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,[email protected] 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
ωk∧ωm−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 ≤ k ≤ m−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
Calculations infer that the endomorphismAwrites
A:T1,0−→T1,0,
Ui∂i−→AjiUi∂jgjgiUi∂j.
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/∀1 ≤ j ≤ k, 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 m ≥ 3 with nonnegative holomorphic bisectional curvature, and let f : M → Rbe a function of classC∞satisfying
Mefωm mk
Mωm. There exists a unique functionϕ:M → R of classC∞such that
1
M
ϕ ωm0, 1.3
2ωk∧ωm−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
Mωk∧ωm−k
Mωm. Proof. See9, page 44.
Let us writeEkdifferently.
Lemma 1.4. Considerωk∧ωm−k σkλg−1g/ mkωm.
Proof. LetP ∈ M. 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,ωidza∧dzaand
ωiλa0dza∧dza. Thus
ωk∧ωm−k
a
iλa0dza∧dza k
∧
b
idzb∧dzb m−k
a1,..., ak∈{1,..., m}
distinct integers b1,..., bm−k∈{1,..., m}\{a1,..., ak}
distinct integers
imλa10· · ·λak0
dza1∧dza1 ∧ · · · ∧
dzak∧dzak ∧
dzb1∧dzb1 ∧ · · · ∧
dzbm−k∧dzbm−k .
1.4
Now a1, . . . , ak, b1, . . . , bm−k are m distinct integers of {1, . . . , m} and 2-forms commute therefore,
ωk∧ωm−k
⎛
⎜⎜
⎜⎜
⎜⎜
⎜⎜
⎝
a1,..., ak∈{1,..., m}
distinct integers b1,..., bm−k∈{1,..., m}\{a1,..., ak}
distinct integers
λa10· · ·λak0
⎞
⎟⎟
⎟⎟
⎟⎟
⎟⎟
⎠
im
dz1∧dz1 ∧ · · · ∧
dzm∧dzm
ωm m!
⎛
⎜⎜
⎝
a1,..., ak∈{1,..., m}
distinct integers
m−k! λa10· · ·λak0
⎞
⎟⎟
⎠ωm m!
ωk∧ωm−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. AtP ∈Min a chartz, we have Mat∂1,...,∂mAP Aijz1≤i,j≤m, thus
σkλAP σkλAijz1≤i, j≤m. In addition,Aji gjgigjgi∂iϕ δijgj∂iϕ, so the equation writes locally:
σk
λ
δijgj∂iϕ
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 gj∂iϕ
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
k2.1. Calculation of the Derivatives at a Diagonal Matrix
The first derivatives of the symmetric polynomial σk are given by the following: for all 1 ≤ i ≤ m,∂σk/∂λiλ σk−1, iλ whereσk−1, iλ : σk−1|λi0. For 1 ≤ i /j ≤ m, 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
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
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 u◦λ and more generallyu◦λB 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∂BijBζjiζsr ≤0. 2.9
LetB ∈ λ−1Γk,ζ ∈ HmC, andU ∈UmCsuch that tUBU diagλ1, . . . , λm. We have λ λ1, . . . , λm∈Γk. Let us denoteζ tUζU∈ HmC:
S: m
i, j, r, s1
∂2Fk
∂Bsr∂BjiBζjiζsr
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
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 cij ∂2lnσk/∂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 functionϕttϕ1 1−tϕ0 ϕ0tϕwithϕϕ1−ϕ0. LetP ∈ M, 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
:αtijP
∂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 pointm0∈M, 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 ≤ q ≤ k, we have σqλg−1gϕtm0−σqλg−1gϕ0m0 hmq0t−hmq00 t
0hmq0sds. Let us prove that
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
δjigjm0∂iϕsm0 gjm0∂iϕm0
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 ≤ i ≤ m. Besides,∂iiϕm0 ≥ 0 since the matrix ∂ijϕm01≤i, j≤m is positive-semi- definite. Therefore, we infer thathmq0s≥0. Consequently, we obtain thatσqλg−1gϕtm0≥ σqλg−1gϕ0m0>0sinceϕ0isk-admissible. This holds for all 1≤q≤k; 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 mapP →aijP 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 allP ∈Bm0 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 atm0∈Bm0, and satisfiesLϕ0; then by the Hopf maximum principle 12, we deduce thatϕP ϕm0for allP ∈Bm0. 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 allP ∈Bm ϕP ϕmso for allP ∈BmϕP ϕm0thenBm ⊂ S, which proves thatSis an open set. But the manifoldMis connected; thenS M, namely,ϕP ϕm0 for allP∈M. Besides
Mϕωm0, therefore we deduce thatϕ≡0 onMnamely thatϕ1≡ϕ0
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
δjigj∂iϕt
1≤i, j≤m
tfln
⎛
⎜⎝mk
Mωm
Metfωm
⎞
⎟⎠
At
. Ek,t
The functionϕ0 ≡ 0 is ak-admissible solution of Ek,0:σkλδjigj∂iϕ01≤i, j≤m mk and satisfies
Mϕ0ωm0. Fort1,A11 soEk,1corresponds to Ek
.
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 solutionϕ∈Cl, 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:
ϕ∈Cl, αM,
M
ϕωm0
"
, Sl, α:#
ϕ∈Sl, 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,α → Cl−2, αM, ϕ → Fkϕ Fkδijgj∂iϕ1≤i, j≤m, is differentiable, and its differential at a pointϕ∈Sl, α,dFkϕ ∈ LSl, α, Cl−2, αMis equal to
dFkϕ·ψ m
i, j1
∂Fk
∂Bji
δji gj∂iϕ gj∂iψ ∀ψ ∈Sl,α. 4.3
Proof. See9, page 60.
Proposition 4.2. The nonlinear operatorFkis elliptic onSl, α.
Proof. Let us fix a functionϕ∈Sl, 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:
dFkϕ·v m
i, o1
⎛
⎝m
j1
∂Fk
∂Bji
δji gj o∂ioϕ
1≤i, j≤m×gj o
⎞
⎠∂i ov. 4.4
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,
dFkϕ·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
δijgj∂iϕ
1≤i, j≤m
. 4.6
AsFk, the operatorFk:Sl, α → Cl−2, αMis differentiable and elliptic onSl, αof differential dFkϕ·ψ m
i, j1
∂fk
∂Bij
δji gj∂iϕ gj∂iψ ∀ψ∈Sl, α. 4.7
Let us denoteaϕthe matrixδji gj∂iϕ1≤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
dFkϕ·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
gj1s∂i1sv aϕ
j2
i2· · · aϕ
jk
ik
1 k!
1≤i1,..., ik, j1,..., jk≤m
εij1···ik
1···jk
aϕj1
i1
gj2s∂i2sv · · · aϕjk
ik
· · · 1 k!
1≤i1,..., ik, j1,..., jk≤m
εij1···ik
1···jk
aϕ
j1
i1· · · aϕ
jk−1
ik−1
gjks∂iksv
1 k−1!
1≤i1,..., ik, j1,..., jk≤m
εij1···ik
1···jk
aϕj1
i1· · · aϕjk−1
ik−1
gjks∂iksv
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:
dFkϕ∇i
Cij aϕ
∇j . 4.10
Proof. By4.9we have dFkϕ·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−2∇i aϕjk−1
ik−1 . 4.12
But∇iaϕjik−1k−1 ∇iδjik−1k−1∇jik−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−2∇jiik−1k−1ϕ. 4.13
Besides, the quantity∇jiik−1k−1ϕis symmetric ini, ik−1indeed,∇jiik−1k−1ϕ− ∇jik−1k−1iϕ Rjsiik−1
k−1∇sϕand 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
i1∇iCijaϕ 0, consequentlydFkϕ·vm
i1∇im
j1Cijaϕ∇jv.
FromProposition 4.3, we infer easily9, page 62the following corollary.
Corollary 4.4. The mapF:Sl,α → Sl−2, α, ϕ→Fϕ Fkϕ−mkis well defined and differentiable and its differential equalsdFϕ dFkϕ∇iCijaϕ∇j∈ LSl, α,Sl−2, α.
Now, let t0 ∈ Tl, α and let ϕ0 ∈ Sl, α be a solution of the corresponding equation Ek,t0:Fϕ0 et0fAt0−mk.
Lemma 4.5. dFϕ0 :Sl, α → Sl−2, αis an isomorphism.
Proof. Letψ ∈Cl−2,αMwith
Mψvg0. Let us consider the equation
∇i
Cij aϕ0
∇ju ψ. 4.14
We have Cijaϕ0∈Cl−2,αM and the matrix Cijaϕ01≤i, j≤m ∂fk/∂Bijδij gj∂iϕ01≤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 u∈Cl, αMsatisfying
Muvg0 which is solution of4.14and then solution ofdFϕ0uψ.
Thus, the linear continuous mapdFϕ0 :Sl, α → Sl−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 Fϕ0 such that F : U → V 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
etfAt−mk−et0fAt0−mkCl−2, αMis small enough so thatetfAt−mk ∈V, thus there
existsϕ∈U⊂Sl,αsuch thatFϕ etfAt−mkand consequently there existsϕ∈Cl, 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 gj∂iϕts
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,Rtoϕτ ∈C2M,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 ϕτ ∈C∞M,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