Università di Roma ”La Sapienza”

Eigenvalue, maximum principle and regularity for fully non linear homogeneous operators. Abstract The main scope of this article is to define the concept of principal eigen-value for fully non linear second order operators in bounded domains that are elliptic, homogenous with lower order terms. In particular we prove maximum and comparison principle, Hölder and Lipschitz regularity. This leads to the existence of a first eigenvalue and eigenfunction and to the existence of solutions of Dirichlet problems within this class of operators.


Introduction
In [5], inspired by the acclaimed work of Berestycki, Nirenberg and Varadhan [3], we extended the definition of principal eigenvalue to Dirichlet problems for fully-non linear second order elliptic operators.
(II)Furthermore, suppose that λ <λ. If f ≤ 0 is bounded and continuous in Ω, then there exists u non-negative, viscosity solution of If moreover f is negative in Ω, the solution is unique.
Henceλ was denoted principal eigenvalue of −F in Ω.
In the case α = 0, and for F a linear uniformly elliptic second order operator, these results are included in [3]; when F is one of the Pucci operators the problem has been treated by Quaas [25] and Busca, Esteban and Quaas [8]. Their papers give a more complete description of the spectrum and also treat bifurcation problems. In [5] and in this note the situation is complicated by the fact that there are no known results about the regularity of the solution, or the existence of the solution, even without the zero order term.
The scope of this article is to complete the results of [5]; indeed we consider operators that depend explicitly on x, we include lower order terms, moreover we defineλ in a more suitable way i.e. without requiring that super-solutions are positive up to the boundary. Precisely we shall study existence of solutions, eigenvalue problems and regularity of the solutions for operators of the following type: G(x, u, ∇u, D 2 u) := F (x, ∇u, D 2 u) + b(x).∇u|∇u| α + c(x)|u| α u where F satisfies assumptions as in [4] i.e. the above assumption (H1) and (H2), plus some continuity with respect to the x variable. See e.g. [16] for similar conditions. Because of the new setting the proofs differ in nature from [5] .
The hypothesis on b and c are quite standard and they will be described in the next section.
The main aim of this paper is to prove the two following existence results: Suppose that f ≤ 0, bounded and continuous, that λ <λ, then there exists a nonnegative solution of G(x, u, ∇u, D 2 u) + λu 1+α = f in Ω, u = 0 on ∂Ω.
If G(x, u, p, X) = −F (x, p, −X)+b(x)·p|p| α +c(x)|u| α u then λ =λ(G). Furthermore if F satisfies (H2) then so does F (x, p, X) = −F (x, p, −X). Hence it is possible to prove for λ the same results than forλ. It is important to remark that in general G = G and hence λ can be different fromλ.
While we were completing this paper, we received a very interesting preprint of Ishii and Yoshimura [18] where similar results are obtained in the case α = 0. Let us mention that they call the eigenvalue a demi-eigenvalue as in the paper of P.L. Lions [23], and they characterize it as the supremum of those λ ∈ IR for which there is a viscosity supersolution u ∈ C(Ω) of F [u] = λu + 1 in Ω which satisfies u ≥ 0 in Ω. " (their F is our −G).
In the next section we state precisely the conditions on G and the definition of viscosity solution in this setting. In section 3 we prove a comparison principle and some boundary estimates that allow to prove that for λ <λ the maximum principle holds. This will be done in the fourth section, where we also provide some estimates onλ and a further comparison principle when λ <λ. In section 5, using Ishii-Lions technique we prove regularity results, these in particular give the required relative compactness for the sequence of solutions that are used to prove the main existence's results in the last section.

Main assumptions and definitions.
In this section, we state the assumptions on the operators G(x, u, ∇u, D 2 u) = F (x, ∇u, D 2 u) + b(x).∇u|∇u| α + c(x)|u| α u treated in this note and the notion of viscosity solution.
The operator F is continuous on IR N × (IR N ) ⋆ × S, where S denotes the space of symmetric matrices on IR N .
(H4) There exists a continuous function ω with ω(0) = 0, such that if (X, Y ) ∈ S 2 and ζ ∈ IR satisfy and I is the identity matrix in IR N , then for all (x, y) ∈ IR N , x = y The condition (H2), usually called uniformly elliptic condition, will be in some cases replaced by the much weaker condition

Remark 1
The assumption (H2) and the fact that F (x, p, 0) = 0 implies that where M = M + − M − is a minimal decomposition of M into positive and negative symmetric matrices.
We now assume conditions for the lower order terms i.e. we shall suppose that b : Ω → IR N and c : Ω → IR are continuous and bounded.
We shall sometimes require (for example for the comparison and the maximum principle) that b satisfies: (H5) -Either α < 0 and b is Hölderian of exponent 1 + α, -or α ≥ 0 and, for all x and y, Let us recall what we mean by viscosity solutions, adapted to our context. It is well known that in dealing with viscosity respectively sub and super solutions one works with u ⋆ (x) = lim sup y,|y−x|≤r u(y) and u ⋆ (x) = lim inf y,|y−x|≤r u(y).

Definition 1
Let Ω be a bounded domain in IR N , then v, bounded on Ω is called a viscosity super solution of G(x, ∇u, Of course u is a viscosity sub solution if for all x 0 ∈ Ω, -Either there exists a ball B(x 0 , δ), δ > 0 on which u = cte = c and 0 ≥ g(x, c), for all x ∈ B(x 0 , δ) -Or ∀ϕ ∈ C 2 (Ω), such that u ⋆ − ϕ has a local maximum on x 0 and ∇ϕ(x 0 ) = 0, one has A viscosity solution is a function which is both a super-solution and a subsolution.
In particular we shall use this definition with G(x, p, X) = F (x, p, X)+b(x)·p|p| α . See e.g. [10] for similar definition of viscosity solution for equations with singular operators.
For convenience we recall the definition of semi-jets given e.g. in [11] In the definition of viscosity solutions the test functions can be substituted by the elements of the semi-jets in the sense that in the definition above one can restrict to the functions φ defined by Remark 2 In all the paper we shall consider that Ω is a bounded C 2 domain. In particular we shall use several times the fact that this implies that the distance to the boundary: d(x, ∂Ω) := d(x) := inf{|x − y|, y ∈ ∂Ω} satisfies the following properties: 3 A comparison principle and some boundary estimates.
We start by establishing a comparison result which is a sort of extension of the comparison Theorem 2.1 in [4].
Theorem 1 Suppose that F satisfies (H1), (H2'), and (H4), that b is continuous and bounded and b satisfies (H5). Let f and g be respectively upper and lower semi continuous.
Suppose that β is some continuous function on IR + such that β(0) = 0. Suppose that φ > 0 in Ω lower semicontinuous and σ upper semicontinous, satisfy, respectively, in the viscosity sense, Suppose that β is increasing on IR + and f ≤ g, or β is nondecreasing and f < g.
Before starting the proof, for convenience of the reader, let us recall the following lemmata proved in [4], the first one being an extension of Ishii's acclaimed result.

Lemma 1
Let Ω be a bounded open set in IR N , which is piecewise C 1 . Let u upper semi-continous in Ω, v lower semicontinous in Ω, (x j , y j ) ∈ Ω 2 , x j = y j , and q ≥ sup{2, α+2 α+1 }. We assume that the function Lemma 2 Under the previous assumptions on F , let v be a lower semicontinuous, viscosity supersolution of for some functions (f, β) upper semi continuous in Ω. Suppose thatx is some point wherex is a strict local minimum of the left hand side and v is not locally constant aroundx. Then, −β(v(x)) ≤ f (x).
Remark: This Lemma was stated and proved for continuous super solutions in [4]. The proof is similar but it is adjusted to lower semi continuous super solutions and is given here for the convenience of the reader.
Proof Without loss of generality we can suppose thatx = 0. Since the infimum is strict, for ǫ > 0, there exists N such that for any n > N We take in the following also N large enough in order that and such that C(diamΩ + 1) q−1 q N < ǫ/4. Since v is not locally contant for any n , there exists (t n , z n ) in We prove in what follows that the infimum is achieved in B(0, 1 n ) and that it is not achieved on t n .
Let us observe indeed the infimum on |x| ≤ 1/n cannot be achieved on t n . Moreover Since the infimum cannot be achieved on t n , let y n , |y n | ≤ 1 n be a point such that the infimum is achieved on y n , then is a test function for v on y n with a gradient = 0 on that point. Since v is a supersolution one gets Let us observe that v(y n ) → v(0). Indeed one has by the lower semicontinuity of v v(0) ≤ lim inf v(y n ) and using v(0) + C|t n | q ≥ v(y n ) + C|y n − t n | q one has the reverse inequality. Then by the uppersemicontinuity of f and β one gets that which is the desired conclusion.
Proof of Theorem 1. Suppose by contradiction that max (σ − φ) > 0 in Ω. Since σ ≤ φ on the boundary, the supremum can only be achieved inside Ω.
Let us consider for j ∈ N and for some q > max(2, α+2 α+1 ) Suppose that (x j , y j ) is a maximum for ψ j . Then (i) from the boundedness of σ and φ one deduces that |x j − y j | → 0 as j → ∞.
Claim: For j large enough, there exist x j and y j such that (x j , y j ) is a maximum pair for ψ j and x j = y j . Indeed suppose that x j = y j . Then one would have and then x j would be a local maximum for and similarly a local minimum for We first exclude that x j is both a strict local maximum and a strict local minimum. Indeed in that case, by Lemma 2 This is a contradiction because either β is increasing Hence x j cannot be both a strict minimum for Φ and a strict maximum for Σ. In the first case there exist δ > 0 and R > δ such that B(x j , R) ⊂ Ω and Then if y j is a point on which the minimum above is achieved, one has and (x j , y j ) is still a maximum point for ψ j since for all (x, y) ∈ Ω 2 This concludes the Claim. In the other case, similarly, one can replace x j by a point y j near x j with and (y j , x j ) is still a maximum point for ψ j .
We can now conclude. By Lemma 1 there exist X j and Y j such that and We can use the fact that σ and φ are respectively sub and super solution to obtain: Passing to the limit and using the fact that g and f are respectively lower and upper semi continuous and β is continuous, we obtain which contradicts our hypotheses in all cases and σ ≤ 0 in Ω. This ends the proof.
As an application of the comparison theorem we will state bounds for sub and super solutions near the boundary. The conclusions given in Proposition 1 and Corollary 2 will be used in the proof of the maximum principle Theorem 3.
Proposition 1 Suppose that F satisfies (H1) and (H2), and that b is bounded.
Let u be uppersemicontinuous subsolution of Then there exists δ > 0 and some constant C 3 that depends only on the structural data such that if the distance to the boundary d(x, ∂Ω) < δ.
Proof Let d(x) = d(x, ∂Ω). First let us observe that one can assume that there exists d 0 such that Ω d0 = {x ∈ Ω such that d(x) < d 0 } the supremum of u is positive because otherwise there is nothing to prove. We recall that Ω is a bounded C 2 domain and using the properties of the distance function stated in Remark 2 we know that D 2 d ≤ C 1 Id for some constant C 1 that depends on Ω. Let Ω δ = {x ∈ Ω; d(x) < δ}. Suppose that δ < a 4(C1(A+a)N +|b|∞) . For some constants γ and C 2 that will be chosen later we introduce We use the inequalities With this choice of constants we have obtained that and furthermore u ≤ ψ on ∂Ω δ .
since u ≤ ψ in Ω δ . This ends the proof.
The comparison principle in [4] allows also to establish a strict maximum principle: Theorem 2 Suppose that F satisfies (H2), b and c are continuous and bounded and b satisfies (H5). Let u be a viscosity non-negative lowersemicontinuous super solution of Then either u ≡ 0 or u > 0 in Ω.
Remark: Other strong maximum principles and strong minimal principles have been established in [2] for a more general class of fully nonlinear operators that are "proper" .
Proof. Using the inequality in (H2), let us recall, using Remark 1, that Hence it is sufficient to prove the proposition when u is a super solution of H does not depend on x and it satisfies the hypothesis of Theorem 1.
Moreover one can assume that c is some negative constant. Indeed, suppose that we have proved that for any u ≥ 0 super solution of we have that u > 0 in Ω. Then if for some u ≥ 0 we have that u is a non negative super solution of (5) and then u > 0 and that would conclude the proof. Hence we suppose by contradiction that x 0 is some point inside Ω on which u(x 0 ) = 0. Following e.g. Vazquez [27], one can assume that on the ball |x − Then take k such that If k is as above, let m be chosen such that The function v is a strict subsolution in the annulus, in the sense that it satisfies H(∇v, Hence u ≥ v everywhere on the boundary of the annulus. In fact u ≥ v everywhere in the annulus, since we can use the comparison principle Theorem 1 for the Then v is a test function for u at x 0 . Then, since u is a super solution and ∇v(x 0 ) = 0: which clearly contradicts the definition of v. Finally u cannot be zero inside Ω. This ends the proof.

Corollary 1 (Hopf )
Let v be a viscosity continuous super solution of Suppose that v is positive in a neighborhood of x o ∈ ∂Ω and v(x o ) = 0 then there exist C > 0 and δ > 0 such that To prove this corollary just proceed as in the proof of Theorem 2 and remark that e −kρ − e −kR ≥ C(R − ρ).
In fact, one can get a better estimate about supersolutions near the boundary i.e. some sort of limited expansion at the order two. We still denote by d(x) the distance to the boundary of Ω and, for d > 0, Proof. We start by proving the following Claim: For some constant C > 0 large enough, there exists a neighborhood of ∂Ω such that Let d 0 be such that in Ω d0 := {x ∈ Ω : d(x) < d 0 } the distance is smooth and there exists C 1 such that |D 2 d| ∞ ≤ C 1 as seen in Remark 2. Note that this implies that tr( We compute the two first derivatives of ϕ : and then 1 ≤ |∇ϕ| ≤ 2 In particular These imply that Hence we obtain This ends the proof of the Claim.
To conclude the proof of the proposition we choose C and d 0 as in the claim, This ends the proof.

Maximum principle.
We can now state and prove the following Maximum principle: Let Ω be a bounded domain of IR n . Suppose that F satisfies (H1), (H2'), (H4), that b and c are continuous and b satisfies (H5). Suppose that τ <λ and that u is a viscosity sub solution of

Remark:
Similarly it is possible to prove that if τ < λ and v is a super solution of Proof. Let λ ∈]τ,λ[, and let v be a super solution of satisfying v > 0 in Ω, which exists by definition ofλ.
We assume by contradiction that sup u(x) > 0 in Ω. Claim : sup u v < +∞. Near the boundary this holds true since from Proposition 1 and Corollary 1 there exists δ > 0 such that In the interior we just use the fact that We now define γ ′ = sup x∈Ω u v and w = γv, where 0 < γ < γ ′ and γ is sufficiently The supremum of u − w is strictly positive, and it is necessarily achieved on x ∈ Ω since on the boundary u − w ≤ 0. One has On the other hand and then w(x) ≥ γ γ ′ u(x). As in the comparison principle, we consider, for j ∈ N and for some q > max(2, α+2 α+1 ): Since sup(u − w) > 0, the supremum of ψ j is achieved in (x j , y j ) ∈ Ω 2 . For j large enough, ψ j achieves its positive maximum on some couple (x j , y j ) ∈ Ω 2 such that 1) x j = y j for j large enough, (this uses lemma 2 and the definition of γ).
2) (x j , y j ) → (x,x) which is a maximum point for u − w and it is an interior and The proof of these facts proceeds similarly to the one given in Theorem 1.

Condition (H4) implies that
Then, using the above inequality, the properties of the sequence (x j , y j ), the condition on b -with C b below being either its Hölder constant or 0-, and the homogeneity condition (H1), one obtains By passing to the limit when j goes to infinity, since c is continuous one gets If c(x) + λ > 0 one obtains that This contradicts the hypothesis that If c(x) + λ = 0 then τ < λ implies that once more a contradiction since τ < λ. This ends the proof.

Bounds onλ
We deal with the particular case of the dimension 1. In that case we shall use variational techniques and weak solutions to estimate the first eigenvalue, this being justified by the following lemma R], and that b is continuous and bounded. Suppose that g is continuous, then the weak solutions (in W 1,2+α (] − R, R[)) and the viscosity solutions of Proof. Suppose first that u ∈ W 1,2+α is a weak solution.
The previous equation can also be written as We now prove that u is a viscosity solution.
Suppose that ϕ is some test function by above for u onx, again we are only intrested in the case ϕ ′ (x) = 0 which implies that u ′ cannot be zero, and ϕ ′′ (x) ≥ u ′′ (x) Then since on those points u is a solution in the classical sense This implies which implies that u is a sub solution .
We prove that the viscosity solutions are weak solutions, in the one dimensional case.
Let v be a weak solution of v = 0 on the boundary. Let now u be a viscosity solution of the same equation. We want to prove that u = v. For that aim let ǫ and let v ǫ be the weak solution of , v ǫ = 0 on the boundary, and v ǫ be the weak solution of , v ǫ = 0 on the boundary. By the previous part v ǫ and v ǫ are viscosity solutions and by the comparison theorem 1 gets Moroever by passing to the limit for weak solutions (for example using variational technics) it is easy to prove that v ǫ and v ǫ tend to v weakly in W 1,2+α and then in particular uniformyl on [−R, R]. We obtain that v = u.
This ends the proof.
and b bounded, then there exist some constants C 1 > 0, C 2 > 0 which depend on a and the bound of b such that Then it is easy to show that λ ≥ λ 1 := inf Indeed, the infimum is achieved and u, a function achieving the infimum, is a weak solution of Due to the previous lemma u is also a viscosity solution. One can assume that u ≥ 0, so u > 0 in Ω, using strong maximum principle of Vazquez. Hence, by definition,λ ≥ λ 1 . But one has, for some universal constant C which is the desired result. This ends the proof.

Proof of Proposition 3.
Suppose that Ω is contained in [−R, R] × IR N −1 , let us define where q = 2.3 q R|b1|∞ a + 2, then |∂ x1 u| ≥ qR q−1 and Finally, using also u(x) ≤ 3 q R q , one gets and, by definition,λ Using the expression of q in function of b 1 one gets the announced estimate.
Remark 3 Let us note that in the case b = cte or when there exists some direction e 1 such that b(x).e 1 = cte and Ω is bounded in this direction one has a better estimate.
Indeed, similarly to [5] one can consider This function is positive on x 1 ∈] − R, R], its gradient is never zero. Hence one has, for some constant C: This implies that which is a more accurate lower bound than in the general case.
Proposition 5 Suppose that R is the radius of the largest ball contained in Ω and suppose that F satisfies assumption (H1) and (H2).
Furthermore let b and c be bounded functions. Then, there exists some constant C 1 which depends only on N , Ω α, a and A, such that Proof. Without loss of generality we can suppose that the largest ball contained in Ω is B R (0). Let σ be defined as Let σ(x) = g(r), for r = |x|. Clearly g ′ (r) = r 2q−1 − r q−1 R q and Furthermore g ′ ≤ 0 while g ′′ ≤ 0 for r ≤ q−1 2q−1 1 q R and positive elsewhere.
Hence by condition (H2) and using the fact that for radial functions the eigenvalues of the Hessian are g ′ r with multiplicity N-1 and g ′′ (see [?]) we get where B 1 = a(2q − 1) + A(N − 1) and B 2 = a(q − 1) + A(N − 1). Let R 1 be defined as Hence the supremum is achieved for r ≤ R 1 . On that set one can use an upper bound for | − F (x, ∇σ, D 2 σ) + b.∇σ|∇σ| α | and a lower bound for σ e.g.
More precisely for r ≤ R 1 and for some constants C 1 , C 2 , C ′ 1 C ′ 2 depending on a, A, N and |b| ∞ one has: Then σ is a subsolution in B R (0) of R α+2 + C2 R α+1 + |c| ∞ , and then according to the maximum principle, Theorem 3, one should have that σ ≤ 0 in B R (0), a contradiction. This ends the proof.

Comparison theorem for λ <λ
Theorem 4 Suppose that F satisfies (H1), (H2'), and (H4), that b and c are continuous and bounded and b satisfies (H5). Suppose that τ <λ, f ≤ 0, f is upper semi-continuous and g is lower semi-continuous with f ≤ g.
Suppose that there exist σ upper semi continuous , and v non-negative and lower semi continuous , satisfying Then σ ≤ v in Ω in each of these two cases: 1) If v > 0 on Ω and either f < 0 in Ω, or g(x) > 0 on every pointx such that f (x) = 0, 2) If v > 0 in Ω, f < 0 and f < g on Ω Proof. We act as in the proof of Theorem 3.6 in [4]. 1) We assume first that v > 0 on Ω. Suppose by contradiction that σ > v somewhere in Ω. The supremum of the function σ v on ∂Ω is less than 1 since σ ≤ v on ∂Ω and v > 0 on ∂Ω, then its supremum is achieved inside Ω. Letx be a point such that We define For j large enough, this function achieves its maximum which is greater than 1, on some couple (x j , y j ) ∈ Ω 2 . It is easy to see that this sequence converges to (x,x), a maximum point for σ v . Since on test functions that have zero gradient the definition of viscosity solutions doesn't require to test equation, we need to prove first that x j , y j can be chosen such that x j = y j for j large enough.
Indeed, if x j = y j one would have for all x ∈ Ω which implies that This means that σ has a local maximum on the point x j . We argue as it is done in the proof of Theorem 1 : If x j is not a strict local maximum then x j can be replaced by x ′ j close to it and then (x ′ j , x j ) is also a maximum point for ψ j . If the maximum is strict using Lemma 2 one gets that hence x j is a local minimum for v, and if it is not strict, there exists x ′ j which is different from x j such that (x ′ j , x j ) is also a maximum point for ψ j . If the minimum is strict using once more Lemma 2 one would have This is a contradiction for j large enough. Indeed, passing to the limit one would get Since σ(x) > v(x) this implies that Now there are two cases either f (x) < 0 or f (x) = 0 and the above inequality is strict. In both cases it contradicts We can take x j and y j such that x j = y j .
Moreover there exist X j and Y j such that and We can use the fact that σ and v are respectively sub and super solution to obtain: Passing to the limit, since c is continuous, we get: Either f (x) = 0 and then we have reached a contradiction because, in that case, by hypothesis g(x) > 0, or f (x) < 0, and then we get This concludes the proof of the first part.
2) For the second part, let m be such that f − g ≤ −m < 0, and f < − m 2 . Let ǫ be given such that by the uniform continuity of the function ( We are now in a position to use the first part of the theorem, since and then u ≤ v + ǫ in Ω. Letting ǫ go to zero we get the required conclusion. This ends the proof.

Regularity results
In this section we shall prove that the viscosity solutions are Hölder continuous.
Since the Hölder estimates depend only on the bounds of f and the structural constants, this Hölder continuity will allow us to have a compactness criteria that will be useful in the next section. Let us note that we state all the results with c = 0. Indeed, one can consider c(x)|u| α u in the right hand side since it is bounded, and get the same regularity results.
Proposition 6 Suppose that F satisfies (H1), (H2), (H3). Let f be a bounded function in Ω. Let u be a viscosity non-negative bounded solution of Then if b is bounded, for any γ ∈ (0, 1) there exists some constant C which depends only on |f | ∞ and |b| ∞ such that for any (x, y) ∈Ω 2 An immediate consequence of the above Proposition is the Corollary 2 Suppose that F satisfies (H1), (H2) and (H3). Suppose that f n is a sequence of continuous and uniformly bounded functions, and u n is a sequence of bounded viscosity solutions of F (x, ∇u n , D 2 u n ) + b(x).∇u n |∇u n | α = f n (x) with b bounded, u n = 0 on ∂Ω. Then the sequence u n is relatively compact in C(Ω).
Proof. The proof relies on ideas used to prove Hölder and Lipschitz estimates in [17], as it is done in [5]. We use Proposition 1 in section 3 which implies in particular that there exists for d(x) := d(x, ∂Ω) ≤ δ.
We construct a function Φ as follows: Let M o and γ be as in (7), M = sup(M o , 2 sup u δ γ ) and Φ(x) = M |x| γ . We also define If the Claim holds this completes the proof, indeed taking x = y we would get that u ⋆ = u ⋆ and then u is continuous. Therefore, going back to (8), for (x, y) ∈ ∆ δ which is equivalent to the local Hölder continuity.

Now we consider interior points. Suppose by contradiction that
Clearlyx =ȳ. Then using Ishii's Lemma [16], there exist X and Y such that We need a more precise estimate, as in [17]. For that aim let : Using −(X + Y ) ≥ 0 and (I − P ) ≥ 0 and the properties of the symmetric matrices one has tr(X + Y ) ≤ tr(P (X + Y )).
For completeness sake we shall now prove some Lipschitz regularity of the solution. To get Lipschitz regularity we need a further assumption as it was done in [5]. Let us remark that Lipschitz regularity is not necessary to prove the existence results, hence this further assumption will be used only in the present part of the paper.
(H7)There exists ν > 0 and κ ∈]1/2, 1] such that for all |p| = 1 , |q| ≤ 1 2 , B ∈ S which implies by homogeneity that for all p = 0 , |q| ≤ |p| One has, then, the following regularity result: Proof of Theorem 5. The proof proceeds similarly to the proof given by Ishii and Lions in [17] and as it is required in that paper, we use the fact that we already know that u is Hölder continuous, together with the additional assumption (H7) .
To simplify the calculation but, without loss of generality we shall suppose that in hypothesis (H2) a = A = 1. Let γ be in ] 1 2κ , 1[ and c γ such that by the Hölder's continuity proved before Let µ be an increasing function such that µ(0) = 0, and µ(r) ≥ r, let l(r) = r 0 ds s 0 µ(σ) σ dσ, let us note that since µ ≥ 0, for r > 0: Let r 0 be such that l ′ (r 0 ) = 1 2 . Let also δ > 0 be given, K = r0 δ , and z be such that d(z, ∂Ω) ≥ 2δ. We We shall now choose all the constants above.
-k is such that k = and L = c γ δ γ−k , using the Hölder continuity of u, one has u(x) − u(y) ≤ ϕ(x, y) on ∂∆ z . Indeed, the assumption on r 0 implies that Φ(x) ≥ M K |x| 2 for |x| ≤ r 0 and then if |x − y| = δ, Suppose by contradiction that for some point (x,ȳ) one has Proceeding as in the previous proof, there exist X, Y such that where the matrices X and Y satisfy with B = D 2 Φ(x −ȳ) and Let us note that similarly to the Hölder case, (11) implies that X + Y −L ≤ 4B and then This gives: Let us note that and From this we get in particular that for δ > 0 small enough (or K large enough) , from which we derive that for K large enough tr(X + Y ) ≤ 0 and for some > 0 universal constant C, and |L| ≤ |tr(X + Y )| for K large enough.
In the following we shall need a bound from above for |X|. In order to make the reading easier the constants C or c will be constants which depend only on the data, and they may vary from one line to another. Remark that the lemma III.1 in [17] ensures the existence of some universal constant such that |X −L| + |Y | ≤ C |B| Let us note that and then with the assumptions on µ, |tr(X + Y )| ≥ C ≥ K|tr(X + Y )| 1 2 from which one derives that |X| ≤ |tr(X + Y )|(1 + 1 ).
We need to prove that For that aim we write We now obtain using assumption (H2) and (H3 ) concerning F From this one gets a contradiction for K large. We have proved that for all x such that d(x, ∂Ω) ≥ 2δ and for y such that |x − y| ≤ δ Recovering the compact set Ω by a finite number of C 2 sets Ω i , Ω i ⊂ Ω i+1 such that d(∂Ω i , ∂Ω i+1 ) ≤ 2δ, the local Lipschitz continuity is proved.

Existence's results
We now prove the existence of non negative solutions of where f is a given positive function. The steps are the following : Step 1: Exhibit a sub and a super solution of the equation when the coefficient of the zero order is non positive and f is constant.
Step 2: Under the same conditions on the zero order term, use Perron's method to solve the equation for any negative function −f .
Step 3: From the previous steps we construct a solution of the above Dirichlet problem when λ <λ without conditions on the sign of c(x).
Step 4: This will also allow to prove the existence of the associated eigenvalue. The first step is obtained by remarking that 0 is a sub solution and establishing the following Proposition 7 Suppose that F satisfies (H1) and (H2), b and c are bounded; furthermore let c be non-positive in Ω. Then there exists a function u which is a nonnegative viscosity super solution of Proof. Let d be the distance function to ∂Ω, which is well defined in Ω and satisfies the properties stated in Remark 2. Let K > diamΩ. Then d ≤ K. Let γ ∈]0, 1[ and let k be a large enough constant to be chosen later. Let u be defined as Clearly u = 0 on the boundary.
Suppose that ψ is a C 2 function such that (u − ψ)(x) ≥ (u − ψ)(x) = 0, for all x in a small neighbourhood ofx. Then J 2,− u(x) = ∅ and then the function φ defined as This implies that J 2,− d(x) = ∅. According to some of the properties of d recalled in the introduction, on such a point d is differentiable and then ∇φ(x) = ∇d(x) has modulus 1.
We use Remark 2 on the distance function and the following inequalities on symmetric matrices Using these with X = D 2 φ and Y = Dφ ⊗ Dφ, and condition (H2) we obtain The function d γ(α+1)−2−α (1+d γ ) k(α+1)+α+2 is decreasing hence it is greater than We shall use this later.
We have obtained that there exists a constant C = C(A, a, |b| ∞ , N ) such that Clearly since γ < 1 we can choose k large enough in order that which gives the result. This ends the proof.
We are now in a position to solve step 2: Theorem 6 Suppose that F satisfies (H1) and (H2), that b and c are continuous with c ≤ 0.
1. If f is continuous, bounded and f ≤ 0 on Ω, then there exists u a nonnegative viscosity solution of Remark 5 In a forthcoming paper, [6] we prove existence's results with general data.

Proof.
Let u 2 be the viscosity super solution given in Proposition 7 (see Remark 4), of We use Perron's method, see Ishii's paper [16]. We define Let v(x) = sup u∈M u(x). We prove that v is both a sub and a super solution.
We use the same process as in [5] to prove that v ⋆ is a sub solution.
We now prove that v ⋆ is a super solution. If not, there would existx ∈ Ω, r > 0 and ϕ ∈ C 2 (B(x, r), with ∇ϕ(x) = 0, satisfying We prove that ϕ(x) < v 2 (x). If not one would have ϕ( hence since v 2 is a super solution and ϕ is a test function for v 2 onx, a contradiction. Then ϕ(x) < v 2 (x). We construct now a sub solution which is greater than v ⋆ and less than v 2 .
Let ε > 0 be such that and let δ be such that for |x −x| ≤ δ: Then One can assume that We take r < δ 4 and such that 0 < r < inf |x−x|≤δ (v 2 (x) − ϕ(x)), and define w is LSC as it is the supremum of two LSC functions. One has w(x) = ϕ(x) + r, and w = v ⋆ for r < |x −x| < δ. w is a sub solution, since when w = ϕ + r one can use ϕ + r as a test function, and using the continuity of c, Elsewhere w = v ⋆ , hence it is a sub solution. Moreover w ≥ v ⋆ , w = v ⋆ and w ≤ g. This contradicts the fact that v ⋆ is the supremum of the sub solutions. Using Hölder regularity we get that v ⋆ is Hölder. For the proof of the second statement, it is enough to remark that u = c o is a sub solution of (13) and then proceed as above with u 2 the solution, given in Proposition 7 (see Remark 4), of G(x, u 2 , ∇u 2 , D 2 u 2 ) ≤ −|f | ∞ , u 2 = c o on ∂Ω. This ends the proof.
We now prove an existence result for λ <λ i.e. step 3.
Proof. We define a sequence by induction with u 1 = 0 and u n+1 as the solution of which exists by the previous theorem.
We have obtained that the sequence u n is bounded. Letting n go to infinity, and using the compactness result (Corollary 2), the sequence being in addition monotone, it converges in its whole to u which is a solution.
The solution is unique if f ≤ −m < 0 on Ω. Indeed suppose that u and v are two solutions then v(1 + ǫ) is a solution with f (1 + ǫ) 1+α in the right hand side. Since it is strictly less than f one gets by the comparison principle 4 that v(1 + ǫ) ≥ u and since ǫ is arbitrary v ≥ u. One can of course exchange u and v and obtain that u = v This ends the proof.
By Theorem 6 the sequence u n is well defined. We shall prove that (u n ) is not bounded. Indeed suppose by contradiction that it is. Then by the Hölder's estimate and the compactness result (Corollary 2), one would have that a subsequence, still denoted u n , tends uniformly to a nonnegative continuous function u which would be a viscosity solution of F (x, ∇u, D 2 u) + b(x).∇u|∇u| α + (c(x) +λ)u 1+α = −1 in Ω u = 0 on ∂Ω.
The boundary condition is given by the uniform convergence. Clearly w is Hölder continuous, and if F satisfies the assumption (H7), then it is also locally Lipschitz continuous. This ends the proof.