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Eigenvalue pinching and application to the stability and the almost umbilicity of hypersurfaces
Jean-Francois Grosjean, Julien Roth
To cite this version:
Jean-Francois Grosjean, Julien Roth. Eigenvalue pinching and application to the stability and the
almost umbilicity of hypersurfaces. Mathematische Zeitschrift, Springer, 2012, 271 (no. 1-2), pp.469-
488. �hal-00170114v3�
THE ALMOST UMBILICITY OF HYPERSURFACES
J.-F. GROSJEAN, J.ROTH
Abstract. In this paper we give pinching theorems for the first nonzero eigenvalue of the Laplacian on compact hypersurfaces of ambient spaces with bounded sectional curvature. As an application we deduce a rigidity result for stable constant mean curvature hypersurfacesM of these spaces N. Indeed, we prove that if M is included in a ball of radius small enough then the Hausdorff-distance betweenM and a geodesic sphereS ofN is small. MoreoverM is diffeomorphic and quasi-isometric toS. As other application, we obtain rigidity results for almost umbilic hypersurfaces.
Contents
1. Introduction 1
2. Preliminaries 6
3. An L
2-approach 8
4. Proof of the diffeomorphism 12
5. Application to the stability 17
6. Application to the almost umbilic hypersurfaces 19
References 20
1. Introduction
One way to show that the geodesic spheres are the only stable constant mean curvature hypersurfaces of classical space forms (i.e. Euclidean space, spherical space and hyperbolic space) is to prove that there is equality in the well-known Reilly’s inequality. One of the main points of the present paper is to obtain new stability results for hypersufaces immersed in more general ambient spaces by using a Reilly’s inequality proved by Heintze ([9]). Note that the isoperimetric problem is a particular case of these stability results proved here. Indeed compact stable constant mean curvature hypersurfaces bounding a domain appear as solution of the isoperimetric problem. We know that solutions of this problem exist on any compact Riemannian manifold and are smooth possibly up to a singular set of codimension at least 8 (see theorem 1 of [18], see also [13] and [15]). Moreover, in any dimension, smooth solutions exist in a neighborhood of non-degenerate critical point of the scalar curvature ([24]).
First, let us recall Reilly’s inequality. Let (M
m, g) be a compact, connected and oriented m-dimensional Riemannian manifold without boundary isometrically
Date: 14th February 2011.
2000Mathematics Subject Classification. 53A07, 53C21.
Key words and phrases. Spectrum, Laplacian, pinching results, hypersurfaces.
1
immersed by φ in the simply connected space form N
n+1(c) (c = 0, 1 , − 1 respect- ively for Euclidean space, sphere or hyperbolic space). Reilly’s inequality gives an extrinsic upper bound for the first nonzero eigenvalue λ
1(M ) of the Laplacian of (M
m, g) in term of the square of the length of the mean curvature H. Indeed we have
λ
1(M ) 6 m V (M )
Z
M
( | H |
2+ c)dv (1)
where dv and V (M ) denote respectively the Riemannian volume element and the volume of (M
m, g ). Moreover in the case of hypersurfaces (i.e. m = n), equality holds if and only if (M
n, g) is immersed as a geodesic sphere of N
n+1(c). For c = 0 this inequality was proved by Reilly ([16]) and can easily be extended to the spherical case c = 1 by considering the canonical embedding of S
nin R
n+1. For c = − 1 it has been proved by El Soufi and Ilias in [8].
In the sequel we will consider a weaker inequality due to Heintze ([9]) which generalizes the previous one for the case where (M
m, g) is isometrically immersed by φ in a (n+1)-dimensional Riemannian manifold (N
n+1, h) whose sectional curvature K
Nis bounded above by δ. Indeed if φ(M ) lies in a convex ball and if the radius of this ball is
4√πδin the case δ > 0, we have
λ
1(M ) 6 m( k H k
2∞+ δ) (2)
where k H k
∞denotes the L
∞-norm of the mean curvature. Now for m = n if we assume that K
Nis bounded below by µ and M has a constant mean curvature H and is stable (see section 5) we have
n(H
2+ µ) 6 λ
1(M) 6 n(H
2+ δ)
Consequently we see that if N is not of constant sectional curvature we can’t con- clude as in the case of space forms. However, the above inequality is a kind of pinching on Reilly’s inequality, that is a condition of almost equality. Such con- ditions have been studied for Reilly’s inequality in Euclidean space in [7]. In the present paper we will generalize the results of [7] to the inequality (2) for hypersur- faces (i.e. m = n) of ambient spaces with non constant sectional curvature. That amounts to finding conditions on geometric invariants so that if we have
(Λ
ε) n( k H k
2∞+ δ) < λ
1(M )(1 + ε) then M is close to a sphere in a certain sense.
This problem is a particular case of a pinching concerning the moment of inertia J
p(M) of M with respect to a point p. It is defined by
J
p(M) := k X k
2where X
x:= s
δ(r(x)) ∇
Nr |
x, r(x) is the geodesic distance between x and p, k · k
qis the L
q-norm on C
∞(M ) defined by k f k
qq=
V(M)1Z
M
| f |
qdv and s
δis the function
defined by
s
δ(r) =
√1 δ
sin √
δr if δ > 0
r if δ = 0
√
1|δ|
sinh p
| δ | r if δ < 0, The invariant J
p(M ) satisfies the inequality
1 6 ( k H k
2∞+ δ)J
p(M)
2(3)
where φ(M ) is contained in the ball of center p and of radius
4√πδif δ > 0. We associate to this inequality the pinching
(I
p,ε) ( k H k
2∞+ δ)J
p(M )
26 1 + ε
If (Λ
ε) holds and ε is small enough (ε < 1/4) then for the center of mass p
0of M, (I
p0,6ε) is satisfied (see proposition 2.1).
In fact, the pinching (I
p,ε) is more general than this one concerning the extrinsic radius R(M ) defined as the radius of the smallest ball containing φ(M). We recall that we have the following lower bound of the radius (see [3] for instance)
s
δ(R(M))
c
δ(R(M )) > 1 k H k
∞(4)
where c
δ= s
′δor equivalently
1 6 ( k H k
2∞+ δ)s
δ(R(M))
2(5)
In the case of hypersurfaces of the space form of curvature δ, equality in (5) cha- racterizes geodesic spheres. The associated pinching
(R
ε) ( k H k
2∞+ δ)s
δ(R(M))
26 (1 + ε).
has been treated for hypersurfaces of ambient spaces with constant sectional curvature in [19]. It is easy to see that if (R
ε) holds, then (I
p0,ε) is satisfied for the center p
0of the ball of radius R(M ) containing φ(M).
Before giving the main theorems, we set some notations which will be more convenient. Throughout the paper, we will let h = ( k H k
2∞+ δ)
1/2and use B to denote the second fundamental form. Moreover we will let B(p, R) the geodesic ball in N of center p and radius R.
We will need two hypotheses on the volume of M and on the injectivity radius i(N ) coming from hypotheses assumed in a result on a Sobolev inequality due to Hoffman and Spruck ([10] and [11]). Indeed we will assume that i(N ) >
√πδ
if δ > 0 and we will consider H
V(n, N ) the space of all Riemannian compact, connected and oriented n-dimensional Riemannian manifolds without boundary isometrically immersed by φ in (N
n+1, h) which satisfy the following hypothesis on the volume : V (M ) 6
cωnδn/2
if δ > 0 and V (M) 6 cω
ni(N )
nif δ 6 0 for some constant c.
For convenience we take 1/ √
δ = + ∞ if δ 6 0.
Let us state the first main theorem.
Theorem 1.1. Let (N
n+1, h) be a n + 1-dimensional Riemannian manifold whose sectional curvature K
Nsatisfies µ 6 K
N6 δ and i(N) >
√πδ
if δ > 0. Let M ∈ H
V(n, N ) and p be a point of N such that φ(M) ⊂ B
p, min
π 4√
δ
, i(N )
. Let
ε < 1, q > n and A > 0. Let us assume that max(V (M )
1/nk H k
∞, V (M )
1/nk B k
q) 6 A for δ > 0 (resp. max(V (M )
1/nk H k
∞,
kHhk∞, V (M )
1/nk B k
q) 6 A for δ < 0).
Then there exist positive constants C := C(n, q, A), α := α(q, n) such that if (I
p,ε) holds, ε
α< 1/C and φ(M ) is contained in the ball B
p, s
−δ1q
ε δ−µ
then
d
Hφ(M ), S
p, s
−δ11
h
6 C h ε
αwhere d
Hdenotes the Hausdorff distance. Moreover M is diffeomorphic and ε
α- quasi-isometric to S(p, s
−δ1 1h). Namely there exists a diffeomorphism from M into S(p, s
−δ1 1h) so that
| dF
x(u) |
2− 1
6 Cε
αfor any x ∈ M, u ∈ T
xM and | u | = 1.
We recall that the Hausdorff distance between two compact subsets A and B of a metric space is given by
d
H(A, B) = inf { A ⊂ V
η(B) and B ⊂ V
η(A) }
where for any subset A, V
η(A) is the tubular neighborhood of A defined by V
η(A) = { x | d(x, A) < η } .
As in the euclidean case (see [7]) for the pinching of λ
1(M ) or as in the hyperbolic case or spherical case ([19]) for the pinching of extrinsic radius we can obtain the Hausdorff proximity strictly with a dependence on k H k
∞.
Obviously we have the following corollary
Corollary 1.1. Let (N
n+1, h) be a n + 1-dimensional Riemannian manifold whose sectional curvature K
Nsatisfies µ 6 K
N6 δ and i(N ) >
√πδ
if δ > 0. Let M ∈ H
V(n, N ). Let us assume that φ(M ) lies in a convex ball of radius min
π 8√
δ
,
i(N)2. Let p
0be the center of mass of M . Let ε < 1/6 , q > n and A > 0. Let us assume that max(V (M )
1/nk H k
∞, V (M )
1/nk B k
q) 6 A for δ > 0 (resp. max(V (M )
1/nk H k
∞,
kHhk∞, V (M )
1/nk B k
q) 6 A for δ < 0).
Then there exist positive constants C := C(n, q, A), α := α(q, n) such that if (Λ
ε) holds, ε
α< 1/C and φ(M ) is contained in the ball B
p
0, s
−δ1q
ε δ−µ
then
d
Hφ(M ), S
p
0, s
−δ11
h
6 C h ε
αand M is diffeomorphic and ε
α-quasi-isometric to S(p
0, s
−δ1 h1).
Theorem 1.1 allows to obtain an application for the stable constant mean curvature hypersurfaces. Indeed we have the following stability theorem
Theorem 1.2. Let (N
n+1, h) be a n + 1-dimensional Riemannian manifold whose sectional curvature K
Nsatisfies µ 6 K
N6 δ and i(N ) >
√πδ
if δ > 0 and let M ∈ H
V(n, N ). Let us assume that φ(M ) lies in a convex ball of radius min
π 8√
δ
,
i(N2).
Let p
0be the center of mass of M. Let ε < 1/6, q > n and A > 0. Then there
exist positive constants C := C(n, q, A), α := α(q, n) and R(δ, µ, ε) such that if φ is of constant mean curvature H and stable, V (M )
1/nk B k
q6 A for δ > 0 (resp.
max(
Hh, V (M )
1/nk B k
q) 6 A for δ < 0), ε
α< 1/C and φ(M ) is contained in a convex ball of radius
12s
−δ1q
ε 2(δ−µ)
then
d
Hφ(M ), S
p
0, s
−δ11
h
6 C h ε
αand M is diffeomorphic and ε
α-quasi-isometric to S(p
0, s
−δ1 h1).
Remark 1.1. Note that from ([14]) we know that stable constant mean curvature embedded hypersurfaces bounding a small volume are nearly round spheres. In our corollary we consider the more general case of immersed hypersurfaces. Moreover we give a proximity with a geodesic sphere of the ambient space with explicite center and radius.
On the other hand let us recall a result concerning the topology of isoperimetric hypersurfaces (the embedded case). For instance if n = 2 and the Ricci curvature Ric
Nof N is bounded below by 2, Ros proved that if the volume of M is large enough then M is homeomorphic either to a sphere or a torus ([17]).
As another application of theorems 1.1 we have results for the almost umbilic hypersurfaces of space forms. These theorems are to be compared with results of Shiohama and Xu ([21] and [22]) who obtain conditions on the Betti numbers.
Theorem 1.3. Let (N
n+1, h) be a (n + 1)-dimensional Riemannian manifold with constant sectional curvature δ 6 = 0 and let M ∈ H
V(n, N ). Let us assume that φ(M ) lies in a convex ball of radius
8√πδ. Let p be the center of mass of M. Let ε < 1, r, q >
n and A > 0. Moreover let us assume that max(V (M )
1/nk H k
∞, V (M)
1/nk B k
q) 6 A for δ > 0 (resp. max(V (M )
1/nk H k
∞,
kHhk∞, V (M )
1/nk B k
q) 6 A for δ < 0).
Then there exist positive constants C := C(n, q, A) , α := α(q, n) such that if ε
α6 1/C and
(1) k τ k
r6 k H k
rε.
(2) k H
2− k H k
2∞k
r/26 k H k
2rε.
Then M is ε-Hausdorff close, diffeomorphic and ε-quasi-isometric to S(p, s
−δ1 1h).
Remark 1.2. The dependence on k B k
qis not necessary for the Hausdorff proximity.
In the Euclidean case, using the pinching theorem proved in [7] we can improve the condition 2)
Theorem 1.4. Let (M
n, g) be a compact, connected and oriented n-dimensional Riemannian manifold without boundary isometrically immersed by φ in R
n+1. Let p be the center of mass of M . Let ε < 1, r, q > n, s > r and A > 0. Let us assume that V (M )
1/nk H k
q6 A. Then there exist positive constants C := C(n, q, A), α :=
α(q, n) such that if ε
α6 1/C and (1) k τ k
r6 k H k
rε.
(2) k H
2− k H k
2sk
r/26 k H k
2rε.
Then M is ε-Hausdorff close to S p,
kH1k2
. Moreover if V (M )
1/nk B k
q6 A then M is diffeomorphic and ε-quasi-isometric to S
p,
kH1k2. 2. Preliminaries
Let (M
n, g) be a compact, connected n-dimensional Riemannian manifold iso- metrically immersed by φ in an (n + 1)-dimensional Riemannian manifold (N
n+1, h) whose sectional curvature is bounded by δ. For any point p ∈ N let us consider exp be the exponential map at this point. Locally we consider (x
i)
16i6nthe normal coordinates of N centered at p and for all x ∈ N, we denote by r(x) = d(p, x), the geodesic distance between p and x on (N
n+1, h).
We recall that the function c
δis defined by c
δ= s
′δ. Obviously, we have c
2δ+δs
2δ= 1 and c
′δ= − δs
δ.
The gradient of a function u defined on N with respect to h will be denoted by
∇
Nu and the gradient with respect to g of the restriction of u on M will be denoted by ∇
Mu.
Now considering the vector field on M , X = s
δ∇
Nr we recall that Heintze proved that
div (X
T) > nc
δ− nH h X, ν i (6)
Then using this identity Z
M
(n − δ | X
T|
2)dv = Z
M
(n − div (X
T)c
δ)dv 6
Z
M
(n − nc
2δ+ nH h X, ν i c
δ)dv
= Z
M
(nδs
2δ+ nH h X, ν i c
δ)dv 6
Z
M
nδs
2δdv + k H k
∞Z
M
ns
δc
δdv and using again (6) we get
Z
M
(n − δ | X
T|
2)dv 6 nδ Z
M
| X |
2dv + k H k
∞Z
M
(nH h X, ν i s
δ+ div (X
T)s
δ)dv
= nδ Z
M
| X |
2dv + k H k
∞Z
M
(nH h X, ν i s
δ− c
δs
δ|∇
Mr |
2)dv 6 nδ
Z
M
| X |
2dv + n k H k
2∞Z
M
|h X, ν i|| X | dv − k H k
∞Z
M
c
δs
δ|∇
Mr |
2dv 6 nδ
Z
M
| X |
2dv + n k H k
2∞Z
M
| X |
2dv − k H k
∞Z
M
c
δs
δ|∇
Mr |
2dv 6 n( k H k
2∞+ δ)
Z
M
| X |
2dv − k H k
∞Z
M
c
δs
δ|∇
Mr |
2dv Finally
1 6 ( k H k
2∞+ δ) k X k
22+ 1 nV (M )
Z
M
(δs
2δ− c
δs
δk H k
∞) |∇
Mr |
2dv
Now if δ 6 0 the last term is nonpositif. If δ > 0, since we have assumed that φ(M ) is in the ball B
p,
π4√ δ
, it follows that
sδπ
4√ δ
cδ
π 4√ δ
>
1kHk∞
and then δ
k H k
2∞6 1.
(7)
It follows that δs
2δ− c
δs
δk H k
∞6 0. This completes the proof of (3).
Now let us recall briefly the proof of Heintze. We will use
sδr(r)x
ias test functions in the variational characterization of λ
1(M ). But these functions must be L
2- orthogonal to the constant functions. For this purpose, we use a standard argument used by Chavel and Heintze ([6] and [9]). Indeed, if φ(M ) lies in a convex ball B the vector field Y defined in a neighborhood of B by
Y
q= Z
M
s
δ(d(q, x))
d(q, x) exp
−q1(x)dv(x) ∈ T
qN, q ∈ M ,
has necessarily a zero in B at a point p called the center of mass of M . Consequently, for a such p,
Z
M
s
δ(r)
r x
idv = 0. For δ > 0, we assume in addition that φ(M ) is contained in a ball of radius
4√πδ. Indeed, in this case φ(M ) lies in a ball of center p (the point p so that Y
p= 0) with a radius less or equal to
2√πδand c
δis then a nonnegative function. First note that the coordinates of X in the normal local frame are
sδ(r) r
x
i16i6n
. Moreover Heintze has proved that
n+1
X
i=1
∇
Ms
δ(r) r x
i2
6 n − δ | X
T|
2. Then from the variational characterization of the first eigenvalue and the previous proof we get
λ
1(M) k X k
226 1 V (M )
Z
M
(n − δ | X
T|
2)dv 6 n( k H k
2∞+ δ) k X k
22− k H k
∞V (M ) Z
M
c
δs
δ|∇
Mr |
2dv.
We end this section by the following proposition.
Proposition 2.1. Let (N
n+1, h) be a (n + 1)-dimensional Riemannian manifold whose sectional curvature K
Nsatisfies K
N6 δ. If φ(M ) lies in a convex ball of radius min
i(N ),
π4√ δ
then for any ε < 1/4, (Λ
ε) implies (I
p0,6ε) where p
0is the center of mass of M .
Proof. If (Λ
ε) holds then
n( k H k
2∞+ δ)J
p0(M )
26 (1 + ε)λ
1(M)J
p0(M )
26 (n − δ k X
Tk
22)(1 + ε) If δ > 0 then (I
p0,ε) is satisfied. If δ < 0 from the proof of the inequality of Reilly we have
λ
1(M)J
p0(M )
26 n( k H k
2∞+ δ)J
p0(M )
2− k H k
∞V (M ) Z
M
c
δs
δ| X
T|
22dv
6 λ
1(M )J
p0(M )
2(1 + ε)
It follows that p
| δ |k H k
∞k X
Tk
226 λ
1(M )J
p0(M )
2ε. Therefore n( k H k
2∞+ δ)J
p0(M )
26 (n − δ k X
Tk
22)(1 + ε)
6 n + p | δ | k H k
∞λ
1(M )J
p0(M)
2ε
!
(1 + ε) Now noting that
kH|δk|2∞
6 1, λ
1(M) 6 n( k H k
2∞+ δ) and ε < 1, we get ( k H k
2∞+ δ)J
p0(M)
2(1 − 2ε) 6 (1 + ε)
and if ε < 1/4, then (I
p0,6ε) is satisfied. Note that if δ > 0, it is not necessary to suppose that φ(M ) lies in a ball or radius
4√πδto prove that (Λ
ε) implies (I
p0,6ε).
3. An L
2-approach Throughout the paper we assume that φ(M ) ⊂ B
p,
4√πδfor δ > 0.
Lemma 3.1. If the pinching condition (I
p,ε) holds then k X
Tk
226
2εkHk2∞
. Proof. We have
k X
Tk
22= 1 V (M )
Z
M
| X |
2dv − Z
M
h X, ν i
2dv
6 2
V (M) Z
M
( | X |
2− |h X, ν i|| X | )dv 6 2ε k H k
2∞where the last inequality is coming from the proof of (3) recalled in the preliminaries.
Lemma 3.2. Let Y = nHc
δν − n k H k
2∞X. If (I
p,ε) holds then k Y k
226 4n
2k H k
2∞ε.
Proof. Using again (6) and the previous lemmas we have k Y k
22= n
2V (M) Z
M
H
2c
2δdv − 2 n
2V (M) k H k
2∞Z
M
H h X, ν i c
δdv + n
2k H k
4∞k X k
226 n
2V (M )
Z
M
H
2c
2δdv + 2n k H k
2∞V (M )
Z
M
(div (X
T)c
δ− nc
2δ)dv + n
2k H k
4∞k X k
22= n
2V (M)
Z
M
H
2c
2δdv + 2nδ k H k
2∞k X
Tk
22− 2n
2k H k
2∞V (M )
Z
M
c
2δdv + n
2k H k
4∞k X k
226 − n
2V (M) k H k
2∞Z
M
c
2δdv + n
2k H k
4∞k X k
22+ 2nδ k H k
2∞k X
Tk
22= − n
2k H k
2∞+ n
2k H k
2∞( k H k
2∞+ δ)J
p(M )
2+ 2nδ k H k
2∞k X
Tk
22= n
2k H k
2∞ε + 2nδ k H k
2∞k X
Tk
22Now we conclude by applying the lemma 3.1 and (7).
Lemma 3.3. Let W = | X |
1/2δX + Hc
δν − h
|XX|. If (I
p,ε) holds and if δ > 0, then k W k
226 6hε. If δ < 0 and V (M )
1/nk H k
∞6 A then k W k
226
2h +
C(n)khHk2∞A
n/2ε.
Proof. First we have k W k
226 1
V (M) Z
M
| X || δX + Hc
δν |
2− 2h h δX + Hc
δν, X i + h
2| X | dv
6 1
V (M) Z
M
| X || δX + Hc
δν |
2− 2h h δX + Hc
δν, X i
dv + h
2k X k
2(8)
Let us compute the first term 1
V (M ) Z
M
| X || δX + Hc
δν |
2dv = 1 V (M )
Z
M
| X | δ
2| X |
2+ 2δc
δH h X, ν i + H
2c
2δdv
= 1
V (M ) Z
M
| X | δ(1 − c
2δ) + H
2(1 − δs
2δ)H
2+ 2δc
δH h X, ν i dv
= 1
V (M ) Z
M
| X | (H
2+ δ − δ | HX − c
δν |
2)dv 6 h
2k X k
2− δ
V (M) Z
M
| X || HX − c
δν |
2dv (9)
Now let us compute the last two terms of (8)
− 2h V (M )
Z
M
h δX + Hc
δν, X i dv + h
2k X k
26 − 2δh
V (M ) Z
M
s
2δdv + 2h nV (M )
Z
M
div (X
T)c
δdv − 2h V (M)
Z
M
c
2δdv + h
2k X k
2= − 2h + 2hδ
n k X
Tk
22+ h
2k X k
2Therefore reporting this and (9) in (8), we get k W k
226 2hε + 2hδ
n k X
Tk
22− δ V (M )
Z
M
| X || HX − c
δν |
2dv Now if δ > 0 then k W k
226 6hε. If δ < 0 we have
k W k
226 2hε + | δ |
V (M) k X k
∞Z
M
| HX − c
δν |
2dv Moreover,
Z
M
| HX − c
δν |
2dv 6 k H k
2∞Z
M
s
2δdv − 2 Z
M
H h X, ν i c
δdv + Z
M
c
2δdv
Now from the proof of (3) recalled in the preliminaries and the pinching condition we have
k H k
2∞Z
M
s
2δdv − Z
M
H h X, ν i c
δdv 6 nh
2k X k
22C 6 nV (M )ε and
Z
M
c
2δdv − Z
M
H h X, ν i c
δdv 6 1 n
Z
M
div (X
T)c
δdv = δV (M) k X
Tk
22Then we have proved that if δ < 0 then
k W k
226 (2h + n | δ |k X k
∞)ε
(10)
Now the researched inequality is a straightforward consequence of the following lemma
Lemma 3.4. If V (M)
1/nk H k
∞6 A then k X k
∞6 K (n)A
n/2k X k
2.
The proof of the lemma 3.4 uses a Nirenberg-Moser type of proof (see [7]) based on a Sobolev inequality due to Hoffman and Spruck (see [10], [11] and [12]) which is available under the conditions on the injectivity radius of N and the volume of M contained in the definition of H
V(n, N ).
Proof of the lemma 3.4: Let us put ϕ = | X | . An easy computation shows that | dϕ
2α| 6 2αϕ
2α−1c
δ. Then if δ > 0, | dϕ
2α| 6 2αϕ
2α−1. If not we have
| dϕ
2α| 6 2αϕ
2α−1p
1 − δs
2δ6 2α(1 + p
| δ |k ϕ k
∞)ϕ
2α−16 2α(1 + k H k
∞k ϕ k
∞)ϕ
2α−1. Moreover since 1 6 ( k H k
2∞+ δ) k X k
226 2 k H k
2∞k X k
2∞we deduce that for δ < 0 we have | dϕ
2α| 6 4αϕ
2α−1k H k
∞k ϕ k
∞. Hence, using the Sobolev inequality (see [10], [11] and [12])
k f k
nn−16 K (n)V (M)
n1k df k
1+ k Hf k
1(11)
we get for any α > 1 and f = ϕ
2αk ϕ k
2α2αnn−1
6 K(n)V (M )
1/n2α k H k
∞k ϕ k
∞k ϕ k
2α2α−−11Then putting ν =
n−n1and α =
ap2+1where a
p+1= (a
p+ 1)ν and a
0= 2 we have k ϕ k
aap+1p+1ν6 K(n)V (M)
n1(a
p+ 1) k H k
∞k ϕ k
∞k ϕ k
aapp6 K(n)V (M)
n1a
pk H k
∞k ϕ k
∞k ϕ k
aappThen by iterating we find k ϕ k
ap+1 νp+1
ap+1
6
K(n)V (M )
1na
pk H k
∞k ϕ k
∞ 1/νpk ϕ k
ap
aνpp
6
p
Y
i=0
a
1
iνi
!
K(n)V (M )
1nk H k
∞k ϕ k
∞ n(
1−νp1+1) k ϕ k
aa00Now since
aνppconverges to a
0+ n and a
0= 2 we get k ϕ k
2∞6 C(n)
V (M )
n1k H k
∞ nk ϕ k
226 C(n)A
nk ϕ k
22Let’s introduce now the function ψ = | X |
1/2| X | −
1h= | X |
1/2X −
1h|XX|. We will give an L
2-estimate of ψ.
Lemma 3.5. If (I
p,ε) holds and δ > 0 then k ψ k
16
Ch3/2
ε
1/2. If δ < 0 and max(V (M )
1/nk H k
∞,
kHhk∞) 6 A then k ψ k
16
C(n)A1+n/4h3/2
ε
1/2.
Proof. First we have ψ = | X |
1/21
h
2(h
2X − δX − Hc
δν) + 1 h
2δX + Hc
δν − h X
| X |
6 | X |
1/2nh
2| Y | + 1 h
2| W | Then by H¨older inequality we get
k ψ k
16 1 h
21
n k X k
1/22k Y k
2+ k W k
2From Lemmas 3.2 and 3.3, we deduce easily the inequality for δ > 0 and for δ < 0 we get
k ψ k
16 C(n) h
2k H k
∞h
1/2+ h
1/2+ k H k
∞h
1/2A
n/4ε
1/26 C(n)
h
2(Ah
1/2+ h
1/2+ A
1+n/4h
1/2)ε
1/2Now from (11) by taking f = 1 we see that
K(n) 6 V (M )
1/nk H k
∞6 A (12)
This allows us to obtain the desired inequality for δ < 0.
Lemma 3.6. Let ε < 1 be a positive real number and let us assume that V (M )
1/nk H k
∞6 A (resp. max(V (M )
1/nk H k
∞,
kHhk∞) 6 A for δ < 0). Then there exist constants C := C(n) and α := α(n) so that if (I
p,ε) holds then
k ψ k
∞6 CA
αh
3/2ε
2(2n+1)1Proof. Let α > 1 then
| dψ
2α| = αψ
2α−2| dψ
2|
= αψ
2α−2| X | − 1 h
3 | X | − 1 h | d | X ||
6 3αψ
2α−2k X k
∞+ 1 h
2c
δProceeding as in the proof of Lemma 3.4 we find that | dψ
2α| 6 αEψ
2α−2where E = 3 k X k
∞+
h12if δ > 0 and E = 3 k X k
∞+
1h2(1 + k H k
∞k X k
∞) if not. It follows that
k ψ k
2αn2αn−1
6 K (n)V (M)
1/n(αE + k ψ k
2∞k H k
∞) k ψ k
2α2α−−22Now we know that for δ > 0, k H k
∞6 h and
kHδk2∞
6 1. Moreover, h 6 k H k
∞for δ < 0. From these facts we deduce that
k ψ k
2αn2αn−1
6 K(n)AαE
′k ψ k
2α2α−−22where E
′=
Eh+ k X k
∞k X k
∞+
h12. Now we put a
p+1= (a
p+ 2)ν with ν =
nn−1, a
0= 1 and α =
ap2+2. Then noting that
aνppconverges to a
0+ 2n, the end of the proof is similar to that Lemma 3.4 and we find
k ψ k
1+2n∞6 K(n)(AE
′)
nk ψ k
1Now Lemma 3.4 and 3.5 combining with (12) allow us to conclude that k ψ k
∞6
K(n)
Ahα(n)3/2ε
2(2n+1)1.
Lemma 3.7. Let ε < 1 be a positive real and let us assume that V (M )
1/nk H k
∞6 A (resp. max(V (M )
1/nk H k
∞,
kHhk∞) 6 A for δ < 0). Then there exist constants C := C(n) and α := α(n) so that if ε
2n+116
116(CAα)2
and (I
p,ε) holds then
| X | − 1 h
6 CA
αh ε
2(2n+1)1and
r − s
−δ11
h
6 CA
αh ε
2(2n+1)1Proof. Consider the function f(t) = t t −
h12which is increasing on [0,
3h1] and [
1h, + ∞ ) and decreasing on [
3h1,
h1]. Then k ψ k
2∞= k f ( | X | ) k
∞6
(CAh3α)2ε
2n+11. If (CA
α)
2ε
2n+116
271then f ( | X | ) 6
27h13< f
3h1. Now since k X k
22>
h12there exists x
0∈ M so that | X
x0| >
12h
>
3h1and by connectedness of M , it follows that | X | >
3h1over M. Then
| X | −
1h6
√3CAh αε
2(2n+1)1. Moreover assume that (CA
α)
2ε
2n+116
481in order to have
√hδ< 1 for δ > 0. We have
r − s
−δ11
h
6 sup
I
1 p 1 − δy
2!
| X | − 1 h
6 3 √ 3CA
αh ε
2(2n+1)1where I = R
+for δ 6 0 and I = [0,
3√42δ] for δ > 0. We obtain the desired result by choosing the new constant C
′= 3 √
3C.
4. Proof of the diffeomorphism
From now we will need a dependence on the second fundamental form in order to prove the diffeomorphism and the quasi-isometry.
Let us consider F : M −→ S p, s
−δ1 h1x 7−→ exp
ps
−δ1 1h Y|Y|
, where Y = exp
−p1(x). For more convenience we will put ̺ = s
−δ1 h1 Y|Y|
.
Lemma 4.1. Let u ∈ T
xM so that | u | = 1 and v = u − h u, ∇
Mr i∇
Nr. We have 1
h
2s
µ(r)
2| v |
26 | dF
x(u) |
26 s
µs
−δ1 1h2s
δ(r)
2| v |
2Proof. An easy computation shows that
d Y
| Y |
|
x(u) = 1
r d exp
−p1|
x(u) − dr(u)
r
2exp
−p1(x)
Then we deduce that dF
x(u) = d exp
p|
̺s
−δ11 h
d
Y
| Y |
|
x(u)
= s
−δ1 h1r d exp
p|
̺d exp
−p1|
x(u)
− s
−δ1 h1dr(u)
r
2d exp
p|
̺(exp
−p1(x))
= s
−δ1 h1r d exp
p|
̺d exp
−p1|
x(u)
− s
−δ1 1hdr(u) r ∇
Nr
F(x)Now let us compute the norm of dF
x(u). We have
| dF
x(u) |
2= s
−δ1 1h2r
2h
d exp
p|
̺d exp
−p1|
x(u)
2
− 2 h d exp
p|
̺d exp
−p1|
x(u)
, ∇
Nr i
F(x)dr(u) + dr(u)
2Now since exp
pis a radial isometry (see for instance [20]), we have
h d exp
p|
̺d exp
−p1|
x(u)
, ∇
Nr i
F(x)= h d exp
−p1|
x(u) , Y
| Y | i = h u, ∇
Nr i
xand it follows that
| dF
x(u) |
2= s
−δ1 h12r
2h
d exp
p|
̺d exp
−p1|
x(u)
2
− h∇
Mr, u i
2i (13)
Now
d exp
p|
̺d exp
−p1|
x(u)
2
=
d exp
p|
̺(d exp
−p1|
x(v) ) + h u, ∇
Mr i d exp
p|
̺d exp
−p1 x( ∇
Nr)
2
where v = u − h u, ∇
Mr i∇
Nr. Expending this expression we get
d exp
p|
̺d exp
−p1|
x(u)
2
=
d exp
p|
̺(d exp
−p1|
x(v ) )
2
+ h u, ∇
Mr i
2d exp
p|
̺d exp
−p1 x( ∇
Nr)
2
+ 2 h u, ∇
Mr ih d exp
p|
̺(d exp
−p1|
x(v )), d exp
p|
̺d exp
−p1 x( ∇
Nr) i
=
d exp
p|
̺(d exp
−p1|
x(v))
2
+ h u, ∇
Mr i
2where in the last equality we have used again the radial isometry property of the exponential map. And reporting this in (13) we obtain
| dF
x(u) |
2= s
−δ1 h12r
2d exp
p|
̺(d exp
−p1|
x(v) )
2
Since µ 6 K
N6 δ the standard Jacobi field estimates (see for instance corollary 2.8, p 153 of [20]) say that for any vector w orthogonal to ∇
Nr at y we have
| w |
2r
2s
µ(r)
26 | d exp
−p1|
y(w) |
26 | w |
2r
2s
δ(r)
2This gives s
δ(s
−δ1 1h)
2r
2| d exp
−p1|
x(v) |
26 | dF
x(u) |
26 s
µ(s
−δ1 h1)
2r
2| d exp
−p1|
x(v) |
2and applying again the standard Jacobi field estimates we obtain the desired in-
equalities of the lemma.
From now we denote by D any constant of the form D := c if µ > 0 and D := cc
µ ch
for some positive constant c.
Lemma 4.2. Let u ∈ T
xM so that | u | = 1.
1 − k∇
Mr k
2∞h
2s
2δ(r) 6 | dF
x(u) |
26 1 h
2s
2δ(r)
1 + D
δ − µ h
2 2Proof. Let r > 0. For t ∈ ( −∞ ,
9rπ22], consider the function σ
r(t) = s
t(r). An easy check yields that σ
ris C
1on ( −∞ ,
9rπ22] and
σ
r′(t) =
r3ct(r) 2
√tr−tan(√ tr) (√
tr)3
if t ∈ (0,
9rπ22]
−
r63if t = 0
r3ct(r) 2
−√−tr+tanh(√
−tr) (√
−tr)3
if t ∈ ( −∞ , 0)
It follows that σ
ris decreasing on ( −∞ ,
9rπ22] and that there exists a constant E so that | σ
r′(t) | 6 Er
3c
t(r), for any t ∈ ( −∞ ,
9rπ22]. From this we deduce that
0 6 s
µ(r) − s
δ(r) 6 Er
3c
µ(r)(δ − µ) (14)
From the lemma 3.7 and from the fact that
h16 s
−δ1 1hand φ(M ) ⊂ B
p,
4√πδfor δ > 0, we see that s
−δ1 1h6
π3√
δ
. Then applying (14) to s
−δ1 1hwe obtain that s
µs
−δ11 h
6 1
h
1 + D
δ − µ h
2(15)
From (14), (15) and the lemma 4.1 we get the desired result.
Lemma 4.3. Let ε < 1 , q > n and A be positive real numbers. Then there exist constants C := C(q, n) , α := α(q, n) and β := β(q, n) so that if max(V (M )
1/nk H k
∞, V (M )
1/nk B k
q) 6 A for δ > 0 (resp.
max(V (M )
1/nk H k
∞,
kHhk∞, V (M)
1/nk B k
q) 6 A for δ < 0), ε
β6
CA1αand (I
p,ε) holds then
k X
Tk
∞6 CA
αh D
αε
βProof. Put χ = | X
T| . Then | dχ
2α| = 2αχ
2α−1(c
δ(r) |∇
Mr |
2+ s
δ(r) | d |∇
Mr || ). Let us
estimate | d |∇
Mr || at a point x. For this, consider (e
i)
16i6nan orthonormal basis at
x. We have
| d |∇
Mr ||
2= 1
4 |∇
Mr |
2| d h∇
Nr, ν i
2|
2= h∇
Nr, ν i
2|∇
Mr |
2n
X
i=1
(e
ih∇
Nr, ν i )
2= h∇
Nr, ν i
2|∇
Mr |
2n
X
i=1
( ∇
Ndr(e
i, ν) + B (e
i, ∇
Mr))
26 2
|∇
Mr |
2n
X
i=1
∇
Ndr(e
i, ν)
2+ | B |
2|∇
Mr |
2!
Now
n
X
i=1
∇
Ndr(e
i, ν)
26 |∇
Ndr |
26
n+1
X
i=1
∇
Ndr(u
i, u
i) where (u
i)
16i6n+1is an or- thonormal basis which diagonalizes ∇
Ndr. From the comparison theorems (see for instance [20] p 153) we deduce that
n+1
X
i=1
∇
Ndr(u
i, u
i)
26 c
µs
µ 2n+1X
i=1
| u
i− h u
i, ∇
Nr i∇
Nr |
2= n c
µs
µ 2It follows that | d |∇
Mr ||
26 2n
|∇
Mr |
2c
µs
µ 2+ 2 | B |
2and
| dχ
2α| 6 2αχ
2α−1C(n)
c
δ+ s
δ|∇
Mr | c
µs
µ+ s
δ| B |
Now it is easy to see that
ssδµ
is bounded by a constant. Then
| dχ
2α| 6 2αχ
2α−1C(n)
c
δ+ c
µ|∇
Mr | + k X k
∞| B |
6 2αχ
2α−1C(n) k X k
∞c
δ+ c
µχ + | B |
6 2αχ
2α−2C(n) k X k
∞(c
δ+ c
µ+ k χ k
∞| B | ) 6 2αC(n) k X k
∞(E + k X k
∞| B | )χ
2α−2where E = 1 if µ > 0 and E = c
µ(r) if not. Now let us assume that α > 1. Then k χ k
2α2αnn−1
6 K (n)V (M)
1/n2α( k X k
∞E k χ k
2α2α−−22+ k X k
2∞k B k
qk χ k
2α(2α−−22)q q−1) 6 K (n)A2α
k X k
∞h E + k X k
2∞k χ k
2α(2α−−22)q q−1Now we put ν :=
n(q(n−−1)q1), a
p+1:= a
pν +
n2n−1, a
0= 2 and α :=
12q−1 q
a
p+ 1. Then a
p+1=
n2αn−1and
k χ k
ap+1 νp+1
ap+1
6
K (n)Aa
p+1k X k
∞h E + k X k
2∞ nn−1νp1+1k χ k
ap
aνpp
6
p+1
Y
i=1
a
1
iνi
!
n−1nK(n)A
k X k
∞h E + k X k
2∞n n−1
p+1
X
i=1
1 ν
ik χ k
aa00Now noting that
νappconverges to a
0+
q2nq−nwe get k χ k
∞6 C(n, q)
A
k X k
∞h E + k X k
2∞ 2(1+γ)γk χ k
1 1+γ
2
where γ :=
qnq−n. Now combining Lemmas 3.4 and 3.1 we obtain that k χ k
∞6 C(q, n) A
α(q,n)(E + 1)
α(q,n)h ε
β(q,n)Now we conclude the proof by noting that if µ < 0, then from Lemma 3.7 we have E = c
µ(r) 6 cosh
√ µ
s
−δ11
h
+ CA
αh ε
2(2n+1)16 cosh √ µ
s
−δ11 h
+ 1
4h
6 c
µc h
We can now give the proof of Theorem 1.1.
Proof of Theorem 1.1: From the lemma 3.7 we have 1
h
2s
2δ(r)
1 + D
δ − µ h
2 2− 1 6 1 + D
δh−2µ 2(1 − CA
αε
2(2n+1)1)
2− 1 6 4
3
2 + D
δ − µ h
2− CA
αε
2(2n+1)1D
δ − µ h
2+ CA
αε
2(2n+1)16 DCA
α1 + δ − µ εh
2 2ε
2(2n+1)1On the other hand from Lemmas 3.4 and 4.3 we have k∇
Mr k
2∞6 h
2(1 − CA
αε
2(2n+1)1)
2k X
Tk
2∞6 16
9 (CA
α)
2D
2αε
2βThen we deduce that 1 − k∇
Mr k
2∞h
2s
2δ(r) − 1 > − 2(CA
α)
2D
2αε
2β+ (1 + CA
αε
2(2n+1)1)
2− 1 (1 + CA
αε
2(2n+1)1)
2> −
2(CA
α)
2D
2αε
2β+ 2CA
αε
2(2n+1)1+ (CA
α)
2ε
2n+11> − CA
α′(D
2α+ 1)ε
β′It follows from this, the previous inequality and Lemma 4.2 that
|| dF
x(u) |
2− 1 | 6 CA
α′max (D
2α+ 1)ε
β′,
1 + δ − µ εh
2 2ε
2(2n+1)1!
6 CA
α′D
α′ε
β′′1 + δ − µ εh
2 2From (5) we deduce that if φ(M ) lies in a ball of center p and radius s
−δ1q
ε δ−µ
then || dF
x(u) |
2− 1 | 6 CA
α′D
α′ε
β′′. Moreover it is easy to see that for δ > 0 then D 6 c cosh c. For δ < 0, we conclude by noting that
p | µ |
hc26 c
2 δεh−µ2+ c
2|hδ2|6 c
2+ c
2kH|δk|2∞
A
26 c
2(1 + A
2).
Proof of Corollary 1.1: Since φ(M ) lies in a convex ball of radius min
π 8√
δ
,
i(N)2, we deduce that φ(M ) ⊂ B(p
0, min
π 4√
δ
, i(N )
where p
0is the center of mass of M . From Proposition 2.1 and the fact that fact that ε < 1/6 we know that (I
p0,6ε) holds. Then we can apply Theorem 1.1.
5. Application to the stability
Briefly, we recall the problem of the stability of hypersurfaces with constant mean curvature (see for instance [4]).
Let (M
n, g) be an oriented compact n -dimensional hypersurface isometrically immersed by φ in a n + 1-dimensional oriented manifold (N
n+1, h). We assume that M is oriented by the global unit normal field ν so that ν is compatible with the orientations of M and N . Let F : ( − ε, ε) × M −→ N be a variation of φ so that F (0, .) = φ. We recall that the balance volume is the function V : ( − ε, ε) −→ R defined by
Z
[0,t]×M
F
⋆dv
hwhere dv
his the element volume associated to the metric h. It is well known that V
′(0) =
Z
M
f dv
where f(x) = h
∂F∂t(0, x), ν i . Moreover the area function A(t) = Z
M
dv
Ft⋆hsatisfies A
′(0) = − n
Z
M