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# Comparison Theorems for Manifold with Mean Convex Boundary

Jian Ge

###### Abstract.

Let $M^{n}$ be an $n$-dimensional Riemannian manifold with boundary $\partial
M$. Assume that Ricci curvature is bounded from below by $(n-1)k$, for
$k\in\mathds{R}$, we give a sharp estimate of the upper bound of
$\rho(x)=\operatorname{d}(x,\partial M)$, in terms of the mean curvature bound
of the boundary. When $\partial M$ is compact, the upper bound is achieved if
and only if $M$ is isometric to a disk in space form. A Kähler version of
estimation is also proved. Moreover we prove a Laplace comparison theorem for
distance function to the boundary of Kähler manifold and also estimate the
first eigenvalue of the real Laplacian.

##  0\. Introduction

Let $M^{n}$ be an $n$-dimensional Riemannian manifold with boundary $\partial
M$, $\rho(x)=\operatorname{d}(x,\partial M)$ be the distance function to the
boundary. In a recent paper [Li12], the author proved $\max_{x\in
M}\rho(x)\leq 1/k$ under the assumptions that the Ricci curvature
$\operatorname{Ric}\geq 0$ and the mean curvature $H$ of the boundary
satisfies $H\geq(n-1)k$ for $k>0$, the equality holds if and only if $M$
isometric to the Euclidean ball of radius $1/k$. The argument is essentially
the well known Jacobi field estimates. The idea can be traced back to [HK78],
although estimations of volume is main topic there instead of the distance
functions, also they treated compact manifold only. Under a stronger
assumption that the sectional curvatures bounded from below, Dekster estimated
$\rho$ and lenght of more general curves in [Dek77]. Alexander and Bishop
studied more general Alexandrov spaces with certain convexity condition on the
boundary in [AB10], among many other things, one estimate of the upper bound
of $\rho$ has also been derived for such spaces. In this note we generalize
the above mentioned theorem to Riemannian manifold with lower Ricci curvature
bound $k\in\mathds{R}$ and give an unified proof for all $k$. i.e. we prove
the following

######  Theorem 0.1.

Let $M^{n}$ be a complete $n$-dimensional Riemannian manifold with lower Ricci
curvature bound $(n-1)k$ and boundary $\partial M$. Assume the mean curvature
$H$ of the boundary satisfies $H\geq(n-1)h$. For the case $k\leq 0$ we assumer
further that $h>\sqrt{-k}$. Let $\rho(x)=\operatorname{d}(x,\partial M)$ be
the distance to the boundary. Then

$\rho(x)\leq\left\\{\begin{aligned}
\frac{1}{\sqrt{|k|}}&\coth^{-1}\Big{(}\frac{h}{\sqrt{|k|}}\Big{)}\quad&{\rm
for}\quad k<0\\\ &\frac{1}{h}\quad&{\rm for}\quad k=0\\\
\frac{1}{\sqrt{k}}&\cot^{-1}\Big{(}\frac{h}{\sqrt{k}}\Big{)}\quad&{\rm
for}\quad k>0\\\ \end{aligned}\right.$ (0.1)

If we assume further that $\partial M$ is bounded, then the upper bound in
(0.1) implies that $M$ is compact. Note also that the condition $h>\sqrt{-k}$
for $k\leq 0$ is also sharp. Since the distance function $\rho(x)$ defined on
the warp product $[0,\infty)\times_{e^{-t}}\SS^{n-1}$ is unbounded and
satisfies the mean convex condition for $h=\sqrt{-k}=1$. Note one can also
give a Laplace comparison of $\rho$ for manifold with lower Ricci curvature
bound and mean convex boundary, which is implicitly proved in [HK78]. Hence we
will only list theorem here and omitted the proof. However, the proof is
similar as the Kähler version comparison proved in section 2. Define

$\textsf{H}_{k}(r)=(n-1)\textbf{{c}}_{k}(r)/\textbf{{s}}_{k}(r),$

which is the mean curvature of the geodesic sphere of radius $r$ in simply
connected space form with constant sectional curvature $k$, for
$r\in(0,\pi/\sqrt{k})$ if $k>0$. See (1.2) and (1.3) for the definition of
$\textbf{{s}}_{k}$ and $\textbf{{c}}_{k}$. Hence we have

######  Theorem 0.2 (Implicitly given in [HK78]).

Let $M^{n}$ be a complete $n$-dimensional Riemannian manifold with lower Ricci
curvature bound $(n-1)k$ and boundary $\partial M$. Assume the mean curvature
$H$ of the boundary satisfies $H\geq\textsf{H}_{k}(h)>0$. Let
$\rho(x)=\operatorname{d}(x,\partial M)$ be the distance to the boundary. Then

$\Delta\rho(x)\leq-\textsf{H}_{k}(h-\rho(x)).$

Let $\SS^{n}_{k}$ be the simply connected $n$-dimensional space form of
constant sectional curvature $k$. Let $D_{X}(p,r)$ denote the close disk of
radius $r$ centered at point $p\in X$. i.e. $D_{X}(p,r)=\\{x\in
X|\operatorname{d}(x,p)\leq r\\}$. The proof of the rigidity part in [Li12]
for $\operatorname{Ric}\geq 0$ does not extend to general lower curvature
bound, hence we use Theorem 0.2 and the idea in Cheeger-Gromoll’s proof of
splitting theorem [CG71], one can prove the following rigidity theorem for
Theorem 0.1.

######  Theorem 0.3.

Let $M^{n}$ be a complete $n$-dimensional Riemannian manifold with lower Ricci
curvature bound $(n-1)k$ and boundary $\partial M$. Assume the mean curvature
$H$ of the boundary satisfies $H\geq\textsf{H}_{k}(h)$. Let
$\rho(x)=\operatorname{d}(x,\partial M)$ be the distance to the boundary. Let
$\ell=\max_{x\in M}\rho(x)$, then $\ell=h$ if and only if $M$ is isometric to
$D_{\SS^{n}_{k}}(h)$.

It is well known that results related to Riemannian curvatures are sometimes
also hold for Kähler manifold when hypothesis are suitably phrased in terms of
bi-sectional or holomorphic sectional curvature. In fact the idea of the proof
of Theorem 0.1 can be used to prove the following estimate for Kähler
manifold. It suits our purposes well in this note to avoid complex vector
spaces. In fact we will treat Kähler manifold as a Riemannian manifold with
metric $g$ admitting a parallel skew-symmetric linear transformation $J$ on
the tangent bundle such that $J^{2}=-I$, where $I$ is the identity
transformation of $TM$. Let $R$ be the Riemannian curvature tensor of $g$ with
the convention

$R(X,Y,Z,W)=\langle-\nabla_{X}\nabla_{Y}Z+\nabla_{Y}\nabla_{X}Z+\nabla_{[X,Y]}Z,W\rangle.$

Hence the sectional curvature of a plane $\sigma$ is $R(X,Y,X,Y)$, where $X,Y$
are an orthonormal basis of $\sigma$. Let $\sigma_{1}$ and $\sigma_{2}$ are
two planes in $T_{x}M$ each invariant under $J$. Let $X$ and $Y$ be unit
vectors in $\sigma_{1}$ and $\sigma_{2}$ respectively. Then the holomorphic
bisectional curvature of $\sigma_{1}$ and $\sigma_{2}$ is defined by:

${\rm bisec}(\sigma_{1},\sigma_{2})=R(X,JX,Y,JY).$

By the Bianchi identity we have

${\rm bisec}(\sigma_{1},\sigma_{2})=R(X,Y,X,Y)+R(X,JY,X,JY)$

Notice that our definition of bi-sectional curvature differs by factor $2$ to
the one gave in [LW05]. By writing the inequality ${\rm bisec}\geq 2k$ we mean

$R(X,Y,X,Y)+R(X,JY,X,JY)\geq 2k(|X|^{2}|Y|^{2}+\langle X,Y\rangle^{2}+\langle
X,JY\rangle^{2}),$

for any $X,Y\in T_{x}M$ and any $x\in M$. Note that the complex projective
space $\mathds{CP}^{n}$ with the Fubini-study metric is normalized to have
constant holomorphic sectional curvature $4$. The condition ${\rm bisec}\geq
2k$ are stronger than the condition $\operatorname{Ric}\geq(2n+2)k$, however
the conclusions one can get are also stronger. Define

$\textsf{KH}_{k}(r)=\textbf{{c}}_{4k}(r)/\textbf{{s}}_{4k}(r)+(2n-2)\textbf{{c}}_{k}(r)/\textbf{{s}}_{k}(r),$

which is the mean curvature of the geodesic sphere of radius $r$ in simply
connected complex space form with constant holomorphic sectional curvature
$4k$. One easily verify that $\textsf{KH}_{k}(r)$ is monotone decreasing in
it’s natural domain of definition. We have the following Kähler version of
Theorem 0.1

######  Theorem 0.4.

Let $M^{n}$ be a Kähler manifold of complex dimension $n$ with bisectional
curvature ${\rm bisec}\geq 2k$, for $k\in\mathds{R}$. If $\partial M$ is mean
convex with mean curvature $H\geq\textsf{KH}_{k}(h)$, if $k=1$ we assume
$0<h<\pi/2$. Let $\rho(x)=\operatorname{d}(x,\partial M)$ be the distance to
the boundary. Then for all $x\in M$,

$\rho(x)\leq h.$

In fact if we put a stronger assumption on the convexity of $\partial M$, then
Theorem 0.4 can also be viewed as a consequence of the following Laplacian
comparison theorem for $\rho(x)$:

######  Theorem 0.5.

Let $M^{n}$ be a Kähler manifold of complex dimension $n$ with bisectional
curvature ${\rm bisec}\geq 2k$, for $k\in\mathds{R}$. If the second
fundamental form of $\partial M$ with respect to inner normal direction $\nu$
satisfy
$\operatorname{II}(J\nu,J\nu)\geq\textbf{{c}}_{4k}(h)/\textbf{{s}}_{4k}(h)$
and $\operatorname{II}(V,V)+\operatorname{II}(JV,JV)\geq
2\textbf{{c}}_{k}(h)/\textbf{{s}}_{k}(h)$ for any $V\perp J\nu$, if $k=1$ we
assume $0<h<\pi/2$. Let $\rho(x)=\operatorname{d}(x,\partial M)$ be the
distance to the boundary. Then

$\Delta\rho(x)\leq-\textsf{KH}_{k}(h-\rho(x))\leq-\textsf{KH}_{k}(h).$

The Laplace comparison for distance function to a point with arbitrary
holomorphic bisectional curvature lower bound is given by Li-Wang [LW05].

######  Theorem 0.6 (Li-Wang’s Laplace Comparison, [LW05]).

Let $M^{n}$ be a Kähler manifold of complex dimension $n$ with holomorphic
bisectional curvature ${\rm bisec}\geq 2k$, for $k\in\mathds{R}$. Let
$r(x)=\operatorname{d}(x,o)$, where $o\in M$ is a fixed point in $M$. Then

$\Delta r(x)\leq\textsf{KH}_{k}(r(x)).$

We also note that a complex Hessian comparison theorem for the Busemann
function on complete Kähler manifold with non-negative holomorphic bisectional
curvature was proved by Greene-Wu [GW78]. Cao-Ni [CN05] proved the complex
Hessian comparison theorem for the distance function on Kähler manifold with
non-negative holomorphic bisectional curvature. Our techniques used in the
proofs of Theorem 0.4 and Theorem 0.5 are inspired by [GW78] and [CS05]. The
Laplacian comparison of distance function to the boundary is a natural
complement of the comparison theorems given in [LW05]. As an application of
the Laplace comparison, we estimate the lower bound of the first eigenvalue of
the Laplacian for Kähler manifold with boundary using the idea of Song-Ying Li
and Xiaodong Wang [LW12]. They gave an estimation in the similar settings with
a fixed second fundamental form $2n$.

######  Theorem 0.7.

Assume $M^{n}$ satisfies the condition in Theorem 0.5 with
$\textsf{KH}_{k}(h)\geq 0$ and also assume $M$ is compact. Denote the first
Dirichlet eigenvalue of the Laplacian by $\lambda_{1}$. Then

$\lambda_{1}\geq\Big{(}\frac{\textsf{KH}_{k}(h)}{2}\Big{)}^{2}.\\\ $

It is my pleasure to thank Werner Ballmann for useful comments.

##  1\. Proof of the Theorem 0.1 and Theorem 0.3

In this note, all proofs are given in an uniform flavor, i.e. independent of
the sign of $k$. Hence let’s recall some definitions.

######  Definition .

Given a real constant $k$, we let $\textbf{{s}}_{k}$ denote the solution to
the ordinary differential equation

$\left\\{\begin{aligned} \phi^{\prime\prime}+k\phi&=0,\\\
\phi(0)=0,\quad&\phi^{\prime}(0)=1\end{aligned}\right.$ (1.1)

Setting $\textbf{{c}}_{k}(t)=\textbf{{s}}^{\prime}_{k}(t)$, we clearly get
$\textbf{{c}}^{\prime}_{k}(t)=-k\textbf{{s}}_{k}(t)$ and $\textbf{{c}}_{k}$
satisfies $\textbf{{c}}_{k}^{\prime\prime}+k\textbf{{c}}_{k}=0$ , with initial
condition $\textbf{{c}}_{k}(0)=1,\textbf{{c}}^{\prime}_{k}(0)=0$.

The explicit expressions are given by

$\textbf{{s}}_{k}(t)=\left\\{\begin{aligned}
&\frac{\sinh{(\sqrt{|k|}t)}}{\sqrt{|k|}}\quad&{\rm for}\quad k<0,\\\
&t\quad\quad&{\rm for}\quad k=0,\\\
&\frac{\sin{(\sqrt{k}t)}}{\sqrt{k}}\quad&{\rm for}\quad k>0,\\\
\end{aligned}\right.$ (1.2) $\textbf{{c}}_{k}(t)=\left\\{\begin{aligned}
&\cosh{(\sqrt{|k|}t)}\quad&{\rm for}\quad k<0\\\ &1\quad\quad&{\rm for}\quad
k=0\\\ &\cos{(\sqrt{k}t)}\quad&{\rm for}\quad k>0\\\ \end{aligned}\right.$
(1.3)

We sum up several basic formulas of $\textbf{{s}}_{k}$ and $\textbf{{c}}_{k}$,
the proof is straightforward calculations.

######  Proposition .

$\displaystyle\textbf{{s}}_{k}$ $\displaystyle{}^{\prime}=\textbf{{c}}_{k};$
(1.4) $\displaystyle\textbf{{c}}_{k}^{\prime}$
$\displaystyle=-k\textbf{{s}}_{k};$ $\displaystyle\textbf{{c}}_{k}(A+B)$
$\displaystyle=\textbf{{c}}_{k}(A)\textbf{{c}}_{k}(B)-k\textbf{{s}}_{k}(A)\textbf{{s}}_{k}(B);$
$\displaystyle\textbf{{s}}_{k}(A+B)$
$\displaystyle=\textbf{{s}}_{k}(A)\textbf{{c}}_{k}(B)+\textbf{{c}}_{k}(A)\textbf{{s}}_{k}(B).$

###### Proof of the Theorem 0.1.

For any fixed $\ell>0$ (if $k\geq 0$ we also assume $\ell<\pi/\sqrt{k}$), we
consider the differential equation

$\left\\{\begin{aligned} f^{\prime\prime}(s)+kf(s)&=0\\\ f(0)&=0\\\
f(\ell)&=1\\\ \end{aligned}\right.$ (1.5)

One easily verified

$f(s)=\frac{\textbf{{s}}_{k}(s)}{\textbf{{s}}_{k}(\ell)}$ (1.6)

is the solution. For any point $x\in M$, let $p\in\partial M$ such that
$\operatorname{d}(p,x)=\rho(x)$. Let $\gamma:[0,\ell]\to M$ be a shortest
geodesic joining $x$ to $p$, i.e. $\gamma(0)=x,\gamma(\ell)=p$ and
$\ell=\rho(x)$. By the first variation formula, $\gamma^{\prime}(\ell)\perp
T_{p}(\partial M)$. For and vector field $V$ normal to $\gamma$, the second
variational formula of arc-length gives

$\displaystyle L^{\prime\prime}(V,V)$
$\displaystyle=\int_{0}^{\ell}\Big{(}|V^{\prime}(s)|^{2}-|V(s)|^{2}K(\gamma^{\prime}(s),V(s))\Big{)}ds$
(1.7)
$\displaystyle\quad-\langle\nabla_{V(\ell)}(\gamma^{\prime}(\ell)),V(\ell)\rangle$
$\displaystyle\geq 0,$

where $K(v,w)$ denotes the sectional curvature of the plane spanned by $v$ and
$w$. Choose an orthonormal basis $E_{1},\cdots,E_{n-1}$ of $T_{p}(\partial
M)$. Let $E_{i}(s)$ be the parallel transportation of $E_{i}$ along $\gamma$.
Define

$V_{i}(s)=f(s)E_{i}(s)$

Let $V=V_{i}$ in (1.7), sum over $i$ for $1$ to $(n-1)$ and apply the boundary
condition of (1.5), one get

$\int_{0}^{\ell}\Big{(}(n-1)f^{\prime
2}(s)-f^{2}(s)\operatorname{Ric}(\gamma^{\prime}(s))\Big{)}ds-H_{p}\geq 0.$

where $\operatorname{Ric}(\gamma^{\prime}(s))$ denotes the Ricci curvature
along the direction $\gamma^{\prime}(s)$ and $H_{p}$ denotes the mean
curvature of $\partial M$ at $p$ w.r.t. the inner normal direction
$-\gamma^{\prime}(\ell)$:

$H_{p}=\sum_{i=1}^{n-1}\langle\nabla_{V_{i}(\ell)}\gamma^{\prime}(\ell),V_{i}(\ell)\rangle.$

Integration by parts for the term $f^{\prime 2}$ and make use of the
differential equation (1.5), one get:

$(n-1)[ff^{\prime}]|_{0}^{\ell}-H_{p}\geq 0.$

Since $H\geq(n-1)h>0$, we have:

$f^{\prime}(\ell)\geq h.$ (1.8)

That is $\textbf{{c}}_{k}(\ell)/\textbf{{s}}_{k}(\ell)\geq h$. Hence the
theorem follows from the explicit expressions (1.2) and (1.3). ∎

######  Remark .

It can be seen from the proof that the condition $h>\sqrt{|k|}$ is essential
to get an upper bound when solving (1.8). For positive $k$, the estimates also
holds for mean curvature bounded by a negative constant.

In [CG71], Cheeger and Gromoll proved the celebrate splitting theorem for open
manifold with non-negative Ricci curvature, the key step is that the sum two
Busemann functions constructed out of a line is super-harmonic and achieves a
minimum along this line, hence the minimal principle of super-harmonic
function implies it must be constant. Our proof of the Rigidity Theorem 0.3 is
similar to Cheeger-Gromoll’s proof. Note Li’s proof of the rigidity for $k=0$
given in [Li12] is along a slightly different line, where a rigidity theorem
for isoperemetric inequality plays an important role.

###### Proof of Rigidity Theorem 0.3.

Since $\partial M$ is compact, Theorem 0.1 implies that $M$ itself is compact.
Hence there exists $x_{0}\in M$ such that $\rho(x_{0})=h=\max_{M}(\rho(x))$.
Define

$r(x)=\operatorname{d}(x,x_{0}).$

The classical Laplacian comparison theorem implies that

$\Delta r(x)\leq\textsf{H}_{k}(r(x)).$ (1.9)

Theorem 0.2 implies

$\Delta\rho(x)\leq-\textsf{H}_{k}(h-\rho(x)).$ (1.10)

Follow the idea of Cheeger-Gromoll mentioned above, one define:

$F(x)=r(x)+\rho(x)-h.$

By triangle inequality, we clearly have $F(x)\geq 0$ for $x\in M$. Let
$\gamma:[0,h]\to M$ be a length minimizing geodesic joining $x_{0}$ and
$p\in\partial M$. Hence $F(\gamma(t))\equiv 0$. One calculate

$\displaystyle\Delta F(x)$ $\displaystyle=\Delta r(x)+\Delta\rho(x)$ (1.11)
$\displaystyle\leq\textsf{H}_{k}(r(x))-\textsf{H}_{k}(h-\rho(x))$
$\displaystyle\leq\textsf{H}_{k}(r(x))-\textsf{H}_{k}(r(x))$
$\displaystyle\leq 0$

where we used (1.9) and (1.10) for the first inequality. The second inequality
follows by the triangle inequality $h-\rho(x)\leq r(x)$ and the fact that
$\textsf{H}_{k}(t)$ is monotone decreasing. Therefore $F$ is a nonnegative
super-harmonic function on $M$ and achieves the minimal $0$ at some interior
point $\gamma(s)$ for $s\in(0,h)$. Hence $F$ must be identically $0$. This
shows the smoothness of the $r$ and $\rho$ in $M-\\{x_{0},\partial M\\}$. In
fact if $x\in M-\\{x_{0},\partial M\\}$, then $x$ can be jointed to both
$x_{0}$ and $\partial M$ by geodesic segments $\sigma_{1}$ and $\sigma_{2}$.
If we put this two segment together then it has length
$h=\operatorname{d}(x_{0},\partial M)$, such a segment must be smooth. Hence
$r$ is smooth and $\rho=h-r$ is also smooth. Also note that the geodesic never
bifurcate, which implies the uniqueness of such geodesic connecting $x_{0}$ to
$\partial M$ passing through $x$. Hence $M=D_{M}(x_{0},h)$, $\partial
M=\partial D_{M}(x_{0},h)$ and moreover

$\exp:B_{T_{x_{0}}M}(o,h)\to M$

is a diffeomorphism. The equality $\Delta F=0$ also implies

$\Delta r=H_{k}(r)=(n-1)\frac{\textbf{{c}}_{k}(r)}{\textbf{{s}}_{k}(r)},$

Taking derivative of $\textsf{H}_{k}(r)$ with respect to the direction
$\partial_{r}:=\partial/\partial r$ one get

$\displaystyle-(n-1)k$ $\displaystyle=\partial_{r}(\Delta r)+\frac{(\Delta
r)^{2}}{n-1}$ (1.12) $\displaystyle\leq\partial_{r}(\Delta
r)+|\operatorname{Hess}r|^{2}$
$\displaystyle=-\operatorname{Ric}(\partial_{r},\partial_{r})$
$\displaystyle\leq-(n-1)k,$

where the first inequality is Cauchy-Schwartz and the second equality is
Weitzenböck formula applied for distance function $r$. Hence all inequalities
in (1.12) are equalities. In particular equality in Cauchy-Schwartz implies
that Hessian is diagonal:
$\operatorname{Hess}r=\textbf{{c}}_{k}(r)/\textbf{{s}}_{k}(r)g_{r}$ for
$0<r<h$. Therefore the metric can be written in polar coordinate as

$g=dr^{2}+\textbf{{s}}_{k}^{2}(r)ds_{n-1}^{2}$

where $ds_{n-1}^{2}$ is the standard metric on $(n-1)$-dimensional sphere of
sectional curvature 1. Hence $M$ is isometric to the disk of radius $h$ in
space form $\SS^{n}_{k}$. ∎

##  2\. Comparison Theorems for Kähler Manifolds

We first give a proof similar to the proof of Theorem 0.1.

###### Proof of Theorem 0.4.

Fix $x\in M$. Let $p\in\partial M$ be a point such that
$\operatorname{d}(x,p)=\rho(x)$. It is know that under the assumption ${\rm
bisec}\geq 2$ the diameter of $M$ is less than $\pi/2$ see [LW05]. Hence we
can assume $\ell\leq\pi/2$ for the case $k=1$. Let $\gamma:[0,\ell]\to M$ be a
geodesic such that $\gamma(0)=x,\gamma(\ell)=p$ and $\ell=\rho(x)$. Choose an
orthonormal basis of $T_{p}M$:

$\\{E_{1},E_{2},\cdots,E_{2n-1},E_{2n}\\},$

satisfying $E_{2k}=JE_{2k-1}$ for $k=1,\cdots,(n-1)$ and
$E_{2n}=\gamma^{\prime}(\ell)$. Denote by $E_{i}(s)$ the parallel translation
of $E_{i}$ along $\gamma$. Define:

$f(s)=\frac{\textbf{{s}}_{k}(s)}{\textbf{{s}}_{k}(\ell)},\quad
g(s)=\frac{\textbf{{s}}_{4k}(s)}{\textbf{{s}}_{4k}(\ell)}.$

and

$V_{i}(s)=f(s)E_{i}(s),\ \text{for }i=1,\cdots,2n-2,\quad
V_{2n-1}(s)=g(s)E_{2n-1}(s).$

By the second variational formula one have

$\begin{split}0&\leq\sum_{i=1}^{2n-1}\delta^{2}(V_{i},V_{i})\\\
&=\int_{0}^{\ell}\Big{(}(2n-2)f^{\prime
2}(s)-\sum_{i=1}^{2n-2}f^{2}(s)K(\gamma^{\prime},E_{i})\\\
&\quad\quad+g^{\prime
2}(s)-g^{2}(s)K(\gamma^{\prime},J\gamma^{\prime})\Big{)}ds-H_{p}\\\
&=(2n-2)\textbf{{c}}_{k}(\ell)/\textbf{{s}}_{k}(\ell)+\textbf{{c}}_{4k}(\ell)/\textbf{{s}}_{4k}(\ell)-H_{p}\\\
\end{split}$ (2.1)

i.e.

$\textsf{KH}_{k}(\ell)\geq\textsf{KH}_{k}(h).$

By the monotonicity of $\textsf{KH}_{k}$, we have $\ell\leq h$. ∎

The distance estimate can also be derived from the following Hessian
comparison theorem. But let us recall first some basic definitions, which can
be found in [BC64]. Let $N\in M$ a submanifold, then the $N$-Jacobi field
along a geodesic $\gamma:[0,b]\to M$ with $\gamma(0)=p\in N$ and
$\gamma^{\prime}(0)\perp T_{p}(N)$ is the unique Jacobi field $J$ satisfying:

$J(0)\in T_{p}(N),\nabla_{\gamma^{\prime}(0)}J(0)-S_{\xi}J(0)\in
T_{p}^{\perp}N,$

where $S_{\xi}$ denotes the shape operator of $N$ with respect to the normal
vector $\xi$, i.e. $\langle S_{\xi}V,W\rangle=\langle\nabla_{V}\xi,W\rangle$
and $T_{p}^{\perp}N$ the normal bundle of $N$ in $M$. Recall also that the
index form $I_{\gamma}(,)$ associated with $\gamma$, for any vector field
$V\perp\gamma^{\prime}$ along $\gamma$, it is defined by:

$I_{\gamma}(V,V)=\langle
S_{\gamma^{\prime}(0)}V(0),V(0)\rangle+\int_{0}^{\ell}\Big{|}V^{\prime}|^{2}+|V|^{2}K(\gamma^{\prime},V)\Big{)}ds.$

Note that the second fundamental form with respect to the inner normal
direction of $\partial M$ is defined by $\operatorname{II}(V(0),V(0))=-\langle
S_{\gamma^{\prime}(0)}V(0),V(0)\rangle$. We have the following

######  Proposition  (The basic inequality. cf. [BC64]).

Suppose there is no focal point of $N$ on $\gamma(0,\ell]$. For all vector
field $V$ along $\gamma$ with $V(0)\in T_{p}N$, there is an unique $N$-Jacobi
field $J$ such that $J(\ell)=V(\ell)$. Moreover $I_{\gamma}(V)\geq
I_{\gamma}(J)$ and equality occurs if and only if $V=J$.

Theorem 0.5 is a consequence of the following complex Hessian comparison
theorem.

######  Theorem 2.1.

Let $M^{n}$ be a Kähler manifold of complex dimension $n$ with holomorphic
bisectional curvature ${\rm bisec}\geq 2k$, where $k=1,0$ or $-1$. If the
second fundamental form of $\partial M$ with respect to inner normal direction
$\nu$ satisfy
$\operatorname{II}(J\nu,J\nu)\geq\textbf{{c}}_{4k}(h)/\textbf{{s}}_{4k}(h)$
and $\operatorname{II}(V,V)+\operatorname{II}(JV,JV)\geq
2\textbf{{c}}_{k}(h)/\textbf{{s}}_{k}(h)$ for any $V\perp J\nu$, if $k=1$ we
assume $0<h<\pi/2$. Let $\rho(x)=\operatorname{d}(x,\partial M):=\ell$ be the
distance to the boundary. Then for all $\ell<h$, we have

$\displaystyle\operatorname{Hess}\rho(J\nu,J\nu)$
$\displaystyle\leq-\frac{\textbf{{c}}_{4k}(h-\ell)}{\textbf{{s}}_{4k}(h-\ell)}$
(2.2)
$\displaystyle\operatorname{Hess}\rho(V,V)+\operatorname{Hess}\rho(JV,JV)$
$\displaystyle\leq-2\frac{\textbf{{c}}_{k}(h-\ell)}{\textbf{{s}}_{k}(h-\ell)}$

In particular,

$\Delta(\rho)(x)\leq-\textsf{KH}_{k}(h-\rho(x))\leq-\textsf{KH}_{k}(h)$ (2.3)

###### Proof.

For any $x\in M$ let $p\in\partial M$ be a point such that
$\rho(x)=\operatorname{d}(x,\partial M)=:\ell$. Let $\gamma:[0,\ell]\to M$ be
a geodesic from $p$ to $x$. By Theorem 0.4, $\ell\leq h$. It suffices to
estimate the Laplacian for non cut point $x$. Let $f$ be the solution of the
differential equation:

$\left\\{\begin{aligned} f^{\prime\prime}+4kf&=0,\\\
f(0)=1,\quad&f^{\prime}(0)=-h_{1},\end{aligned}\right.$ (2.4)

where $h_{1}=\textbf{{c}}_{4k}(h)/\textbf{{s}}_{4k}(h)$. The explicit solution
is

$f(s)=\textbf{{c}}_{4k}(s)-h_{1}\textbf{{s}}_{4k}(s).$

Also let $g$ be the solution of

$\left\\{\begin{aligned} g^{\prime\prime}+kg&=0,\\\
g(0)=1,\quad&g^{\prime}(0)=-h_{2},\end{aligned}\right.$ (2.5)

where $h_{2}=\textbf{{c}}_{k}(h)/\textbf{{s}}_{k}(h)$. The explicit solution
is

$g(s)=\textbf{{c}}_{k}(s)-h_{2}\textbf{{s}}_{k}(s).$

Since $\ell<h$, one easily verify that $f(\ell)$ and $g(\ell)$ are non-zero,
hence we can define

$\tilde{f}(s)=f(s)/f(\ell),\quad\tilde{g}(s)=g(s)/g(\ell).$

and let $\\{E_{1},\cdots,E_{2n}\\}$ be an orthonormal basis of $T_{p}M$ such
that $JE_{2i-1}=E_{2i}$ for $i=1,\cdots,(n-1)$ and
$E_{2n}=\gamma^{\prime}(0)$. Parallel translate the $E_{i}$ along $\gamma$,
one get $E_{i}(s)$ and define

$V_{i}(s)=\tilde{g}(s)E_{i},\ \text{for }\ i=1,\cdots,2n-1,\ \text{and }\
V_{2n-1}(s)=\tilde{f}(s)E_{2n-1}.$

Let $J_{i}$ be the unique $\partial M$-Jacobi field with
$J_{i}(\ell)=E_{i}(\ell)$. Hence

$\nabla^{2}\rho(E_{i}(\ell),E_{i}(\ell))=\nabla^{2}\rho(J_{i}(\ell),J_{i}(\ell))=\langle\nabla_{\gamma^{\prime}(\ell)}J_{i},J_{i}\rangle=I_{\gamma}(J_{i},J_{i}),$

By Section 2, one have $\nabla^{2}\rho(E_{i}(\ell),E_{i}(\ell))\leq
I(V_{i},V_{i})$. Hence it suffices to estimate $I(V_{i},V_{i})$ from above. In
fact we have

$\begin{split}I_{\gamma}(V_{2n-1},V_{2n-1})&=-\tilde{f}^{2}(0)\operatorname{II}(E_{2n-1},E_{2n-1})\\\
&\quad+\int_{0}^{\ell}\Big{(}(\tilde{f}^{\prime})^{2}-(\tilde{f})^{2}K(\gamma^{\prime},E_{2n-1})\Big{)}ds\\\
&\leq-\tilde{f}^{2}(0)h_{1}+\int_{0}^{\ell}\Big{(}(\tilde{f}^{\prime})^{2}-4k(\tilde{f})^{2}\Big{)}ds\\\
&=-\tilde{f}^{2}(0)h_{1}+\tilde{f}^{\prime}\tilde{f}|_{0}^{\ell}\\\
&=f^{\prime}(\ell)/f(\ell)\\\
&=\frac{-4k\textbf{{s}}_{4k}(\ell)-\textbf{{c}}_{4k}(\ell)\textbf{{c}}_{4k}(h)/\textbf{{s}}_{4k}(h)}{\textbf{{c}}_{4k}(\ell)-\textbf{{s}}_{4k}(\ell)\textbf{{c}}_{4k}(h)/\textbf{{s}}_{4k}(h)}\\\
&=-\frac{\textbf{{c}}_{4k}(h-\ell)}{\textbf{{s}}_{4k}(h-\ell)}\end{split}$
(2.6)

Similar calculation shows that

$I_{\gamma}(V_{2i-1},V_{2i-1})+I_{\gamma}(V_{2i},V_{2i})\leq
2g^{\prime}(\ell)/g(\ell)=-2\frac{\textbf{{c}}_{k}(h-\ell)}{\textbf{{s}}_{k}(h-\ell)}$
(2.7)

Sum up for $i=1$ to $2n-1$, one get

$\begin{split}\Delta\rho&\leq
I_{\gamma}(V_{2n-1},V_{2n-1})+\sum_{i=1}^{n-1}I_{\gamma}(V_{2i-1},V_{2i-1})+I_{\gamma}(V_{2i},V_{2i})\\\
&\leq f^{\prime}(\ell)/f(\ell)+2(n-1)g^{\prime}(\ell)/g(\ell)\\\
&\leq-\textsf{KH}_{k}(h-\ell).\end{split}$ (2.8)

The second inequality in (2.3) follows from the Theorem 0.4 and monotonicity
of $\textsf{KH}_{k}$. ∎

Using the same idea of [LW12], we can estimate $\lambda_{0}$ as follows.

###### Proof of Theorem 0.7.

Let $f$ be the eigenfunction correspond to $\lambda_{0}$. i.e. $f$ satisfies

$\left\\{\begin{aligned} \Delta f+\lambda_{0}f&=0,\quad&\text{on }M,\\\
f&=0,\quad&\text{on }\partial M.\end{aligned}\right.$ (2.9)

We can assume $f>0$ on $M$. For any $C^{2}$ function $g:M\to\mathds{R}$, we
let

$F(x)=f(x)e^{g(x)}.$

Clearly $F\geq 0$ on $M$ and $F|_{\partial M}=0$. Hence by compactness of $M$,
$F$ reaches its maximum at an interior point, say $p_{0}\in M$. Hence at
$p_{0}$ we have

$\nabla F(p_{0})=(\nabla f\cdot e^{g}+fe^{g}\nabla g)|_{p_{0}}=0;$

where $\nabla f$ denotes the gradient vector of $f$. Hence at $p_{0}$:

$\nabla f=-f\nabla g.$

One calculates at $p_{0}$:

$\displaystyle 0\geq\Delta F$ $\displaystyle=\Delta(fe^{g})$ (2.10)
$\displaystyle=e^{g}(\Delta f+2\langle\nabla f,\nabla g\rangle+f\Delta
g+f\langle\nabla g,\nabla g\rangle)$
$\displaystyle=e^{g}(-\lambda_{0}f+2\langle-f\nabla g,\nabla g\rangle+f\Delta
g+f\langle\nabla g,\nabla g\rangle)$
$\displaystyle=e^{g}f(-\lambda_{0}-|\nabla g|^{2}+\Delta g).$

Therefore

$\lambda_{0}\geq(\Delta g-|\nabla g|^{2})|_{p_{0}}.$ (2.11)

for and $g\in C^{2}(M)$ and $p_{0}$ depends on $g$. Hence to get a lower bound
for $\lambda_{0}$ one let $g(x)=-c\rho(x)$, where
$\rho(x)=\operatorname{d}(x,\partial M)$ and $c$ is some positive constant to
be determined. Hence by Theorem 0.5, the following inequalities hold in
barrier sense

$\Delta g\geq c\textsf{KH}_{k}(h),\quad|\nabla g|\leq c$

Hence by (2.11), the following inequality holds for all $c\geq 0$

$\lambda_{0}\geq c(\textsf{KH}_{k}(h)-c).$ (2.12)

One easily see the maximum of the right hand side of (2.12) is

$\Big{(}\frac{\textsf{KH}_{k}(h)}{2}\Big{)}^{2},$

which is achieved when $c=\textsf{KH}_{k}(h)/2\geq 0$. ∎

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Jian Ge
Max Planck Institute for Mathematics
Vivatsgasse 7
53111 Bonn, Germany
<EMAIL_ADDRESS>