∈ W 1,2

0 (R2,Rm) be the extension

with compact support of u − u. Then we have that ∇u = ∇u on B 12. Moreover we use

Poincare's inequality to get

||∇u||L2(R2) ≤ c||u− u||W 1,2(B 12

) ≤ c||∇u||L2(B).

We extend B in the same way and denote the resulting map by B ∈ W 1,20 (R2,M(m)). Then

we let α be the solution of

∆α = −∇⊥B∇u (5.40)

on R2. Since by Corollary ?? −∇⊥B∇u ∈ H1(R2) with

|| − ∇⊥B∇u||H1(R2) ≤ c||∇B||L2(B)||∇u||L2(B)

we can apply Theorem 4.3.8 to get that α ∈W 2,1(R2). Since

∆(α− α) = 0

on B 12we get that α ∈ W 2,1(B 1

4) (harmonic functions are smooth). Next we observe that β

solves

∆β = ∇⊥A∇u (5.41)

and hence we can argue as before to get that β ∈ W 2,1(B 14). Therefore we see from (5.38)

that

A∇u ∈W 1,1(B 14) (5.42)

and therefore (using (5.8))

∇u ∈W 1,1(B 14) or u ∈W 2,1(B 1

4). (5.43)

Combining this with Corollary ?? we nish the proof of the Theorem.

With the following counterexamples of Frehse [16] we show that one can not drop thecondition that Ω has to be antisymmetric.

Remark 5.2.4. Let u = (u1, u2) ∈W 1,2(B,S1 ⊂ R2) be dened by

u1(x) = sin ln ln2

|x|,

u2(x) = cos ln ln2

|x|


then it is easy to check that u solves the elliptic system −∆u = Ω∇u with

Ω =

((u1 + u2)∇u1 (u1 + u2)∇u2

(u2 − u1)∇u1 (u2 − u1)∇u2

).

So in this case Ω is not antisymmetric and u is bounded but not continuous.For u ∈W 1,2(B,R2) given by

u1(x) = ln ln2

|x|,

u2(x) = ln ln2

|x|

we have that u solves the elliptic system −∆u = Ω∇u with

Ω =

(∇u1 0

0 ∇u2

).

Here we even don`t have that u is bounded.

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