A prescribed scalar and boundary mean curvature problem on compact manifolds with boundary
aa r X i v : . [ m a t h . DG ] F e b A prescribed scalar and boundary mean curvatureproblem on compact manifolds with boundary
Vladmir Sicca and Gantumur Tsogtgerel
McGill University
Abstract
We consider the problem of finding a metric in a given conformal class with prescribed nonpositive scalarcurvature and nonpositive boundary mean curvature on a compact manifold with boundary, and establish anecessary and sufficient condition in terms of a conformal invariant that measures the zero set of the targetcurvatures.
Let (
M, g ) be a compact Riemannian manifold with boundary. Suppose that R ′ is a nonpositivefunction in M , and H ′ is a nonpositive function on ∂M . Then we ask the question: Can weconformally transform the metric g so that the resulting metric has the scalar curvature equal to R ′ , and the boundary mean curvature equal to H ′ ? Setting the dimension of M to be three for thesake of this introduction, this is equivalent to solubility of the equation ( − u + Ru = R ′ u in M, ∂ ν u + Hu = H ′ u on ∂M, (1)for a positive function u , where R and H are the scalar curvature and the boundary mean curvatureof g , respectively, and ∂ ν is the outward normal derivative. Apart from its intrinsic importance,we are led to this problem by the study of the Einstein constraint equations. More specifically, inthe negative Yamabe case, solubility of an important class of Lichnerowicz equations on compactmanifolds with boundary is shown to be equivalent to a certain instance of (1), cf. [5].The analogue of this problem on closed manifolds was first studied by Rauzy in [7], and morerecently, the analysis has been extended to asymptotically flat manifolds by Dilts and Maxwell in[6], and to asymptotically hyperbolic manifolds by Gicquaid in [3]. Restricting ourselves to theboundaryless case, those works show that a generalization of the Yamabe invariant to subsets of themanifold plays an important role, in that (1) has a positive solution if and only if this generalizedYamabe invariant for the zero set of R ′ is positive. The positivity of the Yamabe invariant is ina certain sense a measure of the smallness of the subset, as observed for asymptotically Euclideanmanifolds in Lemma 3.15 of [6].In this work, we adapt these ideas – particularly the ones in [6] – to the framework of manifoldswith boundary. To do so we extend the Yamabe invariant of manifolds with boundary as defined byEscobar to pairs (Ω , Σ), where Ω ⊂ M and Σ ⊂ ∂M , which we call the relative Yamabe invariant ,by looking at test functions supported only on Ω ∪ Σ and develop some basic properties of theinvariant. In particular, we relate this invariant to a relative version of the first eigenvalue of theLaplacian of the pair (Ω , Σ), allowing us to use results related to the linear version of problem (1).1
The relative Yamabe invariant 2
In the end, we can prove that, in the Yamabe negative case, (1) has a positive solution if and onlyif the relative Yamabe invariant of the pair ( { R ′ = 0 } , { H ′ = 0 } ) is positive. The paper is structured as follows. In section 2 we take the necessary steps to define the relativeYamabe invariant in the framework of manifolds with boundary and establish its relation to thesubcritical problem corresponding to (1). In section 3 we develop the theory of the linearizedproblem by defining the relative eigenvalue of the Laplacian and showing that it can be used in thestudy of the relative Yamabe invariant since both have the same sign. In section 4 we bring bothideas together in using variational techniques to solve the prescribed curvature problem for Yamabenegative manifolds with boundary when R ′ ≤ H ′ ≤
0. Finally, in section 5 we put our resultsin context with some immediate consequences and examples.
Let M be a smooth, connected, compact manifold with boundary and dimension n ≥
3. Assumethat M is equipped with a Riemannian metric g ∈ W s,p , where sp > n and s ≥
1. We denoteby R ∈ W s − ,p ( M ) the scalar curvature of ( M, g ), and by H ∈ W s − − p ,p ( ∂M ) the mean extrinsiccurvature of the boundary ∂M , with respect to the outer normal. Let Ω ⊂ M and Σ ⊂ ∂M be(relatively) measurable sets, and consider the functional E : W , ( M ) → R defined by E ( ϕ ) = Z Ω |∇ ϕ | dV g + n − n − Z Ω Rϕ dV g + n − Z Σ H ( γϕ ) dσ g , (2)where γ : W , ( M ) → W , ( ∂M ) is the trace map and dV g and dσ g are the volume forms inducedby g on M and on ∂M respectively. By using the assumptions sp > n and s ≥
1, one can showthat E ( ϕ ) is finite for each ϕ ∈ W , ( M ).Let ¯ q = nn − . Then for 2 ≤ q ≤ q , and 2 ≤ r ≤ ¯ q + 1 with q ≥ r , and for b ∈ R , we define Y q,rb (Ω , Σ) = inf ϕ ∈ B q,rb (Ω , Σ) E ( ϕ ) , (3)where B q,rb (Ω , Σ) = { ϕ ∈ W , (Ω , Σ) : k ϕ k qL q (Ω) + b k γϕ k rL r (Σ) = 1 } , (4)and W , (Ω , Σ) = (cid:8) ϕ ∈ C (Ω) ∩ C (Ω) : ϕ | M \ Ω ≡ γϕ ) | ∂M \ Σ ≡ (cid:9) , (5)with the closure taken in W , ( M ).Note that q = 2¯ q is the critical exponent of the embedding W , ( M ) ֒ → L q ( M ), while r = ¯ q + 1is the critical exponent for the continuous trace operator W , ( M ) ֒ → L r ( ∂M ).A primary aim of this section is to establish that the sign of Y q,rb (Ω , Σ) is a conformal invariantand does not depend on the indices. To this end, we start by stating some lemmata.
Lemma 2.1 ([1]) . Let q > r > , a > and b be constants (if b > − a we can have q = r ), and let f b ( x ) = ax q + bx r , (6) where b ∈ R is a parameter. Then we have the following.(a) The equation f b ( x ) = 1 has a unique positive solution x b > . The relative Yamabe invariant 3 (b) The correspondence b x b is continuous. The next lemma is adapted from Proposition 2.3 of [1].
Lemma 2.2.
Let ≤ q ≤ q , and ≤ r ≤ ¯ q + 1 with q > r , and let b ∈ R . Then for any ε > ,there exists a constant C ε ≥ such that k ϕ k L (Ω) ≤ ε k∇ ϕ k L (Ω) + C ε , (7) and k γϕ k L (Σ) ≤ ε k∇ ϕ k L (Ω) + C ε , (8) for all ϕ ∈ B q,rb (Ω , Σ) .Proof. We will prove that the inequalities hold for all ϕ satisfying ϕ ∈ C (Ω) ∩ C (Ω) , supp ϕ ⊂ Ω ∪ Σ , k ϕ k qL q (Ω) + b k γϕ k rL r (Σ) =: F ( ϕ ) < . Let us first prove (7). We start by using H¨older’s inequality Z Ω ϕ dV ≤ (cid:20)Z Ω | ϕ | q dV (cid:21) q (cid:20)Z Ω dV (cid:21) q − q = (cid:20)Z Ω | ϕ | q dV (cid:21) q vol(Ω) q − q ≤ (cid:20) − b Z Σ | γϕ | r dσ (cid:21) q vol(Ω) q − q ≤ (cid:20) | b | Z Σ | γϕ | r dσ (cid:21) q vol(Ω) q − q . Since q ≤
1, we have ( a + b ) q ≤ a q + b q for a, b ≥
0, and so (cid:20) | b | Z Σ | γϕ | r dσ (cid:21) q vol(Ω) q − q ≤ " q + | b | q (cid:20)Z Σ | γϕ | r dσ (cid:21) q vol(Ω) q − q ≤ q vol(Ω) q − q + 2 q vol(Ω) q − q | b | q (cid:20)Z Σ | γϕ | r dσ (cid:21) q , yielding (cid:20) | b | Z Σ | γϕ | r dσ (cid:21) q vol(Ω) q − q ≤ C + C | b | q (cid:20)Z Σ | γϕ | r dσ (cid:21) q , (9)where C = 2 q vol(Ω) q − q .Now, the trace inequality says that there is a positive constant C = C ( M, g, r ) such that (cid:20)Z Σ | γϕ | r dV (cid:21) r = (cid:20)Z ∂M | γϕ | r dV (cid:21) r ≤ C Z M (cid:0) |∇ ϕ | + ϕ (cid:1) dV = C Z Ω (cid:0) |∇ ϕ | + ϕ (cid:1) dV, The relative Yamabe invariant 4 which leads to (cid:20)Z Σ | γϕ | r dV (cid:21) q ≤ C rq (cid:20)Z Ω |∇ ϕ | dV + Z Ω ϕ dV (cid:21) rq ≤ C rq (cid:18)Z Ω |∇ ϕ | dV (cid:19) rq + C rq (cid:18)Z Ω ϕ dV (cid:19) rq , (10)since rq ≤
1. Plugging (10) into (9) we get Z Ω ϕ dV ≤ C + C | b | q (cid:20)Z Σ | γϕ | r dσ (cid:21) q ≤ C + C | b | q C rq "(cid:18)Z Ω |∇ ϕ | dV (cid:19) rq + (cid:18)Z Ω ϕ dV (cid:19) rq = C + C "(cid:18)Z Ω |∇ ϕ | dV (cid:19) rq + (cid:18)Z Ω ϕ dV (cid:19) rq , (11)where we introduced C = C | b | q C rq > ǫ >
0, there is C ( ǫ ) such that t rq ≤ ( ε /C ) t + C for t >
0. Applying this with t = R Ω ϕ dV , we get R Ω ϕ dV ≤ C + C (cid:0)R Ω |∇ ϕ | dV (cid:1) rq + ǫ C C R Ω ϕ dV + C . Then with t = R Ω |∇ ϕ | dV , we have R Ω ϕ dV ≤ C + ǫ R Ω |∇ ϕ | dV + C + ǫ R Ω ϕ dV + C ⇒ (1 − ǫ ) R Ω ϕ dV ≤ ǫ R Ω |∇ ϕ | dV + C + 2 C ⇒ R Ω ϕ dV ≤ ǫ − ǫ R Ω |∇ ϕ | dV + C where C := C +2 C − ǫ . This gives (7).Now, again by the Sobolev trace inequality, there is C > Z Σ ( γϕ ) dσ = Z ∂M ( γϕ ) dσ ≤ C (cid:18)Z M ϕ dV + Z M (cid:12)(cid:12) ∇ (cid:0) ϕ (cid:1)(cid:12)(cid:12) dV (cid:19) = C (cid:18)Z Ω ϕ dV + Z Ω (cid:12)(cid:12) ∇ (cid:0) ϕ (cid:1)(cid:12)(cid:12) dV (cid:19) . However, if ǫ > |∇ ( u ) | = | u ∇ u | ≤ | u ||∇ u | ≤ ǫ |∇ u | + u ǫ , (12)because (cid:16) √ ǫ |∇ u | − u √ ǫ (cid:17) ≥ C such that Z Ω ϕ dV ≤ ǫ Z Ω |∇ ϕ | dV + C , (13)which combined with (12) and (13) implies that R Σ ( γϕ ) dσ ≤ C ǫ R Ω |∇ ϕ | dV + C C + C ǫ R Ω |∇ ϕ | dV + C ǫ R Ω ϕ dV ≤ C ( ǫ + 2 ǫ ) R Ω |∇ ϕ | dV + C C + C C ǫ , again by (13). The estimate (8) follows from taking ǫ = C ( ǫ + 2 ǫ ) and C ǫ = max { C C + C C ǫ , C ( ǫ ) } . The relative Yamabe invariant 5
Later we are going to use the fact that if C decreases, C may decrease as well and thus C ǫ may decrease. That is, if we consider C ǫ = C ǫ ( b ), we can assume that | b | > | b | ⇒ C ǫ ( b ) ≥ C ǫ ( b ).The following result will be important. Lemma 2.3.
Let ≤ q ≤ q , and ≤ r ≤ ¯ q + 1 with q ≥ r , and let b ∈ R . Then we have thefollowing.(a) Given ǫ > , there exists K ǫ > such that (cid:12)(cid:12)(cid:12)(cid:12)Z Ω Rϕ dV g (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ || ϕ || W , (Ω) + K ǫ || ϕ || L (Ω) , (14) and that (cid:12)(cid:12)(cid:12)(cid:12)Z Σ H ( γϕ ) dσ g (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ || γϕ || W , (Σ) + K ǫ || γϕ || L (Σ) . (15) (b) For q > r , if B q,rb (Ω , Σ) = ∅ , Y q,rb (Ω , Σ) is finite.(c) Again, if q > r , there are constants C and K such that k∇ ϕ k L (Ω) ≤ KE ( ϕ ) + C, (16) for all ϕ ∈ B q,rb (Ω , Σ) .(d) The quantity Y g (Ω , Σ) := Y q,r (Ω , Σ) is a conformal invariant, that is, Y g (Ω , Σ) = Y ˜ g (Ω , Σ) for any two metrics ˜ g ∼ g of class W s,p . We refer to Y g (Ω , Σ) as the relative Yamabe invariantof (Ω , Σ) .(e) The quantity Y q, ¯ q +1 b (Ω , Σ) is also a conformal invariant.Proof. To prove (a) we first refer to the discussion on page 9 of [5] to justify that we can bound thecurvature terms of E ( ϕ ):If ϕ ∈ W (1 − δ ) , ( M ) for a sufficiently small δ >
0, we have ϕ ∈ W − s,p ′ ( M ) = (cid:2) W s − ,p ( M ) (cid:3) ′ for sp > n and s ≥
1, with p ′ the H¨older conjugate of p . In particular, if ϕ ∈ W , (Ω , Σ), there is K such that (cid:12)(cid:12)(cid:12)(cid:12)Z Ω Rϕ dV g (cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)Z M Rϕ dV g (cid:12)(cid:12)(cid:12)(cid:12) ≤ K || R || W s − ,p ( M ) || ϕ || W − s,p ′ ( M ) . (17)However the pointwise multiplication is bounded as a map W (1 − δ ) , ( M ) ⊗ W (1 − δ ) , ( M ) → W − s,p ′ ( M ),so there is K a constant providing || ϕ || W − s,p ′ ( M ) ≤ K || ϕ || W (1 − δ ) , ( M ) , (18)for a suitably small δ .It follows, by an interpolation argument, that one gets for all ǫ > K ǫ satisfying (cid:12)(cid:12)(cid:12)(cid:12)Z Ω Rϕ dV g (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ || ϕ || W , (Ω) + K ǫ || ϕ || L (Ω) , (19)proving the first estimate in (a).If q > r , we can combine this with (7) and get that for all ǫ >
0, there is L ǫ > (cid:12)(cid:12)(cid:12)(cid:12)Z Ω Rϕ dV g (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ ||∇ ϕ || L (Ω) + L ǫ . (20) The relative Yamabe invariant 6
Since ϕ ∈ W , (Ω) implies ( γϕ ) ∈ W , ( ∂M ), Corollary A.5 of [5] says ( γϕ ) ∈ W p +1 − s,p ′ ( ∂M ) if ( σ = p + 1 − s ≤ σ − n − p ′ = p + 1 − s − np ′ + p ′ = 2 − s − np ′ ≤ − n − which can be rewritten as ( ps ≥ p + 1 psn ≥ pn − p . Now, since sp > n >
2, the second condition is satisfied since psn > > pn − p . Adding that s ≥ p < ps ≤ ps psn < ps. So, as in the derivation of equation (19) above, there is ˜ K ǫ > (cid:12)(cid:12)(cid:12)(cid:12)Z Σ H ( γϕ ) dσ g (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ || γϕ || W , (Σ) + ˜ K ǫ || ϕ || L (Σ) , (21)which is the second estimate in (a).Using the trace inequality followed by (8), if q > r we get that given ǫ >
0, there is L ǫ > (cid:12)(cid:12)(cid:12)(cid:12)Z Σ H ( γϕ ) dσ g (cid:12)(cid:12)(cid:12)(cid:12) ≤ ǫ ||∇ ϕ || W , (Ω) + L ǫ . (22)We can finally prove (c), which is a lower regularity version of proposition 2.4 in [1]: ||∇ ϕ || L (Ω) = E ( ϕ ) − n − n − R Ω Rϕ dV g − n − R Σ H ( γϕ ) dσ g ≤ E ( ϕ ) + 2 ǫ ||∇ ϕ || L (Ω) + 2 L ǫ , for a suitable choice of ǫ and L ǫ . Thus, making sure ǫ < , we get ||∇ ϕ || L (Ω) ≤ − ǫ E ( ϕ ) + 21 − ǫ L ǫ and the result follows by choosing K = − ǫ and C = − ǫ L ǫ .With this in hand, it is simple to prove (b), since for all ϕ : E ( ϕ ) ≥ K ||∇ ϕ || L (Ω) − CK ≥ − CK .
For (d) and (e), let ˜ g = φ n − g and let us see how each element of the expression of E ( ϕ )transforms between the two metrics: dV ˜ g = φ nn − dV g = φ q dV g dσ ˜ g = φ n − n − dσ g = φ ¯ q +1 dσ g R ˜ g = φ − n +2 n − (cid:16) − n − n − ∆ g φ + R g φ (cid:17) H ˜ g = φ − nn − (cid:16) H g φ + n − ∂φ∂η (cid:17) |∇ ϕ | g = φ − n − |∇ ϕ | g The relative Yamabe invariant 7
The next step is to prove that E ˜ g ( ϕ ) = E g ( φϕ ): E ˜ g ( ϕ ) = R Ω (cid:16) |∇ ϕ | g + n − n − R ˜ g ϕ (cid:17) dV ˜ g + n − R Σ H ˜ g ( γφ ) dσ ˜ g = R Ω (cid:16) φ − n − |∇ ϕ | g + n − n − φ − n +2 n − (cid:16) − n − n − ∆ g φ + R g φ (cid:17) ϕ (cid:17) φ nn − dV g + n − R Σ φ − nn − (cid:16) H g φ + n − ∂φ∂η (cid:17) ( γϕ ) φ n − n − dσ ˜ g = R Ω (cid:16) φ |∇ ϕ | g − φϕ ∆ g φ + n − n − R g ( φϕ ) (cid:17) dV g + R Σ n − H g ( φ ( γϕ )) + φ ( γϕ ) ∂φ∂η dσ ˜ g = E g ( φϕ ) + R Ω (cid:0) −|∇ ( φϕ ) | g + φ |∇ ϕ | g − φϕ ∆ g φ (cid:1) dV g + R Σ φ ( γϕ ) ∂φ∂η dσ ˜ g = E g ( φϕ ) + R Ω (cid:0) −|∇ ( φϕ ) | g + φ |∇ ϕ | g + ∇ ( φϕ ) · ∇ φ (cid:1) dV g where we used Green’s first identity in the last equality. Then if we use the identities |∇ ( φϕ ) | g = φ |∇ ϕ | g + ϕ |∇ φ | g + 2 φϕ ∇ φ · ∇ ϕ, (23)and (cid:0) ∇ ( φϕ ) (cid:1) · ∇ φ = ϕ |∇ φ | g + 2 φϕ ∇ ϕ · ∇ φ, (24)we get that E ˜ g ( ϕ ) = E g ( φϕ ) . (25)Finally, the transformation of the respective measures under the conformal change of the metricimplies that Z Ω | ϕ | q dV ˜ g = Z Ω | φϕ | q dV g and Z Σ | γϕ | ¯ q +1 dσ ˜ g = Z Σ | φ ( γϕ ) | ¯ q +1 dσ g so ϕ ∈ B q,r , ˜ g (Ω , Σ) if, and only if, φϕ ∈ B q,r ,g (Ω , Σ), proving (c), and ϕ ∈ B q, ¯ q +1 b, ˜ g (Ω , Σ) if, and onlyif, φϕ ∈ B q, ¯ q +1 b,g (Ω , Σ), proving (d).
Theorem 2.4.
Let M be a smooth connected Riemannian manifold with dimension n ≥ and witha metric g ∈ W s,p , where we assume sp > n and s ≥ . Let q ∈ [2 , q ) , and r ∈ [2 , ¯ q + 1) with q > r . Let also b ∈ R and let Ω and Σ be relatively open sets. Then, there exists a strictly positivefunction φ ∈ B q,rb (Ω , Σ) ∩ W s,p ( M ) , such that ( − ∆ φ + n − n − Rφ = λqφ q − , in Ω γ∂ ν φ + n − Hγφ = λrb ( γφ ) r − , in Σ (26) where the sign of λ is the same as that of Y q,rb (Ω , Σ) defined above.Proof. The above equation is the Euler-Lagrange equation for the functional E over positive func-tions with the Lagrange multiplier λ , so it suffices to show that E attains its infimum Y q,rb (Ω , Σ)over B q,rb (Ω , Σ) at a positive function φ ∈ W s,p ( M ) (notice that the existence of a minimizer doesnot depend on the sets Ω and Σ being open).Let { φ i } i ⊂ B q,rb (Ω , Σ) be a sequence satisfying E ( φ i ) → Y q,rb (Ω , Σ). Since E ( ϕ ) = E ( | ϕ | ), forall ϕ , we can assume φ i ≥ , ∀ i . If ϕ ∈ B q,rb (Ω , Σ) satisfies the bound E ( ϕ ) ≤ Λ, then inequality(7) and item (c) of Lemma 2.3 tell us there is C (Λ) > k ϕ k W , (Ω) ≤ C (Λ). Now, since Y q,rb (Ω , Σ) is finite, we conclude that { φ i } i is bounded in W , (Ω). By the reflexivity of W , (Ω),the compactness of W , (Ω) ֒ → L q (Ω), and the compactness of the trace map γ : W , (Ω) ֒ → L r (Σ),there exists an element φ ∈ W , (Ω) and a subsequence { φ ′ i } ⊂ { φ i } such that, for a fixed small δ > The relative Yamabe invariant 8 (i) φ ′ i ⇀ φ in W , (Ω),(ii) φ ′ i → φ in W − δ, (Ω),(iii) φ ′ i → φ in L q (Ω),(iv) φ ′ i → φ in L (Ω),(v) γφ ′ i → γφ in W − δ, ,(vi) γφ ′ i → γφ in L r (Σ).Items (iii) and (vi) imply φ ∈ B q,rb (Ω , Σ), hence φ ≥ E ( φ ) ≤ lim E ( φ ′ i ) = Y q,rb (Ω , Σ), because then E ( φ ) = Y q,rb (Ω , Σ) and thus φ satisfies (26).Indeed, the proof of Lemma 2.3 (a) shows that ϕ
7→ h
R, ϕ i Ω is continuous in W − δ, (Ω) and ϕ
7→ h
H, ϕ i Σ is continuous in W − δ, (Σ). So the maps ϕ
7→ h
R, ϕ i Ω and ϕ
7→ h
H, ϕ i Σ have,respectively, the same regularity. Therefore, by (ii) and (v), we have E ( φ ) − Y q,rb (Ω , Σ) = E ( φ ) − lim E ( φ ′ i ) = k∇ φ k L (Ω) − lim k∇ φ ′ i k L (Ω) (27)and so by (i) and (iv), E ( φ ) ≤ Y q,rb (Ω , Σ) as expected.Corollary B.4 of [5] implies that φ ∈ W s,p ( M ), as is detailed in the proof of Lemma 4.5 below.Also, since φ ∈ B q,rb (Ω , Σ), we have φ
0. But, if we call N = ( γφ ) − ( { } ), Lemma B.7 of [5] tellsus that φ > φ and integrating by parts, we conclude that the sign of the Lagrangemultiplier λ is the same as that of Y q,rb (Ω , Σ).Under the conformal scaling ˜ g = ϕ q − g , the scalar curvature and the mean extrinsic curvaturetransform as ( ˜ R = ϕ − q ( − n − n − ∆ ϕ + Rϕ ) , ˜ H = ( γϕ ) − ¯ q ( n − γ∂ ν ϕ + Hγϕ ) , (28)so assuming the conditions of the above theorem we infer that any given metric g ∈ W s,p can betransformed to the metric ˜ g = φ q − g with the continuous scalar curvature ˜ R = λq ( n − n − φ q − q inΩ, and the continuous mean curvature ˜ H = λbrn − ( γφ ) r − ¯ q − on Σ, where the conformal factor φ is asin the theorem. In other words, given any metric g ∈ W s,p , there exist φ ∈ W s,p ( M ) with φ > R ∈ W s,p ( M ) and ˜ H ∈ W s − p ,p (Σ), having constant sign , such that ( − n − n − ∆ φ + Rφ = ˜ Rφ q − , n − γ∂ ν φ + Hγφ = ˜ H ( γφ ) ¯ q . (29)We will prove below that the conformal invariant Y g of the metric g completely determines the signof ˜ R , independent of the indices q , r or b , giving rise to a Yamabe classification of metrics in W s,p .Note that the sign of the boundary mean curvature can be controlled by the sign of the parameter b ∈ R , unless of course ˜ R ≡
0, in which case we are forced to have ˜ H ≡ Lemma 2.5. If q ∈ [2 , q ] , r ∈ [2 , ¯ q + 1] , q > r , then the map b
7→ Y q,rb (Ω , Σ) is non-increasing. The relative Yamabe invariant 9
Proof.
Let b ′ < b and ϕ ∈ B q,rb (Ω , Σ). Then,as a consequence of Lemma 2.1, there exists k ϕ > k ϕ ϕ ∈ B q,rb ′ (Ω , Σ). That means1 = k qϕ || ϕ || qL q (Ω) + k rϕ b ′ || γϕ || rL r (Σ) = || ϕ || qL q (Ω) + b || γϕ || rL r (Σ) = || ϕ || qL q (Ω) + b ′ || γϕ || r Σ ,r + ( b − b ′ ) || γϕ || rL r (Σ) , and so, since b − b ′ >
0, we get k qϕ || ϕ || qL q (Ω) + k rϕ b ′ || γϕ || rL r (Σ) ≥ || ϕ || qL q (Ω) + b ′ || γϕ || rL r (Σ) , implying k ϕ ≥ E ( k ϕ ϕ ) = k ϕ E ( ϕ ) ≥ E ( ϕ ). Hence Y q,rb ′ (Ω , Σ) ≥ Y q,rb (Ω , Σ), and thus the map isnon-increasing.
Theorem 2.6.
Let ( M, g ) be a smooth, compact, connected, n -dimensional Riemannian manifoldwith boundary, where we assume that the components of the metric g are (locally) in W s,p , with sp > n , s ≥ , and n ≥ . Let Ω ⊂ M and Σ ⊂ ∂M be (relatively) closed sets. Then, the sign of Y q,rb (Ω , Σ) is independent of q ∈ [2 , q ] , r ∈ [2 , ¯ q + 1] , and b ∈ R , as long as q > r . In particular,this sign is a conformal invariant.Proof. Let q, q ′ ∈ [2 , q ], r, r ′ ∈ [2 , ¯ q + 1], and b, b ′ ∈ R , with q > r and q ′ > r ′ .Suppose Y q,rb (Ω , Σ) <
0. Then E ( ϕ ) < ϕ ∈ B q,rb (Ω , Σ). By scaling, there is some k > kϕ ∈ B q ′ ,r ′ b ′ (Ω , Σ). Now since E ( kϕ ) = k E ( ϕ ), we infer that Y q ′ ,r ′ b ′ (Ω , Σ) < Y q,rb (Ω , Σ) ≥
0. Then E ( ϕ ) ≥ ϕ ∈ B q,rb (Ω , Σ). On the other hand, for any ψ ∈ B q ′ ,r ′ b ′ (Ω , Σ), there is some k > kψ ∈ B q,rb (Ω , Σ), and hence E ( ψ ) = k − E ( kψ ) ≥ Y q ′ ,r ′ b ′ (Ω , Σ) ≥ Y q,rb (Ω , Σ) > Y q ′ ,r ′ b ′ (Ω , Σ) >
0. Since the parametersare arbitrary, in light of the preceding paragraph, this would also establish the equivalence between Y q,rb (Ω , Σ) = 0 and Y q ′ ,r ′ b ′ (Ω , Σ) = 0.Let us start by fixing q and r . Assume that Y q,rb (Ω , Σ) = α > b <
0. Let ψ ∈ B q,rb ′ (Ω , Σ). Then there exists some k > kψ ∈ B q,rb (Ω , Σ), and E ( kψ ) = k E ( ψ ) ≥ α. (30)The value of k comes from the equation k q k ψ k qL q (Ω) + bk r k γψ k rL r (Σ) = 1 . (31)Let us focus first at b ′ = 0. Then || ϕ || qq = 1 for all ϕ and k q − r = k − r − b k γψ k rL r (Σ) (32)If Y q,r (Ω , Σ) = 0, there is a minimizing sequence { ψ m } m ⊂ B q,r (Ω , Σ) such that E ( ψ m ) →
0. But, if k m ψ m ∈ B q,rb (Ω , Σ), constraint (30) implies k m → ∞ and thus, by equation (32), k γψ m k rL r (Σ) → ∞ for b < E ( ψ m ) → M > E ( ψ m ) < M forall m and Lemma 2.3 (c) implies there are K and C such that k∇ ψ m k L (Ω) ≤ KM + C . But by(7), given ǫ >
0, there exists C ǫ > k ψ m k L (Ω) ≤ ǫ k∇ ψ m k L (Ω) + C ǫ (33) The relative Yamabe invariant 10 and hence the sequence n k ψ m k W , (Ω) o m is bounded and the continuity of the trace map implies that (cid:26) k γψ m k W , (Σ) (cid:27) m is bounded as well. But since r < ¯ q + 1 the Sobolev embedding theorem implies W , (Σ) ֒ → L r (Σ) and there is a uniform bound on k γψ m k rL r (Σ) . Contradiction. So Y q,r (Ω , Σ) = ǫ for some ǫ > b ′ >
0. Assume Y q,rb ′ (Ω , Σ) = 0 and let { ψ m } m ⊂ B q,rb ′ (Ω , Σ)be a minimizing sequence. If you take b > k m > k m ψ m ∈ B q,rb (Ω , Σ), then E ( k m ψ m ) = k m E ( ψ m ). So either k m → ∞ or E ( k m ψ m ) → Y q,rb (Ω , Σ) = 0. But k m sat-isfies k qm k ψ m k qL q (Ω) + k rm b k γψ m k rL r (Σ) = 1 (34)as a consequence, since b > k m → ∞ implies k ψ m k qL q (Ω) → k γψ m k rL r (Σ) →
0. But thatwould contradict the fact that ψ m ∈ B q,rb ′ (Ω , Σ), for all m . Hence k m is bounded and Y q,rb (Ω , Σ) = 0,for all b > Y q,rb ′ (Ω , Σ) = 0 for some b ′ >
0, let { b m } m be a decreasing sequenceof strictly positive numbers such that b m →
0. Then Y q,rb m (Ω , Σ) = 0 , ∀ m , and for each m , there is ψ m ∈ B q,rb m (Ω , Σ) such that E ( ψ m ) < m . In particular, E ( ψ m ) ≤ , ∀ m . Now, since the sequence ofthe b m ’s is bounded, the remark after the proof of lemma 2.2 allows us to use theorem 2.3, item (c)to get constants K, C > k∇ ψ m k L (Ω) ≤ K + M (35)which, just as above, implies n k γψ m k rL r (Σ) o m is uniformly bounded.Finally, for each m let l m > l m ψ m ∈ B q,r (Ω , Σ). Then E ( l m ψ m ) = l m E ( ψ m ) and l qm k ψ m k qL q (Ω) = 1 (36)so l m → ∞ iff k ψ m k L q (Ω) →
0. However k ψ m k qL q (Ω) + b m k γψ m k rL r (Σ) = 1 . (37)It follows, since b m →
0, that k ψ m k qL q (Ω) → k γψ m k rL r (Σ) → ∞ , which cannot be. As aconsequence, { l m } m is uniformly bounded and E ( l m ψ m ) →
0, which implies that Y (Ω , Σ) = 0.Summing up, we have that if there is b < Y q,rb (Ω , Σ) > Y q,r (Ω , Σ) >
0, while ifthere is b ′ > Y q,rb ′ (Ω , Σ) = 0, Y q,r (Ω , Σ) = 0, and we can’t have both at the same time. So thesign of Y q,rb (Ω , Σ) does not depend on b .Now, fixing b = 0, by definition Y q,r (Ω , Σ) does not depend on r , and what we said about b holds for r as well.To prove that the sign of the relative Yamabe invariant does not depend on q either we willalso argue that the sign of Y q,r (Ω , Σ) is independent of q . If q > q ′ , there exists C q > k ψ k L q ′ (Ω) ≤ C q k ψ k L q (Ω) . On the other hand, if ψ ∈ B q,r (Ω , Σ), let k > kψ ∈ B q ′ ,r (Ω , Σ), so 1 = k q ′ k ψ k q ′ L q ′ (Ω) ≤ k q ′ C qq k ψ k qL q (Ω) = k q ′ C qq and that implies k q ′ = 1 k ψ k q ′ L q ′ (Ω) ≥ C qq . The relative Yamabe invariant 11
Hence, E ( kψ ) = k E ( ψ ) ≥ E ( ψ ) C qq and Y q,r (Ω , Σ) > Y q ′ ,r (Ω , Σ) > Y q ′ ,r (Ω , Σ) > Y q,r (Ω , Σ) = 0. Then we have Y q ′ ,r (Ω , Σ) = ǫ > Y q,r (Ω , Σ) = 0.Let { ψ m } m ⊂ B q,r (Ω , Σ) be a minimizing sequence and k m > k m ψ m ∈ B q ′ ,r (Ω , Σ).So E ( k m ψ m ) = k m E ( ψ m ) ≥ ǫ implies k m → ∞ . But k m satisfies k q ′ m || ψ m || q ′ L q ′ (Ω) + k rm k γψ m k rL r (Σ) = 1,thus k ψ m k L q ′ (Ω) → k γψ m k L r (Σ) →
0. Also, q ′ ≥ , r ≥
2, so ( k ψ m k L (Ω) → k γψ m k L (Σ) → . (38)On the other hand, k γψ m k L r (Σ) → k ψ m k L q (Ω) → q ≤ q implies thereis K q > k ψ m k W , (Ω) ≥ k ψ m k Lq (Ω) K q :lim k ψ m k W , (Ω) ≥ K q ⇒ lim k∇ ψ m k L (Ω) ≥ K q (39)Finally, since E ( ψ m ) →
0, item (c) of theorem 2.3 implies k∇ ψ m k L (Ω) is bounded and then,along with (38), item (a) of the same theorem implieslim (cid:12)(cid:12)(cid:12)(cid:12)Z Ω Rψ m dV g (cid:12)(cid:12)(cid:12)(cid:12) = lim (cid:12)(cid:12)(cid:12)(cid:12)Z Σ H ( γψ m ) dσ g (cid:12)(cid:12)(cid:12)(cid:12) = 0 (40)thus, by definition of E : lim k∇ ψ m k L (Ω) = lim E ( ψ m ) = 0 (41)which contradicts (39). So Y q,r (Ω , Σ) > Y q,rb (Ω , Σ) does not depend on q .We finish this section by proving some general properties of Y q,rb (Ω , Σ) that will not be necessaryfor the rest of the paper.
Lemma 2.7 (Monotonicity) . If Ω ⊂ Ω and Σ ⊂ Σ , then we have Y q,rb (Ω , Σ ) ≥ Y q,rb (Ω , Σ ) . Proof.
It follows straight from the definition since we expand the set of test functions ϕ . Lemma 2.8 (Continuity from above) . Let Ω ⊂ M , Σ ⊂ ∂M , q , r and b as in theorem 2.4. Let Ω k be a decreasing sequence of relatively open sets and let Σ k be a decreasing sequence of relativelyclosed sets such that (T k Ω k = Ω T k Σ k = Σ . Then we have lim k Y q,rb (Ω k , Σ k ) = Y q,rb (Ω , Σ) . (42) Proof.
From monotonicity, (Ω k ∪ Σ k ) ⊃ (Ω k +1 ∪ Σ k +1 ) implies Y q,rb (Ω k , Σ k ) ≤ Y q,rb (Ω k +1 , Σ k +1 ),so the sequence of the Y k := Y q,rb (Ω k , Σ k ) is non-decreasing. In addition, again by monotonicity, Y := Y q,rb (Ω , Σ) ≥ Y k , for all k . Hence, we can define Λ = lim k Y k ≥ Y k , for all k . We also have Y ≥ Λ.If Λ = + ∞ , Y = + ∞ as well and the result holds automatically. Eigenvalue of the Laplacian 12
On the other hand, assume Λ, and thus the Y k , are all finite. So, by Theorem 2.4, for each k there is u k ∈ B q,rb (Ω k , Σ k ) satisfying E ( u k ) = Y k ≤ Λ. Then, by lemmata 2.3 and 2.2, there is
K > k u k k W , ( M ) ≤ K, ∀ k . Now, all u k are in W , (Ω , Σ ), implying that the sequenceof the u k is bounded in W , (Ω ) and hence, in an argument similar to the one in theorem 2.4,there is u ∈ W , (Ω , Σ ) such that:(i) u k ⇀ u in W , (Ω , Σ ),(ii) u k → u in L q (Ω ),(iii) γu k → γu in L r (Σ ),(iv) u k → u a.e. in Ω ,(v) γu k → γu a.e. in Σ .By (iv) and (v), u ∈ W , (Ω , Σ). In fact, by (ii) and (iii), u ∈ B q,rb (Ω , Σ). So E ( u ) ≥ Y . But,just as in the proof of theorem 2.4, we have E ( u ) ≤ lim inf E ( u k ) = Λ. Thus we conclude thatΛ = Y .As a remark, the result is not true if we drop the hypothesis on the indices being subcritical, asexplained after the proof of Lemma 3.14 in [6]. To study the more general problem of prescribing scalar curvature and mean extrinsic curvaturewe try to adapt the strategy of the paper [6] to the case of manifolds with boundary. In order todo so we have to study a specific instance of Y q,rb (Ω , Σ) not covered by the results in the previoussection, with q = r = 2, which corresponds to the first eigenvalue of the Laplacian relative to thepair (Ω , Σ).First we define the Raileigh quotient for functions in W , ( M, ∂M ) as Q g ( ϕ ) = E ( ϕ ) || ϕ || L (Ω) + || γϕ || L (Σ) (43)and the first eigenvalue of the Laplacian for the set Ω ∪ Σ as λ q,rb (Ω , Σ) := inf B q,rb (Ω , Σ) Q g ( ϕ ) (44)Notice that, since Q g ( kϕ ) = Q g ( ϕ ), for all k >
0, Lemma 2.1 implies that λ q,rb (Ω , Σ) does notreally depend on q , r or b , so we can drop the indices. Proposition 3.1. If ( M, g ) , Ω , Σ are as before, there is u ∈ W , (Ω , Σ) such that λ (Ω , Σ) = Q g ( u ) . Proof.
We want to be able to simply take q = r = 2, b = 1 in theorem 2.4. Checking the proof, q > r is a relevant hypothesis just to show that a minimizing sequence { φ i } i ⊂ B , (Ω , Σ) for
Eigenvalue of the Laplacian 13 E ( φ i ) → λ (Ω , Σ) will be bounded in W , (Ω , Σ). Now, if φ i ∈ B , (Ω , Σ), k φ i k L (Ω) ≤
1. Also,Proposition 2.3 (a) does not require q > r , so for any ǫ >
0, there is K ǫ > R Ω |∇ φ i | dV g = E ( φ i ) − n − n − R Ω Rφ i dV g − n − R Σ H ( γφ i ) dσ g ≤ E ( φ i ) + ǫ k φ i k W , (Ω) + K ǫ k φ i k L (Ω) + ǫ k γφ i k W , (Σ) + K ǫ k γφ i k L (Σ) ≤ E ( φ i ) + ǫ (1 + C ) R Ω |∇ φ i | dV g + ǫ + ǫC + 2 K ǫ , with C the constant associated to the trace inequality. So, if we choose ǫ small enough, for any i k φ i k W , (Ω) ≤ − ǫ (1 + C ) ( E ( φ i ) + ǫ (1 + C ) + 2 K ǫ ) + 1 (45)and the proof follows as in theorem 2.4.Instead of dealing with the relative Yamabe invariant of a set we want to work with its linearcounterpart, the first eigenvalue of the Laplacian. For our purposes, we only need λ (Ω , Σ) to coincidewith Y q,rb (Ω , Σ) in sign. In views of Theorem 2.6, it is enough to prove that the signs agree for b = 1. Theorem 3.2. If q > r , we have sign( λ (Ω , Σ)) = sign( Y q,r (Ω , Σ)) . Proof.
Let us omit the sets to simplify notation.i. Y q,r < ϕ ∈ B q,r (Ω , Σ) such that E ( ϕ ) <
0, which is itself true if, andonly if, λ < Y q,r = ǫ >
0. Since r, q ≥
2, there is C q,r > || ϕ || L (Ω) ≤ C q,r || ϕ || qL q (Ω) and || γϕ || L (Σ) ≤ C q,r || γϕ || rL r (Σ) , ∀ ϕ ∈ B q,r (Ω , Σ). So, if ϕ ∈ B q,r (Ω , Σ): || ϕ || L (Ω) + || γϕ || L (Σ) ≤ C q,r . ⇒ E ( ϕ ) || ϕ || L (Ω) + || γϕ || L (Σ) ≥ E ( ϕ ) C q,r , and λ ≥ Y q,r C q,r = ǫC q,r > λ > Y q,r ≤
0, hence Y q,r ≤
0. Let ϕ k be a minimizing sequence for Y q,r .Then we can assume, up to a subsequence, E ( ϕ k ) < k , ∀ k .So we have that, since λ > E ( ϕ k ) = Q g ( ϕ k ) (cid:16) || ϕ k || L (Ω) + || γϕ k || L (Σ) (cid:17) < k ⇒ lim k →∞ (cid:16) || ϕ k || L (Ω) + || γϕ k || L (Σ) (cid:17) = 0 . (46)Also E ( ϕ k ) ≤ k ⇒ ||∇ ϕ k || L (Ω) ≤ k − n − n − Z Ω Rϕ k dV g − n − Z Σ H ( γϕ k ) dσ g (47)and || ϕ k || L q (Ω) = 1 , q ≤ q . As a consequence, there is K q > || ϕ k || W , (Ω) ≥ K q and,by (46) lim k →∞ ||∇ ϕ k || L (Ω) ≥ K q . (48)Turning back to equations (14) and (15) we see that, given ǫ >
0, there exists K ǫ > ( − n − n − R Ω Rϕ dV g ≤ n − n − ǫ ||∇ ϕ k || L (Ω) + n − n − K ǫ || ϕ k || L (Ω) − n − R Σ H ( γϕ k ) dσ g ≤ n − ǫ ||∇ ϕ k || L (Ω) + n − K ǫ || γϕ k || L (Σ) . (49) The Prescribed Scalar Curvature and Mean Curvature Problem 14
But also, by (c) in lemma 2.3 ||∇ ϕ k || L (Ω) ≤ KE ( ϕ k ) + C. Plugging that into (49) we get ( − n − n − R Ω Rϕ dV g ≤ n − n − ǫ ( KE ( ϕ k ) + C ) + n − n − K ǫ || ϕ k || L (Ω) − n − R Σ H ( γϕ k ) dσ g ≤ n − ǫ ( KE ( ϕ k ) + C ) + n − K ǫ || γϕ k || L (Σ) and if we take the limit in (47) and use the result (46) we end up finding that there is a constant L = ( KE ( ϕ k ) + C ) (cid:16) n − n − + n − (cid:17) such thatlim k →∞ ||∇ ϕ k || L (Ω) ≤ ǫL, for all ǫ >
0. Contradiction.iv. Gathering the results i, ii and iii we find that Y q,r = 0 if, and only if, λ = 0. Let’s define the following functional over W , ( M, ∂M ) F q,r ( u ) = E ( u ) − Z M n − q ( n − R ′ | u | q dV g − Z ∂M n − r H ′ | γu | r dσ g (50)and prove some of its important properties. But first we need a technical lemma. Lemma 4.1.
Let
A, B, C > be real constants, x, y, z ≥ be positive numbers and assume y + z > Cx. (51) If q, r, p are exponents satisfying q ≥ q > , r ≥ r > and r , q > p > , there is J ( q , r , p ) > such that if x ≥ L f ( y, z ) = Ay q + Bz r ≥ x p . Proof. If Ay q ≥ x p , there is nothing to do. So assume y < A − q x pq . Then z − B − r x pr > Cx − y − B − r x pr > Cx − A − q x pq − B − r x pr (52)Since pq , pr <
1, there is
L > x ≥ L then Cx − A − q x pq − B − r x pr > z > B − r x pr and Bz r > x p .For simplicity, for the rest of this section, we will use the notation k u k L ( M,∂M ) = k u k L ( M ) + k γu k L ( ∂M ) , which defines a norm in W , ( M ).Also, we will denote Z := { s ∈ M | R ′ ( s ) = 0 } and Z ∂ = { p ∈ ∂M | H ′ ( p ) = 0 } the zero sets of R ′ and H ′ respectively.Finally, we can prove our first important result. The Prescribed Scalar Curvature and Mean Curvature Problem 15
Lemma 4.2 (Coercivity) . Let q , r be subcritical (as in theorem 2.4), R ′ ≤ , H ′ ≤ .If ( Z, Z ∂ ) is Yamabe positive (i.e. Y p,qb ( Z, Z δ ) > ), ∀ B ∈ R , ∃ K ( q , r , B ) > such that if q, r are subcritical, q ≥ q , r ≥ r , q > r , u ∈ W , ( M, ∂M ) and || u || L ( M,∂M ) ≥ K , then F q,r ( u ) ≥ B .Proof. Adapting the proof of proposition 4.5 of [6].For each ǫ >
0, let’s define A ǫ = (cid:8) u ∈ W , ( M, ∂M ) (cid:12)(cid:12)R M | R ′ | u dV g + R ∂M | H ′ | ( γu ) dσ g ≤ ǫ || u || L ( M,∂M ) (cid:0)R M | R ′ | dV g + R ∂M H ′ dσ g (cid:1) o . (54)Now, since ( Z, Z ∂ ) is Yamabe positive, λ ( Z, Z ∂ ) >
0. So we can fix L ∈ (0 , λ ( Z, Z ∂ )). Claim:
There is ǫ < u ∈ A ǫ E ( u ) ≥ L || u || L ( M,∂M ) . (55)Assume that is not true. So if ǫ k is a sequence such that ǫ k →
0, we can choose v k ∈ A ǫ k , || v k || L ( M,∂M ) = 1 violating equation (55): E ( v k ) < L, ∀ k. (56)Putting this together with item 3 of lemma 2.3 we have that { v k } is a bounded sequence in W , ( M ).So, just as in the proof of theorem 2.4, there is v ∈ W , ( M ) such that v k ⇀ v in W , ( M ) and v k → v in L ( M, ∂M ), and hence || v || L ( M,∂M ) = 1.Also, as in the proof of theorem 2.4, changing R by R ′ , we have that Z M | R ′ | v k dV g → Z M | R ′ | v dV g (57)and Z ∂M | H ′ | ( γv k ) dσ g → Z ∂M | H ′ | ( γv ) dσ g (58)But, by choice of the v k ’s0 ≤ Z M | R ′ | v k dV g + Z ∂M | H ′ | ( γv k ) dσ g ≤ ǫ k || v k || L ( M,∂M ) (cid:18)Z M | R ′ | dV g + Z ∂M | H ′ | dσ g (cid:19) → Z M | R ′ | v dV g + Z ∂M | H ′ | ( γv ) dσ g = 0 (60)So v ≡ Z ∪ Z ∂ and v ∈ W , ( Z, Z ∂ ). But, since v k ⇀ v in W , , v k → v in L and fromthe last equality E ( v ) ≤ E ( v k ) ≤ L < λ ( Z, ∅ ) . Contradiction.Now we divide the proof of the theorem itself in two cases, taking ǫ as in the claim: Case 1: ( u A ǫ ) If u A ǫ Z M | R ′ | u dV g + Z ∂M | H ′ | ( γu ) dσ g > ǫ || u || L ( M,∂M ) (cid:18)Z M | R ′ | dV g + Z ∂M | H ′ | dσ g (cid:19) (61)Now, if R ′ , H ′ ≤ F q,r ( u ) = E ( u ) + Z M n − q ( n − | R ′ || u | q dV g + Z ∂M n − r | H ′ || γu | r dσ g (62) The Prescribed Scalar Curvature and Mean Curvature Problem 16
Now, using H¨older inequality on the higher power terms Z M | R ′ || u | dV g = Z M | R ′ | − q | R ′ | q | u | dV g ≤ (cid:20)Z M | R ′ | dV g (cid:21) − q (cid:20)Z M | R ′ || u | q dV g (cid:21) q (63)and similarly Z ∂M | H ′ || γu | dσ g ≤ (cid:20)Z ∂M | H ′ | dσ g (cid:21) − r (cid:20)Z ∂M | H ′ || γu | r dσ g (cid:21) r (64)On the other hand, from lemma 2.3 (a) and (b), given η >
0, there are D ( η ) , D ( η ), independentof q or r such that E ( u ) = R M |∇ u | dV g + n − n − R M Ru dV g + n − R ∂M H ( γu ) dσ g ≥ R M |∇ u | dV g − n − n − (cid:16) η k u k W , ( M ) + D k u k L ( M ) (cid:17) − n − (cid:16) η k γu k W , ( ∂M ) + D k γu k L ( ∂M ) (cid:17) ≥ (cid:16) − (cid:16) n − n − − n − D (cid:17) η (cid:17) R M |∇ u | dV g − (cid:16) n − n − ( η + D ) + n − ( ηD ) (cid:17) k u k L ( M ) − n − D k γu k L ( ∂M ) with D the constant associated to the trace inequality. By choosing η small enough that thecoefficient of R M k∇ u k dV g is positive and D the largest value between n − n − ( η + D ) + n − ηD and n − D we have E ( u ) ≥ − D k u k L ( M,∂M ) . (65)Plugging this together with equations (63) and (64) in (62) we have F q,r ( u ) ≥ − D k u k L ( M,∂M ) + A ( q ) (cid:20)Z M | R ′ || u | dV g (cid:21) q + A ( r ) (cid:20)Z ∂M | H ′ || γu | dσ g (cid:21) r (66)with A ( q ) = n − q ( n − (cid:2)R M | R ′ | dV g (cid:3) − q and A ( r ) = n − r (cid:2)R ∂M | H ′ | dσ g (cid:3) − r positive constants thatdepend on q and r respectively. Since both A ( q ) and A ( r ) are continuous on q and r , we candefine A := min { A ( q ) | q ∈ [ q , q ] } and A := min { A ( r ) | r ∈ [ r , ¯ q + 1] } . Notice that they cannotbe both zero otherwise inequality (61) is not satisfied. We can then drop the dependency on q and r of the constants in the previous inequality and have F q,r ( u ) ≥ − D k u k L ( M,∂M ) + A (cid:20)Z M | R ′ || u | dV g (cid:21) q + A (cid:20)Z ∂M | H ′ || γu | dσ g (cid:21) r (67)Now, A = 0 if, and only if, H ′ ≡
0. In this case the right-hand side of inequality (67) becomes F q,r ( u ) ≥ − D ( k u k L ( M,∂M ) ) + A (cid:20)Z ∂M | R ′ | u dV g (cid:21) q (68)and by condition (61) we have F q,r ≥ − D k u k L ( M,∂M ) + A (cid:20) ǫ Z M | R ′ | dV g (cid:21) q || u || qL ( M,∂M ) (69)As said, R M | R ′ | dV g >
0, thus the coefficient of || u || qL ( M,∂M ) is larger than zero and, since q ≥ q >
2, there is K satisfying the hypothesis of the theorem. In fact, if K satisfies the hypothesisfor q = q , the same K satisfies for q > q . A similar thing happens if A = 0 instead. The Prescribed Scalar Curvature and Mean Curvature Problem 17
Finally, if we assume A , A >
0, since q > r >
1, by lemma 4.1, there is p >
J > q, r such that k u k L ( M,∂M ) ≥ J implies, as a consequence of condition (61) A (cid:20)Z M | R ′ || u | dV g (cid:21) q + A (cid:20)Z ∂M | H ′ || γu | dσ g (cid:21) r ≥ k u k pL ( M,∂M ) (70)and replacing in (67) F q,r ( u ) ≥ − D k u k L ( M,∂M ) + k u k pL ( M,∂M ) (71)and because p > K > J satisfying the theorem.
Case 2: ( u ∈ A ǫ ) In that case, if R ′ , H ′ ≤ F q,r ( u ) ≥ E ( u ) ≥ L || u || L ( M,∂M ) . (72)And the result follows immediately.Notice that in the case of inequality (72), F q,r ≥
0, while in the case of inequalities (69) and(71) the right hand sides also have lower bounds, independent of u . Putting all together we see that F q,r has a lower bound independent of u . It is natural, then, to look for minimizers. Proposition 4.3.
Under the hypothesis of the previous lemma, fixing q, r subcritical, assume R ′ and H ′ are bounded. Assume also that n − n − Z M RdV g + n − Z ∂M Hdσ g < Then there is u q,r > a function that minimizes F q,r and is a weak solution to ( − ∆ u q,r + n − n − Ru q,r = n − n − R ′ u q − q,r , in Ω γ∂ ν u q,r + n − Hγu q,r = n − H ′ ( γu q,r ) r − , in Σ . (74) Proof.
Let { u k } k be a minimizing sequence for F q,r in W , ( M, ∂M ). Since F q,r ( | u | ) = F q,r ( u ), wecan assume u k ≥ k .Now, fixing a constant function ˜ u ≡ A ∈ W , ( M, ∂M ) we have F q,r (˜ u ) = A (cid:20) n − n − Z M RdV g + n − Z ∂M Hdσ g (cid:21) − A q q Z M R ′ dV g − A r r Z ∂M H ′ dσ g := B (75)So, since { u k } k is minimizing, we can assume F q,r ( u k ) ≤ B, ∀ k and, by the previous lemma, ∃ K B > k u k k L ( M,∂M ) < K B . Now, as seen in the proof of the previous lemma, F q,r ( u ) ≥ E ( u ) , ∀ u ,so this, together with item a of lemma 2.3, shows that k u k k W , ( M ) = E ( u k ) − n − n − R M Ru k dV g − n − R ∂M H ( γu k ) dσ g + k u k k L ( M ) ≤ B + n − n − | R M Ru k dV g | + n − | R ∂M H ( γu k ) dσ g | + K B ≤ B + n − n − (cid:16) ǫ k u k k W , ( M ) + K ǫ k u k k L ( M ) (cid:17) + n − (cid:18) ǫ k γu k k W , ( ∂M ) + K ǫ k γu k k L ( ∂M ) (cid:19) + K B ≤ B + K B (cid:16) n − n − K ǫ + n − K ǫ (cid:17) + ǫ (cid:16) n − n − + n − C (cid:17) k u k k W , ( M ) The Prescribed Scalar Curvature and Mean Curvature Problem 18 C the constant given by the trace inequality. So, choosing ǫ small enough we finally have k u k k W , ( M ) ≤ B + K B (cid:16) n − n − K ǫ + n − K ǫ (cid:17) − ǫ (cid:16) n − n − + n − C (cid:17) (76)so { u k } k is a bounded sequence in W , ( M ) and we can use the same argument to say that thereis u q,r in W , ( M ) such that:(i) u k ⇀ u q,r in W , ( M );(ii) u k → u q,r in L p ( M ), p ∈ [2 , q );(iii) γu k → γu q,r in L s ( ∂M ), s ∈ [2 , ¯ q + 1).Now, as in the proof of theorem 2.4, we have that E ( u q,r ) ≤ lim inf E ( u k ). Also, since R ′ , H ′ arebounded and u k → u q,r in L q ( M ) and γu k → γu q,r in L r ( ∂M ): (R M R ′ u qk dV g → R M R ′ u qq,r dV g R M H ′ ( γu k ) r dV g → R M H ′ ( γu q,r ) r dV g (77)So F q,r ( u q,r ) ≤ lim inf F q,r ( u k ) and, since u k is a minimizing sequence, u q,r is a minimizer of F q,r and hence a weak solution to equation (74).Now, looking back at equation (75), we see that if condition (73) is true, since q > r ≥ A small enough to guarantee that B <
0, hence F q,r ( u q,r ) < u q,r
0. Thus, bylemma B.7 of [5], u q,r > k u q,r k W , ( M ) , which, of course, gives an uniformbound on k u q,r k L q ( M ) . We want to use this fact to get regularity results for the minimizers byresorting to a bootstrap argument, but we need first to increase this regularity a little. Lemma 4.4.
In the context of the previous lemma, there is
Q > q and C > independent of q and r such that k u q,r k L Q ( M ) ≤ C. (78) Proof.
Let u q,r be a minimizer of F q,r as before and define for some δ > ( w = u δq,r ,ν = u δq,r . (79)Since u q,r is a weak solution to (74), we can test it against ν to have, defining C δ = δ (1+ δ ) : C δ k∇ w k L ( M ) = R M h∇ u q,r , ∇ ν i dV g = n − n − R M (cid:16) R ′ u q − q,r − Ru q,r (cid:17) νdV g + n − R ∂M (cid:0) H ′ ( γu q,r ) r − − Hγu q,r (cid:1) ( γν ) dσ g ≤ − (cid:16) n − n − R M RuνdV g + n − R ∂M H ( γu )( γν ) dσ g (cid:17) ≤ n − n − (cid:12)(cid:12)R M RuνdV g (cid:12)(cid:12) + n − (cid:12)(cid:12)R ∂M H ( γu )( γν ) dσ g (cid:12)(cid:12) = n − n − (cid:12)(cid:12)R M Rw dV g (cid:12)(cid:12) + n − (cid:12)(cid:12)R ∂M H ( γw ) dσ g (cid:12)(cid:12) , the first inequality being true because R ′ ≤ H ′ ≤
0. So, using the estimates from Lemma2.3a we have that for ǫ >
0, there is K ǫ > C δ k∇ w k L ( M ) ≤ ǫ k w k W , ( M ) + K ǫ k w k L ( M ) + ǫ k γw k W , ( ∂M ) + K ǫ k γw k L ( ∂M ) (80) The Prescribed Scalar Curvature and Mean Curvature Problem 19 and hence, if L is the constant associated to the trace inequality, rearranging the terms we have( C δ − ǫ − ǫL ) k∇ w k L ( M ) ≤ ( ǫ + K ǫ + ǫL ) k w k L ( M ) + K ǫ k γw k L ( ∂M ) . (81)If we choose δ small enough to guarantee the right hand side is controlled by k u q,r k L q ( M ) , and then ǫ > C ′ > k w k L q ( M ) ≤ C ′ (82)and the result follows by taking Q = 2¯ q (1 + δ ).Finally, we can prove the main regularity result. Lemma 4.5 (Regularity) . Assume
R, R ′ ∈ L ∞ ( M ) , H, H ′ ∈ W s,t ( ∂M ) with t > , s ≥ and st > n − . If u q,r is a minimizer of F q,r as in the previous lemma, then u q,r ∈ W , n ( M ) , γu q,r ∈ W , n − ( ∂M ) and there are constants C, K independent of q and r such that k u q,r k W , n ( M ) ≤ C k γu q,r k W , n − ( ∂M ) ≤ K (83) Proof.
Both the regularity and the uniform bounds come from a bootstrap procedure throughapplications of corollary B.4 of [5]. Taking D = ∅ , ∃ C p i > C p i k u q,r k W ,pi ( M ) ≤ k ∆ u q,r k W ,pi ( M ) + k γ∂ ν u q,r k W − pi ,pi ( ∂M ) + k u q,r k W ,pi ( M ) Then, doing more straightforward calculations we get C p i k u q,r k W ,pi ( M ) ≤ n − n − k R ′ u q − q,r − Ru q,r k L pi ( M ) + n − k H ′ ( γu q,r ) r − − Hγu q,r k W − pi ,pi ( ∂M ) + k u q,r k L pi ( M ) because u q,r is a solution to equation (74). Hence C p i k u q,r k W ,pi ( M ) ≤ n − n − max | R ′ |k u q − q,r k L pi ( M ) + (cid:16) n − n − max | R | + 1 (cid:17) k u q,r k L pi ( M ) + n − n − (cid:18) k H ′ ( γu q,r ) r − k W − pi ,pi ( ∂M ) + k Hγu q,r k W − pi ,pi ( ∂M ) (cid:19) and by corollary A.5 of [5] there is D > D > max {k H ′ k W s,t ( ∂M ) , k H k W s,t ( ∂M ) } , C p i k u q,r k W ,pi ( M ) ≤ n − n − max | R ′ |k u q − q,r k L pi ( M ) + (cid:16) n − n − max | R | + 1 (cid:17) k u q,r k L pi ( M ) + n − n − D D (cid:18) k ( γu q,r ) r − k W − pi ,pi ( ∂M ) + k γu q,r k W − pi ,pi ( ∂M ) (cid:19) ≤ n − n − max | R ′ |k u q − q,r k L pi ( M ) + (cid:16) n − n − max | R | + 1 (cid:17) k u q,r k L pi ( M ) + n − n − D D D (cid:16) k u r − q,r k W ,pi ( M ) + k u q,r k W ,pi ( M ) (cid:17) , with D given by the trace inequality.Now, of course k u q − q,r k L pi ( M ) = k u q,r k q − L pi ( q − ( M ) . We can find a similar estimate for k u r − q,r k W ,pi ( M ) .Using the fact that ∇ (cid:0) u r − (cid:1) = ( r − u r − ∇ u , for any u : k u r − k p i W ,pi ( M ) = R M | u | p i ( r − ( | u | p i + r |∇ u | p i ) dV g ≤ r R M | u | p i ( r − ( | u | p i + |∇ u | p i ) dV g , The Prescribed Scalar Curvature and Mean Curvature Problem 20 because r >
1. Combining this with H¨older’s inequality we find that k u r − k p i W ,pi ( M ) ≤ r k u p i ( r − k L q − q − ( M ) k| u | p i + |∇ u | p i k L q − q ( M ) . (84)So, since r ≤ ¯ q + 1, there is D > k u r − k W ,pi ( M ) ≤ D k u k r − L pi (2¯ q − ( M ) k u k W , pi (2¯ q − q ( M ) . (85)Now, coming back to the main flow of the proof, we have that k u q,r k W ,pi ( M ) ≤ C (1) p i +1 k u q,r k q − L pi (2¯ q − ( M ) + C (2) p i +1 k u q,r k L pi ( M ) + C (3) p i +1 k u q,r k W ,pi ( M ) + C (4) p i +1 k u q,r k r − L pi (2¯ q − ( M ) k u q,r k W , pi (2¯ q − q ( M ) , (86)if C (1) p i +1 = n − n − max | R ′ | [ V g ( M )] q − pi ( q − − q − pi (2¯ q − , C (2) p i +1 = (cid:16) n − n − max | R | + 1 (cid:17) , C (3) p i +1 = n − n − D D D and C (4) p i +1 = n − n − D D D D . Now, if k u q,r k L pi (2¯ q − ( M ) ≤
1, its corresponding terms are con-trolled. If not k u q,r k W ,pi ( M ) ≤ C (1) p i +1 k u q,r k q − L pi (2¯ q − ( M ) + C (2) p i +1 k u q,r k L pi ( M ) + C (3) p i +1 k u q,r k W ,pi ( M ) + C (4) p i +1 k u q,r k ¯ q − L pi (2¯ q − ( M ) k u q,r k W , pi (2¯ q − q ( M ) . (87)Finally, by Sobolev embedding theorems, there are K (1) i +1 , K (2) i +1 , K (3) i +1 > K (4) i +1 such that k u q,r k W ,pi ( M ) ≤ K (1) i +1 k u q,r k q − W ,p (1) i +1 ( M ) + K (2) i +1 k u q,r k W ,p (2) i +1 ( M ) + K (3) i +1 k u q,r k W ,p (3) i +1 ( M ) + K (4) i +1 k u q,r k ¯ q − W ,p (1) i +1 ( M ) k u q,r k W ,p (4) i +1 ( M ) (88)with p (1) i +1 = np i (2¯ q − n +2 p i (2¯ q − ≥ p (2) i +1 and p (4) i +1 = np i (2¯ q − n ¯ q + p i (2¯ q − ≥ p (3) i +1 . But we also have p (1) i +1 > p (4) i +1 ! So,since Q > q , the usual bootstrap arguments can be used to give a constant D >
0, independent of q , r subcritical such that k u q,r k W , n ( M ) ≤ D k u q,r k L Q ( M ) (89)and Lemma 4.4 says there is C >
0, independent of q, r such that k u q,r k W , n ( M ) ≤ C. (90)The bound on the traces come straight from the trace inequality.We can finally prove our main theorem Theorem 4.6.
Let ( M, ∂M ) be a smooth, Yamabe-negative manifold with boundary, g a W s,p Riemannian metric, sp > n , s ≥ with bounded scalar curvature R and mean extrinsic curvature H ∈ W r,t ( ∂M ) along the boundary, t > , r ≥ and rt > n − . If R ′ , is a bounded non-positivefunction, H ′ ∈ W r,t ( ∂M ) , then there is a function u ∈ W , n ( M ) ∩ L ∞ ( ∂M ) such that the metric g ′ = u q − g has scalar curvature R ′ inside M and mean extrinsic curvature H ′ along the boundary ∂M if, and only if, ( Z, Z ∂ ) is Yamabe positive. Examples and Consequences 21
Proof.
First, since (
M, ∂M ) is Yamabe-negative, equation (28) and theorem 2.4 give φ > M such if ˜ g = φ q − g , R ˜ g , H ˜ g < g . So we proceed with ( M, ∂M, ˜ g ) instead of ( M, ∂M, g ).Now, if (
Z, Z ∂ ) is Yamabe positive, let ( q n , r n ) be a sequence, increasing on each coordinate,converging to (2¯ q, ¯ q + 1). Then we have a sequence { u q n ,r n } n of minimizers of F q n ,r n . Now, thissequence is bounded in W , n ( M ), so there is u ∈ W , n ( M ) such that, up to subsequences, u q n ,r n → u in W , ( M ) and u q n ,r n → u uniformly in ¯ M . So u is a solution to ( − ∆ u + n − n − R ˜ g u = n − n − R ′ u q − , in Mγ∂ ν u + n − H ˜ g γu = n − H ′ ( γu ) ¯ q , in ∂M . (91)On the other hand, assume there is such solution u . Then we can calculate the Yamabe invariantof ( Z, Z ∂ ) with respect to the metric g ′ = u q − ˜ g . If B , ( Z, Z ∂ ) is empty, Y , ( Z, Z ∂ ) = + ∞ bydefinition. Otherwise, there is a minimizer ¯ u ∈ B , ( Z, Z ∂ ) such that Y , ( Z, Z ∂ ) = E (¯ u ).But if ¯ u ∈ W , ( Z, Z ∂ ), ¯ u is supported in Z ∪ Z ∂ , so Z Z R ′ ¯ u dV g = Z Z ∂ H ′ ( γ ¯ u ) dσ g = 0and E (¯ u ) = R Z |∇ ¯ u | dV g ≥
0. We just have to rule out E ( u ) = 0.If Z has positive measure, E (¯ u ) = 0 if, and only if, ¯ u is a constant. But ¯ u is supported in( Z, Z ∂ ), so either ¯ u ≡
0, which is not possible because ¯ u ∈ B , ( Z, Z ∂ ) or ( Z, Z ∂ ) = ( M, ∂M ) andthe manifold is Yamabe zero, which is not the case.Now, if Z has measure zero, and ¯ u is a function in W , ( M ) supported in Z , γ ¯ u ≡
0. So B , ( Z, Z ∂ ) = ∅ , which is not the case for us here. So E (¯ u ) > With our main result stated, it seems worthwhile to study some specific cases in order to make abetter sense of the tools we just developed.In the next two corollaries we show some cases where we can guarantee, under weaker conditions,that Y ( Z, Z ∂ ) is positive. Behind both results is the fact that test functions in W , (Ω , Σ) in generalcannot be concentrated only on the boundary. That is, if u ∈ W , (Ω , Σ) is non-zero in a point p ∈ Σ in the boundary, it must be non-zero in nearby points inside the manifold. But in that case, u ≡ ∪ Σ, so that is only possible if p lies in the closure of Ω. As a result we can seethat, in fact, only the set ¯Ω counts towards the evaluation of Y (Ω , Σ).
Corollary 5.1.
In the context of Theorem 4.6, if Z ∂ ∩ ¯ Z = ∅ , one can find the metric g ′ if, andonly if, Y ( Z, ∅ ) (which equals Y ( Z ) in the sense of [6]) is positive.Proof. In this case, u ∈ W , ( Z, Z ∂ ) implies ( γu ) | Z ∂ ≡ ∪ Σ has no interior in ¯ M , W , (Ω , Σ) is trivial and B q,rb (Ω , Σ) is empty regardless of q , r or b . As a consequence, if Ω hasmeasure zero, Y (Ω , Σ) = + ∞ . Corollary 5.2.
In the context of Theorem 4.6, if Z has measure zero, one can find the metric g ′ regardless of H ′ , as long as H ′ ≤ . Examples and Consequences 22 R ⋉ A set of cases for which we can calculate the Yamabe invariant - and compare with Escobar resultsin [2] for example - is when M is an open bounded subset of R n . In that case, R ≡ E ( u ) = Z Ω k∇ u k dV g + n − Z Σ H ( γu ) dσ g . (92)In particular, if Σ = ∅ , E ( u ) ≥ u ∈ W , (Ω , ∅ ), with equality if, and only if, u ≡
0, sincethere are no other constants in W , (Ω , ∅ ). As a consequence, all open sets in the interior of flatmanifolds are Yamabe positive. In particular, Y ( M, ∅ ) > H ′ < R ′ ≤ M is a bounded open subset of R n and Y ( M, ∂M ) <
0, there is a metric conformal to the Euclidean metric on (
M, ∂M ) with R ′ asits scalar curvature inside the manifold and H ′ its mean curvature over ∂M . With the appropriatehypothesis, this implies the negative case for Theorem 1 in [2]. We do not have restrictions on thedimension for the Yamabe-negative case though.Escobar’s classification also shows that the Yamabe-negative condition cannot be dropped inTheorem 4.6, since it is not true that one can realize R ′ ≡ H ′ < R n . For example, a straightforward application of thedivergence theorem shows there is no solution to ( − ∆ u ≡ , in D n ,γ∂ ν u + n − γu = − n − ( γu ) ¯ q , in S n − , (93)what does not contradict our theorem because the unit ball with its boundary is Yamabe-positive,since for u ∈ W , ( D n , S n − ) E ( u ) = Z D n k∇ u k dV g + Z S n − ( γu ) dσ g (94)which is non-negative and vanishes if, and only if, u ≡ The Lichnerowicz equation arises in General Relativity in the formulation of the initial value problemfor Einstein equations. Briefly, one wants to find a metric in a given conformal class that satisfies thegeometric constraint equations for a spacetime that solves Einstein’s equations with given matterfields. The Lichnerowicz equation then is a scalar condition on the conformal factor. A derivationcan be found, for example, in section 1.3 of [4].In [5] the authors discuss the Lichnerowicz equation in the context of manifolds with boundaryand find existence conditions to a wide class of problems that can be stated as boundary valueformulations of the Lichnerowicz equation. In their theorem 6.2, the authors exactly deal withYamabe negative manifolds with boundary and prove that, in that case, the existence of solutionsto the Lichnerowicz equation corresponds to the existence of solutions to a prescribed curvatureproblem. Our Theorem 4.6, then, solves the problem to a class of those equations.To state it precisely, consider the Lichnerowicz equation ( − ∆ u + n − n − Ru = n − n − R ′ u q − + a w u − q − , in Mγ∂ ν u + n − Hγu = n − H ′ ( γu ) ¯ q − b w u − ¯ q , in ∂M, (95)notice that there is no part of the boundary where the initial condition specified is a Dirichlet-typecondition. In that case, [5] shows that, if ( M, ∂M ) is Yamabe negative, the existence of solutions
Examples and Consequences 23 to (95) is equivalent to the existence of solutions to (91) if a w ≥ b w ≤ R ′ ≤ H ′ ≤ a w ≥ b w ≤ References [1] J. F. Escobar. “Conformal deformation of a Riemannian metric to a constant scalar curvaturemetric with constant mean curvature on the boundary”. In:
Indiana Univ. Math. J.
Annals of Mathematics
Prescribed non positive scalar curvature on asymptotically hyperbolic man-ifolds with application to the Lichnerowicz equation . 2019. arXiv: .[4] Emmanuel Hebey.
Compactness and Stability for Nonlinear Elliptic Equations . Zurich Lecturesin Advanced Mathematics. Zurich: European Mathematical Society, 2014.[5] Michael Holst and Gantumur Tsogtgerel. “The Lichnerowicz equation on compact manifoldswith boundary”. In:
Classical Quantum Gravity
Communications in Analysis and Geometry