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Volume 2012, Article ID 350183,48pages doi:10.1155/2012/350183

Research Article

Complex Hessian Equations on Some Compact K ¨ahler Manifolds

Asma Jbilou

Universit´e Internationale de Rabat (UIR), Parc Technopolis, Rocade de Rabat, Sal´e, 11 100 Sala Al Jadida, Morocco

Correspondence should be addressed to Asma Jbilou,asmajbilou@yahoo.fr Received 29 March 2012; Revised 22 July 2012; Accepted 31 July 2012 Academic Editor: Aloys Krieg

Copyrightq2012 Asma Jbilou. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

On a compact connected 2m-dimensional K¨ahler manifold with K¨ahler formω, given a smooth functionf :M → Rand an integer 1< k < m, we want to solve uniquely inωthe equation

ωkωm−kefωm, relying on the notion ofk-positivity forω ∈ω the extreme cases are solved:

kmbyYau in 1978, andk1 trivially. We solve by the continuity method the corresponding complex elliptickth Hessian equation, more difficult to solve than the Calabi-Yau equationkm, under the assumption that the holomorphic bisectional curvature of the manifold is nonnegative, required here only to derive an a priori eigenvalues pinching.

1. The Theorem

All manifolds considered in this paper are connected.

LetM, J, g, ωbe a compact connected K¨ahler manifold of complex dimensionm≥3.

Fix an integer 2 ≤ km−1. Letϕ : M → Rbe a smooth function, and let us consider the1,1-formω ωi∂∂ϕand the associated 2-tensorgdefined bygX, Y ωX, JY . Consider the sesquilinear formshandhonT1,0defined byhU, V gU, VandhU, V

gU, V. We denote byλg−1gthe eigenvalues ofhwith respect to the Hermitian formh. By definition, these are the eigenvalues of the unique endomorphismAofT1,0satisfying

hU, V hU, AV ∀U, V ∈T1,0. 1.1

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Calculations infer that the endomorphismAwrites

A:T1,0−→T1,0,

Uii−→AjiUijgjgiUij.

1.2

A is a self-adjoint/Hermitian endomorphism of the Hermitian space T1,0, h, therefore λg−1g ∈ Rm. Let us consider the following cone:Γk {λ ∈ Rm/∀1jk, σjλ > 0}, whereσjdenotes thejth elementary symmetric function.

Definition 1.1. ϕis said to bek-admissible if and only ifλg−1g ∈Γk. In this paper, we prove the following theorem.

Theorem 1.2theσkequation. LetM, J, g, ωbe a compact connected K¨ahler manifold of complex dimension m3 with nonnegative holomorphic bisectional curvature, and let f : M → Rbe a function of classCsatisfying

Mefωm mk

Mωm. There exists a unique functionϕ:M → R of classCsuch that

1

M

ϕ ωm0, 1.3

2ωkωm−k ef

mk

ωm. Ek

Moreover the solutionϕisk-admissible.

This result was announced in a note in the Comptes Rendus de l’Acad´e-mie des Sciences de Paris published online in December 20091. The curvature assumption is used, inSection 6.2only, for an a priori estimate onλg−1g as in2, page 408, and it should be removedas did Aubin for the casekmin3, see also4for this case. For the analogue ofEkonCm, the Dirichlet problem is solved in5,6, and a Bedford-Taylor type theory, for weak solutions of the corresponding degenerate equations, is addressed in7. Thanks to Julien Keller, we learned of an independent work8aiming at the same result as ours, with a different gradient estimate and a similar method to estimateλg−1g, but no proofs given for theC0and theC2estimates.

Let us notice that the functionf appearing in the second member of Ek satisfies necessarily the normalisation condition

Mefωm mk

Mωm. Indeed, this results from the following lemma.

Lemma 1.3. Consider

Mωkωm−k

Mωm. Proof. See9, page 44.

Let us writeEkdifferently.

Lemma 1.4. Considerωkωm−k σkλg−1g/ mkωm.

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Proof. LetPM. It suffices to prove the equality at P in a g-normalg-adapted chart z centered atP. In such a chartgij0 δijandgij0 δijλi0, so atz0,ωidzadzaand

ωiλa0dzadza. Thus

ωkωm−k

a

a0dzadza k

b

idzbdzb m−k

a1,..., ak∈{1,..., m}

distinct integers b1,..., bm−k∈{1,..., m}\{a1,..., ak}

distinct integers

imλa10· · ·λak0

dza1dza1 ∧ · · · ∧

dzakdzak

dzb1dzb1 ∧ · · · ∧

dzbm−kdzbm−k .

1.4

Now a1, . . . , ak, b1, . . . , bm−k are m distinct integers of {1, . . . , m} and 2-forms commute therefore,

ωkωm−k

⎜⎜

⎜⎜

⎜⎜

⎜⎜

a1,..., ak∈{1,..., m}

distinct integers b1,..., bm−k∈{1,..., m}\{a1,..., ak}

distinct integers

λa10· · ·λak0

⎟⎟

⎟⎟

⎟⎟

⎟⎟

im

dz1dz1 ∧ · · · ∧

dzmdzm

ωm m!

⎜⎜

a1,..., ak∈{1,..., m}

distinct integers

m−k! λa10· · ·λak0

⎟⎟

ωm m!

ωkωm−k m−k!

m! k!σkλ10, . . . , λm0ωm σk λ

g−1g mk ωm.

1.5

Consequently,Ekwrites:

σk λ

g−1g

ef. Ek

Let us remark that Em corresponds to the Calabi-Yau equation detg/ detg ef, when E1is just a linear equation in Laplacian form. Since the endomorphismAis Hermitian, the spectral theorem provides anh-orthonormal basis forT1,0 of eigenvectorse1, . . . , em:Aei λiei,λ λ1, . . . , λm∈Γk. AtPMin a chartz, we have Mat1,...,∂mAP Aijz1≤i,j≤m, thus

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σkλAP σkλAijz1≤i, j≤m. In addition,Aji gjgigjgiiϕ δijgjiϕ, so the equation writes locally:

σk

λ

δijgjiϕ

1≤i, j≤m

ef. Ek

Let us notice that a solution of this equation Ek

is necessarilyk-admissible 9, page 46.

Let us definefkB σkλBandFkB lnσkλBwhereB Bji1≤i, j≤mis a Hermitian matrix. The functionfk is a polynomial in the variables Bji, specifically fkB

|I|kBII sum of the principal minors of orderkof the matrixB. Equivalently

Ek writes:

Fk

δji gjiϕ

1≤i, j≤m

f.

Ek

It is a nonlinear elliptic second order PDE of complex Monge-Amp`ere type. We prove the existence of ak-admissible solution by the continuity method.

2. Derivatives and Concavity of F

k

2.1. Calculation of the Derivatives at a Diagonal Matrix

The first derivatives of the symmetric polynomial σk are given by the following: for all 1 ≤ im,∂σk/∂λiλ σk−1, iλ whereσk−1, iλ : σk−1|λi0. For 1 ≤ i /jm, let us denoteσk−2, ijλ : σk−2|λiλj0and σk−2, iiλ 0. The second derivatives of the polynomial σk are given by∂2σk/∂λi∂λjλ σk−2, ijλ. We calculate the derivatives of the function fk:HmC → R, whereHmCdenotes the set of Hermitian matrices, at diagonal matrices using the formula:

fkB

1≤i1<···<ik≤m

σ∈Sk

εσBiiσ11 · · ·Biiσk

k

1 k!

1≤i1,...,ik,j1,...,jk≤m

εij1···ik

1···jkBji1

1· · ·Bjik

k,

2.1

where

εji1···ik

1···jk

⎧⎨

1 ifi1, . . . , ik distinct andj1, . . . , jk even permutation ofi1, . . . , ik,

−1 ifi1, . . . , ikdistinct andj1, . . . , jk odd permutation ofi1, . . . , ik, 0 else.

2.2

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These derivatives are given by9, page 48

∂fk

∂Bji

diagb1, . . . , bm

0 ifi /j, σk−1, ib1, . . . , bm ifij, ifi /j 2fk

∂Bjj∂Bii

diagb1, . . . , bm

σk−2,ijb1, . . . , bm

2fk

∂Bji∂Bji

diagb1, . . . , bm

−σk−2,ijb1, . . . , bm,

2.3

and all the other second derivatives offkat diagb1, . . . , bmvanish.

Consequently, the derivatives of the function Fk lnfk : λ−1Γk ⊂ HmC → R at diagonal matrices diagλ1, . . . , λm with λ λ1, . . . , λm ∈ Γk, where λ−1Γk {B ∈ HmC/λB∈Γk}, are given by

∂Fk

∂Bij

diagλ1, . . . , λm

⎧⎨

0 ifi /j, σk−1, iλ

σkλ if ij, 2.4

if i /j 2Fk

∂Bji∂Bji

diagλ1, . . . , λm

σk−2, ijλ σkλ

2Fk

∂Bjj∂Bii

diagλ1, . . . , λm

σk−2, ijλ

σkλ − σk−1, iλσk−1, jλ σkλ2

2Fk

∂Bii∂Bii

diagλ1, . . . , λm

−σk−1, iλ2 σkλ2

2.5

and all the other second derivatives ofFkat diagλ1, . . . , λmvanish.

2.2. The Invariance ofFk and of Its First and Second Differentials The functionFk:λ−1Γk → Ris invariant under unitary similitudes:

∀B∈λ−1Γk, ∀U∈UmC, FkB Fk

tUBU . 2.6

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Differentiating the previous invariance formula 2.6, we show that the first and second differentials ofFkare also invariant under unitary similitudes:

∀B∈λ−1Γk, ∀ζ∈ HmC, ∀U∈UmC, dFkB·ζ dFktUBU·

tUζU ,

2.7

∀B∈λ−1Γk, ∀ζ∈ HmC, ∀Θ∈ HmC, ∀U∈UmC,

d2Fk

B·ζ,Θ d2Fk

tUBU·

tUζU,tUΘU . 2.8

These invariance formulas are allowed to come down to the diagonal case, when it is useful.

2.3. Concavity ofFk

We prove in9the concavity of the functions uλ and more generallyuλB whenu ∈ Γ0Rmand is symmetric9, Theorem VII.4.2, which in particular gives the concavity of the functionsFklnσkλ9, Corollary VII.4.30and more generally lnσkλB9, Theorem VII.4.29.

In this section, let us show by an elementary calculation the concavity of the functionFk. Proposition 2.1. The functionFk:λ−1Γk → R, B→FkB lnσkλBis concave (this holds for allk∈ {1, . . . , m}).

Proof. The functionFkis of classC2, so its concavity is equivalent to the following inequality:

∀B∈λ−1Γk, ∀ζ∈ HmC m

i, j, r, s1

2Fk

∂Brs∂Bijjiζsr ≤0. 2.9

LetBλ−1Γk,ζ ∈ HmC, andUUmCsuch that tUBU diagλ1, . . . , λm. We have λ λ1, . . . , λm∈Γk. Let us denoteζ tUζU∈ HmC:

S: m

i, j, r, s1

2Fk

∂Bsr∂Bjijiζsr

d2Fk

B·ζ, ζ so by the invariance formula2.8

d2Fk

tUBU·

tUζU,tUζU

m

i,j,r,s1

2Fk

∂Brs∂Bij

diagλ1, . . . , λmζjiζsr

m

i /j1

σk−2,ijλ σkλ ζji ζij

ζji

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m

i /j1

σk−2, ijλ

σkλ −σk−1, iλσk−1, jλ σkλ2

:cij

ζiiζjjm

i1

−σk−1, iλ2 σkλ2

ζii 2

m

i,j1

σk−2, ijλ

σkλ ζji2 m

i, j1

cijζiiζjj.

2.10

But cij2lnσk/∂λi∂λjλ, and ζii ∈ R, so m

i,j1cijζiiζjj ≤ 0 by concavity of lnσk

at λ ∈ Γk 10, page 269. In addition, σk−2,ijλ > 0 since λ ∈ Γk 11, consequently m

i,j1−σk−2,ijλ/σkλ|ζji|2≤0, which shows thatS≤0 and achieves the proof.

3. The Proof of Uniqueness

Letϕ0andϕ1be two smoothk-admissible solutions of Ek

such that

Mϕ0ωm

Mϕ1ωm 0. For allt∈0,1, let us consider the functionϕt1 1−0 ϕ0withϕϕ1ϕ0. LetPM, and let us denotehPkt fkδji gjP∂iϕtP. We havehPk1−hPk0 0 which is equivalent to1

0hPktdt0. But

hPkt m

i, j1

m

1

∂fk

∂Bi

δijgjP∂iϕtP gjP

tijP

ijϕP.

3.1

Therefore we obtain

LϕP: m

i, j1

aijP∂ijϕP 0 withaijP 1

0

αtijPdt. 3.2

We show easily that the matrixaijP1≤i, j≤mis Hermitian9, page 53. Besides the function ϕis continuous on the compact manifoldMso it assumes its minimum at a pointm0M, so that the complex Hessian matrix ofϕat the pointm0, namely,∂ijϕm01≤i, j≤2m, is positive- semidefinite.

Lemma 3.1. For allt∈0,1,λg−1gϕtm0∈Γk; namely, the functionsϕtt∈0,1arek-admissible atm0.

Proof. Let us denoteW:{t∈0,1/λg−1gϕtm0∈Γk}. The setWis nonempty, it contains 0, and it is an open subset of0,1. Lettbe the largest number of0,1such that0, t⊂ W.

Let us suppose thatt < 1 and show that we get a contradiction. Let 1 ≤ qk, we have σqλg−1gϕtm0σqλg−1gϕ0m0 hmq0t−hmq00 t

0hmq0sds. Let us prove that

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hmq0s≥0 for alls∈0, t. Fixs∈0, t; the quantityhmq0sis intrinsic so it suffices to prove the assertion in a particular chart atm0. Now atm0in ag-unitarygϕs-adapted chart atm0

hmq0s m

i, j, 1

∂fq

∂Bji

δjigjm0iϕsm0 gjm0iϕm0

m

i1

∂σq

∂λi

λ

g−1gϕs m0 iiϕm0.

3.3

But λg−1gϕsm0 ∈ Γk ⊂ Γq since s ∈ 0, t⊂ W, then ∂σq/∂λiλg−1gϕsm0 > 0 for all 1 ≤ im. Besides,∂iiϕm0 ≥ 0 since the matrix ∂ijϕm01≤i, j≤m is positive-semi- definite. Therefore, we infer thathmq0s≥0. Consequently, we obtain thatσqλg−1gϕtm0σqλg−1gϕ0m0>0sinceϕ0isk-admissible. This holds for all 1qk; we deduce then thatλg−1gϕtm0∈ Γkwhich proves thatt ∈ W. This is a contradiction; we infer then that W 0,1.

We check easily that the Hermitian matrixaijm01≤i, j≤mis positive definite9, page 54and deduce then the following lemma since the mapPaijP 1

0m

1∂fk/∂Biδji gjP∂iϕtPgjPdtis continuous on a neighbourhood ofm0.

Lemma 3.2. There exists an open ballBm0 centered atm0 such that for allPBm0 the Hermitian matrixaijP1≤i, j≤mis positive definite.

Consequently, the operatorLis elliptic on the open set Bm0. But the map ϕ isC, assumes its minimum atm0Bm0, and satisfiesLϕ0; then by the Hopf maximum principle 12, we deduce thatϕP ϕm0for allPBm0. Let us denoteS:{P ∈M/ϕP ϕm0}.

This set is nonempty and it is a closed set. Let us prove thatSis an open set: letmbe a point ofS, soϕm ϕm0, then the mapϕassumes its minimum at the pointm. Therefore, by the same proof as for the pointm0, we infer that there exists an open ballBmcentered atmsuch that for allPBm ϕP ϕmso for allPBmϕP ϕm0thenBm ⊂ S, which proves thatSis an open set. But the manifoldMis connected; thenS M, namely,ϕP ϕm0 for allPM. Besides

Mϕωm0, therefore we deduce thatϕ≡0 onMnamely thatϕ1ϕ0

onM, which achieves the proof of uniqueness.

4. The Continuity Method

Let us consider the one parameter family ofEk,t,t∈0,1

Fk ϕt!

:Fk

δjigjiϕt

1≤i, j≤m

tfln

⎜⎝mk

Mωm

Metfωm

⎟⎠

At

. Ek,t

The functionϕ0 ≡ 0 is ak-admissible solution of Ek,0:σkλδjigjiϕ01≤i, j≤m mk and satisfies

Mϕ0ωm0. Fort1,A11 soEk,1corresponds to Ek

.

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Let us fixl ∈N,l≥5 and 0< α <1, and let us consider the nonempty setcontaining 0:

Tl, α:

t∈0,1/Ek,thave ak-admissible solutionϕCl, αM such that

M

ϕωm0

"

.

4.1

The aim is to prove that 1 ∈ Tl, α. For this we prove, using the connectedness of0,1, that Tl, α 0,1.

4.1.Tl, α Is an Open Set of0,1

This arises from the local inverse mapping theorem and from solving a linear problem. Let us consider the following sets:

Sl, α:

ϕCl, αM,

M

ϕωm0

"

, Sl, α:#

ϕSl, α, k-admissible forg$ ,

4.2

whereSl, αis a vector space andSl, αis an open set ofSl, α. Using these notations, the setTl, α

writesTl, α:{t∈0,1/∃ϕ∈Sl, αsolution ofEk,t}.

Lemma 4.1. The operatorFk : Sl,αCl−2, αM, ϕ → Fkϕ Fkδijgjiϕ1≤i, j≤m, is differentiable, and its differential at a pointϕSl, α,dF ∈ LSl, α, Cl−2, αMis equal to

dF·ψ m

i, j1

∂Fk

∂Bji

δji gjiϕ gjiψ ∀ψ ∈Sl,α. 4.3

Proof. See9, page 60.

Proposition 4.2. The nonlinear operatorFkis elliptic onSl, α.

Proof. Let us fix a functionϕSl, αand check that the nonlinear operatorFkis elliptic for this functionϕ. This goes back to show that the linearization atϕof the nonlinear operatorFkis elliptic. ByLemma 4.1, this linearization is the following linear operator:

dF·v m

i, o1

m

j1

∂Fk

∂Bji

δji gj oioϕ

1≤i, j≤m×gj o

i ov. 4.4

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In order to prove that this linear operator is elliptic, it suffices to check the ellipticity in a particular chart, for example, at the center of ag-normalgϕ-adapted chart. At the center of such a chart,

dF·v m

i, o1

∂Fk

∂Bio

diagλ

g−1g

i ovm

i1

σk−1, iλ g−1g σkλ

g−1g iiv. 4.5 But for alli∈ {1, . . . , m}we haveσk−1,iλg−1g/σ kλg−1g>0 onMsinceλg−1g ∈Γk11, which proves that the linearization is elliptic and achieves the proof.

Let us denoteFkthe operator

Fk ϕ!

:fk

δijgjiϕ

1≤i, j≤m

. 4.6

AsFk, the operatorFk:Sl, αCl−2, αMis differentiable and elliptic onSl, αof differential dF·ψ m

i, j1

∂fk

∂Bij

δji gjiϕ gjiψ ∀ψ∈Sl, α. 4.7

Let us denoteaϕthe matrixδji gjiϕ1≤i, j≤mand calculate this linearization in a different way, by using the expression2.1offk:

Fk ϕ! fk

aϕ 1

k!

1≤i1,..., ik, j1,..., jk≤m

εij1···ik

1···jk

aϕj1

i1· · · aϕjk

ik. 4.8

Thus

dF·v d dt

Fk ϕtv!

|t0

d dt

⎝1 k!

1≤i1,..., ik, j1,..., jk≤m

εji1···ik

1···jk

aϕtvj1

i1· · · aϕtvjk

ik

|t0

1 k!

1≤i1,..., ik, j1,..., jk≤m

εij1···ik

1...jk

gj1si1sv aϕ

j2

i2· · · aϕ

jk

ik

1 k!

1≤i1,..., ik, j1,..., jk≤m

εij1···ik

1···jk

aϕj1

i1

gj2si2sv · · · aϕjk

ik

· · · 1 k!

1≤i1,..., ik, j1,..., jk≤m

εij1···ik

1···jk

aϕ

j1

i1· · · aϕ

jk−1

ik−1

gjksiksv

1 k−1!

1≤i1,..., ik, j1,..., jk≤m

εij1···ik

1···jk

aϕj1

i1· · · aϕjk−1

ik−1

gjksiksv

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by symmetry

m

i, j1

⎝ 1 k−1!

1≤i1,..., ik−1, j1,..., jk−1≤m

εji1···ik−1i

1···jk−1j

aϕj1

i1· · · aϕjk−1

ik−1

:Cijaϕ

jiv.

4.9

We infer then the following proposition.

Proposition 4.3. The linearizationdFkof the operatorFkis of divergence type:

dFi

Cij aϕ

j . 4.10

Proof. By4.9we have dF·v m

i, j1

Cij aϕ

jiv

m

i1

i

m

j1

Cij aϕ

jv

⎠−m

j1

m

i1

i

Cij aϕ

jv.

4.11

Moreover m

i1

i

Cij

aϕ 1

k−2!

m i1

1≤i1,..., ik−1, j1,..., jk−1≤m

εij1···ik−1i

1···jk−1j

aϕj1

i1· · · aϕjk−2

ik−2i aϕjk−1

ik−1 . 4.12

But∇iaϕjik−1k−1iδjik−1k−1jik−1k−1ϕ jiik−1k−1ϕ, then m

i1

i

Cij

aϕ 1

k−2!

m i1

1≤i1,..., ik−1, j1,..., jk−1≤m

εji1···ik−1i

1···jk−1j

aϕj1

i1· · · aϕjk−2

ik−2jiik−1k−1ϕ. 4.13

Besides, the quantity∇jiik−1k−1ϕis symmetric ini, ik−1indeed,∇jiik−1k−1ϕ− ∇jik−1k−1iϕ Rjsiik−1

k−1sϕand Rjsiik−1

k−1 0 since g is K¨ahler, and εij1···ik−1i

1···jk−1j is antisymmetric in i, ik−1; it follows then that m

i1iCijaϕ 0, consequentlydF·vm

i1im

j1Cijaϕjv.

FromProposition 4.3, we infer easily9, page 62the following corollary.

Corollary 4.4. The mapF:Sl,αSl−2, α, ϕFkϕ−mkis well defined and differentiable and its differential equalsdFϕ dFiCijaϕj∈ LSl, α,Sl−2, α.

Now, let t0 ∈ Tl, α and let ϕ0Sl, α be a solution of the corresponding equation Ek,t0:0 et0fAt0mk.

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Lemma 4.5. dFϕ0 :Sl, αSl−2, αis an isomorphism.

Proof. LetψCl−2,αMwith

Mψvg0. Let us consider the equation

i

Cij aϕ0

ju ψ. 4.14

We have Cijaϕ0Cl−2,αM and the matrix Cijaϕ01≤i, j≤m ∂fk/∂Bijδij gjiϕ01≤i, j≤mis positive definitesince Fk is elliptic atϕ0; then by Theorem 4.7 of13, p. 104on the operators of divergence type, we deduce that there exists a unique function uCl, αMsatisfying

Muvg0 which is solution of4.14and then solution ofdFϕ0uψ.

Thus, the linear continuous mapdFϕ0 :Sl, αSl−2, αis bijective, and its inverse is continuous by the open map theorem, which achieves the proof.

We deduce then by the local inverse mapping theorem that there exists an open set U of Sl, α containing ϕ0 and an open set V of Sl−2, α containing 0 such that F : UV is a diffeomorphism. Now, let us consider a real number t ∈ 0,1 very close to t0 and let us check that it belongs also to Tl, α: if |t−t0| ≤ ε is sufficiently small then

etfAtmk−et0fAt0mkCl−2, αMis small enough so thatetfAtmkV, thus there

existsϕUSl,αsuch thatFϕ etfAtmkand consequently there existsϕCl, αMof vanishing integral forgwhich is solution ofEk,t. Hencet∈ Tl, α. We conclude therefore that Tl, αis an open set of0,1.

4.2.Tl,α Is a Closed Set of0,1: The Scheme of the Proof

This section is based on a priori estimates. Finding these estimates is the most difficult step of the proof. Lettss∈Nbe a sequence of elements ofTl, αthat converges toτ ∈0,1, and let ϕtss∈Nbe the corresponding sequence of functions:ϕtsisCl, α,k-admissible, has a vanishing integral, and is a solution of

Fk

δji gjiϕts

1≤i, j≤m

tsflnAts. Ek,ts

Let us prove thatτ∈ Tl, α. Here is the scheme of the proof.

1Reduction to aC2, βMestimate: ifϕtss∈Nis bounded in aC2, βMwith 0< β <1, the inclusionC2, βM ⊂C2M,Rbeing compact, we deduce that after extraction ϕtss∈Nconverges inC2M,RtoϕτC2M,R. We show by tending to the limit thatϕτis a solution ofEk,τ it is then necessarilyk-admissibleand of vanishing integral forg. We check finally by a nonlinear regularity theorem14, page 467that ϕτCM,R, which allows us to deduce thatτ ∈ Tl, αsee9, pages 64–67for details.

2We show that ϕtss∈N is bounded inC0M,R: first of all we prove a positivity Lemma 5.4forEk,t, inspired by the ones of15, page 843 fork m, but in a very different way, required since thek-positivity ofωts is weaker withk < min this case, some eigenvalues can be nonpositive, which complicates the proof, using a polarization method of7, page 1740 cf. 5.2and a G˚arding inequality 5.3; we

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