EXTREMAL FUNCTIONS FOR MOSER-TRUDINGER TYPE INEQUALITY ON COMPACT CLOSED 4-MANIFOLDS YUXIANG LI, CHEIKH BIRAHIM NDIAYE
Abstract. Given a compact closed four dimensional smooth Riemannian manifold, we prove existence of extremal functions for Moser-Trudinger type inequality. The method used is Blow-up analysis combined with capacity techniques.
Keywords: Moser-Trudinger inequality, Blow-up analysis, Capacity, Extremal function, Green function. 2000 Mathematics Subject Classification: 46E35, 26D10
Contents 1. Introduction 2. Notations and Preliminaries 3. Proof of Theorem 1.1 3.1. Concentration behavior and profile of uk 3.2. Pohozaev type identity and application 3.3. Blow-up analysis 3.4. Capacity estimates 3.5. The test function 3.6. Proof of Theorem 1.1 4. Proof of Theorem 1.2 References
1 4 6 6 10 13 17 25 27 27 28
1. Introduction It is well-known that Moser-Trudinger type inequalities are crucial analytic tools in the study of partial differential equations arising from geometry and physics. In fact, much work has been done on such inequalities and their applications in the last decades, see for example, [1], [3], [4], [6], [8], [18], [22], and the references therein. There are two important objects in the study of Moser-Trudinger type inequalities: one is to find the best constant and the other is to determine whether there exist extremal functions. For the best constant there are the celebrated work of Moser[19] and the extension to higher order derivatives by Adams [1] on flat spaces. In the context of curved spaces Fontana has extended the results of Adams, see [9]. To mention results about extremal functions, we cite the results of Carleson and Chang [5], Flucher [10] and Lin [16] in the Euclidean case and the results of Li [14], [15] in the curved one. In [14] and [15] the author have proved the existence of an extremal function for the classic MoserTrudinger inequality on a compact manifold under a constraint involving only the first derivatives. In this paper, we will extend the results of Li to a compact closed four dimensional smooth Riemannian manifold under a constraint involving the Laplacian. More precisely we prove the following Theorems Date: October 30, 2007. 1
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englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
Theorem 1.1. Let (M, g) be a compact closed smooth 4-dimensional Riemannian manifold. Then setting Z H1 = {u ∈ H 2 (M ) : u = 0,
|∆g u|2 dVg = 1}
M
we have that
Z sup u∈H1
e32π
2
u2
dVg
M
is attained. R On the 4−dimensional manifold (M, g) , the so-called Paneitz operator, which is defined in terms of the scalar curvature Rg and the Ricci tensor Ricg as 2 Pg4 u = ∆2g u + divg ( Rg g − 2Ricg )du u ∈ C ∞ (M ), 3 plays an important role in conformal geometry see [4], [6], [7], [8], [11], [20], [21]. In particular, the relation between the Paneitz operator and the Q-curvature, which is defined as 1 (1.1) Qg = − (∆g Rg − Rg2 + 3|Ricg |2 ), 12 is of great interest. It is well-known that Moser-Trudinger inequalities involving Pg4 play an important role in the problem of prescribing constant Q-curvature see [8], [12], [20]. Therefore it is worth having an extension of Theorem 1.1 concerning the Paneitz operator as well. Our next result goes in this direction. More precisely we have the following. Theorem 1.2. Let (M, g) be a compact closed smooth 4-dimensional Riemannian manifold. Assuming that Pg4 is non-negative and kerPg4 ' R, then setting H2 = {u ∈ H 2 (M ) : u = 0, < Pg4 u, u >= 1} we have
Z sup u∈H2
e32π
2
u2
dVg
M
is attained. Remark 1.3. Since the leading term of Pg4 ( for the definition see the Section 2) is ∆2g then the two Theorems are quite similar. We point out that the same proof is valid for both except some trivial adaptations, hence we will give a full proof of Theorem 1.1 only and sketch the proof of Theorem 1.2 in the last section. Remark 1.4. R We mention that due to a result by Gursky, see [11] if both the Yamabe class Y (g) and M Qg dVg are non-negative, then we have that Pg4 is non-negative and kerPg4 ' R. We are going to describe our approach to prove Theorem 1.1. We will use Blow-up analysis. First of all we take a sequence (αk )k such that αk % 32π 2 , and by using Direct Methods of the Calculus of variations we can find uk ∈ H1 such that Z Z 2 αk u2k e eαk v dVg . dVg = sup v∈H1
M
M
see Lemma 3.1. Moreover using the Lagrange multiplier rule we have that (uk )k satisfies the equation: uk αk u2k ∆2g uk = e − γk (1.2) λk for some constants λk and γk . R 2 Now it is easy to see that if there exists α > 32π 2 such that M eαuk dVg is bounded, then by using Lagrange formula, Young’s inequality and Rellich compactness Theorem, we obtain that the weak limit of uk becomes an extremizer. On the other hand if ck = max |uk | = |uk |(xk ) M
englishM-T PROBLEM
3
is bounded, then from standard elliptic regularity theory uk is compact, thus converges uniformly to an extremizer. Hence assuming that Theorem 1.1 does not hold, we get 1) Z 2 ∀α > 32π 2 lim eαuk dVg → +∞ k→+∞
M
2) ck → +∞ We will follow the same method as in [14] up to some extents. In [14], the function sequence we studied is the following: uk α0k u2k e −∆g uk = − γk , λk R 0 2 where αk0 % 4π, and uk attains R sup eαk u dVg . We also assumed ck → +∞. Then M M
|∇g u|2 dVg =1,¯ u=0
we have 2αk ck (uk (xk + rk x) − ck ) → −2 log(1 + π|x|2 ) for suitable choices of rk , xk . Next we proved the following Z 1 |∇g uk |2 dVg = ∀A > 1, lim k→+∞ {u ≤ ck } A k A
(1.3)
(1.4)
which implies that Z
2
eαk uk dVg = µ(M ) + lim
lim
k→+∞
M
k→+∞
λk , c2k
and that ck uk converges to some Green function weakly. In the end, we got an upper bound of via capacity.
λk c2k
Remark 1.5. (1.3) was first discovered by Struwe in [23]. Remark 1.6. (1.4) also appeared in [2]. However there are two main differences between the present case and the one in [14]. One is that there is no direct maximum principle for equation (1.2) and the other one is that truncations are not allowed in the space H 2 (M ) . Hence to get a counterpart of (1.3) and (1.4) is not easy. To solve the first difficulty, we replace ck (uk (xk +rk x)−ck ) with βk (uk (expxk (rk x))−ck ), where Z |uk | αk u2k e dVg . 1/βk = M λk By using the strength of the Green representation formula, we get that the profile of uk is either a constant function or a standard bubble. The second difficulty will be solved by applying capacity and Pohozaev type identity. In more detail we will prove that βk uk * G (see Lemma 3.6) which satisfies 2 R∆g G = τ (δx0 − V olg (M )) G = 0. M for some τ ∈ (0, 1]. Then we can derive from a Pohozaev type identity (see Lemma 3.7) that Z 2 λk eαk uk dVg = V olg (M ) + lim τ 2 2 . lim k→+∞ M k→+∞ βk In order to apply the capacity, we will follow some ideas in [12]. Concretely, we will show that up to a small term the energy of uk on some annulus is bounded below by the Euclidean one (see Lemma 3.10). Moreover one can prove the existence of Uk (see Lemma 3.11) such that the energy of Uk is comparable to the Euclidean energy of uk , and the Dirichlet datum and Neumann datum of Uk at the boundary of the annulus are also comparable to those of uk . In this sense,
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englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
we simplify the calculation of capacity in [15]. Now using capacity techniques we get dτ = 1, see Proposition 3.12. Furthermore we have that lim τ 2
k→+∞
ck βk
→ d and
λk π 2 5 +32π2 S0 ≤ e3 . 2 βk 6
Hence we arrive to Z sup u∈H1
M
e32π
2
u2
dVg ≤ V olg (M ) +
π 2 5 +32π2 S0 e3 . 6
(1.5)
In the end, we will find test functions in order to contradict (1.5). We will simplify the arguments in [14]. Indeed we use carefully the regular part of G to avoid cut-off functions and hence making the calculations simpler. The plan of the paper is the following: In Section 2 we collect some preliminary results regarding the existence of the Green functions for ∆2g and Pg4 , and associated Moser-Trudinger type inequality. In Section 2 we prove Theorem 1.1. This Section is divided into six subsections. In the first one, we deal with concentration behavior and the profile of the blowing-up sequence. The second one is concerned about the derivation of a Pohozaev type identity and its application. In subsection 3 we perform the Blow-up analysis to get either the zero function or a standard bubble in the limit. In the subsection 4, we deal with the capacity estimates to get an upper bound. And in the subsection 5, we construct test functions . In the last subsection we show how to reach a contradiction. The last Section is concerned about the sketch of the proof of Theorem 1.2. Acknowledgements The second author has been supported by M.U.R.S.T within the PRIN 2004 Variational methods and nonlinear differential equations. 2. Notations and Preliminaries In this brief section we collect some useful notations, and state a lemma giving the existence of the Green functions of ∆2g and of the operator Pg4 with the asymptotics near the singularity. We also give a version of Adams inequality on the a manifold due to Fontana and an analogue of the well-known Moser-Trudinger inequality for the operator Pg4 when it is non-negative. In the following, Br (x) stands for the metric ball of radius r and center x in M , B r (p) and stands for the Euclidean ball of center p and radius r. We also denote with dg (x, y) the metric distance between two points x and y of M . H 2 (M ) stands for the usual Sobolev space of functions on M , i.e functions which are in L2 together with their first and second derivatives. W 2,q (M ) denotes the usual Sobolev spaces of functions which are in Lq (M ) with their first and second derivatives. Large positive constants are always denoted by C, and the value of C is allowed to vary from formula to formula and also within the same line. M 2 stands for the cartesian product M ×M , while Diag(M ) is the diagonal of M 2 . Given a function u ∈ RL1 (M ), u ¯ denotes −1 R its average on M , that is u ¯ = (V olg (M )) u(x)dV (x) where V ol (M ) = dV . g g g M M Ak = ok (1) means that Ak → 0 as the integer k → +∞. Aδ = oδ (1) means that Aδ → 0 as the real number δ −→ 0. Ak,δ = ok,δ (1) means that Ak,δ → 0 as k → +∞ first and after the real number δ −→ 0. Ak = O(Bk ) means that Ak ≤ CBk for some fixed constant C. injg (M ) stands for the injectivity radius of M . dVg denotes the Riemannian measure associated to the metric g. dSg stands for the surface measure associated to g. Given a metric g on M , and x ∈ M , |g(x)|, stands for determinant of the matrix with entries gi,j (x) where gi,j (x) are the components of g(x) in some system of coordinates. ∆0 stands for the Euclidean Laplacian and ∆g the Laplace-Beltrami with respect to the background metric g. As mentioned before we begin by stating a lemma giving the existence of the Green function of ∆2g and Pg4 , and their asymptotics near the singularities.
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5
Lemma 2.1. We have that the Green function F (x, y) of ∆2g exists in the following sense : a) For all functions u ∈ C 2 (M ), we have Z u(x) − u ¯= F (x, y)∆2g u(y)dVg (y) x 6= y ∈ M M
b) F (x, y) = H(x, y) + K(x, y) is smooth on M 2 \ Diag(M 2 ), K extends to a C 1+α function on M 2 and H(x, y) =
1 1 f (r) log 8π 2 r
where, r = dg (x, y) is the geodesic distance from x to y; f (r) is a C ∞ positive decreasing function, f (r) = 1 in a neighborhood of r = 0 and f (r) = 0 for r ≥ injg (M ). Moreover we have that the following estimates holds |∇g F (x, y)| ≤ C
1 1 |∇2g F (x, y)| ≤ C . dg (x, y) dg (x, y)2
Proof. For the proof see [6] and the proof of Lemma 2.3 in [17].
Lemma 2.2. Suppose KerPg4 ' R. Then the Green function Q(x, y) of Pg4 exists in the following sense : a) For all functions u ∈ C 2 (M ), we have Z u(x) − u ¯= Q(x, y)Pg4 u(y)dVg (y) x 6= y ∈ M M
b) Q(x, y) = H0 (x, y) + K0 (x, y) 2
2
is smooth on M \ Diag(M ), K extends to a C 2+α function on M 2 and H(x, y) =
1 1 f (r) log 8π 2 r
where, r = dg (x, y) is the geodesic distance from x to y; f (r) is a C ∞ positive decreasing function, f (r) = 1 in a neighborhood of r = 0 and f (r) = 0 for r ≥ injg (M ). Proof. For the proof see Lemma 2.1 in [20].
Next we state a Theorem due to Fontana[9]. Theorem 2.3. ([9]) There exists a constant C = C(M ) > 0 such that the following holds Z Z 2 2 e32π u dVg ≤ C for all u ∈ H 2 (M ) such that |∆2g u|dVg = 1. M
M
Moreover this constant is optimal in the sense that if we replace it by any α bigger then the integral can be maken as large as we want. Next we state a Moser-Trudinger type inequality corresponding to Pg4 when it is non-negative. The proof can be found in [20] where it is proven for every Pgn (where Pgn stands for higher order Paneitz operator). Proposition 2.4. Suppose that Pg4 is non-negative and that kerPg4 = R, then there exists a constant C = C(M ) > 0 such that Z
2 2 e32π u dVg ≤ C for all u ∈ H 2 (M ) such that Pg4 u, u = 1. M
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englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
3. Proof of Theorem 1.1 Lemma 3.1. Let αk be an increasing sequence converging to 32π 2 . Then for every k there exists uk ∈ H1 such that Z Z 2
2
eαk u dVg .
eαk uk dVg = sup
u∈H1
M
M
Moreover uk satisfies the following equation 2 1 uk eαk uk − γk λk
∆2g uk =
(3.1)
where Z λk =
2
u2k eαk uk dVg
M
and γk =
1 λk V olg (M )
Z
2
uk eαk uk dVg .
M
∞
Moreover we have uk ∈ C (M ). Proof. First of all using the inequality in Theorem 2.3, one can check easily that the functional Z 2 Ik (u) = eαk u dVg ; M
is weakly continous. Hence using Direct Methods of the Calculus of Variations we get the existence of maximizer say uk . On the other hand using the Lagrange multiplier rule one get the equation (3.1). Moreover integrating the equation (3.1) and after multipling it by uk and integrating again, we get the value of γk and λk respectively. Moreover using standard elliptic regularity we get that uk ∈ C ∞ (M ). Hence the Lemma is proved. Now we are ready to give the proof of Theorem 1.1. ¿From now on we suppose by contradiction that Theorem 1.1 does not hold. Hence from the same considerations as in the Introduction we have that : 1) ∀α > 32π
2
Z lim
k→+∞
2
eαuk dVg → +∞
(3.2)
M
2) ck = max |uk | = |uk |(xk ) → +∞ M
We will divide the reminder of the proof into six subsections. 3.1. Concentration behavior and profile of uk . This subsection is concerned about two main ingredients. The first one is the study of the concentration phenomenon of the energy corresponding to uk . The second one is the description of the profile of βk uk as k → +∞, where βk is given by the relation Z |uk | αk u2k e dVg . 1/βk = M λk We start by giving an energy concentration lemma which is inspired from P.L.Lions’work. Lemma 3.2. uk verifies : uk * 0 in H 2 (M ); and |∆g uk |2 δx0 for some x0 ∈ M .
englishM-T PROBLEM
7
Proof. First of all from the fact that uk ∈ H1 we can assume without loss of generality that uk * u0 in H 2 (M ). Now let us show that u0 = 0. We have the trivial identity Z Z |∆g (uk − u0 )|2 dVg = M
|∆g uk |2 dVg +
M
Z
(3.3)
|∆g u0 |2 dVg − 2
Z
M
R Hence using the fact that M |∆g uk |2 dVg = 1 we derive Z Z Z 2 2 |∆g (uk − u0 )| dVg = 1 + |∆g u0 | dVg − 2 M
∆g uk ∆g u0 dVg . M
M
∆g uk ∆g u0 dVg
M
So using (3.3) we get Z
|∆g (uk − u0 )|2 dVg 1 −
lim
k→0
Z
M
∆g u0 ∆g u0 dVg M
Now suppose that u0 6= 0 and let us argue for a contradiction. Then there exists some β < 1 such that for k large enough the following holds Z |∆g (uk − u0 )|2 dVg < β. M
Hence using Fontana’s result see Theorem 2.3 we obtain that Z 2 eα1 (uk −u0 ) dVg ≤ C for some α1 > 32π 2 . M
Now using Cauchy inequality one can check easily that Z 2 eα2 uk dVg ≤ C for some α2 > 32π 2 . M
Hence reaching a contradiction to (3.2). On the other hand without lost of generality we can assume that |∆g uk |dVg * µ. Now suppose µ 6= δp for every p ∈ M and let us argue for a contradiction to (3.2) again. First of all let us take a cut-off function η ∈ C0∞ (Bδ (x)) , η = 1 on Bx ( 2δ ) where x is a fixed point in M and δ a fixed positive and small number. We have that Z |∆g uk |2 dVg < 1.
lim sup k→+∞
Bδ (x)
Now working in a normal coordinate system around x and using standard elliptic regularity theory we get Z Z B δ (˜ x)
|∆0 ηg uk |2 dVg ≤ (1 + oδ (1))
|∆g uk |2 dVg ;
Bδ (x)
where x ˜ is the point corresponding to x in R4 and ηg uk the expression of ηuk on the normal coordinate system. Hence for δ small we get Z |∆0 ηg uk |2 dVg < 1 B δ (˜ x)
Thus using the Adams result see [1] we have that Z ^ 2 ˜ ηu k) eα( dx ≤ C for some α ˜ > 32π 2 . B δ (˜ x)
Hence using a covering argument we infer that Z ¯ 2k eαu dVg ≤ Cfor someα ¯ > 32π 2 , M
so reaching a contradiction. Hence the Lemma is proved.
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englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
Lemma 3.3. We have the following hold: lim λk = +∞,
lim γk = 0.
k→+∞
k→+∞
Proof. Let N > 0 be large enougth. By using the definition of λk we have that Z Z Z Z 2 2 2 2 λk = u2k eαk uk dVg ≥ N 2 eαk uk dVg = N 2 ( eαk uk dVg − eαk uk dVg ). {uk ≥N }
M
On the other hand Z lim k→+∞
αk u2k
e
!
Z
αk u2k
dVg −
e
dVg
Z = lim
k→+∞
{uk ≤N }
M
{uk ≤N }
M
2
eαk uk dVg − V olg (M ).
M
Hence using the fact that Z lim
k→+∞
2
eαk uk dVg = sup u∈H1
M
Z
e32π
2
u2
dVg > V olg (M )
M
we have that 1) holds. Now we prove 2). using the definition of γk , we get 2 2 1 1 N N e32π N + . |γk | ≤ λk V olg (M ) N Hence by using point 1 and letting k → +∞ and after N → +∞ we get point 2. So the Lemma is proved. Next let us set
Z τk = M
βk uk αk u2k e . λk
One can check easily the following Lemma 3.4. With the definition above we have that 0 ≤ βk ≤ ck , |τk | ≤ 1 and βk γk is bounded. Moreover up to a subsequence and up to changing uk to −uk τk → τ ≥ 0. The next Lemma gives some Lebesgue estimates on Ball in terms of the radius with constant independent of the ball. As a corollary we get the profile of βk uk as k → +∞. Lemma 3.5. There are constants C1 (p),and C2 (p) depending only on p and M such that, for r sufficiently small and for any x ∈ M there holds Z |∇2g βk uk |p dVg ≤ C2 (p)r4−2p ; Br (x)
and
Z
|∇g βk uk |p dVg ≤ C1 (p)r4−p
Br (x)
where, respectively, p < 2, and p < 4. Proof. First of all using the Green representation formula we have Z uk (x) = F (x, y)∆2g uk dVg (y) ∀x ∈ M. M
Hence using the equation we get Z Z 1 αk u2k uk (x) = F (x, y) uk e dVg (y) − F (x, y)γk dVg (y). λk M M Now by differentiating with respect to x for every m = 1, 2 we have that Z Z 1 m m αk u2k |∇g uk (x)| ≤ |∇g F (x, y)| |uk |e dVg (y) + |∇m g F (x, y)| |γk | . λk M M Hence we get Z Z 1 αk u2k m m |∇m |uk |e dVg (y) + |∇g (βk uk (x))| ≤ |∇g F (x, y)|βk g F (x, y)|βk |γk | . λk M M
englishM-T PROBLEM
9
Taking the p-th power in both side of the inequality and using the basic inequality (a + b)p ≤ 2p−1 (ap + bp ) for a ≥ 0 and b ≥ 0 we obtain p |∇m g (βk uk (x))|
p−1
Z
≤2
|∇m g F (x, y)|βm
M
+2p−1
p |uk |e dVg (y) p |∇m F (x, y)|β |γ | k k g
1 λk
Z
αk u2k
M
Now integrating both sides of the inequality we obtain p Z Z Z 1 p−1 m αk u2k dVg (z) |∇m (β u (z))|dV (z) ≤ 2 |∇ F (z, y)|β |u |e dV (y) k k g k k g g g λk Br (x) Br (x) M Z p Z p−1 m +2 |∇g F (z, y)|βk |γk | dVg (z). Br (x)
M
First let us estimate the second term in the right hand side of the inequality p Z Z Z 1 dV (z) ≤ C |γ | sup |∇m F (z, y)|β dVg (z) ≤ C(M )r4−mp g k k g pm (z, y) d g y∈M Br (x) Br (x) M Thanks to the fact that βk γk is bounded, to the asymptotics of the Green function and to Jensen’s inequality. Now let us estimates the second term. First of all we define the following auxiliary measure 2 1 |uk |eαk uk dVg mk = βk λk We have that mk is a probability measure. On the other hand we can write p Z Z 1 αk u2k m |uk |e dVg (y) dVg (z) |∇g F (z, y)|βk λk Br (x) M (3.4) Z p Z = |∇m F (z, y)|dm (y) dV (z). k g g Br (x)
M
Now by using Jensen’s inequality we have that Z p Z m |∇g F (z, y)|dmk (y) ≤ M
p |∇m g F (z, y)| dmk (y)
M
Thus with the (3.4) we have that p Z Z 1 αk u2k |u |e dV (y) dVg (z) ≤ |∇m F (z, y)|β k k g g λk Br (x) M Z Z m p |∇g F (z, y)| dmk (y) dVg (z). Br (x)
M
Now by using again the same argument as in the first term we obtain Z Z m p |∇g F (z, y)| dmk (y) dVg (z) ≤ C(M )r4−mp . Br (x)
M
Hence the Lemma is proved.
Next we give a corollary of this Lemma. Corollary 3.6. We have βk uk * G W 2,p (M ) for p ∈ (1, 2), βk uk * G smoothly in M \Bδ (x0 ) where δ is small and G satisfies 2 ∆g G = τ (δx0 − V olg1(M ) ) in M ; G=0 Moreover G(x) =
1 τ log + τ S(x) 8π 2 r
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englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
with r = dg (x, x0 ). S = S0 + S1 (x) , S0 = S(x0 ) and S ∈ W 2,q (M ) for every q ≥ 1. Proof. By Lemma 3.5 we have that βk uk * G W 2,p (M ) p ∈ (1, 2) 2
On the other hand using Lemma 3.2 we get eαk uk is bounded in Lp (M \ Bδ (x0 )). Hence the standard elliptic regularity implies that βk uk → G smoothly in M \ Bδ (x0 ).
(3.5)
So to end the proof of the proposition we need only to show that 2 βk uk eαk uk * τ δx0 . λk
(3.6)
To do this let us take ϕ ∈ C ∞ (M ) then we have Z Z Z 2 βk βk βk αk u2k αk u2k dVg = dVg + ϕ uk e ϕ uk e ϕ uk eαk uk dVg λ λ λ k k k M M \Bδ (x0 ) Bδ (x0 ) Using (3.5) we have that Z ϕ M \Bδ (x0 )
2 1 βk uk eαk uk dVg = O( ). λk λk
On the other hand, we can write inside the ball Bδ (x0 ) Z R βk αk u2k ϕ u e dVg = (ϕ(x0 ) + oδ (1)) Bδ (x0 ) λk k =
2 βk uk eαk uk dVg λ Bδ (x0 ) k ! Z 2 βk (ϕ(x0 ) + oδ (1)) τ − uk eαk uk dVg M \Bδ (x0 ) λk
Now using again (3.5) we derive Z M \Bδ (x0 )
2 βk 1 uk eαk uk = O( ). λk λk
Hence we arrive to Z ϕ Bδ (x0 )
2 βk uk eαk uk dVg = τ ϕ(x0 ) + ok,δ (1). λk
Thus we get Z ϕ M
2 1 βk uk eαk uk dVg = O( ) + τ ϕ(x0 ) + ok,δ (1). λk λk
Hence from Lemma 3.3 we conclude the proof of claim (3.6) )and of the Corollary too.
3.2. Pohozaev type identity and application. As it is already said in the introduction this subsection R deals 2with the derivation of a Pohozaev type identity. And as corollary we give the limit of M eαk uk dVg in terms of V olg (M ), λk , βk and τ Lemma 3.7. Setting Uk = ∆g uk we have the following holds Z Z Z Z 2 δ ∂uk αk u2k 2 − e dVg = − U dSg − δ ∇g uk ∇g Uk dVg + 2 Uk αk λk Bδ (xk ) 2 ∂Bδ (xk ) k ∂r ∂Bδ (xk ) ∂Bδ (xk ) Z Z ∂Uk ∂uk +2δ dSg + O(r2 )∇g uk ∇g Uk dVg ∂Bδ (xk ) ∂r ∂r Bδ (xk ) Z Z Z 2 2 δ δ + O(r2 )Uk2 dVg + eαk uk O(r2 )dVg − eαk uk dVg + O( 2 ). 2λk αk ∂Bδ (xk ) βk Bδ (xk ) Bδ (xk ) where δ is small and fixed real number.
englishM-T PROBLEM
11
Proof. The proof relies on the divergence formula and the asymptotics of the metric g in normal coordinates around xk . By the definition of Uk we have that ( ∆g uk = Uk uk αk u2k λk e
∆g Uk = The first issue is to compute On one side we obtain Z Z ∂Uk ∆g uk dVg = − r Bδ (xk ) ∂r
R Bδ (xk )
k r ∂U ∂r ∆g uk in two different ways, where r(x) = dg (x, xk ).
(∇g Uk ∇g uk + r Bδ (xk )
On the other side we get Z ∂Uk ∆g uk dVg r Bδ (xk ) ∂r
− γk .
∂∇g Uk ∇g uk )dVg + ∂r
∂Bδ (xk )
r ∂Bδ (xk )
∂Uk ∂uk dSg . ∂r ∂r
Z
∂Uk Uk dVg ∂r Z δBδ (xkZ) ∂Uk p Uk |g|r4 dSdr = 2π 2 ∂r ∂Br (xk ) Z0 Z = 2δ Uk2 dSg − 2 Uk2 (1 + O(r2 ))dVg . =
r
∂Bδ (xk )
Thus we have Z Z 2 δ U dS − 2 g k 2
Z
Uk2 dVg Bδ (xk )
Z = − Z +
Bδ (xk )
(∇g Uk ∇g uk + r Bδ (xk )
r ∂Bδ (xk )
∂∇g Uk ∇g uk )dVg Z∂r
∂Uk ∂vk dSg + ∂r ∂r
O(r2 )Uk2 dVg
Bδ (xk )
In the same way we obtain Z Z 2 2 αk u2k δ eαk uk (1 + O(r2 ))dVg e dSg − 2λk αk λ α k k Bδ (xk ) ∂B Z Z δ (xk ) ∂Uk ∂uk δ ∂∇g uk ∇g Uk )dVg + r dSg + O( 2 ). =− (∇g Uk ∇g uk + r ∂r ∂r ∂r β ∂Bδ (xk ) Bδ (xk ) k Hence by summing this two last lines we arrive to Z Z Z Z 2 δ αk u2k αk u2k 2 δ e − e dV + U dS − 2 Uk2 dVg dS g g g k 2λk αk λ α 2 k k ∂B (x ) B (x ) ∂B (x ) B (x ) δ k δ k δ k δ k Z Z ∂ ∂Uk ∂uk =− (2∇g Uk ∇g uk + r ∇g uk ∇g Uk )dVg + 2 r dSg ∂Bδ (xk ) ∂r ∂r Z Bδ (xk ) Z ∂r 2 δ + O(r2 )Uk2 dVg + eαk uk O(r2 )dVg + O( 2 ). βk Bδ (xk ) Bδ (xk ) (3.7) On the other hand using the same method one can check easily that Z Z Z ∂ r ∇g uk ∇g Uk dVg = δ ∇g uk ∇g Uk dVg − 4 ∇g uk ∇g Uk dVg Bδ (xk ) ∂r Bδ (xk ) Z ∂Bδ (xk )
(3.8)
O(r2 )∇g uk ∇g Uk dVg
+ Bδ (xk )
and Z ∇g Uk ∇g uk dVg Bδ (xk )
Z = − Z = −
Z Uk ∆g uk dVg + Bδ (xk ) Z Uk2 dVg + Bδ (xk )
Uk ∂Bδ (xk )
∂Bδ (xk )
Uk
∂uk dSg ∂r
∂uk dSg , ∂r
(3.9)
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englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
So using (3.7),(3.8) and (3.9) we arrive to Z Z Z Z 2 ∂uk − αk2λk eαk uk dVg = − 2δ Uk2 dSg − δ ∇g uk ∇g Uk dVg + 2 Uk ∂r Bδ (xk ) ∂Bδ (x ∂Bδ (xk ) Z ∂Bδ (xk ) Z k) ∂Uk ∂uk 2 +2δ dSg + O(r )∇g uk ∇g Uk dVg Bδ (xk ) Z ∂Bδ (xk ) ∂r ∂r Z 2 + O(r2 )Uk2 dVg + eαk uk O(r2 )dVg Bδ (xZk ) Bδ (xk ) δ αk u2k δ − 2λk αk e dVg + O( 2 ). β ∂Bδ (xk ) k Thus the Lemma is proved
Corollary 3.8. We have that Z lim
k→+∞
λk 2. k→+∞ βk
2
eαk uk = V olg (M ) + τ 2 lim
M
Moreover we have that τ ∈ (0, 1]. Proof. First of all we have that the sequence ( λβk2 )k is bounded. Indeed using the definition of k βk we have that Z 2 λk 1 = ( |uk |eαk uk dVg )2 . 2 βk λk M Hence using Jensen’s inequality we obtain Z Z 2 1 λk αk u2k e u2k eαk uk dVg . ≤ dVg βk2 λk M M Thus using the definition of λk we have that Z 2 λk ≤ eαk uk dVg . βk2 M On the other hand one can check easily that Z Z 2 lim eαk uk dVg = sup k→+∞
u∈H1
M
Hence we derive that ( λβk2 )k is bounded.
e32π
2
u2
dVg < ∞.
M
So we can suppose without lost of generality that
k
( λβk2 )k k
converges. Now from Lemma 3.7 we have that Z Z 2 (βk Uk )2 dSg lim eαk uk dVg = 16π 2 lim λβk2 ( 2δ k→+∞ k→+∞ k ∂B (x ) Bδ (xk ) δ k Z Z ∂(βk uk ) ∇g (βk uk )∇g (βk Uk )dSg − 2 (βk Uk ) +δ ∂r ∂Bδ (xk ) Z∂Bδ (xk ) ∂(βk Uk ) ∂(βk uk ) −2δ dSg + O(δ)). ∂r ∂r ∂Bδ (xk ) So using Lemma 3.6 we obtain Z 2 lim eαk uk dVg k→+∞
Bδ (xk )
=
lim λk2 ( δ k→+∞ βk 2
Z
|∆g G|2 dSg 16π ∂Bδ (x0 ) Z Z ∂G +δ ∇g G∇g (∆g G)dSg − 2 ∆g G ∂r ∂Bδ (x0 ) Z∂Bδ (x0) ∂∆g G ∂G −2δ dSg + O(δ)). ∂r ∂Bδ (x0 ) ∂r
Moreover by trivial calculations we get Z ∂Bδ (x0 )
2
|∆g G|2 dSg =
τ2 + O(1); 8π 2 δ
englishM-T PROBLEM
τ2 + O(1); 8π 2 δ
Z ∇g G∇g (∆g G)dSg = − ∂Bδ (x0 )
Z ∆g G ∂Bδ (x0 )
13
τ2 ∂G = + O(δ); ∂r 16π 2
and ∂∆g G ∂G τ2 dSg = − 2 + O(1) ∂r ∂r 8π δ
Z ∂Bδ (x0 )
Hence with this we obtain Z
2
eαk uk dVg = τ 2 lim
lim
k→+∞
k→+∞
Bδ (xk )
On the other hand we have that Z Z αk u2k e dVg = M
αk u2k
e
Z
2
eαk uk dVg
dVg + M \Bδ (xk )
Bδ (xk )
Moreover by Lemma 3.2 we have that Z
λk + O(δ). βk2
2
eαk uk dVg = V olg (M ) + ok,δ (1).
M \Bδ (xk )
Thus we derive that Z
2
eαk uk dVg = V olg (M ) + τ 2 lim
lim
k→+∞
k→+∞
M
λk + oδ (1). βk2
Hence letting δ → 0 we obtain Z
2
eαk uk dVg = V olg (M ) + τ 2 lim
lim
k→+∞
k→+∞
M
λk . βk2
Now suppose τ = 0 then we get Z
2
eαk uk dVg = V olg (M ).
lim
k→+∞
M
On the other hand we have that Z Z αk u2k lim e dVg = sup k→+∞
u∈H1
M
e32π
2
u2
dVg > V olg (M );
M
hence a contradiction. Thus τ 6= 0 and the Corollary is proved.
3.3. Blow-up analysis. In this subsection we perform the Blow-up analysis and show that the asymptotic profile of uk is either the zero function or a standard Bubble. First of all let us introduce some notations. We set λk −αk c2k rk4 = e . βk ck −1
Now for x ∈ B rk
δ
(0) with δ > 0 small we set wk (x) = 2αk βk (uk (expxk (rk x)) − ck ) ; vk (x) =
1 uk (expxk (rk x)); ck
gk (x) = (exp∗xk g)(rk x). Next we define dk =
ck βk
d = lim dk . k→+∞
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englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
Proposition 3.9. The following hold: We have 1 4 2 in Cloc q (R4 ); if d < +∞ then wk → w(x) := log d d 2 1+ |x| 6
and if d = ∞ then
2 wk → w = 0 in Cloc (R4 ).
Proof. First of all we recall that 2 gk → dx2 in Cloc (R4 ).
Since ( λβk2 ), ( βckk ) are bounded and ck → +∞ , then we infer that k
rk → 0 as k → 0. Now using the Green representation formula for ∆2g (see Lemma 2.1) we have that Z uk (x) = F (x, y)∆2g uk dVg (y) ∀x ∈ M. M
Now using equation and differentiating Z m |∇g uk (x)| ≤
with respect to x we obtain that for m = 1, 2 m ∇g F (x, y) uk eαk u2k − γk dVg (y). λk M
Hence from the fact that βk γk is bounded see Lemma 3.4 we get Z m u k α u2 −1 k k e |∇m u (x)| ≤ ∇ F (x, y) g k g λk dVg (y) + O(βk ). M Now for yk ∈ BLrk (xk ), L > 0 fixed we write that ! Z Z m |uk | αk u2k ∇g F (yk , y) |uk | eαk u2k dVg (y) = O r−m e dVg (y) k λk M M \BLrk (yk ) λk ! Z ck αk c2k −m +O e dg (yk , y) dVg (y) λk BLr (yk ) k
= O(rk−m βk−1 ). thanks to the fact that |uk | ≤ ck to the definition of rk . Now it is not worth remarking that ck = uk (xk ) since we have taken τ ≥ 0 (see Lemma 3.4). Hence we have that wk (x) ≤ wk (0) = 0 ∀ x ∈ R4 . So we get from the estimate above that wk is uniformly bounded in C 2 (K) for every compact subset K of R4 . Thus by Arz´ela-Ascoli Theorem we infer that 1 wk −→ w ∈ Cloc (R4 ).
Clearly w is a Lipschitz function since the constant which bounds the gradient of wk is independent of the compact set K. On the other hand from the Green representation formula we have for x ∈ R4 fixed and for L big enough such that x ∈ B L (0) Z uk (expxk (rk x)) = F (expxk (rk x), y)∆2g uk (y)dVg (y). M
Now remarking that uk (xk ) = uk (expxk (rk 0)); we have that Z uk (expxk (rk x)) − uk (xk ) = M
(F (expx (rk x), y) − F (expxk (0), y)) ∆2g uk (y)dVg (y).
englishM-T PROBLEM
15
Hence using (3.1) we obtain Z uk (expxk (rk x)) − uk (xk ) =
2 uk (F (expxk (rk x), y) − F (expxk (0), y)) eαk uk dVg (y) λk M Z − (F (expxk (rk x), y) − F (expxk (0), y)) (γk )dVg (y).
M
Now setting Z (F (expxk (rk x), y) − F (expxk (0), y))
Ik (x) = BLrk (xk )
uk αk u2k e dVg (y); λk
Z (F (expxk (rk x), y) − F (expxk (0), y))
IIk (x) = M \BLrk (xk )
and
uk αk u2k e dVg (y) λk
Z (F (expxk (rk x), y) − F (expxk (0), y)) (γk )dVg (y);
IIIk (x) = M
we find uk (expxk (rk x)) − uk (xk ) = Ik (x) + IIk (x) + IIIk (x). So using the definition of wk we arrive to wk = 2αk βk (Ik (x) + IIk (x) + IIIk (x)) . Now to continue the proof we consider two cases: Case 1: d < +∞ First of all let us study each of the terms 2αk βk Ik (x), 2αk βk IIk (x), 2αk βk IIIk (x) separately. Using the change of variables y = expxk (rk z) we have Z 2αk βk Ik (x) = (F (expxk (rk x), expxk (rk z)) − F (expxk (0), expxk (rk z))) B L (0)
2αk βk uk (expxk (rk z)) αk u2 (expx (rk z) 4 k e rk dVgk (z). λk and vk one can check easily that the following holds
Hence using the definition of rk Z 2αk βk Ik (x) = 2αk (G(expx (rk x), expx (rk z)) − G(expx (0), expx (rk z))) vk (z) B L (0)
e
dk 2
(wk (z)(1+vk )
dVgk (z).
Moreover from the asymptotics of the Green function see Lemma 2.1 we have that Z dk 1 |z| 2αk βk Ik (x) = 2αk log + K (x, z) vk (z)e 2 (wk (z)(1+vk (z))) dVgk (z). k 2 8π |x − z| B L (0) where Kk (x, z) = [K(expxk (rk x), expxk (rk z)) − (K(expxk (0), expxk (rk z)] . 2 Hence since K is of class C 1 on M 2 and gk → dx2 in Cloc (R4 ) and vk → 1 then letting k → +∞ we derive Z |z| dw(z) lim 2αk βk Ik (x) = 8 log e dz. k→+∞ |x − z| L B (0)
Now to estimate αk βk IIk (x) we write for k large enough Z 1 dg (expxk (0), y) 2αk βk uk αk u2k log e αk βk IIk (x) = dVg (y) 2 dg (expxk (rk x), y) λk M \BLrk (xk ) 8π Z ¯ k (x, y) 2αk βk uk eαk u2k dVg (y), + K λk M \BLr (xk ) k
where ¯ k (x, y) = (K(expx (rk x), y) − K(expx (0), y)) . K k k
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englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
Taking the absolute value in both sides of the equality and using the change of variable y = expxk (rk z) and the fact that K ∈ C 1 we obtain, Z d |z| |vk| (z)e 2k (wk (z)(1+vk (z))) dVg (z) |2αk βk IIk (x)| ≤ 8 log k |x − z| R4 \B L (0) Z 2αk βk uk αk u2k +Lrk e dVg (y). λk M \BLr (xk ) k
Hence letting k → +∞ we deduce that lim sup |2αk βk IIk (x)| = oL (1). k→+∞
Now using the same method one proves that 2αk βk IIIk (x) → 0 as k → +∞. So we have that
Z
|z| |x − z|
8 log
w(x) = B L (R)
edw(z) dz + lim 2αk βk IIk (x). k→+∞
Hence letting L → +∞ we obtain that w is a solution of the following integral equation Z |z| edw(z) dz. (3.10) w(x) = 8 log |x − z| R4 Now since w is Lipschitz then the theory of singular integral operator gives that w ∈ C 1 (R4 ). Since Z Z 2αk βk uk αk u2k e edw(x) dx. lim dVg = 64π 2 k→+∞ B λk B L (0) Lr (xk ) k
and
Z
2αk βk uk αk u2k e dVg ≤ 64π 2 , λk
BLrk (xk )
then we get Z
dw(x)
lim
L→+∞
e
Z dx =
B L (0)
edw(x) dx ≤ 1.
R4
Now setting d 1 8π 2 d w(x) + log( ); 4 4 3 we have that w ˜ satisfies the following conformally invariant integral equation Z 6 |z| 1 8π 2 d w(z) ˜ w(x) ˜ = log e dz + log( ), 2 |x − z| 4 3 R4 8π w(x) ˜ =
and
Z
˜ e4w(x) dx < +∞.
R4
Hence from the classification result by X.Xu see Theorem 1.2 in [25] we derive that 2λ w(x) ˜ = log λ2 + |x − x0 |2 for some λ > 0 and x0 ∈ R4 . ¿From the fact that w(x) ≤ w(0) = 0 ∀x ∈ R4 ; we obtain w(x) ˜ ≤ w(0) ˜ =
1 8π 2 d log( ) ∀x ∈ R4 . 4 3
Then we derive x0 = 0, λ = 2(
8π 2 d − 1 ) 4 3
(3.11)
englishM-T PROBLEM
17
Hence by trivial calculations we get
w(x) =
1 4 . q log d 1 + d |x|2 6
Case 2: d = +∞. In this case using the same argument we get lim sup |αk βk IIk (x)| = oL (1); k→+∞
and αk βk IIIk (x) = ok (1), Now let us show that αk βk Ik (x) = ok (1) By using the same arguments as in Case 1 we get Z 1 |z| αk βk Ik (x) = log + K (x, z) vk (z)edk (wk (z)(1+vk (z))) dVgk (z) k 8π 2 |x − z| B L (0) Now since K is C 1 we need only to show that Z |z| 1 log vk (z)edk (wk (z)(1+vk (z))) dVgk (z) = ok (1). 2 8π |x − z| L B (0) By using the trivial inequality Z BLrk (xk )
u2k αk u2k e dVg ≤ 1; λk
and the change of variables as above, we obtain Z 1 vk2 (z)edk (wk (z)(1+vk (z))) dVgk (z) = O( ) = ok (1). d L k B (0) On the other hand using the property of vk one can check easily that Z Z dk (wk (z)(1+vk (z))) vk (z)e dVgk (z) = vk2 (z)edk (wk (z)(1+vk (z))) dVgk (z) + ok (1). B L (0)
B L (0)
Thus we arrive to Z B L (0)
1 |z| log vk (z)edk (wk (z)(1+vk (z))) dVgk (z) = ok (1) 8π 2 |x − z|
So we get αk βk Ik (x) = ok (1) Thus letting k → +∞, we obtain w(x) = 0 ∀x ∈ R4 . Hence the Proposition is proved.
3.4. Capacity estimates. This subsection deals with some capacity-type estimates which allow us to get an upper bound of τ 2 limk→+∞ λβk2 . We start by giving a first Lemma to show that we k can basically work on Euclidean space in order to get the capacity estimates as already said in the Introduction. Lemma 3.10. There is a constant B which is independent of k, L and δ s.t. Z Z J1 (k, L, δ) |(1 − B|x|2 )∆0 u ˜k |2 dx ≤ |∆g uk |2 dVg + , βk2 B δ (0)\B Lrk (0) Bδ (xk )\BLr (xk ) k
where u ˜(x) = uk (expxk (x)). Moreover we have that lim lim J1 (k, L, δ) = 0.
δ→0 k→+∞
18
englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
Proof. First of all by using the definition of ∆g ie p 1 ∆g = p ∂r ( |g|g rs ∂s ); |g| we get 2 ˜k |∆g βk uk |2 = |g rs βk ∂x∂ r u∂x ˜k |)|2 s + O(|∇βk u 2
˜k 2 2 = |g rs βk ∂x∂ r u∂x ˜k ||∇βk u ˜k |)) + O((|∇βk u ˜k |)2 ) s | + O(|∇ βk u On the other hand using the fact that (see Corollary 3.6)) ˜ in W 2,p (M ); βk u ˜k * G
˜ where p ∈ (1, 2); and G(x) = G(expx0 (x)); we obtain Z O(|∇2 βk u ˜k ||∇βk u ˜k |) + O((|∇βk u ˜k |)2 ) B δ (0)\B Lrk (0)
˜ 1,2 δ ≤ C||G|| W (B (0)\B Lrk (0)) = J2 (k, L, δ), and it is clear that lim lim J2 (k, L, δ) = 0
δ→0 k→+∞ 2
˜k 2 To do this, we first write the inverse of the Now let us estimate B δ (0)\B Lrk (0) |g rs βk ∂x∂ r u∂x s| . metric in the following form g rs = δ rs + Ars with |Ars | ≤ C|x|2 . We can write X X ∂2u ˜k ∂2u ˜k ∂ 2 u ˜k ∂2u ˜k Apq ∆0 u ˜k p q + Ars Apq r s p q ˜k |2 + 2 |g rs r s |2 |∆0 u ∂x ∂x ∂x ∂x ∂x ∂x ∂x ∂x p,q r,s,p,q
R
Furthermore we derive Z X Z X ∂2u ˜k ∂2u ˜k |Apq ∆0 u ˜k p q |dVg ≤ C 2 |x|2 | p q |2 )dx (|x|2 |∆0 u ˜k |2 + Lrk Lrk ∂x ∂x ∂x ∂x δ δ (0) B (0)\B B (0)\B (0) p,q p,q On the other hand we have that XZ
∂2u ˜k ∂ 2 u ˜k dx s s p ∂x ∂x ∂x ∂xp B δ (0)\B Lrk (0) ∂u ˜k ∂ 2 u ˜k ∂ ∂ |x|2 q p q , dS ∂x ∂x ∂x ∂xp ∂r ∂(B δ (0)\B Lrk (0)) Z ˜k ∂ 2 u ˜k ∂ ∂ 2 ∂u + |x| , dS. ∂xq ∂xp ∂xp ∂xq ∂r ∂(B δ (0)\B Lrk (0))
∂2u ˜k 2 | dx p Lr ∂x ∂xq δ p,q B (0)\B k (0) Z O(|∇˜ uk ||∇2 u ˜k |)dx +
Z +
|x|2 |
B δ (0)\B Lrk (0)
Z
|x|2
So setting J3 (k, L, δ) = βk2
Z
Z
2
O(|∇˜ uk ||∇ u ˜k |)dx + ∂(B δ (0)\B Lrk (0))
B δ (0)\B Lrk (0)
Z
|x|2
+ ∂(B δ (0)\B Lrk (0))
We obtain XZ p,q
B δ (0)\B Lrk (0)
|x|2 |
∂2u ˜k 2 | = ∂xp ∂xq
Z B δ (0)\B Lrk (0)
|x|2
∂u ˜k ∂ 2 u ˜k ) ∂xq ∂xp ∂xp
∂ ∂ , ∂xp ∂r
∂ ∂ , ∂xq ∂r
∂2u ˜k ∂ 2 u ˜k J3 (k, L, δ) dx + . q ∂x ∂xq ∂xp ∂xp βk2
Moreover we have that lim lim J3 (k, L, δ) = 0.
δ→0 k→+∞
∂2u ˜k p ∂x ∂xq
∂u ˜k |x|2 q ∂x
dS dS
englishM-T PROBLEM
Hence we get XZ 2 p,q
|Apq
B δ (0)\B Lrk (0)
∂2u ˜k ∂ 2 u ˜k |≤C s s p ∂x ∂x ∂x ∂xq
Z
19
|x|2 |∆0 u ˜k |2 dx +
B δ (0)\B Lrk (0)
J4 (k, L, δ) βk2
with lim lim J4 (k, L, δ) = 0.
δ→0 k→+∞
On the other hand using similar arguments we get Z Z X J5 (k, L, δ) ˜k ∂2u ˜k ∂ 2 u |x|4 |∆0 u ˜k |2 dx + . Ars Apq r s p q ≤ C ∂x ∂x ∂x ∂x βk2 B δ (0)\B Lrk (0) B δ (0)\B Lrk (0) r,s,p,q with lim lim J5 (k, L, δ) = 0.
δ→0 k→+∞
So we arrive to Z
|∆g uk |2 dVg ≤
Z
(1 + C|x|2 + C|x|4 )|∆0 u ˜k |2 dx +
B δ (0)\B Lrk (0)
Bδ (xk )\BLrk (xk )
J6 (k, L, δ) ; βk2
with lim lim J6 (k, L, δ) = 0
δ→0 k→+∞
Hence we can find a constant B1 independent of k, L and δ s.t Z Z J7 (k, L, δ) . |∆g uk |2 dVg ≥ (1 − B1 |x|2 )|∆0 u ˜k |2 dx + Lrk βk2 δ Bδ (xk )\BLr B (0)\B (0) k
So setting J1 (k, L, δ) = −J7 (k, L, δ) and B = B1 we have the proved the Lemma. Next we give a technical Lemma Lemma 3.11. There exists a sequence of functions Uk ∈ W 2,2 (B δ (0) \ B Lrk (0)) s.t Uk |∂B δ (0) = τ and
1 − 16π w(L) 2 log δ + S0 , Uk |∂B Lrk (0) = + ck ; βk 2αk βk
∂Uk ∂Uk τ w0 (L) , . |∂Bδ (0) = − 2 |∂B Lrk (0) = ∂r 8π δβk ∂r 2αk βk rk
Moreover there holds Z lim lim βk2 ( δ→0 k→+∞
|∆0 (1 − B|x|2 )Uk |2 dx −
B δ (0)\B Lrk (0)
Z
|(1 − B|x|2 )∆0 u ˜k |2 dx) = 0.
B δ \B Lrk (0)
Proof. First of all let us set hk (x) = uk (expxk (rk x)). and
u0k
to be the solution of 2 0 ∆0 uk = ∆20 hk ∂u0k ∂hk 0 ∂n |∂B2L ∂n |∂B2L , uk |∂B 2L (0) = hk |∂B 2L (0) ∂u0k 1 ∂w w 0 ∂n |∂B L (0) 2αk βk ∂n |∂B L (0) , uk |∂B L (0) = 2αk βk |∂B L (0) .
Next let us define ( Uk0
=
u0k ( rxk ) Lrk ≤ |x| ≤ 2Lrk u ˜k (x)
2Lrk ≤ |x|.
Clearly we have that Z lim
k→+∞
B 2Lrk (0)\B Lrk (0)
(1 − B|x|2 )(|∆0 Uk0 |2 − |∆0 u ˜k |2 )dx = 0,
20
englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
and lim |Uk0 − u ˜0k |C 0 (B 2Lrk (0)\B Lrk (0)) = 0.
k→+∞
Now let η be a smooth function which satisfies (
1
t ≤ 1/2
0
t > 2/3
η(t) = and set Gk = η(
|x| τ τ )(˜ uk − τ S0 + 2 log |x|) − 2 log |x| + τ S0 . δ 8π 8π
Then we have that Gk → −
|x| τ log |x| + τ S0 + τ η( )S˜1 (x); 8π 2 δ
where S˜1 (x) = S1 (expx0 (x)) . Furthermore we obtain βk u ˜ k − Gk → τ
|x| 1 − η( ) S1 (x), δ
then Z lim |
→0
Z
2
|∆0 Gk |2 dx| ≤ Σ.
|∆0 βk u ˜k | dx − B δ (0)\B δ/2 (0)
B δ (0)\B δ/2 (0)
where Σ
=
qR B δ (0)\B δ/2 (0)
≤ Cδ
p
2 ˜ |∆0 (1 − η( |x| δ ))S1 (x)| dx
R B δ (0)\B δ/2 (0)
˜− |∆0 (G
1 8π 2
2 ˜ log |x| + η( |x| δ )S1 (x))| dx
| log δ|.
So we get Z lim |
→0
|∆0 βk u ˜k |2 dx −
Z
B δ (0)\B δ/2 (0)
|∆0 Gk |2 dx| ≤ Cδ
p
| log δ|.
B δ (0)\B δ/2 (0)
Hence setting ( Uk =
Uk0 (x)
|x| ≤
δ 2
Gk (x) δ/2 ≤ |x| ≤ δ
we have proved the Lemma.
Proposition 3.12. We have the following holds π 2 5 +32π2 S0 λk ≤ e3 ; 2 k→+∞ βk 6
τ 2 lim and
dτ = 1. Proof. First using Lemma 3.10 and Lemma 3.11 we get R R Z |∆w|2 + M \Bδ (x0 ) |∆G|2 + J0 (k, L, δ) BL (x0 ) 2 2 |∆0 (1 − B|x| )Uk | dx ≤ 1 − . (3.12) βk2 B δ (0)\B Lrk (0) with lim lim J0 (k, L, δ) = 0. R Next we will apply capacity to give a lower boundary of B δ (0)\B Lrk (0) |∆0 (1 − B|x|2 )Uk |2 dx. Hence we need to calculate Z |∆0 Φ|2 dx, inf δ→0 k→+∞
∂Φ r Φ|∂B r (0) =P1 ,Φ|∂B R (0) =P2 , ∂Φ ∂r |∂B (0) =Q1 , ∂r |∂B R (0) =Q2
B R (0)\B r (0)
englishM-T PROBLEM
21
where P1 , P2 , Q1 , Q2 are constants. It is obvious that the infimum is attained by the function Φ which satisfies ( 2 ∆0 Φ = 0 ∂Φ r Φ|∂B r (0) = P1 , Φ|∂B R (0) = P2 , ∂Φ ∂r |∂B (0) = Q1 , ∂r |∂B R (0) = Q2 .
Moreover we can require the function Φ to be of the form C + D, r2 where A, B, C, D are all constants which satisfies the following linear system of equations A log r + Br2 + rC2 + D = P1 A log R + BR2 + C + D = P Φ = A log r + Br2 +
2
R2
2 rC3
A r
+ 2Br −
A R
+ 2BR − 2 RC3 = Q2
= Q1
Now by straightforward calculations we obtain the explicit expression of A and B % P1 −P2 + % 2 rQ1 + 2 RQ2 A= log r/R+% 2
2
B= Where % = Z
2
2
R −r R2 +r 2 .
−2P1 +2P2 −rQ1 (1+ R22r−r2 log r/R)+RQ2 (1+ R2R 2 −r 2 log r/R) 4(R2 +r 2 )(log r/R+%)
Furthermore we have
|∆0 Φ|2 dx = −8π 2 A2 log r/R + 32π 2 AB(R2 − r2 ) + 32π 2 B 2 (R4 − r4 )
B R (0)\B r (0)
In our case in which we have that r = Lrk ,
R=δ P1 = ck +
− τ 2 log δ + τ S0 + O(δ log δ) w(L) + O(rk ck ) P2 8π 2αk βk βk
τ + O(δ log δ) w0 (L) + O(rk ck ) Q2 = − . 2αk βk rk 8π 2 βk δ Then by the formula giving A we obtain by trivial calculations Q1 =
ck +
A=
− log δ + log L + where Nk =
τ Nk + 8π 2 log δ βk
log
λk βk ck
−αk c2k
4
+ 1 + O(rk2 )
w(L) w0 (L)L τ + O(δ log δ) + O(rk c2k ). − τ S0 + − 2αk 4αk 16π 2
Moreover using the the fact that the sequence ( λβk2 )k is bounded it is easily seen that k
A = O(
1 ). ck
Furthermore using the formula of B we get still by trivial calculations B=
αk c2k τ 1 8π 2 βk 2 + O( βk ) . δ 2 (−αk c2k + log βλkkck )
−2ck +
and then B = O(
1 1 ) . βk δ 2
(3.13)
22
englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
Now let compute 8π 2 A2 log r/R. By using the expression of A, r and R , we have that r −8π A log( ) = −8π 2 ( R 2
ck +
2
τ Nk + 8π 2 log δ βk
log
− log δ + log L +
λk βk ck
−αk c2k
4
)2 (
log
λk βk ck
− αk c2k
4
+ 1 + O(rk2 )
−log δ+log L)
Now using the relation 2 αk c2k 2 1 λk 2 ( ) 1− (−4 log δ + 4 log L + log + 4 + O(rk )) = 4 αk c2k βk ck !2 log βλkkck − αk c2k 2 − log δ + log L + + 1 + O(rk ) 4 we derive ck + r −8π 2 A2 log( ) = −8π 2 ( R
τ Nk + 8π 2 log δ βk )2 αk c2k 4
−2 λk 1 2 (−4 log δ + 4 log L + log + 4 + O(rk )) 1− αk c2k βk ck ×(
log
λk βk ck
− αk c2k
4
− log δ + log L).
On the other hand using Taylor expansion we have the following identity −2 log βλkkck + 4 − 4 log δ + 4 log L λk 1 2 + 4 + O(r (−4 log δ + 4 log L + log 1− )) = 1 + 2 k αk c2k βk ck αk c2k +O(
log2 ck ); c4k
hence we get τ Nk + 8π 2 log δ log βλkkck − αk c2k βk 2 ) ( − log δ + log L) 2 αk ck 4 4 log βλkkck + 4 − 4 log δ + 4 log L log2 ck + O( )) +2 αk c2k c4k
ck + r −8π 2 A2 log( ) = −8π 2 ( R ×(1
On the other hand using the relation 2
−8π (
ck +
32π 2 1 (ck + αk c2k
τ Nk + 8π 2 log δ log βλkkck − αk c2k βk 2 − log δ + log L) = ) ( αk c2k 4 4 log βλkkck − 4 log δ + 4 log L Nk + 8πτ 2 log δ 2 ) (1 − ) βk αk c2k
we obtain −8π 2 A2 log(
log Nk + 8πτ 2 log δ 2 32π 2 1 r )= (c + ) (1 + 2 k R αk c2k βk
λk βk ck
+ 4 − 4 log δ + 4 log L αk c2k ×(1 −
log
λk βk ck
+ O(
log2 ck )) c4k
− 4 log δ + 4 log L αk c2k
Moreover using again the trivial relation (1 + 2
log
λk βk ck
+ 4 − 4 log δ + 4 log L αk c2k
log log2 ck + O( ))(1 − 4 ck
(1 +
log
λk βk ck
λk βk ck
− 4 log δ + 4 log L αk c2k
+ 8 − 4 log δ + 4 log L αk c2k
+ O(
)=
log2 ck )) c4k
)
englishM-T PROBLEM
23
we arrive to −8π 2 A2 log(
log Nk + 8πτ 2 log δ 2 r 32π 2 1 )= (ck + ) (1 + 2 R αk ck βk
λk βk ck
+ 8 − 4 log δ + 4 log L αk c2k
+ O(
log2 ck )) c4k
On the other hand one can check easily that the following holds Nk +
(ck + log
c2k +
τ 8π 2
log δ
βk
λk βk ck
2
) (1 +
log
+ 8 − 4 log δ + 4 log L αk
λk βk ck
+ 2ck
+ 8 − 4 log δ + 4 log L αk c2k Nk + 8πτ 2 log δ βk
log2 ck )) = c4k ! log ck 1 + O( 2 ) + O( 2 ) ; ck βk + O(
thus we obtain r 32π 2 1 −8π A log( ) = R αk c2k 2
2
c2k
log
+
λk βk ck
+ 8 − 4 log δ + 4 log L αk 32π 2 1 + αk c2k
+ 2ck
τ 8π 2
Nk +
log δ
!
βk
log ck 1 O( 2 ) + O( 2 ) ck βk
Furthermore using the relation λk βk ck
! log c 1 k c2k + + 2ck + O( 2 ) + O( 2 ) = αk βk ck βk λk 4 1 8 4 log L 1 log − log δ + 2 dk τ log δ + 2dk Nk + + + ok (1) c2k + αk βk ck αk 4π αk αk log
+ 8 − 4 log δ + 4 log L
Nk +
τ 8π 2
log δ
we get r 32π 2 1 −8π A log( ) = R αk2 c2k 2
2
c2k
λk 4 1 4 log L 8 1 log − log δ + 2 dk τ log δ + 2dk Nk + + + αk βk ck αk 4π αk αk +
32π 2 1 ok (1) αk2 c2k (3.14)
R Next we will evaluate M \Bδ (x0 ) ∆g G∆g GdVg . We have that by Green formula Z Z Z Z ∂G ∂∆g G ∆g G∆g GdVg = G∆2g GdVg − ∆g G + G . ∂r M \Bδ (x0 ) M \Bδ (x0 ) ∂Bδ (x0 ) ∂r ∂Bδ (x0 ) Thus using the equation solved by G we get Z Z Z τ ∂(− log r) τ2 ∆g G∆g GdVg = − GdVg − ∆0 (− log r) 4 µ(M ) M \Bδ (p) 64π ∂Bδ (x0 ) ∂r M \Bδ (x0 ) Z ∂∆0 (− 8πτ 2 log r) τ + O(δ log δ) + (− 2 log r + S0 ) 8π ∂r ∂Bδ (x0 ) Hence we obtain Z ∆g G∆g GdVg = − M \Bδ (x0 )
Now let us set
Z P (L) =
τ2 τ2 − 2 log δ + τ 2 S0 + O(δ log δ), 2 16π 8π |∆0 w|2 dx/(2 × 32π 2 )2 .
B L (0)
Hence using (3.12), (3.13), (3.14), we derive that 32π 2 1 λk 4 1 4 log L 8 2 ck + log − log δ + 2 dk τ log δ + 2dk Nk + + αk αk βk ck αk 4π αk αk ≤
c2k (1
−
P (L) −
τ2 16π 2
−
τ2 8π 2
log δ + τ S0 + O(δ log δ) + ok,δ (1) ) + δ 2 O(c2k AB) + δ 4 O(c2k B 2 ). βk2
24
englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
Moreover by isolating the term we get
32π 2 α2k
log
λk βk ck
in the left and transposing all the other in the right
32π 2 λk 1 64 32π 2 32π 2 4 log L 8 2 2 log ≤ (d τ − d τ + ( ) ) log δ − (2dk Nk + + ) k k 2 αk βk ck 8π 2 αk αk αk αk αk τ2 + τ S0 − + O(δ log δ) + ok (1)) + δ 2 O(c2k AB) + δ 4 O(c2k B 2 ). 16π 2 Hence using the trivial identity
(3.15)
−d2k (P (L)
log
λk λk = log + log dk βk2 βk ck
we get λk 1 64 32π 2 32π 2 2 + 4 log L 2 32π 2 2 2 log 2 ≤ (d τ − d τ + ( ) ) log δ − (2dk Nk + + ) k k 2 αk βk 8π 2 αk αk αk αk αk −d2k (P (L) + τ S0 −
32π 2 τ2 + O(δ log δ) + o (1)) + log dk + O(d2k ). k 16π 2 αk2
Now suppose d = +∞, letting δ → 0, then we have that lim log
k→+∞
λk = −∞, βk2
thus we derive λk =0 βk2 Hence using Corollary 3.8 we obtain a contradiction. So d must be finite. On the other hand one can check easily that the following holds lim
k→+∞
λk 1 32π 2 2 32π 2 log ≤ (d τ − ) log δ + O(1)(d2k + dk + log dk ) + O(1). k αk2 βk2 8π 2 αk Hence we derive dk τ → 1; otherwise we reach the same contradiction. So we have that dτ = 1. Hence by using this we can rewrite B as follows B=
−2ck + δ(− 8π21ck δ 2
−αk c2k ) 4
δ 2 (−αk c2k )
+ O(1/ck )
+ O(1)
=
ok (1) . ck
Thus we obtain 32π 2 AB(R2 − r2 ) + 32π 2 B 2 (R4 − r4 ) =
ok (1) . c2k
On the other hand since d < +∞, we have that by Lemma 3.9 q 4 log(1 + d6 π|x|2 ) w=− . d Moreover by trivial calculations we get q d 2 log(1 + 1 6 πL ) P (L) = + . 96d2 π 2 16d2 π 2 Furthermore by taking the limit as k → +∞ in (3.15) we obtain q r d 2 4 25 d 2 λk 6 πL 2 2 2 q ≤ − + 4dτ + 2d τ + 32π S0 + + 2 log(1 + πL ) − 4 log L lim log k→+∞ βk ck 3 6 1 + d πL2 6
englishM-T PROBLEM
25
Now letting L → +∞, we get λk 5 ≤ − log 6 + log π 2 + log d. βk ck 3 Hence by remarking the trivial identity λk 1 λk lim lim 2 k→+∞ βk ck d k→+∞ βk lim log
k→+∞
we get λk π 2 5 +32π2 S0 ≤ e3 . 2 k→+∞ βk 6 So the proof of the proposition is done. τ 2 lim
3.5. The test function. This Subsection deals with the construction of some test functions in order to reach a contradiction. Now let > 0, c > 0, L > 0 and set “ ” dg (x,x0 ) 2 2 c + Λ+Bdg (x,x0 ) −4 log 1+λ( ) + S(x) d (x, x ) ≤ L g 0 64π 2 c c f (x) = G(x) d (x, x ) > L g
c
where
0
π 4 λ= √ , B=− 2 2 L (1 + λL2 ) 6
and Λ = −64π 2 c2 − BL2 2 − 8 log(L) + 4 log(1 + λL2 ).
(3.16)
Proposition 3.13. We have that for small, there exist suitable c and L such that Z |∆g f |2 dVg = 1; M
and Z
e32π
lim sup →0
2
(f −f¯ )2
dVg > V ol(M ) +
M
π 2 5 +32π2 S0 e3 . 6
Proof. First of all using the expansion of g in normal coordinates we get Z Z Z 2 2 2 ˜ |∆g f | dVg |∆0 f | (1 + O(L) )dx + O(r2 |∇0 f˜ |2 )dx B L (0)
BL (x0 )
B L (0)
where f˜ (x) = f (expx0 (x)). On the other hand by direct calculations owe obtain Z 12 + λL2 (30 + λL2 (21 + λL2 )) + 6(1 + λL3 )3 log(1 + λL2 ) |∆0 f˜ |2 dx = 96c2 (1 + λL2 )3 π 2 B L (0) Hence we arrive to R |∆g f |2 dVg BL (x0 )
= =
2
(1 + O(L)2 ) 12+λL
(30+λL2 (21+λL2 ))+6(1+λL3 )3 log(1+λL2 ) 96c2 (1+λL2 )3 π 2
2 2 1 1 3 +4 log(1+λL )+O( L2 )+O((L) 32c2 π 2
log L)
Furthermore, by direct computation, we have Z L4 4 r2 |∇0 f˜ |2 dx = O( 2 ). c B L (0) Moreover using Green formula we get Z Z |∆g G|2 dVg = M \BL (x0 )
Z
GdVg −
M \BL (x0 ) 1 = − 16π 2 + S0 −
log L 8π 2
∂G ∆g GdSg + ∂BL (x0 ) ∂r + O(L log L)
Z G ∂BL
∂∆g G dSg ∂r
26
englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
R Now let us find a condition to have M |∆g f |2 dVg = 1. By trivial calculations we can see that it is equivalent to 1 5 1 2 2 − + 2 log(1 + λL ) + 32π S − 4 log L + O( ) + O(L log L) = 1. 0 32π 2 c2 3 L2 i.e.
5 1 32π 2 c2 = − + 2 log(1 + λL2 ) + 32π 2 S0 − 4 log L + O( 2 ) + O(L log L). 3 L Hence by (3.16) Λ take the following form 10 1 − 64π 2 S0 + O( 2 ) + O(L log L). 3 L On the other hand it is easily seen that Z f dVg = O(c(L)4 ); Λ=
BL (x0 )
and Z
Z f dVg = −
M \BL (x0 )
BL
(L)4 log L G = O( ). c c
hence f¯ = O(c(L)4 ). Furthermore by trivial calculations one gets that in BL (x0 ) 2 (f − f¯ )2 ≥ c2 + 64π Λ + Br2 − 4 log(1 + λ( r )2 ) + 64π 2 S0 + O(L) + O(c2 (L)4 ) 2 = c2 + hence Z
32π (f −f¯ )2
e
2
dVg
5 48π 2
−
log(1+λ(r/)2 ) 8π 2
« „ log(1+λ(r/)2 5 32π 2 c2 + 48π +O( L12 )+O(L log L)+O(c2 (L)4 ) 2− 8π 2
Z 2 ≥ (1 + O(L) )
BL (x0 )
= 4 e
10
= on the other hand Z 2 ¯ 2 e32π (f −f ) dVg
e
BL (x0 ) 2 4 2 2 1 10 3 +32π c +O( L2 )+O(L log L)+O(c (L) )
= 4 e 3 +32π
M \BL (x0 )
+ O( L12 ) + O(L log L) + O(c2 (L)4 );
2 2
c
6
L 2 π 2 1+λL 6 + O(L)
π 2 (1 + O( L12 ) + O(L log L) + O(L)2 )
π 2 53 +32π 2 S0 (1 6 e
+ O(L log L) + O( L12 ) + O(c2 (L)4 )).
Z
(1 + 32π 2 (f − f¯ )2 )dVg Z 32π 2 G2 dVg + O(c(L)4 ) M \BL (x0 ) ≥ V ol(M \ BL (x0 )) + c2 Z 32π 2 G2 dVg + O(L)4 log L = V ol(M ) + M c2 ≥
M \BL (x0 )
Thus we arrive to R Z 32π 2 G2 dVg π 2 5 +32π2 S0 M \BL (x0 ) 32π 2 (f −f¯ )2 e dVg ≥ V ol(M ) + e3 + 6 c2 M ; 1 2 4 +O(L log(L)) + O( 2 ) + O(c (L) ) L and factorizing by c12 we get Z 2 2 2 5 ¯ 2 e32π (f −f ) dVg ≥ V ol(M ) + π6 e 3 +32π S0 M Z c2 2 2 2 4 4 1 + c2 32π G dVg + O(c L log(L)) + O( 2 ) + O(c (L) ) . L M
dx
englishM-T PROBLEM
27
On the other hand setting L = log
1
we get O(c2 L log(L)) + O(
c2 ) + O(c4 (L)4 ) → 0 as → 0. L2
Hence the Proposition is proved. 3.6. Proof of Theorem 1.1. This small subsection is concerned about the proof of Theorem 1.1. First of all by corollary we have that Z 2 λk lim eαk uk = V olg (M ) + τ 2 lim 2 k→+∞ M k→+∞ βk with τ = 6 0. On the other hand from Proposition 3.12 we get λk π 2 5 +32π2 S0 ≤ e3 . 2 k→+∞ βk 6
τ 2 lim Hence we obtain Z
2
eαk uk ≤ V olg (M ) +
lim
k→+∞
M
π 2 5 +32π2 S0 e3 . 6
Thus using the relation Z
2
eαk uk dVg = sup
lim
k→+∞
u∈H1
M
Z
e32π
2
u2
dVg .
M
we derive
π 2 5 +32π2 S0 e3 . 6 u∈H1 M On the other hand from Proposition 3.13 we have the existence of a family of function f such that Z |∆g f |2 dVg = 1; Z
sup
e32π
2
u2
dVg ≤ V olg (M ) +
M
and
Z
2 1 5 dVg > V ol(M ) + e 3 +32π S0 π 2 . 6 →0 M Hence we reach a contradiction. So the proof of Theorem 1.1 is completed.
lim sup
e32π
2
(f −f¯ )2
4. Proof of Theorem 1.2 As already said in the Introduction, in this brief Section we will explain how the proof of Theorem 1.1 remains valid for Theorem 1.2. First of all we remark that all the analysis above have been possible due to the following facts 1) R |∆g u|2 dVg is an equivalent norm to the standard norm of H 2 (M ) on H1 . M 2) The existence of the Green function for ∆2g . 3) The result of Fontana.
On the other hand we have a counterpart of 2) and 3). Moreover it is easy to see that Pg4 u, u is also an equivalent norm to the standard norm of H 2 (M ) on H2 . Notice that for a blowing-up sequence uk we have that Z
4 Pg u k , u k = |∆g uk |2 dVg + ok (1); (4.1) M
then it is easy to see that the same proof is valid up to the subsection of test functions. Notice that (4.1) holds for the test functions f , then it is easy to see that continuing the same proof we get Theorem 1.2.
28
englishYUXIANG LI, CHEIKH BIRAHIM NDIAYE
References [1] Adams D.R., A sharp inequality of J.Moser for higher order derivatives, Ann. math 128 (1988) 385-398. [2] Adimurthi and O.Druet, Blow up analysis in dimension 2 and a sharp form of Trudinger Moser inequality, Comm. PDE 29 No 1-2, (2004), 295-322. [3] Beckner W., Sharp Sobolev inequalities on the sphere and the Moser-trudinger inequality, Ann. Math 138(1) (1993) 213-242. [4] Chang S.Y.A., The Moser-Trudinger inequality and applications to some problems in conformal geometry,in Nonlinear Partial differential equations in Differential Geometry (Park City, UT, 1992) IAS/Park City Mathematics series, Vol. 2(American Mathematical society, Providence, rI, 1996), pp. 65-125. [5] Carleson L, Chang S.Y.A., On the existence of an extremal function for an inequality of J.Moser, Bull. Sci. Math 110 (1986) 113-127. [6] Chang S.Y.A., Yang P.C.,Extremal metrics of zeta functional determinants on 4-manifolds, ann. of Math. 142(1995), 171-212. [7] Chang S.Y.A., Gursky M.J., Yang P.C.,A conformally invariant sphere theorem in four dimensions, Publ. Math. Inst. Hautes etudes Sci. 9892003), 105-143. [8] Djadli Z., Malchiodi A., Existence of conformal metrics with constant Q-curvature, to appear in ann. of Math. [9] Fontana L., Sharp bordeline sobolev inequalities on compact Riemmanian manifolds, comment. Math.Helv. 68 (1993) 415-454. [10] Flucher M., Extremal functions for Trudinger-Moser type inequality in 2 dimensions, Comment. math.Helv. 67 (1992) 471-497. [11] Gursky M., The principal eigenvalue of a conformally invariant differential operator, with an application to semilinear elliptic PDE, Comm. Math. Phys. 207-1 (1999), 131-143. [12] Li J., Li Y., Liu P., The Q-curvature on a 4-dimensional Riemannian manifold (M, g) with preprint.
R M
QdVg = 8π 2 ,
[13] Li Y., Liu P., Moser-Trudinger inequality on the boundary of compact of compact Riemannian surface, Math. Z. 250 (2) (2005) 363-386. [14] Li Y., Moser-Trudinger inequality on compact Riemannian manifolds of dimension two, J.Partial Differential equations 14(2) (2001) 1289-1319. [15] Li Y., The extremal functions for Moser-trudinger inequality on compact Riemannian manifolds, Sci. China Series A. Math. 48 (2005) 18-648. [16] Lin K.C.,Extremal functions for moser’s inequality, trans. Amer. Math. Soc. 348 (1996) 2663-2671. [17] Malchiodi A., Compactness of solutions to some geometric fourth-order equations, J. Reine Angew. Math., to appear. [18] Malchiodi A., Ndiaye C.B., Some existence results for the Toda system on closed surfaces, preprint, 2005 [19] Moser J., A Sharp form of an inequality of Trudinger, Ind. univ. math. j. 20 (1971) 1077-1091. [20] Ndiaye C.B., Constant Q-curvature metrics in arbitrary dimension preprint 2006. [21] Paneitz S., A quartic conformally covariant differential operator for arbitrary pseudo-Riemannian manifolds, preprint, 1983. [22] Pohozaev S.I.,: The Sobolev embedding in the case pl = n. Proceedings of the Technical Scientific Conference on Advances of Scientific Research 1964-1965, Mathematics Section, 158-170, Moskov. Energet. Inst., Moscow, 1965 [23] Struwe M.,: Critical points of embedding of H01,n into Orlicz space, Ann. Inst. Henri., 5(5):425-464, 1988.
englishM-T PROBLEM
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[24] Trudinger N.S.,on embedding into orlicz space and some applications, J. Math. mech. 17 (1967) 473-484. [25] Xu Xingwang., Uniqueness and non-existence theorems for conformally invariant equations, Journal of Functional Analysis. 222(2005) 1-28. (Yuxiang Li) ICTP, Mathematics Section, Strada Costiera 11, 34014 Trieste, Italy E-mail address, yuxiang li:
[email protected] (Cheikh Birahim Ndiaye) S.I.S.S.A., Via Beirut 2-4, 34014, Trieste, Italy E-mail address, Cheikh Birahim Ndiaye:
[email protected] 
