Around the Nash-Moser theorem

David Gerard-Varet

February 10, 2019

Preamble

These lecture notes correspond to the contents of a graduate course taught in UniversityParis Diderot in 2018. The general concern of these notes is the use of perturbative methodsfor solving equations in infinite dimension. Typically, we want to solve an equation of theform

(1) F (u) = f,

where F : X → Y is a smooth function between two Banach spaces X and Y such thatF (0) = 0, and where ‖f‖Y is small enough. In many applications that we will consider, themapping F will be a (nonlinear) partial differential operator, and X and Y will always bespaces of functions.

In the case where the differential at 0, denoted F ′(0), is an isomorphism from X to Y , thelocal inversion theorem provides us with a unique solution u of (1) close to 0 in X, attendingthat f is close enough to 0 in Y . The main tool behind the local inversion theorem is thePicard fixed point theorem: namely, a key idea is to reformulate equation (1) under the form

u = F ′(0)−1 (f − F (u) + F ′(0)u)

and show that it is a contraction from a small ball in X to itself. In this way, the solutionu can be obtained as the limit of the iteration

(2) un+1 = F ′(0)−1 (f − F (un) + F ′(0)un) .

The convergence typically holds at geometric rate, as

‖un+1 − un‖X ≤ κ‖un − un−1‖X

for some fixed κ ∈ (0, 1), which implies ‖un − u‖X ≤ Cκn for some C > 0.

The Picard iteration (2) can be compared to another famous iterative method, namelyNewton’s method. In our setting, the Newton-Raphson iteration can be written as:

(3) un+1 = F ′(un)−1 (f − F (un) + F ′(un)un) .

On one hand, the Newton’s iteration is more demanding: at each step, one must solve adifferent linearized equation, involving the operator F ′(un) rather than F ′(0). On the otherhand, if ‖u0‖X and ‖f‖Y are small enough, the Newton’s scheme converges quadraticallyrather than geometrically: one has

(4) ‖un+1 − un‖X ≤ c‖un − un−1‖2X

which implies ‖un − u‖X ≤ Cκ2n for some C > 0 and κ = c‖u1 − u0‖X .

Overall, the common point between these methods is an assumption of invertibility of thelinearized operator, with an estimate of the type:

F ′(u)v = g ⇒ ‖v‖X ≤ C‖g‖Y

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either for u = 0, or for u in a small neighborhood of 0. However, there are many interestingand important situations where such nice invertibility assumption does not hold. What mayfor instance happen is that the linearized equation is solvable, but only in a functional spacewhich is larger than X. This larger space often corresponds to a loss of regularity. Typically,one can think of X = Xs0 and Y = Ys0 as elements of two families of Banach spaces (Xs)s∈Nand (Ys)s∈N, where s denotes a regularity index: for instance, Xs = Cs

b (Rd) (space of s-timescontinuously differentiable functions with bounded derivatives), or Xs = Hs(Rd) (Sobolevspace of index s). Then, the bad situations are those in which F sends Xs to Ys, but theinverse linearized operator F ′(u)−1 is only defined from Ys to Xs−σ, for some σ > 0:

(5) F ′(u)v = g ⇒ ‖v‖Xs−σ ≤ Cσ‖g‖Ys

Obviously, such feature plagues the use of the standard iterative methods (2) or (3): at eachstep of the iteration, the regularity of the approximation un is lowered by σ, so that no moreregularity is available after a finite number of steps.

This phenomenon of loss of derivatives will be the central topic of these lecture notes. Itappears in many different physical and mathematical contexts: celestial mechanics (in con-nection to the so-called KAM theory and small divisor problems), riemannian geometry (inconnection to the so-called isometric embedding problem), or in various PDE problems (forinstance in the recent work by Mouhot and Villani on Landau damping [13]). We will ofcourse examine several examples in these notes, and present the mathematical tools that areused to handle this kind of problems. These tools were mostly developped in the 50’s and60’s, through the works of Kolmogorov, Arnold, Nash and Moser.

The bottom line of these works is that, while the Picard iteration becomes useless, a properadaptation of Newton’s iteration can still be used to solve (1): the very fast convergenceof the Newton’s scheme can somehow compensate for the loss of derivatives. Obviously, totake advantage of the former over the latter, both must be put onto a common ground: oneneeds to make the information about the loss of derivatives more quantitative. One wayto proceed is to work in the framework of analytic functions, meaning that Xs or Ys arespaces of analytic functions (with s being more or less the radius of analyticity). In suchframework, the Cauchy formula allows to quantify the loss of derivatives in terms of lossof the radius of analyticity. This information can be implemented in the estimate (5), andthen in an analogue of (4). This is in short the path followed by Kolmogorov and Arnoldin the treatment of small divisor problems (see chapter 2). Another way to quantify theloss of derivatives is the Fourier transform: if a function has frequencies in a ball of radiusN , a derivative ”costs at most a factor N”. This idea is at the basis of the Nash-Mosertheorem, which allows to overcome the problem of the loss of derivatives in finite regularity(see chapter 3).

The outline of the lectures is as follows. The first chapter is a reminder on iterative methodsfor solving nonlinear equations (Picard, Newton). A simple example illustrating the phe-nomenon of loss of derivatives is provided, giving the motivation for the next chapters. Thesecond chapter is devoted to circle diffeomorphisms. The goal of this chapter is to establish

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a theorem of Arnold about the analytic conjugacy of analytic diffeomorphisms of the cir-cle. This theorem exhibits a first way to circumvent the loss of derivatives, by combiningthe properties of analytic functions together with the Newton’s scheme. The third chapteris devoted to a discussion and proof of the famous Nash-Moser theorem, which allows tosolve (1) in finite regularity. For the sake of clarity, we restrict in this chapter to a simplestatement inspired by Moser’s paper [14] (see also [24]). Chapter 4 is about the applicationof the Nash-Moser theorem to the isometric embedding problem. Chapter 5 is about somemodern tools of Fourier analysis (Littlewood-Paley decomposition, paraproduct), which haverevealed a good substitute to the Nash-Moser linearization technique in some PDE contexts(see for instance [3] on the Landau damping problem). Eventually, we conclude these lecturenotes by discussing the famous De Giorgi-Nash-Moser theorem, on the Holder regularity ofsolutions of linear scalar elliptic equations with measurable coefficients. We follow here theapproach of De Giorgi [7, 5]. Although this last topic may seem a bit disconnected fromthe rest of the lectures, it turns out that one of the core arguments borrows to the samephilosophy: the gain in regularity due to the ellipticity competes with the shrinking of thedomain where interior elliptic estimates can be derived, a bit like the gain in the error ofconvergence competes with the loss of derivatives in the previous settings. Furthermore, onekey idea is to express the gain due to elliptic regularity ”in a nonlinear manner”, to echo thenonlinear nature of the Newton’s estimate (4)

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Contents

1 Iterative methods for solving nonlinear equations 61.1 Picard iteration . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.2 Newton’s method . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.3 Loss of derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

2 Conjugacy of circle diffeomorphisms 112.1 Circle maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 112.2 Circle homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 122.3 Conjugacy of analytic diffeomorphisms . . . . . . . . . . . . . . . . . . . . . 19

2.3.1 Setting of the problem . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3.2 A reminder about analytic functions . . . . . . . . . . . . . . . . . . 212.3.3 Arnold’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

2.4 Proof of Arnold’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3 The Nash-Moser theorem 303.1 Statement of the Nash-Moser theorem . . . . . . . . . . . . . . . . . . . . . 313.2 Proof of the Nash-Moser theorem . . . . . . . . . . . . . . . . . . . . . . . . 33

4 The isometric embedding of riemannian manifolds 374.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 384.2 Use of the Nash-Moser theorem . . . . . . . . . . . . . . . . . . . . . . . . . 41

5 Littlewood-Paley decomposition 445.1 The space of temperate distributions S ′(Rd) . . . . . . . . . . . . . . . . . . 44

5.1.1 Definitions, examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 445.1.2 Link with the Fourier transform . . . . . . . . . . . . . . . . . . . . . 46

5.2 Littlewood-Paley decomposition . . . . . . . . . . . . . . . . . . . . . . . . . 475.3 Link with Sobolev and Holder spaces . . . . . . . . . . . . . . . . . . . . . . 505.4 Paraproduct and tame estimates . . . . . . . . . . . . . . . . . . . . . . . . . 585.5 Connection to the Nash-Moser methodology . . . . . . . . . . . . . . . . . . 61

6 The regularity theorem of De Giorgi for scalar elliptic equations 646.1 Weak solutions of elliptic equations . . . . . . . . . . . . . . . . . . . . . . . 65

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6.2 Regularity of the weak solution . . . . . . . . . . . . . . . . . . . . . . . . . 686.2.1 First step: from L2 to L∞. . . . . . . . . . . . . . . . . . . . . . . . . 686.2.2 Holder regularity with no source . . . . . . . . . . . . . . . . . . . . . 716.2.3 Holder regularity with source . . . . . . . . . . . . . . . . . . . . . . 75

A Complement to the proof of the local inversion theorem 81

B Complement to the proof of the isometric embedding 83

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Chapter 1

Iterative methods for solvingnonlinear equations

1.1 Picard iteration

Let X, Y Banach spaces, and F : X → Y , F C1 with F (0) = 0.

Theorem 1.1. (local inversion theorem)

If the differential at 0, denoted F ′(0), is an isomorphism from X to Y , there exists an openneighborhood U × V of (0, 0) in X × Y such that F is a C1 diffeomorphism from U to V .

Proof. The idea is to use the following global result.

Lemma 1.1. Let A an isomorphism from X to Y , ϕ : X → Y a Lipschitz map with

Lip(ϕ) <1

||A−1||L(Y,X)

.

Then F = A+ϕ is a bijection from X to Y , with Lipschitz inverse. Moreover, if F is C1 inan open set U of X and if F ′(u) in an isomorphim for all u ∈ U , then F−1 is C1 on F (U).

Proof of the lemma.1 To show that F is a bijection, we have to show that for any f ∈ Y ,there exists a unique u ∈ X such that F (u) = f . We reformulate this equation as

u = A−1f − A−1ϕ(u).

Hence, we have to show that for all f ∈ Y , the map Gf (u) = A−1f −A−1ϕ(u) has a uniquefixed point. This is a consequence of the Banach fixed point theorem: for all u, v ∈ X

‖Gf (u)−Gf (v)‖X = ‖A−1ϕ(u)− A−1ϕ(v)‖X ≤ ‖A−1‖ ‖ϕ(u)− ϕ(v)‖Y≤ ‖A−1‖ Lip(ϕ) ‖u− v‖X

1This proof is borrowed to unpublished lecture notes of Albert Fathi, taught in ENS Lyon in 1997.

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so thatG is a contraction. For the Lipschitz regularity of F−1, we take two elements f = F (u)and f ′ = F (u′) in Y and write

u− u′ = A−1(f − f ′) + A−1(ϕ(u)− ϕ(u′))

so that‖u− u′‖X ≤ ‖A−1‖ ‖f − f ′‖Y + ‖A−1‖Lip(ϕ)‖u− u′‖X ,

and eventually

‖F−1(f)− F−1(f ′)‖X = ‖u− u′‖X ≤‖A−1‖

1− ‖A−1 ‖Lip(ϕ)‖f − f ′‖Y .

Hence, F is bi-Lipschitz. The second part of the lemma follows then easily from LemmaA.1, Appendix A.

Back to the proof of the local inversion theorem. We write A = F ′(0), ϕ = F − A.The application ϕ is C1, with ϕ′(0) = 0. By continuity, there exists r > 0 such thatsupu∈Br ‖ϕ

′(u)‖ < 14‖A−1‖ , where Br (resp. Br) is the open (resp. closed) ball of radius r in

X. This of course implies that over Br, Lip(ϕ) < 14‖A−1‖ . Let ρr : X → Br, defined by

ρr(x) = x, x ∈ Br, ρr(x) = rx

‖x‖, x /∈ Br.

The proof of the following lemma is given in Appendix A:

Lemma 1.2. For any r > 0, ρr is a Lipschitz map, with Lip(ρr) ≤ 2.

We now set ϕr = ϕ ρr. One has ϕr Lipschitz in X, with Lip(ϕr) <1

2‖A−1‖ . By Lemma

1.1, Fr = A+ ϕr is a Lipschitz bijection with Lipschitz inverse. As Fr = F on Br, one findsthat F (Br) is open in F , and F : Br → F (Br) is a Lipschitz bijection with Lipschitz inverse.Moreover, up to take a smaller r, we can always assume that F ′(u) is an isomorphism for allu ∈ Br. By the second part of Lemma 1.1, it yields the C1 regularity of F−1.

Remark 1.1. The previous proof of the local inversion theorem emphasizes the key role ofPicard fixed point theorem. The theorem itself relies on an iteration to construct the fixedpoint. In our context, the iterative process takes the form:

un+1 = A−1f − A−1ϕ(un) ⇔ F ′(0) (un+1 − un) + F (un) = f.

As is well-known, the convergence if of geometric type, that is ‖un− u‖X ≤ Cκn, κ ∈ (0, 1).

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1.2 Newton’s method

Besides Picard iteration, another famous iteration for solving equations is given by Newton’s(or Newton-Raphson’s) method:

F ′(un)(un+1 − un) + F (un) = f.

To study its convergence, we introduce the ”error” εn = ‖F (un) − f‖Y . In particular,taking u0 = 0, we have ε0 = ‖f‖Y . Let V a neighborhood of 0, and c > 0 such thatsupu∈V ‖F ′(u)−1‖L(Y,X) ≤ c. Then, assuming that un ∈ V for all n, we deduce:

(1.1) ‖un+1 − un‖X ≤ cεn

Then,

εn+1 = ‖F (un+1)− f‖Y≤ ‖F (un+1)− F ′(un)(un+1 − un)− F (un)‖Y + ‖F ′(un)(un+1 − un) + F (un)− f‖Y= ‖F (un+1)− F ′(un)(un+1 − un)− F (un)‖Y

Under the assumption that: for all u, v ∈ V ,

(1.2) ‖F (v)− F (u)− F ′(u)(v − u)‖Y ≤ C‖v − u‖2

we obtain

(1.3) εn+1 ≤ C‖un+1 − un‖2 ≤ Cc2ε2n

where the last inequality is deduced from (1.1). Setting ε′n = Cc2εn, the last inequality canbe written as ε′n+1 ≤ (ε′n)2, so that ε′n ≤ (ε′0)2n . Hence, the error converges very fast to zeroif ε′0 < 1, that is if ε0 < 1/(Cc2).

Exercice 1.1. Complete the proof: show rigorously that un ∈ V for all n and all estimatesabove.

1.3 Loss of derivatives

The previous iterations rely crucially on the fact that F ′(0)−1 (or F ′(u)−1 for u near 0) is acontinuous linear operator from Y to X. However, for many functionals F , notably many(nonlinear) partial differential operators, this property is not satisfied. We present here acrude example, as more interesting and involved ones will appear in the next chapters. Weconsider

F (u) = ∂tu+ ∂xu+ u∂xu

acting on functions u = u(t, x), with t ∈ [0, T ], x ∈ R. More precisely, we see F as actingfrom

X = u ∈ C2b ([0, T ]× R), u|t=0 = 0 to Y = C1

b ([0, T ]× R).

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We remind that Ckb ([0, T ]×R) is the space of functions of class Ck over [0, T ]×R such that

‖u‖Ckb := supα∈N2,|α|≤k

sup[0,T ]×R

| ∂αu| < +∞

Clearly, F ′(0) is given by F ′(0)v = ∂tv + ∂xv, and defines a continuous linear operator fromX to Y . As regards its invertibility, for all f ∈ Y , it is easy to verify that

v(t, x) =

∫ t

0

f(s, x− (t− s)) ds

is the unique solution in Y satisfying v|t=0 = 0, and if we denote by F ′(0)−1 the mappingf → v, one has ‖F ′(0)−1f‖Y ≤ C‖f‖Y . But clearly, the lack of regularization in x preventsF ′(0)−1 to send Y into X. This phenomenon is called a loss of derivatives.

As mentioned in the preamble, many situations where the loss of derivatives appearinvolve families of Banach spaces Xs and Ys, where s denotes a regularity index. One canthink of Xs = Cs(T) (1-periodic functions of class Cs), or Xs = Hs(Rn) (Sobolev functionsof index s). The typical pathology corresponds to functionals F : Xs → Ys, such thatF ′(0) : Xs → Ys, but the inverse F ′(0)−1 only exists as a linear operator from Xs to Xs−rfor some r > 0. This loss of regularity in inverting the linearized equation is incompatiblewith the classical Picard or Newton’s schemes: regularity is lost after a finite number ofiterations.

The goal of the following chapters will be to illustrate methods that allow to overcomethis problem. The basic idea is that with a suitable modification, the Newton’s scheme canresist this phenomenon : the quadratic gain in the error can win over the loss of derivatives.Of course, to have a competition between these two mechanisms, one must ”quantify the lossof derivatives”, and manage to encode it into an error estimate like (1.3). We will considertwo cases:

• the case where the index loss r can be taken arbitrarily small, with estimates of thetype: for all s > 0,

‖F ′(0)−1f‖Xs−r ≤ Cr,s‖f‖Y s , limr→0

Cr,s = +∞

Such case is typical of analytic settings, and will be discussed in detail in Chapter 2in the context of circle diffeomorphisms. Roughly, the Xs will be spaces of analyticfunctions with radius of analyticity s, and the Cauchy formula will allow to quantifythe loss of derivatives through estimates of the type:

∀r > 0, ‖η′‖Xs−r ≤1

r‖η‖Xs .

• the case where the index loss r is fixed. This case enters the scope of the celebratedNash-Moser theorem, see Chapter 3. Roughly, the idea behind this theorem is to

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modify the Newton’s scheme by introducing truncations in frequencies. For instance,if we work with Xs = Cs

2π (the space of 2π-periodic functions of class Cs), and definefor N ≥ 0 the truncation operator:

SNf(x) =∑|k|≤N

f(k)eikx, f(k) =1

2π

∫ 2π

0

f(t)e−itdt

we can for instance take advantage of the Bernstein inequality [21], which implies

‖(SNf)′‖L∞ ≤ N‖SN f‖L∞

and in turn∀N ≥ 1, ‖(SNf)′‖Xs ≤ N‖SN f‖Xs

to quantify the loss of derivatives.

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Chapter 2

Conjugacy of circle diffeomorphisms

2.1 Circle maps

Let S1 = z, |z| = 1 the unit circle, with the metric

d(z, z′) = inf|t− t′|, t, t′ ∈ R, z = e2iπt, z′ = e2iπt′

Let Π : R→ S1, t→ e2iπt. Note that it is continuous, and that

Πs : (s− 1

2, s+

1

2)→ S1 \ −e2iπs, Πs(t) = Π(t)

is a homeomorphism for all s ∈ R.

Definition 2.1. (Lift of a circle map)

Let f : S1 → S1. A lift of f is a map F : R→ R such that Π F = f Π.

Theorem 2.1. Any continuous circle map f has a continuous lift. Moreover, if F1 and F2

are two continuous lifts, there exists k ∈ Z such that F1 = F2 + k.

Proof. Uniqueness. If F1 and F2 are two continuous lifts of the same f , the relation ΠF1 =π F2 shows that F1 − F2 takes integer values. But as it is continuous, it is constant (andequal to a fixed integer).

Existence. It is enough to show that there exists F : [0, 1] → R continuous and satisfyingΠF = f Π over [0, 1]. Indeed, one can check that F (t) = F (t−E(t)) +E(t)(F (1)−F (0))is then a continuous lift of f . The existence of such an F is deduced easily from the followingproperty:

There exists ε > 0 such that for all x ∈ [0, 1], if there exists an F : [0, x]→ R continuous andsatisfying Π F = f Π over [0, x], then there exists Fε : [0,min(x + ε, 1)]→ R continuousand satisfying Π Fε = f Π over [0,min(x+ ε, 1)].

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To prove this property, we first express that f is uniformly continuous on S1: there existsε > 0, such that d(z, z′) ≤ ε ⇒ d(f(z), f(z′)) < 1

4. Then, for F as above, the function Fε

defined by

Fε(t) = F (t) on [0, x], Fε(t) = Π−1F (x) f Π(t) on [x,min(x+ ε, 1)]

is suitable.

In what follows, all circle maps will be continuous, and the word lift will refer automaticallyto a continuous lift.

Definition 2.2. (Degree of a circle map)

Let f be a continuous circle map, F a lift of f . The degree of f is deg(f) = F (1)− F (0).

Remark 2.1. (Properties of the degree)

• By the uniqueness part of Theorem 2.1, deg(f) depends only on f , not on the lift.

• By the definition of the lift, e2iπF (1) = f(e2iπ) = f(1) = e2iπF (0) so that deg(f) is aninteger.

• One can show that for any lift F , any x ∈ R and any integer k ∈ Z,

F (x+ k) = F (x) + k deg(f)

Indeed, as F (·+ k) is a lift, there exists nk ∈ Z such that F (·+ k) = F + nk. Clearly,n0 = 0, while nk−nk−1 = F (k)−F (k−1) = F (·+k−1)|x=1−F (·+k−1)|x=0 = deg(f)by definition, because F (·+ k − 1) is also a lift. The formula follows.

• From the previous formula, it is easy to see that deg(f g) = deg(f)deg(g).

Example 2.1. f(z) = zn, n ∈ Z is a circle map, with lift F (t) = nt. Hence, deg(f) = n.

2.2 Circle homeomorphisms

We consider in this section circle homeomorphisms (which, as the circle is compact, areexactly the continuous and bijective circle maps). We say that a homeomorphism f isorientation-preserving (resp. orientation-reversing) if the image by f of any triplet of pointsin anti-clockwise order is in anti-clockwise order (resp. clockwise order). One can see that thisis equivalent to f preserving (resp. reversing) the orientation of one triplet only, cf the proofof the next proposition. In other words, there are only two classes of circle homeomorphisms:those which preserve orientation and those which reverse orientation.

Proposition 2.1. If f is an orientation-preserving circle homeomorphism, then deg(f) = 1,and any lift of F is an increasing homeomorphism from R to R.

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Proof. As f is one-to-one, any lift F is one-to-one over (0, 1), and therefore monotonicover [0, 1]. Moreover, if there exists t ∈ (0, 1) such that F (t) − F (0) ∈ −1,+1, one hasf(e2iπt) = f(1), which again contradicts the fact that f is one-to-one. It follows that eitherF ((0, 1)) ⊂ (F (0), F (0)+1) or F ((0, 1)) ⊂ (F (0)−1, F (0)). As f preserves orientation, F |[0,1]

is increasing and F (1) = F (0)+1, or deg(f) = 1. As F (x+ 1) = F (x) + 1 (see Remark 2.1),it follows that F is an increasing homeomorphism from R to R.

Proposition 2.2. Let f be an orientation-preserving circle homeomorphism, and F a liftof f . Then, the number ρ0(F ) = limn→+∞

Fn(x)n

exists and does not depend on x. Moreover,ρ0(F + k) = ρ0(F ) + k for all k ∈ Z.

Proof. From the property F (x + k) = F (x) + k, see Remark 2.1, it is easily deduced thatF n(x+ k) = F n(x) + k (for all x, for all n, k ∈ Z).

Let us first prove that limnFn(0)n

exists. We remind the following well-known lemma:

Lemma 2.1. Let (an)n∈N be a real sequence satisfying an+m ≤ an + am for all n,m. Then,limn→+∞

ann

= infn∈Nann

exists in R ∪ −∞.

Let x, y ∈ R, and k ∈ Z such that x+ k ≤ y ≤ x+ k + 1. Applying the increasing functionF n, we obtain

F n(x+ k) ≤ F n(y) ≤ F n(x+ k + 1) ⇔ F n(x) + k ≤ F n(y) ≤ F n(x) + k + 1.

By substracting the double inequality x+ k ≤ y ≤ x+ k + 1, we end up with

(2.1) F n(x) + y − x− 1 ≤ F n(y) ≤ F n(x) + y − x+ 1

Taking x = 0, y = Fm(0), and setting an = −F n(0) + 1, we get an+m ≤ an + am. By the

lemma, this shows that limn→+∞Fn(0)n

exists in R ∪ +∞. Moreover, as F k is increasingfor all k ∈ N,

F n(0) = F n−1(F (0)) = F n−1(F (0)− E(F (0))

)+ E(F (0))

≤ F n−1(1) + E(F (0)) ≤ F n−1(0) + 1 + E(F (0)).

It implies that F n(0) ≤ C(n + 1) for some C and for all n, which implies in turn that

limn→+∞Fn(0)n

exists (and is finite).

To conclude the proof of the proposition, we write F n(x) = F n(x− E(x)) + E(x), so that

F n(0) ≤ F n(x− E(x)) ≤ F n(1) = F n(0) + 1.

We getE(x)

n≤ F n(x)

n− F n(0)

n≤ E(x) + 1

n.

Hence for all x, limn→+∞Fn(x)n

exists and is equal to limn→+∞Fn(0)n

. Finally, the relationρ0(F + k) = ρ0(F ) + k is deduced easily from the relation F n(x+ k) = F n(x) + k.

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Exercice 2.1. Show that for any orientation preserving homeomorphism f and continuouslift F of f , the quantity Fn(x)−x

nconverges uniformly in x to ρ0(F ) as n→ +∞.

The previous proposition allows to set the following

Definition 2.3. (rotation number) Let f be an orientation-preserving circle homeomor-phism, F a lift of f . Its rotation number is defined as ρ(f) = ρ0(F ) (mod 1).

Example 2.2. Let α ∈ R, Rα(z) = e2iπαz. A lift of Rα is F (t) = t+ α. Hence, ρ0(F ) = α,and ρ(f) = α mod 1.

Exercice 2.2. Let f a circle homeomorphism that reverses orientation.

1. Show that any lift F of f is decreasing from [0, 1] to [F (0)− 1, F (0)].

2. Show that there exists a lift F of f having a fixed point over [0, 1].

3. Show that for all x, limn→+∞Fn(x)n

= 0.

Proposition 2.3. (A few properties of the rotation number)

• For all n ∈ Z, ρ(fn) = nρ(f).

• f → ρ(f) is continuous from the set of orientation-preserving circle homeomorphisms(endowed with the sup norm) to R/Z.

• ρ(f) ∈ Q if and only if f has a periodic point.

• Let h : S1 → S1 a continuous circle map with deg(h) = 1, and f, g two orientation-preserving homeomorphisms such that h f = g h. Then, ρ(f) = ρ(g). In particular,if h is an orientation-preserving homeomorphism, ρ(h f h−1) = ρ(f).

Proof. Let F be a lift of f . The first item follows from the series of equalities:

ρ0(F n) = limk→+∞

(F n)k(0)

k= n lim

k→+∞

F (nk)(0)

nk= nρ0(F )

For the second item, we fix ε > 0 and n such that 1n< ε. From (2.1), we deduce that

(2.2) F n(0)− 2 < F n(t)− t < F n(0) + 2 ∀t ∈ R.

For homeomorphisms g close enough to f in C0 topology, and for a proper choice of of thelift G of g, one can make ‖F n −Gn‖∞ arbitrarily small. Hence, we can ensure that

(2.3) F n(0)− 2 < Gn(t)− t < F n(0) + 2 ∀t ∈ R.

14

Using that for any k ∈ N:

F nk(0) = F nk(0)− 0 =k−1∑j=0

F n F (jn)(0)− F (jn)(0)

we can deduce from (2.2) that

k(F n(0)− 2) < F nk(0) < k(F n(0) + 2)

and similarly, we can deduce from (2.3)

k(F n(0)− 2) < Gnk(0) < k(F n(0) + 2)

Dividing by nk and letting k go to infinity, we get

F n(0)− 2

n≤ ρ0(F ) ≤ F n(0) + 2

n,

F n(0)− 2

n≤ ρ0(G) ≤ F n(0) + 2

n

which implies |ρ0(F )− ρ0(G)| ≤ 4n≤ 4ε.

We now focus on the third item. If f has a periodic point z = e2iπx of period q, thenΠ F q(x) = Π(x), so that there exists p ∈ Z such that F q(x) = x + p. Inductively, we getthat F nq(x) = x+ np, and ρ0(F ) = p

q.

Conversely, if ρ0(F ) = pq, then, using the first item of the proposition, G = F q − p is a lift of

f q that satisfies ρ0(G) = 0.

• if G(0) > 0, one obtains that (Gn(0))n∈N is an increasing sequence, because G isincreasing. Then, either it is is bounded from above by 1 and converges to a fixedpoint of G, as expected, or there exists k ≥ 1 such that Gk(0) > 1. In the lattercase, we get G2k(0) > Gk(1) > Gk(0) + 1. Inductively, one shows that for all n ≥ 1,Gnk(0) > n, so that ρ0(G) ≥ 1

k, reaching a contradiction.

• if G(0) < 0, one can proceed similarly with the decreasing sequence (Gn(0))n∈N.

As regards the last item, let F,G,H lifts of f, g, h respectively. One can check easily thatH F and G H are two lifts of h f = g h, so that there exists N ∈ Z such thatH F = G H, where G = G+N . This implies H F n = Gn H for all n ∈ Z. Now,

ρ0(G) = limn

Gn(H(x))

n= lim

n

H(F n(x))

n= lim

n

H(F n(x)− E(F n(x)))

n+E(F n(x))

n

where E is the integer part, and the last equality comes from the property: H(x + k) =H(x) + k, see Remark 2.1. As H is bounded on [0, 1], the first term at the right-hand sidegoes to zero, while the second one goes to ρ0(F ). This concludes the proof.

15

Exercice 2.3. Let f be an orientation preserving homeomorphism, with ρ(f) = pq, p ∈ Z∗,

q ∈ N∗, p ∧ q = 1. Show that f has a periodic point of period q, but not of period q′ with0 ≤ q′ < q.

Exercice 2.4. Let f be a circle homeomorphism. For all x ∈ S1, we introduce the ω-limitset of x (under f): ω(x, f) = ∩n∈Nfk(x), k ≥ n.

1. Show that if f is orientation preserving and has a periodic point x0 of period q, thenfor any x ∈ S1, fnq(x) converges to a fixed point of f q as n → +∞ (which is notnecessarily x0). Show that such a point is a periodic point of f of period q.

2. Deduce from the previous question and the previous exercise that if ρ(f) = pq, p ∈ Z∗,

q ∈ N∗, p ∧ q = 1, then for all x ∈ S1, ω(x, f) is a periodic orbit of period q.

We see from the last point of the previous proposition that the rotation number is invariantby conjugacy. It is then natural to wonder if the converse holds, that is if two (orientation-preserving) homeomorphisms having the same rotation number are in the same conjugacyclass. In other words, one can ask if a homeomorphism with rotation number α is conjugatedto the rotation Rα. In the case where α ∈ Q (mod 1), this is clearly not enough: for instance,any orientation-preserving homeomorphism which fixes 1 has rotation number 0, but is notnecessarily the identity. In the case where α is irrational, the problem is deeper. What isalways true is the following weaker result:

Proposition 2.4. Let f be an orientation-preserving diffeomorphism with irrational rotationnumber α. Then, f is semi-conjugated to Rα, which means that there exists a continuouscircle map h, with deg(h) = 1 and a non-decreasing lift (but not necessarily one-to-one),such that h f = Rα h.

Proof. 1 By the Krylov-Bogolyubov theorem [22, p10], there exists a Borel probability mea-sure µ on S1 which is invariant by f . As f is a homeomorphism, this amounts to the identityµ(f(A)) = µ(A) for any Borel set A. As the rotation number of f is irrational, f has noperiodic point, cf Proposition 2.3. Hence, for any a ∈ S1, the fn(a), n ∈ Z, are all distinct.This implies

1 = µ(S1) ≥∑n∈Z

µ(fn(a)) =∑n∈Z

µ(a), so that µ(a) = 0.

Hence, µ has no atom. It follows that the circle map h(x) = Π µ([1, x]), where [1, x] is thearc joining 1 to x anticlockwise, is continuous. It is also orientation-preserving and satisfiesdeg(h) = 1. Note that a lift H of h is given by H(y) = µ([1,Π(y)]) + E(y). Now, for allx ∈ S1, taking A = [1, x] and noticing that f([1, x]) = [f(1), f(x)] because f is orientation-preserving, we deduce that µ([f(1), f(x)]) = µ([1, x]), which implies zh f(x) = h(x), wherez = Π µ([(f(1), 1]). This means that there exists β such that h f(x) = Rβ h(x). But,by Proposition 2.3, ρ(f) = ρ(Rβ), which means α = β, and the result follows.

1This proof borrows to unpublished lecture notes of Albert Fathi, taught in ENS Lyon in 1998.

16

Exercice 2.5. (Proof of the Krylov-Bogolyubov theorem)

Let K a compact metric space, T : K → K continuous. Let ν0 any finite borelian measureon K.

1. For all k ∈ N, we introduce νk = T k∗ ν0, defined by νk(A) = ν0(T−k(A)). Explain whythe sequence of averaged measures (µk = 1

l

∑k−1l=0 νl)k≥1 has a subsequence (µϕ(k))k≥1

that weakly converges to some finite borelian measure µ.

2. Deduce from there that for all continuous function f on K, 1ϕ(k)

∑ϕ(k)−1l=0

∫Kf T l+1dν0

converges to∫Kf Tdµ as k → +∞.

3. Show that it also converges to∫Kfdµ, so that µ is an invariant measure for T .

To determine if f is conjugated to Rα, one can examine under which conditions the circlemap h given by the previous proposition is one-to-one. Note that as α is irrational, Rn1(x) 6=Rn2(x) for all x and all n1 6= n2. It follows that h fn1(x) 6= h fn2(x) for all n1 6= n2. Ifwe can show that f has a dense orbit, then we will find x, n1 6= n2 such that for any z 6= z′

z < fn1(x) < fn2(x) < z′ (the symbol < refering to the trigonometric order). As h preservesorientation, it will follow that h(z) 6= h(z′), proving conjugacy of f to a rotation. We willsee that this is the case when f is regular enough.

Definition 2.4. (Circle diffeomorphism) We say that a circle map f is Ck, for k ∈N∗ ∪ ∞ ∪ ω2, if it has a lift which is Ck over R. It is a circle diffeomorphism of classCk if it is Ck, bijective, with a Ck inverse.

Theorem 2.2. (Denjoy’s theorem) Let f be a circle diffeomorphism of class C2, withirrotational rotation number α. Then, f is topologically conjugated to Rα (that is conjugatedby an orientation-preserving homeomorphism).

Remark 2.2. If h1 and h2 are two homeomorphisms conjugating f to Rα, then

h1 Rα h−11 = h2 Rα h−1

2

so that h = h−11 h2 satisfies Rα h = h Rα, that is : ∀x, e2iπαh(x) = h(e2iπαx). From

there:∀n ∈ Z, e2iπnαh(1) = h(e2iπnα).

By density of (e2iπnα)n∈Z in S1, it follows that h(z) = h(1)z for all z ∈ S1, so that h1 = h2Rβ

for some β. Hence, the conjugacy is ”almost unique”.

Proof. 3 Let h given by Proposition 2.4. Let us assume that h is not one-to-one. Then thereexists a non-trivial arc I on which h is constant. From the relation hf = Rα h, we deducethat h fn = Rn

α h. In particular, h is constant on fn(I). Moreover, as Rn1α (x) 6= Rn2

α (x)

2the notation Cω refers to analytic functions.3This proof borrows to unpublished lecture notes of Albert Fathi, taught in ENS Lyon in 1998.

17

for any x as soon as n1 6= n2, we see that the arcs fn(I), n ∈ Z, must be disjoint. DenotingIn = fn(I) and ln the length of In, this implies in turn that ln → 0 as n→ ±∞. Now,

ln =

∫Π−1(In)∩[0,1]

ds =

∫F−n(Π−1(In)∩[0,1])

(F n)′(t)dt

=

∫F−n(Π−1(In))∩[F−n(0),F−n(1)]

(F n)′(t)dt =

∫Π−1(I)∩[0,1]

(F n)′(t)dt

where we used the 1-periodicity of (F n)′ and the identity : F−n(Π−1(In)) = Π−1(I). Hence,we find that for some θn ∈ Π−1(I) ∩ [0, 1] (with lI the length of I):

ln = (F n)′(θn) lI =n−1∏i=1

F ′(F i(θn)) lI .

With a similar calculation, we find that for some θ′n ∈ Π−1(I−n) ∩ [0, 1],

lI = (F n)′(θ′n) l−n =n−1∏i=1

F ′(F i(θ′n)) ln.

As F ′ > 0, this leads to

ln(ln l−n) =n−1∑i=0

(lnF ′(F i(θn))− lnF ′(F i(θ′n))

)+ 2 ln lI

=n−1∑i=0

(lnF ′

(F i(θn)− E(F i(θn))

)− lnF ′

(F i(θ′n)− E(F i(θ′n))

))+ 2 ln lI

where the last equality comes from the 1-periodicity of F ′.

We now claim that : there exists a strictly increasing sequence of integers nl, such that foreach l ≥ 0, the arcs [f i Π(θnl), f

i Π(θ′nl)], i = 0, . . . , nl−1 are disjoint.

Let us assume temporarily that this claim holds. For all i, for all n,

Π−1[f i Π(θn), f i Π(θ′n)] ∩ [0, 1]

is either a single interval, of the type

[F i(θn)− E(F i(θn)), F i(θ′n)− E(F i(θ′n))]

(possibly switching θn and θ′n), or a union of two disjoint intervals, of the type

[0, F i(θn)− E(F i(θn))] ∪ [F i(θ′n)− E(F i(θ′n)), 2π].

By the claim, when n = nl, and i varies between 0 and nl− 1, we obtain a family of disjointsubintervals of [0, 1]. Moreover, lnF ′ is 1-periodic, and it is C1 (therefore with boundedvariation) over [0, 1]. From there, writing

lnF ′(F i(θn))− lnF ′(F i(θ′n)) = lnF ′(F i(θn)− E(F i(θn))

)− lnF ′

(F i(θ′n)− E(F i(θ′n))

)18

in the case of a single interval and

lnF ′(F i(θn))− lnF ′(F i(θ′n)) = lnF ′(F i(θn)− E(F i(θn))

)− lnF ′(0)

+ lnF ′(1)− lnF ′(F i(θ′n)− E(F i(θ′n))

)in the case of a union of two intervals, we get that the right-hand side of the previous equalityis bounded uniformly in n = nl. But the left hand-side ln(lnl l−nl) → −∞ as l → +∞. Wereach in this way a contradiction.

It remains to prove the claim. We remind once more that h f i = Riα h for all i ∈ Z. As h

is constant on I (say equal to z), we have h Π(θnl) = z and h Π(θ′nl) = R−nlα (z). It followsthat

h([f i Π(θnl), f

i Π(θ′nl)])

= [Riα(t), Ri−nl

α (t)]

It is then enough to show that the intervals at the right hand-side are disjoint for some goodsequence nl and i = 0, . . . , nl−1. There is no loss of generality in taking t = 1. We thenintroduce the following sequence:

n0 = 1, nl+1 = infn ≥ nl, d(1, Rnα(1)) < d(1, Rnl

α (1)).

Note that the infimum is well-defined, by the density of the orbits of Rα (α is irrational).The sequence nl is strictly increasing and goes to infinity. Moreover, for all p1 < p2, p1, p2

in 0, ..., nl − 1, one has

d(Rp1α (1), Rp2

α (1)) = d(1, Rp2−p1α (1)) > d(1, Rnl

α (1)).

Thanks to this property, one can show that the arcs [Riα(1), Ri−nl

α (1)] are disjoint, whichconcludes the proof.

Let us point out that the smoothness assumption on f in Denjoy’s theorem is crucial:Denjoy himself exhibited a counterexample with C1 regularity. On the other hand, theconjugacy h of the theorem is only shown to be continuous. A natural question is then: if fis smooth, is the conjugacy h itself smooth ? We shall in the next sections discuss the specialcase of a perturbative and analytic framework.

2.3 Conjugacy of analytic diffeomorphisms

2.3.1 Setting of the problem

Let f be an analytic diffeomorphism of the circle, orientation-preserving. We want to studythe regularity properties of a map h that conjugates f to the rotation Rα, where α = ρ(f)is in R \Q. We will restrict to the case where f is a small perturbation of Rα. Roughly, wewill assume that f has a lift F of the form F (x) = x+α+ η(x), where η will be small in ananalytic norm (to be specified later on). Let us note that the relation F (x+ 1) = F (x) + 1implies the relation η(x+ 1) = η(x), so that η is 1-periodic.

19

Remark 2.3. We know from the definition of the rotation number that ρ0(F ) = α + k forsome k ∈ Z. Moreover, for η small enough in analytic norm, one can ensure that |ρ0(F )−ρ0(x→ x+ α)| ≤ 1

2, see the proof of the second point in Proposition 2.3. Hence, ρ0(F ) = α,

which puts some constraint on η. Namely, one has F n(x) = x+nα+∑n−1

i=0 η(F i(x)), so that

limn→+∞

1

n

n−1∑i=0

η(F i(x)) = 0.

This obviously implies that η vanishes at some point, a property that will be used later.

As f is a small perturbation of Rα, it is natural to look for the conjugacy h as a smallperturbation of the identity. More precisely, we will look for a lift H under the form H(x) =x + U(x), where the 1-periodic function U will also be small in some appropriate analyticnorm.

Formally, the conjugacy relation writes: for all x ∈ R,

F H(x) = H(x+ α)⇔ H(x) + α + η(H(x)) = H(x+ α)

⇔ U(x+ α)− U(x) = η(x+ U(x))

A natural approach to build a solution U to the last equation is to look for successiveapproximations, taking as a first approximation the solution of the linear equation

(2.4) ∀x ∈ R, U(x+ α)− U(x) = η(x)

As U and η are 1-periodic, it is natural to solve them using Fourier series:

U(x) =∑n∈Z

U(n)e2iπnx, η(x) =∑n∈Z

η(n)e2iπnx.

We find for all n ∈ Z:

(2.5) (e2iπnα − 1)U(n) = η(n).

Two difficulties arise from this linearized approach. First, in the case n = 0, the compatibilitycondition η(0) = 0 is required. More importantly, one faces a small divisor problem whenwriting

U(n) =η(n)

e2iπnα − 1∀n 6= 0.

Indeed, as α is irrational, the denominator does not vanish for n 6= 0, but it can still takearbitrarily small values, which can prevent the convergence of the Fourier series which issupposed to define U .

To control the smallness of the denominator, it is then natural to put some condition on α,which expresses the fact that α can not be approximated too well by rational numbers. A

20

convenient condition is the following diophantine condition: there exists ν > 2, K > 0 suchthat

(HK,ν) ∀(m,n) ∈ Z× Z∗, |α− m

n| ≥ K

|n|ν

As is well-known, this property is generic: more precisely, one has the following

Lemma 2.2. For all ν > 2, the set of α satisfying (HK,ν) is of full measure in R.

Under assumption (HK,ν), the inequality

(2.6) ∀n 6= 0, |e2iπnα − 1| ≥ 4K

|n|ν−1

holds. Indeed, for m ∈ Z such that |nα−m| ≤ 12, one can write

|e2iπnα − 1| = |e2iπ(nα−m) − 1| = 2| sin(π(nα−m))| ≥ 4|nα−m|

using the inequality sin θ ≥ 2πθ, valid for 0 ≤ θ ≤ π

2. The inequality follows.

Still, in spite of this control, the resolution of (2.4) exhibits the phenomenon of loss ofderivatives mentioned in the preamble to these lecture notes. Introducing for instance theperiodic Sobolev norm over 1-periodic functions with zero mean:

‖F‖Hsper

=

(∑n∈Z

|n|2s|F (n)|2)1/2

, s ≥ 0

the inequality (2.6) yields

‖U‖Hsper≤ 1

4K‖η‖Hs+ν−1

per.

As explained in the preamble to the lecture notes, this loss is incompatible with a standardresolution through linearization. Up to our knowledge, this kind of small divisor problemswas first noticed in the context of celestial mechanics. We will see here how to overcomesuch problems in the analytic setting.

2.3.2 A reminder about analytic functions

We will consider here functions that are 1-periodic and analytic over the real line. Moreprecisely, we introduce for all σ > 0:

Bσ = x+ iy, x ∈ R, |y| < σ,Xσ = η C∞, 1-periodic over R,

such that η has an holomorphic extension over Bσ, continuous over Bσ.

21

Note that the extension of η ∈ Xσ is unique by analytique continuation. To lighten notations,we will still denote by η = η(z) such extension. Note also that

(2.7) η(x) = η(x+ 1) ∀x ∈ R ⇒ η(z) = η(z + 1) ∀z ∈ Bσ

by analytic continuation. We infer from this property that functions in Xσ are bounded.More precisely,

‖η‖σ = maxz∈Bσ|η(z)|

makes Xσ a Banach space (using Weierstrass theorem).

We now state a version of the Paley-Wiener theorem adapted to our 1-periodic functions:

Lemma 2.3. Let η ∈ Xσ. Then, for all n ∈ Z,

|η(n)| ≤ e−2πσ|n|‖η‖σ

Proof. We introduce the closed and rectangular oriented contour γ− defined as the concate-nation of the segments [0, 1], [1, 1− iσ], [1− iσ,−iσ] and [−iσ, 0]. By Cauchy’s theorem, wehave for all n ≥ 0: ∫

γ−

η(z)e−2iπnz dz = 0.

By 1-periodicity of η, see (2.7), the integrals over the two vertical boundaries of γ− cancelout. We are left with ∫

[0,1]

η(z)e−2iπnzdz =

∫[−iσ,1−iσ]

η(z)e−2iπnzdz.

The left hand-side is exactly η(n), while the right-hand side has its modulus bounded by‖η‖σe−2πn|σ|. This gives the desired inequality for all n ≥ 0. The inequality for n < 0 isobtained using the contour γ+, symmetric of γ− with respect to the x-axis.

Remark 2.4. Conversely, if η is a 1-periodic function satisfying |η(n)| ≤ C e−2πσ|n| for alln, it can be shown that it is an element of Xσ′ for all 0 < σ′ < σ.

The next lemma will be crucial to overcome the loss of derivatives.

Lemma 2.4. Let η ∈ Xσ. For all 0 < δ < σ, η′ ∈ Xσ−δ, with the estimate

(2.8) ‖η′‖σ−δ ≤1

δ‖η‖σ

Proof. This is a consequence of the Cauchy formula. For z ∈ Bσ−δ, and γ = C(z, δ), therelation

η′(z) =1

2iπ

∫γ

η(ξ)

(ξ − z)2dξ

implies the formula.

22

2.3.3 Arnold’s theorem

The main theorem of the chapter is

Theorem 2.3. Let ν > 2, K > 0 and α satisfying (HK,ν). Let f be an orientation-preservinghomeomorphism with ρ(f) = α. For all σ > 0, there exists ε = ε(K, ν, σ) such that if f hasa lift F satisfying:

F (x) = x+ α + η(x) with ‖η‖σ ≤ ε

then f is conjugated to Rα by an analytic orientation-preserving diffeomorphism.

The next section will be devoted to the detailed proof of the theorem. The general pictureis as follows. The aim is to construct the analytic conjugacy h (more precisely its lift H)through an iterative process. One first sets F0(x) = F (x) = x + α + η(x). Then, givenFn(x) = x+ α + ηn(x), one builds

• The solution Un of Un(x+ α)− Un(x) = ηn(x)− ηn(0).

• Hn(x) = x+ Un(x)

• Fn+1 = H−1n Fn Hn = (H1 · · · Hn)−1 F (H1 · · · Hn).

Roughly, the goal is then to show that Hn = H1 · · · Hn converges to an analytic diffeo-morphism H, while Fn converges to the rotation, meaning that the error ηn converges to 0.A keypoint in establishing this convergence is to show an estimate of the type

‖ηn+1‖σn−δn+1 ≤C

δMn‖ηn‖2

σn

which encodes both the loss of derivatives through the denominator δMn and the quadraticconvergence of the error. Roughly, the idea will be to take δn = δ0

n2 , δ0 1, so that losingδ0 + δ1 + ... in the regularity index through the iteration process will leave us with someregularity. Meanwhile, we expect as in the Newton’s scheme a decay of the error like ε2n

0 . Inparticular, such fast decay will win over the growth induced by 1

δMn. Obviously, the exponent

2 (more generally the fact that the exponent is greater than 1) at the right hand-side iscrucial: an estimate like ‖ηn+1‖σn−δn+1 ≤ C

δMn‖ηn‖σn has no chance to yield a decay of ηn.

2.4 Proof of Arnold’s theorem

The proof presented here follows the lines of paper [25], which is a nice introduction to KAMtheory from an analytic viewpoint. We first gather all estimates that will be used at eachstep of the iteration alluded to above.

Proposition 2.5. Let σ > 0, η ∈ Xσ, δ ∈ (0,min(σ6, 1)). Let F (x) = x + α + η(x), with

ρ0(F ) = α.

23

1. The equationU(·+ α)− U = η − η(0)

has a unique solution with zero mean in Xσ−δ, satisfying

(2.9) ‖U‖σ−δ ≤Cν,Kδν‖η‖σ

2. There exists c = cν,K such that under the assumption ‖η‖σ ≤ cδν+1, the followingstatements hold:

• H(z) = z + U(z) is an analytic diffeomorphism from Bσ−2δ to its range. Thisrange contains Bσ−3δ.

• V (z) = H−1(z)− (z − U(z)) is in Xσ−4δ, with estimate

(2.10) ‖V ‖σ−4δ ≤Cν,Kδ2ν+1

‖η‖2σ.

• F = H−1 F H can be written as F (x) = x+ α + η(x) with η ∈ Xσ−6δ and

(2.11) ‖η‖σ−6δ ≤Cν,Kδ2ν+1

‖η‖2σ.

Proof.

1. Following the end of subsection 2.3.1, we know that if U =∑

n∈Z∗ U(n)e2iπnx is a

solution of (2.4), then for all n ∈ Z∗, U(n) = η(n)e2iπnα−1

. By the lower bound (2.6) onthe denominator, and the upper bound on the numerator given by Lemma 2.3, we get,for all z ∈ Bσ−δ:

| U(n)e2iπnz| ≤ |η(n)||e2iπnα − 1|

e2π|nImz| ≤ ‖η‖σe−2π|n| δ

4K|n|ν−1.

This shows that the Fourier series converges absolutely overBσ−δ. It provides a solutionto (2.4), with estimate

‖U‖σ−δ ≤‖η‖σ4K

∑n∈Z∗

e−2π|n| δ|n|ν−1 =‖η‖σ

4Kδν−1

∑n∈Z∗

e−2π|n| δ|δn|ν−1

One can check that there exists M = Mν such that the function gν : x→ e−2πxxν−1 isincreasing for 0 ≤ x ≤ Mν and decreasing for x ≥ Mν . We can then split the sum atthe right hand-side between indices n with |nδ| ≤Mν and those with |nδ| ≥Mν . Thefirst sum can be bounded through:∑

n, |n δ| ≤Mν

e−2π|n| δ|δn|ν−1 ≤∑

n, |n δ| ≤Mν

gν(Mν) ≤2Mν

δgν(Mν)

while the second sum can be compared to the integral∫|xδ|≥Mν

e−2πxδ|δx|ν−1 dx = Cνδ

.

Estimate (2.9) follows.

24

2. To prove that H(z) = z+U(z) is an analytic diffeomorphim from Bσ−2δ onto its range,it is enough to show that |U ′(z)| < 1 for all z ∈ Bσ−2δ: indeed, it follows easily fromthis property that H is one-to-one, and a local analytic diffeomorphism (by the localinversion theorem in its analytic version). By (2.8),

‖U ′‖σ−2δ ≤1

δ‖U‖σ−δ ≤

Cν,Kδν+1‖η‖σ

where the last inequality comes from (2.9). If ‖η‖σ ≤ δν+1

2Cν,K, then ‖U‖σ−δ ≤ δ

2, and

‖U ′‖σ−2δ ≤ 12.

To show that Bσ−3δ ⊂ H(Bσ−2δ), it is enough to show that for all y ∈ Bσ−3δ, thefunction z → y−U(z) has a fixed point in Bσ−2δ. As ‖U‖σ−δ ≤ δ

2, this mapping sends

Bσ−2δ to itself, and the inequality ‖U ′‖σ−2δ < 1 implies that it is a contraction overthis strip. The result follows from Picard fixed point theorem.

Let now V (z) = H−1(z)− (z − U(z)), which is well-defined on Bσ−3δ by the previousitem. We write

z = H−1 H(z) = H−1(z + U(z)) = z + U(z)− U(z + U(z)) + V (z + U(z)),

so that

V (z + U(z)) = U(z + U(z))− U(z) =

∫ 1

0

U ′(z + sU(z))U(z)ds.

Setting ξ = z + U(z) = H(z), we find

V (ξ) =

∫ 1

0

U ′(H−1(ξ) + sU(H−1(ξ))

)U(H−1(ξ))ds.

Note that, in the same way that Bσ−3δ ⊂ H(Bσ−2δ), one has Bσ−4δ ⊂ H(Bσ−3δ), sothat the previous formula for V (ξ) holds for all ξ ∈ Bσ−4δ. Moreover, H−1(ξ) ∈ Bσ−3δ,and H−1(ξ) + sU(H−1(ξ)) ∈ Bσ−2δ. Eventually,

|V (ξ)| ≤ ‖U ′‖σ−2δ‖U‖σ−3δ ≤ ‖U ′‖σ−2δ‖U‖σ−δ ≤C ′ν,Kδ2ν+1

‖η‖2σ.

The estimate (2.10) follows.

25

To conclude the proof, we compute

F (x) = H−1 F H(x) = H−1(x+ U(x) + α + η(x+ U(x))

)= x+ U(x) + α + η(x+ U(x))− U

(x+ U(x) + α + η(x+ U(x))

)+ V

(x+ U(x) + α + η(x+ U(x))

)= x+ α + η(x) +

(U(x)− U(x+ α)

)+(η(x+ U(x))− η(x)

)+(U(x+ α)− U

(x+ α + U(x) + η(x+ U(x))

))+ V

(x+ U(x) + α + η(x+ U(x))

).

Hence, we can write F (x) = x+ α + η(x), where

η(z) = η(0) +(η(z + U(z))− η(z)

)+(U(z + α)− U

(z + α + U(z) + η(z + U(z))

))+ V

(z + U(z) + α + η(z + U(z))

)= η(0) + I(z) + II(z) + III(z).

It remains to show that the right-hand side is well-defined for z ∈ Bσ−6δ, holomorphicin the interior and continuous on the closure, with sup norm bounded by

Cν,Kδ2ν+1‖η‖2

σ.

• We have seen that under the condition ‖η‖σ ≤ δν+1

2Cν,K, one has ‖U‖σ−δ ≤ δ

2. It

follows easily that I is defined over Bσ−δ.Moreover,

I(z) =

∫ 1

0

η′(z + sU(z))U(z) ds

so that

‖I‖σ−2δ ≤ ‖η′‖σ−δ‖U‖σ−2δ ≤Cν,Kδν+1‖η‖2

σ

by (2.8) and (2.9).

• Using that ‖U‖σ−δ ≤ δ2

and taking ‖η‖σ ≤ min(

12Cν,K

, 1)δν+1 ≤ δ, we see easily

that II is defined over Bσ−3δ. Moreover,

II(z) =

∫ 1

0

U ′(z + α + s

(U(z) + η(z + U(z))

)) (U(z) + η(z + U(z))

)ds

so that

‖II‖σ−4δ ≤ ‖U ′‖σ−2δ (‖U‖σ−4δ + ‖η‖σ−3δ)

≤ Cν,Kδν+1

(Cν,Kδν

+ 1

)‖η‖2

σ ≤C ′ν,Kδ2ν+1

‖η‖2σ.

26

• With similar arguments as before, for z ∈ Bσ−6δ, z + α + U(z) + η(z + U(z)) ∈Bσ−4δ, so that III is well-defined in Bσ−6δ, and satisfies

‖III‖σ−6δ ≤ ‖V ‖σ−4δ ≤Cν,Kδ2ν+1

‖η‖2σ

by (2.10).

• It remains to evaluate η(0). To that purpose, we notice that ρ0(F ) = ρ0(F ) = α(see the proof of the last point in Proposition 2.3). By Remark 2.3, there existsx0 such that η(x0) = 0. It follows that

η(0) = −I(x0)− II(x0)− III(x0)

and by the previous estimates, |η(0)| ≤ Cν,Kδ2ν+1‖η‖2

σ.

Putting together all estimates yields (2.11).

Thanks to Proposition 2.5, we are now in order to prove Arnold’s theorem. As sketched inParagraph 2.3.3, conjugacy to the rotation is obtained as the limit of an induction process,starting from F0 = F , η0 = η: F0(x) = x+α+η0(x). The induction relation is the following:given Fn(x) = x+ α + ηn(x), we solve

Un(·+ α)− Un = ηn − ηn(0)

and then set

Hn(x) = x+ Un(x), Fn+1(x) = H−1n Fn Hn(x) = x+ α + ηn+1(x).

More precisely, let δn = min(σ,1)2π2n2 , n ≥ 1, so that in particular

∑n≥1 δn ≤

min(σ,1)12

. Let

εn = ε( 3

2)n

0 , n ≥ 0, ε0 > 0. We shall show inductively that for a good choice of ε0 (dependingon ν,K) and a smallness assumption on ‖η‖σ, the following property holds for all n ≥ 0:

(Pn)ηn defines an element of Xσn where σn = σ − 6

n∑k=1

δk, ‖ηn‖σn ≤ εn,

and ‖ηn‖σn ≤ c δν+1n+1 (see Proposition 2.5 for the definition of c).

First, P0 is clearly satisfied under the smallness assumption ‖η‖σ ≤ ε0 for any choice ofε0. Now, assume Pn. To establish Pn+1, we first notice that our choice of δn ensuresthat δn+1 ≤ min

(σn6, 1). This, together with inequality ‖ηn‖σn ≤ c δν+1

n+1, allows to applyProposition 2.5 with σ = σn and δ = δn+1: it shows that the function ηn+1 introduced aboveis well-defined as an element of Xσn−6δn+1 = Xσn+1 , and that

(2.12) ‖ηn+1‖σn+1 ≤Cν,K

δ2ν+1n+1

‖ηn‖2σn .

27

As ‖ηn‖σn ≤ εn, this last inequality yields:

‖ηn+1‖σn+1 ≤Cν,K

δ2ν+1n+1

(εn)2.

In particular, ‖ηn+1‖σn+1 ≤ εn+1 holds wheneverCν,K

δ2ν+1n+1

(εn)2 ≤ εn+1, which can be written in

the formC ′ν,K(n+ 1)4ν+2ε

12

( 32

)n

0 ≤ 1

and is guaranteed for ε0 small enough (depending only on ν,K). Moreover, still by (Pn) and(2.12), we have

‖ηn+1‖σn+1 ≤Cν,K

δ2ν+1n+1

(cδν+1n+1

)εn

so that we obtain ‖ηn+1‖σn+1 ≤ c δν+1n+2 whenever

Cν,Kδνn+1

εn ≤ δν+1n+2, which can be again guaran-

teed for ε0 small enough (depending on ν,K). This proves Pn+1.

Conclusion of the proof of Arnold’s theorem.

From Proposition 2.5 and the properties of ηn, one can check that Hn is analytic on Bσn−2δn+1

with range included in Bσn−δn+1 ⊂ Bσn−1−2δn . Hence, if we set Hn = H1 · · · Hn,then Hn

is well-defined and analytic over Bσn−2δn+1 , and by construction: Fn+1 = H−1n F Hn. We

compute

Hn(z) = H1 · · · Hn−1(z + Un(z))

= H1 · · · Hn−2(z + Un(z) + Un−1(z + Un(z)))

= z + Un(z) + Un−1(z + Un(z)) + Un−2(z + Un(z) + Un−1(z + Un(z))) + . . .

We deduce from the previous writing that

‖Hn − z‖σn−δn+1 ≤n∑k=0

‖Uk‖σk−2δk+1

≤ Cν,K

n∑k=0

‖ηk‖σkδνk+1

≤ Cν,K

n∑k=0

εkδνk+1

≤ C ′ν,Kε0.

Moreover, for all n,

Hn+1(z)−Hn(z) = Hn Hn+1(z)−Hn(z) =

∫ 1

0

H′n(z + sUn+1(z))Un+1(z)ds

so that

‖Hn+1 −Hn‖σn+1−2δn+2 ≤ ‖H′n‖σn+1−δn+2‖Un+1‖σn+1−2δn+2

≤ ‖H′n‖σn−3δn+1 ‖Un+1‖σn+1−δn+2

≤ (‖(Hn − z)′‖σn−3δn+1 + 1) ‖Un+1‖σn+1−δn+2 ≤(C ′ν,Kε0

δn+1

+ 1

)Cν,Kδνn+2

εn.

28

This implies that (Hn)n∈N is a Cauchy sequence in Xσ∗ , with σ∗ = limn σn, therefore con-verging to an element H in Xσ∗ . Moreover, the estimate on Hn − z shows that

‖H − z‖σ∗ ≤ C ′ν,Kε0

By (2.8), it follows that

‖H′ − 1‖σ∗−δ∗ ≤C ′ν,Kδ∗

ε0,

where (for instance) δ∗ = σ∗4

. Reasoning as in Proposition (2.5) (to establish the propertiesof H), we can deduce from the two previous inequalities that for ε0 small enough (dependingon ν,K, and also on σ∗ therefore on σ), H is invertible from Bσ∗−δ∗ onto its range, and thatthis range contains Bσ∗−2δ∗ . Eventually:

F H(x) = limnF Hn(x) = lim

nHn Fn(x) = lim

nHn(x+ α + ηn(x)) = H(x+ α)

for all x ∈ R. This concludes the proof.

29

Chapter 3

The Nash-Moser theorem

The previous chapter was dedicated to the phenomenon of loss of derivatives in an analyticframework. The main point in solving the problem was to establish an error estimate of theform

‖ηn+1‖σ−δ ≤C

δM‖ηn‖2

σ,

for arbitrary small δ. The fact that δ could be arbitrarily close to zero was used by takingδ = δn with limn→+∞ δn = 0. In this chapter, we are rather interested to spaces of functionswith finite regularity, where the error estimate takes the form

(3.1) ‖ηn+1‖s−r ≤ Cs,r‖ηn‖2s, for some fixed r.

As one can not play on the index loss r anymore, another strategy is needed to handlethe loss of derivatives. We will describe the famous approach introduced by Nash [18]and further developped by Moser [14, 15, 16]. The crude idea behind this approach is tomodify the Newton’s scheme by adding some frequency truncation operators in the iterationformula. The goal of these truncations is to control at iteration n the support of the Fouriertransform of ηn. Note that we will consider only 1-periodic functions: the Fourier transformf(n), n ∈ Z, will refer to the Fourier coefficient n of the 1-periodic function f .

To explain a little more the idea, let us assume that, for all n, the error ηn is a 1-periodicfunction with Fourier support included in a ball of radius Nn:

ηn(x) =∑|k| ≤Nn

ηn(k)e2iπk·x, x ∈ Rd.

Introducing the Sobolev norm

‖η‖s =

(∑k∈Z

〈k〉2s|η(k)|2)1/2

< +∞ (s ∈ R, 〈x〉 =√

1 + |x|2)

one has easily: ‖ηn‖s ≤ 〈Nn〉r‖ηn‖s−r, so that an estimate like (3.1) turns into

‖ηn+1‖s−r ≤ Cs,r〈Nn〉2r‖ηn‖2s−r.

30

Hence, if the scheme under consideration (itself related to the equation that we try to solve)ensures that the radius Nn grows subquadratically, typically if Nn+1 . Nµ

n for some µ < 2,then the error will converge to zero for η0 small enough.

3.1 Statement of the Nash-Moser theorem

The Nash-Moser theorem reflects a methodology introduced by Nash and Moser for theresolution of perturbative problems: the goal is to solve an equation of the form F (u) = f ,with F (0) = 0 and f close to zero.

Notation. In what follows, for any two families of norms ‖ ‖si and ‖ ‖tk , we shall use thenotation

‖F1(u1, . . . um)‖s1 + · · ·+ ‖Fk(u1, . . . um)‖sk . ‖G1(v1, . . . vn)‖t1 + · · ·+ ‖Gl(v1, . . . vn)‖tl

whenever

‖F1(u1, . . . um)‖s1 + · · ·+ ‖Fk(u1, . . . um)‖sk ≤ C(‖G1(v1, . . . vn)‖t1 + · · ·+ ‖Gl(v1, . . . vn)‖tl

)for a constant C that depends only on s1, . . . , sk, t1, . . . , tl.

Functional framework.

We consider a family of Banach spaces (Xs, ‖ ‖s)s≥0 with the following properties:

i) (Xs)s≥0 is decreasing, with ‖ · ‖s ≤ ‖ · ‖s′ for s ≤ s′.

ii) There is δ > 0, and a sequence (TN)N≥1 of linear operators from X0 to X∞ = ∩s≥0Xs

such that:

‖TNu‖s′ . N s′−s+δ‖u‖s ∀s′ ≥ s,∀u ∈ Xs,

‖(Id− TN)u‖s′ . N s′−s+δ‖u‖s ∀s′ < s,∀u ∈ Xs,

Besides (Xs, || ||s), we consider another family of decreasing Banach spaces (Ys, | |s)s≥0, with| · |s ≤ | · |s′ for s ≤ s′.

Remark 3.1. One can consider families (Xs) and (Ys) either indexed by s ∈ R+, or by s ina subset of R+, typically N.

Examples of families satisfying i) and ii) are (Hs(Td))s≥0, see the exercise below, or (Hs(Rd))s≥0:see chapter 5 for more. In those examples, one can take δ = 0 in i) and ii). The family(Cs(Td))s∈N is also adapted, with for instance δ = d (exercise).

Assumptions on F .

Let k ≥ 2r ≥ 0, and F : Bk → Yk−r of class C1, where Bk is an open ball centered at 0 inXk. We make the folllowing assumptions

31

1. (C1,1 condition)

For all u, v ∈ Xk such that u, u+ v ∈ Bk,

| F (u+ v)− F (u)− F ′(u)v|k−r . ‖v‖2k.

2. (C1 condition)

For all u, v ∈ Xk such that u ∈ Bk,

| F ′(u)v|k−r . ‖v‖k

3. (C0 condition)

For all s ≥ k, for all u ∈ Xs ∩Bk,

| F (u)|s−r . ‖u‖s

4. (right invertibility condition)

For all u ∈ X∞ ∩Bk, there exists L(u) such that: for all s ≥ k − r,

L(u) : Ys → Xs−r, F ′(u)L(u)|Ys+2r = Id|Ys+2r

and such that for all g ∈ Ys,

‖L(u)g‖s−r . | g|s + | g|k−r‖u‖s

Remark 3.2. Note that conditions 1. and 2. must hold only for the index k, while conditions3. and 4. hold for all s ≥ k.

As will be clear from the proof, the keypoint in conditions 3. and 4. is the fact that theinequalities are subquadratic (and even linear) in the high norm: for instance, there is noterm like ‖u‖2

s in 3. or no term like ‖u‖s| g|s in 4. Such estimates, linear in the high norm,are called tame estimates. We shall come back to this notion in the next chapters.

Theorem 3.1. Under the assumptions 1. to 4. above, there exists k′ ≥ k − r, such thatfor all f ∈ Yk′ with | f |k−r small enough (the smallness condition depending on |f |k′), thereexists a solution u in Bk of F (u) = f .

By depending on |f |k′ , we mean that for all M > 0, one can find ε = ε(M) > 0 such that:for all f ∈ Yk′ with | f |k′ ≤M and | f |k−r ≤ ε, there exists a solution.

Remark 3.3. This statement of the Nash-Moser theorem and its proof follow the article [14]by J. Moser, see also [24]. A refined statement (due to L. Hormander) can be found in [1].See also [10].

32

Exercice 3.1. (Sobolev spaces of periodic functions)

Let s ∈ R+. We define

Hs(Td) = u ∈ L2loc(Rd), 1-periodic, such that

∑n∈Zd〈n〉2s|u(n)|2 < +∞

where 〈x〉 =√

1 + |x|2, and u(n) =∫

[0,1]du(t)e−2iπn·tdt, ∀n ∈ Zd.

1. Show that Hs(Td) is a Banach space with the norm ‖u‖s =(∑

n∈Zd〈n〉2s|u(n)|2)1/2

.

2. Show that for s ∈ N, Hs(Td) is the completion of C∞(Td) for the norm

‖u‖Hs =

∑α∈Nd,|α|≤s

‖∂αu‖2L2

.

Show that ‖ · ‖s and ‖ · ‖Hs are equivalent.

3. Show that the operators TN , N ≥ 1 defined by

TNf =∑

n∈Zd,|n|≤N

u(n)e2iπn·x

satisfy the properties ii) mentioned in the functional framework above, with δ = 0.

4. Show that for all s ∈ N and any ε > 0, Hs+ d2

+ε is continuously embedded in Cs(Td).

3.2 Proof of the Nash-Moser theorem

To lighten a bit the notations, we shall only consider the case where the exponent δ in thebounds satisfied by (TN)N≥1 is zero. We start by collecting a few lemmas:

Lemma 3.1. Let µ ≥ 6r, and Λ > 15r + 92µ+ 3. Let c > 0, n∗ ∈ N ∪ +∞, N0 > 0.

Let (un)n∈N be a sequence in X∞ such that for all n ∈ [|1, n∗|]:

(3.2) ‖un+1 − un‖k ≤ c(N3rn ‖un − un−1‖2

k +N5r−Λn NΛ

n−1

), Nn = N

( 32

)n

0 .

If N0 ≥ 4c2 and Nµ1 ‖u1 − u0‖k ≤ 1

2c, then

Nµn+1‖un+1 − un‖k ≤

1

2c, ∀n ∈ [|0, n∗|].

Proof. We set δn = Nµn‖un − un−1‖k for n ∈ [|1, n∗|]. Inequality (3.2) can be written as

δn+1 ≤ c(N

3r−µ2

n δ2n +N

152r+ 9

4µ−Λ

2n−1

)33

Note that by our choice of µ, 3r− µ2≤ 0, and by our choice of Λ, 15

2r+ 9

4µ− Λ

2< −3

2. Thus,

we obtain

δn+1 ≤ c

(δ2n +

1

Nn

).

Under the assumptions N0 ≥ 4c2 and δ1 <12c

, an easy induction shows that δn+1 <12c

for alln ∈ [1, n∗], which concludes the proof.

Lemma 3.2. Let Λ ∈ N∗, M > 0, and f ∈ Yk−r+Λ, |f |k−r+Λ ≤ M . Then, there exists apositive constant a (depending on Λ and M) such that: for all N ≥ 1, for all u ∈ X∞:

∀λ ∈ [|0,Λ|], ‖u‖k+λ ≤ a−1Nλ ⇒ ∀λ ∈ [|0,Λ|], ‖L(u)(F (u)− f)‖k−2r+λ ≤ aNλ

Proof. Let u as in the statement of the Lemma.Taking λ = 0 and a > 1, we ensure thatu ∈ Bk, so that L(u) is well-defined. By the properties 3. and 4. satisfied by F , we get

‖L(u)(F (u)− f)‖k−2r+λ . | F (u)− f |k−r+λ + | F (u)− f |k−r‖u‖k−r+λ. ‖u‖k+λ + |f |k−r+λ + (‖u‖k + |f |k−r)‖u‖k+λ.

By the assumption on u, we find that

‖L(u)(F (u)− f)‖k−2r+λ ≤ Cλ(a−1Nλ + |f |k−r+λ + a−1

(a−1 + |f |k−r

)Nλ)

≤ aNλ

taking a large enough so that:(maxλ∈[|0,Λ|]

Cλ

)(a−1 + M + a−1

(a−1 +M

))≤ a.

We now turn to the core of the proof of the Nash-Moser theorem. As mentioned earlier, itconsists in a regularized Newton’s scheme, involving truncation operators. More precisely,

we introduce Nn = N( 3

2)n

0 , N0 to be specified later, and then consider the following scheme:

(3.3) u0 = 0, un+1 = un − TNn+1L(un) (f − F (un))

Note that due to the application of the truncation operator, un will be in the range of TNnand therefore in X∞. The scheme will be well-defined as long as un ∈ Bk. We will show thatunder suitable assumptions on f , and a suitable choice of N0, it will be globally well-definedand will converge to a solution u of F (u) = f .

One main step is to show that the sequence (un) satisfies an inequality of the type (3.2). Wefix Λ ∈ N∗ as in Lemma 3.1 and then a as in Lemma 3.2. We introduce n∗ the biggest index(possibly infinite) such that: for all n ∈ [|0, n∗|]

∀λ ∈ [0,Λ], ‖un‖k+λ ≤1

2a−1Nλ

n .

34

From (3.3), we deduce: for all n ∈ [|0, n∗|],

(3.4)‖un+1 − un‖k = ‖TNn+1L(un) (f − F (un)) ‖k . N2r

n+1‖L(un) (f − F (un)) ‖k−2r

. N2rn+1| f − F (un)|k−r

where the last inequality comes from assumption 4. To evaluate | f − F (un)|k−r, we write,for n ∈ [|1, n∗|]

| F (un)− f |k−r ≤ | F (un)− F (un−1)− F ′(un−1)(un − un−1)|k−r+ |F (un−1) + F ′(un−1)(un − un−1)− f |k−r. ‖un − un−1‖2

k + | F ′(un−1)(Id− TNn)L(un−1)(f − F (un−1))|k−r.

where the first term at the right hand-side of the last inequality comes from assumption 1.,while the second term comes from formula (3.3). By assumption 3. we get:

| F (un)− f |k−r . ‖un − un−1‖2k + ‖(Id− TNn)L(un−1)(f − F (un−1))‖k.

To handle the second term, we proceed as follows: taking advantage of the properties of thetruncation operator, cf property ii), we write

∀λ, ‖(Id− TNn)L(un−1)(f − F (un−1))‖k . N2r−λn ‖L(un−1)(f − F (un−1))‖k−2r+λ

for λ ≥ 2r. By the definition of n∗ and Lemma 3.2 we get:

∀λ ∈ [2r,Λ], ‖(Id− TNn)L(un−1)(f − F (un−1))‖k ≤ Cλ aN2r−λn Nλ

n−1.

Hence, for all n ∈ [|1, n∗|], for λ = Λ,

‖un+1 − un‖k ≤ c(N2rn+1‖un − un−1‖2

k +N5r−Λn NΛ

n−1

)≤ c

(N3rn ‖un − un−1‖2

k +N5r−Λn NΛ

n−1

)where the constant c depends on Λ and a.

We can apply Lemma 3.1, which yields: if N0 ≥ 4c2, and Nµ1 ‖u1 − u0‖k = Nµ

1 ‖u1‖k ≤ 12c

,then for all n ∈ [|0, n∗|],

(3.5) ‖un+1 − un‖k ≤1

2cN−µn+1

Note that by inequality (3.4) with n = 0,

‖u1‖k ≤ CN2r1 | f |k−r

35

so that for any N0, the smallness condition will be satisfied by taking | f |k−r is small enough.To conclude we notice that : for all finite n ∈ [|0, n∗|]

‖un+1‖k+λ ≤n∑j=0

‖uj+1 − uj‖k+λ

≤ Cλ

n∑j=0

Nλj+1 ‖uj+1 − uj‖k

≤ Cλ2cNλn+1

n∑j=0

N−µj+1 ≤ aNλn+1

if we take N0 large enough so that maxλCλ2c

∑nj=1 N

−µj+1 < a.

This shows that n∗ = +∞, and that the estimate (3.5) holds for all n. From there, deducingthe convergence of un to a solution of F (u) = f is easy, and left to the reader.

36

Chapter 4

The isometric embedding ofriemannian manifolds

We assume here that the reader has a basic knowledge of riemannian geometry (essentiallythat he knows the classical definitions). If not, he can jump to section 4.2, where the nonlinearPDE (4.1) below is solved thanks to the Nash-Moser theorem, without any reference togeometrical objects.

The general problem is the following: given a smooth manifold M of dimension d, equippedwith a riemannian metrics of class Ck, can we find some integer n and a Ck isometricembedding u : M → Rn ?

We remind a few definitions to clarify this statement:

Definition 4.1. (immersion, embedding)

A C1 mapping u : M → Rn is an immersion if for any x in M , the differential u′(x) :TxM → Rn is injective. We say that u is an embedding if it is an immersion and if it is anhomeomorphism onto its image.

The fact that u is an immersion ensures that u is a local embedding, but not necessarilyglobal. In other words, an embedding excludes self-intersection, while an immersion allowsfor it. Note that in the case where M is a compact manifold, an injective immersion is anembedding.

Definition 4.2. For (M, g) and (N, h) two riemannian manifolds, a C1 mapping u : M → Nis an isometry if u∗h = g (the pullback of h by u is g). In local coordinates, this reads

h

(∂u

∂xi,∂u

∂xj

)= g

(∂

∂xi,∂

∂xj

)(= gij), ∀i ≤ j = 1...d

Note that when (N, h) = (Rn, e) with e the usual euclidean scalar product, the previousequation reads

(4.1)∂u

∂xi· ∂u∂xj

= gij, 1 ≤ i ≤ j ≤ d.

37

The isometric embedding problem enters a more general framework, that is the correspon-dence between abstract manifolds and submanifolds of Rn. In this regards, we remind thefamous theorem due to Whitney:

Theorem 4.1. A smooth manifold of dimension d can be smoothly embedded in R2d.

The exponent 2d in the theorem of Whitney can be compared to the dimension suggestedby the isometry constraints (4.1). Namely, (4.1) is a system of sd = d(d+1)

2equations in n

unknowns (the components of u), so that it is very natural that n ≥ sd.

The isometric embedding problem was formulated by Schlafli in 1871, and has a long history.We refer for instance to the survey [27], or [23]. In these lecture notes, we want to give someinsight into the breakthrough result of Nash (1956):

Theorem 4.2. For k ≥ 3, and n ≥ 3sd + 4d, any smooth compact manifold of dimension dwith Ck metrics can be Ck isometrically embedded in Rn.

The exponent was later improved by Gromov and Rohlin, down to n ≥ sd + d + max(d, 5),and the case of non-compact manifold was treated as well.

In the rest of the chapter, we will establish a weaker version, not caring about the lowerbounds on k and n. Namely, for a given k, we will show the existence of n and k′ ≥ k suchthat for any g in Ck′, there exists a Ck isometric embedding u from M to Rn. The firstsection is dedicated to preliminary steps, that will allow to go from the global embeddingproblem to the resolution of (4.1) over a torus, in a perturbative regime. In the secondsection, we will solve this PDE through the use of the Nash-Moser theorem.

4.1 Preliminaries

In all what follows, k ≥ 1 is fixed, M is smooth and compact, and g is Ck′ on M , k′ ≥ k.

Step 1: Reduction to the case M = Td.We show here that we can always restrict to the case where M is a torus, with a Ck′

riemannian metrics g. This will allow us to benefit from global coordinates on M . Inparticular, the equation (4.1) will be set over the whole of M . To restrict to this case, wefirst invoke Whitney’s theorem: by replacing M by uW (M) and g by its pushforward (uW )∗g,where uW : M → R2d is the Whitney’s embedding, we can always consider the case whereM is a smooth compact submanifold of R2d. We then rely on

Proposition 4.1. For any smooth compact submanifold M of RN , any Ck′ riemannianmetrics g on M , and for any bounded neighborhood U of M , there exists a Ck′ riemannianmetrics G on RN with

• G = e in RN \ U , e the euclidean scalar product.

• G|M = g, in the sense that for all x ∈M , for all v1, v2 ∈ TxM , Gx(v1, v2) = gx(v1, v2).

38

This proposition will be proved in Appendix B. As g is constant outside a compact set, wecan always consider its restriction to a large cube and periodize it so as to obtain a smoothmetrics on a torus, as claimed.

Step 2: From an embedding to an immersion.

We now show that the requirement that u be one-to-one can be removed. In particular, thismeans that we will only need to find a u satisfying the relation (4.1), as it automaticallyimplies that u is an immersion. This simplification is made possible thanks to the followingproperty:

Lemma 4.1. Let ui : M → Rni, gi = (ui)∗ ei, ti > 0, i = 1, 2. Then,

u = (t1u1, t2u2) : M → Rn1+n2

satisfies u∗e = t21g1 + t22g2.

Here, ei, resp. e, denotes the scalar product on Rni , resp. Rn. The proof of this lemma iseasy and left to the reader. Now, assume that we know:

i) a smooth embedding u1 : M → Rn1 (for instance given by Whithney’s theorem).

ii) for some k′ ≥ k and any Ck′ metrics g2 on M , the existence of u2 : M → Rn2 of classCk such that g2 = u∗2e.

Then, given a Ck′ riemannian metrics g, with k′ as in point ii), we set g1 = u∗1e, g2 = g−λ2g1,λ > 0. As M is compact, for λ > 0 small enough, g2 is definite positive, hence a Ck′

riemannian metrics. By ii), there exists u2 : (M, g2) → Rn2 of class Ck with g2 = u∗2e.By the lemma, it follows that u = (λu1, u2) satisfies u∗e = g, and as u1 is one-to-one, u isone-to-one as well. We get in this way a Ck embedding from (M, g) to some Rn. Therefore,we only need to focus on ii).

Step 3: Density of representable metrics and reduction to a perturbative problem.

We call representable any metrics g on M of the form u∗e, for an immersion u : M → Rn.In a local chart (or even global in the case of a torus), this reads

g = (∇u)(∇u)t, ∇u = (∂iuj)1≤i≤d,1≤j≤n, see (4.1).

A key ingredient to reduce the problem to a perturbative setting is the following densityresult of representable metrics.

Proposition 4.2. Let k′ ≥ 1, g a Ck′ riemannian metrics on M . One can find a family ofCk′−1 riemannian metrics (gε)ε>0 such that

• ‖gε − g‖Ck′−1 ≤ ε.

• gε = u∗εe, for some uε : M → Rnε of class Ck′.

39

The proof of this proposition is postponed to Appendix B. Thanks to this proposition, it isenough to prove the following claim:

Claim: There exists a smooth riemannian metrics g0 on M , k′ ≥ k and ε > 0 such that: forall Ck′ riemannian metrics g with ‖g − g0‖Ck′ ≤ ε, g is representable, i.e. there exists a Ck

immersion u : M → Rn for some n with u∗e = g.

Indeed, assume that this claim holds and let g be an arbitrary Ck′+1 riemannian metricson M . We can always assume that g′ = g − g0 is also a riemannian metrics: indeed, wecan always replace g0 by c0g0 for a small enough positive constant c0, which amounts toreplace ε by c0ε in the claim. For the ε of the claim, by the density lemma, there exists aCk′ representable metrics gε (with a Ck′ immersion) such that ‖g′ − gε‖Ck′ ≤ ε. This lastinequality reads ‖g − g0‖Ck′ ≤ ε, where g = g − gε. It follows from the claim that g − gεis representable (with a Ck immersion), but as gε is also representable, Lemma 4.1 impliesthat the sum is representable (with a Ck immersion). This proves that any regular enoughriemannian metric g is representable, as expected. Thus, it remains to prove the claim, whichis a perturbative result.

Step 4: Choice of the reference solution.

The last step of these preliminaries consists in selecting a good g0 for which one is able tosolve (4.1) for g near g0. We take a metrics g0 that can be written in local coordinates asg0 = (∇u0)(∇u0)t, where the vectors ∂iu0(x), ∂2

iju0(x), 1 ≤ i ≤ j ≤ d are independent for allx ∈M . Such immersion will be called free. One can check that the notion of free immersionis independent of the choice of the local chart (exercise). Note that, as we restricted inStep 1 to the case of a torus, we could take global coordinates, but this is not useful here.Obviously, before picking up a free immersion, one must check that free immersions exist !This can be shown as follows. By Whitney’s theorem, we can always assume that M is asubmanifold of R2d of dimension d. Moreover, locally, M can be expressed as a graph: forany x ∈ M , after possible reindexing of the coordinates, there exists an open neighborhoodU of x in R2d such that all points x in U ∩M take the form

x = (x1, . . . , xd, ψ(x1, . . . , xd)) , ψ with values in Rd.

Letφ : R2d → R2d+d(2d+1), φ(x1, . . . , x2d) = (xi, xk xl)1≤i≤2d,1≤k≤l≤2d.

It is a simple verification that φ|M is a free immersion of M in R2d+d(2d+1). Indeed, one cancheck that

Φ(x1, . . . , xd) = φ (x1, . . . , xd, ψ(x1, . . . , xd))

satisfies:

∂iΦ(x1, . . . , xd), ∂k∂lΦ(x1, . . . , xd), 1 ≤ i ≤ d, 1 ≤ k ≤ l ≤ d, are independent vectors.

Conclusion. After this simplification procedure, it remains to show the following

40

Theorem 4.3. Let M = Td a d-dimensional torus, g0 = (∇u0)(∇u0)t a smooth riemannianmetrics given by a smooth and free immersion u0 : M → Rn0. There exists k′ ≥ k, ε > 0,such that for any riemannian metrics g with ‖g − g0‖Ck′ ≤ ε, there exists a Ck immersionu : M → Rn0 (close to u0) with u∗ e = g.

The next section is dedicated to the proof of this result.

4.2 Use of the Nash-Moser theorem

The proof of Theorem 4.3 comes down to the resolution of equation (4.1), for g close tog0. This resolution can be tackled using the Nash-Moser theorem 3.1. We shall consider

the family of spaces Xs = (Cs(Td))n0 , see Remark 3.1, while Ys = (Cs(Td))d(d+1)

2 . Settingf = g − g0, u = u− u0, we rewrite (4.1) as

F (u) = f, Fij(u) =∂u0

∂xi· ∂u∂xj

+∂u

∂xi· ∂u0

∂xj+∂u

∂xi· ∂u∂xj

, 1 ≤ i ≤ j ≤ d.

We fix an integer k > 4, and want to verify the assumptions 1. to 4. in chapter 3. Clearly,F is C1 from Bk to Yk−1 for any ball Bk in Xk, and for all u ∈ Bk, v ∈ Xk:

F ′(u)v =∂(u0 + u)

∂xi· ∂v∂xj

+∂v

∂xi· ∂(u0 + u)

∂xj

while for all u, v ∈ Bk

F (u + v)− F (u)− F ′(u)v =∂v

∂xi· ∂v∂xj

Assumptions 1. and 2. hold for any k ≥ r ≥ 1.

To check the so-called tame estimate 3., we rely on the following product estimate

Lemma 4.2. For all t ∈ N, and f, g ∈ Ct(Td)

(4.2) ‖fg‖Ct . ‖f‖C0 ‖g‖Ct + ‖f‖Ct ‖g‖C0

Proof. From the classical formula ‖h′‖C0 ≤√

2‖h‖C0‖h′′‖C0 valid for any periodic function

h = h(t) of one real variable, we deduce that for all i = 1 . . . , d, ‖∂if‖C0 ≤ 2‖f‖1/2

C0 ‖∂2i f‖

1/2

C0 .By a classical induction argument, it follows that for all t, for all α ∈ Nd with |α| ≤ t, wehave

‖∂αf‖C0 . ‖f‖1− |α|t

C0 ‖f‖|α|t

Ct .

To establish (4.2), it is enough to show by induction that: for all t, for all α, β ∈ Nd with|α + β| ≤ t,

‖∂αf‖C0 ‖∂βg‖C0 . ‖f‖C0 ‖g‖Ct + ‖f‖t ‖g‖C0

41

The case t = 0 is obvious. Let us assume that the inequality holds for all α, β ∈ Nd with|α+β| ≤ t. As ‖ ‖Ct ≤ ‖‖Ct+1 , the inequality still holds with t replaced by t+1 if |α+β| ≤ t.For |α + β| = t+ 1, we use the interpolation inequality, that leads to:

‖∂αf‖C0 ‖∂βg‖C0 .

(‖f‖1− |α|

t+1

C0 ‖g‖|β|t+1

Ct+1

) (‖f‖

|α|t+1

Ct+1‖g‖1− |β|

t+1

C0

)The desired inequality then folllows from the convexity inequality : ab ≤ 1

pap + 1

qbq for

a, b > 0, 1p

+ 1q

= 1, applied with p = 1

1− |α|t+1

= t+1β

and q = 1

1− |β|t+1

= t+1α

By taking t = s− 1 with s ≥ k, the lemma implies easily assumption 3. for any r ≥ 1.

The last piece of the proof is to check assumption 4. Here, we face the difficulty that thelinearized system F ′(u)v = g, which can be written as

(4.3)∂(u0 + u)

∂xi· ∂v∂xj

+∂v

∂xi· ∂(u0 + u)

∂xj= gij, ∀1 ≤ i ≤ j ≤ d

is underdetermined. A nice idea of Nash is to add constraints on v, namely

(4.4)∂(u0 + u)

∂xk· v = 0, ∀1 ≤ k ≤ d

By taking k = i (resp. k = j) in the previous relation, differentiating with respect to xj(resp. to xi), we infer that

(4.5)∂2(u0 + u)

∂xixj· v =

∂(u0 + u)

∂xi· ∂v∂xj

=∂(u0 + u)

∂xj· ∂v∂xi

so that (4.3) becomes

(4.6) 2∂2(u0 + u)

∂xixj· v = gij, ∀1 ≤ i ≤ j ≤ d.

As u0 is a free immersion, under the assumption that the radius of Bk is small enough,one has that for any u ∈ X∞ ∩ Bk, u0 + u is a free immersion as well. Hence, the system(4.4)-(4.6) (which is equivalent to (4.4)-(4.3)) is pointwise solvable as soon as n0 > d+ d(d+1)

2.

To build a right inverse, we can for instance proceed in the following way. For all x ∈M , we

orthonormalize the family ∂2(u0+u)∂xixj

(x), ∂(u0+u)∂xk

(x), resulting in an orthonormal family e`(x),

` = 1, . . . , d+ d(d+1)2

, with explicit dependence on the vectors of the former family, and thatspans the same vector space. We can then compute from (4.4) and (4.6) the values of v · e`for all `, and set v =

∑`(v · e`)e`. Note that the dependence on v is explicit and smooth

on ∂2(u0+u)∂xixj

(x), ∂(u0+u)∂xk

(x), and it is smooth and linear in the coefficients of the riemannian

metrics g(x). Hence, by Lemma 4.2, the right inverse v = L(u)g satisfies for all t

(4.7) ‖L(u)g‖t ≤ ‖F(∂2ij(u0 + u), ∂k(u0 + u)

)‖C0 | g|t + ‖F

(∂2ij(u0 + u), ∂k(u0 + u)

)‖Ct | g|0.

where F is a smooth vector valued function of its arguments. We then use the

42

Lemma 4.3. Let G : Ω→ R a smooth function, Ω an open set of Rn. Then, for all t ∈ N,for all U ∈ Ct(Td, K) with K b Ω,

‖G(U)‖Ct . 1 + C(‖U‖L∞)‖U‖t.

for an increasing function C.

For brevity, we skip the proof of this lemma, that can be obtained by writing ∂αG(u) as alinear combination of products of some G(k)(u) and some derivatives of u. These productscan then be controlled using Lemma 4.2.

Applying this lemma in (4.7), we see that the assumption 4. of the Nash-Moser theorem issatisfied for r = 2. The Nash-Moser theorem therefore provides a solution to the embeddingproblem.

43

Chapter 5

Littlewood-Paley decomposition

We have encountered in chapter 3 the notion of tame estimate, see Remark 3.2 or Lemma4.2. We shall see here that tame estimates are very natural, notably in a PDE context: inmost functional spaces used in the analysis of PDEs, products and compositions by smoothfunctions obey tame estimates, similar to those encountered in Lemma 4.2 or Lemma 4.3.To establish these estimates, we shall rely on the so-called Littlewood-Paley decomposition,which is an important tool of Fourier analysis. This decomposition, and the related notionof paraproduct have revealed efficient in the study of nonlinear PDE’s, and sometimes usedas a substitute to Nash-Moser techniques ([11, 3]). This will be a bit discussed at the endof the chapter.

For a detailed description of Littlewood Paley analysis and its applications, we refer thereader to the nice monograph [2]. We rather borrow here to the lighter presentation in [1],see also the last chapter in [6]. We start with reminders on temperate distributions.

5.1 The space of temperate distributions S ′(Rd)

5.1.1 Definitions, examples

Definition 5.1. (the Schwartz space S(Rd))

The Schwartz space or space of rapidly decreasing functions on Rd is

S(Rd) =ϕ ∈ C∞(Rd,C), ∀k ∈ N, ‖ϕ‖k < +∞

, ‖ϕ‖k = sup

x∈Rd,|α|≤k(1 + |x|)k|∂αϕ(x)|.

Proposition 5.1. For all k ∈ N, ‖ϕ ‖k = supx∈Rd,|α|≤k(1 + |x|)k|∂αϕ(x)| defines a norm on

S(Rd), and

dS(ϕ, ψ) =∑k∈N

min(1, ‖ϕ− ψ‖k)2k

defines a distance on S(Rd) which makes it a complete metric space.

Remark 5.1. One can check that dS(ϕn, ϕ)→ 0 if and only if ‖ϕn − ϕ‖k → 0 for all k.

44

Definition 5.2. (the space of temperate distributions S(Rd))

The space of temperate distributions S ′(Rd) is the space of all continuous linear forms onS(Rd)

Remark 5.2. By continuous on S(Rd), we mean continuous on (S(Rd), dS). One can showthat u : S(Rd)→ R is continuous if and only if there exists k ∈ N and Ck > 0 such that

∀ϕ ∈ S(Rd), |u(ϕ)| ≤ Ck‖ϕ‖k.

For u ∈ S ′(Rd) and ϕ ∈ S(Rd), we will often note 〈u, ϕ〉 (duality bracket) for u(ϕ).

We remind here a few examples of temperate distributions:

• If u ∈ L1loc(Rd) grows at most polynomially, i.e. there exists C, k,M > 0 such that

|u(x)| ≤ C(1 + |x|)k for a.e. |x| ≥M , then u defines an element Tu of S ′(Rd) throughthe formula

〈Tu, ϕ〉 =

∫Rduϕ.

The same holds if u ∈ Lp(Rd) for some 1 ≤ p ≤ +∞. In both cases, the mappingu → Tu is one-to-one, as a consequence of the following general result: for any openset Ω ⊂ Rd, and any u ∈ L1

loc(Ω),

(5.1)

∫Ω

uϕ = 0 ∀ϕ ∈ C∞c (Ω) ⇒ u = 0 a.e. in Ω

(if two functions are equal in the sense of distributions, they are equal almost every-where). This justifies to identify the function and the distribution, and use notation〈u, ϕ〉 instead of 〈T, ϕ〉.

• If µ is a complex measure on the borelians of Rd, it can be identified to a linear formon Cb(Rd), and µ|S(Rd) belongs to S ′(Rd).

• If χ ∈ C∞(Rd) is a smooth function that grows at most polynomially (see the firstitem), and S ′(Rd) one can define the product uχ ∈ S ′(Rd) (also noted χu) by theformula 〈uχ, ϕ〉 = 〈u, χϕ〉.

• Derivative of a temperate distribution. Given u ∈ S ′(Rd) and α ∈ Nd, one definesthe element ∂αu ∈ S ′(Rd) by the formula: < ∂αu, ϕ >= (−1)|α|〈u, ϕ〉.

Proposition 5.2. (Support of a distribution)

We say that u ∈ S ′(Rd) is zero on an open set O of Rd if 〈u, ϕ〉 = 0 for all ϕ ∈ S(Rd),compactly supported in O. The union of all open sets on which u is zero is still an open seton which u is zero, and we call its support its complementary set.

45

5.1.2 Link with the Fourier transform

Notation. F : L1(Rd) → C00(Rd), Ff(ξ) =

∫Rd e

−2iπξ·xf(x)dx. One shall also use notation

f instead of F .

From the algebraic identities:

∂αf(ξ) = i|α|ξαf(ξ), and xαf(ξ) = i|α|∂αf(ξ)

it is easily seen that F sends the Schwartz space to itself. Combining this with the Fourierinversion formula, it is easily seen that

Proposition 5.3. F is an isomorphism from S(Rd) to itself, with inverse given by F−1ϕ(x) =1

(2π)d

∫Rd e

ix·ξϕ(ξ)dξ.

This proposition allows to define the Fourier transform on the space of temperate distribu-tions, through duality:

Definition 5.3. (Fourier transform of a temperate distribution)

For u ∈ S ′(Rd), the Fourier transform u of u is given by the formula:

〈u, ϕ〉 = 〈u, ϕ〉, ∀ϕ ∈ S(Rd)

The following can be shown:

Proposition 5.4. u→ u is an isomorphism from S ′(Rd) to itself.

Remark 5.3. Continuity of a function F : S ′(Rd) → X with X a topological space isunderstood as follows: if un → u in S ′(Rd), in the sense that 〈un, ϕ〉 → 〈u, ϕ〉 for allϕ ∈ S(Rd), then F (un)→ F (u).

Remark 5.4. For u ∈ S ′(Rd), the inverse Fourier transform of u satisfies:

〈F−1u, ϕ〉 = 〈u,F−1ϕ〉, ∀ϕ ∈ S(Rd)

Remark 5.5. In the case where u ∈ S(Rd), or u ∈ L1(Rd), or u ∈ L2(Rd), the usualdefinition of u coincides with the definition in the sense of S ′ (seeing functions as temperatedistributions, see the first example before paragraph 5.1.2.

Definition 5.4. (Fourier multiplier)

Let χ ∈ C∞(Rd,C) that grows at most polynomially. We define χ(D) : S ′(Rd)→ S ′(Rd) by

the formula: χ(D)u = χu.

In the next paragraph, we will use the convolution between a temperate distribution and afastly decreasing function, defined as follows:

Definition 5.5. For u ∈ S ′(Rd), ϕ ∈ S(Rd), one can define the convolution u ? ϕ by

u ? ϕ(x) = 〈u, ϕ(x− ·)〉

46

Note that again, this definition is consistent with the usual ones when u is more than atemperate distribution. In the same spirit, one has u ? ϕ = ϕ ? u when u is such that theright side is well-defined.

Proposition 5.5. u ? ϕ is a smooth function with at most polynomial growth, and for anyα ∈ Nd, ∂α(u ? ϕ) = ∂αu ? ϕ = u ? ∂αϕ. In particular, u ? ϕ ∈ S ′(Rd) (see the first exampleafter paragraph 5.1.2).

Proposition 5.6. For u ∈ S ′(Rd), ϕ ∈ S(Rd), u ? ϕ = u ϕ.

Note that the right-hand side is well-defined as the product of the temperate distributionu and the fastly decreasing function ϕ (see the third example after paragraph 5.1.2). Notealso that the identity in the proposition implies that a temperate distribution whose Fouriertransform is compactly supported is a smooth function with polynomial growth. Indeed, onecan take φ ∈ S(Rd) such that φ = 1 near the support of u. We then have u = uφ so thatu = u ? g with g = φ. The result follows from Proposition 5.5.

5.2 Littlewood-Paley decomposition

The Littlewood-Paley decomposition is a decomposition of functions (or temperate distri-butions) where each term in the decomposition has a localized frequency support, either ina ball B(0, R) or in an annulus C(0, R,R′) (with small radius R, large radius R′). Amongits possible interests, such decomposition allows for more precise estimates: the underlyingingredient is the following

Proposition 5.7. (Bernstein inequalities) Let r1, r2 > 0. There exists C such that forall k ≥ 1, for all λ > 0 u ∈ Lp(Rd) (1 ≤ p ≤ +∞):

(5.2) supp u ⊂ B(0, r1λ) ⇒ sup|α|=k‖∂αu‖Lp ≤ Ckλk‖u‖Lp

(5.3) supp u ⊂ C(0, r1λ, r2λ) ⇒ C−kλk‖u‖Lp ≤ sup|α|=k‖∂αu‖Lp ≤ Ckλk‖u‖Lp

Proof. By considering uλ(x) = u(λx) one can restrict to the case λ = 1. Let φ ∈ C∞c (Rd)such that φ = 1 in a vicinity of B(0, r1), and g ∈ S(Rd) such that g = ϕ. If u is supportedin B(0, r1), then uφ = u, which implies that u = u ? g, and from there ∂αu = u ? (∂αg). ByYoung’s inequality, we get

‖∂αu‖Lp ≤ ‖∂αg‖L1 ‖u‖Lp≤ Cd ‖(1 + |x|2)d∂αg‖L∞ ‖u‖Lp≤ C ′d‖(1−∆)d(ξαφ)‖L1 ‖u‖Lp ≤ C |α|‖ ‖u‖Lp

for C large enough. This shows (5.2), which implies the right-hand side inequality of (5.3).

47

For the remaining inequality, we shall use the identity

|ξ|2k = (ξ21 + · · ·+ ξ2

d)k =

∑|α|=k

Aα ξ2α11 . . . ξ2αd

d = (−1)k∑|α|=k

Aα(iξ)α(iξ)α,

with Aα = k!α1!...αd!

. Note that

(5.4)∑|α|=k

Aα = (1 + · · ·+ 1)k = dk, so that Aα ≤ dk.

We consider φ ∈ C∞c (Rd \ 0) such that φ = 1 near C(0, r1, r2) and set for all α of length k:

φα(ξ) = (−1)kAα(iξ)α

|ξ|2kφ(ξ), gα = F−1φα.

Note that φα ∈ S(Rd): there is in particular no singularity at ξ = 0, due to the vanishing ofφ. By the above algebraic identity, and as uφ = u, we deduce:

u =∑|α|=k

∂αu ? gα, so that ‖u‖Lp ≤∑|α|=k

‖gα‖L1‖∂αu‖Lp

One can show as for (5.2) and taking into account (5.4) that ‖gα‖L1 ≤ Ck for C large enough,which concludes the proof.

We now turn to the definition of the Littlewood-Paley decomposition. Let Ψ a smooth andnon-increasing function on R+ such that Ψ(r) = 1 for 0 ≤ r ≤ 1

2, Ψ(r) = 0 for r ≥ 1. Let

ψ(x) = Ψ(|x|), χ(ξ) = ψ( ξ2) − ψ(ξ). From the definitions, it follows that ψ and χ are non-

negative, that ψ is supported in B(0, 1), that χ is supported in the annulus C = C(0, 12, 2),

and that for all ξ ∈ Rd:

(5.5) 1 = ψ(ξ) +∑p≥0

χ(2−pξ)

Note that the sum above has at most two non-zero terms for each ξ: indeed, ξ → χ(2−pξ)has support in 2pC = C(0, 2p−1, 2p+1), so that 2pC ∩ 2qC = ∅ as soon as |q − p| ≥ 2. One canthen show (exercise) that for all ϕ ∈ S(Rd),

ϕ = ψϕ+∑p≥0

χ(2−p·)ϕ

with convergence of the right-hand side in S(R)d. It follows easily that

(5.6) ∀u ∈ S ′(Rd), u = ψ(D)u+∑p≥0

χ(2−pD)u

48

with convergence of the series in S ′(Rd). We shall note

∆−1u = ψ(D)u, ∆pu = χ(2−pD)u for all p ≥ 0,

and eventually

Spu = ψ(2−pD)u =

p−1∑q=−1

∆qu for all p ≥ 0.

Note thatu = lim

p→+∞Spu, ∀u ∈ S ′(Rd).

Remark 5.6. As ∆pu or Spu has compact frequency support, it is smooth with at mostpolynomial growth.

Lemma 5.1. There exists C > 0 such that for all 1 ≤ q ≤ +∞, for all u ∈ Lq(Rd),

supp≥−1‖Spu‖Lq ≤ C‖u‖Lq , sup

p≥−1‖∆pu‖Lq ≤ C‖u‖Lq

Proof. As Spu = ψ(2−pD)u, it can also be written 2pd(F−1ψ)(2p·) ? u. By Young inequality,it follows that

‖Spu‖Lq ≤ ‖2pd(F−1ψ)(2p·)‖L1 ‖u‖Lq ≤ ‖F−1ψ‖L1‖u‖Lq

The same reasoning applies to ∆pu, replacing ψ by χ.

Lemma 5.2. (Almost orthogonality)

∀ξ ∈ Rd,1

2≤ ψ2(ξ) +

∑p≥0

χ2(2−pξ) ≤ 1.

and for all u ∈ L2(Rd), ∑p≥−1

‖∆pu‖2L2 ≤ ‖u‖2

L2 ≤ 2∑p≥−1

‖∆pu‖2

Proof. The first double inequality follows from the fact that the sum 1 = ψ(ξ)+∑

p≥0 χ(2−pξ)has at most two non-zero terms for each value of ξ. Hence, the result follows from theelementary inequalities a2 + b2 ≤ (a + b)2 ≤ 2 (a2 + b2). The second double inequality isobtained from the first one by multiplying by u(ξ), integrating over Rd, and using Plancherelidentity.

49

5.3 Link with Sobolev and Holder spaces

Definition 5.6. (Sobolev spaces) For all s ∈ R, we define

Hs(Rd) = u ∈ S ′(Rd), ξ → (1 + |ξ|2)s/2u(ξ) ∈ L2(Rd)

Note that for s ≥ 0, u ∈ L2(Rd), so that u ∈ L2(Rd). One can show that it is a Hilbert space,

equipped with the norm ‖u‖Hs =(∫

Rd(1 + |ξ|2)s|u(ξ)|2dξ)1/2

.

In the case when s ∈ N, there is an alternative definition, in terms of weak derivatives in L2:

Definition 5.7. (weak derivatives) Let Ω an open set of Rd, 1 ≤ p ≤ +∞ and α ∈ Nd.For u ∈ L1

loc(Ω), we say that ∂αu ∈ Lp(Ω) if there exists vα ∈ Lp(Ω) satisfying: for allϕ ∈ C∞c (Ω),

(5.7)

∫Ω

u ∂αϕ = (−1)|α|∫

Ω

vα ϕ.

Remark 5.7. One can show that this vα is unique, and we denote it ∂αu. In terms ofclassical distributions, it means that the derivative ∂αu ∈ D′(Ω) is representable by a functionin Lq(Ω).

Definition 5.8. For s ∈ N, 1 ≤ p ≤ +∞, Ω an open set of Rd:

W s,p(Ω) = u ∈ Lp(Ω), ∂αu ∈ Lp(Ω), ∀|α| ≤ s

It is a Banach space, equipped with the norm ‖u‖W s,p =(∑

|α|≤s ‖u‖2Lp

)1/2

(and a Hilbert

space in the case p = 2).

Proposition 5.8. Hs(Rd) = W s,2(Rd), with equivalent norms.

Proof. Here are a few ideas of the proof. If u ∈ Hs(Rd), for all α with |α| ≤ s, the function(iξ)αu belongs to L2, so that vα = F−1 ((iξ)αu) ∈ L2(Rd). One then shows that vα is a weakderivative of u, cf identity (5.7). This identity can be easily proved thanks to the Plancherelformula. Conversely, if u ∈ W s,2(Rd), and vα = ∂αu denoting the weak derivative, the pointis to show that vα = (iξ)αu. In other words, one has to show that this algebraic identity,valid for usual derivatives, extends to the weak derivatives. This again relies on identity(5.7), through Plancherel formula.

Definition 5.9. (Holder spaces) For α > 0, α /∈ N, Cα(Rd) is the space of functionsf ∈ Ck

b (Rd), k = E(α), such that

∀β, |β| ≤ k, supx 6=y

|∂βf(x)− ∂βf(y)||x− y|α−k

< +∞

Cα(Rd) is a Hilbert space equipped with the norm

‖u‖∼Cα = sup|β|≤k‖∂βu‖L∞ + sup

|β|=ksupx 6=y

|∂βu(x)− ∂βu(y)||x− y|α−k

50

We now characterize these functional spaces in terms of properties of the LittlewoodPaley decomposition.

Proposition 5.9. (characterization of Sobolev spaces)

For all s ∈ R, for all u ∈ S ′(Rd), u ∈ Hs(Rd) if and only if∑

p≥−1 22ps‖∆pu‖2L2 < +∞.

Moreover, there exists C > 0 such that for all u ∈ Hs(Rd)

1

C

∑p≥−1

22ps‖∆pu‖2L2 ≤ ‖u‖2

Hs ≤ C∑p≥−1

22ps‖∆pu‖2L2

Proof. If u ∈ Hs(Rd), ‖u‖2s = ‖〈D〉su‖L2 , where 〈ξ〉 =

√1 + |ξ|2. By the second property of

Lemma 5.2, we deduce∑‖∆p〈D〉su‖2

L2 ≤ ‖u‖2Hs ≤ 2

∑‖∆p〈D〉su‖2

L2 .

Then, using Plancherel formula and the definition of ∆p, we have easily that there existsC > 0 such that for all p ≥ −1

1

C2ps‖∆pu‖L2 ≤ ‖∆p〈D〉su‖L2 ≤ C2ps‖∆pu‖L2

so that for some C > 0

(5.8)1

C

∑p≥−1

22ps‖∆pu‖2L2 ≤ ‖u‖2

s ≤ C∑p≥−1

22ps‖∆pu‖2L2

The left inequality yields in particular that∑

p≥−1 22ps‖∆pu‖2L2 < +∞.

Conversely, let u ∈ S ′(Rd) such that∑

p≥−1 22ps‖∆pu‖2L2 < +∞. It implies that for all

p ≥ −1, ∆pu ∈ L2(Rd), so that for any N ≥ 0, SNu belongs to L2(Rd), and even Hs(Rd)as it is compactly supported in frequency. We can apply the reasoning above replacing u bySNu, and the right inequality in (5.8) yields

‖SNu‖2s ≤ C

∑p≥−1

22ps‖∆pSNu‖2L2 ≤

∑p≥−1

22ps‖∆pu‖2L2

Hence, the sequence (SNu)N is bounded in Hs(Rd), so that it has a subsequence weaklyconverging to u ∈ Hs(Rd) . But it also converges to u ∈ S ′(Rd), so that u = u ∈ Hs(Rd).The equivalence of the norms is then given by (5.8).

Proposition 5.10. (characterization of Holder spaces)

For all α ∈ R+ \N, for all u ∈ S ′(Rd), u ∈ Cα(Rd) if and only if supp≥−1 2pα‖∆pu‖L∞ <∞.Moreover, there exists C > 0 such that for all u ∈ Cα(Rd)

1

Csupp≥−1

2pα‖∆pu‖L∞ ≤ ‖u‖∼2Cα ≤ C sup

p≥−12pα‖∆pu‖L∞

51

Proof. We focus on the first part of the equivalence. Let u ∈ Cα(Rd), and k = E(α). ByLemma 5.1, ‖∆−1u‖L∞ ≤ ‖u‖L∞ ≤ ‖u‖α. By Bernstein inequality (5.3), we find that for allp ≥ 0,

2pα‖∆pu‖L∞ = 2p(α−k)2pk‖∆pu‖L∞≤ Ck2p(α−k) sup

|β|=k‖∂β∆pu‖L∞ = Ck2p(α−k) sup

|β|=k‖∆p∂

βu‖L∞

As ∂βu ∈ Cα−k(Rd), the problem reduces to proving the first part of the equivalence forα ∈ (0, 1). In this case, we proceed as follows. We introduce hχ = 2pdF−1χ(2p·), so that∆pu = hχ ? u. As χ(0) = 0,

∫Rd hχ = 0, which implies

∀x, ∆pu(x) = ∆pu(x)−(∫

Rdhχ

)u(x) =

∫Rdhχ(y)(u(x− y)− u(x))dy.

We deduce :

| ∆pu(x)| ≤∫Rdhχ(y)|y|αdy ‖u‖Cα ≤ 2−pα‖y → hχ(y)2p|y|α‖L1‖u‖Cα

≤ 2−pα‖y → F−1χ(y) |y|α‖L1 ‖u‖Cα

Hence, 2pα‖∆pu‖L∞ is bounded uniformly in p by C‖u‖Cα .

Conversely, let us assume that Mα = supp≥−1 2pα‖∆pu‖L∞ < +∞. We set again k = E(α).We write, for any |β| ≤ k:

‖∂βSNu− ∂βSMu‖L∞ ≤N−1∑p=M

‖∂β ∆pu‖L∞ ≤ Ck

N−1∑p=M

2pk‖∆pu‖L∞

≤ CkMα

N−1∑p=M

2p(k−α)

It follows that (SNu)N is a Cauchy sequence in Ckb (Rd), converging to some u. But as (SNu)

converges to u in S ′(Rd), u = u ∈ Ckb (Rd). By taking SNu instead of (SN − SM)u in the

inequalities above, and letting N → +∞, we get:

‖∂βu‖L∞ ≤ C supp≥−1

2pα‖∆pu‖L∞ .

It remains to show that for all β with |β| = k,

‖∂βu‖Cα−k ≤ C supp

2pα‖∆pu‖L∞

By Bernstein inequality, ‖∂β∆pu‖L∞ ≤ Ck2pk‖∆pu‖L∞ , so that it is enough to show that

‖∂βu‖Cα−k ≤ C supp

2p(α−k)‖∂β∆pu‖L∞ = C supp

2p(α−k)‖∆p ∂βu‖L∞ .

52

Setting v = ∂βu, we see that we can restrict to the case β = 0, α ∈ (0, 1) (and u continuous).We write

u(x) = Spu(x) +Rpu(x), Rpu =∑q≥p

∆qu.

We find‖Rpu‖L∞ ≤

∑q≥p

‖∆qu‖L∞ ≤Mα

∑q≥p

2−qα ≤ CMα2−pα.

Also,

|Spu(x)− Spu(y)| ≤ |x− y|p−1∑q=1

‖∇∆qu‖L∞ .

But by (5.2), ‖∇∆qu‖L∞ ≤ C2q‖∆qu‖L∞ ≤ C ′Mα2q(1−α). Collecting the previous bounds,we obtain:

|u(x)− u(y)| ≤ CMα

(|x− y|2p(1−α) + 2−pα

)One can then optimize in p, taking for p the biggest value such that |x− y|2p ≤ 1. It followsthat

|u(x)− u(y)| ≤ C ′Mα|x− y|α

which concludes the proof.

This last proposition suggests to extend the definition of the Holder spaces through thefollowing

Definition 5.10. For any α ∈ R, we denote

Cα(Rd) = u ∈ S ′(Rd), supp≥−1

2pα‖∆pu‖L∞ < +∞

which is a Banach space under the norm ‖u‖Cα = supp≥−1 2pα‖∆pu‖L∞.

Remark 5.8. Warning ! One can show that for α ∈ N, Cα(Rd) does not coincide withthe space Cα

b (Rd). To emphasize the difference, it is customary to denote Cα∗ (Rd) the space

Cα(Rd) in the special case where α ∈ N.

Exercice 5.1. (convexity inequalities)

1. Show that if s = θs0 + (1− θ)s1, with 0 ≤ θ ≤ 1, s0, s1 ∈ R,

‖u‖Hs ≤ ‖u‖θHs0 ‖u‖1−θHs1 .

2. Show that If α = θα0 + (1− θ)α1, with 0 ≤ θ ≤ 1, α0, α1 ∈ R,

‖u‖Cα ≤ ‖u‖θCα0 ‖u‖1−θCα1 .

Proposition 5.11. (first Sobolev imbedding)

For all s ∈ R, Hs(Rd) is embedded continuously in Cs− d2 (Rd).

53

Proof. Let u ∈ Hs(Rd). As a byproduct of Proposition 5.9, we know that ∆pu ∈ L2(Rd).Thus, its Fourier transform is in L2(Rd) and also compactly supported, so that it is in L1(Rd).

We write the Fourier inversion formula: for p ≥ −1, ∆pu(x) = 1(2π)d

∫Rd e

ix·ξ∆pu(ξ)dξ, whichimplies

‖∆pu‖L∞ ≤ ‖∆pu‖L2 |B(0, C2p)|1/2 ≤ C ′2p(d2−s) sup

p2ps‖∆pu‖L2

≤ C ′′ 2p(d2−s)(∑p

2ps‖∆pu‖2L2)1/2

By Propositions 5.9 and 5.10, we get that ‖u‖Cs−

d2≤ C‖u‖Hs .

Corollary 5.1. For s > d2, Hs(Rd) is embedded continuously in Cb(Rd).

Proposition 5.12. (second Sobolev imbedding)

For 0 < s < d2, Hs(Rd) is continuously embedded in L

2dd−2s (Rd). More precisely, there exists

C > 0 such that

‖u‖L

2dd−2s≤ C‖u‖Hs , ‖u‖Hs =

(∫Rd|ξ|2s|u(ξ)|2 dξ

)1/2

.

Remark 5.9. The exponent 2dd−2s

can be deduced from a homogeneity argument. Indeed,suppose that we have an inequality of the type

‖u‖Lp ≤ C

(∫Rd|ξ|2s|u(ξ)|2 dξ

)1/2

for some p. Given some non-zero u, we consider the family uλ(x) = u(λx), λ > 0. Asimple calculation shows that ‖uλ‖Lp = λ−d/p‖u‖Lp, while, as uλ = λ−d u(λ−1·), we get

‖uλ‖Hs = λs−d2‖u‖Hs. Taking uλ in the previous inequality, we deduce

λ−d/p‖u‖Lp ≤ Cλs−d2‖u‖Hs .

Sending λ to 0, resp. λ to +∞, we find −dp− s + d

2≥ 0, resp. −d

p− s + d

2≤ 0. We get

p = 2dd−2s

.

Remark 5.10. In the case s ∈ N∗, with the identification Hs(Rd) = W s,2(Rd) (see Proposi-tion 5.8 and its proof), the inequality can be written equivalently as:

(5.9) ‖u‖L

2dd−2s≤ C sup

|α|=s‖∂αu‖L2 .

Proof. 1 We denote p = 2dd−2s

. We start from a famous identity for the Lp norm, namely

‖f‖pLp = p

∫ +∞

0

λp−1Leb(|f | > λ) dλ.

1This proof is borrowed from the lecture notes [6], and close in spirit to the proof of the famousMarcinkiewicz interpolation theorem, see [26, 187-194].

54

We decompose f in low and high frequencies, setting

f = f1,A + f2,A f1,A = 1B(0,A)(D)f, f2,A = 1B(0,A)c(D)f.

If s < d2, ξ → |ξ|−s1B(0,A)(ξ) belongs to L2, so that f1,A(ξ) = |ξ|−s1B(0,A)(ξ)|ξ|sf belongs to

L1, and

‖f1,A‖L∞ ≤1

(2π)d‖f1,A‖L1 ≤ 1

(2π)d

(∫B(0,A)

|ξ|−2sdξ

)1/2

‖f‖Hs ≤ CsAd2−s‖f‖Hs .

Now, we notice that

|f | > λ ⊂ |f1,A| >λ

2 ∪ |f2,A| >

λ

2

which implies

Leb(|f | > λ) ≤ Leb(|f1,A| >λ

2) + Leb(|f2,A| >

λ

2)

We take A = Aλ =(

λ4Cs‖f‖Hs

) pd, so that Leb(|f1,A| > λ

2) = 0. Hence,

‖f‖pLp = p

∫ +∞

0

λp−1Leb(|f2,A| >λ

2)dλ.

For the high-frequency part, we use the Tchebytchev inequality, and then Plancherel identity,which leads to

Leb(|f2,A| >λ

2) ≤ 4

‖f2,A‖2L2

λ2≤ C

λ2‖f2,A‖2

L2

We finally get

‖f‖pLp ≤ C

∫ +∞

0

∫Rdλp−31|ξ|≥Aλ|f(ξ)|2dξ

Note that, by definition of Aλ, |ξ| ≥ Aλ is equivalent to λ ≤ 4Cs‖f‖Hs|ξ|dp := Cf |ξ|

dp . By

Fubini’s theorem:

‖f‖pLp ≤ C

∫Rd|f(ξ)|2

∫ Cf |ξ|dp

0

λp−3dλdξ ≤ C ′Cp−2f

∫Rd|ξ|

d(p−2)p |f(ξ)|2dξ = C ′′‖f‖p

Hs.

Corollary 5.2. Hd/2(Rd) is continuously embedded in Lq(Rd) for all q ≥ 2.

Proof. Let sn = d2− 1

n. By the previous proposition Hsn is continuously embedded in Lpn(Rd)

(∩L2(Rd)), with pn = 2dd−2sn

. As pn → +∞ as n→ +∞, the result follows.

55

We conclude this paragraph with three lemma, that express some robustness with respect tothe choice of the frequency decomposition in the Littlewood-Paley approach. The first twoare concerned with decompositions of the form u =

∑p up, with the spectrum of up localized

in either a ball of radius O(2p), or an annulus of typical radius O(2p). We show that theSobolev and Holder characterizations in terms of the family (∆pu)p extend at least partiallyto the family (up)p. The third lemma treats the case of a decomposition u =

∑up where up

mimics a localization frequency, in the sense that one derivative costs O(2p).

Lemma 5.3. Let (up)p≥−1 such that supp up ⊂ B(0, R2p), for some R > 0, for all p ≥ −1.

• If supp 2pα‖up‖L∞ < +∞ (with α > 0), then u =∑up belongs to Cα(Rd), and

‖u‖Cα ≤ C supp

2pα‖up‖L∞ .

• If∑

p 22ps‖up‖2L2 < +∞ (with s > 0), then u =

∑up belongs to Hs(Rd), and

‖u‖2Hs ≤ C

∑p

22ps‖up‖2L2 .

Remark 5.11. Note the limitation s > 0 in the last statement.

Proof. We focus on the second item, the first one is very similar and slighty easier. Letq ≥ −1. By the support condition on (up), there exists N > 0 such that for all q ≥ −1,

(5.10) ∆qu =∑

p≥q−N

∆qup.

It follows that

2qs‖∆qu‖L2 ≤ 2qs∑

p≥q−N

‖∆qup‖L2 ≤ 2qs∑

p≥q−N

‖up‖L2 ≤∑

p≥q−N

2ps‖up‖L22(q−p)s ≤∑p∈Z

ap bq−p

where:

• ap = 2ps‖up‖L2 for p ≥ −1, ap = 0 otherwise.

• br = 2rs for r ≤ N , br = 0 otherwise.

As s > 0, (br)r∈Z is sommable, the discrete Young convolution inequality yields

‖2ps‖∆pu‖L2‖l2 ≤ ‖ap‖l2 ‖bp‖l1

so that ∑p≥−1

22ps‖∆pu‖2L2 ≤ C

∑p≥−1

22ps‖up‖2L2

and we conclude thanks to Proposition 5.9.

56

Lemma 5.4. Let (up)p≥0 such that supp up ⊂ C(0, c2p, C2p), for some C > c > 0, for allp ≥ −1.

• If supp 2pα‖up‖L∞ < +∞ (with α ∈ R), then u =∑up belongs to Cα(Rd), and

‖u‖Cα ≤ C supp

2pα‖up‖L∞ .

• If∑

p 22ps‖up‖2L2 < +∞ (with s ∈ R), then u =

∑up belongs to Hs(Rd), and

‖u‖2s ≤ C

∑p

22ps‖up‖2L2 .

Proof. The proof is very similar to the proof of the previous statement. The difference isthat the relation (5.10) is replaced by

∆qu =∑|p−q|≤N

∆qup.

We arrive this time at the inequality

2qs‖∆qu‖L2 ≤∑p∈Z

ap bq−p

where:

• ap = 2ps‖up‖L2 for p ≥ −1, ap = 0 otherwise.

• br = 2rs for |r| ≤ N , br = 0 otherwise.

so that the sequence (br)r∈Z is summable without constraint on the sign of s (or α).

Lemma 5.5. Let s > 0, n ∈ N, n > s. There exists C > 0 such that for all family up(p ∈ Z) in Hn(Rd), with

‖∂αuk‖L2 ≤ 2k(|α|−s)εk, with (εk)k∈Z ∈ l2(Z),

the sum u =∑uk ∈ Hs(Rd), and ‖u‖2

Hs ≤ C∑ε2p.

Proof. We first note that∑fj converges in L2, so that f is well-defined as an element of

L2. We write2js‖∆jf‖L2 ≤

∑k≥j

2js‖∆jfk‖L2 +∑k<j

2js‖∆jfk‖L2 .

To bound the first sum, we shall use the Bernstein inequality and the assumption of thelemma, which yields

‖∆jfk‖L2 ≤ C‖fk‖L2 ≤ C2−ksεk, ∀j ≥ −1

57

while to bound the second sum, we shall use

‖∆jfk‖L2 ≤ C2−nj∑|α|=n

‖∂α∆jfk‖L2 ≤ C2−nj‖fk‖Hn ≤ C2−(j−k)n2−ksεk, ∀j ≥ 0.

We find

2js‖∆jf‖L2 ≤ C∑k≥j

2(j−k)sεk + C∑k<j

2(j−k)(s−n)εk = C(a ? ε)j

where ? refers to a discrete convolution, ε = (εk)k∈Z, and a = (al)l∈Z is defined by

al = 2ls for l ≤ 0, al = 2l(s−n) for l > 0.

As a ∈ l1(Z), the result follows from Young’s inequality.

5.4 Paraproduct and tame estimates

We will now use the Littlewood Paley tools to obtain a refined decomposition of a productof two functions, due to Bony [4]. Formally, the idea is to write

uv =∑k,j

∆ku∆jv =∑j≥2

Sj−2u∆jv +∑k≥2

Sk−2v∆ku+∑|k−j|≤2

∆ku∆jv

= Tuv + Tvu+R(u, v)

where

• Tuv describes the part of the interaction in which the frequencies of u are much lowerthan the frequencies of v.

• Tvu describes the part of the interaction in which the frequencies of v are much lowerthan the frequencies of u.

• R(u, v) describes the part of the interaction in which the frequencies of u and v arecomparable.

The term Tuv (resp. Tvu) is called the paraproduct of v by u (resp. of u by v), while R(u, v)is the remainder.

Remark 5.12. One can show that, for any u, v ∈ S ′(Rd), the series defining Tuv or Tvu areconverging in S ′(Rd) (see the exercise below). The third series, that has to be understood as

R(u, v) =∑j≥1

∑k∈j,j±2,j±1

∆ku∆jv

is not always converging (which is consistent with the fact that the product uv is not alwayswell-defined for two temperate distributions u and v). We shall give sufficient conditionsbelow that ensure the convergence of the three series.

58

Exercice 5.2. Let u, v ∈ S ′(Rd).

1. Show that for all ξ ∈ C(0, 2j−1, 2j+1), for all ξ′ ∈ B(0, 2j−2), for all α ∈ Nd, for allK > 0, for all ϕ ∈ S(Rd), for all j ≥ 2,

|∂αξ ϕ(ξ − ξ′)| ≤ CK,α 2−jK .

2. Deduce from the previous question that for all u ∈ S ′(Rd), for all ϕ ∈ S(Rd), the series∑j≥2 ∆j (Sj−2uϕ) converges in S(Rd).

3. Deduce that the paraproduct Tuv is well-defined for all u, v ∈ S ′(Rd).

The main proposition of this section is

Proposition 5.13. (estimates for paraproduct and remainder)

• For all s ∈ R, u ∈ L∞, v ∈ Hs,

‖Tuv‖s ≤ Cs‖u‖L∞‖v‖s.

• For all α ∈ R, u ∈ L∞, v ∈ Cα,

‖Tuv‖Cα ≤ Cα‖u‖L∞‖v‖Cα .

• For all r, s with r + s > 0, u ∈ Cr, v ∈ Hs,

‖R(u, v)‖Hr+s ≤ Cr,s‖u‖Cr ‖v‖Hs .

• For all α, α′ with α + α′ > 0, u ∈ Cα, v ∈ Cα′,

‖R(u, v)‖Cα+α′ ≤ Cα,α′‖u‖Cα‖v‖Cα′ .

Remark 5.13. The bottom line of this proposition is that the regularity of Tuv is mainlydetermined by v, while the regularities of u and v add to provide the regularity of R(u, v).

Proof. We remark that Tuv has a decomposition that obeys the assumption of Lemma5.4. Indeed, Tuv =

∑p≥2 Sp−2u∆pv, and the Fourier transform of Sp−2u∆pv is included

in B(0, 2p−2) + C(0, 2p−1, 2p+1) ⊂ C(0, c2p, C2p) for some C > c > 0. As regards the firstitem, we write:∑

p

22ps‖Sp−2u∆pv‖2L2 ≤

∑p

22ps‖Sp−2u‖2L∞‖∆pv‖2

L2 ≤ C‖u‖2L∞

∑p

22ps‖∆pv‖2L2

≤ C ′‖u‖2L∞‖u‖2

Hs .

where the last inequality comes from Proposition 5.9. The inequality ‖Tuv‖s ≤ Cs‖u‖L∞‖v‖sis then deduced from the first item of Lemma 5.4. The second item of the proposition canbe handled along the same lines, using this time the second item of Lemma 5.4. Eventually,the third and fourth items rely on the same kind of arguments together with Lemma 5.3.We leave the details to the reader.

59

Corollary 5.3. (Tame estimates for the product)

• For any s > 0, for all u, v ∈ L∞ ∩Hs(Rd),

‖uv‖s ≤ C (‖u‖L∞‖v‖s + ‖v‖L∞‖u‖s)

• For any α ∈ R+ \ N, for all u, v ∈ Cα(Rd),

‖uv‖Cα ≤ C (‖u‖L∞‖v‖Cα + ‖v‖L∞‖u‖Cα)

Proof. This is a simple consequence of the estimates on the paraproduct and the remainder,taking into account that L∞ is included in C0

∗ .

Proposition 5.14. (Tame estimate for the composition)

Let F a C∞ function on R, with F (0) = 0. If u ∈ Hs(Rd) ∩ L∞(Rd) with s ≥ 0, thenF (u) ∈ Hs(Rd) and

‖F (u)‖Hs ≤ Cs,F,‖u‖L∞‖u‖Hs .

Proof. For s = 0, it is enough to write F (u) = G(u)u with G ∈ C∞(R): it follows that‖F (u)‖L2 ≤ sup |G(t)|, t ∈ [−‖u‖L∞ , ‖u‖L∞ ] ‖u‖L2 .

For s > 0, we want to write

F (u) = F (S0u) ++∞∑p=0

F (Sp+1u)− F (Spu).

Let us show that this series converges in L2, that is F (Snu)→ F (u) in L2 as n→ +∞. Weknow that Snu→ u in L2, using for instance the almost orthogonality of the family (∆ku),see Lemma 5.2. Moreover, ‖Spu‖L∞ ≤ C‖u‖L∞ . It follows that

‖F (Snu)− F (u)‖L2 ≤ C supt∈[0,1]

‖F ′(tSnu+ (1− t)u)‖L∞‖Snu− u‖L2 → 0.

Now, we write this sum as

F (u) =∑p≥−1

mp ∆pu, mp =

∫ 1

0

F ′(Spu+ t∆pu)dt

with the convention that S−1u = 0. For any β ∈ Nd,

‖∂β(Spu+ t∆pu)‖L∞ ≤ Cβ2|β|p‖u‖L∞ .It follows by Leibnitz rule that for all α ∈ Nd,

‖∂αmp‖L∞ ≤ Cα,F,‖u‖L∞2|α|p

Using that u ∈ Hs, we also have for all α

‖∂α∆pu‖L2 ≤ Cα2|α|(p−s)εp,∑

ε2p = ‖u‖2

Hs .

Combining previous inequalities, we find for all tα,

‖∂α(mp∆pv)‖L2 ≤ Cα,F,‖u‖L∞2|α|(p−s)εp.

We conclude by applying Lemma 5.5

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5.5 Connection to the Nash-Moser methodology

This last paragraph is a very informal discussion on how the paraproducts (and more gener-ally paradifferential calculus) can substitute to the Nash-Moser technique. We borrow herefrom the article [11]. For much more on paradifferential calculus, we refer to [4] and to thenice monograph [12].

As seen from Proposition 5.13, the paraproduct Tuv has essentially the regularity of v:the derivatives almost commute to the paraproduct. Hence, in PDE problems with lossof derivatives, when the loss comes from the coefficients of the right inverse g → L(u)g ofthe linearized operator v → F ′(u)v, it can be very useful to substitute to a linearization aparalinearization. We note that this approach, that we will discuss here in the context ofthe isometric embedding problem, relies on the paraproduct, that is on removing certainhigh frequencies. To this respect, it relates to the frequency truncation in the Nash-Moserapproach.

The starting point of the paralinearization approach is the following theorem, due to Bony:

Theorem 5.1. (Paralinearization Theorem)

Let F = F (x, u) ∈ C∞(Rd × Rm), F and its derivatives bounded on Rd ×K for all boundedK ⊂ Rm. Assume F (x, 0) = 0. Denote F ′u the fonction x→ ∂

∂uF (x, u(x)).

• If u ∈ Hs(Rd) with ρ := s− d2> 0, then

F (·, u)− TF ′uu ∈ Hs+ρ(Rd).

• If u ∈ Cr(Rd) with r > 0, then

F (·, u)− TF ′uu ∈ C2r(Rd).

See [4] or [12] for a proof. Another property of the paraproduct that is often used is

Proposition 5.15. Let r > 0. Let u, v ∈ Cr(Rd). Then,

TuTv − Tuv ∈ L(Hs(Rd), Hs+r(Rd)) ∀s ∈ R,TuTv − Tuv ∈ L(Cs(/Rd), Cs+r(Rd)) ∀s ∈ R.

See again [4] or [12] for a proof.

We can now introduce the notion of paradifferential operator, in the restricted context ofclassical differential operators. For a much wider framework, cf [12].

Definition 5.11. For a differential operator of order m over Rd, A =∑|α|≤m aα(x)∂α, the

paradifferential operator associated to A is the operator TA =∑|α|≤m Taα∂

α.

61

For a general nonlinear partial differential operator F (x, (∂αu)|α|≤m), with F = F (x, U)satisfying the assumptions of Theorem 5.1, the application of the theorem to the functionF (x, U(x)), U(x) = (∂αu)|α|≤m(x), shows the following: for all u ∈ Cr+m(Rd),

F (·, (∂αu)|α|≤m))− TF′uu ∈ C2r(Rd),

where this time TF′u is the paradifferential operator associated to

F′u =∑|α|≤M

∂

∂UαF (x, U(x))∂α.

Similarly, one can from Proposition 5.15 deduce the following result: for two differentialoperators A =

∑α≤m aα∂

α and B =∑|β|≤m′ bβ∂

β, if the coefficients aα, bβ belong to Cr(Rd)with r > m, then

TATB − TAB ∈ L(Cm+m′+σ(Rd), Cσ+r(Rd)), ∀σ ∈ R.

To apply these results and illustrate the paralinearization approach, we now come back tothe isometric problem from Chapter 4. We will give indications on how to solve it, basedon the article [11]. We refer to this article for all necessary additional details. As seen inChapter 4, the heart of the proof is to be able to solve an equation of the type

(5.11) Φ(u0 + u) = Φ(u0) + f, x ∈ Td, Φ(u) =

(∂u

∂xi· ∂u∂xj

)1≤i≤j≤d

.

Here u0 is a free immersion, C∞, with values in Rn. Ideally, one would like to find for

f ∈ Cr+1(Td,Rd(d+1)

2 ) small enough, a solution u ∈ Cr+1(Td,Rn) (we will explain how todo so for any r > 2). As u0 is a free immersion, one can build for ‖u‖Cr+1 small enough aright-inverse g → L(u0 + u)g to the linearized operator v → Φ′(u0 + u)v. We remind thatL(u0 + u) is a zero order operator (matrix product), with coefficients involving second orderderivatives of u0 + u, which is where the strongest loss of derivatives occurs.

If L was not involving any loss of derivatives, one could proceed as follows to solve (5.11).One could rewrite the equation (5.11) as

Φ′(u0)u+R(u) = f,

look for u under the form u = u(g) = L(u0)g, and solve the equation with unknown g:

Φ′(u0)u(g) +R(u(g)) = f, that is g +R(u(g)) = f.

Solvability would be possible for small f by the implicit function theorem.

Here, to treat the loss of derivatives, the idea is the following. We start by rewriting (5.11)as

(5.12) T ′Φ(u0 + u) +R(u) = f

62

where for u ∈ Cr+1,

R(u) = Φ(u0 + u)− Φ(u0)− T ′Φ(u0 + u) ∈ C2r(Td)

by the paralinearization result for nonlinear partial differential operators (here, the operatoris first order). Note that all notions and results of this chapter were stated in Rd, butstraightforward analogues are available on Td. In particular, for 2r ≥ r + 1, that is r ≥ 1,the functional R sends Cr+1 to Cr+1. Moreover, one can show that this functional is smooth,see [11] for details.

The next step is to look at the equation

(5.13) u− TL(u0+u)g − L(u0)g + TL(u0)g = 0.

of unknown u, with data g ∈ Cr+1. Using Proposition 5.13, one can check that as u0 isC∞, −L(u0)g + TL(u0)g is in Cσ for all σ. By the same proposition, TL(u0+u)g belongs toCr+1. Eventually, one can show that this equation is of the form G(u, g) = 0 with a smoothG : Cr+1×Cr+1 → Cr+1, satisfying ∂uG(0, 0) = Id. It follows from the usual implicit functiontheorem that for ‖g‖Cr+1 small enough, (5.13) has a unique solution u = u(g) ∈ Cr+1.

The last step is to solve the equation

Φ(u0 + u(g)) = Φ(u0) + f

with unknown g ∈ Cr+1. Taking into account (5.12) and (5.13), we can write it

TΦ′(u0+u(g))TL(u0+u(g))g + TΦ′(u0+u(g))

(L(u0)g − TL(u0)g

)+R(u(g)) = f

The second and last terms at the left-hand side belong to Cr+1. As regards the first term,we use the result on the commutation of paradifferential operators: as

Φ′(u0 + u)L(u0 + u) = Id,

taking into account that Φ′(u0 + u) and L(u0 + u) are respectively first order and zeroorder operators, and that their coefficients are respectively in Cr and Cr−1, we get that, forr − 1 > 1,

TΦ′(u0+u)TL(u0+u) ∈ L(Cσ+1, Cσ+r−1).

Taking σ = r > 2, we find that TΦ′(u0+u(g))TL(u0+u(g))g belongs to Cr+1. Eventually, one cancheck that the function

g → Φ(u0 + u(g))− Φ(u0)

(which sends Cr+1 to Cr+1 as we just established), is smooth and that its differential at 0 is0. We conclude by the implicit function theorem.

63

Chapter 6

The regularity theorem of De Giorgifor scalar elliptic equations

We will present here a celebrated result on the regularity of weak solutions to scalar ellipticequations. This result was established independently by De Giorgi [7] and Nash [19], andrevisited later by Moser [17]. We shall present here the approach of De Giorgi. Our mainreferences for this chapter are [20] and [5]. We shall emphasize some proximity with thearguments encountered in Chapter 2, used to overcome losses of derivatives.

The PDEs under consideration are diffusion equations of the form:

(6.1) −div (A∇u) = f,

or in coordinates−∑i,j

∂i(aij∂ju) = f.

Such equations are considered in a domain (that is an open and connected set) Ω of Rd,d ≥ 2:

u, f : Ω→ R, A = (aij)1≤i,j≤d : Ω→Md(R).

The diffusion matrix satisfies the following fundamental assumptions:

(H1) (L∞ bound): aij ∈ L∞(Ω) for all 1 ≤ i, j ≤ d.

(H2) (uniform ellipticity): There exists λ > 0 such that for a.e. x ∈ Ω, and all ξ ∈ Rd,

A(x)ξ · ξ ≥ λξ · ξ.

The analysis of equation (6.1) has a long history, starting from the study of harmonic func-tions ∆u = 0 and of the Poisson equation −∆u = f . As regards harmonic functions, itis a well-known result that if a distribution u ∈ D′(Ω) satisfies ∆u = 0 in Ω, it is in factsmooth (and even analytic). This in turn allows to establish various regularity results forthe Poisson equation. As an example, we state

64

Proposition 6.1. (Holder regularity for Poisson equation)

Let α ∈ (0, 1), Ω an open set of Rd. If f is of class Cα on Ω, any distributional solution uof −∆u = f in Ω is of class C2+α on Ω.

Proof. Let x0 ∈ Ω, and χ ∈ C∞c (Ω) such that χ = 1 near x0. Let uχ the solution in S ′(Rd)of

−∆uχ = χf

which is obtained as the convolution of the fundamental solution and χf (seen as a functionover Rd after extension by zero outside Ω). We note that u−uχ is harmonic near x0, so thatit is smooth near x0. As ∆−1uχ is also smooth, it is then enough to show that

∑p≥0 ∆puχ

belongs to C2+α(Rd), given that χf ∈ Cα(Rd) (after extension by zero outside Ω). As

−∆∆puχ = −∆p∆uχ = −∆p(χf)

we deduce from (5.3) that for all p ≥ 0,

22p‖∆puχ‖L∞ ≤ C‖∆p(χf)‖L∞ ≤ C ′2−pα‖χf‖Cα

The conclusion follows from Lemma 5.3

How is the situation modified in the case of heterogeneous diffusions (that is of non-constant matrices) ? When the coefficients aij have C0,α regularity, one can show through aperturbative approach that the conclusion for constant coefficients equations remains: if thedata has Holder continuity Cα, the solution of (6.1) has regularity C2+α. Such result, togetherwith a complete existence theory and set of estimates, was established by J. Schauder inthe 1930’s. For a description of Schauder estimates, we refer the reader to the textbook[9, chapter 6]. Still, with regards to applications, a Holder continuity assumption on thecoefficients is restrictive, and one would like to have some existence and regularity theoryunder the mere assumptions (H1) and (H2). We remind in the next chapter the standardnotion of weak solutions, before turning to the De Giorgi-Nash-Moser theorem on the Holderregularity of weak solutions.

6.1 Weak solutions of elliptic equations

In this whole section, Ω denotes a domain of Rd (open connected set). The Sobolev spaceW s,p(Ω) was introduced in Definition 5.8. For an extensive study of Sobolev spaces (approx-imation by smooth functions, extension to the whole space, traces, compactness...), we referto [8, chapter 5]. We shall focus here on the space H1(Ω) = W 1,2(Ω). We shall notably use

Theorem 6.1. (Rellich compactness theorem)

If Ω is of class C1 and bounded, the embedding of H1(Ω) into L2(Ω) is compact: any boundedsequence in H1(Ω) has a subsequence that converges in L2(Ω).

65

Remark 6.1. Let (un) be a bounded sequence in H1(Ω), strongly converging in L2(Ω) tosome u. This sequence being bounded in the Hilbert space H1(Ω), it also has a subsequencethat converges weakly in H1(Ω) to some u. By comparison of the limits in the distributionalsense, we obtain u = u. Hence, combining Rellich theorem and this argument, we see thatone can extract from a bounded sequence in H1(Ω) a subsequence that converges strongly inL2 and weakly in H1 to some u.

For later use, we also single out the following composition result (see [8, chapter 5, problem16]):

Proposition 6.2. Let G ∈ C1(R) such that G(0) = 0 and G′ bounded. For all u ∈ H1(Ω),G(u) ∈ H1(Ω), with for all 1 ≤ i ≤ d, ∂iG(u) = G′(u)∂iu.

Exercice 6.1. ([8, chapter 5, problem 17])

Let u ∈ H1(Ω), u+ = max(u, 0), u− = max(−u, 0). Show that u± ∈ H1(Ω),

∇u+ = 1u>0∇u , ∇u− = −1u<0∇u.

Hint: use the previous proposition with the function F ε(z) = (z2 + ε2)1/2 − ε for z > 0,F ε(z) = 0 for z < 0.

In the resolution of elliptic equations, a crucial role is played by a subspace of H1(Ω). Namely,one denotes by H1

0 (Ω) the closure of C∞c (Ω) in H1(Ω). A key property of the space H10 (Ω)

is the so-called

Proposition 6.3. (Poincare inequality) If Ω is bounded, there exists C > 0 (that can betaken proportional to the diameter of Ω) such that for all u ∈ H1

0 (Ω):

‖u‖L2 ≤ C‖∇u‖L2 .

Hence, ‖u‖H10 (Ω) :=

(∫Ω|∇u|2

)1/2is a norm on H1

0 (Ω), equivalent to the usual one.

Remark 6.2. When Ω is C1 and bounded (so that Rellich theorem applies), one can provethe inequality

‖u‖L2 ≤ C‖∇u‖L2 , C = C(Ω),∀u ∈ V

for V any closed subspace of H1(Ω) such that the only constant function in V is zero. Thisapplies for instance to the functions of H1(Ω) that have zero mean. See also Lemma 6.3

It is easily seen that, given u ∈ H10 (Ω), its extension u by 0 outside Ω belongs to H1(Rd).

This allows to apply the results of Chapter 5, notably the results on the Sobolev embeddings.We find (combined with Poincare inequality)

Proposition 6.4. (Sobolev embeddings)

66

• If d > 2, the space H10 (Ω) embedds continuously in L

2dd−2 (Ω), with

‖u‖L

2dd−2≤ C‖∇u‖L2

for some C that depends on d and Ω (and can be taken proportional to the diameter ofΩ).

• If d = 2, H10 (Ω) embedds continuously in Lq(Ω) for all q ∈ [2,+∞) (q ∈ [1,+∞) for

bounded Ω), with‖u‖Lq ≤ C‖∇u‖L2

Remark 6.3. In the whole space case, the second embedding involves the non-homogeneousH1 norm of u, not only the L2 norm of the gradient. But in the case of H1

0 (Ω), the Poincareinequality allows to replace the H1 norm by the H1

0 norm.

Remark 6.4. These embeddings still hold when H10 (Ω) is replaced wth H1(Ω), under the

assumption that Ω is of class C1, and if the H10 norm at the right-hand side is replaced by

the (inhomogeneous) H1 norm.

The space H10 (Ω) arises naturally in the notion of weak solutions for Poisson type systems:

(6.2) −div (A∇u) = f in Ω, u|∂Ω = 0.

Definition 6.1. (weak solution of the Poisson equation)

For any f ∈ H−1(Ω) = (H10 (Ω))′, we say that u ∈ H1

0 (Ω) is a weak solution of (6.2) if:

(6.3)

∫Ω

∇u · ∇ϕ = f(ϕ), ∀ϕ ∈ H10 (Ω).

Note that if Ω is bounded and if f ∈ Lq(Ω), with q ≥ 2dd+2

if d > 2 or q > 1 if d = 2, then f

can be seen as an element of H−1(Ω) through the mapping ϕ→∫

Ωfϕ. For instance, in the

case d > 2:

|∫

Ω

fϕ| ≤ ‖f‖L

2dd+2 (Ω)

‖ϕ‖L

2dd−2 (Ω)

≤ C‖f‖L

2dd+2 (Ω)

‖f‖H10 (Ω) ≤ C ′‖f‖Lq(Ω)‖f‖H1

0 (Ω)

where we have used Proposition 6.4 for the second inequality.

The notion of weak solution is extremely convenient, as it provides an easy framework forunique solvability of (6.2). This is the sense of

Proposition 6.5. (Existence and uniqueness of weak solutions)

Let Ω be a bounded domain. For any f ∈ H−1(Ω), there is a unique weak solution of (6.2).

Proof. This is a consequence of the following abstract result:

Lemma 6.1. (Lax-Milgram Lemma) Let H be a Hilbert space, a : H ×H → R bilinear,continuous, and coercive in the sense that a(u, u) ≥ αu for some α > 0, for all u ∈ H.Then, for all f ∈ H ′, there exists a unique solution u ∈ H of a(u, v) = f(v).

For a proof of this lemma, see [8, chapitre 6]. The theorem follows, with H = H10 (Ω),

a(u, v) =∫

ΩA(x)∇u(x) · ∇v(x)dx.

67

6.2 Regularity of the weak solution

A natural question is: how regular is the weak solution provided by Theorem 6.5 ? We focushere on interior regularity (regularity up to the boundary would require more assumptionson Ω). The main goal of this chapter is proving Holder regularity. We introduce, for an openset Ω, the norm

‖u‖Cα(Ω) = ‖u‖L∞(Ω) + supx 6=y∈Ω

|u(x)− u(y)||x− y|α

.

Theorem 6.2. [7, 19]

Suppose that A = (aij) satisfies assumptions (H1)-(H2). Let q > d2, f ∈ Lq(Ω). Suppose that

u ∈ H1(Ω) satisfies ∫Ω

A∇u · ∇ϕ =

∫Ω

fϕ, ∀ϕ ∈ H10 (Ω).

Then, there exists α = α(d, λ, ‖A‖L∞ , q) ∈ (0, 1) such that for all domain Ω b Ω,

(6.4) ‖u‖Cα(Ω) ≤ C(‖f‖Lq(Ω) + ‖u‖H1(Ω)

),

where C depends on d, λ, ‖A‖L∞ , q and Ω.

Note that by a covering argument, it is enough to establish the bound

‖u‖Cα(Br(x0)) ≤ C(‖f‖Lq(B4r(x0)) + ‖u‖H1(B4r(x0))

),

for any x0 ∈ Ω and r > 0 such that B(x0, 4r) b Ω, where C is allowed to depend ond, λ,Λ, q, r. Furthermore, by dilation and translation, we can always assume x0 = 0, r = 1

4.

Eventually, it is enough to prove the theorem in the special case Ω = B(0, 1), Ω = B(0, 1/4).

6.2.1 First step: from L2 to L∞.

Proposition 6.6. Suppose that A = (aij) is bounded measurable in B(0, 1) and satisfies forsome λ > 0:

λξ · ξ ≤ A(x)ξ · ξ, for almost every x ∈ B(0, 1), ∀ξ ∈ Rd.

Let q > d2, f ∈ Lq(B(0, 1)) and u ∈ H1(B(0, 1)) a subsolution of the Poisson equation in B1,

namely, ∫B(0,1)

A∇u · ∇ϕ ≤∫

Ω

fϕ ∀ϕ ∈ H10 (B(0, 1)), ϕ ≥ 0 in B(0, 1)

Then u+ ∈ L∞(B(0, 1/2)), with

‖u+‖L∞(B(0,1/2)) ≤ C(‖f‖Lq(B(0,1)) + ‖u+‖L2(B(0,1))

)for a constant C that depends only on d, q, λ, ‖A‖L∞.

68

Proof. We follow here the approach by De Giorgi and assume d ≥ 3 (the case d = 2 resuiresvery minor modifications). We consider

• a sequence of balls Bk = B(0, rk) where r0 = 1 and rk is decreasing towards 12.

• a sequence of smooth functions (χk)k≥1, with values in (0, 1), such that χk = 1 in Bk,χk = 0 in Bc

k−1, with |∇χk| ≤ Crk−rk−1

.

• a sequence of increasing constants Ck going to 1.

The goal is to show that there exists δ > 0 such that∫B(0,1/2)

((u− 1)+)2 = limk→+∞

∫B(0,1)

χk((u− Ck)+)2 = 0

under the assumption ‖f‖Lq(B(0,1)) + ‖u+‖L2(B(0,1)) ≤ δ. Then the general case follows by

considering (uδ, fδ) = δ (u,f)‖f‖Lq(B(0,1))+‖u+‖L2(B(0,1))

.

We denote uk = (u− Ck)+. The first ingredient of the proof is sometimes called Cacciopoliinequality. Taking ϕ = χ2

k+1uk+1 as a test function, we find∫B(0,1)

A∇u · ∇(χ2k+1uk+1

)≤∫

Ω

fχ2k+1uk+1

Expanding the gradient at the left-hand side, we find∫B(0,1)

χ2k+1A∇u · ∇uk+1 ≤

∫Ω

fχ2k+1uk+1 − 2

∫B(0,1)

(A∇u · ∇χk+1)χk+1uk+1

We note that A∇u · ∇uk+1 = A∇uk+1 · ∇uk+1 ≥ λ|∇uk+1|2, and that

− 2

∫B(0,1)

(A∇u · ∇χk+1)χk+1uk+1 = −2

∫B(0,1)

(A∇uk+1 · ∇χk+1)χk+1uk+1

≤ λ

2

∫B(0,1)

χ2k+1|∇uk+1|2 + C

∫B(0,1)

|∇χk+1|2|uk+1|2

≤ λ

2

∫B(0,1)

χ2k+1|∇uk+1|2 +

C ′

(rk − rk+1)2

∫B(0,1)

|χkuk|2

where C,C ′ depend on λ and Λ = ‖A‖L∞ . We have used here that χk = 1 on the support of∇χk+1 and that uk+1 ≤ uk. Hence,∫

B(0,1)

χ2k+1|∇uk+1|2 ≤ C

(1

(rk − rk+1)2

∫B(0,1)

|χkuk|2 +

∫Ω

fχ2k+1uk+1

).

This implies∫B(0,1)

|∇(χk+1uk+1)|2 ≤ C ′(

1

(rk − rk+1)2

∫B(0,1)

|χkuk|2 +

∫Ω

fχ2k+1uk+1

).

69

The second ingredient of the proof is the Sobolev imbedding. More precisely, we write, forp = 2d

d−2:∫

Ω

fχ2k+1uk+1 ≤ ‖f‖Lq Leb(χk+1uk+1 > 0)1− 1

p− 1q ‖χk+1uk+1‖Lp

≤ C ‖f‖Lq(B(0,1)) Leb(χk+1uk+1 > 0)1− 1p− 1q ‖∇(χk+1uk+1)‖L2

≤ δ ‖∇(χk+1uk+1)‖2L2 + Cδ ‖f‖2

Lq(B(0,1)) Leb(χk+1uk+1 > 0)2− 2p− 2q

We also have∫B(0,1)

|χk+1uk+1|2 ≤ C

(∫B(0,1)

|χk+1uk+1|p)2/p

Leb(χk+1uk+1 > 0)2/d

≤ C

(∫B(0,1)

|∇(χk+1uk+1)|2)Leb(χk+1uk+1 > 0)2/d

Combining, we get∫B(0,1)

|χk+1uk+1|2 ≤ C( 1

(rk − rk+1)2

(∫B(0,1)

|χkuk|2)Leb(χk+1uk+1 > 0)2/d

+ ‖f‖2Lq Leb(χk+1uk+1 > 0)1+ε

)where ε = 1− 2

p− 2

q+ 2

d> 0. To evaluate the Lebesgue measure of the set χk+1uk+1 > 0,

we use the fact thatχk+1uk+1 > 0 ⇒ χkuk > Ck+1 − Ck

which by Bienaime-Tchebytchev yields

Leb(χk+1uk+1 > 0) ≤ 1

(Ck+1 − Ck)2

∫B(0,1)

|χkuk|2.

Setting Uk =∫B(0,1)

|χkuk|2, we end up with the inequality

(6.5) Uk+1 ≤ C

(1

(rk − rk+1)2(Ck+1 − Ck)2/dU

1+2/dk +

1

(Ck+1 − Ck)2(1+ε)‖f‖LqU1+2ε

k

)Taking rk = 1

2+ 1

2k+1 , Ck = 1− 12k

, we get for some fixed N :

(6.6) Uk+1 ≤ C2Nk(U1+2/dk + ‖f‖2

LqU1+2εk )

From there, one can show by induction that for ‖f‖Lq ≤ 1, for δ and ε > 0 small enough andU0 =

∫B(0,1)

|u+|2 ≤ δ, one has Uk ≤ δ(1+ε)k for all k. This implies that Uk → 0 as k → +∞,

which ends the proof.

70

Corollary 6.1. Under the same assumptions on A and f as in the previous proposition, ifu ∈ H1(B(0, 1)) satisfies∫

B(0,1)

A · ∇u · ∇ϕ =

∫B(0,1)

fϕ, ∀ϕ ∈ H10 (B(0, 1))

then u ∈ L∞(B(0, 1/2)) and

‖u‖L∞(B(0,1/2)) ≤ C(‖f‖Lq(B(0,1)) + ‖u‖L2(B(0,1))

)for a constant C that depends only on d, q, λ, ‖A‖L∞.

Remark 6.5. There is a strong analogy between the method used to prove Proposition 6.6and the one encountered in Chapter 2. In the former, the loss of derivatives was resultingin negative powers of δ, where δ corresponded to a loss in the radius of analyticity. Here,the loss is due to the shrinking of the domain on which regularity estimates are obtained(negative powers of rk − rk+1), and on negative powers of Ck+1 −Ck. In both contexts, whatcompensates this loss is the nonlinear nature of the iterative process. Note that in the proof ofProposition 6.6, this nonlinearity is made possible thanks to the use of the Sobolev imbedding,despite the linear nature of the equation.

6.2.2 Holder regularity with no source

We now turn to the proof of De Giorgi-Nash-Moser theorem. We first focus on the homoge-neous case, namely on solution of

(6.7) −div (A∇u) = 0.

Definition 6.2. Let Ω ⊂ Rd an open set. u ∈ H1(Ω) is a subsolution (resp. supersolution)of (6.7) in Ω if ∫

Ω

A∇u · ∇ϕ ≤ 0 (resp. ≥ 0) ∀ϕ ∈ H10 (Ω), ϕ ≥ 0 in Ω

Lemma 6.2. Let φ ∈ C1(R) a convex function with φ(0) = 0 and φ′ bounded. Let u ∈H1(B(0, 1)). Then,

i) If u is a subsolution and φ′ ≥ 0, then v = φ(u) is also a subsolution.

ii) If u is a supersolution and φ′ ≤ 0, then v = φ(u) is a subsolution.

Remark 6.6. One could replace B(0, 1) by any smooth bounded domain.

Proof. Note that by Proposition 6.2, φ(u) ∈ H1(B(0, 1)). We prove the first point, in thecase φ ∈ C2

b (R): the general case follows by considering φε = φ ? ρε with ρε a standardmollifier and passing to the limit into the relation∫

B(0,1)

A∇φε(u) · ∇ϕ =

∫B(0,1)

φ′ε(u)∇u · ∇ϕ ≤ 0.

71

Let ϕ ∈ C∞c (B(0, 1)) with ϕ ≥ 0. We compute∫B(0,1)

ADv ·Dϕ =

∫B(0,1)

φ′(u)ADu ·Dϕ

=

∫B(0,1)

ADu ·D (φ′(u)ϕ)−∫B(0,1)

ADu ·Duφ′′(u)ϕ ≤ 0

where the first term is non-positive as u is a subsolution, and the second term is non-positiveby the convexity of Φ. Note that the test function φ′(u)ϕ in the second integral belongs toH1

0 (B1), as φ′(u) ∈ H1(B(0, 1)) (see Proposition 6.2) and ϕ ∈ C∞c (B(0, 1)).

To conclude, we leave to the reader to show that any non-negative function ϕ in H10 (B(0, 1))

is the limit in H1 of a sequence (ϕn) of non-negative functions in C∞c (B(0, 1)). By theprevious lines, for all n,

∫B(0,1)

ADv ·Dϕn ≤ 0, so that∫B(0,1)

ADv ·Dϕ ≤ 0.

We now state a variation of Poincare inequality:

Lemma 6.3. For any ε > 0, there exists C = C(ε, d) such that, for all u ∈ H1(B(0, 1)) withLeb(u = 0) ≥ ε, we have ∫

B(0,1)

|u|2 ≤ C

∫B(0,1)

|∇u|2

Remark 6.7. One could replace B(0, 1) by any smooth bounded domain.

Proof. If the statement does not hold, one can find ε > 0 and a sequence (un) in H1 suchthat

Leb(un = 0) ≥ ε, ‖un‖L2 = 1, ‖∇un‖L2 = O(1/n).

By the Rellich theorem, see [8], one can up to a subsequence assume that un converges weaklyin H1 and strongly in L2 to some u ∈ H1. By the gradient bound, we find that ∇u = 0, sothat u = u is constant over B(0, 1). As ‖u‖L2 = limn ‖un‖L2 = 1, the constant u is non-zero.Eventually, we get

ε ≤ Leb(un = 0) ≤ Leb(|un − u| ≥ |u|) ≤1

|u|2

∫B(0,1)

|un − u|2

where the last inequality is the Bienaime-Tchebytchev inequality. As the right-hand sidegoes to zero when n goes to infinity, we reach a contradiction.

We now prove:

Proposition 6.7. Suppose that u is a non-negative supersolution in B(0, 2), and

Leb(x ∈ B(0, 1), u ≥ 1) ≥ ε, ε > 0.

Then there exists a constant c > 0 depending on ε, d, λ, Λ = ‖A‖L∞ such that

infB(0,1/2)

u ≥ c.

72

Proof. We may assume that u ≥ δ > 0, then the result for any non-negative u can beobtained by considering uδ = u+ δ and letting δ go to 0+.

We then consider v = (ln(u))− = (− lnu)+. We claim that v a non-negative subsolution of(6.7) in B(0, 2). This can be shown by considering the function Fε (− lnu) where Fε wasdefined in Exercice 6.1, applying Proposition 6.2 and sending ε to 0. Moreover, we have

Leb(x ∈ B(0, 1), u ≥ 1) = Leb(x ∈ B(0, 1), v = 0)

From the previous proposition, we can apply the Poincare inequality:∫B(0,1)

|v|2 ≤ C

∫B(0,1)

|∇v|2

Moreover, we can also apply Proposition 6.6 to get

‖v‖L∞(B(0,1/2)) ≤ C‖v‖L2(B(0,1)).

Combining both, we get

‖v‖L∞(B(0,1/2)) ≤ C

∫B(0,1)

|∇v|2.

To prove that the right-hand side is bounded uniformly in v, we take ϕ = χ2

u, χ smooth and

compactly supported in B(0, 2) in the definition of the supersolution. We find:

0 ≤∫B(0,2)

A∇u · ∇(χ2

u

)= −

∫B(0,1)

χ2

u2A∇u · ∇u+ 2

∫B(0,2)

A∇u · ∇χχu

which implies after a few manipulations that∫B(0,2)

χ2|∇ ln(u)|2 ≤ C

∫B(0,2)

|∇χ|2

Taking χ = 1 in a vicinity of B(0, 1), we deduce∫B(0,1)

|∇ ln(u)|2 ≤ C

Eventually, ‖v‖L∞(B(0,1/2)) ≤ C, so that infB(0,1/2) u ≥ c = e−C .

Proposition 6.8. (Control of the oscillations)

Assume that u ∈ H1(B(0, 2)) ∩ L∞(B(0, 2)) satisfies∫B(0,2)

A∇u · ∇ϕ = 0, ∀ϕ ∈ H10 (B(0, 2)).

Then, there exists a γ ∈ (0, 1) depending on d, λ, ‖A‖L∞ such that

oscB(0,1/2)u ≤ γoscB(0,2)u

73

Proof. We set α = infB(0,2) u, β = supB(0,2) u, and α′ = infB(0,1/2) u, β′ = supB(0,1/2) u. Notethe following equivalences:

u ≥ 1

2(α + β) ⇔ 2

u− αβ − α

≥ 1

u ≤ 1

2(α + β) ⇔ 2

β − uβ − α

≥ 1

Hence, there are two cases:

1. Leb(x ∈ B(0, 1), 2u−αβ−α ≥ 1 ≥ 1

2Leb(B(0, 1)). Then we apply the previous proposition

to 2u−αβ−α which is a non-negative supersolution in B(0, 2). It follows that: for some

C > 1,

infB(0,1/2)

2u− αβ − α

>1

C,

which amounts to

α′ ≥ α +1

2C(β − α)

so that β′ − α′ ≤ (1− 12C

)(β − α).

2. Leb(x ∈ B(0, 1), 2 β−uβ−α ≥ 1) ≥ 1

2Leb(B(0, 1)). This case can be handled similarly.

Corollary 6.2. Theorem 6.2 holds in the case f = 0.

Proof. As explained after the statement of Theorem 6.2, it is enough to consider the caseΩ = B(0, 1/4), Ω = B(0, 1). By Corollary (6.1), we already have the L∞ control (remindthat f = 0 here):

‖u‖L∞(B(0,1/2)) ≤ C‖u‖L2(B(0,1)).

We then consider, for an arbitrary x0 ∈ B(0, 1/4):

u1(y) = u(x0 + y/8), un+1(y) = un(y/8) = u(x0 + y/8n+1).

It is easy to see that for all n ≥ 1, un ∈ H1(B(0, 2)) ∩ L∞(B(0, 2)) (u being bounded inB(0, 1/2)) and that it satisfies∫

B(0,2)

An∇un · ∇ϕ = 0, ∀ϕ ∈ H10 (B(0, 2))

where An(y) = A(x0 + y8n

) satisfies the same assumptions as the original A. We can applythe previous proposition, that yields

oscB(0,1/2)un+1 ≤ γoscB(0,2)un+1 ≤ γoscB(0,1/4)un ≤ γoscB(0,1/2)un.

74

This leads tooscB(0,1/2)un ≤ γnoscB(0,1/2)u1

which impliessup

|x−x0|≤ 18n

|u(x)− u(x0)| ≤ 2γn‖u‖L∞(B(0,1/2)).

We claim that this relation implies that there exists C > 0 such that for all x ∈ B(0, 1/4),

|u(x)− u(x0)| ≤ C|x− x0|α‖u‖L∞(B(0,1/2)), with α = − ln γ

ln 8.

Indeed, it is enough to show that there exists C > 0 such that this inequality holds forC|x−x0|α ≤ 2, otherwise no matter the value of C the inequality is obvious. Taking C largeenough so that (2/C)1/α ≤ 1/8, there is a biggest n for which |x − x0| ≤ 1

8n, which implies

by the previous bound:|u(x)− u(x0)| ≤ 2γn‖u‖L∞(B(0,1/2))

One can then use the inequality 18n+1 ≤ |x − x0| to have an upper bound on n in terms of

|x − x0|, and show the desired inequality for a good C. We leave the last details to thereader.

6.2.3 Holder regularity with source

After the treatment of the homogeneous equation (6.7) performed in the previous paragraph,we now consider the equation (6.1) with general source term f . We will use a very usefulcharacterization of Holder spaces due to Campanato, in terms of the growth of local integrals.

For Ω a bounded domain of Rd, u ∈ L1(Ω), and x, r such that B(x, r) ⊂ Ω, we define

ux,r =1

Leb(B(x, r))

∫B(x,r)

u.

Theorem 6.3. (Campanato’s characterization of Holder spaces)

Suppose that u ∈ L2(Ω) satisfies for all B(x, r) ⊂ Ω:∫B(x,r)

|u− ux,r|2 ≤M2rd+2α

for some α ∈ (0, 1). Then u is of class Cα in Ω, and for all Ω b Ω,

‖u‖Cα(Ω) ≤ C(M + ‖u‖L2(Ω))

for some C depending on d, α, Ω, Ω′.

Exercice 6.2. 1. Show that for all c ∈ R, for all u ∈ L2(Ω) and all B(x, r) ⊂ Ω,∫B(x,r)

|u− ux,r|2 ≤∫B(x,r)

|u− c|2.

75

2. Show that if u is of class Cα in a vicinity Ω of B(x, r),∫B(x,r)

|u− ux,r|2 ≤ ‖u‖Cα(Ω)rd+2α.

Proof. Denote R0 = dist(Ω,Ωc). For any x0 ∈ Ω and any 0 < r1 < r2 < R0,

|ux0,r1 − ux0,r2|2 ≤ 2(|u(x)− ux0,r1|2 + |u(x)− ux0,r2|2

)so that after integration over B(x0, r1):

|ux0,r1 − ux0,r2|2 ≤2

ωdrd1

(∫B(x0,r1)

|u− ux0,r1|2 +

∫B(x0,r2)

|u− ux0,r2|2).

Here, ωd is the volume of the unit ball in Rd. Hence,

(6.8) |ux0,r1 − ux0,r2 |2 ≤ CM2r−d1

(rd+2α

1 + rd+2α2

).

For any r < R0, set r1 = r2i+1 , r2 = r

2i, we obtain

|ux0,2−(i+1)r − ux0,2−ir| ≤ C ′M2−(i+1)αrα.

Summing from i = m− 1 to n (with m < n), we find easily

|ux0,2−mr − ux0,2−nr| ≤C ′′

2mαMrα

so that (ux0,2−ir)i is a Cauchy sequence, hence converging, for any r > 0. This limit isindependent of r, as can be seen easily from taking r1 = 2−ir and r2 = 2−ir′, for 0 < r 6=r′ ≤ R0. We denote by u(x0) this limit. Taking m = 0, n = +∞, we find

|ux0,r − u(x0)| ≤ C ′′Mrα, ∀r ≤ R0.

In particular, the convergence as r → 0 is uniform over Ω, and as x0 → ux0,r is continuous,u is continuous as well.

Now, invoking Lebesgue’s differentiation theorem, we know that for a.e. x0, ux0,r → u(x0)as r → 0, which means that u = u a.e. Hence, u is continuous (in the sense that it has acontinuous representative), and from the previous inequality:

(6.9) |ux,r − u(x)| ≤ C ′′Mrα, ∀r ≤ R0.

With r = R0, we obtain:

(6.10) supΩ

|u| ≤ C ′′MRα0 + sup

x∈Ω′ux,R0 ≤ c

(M + ‖u‖L2(Ω)

)

76

As regards the Holder continuity, let x 6= y ∈ Ω, and set r = |x− y|. If r < R0

2, we write

|u(x)− u(y)| ≤ |u(x)− ux,2r|+ |u(y)− uy,2r|+ |ux,2r − uy,2r|.

The first two terms are controlled by (6.9). For the last one, we split again:

|ux,2r − uy,2r|2 ≤ 2(|u(z)− ux,2r|2 + |u(z)− uy,2r|2

)and integrate with respect to z ∈ B(x, 2r) ∩B(y, 2r) ⊃ B(x, r). Thus,

|ux,2r − uy,2r|2 ≤2

Leb(B(x, r))

(∫B(x,2r)

|u− ux,2r|2 +

∫B(y,2r)

|u− uy,2r|2)≤ CM2r2α

We get, if r = |x− y| ≤ R0

2,

|u(x)− u(y)| ≤ CM |x− y|α.

On the other hand, if |x− y| ≥ R0

2

|u(x)− u(y)| ≤ 2 supΩ

|u| ≤ 2α+1

Rα0

|x− y|α ≤ C(M + ‖u‖L2(Ω))|x− y|α

by (6.10).

Corollary 6.3. Assume u ∈ H1loc(Ω) satisfies∫

B(x,r)

|∇u|2 ≤M2rd−2+2α for any B(x, r) ⊂ Ω

for some α ∈ (0, 1). Then u is of class Cα in Ω, and for all Ω b Ω,

‖u‖Cα(Ω) ≤ C(M + ‖u‖L2(Ω))

for some C depending on d, α, Ω, Ω′.

Proof. By the Poincare inequality for functions with zero mean, see Remark 6.2, we obtain∫B(x,r)

|u− ux,r|2 ≤ Cr2

∫B(x,r)

|∇u|2 ≤ C ′M2rd+2α

and the result follows from the previous theorem.

We will now prove Theorem 6.2 in the case f 6= 0. x0 ∈ Ω, and ρ, r such that 0 < r < R0,R0 = dist(Ω,Ωc). We can with no loss of generality assume that R0 < 1. We decomposeu = v + w, where v ∈ H1

0 (B(x, r)) satisfies the homogeneous equation:∫Ω

A∇v · ∇ϕ =

∫Ω

fϕ, ∀ϕ ∈ H10 (B(x, r))

77

(and w = u− v). Taking ϕ = v, we get∫B(x,r)

|∇v|2 ≤ C

∫B(x,r)

|f v| ≤ C‖f‖L

2dd+2 (B(x,r))

‖v ‖L

2dd−2 (B(x,r))

≤ C ′‖f‖L

2dd+2 (B(x,r))

‖∇v‖L2(B(x,r)) ≤ C ′′‖f‖Lq(Ω)r12

(d+2− 2dq

)‖∇v‖L2(B(x,r))

where the second inequality comes from the Sobolev embedding, while the third comes fromHolder inequality. We end up with(6.11)∫

B(x,r)

|∇v|2 ≤ C‖f‖2Lq(B(0,1))r

d+2− 2dq ‖∇v‖L2(B(x,r)) = C ‖f‖2

Lq(B(0,1))rd−2+2α1‖∇v‖L2(B(x,r))

with α1 = 2− dq> 0.

As regards the estimate for w, we state:

Lemma 6.4. Let w ∈ H1(B(x0, r)) satisfying∫B(x0,r)

A∇w · ∇ϕ = 0, ∀ϕ ∈ H10 (B(x0, r)).

There exists α2 ∈ (0, 1) such that for any ρ < r we have:

(6.12)

∫B(x0,ρ)

|∇w|2 ≤ C(ρr

)d−2+2α2∫B(x0,r)

|∇w|2

where C only depends on d, λ, ‖A‖L∞.

Let us postpone the proof of this lemma. On the basis of estimates (6.11) and (6.12), wecan write ∫

B(x0,ρ)

|∇u|2 ≤ 2

(∫B(x0,ρ)

|∇w|2 +

∫B(x0,ρ)

|∇v|2)

≤ C

((ρr

)d−2+2α2∫B(x0,r)

|∇w|2 +

∫B(x0,r)

|∇v|2)

≤ C ′((ρ

r

)d−2+2α2∫B(x0,r)

|∇u|2 +

∫B(x0,r)

|∇v|2)

≤ C ′′((ρ

r

)d−2+2α2∫B(x0,r)

|∇u|2 + ‖f‖2Lq(B(0,1))r

d−2+2α1

).

We then state another lemma:

Lemma 6.5. Let φ a non-negative and non-decreasing function on [0, R]. Assume that

φ(ρ) ≤ A(ρr

)γφ(r) +Brβ,

78

for any 0 < ρ ≤ r ≤ R, with A,B, γ, β ≥ 0, and β < α. Then, for any δ ∈ (β, γ), thereexists c depending on d, β, γ, δ such that

φ(r) ≤ c

(Φ(R)

Rδrδ +Brβ

).

Applying this lemma with R = R0

2, γ = d−2+α2, B = ‖f‖Lq(B(0,1)), β = min(d−2+α1, d−

2 + 2α2), and φ(r) =∫B(x0,r)

|∇u|2, we obtain the existence of α > 0 such that∫B(x0,r)

|∇u|2 ≤ C0

(∫B(x0,

R02

)

|∇u|2 + ‖f‖2Lq(Ω)

)rd−2+2α

As a last step, one must establish that∫B(x0,

R02

)

|∇u|2 ≤ C(‖u‖2

L2(Ω) + ‖f‖2Lq(Ω)

)which is classical: one can for instance test the weak formulation against χ2u, where χ is asmooth truncation that is compactly supported in Ω, and satisfying χ = 1 near B(x0,

R0

2).

See the proof of Proposition 6.6 for similar (but more involved) estimates. Eventually, weget ∫

B(x0,r)

|∇u|2 ≤ C(‖u‖2

L2(Ω) + ‖f‖2Lq(Ω)

)rd−2+2α

and we can conclude by application of Corollary 6.3.

We still have to prove the last two lemmas.

Proof of Lemma 6.4.

By translation and dilation, one can restrict to x0 = 0, r = 2. One can also restrict to ρ ≤ 14,

otherwise the statement is trivial. Furthermore, we can assume that u has zero mean overB(0, 2), because the statement of the lemma is insensitive to the addition of a constant. ByPoincare inequality for functions with zero mean, see Remark 6.2:∫

B(0,2)

w2 ≤ C

∫B(0,2)

|∇w|2.

Hence, Corollary 6.2 yields: for all x ≤ 12,

|w(x)− w(0)|2 ≤ C|x|2α∫B(0,2)

|∇w|2.

We introduce χρ a smooth truncation function that is compactly supported in B(0, 2ρ),χρ = 1 in B(0, ρ), |∇χρ| ≤ 2

ρ. We then take as a test function: ϕ = χ2

ρ(w − w(0)).Straightforward manipulations yield∫

B(0,ρ)

|Dw|2 ≤∫χ2ρ|Dw|2 ≤

C

ρ2

∫B(0,2ρ)

|w − w(0)|2 ≤ C ′ρd−2+2α

∫B(0,2)

|Dw|2

79

see the proof of Proposition 6.6 for similar (but more involved) estimates. We end up with∫B(0,ρ)

|Dw|2 ≤ C ′ρd−2+2α

∫B(0,2)

|Dw|2

Proof of Lemma 6.5.

Let δ ∈ (β, γ), r < R. Choose τ ∈ (0, 1) such that Aτ γ = τ δ. We find

φ(τr) ≤ τ γφ(r) +Brβ.

For all integers k > 0,

φ(τ k+1r) ≤ τ γφ(τ kr) +Bτ kβrβ

τ (k+1)γ +Bτ kβrβk∑j=0

τ j(γ−β)

τ (k+1)γ +Bτ kβrβ

1− τ γ−β

Choosing k such that τ k+2r < ρ ≤ τ k+1r, this inequality gives

φ(ρ) ≤ 1

τ γ

(ρr

)γφ(r) +

Bρβ

τ 2β(1− τ γ−β.

80

Appendix A

Complement to the proof of the localinversion theorem

We show first the following lemma, which easily implies the second part of Lemma 1.1.

Lemma A.1. Let F : X → Y a bi-Lipschitz map. Let U be an open set of X, such thatF is C1 on U and F ′(u) is an isomorphism from X to Y for all u ∈ U . Then, F is a C1

diffeomorphism from U to F (U).

Proof. As F is an homeomorphism fromX to Y , F (U) is an open set. To prove that F−1 is C1

over F (U), it is enough to show that F−1 is differentiable: indeed, one can then differentiatethe identity F F−1 = Id in F (U), and see that necessarily (F−1)′(F (u)) = F ′(u)−1 and istherefore continuous in u.

To prove the differentiability of F−1 at any point f0 = F (u0) ∈ F (U), we first write for anyf = F (u) ∈ F (U):

f − f0 = F (u)− F (u0) = F ′(u0)(u− u0) + ‖u− u0‖ε(u), limu→u0

ε(u) = 0

We apply F ′(u0)−1 to this identity, and replace u by F−1(f), u0 by F−1(f0) to get

F−1(f)− F−1(f0) = F ′(u0)−1(f − f0)− ‖F−1(f)− F−1(f0)‖ε′(f)

where ε′(f) = F ′(u0)−1ε(F−1(f)) goes to zero as f goes to f0 (by the continuity of F−1 atf0). Also, as F−1 is Lipschitz, we can write

‖F−1(f)− F−1(f0)‖ = M(f)‖f − f0‖

where the function M is bounded by Lip(F−1). Eventually,

F−1(f)− F−1(f0) = F ′(u0)−1(f − f0) + ‖f − f0‖ε′′(f)

where ε′′ = M ε′ goes to zero as f goes to f0. This proves the differentiability.

81

[7, 19]

We then give the proof of:

Lemma 1.2. Let r > 0, ρr : x→ Br such that

ρr(x) = x, x ∈ Br, ρr(x) = rx

‖x‖, x /∈ Br

Then, ρr is Lipschitz, with Lip(ρr) ≤ 2.

Proof. We split following the values of ‖x‖ and ‖y‖.

If ‖x‖ ≤ r, ‖y‖ ≤ r, ‖ρr(x)− ρr(y)‖ = ‖x− y‖.

If ‖x‖ ≤ r ≤ ‖y‖,

‖ρr(x)− ρr(y)‖ = ‖x− r y

‖y‖‖ =

1

‖y‖‖ ‖y‖x− ry‖ =

1

‖y‖‖ (‖y‖x− ‖y‖ y) + (‖y‖ y − ry) ‖

≤ ‖x− y‖+ ‖y‖ − r ≤ ‖x− y‖+ ‖y‖ − ‖x‖ ≤ 2‖x− y‖.

If ‖x‖ ≥ r, ‖y‖ ≥ r,

‖ρr(x)− ρr(y)‖ = r‖ x

‖x‖− y

‖y‖‖ ≤ r

‖x‖ ‖y‖‖ ‖y‖x− ‖x‖y ‖

=r

‖x‖ ‖y‖‖(‖y‖x− ‖y‖y) + (‖y‖y − ‖x‖y)‖

≤ r

‖x‖‖x− y‖ +

r

‖x‖| ‖y‖ − ‖x‖| ≤ 2‖x− y‖.

82

Appendix B

Complement to the proof of theisometric embedding

Proposition 4.1.

For any smooth compact submanifold M of RN , any Ck′ riemannian metrics g on M , andfor any bounded neighborhood U of M , there exists a Ck′ riemannian metrics G on RN with

• G = e in RN \ U , e the euclidean scalar product.

• G|M = g, in the sense that for all x ∈M , for all v1, v2 ∈ TxM , Gx(v1, v2) = gx(v1, v2).

Proof. Let (Uj)j∈J be a covering of a neighborhood of M by local charts, meaning that thereare smooth diffeomorphisms ϕj from Uj ⊂ RN to open sets Ωj ⊂ RN such that

ϕj(Uj ∩M) = Ωj ∩ Rd × 0

where d is the dimension of M . We can always assume that ∪j Uj ⊂ U . Let (ψj)j∈J apartition of unity such that the support of ψj is included in Uj, and

∑j ψj = 1 in an open

neighborhood U of M . Let gj = g|Uj∩M , and hj = (ϕj)∗gj, that is the push forward of gj byϕj. It defines a riemannian metrics on Ωj ∩Rd×0, with for all y = (y′, 0) ∈ Ωj ∩Rd×0,hj|y is a scalar product on Rd × 0. We extend it as a metrics over Ωj by setting: for ally = (y′, y′′) ∈ Ωj, for all v = (v′, v′′) ∈ RN ,

hj|y(v1, v2) = hj|(y′,0) ((v′1, 0), (v′2, 0)) + v′′1 · v′′2

where the second term at the right-hand side refers to the scalar product in RN−d. Eventually,we define for all y ∈ Ωj:

Hj|y = χ(y′′)hj|(y′,0) + (1− χ(y′′))(ϕj)∗e

where e is the euclidean scalar product on RN , and χ is a smooth function with values in(0, 1), χ = 1 near 0, χ = 0 outside some open neighbourhood U ′′ of 0. More precisely, wetake U ′′ so that

Ωj ∩ (Rd × U ′′) ⊂ ϕ(Uj ∩ U).

83

Eventually, we define Gj = (ϕj)∗Hj in Uj, Gj = e outside Uj, and set

G =∑

ψjGj in U , G = e outside U .

Proposition 4.2.

Let k′ ≥ 1, g a Ck′ riemannian metrics on M . One can find a family of Ck′−1 riemannianmetrics (gε)ε>0 such that

• ‖gε − g‖Ck′−1 ≤ ε.

• gε = u∗εe, for some uε : M → Rnε of class Ck′ .

Proof. First, we prove that there exists a radial, smooth, and compactly supported functionχ such that

(B.1)

∫Rd

(∇χ)(∇χ)t(x) dx = Id

Taking χ(x) = ψ(|x|), we compute:∫Rd

(∇χ)(∇χ)t(x)dx =

∫ +∞

0

ψ′(r)rn−1

∫Sd−1

er × erdσdr.

We need to show that A =∫Sd−1 er × er dσ is proportional to the identity matrix: indeed,

afterwards, one can take ψ = cψ for ψ a smooth compactly supported function such that∫ +∞0

ψ′(r)rn−1dr 6= 0 and c a proper constant. Therefore, we notice that

A =

∫B(0,1)

(x

|x|⊗ x

|x|

)dx

which implies easily that Ax · x is constant on the unit sphere of Rd. From there, we deducethat Ax is proportional to x for all x (Lagrange multiplier theorem), and a classical argumentimplies that A is a scalar endomorphism.

Now, for a ∈ C∞c (Rd), A ∈ C∞(Rd, GLd(R)), and y ∈ Rd, we define uδ,y : Rd → R by

uδ,y = | det(A(x))|1/2δ1−d/2χ

(A(x)

x− yδ

)a(x).

A direct calculation shows that

∇xuδ,y = | det(A(x))|1/2δ1−d/2a(x)

(At

δ+x− yδ· ∇At(x)

)∇χ

(A(x)

x− yδ

)+O(δ1−d/2)

84

which implies (after the change of variable z = A(x)x−yδ

):∫Rd

(∇xuδ,y)(∇xuδ,y)t dy = a(x)2A(x)t

(∫Rd

(∇χ(z))(∇χ(z))tdz

)A(x) + O(δ)

= a(x)2A(x)tA(x) +O(δ)

where the O(δ) is measured in Ck′−1 if A and a are Ck′ .

To conclude the proof of the proposition, we introduce a partition of unity (ψj)j=1...J asso-ciated to a covering (Uj)j=1...J of the manifold M by local charts. We can then decomposeg =

∑j ψjg and remark that, expressed in local coordinates, each ψjg is of the form a(x)A(x)

where a ∈ C∞c (Rd), while A is Ck′ and has values in the set of symmetric definite positivematrices. Hence, as above, one can construct ujδ,y defined on M such that∫

Rd(∇ujδ,y)(∇u

jδ,y)

t dy → ψjg in Ck′−1, as δ → 0.

By approximating the integral in y by a discrete sum of the form

l∑i=1

λ2i (∇u

jδ,yi

)(∇ujδ,yi)t =

l∑i=1

(∇λi∇ujδ,yi)(∇λiujδ,yi

)t

we find eventually that for each ε > 0, for each j, there exists uj1, . . . , ujl , l = l(ε, j), of class

Ck′ , supported in the local chart Uj, such that

‖k(ε,j)∑i=1

(∇uji )(∇uji )t − ψjg‖Ck′−1 ≤

ε

J

so that

‖J∑j=1

k(ε,j)∑i=1

(∇uji )(∇uji )t − ψjg‖Ck′−1 ≤ ε.

By Lemma 4.1, we deduce that for all ε > 0, there exists n = n(ε) and uε = (uji ) of class Ck′

from M to Rn such that ‖u∗ε e− g‖Ck′−1 ≤ ε.

85

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