Equivariant Yamabe problem with boundary

As a generalization of the Yamabe problem, Hebey and Vaugon considered the equivariant Yamabe problem: for a subgroup G of the isometry group, find a G-invariant metric whose scalar curvature is constant in a given conformal class. In this paper, we study the equivariant Yamabe problem with boundary.


Introduction
Let M be a closed (that is, compact without boundary) n-dimensional manifold with n ≥ 3, and g a Riemannian metric on M. As a generalization of the Uniformization Theorem for surfaces, the Yamabe problem [22] is to find a metric conformal to g such that its scalar curvature is constant. This was solved by Trudinger [20], Aubin [2], and Schoen [17].
In 1993, a question was raised by Hebey and Vaugon in [10] as to whether the Yamabe problem could be extended under the control of its isometry group. This problem is called the equivariant Yamabe problem (or Hebey-Vaugon Conjecture). To state it, we need the following definitions. Denote [g] the conformal class of g. The group of all conformal transformations of (M, g), C (M, g), is defined as (1.1) The group of all isometries in (M, g), I (M, g), is defined as where E(u) = M ( 4(n−1) n−2 |∇ g u| 2 + R g u 2 )dV g ( M u 2n n−2 dV g ) In [10], Hebey and Vaugon proved the following: [10]) There always holds where O G (x) denotes the orbit of x ∈ M under G, i.e. O G (x) = {σ (x) : σ ∈ G}, and Y (S n , g S n ) is the Yamabe constant of the standard n-dimensional unit sphere (S n , g S n ).

Theorem 1.3 (Theorem C in [10]) If the strict inequality holds in (1.3), then the infimum on the left hand side is achieved and the equivariant Yamabe problem is solvable.
On the other hand, Hebey and Vaugon proved in [10,Theorem B] that the strict inequality holds in (1.3), and as a result the Conjecture 1.1 holds, if one of the following conditions is satisfied: (i) The action of G on M is free. (ii) 3 ≤ dim M ≤ 11. (iii) There exists a point p ∈ M with finite minimal orbit under G such that ω( p) := inf{i ∈ N : ∇ i W g ( p) = 0} > (n − 6)/2 or ω( p) ∈ {0, 1, 2}.
See also [11,14,15,19] for results related to the equivariant Yamabe problem. The Yamabe problem can be formulated for manifolds with boundary. Let M be a compact n-dimensional manifold with smooth boundary ∂ M, where n ≥ 3, and g is a Riemannian metric on M. There are two types for the Yamabe problem with boundary: The Yamabe problem with boundary (I). Find a conformal metric g such that its scalar curvature R g is constant in M and its mean curvature H g is zero on ∂ M. The Yamabe problem with boundary (II). Find a conformal metric g such that its scalar curvature R g is zero in M and its mean curvature H g is constant on ∂ M.
Inspired by Conjecture 1.1, we study in this paper the equivariant Yamabe problem with boundary. We are interested in the following two conjectures which are the corresponding versions of the Hebey-Vaugon Conjecture on manifolds with boundary: Conjecture 1.4 (Equivariant Yamabe problem with boundary (I)) Given a compact subgroup G of I(M, ∂ M, g), there exists a G-invariant metric conformal to g such that its scalar curvature is constant and its mean curvature vanishes.

Conjecture 1.5 (Equivariant Yamabe problem with boundary (II)) Given a compact subgroup G of I(∂ M, g| ∂ M ), there exists a G-invariant metric conformal to g such that its scalar curvature vanishes and its mean curvature is constant.
We remark that the Yamabe problem with boundary (I) is the special case of Conjecture 1.4 when G = {id M } and the Yamabe problem with boundary (II) is the special case of Conjecture 1.5 when G = {id ∂ M }. We also remark that one may modify Conjecture 1.5 as follows:

Conjecture 1.6 Given a compact subgroup G of I (M, ∂ M, g), there exists a G-invariant metric conformal to g such that its scalar curvature vanishes and its mean curvature is constant.
Since f ∈ I (M, ∂ M, g) implies that f | ∂ M ∈ I (∂ M, g| ∂ M ), Conjecture 1.6 in fact implies Conjecture 1.5. But we find that it is easier to consider Conjecture 1.5 than the more restrictive Conjecture 1.6.
Let g S n + be the standard metric in S n + , and the boundary of S n + is given by The following is the corresponding version of Theorem 1. The following is the corresponding version of Theorem 1.3:

Theorem 1.8 If the strict inequality holds in (1.6), then the infimum on the left hand side is achieved and the equivariant Yamabe problem with boundary (I) is solvable.
We are able to prove that the inequality in (1.6) is strict for some cases. More precisely, we have the following: Theorem 1.9 Let (M, g) be a compact n-dimensional Riemannian manifold with boundary ∂ M. Assume that ∂ M is umbilic and that M is not conformally diffeomorphic to S n + . If either M is locally conformally flat or 3 ≤ n ≤ 5, then the strict inequality holds in (1.6). In particular, equivariant Yamabe problem with boundary (I) is solvable for such (M, ∂ M, g).
Similar results are also obtained for the equivariant Yamabe problem with boundary (II). Define (1.7) The generalized Yamabe constant of (M, ∂ M, g) is defined as Then D n equipped with the standard flat metric g D n is an n-dimensional Riemannian manifold with boundary Similar to Theorem 1.7, we have the following: Theorem 1.10 There always holds The plan of this paper is as follows: In Sect. 2, we prove some basic facts related to the equivariant Yamabe problem with boundary (I), and then prove Theorem 1.7 and Theorem 1.8. In Sect. 3, we study the locally conformally flat case of Theorem 1.9. In Sect. 4, we prove the remaining cases of Theorem 1.9, namely, when 3 ≤ dim M ≤ 5. In Sect. 5, after proving some basic facts related to the equivariant Yamabe problem with boundary (II), we then prove Theorem 1.10 and Theorem 1.11. The locally conformally flat case of Theorem 1.12 is proved in Sect. 6, while the case of Theorem 1.12 when 3 ≤ dim M ≤ 5 is proved in Sect. 7.

Notations and preliminary results for type (I)
Let (M, ∂ M, g) be a compact n-dimensional Riemannian manifold with smooth boundary ∂ M. By conformal change of g, we may assume that H g = 0 on ∂ M (see Lemma 1.1 in [9]). It is known that if Y (M, ∂ M, g) ≤ 0, then the Equivariant Yamabe problem with boundary (I) is always solvable; for one can seek the solution to the subcritical case, pass to the limit and get the solution to the critical case, as done by Trudinger for the classical Yamabe problem.
More precisely, for all 2 < q ≤ 2n n − 2 , there exists u q > 0 such that u q is G-invariant and , where u is a solution to the Equivariant Yamabe problem with boundary (I).
Hence, we assume that Y (M, ∂ M, g) > 0. By conformal change of g, we may assume without loss of generality (see Lemma 1.1 in [9]), i.e. R g > 0 in M and H g = 0 on ∂ M. (2.1) It follows from Sobolev embedding that for 0 ≡ u ∈ (M). In view of (2.1), we have E 2n which is well-defined, since (M) ⊆ L q (M) for any 2 ≤ q ≤ 2n n − 2 (see Theorem 2.30 in [3] for example).
Proof Consider a sequence {u i } of positive functions in (M) which is G-invariant and satisfies (c.f. Theorem 2.34 in [3]), by passing to a subsequence, we can suppose (2.7) It follows from (2.4) that Combining this with (2.7), we obtain This implies that u q realizes the minimum of E q (u) in the space of u ∈ (M) which are G-invariant. The Euler-Lagrange equation then implies that for all G-invariant functions v ∈ (M). It also follows from (2.4) that Let L g be the conformal Laplacian of g: Since R g > 0 by (2.1), the conformal Laplacian L g defined in (M) is invertible. Hence, let w ∈ (M) be the unique solution of the equation Since u q is G-invariant, the uniqueness of w implies that w is also G-invariant. On the other hand, it follows from (2.9) and (2.11) that where we have used (2.10) and integration by parts. Since L g is invertible, u q = w, i.e. u q is a weak solution of A regularity theorem of Cherrier [6] asserts that u q ∈ C ∞ (M). This proves the assertion. Proof First of all, we note that To see this, it suffices to consider 0 < v ∈ C ∞ (M) which is G-invariant and satisfies E 2n Without loss of generality, we suppose now that the subsequence {u q } satisfies lim u q = u , where the second inequality follows from M u q q dV g = 1, which implies that M u 2n n−2 dV g ≤ 1. (2.14) Multiplying the second equation in (2.13) by u, integrating over M and using the first equation in (2.13), we get μ 2n for all δ > 0 and for each i = 1, 2..., k.
The proof of Lemma 2.3 will be given in Appendix. Now we are ready to prove Theorem 1.8.
Proof of Theorem 1.8 By the assumption in Theorem 1.8, μ 2n and it is about to show that the situation described We now use the invariance of u q under G and the property M u q q dV g = 1. (2.17) Two cases arise. In the first case, the orbit of It follows from (2.17) that we can choose δ > 0 small enough to get In the second case, card O G (x i ) = ∞. It is then easy to see that, for all > 0, there exists We remark that we just showed that if the orbit of any point in G is infinite, then μ 2n is attained.
Proof of Theorem 1. 7 If the orbit of any point in G is infinite, there is nothing to prove. Suppose that G ⊆ I (M, ∂ M, g) possesses a finite orbit. Let x 1 be a point of minimal orbit.
Consider the case when x 1 ∈ M. If x 1 ∈ M, then by conformally changing g, we may assume that the mean curvature of g vanishes and the scalar curvature of g at x 1 is zero, i.e. H g = 0 on ∂ M and R g (x 1 ) = 0. To see this, suppose that f ∈ C ∞ (M) such that f is G-invariant and is strictly positive in at least one point of M which satisfies f (x 1 ) = 0. Since the orbit of x 1 in G is finite, it is always possible to construct such a function f . Let 0 < u ∈ C ∞ (M) be the unique solution of (2.20). In particular, we have R g ( The function u is therefore G-invariant and since E 2n . Then we can construct a Riemannian metric g conformal to g and Ginvariant which satisfies Rc(g ) = 0 and H (g ) = 0 at any point of a minimal orbit under G (See Lemma 4.1).

Lemma 2.5 If x is a point of S n
We now consider g a metric on S n + conformal to g S n + , and I x (S n + , ∂ S n is a compact subgroup of conformal transformation of S n + . We only deal with the case of x ∈ ∂ S n + here, and the case of x ∈ S n + can be proved in the same way. Suppose x ∈ ∂ S n + . If ϕ : S n + − {x} → R n + is the stereographic projection of pole x, then ϕ| ∂ S n + −{x} is a map from ∂ S n + to R n−1 . Then ϕ| ∂ S n + −{x} I x (S n + , ∂ S n + , g)(ϕ| ∂ S n + −{x} ) −1 becomes a compact group of conformal transformations of R n−1 , a subgroup of the group of isometries of a metric conformal to the Euclidean metric, and thus makes it a subcompact group of the affine group. Consequently, ϕ| ∂ S n + −{x} I x (S n + , ∂ S n + , g)(ϕ| ∂ S n + −{x} ) −1 necessarily admits a fixed point (which we denote by y ∈ R n−1 = ∂R n + ). Now, if t y , t −y : R n + → R n + represent the translations of respective vectors y and −y, then t −y (ϕ I x (S n + , ∂ S n + , g)ϕ −1 )t y becomes a group of vector isometries of R n + . So, one can see that the conformal transformation τ we want is given by τ = ϕ −1 t y ϕ.

Locally conformally flat case for type (I)
In this section, we prove the first case of Theorem 1.9, i.e. we will show that if the manifold is locally conformally flat, then the strict inequality holds in (1.6). Let (M, ∂ M, g) be a compact Riemannian manifold with boundary and G ⊂ I (M, ∂ M, g) be a subgroup of the group of isometries of g. Assume that ∂ M is umbilic and that M is not conformally diffeomorphic to S n + . We suppose that G has finite orbits, and we denote by O G (x 1 ) = {x 1 , · · · , x k } a minimal orbit under G. First, we consider the case of Suppose that (M, ∂ M, g) is locally conformally flat. By conformal change of g if necessary, we can assume that ∂ M is minimal with respect to g (see Lemma 2.1 in [4]), i.e. H g = 0. Since ∂ M is minimal and umbilic, ∂ M is totally geodesic. We have the following:

Lemma 3.1 If (M, ∂ M, g) is locally conformally flat, then for all x ∈ ∂ M, there exits a metric g x conformal to g and I x (M, ∂ M, g)-invariant which is Euclidean in the neighborhood of x.
Proof Let be an open neighborhood of x. Since (M, ∂ M, g) is locally conformally flat and ∂ M is totally geodesic, we can double M along ∂ M to obtainM, which is locally conformal flat, closed manifold. Then we can a conformal immersionφ from a neighborhood of x ∈M to S n (c.f. [12]), which gives a conformal immersion ϕ : . Therefore it is also a subgroup of I p (S n + , ∂ S n + , g ) for a metric g conformal to g S n + . Then, by Lemma 2.5, there exists τ ∈ C p (S n + , ∂ S n We set ψ = τ • ϕ and let g 0 be the Euclidean metric on S n + − {p} conformal to g S n + , which is obtained by reciprocal image of the Euclidean metric of R n + under the stereographic projection of pole − p. It is easy to see Then we can see that the metric g is G-invariant and Euclidean in the neighborhood of each point of a minimal orbit under G. Then, we can assume that near each x i ∈ ∂ M the metric is flat and the boundary is minimal. Let (y i 1 , · · · , y i n ) be rectangular coordinates around x i ∈ ∂ M. Since ∂ M is umbilic and minimal, it is a hyperplane. So we can assume that ∂ M is given by In the neighborhood of x i , G i admits an expansion of the type where A is a constant, α i (x) is a smooth harmonic function near x i with α(x i ) = 0, and For each x i , we now consider the function u i δ, (like [9]) defined by In order for the function u i δ, to be continuous across ∂ B δ (x i ) we must require 0 to satisfy To simplify the notations, we will write u = u δ, and u i = u i δ, . We have and with the invariant under G, we find By (4.12) in [9], we have On the other hand, on ∂ B 2δ (x 1 ), we have and with the definition of u i , we see that We therefore obtain Combining all these estimates , we can conclude that It now remains to evaluate u 2 . To this end, we compute where the second inequality follows from (a + b) N ≥ a N + N a n−1 b, and the third inequality follows from the invariance under G. From the definition of u 1 , we have We can see that (n−2)/2 is equivalent to 0 from (3.2), and 0 where C 0 and C 1 are strictly positive constants independent of δ and . Finally it remains to notice that we always have , we obtain two strictly positive constants C 0 and C 2 , independent of δ and , with In particular, we can find δ and sufficiently small so that E 2n By the positive mass theorem, we have A > 0 (cf. [9,Appendix]). Therefore we can conclude that μ 2n Then, since M is locally conformally flat, we can follow the same argument as in the above and eventually obtain the Eq. (3.3). What makes this case different from the cases discussed earlier is the fact that, when M = S n + , we have A = 0 (cf. [9,Appendix]). Since C 2 is a positive constant, so if k = 1, we have μ 2n admits a fixed point (denote by x). Since x is a fixed point, G becomes a subgroup of I x (S n + , ∂ S n + , g). Then, by Lemma 2.5, there exist τ ∈ C x (S n + , ∂ S n + , g S n + ) so that G ⊂ τ −1 I x (S n + , ∂ S n + , g S n + )τ . Let f , ϕ > 0 be smooth functions such that (τ (x)). First, we show that u is G-invariant. To see this, consider σ ∈ G and i ∈ I x 1 (S n + , ∂ S n + , g S n + ) which satisfies σ = τ −1 iτ . Then . This shows that the invariance of u under σ , hence under G. Since ϕu = 1/( f •τ ) and τ * g S n + = ( f •τ ) −4/(n−2) g S n + , we can easily verify that E 2n n−2 (u) = Y (S n + , ∂ S n + , g S n + ). This shows that μ 2n n−2 (G) = Y (S n + , ∂ S n + , g S n + ). Now we consider the case of x 1 ∈ M. Even in this case, we can show that there is a metric g that is G-invariant and Euclidean in the neighborhood of each point of a minimal orbit through the same method as before. Then, near x i , the Green's function G i of the conformal Laplacian L g admits an expansion of the type where A is a constant, α i (x) is a smooth harmonic function near x i with α(x i ) = 0, and For each x i , we now consider the function u i δ, defined by where δ is chosen small enough so that B 4δ (x i ) ∩ ∂ M = ∅, and the η is a smooth cut-off function such that 0 Then we can get the following using the arguments used above: Then, once again by the positive mass theorem (cf. [9, Appendix]) we obtain μ 2n

The case of dimension 3, 4, and 5 for type (I)
In this section, we prove Theorem 1.9 when 3 ≤ n ≤ 5. Let (M, g) be a compact ndimensional Riemannian manifold with boundary ∂ M. First assume that ∂ M is umbilic and that M is not conformally diffeomorphic to S n + . We also suppose that G has finite orbits, and we denote by O G (x 1 ) = {x 1 , · · · , x k } a minimal orbit under G. In this section we only discuss the case of x 1 ∈ ∂ M. However, in the case of x 1 ∈ M, one can obtain the desired result in the same way as x 1 ∈ ∂ M, as it was in the previous section.
Unlike the previous section, g is no longer assumed to be Euclidean in the neighborhood of x i , and as a result, some additional terms appear in the estimates. But we will see that the additional terms are always negligible in lower dimensions.
Set ϕ 1 to be the first eigenfunction for the conformal Laplacian with respect to the boundary And O is defined similarly. We define u i δ, , and u δ, as in the previous section. It follows from [9, section 4] that where I 3 is the error term satisfying As before, we can show that where C 0 and C 2 are positive constants. And the positive mass theorem (cf. [9,Appendix]) ). To prove the cases for n = 4 and n = 5, we need the following: Proof The dimension has in fact nothing to do here. We will first show that given a compact Riemannian manifold (M, g) with boundary ∂ M and x a point of ∂ M, we can always find a I x (M, ∂ M, g)-invariant metric g x conformal to g, which satisfies H g x (x) = 0 and Rc(g x )(x) = 0.
We consider an open neighborhood of 0 in  , g(x)). We now consider Rc(g)(x), which is a symmetrical bilinear form over T x (X ). Let (e 1 , · · · , e n ) be an orthonormal basis of T x M which diagonalizes Rc(g)(x), and E i , i = 1, · · · , n the associated eigenspace. Then the invariance of Rc(g)(x) under I x (M, ∂ M, g) imposes the invariance of E i under the same group. Let r i be the distance from the origin of T x M measured on E i and a 1 , · · · , a n real numbers. Then we consider f : T x M → R defined byf (e i ) = a i r 2 i . The functionf is clearly I x (M, ∂ M, g)-invariant so that the function f : exp x ( ) → R defined by f =f • exp −1 x also becomes I x (M, ∂ M, g)invariant. We now show that it is possible to choose the a i (in fact in a unique way) so that f satisfies Rc(e f g)(x) = 0.
One can easily see that f (x) = 0 and ∇ g f (x) = 0, so we have and in the normal map (exp x ( ), exp −1 x ) associated with (e 1 , · · · , e n ), we can see that It is clear that this system always admits a solution (which is unique). Now we set g x = η(e f g) + (1 − η)g. This proves the first part of Lemma 4.1.
For the second part, it remains to show that we can also have Sym∇ Rc(x) = 0. Although it is more technical, basically we can adopt the same method as in the first part. Starting from the metric g x which has just been obtained, it will be a question of constructing a homogeneous polynomialf of third degree so that Sym∇ Rc(e f g x )(x) = 0.
In order not to overload the drafting, we are leaving this point aside. Now we consider the metric Then we can see that g is conformal to g, G-invariant, and satisfies Rc(g ) = 0 (resp. Rc(g ) = 0 and ∇ Rc(g ) = 0) and H (g ) = 0 at any point of a minimal orbit under G. Let us point out that, by the same reasoning of recurrence of this type, it is possible to prove the existence of conformal normal G-invariant coordinates. More precisely, we have the following:

where r represents the distance to point x. In particular, if G is a subgroup of I (M, ∂ M, g) and if x ∈ ∂ M has a finite orbit under G, then for any positive integer m, we can associate a G-invariant metric g conformal to g and so that
in a normal geodesic coordinates at any point in the orbit of x under G.
By Lemma 4.1, when n = 4, the Green function's G i for the conformal Laplacian with boundary conditions ∂G i ∂ν g = 0 has the expansion for x i small as where A is a constant. We define u i δ, and u δ, as in section 4. It follows from [9, section 4] that where I 4 is the error term satisfying

Then we can show that
where C 0 and C 2 are positive constants. And the positive mass theorem guarantees that ).
Suppose n = 5. Let (y i 1 , · · · , y i n ) denote a normal rectangular coordinate system centered at x i . Let r i = |y i |, ζ = y i /|y i |. Then the metric g of M can be locally written as where h r i is a metric on S n−1 + with h 0 being the standard metric on S n−1 + . Given ρ > 0 with ρ small, let ξ(r i ) be a smooth nonincreasing function satisfying ξ(r i ) = 1 for r i ≤ ρ, and ξ(r i ) = 0 for r i ≥ 2ρ; let |ξ (r i )| ≤ cρ −1 and let |ξ (r i )| ≤ cρ −2 for all r i > 0. We define the metric ρ g on M by setting ρ g = 0 on M − B + 2ρ and Thus ρ g is Euclidean in B + ρ and coincides with g outside B + 2ρ . It is easy to check that the curvature tensor of ρ g is bounded, independent of ρ. It follows from [8, Theorem 7.1] that the boundary ∂ M is totally geodesic with respect to the metric ρ g.
Let L ρ and B ρ denote the following linear operators taken in terms of ρ g: Let λ ρ denote the lowest eigenvalue of L ρ , and λ the lowest eigenvalue of L. Then since g is conformally related to a metric with positive scalar curvature, where the boundary is minimal, we have λ > 0. Let G i be the multiple of the Green's function of (L, B) with pole at x i and normalize so that lim |y i |→0 |y i | 3 G i (y i ) = 1. Note that ρ g converges in the C 1 -norm to g as ρ 0 and R ρ is uniformly bounded. Thus, for ρ sufficiently small, λ ρ > 0. Hence we can consider G i,ρ to be the positive Green's function (L ρ , B ρ ). Normalize the function G i,ρ such that lim |y i |→0 |y i | 3 G i,ρ = 1. Consider the double of M,M = M ∪ ∂ M ∪ M, equipped with the metric induced from M. Then we can show that the function G i,ρ converges to G i as ρ → 0 in the C 2 -norm on compact subsets of M − {x i } by applying Lemma 1 in [17]. Since the metric ρ g are Euclidean in B ρ (x i ), the function G i,ρ is harmonic in B ρ and hance has an expansion for |y i | small, We define u i δ, , and u δ, as in the previous section. It follows from (2.8) in [17] that where c depends on ρ but c 1 does not. Then we can show that where C 0 and C 2 are positive constants. Since we assumed that M is not conformally equivalent to S 5 + , it follows from [9,Section4] that So, fixing ρ small, then fixing ρ 0 , and finally choosing 0 sufficiently small, we prove μ 10 ).

Notations and preliminary results for type (II)
Let (M, ∂ M, g) be a compact n-dimensional Riemannian manifold with smooth boundary ∂ M. By conformal change of g, we may assume that R g = 0 in M (see Proposition 1.4 in [8]). It is known that if Q(M, ∂ M, g) ≤ 0, then the Equivariant Yamabe problem with boundary (II) is always solvable; for one can seek the solution to the subcritical case, pass to the limit and get the solution to the critical case, as done by Trudinger for the classical Yamabe problem. More precisely, for all 2 < q ≤ 2(n − 1) n − 2 , there exists u q > 0 such that where μ q (G) is defined below. By the assumption Y (M, ∂ M, g) ≤ 0, u q converges to u as where u is a solution to the Equivariant Yamabe problem with boundary (II).
Hence, we assume that Q(M, ∂ M, g) > 0. By a conformal change of g, we may assume that the scalar curvature of g vanishes in M (see Proposition 1.4 in [8]), i.e.
where 0 ≡ u ∈ H 2 1 (M). Note that Q 2(n−1) n−2 (u) = Q(u) defined in (1.7) thanks to (5.1). For Proof It follows from (5.5) that u q is positive almost everywhere. Since It also follows from (5.2) and (5.3) that This implies that u q realizes the minimum of Q q (u) on the space of u ∈ H Let B g be the boundary operator of g: Since H g > 0 by (5.1), the boundary operator B g defined in H 1 2 (M) is invertible. Hence, let w ∈ H 1 2 (M) be the unique solution of the equation On the other hand, it follows from (5.9) and (5.11) that where we have used (5.10) and integration by parts. Since B g is invertible, u q = w, i.e. u q is a weak solution of A regularity theorem of Cherrier [6] asserts that u q ∈ C ∞ (M). This proves the assertion. Proof First of all, we note that lim sup To see this, it suffices to consider 0 < v ∈ C ∞ (M) such that v| ∂ M is G-invariant and satisfies Q 2(n−1) (5.14) Multiplying the second equation in (5.13) by u, integrating over ∂ M and using the first equation in (5.13), we get where the last inequality follows from (5.14). Combining (5.12), (5.14) and (5.15), we can conclude that μ = μ 2(n−1) n−2 (G) and  , d g (x, y) is the distance between x and y in M with respect to the Riemannian metric g.

Lemma 5.3 In the situation described above, there exists a finite number of points
for all δ > 0 and for each i = 1, 2..., k. Furthermore, for all p ∈ N, the sequence {u q } 2≤q< 2(n−1) n−2 converges in C p (M) to zero on any compact set of M − {x 1 , · · · , x k }.
Proof It follows from Lemma 4.1 in [21].
Proof of Theorem 1.11 By the assumption in Theorem 1.11, μ 2(n−1) and it is about to show that the situation described above cannot occur (so that Lemma 5.2 applies).
Suppose that for any subsequence of {u q } 2≤q≤ 2(n−1) n−2 converges to zero. Then by Lemma 5.3 there exists x 1 , · · · , x k ∈ ∂ M such that {u q } 2≤q< 2(n−1) n−2 converges to zero in M −{x 1 , · · · , x k } and for each i = 1, ..., k and for all δ > 0 Two cases arise. In the first case, the orbit of It follows from (5.17) that we can choose δ > 0 small enough to get , which gives lim sup In the second case, card O G (x i ) = ∞. It is then easy to see that, for all > 0, there exists 0 < δ 1 with lim sup But if is sufficiently small, this equality contradicts (5.16). Therefore, for all i = 1, ..., k, the orbit of x i in G is finite and by (5.16) and (5.18). The hypothesis that the strict inequality holds in (1.9), i.e.
We remark that we just showed that if the orbit of any point in G is infinite, then μ 2(n−1) is attained. card O G (x). By conformally changing g, we may assume that the scalar curvature of g vanishes and the mean curvature of g at x 1 is zero, i.e. R g = 0 in M and H g (x 1 ) = 0. To see this, suppose that f ∈ C ∞ (∂ M) such that f is G-invariant and is strictly positive in at least one point of ∂ M which satisfies f (x 1 ) = 0. Since the orbit of x 1 in G is finite, it is always possible to construct such a function f . Let 0 < u ∈ C ∞ (M) be the unique solution of by (5.1) and (5.20). In particular, we have H g ( n and the second fundamental form of ∂ M at x 1 has a diagonal form. By the invariant under G, we have normal coordinates (y i 1 , · · · , y i n ) around each x i ∈ ∂ M which satisfy the properties described above. Let δ > 0 be chosen such that where |y i | 2 = (y i 1 ) 2 + · · · + (y i n−1 ) 2 , and ψ(x) be a piecewise smooth function which satisfies ψ(s) = 1 for |s| ≤ δ, ψ(s) = 0 for |s| ≥ 2δ, and |ψ (s)| ≤ δ −1 for δ ≤ |s| ≤ 2δ. For i = 1, ..., k, we define The function u | ∂ M is therefore G-invariant and since Q 2(n−1) From the proof of Theorem 1.10, we have the following: For any metric g conformal to g D n , we can find τ ∈ C x (D n , ∂ D n , g D n ) such that τ −1 I x (D n , ∂ D n , g)τ ⊂ I x (D n , ∂ D n , g D n ).
Proof Note that I x (D n , ∂ D n , g) is a compact subgroup of conformal transformation of D n . Let ψ : S n + → D n be the projection from the upper hemisphere to the unit disc. If ϕ : S n + − {x} → R n + is the stereographic projection of pole x, ϕψ I x (D n + , ∂ D n , g)ψ −1 ϕ −1 becomes a compact group of conformal transformations of R n + , a subgroup of the group of isometries of a metric conformal to the Euclidean metric, and thus makes it a subcompact group of the affine group. Therefore, ϕψ I x (D n , ∂ D n , g)ψ −1 ϕ −1 necessarily admits a fixed point. Then we can follow the argument in Lemma 2.5 to finish the proof.

Locally conformally flat case for type (II)
In this section, we prove Theorem 1.12 when (M, ∂ M, g) is locally conformally flat, i.e. we will show that if (M, ∂ M, g) is locally conformally flat, then the strict inequality holds in (1.9). Let (M, g) be a compact Riemannian manifold with boundary ∂ M, and G be a subgroup of I (M, ∂ M, g), the group of isometries of g. Assume that ∂ M is umbilic and M is not conformally diffeomorphic to D n . We suppose that G has a finite orbit, and we denote by O g (x 1 ) = {x 1 , · · · , x k } a minimal orbit under G.
Suppose that (M, ∂ M, g) is locally conformally flat. By Lemma 3.1, we can assume that g is Euclidean in the neighborhood of each the x i . By umbilicity, we can assume that the boundary is a hyperplane. Let (y i 1 , · · · , y i n ) be rectangluar coordinates around x i ∈ ∂ M such that ∂ M is defined by y i n = 0 locally. Let G i be the positive solution of g G i = 0 on M − {x i } and B g G i = 0 on ∂ M. Then G i has the following asymptotic expansion for r i small: Let η i be a smooth cut-off function as in (3.4). For each x i , we now consider (like [8]) the function u i δ, defined by where Observe that the function u i δ, is continuous across ∂ B δ and ∂ B 2δ .
we have ∂G 1 ∂ν ≤ C and G i ≤ Cδ 2−n , which gives On the other hand, on ∂ B 2δ (x 1 ), we have We thus obtain and with the definition of u i , we see that Hence, we obtain Combining all these estimates yields . To this end, we compute where the second inequality follows from (a + b) N ≥ a N + N a n−1 b, and the third inequality follows from the invariant under G. From the definition of u 1 , we have where the second equality follows from by the definition of W i in (6.1). Therefore, as in Sect. 3, we obtain where C 0 and C 2 are strictly positive constants. In particular, we can find δ and sufficiently small so that Q 2(n−1) The positive mass theorem guarantees that A > 0 (cf. [9,Appendix]. Therefore we can conclude that μ 2(n−1) . Now we suppose M = D n . Then, since M is locally conformally flat, we can follow the above argument and eventually obtain (6.2). But, in this case, the constant A is zero. If k ≥ 2, we still have Therefore, we suppose k = card O G (x 1 ) = 1, i.e. G admits a fixed point (denote by x). Since x is a fixed point, G becomes a subgroup of I x (D n , ∂ D n , g). Then, by Lemma 5.5, We set u(x) = 1/ϕ(x) f (τ (x)). First, we show that u is G-invariant. To see this, consider σ ∈ G and i ∈ I x 1 (D n , ∂ D n , g D n ) which satisfies σ = τ −1 iτ . Then . This shows that the invariance of u under σ , hence under G. Since ϕu = 1/( f • τ ) and τ * g D n = ( f • τ ) −4/(n−2) g D n , we can easily verify that Q 2(n−1) n−2 (u) = Q(D n , ∂ D n , g D n ). This shows that μ 2(n−1)

The case of dimension 3, 4 and 5 for type (II)
In this section, we prove Theorem 1.12 when 3 ≤ n ≤ 5. Basically, it can be proved by the same argument as Type I. Therefore, we will only consider the case when n = 5.
Suppose that n = 5. Let (y i 1 , · · · , y i n ) denote a normal rectangular coordinate system centered at x i . Let r i = |y i |, ζ = y i /|y i |. Then the metric g of M can be locally written as where h r i is a metric on S n−1 + with h 0 being the standard metric. Given ρ > 0 with ρ small, let ξ(r i ) be a smooth nonincreasing function satisfying ξ(r i ) = 1 for r i ≤ ρ, and ξ(r i ) = 0 for r i ≥ 2ρ; let |ξ (r i )| ≤ cρ −1 and let |ξ (r i )| ≤ cρ −2 for all r i > 0. We define the metric ρ g on M by setting ρ g = 0 on M − B + 2ρ and Thus ρ g is Euclidean in B + ρ and coincides with g outside B + 2ρ . It is easy to check that the curvature tensor of ρ g is bounded, independent of ρ. It follows from [8,Theorem 7.1] that the boundary ∂ M is totally geodesic with respect to the metric ρ g.
Let L ρ and B ρ be the following linear operator taken in terms of ρ g: Let λ ρ denote the lowest eigenvalue of L ρ , and λ the lowest eigenvalue of L. Then since g is conformally related to a metric with positive scalar curvature, where the boundary is minimal, we have λ > 0. Let G i be the multiple of the Green's function of (L, B) with pole at x i and normalize so that lim |y i |→0 |y i | 3 G i (y i ) = 1. Note that ρ g converges in the C 1 -norm to g as ρ 0 and R ρ is uniformly bounded. Thus, for ρ sufficiently small, λ ρ > 0. Hence we can consider G i,ρ to be the positive Green's function (L ρ , B ρ ). Normalize the function G i,ρ such that lim |y i |→0 |y i | 3 G i,ρ = 1. ConsiderM = M ∪∂ M ∪ M as the double of M with the standard metric induced from M. Then we can show that the function G i,ρ converges to G i as ρ → 0 in the C 2 -norm on compact subsets of M − {x i } by applying Lemma 1 in [17]. Since the metric ρ g are Euclidean in B ρ (x i ), the function G i,ρ is harmonic in B ρ and hance has an expansion for |y i | small, We define u i δ, , and u δ, as in the previous section. Then it follows from (7.16) in [8] that where c depends on ρ but c 1 does not. Then we can show that where C 0 and C 2 are positive constants. Since M is not conformally equivalent to S 5 + , if follows from [8,Section7] Fixing ρ small, then fixing ρ 0 , and finally choosing 0 sufficiently small, we thus have (i) For any σ < Y (S n + , ∂ S n + , g S n + ), there exist constants q 0 > 2n n−2 , r 0 > 0 such that, whenever for some r < r 0 and some x 0 ∈ M there holds P L n/2 (B 2r (x 0 )) ≤ σ , then (ii) For any q > 2n n−2 , any r > 0, there exist a constant C = C(q, r ) > 0 such that , whenever there holds P L n/2 (B 4r (x 0 )) < 2nY (S n + , ∂ S n + , g S n + )/((n − 2)q).
Proof (i) This assertion will follow by performing one step of a Moser iteration. Choose some smooth cut-off function η ∈ C 1 0 (B 2r (x 0 )) satisfying η(x) = 1 in B r (x 0 ) and |∇ g η| < C/r . For suitable p ≥ 1, we then let v = u 2 p−1 η 2 . Multiplying the first equation of (8.1) by v, using integrating by parts and the second equation of (8.1), we have Let w = u p η. Then |w| 2 = uv, and Young's inequality implies that It follows from (8.2), (8.3), and the Hölder's inequality that .
Proof Fix a point x 0 ∈ M and assume that for some r > 0 there holds sup k B 2r (x 0 )∩M |R g k | n 2 dV g k 2 n ≤ σ < Y (S n + , ∂ S n + , g S n + ). (8.10) Writing the Yamabe equation for u k as − 4(n − 1) n − 2 g u k + R g u k = R g k u n+2 n−2 k = P k u k , we have P k = R g k u ≤ σ < Y (S n + , ∂ S n + , g S n + ).
Clearly, we may assume that r ≤ r 0 (σ ) as determined in Lemma 8.1. Lemma 8.1 and the assumption Vol(M, g k ) = 1 then imply that (u k ) is bounded in L q 0 (B r (x 0 ) ∩ M) for some q 0 > 2n n−2 . In particular, Vol(B r (x 0 ), g k ) → 0 as r → 0, (8.11) uniformly in k ∈ N.
Choose q > n 2 such that 1 By Hölder's inequality, for any r > 0 we have Going back to the previous equation, we then find that | g u k | ≤ C|R g k − R g k |u n+2 n−2 k + R g k u n+2 n−2 k + R g u k is bounded in L p (B 2r (x 0 ) ∩ M). Thus, by the Calderón-Zygmund inequality, u k ∈ W 2, p (B r (x 2 ) ∩ M) with u k W 2, p (B r (x 0 )∩M) ≤ C = C(r ). (8.13) Now assume that condition (8.10) is satisfied for every x ∈ M and some radius r = r (x) > 0. The cover (B r (x) (x)) x∈M of M admits a finite subcover B r i (x), where r i = r (x i ), 1 ≤ i ≤ I . From (8.13) we then obtain the desired uniform bound u k W 2, p (M) ≤ I max 1≤i≤I C(r i ).
If (8.10) does not hold for every x with some r = r (x) > 0, we iteratively determine points x l , l ∈ N, and a subsequence (u k ) such that for any r > 0 and any l condition (8.9) is valid. This iteration terminates after finitely many steps. Indeed, given x 1 , · · · , x L , choose 0 < r < min i = j dist(x i , x j )/2. Then (8.9) yields the bound L · Y (S n + , ∂ S n + , g S n + ) By a covering argument as above we then obtain that (u k ) is bounded in W 2, p → L ∞ on any compact subset of M \ {x 1 , · · · , x L }, as claimed.