Introduction to the calculus of variations and -convergenceantonin/CaVa/CV1.pdf · 2020. 4. 9. ·...

Introduction to the calculus of variations and

Γ-convergence

Antonin Chambolle, CMAP, Ecole Polytechnique, CNRS

April 15, 2015

Contents

1 Introduction 5

2 Characterization of the critical points 7

2.1 Differentiability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.2 First variation of a functional . . . . . . . . . . . . . . . . . . . . . . . . 8

2.3 Boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.3.1 The prescribed boundary conditions . . . . . . . . . . . . . . . . 10

2.3.2 The free boundary conditions . . . . . . . . . . . . . . . . . . . . 12

2.4 A remark on “Null Lagrangians” . . . . . . . . . . . . . . . . . . . . . . 13

2.5 Constrained problems, obstacles . . . . . . . . . . . . . . . . . . . . . . . 14

2.5.1 Convex constraints . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.5.2 Differentiable constraints . . . . . . . . . . . . . . . . . . . . . . 15

3 Existence of minimizers 17

3.1 Lower semicontinuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.2 The direct method in the calculus of variations . . . . . . . . . . . . . . 18

3.2.1 Basic principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2.2 An example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

3.3 Compactness results in infinite dimension . . . . . . . . . . . . . . . . . 20

3.3.1 Ascoli-Arzela . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.4 Basic lower-semicontinuity criteria . . . . . . . . . . . . . . . . . . . . . 23

3.4.1 Convex functions . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3.4.2 Non-convex cases . . . . . . . . . . . . . . . . . . . . . . . . . . . 24

3.4.3 Necessity of the convexity . . . . . . . . . . . . . . . . . . . . . . 24

4 Regularity for minimizers (elliptic case) 27

4.1 Basic regularity estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.1.1 Obvious estimates . . . . . . . . . . . . . . . . . . . . . . . . . . 27

4.1.2 Higher derivability . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.2 The De Giorgi-Nash-Moser Thm . . . . . . . . . . . . . . . . . . . . . . 31

4.2.1 Towards more regularity . . . . . . . . . . . . . . . . . . . . . . . 31

4.2.2 Proof: L∞ control . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.2.3 Harnack-type estimate . . . . . . . . . . . . . . . . . . . . . . . . 36

4.2.4 Application: control of the oscillation of a solution . . . . . . . . 38

3

4 CONTENTS

4.2.5 Conclusion: u is Holder . . . . . . . . . . . . . . . . . . . . . . . 38

5 Vector-valued minimization problems 41

6 Variational convergence 45

6.1 Why? . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45

6.1.1 Necessity of the axioms . . . . . . . . . . . . . . . . . . . . . . . 46

6.1.2 Properties and comments . . . . . . . . . . . . . . . . . . . . . . 47

6.1.3 Compactness of Γ-convergence . . . . . . . . . . . . . . . . . . . 48

6.2 Application: discrete to continuum limits . . . . . . . . . . . . . . . . . 50

6.2.1 Higher dimension . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.3 Application: periodic homogenization . . . . . . . . . . . . . . . . . . . 54

6.3.1 Proof of the homogenization result . . . . . . . . . . . . . . . . . 57

6.3.2 A remark in the convex case . . . . . . . . . . . . . . . . . . . . 63

6.3.3 Phase transitions, singular limits . . . . . . . . . . . . . . . . . . 64

Chapter 1

Introduction

The calculus of variations is the study of the minimizers or critical points of “function-

als”, which are functions defined in spaces of infinite dimensions, typically functional

spaces.

This needs to adapt the notions of differential calculus. The stationarity of a

functional E(u) is “simply” characterized by the equation

E ′(u) = 0 (1.1)

which, in general, will be a partial differential equation (PDE) in u (or something more

general).

A goal of the calculus of variations is to “solve” such PDEs: more precisely, to show

that they actually have one or several solutions (or none...), study their properties,

and possibly design numerical methods to compute these solutions or approximations.

A first important observation is that not all PDEs will be solve by a variational

analysis: only the PDEs which are “variational”, meaning that their equation is pre-

cisely of the form (1.1) for a particular E .

Why is it interesting?

• it provides sometimes a very simple tool for showing existence of (weak) solutions

to a problem;

• many PDEs come from problems in physics, mechanics, etc, and precisely from

“variational” principles and are therefore (often minimizing) critical points of

some physical energy.

• many problems in the industry (or finance, etc) are designed as finding the “best”

state according to some criterion, and their solution is precisely a minimizer, or

maximizer, of this criterion (“optimization”).

Standard examples:

1. Laplace equation −∆u = 0 characterizes the critical points of the “Dirichlet

energy”∫|∇u|2dx. In this case, since the energy is convex, critical points and

minimizers are the same.

5

6 CHAPTER 1. INTRODUCTION

2. Horizontal elastic membrane subject to a vertical force f :

minu=0 on ∂Ω

1

2

∫Ω

|∇u|2dx−∫

Ω

fudx,

in this case equation (1.1) reads−∆u = f in Ω,

u = 0 on ∂Ω.

Variant with a nonlinear potential energy:

minu=0 on ∂Ω

1

2

∫Ω

|∇u|2dx+

∫Ω

F (u)dx,

then the equation becomes −∆u+ F ′(u) = 0.

3. Nonlinear elasticity: E(u) =∫

ΩW (∇u)dx where now u : Ω ⊂ R3 → R3 is

vectorial valued. W might depend on ∇uT∇u, det∇u... Equation (1.1) becomes

now a system.

4. Minimal surfaces, geodesics: E(u) =∫

Ω

√1 + |∇u|2dx (area of the graph of u),

E(u) =∫ 1

0|u′(t)|2 dt with u(0) = x0, u(1) = x1, u(x) ∈M a manifold for each x.

For each of these problems, the natural questions are: is there existence of a solu-

tion? Uniqueness? How can it be characterized? How can it be computed?

Chapter 2

Characterization of the

critical points

This chapter addresses the issue of the meaning of equation (1.1).

2.1 Differentiability

Definition 1. Let X be a Banach space and U an open subset. E : U ⊂ X → R is

“Frechet”-differentiable at u ∈ U if there exists a continuous linear form DE(u) ∈ X∗,called “differential”, such that

limv→0

|E(u+ v)− E(u)−DE(u) · v|‖v‖X

= 0

which can also be written (using Hardy’s “o” notation)

E(u+ v) = E(u) +DE(u) · v + o(‖v‖X).

E is of class C1(U) if u 7→ DE(u) is continuous.

Definition 2. E is said to be Gateaux-differentiable at u ∈ U in the direction v ∈ Xif

DvE(u) =d

dtE(u+ tv)|t=0

exists. It is said to be Gateaux-differentiable at u if it exists for all v.

Clearly, if E is (Frechet-)differentiable at u, then it is Gateaux-differentiable, and

DvE(u) = DE(u) ·v for all v. The converse is not true. Examples for functions defined

in X = R2:

1. u(x, y) = x3

x2+y2 (and u(0) = 0): the Gateaux derivative in direction (a, b) 6= 0 is

the function a3/(a2 + b2) which is not linear in (a, b)...

7

8 CHAPTER 2. CHARACTERIZATION OF THE CRITICAL POINTS

2. u(x, y) = x2yx4+y2

√x2 + y2 ≤ 1

2

√x2 + y2 (and 0 in 0): the Gateaux derivative is

0, however, if (x, y) = (t, t2)→ 0 as t→ 0,

u(x, y)√x2 + y2

=t4

2t4=

1

26→ 0.

The Frechet-differential DE(u) is also denoted E ′(u), and by definition, a critical

point u is a point where (1.1) holds.

2.2 First variation of a functional

In practice, to derive the stationarity conditions of an energy, and in particular the

minimality conditions, it is enough to know how to compute directionals derivatives for

all “admissible” v. Indeed, assume that u is a minimizer of E over a set K ⊂ X: then

given v, provided u+ tv ∈ K for t > 0 small (which is the meaning of an “admissible

variation” v) one always have E(u+ tv) ≥ E(u) so that

limt→0,t>0

E(u+ tv)− E(u)

t≥ 0.

Then, if E is differentiable at u, one recovers that DE(u) · v ≥ 0. If both v and −v are

admissible, one recovers DE(u) · v = 0.

For a general theory, we will restrict ourselves to functionals of the form

E(u) =

∫Ω

L(x, u(x), Du(x) dx (2.1)

where Ω ⊂ Rn is a bounded open set, and u : Ω → Rm is a possibly vectorial-valued

function (in some functional space). Here “Du” is the differential, identified to an m×nmatrix with entries ∂ui/∂xα, i = 1, . . . ,m, α = 1, . . . , n (but sometimes, especially for

scalar functions, we will also use the Gradient ∇u which is a vector). The function

L : Ω× Rm × Rm×n → R(x, u, p) 7→ L(x, u, p)

is called the “Lagrangian”, and is assumed to be smooth.

Then, given u, v and t > 0 small, one has

E(u+ tv) = E(u) + t

∫Ω

∑i

∂L∂ui

(x, u,Du)vi +∑i,α

∂L∂piα

(x, u,Du)∂vi

∂xαdx+ o(1),

yielding

d

dtE(u+ tv)|t=0 =

∫Ω

∑i

∂L∂ui


∂L∂piα

(x, u,Du)∂vi

∂xαdx.

Hence at a critical point, one should have for all v admissible:∫Ω

∑i

∂L∂ui


∂L∂piα

(x, u,Du)∂vi

∂xαdx = 0 (2.2)

2.2. FIRST VARIATION OF A FUNCTIONAL 9

If all smooth v = (v1, . . . , vm) with compact support are admissible, one can integrate

by parts the last term, yielding the following form for equation (1.1)

∂L∂ui

(x, u,Du)−n∑α=1

∂

∂xα

(∂L∂piα

(x, u,Du)

)= 0

for all i = 1, . . . ,m, at least in the distributional sense (in D′ [13]).

Definition 3. The Euler-Lagrange equation (or system) associated to E, given by (2.1),

is∂L∂ui

(x, u,Du) =

n∑α=1

∂

∂xα

(∂L∂piα

(x, u,Du)

)= div

(∂L∂piα

(x, u,Du)

)(2.3)

for all i = 1, . . . ,m.

One observes that this equation, which characterizes the (smooth enough) critical

points of E , is in divergence form. In particular, non-divergence equations with non

constant coefficients such as ∑i,j

ai,j∂2u

∂xi∂xj= 0

are in general not variational.

Examples

1. E(u) = 12

∫Ω|∇u|2dx +

∫ΩF (u)dx: here L(x, u, p) = |p|2/2 + F (u) and one re-

covers the equation

div∇u = ∂uF (u)

(or ∆u = F ′(u)). In practice, the steps to derive these equations are as follow:

assume u is a minizer of E in H10 (Ω) (that is, with Dirichlet boundary conditions

u = 0 on ∂Ω. Then, any variation v ∈ H10 (Ω) is admissible: for any t > 0 (or

t ∈ R), u+ tv ∈ H10 (Ω) and one can write

E(u+ tv) ≥ E(u).

Hence,

0 ≤ E(u+ tv)− E(u)

t

=

∫Ω

∇u · ∇v dx+t

2

∫Ω

|∇v|2dx+

∫Ω

F (u+ tv)− F (u)

tdx.

The last term is ∫Ω

1

t

∫ t

0

F ′(u+ sv)v dx

so that if F is C1, it is easy to justify (using for instance Lebesgue’s convergence

theorem) that its limit as t→ 0 is precisely∫

ΩF ′(u)v dx. Hence,∫

Ω

∇u · ∇v + F ′(u)v dx = 0


for all v ∈ H10 (Ω) (or any v ∈ C∞c (Ω) would be enough to conclude). This is

precisely the weak form or variational form for ∆u = F ′(u) (and it is standard

that if u has two derivatives in L2, the two forms are equivalent).

2. For m = 1, given a matrix A = (aα,β)1≤α,β≤n, we let L(x, u, p) =∑α,β aα,βpαpβ .

This is the Lagrangian for∫

Ω(A∇u) · ∇udx: then the equation is

divA∇u = 0.

3. L(x, u, p) = (1/q)|p|q, 1 < q < +∞: the equation is the q-Laplace equation

div |∇u|q−2∇u = 0.

For q = 1, the Lagrangian is not regular enough and the signification of the limit

equation

“div∇u|∇u|

= 0′′

must be clarified when ∇u = 0...

4. L(x, u, p) =√

1 + |p|2: the equation is the minimal surface (or graph) equation

div∇u√

1 + |∇u|2= 0.

2.3 Boundary conditions

The Euler-Lagrange equations we have derived up to now are not complete. Indeed, if

one considers for instance the equation ∆u = 0 (which characterizes the critical points

of∫

Ω|∇u|2dx), it has an infinite number of solutions: any harmonic function in Ω is a

solution. However, if we are considering the problem

minu∈X

∫Ω

|∇u|2dx,

for X = H1(Ω) (such that the functional is finite) or X = H10 (Ω) (that is, if we require

in addition that u vanishes on the boundary), then most harmonic solutions are not a

solution of the problem (neither a minimizer nor a critical point, which in this case are

the same). Indeed, the minimal value for this problem is 0, reached for u = 0, while

in general if u is harmonic, the integral will not vanish...

Remarks: if X = H1(Ω), there are still infinitely many solutions, which ones? Given

u a harmonic function with finite energy, what minimization problem does it solve?

2.3.1 The prescribed boundary conditions

The “Dirichlet” or prescribed boundary conditions are the conditions which come

from restrictions imposed in the space of definition of the functional E . They are thus,

2.3. BOUNDARY CONDITIONS 11

strictly speaking, not part of the Euler-Lagrange equation “E ′(u) = 0” (1.1), but in

the condition “u ∈ X”. Let us consider two examples: the first is the problem

minu∈g+H1

0 (Ω)

∫Ω

|∇u|2dx, (2.4)

where g ∈ H1(Ω) (which is necessary to guarantee that there exists at least one

function u with finite energy, otherwise the problem cannot have a solution!)

In this case, the equation which is solved by a minimier is−∆u = 0 in Ω

u = g on ∂Ω.(2.5)

The first line in this equation comes from the stationarity of E at u, while the second

is because we have prescribed that u ∈ g +H10 (Ω).

In this case, the admissible variations v are the functions v ∈ H10 (Ω) (or the dense

subclass v ∈ C∞c (Ω). Indeed for such functions, u + tv is not modified on ∂Ω. Note

that the admissible variations do not satisfy v = g, but v = 0, on the boundary!

Warning: if the Dirichlet condition u = g is not the trace of a H1 function on ∂Ω,

it will precisely mean that for any u with u = g ∈ Ω,∫

Ω|∇u|2dx = +∞. Hence

the variational problem (2.4) cannot be solved. However, this does not mean that

(2.5) is not solvable! For instance, there exists in the open planar disk B1 a function

u ∈ C∞(B1) (even analytic) such that ∆u = 0 in B1, u(cos θ, sin θ) = sin(θ/2),

−π ≤ θ ≤ π. But there is no function in H1(B1) with a jump discontinuity on the

boundary ∂B1.

A second example is illustrated by the following problem

minu∈g+H2

0 (Ω)

∫Ω

|∆u|2dx

for a given g ∈ H2(Ω). We recall that H20 (Ω) is the closure of C∞c (Ω) in H2(Ω), hence

for the norm u 7→√∫

u2 + |Du|2 + |D2u|2dx.

The Euler-Lagrange equation for this problem is∆2u = 0 in Ω

u = g on ∂Ω

∇u = ∇g on ∂Ω.

The latter condition follows because functions in v ∈ H20 (Ω) have a vanishing gradient

on the boundary (if this boundary is smooth enough). Since the tangential gradient is

obviously vanishing (because v = 0 on ∂Ω), it is equivalent to write that the normal

gradient ∂v/∂ν = ∇v · ν = 0. Hence one could replace the last condition in the

equation by∂u

∂ν=∂g

∂νon ∂Ω,


however, this is still (in my opinion) a “Dirichlet” condition: it is a restriction on the

space of minimization (which makes sense because functions with∫|∆u|2 <∞ have a

well-defined gradient on a regular boundary), it is not coming from the first variation

equation E ′(u) = 0!

2.3.2 The free boundary conditions

On the contrary, the “Neumann” or free boundary conditions are the conditions which

do not arise from restrictions on u but from the equilibrium equation E ′(u) = 0,

written in the variational sense DE(u) · v = 0, and the fact that the variations (or test

functions) v are allowed to vary freely on a part of the boundary, so that this equation

contains, in fact, an information on the behaviour of u on the boundary.

Examples

Let us start with a few examples: assume u is a solution of

minu∈H1(Ω)

∫Ω

|∇u|2 + (u− g)2dx

Then considering variations u+ tv, v ∈ H1(Ω), t small, one easily finds that∫Ω

∇u · ∇v + (u− g)v dx = 0, (2.6)

this for any v ∈ H10 (Ω). If we consider first v ∈ C∞c (Ω), then one can integrate by

parts the first half and find

−∫

Ω

∆u v dx+

∫Ω

(u− g)v dx = 0

and since this must be true for all v ∈ C∞c (Ω), we must have

−∆u+ u = g in Ω.

Now, consider v ∈ C∞(Ω). Then, integrating by parts (2.6) again one finds∫∂Ω

v∇u · ν dσ +

∫Ω

(−∆u+ u− g)v dx = 0,

however we already know that −∆u+ u− g = 0, so that this becomes simply∫∂Ω

v∇u · ν dσ = 0.

Since this must hold for all smooth v, we find that ∂v/∂ν = 0. Hence the Euler-

Lagrange equation is now: −∆u+ u = g in Ω∂u∂ν = 0 on ∂Ω.

The important point is that now, the second equation does not come from restrictions

on the function space where the functional is minimized. It is part of the condition

E ′(u) = 0, and comes, on the contrary, from the fact that admissible variations v are

free to vary on the boundary.

2.4. A REMARK ON “NULL LAGRANGIANS” 13

Example: consider now a (smooth) solution to the problem

minu∈H1(Ω)

∫Ω

|∇u|2 dx− 2

∫∂Ω

ufdσ.

Can you show that (i) −∆u = 0 in Ω? (ii) ∂u/∂ν = g on ∂Ω?

General situation

For the general form (2.1), if we look for a critical point with “free” boundary con-

ditions (that is, variations v ∈ C∞(Ω)m, which do not vanish on the boundary, are

admissible), we find in addition to (2.3) (which is obtained precisely considering the

variations v ∈ C∞c (Ω)m) that, integrating (2.2) by parts

0 =

∫∂Ω

vin∑α=1

(∂L∂piα

(x, u,Du)

)να dσ

+

∫Ω

(∂L∂ui

(x, u,Du)−n∑α=1

∂

∂xα

(∂L∂piα

(x, u,Du)

))vi dx

for all i = 1, . . . ,m (here ν = (να)nα=1 is the outer normal to the boundary of ∂Ω).

Since (2.3) holds, the second integral vanishes, and since the first must then vanish for

all v, we deduce that for i = 1, . . . ,m,(∂L∂piα

(x, u,Du)

)nα=1

· ν = 0. (2.7)

When ∂Ω or u is not regular enough to justify such computations, we say that the

condition (2.7) holds “in the weak sense”.

Remark: you can also check that “Robin” or “oblique” boundary conditions, for

critical points of first order functionals, fall into this category (they arise when the

energy has nonlinear terms on the boundary).

2.4 A remark on “Null Lagrangians”

Consider again the energy in the standard form (2.1).

Definition 4. L is called a “Null Lagrangian” if for any u : Ω→ Rm smooth enough,

(2.3) is satisfied.

It means that any smooth u is a critical point of the energy. In this case, given u, v

regular enough with u = v on the boundary ∂Ω, one has E(u) = E(v)!

Proof. We let for t ∈ [0, 1] f(t) = E(tu+ (1− t)v). We then compute (to simplify we

denote ut = tu+ (1− t)v):

f ′(t) =

∫Ω

∑i

∂L∂ui

(x, ut, Dut)(ui − vi) +

∑i,α

∂L∂piα

(x, ut, Dut)∂(ui − vi)

∂xαdx.


Integrating by parts (we recall that u − v = 0 on the boundary) and using (2.3), it

follows f ′(t) = 0, hence f is constant and f(0) = f(1).

But: does it exist? (But, of course, for the trivial case L = 0.) If m = 1 one

can show that the only Null-Lagrangians are of the form L(x, u, p) = a · p+ b(x), a a

constant vector, or more generally L(x, u, p) = φ(x) · p+ udivφ(x) + b(x) (for φ a C1

vector field). In this case,∫Ω

L(x, u,Du)dx =

∫Ω

div (φu) + b(x)dx =

∫∂Ω

uφ · ν dσ +

∫Ω

b(x)dx

and depends only on the boundary values of u. For m > 1 we will see in Chapter 5

that there are less trivial examples.

2.5 Constrained problems, obstacles

When there are constraints for a minimization problem, the Euler-Lagrange equation

is not valid anymore, at least in general. Let us just illustrate this with two important

examples: the convex constraints, and the differentiable constraints.

2.5.1 Convex constraints

Consider the obstacle problem

minu≥ψ

∫Ω

|∇u|2dx :

where the function u is constrained to belong to the convex set C = u ∈ H10 (Ω) , u ≥

ψ. In this case, instead of considering a variation of the form u + tv (which would

raise the difficult issue of how to choose v) it is more convenient to consider another

candidate v ∈ C, and variations of the form tv + (1 − t)u ∈ C (as tv + (1 − t)u =

u+ t(v−u), it corresponds to nothing else than choosing a variation of the form v−u,

v ∈ C). It easily follows, for a minimizer, the equation

DE(u) · (v − u) ≥ 0

for all v ∈ C, called a “variational inequality”. In the example above, this reads∫Ω

∇u · ∇(v − u)dx ≥ 0

for all v with v ≥ ψ. Now, in the set u > ψ (or at least in its interior if u is not

continuous), one can consider a v of the form u ± tφ, φ ∈ C∞c (u > ψ), which also

satisfies u± tφ ≥ ψ a.e. for t small enough. It follows that∫Ω

∇u · ∇φdx = 0

for all such φ, in other words, −∆u = 0 as in the unconstrained case, in the set where

the constraint is not active.

2.5. CONSTRAINED PROBLEMS, OBSTACLES 15

On the other hand, a function v ≥ u always satisfies v ∈ C. This means that for

any φ ≥ 0, φ ∈ H10 (Ω), one will have∫

Ω

∇u · ∇φdx ≥ 0.

In other words, −∆u ≥ 0, which means u is “superharmonic”. One can show (since

−∆u is a signed distribution) that it is a nonnegative Radon measure, and write the

Euler-Lagrange equation, in this case−∆u = µ µ nonnegative measure,

u = 0 on ∂Ω,

µ(u > ψ) = 0.

A second typical example is the following variational problem

minµ∈P(Ω)

∫Ω×Ω

W (x, y)dµ(x)dµ(y)

where P(Ω) is the set of probability measures on Ω and W a nonnegative, symmetric

weight. If µ is minimizing, one finds that for any ν∫Ω×Ω

W (x, y)dµ(x)d(ν − µ)(y) ≥ 0

and one can deduce that there exists a constant c such that∫

ΩW (x, y)dµ(y) = c µ-a.e.

2.5.2 Differentiable constraints

Formally, one solves a problem of the form

minF(u)=0

E(u)

(or one searchs for a critical point). A minimizer should satisfy that DE(u) · v = 0

in all directions which is tangent to the manifold u : F(u) = 0, that is, with

DF(u) · v = 0. Hence the equation should read formally “DE(u) ‖ DF(u)”. In other

words, there exists a “Lagrange multiplier” λ ∈ R such that

DE(u) = λDF(u). (2.8)

Example: we consider the problem min∫Ωu2=1

∫Ω|∇u|2dx, for u ∈ H1

0 (Ω). For a

general v ∈ H10 (Ω), there is no reason for which one would have

∫Ω

(u+ tv)2dx = 1 for

any t small enough (it is even impossible unless v = 0). However, one may consider a

path through u on the variety∫

Ωu2dx = 1 with tangent v at u, such as

t 7→ u+ tv

‖u+ tv‖L2

The minimality of u yields∫Ω|∇u+ t∇v|2dx∫Ω|u+ tv|2dx

≥∫

Ω

|∇u|2dx


for all v ∈ H10 and any t small (or not). Developping, we obtain∫

Ω|∇u|2dx+ 2t

∫Ω∇u · ∇v dx+ t2

∫Ω|∇v|2dx

1 + 2t∫

Ωuv dx+ t2

∫Ωv2dx

≥∫

Ω

|∇u|2dx.

This means that the left-hand side of this inequality is minimized for t = 0, hence that

its derivative at t = 0 should vanish. This derivative is precisely∫Ω

∇u · ∇v dx−∫

Ω

|∇u|2dx∫

Ω

uv dx.

Hence the Euler-Lagrange equation for this problem is

−∆u = λu

and has the form (2.8), with λ =∫

Ω|∇u|2dx.

Chapter 3

Existence of minimizers

A.k.a. “The Direct Method in the Calculus of Variations”.

3.1 Lower semicontinuity

Recall that “a continuous function on a compact set reaches its bounds”, i.e., if f : C →R is continuous and C is compact, there exists, x, y ∈ C such that f(x) = maxC f and

f(y) = minC f . Now, if we only need to minimize f , is the continuity really needed?

In fact, only half of it is necessary.

Definition 5. Let U ⊂ X be an open subset of a metric vector space X. We say that

f : U → R is lower semicontinuous (lsc) at x ∈ U if for any sequence xn → x, one has

f(x) ≤ lim infn→∞

f(xn). (3.1)

f is lsc in U if it is lsc at any x ∈ U . We say that f is upper semicontinuous (usc) if

−f is lsc.

(In a general topological space, this definition defines the “sequential lower semiconti-

nuity”.) Here, R = R ∪ −∞,+∞, the extended real line.

Proposition 1. Let f : C ∈ R ∪ +∞ be a lsc function on a nonempty compact

metric set C, and assume f 6≡ +∞. Then f reaches its minimum: there exists x ∈ Csuch that f(x) = minC f .

Proof. Since ∃x ∈ C with f(x) < +∞, we can consider a minimizing sequence (xn),

that is a sequence such that limn f(xn) = infC f . Since C is compact, there exists a

subsequence (xnk)k converging to a point x ∈ C. Since f is lsc, one has

f(x) ≤ lim infk→∞

f(xnk) = limn→∞

f(xn) = infCf.

And since x ∈ C, it follows f(x) = minC f .

17

18 CHAPTER 3. EXISTENCE OF MINIMIZERS

Proposition 2. Let (fi)i∈I be a family of (extended) real-valued lsc functions. Then

f(x) := supi∈I fi(x) is lsc.

Proof. Indeed, if xn → x, for any i, we have fi(x) ≤ lim infn→∞ fi(xn) ≤ lim infn→∞ f(xn).

Hence, f(x) = supi fi(x) ≤ lim infn→∞ f(xn).

Important examples of lsc functionals are defined by duality: for instance if

E(u) = sup

∫Ω

udivφdx : φ ∈ C∞c (Ω;Rn), ‖φ‖L2 ≤ 1

(3.2)

then by definition it is obviously lsc in L2(Ω), thanks to Proposition 2 (or in L1(Ω), or

even, in fact, in the distributional sense—sequentially). On the other hand, one can

show that for any u ∈ L2(Ω),

E(u) =

‖∇u‖L2 if u ∈ H1(Ω),

+∞ else.

Proposition 3. f is lsc if and only if for all a ∈ R, the sets f ≤ a are closed, or if

and only if for all a ∈ R, the sets f > a are open.

3.2 The direct method in the calculus of variations

3.2.1 Basic principle

Given a minimization problem

minu∈XE(u),

the “direct method in the calculus of variations” consists in

• showing that minimizing sequences belong to a compact set;

• showing that E is lsc on this set.

However, an important remark here is that in infinite dimension, not all topologies

coincide, and it is “enough” to find a topology in which minimizing sequences are

compact and E is lsc. Observe however that

• The weaker the topology, the easier it will be for sequences to converge, and in

particular it will be more likely that a minimizing sequence is compact;

• However, the stronger the topology, the easier it is for a function to be continuous

or lsc, since there are less converging sequences on which the continuity or lower

semicontinuity needs to be checked... (for instance, the example (3.2) is obviously

continuous in the strong H1 topology, but only lsc for the L2 topology.)

Hence, more precisely the technique of proof follows these steps:

1. one checks that u ∈ X : E(u) < +∞ 6= ∅ (otherwise there will not exist

minimizing sequences);

3.2. THE DIRECT METHOD IN THE CALCULUS OF VARIATIONS 19

2. one considers a minimizing sequence (un)n, that is, such that E(un)→ infX E as

n → ∞. This always exists, by definition of the “inf”, as soon as the previous

condition is satisfied:

∀n ∈ N ∃un s.t. E(un) ≤ max−n, infXE + 1/n;

3. one looks for a topology on X for which such a sequence is [sequentially] pre-

compact, that is, for which there will be subsequences (unk) which converge to

some limit u;

4. one tries to check that E is lsc in this topology, yielding

E(u) ≤ lim infk→∞

E(unk).

3.2.2 An example

We consider the problem

minu∈XE(u) :=

∫Ω

|∇u|2 − 2fu dx (3.3)

for a given function f ∈ L2(Ω), where Ω is a bounded open subset of Rn, and X is

either H10 (Ω) or H1(Ω).

The first important point is to observe that there exists u with E(u) < ∞: for

instance, E(0) = 0. This allows to consider a minimizing sequence (uk)k, with E(uk)→infX E .

We now want to check whether (uk)k is (pre)compact. The standard way is to start

showing bounds on the sequence. We can for instance assume that we have chosen

each uk in such a way that E(uk) ≤ E(0) = 0, hence for all k,∫Ω

|∇uk|2dx ≤ 2

∫Ω

fuk dx ≤ 2‖f‖L2‖uk‖L2 .

If X = H10 (Ω), we can use the Poincare inequality to deduce that

‖uk‖2L2 ≤ C∫

Ω

|∇uk|2dx ≤ 2C‖f‖L2‖uk‖L2 ,

hence ‖uk‖L2 ≤ 2C‖f‖L2 and it follows that for some constant C > 0,

‖uk‖H1 ≤ C‖f‖L2 ,

the sequence is uniformly bounded in H1(Ω).

If X = H1(Ω), assuming Ω is regular (for instance with Lipschitz boundary), the

Poincare-Wirtinger inequality establishes that there exists a constant C such that for

each k, there is a constant ck (for instance, ck =∫

Ωuk dx/|Ω|) such that

‖uk − ck‖2L2 ≤ C‖∇uk‖2L2 ≤ 2C

∫Ω

fuk dx

= 2C

∫Ω

f(uk − ck) dx+ 2Cck

∫Ω

f dx

≤ 2C‖f‖L2‖uk − ck‖L2 + 2Cck

∫Ω

f dx (3.4)


and one cannot hope to control this norm, unless∫

Ωf dx = 0. However, this is good

news if one observes that if∫

Ωf dx 6= 0, for any constant function c ∈ R, E(c) =

−2c∫

Ωf dx and the infimum over constants is −∞. Hence for X = H1(Ω), there

cannot be a solution (and minimizing sequences cannot be compact) unless∫

Ωf dx = 0.

In this case, (3.4) yields

‖uk − ck‖L2 ≤ 2C‖f‖L2

and again one can deduce a global bound in H1 for the modified sequence uk = uk−ck.

How do we use these bounds? We need compactness results ensuring that bounded

functions in H1 will be compact in some topology.

3.3 Compactness results in infinite dimension

We will gather here three essential compactness results which are useful in the study

of minimizing sequences.

3.3.1 Ascoli-Arzela

Theorem 1 (Ascoli Arzela). Let X be a compact metric space and (uk)k a sequence

of uniformly bounded:

supx∈X

supk≥0|uk(x)| < +∞

and uniformly equicontinuous functions:

∀ε > 0,∃η > 0,∀x, y,dist (x, y) < η ⇒ |uk(x)− uk(y)| ≤ ε. (3.5)

Then, there exists u ∈ C0(X) and a subsequence ukl such that ukl → u uniformly as

l→∞.

Proof. Consider (xn) a countable dense sequence in X. One can finds ukl such that

ukl(xn) has a limit, denoted u(xn), for all n. This is a classical diagonal procedure: first

extract uφ1(k)(x1) converging to u(x1). Then, from the sequence uφ1(k)(x2), extract

uφ2(k)(x2) which converges to u(x2). Etc... Eventually let kl := φl(l).

Now, one checks easily that the uniform continuity of uk passes to the limit:

∀ε > 0,∃η > 0,∀xn, xm,dist (xn, xm) < η ⇒ |u(xn)− u(xm)| ≤ ε. (3.6)

Use this to show that u can be extended to all of X, into a well-defined continuous

function u. Idea: for x ∈ X, consider xni → x, then u(xni) is a Cauchy sequence

thanks to (3.6) and has a limit, which we denote u(x), once one has checked that this

limit was not depending on the particular subsequence (xni). Again, u satisfies (3.6)

but now everywhere in X and not only on the dense subset (xn) : n ≥ 1.Eventually, by compactness of X, given ε and η as in the equi-continuity statement,

one can cover X by finitely many balls B(xni , η), i = 1, . . . , N . If l is large enough,

|ukl(xni) − u(xni)| ≤ ε for all i = 1, . . . , N . Then if x ∈ X, there is i such that

x ∈ B(xni , η), and

|ukl(x)− u(x)| ≤ |ukl(x)− ukl(xni)|+ |ukl(xni)− u(xni)|+ |u(xni)− u(x)| ≤ 3ε

and the proof is complete.

3.3. COMPACTNESS RESULTS IN INFINITE DIMENSION 21

Application: In dimension 1, if Ω = (0, 1) in (3.3), we obtain from the bound∫Ω|u′k|2 dx ≤ c <∞ that the uk are uniformly equicontinuous: indeed, for each k and

x, y ∈ Ω,

|uk(y)− uk(x)| =

∣∣∣∣∣∫

[x,y]

u′k(s) ds

∣∣∣∣∣ ≤√|x− y|√c (3.7)

from which (3.5) is easily deduced. (Rigorously, one should first approximate each ukby smooth functions before writing inequality (3.7).) It follows from Ascoli’s theorem

that up to a subsequence, the uk (or the modified uk) converge uniformly to a limit u.

Exercise: One can deduce that u is a minimizer because in L2(Ω), for instance,

v 7→

12

∫Ω|∇v|2dx if u ∈ H1(Ω),

+∞ else.

is lower semicontinuous, as one can check it coincides with the sup of continuous

functionals:

sup

∫Ω

udivφdx − 1

2

∫Ω

|φ|2dx : φ ∈ C∞c (Ω;Rn), ‖φ‖L2 ≤ 1

(using Riesz’ representation theorem).

However, the use of Ascoli-Arzela’s theorem is restricted to few situations, where

one can show uniform equi-continuity of a sequence of functions (mostly, in 1D, or

when the sequence solves higher order problems).

The next important compactness theorem in functional analysis is Rellich’s theo-

rem:

Theorem 2 (Rellich). Let Ω ⊂ Rn be a bounded, regular1 open set. Let (uk)k≥1 be a

bounded sequence in W 1,p(Ω), 1 ≤ p ≤ +∞2. Then there exists u ∈ Lp(Ω) and (ukl) a

subsequence which converges to u in Lp(Ω). If p > 1 one has in addition that W 1,p(Ω).

Proof. The proof relies on three ingredients:

1. A continuous extension operator T : W 1,p(Ω) → W 1,p0 (B), where B is a large

ball containing Ω, such that ‖Tu‖W 1,p(Rn) ≤ C‖u‖W 1,p(Ω) and Tu|Ω = u. This

is the reason for which the domain needs to be bounded and regular, one can

“easily” build such an operator for Lipschitz domains [8]. (To simplify, we now

denote uk the sequence Tuk.)

2. An estimate of the Lp-distance between the smoothing uεk := ρε∗uk of uk and ukitself, where ρε(z) = ε−nρ(z/ε) is a smooth mollifying kernel. This is obtained

1for instance with Lipschitz boundary.2as soon as p > n, one can invoke Ascoli’s theorem, thanks to Sobolev/Morrey’s injections.


as follows:

uεk(x)− uk(x) =

∫Rnρε(z)[uk(x− z)− uk(x)]dz

=

∫Rnρε(z)

∫ 1

0

−∇uk(x− tz) · z dtdz

= −ε∫ 1

0

∫Rn

[(tε)−nρ

(z′

tε

)z′

tε

]· ∇uk(x− z′)dz′dt,

which yields (using ‖f ∗ g‖p ≤ ‖f‖1‖g‖p)

‖uεk − uk‖Lp ≤ ε‖zρ(z)‖L1‖∇uk‖Lp .

3. Ascoli’s theorem applied to the sequences (u1/ik )k≥1 for all i. By a diagonal

argument, one can find a subsequence ukl such that for each i, u1/ikl

converges

uniformly as l→∞. Then, one uses

‖ukm − ukl‖Lp ≤ ‖ukm − u1/ikm‖Lp + ‖u1/i

km− u1/i

kl‖Lp + ‖u1/i

kl− ukl‖Lp

≤ C/i+ ‖u1/ikm− u1/i

kl‖Lp

by the previous step, showing that ukm is a Cauchy sequence in Lp, which ends

the proof.

For p > 1, one also can show (for instance by duality) that ‖∇u‖Lp is lower

semicontinuous in Lp.

Application: if in problem (3.3), the domain is bounded and regular, we immedi-

ately get compactness in L2 of the sequence (uk or uk). Semicontinuity is as before.

In more general situation, one can “almost always” rely on the Banach-Alaoglu-

Bourbaki Thm [5, 4, 12, 11, 14]:

Theorem 3 (Banach-Alaoglu-Bourbaki). Let X be a Banach space and X∗ (or X ′)

its topological dual, that is the set of continuous linear forms, endowed with the norm

‖l‖∗ = sup‖x‖≤1 〈l, x〉. Then BX∗ = l ∈ X∗ : ‖l‖∗ ≤ 1 is compact for the weak-∗topology σ∗(X∗, X), defined as the weakest topology such that l 7→ 〈l, x〉 is continuous

for all x ∈ X:

ln∗ l⇔ ∀x ∈ X, 〈ln, x〉 → 〈l, x〉 .

Remarks 1. If X is reflexive (Lp, W 1,p with 1 < p < +∞...) then a result is that

the unit ball of X is weakly compact. This is not true in general (ex: X = L1(Ω)).

2. If X is separable, that is, when there exists (xn)n≥1 a dense sequence in X (that is,

quite often), then

dist (f, g) =∑n≥1

2−n| 〈f − g, xn〉 |

is a distance on bounded sets of X∗, which induces the weak-∗ topology. It implies

that from any bounded sequence of X∗, one can extract converging subsequences.

3.4. BASIC LOWER-SEMICONTINUITY CRITERIA 23

Proof. In general, the proof relies on the fact that a product of compact spaces is

compact, see for instance [14]. We can give a very simple “constructive proof” in the

simplified setting where X is separable, with a dense sequence (xn)n≥1. In this simpler

case, it is equivalent to show the compactness of bounded sequences (yk)k≥1 of points

of X∗. Assume c = supk ‖yk‖∗ < +∞, then since 〈yk, xn〉 is bounded for each n, one

can extract a subsequence ykl such that 〈ykl , xn〉 converges to a value l(xn) for each n

(by a diagonal argument). Then we observe that l(xn)− l(xm) = liml 〈ykl , xn − xm〉 ≤c‖xn − xm‖, hence l can be extended by continuity to all of X. (Since if xnm → x,

then l(xnm) is a Cauchy sequence thanks to the previous inequality, and converges to

a value l(x) independent of the sequence (xnm).) One checks that l(x) = liml 〈ykl , x〉for all x, indeed if xnm → x, then for all l,m

|l(x)− 〈ykl , x〉 | ≤ |l(x)− l(xnm)|+ |l(xnm)− 〈ykl , xnm〉 |+ | 〈ykl , xnm − x〉 |≤ 2c‖xnm − x‖+ |l(xnm)− 〈ykl , xnm〉 |

which is arbitrarily small if m is large enough and then l large enough. One deduces

that l is linear, continuous, and is the weak-∗ limit of ykl .

Remark: This proof also shows that ‖l‖∗ ≤ c, and more precisely, one easily deduces

that ‖l‖∗ ≤ lim inf l→∞ ‖yml‖∗. In general, if yk∗ y, then the Banach-Steinhaus

theorem ensures that ‖y‖∗ ≤ lim infk ‖yk‖∗, see for instance [5, 4].

Application: In problem (3.3), it follows from Banach-Alaoglu-Bourbaki’s theorem

that uk (or uk) is bounded in H1(Ω), hence admits weakly converging subsequences.

Semicontinuity is as before.

3.4 Basic lower-semicontinuity criteria

A lot of the theory of minimizers (see for instance [6]) is devoted to the study of criteria

for the lower-semicontinuity of functionals. We give here only an elementary one (how-

ever almost “necessary” for integrands of the first order of scalar-valued functions).

3.4.1 Convex functions

Theorem 4. Consider X a Banach space. Let C ⊂ X be a convex set: then C is

closed for the strong topology (induced by the norm) if and only if it is closed for the

weak topology σ(X,X∗) (defined by xn x if and only if 〈l, xn〉 → 〈l, x〉 for all l ∈ X∗.Let f : X → R ∪ +∞ be a convex function: then f is lsc for the strong topology

if and only if it is lsc for the weak topology.

The second statement is a consequence of the first, in the particular case where

the space is X × R and the convex set C is the epigraph (x, t) : t ≥ f(x), which is

closed if and only if f is lsc.

The first statement is a consequence of the Hahn-Banach separation theorem, see [5,

4]. The idea is that if f is strongly lsc, then it can be written as a supremum of


continuous linear form:

f(x) = supl∈X∗

〈l, x〉 − f∗(l)

which is weakly lsc.

Application: returning to problem (3.3), one can check easily that u 7→ E(u) is

convex and continuous in the strong H1 topology, hence it is also weakly lsc in H1.

(Observe that this point of view only requires f ∈ L2 but no assumption on the domain

Ω.)

In general, one cannot hope more. If uk → u in L2 or weakly in H1, one has not

in general∫

Ω|∇uk|2dx→

∫Ω|∇u|2dx. For instance let

uε(x) =

∫ x

0

sign(sε

)ds

which is a kind of sawtooth. Then |u′ε| = 1 a.e.,∫ 1

0|u′ε|2dx = 1, but uε → u = 0

uniformly, which satisfies∫ 1

0|u′|2dx = 0 < 1.

Remark: observe that the previous function uε is a minimizing sequence for the

problem

minu∈H1(0,1)

∫ 1

0

|1− u′2|+ u2dx.

which is not convex in u′.

3.4.2 Non-convex cases

It is not necessary to be convex for a functional to be lsc: for instance

E(u) := 12

∫Ω

|∇u|2 + F (u)dx

for F continuous, bounded from below, is weakly lsc in H1(Ω). However, we now will

see that for scalar integrands of the form∫L(x, u,Du), convexity in the last variable

is essentially necessary.

3.4.3 Necessity of the convexity

Lemma 1. Consider f : R→ R a continuous function such that

u 7→∫ 1

0

f(u(x))dx

is lsc for the weak convergence in Lp(0, 1), 1 < p ≤ ∞ (the weak-∗ convergence for

p =∞). Then f is convex.

Proof. Let a, b ∈ R and t ∈ [0, 1] and consider a 1-periodic function ψ(x) = a if

0 ≤ x < t, ψ(x) = b if t ≤ x < 1. Let then uk(x) = ψ(kx) for k ≥ 1. Then

uk ta+ (1− t)b

3.4. BASIC LOWER-SEMICONTINUITY CRITERIA 25

in Lp. Indeed, up to a subsequence, uk u (Banach-Alaoglu-Bourbaki). Then, one

can identify u by testing against smooth (or just continuous) functions: if g ∈ C(0, 1),

∫ 1

0

uk(x)g(x)dx =1

k

∫ k

0

ψ(y)g( yk )dy =1

k

k∑l=1

∫ l

l−1

ψ(y)g( yk )dy

≈ 1

k

k∑l=1

ta g(l−1k

)+ (1− t)b g

(lk

) k→∞−→ (ta+ (1− t)b)∫ 1

0

g(s)ds

Then, by semicontinuity,∫ 1

0

f(ta+ (1− t)b)dx ≤ lim infk→∞

∫ 1

0

f(uk)dx = tf(a) + (1− t)f(b).

Lemma 2. Let m,n ≥ 1 and f : Rmn → R such that

u 7→∫ 1

0

f(∇u)dx

is weakly lsc in W 1,p(Ω;Rm), where Ω ⊂ Rn. Then if n = 1 or m = 1, f is convex.

The proof is essentially the same. It FAILS for both n > 1 and m > 1. One always

have:

Lemma 3. Let m,n ≥ 1 and f : Rmn → R such that

u 7→∫ 1

0

f(∇u)dx

is weakly lsc in W 1,p(Ω;Rm), where Ω ⊂ Rn. Then f is “quasiconvex” in the sense of

Morrey, that is: for any φ ∈W 1,∞0 (Ω;Rm) and any A ∈ Rm×n,

|Ω|f(A) ≤∫

Ω

f(A+Dφ) dx.

Proof. First, observe that it is equivalent to require that

|Ω′|f(A) ≤∫

Ω′f(A+Dφ) dx

for any φ ∈ W 1,∞(Ω′;Rm) where Ω′ is any other bounded open set of Rn. Indeed,

choosing z, λ such that z + λΩ′ ⊂ Ω, for such φ,

φ′ : x 7→

λφ(x−zλ

)if x ∈ z + λΩ′

0 else∈W 1,∞

0 (Ω;Rm).

Then∫Ω

f(A+Dφ′)dx = |Ω \ (z + λΩ′)|f(A) +

∫z+λΩ′

f(A+Dφ

(x−zλ

))dx

= |Ω \ (z + λΩ′)|f(A) + λn∫

Ω′f(A+Dφ)dx


and since |Ω|f(A) ≤∫

Ωf(A+Dφ′)dx, it follows

|Ω′|f(A) = λ−n|z + λΩ′|f(A)

= λ−n(|Ω| − |Ω \ (z + λΩ′)|)f(A) ≤∫

Ω′f(A+Dφ)dx.

Then, we consider φ ∈W 1,∞(Q;Rm) where Q = (0, 1)n is the unit cube. We then

consider the cubes Qz = z + εQ ⊂ Ω, for z ∈ Zn, and the function

φε :=

εφ(x−zε

)if x ∈ z + εQ ⊂ Ω

0 else

which is in W 1,∞(Ω), and goes uniformly to 0 as ε → 0. Let u(x) := Ax, uε(x) :=

Ax+ φε(x): one easily shows that uε u in W 1,p(Ω;Rm). Hence

|Ω|f(A) =

∫Ω

f(Du)dx ≤ lim infε→0

∫Ω

f(Duε)dx.

The last integral is

∑z∈Zn

z+εQ⊂Ω

∫z+εQ

f(A+Dφ

(x−zε

))dx

= εn]z ∈ Zn : z + εQ ⊂ Ω∫Q

f(A+Dφ)dxε→0−→ |Ω|

∫Q

f(A+Dφ)dx

and it follows (since |Q| = 1)

|Q|f(A) ≤∫Q

f(A+Dφ)dx

establishing the quasiconvexity of f .

Remark: if n = 1 or m = 1, then the quasiconvexity is equivalent to the convexity

(why?)

Chapter 4

Regularity for minimizers

(elliptic case)

Once one has shown existence of the solution to a problem such as (3.3), a natural

question is in what sense the Euler-Lagrange equation

−∆u = f

is valid? A priori, the meaning of this equation is that “the Distributional Laplacian”

of u ∈ H1(Ω) (or the distributional divergence of ∇u ∈ L2(Ω)) is represented by a

L2 function. However, can we say more on the regularity of u? If f is continuous, is

this equation “classical”, that is, is u a C2 function? And before this, does one have

that ∆u is the trace of a L2 matrix D2u? The regularity theory aims at finding the

conditions under which the weak (variational) Euler-Lagrange equations are, in fact,

true a.e., or pointwise, in a stronger sense.

The answer to the second question is easily deduced by the “translation method”

of Nirenberg, which we describe in the next section.

4.1 Basic regularity estimates

4.1.1 Obvious estimates

A first remark is that given a PDE in weak/variational form, it is always wise to

check first from obvious estimates which could follow from taking u, if admissible, or

admissible functions of u as a test function...

Choice of a test function For instance, if u ∈ H10 (Ω) and∫

Ω

∇u · ∇v dx =

∫Ω

fv dx

for all v ∈ H10 (Ω) (Pb. (3.3)) then obviously∫

Ω

|∇u|2 dx ≤ ‖f‖L2‖u‖L2 ,

27

28 CHAPTER 4. REGULARITY FOR MINIMIZERS (ELLIPTIC CASE)

showing that the H10 norm of u is controlled by ‖f‖2. (More precisely, one could bound

it by the “H−1” norm of f , ‖f‖H−1 = supu∈H10 (Ω),‖u‖

H10≤1

∫Ωfu dx.)

Nonlinear variant A variant consists in using nonlinear transformations of u: as

an example, if u minimizes

E(u) =

∫Ω

|∇u|2 + (u− f)2dx

over X = H10 (Ω) or H1(Ω), it satisfies for any v ∈ X∫

Ω

∇u · ∇v + (u− f)v dx = 0.

Taking v = u in the equation, one finds that

‖u‖H1 ≤ ‖f‖L2 .

Taking v = ψ(u) for ψ nondecreasing, with ψ(0) = 0, one find∫Ω

ψ′(u)|∇u|2 + uψ(u)dx =

∫Ω

fψ(u)dx.

Now if ψ(t) = (t+)p−1, p ≥ 1, it follows (by Holder’s inequality)∫Ω

(u+)pdx ≤∫

Ω

f(u+)p−1dx ≤ ‖f+‖Lp‖u+‖p−1Lp

so that ‖u+‖p ≤ ‖f+‖p. Sending p→∞ one also deduces that supu ≤ sup f+.

Truncation The last statement can be derived in a simpler way: if c ≥ sup f (and

c ≥ 0 in case X = H10 (Ω)), then v = minu, c is an admissible candidate in the

energy: E(u) ≤ E(v).

However

E(v) =

∫u<c

|∇u|2 + (u− f)2dx+

∫u>c

(c− f)2dx ≤ E(u),

so that v is a minimizer, which is a contradiction unless u > c has zero measure. It

follows that u ≤ c a.e.

Exercise: assume X = H1(Ω) and v satisfies, in the variational sense in Ω,

−∆v + v ≥ f

and ∇v · ν = 0 on ∂Ω. Show that v ≥ u.

4.1. BASIC REGULARITY ESTIMATES 29

4.1.2 Higher derivability

Theorem 5. Let u ∈ H1(Ω) satisfy −∆u ∈ L2loc. Then u ∈ H2

loc(Ω). More generally

if −divA∇u ∈ L2 with A(x) ∈ Rn×n Lipschitz-continuous and uniformly elliptic:

(Aξ) · ξ ≥ γ|ξ|2

for all ξ ∈ Rn, then u ∈ H2loc(Ω).

Proof. Assume first u ∈ H1(Rn), and f = −∆u ∈ L2(Rn): by defintion this means∫Rn∇u · ∇v dx =

∫Rnfv dx

for all v ∈ H1(Ω). In particular, we can consider for e a unit vector and h ∈ R small

the test function

v = −D−hDhu ∈ H1(Rn)

where Dhψ := (ψ(x+ he)− ψ(x))/h. It follows

−∫Rn∇u · ∇D−hDhu dx = −

∫RnfD−hDhu dx.

Since∫RngDhf dx =

1

h

∫Rng(x)f(x+ he)− g(x)f(x) dx

=1

h

∫Rng(x− he)f(x)− g(x)f(x) dx = −

∫RnD−hgf dx

we obtain ∫Rn|Dh∇u|2dx ≤ ‖f‖L2‖D−hDhu‖L2 .

On the other hand for any ψ ∈ H1(Rn), by Cauchy-Schwartz inequality (and, rigor-

ously, approximating first ψ with a smooth function and then passing to the limit),

∫Rn|Dhψ|2dx =

∫Rn

1

h2

∣∣∣ ∫ h

0

∇ψ(x+ se) · e ds∣∣∣2dx

≤∫Rn

1

h

∫ h

0

|∇ψ(x+ se)|2dsdx = ‖∇ψ‖2L2

so that we can deduce ∫Rn|Dh∇u|2dx ≤ ‖f‖L2‖Dh∇u‖L2 .

which shows that ‖Dh∇u‖L2 ≤ ‖f‖L2 . Hence (Banach-Alaoglu-B.) there exists ξ ∈L2(Rn;Rn) and hk ↓ 0, such that as k →∞,

Dhk∇u ξ


weakly in L2. Now if ψ ∈ C∞c (Rn;Rn) it is easy to see that limkD−hkψ = ∇eψ =∑i ∂iψei, uniformly. Hence∫Rnξ · ψdx = lim

k

∫RnDhk∇u · ψdx = − lim

k

∫Rn∇u ·D−hkψdx = −

∫Rn∇u · ∇eψdx

showing that ξ = D2u e. Since this is valid for any unit vector e, it follows that

D2u ∈ L2(Rn;Rn×n), moreover for any e, ‖D2ue‖2 ≤ ‖f‖2 = ‖∆u‖2. This shows that

one can control all the second derivatives of u in L2 from a control on just the trace

of the Hessian.

Now, if one only knows that ∆u ∈ L2loc(Ω), for Br a ball contained in Ω we can

consider a “cut-off” η ∈ C∞c (Ω; [0, 1]) with η = 1 in Br. Then, ηu is a well defined

function inH1(Rn) (∇(ηu) = η∇u+u∇η), and ∆ηu = η∆u+2∇η·∇u+u∆η ∈ L2(Rn).

It follows that D2(ηu) ∈ H2(Rn) and in particular, D2u ∈ L2(Br), with

‖D2u‖Br ≤ C(‖u‖H1(Br′ )+ ‖∆u‖L2(Br′ )

)

where r′ > r, Br′ ⊂ Ω.

Exercise: show the same result for −divA∇u ∈ L2, A = (ai,j(x))i,j Lipschitz-

continuous and uniformly elliptic.

Extensions. More generally, one can consider integrands of the form u 7→∫

ΩF (x,∇u)dx,

with F (x, p) convex and “uniformly elliptic” in p, meaning now that there exists γ > 0

such that

(D2pF (x, p))ξ · ξ =

∑i,j

∂2F (x, p)

∂pi∂pjξiξj ≥ γ‖ξ‖2

for all ξ ∈ Rn, and check whether the result is still valid. The associated equation is

−div∇pF (x,∇u) = · · ·

and it is natural to wonder whether u has second derivatives in L2 when this quantity

belongs to L2.

If the ambient space is Rn and F does not depend on x, this is almost the same

proof. One uses that

Dh (∇pF (∇u)) (x) =∇pF (∇u(x+ he))−∇pF (∇u(x))

h

=

∫ 1

0

D2pF (∇u(x) + s(∇u(x+ he)−∇u(x)))

∇u(x+ he)−∇u(x)

hds

=

(∫ 1

0

D2pF (∇u(x) + s(∇u(x+ he)−∇u(x)))ds

)Dh∇u(x)

so that

−∫Rn∇pF (∇u) · ∇D−hDhu dx =

∫RnDh (∇pF (∇u)) ·Dh∇u dx

≥ γ‖Dh∇u‖2L2 ,

4.2. THE DE GIORGI-NASH-MOSER THM 31

then one can proceed as before. If F depends on x, one needs to assume that

|Fp(x, p)− Fp(y, p)| ≤ L|x− y|(1 + |p|)

for some constant L.

The localization is more tricky... A way is to take v = −D−hη2Dhu as a test

function in the equation, where η is as before a cut-off. We obtain∫Ω

Dh∇pF (∇u)(2η∇ηDhu+ η2Dh∇u)dx = −∫

Ω

f(D−hη2Dhu)dx.

While from the left hand side we quite easily get, as before (assuming also D2pF ≤ C)

a bound of the form

γ‖ηDh∇u‖2L2 ≤ 2C‖ηDh∇u‖L2‖∇ηDhu‖L2 + terms from right hand side.

Bounding the right hand side is a bit more technical, one may for instance write,

denoting g(x) = Dhu(x):

f(x)(D−hη2g)(x)

= f(x)(D−hη)(x)(ηg)(x− he)) + f(x)η(x)η(x− he)g(x− he)− η(x)g(x)

−h

= f(x)(D−hη)(x)(ηg)(x− he)) + f(x)η(x) 1h

∫ h

0

∇e(η(x− se)g(x− se))ds (4.1)

showing that (remember η is smooth)

‖fD−hη2g‖L2 ≤ c‖f‖L2‖ηg‖L2 + c‖ηf‖L2‖g‖L2 + ‖ηf‖L2‖η∇g‖L2 .

(here all the norms can be taken on a small neighborhood of the support of η). The

proof can be finished as before...

4.2 The De Giorgi-Nash-Moser Thm

4.2.1 Towards more regularity

If F : Rn → R is a strongly convex function with for all p,

λ ≤ D2F (p) ≤ Λ

(that is,

λ|ξ|2 ≤ (D2F (p)ξ) · ξ ≤ Λ|ξ|2

for all ξ ∈ Rn), a minimizer of ∫Ω

F (∇u)− fu dx (4.2)

in H1(Ω) (with any kind of boundary condition), with f ∈ L2(Ω), will satisfy the

equation ∫Ω

∇pF (∇u) · ∇v dx =

∫Ω

fv dx


for all v ∈ H10 (Ω). From the previous section, it follows v ∈ H2

loc(Ω). Taking v = ∂kψ

for ψ ∈ C∞c (Ω), and integrating by parts, one also finds

−∫

Ω

∑i,j

∂2i,jF (∇u)∂2

j,ku∂iψdx =

∫Ω

f∂kψ dx

We know that v = ∂ku ∈ H1loc. The above equation shows that in the weak (Distribu-

tional) sense,

−divA(x)∇v = ∂kf

whereA(x) = D2pF (∇u(x)). The only information on the matrixA(x) = (ai,j(x))1≤i,j≤n

is that it is measurable, and its coefficients satisfy

λ ≤ A(x) ≤ Λ

a.e. in Ω. We will now show the following theorem, concerning the case f = 0.

Theorem 6 (De Giorgi-Nash-Moser). Let v ∈ H1(B1) satisfy −divA∇v = 0, that is,∫B1

∑i,j

ai,j(x)∂iv(x)∂jφ(x)dx = 0 (4.3)

for all φ ∈ H10 (B1). Then v ∈ C0,α(B1/2).

The result is still valid if f 6= 0 provided f ∈ Ln+ε, ε > 0, and also if the right

hand side is of the form Df + g with both f and g in Ln+ε, see [9, Thm 8.22]. For u

minimizing (4.2) it implies that u ∈ C1,α. Hence, the matrix A(x), in fact, is C0,α.

Then, more regularity is obtained from the classical theory of Schauder. The

equation for u becomes

−div∇F (u) = −∑i,j

∂2F

∂pi∂pj(∇u)∂2

i,ju = f,

and if F is smooth enough this is of the form

A(x) : D2u = f

where A is C0,α (since it is a smooth function of the gradient of u). Then if also f is

C0,α, it follows [9] that u is C2,α. But then A is C1,α, etc.

4.2.2 Proof: L∞ control

Definition 6. We say that u ∈ H1(B1) is a subsolution of (4.3) if for any φ ∈ H10 (B1)

with φ ≥ 0, ∫B1

ai,j(x)∂iu(x)∂jφ(x) dx ≤ 0 (4.4)

(where we assume that repeated indices are summed from 1 to n).

Lemma 4. Let u ∈ H1(B1) be a subsolution: then

‖u+‖L∞(B1/2) ≤ c‖u‖L2(B1)

where c = c(n,Λ/λ).

We prove this lemma in several steps.


Choice of a test function First we assume that u+ ∈ L∞ (a more careful choice

of the truncation below could deal with this assumption, we will show a different

strategy).

Given 0 < r < R < 1 we consider a cut-off function η(x) = min1, (R−|x|)+/(R−r) (it is 1 in Br, 0 out of BR, and |∇η| ≤ 1/(R− r)). We use η2u+ ∈ H1

0 (B1;R+) as

a test function in the equation:∫BR

ai,j∂iu∂j(η2u+)dx ≤ 0.

Observe that

ai,j∂iu∂j(η2u+) = ai,j∂iu(∂jη)ηu+ + ai,j∂iu(∂j(ηu

+))η

= ai,j∂iu+(∂jη)ηu+ + ai,j∂i(ηu

+)∂j(ηu+)− ai,j∂iη(∂j(ηu

+))u+

= ai,j∂i(ηu+)∂j(ηu

+)− ai,j∂iη(∂jη)(u+)2

where we have used the fact that where u+ > 0 or ∂iu+ 6= 0, u = u+, and the symmetry

of ai,j .

It follows that∫BR

ai,j∂i(ηu+)∂j(ηu

+)dx ≤∫BR

(u+)2ai,j∂iη∂jηdx,

hence

λ‖∇(ηu+)‖2L2(BR) ≤Λ

(R− r)2‖u+‖2L2(BR).

We now use Sobolev’s inequality for the compactly supported function ηu+: there

exists c = c(n) such that

‖ηu+‖L2∗ ≤ c‖∇(ηu+)‖L2 ,

where we recall that 2∗ = 2n/(n− 2), and it follows

‖u+‖L2∗ (Br) ≤ c√

Λ

λ

1

R− r‖u+‖L2(BR). (4.5)

This shows that we get a control of u+ with a better exponent in a smaller ball. We

will now show how to “iterate” this inequality, this trick is due to Moser (J. K. Moser,

1928-1999).

“Stability” of subsolutions

Lemma 5. Let u ∈ H1(B1) be a subsolution and assume that Φ : R → R is convex,

nondecreasing, Lipschitz-continuous (so that Φ(u) ∈ H1(B1)), then Φ(u) is also a

subsolution.

Proof. If Φ is smooth one can write, for φ a non-negative test function∫B1

ai,j∂i(Φ(u))∂jφdx =

∫B1

ai,j∂iu(Φ′(u)∂jφ)dx

=

∫B1

ai,j∂iu∂j(Φ′(u)φ)−

∫B1

(ai,j∂iu∂ju)Φ′′(u)φ ≤ 0.


We have the right to write this because, in particular, Φ′(u)φ ∈ H10 (B1;R+) (and

can be approximated with smooth non-negative functions with compact support). We

used Φ′,Φ′′ ≥ 0.

Hence when Φ is smooth, ∫B1

ai,j∂i(Φ(u))∂jφdx ≤ 0

now if it is not, one can approximate it by smooth functions Φn (by convolution) and

check that Φn(u)→ Φ(u) at least weakly in H1(B1) (which is sufficient).

In our case, u+ is a bounded subsolution (since it is Φ(u) with Φ(t) = max0, t)and also (u+)p for all p ≥ 1 (observe that Φ(t) = max0, tp is Lipschitz on bounded

sets, so that the Lemma may be applied if u+ is bounded).

We deduce from (4.5) that

‖(u+)p‖L2∗ (Br) ≤ c√

Λ

λ

1

R− r‖(u+)p‖L2(BR).

for any 0 < r < R < 1 and any p ≥ 1. Let now χ = n/(n− 2)1 (so that (tp)2∗ = t2χp).

The tricky point is to let

p0 = 1, p1 = χ, . . . , pk = χk, . . .

R0 = 1, R1 = 34 , . . . , Rk = 1

2 + 12k+1 ,

so that in particular 1/(Rk −Rk+1) = 2k+2, and use recursively (4.5) with the subso-

lution (u+)pk , r = Rk+1, R = Rk:

‖(u+)pk‖L2∗ (BRk+1) ≤ c

√Λ

λ2k+2‖(u+)pk‖L2(BRk ).

This is: (∫BRk+1

(u+)2pk+1dx

) 12χ

≤ c√

Λ

λ2k+2

(∫BRk

(u+)2pkdx

) 12

and taking the power (1/pk) on both sides, we get

‖u+‖L2pk+1 (BRk+1) ≤

(c

√Λ

λ2k+2

) 1χk

‖u+‖L2pk (BRk )

≤k∏l=0

(c

√Λ

λ2l+2

) 1χl

‖u+‖L2(B1)

The point to understand now is whether the product can be uniformly bounded: this

is easy, as taking the logarithm we obtain

ln

k∏l=0

(c

√Λ

λ2l+2

) 1χl

=

k∑l=0

1

χl

(ln 4c

√Λ

λ+ l ln 2

)1If the dimension n = 2, the the proof must be modified a little, for instance taking χ > 1 arbitrary.


which is easily shown to be bounded (since 1/χ = 1− 2/n < 1), by

n

2ln 4c

√Λ

λ+n(n− 2)

4ln 2.

Letting

C = (4c)n2

(Λ

λ

)n4

2n(n−2)

4

which as we see, depends only on n (also through c which is the constant in the

Sobolev inequality) and on the ellipticity constant Λ/λ, we find that for all k (using

also Rk+1 ≥ 1/2),

‖u+‖L2pk+1 (B1/2) ≤ C‖u+‖L2(B1)

and now we can send k →∞:

‖u+‖L∞(B1/2) ≤ C‖u+‖L2(B1)

which almost shows Lemma 4. Indeed, we need to check that this is also true without

the a priory assumption u+ ∈ L∞.

Approximation by bounded subsolutions If we can show that any subsolution

can be approximated in L2 (or H1-weak) by bounded subsolutions uk, then it will be

enough to pass to the limit in

‖u+k ‖L∞(B1/2) ≤ C‖u+

k ‖L2(B1) (4.6)

and observe that u 7→ ‖u‖∞ is lsc with respect to the L2 convergence (this is easy) to

conclude that Lemma 4 is true.

The key observation is that a subsolution is in fact a critical point (or here, since

the energy∫B1ai,j∂iv∂jvdx is convex, a minimizer) of the energy with a constraint

v ≤ u.

It is therefore natural to introduce uk the solution of

min

∫B1

ai,j∂iv∂jv dx : v ∈ H1(B1), v = minu, k on ∂B1, v ≤ minu, k a.e.

which satisfies u+

k ≤ k. If φ ≥ 0 is a smooth, nonnegative compactly supported test

function, we see that uk − εφ is also admissible so that for all ε > 0 small,∫B1

ai,j∂iuk∂jukdx ≤∫B1

ai,j∂i(uk − εφ)∂j(uk − εφ)dx

=

∫B1

ai,j∂iuk∂jukdx− 2ε

∫B1

ai,j∂iuk∂jφdx+ ε2

∫B1

ai,j∂iφ∂jφdx.

It follows that ∫B1

ai,j∂iuk∂jφdx ≤ 0 (4.7)

showing that uk is a subsolution: and hence (4.6) holds. Observe also that

λ‖∇uk‖2L2 ≤∫B1

ai,j∂iuk∂jukdx ≤∫B1∩u<k

ai,j∂iu∂judx


(since minu, k is admissible in the minimization problem) so that uk is uniformly

bounded in H1(B1) (we need to use also the boundary condition, and Poincare’s

inequality, to make sure this is true), and up to a subsequence converges (weakly in

H1, or strongly in L2) to some u = supk uk ≤ u, with u = u on ∂B1. By lsc of the

(convex) energy∫B1

ai,j∂iu∂j u dx ≤ lim infk→∞

∫B1

ai,j∂iuk∂jukdx ≤∫B1

ai,j∂iu∂ju dx.

Passing to the limit in (4.7) we also see that∫B1

ai,j∂iu∂jφdx ≤ 0

for any smooth nonnegative test function φ with compact support, or in fact, by

density, any φ ∈ H10 (B1;R+). In particular taking φ = u − u ∈ H1

0 (B1;R+) in the

subsolution inequation for u∫B1

ai,j∂iu∂ju dx ≤∫B1

ai,j∂iu∂j u dx.

Hence

λ‖∇(u− u)‖2L2 ≤∫B1

ai,j∂i(u− u)∂j(u− u)dx

=

∫B1

ai,j∂iu∂ju dx+ 2

∫B1

ai,j∂iu∂j u dx+

∫B1

ai,j∂iu∂j u dx

≤ 2

(∫B1

ai,j∂iu∂ju dx−∫B1

ai,j∂iu∂j u dx

)≤ 0,

showing that u = u achieving the proof of Lemma 4.

4.2.3 Harnack-type estimate

This part is a bit complicated, and uses nonlinear test functions. It shows:

Theorem 7. Let u ∈ H1(B2) a nonnegative supersolution2 of (4.3), and assume there

exists α ∈ (0, 1) with |u ≥ 1∩B1| ≥ α|B1|. Then there exists C = C(α, n,Λ/λ) such

that

infB1/2

u ≥ C.

This is almost a “Harnack inequality”, which would show that infB1/2u ≥ C if

maxu ≥ 1 (see next section).

Proof. First it is enough to show the result for u+ δ, δ > 0, and then pass to the limit

δ ↓ 0. Hence we assume u ≥ δ > 0.

2meaning that −u is a subsolution.


Then we consider v = (lnu)− = max0,− lnu ≤ − ln δ. Since −u is a subsolution

and t 7→ max0,− ln−t is convex, nondecreasing, and Lipschitz in (−∞,−δ), then v

also is a subsolution (Lemma 5). Hence from Lemma 4, v is bounded from above and

supB1/2

v ≤ c‖v‖L2(B1).

(Here and in the following, “sup” and “inf” are the essential supremum and infimum,

that is, up to Lebesgue-negligible sets.)

We recall that a variant of Poincare’s inequality states that there exists a constant

C depending only on α > 0 such that for w ∈ H1(B1), if

|w = 0| ≥ α|B1|

then ‖w‖L2 ≤ C‖∇w‖L2 . This is easily (and classicaly) proved by contradiction:

if not, there exists wn with |wn = 0| ≥ α|B1| and n‖∇wn‖L2 ≤ ‖wn‖L2 , etc...

Exercise: show then that one may assume ‖wn‖L2 = 1, and that up to subsequences,

wn converges to a constant function. Find a contradiction.

Hence, we have

supB1/2

v ≤ c‖v‖L2(B1) ≤ C‖∇v‖L2(B1) (4.8)

where now C depends on α, n, Λ/λ. The “miraculous” trick is that the right-hand side

can be bounded “universally”. To see this, we use the fact that u is a supersolution in

B2 and consider a test function φ = η2/u where η ∈ C∞c (B2; [0, 1]), η = 1 in B1 (since

u ≥ δ, φ ∈ H10 (B2;R+)). It follows∫

B2

ai,j∂iu

(−η

2

u2∂ju+ 2

η∂jη

u

)dx ≥ 0

Hence, since ∇u/u = ∇ lnu,∫B2

η2ai,j∂i lnu∂j lnu dx ≤ 2

∫B2

ηai,j∂i lnu∂jη dx

≤ 2

√∫B2

η2ai,j∂i lnu∂j lnu dx

√∫B2

ai,j∂iη∂jη dx

because of Cauchy-Schwartz inequality (applied to the nonnegative quadratic form∫B2ai,jξiξjdx), and it follows∫

B2

η2ai,j∂i lnu∂j lnu dx ≤ 4

∫B2

ai,j∂iη∂jη dx.

Choosing well η, the right-hand side can be made as small as 4Λ|B2 \B1|, in any case

it is a constant independent on u: it follows

λ‖∇v‖2L2(B1) ≤∫B1

ai,j∂i lnu∂j lnu dx ≤ C(n)Λ

and we deduce from (4.8) that

supB1/2

v ≤ C

with a constant depending only on n, α,Λ/λ. In other words infB1/2u ≥ e−C > 0,

which is what we needed to prove.


4.2.4 Application: control of the oscillation of a solution

Theorem 8. Let u be a solution in B4. Then there exists a constant γ = γ(n,Λ/λ) ∈(0, 1) such that

oscB1/2u ≤ γoscB2

(u) :

the oscillation of u decreases as we move towards the center of the ball.

Proof. This is a relatively simple consequence of the previous results. By Lemma 4,

u is bounded in B2. Letting α = supB2u, β = infB2

u, α′ = supB1/2u, β′ = infB1/2

u,

we have u− β ≥ 0 and α − u ≥ 0 in B2. At each point, either u− β ≥ (α − β)/2, or

u− β < (α− β)/2 so that α− u > (α− β)/2. Hence one of the two sets2u− βα− β

≥ 1

∩B1 or

2α− uα− β

> 1

∩B1

has a measure larger than |B1|/2. Assume for instance it is the first one, then, since

clearly, 2(u− β)/(α− β) (which is a solution of (4.3)) is a supersolution in B2, there

is a constant C which depends only on n and Λ/λ such that from Theorem 7,

2u− βα− β

≥ C

in B1/2. It follows that β′−β ≥ (C/2)(α−β), so that α′−β′ ≤ α−β′ ≤ (1−C/2)(α−β).

This shows the results with γ = (1− C/2) ∈ (1/2, 1).

Remark: you can show easily that γ is scale-invariant: if u is a solution in B(x, 4r),

then oscB(x,r/2)u ≤ γoscB(x,2r)u. (Exercise...)

4.2.5 Conclusion: u is Holder

Theorem 9. Let u be a solution in B4. Then there exists α > 0 such that

supB1

|u|+ supx,y∈B1

|u(x)− u(y)||x− y|α

≤ C‖u‖L2(B4),

C and α depending on n and Λ/λ.

Exercise: show then that if u is a solution in Ω, for all Ω′ ⊂⊂ Ω (i.e., such that Ω′

is a compact subset of Ω), then ‖u‖C0,α(Ω) ≤ C‖u‖L2(Ω) for some constant C.

Proof. We prove the theorem: first, the fact that |u| is bounded by C‖u‖L2(B4) in

B1 follows from Lemma 4. Then, consider x, y ∈ B1 and let r = |x − y| ≤ 2, z =

(x + y)/2. By definition |u(x) − u(y)| ≤ oscB(z,r/2)u. Hence if r is small enough (so

that B(z, 4r) ⊂ B4),

|u(x)− u(y)| ≤ γoscB(z,2r)u

and by induction,

|u(x)− u(y)| ≤ γ2oscB(z,8r)u ≤ · · · ≤ γkoscB(z,4kr/2)u ≤ Cγk‖u‖L2(B4)


as long as B(z, 4kr/2) ⊂ B2, hence as long as 1 + 4kr/2 ≤ 2 (this is true even for

k = 0, in fact). Choosing the largest k ≥ 0 such that this is true, that is, such that

4k ≤ 2/r and 4k ≥ 1/(2r), we find that (be careful that ln γ < 0)

γk = 4kln γln 4 =

(1

4k

)| ln γln 4 |

≤ (2r)|ln γln 4 |

so that, letting α = − ln γ/ ln 4 ∈ (0, 1/2) (since γ > 1/2),

|u(x)− u(y)| ≤ C|x− y|α‖u‖L2(B4)

which was the thesis we wanted to prove.

Chapter 5

Vector-valued minimization

problems

[maybe this part will be skipped as too close to S. Serfaty’s lecture, if time permits I’ll

write some notes on the subject, otherwise the main references are [6, 3, 10]]

A few results: we recall the definition

Definition 7. f : Rm×n → R is “quasiconvex” in the sense of Morrey if and only if,

given Ω a bounded open set in Rn, for any φ ∈W 1,∞0 (Ω;Rm) and any A ∈ Rm×n,

|Ω|f(A) ≤∫

Ω

f(A+Dφ) dx.

The definition is independent on the choice of Ω.

Some properties:

Proposition 4. Let f be as in definition 7, and assume it is locally bounded (which

means precisely that sup|A|≤R f(A) < +∞ for any R > 0). Then f is rank-one convex:

f(tA+ (1− t)B) ≤ tf(A) + (1− t)f(B)

for t ∈ [0, 1], as soon as rk(B − A) = 1, that is B − A = a ⊗ ν for some a ∈ Rm,

ν ∈ Rn. In particular, if m = 1 or n = 1, f is convex.

Proof. We only give the idea of the proof: without loss of generality (just rotating the

axes) one can assume ν = e1 and Ω = (0, 1)n, and t = 1/2. We let C = (A + B)/2.

Then we consider for ε > 0 the function uε(x) = Ax+ε(k+1)a if 2kε ≤ x1 ≤ (2k+1)ε,

uε(x) = Bx−εka if (2k+1)ε ≤ x1 ≤ (2k+2)ε, and we check that since B−A = a⊗e1, it

is continuous (if x1 = (2k+1)ε, (B−A)x−εka−ε(k+1)a = a(2k+1)ε−ε(2k+1)a = 0,

etc). Then, one checks that uε → Cx uniformly. Given a cut-off η ∈ C∞c ((0, 1)n) (with

0 ≤ η ≤ 1, η = 1 on Qδ = (δ, 1 − δ)n), and letting φ(x) = η(x)(uε(x) − Cx), one has

41

42 CHAPTER 5. VECTOR-VALUED MINIMIZATION PROBLEMS

by definition

f(C) ≤∫Q

f(C +Dφ)dx =

∫Q

f(ηDuε + (1− η)C + (uε − Cx)⊗∇η)dx

≤∫Qδf(Duε)dx+

∫Q\Qδ

f(ηDuε + (1− η)C + (uε − Cx)⊗∇η)dx

Now, if ε is small enough, then since uε → C uniformly, for any x

ηDuε(x) + (1− η)C(x) + (uε(x)− Cx)⊗∇η(x) ∈ M : dist (M, [A,B]) ≤ 1

(or any neighborhood of [A,B]) so that by assumption, f is less than some bound M ,

hence

f(C) ≤∫Qδf(Duε)dx+ |Q \Qδ|M.

Since Duε = A in about half of the domain and Duε = B in the other half, we find,

sending ε → 0, that f(C) ≤ (f(A) + f(B))/2 up to an error which goes to zero as

δ → 0.

It is well known that a convex function which is locally bounded is also locally

Lipschitz. When the function has a polynomial growth, this can be made more precise.

Proposition 5. Let f : Rn → R be convex “with growth p”: there exist α, β > 0 such

that

α(|A|p − 1) ≤ f(A) ≤ β(|A|p + 1)

for any A ∈ Rn. Then there is a constant C > 0 such that

|f(A)− f(B)| ≤ C(1 + |A|p−1 + |B|p−1)|A−B|.

Proof. Let A,B ∈ Rn and set R = max|A|, |B|, 1. If A 6= B, there exists a unique C

with |C−A| = R andB ∈ [A,C]. Observe that |C| ≤ 2R. Letting t = |B−A|/|C−A| =|B −A|/R, we have B = tC + (1− t)A, hence

f(B) ≤ (1− t)f(A) + tf(C)

so that

f(B)−f(A) ≤ |B−A|f(C)− f(A)

R≤ |B−A|β(1 + (2R)p)− α

R≤ |B−A|(1+2pRp−1)

which ends the proof, since Rp−1 ≤ |A|p−1 + |B|p−1 unless this sum is less than 2.

Corollary 1. Assume f is quasiconvex, locally bounded, with growth p. Then again,

there is a constant C > 0 such that

|f(A)− f(B)| ≤ C(1 + |A|p−1 + |B|p−1)|A−B|. (5.1)

43

Proof. It is enough to apply the previous results to f in rank-1 directions. Given

A = (a1, . . . , an), B = (b1, . . . , bn) ∈ Rm×n two matrices, with ai, bi ∈ Rm, one has

A−B = (A− (b1, a2, . . . , an))

+ ((b1, a2, . . . , an)− (b1, b2, a3, . . . , an)) + · · ·+ ((b1, . . . , bn−1, an)−B).

Since f is rank-one convex with p growth, one has

f(A)− f(b1, a2, . . . , an) ≤ C(1 + |A|p−1 + |B|p−1)|b1 − a1|, etc

Summing, we obtain

f(A)−f(B) ≤ C(1+ |A|p−1 + |B|p−1)

n∑i=1

|bi−ai| ≤ C√n(1+ |A|p−1 + |B|p−1)|B−A|.

44 CHAPTER 5. VECTOR-VALUED MINIMIZATION PROBLEMS

Chapter 6

Variational convergence

6.1 Why?

Let En : X → R ∪ −∞,+∞ be functionals and assume we want to understand the

“limit” of the problems

minx∈XEn(x) (6.1)

In other words: if (xn)n is the sequence of minimizers of (6.1), how does it behave as

n→∞?

The best would be to define a notion of convergence for the En which would yield

that “En → E” implies that xn → x which minimizes E . One would also like to have

a bit of “stability”, and be coherent with the values of the energy, that is:

(A) if G is continuous and En → E , then En + G → E + G,

(B) if En → E , infX En → infX E .

Let us introduce the following definition [7, 2]:

Definition 8. We say that the sequence of functionals (En)n : X → R ∪ −∞,+∞,

where X is a metric space, Γ-converges to E in X if for any x ∈ E,

i. when xn → X, then E(x) ≤ lim infn En(xn);

ii. there exists xn → x such that lim supn En(xn) ≤ E(x).

Another name for this notion is epi-convergence [1]. The sequence (xn)n in (ii) is

called a recovery sequence.

Now it is very easy to check that this will satisfy (A) and (B) above. In fact, (A)

is obvious from the definition (since in both axioms i. and ii., one will have G(xn) →G(x)). For (B), we have:

Theorem 10. Let G be continuous and En which Γ-converges to E. Assume that for

each n, xn is a minimizer of En + G, or more generally that there exists εn ↓ 0 such

that for any x ∈ XEn(xn) + G(xn) ≤ En(x) + G(x).

45

46 CHAPTER 6. VARIATIONAL CONVERGENCE

Then if x is a cluster point of (xn)n (that is, xnk → x for some subsequence), x is a

minimizer of E + G.

Proof. This is obvious: if y ∈ X, by (ii) one finds yn → y such that lim supn En(yn) ≤E(y). One has that

E(x) ≤ lim infkEnk(xnk)

(using (i), possibly for the sequence x′n = x if n 6= nk for some k, and xnk else). Hence

E(x) + G(x) ≤ lim infkEnk(xnk) + lim

kG(xnk)

≤ lim infkEnk(xnk) +G(xnk) ≤ lim inf

kEnk(ynk) +G(ynk) + εnk

≤ lim supkEnk(ynk) + lim

kG(ynk) + εnk ≤ E(y) + G(y).

Remark: one also easily deduces, in this case, that E(xnk) + G(xnk)→ minX E + G.

6.1.1 Necessity of the axioms

Hence, it works. It turns out that formally one can check that conditions (i) and (ii)

are necessary in order for (A) and (B) to hold. We assume for simplicity that En, E

are bounded from below (for instance, nonnegative), coercive and lsc.

Consider x ∈ X and xn → x, for ε > 0 small one could define

En(x) +1

2εdist (x, x)

Assume this has a minimizer xεn, and xεn converges to some xε. Then the stability (A)

ensures that xε should minimize x 7→ E(x) + dist (x, x)/(2ε), while (B) yields

limnEn(xεn) +

1

2εdist (xεn, x) = E(xε) +

1

2εdist (xε, x).

Now as ε → 0, xε → x by construction (unless E is infinite in a neighborhood of x),

and (we also assume that E is lsc), E(x) ≤ lim infε→0 E(xε). Since

E(xε) ≤ limnEn(xεn) +

1

2εdist (xεn, x)

≤ lim infn

En(xn) +1

2εdist (xn, x) [by minimality of xεn]

= lim infn

En(xn)

we deduce that (i) must hold.

On the other hand, if E(x) < +∞, by minimality one also have that

E(xε) +1

2εdist (xε, x) ≤ E(x)

so that

limnEn(xεn) +

1

2εdist (xεn, x) ≤ E(x).

Hence given ε > 0, for n large enough En(xεn) ≤ E(x)+ε and dist (xεn, x) ≤ 2ε(E(x)+ε).

One can therefore build a global sequence x′n → x with lim supn En(x′n) ≤ E(x).

6.1. WHY? 47

6.1.2 Properties and comments

Proposition 6. Assume EnΓ→ E. Then E is lsc.

Proof. If xk → x, for each k using (ii) one has xkn → xk with lim supn En(xkn) ≤ E(xk).

Then (letting N(0) = 0), for each k, there exists N(k) > N(k − 1) such that

n ≥ N(k)⇒ En(xkn) ≤ E(xk) +1

k, dist (xkn, x

k) ≤ 1

k.

Let for each n ≥ 1, yn := xkn if N(k) ≤ n ≤ N(k + 1). Then, for such n,

dist (yn, x) ≤ 1

k+ dist (xk, x)

so that yn → x as n→∞. Hence by (i),

E(x) ≤ lim infnEn(yn) ≤ lim inf

n,N(k)≤n≤N(k+1)E(xk) +

1

k= lim inf

kE(xk)

and E is lsc.

In particular, one can show that

Proposition 7. Let E be a functional and let En := E for each n. Then (En)n Γ-

converges to sc−E, the lower-semicontinuous envelope of E.

More precisely, one can introduce the following definitions:

Definition 9. Given En as in Def. 8, one defines the “Γ-lim inf” and “Γ-lim sup” of

En by letting:

(Γ− lim infnEn)(x) = inf

lim inf

nEn(xn) : xn → x

(Γ− lim supnEn)(x) = inf

lim sup

nEn(xn) : xn → x

.

Then, one checks easily that these functionals are always well-defined (while it is

not true for the Γ-limit), and that

• both Γ− lim infn En and Γ− lim supn En are lsc;

• Γ− lim infn En ≤ Γ− lim supn En;

• En Γ-converges if and only if Γ− lim supn En ≤ Γ− lim infn En, and in this case

the Γ-limit is precisely the common value of these two functionals.

Observe that checking (i) in Def. 8 exactly means that E(x) ≤ (Γ − lim infn En)(x),

while (ii) means that (Γ− lim supn En)(x) ≤ E(x).


Comments: The Γ-convergence is different from the pointwise convergence. Exam-

ple, in R: fn(x) = 0 if |x| ≥ 1/n, 1 − |2nx + 1| if −1/n ≤ x ≤ 0, |2nx − 1| − 1 if

0 ≤ x ≤ 1/n. Then fn → 0 pointwise, while fn Γ-converges to f(x) = 0 if x 6= 0,

f(0) = −1.

It is different from the uniform convergence, for instance, given any E , F+1/n→ F

uniformly even when F is not lsc.

Important: observe that En → E does not imply −En → −E ! In fact, one has:

Proposition 8.En

Γ→ E−En

Γ→ −E⇔ En “continuously converges” to E

where En is said to continuously converge to E if for any xn → x, one has En(xn)→E(x) (this is stronger than the uniform convergence on compact sets).

Proof. The implication ⇐ is obvious, for the other direction we just write

E(x) ≤ lim infnEn(xn) ≤ lim sup

nEn(xn)

= − lim infn

(−En(xn)) ≤ −(−E(x)) = E(x).

6.1.3 Compactness of Γ-convergence

Theorem 11. Let (X, d) be a separable metric space and let (En)n be a sequence of

functionals from X → R ∪ −∞,+∞. Then, there exists a functional E : X →R ∪ −∞,+∞ and a subsequence (Enk)k such that Enk Γ-converges to E.

Remark: one may have, here, that E ≡ −∞ or E ≡ +∞.

Before, let us introduce the following definition:

Definition 10. Consider (An)n a sequence of subsets of X. We define the inferior

and superior Kuratowski limits of (An)n as

Lin→∞

An =

x ∈ X : lim sup

ndist (An, x) = 0

= x ∈ X : ∃N, ∀U ∈ V(x), n ≥ N ⇒ U ∩An 6= ∅ ,

Lsn→∞

An =x ∈ X : lim inf

ndist (An, x) = 0

= x ∈ X : ∀U ∈ V(x), ]n ∈ N : U ∩An 6= ∅ =∞ .

We say that An converges to A in the sense of Kuratowski if and only if A = LinAn =

LsnAn.

6.1. WHY? 49

Here V(x) is the set of the neighborhoods of x. Observe that LinAn ⊆ LsnAn,

and that

LinAn = x : ∃xn → x, xn ∈ An∀n , Ls

nAn = x : ∃xnk → x, xnk ∈ Ank∀k .

Let us show first the following

Theorem 12. Given (An)n a sequence of subsets of X (a separable metric space).

Then there exists a subsequence (Ank)k and a closed set A ⊂ X such that Ank → A in

the sense of Kuratowski.

Proof. Let dn(x) := mindist (x,An), 1. One easily shows that it is 1-Lipschitz. As

usual, one uses a diagonal argument to extract a subsequence such that dnk(xi)→ δ(xi)

as k → ∞, where (xi)i is a dense sequence in X. As usual, being δ 1-Lipschitz, it is

extended in a unique way in a Lipschitz function over X and one can easily show that

dnk(x)→ δ(x) for any x ∈ X (see the proof of Ascoli’s theorem).

We let A = δ = 0. Then, if x ∈ A, dnk(x) → 0 hence there exists xnk ∈ Ankwith xnk → x. This shows that A ⊆ Lik Ank . On the other hand, if x ∈ Lsk Ank , by

definition there exists a subsequence xnkl → x with xnkl ∈ Ankl . But then, dnkl (x) ≤dist (x,Ankl ) → 0, hence δ(x) = 0 and x ∈ A: Lsk Ank ⊆ A. Hence Ank → A in the

Kuratowski sense.

Proof of Thm 11. To prove Theorem 11, we now define the sets En = epi En = (x, t) ∈X ×R, E(x) ≤ t. Then there is a subsequence and a closed set E ⊂ X ×R such that

Enk → E in the Kuratowski sense. We first observe that E is the epigraph of an lsc

function: if (x, t) ∈ E, then it is easy to check that (x, s) ∈ E for any s ≥ t (there is

(xnk , tnk) ∈ Enk converging to (x, t), but then, obviously, (xnk , tnk + s− t) ∈ Enk for

all k...)

Hence there is E such that epi E = E (E(x) = infs : (x, t) ∈ E ∈ [−∞,+∞]). To

simplify we now denote the converging subsequence (Enk)k simply (En)n. Now, con-

sider xn → x. Observe that (xn, En(xn)) ∈ En. If lim infn En(xn) < +∞ there exists

a subsequence such that limk Enk(xnk) = lim infn En(xn). Hence, (xnk , Enk(xnk)) →(x, lim infn En(xn)) so that

(x, lim infnEn(xn)) ∈ Ls

nEn = E.

By definition of E , it follows that E(x) ≤ lim infn En(xn), that is (i). (Observe that

the epigraph of the Γ-liminf is thus the upper Kuratowski limit of the epigraphs.)

On the other hand, given x ∈ X, if E(x) < +∞ then for any t ∈ R with t ≥ E(x)

there is (xn, tn) ∈ En such that xn → x, tn → t so that lim supn En(xn) ≤ t. If

E(x) > −∞, choosing t = E(x) shows (ii). Else one must take a sequence tk ↓ −∞and use a diagonal sequence to show (ii).

WARNING: all these beautiful results are absolutely useless (a) if one does not

know how to bound globally En, so that E is not infinite everywhere; (b) if one has a

way to show that minimizing sequences will have converging subsequence, so that the

Γ-convergence gives an information about the limit.


Definition 11. We say that the sequence (En)n which Γ-converges to E also satisfies

the compactness property if for any sequence (xn)n such that supn En(xn) < +∞, there

exists a subsequence and x ∈ X such that xnk → x as k →∞.

In this case, clearly, if (xn) is a minimizer (or almost minimizing in the sense of

Theorem 10), and if one can show that the values En(xn) are bounded, then xn will

have subsequences converging to minimizers of E .

Let us know illustrate on a few examples this notion. We will consider:

1. Discrete-to-continuum limits;

2. Periodic Homogenization;

3. “Singular limits” (phase transition);

4. ?

6.2 Application: discrete to continuum limits

Here we will illustrate on a very basic examples a concept which can become much

more complex in some contexts.

Let us just consider the function, defined for y ∈ RN+1 with y0 = yN = 0,

EN (y) =

N−1∑i=0

kh

2

(yi+1 − yi

h

)2

+ h

N∑i=0

yigNi

where we have let h = 1/N , and gNi = g(i/N) for a given continuous function g :

[0, 1]→ R.

Defining

E(y) =k

2

∫ 1

0

(y′(x))2dx+

∫ 1

0

y(x)g(x)dx

one expects that EN→E . But in what sense is this true, since the functions do not

“live” in the same space?

This is a very common issue in Γ-convergence problems. The solution is to embeds

all variables in a common space (this way is in general not unique).

For instance, here, knowing that (Rellich) E (and hopefully the En) is coercive in

L2 (which will help showing compactness), a good choice is to take X = L2(0, 1). For

N > 0 fixed, to a vector yN = (yNi )Ni=0 ∈ RN+1 (with yN0 = yNN = 0), one can associate

yN (x) =

N∑i=1

yNi χ((i−1)h,ih)(x) ∈ L2(0, 1)

or

yN (x) =

N−1∑i=1

yNi ∆(Nx− i) ∈ H10 (0, 1)

where ∆(x) = (1 − |x|)+ (“P1” approximation or spline of order 1). The advantage

of the latter is that yN ∈ H10 (0, 1). If we can show that it is bounded in H1

0 , then it

6.2. APPLICATION: DISCRETE TO CONTINUUM LIMITS 51

will be compact in L2. However observe that in this case, both approximations will

converge to the same limit, indeed

∫ 1

0

|yN (x)− yN (x)|2dx =

N∑i=1

∫ h

0

|yNi − (yNith + (1− t

h )yNi−1)|2dx

=

N∑i=1

h

∫ 1

0

(1− s)2|yNi − yNi−1|2dt

=h2

3

N∑i=1

h

(yNi − yNi−1

h

)2

=h2

3

∫ 1

0

|yN ′(x)|2dx. (6.2)

In order to talk about a Γ-limit, we extend the functionals as follows, letting for

u ∈ L2(0, 1)

EN (u) =

EN (yN ) if there exists yN ∈ RN+1 with y0 = yN = 0 and u = yN ,

+∞ else.

and

E(u) =

E(u) if u ∈ H1

0 (0, 1),

+∞ else.

Then the result is as follows:

Proposition 9. EN Γ-converges to E in L2(0, 1) as N →∞. Moreover, if supk ENk(uNk) <

+∞ then (uNk) is bounded in H10 (0, 1), hence precompact in L2(0, 1) (Rellich).

Proof. First, the compactness is obvious, as one checks that

N−1∑i=0

kh

2

(yNi+1 − yNi

h

)2

=k

2

∫ 1

0

(yN ′(x))2dx,

while

h

N∑i=0

yigNi ≤ max

[0,1]|g|∫ 1

0

|yN |dx ≤ C‖yN‖L2 ≤ C(‖yN‖L2 +

h√3‖yN ′‖L2

).

We can conclude, using Poincare’s inequality, that if supN EN (yN ) < +∞, then yN is

bounded in H10 (0, 1), hence it has subsequences which converge in L2(0, 1). (And, as

well, (yN ).) Observe that in this case, also Ascoli Arzela applies and the convergence

is, in fact, uniform.

Let us show (i): let uN → u in L2(0, 1). If lim infN EN (uN ) = +∞ there is

nothing to prove. Otherwise, there is a subsequence such that ENk(uNk) < +∞ and

limk ENk(uNk) = lim infN EN (uN ). For each k, by definition there is yNk ∈ RNk+1

such that uNk = yNk and ENk(uNk) = E(yNk). Then, by the previous remark, both

yNk and yNk converge to u in L2(0, 1). One has that

Nk−1∑i=0

kh

2

(yNki+1 − y

Nki

h

)2

=k

2

∫ 1

0

(yNk′(x))2dx


so that in particular

k

2

∫ 1

0

|u′(x)|2dx ≤ lim infk

Nk−1∑i=0

kh

2

(yNki+1 − y

Nki

h

)2

.

On the other hand,

h

N∑i=0

yNi gNi =

∫ 1

0

yN (x)gN (x)dx

and since both yN and gN converge uniformly, respectively to u and g, it converges to∫ 1

0y(x)g(x)dx. This shows (i).

For (ii), the proof is here elementary. Indeed, if u ∈ L2(0, 1) with E(u) < +∞(otherwise there is nothing to prove), so that u ∈ H1

0 (0, 1), u is also continuous and

one can simply let for any N yNi = u(i/N). One then checks that uN := yN → u

uniformly (and in L2(0, 1)), as well as yN . Now, one has that

h

N−1∑i=0

(yNi+1 − yNi

h

)2

=1

h

N−1∑i=0

(∫ (i+1)h

ih

yN ′(x)dx

)2

≤N−1∑i=0

∫ (i+1)h

ih

(y′N (x)

)2dx =

∫ 1

0

|u′N |2dx

while (by uniform convergence)

h

N∑i=0

yNi gNi =

∫ 1

0

yN (x)gN (x)dx→∫ 1

0

u(x)g(x)dx.

As a consequence, minimizers of EN will converge to the (unique) minimizer of E .

Observe however that for this problem, standard numerical analysis will also provide

quite precise error bounds for this convergence.

6.2.1 Higher dimension

Here, we have used to build the “recovery sequence” the fact that u was in H1(0, 1),

hence continuous. How would we do this in higher dimension? Consider for instance,

for U = (ui,j)0≤i,j≤N with u0,j = uN,j = ui,0 = ui,N = 0 for all i, j, the energy (with

still, h = 1/N)

Eh(U) = h2∑i,j

|ui+1,j − ui,j |2

h2+|ui,j+1 − ui,j |2

h2

which in the same way as before, will Γ-converge, in L2((0, 1)2) to

E(u) =

∫(0,1)2

|∇u|2dx

(for u ∈ H10 ((0, 1)2), +∞ else).

6.2. APPLICATION: DISCRETE TO CONTINUUM LIMITS 53

For the “liminf” (i) we do the same as in one dimension, considering the bilinear

“Q1” approximation:

uN (x, y) =∑i,j

uNi,j∆

(x− ihh

)∆

(y − jhh

).

One checks that

∂uN

∂x=∑i,j

(uNi+1,j − uNi,j

h

)χ(ih,(i+1)h)(x)∆

(y − jhh

).

so that for fixed y,

∫ 1

0

(∂uN

∂x

)2

dx =

N−1∑i=0

N−1∑j=1

h

(uNi+1,j − uNi,j

h∆

(y − jhh

))2

≤ hN−1∑i=0

N−1∑j=1

(uNi+1,j − uNi,j

h

)2

∆

(y − jhh

)where we have used the convexity of X 7→ X2. Then

∫ 1

0

∫ 1

0

(∂uN

∂x

)2

dx dy ≤ hN−1∑i=0

N−1∑j=1

(uNi+1,j − uNi,j

h

)2 ∫ 1

0

∆

(y − jhh

)dy

= h2N−1∑i=0

N−1∑j=1

(uNi+1,j − uNi,j

h

)2

.

Hence if uN → u, one will have

E(u) ≤ lim infN

∫[0,1]2

|∇u|2dx dy ≤ lim infNEN (uN )

But: what about the limsup (ii)? We cannot, here, reproduce the same proof, since

given u ∈ H10 (0, 1), u(ih, jh) does not make any sense! We use a very classical impor-

tant trick: first, we approximate an arbitray u ∈ H10 ((0, 1)2) with simpler functions

uk (here, in particular, smooth, or at least continuous) such that

lim supk→∞

E(uk) ≤ E(u).

Here this is easy, as we know that C∞c ((0, 1)2) is dense in H10 ((0, 1)2).

For this smooth uk, we do as in one dimension, approximating uk with the discrete

matrix ((uNk )i,j)i,j given by (uNk )i,j = uk(ih, jh). Then, as before it is quite easy to

show that

lim supN→∞

EN (uNk ) ≤ E(uk).

The conclusion uses a diagonal procedure: we know that for all l, there exists k(l) >

k(l − 1) such that

k ≥ k(l)⇒

‖uk − u‖L2 ≤ 1

l

E(uk) ≤ E(u) + 1l .


Then, we find N(l) ≥ N(l − 1) such that if N ≥ N(l),

N ≥ N(l)⇒

‖uNk(l) − uk(l)‖L2 ≤ 1

l

E(uNk(l)) ≤ E(uk(l)) + 1l ,

so that

N ≥ N(l)⇒

‖uNk(l) − u‖L2 ≤ 2

l

E(uNk(l)) ≤ E(u) + 2l .

Eventually, we let uN = uNk(l) whenever N(l) ≤ N < N(l + 1).

6.3 Application: periodic homogenization

(See [3].) We consider for u ∈W 1,p(Ω;Rm) (Ω ⊂ Rd bounded open set) the functions

Eε(u) :=

∫Ω

f(xε,Du

)dx.

Here f : Rd × Rm×d → [0,+∞) is a function which is periodic in the first variable:

f(y + k,A) = f(y,A) for all y ∈ Rd, A ∈ Rm×d and for all k ∈ Zd. It satisfies in

addition the growth condition

α(|A|p − 1) ≤ f(y,A) ≤ β(|A|p + 1)

for all y,A, where p > 1 is given.

In particular it follows

α

∫Ω

|Du|p − 1dx ≤ Eε(u) ≤ β∫

Ω

1 + |Du|pdx.

By compactness, there exists a sequence (εk) ↓ 0 and E : W 1,p(Ω)→ R+ ∪+∞ such

that Eεk Γ-converges to E in Lp(Ω) as k →∞, where for all u,

α

∫Ω

|Du|p − 1dx ≤ E(u) ≤ β∫

Ω

1 + |Du|pdx.

It turns out that the following result is true

Theorem 13. The Γ-limit E in Lp(Ω;Rm) is given, for u ∈W 1,p(Ω), by

E(u) =

∫Ω

fhom(Du) dx

where

fhom(A) := limT→∞

1

T dinf

∫(0,T )d

f(x,A+Dφ(x))dx : φ ∈W 1,p0 ((0, T )d;Rm)

.

(6.3)

The result is true, still, if one has a Dirichlet condition on part of the domain

(observe that the statement for the Γ-limsup depends on the Dirichlet condition and

could not have been true anymore), moreover in this case the compactness of bounded

sequences is immediate (if Ω is regular) thanks to Rellich’s theorem.

6.3. APPLICATION: PERIODIC HOMOGENIZATION 55

Remark: it is not clear that the limit defining fhom exists. This is a standard fact.

First, we can define

fT (A) =1

T dinf

∫(0,T )d

f(x,A+Dφ(x))dx : φ ∈W 1,p0 ((O, T )d;Rm)

.

Then, observe that at least if T is an integer, one easily sees that fkT (A) ≤ fT (A)

for all k ∈ N, k ≥ 1. Indeed if φ is almost optimal in the definition of fT , one can

use in (0, kT )d the function φ′(x) = φ(y) whenever x = lT + y, y ∈ (0, T )d, l ∈ Zd

(li = bxi/T c, i = 1, . . . , d), showing that fkT (A) ≤ 1Td

∫(0,T )d

f(x,A+Dφ)dx.

Moreover if S is large, choosing kS such that kST ≤ S < (kS + 1)T , one sees that

for φ ∈W 1,p0 ((0, kST )d;Rm),

fS(A) ≤ (kST )d

Sd1

(kST )d

∫(0,kST )d

f(x,A+Dφ(x))dx+|(0, S)d \ (0, kST )d|β(1 + |A|p)

Sd

hence

fS(A) ≤ fkST (A) +C

(1− (kST )d

Sd

)(1 + |A|p) ≤ fT (A) +C

(1− (kST )d

Sd

)(1 + |A|p).

Moreover since (kS + 1)T ≥ S, 1 − (kST )d/Sd ≤ 1 − (1 − T/S)d → 1 as S → ∞, so

that lim supS→∞ fS(A) ≤ fT (A). It follows (taking the liminf as T → ∞) that the

limit exists (at least along integers, but extending this to all real numbers is really

easy since for large values of T fT (A) varies very little with T ), and that it is, in fact,

infT fT (A).

Lemma 6. fhom is continuous.

Proof. Fix T > 2, let A, B two matrices and t small. Then. Let φ ∈ W 1,p0 ((1, T −

1)d;Rm), η ∈ C∞c ((0, T )d; [0, 1]) be a cut-off function such that η ≡ 1 in (1, T − 1)d

and |∇η| ≤ 2.

fhom(A+ tB) ≤ fT (A+ tB) ≤ 1

T d

∫(0,T )d)

f(x,A+ tB +D(φ(x)− tη(x)Bx))dx

=1

T d

∫(0,T )d)

f(x,A+Dφ(x) + t(1− η)B − t(Bx)⊗∇η)dx

≤ 1

T d

∫(1,T−1)d)

f(x,A+Dφ(x))dx+ C1

T d

∫(0,T )d\(1,T−1)d

|A+ tB|p + tp|B|p|x|pdx

≤(1− 2

T

)d 1

(T − 2)d

∫(1,T−1)d)

f(x,A+Dφ(x))dx

+ C(

1−(1− 2

T

)d)(|A|p + |tB|p + tp√dpT p|B|p

).

Given ε > 0, one can assume that T is large enough and φ is such that

1

(T − 2)d

∫(1,T−1)d)

f(x,A+Dφ(x))dx ≤ fhom(A) + ε,

C(

1−(1− 2

T

)d)(|A|p + |B|p) ≤ ε,


and it follows (if t ∈ [0, 1])

fhom(A+ tB) ≤ fhom(A) + 2ε+√dptpT pε ≤ fhom(A) + 3ε

if t is then chosen small enough.

Lemma 7. fhom is convex in along the directions A + R(a⊗ ei) where (ei)di=1 is the

canonical basis of Rd.

Remark: as a consequence of Thm. 13, fhom will also be quasiconvex hence rank-one

convex.

Proof. We give the main ideas. Let A,B with B−A = a⊗ e1 and let C = (A+B)/2.

Let T > 0 an integer and φA, φB such that fhom(A) ≈ fT (A) and fhom(B) ≈ fT (B).

Consider φA, φB almost realizing the min in fT (A), fT (B):

fhom(A) ≈ 1

T d

∫(0,T )d

f(x,A+DφA)dx, fhom(B) ≈ 1

T d

∫(0,T )d

f(x,B +DφB)dx.

We define then (in Rd) a function φ by letting φ(x) = Ax+φA(x) + (k+ 1)Ta−Cx if

2kT ≤ x1 ≤ (2k+1)T and φ(x) = Bx+φB(x)−kTa−Cx if (2k+1)T ≤ x1 ≤ 2(k+1)T .

Then it is continuous accross the hyperplanes x1 = kT , hence locally in W 1,p. For

S = 2nT , n ∈ N, large enough, we introduce the cut-off ηS = 1 on (1, S − 1)d, 0 on

∂(0, S)d with |∇ηS | = 1 on (0, S)d \ (1, S − 1)d. Then we find

fhom(C) ≤ limS→∞

1

Sd

∫(0,S)d

f(x,C +D(φηS))dx.

Then, we have by construction that C + D(φηS) = A + DφA if x ∈ (1, S − 1)d and

2kT ≤ x1 ≤ (2k + 1)T , B +DφB if x ∈ (1, S − 1)d and (2k + 1)T ≤ x1 ≤ 2(k + 1)T ,

while otherwise C +D(φηS) = C + ηSDφ+ φ⊗∇ηS and one checks easily that∫(0,S)d\(1,S−1)d

f(x,C +D(φηS))dx ≤ cSd−1

for a constant c depending only on A,B, φA, φB . It follows

1

Sd

∫(0,S)d

f(x,C +D(φηS))dx /fhom(A) + fhom(B)

2+c

S

and letting S →∞ we find that fhom is convex along the line (A,B).

Corollary 2. Estimate (5.1) holds for fhom.

Proof. It is enough to observe that the proof of this estimate uses only the convexity

in the directions a⊗ ei for (ei) the canonical basis.


6.3.1 Proof of the homogenization result

limsup

We start with the limsup inequality which is simpler. Given u ∈ W 1,p(Ω) we know

that there exists a sequence of piecewise affine functions un which converge to u in

W 1,p(Ω) (assuming the boundary of Ω is Lipschitz) (“P1” finite elements approxima-

tion). Thanks to estimate (5.1) (Corollary 2), one can show that

limn→∞

∫Ω

fhom(Dun)dx =

∫Ω

fhom(Du)dx

(of more generally, u 7→∫

Ωfhom(Du)dx is continuous in W 1,p(Ω)). We leave the proof

as an exercise, it suffices to bound |fhom(Du)− fhom(Dun)| using estimate (5.1).

We then show that when u is piecewise affine, there exists uε which converges to u

such that lim supε Eε(u) ≤ E(u). A diagonal trick, exactly as in Section 6.2.1, allows

to conclude.

To build uε one proceeds independently in each subdomain where u is affine. Hence

one considers an open domain O ⊂ Rd and u = Ax in O. Then given δ > 0 one lets

Oδ =⋃

z∈δZdz+(0,δ)d⊂O

z + (0, δ)d

and one observes that |O \ Oδ| → 0 as δ ↓ 0. Given ε > 0 small (ε << δ) we let

T = b δT c−1. If ε > 0 is small enough we know that we can find φε ∈W 1,p0 ((0, T )d;Rm)

such that1

T d

∫(0,T )d

f(y,A+Dφε(y))dy ≤ fhom(A) + ε.

We then let

uε(x) :=

Ax if x ∈ O \Oδ,Ax+ εφε

(xε − b

zε c)

if x ∈ z + (0, δ)d ⊂ Ω , z ∈ δZd.

In each cube z + (0, δ)d, using the periodicity of f ,∫z+(0,δ)d

f(xε , Duε(x))dx =

∫z+(0,δ)d

f(xε − b

zε c, A+Dφε

(xε − b

zε c))dx

= εd∫

(0,δε )d

f(y + θε, A+Dφε(y + θε))dy

where we have done the change of variables y = (x− z)/ε and let θε = z/ε− bz/εc ∈[0, 1]. By the choice of φε, and using that θε + (0, T )d ⊂ (0, δ/ε)d,∫

(0,δε )d

f(y + θε, A+Dφε(y + θε))dy ≤ Cδ

ε

d−1

|A|p + T d(fhom(A) + ε).

Hence ∫z+(0,δ)d

f(xε , Duε(x))dx ≤ Cδd−1ε|A|p + δd(T

δ/ε

)d(fhom(A) + ε),


and, using |Oδ| = δd]z ∈ Zd : z + (0, δ)d ⊂ O,∫O

f(xε , Duε

)dx ≤ C|O \Oδ||A|p + C|Oδ|ε

δ|A|p + |Oδ|

(T

δ/ε

)d(fhom(A) + ε),

so that

lim supε→0

∫O

f(xε , Duε

)dx ≤ |Oδ|fhom(A) + C|O \Oδ||A|p δ→0−→ |O|fhom(A).

liminf

Let us consider first an easier case where we consider functions uε such that uε → u =

Ax in Lp(B;Rm), where B is a ball (centered in 0), and we wish to show that

|B|fhom(A) ≤ lim infε→0

∫B

f(xε , uε(x))dx. (6.4)

The right-hand side, after a change of variable, is

εd∫B/ε

f(y,Dvε(u))dy

where we have let vε(y) = uε(εy)/ε. We observe that if the liminf in (6.4) is finite

there is a subsequence such that∫Bf(x/εk, Duεk)dx is finite and goes to the liminf.

In particular, Duεk is globally bounded in Lp(B), and therefore converges weakly to

A. (As usual, since uεk → Ax, one writes for any smooth ψ with compact support∫B

Aψdx = −∫B

divψAxdx = − limk

∫B

divψuεkdx = limk

∫B

Duεkψdx

showing that the weak limit of Duεk cannot be anything but A.) In what follows, to

simplify, we denote (uε) the subsequence.

In a first stage, we assume that we work in the unit cube Q = (0, 1)d rather than

in the ball B and we observe that if T = 1/ε,

εd∫Q/ε

f(y,Dvε)dy =1

T d

∫(0,T )d

f(y,A+ (Dvε −A))dy ' fhom(A)

(the inequality would be true if we had vε −Ax = 0 on ∂(0, T )d).

To make this work we consider δ > 0 and a cut-off η ∈ C∞c (Q; [0, 1]) with |∇η| ≤C/δ and η ≡ 1 on Qδ = (δ, 1 − δ)d. Then, we consider uδε = Ax + η(uε − Ax). If we

replace uε with uδε, the inequality above becomes true and we indeed can assert that

fhom(A) ≤∫Q

f(xε , Duδε(x))dx.

We need to understand the error between this and the energy of uε. A basic compu-

tation yields, using Duδε = A+ (uε −Ax)⊗∇η + η(Duε −A),∫Q

f(xε , Duδε(x))dx ≤

∫Q

f(xε , Duε(x))dx

≤ C|Q \Qδ|(1 + |A|p) +C

δp

∫Q\Qδ

|uε −Ax|p dx+ C

∫Q\Qδ

|Duε|pdx.


Clearly, the first term in the right-hand side can be made arbitrarily small choosing δ

small, and the second term goes to zero as ε→ 0 since we have assumed that uε → Ax

strongly in Lp(Q).

However, there is no clear way to bound the last term. A solution (due to Marcellini,

see [6]) consists in considering not one but N cut-offs η1, . . . , ηN , for a given integer

N , such that ηi ∈ C∞c (Q(i−1)δN ), ηi ≡ 1 on Q

iδN , |∇ηi| ≤ CN/δ. We consider for ε the

functions uiε = Ax+ηi(uε−Ax), and then the same computation as before shows that

for each i, using Duiε = Duε + (1− ηi)(Duε −A) + (uε −Ax)⊗∇ηi,

fhom(A) ≤ 1

|Q(i−1)δN |

∫Q

(i−1)δN

f(xε , Duiε(x))dx

≤ 1

(1− 2 (i−1)δN )d

∫Q

(i−1)δN

f(xε , Duε(x))dx

+CNp

δp

∫QiδN \Q

(i−1)δN

|uε −Ax|pdx+

∫QiδN \Q

(i−1)δN

|Duε −A|pdx.

Now, since Duε is bounded in Lp, one has for a constant C that

N∑i=1

∫QiδN \Q

(i−1)δN

|Duε −A|pdx ≤∫Q

|Duε −A|pdx ≤ C

so that there exists, for each ε > 0, an index i(ε) for which∫QiδN \Q

(i−1)δN

|Duε −A|pdx ≤C

N.

Choosing for each ε this “good” index i, we find that

fhom(A) ≤ 1

(1− 2 δN )d

∫Q

f(xε , Duε(x))dx+CNp

δp

∫Q

|uε −Ax|pdx+C

N

so that (using uε → Ax),

fhom(A) ≤ 1

(1− 2 δN )d

lim infε→0

∫Q

f(xε , Duε(x))dx+C

N.

It remains to send first N →∞, then δ → 0, to conclude.

Next step: from Q to B: the idea is elementary, it consists in covering B with

smaller and smaller disjoint cubes, and use the previous result: if (Qi)Ki=1 are disjoint

cubes in B one has

|B|fhom(A) = |B \K⋃i=1

Qi|fhom(A) +

K∑i=1

|Qi|fhom(A)

≤ |B \K⋃i=1

Qi|fhom(A) +

K∑i=1

lim infε→0

∫Qi

f(xε , Duε)dx

≤ |B \K⋃i=1

Qi|fhom(A) + lim infε→0

∫B

f(xε , Duε)dx.

Taking more and more many cubes, one can make the first term in the right-hand side

arbitrarily small.


Localization? Now, we want to consider an arbitrary uε → u in Lp(Ω;Rm), and

show that ∫Ω

fhom(Du(x))dx ≤ lim infε→0

∫Ω

f(xε , Duε(x))dx.

As before we assume that the right-hand side is not +∞ and we consider a subsequence

such that

limk→∞

∫Ω

f( xεk , Duεk(x))dx = lim infε→0

∫Ω

f(xε , Duε(x))dx.

The idea is to “localize” the problem near each point, use Duεk ' Du(x) near each

point x and use the previous result. The classical method is due initially to Fonseca

and Muller. The first observation is that µk = f( xεk , Duεk(x))dx is a sequence of

bounded measures which are uniformly bounded, so that we may assume that in the

sense of measures, µk∗ µ where µ is a nonnegative measure, which means that for

any φ ∈ C0c (Ω),

limk→∞

∫Ω

φ(x)dµk(x) =

∫Ω

φ(x)dµ(x).

We know that

µ(Ω) ≤ lim infk→∞

µk(Ω)

so that the Γ-liminf inequality (ii) will hold if we can show that µ ≥ fhom(Du(x))dx.

By Radon-Nikodym’s theorem, we know that µ = g(x)dx+ ν with ν ⊥ dx, ν ≥ 0,

and

g(x) = limρ→0

µ(Bρ(x))

|Bρ|

(Lebesgue)-almost-everywhere. Hence we need to show that g(x) ≥ fhom(x,Du(x)),

dx-a.e..

An important result is the following:

Lemma 8. Given x ∈ Ω, for a.e. ρ > 0 one has

limk→∞

µk(Bρ(x)) = µ(Bρ(x)).

Proof. If Bρ(x) is the open ball of center x and radius ρ, then

µ(Bρ(x)) ≤ lim infk→∞

µk(Bρ(x)).

An obvious reason is that given ν a nonnegative measure,

ν(Bρ(x)) = sup

∫Bρ(x)

φ(x)dν : φ ∈ C0c (Bρ(x); [0, 1])

and the supremum on the right-hand side is weakly lsc with respect to ν, as a sup of

continuous functions with respect to weak-star convergences of measures.

For the same reason ν 7→ ν(Bρ(x)) is usc, since it is an infimum of continuous

functions:

ν(Bρ(x)) = inf

∫Ω

φ(x)dν : φ ∈ C0c (Ω) , φ ≥ χBρ(x)

.


Hence, if ρ is such that µ(∂Bρ(x)) = 0 (which is true for all but, at most, a

countable number of radii since all these sets are disjoint when ρ varies)1, one has

µ(Bρ(x)) ≤ lim infk→∞

µk(Bρ(x)) ≤ lim supk→∞

µk(Bρ(x)) ≤ µ(Bρ(x)) = µ(Bρ(x)),

showing the Lemma.

Now, for each n ≥ 1, one can find ρn ≤ ρn−1, not in the exceptional (countable)

set of the previous Lemma, such that∣∣∣∣µ(Bρn(x))

|Bρn |− g(x)

∣∣∣∣ ≤ 1

n

and then one can find εkn ≤ εkn−1 such that∣∣∣∣µkn(Bρn(x))

|Bρn |− µ(Bρn(x))

|Bρn |

∣∣∣∣ ≤ 1

n.

We then denote, to simplify, εkn with εn, and we let ε′n = εn/ρn. Since εn is chosen

“after ρn” (for a fixed, given value of ρn), we should have (and we can assume that)

ε′n → 0 as n→∞.

Then, the change of variable z = x+ ρny yields

µkn(Bρn(x))

|Bρn |=

1

|Bρn |

∫ρn

(x)f( zεn, Duεn(z))dz

=1

|B1|

∫B1

f( xεn + yε′n, Duεn(x+ ρny))dy

=1

|B1|

∫B1

f(θn + yε′n, Duεn(x+ ρny))dy =

1

|B1|

∫B1

f(y+ε′nθnε′n

, Dun(y))dy

where θn = x/εn − bx/εnc ∈ [0, 1]d and un(y) := (uεn(x+ ρny)− uεn(x))/ρn.

Assume we can show that un → Du(x) · y in Lp(B1) as n → ∞. Then we are

back to the previously studied case of a sequence of functions converging to an affine

function in a ball (modulo the varying translations ε′nθn which can be easily handled)

and it will follow that

fhom(Du(x)) ≤ lim infn→∞

1

|B1|

∫B1

f(y+ε′nθnε′n

, Dun(y))dy.

Putting everything together, we will deduce fhom(Du(x)) ≤ g(x), which was to prove.

It remains to show that un → Du(x) · y. This will be true for almost all x, and relies

on fine properties of W 1,p functions.

A first observation is that for any ε, ρ > 0,∥∥∥∥uε(x+ ρ ·)− uε(x)

ρ− u(x+ ρ ·)− u(x)

ρ

∥∥∥∥Lp(B1)

≤ 1

ρ

(∫B1

|uε(x+ ρy)− u(x+ ρy)|pdy) 1p

+1

ρ|B1|

1p |uε(x)− u(x)|.

1A simple method to check such a fact is to consider the sets Ii = ρ : µ(∂Bρ) ≥ 1/i and observe

that µ(Ω) ≥ µ(⋃ρ∈Ii ∂Bρ) ≥ (1/i)]Ii so that ]Ii ≤ Ci. Hence ρ : µ(∂Bρ) > 0 =

⋃i Ii is countable.


Since we could have assumed that uε → u a.e. (extracting an appropriate subsequence),

the right-hand side goes to zero as ε → 0: therefore when we have chosen εkn above,

for fixed ρn, we could have required in addition that∥∥∥∥un − u(x+ ρn·)− u(x)

ρn

∥∥∥∥Lp(B1)

≤ 1

n. (6.5)

It remains to see that we could also have chosen ρn so that∥∥∥∥u(x+ ρny)− u(x)

ρn−Du(x) · y

∥∥∥∥Lpy(B1)

≤ 1

n. (6.6)

For fixed ρ > 0 (and assuming first that u ∈ C1), one has∫B1

∣∣∣∣u(x+ ρy)− u(x)

ρ−Du(x) · y

∣∣∣∣p dy=

∫B1

∣∣∣∣1ρ∫ ρ

0

(Du(x+ sy)−Du(x)) · yds∣∣∣∣p dy

≤∫B1

1

ρ

∫ ρ

0

|(Du(x+ sy)−Du(x)) · y|p dsdy

≤ 1

ρ

∫ ρ

0

[1

sd

∫Bs(x)

|Du(z)−Du(x)|pdz

]ds

This quantity goes a.e. to zero as ρ→ 0, since we have the well known following result:

Theorem 14. Let h ∈ Lp(Ω). Then for a.e. x ∈ Ω,

lims→0

1

|Bs|

∫Bs(x)

|h(z)− h(x)|pdz = 0

Remark: x is called a “Lebesgue point” if there exists h(x) such that 1|Bs|

∫Bs(x)

|h(z)−h(x)|pdz → 0, and h = h a.e. is called the “precise representative” of h.

Proof. The result is a consequence of Radon-Nikodym’s theorem: if f ∈ L1loc, since

f(x)dx is absolutely continuous with respect to Lebesgue’s measure, one has f(x)dx =

f(x)dx for f given by the limit, which exists almost everywhere:

f(x) = lims→0

1

|Bs|

∫Bs(x)

f(z)dz.

But clearly, it follows that f = f a.e., in particular, almost everywhere

f(x) = lims→0

1

|Bs|

∫Bs(x)

f(z)dz.

We apply this to the function |h(x)− t|p ∈ L1loc, for all t ∈ Q. Then, for all t, there

exists a set Et with |Et| = 0 such that for x 6∈ Et,

|h(x)− t|p = lims→0

1

|Bs|

∫Bs(x)

|h(z)− t|pdz.


Then we let E =⋃t∈QE

t, so that |E| = 0. One has if x 6∈ E and t ∈ Q,(1

|Bs|

∫Bs(x)

|h(z)− h(z)|pdz

)1/p

≤

(1

|Bs|

∫Bs(x)

|h(z)− t|pdz

)1/p

+ |t− h(x)|

so that

lim sups→0

(1

|Bs|

∫Bs(x)

|h(z)− h(z)|pdz

)1/p

≤ 2|h(x)− t|

Since t is arbitrary, this value can be sent to zero and the thesis follows.

Eventually, (ii) is proved, since we have now all the elements to see that g(x) ≥fhom(Du(x)) a.e.

6.3.2 A remark in the convex case

We assume that f(y, ·) is convex. In this case, fhom can be represented by a “cell

formula” (see [3]). Let us define

fhom(A) = inf

∫(0,1)d

f(y,A+Du(y))dy : u ∈W 1,p] ((0, 1)d;Rm)

where W 1,p] ((0, 1)d) is the set of periodic functions in W 1,p (W 1,p(Rd/Zd)). Observe

that in the convex case, the infimum defining fhom is in fact a min and we can denote

uA a solution. The function uA is called a corrector.

Theorem 15. fhom = fhom.

Sketch of proof. Let T be a large integer. If η is a cut-off in (0, T )d, we have

fhom(A) ≤ 1

T d

∫(0,T )d

f(y,A+D(ηuA))dy

=1

T d

∫(0,T )d

f(y,A+ ηDuA + uA ⊗∇η))dy ≈ fhom(A)

up to an error which vanishes when the cut-off tends to 1.

On the other hand, let T be a large integer and v ∈W 1,p0 ((0, T )d;Rm) such that

fhom(A) ≈ 1

T d

∫(0,T )d

f(y,A+Dv)dy

We can consider v as a TZd-periodic function (letting v(y+kT ) = v(y) for y ∈ (0, T )d

and k ∈ Zd). Then, we let

w(x) =1

T d

∑k∈Zd∩[0,T [d

v(x+ k)

which is now Zd-periodic, so that

fhom(A) ≤∫

(0,1)df(y,A+Dw(y))dy.


Now,

f(y,A+Dw(y)) = f(y,A+

1

T d

∑k∈Zd∩[0,T [d

Dv(y + k))

≤ 1

T d

∑k∈Zd∩[0,T [d

f(y,A+Dv(y + k))

by convexity of f . We deduce

fhom(A) ≤ 1

T d

∑k∈Zd∩[0,T [d

∫(0,1)d

f(y,A+Dv(y + k)) =1

T d

∫(0,T )d

f(y,A+Dv(y))dy.

Remark: the term “correctors” derives from the fact that, in the Γ-limsup, a way

to correct the affine function Ax in order to recover the right energy is to approximate

it with Ax+ εuA(x/ε).

Remark: What has been proved in case f(x,A) depends only on A?

6.3.3 Phase transitions, singular limits

The Cahn-Hilliard energy in dimension 1

Higher dimension

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