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ANNALES
DE LA FACULTÉ
DES SCIENCES
Mathématiques
XIAODONG WANG
Remark on an inequality for closed hypersurfaces in complete manifolds
with nonnegative Ricci curvature
Tome XXXII, no 1 (2023), p. 173–178.
https://doi.org/10.5802/afst.1733
© les auteurs, 2023.
Les articles des Annales de la Faculté des Sciences de Toulouse sont mis
à disposition sous la license Creative Commons Attribution (CC-BY) 4.0
http://creativecommons.org/licenses/by/4.0/
Annales de la faculté des sciences de Toulouse Volume XXXII, no 1, 2023
pp. 173-178
Remark on an inequality for closed hypersurfaces in
complete manifolds with nonnegative Ricci curvature (∗)
Xiaodong Wang (1)
ABSTRACT. — We give a simple proof of a recent result due to Agostiniani, Fo-
gagnolo and Mazzieri [1].
RÉSUMÉ. — Nous donnons une preuve simple d’un résultat récent dû à Agosti-
niani, Fogagnolo and Mazzieri [1].
The following result was proved by Agostiniani, Fogagnolo and Mazz-
ieri [1].
Theorem. — Let (Mn, g) (n ⩾ 3) be a complete Riemannian manifold
with nonnegative Ricci curvature and Ω ⊂ M a bounded open set with smooth
boundary. Then ∫
∂Ω
∣∣∣∣ Hn − 1
∣∣∣∣n−1 dσ ⩾ AVR(g)|Sn−1|, (1)
where H is the mean curvature of ∂Ω and AVR(g) is the asymptotic volume
ratio of M . Moreover, if AVR(g) > 0, equality holds iff M\Ω is isometric to(
[r0, ∞) × ∂Ω, dr2 + ( rr0 )
2g∂Ω
)
with
r0 =
(
|∂Ω|
AVR(g)|Sn−1|
) 1
n−1
In particular, ∂Ω is a connected totally umbilic submanifold with constant
mean curvature.
(*) Reçu le 25 janvier 2021, accepté le 22 février 2021.
(1) Department of Mathematics, Michigan State University, East Lansing, MI 48824,
USA — xwang@math.msu.edu
Article proposé par Gilles Carron.
– 173 –
Xiaodong Wang
The proof in [1] is highly nontrivial. It is based on the study of the solution
of the following problem 
∆u = 0, on M\Ω
u = 1 on ∂Ω
u(x) → 0 as x → ∞,
which exists when AVR(g) > 0. The key step consists of showing that, with
β ⩾ (n − 2)/(n − 1)
Uβ(t) = t−β(
n−1
n−2 )
∫
u=t
|∇u|β+1dσ
is monotone in t ∈ (0, 1]. The geometric inequality (1) then follows by ana-
lyzing the asymptotic behavior of Uβ(t) as t → 0. It is a beautiful argument.
In this short note, we show that this theorem can be proved by standard
comparison methods in Riemannian geometry.
To prove the inequality (1), we assume, without loss of generality, that
Ω has no hole, i.e. M\Ω has no bounded component. In the following we
write Σ = ∂Ω and let ν be the outer unit normal along Σ. For each p ∈ Σ
let γp(t) = expp tν(p) be the normal geodesic with initial velocity ν(p). We
define
τ(p) = sup{L > 0 : γp is minimizing on [0, L]} ∈ (0, ∞].
It is well known that τ is a continuous function on Σ and the focus locus
C(Σ) = {expp τ(p)ν(p) : τ(p) < ∞}
is a closed set of measure zero in M . Moreover the map Φ(r, p) = expp rν(p)
is a diffeomorphism from
E = {(r, p) ∈ Σ × [0, ∞) : r < τ(p)}
onto (M\Ω)\C(Σ). And on E the pull back of the volume form takes the
form dµ = A(r, p)drdσ(p). We will also understand r as the distance function
to Σ and it is smooth on M\Ω away from C(Σ). By the Bochner formula
and nonnegative Ricci curvature condition
0 = 12∆|∇r|
2 = |D2r|2 + ⟨∇r, ∇∆r⟩ + Ric(∇r, ∇r)
⩾
(∆r)2
n − 1 +
∂
∂r
∆r.
In view of the initial condition ∆r|r=0 = H, it is standard to deduce from
the above inequality τ ⩽ n−1H− and
A′
A
= ∆r ⩽ (n − 1)H
n − 1 + Hr
– 174 –
Remark on a geometric inequality
This shows that the function
θ(r, p) = A(r, p)(
1 + H(p)n−1 r
)n−1
is non-increasing in r on [0, τ(p)). As θ(0, p) = 1, we obtain
A(r, p) ⩽
(
1 + H(p)
n − 1r
)n−1
. (2)
The above analysis is by now standard in Riemannian geometry. For more
details, cf. [6, 7] or the original work of Heintze and Karcher [3] where the
analysis is done using Jacobi fields and the index form in a more general
setting.
Therefore for any R > 0
Vol{x ∈ M : d(x, Ω) < R}
= |Ω| +
∫
Σ
∫ min(R,τ(p))
0
A(r, p)drdσ(p)
⩽ |Ω| +
∫
Σ
∫ min(R,τ(p))
0
(
1 + H(p)
n − 1r
)n−1
drdσ(p)
⩽ |Ω| +
∫
Σ
∫ min(R,τ(p))
0
(
1 + H
+(p)
n − 1 r
)n−1
drdσ(p)
⩽ |Ω| +
∫
Σ
∫ R
0
(
1 + H
+(p)
n − 1 r
)n−1
drdσ(p)
= |Ω| + R
n
n
∫
Σ
(
H+(p)
n − 1
)n−1
dσ(p) + O(Rn−1).
Dividing both sides by |Bn|Rn = |Sn−1|Rn/n and letting R → ∞ yields
AVR(g) ⩽ 1
|Sn−1|
∫
Σ
(
H+
n − 1
)n−1
dσ,
which implies (1).
We now analyze the equality case. Suppose
AVR(g) = 1
|Sn−1|
∫
Σ
(
H+
n − 1
)n−1
dσ > 0. (3)
– 175 –
Xiaodong Wang
From the proof of the inequality and the fact that τ is continuous, we con-
clude that τ ≡ ∞ on the open set Σ+ = {p ∈ Σ : H(p) > 0}. We note
from (2) that A(r, p) ⩽ 1 for p ∈ Σ\Σ+. For any R′ < R we have
Vol{x ∈ M : d(x, Ω) < R}
= |Ω| +
∫
Σ+
∫ R
0
A(r, p)drdσ(p) +
∫
Σ\Σ+
∫ min(R,τ(p))
0
A(r, p)drdσ(p)
⩽ |Ω| +
∫
Σ+
∫ R
0
θ(r, p)
(
1 + H(p)
n − 1r
)n−1
drdσ(p) +
∫
Σ\Σ+
∫ R
0
drdσ(p)
⩽ |Ω| +
∫
Σ+
∫ R
R′
θ(r, p)(1 + H(p)
n − 1r)
n−1drdσ(p)
+
∫
Σ+
∫ R′
0
θ(r, p)
(
1 + H(p)
n − 1r
)n−1
drdσ(p) + O(R)
⩽ |Ω| +
∫
Σ+
θ(R′, p)
∫ R
R′
(
1 + H(p)
n − 1r
)n−1
drdσ(p)
+
∫
Σ+
∫ R′
0
θ(r, p)
(
1 + H(p)
n − 1r
)n−1
drdσ(p) + O(R).
Dividing both sides by |Bn|Rn = |Sn−1|Rn/n and letting R → ∞ yields
AVR(g) ⩽ 1
|Sn−1|
∫
Σ+
(
H(p)
n − 1
)n−1
θ(R′, p)dσ(p).
Letting R′ → ∞ yields
AVR(g) ⩽ 1
|Sn−1|
∫
Σ+
(
H(p)
n − 1
)n−1
θ∞(p)dσ(p),
where θ∞(p) = limr→∞ θ(r, p) ⩽ 1. As we have equality (3) we must have
θ∞(p) = 1 for a.e. p ∈ Σ+. It follows that
A(r, p) =
(
1 + H(p)
n − 1r
)n−1
on [0, ∞)
for a.e. p ∈ Σ+. By continuity the above identity holds for all p ∈ Σ+.
Inspecting the comparison argument, we must have on Φ([0, ∞) × Σ+)
Ric(∇r, ∇r) = 0, (4)
D2r = ∆r
n − 1g =
H
n − 1 + Hr g. (5)
– 176 –
Remark on a geometric inequality
As Ric ⩾ 0, it follows from the 1st equation above that Ric(∇r, · ) = 0.
From the 2nd equation above Σ+ is an umbilic hypersurface, i.e. the 2nd
fundamental form Π = Hn−1 gΣ+ . Working with an orthonormal frame {e0 =
ν, e1, . . . , en−1} along Σ+ we have by the Codazzi equation, with 1 ⩽ i, j, k ⩽
n − 1
R(ek, ej , ei, ν) = Πij,k − Πik,j =
1
n − 1(Hkδij − Hjδik).
Taking trace over i and k yields
−n − 2
n − 1Hj = Ric(ej , ν) = 0.
As a result H is locally constant on Σ+. Therefore Σ+ must be the union of
several components of Σ. We know that Φ is a diffeomorphism form [0, ∞)×
Σ+ onto its image and the pullback metric Φ∗g takes the following form
dr2 + hr,
where hr is a r-dependent family of metrics on Σ+ and h0 = gΣ+ . In terms
of local coordinates {x1, . . . , xn−1} on Σ+ the equation (5) implies
1
2
∂
∂r
hij =
H
n − 1 + Hr hij .
Therefore hr = (1+ Hn−1 r)
2gΣ+ . This proves that Φ([0, ∞)×Σ+) is isometric
to ([r0, ∞) × Σ+, dr2 + ( rr0 )
2gΣ+), where r0 = n−1H .
Since M has nonnegative Ricci curvature and Euclidean volume growth,
it has only one end by the Cheeger–Gromoll theorem. Therefore Σ+ is con-
nected and if Σ has other components besides Σ+, they all bound bounded
components of M\Ω.
If we have the stronger identity
AVR(g) = 1
|Sn−1|
∫
Σ
∣∣∣∣ Hn − 1
∣∣∣∣n−1 dσ > 0,
inspecting the proof of the inequality (1) shows that we must have H ⩾ 0 on
Σ. Then Ω is compact Riemannian manifold with mean convex boundary. It
is a classic fact that its boundary ∂Ω must be connected, see [4, 5] or [2] for
an analytic argument. Therefore Σ = Σ+ is connected and M\Ω is isometric
to ([r0, ∞) × Σ, dr2 + ( rr0 )
2gΣ).
Acknowledgements
The author would like to thank Fengbo Hang for helpful discussions.
– 177 –
Xiaodong Wang
Bibliography
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[2] F. Hang & X. Wang, “Vanishing sectional curvature on the boundary and a conjecture
of Schroeder and Strake”, Pac. J. Math. 232 (2007), no. 2, p. 283-287.
[3] E. Heintze & H. Karcher, “A general comparison theorem with applications to
volume estimates for submanifolds”, Ann. Sci. Éc. Norm. Supér. 11 (1978), no. 4,
p. 451-470.
[4] R. Ichida, “Riemannian manifolds with compact boundary”, Yokohama Math. J. 29
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[7] T. Sakai, Riemannian geometry, Translations of Mathematical Monographs, vol. 149,
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