Free boundary minimal hypersurfaces outside of the ball

In this paper we obtain two classification theorems for free boundary minimal hypersurfaces outside of the unit ball (exterior FBMH for short) in Euclidean space. The first result states that the only exterior stable FBMH with parallel embedded regular ends are the catenoidal hypersurfaces. To achieve this we prove a B\^ocher type result for positive Jacobi functions on regular minimal ends in $\mathbb{R}^{n+1}$ which, after some calculations, implies the first theorem. The second theorem states that any exterior FBMH $\Sigma$ with one regular end is a catenoidal hypersurface. Its proof is based on a symmetrization procedure similar to R. Schoen [14]. We also give a complete description of the catenoidal hypersurfaces, including the calculation of their indices.


Introduction
Over the last few years, the study of free boundary minimal hypersurfaces (FBMH for short) has occupied a prominent place in differential geometry, especially the study of FBMH in the Euclidean unit ball B ⊂ R n+1 (see e.g.[1,2,7,8,9,16,17] and the references therein).
In this paper, we deal with FBMH in R n+1 \ B with compact boundary in ∂B, which are called exterior free boundary minimal hypersurfaces.These hypersurfaces are critical for the n-volume functional with respect to deformations that let the boundary on the unit sphere.For such critical points, the second order derivative of the volume functional is given by the so-called stability operator which here has a contribution from the boundary.In our situation, this contribution is nonnegative due to the concavity of the unit sphere with respect to its outside.An interesting question would be to understand the geometry and the topology of these hypersurfaces in terms of their indices.In the ball, this is the study made by L. Ambrozio, A. Carlotto and B. Sharp in [2].A situation where non-compact FBMH have been studied is the case of Schwarzschild space: R. Montezuma [13] and E. Barbosa and J.M. Espinar [3] have looked at some properties of these hypersurfaces.
In R n+1 \ B, important examples of exterior FBMH are the catenoidal hypersurfaces, defined as exterior FBMH invariant by isometries fixing a straight line.In Section 4 we give a complete description of the catenoidal hypersurfaces and calculate their indices: some have index 0 and others index 1.
The aim of this paper is to prove two classification results for catenoidal hypersurfaces.The first classification theorem is the following.Definitions are given in Sections 2 and 3.
Theorem 1.Let Σ be an exterior free boundary minimal hypersurface in R n+1 \B.Let us assume that Σ is stable and has parallel embedded regular ends.Then Σ is a catenoidal hypersurface.
In order to prove Theorem 1, we first prove that saying that Σ is stable is equivalent to saying that there exists a positive Jacobi function u on Σ satisfying the Robin boundary condition ∂ ν u + u = 0 on ∂Σ.This is the content of Proposition 3. Its proof is based on the proof of a classical stability characterization due to D. Fischer-Colbrie and R. Schoen [6, Theorem 1] for manifolds without boundary, where here we make use of the Harnack inequality for positive solutions to ∆u + qu = 0 on Σ with Robin boundary condition ∂ ν u + u = 0 on ∂Σ proved in Appendix A. Second, we obtain a Bôcher type theorem for positive Jacobi functions on regular minimal ends in R n+1 which, together with Proposition 3, implies that Σ is invariant by isometries fixing a straight line, in other words, Σ is a catenoidal hypersurface.
As mentioned above, some of the catenoidal hypersurfaces have index 1, so it would be interesting to know if the index equal to 1 implies that the hypersurface is catenoidal.When n = 2, it would be a result similar to Lopez-Ros result [12] for boundaryless minimal surfaces.For example, it would be interesting to understand if a control on the index gives a control on the number of ends of the hypersurface.However, the positive contribution of the boundary to the stability operator seems to make this not an easy task.
The second classification theorem is the following.
Theorem 2. Let Σ be an exterior free boundary minimal hypersurface.If Σ has one regular end, then Σ is a catenoidal hypersurface.
The proof of Theorem 2 is based on a symmetrization procedure as in Schoen's paper [14].
The paper is organized as follows.In Section 2 we present some preliminaries on FBMH and prove Proposition 3. In Section 3 we state and prove an auxiliary Bôcher type result (Theorem 4) and present the proof of Theorem 1.In Section 4 we introduce the catenoidal hypersurfaces and give a complete description of them, including the calculation of their indices.In Section 5 we prove Theorem 2. Finally, in Appendix A we present a proof of the Harnack inequality for positive functions satisfying a Robin type boundary condition.

Free boundary minimal hypersurfaces in R n+1 \ B
Let Σ be an n-manifold with compact boundary.We say that F : Σ → R n+1 \ B is an exterior proper immersion if F is a proper immersion and F (Σ) ∩ ∂B = F (∂Σ).In this paper, we always consider such exterior proper immersions.We also assume that Σ is orientable so that a unit normal N is well defined along F .Besides, we will often identify Σ with its image F (Σ) and just say that Σ is an exterior hypersurface.
We will consider exterior hypersurfaces that are critical for the n-volume functional with respect to any deformations keeping the boundary on ∂B.Such a hypersurface Σ has vanishing mean curvature and meets ∂B orthogonally: we call them exterior free boundary minimal hypersurfaces.
Basic examples are given by cones over minimal hypersurfaces S ⊂ S n = ∂B: One can also consider exterior FBMH that are invariant by isometries fixing a straight line: they are called catenoidal hypersurfaces.Their complete description is given in Section 4.
Let Σ be an exterior free boundary minimal hypersurface.The free boundary condition implies that, at P ∈ ∂Σ, the outgoing unit normal ν(P ) = −P is a principal direction of the second fundamental form B of Σ.Indeed, for T ∈ T ∂Σ, we have where D is the covariant derivative in R n+1 and B ∂B is the second fundamental form of ∂B.

Figure 1. Exterior FBMH
As in the boundaryless case, we also have a monotonicity formula for exterior free boundary minimal hypersurfaces: where B R is the Euclidean ball centered at the origin of radius R ≥ 1.In fact, let v where d(X) = |X| is the distance function to the origin.On the other hand, because Σ is minimal, div Σ (X ) = n.Therefore, This gives where above we have used that Thus, integrating last equation from 1 to R and using coarea formula, we obtain (2.2).
2.1.The stability operator.Let {F t } be a family of exterior proper immersions of Σ such that F 0 (Σ) is free boundary minimal and ∂ t F t has compact support.Even if the volume of F t (Σ) is infinite its derivatives can be computed since the deformation has compact support.Then the first derivative of the n-volume functional vanishes at t = 0 and the second derivative at t = 0 can be computed in terms of the function u = (∂ t F t | t=0 , N ) by where ∇ Σ and dµ are the gradient and the n-volume measure on Σ, and ds is the (n − 1)-volume measure on ∂Σ, all with respect to the metric induced by F 0 (see [2]).After integration by parts, one has So the quadratic form Q is associated with the Jacobi operator defined by Lu = ∆u + B 2 u.Then, for any bounded domain Ω in Σ, we can consider the associated spectrum of L: a sequence of eigenvalues λ n +∞ and a L 2 -orthonormal sequence of functions u n on Ω such that The number of negative eigenvalues is then called the index of Q on Ω and is denoted by Ind(Q, Ω).If (Ω n ) is an increasing sequence of domains such that ∪Ω n = Σ, then the limit of the increasing sequence (Ind(Q, Ω n )) is called the index of Σ and is denoted by Ind(Σ).When Ind(Σ) = 0, we say that Σ is stable and this is equivalent to Q(u, u) ≥ 0 for any function u with compact support on Σ.Actually, we have an alternative characterization of stability given by the following Fischer-Colbrie and Schoen type result.Proposition 3. Let Σ be an exterior free boundary minimal hypersurface.Then Σ is stable if and only if there exists a positive solution u on Σ to Proof.The proof is very similar to that one of [6, Theorem 1] by Fischer-Colbrie and Schoen.
In fact, in order to adapt their proof, we just need to observe that, if u is as in (2.3), then ∂ ν ln u = −1 on ∂Σ (this is for the part (iii) =⇒ (i)) and use the Harnack inequality given in Proposition 9 in Appendix A (for the part (ii) =⇒ (iii)).
When n = 2, Fischer-Colbrie's result [5,Theorem 2] gives that an exterior free boundary minimal surface Σ has finite index if and only if it has finite total curvature.One difference is that, in our case, the quadratic form Q does not depend only on the Gauss map but also on the conformal factor along the boundary ∂Σ.A second important point is that we assume ∂Σ to be compact.For example, if Σ is stable, we have a solution u to (2.3) which can be lifted to the universal cover Σ.This implies that the associated quadratic form on Σ is nonnegative.However, the universal cover may not have finite total curvature as we are going to see below (see Example 6).Actually, the universal cover is not properly immersed and thus it is not an exterior surface according to our definition.

Regular ends.
The asymptotic of an exterior free boundary minimal hypersurface can be highly complicated.A simple asymptotic is given by regular ends introduced by Schoen in [14].
In order to describe it, we split P ∈ R n+1 as (X, z) ∈ R n × R. Then an end E of an exterior free boundary minimal hypersurface is said to be regular if, after an isometry, a representative of E is given by the graph of a function f of bounded gradient defined on {|X| ≥ R} with the following asymptotic: where A, B ∈ R and C ∈ R n .We notice that the above estimate on f implies similar estimates on its derivatives (see [14]).For example, one sees that ∇f (X) goes to 0 as |X| goes to +∞ and, in particular, there is a well-defined unit normal at ∞ for such an end.
If n = 2 and Σ has finite total curvature, then Σ is conformally equivalent to a compact Riemann surface with boundary minus a finite number of points.Moreover, a properly embedded annular end with finite total curvature is regular [14, Proposition 1].
In the case 3 ≤ n ≤ 6, following the arguments of J. Tysk [19], if we assume that Σ has finite index and finite volume growth in the sense that lim R→+∞ R −n |Σ ∩ B R | < +∞, then Σ has finitely many ends, all of them being regular.

A Bôcher type result for the Jacobi operator.
In this section, we analyze the asymptotic behavior of positive Jacobi functions (i.e.solutions to Lu = 0) on regular ends.Theorem 4. Let E be a regular minimal end in R n+1 (let X ∈ R n be a coordinate associated to the end as in (2.4) and (2.5)) and consider a positive Jacobi function u on E. Then u has the following asymptotic: there exist where v is such that the function Proof.Writing X = e t p with t ∈ R and p ∈ S n−1 , a regular end can be parametrized by [t 0 , +∞) × S n−1 with a metric g having the asymptotic g = e 2t (δ + O(e −2(n−1)t )), where δ is the product metric on R × S n−1 .Moreover, the second fundamental form can be estimated by B 2 = O(e −2nt ).Thus the Jacobi operator can be computed as where ∆ σ is the Laplacian on S n−1 and M (u) is a second order linear operator whose coefficients have C 0,α -norm bounded by Ce −2(n−1)t for some constant C > 0.
Therefore a Jacobi function u satisfies which is a uniformly elliptic equation on [t 0 , +∞) × S n−1 .As a consequence, by Harnack inequality ([10, Corollary 8.21]), there is a constant C > 0 such that, for any p, q ∈ S n−1 and t, s ≥ t 0 + 1 with |t − s| ≤ 1, and any positive Jacobi function u, we have u(t, p) ≤ Cu(s, q).By Schauder's elliptic estimates ([10, Corollary 6.3]), we also have Then, combining with (3.2), there is a constant C > 0 such that . By integrating (3.1) over S n−1 , we obtain that ū solves ūtt +(n−2)ū t +M (u) = 0. Considering first the case n > 2, let a and b be two functions such that Then we have the system Using the above equations, we obtain

Thus
√ a 2 + b 2 and ū stay bounded on [t 0 , +∞).In particular, |M (u for t sufficiently large.Therefore e nt √ a 2 + b 2 cannot converge to 0 at t goes to +∞.We can also solve the system to obtain We notice that if A = B = 0 then lim t→+∞ e nt a = lim t→+∞ e nt b = 0, which is not possible.Then we can be sure that either A or B is nonzero.As ū = a + b is positive and A = lim t→+∞ (a + b), then either A > 0 or A = 0 and B > 0. Observe that ū − A − Be −(n−2)t = O(e −2(n−1)t ).
If n = 2, we notice that Thus ū = O(e  In both cases, we have M (u) = O(te −2(n−1)t ).Now, to conclude, we need to estimate u − ū.Let v i be a L 2 -unit eigenfunction for the Laplace operator on S n−1 associated to a nonzero eigenvalue λ (in particular, λ ≥ n − 1).Let Observe that µ 2 + (n − 2)µ − λ = 0 has two roots: µ + ≥ 1 and µ − ≤ −(n − 1).Then, solving the above equation, we obtain 3) and the fact that ū = O(t), we have u i = O(t) and thus a i = 0. We also have Now, by Cauchy-Schwarz, Thus, by squaring (3.6), we obtain Let us define Using that µ + ≥ 1 and µ − ≤ −(n − 1), we can sum the above inequalities with respect to i to obtain Actually, u − ū solves the equation Then, combining the above L 2 -estimate with (3.4) and (3.5), Schauder's estimates give We have then proved that This gives the expected result after going back to the original coordinate system.

Classification of stable hypersurfaces.
If Σ is an exterior free boundary minimal hypersurface with regular ends, the unit normal to Σ has a well-defined limit at each end.Then we say that such a hypersurface has parallel ends if these limits coincide up to a sign.We notice that, if Σ is embedded, then its ends are always parallel.Now, we are going to use the above Bôcher type theorem in order to give a classification of stable exterior FBMH with parallel regular ends.
Proof of Theorem 1.Consider the (X, z) coordinate system on R n+1 .After an isometry, we can assume that the unit normal to the ends of Σ are given by ±e z .Now, let M ∈ M n (R) be a skew-symmetric matrix and consider the Killing vector field K(X, z) = M X.Notice that K generates isometries fixing the z-axis.Then the scalar product u = (K, N ) is a solution to ∆u + B 2 u = 0 on Σ.Moreover, since K is tangent to ∂B, u satisfies ∂ ν u + u = 0 on ∂Σ.
Each end of Σ can be parametrized by the graph of a function f with the asymptotic given by (2.4) or (2.5) (depending on n).In particular, So the asymptotic of f gives that u = O(|X| −(n−1) ).
On the other hand, since Σ is stable, there is a positive solution v to (2.3).The asymptotic of v is given by Theorem 4. As a consequence, we see that u(X)/v(X) goes to 0 as |X| goes to +∞.Also, the function w = u/v satisfies ∆w + 2(∇ ln v, ∇w) = 0 on Σ, ∂ ν w = 0 on ∂Σ.
Therefore the maximum principle gives that w = u/v is constant and thus equals zero.This implies that u = 0 and then Σ is invariant by the isometries generated by K.So Σ is a catenoidal hypersurface.

Catenoidal hypersurfaces
Theorem 1 gives that stable hypersurfaces are invariant by isometries fixing an axis.In this section, we describe this kind of exterior free boundary minimal hypersurfaces Σ.We fix the axis to be Re z .
Let φ be a primitive of the function r → (r 2(n−1) −1) −1/2 defined on [1, +∞).The hypersurface is a minimal hypersurface invariant by isometries fixing Re z .Actually, any connected piece of a minimal hypersurface invariant by isometries fixing Re z is a subset of λΣ+µe z for some λ, µ ∈ R.
Half of Σ can be parametrized by the map Given α ∈ (0, π 2 ), we look for a rotational exterior free boundary minimal hypersurface with boundary in {z = sin α}.Let R α = (sin α) −1/(n−1) be such that φ (R α ) = tan α.Notice that R α decreases with α from +∞ to 1. Let C α be the hypersurface parametrized by where λ α and µ α are chosen such that C α has the expected boundary: The hypersurface C α has free boundary because of the choice of R α .Thus, for any α ∈ [0, π 2 ), there is exactly one rotational exterior free boundary minimal hypersurface: C α for α = 0 and Therefore C α is an exterior free boundary minimal catenoidal hypersurface and any connected exterior free boundary minimal catenoidal hypersurface is the image of some C α by a linear isometry of R n+1 . 2 ) such that C α is stable for α ∈ [0, ᾱn ), and C α has index 1 for α ∈ (ᾱ n , π 2 ).Actually, ᾱ2 = α 2 = π 4 and ᾱn > α n for n > 2. Proof.We first study some preliminary stability properties of C α .Notice that C α is a graph over part of R n .Therefore C α is stable as a graph with fixed boundary: the stability operator is nonnegative for any test functions that vanish on ∂C α .
Let us consider on R n coordinates (x, Y ) ∈ R × R n−1 .Let K(x, Y, z) = (−z, 0, x) be the Killing vector field generating rotations around {x = 0, z = 0} in R n+1 .Then the scalar product k = (K, N ) defines on C α a solution to (2.3).The boundary condition comes from the fact that K is tangent to ∂B.Actually, one can compute k in the (r, p) coordinates.By (4.1), we have Hence k has constant sign when p x has constant sign.This implies that the half catenoidal hypersurfaces C α ∩ {±x ≥ 0} are stable.
Let us now study the global stability of C α .We have a one-parameter family {C α } of catenoidal hypersurfaces.Therefore the derivative with respect to α gives a deformation field whose scalar product with the unit normal to C α is a function u which solves (2.3).In the F α parametrization and for the upward pointing unit normal, u can be computed as So u depends only on r and is equal to 1 on ∂C α , i.e. at r = R α .In order to study the sign of u close to r = +∞, let us have a look on λ α .By (4.2), we have Thus ∂ α λ α is decreasing.
If n > 2, we have lim r→+∞ φ(r) < +∞ and, by (4.1), This limit is positive when α ≤ α n .Moreover, when α ≥ α n , the limit is decreasing with α and negative for α close to π 2 .Then there exists ᾱn > α n such that the limit is positive for α < ᾱn and negative for α > ᾱn .
Thus, for α < ᾱn , u is positive on ∂C α and close to the infinity.Therefore, if u changes sign on C α , {u < 0} would be a precompact subdomain of C α with u = 0 on its boundary, but this would contradict the stability of C α as a graph.Hence, for α < ᾱn , u is positive and then, by Proposition 3, C α is stable.C ᾱn is also stable as limit of stable minimal hypersurfaces.
When α > ᾱn , u changes sign on C α .Thus there is A > R α such that u is nonnegative on [R α , A] × S n−1 and vanishes on {A} × S n−1 .This implies that C α has index at least 1.We notice that there is no value B > A such that u vanishes on {B} × S n−1 .Indeed, this would contradict that [A, +∞) × S n−1 is stable as a graph.
Let us now prove that, for α > ᾱn , C α has index 1.If it is not the case, then there is B > R α such that the Jacobi operator has index at least 2 on [R α , B] × S n−1 .Let us consider u 2 the eigenfunction associated to the second eigenvalue We are going to prove that u 2 depends only on the r variable.As above, let us consider (x, Y ) coordinates on R n and let S be the symmetry of R n+1 with respect to {x = 0}.C α (B) is invariant by S and then we can consider on it the function v defined by v(p) = u 2 (p) − u 2 (S(p)).v is then a solution to is unstable since λ 2 < 0, which contradicts the stability of C α ∩ {x ≥ 0}.So v ≡ 0 and u 2 is invariant by S. Changing the choice of the x coordinate, we obtain that u 2 is invariant by isometries fixing the z-axis and then depends only on r.
As u 2 is associated to the second eigenvalue, u 2 must change sign.Then there is C ∈ (R α , B) such that u 2 = 0 on {r = C}.As λ 2 < 0, this implies that {C ≤ r ≤ B} is unstable, which contradicts the stability of C α as a graph.Hence C α has index 1 for α > ᾱn .
Example 6.Consider the universal cover of C α for n = 2: Straightforward computations give that the area element and the Gaussian curvature of F α are given by Thus, the total curvature of C α is given by while the total curvature of its universal cover is infinite.For α ≤ π 4 , this example shows that when the boundary is not compact, even if the stability operator is nonnegative, the total curvature can be infinite.

Classification of one-ended examples
This section is devoted to the proof of Theorem 2. The idea of the proof is based on a symmetrization procedure as in Schoen's paper [14].
After a rotation, we can assume that the end of Σ is the graph of a function f over the outside of a compact set with the following asymptotic: with A ≥ 0 and, if A = 0, B ≥ 0, and The first step consists in proving that either Σ = {z = 0} \ B = C 0 or A > 0 and Σ ⊂ {z > ε} for some ε > 0.
Observe that ∂Σ ⊂ {z ≥ −2} and Σ \ K ⊂ {z ≥ −2} for some compact set K ⊂ R n+1 .Then, by the maximum principle, Σ ⊂ {z ≥ −2}.In fact, for each t < 0, we have Σ \ K ⊂ {z ≥ t} (for a possibly different compact set K). Therefore, if Σ∩{z < 0} = ∅, we can start from t = −2 and let t < 0 increase up to finding a first contact point in Σ ∩ {z = t 0 } for some t 0 < 0. We notice that, since Σ is free boundary, the first contact point cannot be at ∂Σ.Then the maximum principle can be applied at the first contact point in order to guarantee that Σ = {z = t 0 } \ B, which is not free boundary.This shows that Σ ⊂ {z ≥ 0}.Then either ∂Σ ⊂ {z > 0} or ∂Σ has a point in {z = 0} and the boundary maximum principle can be applied so that Σ = {z = 0} \ B.
If ∂Σ ⊂ {z > 0}, then we see that Since z is harmonic on Σ, using the asymptotic of f we obtain that 0 for n > 2. The same estimate gives that 0 < 2πA + o (1) for n = 2. Therefore, if n = 2, we have A > 0 and this implies that f (X) > 1 for |X| sufficiently large.If n > 2, we have A < 0 and, since f ≥ 0, this implies that B > 0 and f (X) > B 2 for |X| sufficiently large.In any case, we obtain that Σ ⊂ {z ≥ ε} for small positive ε since there cannot be any first contact point with {z = t} for 0 ≤ t ≤ ε.This ends the first step.
We fix a (x, Y ) coordinate system in R n = R × R n−1 .We want to prove that Σ is symmetric with respect to {x = 0}.In order to do this, we are going to follow a symmetrization procedure.
Proof.In R 2 , let p = (a cos α, a sin α) be a point with a > 0 and 0 ≤ α ≤ θ and R t be the rotation of angle t.If 0 ≤ t ≤ 2(θ − α), then the angle between − −−− → pR t (p) and the vertical z-axis is α − t 2 and then at most θ (see Figure 3).Because of the asymptotic of Σ, the intersection Σ∩∂B ρ can be parametrized by ∂B ρ ∩{z = 0} in the following way: there is a function g such that where g satisfies and for any vector V tangent to ∂B |X| .
This implies that − −−−− → P R t (P ) makes an angle less than C ρ −2 with the horizontal plane {z = 0}.Therefore, if ρ θ is chosen such that this angle is less than π 2 − θ, we obtain a contradiction with R t (P ) ∈ Σ and the lemma is proved.
We are now ready to finish the proof of Theorem 2. Let Since Σ ⊂ {z > 0}, we may choose θ 0 > 0 small enough such that Σ − θ 0 ⊂ R n+1 \ B ρ θ 0 .By Lemma 7, we have [0, θ 0 ] ⊂ T .The set T is then a closed interval of the form [0, θ 1 ].Let us notice that when we symmetrize with respect to Π θ , the image of a point on ∂B ρ stays on ∂B ρ , so that points in ∂Σ cannot be sent to interior points of Σ and interior points of Σ cannot be sent to ∂Σ.
Then, if θ 1 < π 2 , by Lemma 7, there is a point P ∈ Σ − θ 1 such that one of the following occurs: ) is on one side of Σ.
In the first case, if P / ∈ ∂Σ, then maximum principle gives that Σ is symmetric with respect to Π θ 1 .If P ∈ ∂Σ, then, by free boundary hypothesis, Σ and S θ 1 (Σ − θ 1 ) are normal to ∂B and thus tangent, since S θ 1 (Σ − θ 1 ) is on one side of Σ.As a consequence, the boundary maximum principle implies that Σ is symmetric with respect to Π θ 1 .
In the second case, if P / ∈ ∂Σ, the boundary maximum principle implies that Σ is symmetric with respect to Π θ 1 .If P ∈ ∂Σ, then as S θ 1 (Σ − θ 1 ) is on one side of Σ, we can locally parametrize Σ and S θ 1 (Σ − θ 1 ) over a quarter of the tangent plane T P Σ by two functions u and v such that u ≤ v, u(P ) = v(P ), and ∇u(P ) = ∇v(P ).Moreover, at P , the tangent vector P is an eigenvector of the second fundamental form of Σ (see (2.1)).Thus the Hessian of u and v at P coincide.So, applying Serrin's corner maximum principle [15] to v − u, we obtain that u ≡ v and Σ is symmetric with respect to Π θ 1 .
By changing the coordinate system, we obtain that Σ is symmetric with respect to any vertical hyperplane passing through the origin and then invariant by rotation around the vertical z-axis: Σ is a catenoidal surface.where X is smooth vectorfield and q a smooth function.Then, given a compact domain U ⊂ Σ, there exists a constant C > 0 (not depending on u) such that, for any p, q ∈ U , we have u(p) u(q) ≤ C.
No such statement seems to appear in the literature.Similar results appear in [4,20], however they are not directly applicable here because of certain hypotheses.
At p, ∆(φw) ≤ 0, and so for some constants a, b, c and d which depend only on X and q.Thus, combining with (A. for some constants A, B, C and D which depend only on K, X, q and φ.This implies that φ(p)w(p) ≤ M for some constant M = M (A, B, C, D) and thus w(q) ≤ M for q ∈ U , since φ ≥ 1 on U .
As a consequence, we have the following Harnack inequality for the Robin boundary condition.

Proposition 9.
Let Σ be a Riemannian manifold with compact boundary and u be a positive solution to ∆u + qu = 0 on Σ, ∂ ν u + u = 0 on ∂Σ, where q is a smooth function.Then, given a compact domain U ⊂ Σ, there exists a constant C > 0 (not depending on u) such that, for any p, q ∈ U , we have u(p) u(q) ≤ C.So the above proposition applies to v and gives the expected result for u.

Proposition 8 .
Appendix A. Harnack inequality In this paper, we are considering solutions u to some elliptic equations on Σ under the Robin boundary condition ∂ ν u + u = 0. Elliptic regularity theory for this condition can be found in [11, Theorem 2.4.2.6].Besides, one can also remark that, if d is a smooth function on Σ with ∂ ν d = 1 (for example −d could be the distance function to ∂Σ) and v = e d u, then ∂ ν v = (∂ ν d)e d u + e d ∂ ν u = 0 and v solves some elliptic equation.So results for Neumann boundary data can be translated to the Robin boundary condition.In the proof of Proposition 3, we use a Harnack inequality up to the boundary that can be derived from the following one.Let Σ be a Riemannian manifold with compact boundary and u be a positive solution to ∆u + (X, ∇u) + qu = 0 on Σ, ∂ ν u = 0 on ∂Σ,