Prescribed Scalar Curvature problem onComplete manifoldsDavid HolcmanParis VI UniversityJune 1998AbstractConditions on the geometric structure of a complete Riemannian man-ifold are given to solve the prescribed scalar curvature problem in thepositive case. The conformal metric obtained is complete. A minimizingsequence is constructed which converges strongly to a solution. In a sec-ond part, the prescribed scalar curvature problem of zero value is solvedwhich is equivalent to �nd a solution to a linear PDE on a complete man-ifold. This solution helps to obtain a general result on the prescribedscalar curvature problem, in the positive case [11].This paper deals with the study of the prescribed curvature problem on non-compact complete manifolds. On compact manifolds it has been solved underconditions on the function f . The main di�culties in this case is the con-centration phenomenon [19]. On a complete manifold having positive Yamabeinvariant, another phenomena can appear. Conditions must be prescribed onthe function f in order to avoid it.Some results concerning the existence of conformal metrics of negative scalarcurvature manifolds were obtained by Aviles-MacOwen [5]. They proved thatit is possible to solve the Yamabe equation with a constant value equal to -1for the prescribed curvature . In the positive case, Kim [14] solved the Yamabeproblem on a complete manifold assuming that the Yamabe quotient is less thana certain number.(Vn; g) will denote a complete Riemannian manifold of dimension greater orequal to 3 and R(g), the scalar curvature. We use the notationN = 2nn�2 , �R = Rcnwhere cn = 4(n�1)n�2 . We shall assume that the conformal Yamabe invariant ispositive : � = infH1(Vn) RVn(ru)2 + �Ru2(RVn uN )2=N > 0 (1)There are many instances of such manifolds : for example asymptotically at manifolds with positive scalar curvature. We can construct other examples1

starting with a compact manifold (Vn; g). Let P be a point of Vn and GP bethe Green function of the conformal Laplacian with pole at P . Then (Vn �P;G4=(n�2)P g) is a complete manifold with the same Yamabe invariant than(Vn; g) [1].The purpose of this paper is to �nd a solution u of the following problem�gu+ �Ru = f j u jn+2n�2 on Vn (2)u > 0where f is a strictly positive C1 function such that the conformal metric g0 =u4=(n�2)g is complete.One important conclusion of this paper is that the prescribed scalar curva-ture problem can be solved on conformally at manifolds under conditions onthe geometry (see section 4 for the de�nition of bounded geometry) and localconditions at a maximum point P of f plus some assumptions on the decay of f.The mass of the manifold does not enter into the game as it does in [16]. Weyldenotes in the following the Weyl conformal curvature tensor.The main results of this paper are :Theorem 1 Under the following assumptions1. At a point P where f is maximal, Weyl(P ) 6= 0 , �gf(P ) = �2gf(P ) = 0and n > 62. Ricci > �c, where c � 0 is a constant and the injectivity radius is strictlypositive.3. there exists a constant C such thatj rf j� Cf and j �f j� Cf ZVn Rn=2 � C ZVn f < +1 and ZVn f2=N < +1 (3)and f is bounded.4. The scalar curvature is bounded and nonnegative.there exists a positive C1 solution of equation (2).Remarks1. The �rst assumption prevents the concentration phenomena locally. Theothers, the concentration phenomenon "at in�nity".2. More speci�cally assumption 2 insures that the relevant Sobolev inequalityis valid : for any � > 0, there exists some constant A� such that anyfunction u 2 H1(Vn) satis�es(ZVn uN)2=N � (K2 + �) ZVn j ru j2 +A� ZVn u2where K is a constant. We choose the best constant possible [1].2

3. We could use also the exact inequality(ZVn uN)2=N � K2 ZVn j ru j2 +A ZVn u2but this would require additional assumptions on the geometry of (Vn; g).4. Sobolev embbeding theorem holds in the following case� The Ricci curvature is bounded from below and� The injectivity radius is positive and the metric g is complete [17].5. Examples of functions f satisfying assumption 3 are given below.The second theorem is the followingTheorem 2 (Vn; g) denotes a complete Riemannian manifold which is locallyconformally at. Then there exists a positive solution of (2) under the followingassumptions :1. rf(P ) = ::: = rn�2f(P ) = 0 at a maximum point of f and rf and �fare bounded.2. The Sobolev Inequality holds on (Vn; g) and the scalar curvature R is non-negative.3. there exists a nonconstant bounded solution h of the equation �h+ �R(g)h =0 such that RVn fhNdVg is �nite. Condition 3 on the decay at in�nity ofthe function f of theorem 1 is valid.Moreover when h converges at in�nity ( such is the case for asymptotically at manifolds), the metric obtained in theorem 2 g0 = u4=(n�2)g is complete.The conclusion holds also for theorem 1.The basic ideas of the proof are the following. We solve Equation (2) bya variational method. Using results from the compact case, we construct aminimizing sequence [12]. To prove that this sequence converges we have toshow that the concentration phenomenon does not happen. This results fromthe fact that f decays at in�nity. Indeed, since the solution has �nite energy,the number of possible concentration points is �nite. A compact K is chosen,avoiding these points. Then we prove that no concentration occurs, outside Kbecause of the decay condition on f, and inside K because of condition 1 intheorem 1 or theorem 2 respectively.This paper is organized as follows : in the �rst part a sequence of minimiz-ing solutions is constructed. In the second part the concentration at in�nityphenomenon is de�ned and conditions are given to prevent it. In the third part,this conditions are used to prove that concentration points and concentrationat in�nity do not occur. In the fourth part we use a variational method to solvea linear equation (100). To �nish estimates on the Green function are used toobtain the behavior of the solution for the linear equation at in�nity.3

1 Minimizing sequences approximationIn this part, a variational method produces a minimizing sequence. Let k bean exhaustion of Vn by compact manifolds with smooth boundary such thatk � ok+1. The variational problem is� k = infA k I(u) = infA k Zk (j ru j2 + �Ru2) (4)where A k = fu 2 H1(k) j R f j u jN = k and u = 1 on @kg. A minimizer ofI is called a minimizing solution.When the condition R k fhNk < k is satis�ed, the subset A k is not empty.We suppose that this is the case in the following.1.0.1 Construction of the minimizing sequence ukProposition 1 On each k there exists a positive solution uk for (4), whichgives the Euler-Lagrange equation�guk + �Ruk = �kf(uk + hk)N�1 on k (5)uk = 0 on @k (6)The Lagrange multiplier �k is positive and hk is the positive solution of thelinear equation �ghk + �Rhk = 0 on k (7)hk = 1 on @k (8)ProofThe existence and uniqueness of hk follow from classical linear theory, [9]. Sincethe scalar curvature is bounded at in�nity, and the maximum of hk is reachedon the boundary of k, �ghk is uniformly bounded (supk hk = 1 ). Hence thesequence hk is bounded in C1;� topology and it converges, uniformly on eachcompact subset, to a solution of�gh+ �Rh = 0 (9)h is bounded by 1 on Vn by the maximum principleIt has been proved in [12] and [13], that there exists a minimizing solutionof (5) such that Zk f(uk + hk)N = k (10)where k is a number large enough, given below.4

Moreover the sequence uk satis�es uk+hk > 0, since the variational problemremains invariant substituting j u j for u in the expression :� k = infA k I(u) = infA k Zk (j ru j2 + �Ru2) (11)with A k = fu 2 H1(k) j R f j u jN = k and u = 1 on @kg. Using themaximum principle for equation (5), uk > 0 since R > 0.Recall the two conditions obtained in [12] or [13] : on each compact k :K2�( k)(sup f)2=N 1�2=Nk k�Rk f juk+hkjN�2(uk+hk)hk < 1 (12)and K2 k�Rk f(uk+hk)N�1hk�( k)(sup f)2=N ( k + C1 Rk fhNk + (13)C2 Rk fuN�1k hk)1�2=N < 1C1 and C2 are two constants and K is the best constant in the Sobolevembedding.When they are satis�ed, the solution uk is minimizing. Later on we provethat these conditions are satis�ed. Moreover uk is a C1(�k) sequence by thestandard Trudinger argument.The energy of the sequence uk is de�ned by :E(uk) = RVn(j ruk j2 + �Ru2k)(RVn fuNk )2=N (14)This energy is �nite dues to the nonconcentration condititions (12) (13). Whenthe strong convergence will be proved, the energy of the limit u will be also�nite : it is well known that the number of concentration points in this case is�nite [19]. This will be important in the next section.1.0.2 Convergence of ukIn this part, the convergence of the sequence uk is established. The sequence k of energy is supposed to be equal to a large number.Proposition 2 uk converges weakly on every compact K to a solution of�gu+ �Ru = �f(u+ h)N�1 on K (15)u � 0 (16)5

ProofSuppose in this section that the Lagrange multiplier �k is bounded in the equa-tion (5) (this will be demonstrated in section 3.3)�guk + �Ruk = �kf(uk + hk)N�1 (17)then Zk (j ruk j2 + �Ru2k) = �k( � Zk f(uk + hk)N�1hk) < �k (18)Hence on every compact subset K, as k goes to in�nity, there exists a k0,such that for all k � k0 , K � k andZK(j ruk j2 + �Ru2k) � C (19)where C is a constant. Finally uk is bounded in H1(K). To prove that theconvergence is strong, the Lagrange multiplier sequence needs to be studied. Wewill then prove the fact that uk is bounded in LN 0(K), where N 0 is greater thanN, the Sobolev critical exponent. Then by standard elliptic theory, the sequenceuk is bounded in C2;�(K). So by Arzela-Ascoli theorem, the convergence isuniform on each compact subset.2 Strong convergenceIn this section, conditions to obtain the strong convergence are established, andalso estimates on the Lagrange multiplier so that it converges. The followingnotation is adopted �k = uk + hk. This is a function of compact support in kSuppose that K' is a given compact and N 0 > N , the critical Sobolev expo-nent, if RVn�K0 f�N 0k dVg is bounded, then for any � > 0, there exists a compactK such that K 0 � K and such that RVn�K f�Nk dVg � �, due to the fact thatf 2 L1 and H�older inequality. In a naive point of view everything happens as ifthe phenomena at in�nity disappears.The study of the convergence for RVn f�Nk dVg is split into two parts :ZVn f�Nk dVg = ZK f�Nk dVg + ZVn�K f�Nk dVg (20)K is a compact subset.The convergence of the two integrals are proved separately :ZK f�Nk dVg ! ZK f�NdVg (21)ZVn�K f�Nk dVg ! ZVn�K f�NdVg (22)6

2.0.3 Nonconcentration at in�nity conditionThe convergence of the �rst integral in (21) is a consequence of the noncon-centration condition locally. To study the second integral, �rst we recall thede�nition :De�nition 1 The sequence �k (de�ned in the last section) is concentrated atin�nity if there exists a real � > 0 such that for every R > 0limk!+1 ZVn�BP (R) f�Nk > �where P is a given point.The standard de�nition of the concentration isDe�nition 2 Q is called a concentration point for the sequence �p if9� > O such that 8� > O , limk!1 ZBQ(�) f�Nk > �: (23)If the concentration phenomena at in�nity does not occur, the second integralin (21) could be taken as small as possible by choosing a large compact set.To prove that no concentration at in�nity occurs, equation (5) is multipliedby a test function �2f2=N�pk , where p > 1 is �xed. � is a C1 function equal to 1in Vn � 2K, vanishing identically on K, where K � 2K. 2K denotes a compactsubset.It is standard [4], that equation (5) is transformed4p(p+1)2 RVn�K j r(�f1=N�(p+1)=2k ) j2 � 2(p�1)(p+1)2 RVn�K �f1=N�(�f1=N )�p+1k +RVn�K �R�2f2=N�p+1k � 2p+1 RVn�K j r�f1=N j2�p+1k= �p RVn�K �2f2=N+1�N+p�1k (24)Using H�older's inequality the last term becomes :ZVn�K �2f2=N+1�N+p�1k � (ZVn�K f�Nk )1�2=N ( supVn�K f)2=N (ZVn�K �f�(p+1)N=2k )2=N (25)and �p = �p( )RVn f�N�1p up (26)We use the conditions at in�nity on f to bound the other terms ( the existenceof functions f which satisfy the decay conditions is given at the end of this paper).7

Suppose the following decay at in�nity,j rf j� Cf (27)j �f j� Cf (28)where C is a constant. A computation givesf1=N�f1=N = N � 1N2 f2=N�2j rf j2 + f2=N�1�f (29)Hence j f1=N�f1=N j� Cf2=N (30)Moreover,j �f1=N�(�f1=N ) j=j �2f1=N�f1=N � 2�f2=N�1r�rf + f2=N��� j� Cf2=N (31)The computation of the second term in the left hand-side of (24) givesj ZVn�K �f1=N�(�f1=N )�p+1k j� C ZVn�K f2=N�p+1k (32)By de�nition of the sequence �k = uk + hk, the last expression becomesZVn�K f2=N�p+1k � C(ZVn�K f2=Nhp+1k + ZVn�K f2=Nup+1k ) (33)By H�older inequality, the �rst term in the right hand-side is estimated as followZVn�K f2=Nhp+1k � (ZVn�K f2=N )(p+1)=N (ZVn�K f2=NhNk )1�(p+1)=N (34)Since RVn f2=N exists, and hk is a bounded sequence, the two terms in the lastinequality can be arbitrary small.The second term in (33) is evaluated �rst by using H�older inequality:ZVn�K f2=Nup+1k � (ZVn�K f2=NuNk )(p+1)=N (ZVn�K f2=N )1�(p+1)=N (35)Since the Yamabe invariant � is positive, (RVn uNk )2=N � 1� (RVn j ruk j2 +�Ru2k). By construction of the sequence uk, which has a compact support in k,we have ZVn j ruk j2 + �Ru2k = �k � Zk f(uk + hk)N�1hk � �k (36)8

Hence , (ZVn uNk )2=N � 1��k (37)but by assumptions the Lagrange multiplier satis�es �k � 1K2(sup f)2=N 1�2=NFinally, (ZVn uNk )2=N � 1� 1K2(sup f)2=N 2=N (38)which is bounded.Using these estimates, we obtainZVn f2=NuNk � sup f2=N ZVn uNk � 1�N=2 1KN (sup f) sup f2=N (39)This is term is bounded independently of the sequence uk. Finally, RVn�K f2=Nup+1kis bounded.For the same reasons, using the inequality f2=N�2j rf j2 � Cf2=N , we de-duce thatZVn�K j r�f1=N j2�p+1k � C Z2K�K f2=N�p+1 + C 0 ZVn�K(f1=N�1))2j rf j2�p+1k ) � C (40)Using the conditions at in�nity on the decay of f , we obtain that the lastterm of (24) is bounded RVn�K Rn=2 � C RVn�K f � �. Indeed by H�olderinequalityZVn�K R�2f2=N�p+1k � C(ZVn�K Rn=2)2=n(ZVn�K �f� (p+1)N2 )2=NFinally (24) gives the following estimate :ZVn�K j r(�f1=N�p+1=2k ) j2 � Cte+( �kRVn f�N�1k uk 1�2=N ( supVn�K f)2=N + �)(ZVn�K �f�(p+1)N=2k )2=N (41)According to Aubin [1] [2] [3] for any � > 0 there exists a constant A� andK such that for all functions in H1,(ZVn uN )2=N � (K2 + �) ZVn j ru j2 +A� ZVn u2 (42)9

Hence if �kRVn f�N�1k uk 1�2=N ( supVn�K f)2=N < 1 (43)is satis�ed, then RVn�K j r(�f1=N�p+1=2k ) j2 is bounded and alsoZVn�K f�N 0p (44)for N 0 > N is bounded.The �nal condition is :�( ) � RVn f(u+ h)N�1h 1�2=N ( supVn�K f)2=N < 1 (45)where u is the weak limit of (5) and �( ) denotes�( ) = infA I(u) = infA ZVn(j ru j2 + �Ru2) (46)with A = fu 2 H1(Vn) RVn f j u jN = g.2.0.4 Condition for nonconcentration phenomena on a compact setIn a compact set, conditions for nonconcentration phenomena are well known[13]. The nonconcentration conditions for the solution u in a compact set is thesame as the conditions for each element of the converging sequence uk recalledin the relation (12) and is adapted here :�( ) � RK f(u+ h)N�1h 1�2=N (supK f)2=N < 1 (47)Where is a parameter to be selected. This identity is obtained in [12] byusing test functions.K denotes an arbitrary compact. It is chosen, such that @K avoids possibleconcentration points. This is possible since the sequence uk is of �nite energy,so only a �nite number of concentration points can appear. The fact that theenergy is bounded is proved in the next section using geometric conditions.3 Geometric conditions for nonconcentration atin�nityTo construct the minimizing sequence uk and to obtain a minimum solution uon Vn, an in�nite number of conditions are given associate to the sequence ukof the last section. Another condition was found on the weak limit u.10

For each k, (13) and (12) need to be satis�ed plus the two following conditions: If C denotes a compact subset of Vn�( )K2 � RVn f(u+ h)N�1h 1�2=N ( supVn�C f)2=N < 1 (48)and �( )K2 � RVn f(u+ h)N�1h 1�2=N (supC f)2=N < 1 (49)We prove that the conditions insuring the absence of concentration phenom-ena locally and at in�nity are satis�ed by using test functions and the assump-tions of the theorem 1 and 2. Moreover, the sequence k could be chosen equalto a constant , large enough. Two cases appear :3.1 The nonlocally conformally at caseIn the case the manifold is not locally conformally at, we assume that thereexists a maximum point P of f where the Weyl tensor does not vanish. Denoteby r the geodesic distance to P.Consider the following functionu�(r) = 1(�2 + r2)n=2�1 � 1(�2 + �2)n=2�1 in BP (�) (50)= 0 in Vn �BP (r) (51)� is �xed.We will see that the conditions, for the sequence (uk) to be a minimizing oneach k, is satis�ed independently of k when is a su�ciently large constant.First let us choose � = ZVn f(u� + h)N (52)and �( �) = ZVn j ru� j2 + �Ru2� (53)We introduce the following quantities : Q�, P� Qk� and P k� . They are de�nedby Q� = �( �)K2(sup f)2=N � � RVn f(u+ h)N�1h 1�2=N� ( supVn�K f)2=N (54)11

P� = �( �)K2(sup f)2=N � � RVn f(u+ h)N�1h 1�2=N� (supK f)2=N (55)Qk� = �( k)K2(sup f)2=N k � Rk f(uk + hk)N�1hk 1�2=Nk (supk f)2=N (56)P k� = K2 k�Rk f(uk+hk)N�1hk�( k)(sup f)2=N ( k + C1 Rk fhNk +C2 Rk fuN�1k hk)1�2=N (57)where C1 and C2 are two constants.It remains to show that this quantities are strictly less than 1 independentlyof k. Using the computations of [12]. Later on on this paper we should recalland adapt the relevant parts of [12] for the convenience of the reader and weshall point out where the boundedness assumptions on f are used.The expansions are : � = ZVn f j u� + h jN = Z �0 rn�1dr ZS(r) f j u� + h(x) jNd + ZVn�BP fhN (58)where d is the volume of the unit sphere Sn�1. Obviously, the second termin this expression is independent of �. The term k = , de�ned in the previouspart is also developed. By de�nition, = k = Zk f j u� + hk jN = Z �0 rn�1dr ZS(r) f j u� + hk(x) jNd + Zk�BP fhNk (59)The second term in the last expression is always bounded by assumption,since the sequence hk converges to a bounded solution h and RVn fhN <1. �and k have the same expansion in the variable �, . Let us compute only the�rst : The change of variable x = �u gives � = Z �=�0 �nun�1du ZS(r) f(�u)pg(�u)��2nj v + �n�2h(�u) jNd +O(1) (60)with v(r) = ( 1(r2+1)(n�2)=2 � �n�2(�2+�2)(n�2)=2 ) on BP (�) (61)v(P ) = 0 on Vn �BP (62)developing the following expression12

G(u) = ZS(r) f(�u)pg(�u)��2nj v + �n�2hk(�u) jNd (63)using Taylor expansionj v + �n�2h(�u) jN = j v jN +�n�2 R 10 N j v + t�n�2h(�u) jN�2(v + t�n�2h(�u))h(�u)dt (64)Then if R�( ) denotes the rest withR�(u) = ��n�2 RS(r) f(�u)pg(�u) R 10 N j v + t�n�2h(�u) jN�2(v + t�n�2h(�u))h(�u)dtd (65)we obtain R�( ) = Z �=�0 �nun�1duR�(u) (66)and deduce the following boundR�( ) � C��2 Z �=�0 un�1du ZS(u) j f j j v + �n�2 j suph jjN�1j suph jd +O(1)(67)But h is bounded. C is a positive constant, which does not depend on � and k.Finally, � = ̂� +R�( ) (68)with ̂� = RVn fu�N .Moreover ̂� = 0(��n) since : ̂� = ��n Z �=�0 rn�1dr ZS(r�) f(�x)vN� � C��n Z �=�0 rn�1drvN� (69)but Z �=�0 rn�1drvN� � C (70)Similar types of inequalities gives an estimate of a bound for R� :R�( ) � ��2C Z �=�0 un�1duj v� + �n�2j suph j jN�1 � C��2 (71)Finally, � = ̂�(1 +O(�n�2)) (72)Q� = S� + o(�4) (73)13

and S� = (sup f)2=NK2�(u�) ̂(u�)�2=N (74)To �nish the computation S�, � � is expanded ([12]) with Ipq = R +10 (1 +t)�ptqdt : �(u�) = (n�2)2wn�12�n�2 In=2n (75)� (1� �2 R6n + �4(T (n+2)(n+4)(n�6)(n�4) � Rcn ( R6n + �R2nR 4(n�1)(n�6)(n�4)(n�2))) + o(�4))and ̂�2=N� = (wn�1f(P )��n (n� 2)2n In=2n ) 2�nn (1 + �2( �f2nf(P ) + R6n) +�4( (2� n)n An(n+ 2)(n� 4)(n� 2)f(P ) + n� 1n� 2( R6n + �f2nf(P ))2 + o(�4) (76)with A = f(P )T + 1(n+ 2)n(�2f4! +R�f12 � Rilfil4 ) (77)and T = 1(n+ 2)n4!(6�R5 � 2RijRij15 � R2ijkl5 + 2RijRij3 + R23 ) (78)now using the assumptions on the geometry at the point P,Q� = 1 +Q4�4 + o(�4) (79)and Q4 = 112n(n� 4)(n� 6)(�W 2)� �2g0f4!n(n� 4) (80)where g0 is a conformal metric built as in [2] to eliminate the tensors whichare not conformally invariant. Using assumption �2g0f = 0, we are lead toQ4 < 0 and the result Q� < 1: (81)To conclude this section we give the other bounded estimates : on Vn �K,using supVn�K f � supK f it follows that condition (49) is satis�ed. Moreover k could be chosen independently of k in the expression of (56),(57), and thesame conclusion holds : P k� < 1, Qk� < 1.To prove that P� < 1 we use that the sequence uk is bounded in LN and themain part of the denominator in this quantity is given by (�). IndeedZk f(uk + hk)N�1hk � 1�1=Nk (Zk f(hk)N�1)1=N (82)and the expansion in � gives k � Zk f(uk + hk)N�1hk = k(1 +O(�n=N )) (83)14

3.2 The locally conformally at caseWe assume the manifold to be complete and locally conformally at. For eachsmall open set, there exists a function � > 0 such that g = �4=(n�2)�. In thiscase, it has been proved that there exists a minimizing sequence uk such thatuk is a solution of (5) on each k, k 2 N . Moreover we will see that if thesequence of functions hk, de�ned in (7) are strictly positive then the minimalsolution uk exists, independently of the sign of the mass of k.To built the test function in this case, consider the Green function of theconformal Laplacian, by de�nition�Gk + �RGk = �P on k (84)Gk = 0 on @k (85)and we have the following expansion in an Euclidean neighborhood of P:Gk(r) = r2�n +Ak + �k (86)where r is the geodesic distance relative to P and �k is a C1 function suchthat �k(P ) = 0. The sequence of the conformal Green function Gk convergesto G, uniformly on every compact.To check the nonconcentration conditions for the limit u of the sequence uk,we shall use the following test function�� = �l(�2+r2)(n�2)=2 on BP (�) (87)�0(G+ �) on BP (2�)�BP (�)�0G on Vn �BP (2�)G is the minimal Green function solution of�G+ �RG = �P (88)To check the nonconcentration conditions depending on k, we use�p;�(r) = �l(�2+r2)(n�2)=2 on BP (�) (89)�0(Gp + �p) on BP (2�)�BP (�)�0Gp on p �BP (2�)where Gp is de�ned above (84).In order that The test functions (87) and (89) are continuous the followingcondition must be satis�ed �l(�2 + �2)(n�2)=2 = �0(�2�n +A) (90)15

for one case, l, �0 are constant numbers, such that 0 < l < (n� 2)=n, is C1decreasing function in j r j, (x) = 1 on BP (�), (x) = 0 for j x j� 2� andj r j� 2��1on � �j x j� 2�. For the other case�l(�2 + �2)(n�2)=2 = �0p(�2�n +Ap) (91)where �0p > 0The quantities Q� P� Qk� P k� will be evaluated as in the previous case, usingcomputations of [12].We shall evaluate : � = ZVn f(u� + h)N and �( �) = ZVn j ru� j2 + �Ru2� (92)and k = Zk f(u� + h)N and �( k) = Zk j ru� j2 + �Ru2� (93)the computations give [13]Qk� = P k� = 1� 4hk(0)�n�2�lJn 1In=2�1n < 1 (94)J depending on n only andP� = Q� = 1� 4h(0)�n�2�lJn 1In=2�1n < 1 (95)where Jn = R10 rn�1(r2+1)(n+2)=2 drIn the last section, we prove that h > 0 hence it is not identically 0. Moreoverthe sequence hk converges to h uniformly on every compact. To conclude thissection, remark that the mass of the manifold [15] does not appear as in othercases. The function h plays the fundamental role. If h were identically zero,then by [12], Qk� = P k� = 1� 2nIn=2n Ak�n�2 < 1 (96)and Qk� = P k� = 1� 2nIn=2n A�n�2 < 1 (97)16

3.3 Estimation of the Lagrange multiplierTo �nish this section, estimates of the Lagrange multiplier sequence �k will beobtain as a consequence of the previous study of nonconcentration. In [12] it isproved that lim sup�( k) � �( ) when �k converges to . The sequence f kgsatis�es the following inequality, for = k large enough, relation (18) gives0 < �k � �( k) k � Rk f(uk + hk)N�1hk : (98)Using the nonconcentration relation (12), one obtains the estimate 0 < �k �1K2 sup f2=N 1�2=NFinally the sequence �k is bounded. Suppose that �k converges to 0, thenthe sequence �k = uk + hk converges to 0 on each compact set sinceZk (j ruk j2 + �Ru2k) = �k( � Zk f(uk + hk)N�1hk) (99)But since uk converges strongly in LN and since Rk f(uk + hk)N = ,RVn fhN = . This contradict the fact that > RVn fhN when is chosen largeenough.3.4 Remarks and conclusionIf R(g) decays at in�nity as 1rq , then the condition that RVn Rn=2 � C R f <1gives a upper bound for the decay rate q > 2.The conditions on f at in�nity, stated in the two theorems, are satis�ed, forexample, when f is decreasing like power functions : f � r�q , rf � r�q�1 andrrf � r�q�2, with q > Nn=2 > 2. Indeed RVn f2=N < +1, hence 1r1�n+(1�2=n)qis in L1(R ).When h converges to 1 at in�nity (which is the case in asymptotically atmanifolds [6]) then the inequality h + u � h > 0, shows that the metric g0 =(h+ u)4=(n�2)g is complete.In the next section, it is proved that the condition 2 of theorem 2 is rati�edunder small additional assumptions on the geometry. This is done by studyinga linear PDE of the type introduced by E. Witten [20] in the case of Spinormanifolds. Results on this linear PDE are used by Witten to prove in dimension4 the positive mass theorem.4 Conformal metric of zero scalar curvature oncomplete manifoldsThe purpose of this part is to prove the existence of a function h solution of theproblem 17

�gh+ �Rh = 0 on Vn (100)h! C > 0 when r ! +1A point O 2 Vn is given, r is the Riemannian distance r = d(P;O).h converges to C means that for every � > 0, there exists a compact Kcontaining O such that 8P 2 Vn �K, j h(P )� C j� �.The existence of such a function h is used to solve the prescribed scalarcurvature problem for a large class of complete manifolds [11], such as theasymptotically at case or certain conformally at cases.A certain number of assumptions are made on the geometry of the manifold.Let us recall the de�nition of the Yamabe conformal invariant :� = infH1(Vn) RVn(ru)2 + �Ru2(RVn uN )2=N (101)the assumptions are :1. The Yamabe invariant is positive : � > 02. An integral bound on the scalar curvature : RVn RdVg < +13. The Ricci curvature is bounded from below4. The injectivity radius is strictly positiveWhen the last two conditions are satis�ed, (Vn; g) is said to have a boundedgeometry.If (Vn; g) has a bounded geometry, the Sobolev inequality is valid. Theconformal Laplacian has a minimal Green function which can be estimated.This is a consequence of some works on the heat kernel [18] [7] [8].The main theorem presented here is the followingTheorem 3 (Vn; g) is a complete manifold, with Ricci curvature bounded frombelow and injectivity radius strictly positive, bounded away from zero, and withthe Yamabe conformal invariant � > 0 .If R denotes the scalar curvature, it is also assumed that R � 0 and RVn R <+1. Then G the minimal Green function of the Laplacian exists and if moreoverRVn G(P;Q)R(P )dVg goes to zero at in�nity, then there exists a strictly positivesolution of (100) which converges to a positive constant at in�nity.Geometric conditions can be stated which insure that RVn G(P;Q)R(P )dVggoes to zero at in�nity. For example, it will be proved in the last section, thatthis is the case if the scalar curvature R converges to zero at in�nity or if R isbounded in Lq with q > n=2.The geometric interpretation of this result is the following : when the scalarcurvature is positive, then the solution h is strictly positive. This is summarizedin the proposition 18

Proposition 3 On manifolds with a bounded geometry and strictly positiveYamabe invariant whose scalar curvature is positive and either belongs to L1\Lq(q > n=2) or tend to zero at in�nity, there exists a conformal metric h4=(n�2)ghaving zero scalar curvature and the manifold (h4=(n�2)g; Vn) is complete.As a consequence of the main theorem, one obtains :Proposition 4 Let (Vn; g) be a complete manifold. If in addition to the as-sumptions of the previous theorem, RVn j R jn=2dVg is bounded, then the pre-scribed scalar curvature problem can be solved : there exists a positive solutionto the equation �gu+ �Ru = f j u jn+2n�2 on Vn (102)u > 0under the conditions on the function f described in the �rst section. The metricg0 = u4=n�2g obtained is complete.This proposition is proved exactly as the two theorems of the last section butinstead of choosing 1 as the boundary condition for the approximation sequence,we use the restriction of the linear positive solution of (100). In this case, thesequence hk converges exactly to the solution.Previous results on the subject were obtained by Brill-Cantor [6]: theyproved the existence of a positive solution of (100) on asymptotically at man-ifolds with �nite ends using weighted Sobolev spaces. The precise statementis Assume Vn has a �nite number of ends, having neighborhoods N1; ::Nk dif-feomorphic to the complement of a unit ball in <n. The weight Sobolev spacesare de�ned by : Hps;� = ff s.t. Pj�j�s j D�fi1:::ij��+j�j jLp(Nl) < +1 for eachends Nl g where � = p1 + r2The two following propositions are equivalent [6] :� There exists a solution � of (100) on asymptotically at manifolds with1 < p < 2n=n� 2 , �n=p < � < �2 + n(1 � 1=p) and s > n=p+ 2. Themetric g considered as a section of a vector bundle is in Hps;�. Moreoverthe solution � satis�es �� 1 2 Hps;� and � > 0.� for all u 6= 0, u 2 C10 , the following inequality is satis�ed :ZVn j ru j2 + �R(g)u2 > 0; (103)The di�erentiability conditions on �; r; s stated above are very restrictive :s > n=p + 2 > n=2 + 1. There exists (Vn; g) having bounded geometry suchthat g does not belong to Hps;� but such that the scalar curvature R belongs toL1 \ Lq with q > n=2. 19

5 A Variational problemIn this section, a variational problem is constructed to obtain a solution ofequation (100):Consider � = infA I(u) = infA ZVn j ru j2 + ZVn �Ru2 (104)where A = fu 2 H1(Vn) s.t. RVn �Ru = 1g:Lemma 1 The minimum � is achieved by a positive function h which satis�es: ZVn �Rh = 1 (105)� = ZVn j rh j2 + ZVn �Rh2 < +1 (106)�gh+ �Rh = � �R (107)h 2 LN (108)The last fact implies that h is not a constant. Moreover � is strictly positive.Proof :Consider uk a minimizing sequence satisfying : � � RVn j ruk j2+RVn �Ru2k ��+ 1=k with RVn �Ruk = 1uk is bounded in LN(Vn). On every compact set K, the sequence (uk) isbounded in H1(K) : this is a consequence of the fact that on every compact,the �rst eigenvalue of the operator � + �R is positive. This follow from theinequality 0 < � � CK�1(� + �R) where CK is a positive constant dependingonly of K.Choosing a subsequence if need be we can assume that uk converges a.e onevery compact to a function u. Then � = RVn j ru j2 + RVn �Ru2. MoreoverRVn �Ru = 1. To see this all we have to do is to prove that for any � > 0 thereexists a compact set K such that : RVn�K �Ruk � C� where C = sup RVn �Ru2k.In fact choose K so that RVn�K �R � �2 then by the schwarz inequalityZVn�K �Ruk �sZVn�K �Ru2k ZVn�K �R � CsZVn�K �R � C� (109)the last inequality is a consequence of R 2 L1 and RVn Ru2k is bounded, sinceuk is a minimizing sequence.u is a solution of the variational problem. The Euler-Lagrange equationimplies that there exists a Lagrange multiplier �0 such that :8� 2 C10 ; Z r�ru+ �Ru� = �0 Z �R� (110)20

Hence �u+ �Ru = �0 �R, � = �0. Moreover u is not zero since RVn �Ru = 1 and u isnot constant since u belongs to LN . � is not zero because � = RVn j ru j2+ �Ru2.It remains to prove that u tends to zero at in�nity then it follows that thefunction H = � � u will be a solution of equation (100), which converges to apositive constant � at in�nity.6 Behavior of h at in�nityIn the previous section the existence of a solution of (100) was obtained by thevariational method. In this section estimates on the behavior at in�nity of thesolution are obtained by the use of the Green function.The minimal Green function of the conformal Laplacian is the smallest ofthe solutions of �gG+ �RG = �P (111)Such a function exists [12]. �P denotes the Dirac distribution localized at a givenpoint P. The Green function of the Laplacian is the minimal strictly positivesolution de�ned by �G0 = �P (112)6.1 Existence of the Green functionThere exists conditions for the existence of such a function : Varopoulos statesin [18] that if the Ricci curvature is positive then the existence of the Greenfunction is equivalent to the fact that the integral R +11 tV ol(Bt(x0))dt < +1converges.Here we will use the existence of the Green function under weaker conditions.The Heat Kernel theory developed partially in [18] [8] [7], gives the existenceand estimates of such a function.Suppose that the conditions of the bounded geometry are satis�ed : theRicci Curvature bounded from below and the injectivity radius strictly positive.If p(t,x,y) denotes the heat kernel solution of�p = @tp (113)Chavel [7] proves the estimations :0 < p(t; x; x) � Ct�n=2 (114)and � > 0; 8t > 0 and 8x; y 2 Vn (115)0 < p(t; x; y) � C�t�n=2e�d2(x;y)=(4t(1+�)) (116)C� and C are two constants. 21

Finally G(x; y) = R +10 p(t; x; y)dt exists and8x 6= y 2 Vn , 0 < G(x; y) � Cr�n+2 (117)where r is the Riemannian distance from x to y.This is a consequence of a theorem of [7] which we will recall now. Usingthe same notations : is an open subset with compact closure and smoothboundary, the isoperimetric quotient is :I() = A(@)V ()1�1=n (118)where A denotes the n-1 dimensional Riemannian measure and V denotes then-dimensional Riemannian measure. The isoperimetric In constant is de�nedto be the in�mum of I() under all smooth In = inf I() (119)When the Sobolev inequalities are valid, which is the case in bounded ge-ometry, H1p � Lq with 1q = 1� 1p , the following equality holds by a theorem ofFederer-Fleming [7] :In = K1n = inff2C10 R j rf j(R j f j)1�1=n > 0 (120)�nally to sum up the results :Proposition 5 On bounded geometry, In > 0 and the following estimate holds:8t > 0 and 8x; y 2 Vn 0 < p(t; x; y) � Ct�n=2e�d2(x;y)=(4t(1+�)) (121)Hence the solution G0 of �G0 = �P exists and the following estimate on thebehavior of G0 is also valid : 0 < G0(x; y) � Cr�n+26.2 Application to the behavior of the solutionWe can deduce from the results of the last paragraph and the fact that thescalar curvature is positive, that 0 < G < G0. Indeed since G is the minimalGreen function of the conformal Laplacian, the sequence Gi of Conformal Greenfunction with boundary value equal to 0 on i (an exhaustion sequence) satis�esG = limi!+1Gi. Consider the bounded function Gi �G0, solution of �(G �G0) + �R(G�G0) = � �RG0 � 0 it satis�es the maximum principle Gi �G0 < 0on @i so Gi �G0 < 0. Hence 0 < G < G0.Using the Conformal Green function, the solution of equation (2) can bewritten as: 22

u(P ) = � ZVn R(Q)G(P;Q)dVg (122)To prove that u goes to zero at in�nity under geometric conditions, take anumber � > 0. we shall prove that there exists a number R0 such that for allP with d(P; 0) � R0 , 0 � u(P ) � �: To start �x a constant R1 > 0 such thatK = BO(R1) is a compact set, RK R(Q)G(P;Q)dVg � � for d(P; 0) � R2 andRVn�K R � �. Then :u(P ) = � ZVn�K R(Q)G(P;Q)dVg + � ZK R(Q)G(P;Q)dVg (123)To prove that the �rst integral can be bounded from above by any positiveconstant arbitrary small, consider the decompositionRVn�K R(Q)G(P;Q)dVg = RVn�K\fG�1gR(Q)G(P;Q)dVg +RVn�K\fG�1gR(Q)G(P;Q)dVg (124)ZVn�K\fG�1gR(Q)G(P;Q)dVg � ZVn�K\fG�1gR(Q)dVg (125)� ZVn�K R(Q)dVg � � (126)and ZVn�K\fG�1gR(Q)G(P;Q)dVg � (127)ZVn�K\fG�1g\fd(P;Q)�C1=(n�2)gR(Q)G(P;Q)dVg (128)since the inequality Cd(P;Q)n�2 � G(P;Q) � 1 implies : d(P;Q) � C1=(n�2) Atthis stage of the proof, two possibilities arise: �rst if R tends to 0 at in�nitythen ZVn�K\fG�1g\fd(P;Q)�C1=(n�2)gR(Q)G(P;Q)dVg (129)� supVn�KR Zfd(P;Q)�C1=(n�2)gG(P;Q)dVg (130)and Rfd(P;Q)�C1=(n�2)gG(P;Q)dVg � C 0C2=(n�2) where C 0 is a constant. HenceZVn�K\fG�1gR(Q)G(P;Q)dVg � supVn�KRC 0C2=(n�2) � � (131)23

If R tends to 0 at in�nity, the last expression is bounded by � for d(P;O) � R4.The second possibility is thatRVn�K\fG�1gR(Q)G(P;Q)dVg � (RVn�K\fG�1gR(Q)�dVg)1=� �(RVn�K\fG�1gG(P;Q)�dVg)1=� (132)using H�older inequality. Now RVn�K\fG�1gG(P;Q)�dVg is bounded if � < nn�2and since the integral RVn R(Q)�dVg converges for � > n=2 by assumptions, wecan choose K being enough so that(ZVn�K R(Q)�dVg)1=� � � (133)Finally for P large enough (d(P;O) � Ri for i=1..3) the integral de�ning u(P )goes to zero when P converges to in�nity under geometric conditions statesin the theorems. This shows that u(P ) tends to zero at in�nity. The maintheorem is solved by considering the function �� u which is a solution of (100)and converges to the positive constant �.References[1] T. Aubin{ Some Nonlinear problems in Riemannian Geometry, 1998Springer-Verlag.[2] T. Aubin{ Equations di�erentielles non lin�eaires et probl�eme de Yamabeconcernant la courbure scalaire, J. Math. pures et appl. 55, 1976 p269-296.[3] T. Aubin{ Espace de Sobolev sur les vari�et�es Riemanniennes, Bull. Sci.math. 2, 100 1976 p 149-173[4] T. Aubin{ Meilleures constantes dans le th�eor�eme d'inclusion de Sobolevet un th�eor�eme de Fredholm non lin�eaire pour la transformation conformede la courbure scalaire J. Funct. Anal., 32, 1979 p148-174.[5] P. Aviles{ R. McOwen Conformal deformation to constant negativescalar curvature on noncompact Riemannian manifolds. J. Di�erential Ge-ometry, 27, 1988, 225-239[6] M. Cantor D. Brill The Laplacian on asymptotically at manifolds,Compositio Mathe. , 43, 1981 p 317-330.[7] I. Chavel E. Feldman { Modi�ed isoperimetric constants, and largetime heat di�usion in Riemannian manifolds, Duke math. J. 64, 3, 1991 p473-499.[8] E.Davies. Gaussian Upper Bounds for Heat Kernels of Some Second-Order Operators on Riemannian Manifolds Journ. Func. Anal. 80, 1988 p16-32. 24

[9] D. Gilbarg { N. Trudinger { Elliptic Partial di�erentiable equationsof second order; Springer-Verlag 1983, second edition.[10] E. Hebey { M. Vaugon { The best constant problem in The Sobolevembedding theorem for complete riemann Manifolds, Indiana, Vol.79, n1,1995 . p235-273[11] D.Holcman{ Th�ese de Doctorat de l'universit�e Paris VI, 1998.[12] D. Holcman{ Solutions nodales sur les vari�et�es riemanniennes compactes,C.R. Acad. Sci. Paris t.326 SI, p 1205-1208, 1998[13] D. Holcman{ Solutions nodales sur les vari�et�es riemanniennes �a bord,C.R. Acad. Sci. Paris t.326 SI, p 1321-1324, 1998[14] S.Kim The Yamabe problem and applications on noncompact completeRiemannian manifolds and the speci�cation of the scalar curvature Ge-ometriae Dedicata 64: 1997 p373-381[15] R. Schoen Variational theory for the total scalar curvature functionalfor Riemannian metrics and related topics Montecatini notes, 1987 Lect.Notes 1365, Springer Verlag, p120-155.[16] R. Schoen Conformal deformation of a Riemannian metric to constantscalar curvature, J. Di�. Geom. 20, 1984 p 479-495.[17] N.Varopoulos Hardy-Littlewood theory for semigroups, Journal of Func-tional Analysis, 1985, 63, p240-260[18] N. Varopoulos The Poisson Kernel on Positively Curved Manifolds,Journ. Func. Anal. 44, 3, 1981 p 359-380.[19] M. Vaugon { Transformation de la courbure scalaire sur une vari�et�e Rie-mannienne compacte J. Funct. Anal., 71, No. 1, 1987 p 182-194.[20] E. Witten { A new proof of the positive Energy theorem, Communicationin mathematical physics, 80, 1981 ,p 381-402.

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