Hardy and Rellich inequalities for submanifolds in Hadamard spaces

Hardy and Rellich inequalities for submanifolds in Hadamard spaces

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Hardy and Rellich inequalities for submanifolds in Hadamard spaces M. Batista a,1 , H. Mirandola b,∗ , F. Vitório a a Instituto de Matemática, Universidade Federal de Alagoas, Maceió, AL, CEP 57072-970, Brazil b Instituto de Matemática, Universidade Federal do Rio de Janeiro, Rio de Janeiro, RJ, CEP 21945-970, Brazil

Received 31 December 2016

Abstract Some of the most known integral inequalities are the Sobolev, Hardy and Rellich inequalities on regions in Euclidean spaces. In the context of submanifolds, the Sobolev inequality was proved by Michael–Simon [13] and Hoffman–Spruck [12]. Since then, a sort of applications to the submanifold theory has been derived from those inequalities. Years later, Carron [6] obtained a Hardy inequality for submanifolds in Hadamard spaces. In this paper, we prove the general Hardy and Rellich Inequalities for submanifolds in Hadamard spaces. Some applications are given and we also analyse the equality cases. © 2017 Elsevier Inc. All rights reserved. MSC: primary 53C42; secondary 53B25

1. Introduction Over the years, geometers have been interested in understanding how integral inequalities imply geometric and topological obstructions on Riemannian manifolds. Under this purpose, some integral inequalities lead us to study positive solutions to critical singular quasilinear elliptic problems, sharp constants, existence, non-existence, rigidity and symmetry results for extremal * Corresponding author.

E-mail addresses: [email protected] (M. Batista), [email protected] (H. Mirandola), [email protected] (F. Vitório). 1 Current address: Department of Mathematics, Princeton University, Princeton, NJ, 08540, USA. http://dx.doi.org/10.1016/j.jde.2017.07.003 0022-0396/© 2017 Elsevier Inc. All rights reserved.

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functions on subsets in the Euclidean space. About these subjects, a comprehensive material can be found, for instance, in [1,3,6,9,7,8,12–14] and references therein. In the literature, some of the most known integral inequalities are the Sobolev, Hardy and Rellich inequalities. The validity of these inequalities and their corresponding sharp constants on a given Riemannian manifold measures, in some suitable sense, how close that manifold is to an Euclidean space. These inequalities can be applied to obtain results such as comparison for the volume growth, estimates of the essential spectrum for the Schrödinger operators, parabolicity, among others properties (see, for instance, [15,10,19]). In this paper, we obtain the general Hardy and Rellich Inequalities for submanifolds in Hadamard ambient spaces using an elementary and very efficient approach. Furthermore, we analyse the equalities cases. At the end of this paper, we give some applications. This paper is organized as follows. In Section 2 below, we introduce our notation and show some simple facts about isometric immersions that will be useful in the proofs of the present paper. In Section 3 and 4, we prove Hardy and Rellich Inequalities for submanifolds in Hadamard spaces. In Section 5 we study the equality cases and do some blowup analysis to show that all the integral in our inequalities converge. The last section (Section 6) is devoted to some applications. 2. Preliminaries Let us start recalling some basic concepts, notations and properties about submanifolds. First, let M = M k be a k-dimensional Riemannian manifold with (possibly nonempty) boundary ∂M. ¯ n (that means M¯ Assume M is isometrically immersed in a n-dimensional Hadamard space M is a complete simply-connected manifold with non-positive sectional curvature). We will denote by f : M → M¯ the isometric immersion. By abuse of notation, sometimes, we will identify f (x) = x, for all x ∈ M. No restriction on the codimension of f is required. Let ·, · denote the Riemannian metric on M¯ and consider the same notation to the metric induced on M. Associated to these metrics, consider ∇¯ and ∇ the Levi-Civita connections on M¯ and M, respectively. It is well known that ∇Y Z = (∇¯ Y Z) , where  means the orthogonal projection onto the tangent bundle T M. The Gauss equation says ∇¯ Y Z = ∇Y Z + II(Y, Z), where II is a quadratic form named by second fundamental form. The mean curvature vector is defined by H = Tr M II. ¯ let rξ = d ¯ (· , ξ ) be the distance on M¯ from ξ . Since M¯ is a Hadamard space, Fixed ξ ∈ M, M by the Hessian comparison theorem (see Theorem 2.3 page 29 of [16]), we have 1 ¯ v2 ), Hess r (v, v) ≥ (1 − ∇r, r

(1)

¯ denotes for all points in M¯ ∗ = M¯ \ {ξ } and vector fields v : M¯ ∗ → T M¯ with |v| = 1. Here, ∇r ¯ of r. For a vector field Y : M → T M, ¯ the divergence of Y on M is the gradient vector, on M, given by divM Y =

k  ∇¯ ei Y, ei , i=1

where {e1 , · · · , ek } denotes a local orthonormal frame on M. By simple computations, one has

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Lemma 2.1. Let Y : M → T M¯ be a vector field and ψ ∈ C 1 (M). The following items hold (a) divM Y = divM Y  − H, Y ; (b) divM (ψY ) = ψ divM Y + ∇ M ψ, Y , where ∇ M ψ denotes the gradient vector, on M, of ψ. ¯ We see that X is Henceforth, we will denote by X : M → T M¯ the radial vector X = r ∇r. continuous, with |X| = r and differentiable in M \ {ξ }. The following lemma holds Lemma 2.2. Let γ and p be real numbers and ψ ∈ C 1 (M). The isometric immersion f : M → M¯ satisfies: divM (

ψ p X ψp γ |X ⊥ |2 ψ p−1 X M ) ≥ [k − γ + ] +  p∇ ψ + ψH,  |X|γ |X|γ |X|γ |X|2

(2)

and 1 k−γ |X ⊥ |2 X 1 + γ +  γ , H , ( γ −2 ) ≥ γ γ +2 2−γ |X| |X| |X| |X|

(3)

where (·)⊥ means the orthogonal projection on the normal bundle of M. Moreover, for both inequalities, (2) and (3), their corresponding equalities occur if and only if the radial curvature (Krad )ξ = 0 in supp (ψ) and M, respectively. Before to prove Lemma 2.2, let us recall the definition of radial sectional curvature. Let x ∈ M¯ and, since M¯ is complete, let γ : [0, t0 = rξ (x)] → M¯ be a minimizing geodesic in M¯ from ξ to x. For all orthonormal pair of vectors Y, Z ∈ Tx M¯ we define (K¯ rad )ξ (Y, Z) = ¯ R(Y, γ (t0 ))γ (t0 ), Z. Proof of Lemma 2.2. By the Hessian comparison theorem (1), divM X ≥ k and divM X = k if T and only if (Krad )ξ = 0 in M. Furthermore, since |X| = r, the gradient vector ∇ M |X| = X |X| . By Lemma 2.1 (b), and using that |X|2 = |X  |2 + |X ⊥ |2 , one has divM (

X 1 1 k−γ |X ⊥ |2 M ) = div X + ∇ ( ), X ≥ + γ . M |X|γ |X|γ |X|γ |X|γ |X|γ +2

Thus, by Lemma 2.1 (a), one has divM (

XT X X k−γ |X ⊥ |2 X ) = div ( ) + H,  ≥ + γ +  γ , H . M |X|γ |X|γ |X|γ |X|γ |X| |X|γ +2

Hence, using Lemma 2.1 (b), Item (2) follows. XT Now, since ∇ M ( |X|1γ −2 ) = −(γ − 2) |X|−(γ −2)−1 ∇ M (|X|) = −(γ − 2) |X| γ , one has T

X ( |X|1γ −2 ) = −(γ − 2) divM ( |X| γ ). Item (3) follows. Lemma 2.2 is proved.

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3. Hardy inequalities for submanifolds The classical Hardy’s Inequality [5] states that, for p > 1 and 0 ≤ f ∈ Lp (0, ∞), (p − 1)p pp

∞

F (x)p dx ≤ xp

0

∞ f (x)p dx. 0

Moreover, the constant (p − 1)p /p p is optimal and never attained, unless ψ = 0 everywhere. In the high-dimensional setting, the Hardy’s Inequality states that, for all compactly supported 0 ≤ ψ ∈ Cc1 (Rk ), with k ≥ 3, it holds 

(k − 2)2 4

ψ2 ≤ r2

Rk

 |∇ψ|2 .

(4)

Rk

2

Moreover, the constant (k−2) is sharp and never attained, in the sense that it is the infimum of 4 1 all 0 ≤ ψ ∈ Cc () and the equality in (4) holds if and only if ψ = 0 in Rk . For submanifolds, Carron [6] proved the following Hardy-type inequality: Theorem A (Carron). Let M k be a complete non-compact manifold, with k ≥ 3, isometrically ¯ Let r = d ¯ (· , ξ ) be the distance on M¯ from any fixed point immersed in a Hadamard space M. M ¯ ξ ∈ M. Then, for all compactly supported function ψ ∈ Cc1 (M), the following Hardy inequality holds: (k − 2)2 4



ψ2 ≤ r2

M





|∇ M ψ|2 +

(k − 2)|H |ψ 2  . 2r

M

¯ Theorem A For the special case that M is a minimal submanifold in Hadamard manifold M, was generalized by Bianchini, Mari and Rigoli [4], where they proved the constant (k − 2)2 /4 is not sharp if the ambient space has curvature less than a negative constant. Our first result is a Hardy-type inequality for submanifolds that generalizes Theorem A. As a consequence of Theorem 3.1 below, it follows that the equality in Carron’s Theorem occurs if and only if ψ = 0 in M. Theorem 3.1. Let M k be a compact manifold with (possibly nonempty) boundary ∂M. Assume ¯ and let r = d ¯ (· , ξ ) be the distance M is isometrically immersed in a Hadamard manifold M M in M¯ from any fixed point ξ . Let 1 ≤ p < ∞ and −∞ < γ < k. For all 0 ≤ ψ ∈ C 1 (M), it holds (k − γ )p pp



ψ p (k − γ )p−1 + rγ p p−1

M



ψp +  ≤ rγ

M

+

(k

− γ )p−1



p p−1 M



|∇ M ψ|p r γ −p

M

(k − γ )p−1  + rγ p p−1

ψp



 ∂M

ψp ¯ ∇r, ν, r γ −1

(5)

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¯ ⊥ |2 + r∇r, ¯ H . Here, + , − denote the positive and negative parts of , where  = γ |∇r respectively, and ν is the exterior conormal to ∂M. Moreover, if p > 1, the equality above holds if and only if ψ = 0 in M. To see how Theorem 3.1 generalizes Theorem A above, if γ ≥ 0 then ¯ ⊥ |2 + r∇r, ¯ H + ≥ γ |∇r ¯ ⊥ |2 + = γ |∇r ¯ H − ≤ r|∇r, ¯ H |. − = r∇r,   2 2 For p = γ = 2, it holds M ψr 2 − ≤ M ψr |H |, hence Theorem A follows from Theorem 3.1. Furthermore, as a consequence of Theorem 3.1, it holds Corollary 3.1. Let M k be a compact manifold with (possibly nonempty) boundary ∂M. Assume ¯ and let r = d ¯ (· , ξ ) be the distance M is isometrically immersed in a Hadamard manifold M M ¯ Let 1 ≤ p < ∞ and −∞ < γ < k. For all 0 ≤ ψ ∈ C 1 (M), in M¯ from a fixed point ξ ∈ M. it holds (k − γ )p pp



ψ p γ (k − γ )p−1 + rγ p p−1

M

+

(k

− γ )p−1



p p−1



ψp ¯ ⊥ 2 |∇r | ≤ rγ

M



|∇ M ψ|p r γ −p

M

(k − γ )p−1 ¯ |H, ∇r| + r γ −1 p p−1 ψp

M



ψp ¯ ∇r, ν. r γ −1

∂M

Moreover, if p > 1, the equality above holds if and only if ψ = 0 in M. It is worthwhile to observe that, in the present paper, we are assuming M compact with (possibly nonempty) boundary ∂M. Comparing with assumptions in Theorem A, if M were assumed complete and non-compact then, for a given ψ ∈ Cc1 (M), we could consider a compact manifold M with smooth boundary ∂M satisfying supp (ψ) ⊂ M ⊂ M. Thus, ψ ∈ Cc1 (M ), in particular, the gradient vector Dψ = 0 on ∂M . Here, there is no assumption on the gradient vector Dψ along ∂M. Proof of Corollary 3.1. If γ ≥ 0, Corollary 3.1 follows from Theorem 3.1 together with the  ψp ¯ ⊥ | and − ≤ r|∇r, ¯ H |. Now, if γ < 0 then facts + ≥ γ |∇r M r γ exists. Thus, since  = ¯ ⊥ |2 + r∇r, ¯ H , by Theorem 3.1, + − − = γ |∇r (k − γ )p pp



ψ p (k − γ )p−1 + rγ p p−1

M

 ≤

 ψp  ¯ ⊥ 2 ¯ H γ |∇r | + r∇r, γ r

M

|∇ M ψ|p r γ −p

M



(k − γ )p−1 + p p−1



ψp ¯ ∇r, ν. r γ −1

∂M

¯ H  ≤ r|∇r, ¯ H |. The result follows from the triangular inequality −r∇r, Now, we will prove Theorem 3.1.

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Proof of Theorem 3.1. For convenience, we will prove this theorem by considering the radial ¯ Set 0 ≤ ψ ∈ C 1 (M). vector field X = r ∇r. First, assume p > 1. Write γ = α + β + 1 with α, β ∈ R. By Lemma 2.2, divM (

ψ p X ψp γ |X ⊥ |2 ψ p−1 X M ) ≥ [k − γ + ] + p∇ ψ + ψH,  |X|γ |X|γ |X|γ |X|2 =

γ |X ⊥ |2 p∇ M ψ ψ p−1 X ψp ψp [k − γ + ]+ , + H, X. γ α 2 β+1 |X| |X| |X|γ |X| |X|

(6)

Moreover, if the equality in (6) holds then the radial curvature (Krad )ξ = 0 in supp ψ. Thus, using Cauchy–Schwarz Inequality and Young Inequality with > 0, divM (

 γ |X ⊥ |2 ψ p X ψp  k−γ + )≥ + H , X γ γ 2 |X| |X| |X| −

1 |p∇ M ψ|p q ψ (p−1)q − , p p |X|pα q |X|qβ

(7)

p where p > 1 and q = p−1 . Now, we take β satisfying βq = γ , that is, β = (p − 1)(α + 1). Since γ = α + β + 1, we see easily that pα = γ − p. Thus,

h( )

p  |X ⊥ |2  ψp p ψ γ + + H, X |X|γ |X|γ |X|2

≤ p p−1

|∇ M ψ|p ψ p X + p divM ( ), γ −p |X| |X|γ

q

where h( ) = p (k −γ − q ). The function h achieves its maximum at the instant = (k −γ ) with h( ) =

(k−γ )p . p

(8) p−1 p

,

Thus, it holds

(k − γ )p ψ p ψ p X (k − γ )p−1 ψ p  |∇ M ψ|p (k − γ )p−1 + ≤ + divM , p γ γ γ −p p−1 p−1 p |X| |X| |X| |X|γ p p

(9)

⊥ 2

| ¯ ⊥ |2 + r∇r, ¯ H . We observe that (9) remains valid for where  = γ |X + H, X = γ |∇r |X|2 p = 1, just considering (9) with p > 1 and taking p → 1.   p  ψp We claim that M divM ( ψ|X|Xγ ) = ∂M |X| / M, γ X, ν holds even when 0 ∈ M. In fact, if 0 ∈ we can apply the divergence theorem directly. If 0 ∈ M, we consider a small regular value r0 > 0 for the distance function r = |X|. We have

 divM ( M∩[|X|>r0 ]



ψ p X )= |X|γ

[∂M∩[|X|>r0 ]]∪[M∩[|X|=r0 ]]



= [∂M∩[|X|>r0 ]]

ψp X, ν |X|γ

ψp 1−γ X, ν + O(r0 )O(r0k−1 ), |X|γ

(10)

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 p as r0 → 0. Thus, since k − γ > 0 e ∂M |X|ψγ −1 exists (see (28) in Section 5), by the dominated convergence theorem, taking r0 → 0, our claim follows. Using (9) and (10), (k − γ )p pp

 M

+

ψp (k − γ )p−1 + |X|γ p p−1 (k − γ )p−1 p p−1

 M

 M

ψp +  ≤ |X|γ

 M

|∇ M ψ|p |X|γ −p

ψ p − (k − γ )p−1  + |X|γ p p−1

 ∂M

ψp X, ν, |X|γ

(11)

where + and − denotes the positive and negative parts of  and ν is the outward pointing unit normal field of the boundary ∂M. Now, assume the equality in (11) holds. We will prove that ψ = 0 in M. In fact, the equality in (11) implies the radial sectional curvature (K¯ rad )ξ = 0 in supp ψ and also imply the equalities in the Cauchy and Young Inequalities, with q = k − γ , holds in (7). Thus, we obtain ∇ M ψ = −λX, and q ψ (p−1)q p p |∇ M ψ|p = , p αp |X| |X|βq

(12) (13)

where λ ≥ 0 is a continuous function on M, and the constants α and β satisfy αp = γ − p and βq = γ . Since p+q = pq = (k − γ )p , by (13), (k − γ )p ψ p |∇ M ψ|p = . pp |X|γ |X|γ −p

(14)

Hence, from (12), ∇M ψ = −

k−γ ψX. p

(15)

In particular, it holds X T = X and X = 12 ∇ M (|X|2 ). By contradiction, assume that the open subset W = {ψ > 0} is nonempty. Using (15) and the fact X = 12 ∇ M (|X|2 ), one has ∇ M (log ψ + k−γ 2 2p |X| ) = 0, in W . Hence, ψ = De

2 − k−γ 2p |X|

,

(16)

in W , for some constant D > 0. This implies ψ > 0 everywhere in M. Moreover, using that X ⊥ = 0, by (9) (with the equality case) and (14), one also obtains divM (

ψ p XT ) = 0, |X|γ

(17)

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k−γ

everywhere in M. On the other hand, since ψ p = D p e− 2 |X| , and again using X T = X, it follows by (2) in Lemma 2.2 (with equality since the radial curvature (Krad )ξ = 0 in M), and (15), we obtain divM (

2

ψ p XT 1 )= [(k − γ )ψ p + pψ p−1 ∇ M ψ, X] γ |X| |X|γ =

(k − γ )ψ p [1 − |X|2 ]. |X|γ

Thus, by (17), one has ψ p (1 − |X|2 ) = 0 everywhere in M, which is a contradiction, since |X|2 = 1 would imply X T = 0 in M, hence X = 0 everywhere in M. Therefore, ψ = 0 in M. Theorem 3.1 is proved. 2 It is straightforward to verify that with the same steps as in the proof of Theorem 3.1 above, p−1 but applying in (6) Cauchy–Schwarz and Young Inequalities directly to p∇ M ψ + ψH, ψ|X|γX  p−1

instead p∇ M ψ, ψ|X|γX , one can prove the following variance of Theorem 3.1. Theorem 3.2. Under the same hypothesis of Theorem 3.1, for all 0 ≤ ψ ∈ C 1 (M), it holds (k − γ )p pp



ψ p γ (k − γ )p−1 + rγ p p−1

M

 ≤



ψp ¯ ⊥ 2 |∇r | rγ

M

1  M ψH p (k − γ )p−1 ∇ ψ+ + γ −p r p p p−1

M



ψp . r γ −1

∂M

Moreover, if p > 1, the equality holds if and only if ψ = 0 in M. 4. Rellich inequalities for submanifolds The original Rellich’s Inequality [17] states that, for all 0 ≤ ψ ∈ Cc1 (Rk ), with k ≥ 5, it holds k 2 (k − 4)2 16

 Rk

ψ2 ≤ |x|4

 |ψ|2 . Rk

This inequality was firstly published in the proceedings of the Amsterdam ICM in 1954 (see [17]) but proved earlier in the lectures at New York University in 1953 published posthumously in [18]. It is much less studied than Hardy’s inequality. A more general version of the Rellich’s Inequality was proved by Davies and Hinz [11], p (k − γ )p  γ − 2 + (p − 1)(k − 2) 2p p

 Rk

ψp ≤ |x|γ

 Rk

|ψ|p , |x|γ −2p

(18)

for all 0 ≤ ψ ∈ Cc1 (Rk ), with k > γ > 2. Moreover, the constant in (18) is sharp and never attained, unless ψ = 0 in Rk .

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Here, we present the first version of the Rellich Inequality for submanifolds. We obtain the following. Theorem 4.1. Let M k be a compact manifold with (possibly nonempty) boundary ∂M. Assume ¯ Let r = d ¯ (· , ξ ) be the distance M is isometrically immersed in a Hadamard manifold M. M ¯ Let p ≥ 1 and 2 < γ < k. For the special case that M¯ has in M¯ from a fixed point ξ ∈ M. radial curvature (K¯ rad )ξ = 0 (e.g., M¯ = Rn ), we relax the hypothesis about γ by assuming 2 − (p − 1)(k − 2) < γ < k. For all 0 ≤ ψ ∈ C 1 (M), it holds:      ψp ψp + |ψ|p ψp − ψ p−1 A + B  ≤ + B  + W, ν. (19) rγ rγ rγ r γ −2p r γ −2 M

M

M

M

∂M

¯ ⊥ |2 + r∇r, ¯ H , and Here  = γ |∇r p Ep (k − γ )p  γ − 2 + (p − 1)(k − 2) > 0, = p 2p p p p−1  E E p − 1 B = B(k, γ , p) = p−1 + > 0, and k−γ p p ¯ T E p−1 ψ ∇r W = W (k, γ , p) = p−2 ∇ M ψ + B , r p A = A(k, γ , p) =

with E=

 k−γ  γ − 2 + (p − 1)(k − 2) > 0. p

Furthermore, if p > 1, the equality in (19) holds if and only if ψ = 0 in M. A special case occurs when γ = 2p. In this case, it holds Corollary 4.1. Let M be a compact manifold with (possibly nonempty) boundary ∂M. Assume ¯ and let r = d ¯ (· , ξ ) be the distance in M is isometrically immersed in a Hadamard manifold M M ¯ M from a fixed point ξ . Assume k > 2p, for some p > 1. Then, for all 0 ≤ ψ ∈ C 1 (M), it holds      ψp ψp + ψp − ψ p−1 p A + B  ≤ |ψ| + B  + W, ν. r 2p r 2p r 2p r 2(p−1) M

M

M

M

∂M

¯ ⊥ |2 + r∇r, ¯ H , and Here,  = 2p|∇r kp (k − 2p)p (p − 1)p ; p 2p k p−1 B = 2p−1 (k − 2p)p−1 (p − 1)p (k + 1); and p ¯  (k − 2p)p−1 k p−1 (p − 1)p−1  M (p − 1)(k + 1) ψ ∇r W= ∇ ψ+ . 2p−3 2 r p p A=

Moreover, the equality holds if and only if ψ = 0 everywhere in M.

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Now, we will prove Theorem 4.1. Proof of Theorem 4.1. As in the proof of Theorem 3.1, for convenience, we will prove this ¯ By Lemma 2.2, theorem by considering the radial vector field X = r ∇r.  ∇M ψp  ψ p 1  p 1 1 p M = ( ) + div − ψ ∇ ( ) ψ M (2 − γ )|X|γ −2 2 − γ |X|γ −2 |X|γ −2 |X|γ −2 ≥ where Y = we obtain

∇M ψp |X|γ −2

k − γ  1 |X ⊥ |2 X +γ +  γ ,H , divM Y + ψ p γ γ +2 2−γ |X| |X| |X|

+ (γ −2)ψ |X|γ

p XT

. Using that γ > 2 and ψ p = pψ p−2 [ψψ + (p − 1)|∇ M ψ|2 ],

pψ p−1 ψ p(p − 1)ψ p−2 |∇ M ψ|2 + ≤ divM Y |X|γ −2 |X|γ −2 − (γ − 2)

 ψp  γ |X ⊥ |2 k − γ + + X, H  . |X|γ |X|2

(20)

Furthermore, the equality in (20) also remains valid when γ = 2 or when the radial curvature (Krad )ξ = 0. Henceforth, we will assume γ ≥ 2 or (Krad )ξ = 0. So, (20) holds. We write γ = α + β + 1, hence γ − 2 = (α − 1) + β. Using Young Inequality with σ > 0, we have pψ

ψ p−1 1 |pψ|p σ q ψ (p−1)q ≥ − − , pσ p |X|(α−1)p q |X|βq |X|γ −2

p where q = p−1 . As in the proof of Theorem 3.1, we take βq = γ . We obtain β = (p − 1)(α + 1) and p(α − 1) = γ − 2p. Thus,

pψ p−1 ψ 1 |pψ|p σ q ψ p ≥ − − . pσ p |X|γ −2p q |X|γ |X|γ −2

(21)

Applying (21) into (20), one has 

(γ − 2)(k − γ ) −

M 2 σ q  ψp p−2 |∇ ψ| + p(p − 1)ψ q |X|γ |X|γ −2

≤ divY +

 p p−1 |ψ|p ψ p  γ |X ⊥ |2 − (γ − 2) + X, H  . σ p |X|γ −2p |X|γ |X|2

(22)

On the other hand, we write p = a + b + 1. By Lemma 2.2, divM (

ψ p X γ |X ⊥ |2 ψ p−1 X p k−γ M ) ≥ ψ [ + ] + p∇ ψ + ψH,  |X|γ |X|γ |X|γ |X|γ +2 = ψp[

k−γ γ |X ⊥ |2 pψ a ∇ M ψ ψ b X ψp + ] +  ,  + H, X. |X|γ |X|α |X|γ |X|γ +2 |X|β+1

(23)

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By Cauchy–Schwarz Inequality and Young Inequality with > 0, 

pψ a ∇ M ψ ψ b X p 2 ψ 2a |∇ M ψ|2 2 ψ 2b ,  ≥ − − . |X|α 2 |X|2β |X|β+1 2 2 |X|2α

(24)

We take 2a = p − 2, 2b = p, 2β = γ and 2α = γ − 2. By (23) and (24), divM (

ψ p X ψp γ |X ⊥ |2 ψp  ) ≥ [k − γ + ] + H , X |X|γ |X|γ |X|γ |X|2 −

p 2 ψ p−2 |∇ M ψ|2 2 ψ p − . 2 |X|γ 2 2 |X|γ −2

Thus, p 2 ψ p−2 |∇ M ψ|2 h( )ψ p 2 ψ p 2 ψ p X ≥ +  − div ( ), M 2 |X|γ |X|γ |X|γ |X|γ −2 where  = obtain

γ |X ⊥ |2 |X|2

+ X, H  and h( ) = 2 (k − γ −

A

2 2 ).

Plotting this inequality into (22), we

ψp ψp |ψ|p +B  ≤ divM Z + , γ γ |X| |X| |X|γ −2p

(25)

where A = A(σ, ) =

σq σp  2(p − 1)h( )  (γ − 2)(k − γ ) − + ; p−1 q p p

B = B(σ, ) =

2(p − 1) 2  σp  γ − 2 + ; p p p−1

Z = Z(σ, ) =

σp  2(p − 1) 2 ψ p X T  Y + p |X|γ p p−1

=

2(p − 1) 2 ψ p X T  σ p  ∇M ψp

. + γ − 2 + p |X|γ p p−1 |X|γ −2

In order to maximize A(σ, ), note that k − γ , with h( ) = A=

(k−γ )2 2 .

∂A ∂

= 0 if and only if h ( ) = 0, which gives 2 =

Hence,

σp p−1 σq σq σp 2 [(γ − 2)(k − γ ) + − [E − (k − γ ) ] = ], p q q p p−1 p p−1

where E = (k − γ )[γ − 2 +

p−1 (k − γ ) (k − γ )] = [γ − 2 + (p − 1)(k − 2)]. p p

(26)

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q Thus, ∂A ∂σ = 0 if and only if σ = E. Note that if k − γ > 0 then E > 0 if and only if γ > p σ p+q σ pq Ep 2 − (k − 2)(p − 1). So, by hypothesis, E > 0. We obtain A = pσp−1 E p = p p = p p = p p . It is easy to see that max A(σ, ) = E p /p. Hence σ,

A=

p Ep (k − γ )p  γ − 2 + (p − 1)(k − 2) . = p 2p p p

Furthermore, B=

2(p − 1)(k − γ )  E p−1  E E p−1  p − 1 γ −2+ = p−1 + > 0; p−1 p k−γ p p p

Z=

2(p − 1)(k − γ )  ψ p X T  E p−1  pψ p−1 ∇ M ψ  + γ −2+ p−1 γ −2 p |X|γ p |X|

=

ψ p−1  E p−1 M ψ p−1 ψX T  = ∇ ψ + B W, |X|γ −2 p p−2 |X|2 |X|γ −2 T

p−1

where W = E ∇ M ψ + B ψX . p p−2 |X|2 As in the proof of Theorem 3.1, for a small regular value r0 of the geodesic distance r = |X| satisfies   divM Z = Z, ν M∩[|X|>r0 ]

[∂M∩[|X|>r0 ]]∪[M∩[|X|=r0 ]]



 2−γ 1−γ  Z, ν + O(r0 ) + O(r0 ) O(r0k−1 )

= [∂M∩[|X|>r0 ]]



k−γ

Z, ν + O(r0

=

)

[∂M∩[|X|>r0 ]]

   So, one has M divM Z = limr0 →0 M∩[|X|>r0 ] divM Z = ∂M Z, ν. Therefore, using (25),  A M

ψp +B |X|γ

 M

ψp +  ≤ |X|γ

 M

|ψ|p +B |X|γ −2p

 M

ψp −  + |X|γ

 Z, ν.

∂M

Furthermore, if the equality holds, then (Krad )ξ = 0 in supp ψ and the equality holds in (24) with 2 = k − γ . Thus, it holds that ∇ M ψ = − k−γ p ψX. Following the same arguments in the proof of Theorem 3.1 (after equation (15)) we will obtain ψ = 0. Theorem 4.1 is proved. 2 There is some versions for Hardy and Rellich inequalities by considering r = dM¯ (· , K) as ¯ instead {ξ } (see, for instance, [2]). For this kind of distance, the distance from a subset K ⊂ M, the main difference of our computations is in Lemma 2.2, where making X = r∇r, it holds divM X ≥ c(K), where c(K) is a constant depending on the geometry of K.

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13

5. Equality cases and blowup analysis Let us do a blowup analysis near the point ξ in order to show that the integrals in (5) converge. Note that if ξ ∈ / M or γ ≤ 0, then r1γ ∈ C 1 (M) hence all the integrals in (5) and (19) converge. ¯ for a small Assume γ > 0 and ξ ∈ M. Since M is compact and isometrically immersed in M, r0 > 0, the intersection M ∩ [r ≤ r0 ] contains only a finite number of connected components, ¯ T | ≥ a > 0 at M ∩ all of them containing ξ at their images. Furthermore, we may assume |∇r [r ≤ r0 ]. In particular, all 0 < s < r0 are regular values for r. By the coarea formula,  M∩[r≤r0 ]

1 1 = rγ a

 M∩[r≤r0 ]

r0 =

1 1 |∇r T | = rγ a

r0

0

dσ 2 ds M∩[r=s]

0

1 O(s k−1 )ds = O(1) sγ



1 sγ

r0

k−γ

s k−1−γ ds = O(r0

).

(27)

0

The last term follows from the fact that k − γ > 0. By similar arguments, since ∂M is also an ¯ isometrically immersed submanifold in M,  ∂M∩[r≤r0 ]

1 k−γ = O(r0 ). r γ −1

(28)

In this section, we will analyse functions that satisfy the equalities in Theorems 3.1 and 4.1, for the case p = 1. We recall that, for both theorems, a function that satisfies the equality for p > 1 must be identically null. For p = 1, the same conclusion fails, however, interesting conclusions are derived. Let BR ⊂ Rn be a ball of radius R and centre at some point ξ ∈ Rn and let r be the distance function from ξ . We will prove the following Proposition 5.1. Let M k be a compact submanifold with boundary ∂M immersed in Rn . Assume ∂M is contained in the hypersphere ∂BR = r −1 (R). For 0 ≤ ψ ∈ C 1 (M), the equality in Theorem 3.1, with p = 1 and −∞ < γ < k, occurs if and only if ψ = ψ(r), i.e., it depends only on r, and ψ (r) ≤ 0. Proof. For all 0 ≤ ψ ∈ C 1 (M), it holds  (k − γ )

ψ + rγ

M



ψ ≤ rγ

M

 M

|∇ M ψ| + r γ −1



ψ r γ −1

¯ ν, ∇r,

∂M

¯ ⊥ |2 + r∇r, ¯ H . where  = γ |∇r As in (32),  ψ ∂M

 ¯ T ¯ ψp ∇r ∇r , ν = [k − γ + ] + ∇ M ψ, γ −1 . γ γ −1 r r r M

(29)

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We obtain 

¯ |∇ M ψ| ∇r M + ∇ ψ,  ≥ 0. r γ −1 r γ −1

M

¯ hence ψ = ψ(r), i.e., deThus, the equality in (29) holds if and only if ∇ M ψ = −|∇ M ψ|∇r, pends only on r, and ψ (r) ≤ 0. 2 Now, we will analyse Theorem 4.1 for the case p = 1. Proposition 5.2. Let M k be a compact submanifold with boundary ∂M immersed in Rn . Assume ∂M is contained in the hypersphere ∂BR = r −1 (R). For 0 ≤ ψ ∈ C 1 (M), the equality in Theorem 4.1, with p = 1 and 2 < γ < k, occurs if and only if ψ ≤ 0. Proof. Let 0 ≤ ψ ∈ C 1 (M). By Theorem 4.1 with p = 1 and 2 < γ < k,   ψ ψ (γ − 2)(k − γ ) + (γ − 2)  γ r rγ M

 ≤

M

|ψ| + r γ −2

M

Since r = R in ∂M, 

1

r



1 r γ −2

∇ M ψ + (γ − 2)ψ

¯ T ∇r , ν. r

∂M

∇ M ψ, ν = γ −2



1

∇ M ψ, ν =

R γ −2

∂M

1

 ψ,

R γ −2

∂M

∂M



= M



=

(31)

M

¯ and, by Lemma 2.2 (using that X = r ∇r),    ¯ T ∇r XT XT ψ γ −1 , ν = ψ , ν = divM (ψ ) γ |X| |X|γ r ∂M

(30)

(32)

M

ψ γ |X ⊥ |2 X [k − γ + ] + ∇ M ψ + ψH,  γ 2 |X| |X|γ |X| ¯ ψp ∇r M [k − γ + ] + ∇ ψ, . rγ r γ −1

M

Furthermore,  ∇ M ψ, M

 ¯ ∇r 1 ¯ 2−γ   = − ∇ M ψ, ∇r γ −2 r γ −1 M

=−

1 γ −2



[div(r 2−γ ∇ M ψ) − r 2−γ ψ] M

(33)

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=−

=−

R 2−γ γ −2 R 2−γ γ −2

 ∇ M ψ, ν + ∂M



ψ + M

1 γ −2

1 γ −2 

15

 r 2−γ ψ M

r 2−γ ψ. M

By (31), (32) and (33), we obtain  0≤



1

r

(|ψ| + ψ) = γ −2

M

2 r γ −2

(ψ)+

M

Thus, using that |ψ| = (ψ)+ + (ψ)− and ψ = (ψ)+ − (ψ)− , the equality in (30) holds if and only if ψ ≤ 0 in M. 2 6. Applications We recall that an isometric immersion f : M → Rn is said to be self-shrinker if the mean curvature vector satisfies 1 H = − f ⊥. 2 ¯ coincides with f , in particular the support function In this case, the radial vector X = r ∇r r ¯ ⊥ 2 ¯ H, ∇r = − 2 |∇r | . By Theorem 3.1, it follows Proposition 6.1. Let M k be a closed manifold isometrically immersed in a Hadamard mani¯ Let r = d ¯ (· , ξ ) be the distance in M¯ from any fixed point ξ . Assume the mean curvature fold M. M of M satisfies ¯ ≥ − λr |∇r ¯ ⊥ |2 , H, ∇r 2

(34)

for some λ > 0. Then, supM r 2 ≥ 2k/λ. Proof. By contradiction, we assume supM r 2 < 2k/λ and choose γ < k so that supM r 2 < 2γ /λ. Thus, ¯ ⊥ |2 + r∇r, ¯ H  ≥ (γ −  = γ |∇r

λr 2 ¯ ⊥ 2 )|∇r | > 0. 2

So, − = 0. Applying Theorem 3.1 with ψ = 1, we obtain a contradiction. 2 We√remark that Proposition√6.1 is optimal. In fact, the round k-dimensional hypersphere S k (0, 2k) ⊂ Rk+1 of radius 2k and centred at the origin is self-shrinker. So, the equality in (34) holds with λ = 1 and r 2 = 2k. We define the extrinsic diameter D(M) of a submanifold M in M¯ as the infimum of the ¯ that contains M. diameter among all geodesic ball in M

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Theorem 6.1. Let M k be a compact manifold with boundary ∂M = ∅ isometrically immersed in a Hadamard manifold M¯ and let D = D(M) be the extrinsic diameter of M. Then, the first eigenvalue of the Laplacian of M satisfies λ1 (M) ≥

k2 D (1 − sup |H |). k M D2

Proof. Let 0 ≤ ψ ∈ C 1 (M), with ψ ≡ 0 and ψ = 0 on ∂M. Using Theorem 3.1, with γ = 0 and p = 2, we obtain k2 4

 ψ2 + M

k 2



 ψ 2 ≤

M

r 2 |∇ M ψ|2 ≤

D2 4

M

 |∇ M ψ|2 . M

¯ H  ≥ − D supM |H |. Hence, Note that, in this case,  = r∇r, 2 k2 4

 ψ2 − M

Since λ1 (M) = infF on ∂M. We obtain



M|∇ M

kD sup |H | 4 M

M ψ|2

ψ2



 ψ2 ≤

M

r 2 |∇ M ψ|2 ≤ M

D2 4

 |∇ M ψ|2 . M

, where F is the set of all 0 ≤ ψ ∈ C 1 (M), with ψ ≡ 0 and ψ = 0

k 2 kD D2 − sup |H | ≤ λ1 (M). 4 4 M 4 Theorem 6.1 follows.

2

Acknowledgments The authors were partially supported by CNPq and M.B. was partially supported by CAPES. The second author would like to thank Wladimir Neves and Aldo Bazan for introducing him to the subject. The authors also would like to thank Frederico Girão for his kind hospitality at Universidade Federal do Ceará where the main part of this manuscript was written. References [1] M. Badiale, G. Tarantello, A Sobolev–Hardy inequality with applications to a nonlinear elliptic equation arising in astrophysics, Arch. Ration. Mech. Anal. 163 (4) (2002) 259–293. [2] G. Barbatis, Best constants for higher-order Rellich inequalities in Lp (), Math. Z. 255 (2007) 877–896. [3] A. Bazan, W. Neves, A scaling approach to Caffarelli–Kohn–Nirenberg inequality, preprint, arXiv:1314.1823. [4] B. Bianchini, L. Mari, M. Rigoli, Yamabe type equations with a sign-changing nonlinearity, and the prescribed curvature problem, J. Differential Equations 260 (10) (2016) 7416–7497. [5] T. Calerman, G.H. Hardy, Fourier series and analytic functions, Proc. Royal Ser. A 101 (1922) 124–133. [6] G. Carron, Inégalités de Hardy sur les variétés riemanniennes non-compactes, J. Math. Pures Appl. (9) 76 (10) (1997) 883–891. [7] L. Caffarelli, R. Kohn, L. Nirenberg, First order interpolation inequalities with weights, Compos. Math. 53 (3) (1984) 259–275.

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