About Brezis-Merle Problem with holderian condition: the case of three blow-up points
aa r X i v : . [ m a t h . A P ] A p r ABOUT BREZIS-MERLE PROBLEM WITH HOLDERIAN CONDITION: THECASE OF THREE BLOW-UP POINTS.
SAMY SKANDER BAHOURAA
BSTRACT . We consider the following problem on open set Ω of R : ( − ∆ u i = V i e u i in Ω ⊂ R ,u i = 0 on ∂ Ω . We assume that : Z Ω e u i dy ≤ C, and, ≤ V i ≤ b < + ∞ On the other hand, if we assume that V i s − holderian with / < s ≤ , and, Z Ω V i e u i dy ≤ π − ǫ, ǫ > then we have a compactness result, namely: sup Ω u i ≤ c = c ( b, C, A, s, Ω) . where A is the holderian constant of V i .
1. I
NTRODUCTION AND M AIN R ESULTS
We set ∆ = ∂ + ∂ on open set Ω of R with a smooth boundary.We consider the following problem on Ω ⊂ R : ( P ) ( − ∆ u i = V i e u i u i = 0 in Ω ⊂ R , in ∂ Ω . We assume that, Z Ω e u i dy ≤ C, and, ≤ V i ≤ b < + ∞ The previous equation is called, the Prescribed Scalar Curvature equation, in relation withconformal change of metrics. The function V i is the prescribed curvature.Here, we try to find some a priori estimates for sequences of the previous problem.Equations of this type were studied by many authors, see [2, 3, 4, 6, 7, 8, 10, 11, 12, 13, 16].We can see in [4], different results for the solutions of those type of equations with or withoutboundaries conditions and, with minimal conditions on V , for example we suppose V i ≥ and V i ∈ L p (Ω) or V i e u i ∈ L p (Ω) with p ∈ [1 , + ∞ ] .Among other results, we can see in [4], the following important Theorem, Theorem A (Brezis-Merle [4]) . If ( u i ) i and ( V i ) i are two sequences of functions relatively tothe previous problem ( P ) with, < a ≤ V i ≤ b < + ∞ , and without the boundary condition,then, for all compact set K of Ω , up K u i ≤ c = c ( a, b, m, K, Ω) if inf Ω u i ≥ m. A simple consequence of this theorem is that, if we assume u i = 0 on ∂ Ω then, the sequence ( u i ) i is locally uniformly bounded. We can find in [4] an interior estimate if we assume a = 0 ,but we need an assumption on the integral of e u i . We have in [4]: Theorem B (Brezis-Merle [4]) . If ( u i ) i and ( V i ) i are two sequences of functions relatively tothe previous problem ( P ) with, ≤ V i ≤ b < + ∞ , and, Z Ω e u i dy ≤ C, then, for all compact set K of Ω , sup K u i ≤ c = c ( b, C, K, Ω) . If, we assume V with more regularity, we can have another type of estimates, sup + inf . Itwas proved, by Shafrir, see [13], that, if ( u i ) i , ( V i ) i are two sequences of functions solutions ofthe previous equation without assumption on the boundary and, < a ≤ V i ≤ b < + ∞ , thenwe have the following interior estimate: C (cid:16) ab (cid:17) sup K u i + inf Ω u i ≤ c = c ( a, b, K, Ω) . We can see in [7], an explicit value of C (cid:16) ab (cid:17) = r ab . In his proof, Shafrir has used the Stokesformula and an isoperimetric inequality. For Chen-Lin, they have used the blow-up analysiscombined with some geometric type inequality for the integral curvature.Now, if we suppose ( V i ) i uniformly Lipschitzian with A the Lipschitz constant, then, C ( a/b ) =1 and c = c ( a, b, A, K, Ω) , see Br´ezis-Li-Shafrir [3]. This result was extended for H¨olderiansequences ( V i ) i by Chen-Lin, see [7]. Also, we can see in [10], an extension of the Brezis-Li-Shafrir to compact Riemann surface without boundary. We can see in [11] explicit form,( πm, m ∈ N ∗ exactly), for the numbers in front of the Dirac masses, when the solutions blow-up. Here, the notion of isolated blow-up point is used. Also, we can see in [16] refined estimatesnear the isolated blow-up points and the bubbling behavior of the blow-up sequences.In [4], Brezis and Merle proposed the following Problem: Problem (Brezis-Merle [4]) . If ( u i ) i and ( V i ) i are two sequences of functions relatively to theprevious problem ( P ) with, ≤ V i → V in C (Ω) . Z Ω e u i dy ≤ C, Is it possible to prove that: sup Ω u i ≤ c = c ( C, V,
Ω) ?
Here, we assume more regularity on V i , we suppose that V i ≥ is C s ( s -holderian) (1 / . In fact, we extend the result of the author, see [2]. We argueby contradiction and we use some asymptotic estimates for the blow-up functions. Also, we usea term of the Pohozaev identity to conclude to a contradiction.Our main result is: Theorem . Assume that, V i is uniformly s − holderian with / < s ≤ , and, Z B (0) V i e u i dy ≤ π − ǫ, ǫ > , then we have: sup Ω u i ≤ c = c ( b, C, A, s, Ω) . for solutions of the problem ( P ) , here A is the holderian constant of V i .
2. P
ROOF OF THE RESULT :Proof of the theorem:Without loss of generality, we can assume that
Ω = B (0) the unit ball centered on the origin.Here, G is the Green function of the Laplacian with Dirichlet condition on B (0) . We have(in complex notation): G ( x, y ) = 12 π log | − ¯ xy || x − y | , we can write: u i ( x ) = Z B (0) G ( x, y ) V i ( y ) e u i ( y ) dy, We assume that we are in the case of one blow-up point. Following the notation of a previouspaper, see [2], we have: max Ω u i = u i ( x i ) → + ∞ ,δ i = d ( x i , ∂ Ω) → , for ǫ > small enough, and | y | = ǫ , u i ( x i + δ i y ) ≤ C ǫ , and, the sup + inf inequality gives: u i ( x i ) + 4 log δ i ≤ C. lso, we have the following estimates which imply the smallness for a term of the Pohozaevidentity: ||∇ u i || L q ( B ( x i ,δ i ǫ ′ )) = o (1) . ∀ ≤ q < . We have; because V i is s-holderian with / < s ≤ , the following term of the Pohozaevidentity tends to J i = Z B ( x i ,δ i ǫ ′ ) < x i |∇ ( u i − u ) > ( V i − V i ( x i )) e u i dy = o (1) . Now, we set: r i = e − u i ( x i ) / , we write, for | θ | ≤ δ i ǫ ′ r i , < ǫ ′ < , u i ( x i + r i θ ) = Z Ω G ( x i + r i θ, y ) V i ( y ) e u i ( y ) dx == Z Ω − B ( x i , δ i ǫ ′ ) G ( x i , y ) V i e u i ( y ) dy + Z B ( x i , δ i ǫ ′ ) G ( x i + r i θ, y ) V i e u i ( y ) dy = We write, y = x i + r i ˜ θ , with | ˜ θ | ≤ δ i r i ǫ ′ , u i ( x i + r i θ ) = Z B (0 , δiri ǫ ′ ) π log | − (¯ x i + r i ¯ θ )( x i + r i ˜ θ ) | r i | θ − ˜ θ | V i e u i ( y ) r i dy ++ Z Ω − B ( x i , δ i ǫ ′ ) G ( x i + r i θ, y ) V i e u i ( y ) dyu i ( x i ) = Z Ω − B ( x i , δ i ǫ ′ ) G ( x i , y ) V i e u i ( y ) dy + Z B ( x i , δ i ǫ ′ ) G ( x i , y ) V i e u i ( y ) dy Hence, u i ( x i ) = Z B (0 , δiri ǫ ′ ) π log | − ¯ x i ( x i + r i ˜ θ ) | r i | ˜ θ | V i e u i ( y ) r i dy ++ Z Ω − B ( x i , δ i ǫ ′ ) G ( x i , y ) V i e u i ( y ) dy We look to the difference, v i ( θ ) = u i ( x i + r i θ ) − u i ( x i ) = Z B (0 , δiri ǫ ′ ) π log | ˜ θ || θ − ˜ θ | V i e u i ( y ) r i dy + h + h , where, h ( θ ) = Z Ω − B ( x i , δ i ǫ ′ ) G ( x i + r i θ, y ) V i e u i ( y ) dy − Z Ω − B ( x i , δ i ǫ ′ ) G ( x i , y ) V i e u i ( y ) dy, and, h ( θ ) = Z B (0 , δ i ǫ ′ ) π log | − (¯ x i + r i ¯ θ ) y || − ¯ x i y | V i e u i ( y ) dy. Remark that, h and h are two harmonic functions, uniformly bounded.According to the maximum principle, the harmonic function G ( x i + r i θ, . ) on Ω − B ( x i , δ i ǫ ′ ) take its maximum on the boundary of B ( x i , δ i ǫ ′ ) , we can compute this maximum: G ( x i + r i θ, y i ) = 12 π log | − (¯ x i + r i ¯ θ ) y i || x i + r i θ − y i | ≃ π log ( | | x i | ) δ i − δ i (3 ǫ ′ + o (1)) | δ i ǫ ′ ≤ C ǫ ′ < + ∞ ith y i = x i + 2 δ i θ i ǫ ′ , | θ i | = 1 , and | r i θ | ≤ δ i ǫ ′ .We can remark, for | θ | ≤ δ i ǫ ′ r i , that v i is such that: v i = h + h + Z B (0 , δiri ǫ ′ ) π log | ˜ θ || θ − ˜ θ | V i e u i ( y ) r i dy,v i = h + h + Z B (0 , δiri ǫ ′ ) π log | ˜ θ || θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ, with h and h , the two uniformly bounded harmonic functions. Remark:
In the case of 2 or 3 blow-up points, and if we consider the half ball, we havesupplemntary terms, around the 2 other blow-up terms. Note that the Green function of the halfball is quasi-similar to the one of the unit ball and our computations are the same if we considerthe half ball.We assume that, the blow-up limit is and we take: G ( x, y ) = 12 π log | − ¯ xy || x − y | − π log | − xy || ¯ x − y | , . Asymptotic estimates and the case of one, two and three blow-up points :
By the asymptotic estimates of Cheng-Lin see [8], we can see that, we have the followinguniform estimates at infinity:
Lemma 2.1. ∀ ǫ, ǫ ′ > , ∃ k ǫ,ǫ ′ ∈ R + , i ǫ,ǫ ′ ∈ N and C ǫ,ǫ ′ > , such that for i ≥ i ǫ,ǫ ′ and k ǫ,ǫ ′ ≤ | θ | ≤ δ i ǫ ′ r i ( − − ǫ ) log | θ | − C ǫ,ǫ ′ ≤ v i ( θ ) ≤ ( − ǫ ) log | θ | + C ǫ,ǫ ′ , For the proof, we consider the three following sets: A = { ˜ θ, | ˜ θ | ≤ k ǫ } , A = { ˜ θ, | θ − ˜ θ | ≤ | θ | , | ˜ θ | ≥ k ǫ } , and, A = { ˜ θ, | θ − ˜ θ | ≥ | θ | , | ˜ θ | ≥ k ǫ } . where k ǫ is such that; π (1 − ǫ ) ≤ Z B (0 ,k ǫ ) V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ = Z B ( x i ,k ǫ e − ui ( xi ) / ) V i e u i ( y ) dy ≤ π (1 + ǫ ) . In fact, if we assume that we have one blow-up point: Z B (0 , δi ri ) V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ = Z B ( x i , δi ) V i e u i ( y ) dy → π, To have the uniform bounds C ǫ > , we need to bound uniformly the following quantity: A i = Z B (0 , δi ri ) π log | ˜ θ | V i e u i ( y ) r i dy = Z B (0 , δi ri ) π log | ˜ θ | V i e v i (˜ θ ) d ˜ θ. To obtain this uniform bound, we use the CC.Chen and C.S. Lin computations, see [7], to havethe existence of a sequence l i → + ∞ such that: Z B (0 ,l i ) π log | ˜ θ | V i e v i (˜ θ ) d ˜ θ ≤ C, nd, on the other hand, the computations of YY.Li and I. Shafrir, see [11], to have, for l i ≤| ˜ θ | ≤ δ i r i : e v i (˜ θ ) ≤ C | ˜ θ | β +2 , for some < β < .Finaly, A i ≤ C. Remark that, in the estimate of CC.Chen and C.S Lin, see [7], we need the assumption that V i is s − holderian with < s ≤ .To explain more the previous lemma, we write: − πv i + 2 πh + 2 πh = − Z B (0 , δiri ǫ ′ ) ∩ A π log | ˜ θ || θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + , − Z B (0 , δiri ǫ ′ ) ∩ A π log | ˜ θ || θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + − Z B (0 , δiri ǫ ′ ) ∩ A π log | ˜ θ || θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ == − I − I − I . For I , we have: | θ − ˜ θ | ≤ | ˜ θ | , hence, − I ≤ . For I , it is easy to see that: − I ≤ log | θ | Z A V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + C, with C a constant independant of x and i . Here we use the estimates of Chen-Lin.Since, | θ − ˜ θ | ≤ | ˜ θ | + | θ | ≤ | ˜ θ || θ | for | θ | , | ˜ θ | ≥ , we have: − I ≤ log | θ | Z A V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ, Thus, − πv i + 2 πh + 2 πh ≤ log | θ | Z A ∪ A V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + C, Hence, − πv i + 2 πh + 2 πh ≤ (8 π + ˜ ǫ ) log | θ | + C, Thus, v i − h − h ≥ ( − − ǫ ) log | θ | − C. For the rest of the proof, we use the same argument as in Cheng-Lin, see [8].We write: v i − h − h = Z B (0 , δiri ǫ ′ ) ∩ A π log | ˜ θ || θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + , + Z B (0 , δiri ǫ ′ ) ∩ A π log | ˜ θ || θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + Z B (0 , δiri ǫ ′ ) ∩ A π log | ˜ θ || θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ == I + I + I . We have: I ≤ − log | θ | Z A π V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + C, with C a constant independant of x and i . Here we use the estimates of Chen-Lin.For I , we have: I ≤ . For I , we have: I ≤ π Z {| ˜ θ − θ |≤| θ | − σ } log 1 | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + ǫ | θ | , Hence, v i − h − h ≤ ( − ǫ ) log | θ | + 12 π Z {| ˜ θ − θ |≤| θ | − σ } log 1 | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ. As in [8], we can prove that, ( h and h are harmonic and satisfy the mean value theorem): v i − h − h − Z {| ˜ θ − θ | = r = | θ | − σ } ( v i − h − h ) = 12 π Z B r ( x ) log r | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ Z {| ˜ θ − θ | = r = | θ | − σ } ( v i − h − h ) ≤ ( − ǫ ) log | θ | . As in the proof of the theorem 1.1 of [8], we use the Brezis-Merle estimate and the twoprevious estimates to prove that for θ large enough, we have: v i − h − h ≤ ( − ǫ ) log | θ | + C. To see this : (We write v i − h − h = k + q , with q harmonic with the same boundary value as v i , we use Brezis-Merle estimate). Note that, h and h are uniformly bounded. Let Ω = B r ( θ ) ,where r = 2 | θ | − σ we have: ( − ∆ k = V i e k + q in Ω ,k = 0 on ∂ Ω . By the Brezis-Merle estimate: Z Ω e k ≤ C | θ | − σ . We use the fact that q is harmonic to have: q ( θ ) ≤ Cq (0) + ( C − − min Ω q − ) . By the previous computations we have: min Ω q − = min ∂ Ω q − = min ∂ Ω ( v i − h − h ) − ≥ ( − − ǫ ) log | θ | − C, and by the previous mean value estimate, we have: q (0) = Z {| ˜ θ − θ | = r = | θ | − σ } ( v i − h − h ) ≤ ( − ǫ ) log | θ | . Thus, q ( θ ) ≤ C log | θ | . ere C is a constant independant of i and σ .We have by the same computations as in the proof of the theorem 1.1 of [8] to conclude that: Z {| ˜ θ − θ |≤| θ | − σ } e v i ≤ | θ | − σ +2 C , and by Cauchy-Schwarz inequality, we have: π Z {| ˜ θ − θ |≤| θ | − σ } log 1 | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ ! ≤ C, and that, for θ and σ large enough: v i − h − h ≤ ( − ǫ ) log | θ | + C. Now, we extend the previous asymptotic estimates to the first derivatives:we have, after derivation under the integral: ∂ j v i = ∂ j h + ∂ j h + Z B (0 , δiri ǫ ′ ) π θ j − ˜ θ j | θ − ˜ θ | V i e u i ( y ) r i dy, In other words, we have: ∂ j v i = ∂ j h + ∂ j h + Z B (0 , δiri ǫ ′ ) π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ, We have the following lemma:
Lemma 2.2. ∀ ǫ, ǫ ′ > ∃ k ǫ,ǫ ′ ∈ R + , i ǫ,ǫ ′ ∈ N , such that, for i ≥ i ǫ,ǫ ′ and k ǫ,ǫ ′ ≤ | θ | ≤ δ i ǫ ′ r i , ∂ j v i ( θ ) ≃ ∂ j u ( θ ) ± ǫ | θ | + C (cid:18) r i δ i (cid:19) , where u is the solution to: − ∆ u = V (0) e u , in R . For the proof, we consider the three following sets: A = { ˜ θ, | ˜ θ | ≤ k ǫ } , A = { ˜ θ, | θ − ˜ θ | ≤ | θ | , | ˜ θ | ≥ k ǫ } , and, A = { ˜ θ, | θ − ˜ θ | ≥ | θ | , | ˜ θ | ≥ k ǫ } . where k ǫ is such that; π (1 − ǫ ) ≤ Z B (0 ,k ǫ ) V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ = Z B ( x i ,k ǫ e − ui ( xi ) / ) V i e u i ( y ) dy ≤ π (1 + ǫ ) . Remark 1:
In the case of 2 or 3 blow-up points, and if we consider the half ball, we havesupplemntary terms, around the 2 other blow-up terms. Note that the Green function of thehalf ball is quasi-similar to the one of the unit ball. In the case of 3 blow-up points, we havethe following supplementary term ( x i is the principal blow-up point and y i and t i the 2 otherblow-up points): C (cid:18) r i d ( x i , y i ) (cid:19) + C (cid:18) r i d ( x i , t i ) (cid:19) . We assume that, the blow-up limit is and we take: ( x, y ) = 12 π log | − ¯ xy || x − y | − π log | − xy || ¯ x − y | , . In the previous computations, we have considered the unit ball, but by a conformal transfor-mation , we can have the same estimates on the half ball, with a coefficient of the conformaltransformation. We can assume the estimates on the half ball.Now, we consider the following term of the Pohozaev identity J i = Z B ( x i ,δ i ǫ ′ ) < x i |∇ ( u i − u ) > ( V i − V i ( x i )) e u i dy, We want to show that this term tends to as i tends to infinity. We can reduce the problem,after integration by parts, to the following integral: J ′ i = δ i Z B ( x i ,δ i ǫ ′ ) ∂ u i V i e u i dy = δ i Z B ( x i ,δ i ǫ ′ ) ∂ u i ( − ∆ u i ) But, if we take y = x i + r i θ , with, | θ | ≤ δ i ǫ ′ r i , we have: J ′ i = δ i r i Z B (0 , δiri ǫ ′ ) ∂ v i ( − ∆ v i ) == δ i r i Z ∂B (0 , δiri ǫ ′ ) (cid:18) ( ∂ v i ) ν − ( ∂ v i ) ν + ( ∂ v i )( ∂ v i ) ν (cid:19) dσ i , Thus, if we use the uniform asymptotic estimates, we can see that, we reduce the computationto the Pohozaev identity for the limit blow-up function (which equal to ), plus terms in ǫ | θ | and | θ | . First, we tend i to infinity, after ǫ to and finaly , we tend ǫ ′ to .With this method we can have a compactness result for 3 blow-ups points. First, we can seethe case of 3 exteriors blow-up points, then by the previous formulation we have a compactnessresult, it is the case for one of the following cases ( if we set δ i , δ ′ i and δ ′′ i for the radii of eachexterior blow-up) : d ( x i , y i ) δ i → + ∞ and d ( x i , t i ) δ i → + ∞ , or, d ( y i , x i ) δ ′ i → + ∞ and d ( y i , t i ) δ ′ i → + ∞ , or, d ( t i , x i ) δ ′′ i → + ∞ and d ( t i , y i ) δ ′′ i → + ∞ , or,the case when the distance to two exterior blow-up points is of order the radii. In this last case,we divide the region in 3 parts and use the Pohozaev identity directly. In fact, we are reduced tothe case of two blow-up points.In fact, in the case of 3 exterior blow-up points. By the previous formulation around eachexterior blow-up point we look to the one of the 3 first cases. For example, assume the first case.Then we work around the first blow-up. In fact we have, for 3 blow-up points : Lemma 2.3. ∀ ǫ > , ǫ ′ > ∃ k ǫ,ǫ ′ ∈ R + , i ǫ,ǫ ′ ∈ N and C ǫ,ǫ ′ > , such that, for i ≥ i ǫ,ǫ ′ and k ǫ,ǫ ′ ≤ | θ | ≤ δ i ǫ ′ r i , ( − − ǫ ) log | θ | − C ǫ,ǫ ′ ≤ v i ( θ ) ≤ ( − ǫ ) log | θ | + C ǫ,ǫ ′ , and, ∂ j v i ≃ ∂ j u ( θ ) ± ǫ | θ | + C (cid:18) r i δ i (cid:19) | θ | + m × (cid:18) r i δ i (cid:19) + C (cid:18) r i d ( x i , y i ) (cid:19) + C (cid:18) r i d ( x i , t i ) (cid:19) roof of the compactness :By using the lemma we can see that we have the compactness, because:(to understand this, it is sufficient to do the computations for the half ball directly by using theGreen function of the half ball directly).We have after using the previous term of the Pohozaev identity: o (1) = J ′ i = m ′ + C o (1) + C o (1) , ǫ ′ lim ǫ lim i J ′ i = m ′ , which contradict the fact that m ′ > .Proof of the second estimate of the lemma:For the proof, we consider the three following sets: A = { ˜ θ, | ˜ θ | ≤ k ǫ } , A = { ˜ θ, | θ − ˜ θ | ≤ | θ | , | ˜ θ | ≥ k ǫ } , and, A = { ˜ θ, | θ − ˜ θ | ≥ | θ | , | ˜ θ | ≥ k ǫ } . where k ǫ (large enough), is such that; π (1 − ǫ ) ≤ Z B (0 ,k ǫ ) V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ = Z B ( x i ,k ǫ e − ui ( xi ) / ) V i e u i ( y ) dy ≤ π (1 + ǫ ) . We write: ∂ j v i − ∂ j h − ∂ j h = Z B (0 , δiri ǫ ′ ) π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ,∂ j v i − ∂ j h − ∂ j h = Z A π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ + Z A ∪ A π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ Using the estimates of v i , we obtain: ∂ j v i − ∂ j h − ∂ j h = o (1) | θ | + Z A π θ j − ˜ θ j | θ − ˜ θ | V e u (˜ θ ) d ˜ θ + Z A ∪ A π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ Thus, ∂ j v i − ∂ j h − ∂ j h = ∂ j u + o (1) | θ | + Z A ∪ A π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ Finaly, ∂ j v i − ∂ j h − ∂ j h − ∂ j u = o (1) | θ | + Z A ∪ A π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ. For A and A , we have: | Z A π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ | ≤ π Z {| ˜ θ |≥ k ǫ } | θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ ≤ ǫ | θ | , because, Z {| ˜ θ |≥ k ǫ } V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ → , for k ǫ large enough. or θ ∈ A , | ˜ θ | ≥ | θ | and, | ˜ θ | ≥ | θ − ˜ θ | , and we use the estimate of v i to have: | Z A π θ j − ˜ θ j | θ − ˜ θ | V i ( x i + r i ˜ θ ) e v i (˜ θ ) d ˜ θ | ≤ C | θ || θ | − ǫ ≤ ǫ | θ | , for θ large enough and ǫ small enough.Finaly, we have: | ∂ j v i − ∂ j h − ∂ j h − ∂ j u | ≤ ǫ | θ | , for θ large enough.Now, it is easy to see from the definition of h and h that: | ∂ j h − ∂ j h − m r i δ i | ≤ C (cid:18) r i d ( x i , y i ) (cid:19) + C (cid:18) r i d ( x i , t i ) (cid:19) Thus, | ∂ j v i − ∂ j u − m r i δ i | ≤ C (cid:18) r i d ( x i , y i ) (cid:19) + C (cid:18) r i d ( x i , t i ) (cid:19) for θ large enough. R EFERENCES[1] T. Aubin. Some Nonlinear Problems in Riemannian Geometry. 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