# Modulus of Concavity and Fundamental Gap Estimates on Surfaces

Gabriel Khan  Department of Mathematics, Iowa State University, Ames, IA, USA.
<EMAIL_ADDRESS>,  Malik Tuerkoen  Department of Mathematics, University of
California, Santa Barbara, CA, USA<EMAIL_ADDRESS>and  Guofang Wei
Department of Mathematics, University of California, Santa Barbara, CA, USA.
<EMAIL_ADDRESS>
###### Abstract.

The fundamental gap of a domain is the difference between the first two
eigenvalues of the Laplace operator. In a series of recent and celebrated
works, it was shown that for convex domains in $\mathbb{R}^{n}$ and
$\mathbb{S}^{n}$ with Dirichlet boundary condition the fundamental gap is at
least $\frac{3\pi^{2}}{D^{2}}$ where $D$ denotes the diameter of the domain.
The key to these results is to establish a strong concavity estimate for the
logarithm of the first eigenfunction. In this article, we prove corresponding
log-concavity and fundamental gap estimates for surfaces with non-constant
positive curvature via a two-point maximum principle. However, the curvature
not being constant greatly increases the difficulty of the computation.

G. Khan is supported in part by Simons Collaboration Grant 849022

G. Wei is partially supported by NSF DMS 2104704.

##  1\. Introduction

Given a bounded domain $\Omega$ in a Riemannian manifold $(M,g)$, the Laplace
operator has an increasing sequence of Dirichlet eigenvalues
$0<\lambda_{1}<\lambda_{2}\leq\lambda_{3}\leq\ldots\rightarrow\infty$ and
corresponding eigenfunctions $u_{i}:\Omega\rightarrow\mathbb{R}$ satisfying

(1.1) $\displaystyle-\Delta
u_{i}=\lambda_{i}u_{i},\quad\left.{u_{i}}\right|_{\partial\Omega}=0.$

The _fundamental gap_ is the difference between the first and second
eigenvalues

$\displaystyle\Gamma(\Omega)=\lambda_{2}(\Omega)-\lambda_{1}(\Omega)>0.$

Obtaining upper and lower bounds for this quantity is important both in
mathematics and physics, and has been studied extensively (see, e.g. [SWYY85,
YZ86, LW87, Wan00, Ni13, LR13, GLL16], the references in [AC11, DSW19] and the
references in the rest of introduction). In a celebrated paper, Andrews and
Clutterbuck [AC11] proved the _fundamental gap conjecture_ [VdB83, Yau86,
AB89], which states that for all bounded convex domains
$\Omega\subset\mathbb{R}^{n}$ and Schrödinger operators $-\Delta+V$ where $V$
is a convex potential, the fundamental gap satisfies

(1.2) $\Gamma(\Omega)\geq\frac{3\pi^{2}}{D^{2}},$

where $D$ is the diameter of $\Omega$. The quantity on the right hand side of
this estimate corresponds to the fundamental gap of a one-dimensional interval
with vanishing potential and is optimal in Euclidean space (See Appendix A and
Remark A.2 for more details). Their work uses a two-point maximum principle to
prove a strong modulus of concavity estimate for the first eigenfunction.

Several years later, the third named author joint with Dai, He, Seto, and Wang
(in various subsets) extended the fundamental gap estimate (1.2) to the round
sphere [SWW19, HWZ20, DSW21]. The presence of curvature greatly complicates
the analysis, and these proofs relied in an essential way on the fact that the
sectional curvature is _constant_.

For spaces of negative curvature, the behavior of the fundamental gap turns
out to be very different. In fact, the third named author, Bourni,
Clutterbuck, Nguyen, Stancu, and Wheeler [BCN+22] showed that for any $D>0$
and $\epsilon>0,$ there are convex domains $\Omega\subset\mathbb{H}^{n}$ of
diameter $D$ which satisfy

$\Gamma(\Omega)<\frac{\epsilon}{D^{2}}.$

In other words, there is no universal lower bound on the fundamental gap among
convex domains of any given diameter.

This result was then extended by Nguyen and the first named author. They
showed that for any Riemannian manifold with _any negative sectional
curvature_ , it is possible to construct convex domains of small diameter
whose fundamental gap is _arbitrarily small_ [KN22]. As such, obtaining
estimates on the fundamental gap in terms of the diameter alone requires that
the space has positive curvature.

Establishing a lower bound on the fundamental gap for convex domains on
manifolds with non-constant positive curvature remains a challenge, and is
open even for seemingly simple manifolds such as
$\mathbb{S}^{2}\times\mathbb{S}^{2}$ or $\mathbb{CP}^{n}$. In [KNTW22], the
authors joint with Nguyen obtained the first result in this direction for
surfaces with positive curvature using a (one-point) maximum principle,
showing that $\Gamma(\Omega)>\frac{\pi^{2}}{D^{2}}+\underline{\kappa}$, under
some derivative assumption on the curvature $\kappa$, where
$\underline{\kappa}=\inf_{\Omega}\kappa.$ In this paper, we use two-point
maximum principle to study the fundamental gap of surface with non-constant
positive curvature. In order to make the two-point maximum principle work, we
need to overcome several difficulties, which we will discuss in Subsection
1.1. We obtain stronger gap estimates than in our previous work [KNTW22]
(joint with Nguyen) and, in particular, recover the estimate of [DSW21] for
$\mathbb{S}^{2}$.

######  Theorem 1.1.

Suppose $(M^{2},g)$ is a Riemannian manifold with curvature $\kappa$ bounded
by $0<\underline{\kappa}\leq\kappa\leq c$, there is a constant
$\alpha(\underline{\kappa},\overline{\kappa})>0$ such that if

$|\nabla\kappa|_{\infty}\leq\alpha(\underline{\kappa},\overline{\kappa}),\ \ \
-(\inf\Delta\kappa)_{-}\leq\alpha(\underline{\kappa},\overline{\kappa}),$

then for all convex domains $\Omega\subset M^{2}$ with diameter
$D\leq\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$ we have the gap estimate

(1.3)
$\Gamma(\Omega)\geq\frac{3\pi^{2}}{D^{2}}-(12+3\pi)(\overline{\kappa}-\underline{\kappa}).$

The constant $\alpha(\underline{\kappa},\overline{\kappa})$ will be given
explicitly and goes to zero when $\underline{\kappa}$ goes to zero (see (3.30)
for the precise condition).

As the curvature of $M^{2}$ is $0<\kappa\leq\overline{\kappa}$, we have the
injectivity radius of $M^{2}$ is $\geq\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$
(see e.g. [dC92, Page 281]). Hence there is no cut point in the interior of
$\Omega$.

###  1.1. Log-concavity and the maximum principle

Following [AC11, SWW19], the first key step to proving gap estimates of this
size is establishing a super log-concavity estimate for the first
eigenfunction. More precisely, if we take $u$ to be first Dirichlet
eigenfunction111 Throughout the rest of this article, we drop the subscript
$1$ from the first eigenfunction $u_{1}$ so that subscripts can be used to
indicate derivatives of functions. We also will denote $w=\log u.$ on a convex
domain $\Omega$ and define for $w=\log u,$ we show that for
$x,y\in\Omega\times\Omega,$ $\mathcal{Z}(x,y)\leq 0,$ where we set

(1.4) $\displaystyle\mathcal{Z}(x,y)=\langle\nabla
w(y),\gamma^{\prime}_{x,y}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma^{\prime}_{x,y}(\tfrac{-d}{2})\rangle+\frac{\pi}{L}\tan\left(\frac{\pi}{L}{d(x,y)}\right)-\textup{tn}_{\underline{\kappa}}\left(\frac{d(x,y)}{2}\right),$

where $\gamma_{x,y}$ is the unique minimizing geodesic satisfying
$\gamma_{x,y}(-\tfrac{d(x,y)}{2})=x$ and $\gamma_{x,y}(\tfrac{d(x,y)}{2})=y$.
In this expression, $L$ is taken to be slightly larger than $D$ and satisfies
$L=D+O(\overline{\kappa}-\underline{\kappa})$ as
$(\overline{\kappa}-\underline{\kappa})\rightarrow 0^{+}$. Furthemore, we
denote $\textup{tn}_{K}(s)=\sqrt{K}\tan(\sqrt{K}s)$ (for $K\geq 0$).

In other words, we have that

$F(x,y)=\frac{\pi}{L}\tan\left(\frac{\pi}{L}{d(x,y)}\right)-\textup{tn}_{\underline{\kappa}}\left(\frac{d(x,y)}{2}\right)$

is a _modulus of concavity for $w$_, where we call a function $F(x,y)$ a
modulus of concavity for $w$ if $\lim_{d(x,y)\rightarrow 0}F(x,y)=0$ and if
for all $x,y\in\Omega$

$\displaystyle\langle\nabla
w(y),\gamma^{\prime}_{x,y}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma^{\prime}_{x,y}(-\tfrac{d}{2})\rangle\leq F(x,y).$

Proving $\mathcal{Z}\leq 0$ is the most difficult part of the proof. In order
to establish this fact, one is tempted to apply the maximum principle directly
to the function $\mathcal{Z}(x,y)$, which is what was done for
$\mathbb{R}^{n}$ and $\mathbb{S}^{n}.$ However, when the curvature is not
constant, there are additional terms that cannot be controlled and thus
introduce a significant difficulty, see Lemma 3.3. In order to overcome this
difficulty, we instead apply the maximum principle to the function

(1.5) $\displaystyle Z(x,y)=\langle\nabla
w(y),\gamma^{\prime}_{x,y}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma^{\prime}_{x,y}(\tfrac{-d}{2})\rangle-2\phi(\tfrac{d}{2})-\mathcal{C}(x,y),$

where $\phi$ will be specified below. Here, $\mathcal{C}$ is defined in the
following way.

######  Definition 1.2 ($\mathcal{C}(x,y)$).

Suppose we have two points $x,y\in(M^{2},g)$ which are not cut points of each
other. We let $J_{x,y}$ be such that

(1.6) $\displaystyle J^{\prime\prime}(s)+\kappa(\gamma_{x,y}(s))J(s)=0,\quad
J(-\tfrac{d}{2})=J(\tfrac{d}{2})=1.$

$\mathcal{C}(x,y)$ is defined to be the quantity

(1.7)
$\mathcal{C}(x,y)=\frac{1}{2}\left(J_{x,y}^{\prime}(\tfrac{d}{2})-J_{x,y}^{\prime}(-\tfrac{d}{2})\right).$

Note that $2\mathcal{C}(x,y)$ is the index form of the Jacobi field $J$ along
the geodesic $\gamma_{x,y}$ with
$J(-\tfrac{d}{2})=e_{x},J(\tfrac{d}{2})=e_{y}$, where $e_{x}$ is a unit vector
perpendicular to $\gamma^{\prime}_{x,y}(-\tfrac{d}{2})$ and $e_{y}$ is the
parallel transport of $e_{x}$ along $\gamma_{x,y}$. See Lemma 2.9.

The quantity $\mathcal{C}(x,y)$ is also related to the the classic Gaussian
curvature by taking the limit as $y$ goes to $x$. More precisely, from (1.6),
we have

(1.8) $\displaystyle\frac{-\kappa(x)}{2}=\lim_{y\rightarrow
x}\frac{\mathcal{C}(x,y)}{d(x,y)}.$

In our computation, $\mathcal{C}(x,y)$ serves as part of the modulus of
concavity for $w$, which plays the essential role of absorbing some of the
problematic terms (see Proposition 3.4).

###  1.2. Overview of the paper

Because the proof of Theorem 1.1 requires a number of steps and several
involved calculations, in this subsection we provide an overview of the paper
and the main ideas used in the argument. In Section 2, we establish several
fundamental properties for Jacobi fields which will be used throughout the
paper. In particular, we give a formula for the solution of inhomogenous
Jacobi equations given boundary data. This is used several times in Section 6
and 7. We also prove several comparison results and identities for Jacobi
fields which are used in the later sections.

In Section 3 we apply a two-point maximum principle to $Z$ which is given by

$\displaystyle Z(x,y)=\langle\nabla
w(y),\gamma_{x,y}^{\prime}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma_{x,y}^{\prime}(-\tfrac{d}{2})\rangle-F(x,y),$

assuming that it achieves a positive maximum in the interior of
$\Omega\times\Omega\backslash\\{(x,x)~{}|~{}x\in\Omega\\}$. From the maximum
principle we derive a differential inequality for $F(x,y)$ to satisfy, to make
it a modulus of concavity for $w$. We first make no assumptions on the
dimension or geometry of the manifold (see Lemma 3.3).

We then specialize to the two-dimensional case and focus on functions of the
form $F(x,y)=2\phi(\tfrac{d(x,y)}{2})+\mathcal{C}(x,y).$ The choice of the
function $\mathcal{C}$ here is crucial to make the maximum principle work (see
Proposition 3.4). We use our maximum principle computation to show that
$F(x,y)$ is a modulus of concavity for $w,$ i.e. $Z\leq 0,$ if $\phi$
satisfies a collection of differential inequalities (see Theorem 3.6 and
Theorem 3.11). In this derivation, we take for granted several crucial
calculations, which are postponed to Sections 6 and 7.

Having derived these inequalities, in Section 4 we introduce the _perturbed
Euclidean model_ 222This is a perturbation of the model
$\phi^{\prime\prime}+\lambda\phi=0,$ which [DSW21] called _Euclidean model_.

(1.9)
$\displaystyle\overline{\phi}^{\prime\prime}-4\left(\textup{tn}_{\overline{\kappa}}(s)-\textup{tn}_{\underline{\kappa}}(s)\right)\overline{\phi}^{\prime}=-\overline{\lambda}\overline{\phi}\quad\textup{with
}\overline{\phi}(-D/2)=\overline{\phi}(D/2)=0.$

We then consider the function

$\phi=(\log\overline{\phi}_{1})^{\prime}+\textup{tn}_{\underline{\kappa}},$

where $\overline{\phi}_{1}$ the first eigenfunction of (1.9) and show that
this satisfies the differential inequalities required for a modulus of
concavity. Appealing to Theorem 4.5, we find that
$F(x,y)=2(\log\overline{\phi}_{1})^{\prime}(\tfrac{d(x,y)}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})$
is a modulus of concavity for $w$.

In Section 5, we compare the fundamental gap $\Gamma(\Omega)$ to the gap of a
one-dimensional model. To do so, we first apply a Riccati comparison theorem
to conclude that $\mathcal{Z}\leq 0$ and then compare the gap to that of a
Euclidean model on a slightly larger interval of length $L$, where
$L=D+O(\overline{\kappa}-\underline{\kappa})>D$

$\displaystyle\phi^{\prime\prime}+\lambda\phi=0\quad\textup{in
}[-L/2,L/2],\quad\phi(-L/2)=\phi(L/2)=0.$

Doing so, we prove the estimate

$\displaystyle\Gamma(\Omega)\geq\frac{3\pi^{2}}{L^{2}}.$

Estimating $L$ in terms of $D,\overline{\kappa},$ and $\underline{\kappa}$
then gives the final gap estimate (1.3).

In order to finish the proof, we must go back and derive several of the key
results which were used in the maximum principle argument. In Section 6, we
prove the crucial result (Proposition 3.4) which shows how the derivatives of
$\mathcal{C}$ absorb the terms which cannot be controlled and thus allow for
the maximum principle to be applied. The proof relies on a calculation of
Jacobi fields, and uses the computations done in Section 2.

Applying the maximum principle to $Z$ also involves the second derivatives of
$\mathcal{C}(x,y)$, which are computed in Section 7. The calculations are
similar to those of Section 6, but more involved and involve derivatives of
the curvature, Jacobi fields and its derivatives (see Proposition 7.4). With
enough control on the first and second derivatives of the curvature, it is
possible to control all these quantities.

Finally, in the appendix we show that the constant $3\pi^{2}$ in the gap
estimate

$\Gamma(\Omega)\geq\frac{3\pi^{2}}{D^{2}}-(12+3\pi)(\overline{\kappa}-\underline{\kappa})$

cannot be increased, independently of $D.$

######  Remark 1.3.

We use dimension two in two ways. First the Jacobi equation becomes one scalar
ODE instead of system of coupled ODEs. Second, even when the Jacobi equation
is a decoupled system, like in the case of
$\mathbb{S}^{2}\times\mathbb{S}^{2}$ or $\mathbb{CP}^{n}$, one cannot control
some of the terms as it is not a scalar matrix (see Section 3.3 and the
paragraph above that).

##  2\. Jacobi fields and their comparison geometry

In this section we establish some fundamental properties of Jacobi fields.
These results will be used throughout the paper in the variational
calculations.

###  2.1. Notation

Given any $x\not=y\in M$, we denote the minimal unit-speed geodesic between
them as $\gamma_{x,y}$ and use the parametrization where
$\gamma_{x,y}(-\tfrac{d}{2})=x$ and $\gamma_{x,y}(\tfrac{d}{2})=y$ with
$d=d(x,y).$ Throughout the paper, we will make the standing assumption that
$x$ and $y$ are not cut points, which will be justified by our hypothesis on
the diameter of $\Omega$. When unambiguous, will drop the subscripts and
simply denote this geodesic by $\gamma$.

We then set

$e_{n}=\gamma_{x,y}^{\prime}\left(-\tfrac{d}{2}\right)\in T_{x}\Omega$

to be the unit vector $T_{x}M$ pointing along the geodesic (towards $y$). We
can extend this to an orthonormal basis $\left\\{e_{i}\right\\}_{i=1,\dots,n}$
at $x$ and then parallel translate this orthonormal basis along $\gamma_{x,y}$
to obtain an orthonormal frame along the geodesic. Setting

$R_{ij}(t)=\langle R(e_{i},e_{n})e_{n},e_{j}\rangle,$

we consider the matrix-valued solutions $J^{1,0}(t)$ and $J^{0,1}(t)$ which
satisfy the system

(2.1) $\displaystyle\begin{cases}J^{\prime\prime}(t)+R(t)J(t)=0\\\
J^{1,0}(-\tfrac{d}{2})=I_{n},\,J^{1,0}(\tfrac{d}{2})=0_{n}\\\
J^{0,1}(-\tfrac{d}{2})=0_{n},\,J^{0,1}(\tfrac{d}{2})=I_{n}.\end{cases}$

Recall that given a geodesic $\gamma:[0,a]\rightarrow M$ whose endpoints are
not conjugate, there exists a unique Jacobi field along $\gamma$ for any given
boundary values at the endpoints, see [dC92, Page 118, Prop. 3.9]. As $y$ is
not conjugate to $x$ along $\gamma$, the solutions $J^{1,0},J^{0,1}$ exist and
are unique. Moreover, for all $t\in(-\tfrac{d}{2},\tfrac{d}{2})$ they will be
invertible. Furthermore, $(J^{1,0})^{\prime}(\tfrac{d}{2})$ must also be
invertible by the uniqueness of solutions to differential equations.

###  2.2. Inhomogeneous Jacobi equations

Given a smooth path of $n\times n$ matrices $M(t),$ we consider the
inhomogeneous equation

(2.2) $\displaystyle J^{\prime\prime}+RJ=M.$

It is possible to solve such an equation in terms of the homogeneous basis
solutions $J^{1,0},J^{0,1}$ and a particular solution. We can compute a
particular solution by integration.

######  Lemma 2.1.

Suppose that $J$ solves (2.2), then one has that

$\displaystyle J(s)$
$\displaystyle=J^{1,0}(s)J(\tfrac{-d}{2})+J^{0,1}(s)J(\tfrac{d}{2})$
$\displaystyle\quad+J^{1,0}(s)\int_{\tfrac{-d}{2}}^{s}J^{0,1}(t)[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}{M(t)}\,dt+J^{0,1}(s)\int_{s}^{\tfrac{d}{2}}J^{1,0}(t)[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}M(t)\,dt.$

This lemma is analogous to [FRV12, Lemma 3.2], which solved this inhomogeneous
problem in terms of initial conditions instead of boundary conditions.

###### Proof.

Denote

$\displaystyle F(s)$
$\displaystyle=J^{1,0}(s)J(\tfrac{-d}{2})+J^{0,1}(s)J(\tfrac{d}{2})$
$\displaystyle\quad+J^{1,0}(s)\int_{\tfrac{-d}{2}}^{s}J^{0,1}(t)[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}{M(t)}\,dt+J^{0,1}(s)\int_{s}^{\tfrac{d}{2}}J^{1,0}(t)[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}M(t)\,dt.$

Then we see that $F$ coincides with $J$ at the boundary. It thus remains to
verify that $F$ solves equation (2.2). A straightforward computation shows
that

$\displaystyle F^{\prime\prime}(s)$
$\displaystyle=-R(s)J^{1,0}(s)J(\tfrac{-d}{2})-R(s)J^{0,1}(s)J(\tfrac{d}{2})$
$\displaystyle\quad-R(s)J^{1,0}(s)\int_{\tfrac{-d}{2}}^{s}J^{0,1}(t)[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}{M(t)}\,dt+(J^{1,0})^{\prime}(s)J^{0,1}(s)[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}{M(s)}$
$\displaystyle\quad-R(s)J^{0,1}(s)\int_{s}^{\tfrac{d}{2}}J^{1,0}(t)[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}{M(t)}\,dt-(J^{0,1})^{\prime}(s)J^{1,0}(s)[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}{M(s)}.$

Hence,

$\displaystyle F^{\prime\prime}(s)+R(s)F(s)$
$\displaystyle=\Bigl{[}(J^{1,0})^{\prime}(s)J^{0,1}(s)-(J^{0,1})^{\prime}(s)J^{1,0}(s)\Bigr{]}[(J^{1,0})^{\prime}(\tfrac{d}{2})]^{-1}{M(s)}.$

However, note that

$\displaystyle\frac{d}{ds}\Bigl{[}(J^{1,0})^{\prime}(s)J^{0,1}(s)-(J^{0,1})^{\prime}(s)J^{1,0}(s)\Bigr{]}\equiv
0$

and therefore

$\displaystyle F^{\prime\prime}(s)+RF(s)=M(s).$

∎

###  2.3. Two-point Jacobi field comparison

We now present several comparison results concerning Jacobi fields

$\displaystyle J^{\prime\prime}+\kappa J=0\quad\textup{in }(0,l)$

These are the analogues of the Sturm comparison theorem for Jacobi equations
(see, e.g., [dC92, Pages 238-239]) except that we will need to consider
boundary value conditions instead of initial value conditions. We first prove
a comparison for the solutions of the type $J=J^{1,0}$ or $J=J^{0,1}:$

######  Lemma 2.2.

For $i=1,2$, let $J_{i}$ satisfy

$\displaystyle J_{i}^{\prime\prime}+k_{i}J_{i}=0\quad\textup{on}\quad(0,l)$

with $J_{i}(0)=1,$ and $J_{i}(l)=0$. Suppose that $k_{1}\leq k_{2}$ and
$J_{1},J_{2}>0$ in $(0,l)$. Then

$\displaystyle J_{2}\geq J_{1}\quad\textup{in }(0,l).$

###### Proof.

The Jacobi field equations imply that

(2.3)
$J_{2}^{\prime\prime}J_{1}-J_{1}^{\prime\prime}J_{2}+(k_{2}-k_{1})J_{1}J_{2}=0.$

Integrating this equation from $t$ to $l$ gives

$\displaystyle 0$
$\displaystyle=\int_{t}^{l}J_{2}^{\prime\prime}J_{1}-J_{1}^{\prime\prime}J_{2}+(k_{2}-k_{1})J_{1}J_{2}\,ds$
$\displaystyle\geq\int_{t}^{l}[J_{2}^{\prime}J_{1}-J_{1}^{\prime}J_{2}]^{\prime}\,ds$
$\displaystyle=-J_{2}^{\prime}(t)J_{1}(t)+J_{1}^{\prime}(t)J_{2}(t).$

In other words, we have that

$\displaystyle\frac{J_{2}^{\prime}(t)}{J_{2}(t)}\geq\frac{J_{1}^{\prime}(t)}{J_{1}(t)}.$

Integrating this inequality from $t_{0}$ to $t$ implies that

$\displaystyle\log J_{2}(t)-\log J_{2}(t_{0})\geq\log J_{1}(t)-\log
J_{1}(t_{0}).$

Finally, taking the limit as $t_{0}\rightarrow 0,$ we obtain the result. ∎

The same result holds (with a nearly identical proof) if we instead assume
that $J_{i}(0)=0,$ and $J_{i}(l)=1$.

This gives the following Jacobi field comparison result, which will be used
repeatedly throughout the paper.

######  Lemma 2.3.

Let

$\displaystyle J_{1}^{\prime\prime}+k_{1}J_{1}=0\quad\textup{with }\quad
J_{1}(0)=1,\,J_{1}(l)=1$

and let

$\displaystyle J_{2}^{\prime\prime}+k_{2}J_{2}=0\quad\textup{with }\quad
J_{2}(0)=1,\,J_{2}(l)=1.$

Suppose that $k_{1}\leq k_{2}$ and suppose that $J_{1},J_{2}>0$ in $(0,l),$
then one has that

$\displaystyle J_{2}\geq J_{1}\quad\textup{in }(0,l).$

######  Remark 2.4.

This result also provides a comparison for the derivatives at the endpoints.
In particular, we find that

(2.4) $\displaystyle J_{1}^{\prime}(\tfrac{-d_{0}}{2})\leq
J_{2}^{\prime}(\tfrac{-d_{0}}{2})\quad\textup{as well as }\quad
J_{1}^{\prime}(\tfrac{d_{0}}{2})\geq J_{2}^{\prime}(\tfrac{d_{0}}{2}).$

Applying the comparison results when the curvature satisfies
$\underline{\kappa}\leq\kappa\leq\overline{\kappa}$, we have the following
estimates.

######  Proposition 2.5.

For $\underline{\kappa}\leq\kappa\leq\overline{\kappa},$ and
$d<\tfrac{\pi}{\sqrt{\overline{\kappa}}}$ (if $\overline{\kappa}\leq 0,$ for
any $d<\infty),$ one has that for $s\in[-\tfrac{d}{2},\tfrac{d}{2}]$

(2.5)
$\displaystyle\frac{\textup{cs}_{\underline{\kappa}}(s)}{\textup{cs}_{\underline{\kappa}}(\tfrac{d}{2})}$
$\displaystyle\leq J(s)$
$\displaystyle\leq\frac{\textup{cs}_{\overline{\kappa}}(s)}{\textup{cs}_{\overline{\kappa}}(\tfrac{d}{2})}$
(2.6)
$\displaystyle\frac{-\textup{sn}_{\underline{\kappa}}(-\tfrac{d}{2}+s)}{\textup{sn}_{\underline{\kappa}}(d)}$
$\displaystyle\leq J^{1,0}(s)$
$\displaystyle\leq\frac{-\textup{sn}_{\overline{\kappa}}(-\tfrac{d}{2}+s)}{\textup{sn}_{\overline{\kappa}}(d)}$
(2.7)
$\displaystyle\frac{\textup{sn}_{\underline{\kappa}}(\tfrac{d}{2}+s)}{\textup{sn}_{\underline{\kappa}}(d)}$
$\displaystyle\leq J^{0,1}(s)$
$\displaystyle\leq\frac{\textup{sn}_{\overline{\kappa}}(\tfrac{d}{2}+s)}{\textup{sn}_{\overline{\kappa}}(d)}$

where we denote

$\displaystyle\textup{sn}_{K}(s)=\begin{cases}\tfrac{\sin(\sqrt{K}s)}{\sqrt{K}}\quad&\textup{for
}K>0\\\ s\quad&\textup{for }K=0\\\
\tfrac{\sinh(\sqrt{-K}s)}{\sqrt{-K}}\quad&\textup{for }K<0,\end{cases}$

$\textup{cs}_{K}(s)=\textup{sn}_{K}^{\prime}(s)$ and
$\textup{tn}_{K}(s)=K\tfrac{\textup{sn}_{K}(s)}{\textup{cs}_{K}(s)}.$

Combining Remark 2.4 and Proposition 2.5, one also obtains an estimate on the
derivatives at the end points.

######  Proposition 2.6.

Under the same assumptions as in Proposition 2.5, one has that

$\displaystyle\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2})\leq
J^{\prime}(-\tfrac{d}{2})\leq\textup{tn}_{\overline{\kappa}}(\tfrac{d}{2})\quad\textup{and
}\quad-\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2})\geq
J^{\prime}(\tfrac{d}{2})\geq-\textup{tn}_{\overline{\kappa}}(\tfrac{d}{2})$

and that

$\displaystyle\frac{1}{\textup{sn}_{\underline{\kappa}}(d)}\leq(J^{0,1})^{\prime}(-\tfrac{d}{2})\leq\frac{1}{\textup{sn}_{\overline{\kappa}}(d)}\quad\textup{and}\quad\frac{\textup{cs}_{\overline{\kappa}}(d)}{\textup{sn}_{\overline{\kappa}}(d)}\leq(J^{0,1})^{\prime}(\tfrac{d}{2})\leq\frac{\textup{cs}_{\underline{\kappa}}(d)}{\textup{sn}_{\underline{\kappa}}(d)}.$

From these estimates, we have the following comparison for the quantity
$\mathcal{C}$.

######  Lemma 2.7.

A surface $(M^{2},g)$ satisfies

(2.8)
$\displaystyle\mathcal{C}(x,y)\leq-\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})$
$\displaystyle\iff\kappa\geq\underline{\kappa}$
$\displaystyle\mathcal{C}(x,y)\geq-\textup{tn}_{\overline{\kappa}}(\tfrac{d(x,y)}{2})$
$\displaystyle\iff\kappa\leq\overline{\kappa}.$

Note that $2\mathcal{C}$ is the second order derivative of the distance or
index form (see below). This comparison result also follows directly from
Hessian (Index) comparison theorem [SY94, Page 4] in the two-dimensional case.
See [AC13] for the higher dimension version.

Recall the index form of a geodesic $\gamma:[0,a]\rightarrow M$ is given by
(see, e.g., [CE08, Page 17])

$I(V,W)=\int_{0}^{a}\langle\nabla_{\gamma^{\prime}}V,\nabla_{\gamma^{\prime}}W\rangle-\langle
R(V,\gamma^{\prime})\gamma^{\prime},W\rangle,$

where $V,W$ are vector fields along $\gamma$. It is a symmetric bilinear form
and is independent of orientation of $\gamma$.

######  Lemma 2.8.

Suppose $x,y$ are not cut points of each other. Define
$e_{2}=\gamma_{x,y}^{\prime}(-\tfrac{d}{2})\in T_{x}M.$ Choose $e_{1}\perp
e_{2}$ and parallel translate $e_{1}$ along $\gamma_{x,y}.$ Set
$E_{1}=e_{1}(-\tfrac{d}{2})\oplus e_{1}(\tfrac{d}{2})\in T_{x}M\bigoplus
T_{y}M$, and $J$ the Jacobi field with
$J(\pm\tfrac{d}{2})=e_{1}(\pm\tfrac{d}{2})$. Then

(2.9)
$\displaystyle\mathcal{C}(x,y)=\tfrac{1}{2}\nabla^{2}_{E_{1},E_{1}}d(x,y)=\tfrac{1}{2}I(J,J)$

###### Proof.

To compute the derivative of distance, let $\sigma_{1}(r)$ be the geodesic
with $\sigma_{1}(0)=x_{0},\tfrac{\partial}{\partial
r}\sigma_{1}(0)=e_{1}(\tfrac{-d_{0}}{2})$, $\sigma_{2}(r)$ be the geodesic
with $\sigma_{2}(0)=y_{0},\tfrac{\partial}{\partial
r}\sigma_{2}(0)=e_{1}(\tfrac{d_{0}}{2})$, and $\eta(r,s)$,
$s\in[-\tfrac{d_{0}}{2},\tfrac{d_{0}}{2}]$, be the minimal geodesic connecting
$\sigma_{1}(r)$ and $\sigma_{2}(r)$. Since $x,y$ are not cut points of each
other, the variation $\eta(r,s)$ is smooth and $\tfrac{\partial\eta}{\partial
r}=J$. Then by the second variation formula of distance

(2.10) $\displaystyle\frac{\partial^{2}}{\partial
r^{2}}L[\eta(r,s)]\Big{|}_{r=0}$
$\displaystyle=\int^{\frac{d}{2}}_{-\frac{d}{2}}\Big{[}\langle\nabla_{s}\tfrac{\partial\eta}{\partial
r},\nabla_{s}\tfrac{\partial\eta}{\partial r}\rangle-\langle
R(e_{2},\tfrac{\partial\eta}{\partial r})\tfrac{\partial\eta}{\partial
r},e_{2}\rangle\Big{]}ds+\langle e_{2},\nabla_{r}\tfrac{\partial\eta}{\partial
r}\rangle\biggr{|}_{-d/2}^{d/2}$ $\displaystyle=I(J,J)$
$\displaystyle=\int^{\frac{d}{2}}_{-\frac{d}{2}}\Big{[}\langle
J^{\prime},J\rangle^{\prime}-\langle J^{\prime\prime},J\rangle-\langle
R(J,e_{2})e_{2},J\rangle\Big{]}ds$ $\displaystyle=\langle
J^{\prime},J\rangle\big{|}_{-\tfrac{d_{0}}{2}}^{\tfrac{d_{0}}{2}}$
$\displaystyle=2\mathcal{C}(x,y).$

∎

###  2.4. An identity for Jacobi fields

There is one remarkable identity for the derivatives of Jacobi fields at their
endpoints. This will play an essential role throughout the proof, so we
mention it now.

######  Lemma 2.9.

Suppose we have two Jacobi fields $J_{1}$ and $J_{2}$ which satisfy the
following:

(2.11) $\displaystyle J_{1}^{\prime\prime}+kJ_{1}=0\quad$ with $\displaystyle
J_{1}(0)=1,\,J_{1}(l)=0$ (2.12) $\displaystyle
J_{2}^{\prime\prime}+kJ_{2}=0\quad$ with $\displaystyle
J_{2}(0)=0,\,J_{2}(l)=1.$

Then we have that

$\displaystyle J_{2}^{\prime}(0)+J_{1}^{\prime}(l)=0.$

###### Proof.

Multiplying (2.11) with $J_{2}$ and (2.12) with $J_{1},$ subtracting them from
each other and integrating gives that

$\displaystyle 0$
$\displaystyle=\int_{0}^{l}J_{1}^{\prime\prime}J_{2}-J_{2}^{\prime\prime}J_{1}\,ds$
$\displaystyle=\Bigl{[}J_{1}^{\prime}(s)J_{2}(s)-J_{2}^{\prime}(s)J_{1}(s)\Bigr{]}_{s=0}^{s=l}$
$\displaystyle=J_{1}^{\prime}(l)+J_{2}^{\prime}(0).$

∎

######  Remark 2.10.

This argument also applies to the matrices $J^{1,0}$ and $J^{0,1}.$ In other
words, we have that

$\displaystyle(J^{0,1})^{\prime}(-d/2)=-(J^{1,0})^{\prime}(d/2).$

##  3\. Modulus of Concavity of the first eigenfunction

We first apply the two-point maximum principle in a very general setting – on
a $n$-dimensional Riemannian manifold for a general modulus of concavity
function. Then we restrict our attention to two dimensions and choose a
specific function to obtain our result. The computations are very long so we
postpone some proofs to later sections.

To obtain a sufficiently strong concavity estimate of a function $w$ on a
strictly convex domain $\Omega\subset M^{n}$, one considers the two point
function

(3.1) $Z(x,y)=\langle\nabla
w(y),\gamma_{x,y}^{\prime}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma_{x,y}^{\prime}(\tfrac{-d}{2})\rangle-F(x,y).$

and applies a maximum principle to $Z.$ For more background on two-point
maximum principles and their applications in geometry, we refer to the surveys
by Andrews [And14] and Brendle [Bre14]. The goal of this section is to
construct modulus of concavity for $w$. In other words, we want to find
functions $F$ so that $Z\leq 0.$

As discussed in Subsection 1.1, we do not restrict ourselves to radial moduli
of concavity and instead consider more general functions $F(x,y)$, which will
take to be $C^{2}$ function and vanish on the diagonal ($x=y$). With these
assumptions in place, we first show that $Z$ cannot achieve a positive maximum
at the boundary.

###  3.1. Boundary Asymptotics

Following [AC11, SWW19], we show that if $Z$ achieves a positive maximum, it
must occur away from the boundary of
$\widehat{\Omega}=\Omega\times\Omega-\\{(x,x)\ |\ x\in\Omega\\}.$ We start by
applying [SWW19, Lemma 3.4], which states the following.

######  Lemma 3.1.

Let $\Omega$ be a uniformly convex bounded domain in a Riemannian manifold
$M^{n}$, and $u:\bar{\Omega}\rightarrow\mathbb{R}$ a $C^{2}$ function which is
positive on the interior of $\Omega$ and satisfies $u=0$ on $\partial\Omega$.
Furthermore, supposed that $\nabla u\neq 0$ on $\partial\Omega$. There exists
$r_{1}>0$ such that $\left.\nabla^{2}\log u\right|_{x}<0$ whenever
$d(x,\partial\Omega)<r_{1}$ and $N\in\mathbb{R}$ such that
$\left.\nabla^{2}\log u\right|_{x}(v,v)\leq N\|v\|^{2}$ for all $x\in\Omega$.

Following [SWW19, Lemma 3.5], we also have the following.

######  Lemma 3.2.

With the assumptions of Lemma 3.1 in place and further assuming that $F$ is
continuous on $\overline{\Omega}\times\overline{\Omega},$ for any $\beta>0,$
there exists an open set $U_{\beta}\subset M\times M$ containing
$\partial\widehat{\Omega}$ such that $Z(x,y)<\beta$ for all $(x,y)\in
U_{\beta}\cap\widehat{\Omega}.$

###### Proof.

We consider points $(x_{0},y_{0})\in\partial\widehat{\Omega}$ and establish
the estimate within $B_{r}(x_{0},y_{0}),$ where $r$ is some small radius
(which may depend on $(x_{0},y_{0})$). The set $U_{\beta}$ can be taken to be
the union of these balls. We distinguish two cases.

  1. Case 1.

($x_{0}=y_{0})$ For $x,y\in\Omega$ close to $x_{0}=y_{0},$ we have

$\displaystyle Z(x,y)$
$\displaystyle=\int_{-\tfrac{d}{2}}^{\tfrac{d}{2}}\operatorname{Hess}w(\gamma_{x,y}^{\prime}(s),\gamma^{\prime}_{x,y}(s))\,ds-F(x,y)$
$\displaystyle\leq Nd+F(x_{0},y_{0})-F(x,y).$

We then choose $r$ small such that $Nd(x,y)<\tfrac{\beta}{2}$ and (using the
continuity of $F$), $F(x_{0},y_{0})-F(x,y)<\tfrac{\beta}{2}.$

  2. Case 2.

($x_{0}\neq y_{0}$) In this case, at least one of $x_{0}$ and $y_{0}$ lies on
$\partial\Omega.$ Without loss of generality, we take
$x_{0}\in\partial\Omega$. Let $\alpha_{0}=\|\nabla u\|(x_{0})$, which is
strictly positive. By convexity of the domain,
$\eta:=\langle-\gamma_{x_{0},y_{0}}^{\prime}(-\tfrac{d}{2}),\nu(x_{0})\rangle>0$
by convexity, where $\nu(x_{0})$ is the outward pointing normal. Thus we have
the estimate

$\displaystyle-\langle\nabla
w(x),\gamma^{\prime}_{x,y_{0}}(-\tfrac{d}{2})\rangle=-\frac{1}{u(x)}\langle\nabla
u(x),\gamma^{\prime}_{x,y_{0}}(-\tfrac{d}{2})\rangle\rightarrow-\infty\quad\textup{as
}x\rightarrow x_{0},$

which implies the claim. Note that if $y_{0}\in\partial\Omega$ as well, we
also find that

$\displaystyle\langle\nabla
w(y),\gamma_{x_{0},y}^{\prime}(\tfrac{d}{2})\rangle\rightarrow-\infty\quad\textup{as
}y\rightarrow y_{0}.$

Thus we have $Z(x,y)<0$ in a neighborhood of $(x_{0},y_{0}).$

∎

###  3.2. Two-point maximum principle on a general manifold

We now assume that $Z$ achieves a positive maximum $\alpha>0$ in the interior
of $\widehat{\Omega}$ at $(x_{0},y_{0})$. In this case, we have that
$x_{0}\neq y_{0}$, and the maximum principle implies that for any $E\in
T_{x_{0}}\Omega\oplus T_{y_{0}}\Omega$

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\nabla_{E}Z(x_{0},y_{0}),$
$\displaystyle 0$ $\displaystyle\geq$
$\displaystyle\nabla^{2}_{E,E}Z(x_{0},y_{0}).$

We then construct an orthonormal frame $\\{e_{i}\\}$ at $x_{0}$ such that
$e_{n}=\gamma^{\prime}_{x_{0},y_{0}}(-\tfrac{d_{0}}{2})$ and parallel
translate it along $\gamma_{x_{0},y_{0}}$ to obtain a local frame along the
geodesic. In these computations, we use the notation $E_{i}=e_{i}\oplus
e_{i}\in T_{(x_{0},y_{0})}\Omega\times\Omega$ for $1\leq i\leq n-1$, and
$E_{n}=e_{n}\oplus(-e_{n})$.

Since $\nabla_{e_{n}}\gamma^{\prime}_{x_{0},y_{0}}(s)=0$ we immediately have

$\displaystyle\nabla_{E_{n},E_{n}}Z(x_{0},y_{0})=\langle\nabla_{e_{n}}\nabla_{e_{n}}\nabla
w(y_{0}),e_{n}\rangle-\langle\nabla_{e_{n}}\nabla_{e_{n}}\nabla
w(x_{0}),e_{n}\rangle-\nabla^{2}_{E_{n},E_{n}}F(x_{0},y_{0}).$

To compute the derivative $\nabla^{2}_{E_{i},E_{i}}Z(x_{0},y_{0})$ with
$i=1,\dots,n-1$, we introduce variations of $\gamma_{x_{0},y_{0}}(s)$, which
are denoted

(3.2)
$\eta_{i}(r,s):(-\delta,\delta)\times[-\tfrac{d_{0}}{2}.\tfrac{d_{0}}{2}]\rightarrow\Omega.$

To define this variation, we let $\sigma_{1}(r)$ be the geodesic with
$\sigma_{1}(0)=x_{0},\tfrac{\partial}{\partial
r}\sigma_{1}(0)=e_{i}(\tfrac{-d_{0}}{2})$ and $\sigma_{2}(r)$ be the geodesic
with $\sigma_{2}(0)=y_{0},\tfrac{\partial}{\partial
r}\sigma_{2}(0)=e_{i}(\tfrac{d_{0}}{2})$. We define $\eta_{i}(r,s)$,
$s\in[-\tfrac{d_{0}}{2},\tfrac{d_{0}}{2}]$ to be the minimal geodesic
connecting $\sigma_{1}(r)$ and $\sigma_{2}(r)$. Since $\Omega$ is strongly
convex, the variation $\eta(r,s)$ is smooth.

For fixed $r\neq 0$, the curves $\eta_{i}(r,\cdot)$ will not be unit speed
geodesics in general. Hence we define

$\displaystyle T_{i}(r,s):=\frac{\eta_{i}^{\prime}}{\|\eta_{i}^{\prime}\|},$

where we denoted $\partial/\partial s$ by ′, which is a convention we will use
throughout the rest of the paper.

Doing so, we have the following identity.

(3.3)
$\displaystyle\nabla^{2}_{E_{i},E_{i}}Z(x_{0},y_{0})=\frac{d^{2}}{dr^{2}|_{r=0}}Z(\eta_{i}(r,\tfrac{-d_{0}}{2}),\eta_{i}(r,\tfrac{d_{0}}{2}))$
$\displaystyle=\langle\nabla_{e_{i}}\nabla_{e_{i}}\nabla
w(y_{0}),e_{n}\rangle-\langle\nabla_{e_{i}}\nabla_{e_{i}}\nabla
w(x_{0}),e_{n}\rangle$ $\displaystyle\quad+2\langle\nabla_{e_{i}}\nabla
w(y_{0}),\nabla_{r}T_{i}(r,\tfrac{d_{0}}{2})\rangle-2\langle\nabla_{e_{i}}\nabla
w(x_{0}),\nabla_{r}T_{i}(r,\tfrac{-d_{0}}{2})\rangle$
$\displaystyle\quad+\langle\nabla
w(y_{0}),\nabla_{r}\nabla_{r}T_{i}(0,\tfrac{d_{0}}{2})\rangle-\langle\nabla
w(x_{0}),\nabla_{r}\nabla_{r}T_{i}(0,\tfrac{-d_{0}}{2})\rangle-\nabla^{2}_{E_{i},E_{i}}F(x_{0},y_{0})$

We then denote the variation field

(3.4) $J_{i}(r,s)=\tfrac{\partial}{\partial r}\eta_{i}(r,s),$

, which is the Jacobi field along $\eta(s)$ satisfying
$J_{i}(r,-\tfrac{d_{0}}{2})=\sigma_{1}^{\prime}(r),\
J_{i}(r,\tfrac{d_{0}}{2})=\sigma_{2}^{\prime}(r)$. For simplicity, we will
often drop the initial $0$ and denote $J_{i}(s)=J_{i}(0,s)$.

We can simplify this expression using several formulas derived on [SWW19, Page
363]. In particular, they showed that

$\displaystyle\nabla_{r}T_{i}(0,s)$ $\displaystyle=$
$\displaystyle-\langle\gamma^{\prime},J_{i}^{\prime}\rangle
e_{n}+J_{i}^{\prime},$ $\displaystyle\nabla_{r}\nabla_{r}T_{i}(0,s)$
$\displaystyle=$
$\displaystyle\Bigl{(}3\langle\gamma^{\prime},J_{i}^{\prime}\rangle^{2}-\|J_{i}^{\prime}\|^{2}-\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s},e_{n}\rangle\Bigr{)}e_{n}$
$\displaystyle\quad-2\langle\gamma^{\prime},J_{i}^{\prime}\rangle
J_{i}^{\prime}+\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial s}$

As the tangential component of a Jacobi field is linear and
$\langle\gamma^{\prime},J_{i}\rangle=0,$ at the end points, we have
$\langle\gamma^{\prime},J_{i}\rangle=0.$ Therefore
$\langle\gamma^{\prime},J_{i}^{\prime}\rangle=0.$ Hence

(3.5) $\nabla_{r}T_{i}(0,s)=J_{i}^{\prime},\ \ \
\nabla_{r}\nabla_{r}T_{i}(0,s)=-\Bigl{(}\|J_{i}^{\prime}\|^{2}+\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s},e_{n}\rangle\Bigr{)}e_{n}+\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s}.$

Combining Equations (3.5) and (3.3), we have

$\displaystyle\nabla^{2}_{E_{i},E_{i}}Z(x_{0},y_{0})=$
$\displaystyle\langle\nabla_{e_{i}}\nabla_{e_{i}}\nabla
w(y_{0}),e_{n}\rangle-\langle\nabla_{e_{i}}\nabla_{e_{i}}\nabla
w(x_{0}),e_{n}\rangle+2\langle\nabla_{e_{i}}\nabla
w(y_{0}),J_{i}^{\prime}(\tfrac{d_{0}}{2})\rangle$
$\displaystyle-2\langle\nabla_{e_{i}}\nabla
w(x_{0}),J_{i}^{\prime}(\tfrac{-d_{0}}{2})\rangle-\|J_{i}^{\prime}(\tfrac{d_{0}}{2})\|^{2}\langle\nabla
w(y_{0}),e_{n}\rangle+\|J_{i}^{\prime}(\tfrac{-d_{0}}{2})\|^{2}\langle\nabla
w(x_{0}),e_{n}\rangle$
$\displaystyle+\sum_{j=1}^{n-1}\left(\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s},e_{j}\rangle\langle\nabla
w(y_{0}),e_{j}\rangle-\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s},e_{j}\rangle\langle\nabla
w(x_{0}),e_{j}\rangle\right)-\nabla^{2}_{E_{i},E_{i}}F(x_{0},y_{0}).$

We now consider the sum of this expression from $i=1$ to $n$, which yields

$\displaystyle 0$
$\displaystyle\geq\sum_{i=1}^{n}\nabla^{2}_{E_{i},E_{i}}Z(x_{0},y_{0})$
$\displaystyle=\langle\Delta\nabla w(y_{0}),e_{n}\rangle-\langle\Delta\nabla
w(x_{0}),e_{n}\rangle)-\sum_{i=1}^{n}\nabla_{E_{i},E_{i}}^{2}F(x_{0},y_{0})$
$\displaystyle\quad+\sum_{i=1}^{n-1}\left(2\langle\nabla_{e_{i}}\nabla
w(y_{0}),J_{i}^{\prime}(\tfrac{d_{0}}{2})\rangle-2\langle\nabla_{e_{i}}\nabla
w(x_{0}),J_{i}^{\prime}(\tfrac{-d_{0}}{2})\rangle\right)$
$\displaystyle\quad+\sum_{i=1}^{n-1}\left(-\|J_{i}^{\prime}(\tfrac{d_{0}}{2})\|^{2}\langle\nabla
w(y_{0}),e_{n}\rangle+\|J_{i}^{\prime}(\tfrac{-d_{0}}{2})\|^{2}\langle\nabla
w(x_{0}),e_{n}\rangle\right)$
$\displaystyle\quad+\sum_{i,j=1}^{n-1}\left(\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s},e_{j}\rangle\langle\nabla
w(y_{0}),e_{j}\rangle-\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s},e_{j}\rangle\langle\nabla w(x_{0}),e_{j}\rangle\right).$

Our goal now is the simplify this expression as much as possible. The terms in
the first line can be simplified using the equation for $\Delta w$ and
Bochner’s identity:

$\displaystyle-\lambda-\Delta w$ $\displaystyle=$ $\displaystyle\|\nabla
w\|^{2}$
$\displaystyle\Delta\nabla\omega-\operatorname{Ric}(\nabla\omega,\cdot)$
$\displaystyle=$ $\displaystyle\nabla\Delta\omega.$

Doing so, we find that

$\displaystyle 0\geq$ $\displaystyle-e_{n}(\|\nabla
w\|^{2})(y_{0})+e_{n}(\|\nabla w\|^{2})(x_{0})$
$\displaystyle\quad-\sum_{i=1}^{n}\nabla_{E_{i},E_{i}}^{2}F(x_{0},y_{0})+\sum_{i=1}^{n-1}\left(2\langle\nabla_{e_{i}}\nabla
w(y_{0}),J_{i}^{\prime}(\tfrac{d_{0}}{2})\rangle-2\langle\nabla_{e_{i}}\nabla
w(x_{0}),J_{i}^{\prime}(\tfrac{-d_{0}}{2})\rangle\right)$
$\displaystyle\quad+\sum_{i=1}^{n-1}\left(-\|J_{i}^{\prime}(\tfrac{d_{0}}{2})\|^{2}\langle\nabla
w(y_{0}),e_{n}\rangle+\|J_{i}^{\prime}(\tfrac{-d_{0}}{2})\|^{2}\langle\nabla
w(x_{0}),e_{n}\rangle\right)$
$\displaystyle\quad+\operatorname{Ric}(\nabla\omega,e_{n})(y_{0})-\operatorname{Ric}(\nabla\omega,e_{n})(x_{0})$
$\displaystyle\quad+\sum_{i,j=1}^{n-1}\left(\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s},e_{j}\rangle\langle\nabla
w(y_{0}),e_{j}\rangle-\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta_{i}}{\partial
s},e_{j}\rangle\langle\nabla w(x_{0}),e_{j}\rangle\right).$

The first line can be further simplified:

$\displaystyle-e_{n}(\nabla w,\nabla w)(y_{0})+e_{n}(\nabla w,\nabla
w)(x_{0})$ $\displaystyle=$
$\displaystyle-2w_{n}(y_{0})w_{nn}(y_{0})-2\sum_{j=1}^{n-1}w_{j}(y_{0})w_{jn}(y_{0})$
$\displaystyle+2w_{n}(x_{0})w_{nn}(x_{0})+2\sum_{j=1}^{n-1}w_{j}(x_{0})w_{jn}(x_{0}),$

where $w_{j}=\langle\nabla w,e_{j}\rangle$ and $\
w_{ij}=\operatorname{Hess}w(e_{i},e_{j})$. As such, we must compute $w_{jn}$
for $j=1,\dots,n-1$. Since

$\displaystyle 0$ $\displaystyle=\nabla_{e_{j}\oplus 0}Z(x_{0},y_{0})$
$\displaystyle=\langle\nabla
w(y_{0}),(J_{j}^{1,0})^{\prime}(\tfrac{d_{0}}{2})\rangle-
w_{jn}(x_{0})-\langle\nabla
w(x_{0}),(J_{j}^{1,0})^{\prime}(\tfrac{-d_{0}}{2})\rangle-\nabla_{e_{j}\oplus
0}F(x_{0},y_{0}),$

we find that

$\displaystyle w_{jn}(x_{0})=\langle\nabla
w(y_{0}),(J_{j}^{1,0})^{\prime}(\tfrac{d_{0}}{2})\rangle-\langle\nabla
w(x_{0}),(J_{j}^{1,0})^{\prime}(\tfrac{-d_{0}}{2})\rangle-\nabla_{e_{j}\oplus
0}F(x_{0},y_{0}),$

where $J_{j}^{1,0}$ and $J_{j}^{0,1}$ are the $j$-th components of the matrix-
valued Jacobi fields defined by Equation (2.1). Similarly, using
$\nabla_{0\oplus e_{j}}Z(x_{0},y_{0})=0,$ we get

$\displaystyle-w_{jn}(y_{0})=\langle\nabla
w(y_{0}),(J_{j}^{0,1})^{\prime}(\tfrac{d_{0}}{2})\rangle-\langle\nabla
w(x_{0}),(J_{j}^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\rangle-\nabla_{0\oplus
e_{j}}F(x_{0},y_{0}).$

For the $w_{nn}$ terms, we observe that

(3.6) $\displaystyle 0$ $\displaystyle=$ $\displaystyle\nabla_{e_{n}\oplus
0}Z=-w_{nn}(x_{0})-\nabla_{e_{n}\oplus 0}F(x_{0},y_{0}),$ (3.7) $\displaystyle
0$ $\displaystyle=$ $\displaystyle\nabla_{0\oplus
e_{n}}Z=w_{nn}(y_{0})-\nabla_{0\oplus e_{n}}F(x_{0},y_{0}).$

Using the Jacobi matrix identity from Remark 2.10, we find

$\displaystyle 2\sum_{j=1}^{n-1}w_{j}(y_{0})\Bigl{[}\langle\nabla
w(y_{0}),(J_{j}^{0,1})^{\prime}(\tfrac{d_{0}}{2})\rangle-\langle\nabla
w(x_{0}),(J_{j}^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\rangle\Bigr{]}$
$\displaystyle+2\sum_{j=1}^{n-1}w_{j}(x_{0})\Bigl{[}\langle\nabla
w(y_{0}),(J_{j}^{1,0})^{\prime}(\tfrac{d_{0}}{2})\rangle-\langle\nabla
w(x_{0}),(J_{j}^{1,0})^{\prime}(\tfrac{-d_{0}}{2})\rangle\Bigr{]}$
$\displaystyle=$ $\displaystyle
2\sum_{i,j=1}^{n-1}w_{j}(y_{0})w_{i}(y_{0})(J^{0,1}_{ji})^{\prime}(\tfrac{d_{0}}{2})-w_{j}(x_{0})w_{i}(x_{0})(J^{1,0}_{ji})^{\prime}(\tfrac{-d_{0}}{2})$
$\displaystyle+2\sum_{i,j=1}^{n-1}\left(-w_{j}(y_{0})w_{i}(x_{0})(J_{ji}^{0,1})^{\prime}(\tfrac{-d_{0}}{2})+w_{j}(x_{0})w_{i}(y_{0})(J^{1,0}_{ji})^{\prime}(\tfrac{d_{0}}{2})\right)$
$\displaystyle=$ $\displaystyle
2\sum_{i,j=1}^{n-1}w_{j}(y_{0})w_{i}(y_{0})(J_{ji})^{\prime}(\tfrac{d_{0}}{2})-w_{j}(x_{0})w_{i}(x_{0})(J_{ji})^{\prime}(\tfrac{-d_{0}}{2})$
$\displaystyle+2\sum_{i,j=1}^{n-1}(J_{ji}^{0,1})^{\prime}(\tfrac{-d_{0}}{2})[w_{j}(y_{0})-w_{j}(x_{0})][w_{i}(y_{0})-w_{i}(x_{0})].$

Noting that

$\displaystyle\sum_{i,j=1}^{n-1}(J_{ji}^{0,1})^{\prime}(\tfrac{-d_{0}}{2})[w_{j}(y_{0})-w_{j}(x_{0})][w_{i}(y_{0})-w_{i}(x_{0})]$
$\displaystyle=\textup{Tr}\left((J^{0,1})^{\prime}(\tfrac{-d}{2})\circ[\nabla
w(y_{0})-\nabla w(x_{0})]\otimes[\nabla w(y_{0})-\nabla w(x_{0})]\right),$

we arrive at the following lemma.

######  Lemma 3.3.

Suppose the function $Z(x,y)$ defined by Equation 3.1 achieves a positive
maximum $\alpha$ on the interior of $\widehat{\Omega}$. Then at that point,
the following inequality holds.

(3.8) $\displaystyle 0\geq$ $\displaystyle
2\textup{Tr}\Bigl{(}(J^{1,1})^{\prime}(\tfrac{d_{0}}{2})\circ\left[\nabla
w(y)\otimes\nabla w(y)+\operatorname{Hess}w(y)\right]\Bigr{)}$
$\displaystyle-2\textup{Tr}\Bigl{(}(J^{1,1})^{\prime}(\tfrac{-d_{0}}{2})\circ\left[\nabla
w(x)\otimes\nabla w(x)+\operatorname{Hess}w(x)\right]\Bigr{)}$
$\displaystyle+2\textup{Tr}\left((J^{0,1})^{\prime}(\tfrac{-d}{2})\circ[\nabla
w(y)-\nabla w(x)]\otimes[\nabla w(y)-\nabla w(x)]\right)$
$\displaystyle+\sum_{i,j=1}^{n-1}\Bigl{[}\langle\nabla_{r}\nabla_{r}\eta^{\prime}_{i}(0,\tfrac{d_{0}}{2}),e_{j}\rangle-\tfrac{2}{n-1}\nabla_{0\oplus
e_{j}}F(x_{0},y_{0})+\frac{1}{n-1}\operatorname{Ric}_{jn}(y_{0})\Bigr{]}w_{j}(y_{0})$
$\displaystyle-\sum_{i,j=1}^{n-1}\Bigl{[}\langle\nabla_{r}\nabla_{r}\eta^{\prime}_{i}(0,\tfrac{-d_{0}}{2}),e_{j}\rangle+\tfrac{2}{n-1}\nabla_{e_{j}\oplus
0}F(x_{0},y_{0})+\frac{1}{n-1}\operatorname{Ric}_{jn}(x_{0})\Bigr{]}w_{j}(x_{0})$
$\displaystyle+w_{n}(y_{0})\left(\operatorname{Ric}_{nn}(y_{0})-\sum_{i}\|J_{i}^{\prime}(\tfrac{d_{0}}{2})\|^{2}-2\nabla_{0\oplus
e_{n}}F\right)$ $\displaystyle-
w_{n}(x_{0})\left(\operatorname{Ric}_{nn}(x_{0})-\sum_{i}\|J_{i}^{\prime}(-\tfrac{d_{0}}{2})\|^{2}+2\nabla_{e_{n}\oplus
0}F\right)-\sum_{i=1}^{n}\nabla^{2}_{E_{i},E_{i}}F(x_{0},y_{0}).$

This can be seen as a two point version of Theorem 2.1 in [KNTW22], and holds
for general Riemannian manifolds.

When $M=\mathbb{M}^{n}_{K}$ is a space form, we have

$((J^{1,1})_{ij}^{\prime}(\pm\tfrac{d_{0}}{2}))_{i,j=1}^{n-1}=\textup{tn}_{K}(\mp\tfrac{d_{0}}{2})I_{n-1}.$

Hence $\textup{Tr}\Bigl{(}(J^{1,1})^{\prime}\circ\left(\nabla w\otimes\nabla
w+\operatorname{Hess}w\right)\Bigr{)}$ simplifies in such a way that Equation
(3.2) can be used. This term seems to be the major obstacle in using the two-
point maximum principle for higher dimensional manifolds. This is the first
part of the proof where we need two-dimension to proceed with the proof.
Moreover the terms
$\langle\nabla_{r}\nabla_{r}\eta^{\prime}_{i}(0,\tfrac{-d_{0}}{2}),e_{j}\rangle$
vanish (see [SWW19], Appendix). When $K=0$,
$\nabla_{r}\nabla_{r}\eta^{\prime}_{i}(0,\tfrac{-d_{0}}{2})$ is identically
zero, not just the vertical components.

###  3.3. Specifying to the two-dimensional case

When $n=2$, the trace terms in Equation (3.8) have only one term each. In this
case, $\operatorname{Ric}_{p}=\kappa(p)g,$ and $J^{1,1}$ can be written in
terms of a function $J=J_{x_{0},y_{0}}$ (we omit the subscript in the
following computation), which solves the ODE

$\displaystyle J^{\prime\prime}+\kappa J=0,\quad\textup{in
}(-\tfrac{d_{0}}{2},\tfrac{d_{0}}{2})\quad
J(-\tfrac{d_{0}}{2})=J(\tfrac{d_{0}}{2})=1$

Inequality (3.8) becomes

$\displaystyle 0\geq$
$\displaystyle+2J^{\prime}(\tfrac{d_{0}}{2})\left(w_{11}(y_{0})+w_{1}^{2}(y_{0})\right)-2J^{\prime}(\tfrac{-d_{0}}{2})\left(w_{11}(x_{0})+w_{1}^{2}(x_{0})\right)$
$\displaystyle\quad+\left(\kappa(y_{0})-|J^{\prime}(\tfrac{d_{0}}{2})|^{2}-2\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})\right)w_{2}(y_{0})$
$\displaystyle\quad-\left(\kappa(x_{0})-|J^{\prime}(\tfrac{-d_{0}}{2})|^{2}+2\nabla_{e_{2}\oplus
0}F(x_{0},y_{0})\right)w_{2}(x_{0})$
$\displaystyle\quad+\left(\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta}{\partial
s},e_{1}\rangle-2\nabla_{0\oplus e_{1}}F(x_{0},y_{0})\right)w_{1}(y_{0})$
$\displaystyle\quad-\left(\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta}{\partial
s},e_{1}\rangle+2\nabla_{e_{1}\oplus 0}F(x_{0},y_{0})\right)w_{1}(x_{0})$
$\displaystyle\quad+2(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\Bigl{(}w_{1}(y_{0})-w_{1}(x_{0})\Bigr{)}^{2}-\nabla_{E_{1},E_{1}}^{2}F(x_{0},y_{0})-\nabla^{2}_{E_{2},E_{2}}F(x_{0},y_{0}).$

Note that the first term on the final line is symmetric in $x$ and $y$ via
Lemma 2.9.

Using (3.2) and (3.6), (3.7) we have

$\displaystyle
2J^{\prime}(\tfrac{d_{0}}{2})\left(w_{11}(y_{0})+w_{1}^{2}(y_{0})\right)-2J^{\prime}(\tfrac{-d_{0}}{2})\left(w_{11}(x_{0})+w_{1}^{2}(x_{0})\right)$
$\displaystyle=-\lambda\left(2J^{\prime}(\tfrac{d_{0}}{2})-2J^{\prime}(\tfrac{-d_{0}}{2})\right)-2J^{\prime}(\tfrac{d_{0}}{2})\left(w_{22}(y_{0})+w_{2}^{2}(y_{0})\right)+2J^{\prime}(\tfrac{-d_{0}}{2})\left(w_{22}(x_{0})+w_{2}^{2}(x_{0})\right)$
$\displaystyle=-\lambda\left(2J^{\prime}(\tfrac{d_{0}}{2})-2J^{\prime}(\tfrac{-d_{0}}{2})\right)-2J^{\prime}(\tfrac{d_{0}}{2})w_{2}^{2}(y_{0})+2J^{\prime}(\tfrac{-d_{0}}{2})w_{2}^{2}(x_{0})$
$\displaystyle\quad-2J^{\prime}(\tfrac{d_{0}}{2})\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{-d_{0}}{2})\nabla_{e_{2}\oplus
0}F(x_{0},y_{0}).$

We therefore obtain that

(3.9) $\displaystyle 0\geq$
$\displaystyle-\nabla_{E_{1},E_{1}}^{2}F(x_{0},y_{0})-\nabla^{2}_{E_{2},E_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{d_{0}}{2})\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{-d_{0}}{2})\nabla_{e_{2}\oplus
0}F(x_{0},y_{0})$
$\displaystyle\quad+2(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\Bigl{(}w_{1}(y_{0})-w_{1}(x_{0})\Bigr{)}^{2}-\lambda\left(2J^{\prime}(\tfrac{d_{0}}{2})-2J^{\prime}(\tfrac{-d_{0}}{2})\right)$
$\displaystyle\quad+\left(\kappa(y_{0})-|J^{\prime}(\tfrac{d_{0}}{2})|^{2}-2\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})\right)w_{2}(y_{0})-2J^{\prime}(\tfrac{d_{0}}{2})w_{2}^{2}(y_{0})$
$\displaystyle\quad-\left(\kappa(x_{0})-|J^{\prime}(\tfrac{-d_{0}}{2})|^{2}+2\nabla_{e_{2}\oplus
0}F(x_{0},y_{0})\right)w_{2}(x_{0})+2J^{\prime}(\tfrac{-d_{0}}{2})w_{2}^{2}(x_{0})$
$\displaystyle+\left(\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta}{\partial
s},e_{1}\rangle-2\nabla_{0\oplus
e_{1}}F(x_{0},y_{0})\right)w_{1}(y_{0})-\left(\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta}{\partial
s},e_{1}\rangle+2\nabla_{e_{1}\oplus 0}F(x_{0},y_{0})\right)w_{1}(x_{0}).$

###  3.4. Specifying to a special function F

As we mentioned previously, the term
$\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta}{\partial s},e_{1}\rangle$
vanishes in constant curvature, so the function $F$ can be chosen to a radial
function (i.e. only depending on the distance between the two points). When
the curvature is not constant, we will need to introduce a component of $F$ to
deal with the term $\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta}{\partial
s},e_{1}\rangle$. Quite surprising, it turns out that there is a particular
function which allows this to be done.

Let $F$ be given by

$\displaystyle F(x,y)=2\phi(\tfrac{d(x,y)}{2})+\mathcal{C}(x,y)$

for some $C^{2}$ function $\phi,$ where $\mathcal{C}$ is defined in (1.7).

The key property which makes this function useful is the following
proposition.

######  Proposition 3.4.

For $x,y\in\Omega,$ one has that

(3.10) $\displaystyle\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta}{\partial
s}(0,\tfrac{d_{0}}{2}),e_{1}\rangle-2\nabla_{0\oplus
e_{1}}\mathcal{C}(x_{0},y_{0})=\langle\nabla_{r}\nabla_{r}\tfrac{\partial\eta}{\partial
s}(0,\tfrac{-d_{0}}{2}),e_{1}\rangle+2\nabla_{e_{1}\oplus
0}\mathcal{C}(x_{0},y_{0})=0.$

The proof of this proposition can be found by computing the relevant Jacobi
fields, and will be presented in Section 6. Note that in view of (3.10), the
fifth line of (3.9) is zero since by the first variation formula the
derivatives of the distance are zero.

A second difficulty is that, since the curvature is non constant, we cannot
combine the $w_{2}$ terms in the third and fourth lines. We will deal with
this issue below (see Lemma 3.5 and thereafter).

We now reduce Inequality (3.9) to a simpler form, throughout these
computations we write $d=d(x,y)$ and $d_{0}=d(x_{0},y_{0}).$ Furthermore, we
will often drop the argument from
$\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})$ and simply denote this
quantity by $\textup{tn}_{\underline{\kappa}}$ (and similarly for
$\textup{tn}_{\overline{\kappa}}$).

To control the terms in the third and fourth line, we make use of two
properties. First, we use the following lemma (whose proof will be done in
Section 6).

######  Lemma 3.5.

We have

$\displaystyle 2\nabla_{e_{2}\oplus 0}\mathcal{C}$
$\displaystyle=\kappa(x)+|J^{\prime}_{x,y}(-\tfrac{d}{2})|^{2}$ $\displaystyle
2\nabla_{0\oplus e_{2}}\mathcal{C}$
$\displaystyle=-\kappa(y)-|J^{\prime}_{x,y}(\tfrac{d}{2})|^{2}.$

Second, we assume that $u$ is log-concave which implies

(3.11) $\displaystyle w_{2}(y_{0})\leq w_{2}(x_{0}).$

This might seem to be circular, since the entire goal of this argument is to
establish a log-concavity estimate for $u$. However, this assumption will
later be removed by a continuity method (see proof of Proposition 3.6 below).

Using Lemma 3.5, we consider the terms

$\displaystyle\left(\kappa(y_{0})-|J^{\prime}(\tfrac{d_{0}}{2})|^{2}-2\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})\right)w_{2}(y_{0})$
$\displaystyle-\left(\kappa(x_{0})-|J^{\prime}(\tfrac{-d_{0}}{2})|^{2}+2\nabla_{e_{2}\oplus
0}F(x_{0},y_{0})\right)w_{2}(x_{0})$
$\displaystyle=-2\left(w_{2}(y_{0})-w_{2}(x_{0})\right)\phi^{\prime}(\tfrac{d_{0}}{2})$
$\displaystyle+2\kappa(y_{0})w_{2}(y_{0})-2\kappa(x_{0})w_{2}(x_{0}).$

Using the assumption $w_{2}(y_{0})\leq w_{2}(x_{0})$, we break the analysis of
these terms into three cases.

  1. Case 1.

$(0>w_{2}(x_{0})>w_{2}(y_{0}))$ In this case, we have that

$\displaystyle 2\kappa(y_{0})w_{2}(y_{0})-2\kappa(x_{0})w_{2}(x_{0})$
$\displaystyle=2\kappa(y_{0})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)+\varepsilon(x_{0},y_{0})w_{2}(x_{0}),$

where

(3.12)
$\displaystyle\varepsilon(x,y)=2\left(\kappa(y_{0})-\kappa(x_{0})\right).$

Since $J^{\prime}(\tfrac{d_{0}}{2})<0$, the assumption implies the estimate

$\displaystyle-2J^{\prime}(\tfrac{d_{0}}{2})w_{2}^{2}(y_{0})\geq-2J^{\prime}(\tfrac{d_{0}}{2})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)^{2}\geq
2\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)^{2},$

where we have used Proposition 2.6 in the last inequality. Furthermore, when
completing the square, we find that

$\displaystyle
2J^{\prime}(\tfrac{-d_{0}}{2})w_{2}^{2}(x_{0})+\varepsilon(x_{0},y_{0})w_{2}(x_{0})=2J^{\prime}(\tfrac{-d_{0}}{2})\Bigl{(}w_{2}(x_{0})+\frac{\varepsilon(x_{0},y_{0})}{4J^{\prime}(\tfrac{-d_{0}}{2})}\Bigr{)}^{2}-\frac{\varepsilon^{2}(x_{0},y_{0})}{8J^{\prime}(\tfrac{-d_{0}}{2})}.$

Combining this equation, using Lemma (3.5), and Proposition 2.6, together with
the log-concavity assumption (3.11), (3.9) becomes the following.

(3.13) $\displaystyle 0\geq$
$\displaystyle-\nabla_{E_{1},E_{1}}^{2}F(x_{0},y_{0})-\nabla^{2}_{E_{2},E_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{d_{0}}{2})\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{-d_{0}}{2})\nabla_{e_{2}\oplus
0}F(x_{0},y_{0})$
$\displaystyle\quad+2(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\Bigl{(}w_{1}(y_{0})-w_{1}(x_{0})\Bigr{)}^{2}-\lambda\left(2J^{\prime}(\tfrac{d_{0}}{2})-2J^{\prime}(\tfrac{-d_{0}}{2})\right)-2\left(w_{2}(y_{0})-w_{2}(x_{0})\right)\phi^{\prime}(\tfrac{d_{0}}{2})$
$\displaystyle\quad+2\kappa(y_{0})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)+2\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)^{2}-\frac{\varepsilon^{2}(x_{0},y_{0})}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})}.$

  2. Case 2.

$(w_{2}(x_{0})\geq w_{2}(y_{0})>0)$

In this case, we observe that

$\displaystyle 2\kappa(y_{0})w_{2}(y_{0})-2\kappa(x_{0})w_{2}(x_{0})$
$\displaystyle=2\kappa(x_{0})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)+\varepsilon(x_{0},y_{0})w_{2}(y_{0}).$

Since $J^{\prime}(-\tfrac{d_{0}}{2})>0$, the standing assumption implies the
estimate

$\displaystyle 2J^{\prime}(-\tfrac{d_{0}}{2})w_{2}^{2}(x_{0})\geq
2J^{\prime}(-\tfrac{d_{0}}{2})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)^{2}\geq
2\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)^{2},$

where in the last inequality we have again used Proposition 2.6.

When completing the square, we get that

$\displaystyle-2J^{\prime}(\tfrac{d_{0}}{2})w_{2}^{2}(y_{0})+\varepsilon(x_{0},y_{0})w_{2}(y_{0})=-2J^{\prime}(\tfrac{d_{0}}{2})\Bigl{(}w_{2}(y_{0})-\frac{\varepsilon(x_{0},y_{0})}{4J^{\prime}(\tfrac{d_{0}}{2})}\Bigr{)}^{2}+\frac{\varepsilon^{2}(x_{0},y_{0})}{8J^{\prime}(\tfrac{d_{0}}{2})}.$

Similarly to Case 1, we obtain the following inequality:

(3.14) $\displaystyle 0\geq$
$\displaystyle-\nabla_{E_{1},E_{1}}^{2}F(x_{0},y_{0})-\nabla^{2}_{E_{2},E_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{d_{0}}{2})\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{-d_{0}}{2})\nabla_{e_{2}\oplus
0}F(x_{0},y_{0})$
$\displaystyle\quad+2(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\Bigl{(}w_{1}(y_{0})-w_{1}(x_{0})\Bigr{)}^{2}-\lambda\left(2J^{\prime}(\tfrac{d_{0}}{2})-2J^{\prime}(\tfrac{-d_{0}}{2})\right)-2\left(w_{2}(y_{0})-w_{2}(x_{0})\right)\phi^{\prime}(\tfrac{d_{0}}{2})$
$\displaystyle\quad+2\kappa(x_{0})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)+2\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)^{2}+\frac{\varepsilon^{2}(x_{0},y_{0})}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})}$

Note that Inequality (3.14) differs from Inequality (3.13) only in that
several of the terms are evaluated at $x_{0}$ rather than $y_{0}$.

  3. Case 3.

$(w_{2}(x_{0})\geq 0\geq w_{2}(y_{0}))$

In this case, we have

$\displaystyle 2\kappa(y_{0})w_{2}(y_{0})-2\kappa(x_{0})w_{2}(x_{0})$
$\displaystyle\geq 2\overline{\kappa}\left(w_{2}(y_{0})-w_{2}(x_{0})\right)$

Moreover, by Proposition 2.6

$\displaystyle
J^{\prime}(-\tfrac{d_{0}}{2})w_{2}^{2}(x_{0})-2J^{\prime}(\tfrac{d_{0}}{2})w_{2}^{2}(y_{0})$
$\displaystyle\geq
2\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(w_{2}^{2}(x_{0})+w_{2}^{2}(y_{0})\right)$
$\displaystyle\geq\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(w_{2}(x_{0})-w_{2}(y_{0})\right)^{2}.$

In all three cases, we find the following inequality

(3.15) $\displaystyle 0\geq$
$\displaystyle-\nabla_{E_{1},E_{1}}^{2}F(x_{0},y_{0})-\nabla^{2}_{E_{2},E_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{d_{0}}{2})\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{-d_{0}}{2})\nabla_{e_{2}\oplus
0}F(x_{0},y_{0})$
$\displaystyle\quad+2(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\Bigl{(}w_{1}(y_{0})-w_{1}(x_{0})\Bigr{)}^{2}-4\lambda\mathcal{C}(x_{0},y_{0})-2\left(w_{2}(y_{0})-w_{2}(x_{0})\right)\phi^{\prime}(\tfrac{d_{0}}{2})$
$\displaystyle\quad+2\overline{\kappa}\left(w_{2}(y_{0})-w_{2}(x_{0})\right)+\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)^{2}+\frac{\varepsilon^{2}(x_{0},y_{0})}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})}$

where we have rewritten
$\left(2J^{\prime}(\tfrac{d_{0}}{2})-2J^{\prime}(\tfrac{-d_{0}}{2})\right)=4\mathcal{C}(x_{0},y_{0}).$

We now estimate all the terms of (3.15), in terms of $\phi$ and $\mathcal{C}.$

Since we have assumed that $Z(x_{0},y_{0})=\alpha$, we have that

$\displaystyle
w_{2}(y_{0})-w_{2}(x_{0})=F(x_{0},y_{0})+\alpha=2\phi(\tfrac{d_{0}}{2})+\mathcal{C}(x_{0},y_{0})+\alpha.$

This gives

$\displaystyle-2\left(w_{2}(y_{0})-w_{2}(x_{0})\right)\phi^{\prime}(\tfrac{d_{0}}{2})$
$\displaystyle=$
$\displaystyle-2\alpha\phi^{\prime}(\tfrac{d_{0}}{2})-2\phi^{\prime}(\tfrac{d_{0}}{2})\mathcal{C}(x_{0},y_{0})-4\phi^{\prime}(\tfrac{d_{0}}{2})\phi(\tfrac{d_{0}}{2}),$
$\displaystyle 2\overline{\kappa}\left(w_{2}(y_{0})-w_{2}(x_{0})\right)$
$\displaystyle=$ $\displaystyle
2\overline{\kappa}\mathcal{C}(x_{0},y_{0})+4\overline{\kappa}\phi(\tfrac{d_{0}}{2})+2\alpha\overline{\kappa},$
$\displaystyle\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(w_{2}(y_{0})-w_{2}(x_{0})\right)^{2}$
$\displaystyle=$
$\displaystyle\alpha\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(\alpha+2\mathcal{C}(x_{0},y_{0})+4\phi(\tfrac{d_{0}}{2})\right)+4\phi^{2}(\tfrac{d_{0}}{2})\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})$
$\displaystyle+2\mathcal{C}(x_{0},y_{0})\Bigl{(}2\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\phi(\tfrac{d_{0}}{2})+\frac{1}{2}\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\mathcal{C}(x_{0},y_{0})\Bigr{)}.$

From Lemma 2.8, we have that

(3.19) $\displaystyle-\nabla^{2}_{E_{1},E_{1}}F(x_{0},y_{0})$ $\displaystyle=$
$\displaystyle-2\mathcal{C}(x_{0},y_{0})\phi^{\prime}(\tfrac{d_{0}}{2})+\nabla_{E_{1},E_{1}}^{2}(-\mathcal{C})(x_{0},y_{0}).$
(3.20) $\displaystyle-\nabla_{E_{2},E_{2}}^{2}F(x_{0},y_{0})$ $\displaystyle=$
$\displaystyle\nabla^{2}_{E_{2},E_{2}}(-\mathcal{C})(x_{0},y_{0})-2\phi^{\prime\prime}(\tfrac{d_{0}}{2}).$

Also by Lemma 3.5

(3.21) $\displaystyle-2J^{\prime}(\tfrac{d_{0}}{2})\nabla_{0\oplus
e_{2}}F(x_{0},y_{0})-2J^{\prime}(\tfrac{-d_{0}}{2})\nabla_{e_{2}\oplus
0}F(x_{0},y_{0})$
$\displaystyle=-2\left(J^{\prime}(\tfrac{d_{0}}{2})-J^{\prime}(-\tfrac{d_{0}}{2})\right)\phi^{\prime}(\tfrac{d_{0}}{2})+J^{\prime}(\tfrac{d_{0}}{2})\kappa(y_{0})+J^{\prime}(\tfrac{d_{0}}{2})^{3}-J^{\prime}(-\tfrac{d_{0}}{2})\kappa(x_{0})-J^{\prime}(-\tfrac{d_{0}}{2})^{3}.$

Finally, since

(3.22) $\displaystyle
2(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\Bigl{(}w_{1}(y_{0})-w_{1}(x_{0})\Bigr{)}^{2}\geq
0,$

this term can be dropped from the analysis. Thus, combining
(LABEL:phi'-terms), (LABEL:w_2-terms),(LABEL:w_2-squared-terms), (3.19),
(3.20), (3.21), and (3.22) into (3.15) gives

(3.23) $\displaystyle 0\geq$
$\displaystyle-2\phi^{\prime\prime}(\tfrac{d_{0}}{2})-4\phi^{\prime}(\tfrac{d_{0}}{2})\phi(\tfrac{d_{0}}{2})-8\phi^{\prime}(\tfrac{d_{0}}{2})\mathcal{C}(x_{0},y_{0})+4\overline{\kappa}\phi(\tfrac{d_{0}}{2})$
$\displaystyle\quad+4\phi^{2}(\tfrac{d_{0}}{2})\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})+4\mathcal{C}(x_{0},y_{0})\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\phi(\tfrac{d_{0}}{2})$
$\displaystyle\quad+\alpha\Bigl{(}2\overline{\kappa}-2\phi^{\prime}(\tfrac{d_{0}}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\left(\alpha+2\mathcal{C}(x_{0},y_{0})+4\phi(\tfrac{d_{0}}{2})\right)\Bigr{)}$
$\displaystyle\quad+2\overline{\kappa}\mathcal{C}(x_{0},y_{0})+\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})\mathcal{C}^{2}(x_{0},y_{0})-4\lambda\mathcal{C}(x_{0},y_{0})$
$\displaystyle\quad+J^{\prime}(\tfrac{d_{0}}{2})\kappa(y_{0})+[J^{\prime}(\tfrac{d_{0}}{2})]^{3}-J^{\prime}(-\tfrac{d_{0}}{2})\kappa(x_{0})-[J^{\prime}(-\tfrac{d_{0}}{2})]^{3}$
$\displaystyle\quad+\nabla_{E_{1},E_{1}}^{2}(-\mathcal{C})(x_{0},y_{0})+\nabla^{2}_{E_{2},E_{2}}(-\mathcal{C})(x_{0},y_{0})-\frac{\varepsilon^{2}(x_{0},y_{0})}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d_{0}}{2})}.$

Computing the term
$\nabla^{2}_{E_{1},E_{1}}(-\mathcal{C})+\nabla^{2}_{E_{2},E_{2}}(-\mathcal{C})$
see Proposition 7.4, we get

$\displaystyle\left(\nabla^{2}_{E_{1},E_{1}}+\nabla_{E_{2},E_{2}}^{2}\right)(-\mathcal{C})$
$\displaystyle=\kappa(x_{0})J^{\prime}(\tfrac{-d_{0}}{2})-\kappa(y_{0})J^{\prime}(\tfrac{d_{0}}{2})+[J^{\prime}(\tfrac{-d_{0}}{2})]^{3}-[J^{\prime}(\tfrac{d_{0}}{2})]^{3}$
$\displaystyle+\mathcal{C}(x_{0},y_{0})\nabla_{e_{2}\oplus(-e_{2})}\mathcal{C}(x_{0},y_{0})+\mathcal{D}(x_{0},y_{0}),$

where $\mathcal{D}$ is defined in Equation (7.14) and satisfies
$|\mathcal{D}(x,y)|\leq
Cd(x,y)\left(|\nabla\kappa|_{\infty}+|\Delta\kappa|_{\infty}\right)$ for some
$C>0$ (to be made precise in Section 7). This cancels the fifth line of
(3.23). Estimating

$\displaystyle\mathcal{C}(x_{0},y_{0})\nabla_{e_{2}\oplus(-e_{2})}\mathcal{C}(x_{0},y_{0})\geq\mathcal{C}(x_{0},y_{0})\left(\overline{\kappa}+\textup{tn}_{\overline{\kappa}}^{2}\right),$

we arrive at the following theorem.

######  Theorem 3.6.

Suppose $(M^{2},g)$ is a complete manifold of positive curvature. Let
$\Omega\subset M^{2}$ be a uniformly convex domain with $w=\log u$ the log of
the first Dirichlet eigenfunction. Suppose further that
$\phi:[0,D/2]\rightarrow\mathbb{R}$ is a $C^{2}$ function with $\phi(0)=0$
which satisfies the following inequalities (for any $x,y\in\Omega$):

(3.24)
$\displaystyle\begin{cases}0\leq&-2\phi^{\prime\prime}-4\phi\phi^{\prime}+4\overline{\kappa}\phi+4\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\phi^{2}+(-4\phi^{\prime}+2\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\phi)2\mathcal{C}(x,y)\\\
&+\Bigr{(}-4\lambda+3\overline{\kappa}+\textup{tn}_{\overline{\kappa}}^{2}(\tfrac{d(x,y)}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\mathcal{C}(x,y)\Bigl{)}\mathcal{C}(x,y)\\\
&-\frac{\varepsilon^{2}(x,y)}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})}+\mathcal{D}(x,y)\\\
0\leq&2\overline{\kappa}-2\phi^{\prime}(\tfrac{d(x,y)}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\left(2\mathcal{C}(x,y)+4\phi(\tfrac{d(x,y)}{2})\right)\\\
0>&\phi^{\prime}(0)-\frac{\underline{\kappa}}{2}\end{cases},$

where $\varepsilon$ is defined (3.12) and $\mathcal{D}$ is defined in (7.14).

Then, for any $x,y\in\Omega$

(3.25) $\langle\nabla
w(y),\gamma_{x,y}^{\prime}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma_{x,y}^{\prime}(\tfrac{-d}{2})\rangle\leq
2\phi(\tfrac{d}{2})+\mathcal{C}(x,y)$

###### Proof.

We consider an increasing family of convex domains $\Omega(t)\subset\Omega$
such that $\Omega(0)$ is a sufficiently close to a very small ball and where
$\Omega(1)=\Omega$. Such a family can be constructed by deforming the boundary
$\partial\Omega$ via curve shortening flow. This process gives a decreasing
family of convex domains which converge to a round point at the time of
singularity [Gag90]. As such, we can stop the flow close to the singular time
(when the domain is very close to a small ball). Doing this process backwards
in time (and reparametrizing so that it occurs in between $t=0$ and $t=1$), we
have the family of interest.

Since $\Omega(0)$ is very close to a small ball (and the metric is nearly
Euclidean in a small neighborhood), elliptic regularity implies that for $t=0$
(and for $t$ sufficiently small), the first eigenfunction of $\Omega(t)$ is
strictly log-concave.

Starting at $\Omega(0)$, we move forwards in time until the first time $t_{0}$
where $\textup{max}\operatorname{Hess}w=0,$ where the maximum ranges over the
unit tangent bundle. By the maximum principle computation above, $Z$ cannot
achieve a positive maximum in the interior and thus $Z$ must be non-positive
(since it is non-positive on the boundary). However, for any unit vector $X\in
T_{p}\Omega$ and geodesic $\gamma$ such that $\gamma^{\prime}(0)=X$, we have
the inequality

$\displaystyle\operatorname{Hess}w(X,X)$ $\displaystyle=\lim_{d\rightarrow
0^{+}}\frac{1}{d}\int_{-\tfrac{d}{2}}^{\tfrac{d}{2}}\operatorname{Hess}w(\gamma^{\prime}(s),\gamma^{\prime}(s))\,ds$
$\displaystyle\leq\lim_{d\rightarrow
0^{+}}\frac{1}{d}\left(2\phi(\tfrac{d}{2})+\mathcal{C}(\gamma(\tfrac{-d}{2}),\gamma(\tfrac{d}{2}))\right)$
$\displaystyle=\phi^{\prime}(0)-\frac{\kappa(p)}{2}<0,$

which induces a contradiction.

This shows that the eigenfunction is log-concave for all $t\in[0,1]$. As such,
we can apply the maximum principle to show that $Z$ must be non-positive at
$t=1.$ In other words,

$\displaystyle\langle\nabla
w(y),\gamma^{\prime}_{x,y}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma^{\prime}_{x,y}(-\tfrac{d}{2})\rangle\leq
2\phi(\tfrac{d}{2})+\mathcal{C}(x,y).$

∎

In Section 7, we estimate $\mathcal{D}$ and in particular obtain the following
estimate to deal with all non-ODE terms in (3.24).

######  Lemma 3.7.

When the diameter satisfies $D\leq\pi/(2\sqrt{\overline{\kappa}})$, we have
the estimate

$\displaystyle\Bigl{(}-4\lambda+3\overline{\kappa}+\textup{tn}_{\overline{\kappa}}^{2}(\tfrac{d(x,y)}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\mathcal{C}(x,y)\Bigr{)}\mathcal{C}(x,y)$
$\displaystyle-\frac{\varepsilon^{2}(x,y)}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})}+\mathcal{D}(x,y)$
$\displaystyle\geq$
$\displaystyle\Bigr{(}-4\lambda+3\overline{\kappa}+(\overline{\kappa}-\underline{\kappa})(1+\frac{\pi}{2})+\sqrt{2}\overline{\kappa}\left(\frac{\overline{\kappa}-\underline{\kappa}}{\underline{\kappa}}\right)^{2}\left(\frac{4{\overline{\kappa}}}{\pi\sqrt{\underline{\kappa}}}+1\right)^{2}\Bigl{)}\mathcal{C}(x,y),$
$\displaystyle\,+2\mathcal{C}(x,y)\left(\frac{-2(\inf\Delta\kappa)_{-}}{\underline{\kappa}}+\left(\frac{1}{\underline{\kappa}^{2}}+\frac{3\pi^{2}\sqrt{8}}{4\overline{\kappa}\underline{\kappa}}\right)|\nabla\kappa|^{2}_{\infty}+\frac{4\pi|\nabla\kappa|_{\infty}}{\underline{\kappa}}\right).$

###  3.5. Explicit modulus of concavity estimates

We are now in the position to starting applying Theorem 3.6 to several choices
of $\phi.$ We start with the simplest choice, which is $\phi\equiv 0.$

######  Theorem 3.8.

Suppose that $(M^{2},g)$ is a surface of positive curvature $\kappa$ such that
$0<\underline{\kappa}\leq\kappa<\overline{\kappa}.$ Let $\Omega\subset M^{2}$
be uniformly and geodesically convex with
$\textup{diam}(\Omega)\leq\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$ and suppose
that

(3.26)
$\displaystyle-2\lambda+\tfrac{3}{2}\overline{\kappa}+(\overline{\kappa}-\underline{\kappa})(\tfrac{1}{2}+\tfrac{\pi}{4})+\sqrt{2}\overline{\kappa}\left(\tfrac{\overline{\kappa}-\underline{\kappa}}{\underline{\kappa}}\right)^{2}\left(\tfrac{4{\overline{\kappa}}}{\pi\sqrt{\underline{\kappa}}}+1\right)^{2}$
$\displaystyle-\tfrac{2}{\underline{\kappa}}(\inf\Delta\kappa)_{-}+\left(\tfrac{1}{\underline{\kappa}^{2}}+\tfrac{3\pi^{2}\sqrt{8}}{4\overline{\kappa}\underline{\kappa}}\right)|\nabla\kappa|^{2}_{\infty}+\tfrac{4\pi}{\underline{\kappa}}|\nabla\kappa|_{\infty}<0.$

Then one has that

(3.27) $\displaystyle\langle\nabla
w(y),\gamma_{x,y}^{\prime}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma_{x,y}^{\prime}(-\tfrac{d}{2})\rangle\leq\mathcal{C}(x,y).$

######  Remark 3.9.

Theorem 3.8 also holds true under the weaker assumption that without any
diameter restriction on $\Omega$ and any $x,y\in\Omega$

(3.28)
$\displaystyle\Bigr{(}-4\lambda+3\overline{\kappa}+\textup{tn}_{\overline{\kappa}}^{2}(\tfrac{d}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2})\mathcal{C}(x,y)\Bigl{)}\mathcal{C}(x,y)-\frac{\varepsilon^{2}(x,y)}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2})}+\mathcal{D}(x,y)>0,$

where $\mathcal{C}$ is defined in (1.7), $\mathcal{D}$ in (7.14) and
$\varepsilon$ in (3.12).

###### Proof of Theorem 3.8.

This follows from Proposition 3.6 with $\phi\equiv 0$, Estimate (3.7) and the
fact that $\mathcal{C}(x,y)<0.$ ∎

######  Remark 3.10.

Note that if $(M^{2},g)$ satisfies the assumptions of Theorem 3.8, one
recovers the Hessian estimate from [KNTW22]. Indeed, dividing (3.27) by $d$
and after passing to the limit as $d\rightarrow 0^{+},$ one recovers

(3.29) $\displaystyle\operatorname{Hess}w\leq-\frac{\kappa}{2}.$

Note that in that paper, we used the function $\frac{\kappa}{2}$ to absorb
problematic term in a one-point maximum principle. Interestingly,
$\mathcal{C}$ helps us to control problematic terms in a similar way and these
two quantities are related by Equation (1.8).

We now apply Theorem 3.11 again, this time with
$\phi=\psi+\textup{tn}_{\underline{\kappa}}$ and derive differential
inequalities for $\psi$. The result below generalizes Theorem 3.6 in [SWW19]
to surfaces of non-constant sectional curvature. Their version of the theorem
provides a parabolic version of this argument. We only state the elliptic
version as the proof of the parabolic version is almost identical.

######  Theorem 3.11.

Suppose the $(M^{2},g)$ is a complete manifold of positive sectional curvature
$\kappa$, such that $0<\underline{\kappa}\leq\kappa\leq\overline{\kappa}$
which satisfies

(3.30)
$\displaystyle-\tfrac{1}{2}\overline{\kappa}+(\overline{\kappa}-\underline{\kappa})(\tfrac{1}{2}+\tfrac{\pi}{4})+\sqrt{2}\overline{\kappa}\left(\tfrac{\overline{\kappa}-\underline{\kappa}}{\underline{\kappa}}\right)^{2}\left(\tfrac{4{\overline{\kappa}}}{\pi\sqrt{\underline{\kappa}}}+1\right)^{2}$
$\displaystyle\quad-\tfrac{2}{\underline{\kappa}}(\inf\Delta\kappa)_{-}+\left(\tfrac{1}{\underline{\kappa}^{2}}+\tfrac{3\pi^{2}\sqrt{8}}{4\overline{\kappa}\underline{\kappa}}\right)|\nabla\kappa|^{2}_{\infty}+\tfrac{4\pi}{\underline{\kappa}}|\nabla\kappa|_{\infty}<0$

Then we have that for any convex $\Omega\subset M,$ with
$\textup{diam}(\Omega)=D\leq\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$

$\displaystyle\langle\nabla
w(y),\gamma_{x,y}^{\prime}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma_{x,y}^{\prime}(\tfrac{-d}{2})\rangle\leq
2\psi(\tfrac{d}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2}),$

where $\psi:[0,D/2]\rightarrow\mathbb{R}$ satisfies

$\displaystyle\begin{cases}0\geq\psi^{\prime\prime}+2\psi\psi^{\prime}-2\textup{tn}_{\underline{\kappa}}\left(\psi^{\prime}+\psi^{2}+\lambda\right)-2\psi\left(\overline{\kappa}-\underline{\kappa}\right)-4\psi^{\prime}\left(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}}\right)\\\
0\geq-2(\overline{\kappa}-\underline{\kappa})+2\psi^{\prime}-4\textup{tn}_{\underline{\kappa}}\psi-2\textup{tn}_{\underline{\kappa}}^{2}+2\textup{tn}_{\underline{\kappa}}\textup{tn}_{\overline{\kappa}}\\\
0>\psi^{\prime}(0)+\frac{\underline{\kappa}}{2}\\\ 0\geq\psi^{\prime}\\\
0=\psi(0).\end{cases}$

###### Proof.

This follows from setting $\phi=\psi+\textup{tn}_{\underline{\kappa}}.$
Indeed, Using $\psi,\psi^{\prime}\leq 0,$ and Lemma 2.7, we have

$\displaystyle(-4\phi^{\prime}+2\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\phi)2\mathcal{C}(x,y)$
$\displaystyle=$
$\displaystyle(-4\psi^{\prime}-4{\underline{\kappa}}+2\textup{tn}_{\underline{\kappa}}\psi-2\textup{tn}_{\underline{\kappa}}^{2})2\mathcal{C}(x,y)$
$\displaystyle\geq$ $\displaystyle
8\psi^{\prime}\textup{tn}_{\overline{\kappa}}+8\underline{\kappa}\textup{tn}_{\underline{\kappa}}-4\textup{tn}_{\underline{\kappa}}^{2}\psi+4\textup{tn}_{\underline{\kappa}}^{3},$

where we used

$\textup{tn}_{\underline{\kappa}}^{\prime}=\underline{\kappa}+\textup{tn}_{\underline{\kappa}}^{2}.$

Hence

(3.31)
$\displaystyle-2\phi^{\prime\prime}-4\phi\phi^{\prime}+4\overline{\kappa}\phi+4\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\phi^{2}+(-4\phi^{\prime}+2\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\phi)2\mathcal{C}(x,y)$
$\displaystyle\geq-2(\psi^{\prime\prime}+2\underline{\kappa}\textup{tn}_{\underline{\kappa}}+2{\textup{tn}_{\underline{\kappa}}^{3}})-4(\psi+\textup{tn}_{\underline{\kappa}})(\psi^{\prime}+\textup{tn}^{2}_{\underline{\kappa}}+\underline{\kappa})+4\overline{\kappa}(\psi+\textup{tn}_{\underline{\kappa}})+4\textup{tn}_{\underline{\kappa}}(\psi+\textup{tn}_{\underline{\kappa}})^{2}$
$\displaystyle\ \ \ \ \ \ \ \ \
+8\psi^{\prime}\textup{tn}_{\overline{\kappa}}+8\underline{\kappa}\textup{tn}_{\underline{\kappa}}-4\textup{tn}_{\underline{\kappa}}^{2}\psi+4\textup{tn}_{\underline{\kappa}}^{3}$
$\displaystyle=-2\psi^{\prime\prime}-4\psi\psi^{\prime}+4\textup{tn}_{\underline{\kappa}}\left(\psi^{\prime}+\psi^{2}\right)+4\psi(\overline{\kappa}-\underline{\kappa})+8\psi^{\prime}\left(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}}\right)+4\overline{\kappa}\textup{tn}_{\underline{\kappa}}.$

Now in order to apply Theorem 3.6, we use (3.31), Lemma 3.7, (3.30) together
with (2.8) to obtain the estimate

$\displaystyle-2\phi^{\prime\prime}-4\phi\phi^{\prime}+4\overline{\kappa}\phi+4\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\phi^{2}+(-4\phi^{\prime}+2\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\phi)2\mathcal{C}(x,y)$
$\displaystyle+\Bigr{(}-4\lambda+3\overline{\kappa}+\textup{tn}_{\overline{\kappa}}^{2}(\tfrac{d(x,y)}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\mathcal{C}(x,y)\Bigl{)}\mathcal{C}(x,y)$
$\displaystyle-\frac{\varepsilon^{2}(x,y)}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})}+\mathcal{D}(x,y)$
$\displaystyle\geq$
$\displaystyle-2\psi^{\prime\prime}-4\psi\psi^{\prime}+4\textup{tn}_{\underline{\kappa}}\left(\psi^{\prime}+\psi^{2}+\lambda\right)+4\psi(\overline{\kappa}-\underline{\kappa})+8\psi^{\prime}\left(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}}\right)$
$\displaystyle\,-\Bigr{(}-\overline{\kappa}+(\overline{\kappa}-\underline{\kappa})(1+\frac{\pi}{2})+\sqrt{2}\overline{\kappa}\left(\frac{\overline{\kappa}-\underline{\kappa}}{\underline{\kappa}}\right)^{2}\left(\frac{4{\overline{\kappa}}}{\pi\sqrt{\underline{\kappa}}}+1\right)^{2}\Bigl{)}\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})$
$\displaystyle\,-2\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\left(-\tfrac{2}{\underline{\kappa}}(\inf\Delta\kappa)_{-}+\left(\tfrac{1}{\underline{\kappa}^{2}}+\tfrac{3\pi^{2}\sqrt{8}}{4\overline{\kappa}\underline{\kappa}}\right)|\nabla\kappa|^{2}_{\infty}+\tfrac{4\pi}{\underline{\kappa}}|\nabla\kappa|_{\infty}\right),$

which is non-negative by assumption. Next, we estimate the second inequality
of (3.24), using $\phi=\psi+\textup{tn}_{\underline{\kappa}}.$ We obtain that

(3.32) $\displaystyle
2\overline{\kappa}-2\phi^{\prime}(\tfrac{d(x,y)}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})\left(2\mathcal{C}(x,y)+4\phi(\tfrac{d(x,y)}{2})\right)$
$\displaystyle\geq
2(\overline{\kappa}-\underline{\kappa})-2\psi^{\prime}+4\textup{tn}_{\underline{\kappa}}\psi+2\textup{tn}_{\underline{\kappa}}^{2}-2\textup{tn}_{\underline{\kappa}}\textup{tn}_{\overline{\kappa}}$

which is positive by assumption. We conclude that the first two inequalities
of (3.24) hold true. Thus, it remains only to verify the last inequality. To
see this, note that

$\displaystyle\phi^{\prime}(0)-\tfrac{\underline{\kappa}}{2}=\psi^{\prime}(0)+\tfrac{\underline{\kappa}}{2}<0,$

by assumption. From Proposition 3.6, we infer that

$\displaystyle
F(x,y)=2\psi(\tfrac{d(x,y)}{2})+2\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})+\mathcal{C}(x,y)$

is a modulus of concavity for $w=\log u.$ Using Lemma 2.7, we infer the claim.
∎

######  Remark 3.12.

Theorem 3.11 holds true under the weaker assumption that for any
$x,y\in\Omega$

$\displaystyle
4\overline{\kappa}\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2})+\Bigr{(}3\overline{\kappa}+\textup{tn}_{\overline{\kappa}}^{2}(\tfrac{d}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2})\mathcal{C}(x,y)\Bigl{)}\mathcal{C}(x,y)-\frac{\varepsilon^{2}(x,y)}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2})}+\mathcal{D}(x,y)>0.$

##  4\. The perturbed Euclidean model

We now apply Theorem 3.11 to find an explicit modulus of concavity for $w=\log
u.$ Note that when $\overline{\kappa}=\underline{\kappa},$ we recover Theorem
3.6 in [SWW19]. Namely, following [DSW21], one obtains that
$\psi(s)=(\log\phi_{e})^{\prime}=\tfrac{\pi}{2D}\tan(\tfrac{2\pi}{D}s)$ is a
modulus of concavity, where $\phi_{e}$ is the first eigenfunction of the
Euclidean model

$\displaystyle\phi^{\prime\prime}=-\lambda\phi\quad\textup{in
}[-D/2,D/2]\quad\textup{with }\phi(-D/2)=\phi(D/2)=0.$

However, since we are interested in the case where the curvature is not
constant, it is necessary to use a different model. Indeed, we consider the
perturbed Euclidean model

(4.1)
$\displaystyle\overline{\phi}^{\prime\prime}-4(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})\overline{\phi}^{\prime}=-\overline{\lambda}\phi\quad\textup{in
}[-{D}/2,{D}/2]\quad\textup{ with
}\,\overline{\phi}(-\tfrac{D}{2})=\overline{\phi}(\tfrac{D}{2})=0.$

We now show that $\psi=(\log\overline{\phi}_{D})^{\prime}$ satisfies the
differential inequalities of Theorem 3.11 where $\overline{\phi}_{D}$ denotes
the first eigenfunction of the problem (4.1). We divide the proof in several
lemma.

We first show that the last two conditions of Theorem 3.11 are satisfied.

######  Lemma 4.1.

Suppose that $\overline{\phi}_{D}$ is the first eigenfunction of the perturbed
Euclidean model (4.1) on $[-{D}/2,{D}/2]$ with Dirichlet boundary conditions.
Choosing $\overline{\phi}_{D}$ to be positive, we find that
$\overline{\phi}_{D}$ and $\psi=(\log\overline{\phi}_{D})^{\prime}$ are
strictly decreasing on $(0,{D}/2).$

###### Proof.

Note that the the first eigenfunction does not switch sign (by the nodal
domain theorem) so can be chosen to be positive. Moreover, since any solution
$\phi$ can be replaced by $\widehat{\phi}(s):=\phi(-s)+\phi(s),$ we find that
$\overline{\phi}_{D}$ is even. We now show that $\overline{\phi}_{D}$ is
strictly decreasing on $(0,{D}/2].$ Using an integrating factor, one finds
that

$\displaystyle\left(\overline{\phi}_{D}^{\prime}\exp(-4\int_{0}^{x}(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})\,ds)\right)^{\prime}=-\overline{\lambda}_{1}\exp(-4\int_{0}^{x}(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})\,ds)\overline{\phi}_{D}<0.$

Since $\overline{\phi}_{D}$ is an even function, we find that
$\overline{\phi}_{D}^{\prime}(0)=0$ and thus integrating this equation yields
the first claim. To show that $\psi$ is strictly decreasing, we calculate

$\displaystyle\psi^{\prime}=\left(\frac{\phi^{\prime}}{\phi}\right)^{\prime}$
$\displaystyle=\frac{\phi^{\prime\prime}\phi-(\phi^{\prime})^{2}}{\phi^{2}}$
$\displaystyle=4(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})\psi-\overline{\lambda_{1}}-\psi^{2}<0.$

∎

We are now concerned with the first conditions from Theorem 3.11.

######  Lemma 4.2.

We have the inequality

$\displaystyle
0\geq\psi^{\prime\prime}+2\psi\psi^{\prime}-2\textup{tn}_{\underline{\kappa}}\left(\psi^{\prime}+\psi^{2}+\overline{\lambda}\right)-2\psi\left(\overline{\kappa}-\underline{\kappa}\right)-4\psi^{\prime}\left(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}}\right)$

in $[0,{D}/2).$

###### Proof.

By Lemma 4.1, $\psi$ is negative. As such, combining this with the first
inequality of Theorem 3.11, we find that

$\displaystyle
4\psi\left(\overline{\kappa}-\underline{\kappa}+\textup{tn}_{\overline{\kappa}}^{2}-\textup{tn}_{\underline{\kappa}}^{2}\right)-8\textup{tn}_{\underline{\kappa}}\psi(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})-2\psi\left(\overline{\kappa}-\underline{\kappa}\right)\leq
0$

if and only if

$\displaystyle
2(\overline{\kappa}-\underline{\kappa})+4\textup{tn}_{\overline{\kappa}}^{2}+4\textup{tn}_{\underline{\kappa}}^{2}-8\textup{tn}_{\underline{\kappa}}\textup{tn}_{\overline{\kappa}}\geq
0.$

The latter of which is true since $\overline{\kappa}\geq\underline{\kappa}.$ ∎

We now show the second condition.

######  Lemma 4.3.

For ${D}<\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$ and
$\overline{\kappa}\leq\tfrac{8+4\pi}{3+4\pi}\underline{\kappa},$ one has

$\displaystyle
2(\overline{\kappa}-\underline{\kappa})-2\psi^{\prime}+4\textup{tn}_{\underline{\kappa}}\psi+2\textup{tn}_{\underline{\kappa}}^{2}-2\textup{tn}_{\underline{\kappa}}\textup{tn}_{\overline{\kappa}}\geq
0.$

###### Proof.

Note that

$\displaystyle
2(\overline{\kappa}-\underline{\kappa})-2\psi^{\prime}+4\textup{tn}_{\underline{\kappa}}\psi+2\textup{tn}_{\underline{\kappa}}^{2}-2\textup{tn}_{\underline{\kappa}}\textup{tn}_{\overline{\kappa}}$
$\displaystyle=$ $\displaystyle
2(\overline{\kappa}-\underline{\kappa})+2\overline{\lambda}_{1}-8(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})\psi+2\psi^{2}+4\psi\textup{tn}_{\underline{\kappa}}+(\textup{tn}_{\underline{\kappa}}-\textup{tn}_{\overline{\kappa}})^{2}-\tfrac{1}{2}{\textup{tn}_{\overline{\kappa}}^{2}}.$

Using the identity,
$2\psi^{2}+4\psi\textup{tn}_{\underline{\kappa}}=2\left(\psi+\textup{tn}_{\underline{\kappa}}\right)^{2}-2\textup{tn}_{\underline{\kappa}}^{2}$,
we find the estimate

(4.2) $\displaystyle
2(\overline{\kappa}-\underline{\kappa})-2\psi^{\prime}+4\textup{tn}_{\underline{\kappa}}\psi+2\textup{tn}_{\underline{\kappa}}^{2}-2\textup{tn}_{\underline{\kappa}}\textup{tn}_{\overline{\kappa}}$
$\displaystyle\geq
2(\overline{\kappa}-\underline{\kappa})+2\overline{\lambda}_{1}-\tfrac{1}{2}\textup{tn}_{\overline{\kappa}}^{2}-2\textup{tn}_{\underline{\kappa}}^{2}.$

The latter is positive by our diameter and curvature pinching assumption. Now
we can bound the first eigenvalue of (4.1) from below by converting it into a
Schrödinger equation with Dirichlet boundary condition:

$\displaystyle-\varphi^{\prime\prime}+\Bigl{(}4(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})^{2}-2(\textup{tn}^{2}_{\overline{\kappa}}-\textup{tn}^{2}_{\underline{\kappa}}+\overline{\kappa}-\underline{\kappa})\Bigr{)}\varphi=\lambda\varphi.$

From this we get

(4.3) $\displaystyle\overline{\lambda}_{1}$ $\displaystyle=\inf_{u\in
H^{1}_{0}(-{D}/2,{D}/2)}\frac{\int_{-{D}/2}^{{D}/2}(u^{\prime})^{2}+\Bigl{(}4(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})^{2}-2(\textup{tn}^{2}_{\overline{\kappa}}-\textup{tn}^{2}_{\underline{\kappa}}+\overline{\kappa}-\underline{\kappa})\Bigr{)}u^{2}\,ds}{\int_{-{D}/2}^{{D}/2}u^{2}\,ds}$
$\displaystyle\geq\frac{\pi^{2}}{{D}^{2}}-(4+\pi)(\overline{\kappa}-\underline{\kappa})$

where we used (7.16). Thus, (4.2) can be estimated as follows

$\displaystyle
2(\overline{\kappa}-\underline{\kappa})+2\overline{\lambda}_{1}-\tfrac{1}{2}\textup{tn}_{\overline{\kappa}}^{2}-2\textup{tn}_{\underline{\kappa}}^{2}$
(4.4)
$\displaystyle\geq-(6+2\pi)(\overline{\kappa}-\underline{\kappa})+8\overline{\kappa}-\tfrac{1}{2}\overline{\kappa}-2\underline{\kappa},$

where we used ${D}<\pi/(2\sqrt{\overline{\kappa}})$. Note that (4) is
nonnegative whenever

$\displaystyle\overline{\kappa}\leq\frac{8+4\pi}{3+4\pi}\underline{\kappa},$

which is our pinch assumption. ∎

######  Lemma 4.4.

For ${D}\leq\pi/(2\sqrt{\overline{\kappa}})$ and
$\overline{\kappa}<\tfrac{7+2\pi}{2\pi}\underline{\kappa},$

$\displaystyle\psi^{\prime}(0)+\frac{\underline{\kappa}}{2}<0.$

###### Proof.

Observe that
$\psi^{\prime}(0)=-\overline{\lambda}_{1}\leq-\tfrac{\pi^{2}}{{D}^{2}}+(4+\pi)(\overline{\kappa}-\underline{\kappa})$
by (4.3). From the diameter restriction
${D}\leq\pi/(2\sqrt{\overline{\kappa}}),$ one finds that

$\displaystyle\psi^{\prime}(0)+\frac{\underline{\kappa}}{2}\leq-4\overline{\kappa}+(4+\pi)(\overline{\kappa}-\underline{\kappa})+\frac{\underline{\kappa}}{2}<0$

by our pinch assumption. ∎

We thus arrived at the following

######  Theorem 4.5.

Suppose the $(M^{2},g)$ is a complete manifold of positive sectional curvature
$\kappa$, such that
$0<\underline{\kappa}\leq\kappa\leq\overline{\kappa}<\tfrac{8+4\pi}{3+4\pi}\underline{\kappa}$
which satisfies (3.30). Suppose that $\Omega\subset M^{2}$ is a geodesically
convex domain such that
$\textup{diam}(\Omega)=D<\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$. Then for any
$x,y\in\Omega$,

$\displaystyle\langle\nabla
w(y),\gamma_{x,y}^{\prime}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma_{x,y}^{\prime}(\tfrac{-d}{2})\rangle\leq
2(\log\overline{\phi}_{D})^{\prime}(\tfrac{d}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2}),$

where $w=\log u$, the log of the first Dirichlet eigenfunction on $\Omega$ and
$\overline{\phi}_{D}$ is the first eigenfunction of the perturbed Euclidean
model (4.1).

###### Proof of Theorem 4.5.

We let ${D^{\prime}}>D$ such that
${D^{\prime}}<\tfrac{\pi}{2\sqrt{\overline{\kappa}}}.$ Consider
$\psi_{D^{\prime}}=(\log\overline{\phi}_{D^{\prime}})^{\prime},$ where
$\overline{\phi}_{D^{\prime}}$ is the solution of the problem

$\displaystyle\overline{\phi}^{\prime\prime}-4(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})\overline{\phi}^{\prime}=-\overline{\lambda}\phi\quad\textup{in
}[-{D^{\prime}}/2,{D^{\prime}}/2]\quad\textup{ with
}\,\overline{\phi}(-\tfrac{D^{\prime}}{2})=\overline{\phi}(\tfrac{D^{\prime}}{2})=0.$

Note also that since ${D^{\prime}}<\tfrac{\pi}{2\sqrt{\overline{\kappa}}},$
all the previous lemmas apply to
$\psi_{D^{\prime}}=(\log\overline{\phi}_{D^{\prime}})^{\prime}$ and hence the
assumptions of Theorem 3.11 are satisfied (since $\psi_{D^{\prime}}$ is
$C^{2}$ in $[0,D/2]$).

As a result, we find that for any $x,y\in\Omega$

$\displaystyle\langle\nabla
w(y),\gamma_{x,y}^{\prime}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma_{x,y}^{\prime}(\tfrac{d}{2})\rangle\leq
2(\log\overline{\phi}_{D^{\prime}})^{\prime}(\tfrac{d(x,y)}{2})+\textup{tn}_{\underline{\kappa}}(\tfrac{d(x,y)}{2}).$

Then letting ${D^{\prime}}\rightarrow D$ gives the claim. ∎

##  5\. Fundamental Gap Comparison

In this section, we compare the fundamental gap to the perturbed Euclidean
model (4.1) in order to obtain the desired lower bound. However, previously
established gap comparison theorems do not apply directly in this case, so
some modifications are needed.

Since $\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}}\geq
0$, we can apply a Riccati comparison theorem (see, e.g., [EH90]).

######  Proposition 5.1.

Suppose we have two functions $\psi_{1},\psi_{2}\leq 0$ which satisfy
$\psi_{1}(0)=\psi_{2}(0)=0$ and

$\displaystyle\psi_{1}^{\prime}+\psi_{1}^{2}+\lambda-4(\textup{tn}_{\overline{\kappa}}-\textup{tn}_{\underline{\kappa}})\psi_{1}$
$\displaystyle=$ $\displaystyle 0,$
$\displaystyle\psi_{2}^{\prime}+\psi_{2}^{2}+\lambda$ $\displaystyle=$
$\displaystyle 0.$

Then we have

$\displaystyle\psi_{1}\leq\psi_{2}.$

Applying this to $\psi_{1}=(\log\overline{\phi}_{1})$ and
$\psi_{2}=(\log\tilde{\phi})^{\prime},$ where $\tilde{\phi}$ satisfies

$\displaystyle\phi^{\prime\prime}=-\lambda\phi\quad\textup{on
}[-L/2,L/2]\quad\textup{ with }\quad\phi(-L/2)=\phi(L/2)=0,$

where $L>0$ is chosen such that
$\tfrac{\pi^{2}}{L^{2}}=\overline{\lambda}_{1}.$ From this, we obtain the our
final log-concavity estimate.

(5.1) $\displaystyle\langle\nabla
w(y),\gamma^{\prime}(\tfrac{d}{2})\rangle-\langle\nabla
w(x),\gamma^{\prime}(\tfrac{-d}{2})\rangle\leq-\frac{\pi}{L}\tan\left(\frac{\pi}{L}{d(x,y)}\right)+\textup{tn}_{\underline{\kappa}}\left(\frac{d(x,y)}{2}\right),$

where $L=\tfrac{\pi}{\sqrt{\overline{\lambda}_{1}}},$ and
$\overline{\lambda}_{1}$ denotes the first eigenvalue of the perturbed
Euclidean model.

Using this estimate, we can compare the fundamental gap of our domain to that
of a one-dimensional interval which is slightly longer, where the length of
the interval depends on the pinch. More precisely, we prove that

$\displaystyle\Gamma(\Omega)\geq\frac{3\pi^{2}}{L^{2}}.$

In order to establish this, we appeal to a previously known gap comparison
theorem.

######  Theorem 5.2 (Theorem 4.6 [SWW19]).

Let $\Omega\subset M^{n}$ be convex and suppose that
$\textup{Ric}_{M}\geq(n-1)K.$ Let $v=v(x,t)$ be a solution to the equation

$\displaystyle\begin{cases}\partial_{t}v=\Delta v+2\langle\nabla v,X\rangle\\\
\partial_{\nu}v=0\quad\textup{on }\partial\Omega\end{cases}$

Suppose that $X$ is a time dependent vector field such that

$\displaystyle\langle X,\gamma^{\prime}(\tfrac{d}{2})\rangle-\langle
X,\gamma^{\prime}(\tfrac{-d}{2})\rangle\leq
2w(\tfrac{d}{2})+(n-1)\textup{tn}_{K}(\tfrac{d}{2}).$

Moreover, assume that
$\varphi:[0,D/2)\times\mathbb{R}_{+}\rightarrow\mathbb{R}$ satisfies

  * $i)$

$\varphi(\cdot,0)$ is a modulus of continuity of $v(\cdot,0),$ i.e.

$\displaystyle|v(y,0)-v(x,0)|\leq 2\varphi(\tfrac{d}{2},0),$

  * $ii)$

$\varphi$ satisfies

$\displaystyle\frac{\partial\varphi}{\partial
t}\geq\varphi^{\prime\prime}+2w\varphi^{\prime}$

  * $iii)$

$\varphi^{\prime}>0$ in $[0,D/2)\times\mathbb{R}_{+}$

  * $iv)$

$\varphi(0,t)\geq 0$ for all $t\geq 0.$

Then we have that

$\displaystyle|v(y,t)-v(x,t)|\leq 2\varphi(\tfrac{d}{2},t)\quad\textup{for all
}t\geq 0.$

With this, we are finally able to give a proof of the main theorem.

######  Theorem 5.3.

Suppose that $(M^{2},g)$ is a complete manifold of positive sectional
curvature $\kappa$ such that
$0<\underline{\kappa}\leq\kappa\leq\overline{\kappa}\leq\tfrac{8+4\pi}{3+4\pi}\underline{\kappa}$
which satisfies (3.30). Then for any geodesically convex $\Omega\subset M^{2}$
such that $\textup{diam}(\Omega)=D<\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$

$\displaystyle\Gamma\geq\frac{3\pi^{2}}{D^{2}}-(12+3\pi)(\overline{\kappa}-\underline{\kappa}).$

###### Proof.

For $C>0$ large enough, a straightforward computation shows that

$\displaystyle\varphi(s,t)=Ce^{-\tfrac{3\pi^{2}}{L^{2}}t}\frac{\cos(\frac{2\pi
s}{L})}{\cos(\frac{\pi s}{L})}$ $\displaystyle
s\in\left(-\frac{L}{2},\frac{L}{2}\right)$ $\displaystyle
v(x,t)=e^{-\Gamma(\Omega)t}\frac{u_{2}(x)}{u_{1}(x)}$ $\displaystyle
x\in\Omega$

satisfy the assumption of Theorem 5.2, together with $X=\nabla\log u_{1}$. As
a result, for all $t>0$ we have the estimate

$\displaystyle
e^{-\Gamma(\Omega)t}\left(\frac{u_{2}(y)}{u_{1}(y)}-\frac{u_{2}(x)}{u_{1}(x)}\right)\leq
Ce^{-\tfrac{3\pi^{2}}{L^{2}}t}\frac{\cos(\frac{d(x,y)\pi}{L})}{\cos(\frac{d(x,y)\pi}{2L})}$

This implies that $\Gamma(\Omega)\geq\frac{3\pi^{2}}{L^{2}}.$ Since
$\lambda_{1}=\pi^{2}/L^{2},$ and by (4.3), we conclude

$\displaystyle\Gamma(\Omega)\geq\frac{3\pi^{2}}{D^{2}}-(12+3\pi)(\overline{\kappa}-\underline{\kappa}).$

∎

##  6\. Proof of Proposition 3.4 and Lemma 3.5

In order to prove Proposition 3.4, we begin by computing the derivatives of
the variation $\eta$, and then the derivatives of $\mathcal{C}$.

###  6.1. The derivatives of the variation $\eta$

Throughout this section, we freely use the variations $\eta_{i}$ defined by
Equation (3.2). We will also use the fact that the Jacobi fields
$J^{1,0},J^{0,1}$ and $J=J^{1,0}+J^{0,1}$ only have an $e_{1}$ component, and
so we also denote their coefficients with the same notation.

######  Lemma 6.1.

We have the following derivative formulas:

$\displaystyle\langle\nabla_{r}\nabla_{r}{\eta^{\prime}},e_{1}\rangle(0,\tfrac{-d_{0}}{2})=\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}\kappa_{1}(t)J^{1,0}(t)(J(t))^{2}\,dt,$

$\displaystyle\langle\nabla_{r}\nabla_{r}{\eta^{\prime}},e_{1}\rangle(0,\tfrac{d_{0}}{2})=-\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}\kappa_{1}(t)J^{0,1}(t)(J(t))^{2}\,dt.$

Here $\kappa$ is the sectional curvature and $\kappa_{1}=e_{1}(\kappa)$.

###### Proof.

In order to establish these identities, we calculate
$\nabla_{r}\nabla_{r}{\eta^{\prime}}(0,\tfrac{d_{0}}{2})$. To do so, we define
an orthonormal frame along $\eta(r,s)$ in the following way:

We parallel transport the vector $e_{1}$ from $x$ along $\sigma_{1}(r)$. In
other words, we take $e_{1}=\tfrac{\partial}{\partial r}\sigma_{1}(r)$. For
$r$ small, $\eta^{\prime}(r,-\tfrac{d_{0}}{2}),e_{1}$ will be a frame along
$\sigma_{1}(r)$. We then define

(6.1)
$T=\tfrac{\eta_{i}^{\prime}(r,-\tfrac{d_{0}}{2})}{\|\eta_{i}^{\prime}(r,-\tfrac{d_{0}}{2})\|},\quad
V(r)=\frac{\widetilde{V}(r)}{\|\widetilde{V}(r)\|},$

where $\widetilde{V}(r)=e_{1}-\langle T,e_{1}\rangle T$. The pair $\\{T,V\\}$
is an orthonormal frame along $\sigma_{1}(r)$ with $V(0)=e_{1},T(0)=e_{2}.$
Finally, we parallel translate $V(r)$ along $\eta(r,\cdot)$, so that
$\\{T,V\\}$ form an orthornormal frame which is parallel along each geodesic
$\eta(r,\cdot)$.

We now establish that the covariant derivative
$\langle\nabla_{r}V,e_{1}\rangle$ vanish at the $x_{0}$ and $y_{0}$.

(6.2) $\langle\nabla_{r}V,e_{1}\rangle(0,\pm\tfrac{d_{0}}{2})=0.$

At $x_{0}$, this follows from the face that $\sigma_{1}(r)$ is a geodesic, the
identity $\langle T,e_{1}\rangle=0$ when $r=0$. For $y_{0}$, note that since

$\overline{V}(r)=\tfrac{\partial}{\partial r}\sigma_{2}(r)-\langle
T(r,\tfrac{d_{0}}{2}),\tfrac{\partial}{\partial r}\sigma_{2}(r)\rangle
T(r,\tfrac{d_{0}}{2})$

is orthogonal to $T(r,\tfrac{d_{0}}{2})$, so the value of $V$ along
$\sigma_{2}(r)$ is also given by
$V(r,\tfrac{d_{0}}{2})=\frac{\overline{V}(r)}{\|\overline{V}(r)\|}$. Then
using the fact that $\sigma_{2}(r)$ is a geodesic, this claim follows from the
identity $\langle T,e_{1}\rangle=0$ at $r=0$.

We now decompose $J$ into terms of the orthonormal frame $\\{T,V\\}$:

$\displaystyle J(r,s)=J_{V}(r,s)V(r,s)+J_{T}(r,s)T(r,s)\quad\textup{for
}s\in(\tfrac{-d_{0}}{2},\tfrac{d_{0}}{2}).$

Doing so, we find that

$\displaystyle
J^{\prime}(r,\pm\tfrac{d_{0}}{2})=J_{V}^{\prime}(r,\pm\tfrac{d_{0}}{2})V(r,\pm\tfrac{d_{0}}{2})+J^{\prime}_{T}(r,\pm\tfrac{d_{0}}{2})T(r,\pm\tfrac{d_{0}}{2}).$

The tangential component of a Jacobi field is linear, so from the boundary
condition we have

$\displaystyle
J_{T}(r,s)=\frac{-1}{d_{0}}(s-\tfrac{d_{0}}{2})\Bigl{\langle}T(r,\tfrac{-d_{0}}{2}),\frac{\partial\sigma_{1}}{\partial
r}(r)\Bigr{\rangle}+\frac{1}{d_{0}}(s+\tfrac{d_{0}}{2})\Bigl{\langle}T(r,\tfrac{d_{0}}{2}),\frac{\partial\sigma_{2}}{\partial
r}(r)\Bigr{\rangle},$

and thus

(6.3) $\displaystyle
J_{T}(0,\pm\tfrac{d_{0}}{2})=J_{T}^{\prime}(0,\pm\tfrac{d_{0}}{2})=0.$

Now we compute

$\displaystyle\nabla_{r}\nabla_{r}\eta^{\prime}(0,\pm\tfrac{d_{0}}{2})$
$\displaystyle=$
$\displaystyle\nabla_{r}\Bigl{[}J^{\prime}_{V}(r,\pm\tfrac{d_{0}}{2})V(r,\pm\tfrac{d_{0}}{2})+J^{\prime}_{T}(r,\pm\tfrac{d_{0}}{2})T(r,\pm\tfrac{d_{0}}{2})\Bigr{]}_{r=0}$
$\displaystyle=$ $\displaystyle\left[\tfrac{\partial}{\partial
r}(J^{\prime}_{V})(0,\pm\tfrac{d_{0}}{2})e_{1}+J^{\prime}_{V}(0,\pm\tfrac{d_{0}}{2})\nabla_{r}V(0,\pm\tfrac{d_{0}}{2})\right]+\nabla_{r}J^{\prime}_{T}(r,\pm\tfrac{d_{0}}{2})e_{2}.$

Using (6.2), we have

(6.4)
$\langle\nabla_{r}\nabla_{r}{\eta^{\prime}},e_{1}\rangle(0,\tfrac{d_{0}}{2})=\tfrac{\partial}{\partial
r}(J^{\prime}_{V})(0,\pm\tfrac{d_{0}}{2}).$

To compute $\frac{\partial}{\partial r}J^{\prime}_{V}$, we differentiate the
Jacobi equation

$J_{V}^{\prime\prime}+J_{V}\langle R(V,T)T,V\rangle\|\eta^{\prime}\|^{2}=0$

with respect to $r$, which gives

(6.5) $\tfrac{\partial}{\partial
r}(J_{V}^{\prime\prime})+\tfrac{\partial}{\partial
r}(J_{V}\kappa(r,s)\|\eta^{\prime}\|^{2})=0,$

where $\kappa(r,s)$ denotes the sectional curvature at the point $\eta(r,s).$
Since the partial derivatives commute and when $r=0$,
$\frac{\partial}{\partial r}\|\eta^{\prime}\|^{2}=\langle
J^{\prime},e_{2}\rangle=0$, evaluating Equation (6.5) at $r=0$ gives

(6.6) $\left(\tfrac{\partial}{\partial
r}J_{V}\right)^{\prime\prime}(0,s)+(\tfrac{\partial}{\partial
r}J_{V})\kappa(0,s)+J_{V}(\tfrac{\partial}{\partial r}\kappa)(0,s)=0.$

Since $\nabla_{\tfrac{\partial}{\partial r}}J(0,\pm\tfrac{d_{0}}{2})=0$, with
(6.3) and (6.2), we have $\tfrac{\partial}{\partial
r}J_{V}(0,\pm\tfrac{d_{0}}{2})=0$.

As a result, this is a non-homogeneous Jacobi field and we can apply Lemma 2.1
to find the solution

(6.7) $\tfrac{\partial}{\partial
r}J_{V}(0,s)=J^{1,0}(s)\int_{-\tfrac{d_{0}}{2}}^{s}J^{0,1}(t)[(J^{1,0})^{\prime}(\tfrac{d_{0}}{2})]^{-1}{m(t)}\,dt+J^{0,1}(s)\int_{s}^{\tfrac{d_{0}}{2}}J^{1,0}(t)[(J^{1,0})^{\prime}(\tfrac{d_{0}}{2})]^{-1}m(t)\,dt,$

where $m(t)=-(\tfrac{\partial}{\partial
r}\kappa)(0,t)J_{V}(0,t)=-(\tfrac{\partial}{\partial r}\kappa)(0,t)J(0,t)$ as
$J_{T}(0,t)=0$. Since $\tfrac{\partial}{\partial r}=Je_{1}$, we write
$\tfrac{\partial}{\partial r}\kappa=Je_{1}(\kappa)=J\kappa_{1}$.

Taking the derivative of (6.7) with respect to $s$ and evaluating this at the
end points gives

(6.8) $\displaystyle\left(\tfrac{\partial}{\partial
r}J_{V}\right)^{\prime}(0,-\tfrac{d_{0}}{2})$ $\displaystyle=$
$\displaystyle-\left(J^{0,1}\right)^{\prime}(-\tfrac{d_{0}}{2})\int_{-\tfrac{d_{0}}{2}}^{\tfrac{d_{0}}{2}}J^{1,0}(t)[(J^{1,0})^{\prime}(\tfrac{d_{0}}{2})]^{-1}\kappa_{1}(t)(J(t))^{2}\,dt$
$\displaystyle=$
$\displaystyle\int_{-\tfrac{d_{0}}{2}}^{\tfrac{d_{0}}{2}}J^{1,0}(t)\kappa_{1}(t)(J(t))^{2}\,dt,$
(6.9) $\displaystyle\left(\tfrac{\partial}{\partial
r}J_{V}\right)^{\prime}(0,\tfrac{d_{0}}{2})$ $\displaystyle=$
$\displaystyle-\int_{-\tfrac{d_{0}}{2}}^{\tfrac{d_{0}}{2}}J^{0,1}(t)\kappa_{1}(t)(J(t))^{2}\,dt,$

where we used Lemma 2.9 in the second equality. Combining Equation 6.8 with
Equation (6.4), we find the desired result.

∎

###  6.2. Normal derivatives of $\mathcal{C}$

We now compute the $e_{1}$ derivative of the function $\mathcal{C}$ defined in
(1.7).

######  Lemma 6.2.

We have

$\displaystyle\nabla_{0\oplus
e_{1}}\mathcal{C}=-\frac{1}{2}\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}\kappa_{1}J^{0,1}J^{2}\,dt$
$\displaystyle\nabla_{e_{1}\oplus
0}\mathcal{C}=-\frac{1}{2}\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}\kappa_{1}J^{1,0}J^{2}\,dt.$

This combines with the previous lemma to prove (3.10).

###### Proof.

We will focus on $\nabla_{0\oplus e_{1}}C,$ but the computation for
$\nabla_{e_{1}\oplus 0}C$ is nearly identical. Let $\sigma(\rho)$ be a curve
such that $\sigma(0)=y,$ $\sigma^{\prime}(0)=e_{1}(\tfrac{d}{2})$ (to indicate
a different variation we use $\rho$ instead of $r$ as before). Doing so, we
have that

$\displaystyle\nabla_{0\oplus
e_{1}}\mathcal{C}(x,y)=\frac{d}{d\rho}_{|_{\rho=0}}\mathcal{C}(x,\sigma(\rho)).$

Now, for each $\rho,$ we have

$\displaystyle\mathcal{C}(x,\sigma(\rho))=\frac{1}{2}\left(J^{\prime}(\rho,\tfrac{d(x,\sigma(\rho))}{2})-J^{\prime}(\rho,-\tfrac{d(x,\sigma(\rho))}{2})\right),$

where $J(\rho,s)$ solves

$\displaystyle\begin{cases}J^{\prime\prime}(\rho,s)+\kappa(\rho,s)J(\rho,s)&=0\\\
J(\rho,-\tfrac{d(x,\sigma(\rho))}{2})=J(\rho,\tfrac{d(x,\sigma(\rho))}{2})&=1.\end{cases}$

Differentiating with respect to $\rho$, commuting the derivatives as before,
and evaluating at $\rho=0$, we find that

(6.10)
$\displaystyle(\tfrac{\partial}{\partial\rho}J)^{\prime\prime}(0,s)+\kappa(0,s)\tfrac{\partial
J}{\partial\rho}(0,s)+\kappa_{\rho}(0,s)J=0.$

To find the boundary values of $\tfrac{\partial J}{\partial\rho},$ we note
that $J$ at the boundary points is constant (i.e. $1$). So by the first
variation formula
$\tfrac{\partial}{\partial\rho}_{|_{\rho=0}}d(x,\sigma(\rho))=0,$ and hence

$\displaystyle 0=$
$\displaystyle\frac{d}{d\rho}_{|_{\rho=0}}J(\rho,\pm\tfrac{d(x,\sigma(\rho))}{2})$
$\displaystyle=$ $\displaystyle\tfrac{\partial
J}{\partial\rho}(0,\pm\tfrac{d(x,y)}{2})+J^{\prime}(0,\pm\tfrac{d(x,y)}{2})\tfrac{d}{d\rho}_{|_{\rho=0}}d(x,\sigma(\rho))$
$\displaystyle=$ $\displaystyle\tfrac{\partial
J}{\partial\rho}(0,\pm\tfrac{d(x,y)}{2}).$

Similarly to the computation in the previous lemma, we apply Lemma 2.1 to
equation (6.10) to solve for $\tfrac{\partial J}{\partial\rho}(0,s)$, then
differentiate this equation with respect to $s$. Evaluating the result at the
endpoints, we find that

$\displaystyle\tfrac{\partial
J^{\prime}}{\partial\rho}(0,-\tfrac{d_{0}}{2})-\tfrac{\partial
J^{\prime}}{\partial\rho}(0,\tfrac{d_{0}}{2})$
$\displaystyle=\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}J^{1.0}\kappa_{\rho}(t)J(t)\,dt-\left(-\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}J^{0,1}\kappa_{\rho}(t)J(t)\,dt\right)$
$\displaystyle=\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}\kappa_{\rho}(t)J^{2}(t)\,dt=\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}\kappa_{1}(t)J^{0,1}(t)J^{2}(t)\,dt$

where we have used $J=J^{1,0}+J^{0,1}$ and
$\tfrac{\partial}{\partial\rho}=J^{0,1}e_{1}$ in the last two equalities. ∎

Theorem 3.4 now follows from the previous lemmas.

###  6.3. Tangential derivatives of $\mathcal{C}$

We now compute the $0\oplus e_{2}$ and $e_{2}\oplus 0$ derivatives of
$\mathcal{C}$ and thus prove Lemma 3.5.

###### Proof of Lemma 3.5.

We first compute $\nabla_{0\oplus e_{2}}\mathcal{C}.$ Note that

$\displaystyle\nabla_{0\oplus
e_{2}}\mathcal{C}=\frac{d}{dt|_{t=0}}\mathcal{C}(x,\gamma(\tfrac{d_{0}}{2}+t)).$

For simplicity, we use the variation $\eta(t,s)=\gamma(s)$, with
$s\in[-\tfrac{d_{0}}{2},\tfrac{d_{0}}{2}+t]$. For $t$ small, let $J(t,s)$
denote the solution to

$\displaystyle\begin{cases}J^{\prime\prime}(t,s)+\kappa(\gamma(s))J(t,s)=0\quad\textup{for
}s\in(-\tfrac{d_{0}}{2},\tfrac{d_{0}}{2}+t)\\\
J(t,-\tfrac{d_{0}}{2})=J(t,\tfrac{d_{0}}{2}+t)=1.\end{cases}$

Then

$2\nabla_{e_{2}\oplus
0}\mathcal{C}=\frac{d}{dt}_{|_{t=0}}\left(J^{\prime}(t,\tfrac{d}{2}+t)-J^{\prime}(t,-\tfrac{d}{2})\right)$

In this formula, we changed the parametrization, but the solution is
independent of the parametrization. Differentiating at $t=0$ gives
$J_{t}(0,s)$ for $s\in(-\tfrac{d_{0}}{2},\tfrac{d_{0}}{2}).$ This satisfies
the same equation, being

(6.11) $\displaystyle J_{t}^{\prime\prime}+\kappa J_{t}=0\quad\textup{in
}(\tfrac{-d_{0}}{2},\tfrac{d_{0}}{2}).$

We have the boundary conditions:

$\displaystyle
0=\tfrac{d}{dt}_{|_{t=0}}J(t,\tfrac{-d_{0}}{2})=J_{t}(0,\tfrac{-d_{0}}{2})$

as well as

$\displaystyle
0=\tfrac{d}{dt}_{|_{t=0}}J(t,\tfrac{d_{0}}{2}+t)=J_{t}(0,\tfrac{d_{0}}{2})+J^{\prime}(\tfrac{d_{0}}{2}).$

In other words, we have that

$\displaystyle J_{t}(0,\tfrac{-d_{0}}{2})=0\quad
J_{t}(0,\tfrac{d_{0}}{2})=-J^{\prime}(\tfrac{d_{0}}{2}).$

Applying Lemma 2.1 to (6.11), we get that

$\displaystyle J_{t}(0,s)$
$\displaystyle=-J^{\prime}(\tfrac{d_{0}}{2})J^{0,1}(s),$

which yields

(6.12) $\displaystyle
J_{t}^{\prime}(\tfrac{d}{2})-J_{t}^{\prime}(-\tfrac{d}{2})$
$\displaystyle=-J^{\prime}(\tfrac{d}{2})\left((J^{0,1})^{\prime}(\tfrac{d}{2})-(J^{0,1})^{\prime}(-\tfrac{d}{2})\right)$
$\displaystyle=-|J^{\prime}(\tfrac{d}{2})|^{2},$

where in the second equality we used Lemma 2.9. We thus conclude that

$\displaystyle 2\nabla_{0\oplus e_{2}}\mathcal{C}$
$\displaystyle=\frac{d}{dt}_{|_{t=0}}\left(J^{\prime}(t,\tfrac{d}{2}+t)-J^{\prime}(t,-\tfrac{d}{2})\right)$
$\displaystyle=J_{t}^{\prime}(0,\tfrac{d}{2})-J_{t}^{\prime}(0,-\tfrac{d}{2})+J^{\prime\prime}(\tfrac{d_{0}}{2})$
$\displaystyle=-|J^{\prime}(\tfrac{d}{2})|^{2}-\kappa(y),$

where we have used (6.11) in the second equality.
We are now computing $\nabla_{e_{2}\oplus 0}\mathcal{C},$ being

$\displaystyle\nabla_{e_{2}\oplus
0}\mathcal{C}=\frac{d}{dt}|_{t=0}\mathcal{C}(\gamma(-\tfrac{d}{2}+t),y).$

With similar arguments as before, we find that $J(t,\cdot)$ is the solution to
the problem

(6.13)
$\displaystyle\begin{cases}J^{\prime\prime}(t,s)+\kappa(\gamma_{x,y}(s))J(t,s)=0\quad\textup{for
}s\in(-\tfrac{d_{0}}{2}+t,\tfrac{d_{0}}{2})\\\
J(t,-\tfrac{d_{0}}{2}+t)=J(t,\tfrac{d_{0}}{2})=1.\end{cases}$

Then, after differentiating the equations (6.13) for fixed
$s\in(-\tfrac{d}{2},\tfrac{d}{2}),$ and applying Lemma 2.2, we find that

$\displaystyle J_{t}(0,s)=-J^{\prime}(-\tfrac{d}{2})J^{1,0}(s).$

From which we get that, using Lemma 2.9

$\displaystyle
J_{t}^{\prime}(\tfrac{d}{2})-J_{t}^{\prime}(-\tfrac{d}{2})=|J^{\prime}(-\tfrac{d}{2})|^{2},$

so that we conclude

$\displaystyle 2\nabla_{e_{2}\oplus 0}\mathcal{C}$
$\displaystyle=\frac{d}{dt}_{|_{t=0}}\left(J^{\prime}(t,\tfrac{d}{2})-J^{\prime}(t,-\tfrac{d}{2}+t)\right)$
$\displaystyle=J_{t}^{\prime}(0,\tfrac{d}{2})-J_{t}^{\prime}(0,-\tfrac{d}{2})-J^{\prime\prime}(-\tfrac{d}{2})$
$\displaystyle=|J^{\prime}(-\tfrac{d}{2})|^{2}+\kappa(x).$

∎

##  7\. Second order Derivatives of $\mathcal{C}$

In this section we compute the second order derivatives of $\mathcal{C}$ and
estimate the quantities in terms of curvature and its derivatives, showing
that (3.7) holds.

Recall

(7.1) $\displaystyle E_{1}=e_{1}\oplus e_{1}\in T_{x}\Omega\oplus
T_{y}\Omega,\quad E_{2}=e_{2}\oplus(-e_{2})\in T_{x}\Omega\oplus T_{y}\Omega.$

###  7.1.  $E_{2}$ derivatives

We first compute the quantity $\nabla_{E_{2},E_{2}}^{2}(-\mathcal{C})$.

######  Lemma 7.1.

$\displaystyle\nabla_{E_{2},E_{2}}^{2}(-\mathcal{C})$
$\displaystyle=\kappa(x_{0})J^{\prime}(\tfrac{-d_{0}}{2})-\kappa(y_{0})J^{\prime}(\tfrac{d_{0}}{2})+\frac{1}{2}\Bigl{[}-\kappa^{\prime}(\tfrac{-d_{0}}{2})+\kappa^{\prime}(\tfrac{d_{0}}{2})\Bigr{]}$
$\displaystyle+[J^{\prime}(\tfrac{-d_{0}}{2})]^{3}-[J^{\prime}(\tfrac{d_{0}}{2})]^{3}-(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\left(J^{\prime}(\tfrac{d_{0}}{2})+J^{\prime}(\tfrac{-d_{0}}{2})\right)^{2}$

###### Proof.

Since we are differentiating $\mathcal{C}$ twice in the direction of the
geodesic, we consider $J(t,\cdot)$ which solve the equation

$\displaystyle\begin{cases}J^{\prime\prime}(t,s)+\kappa(\gamma_{x,y}(s))J(t,s)=0\quad\textup{for
}s\in(-\tfrac{d}{2}+t,\tfrac{d}{2}-t)\\\
J(t,\tfrac{-d_{0}}{2}+t)=J(t,\tfrac{d_{0}}{2}-t)=1.\end{cases}$

Doing so, we find the following:

$\displaystyle\nabla^{2}_{E_{2},E_{2}}\mathcal{C}$ $\displaystyle=$
$\displaystyle\frac{d^{2}}{dt^{2}}_{|_{t=0}}\frac{1}{2}\left(J^{\prime}(t,\tfrac{-d_{0}}{2}+t)-J^{\prime}(t,\tfrac{d_{0}}{2}-t)\right)$
$\displaystyle=$
$\displaystyle\frac{1}{2}\left(\underbrace{J_{tt}^{\prime}(0,\tfrac{-d_{0}}{2})-J_{tt}^{\prime}(0,\tfrac{d_{0}}{2})}_{I}+\underbrace{J^{\prime\prime\prime}(0,\tfrac{-d_{0}}{2})-J^{\prime\prime\prime}(0,\tfrac{d_{0}}{2})}_{II}\right)$
$\displaystyle+\left(\underbrace{J^{\prime\prime}_{t}(0,\tfrac{-d_{0}}{2})+J^{\prime\prime}_{t}(0,\tfrac{d_{0}}{2})}_{III}\right).$

To finish the lemma, we compute the terms $I,II,$ and $III$ on the right hand
side in terms of curvature and its derivatives.

Step 1: Computing $III.$
Observe that $J_{t}(0,\cdot)=\tfrac{\partial J}{\partial t}(0,\cdot)$ solves
the equation

$\displaystyle J^{\prime\prime}+\kappa J=0\quad\textrm{ in
}(-\tfrac{d_{0}}{2},\tfrac{d_{0}}{2})$

with boundary conditions

$\displaystyle
J_{t}(0,\tfrac{-d_{0}}{2})=-J^{\prime}(-\tfrac{d_{0}}{2}),\quad\quad
J_{t}(0,\tfrac{d_{0}}{2})=J^{\prime}(\tfrac{d_{0}}{2}).$

We therefore find

(7.3) $\displaystyle
J_{t}(s)=-J^{\prime}(\tfrac{-d_{0}}{2})J^{1,0}(s)+J^{\prime}(\tfrac{d_{0}}{2})J^{0,1}(s),$

from which it follows that

(7.4)
$J^{\prime\prime}_{t}(0,\tfrac{-d_{0}}{2})+J^{\prime\prime}_{t}(0,\tfrac{d_{0}}{2})=\kappa(x_{0})J^{\prime}(\tfrac{-d_{0}}{2})-\kappa(y_{0})J^{\prime}(\tfrac{d_{0}}{2}).$

Step 2: Computing $II.$
For this term, note that

$J^{\prime\prime\prime}=-(\kappa J)^{\prime}=-\kappa^{\prime}J-\kappa
J^{\prime},$

from which we find that

(7.5)
$J^{\prime\prime\prime}(0,\tfrac{-d_{0}}{2})-J^{\prime\prime\prime}(0,\tfrac{+d_{0}}{2})=-\kappa(x_{0})J^{\prime}(\tfrac{-d_{0}}{2})-\kappa^{\prime}(\tfrac{-d_{0}}{2})+\kappa(y_{0})J^{\prime}(\tfrac{d_{0}}{2})+\kappa^{\prime}(\tfrac{d_{0}}{2}).$

Step 3: Computing $I.$
Observe that $J_{tt}(0,\cdot)$ solves the equation

$\displaystyle J^{\prime\prime}+\kappa J=0.$

Using the identities

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{d^{2}}{dt^{2}}_{|_{t=0}}J(t,\tfrac{-d_{0}}{2}+t)=J_{tt}(0,\tfrac{-d_{0}}{2})+2J_{t}^{\prime}(\tfrac{-d_{0}}{2})-\kappa(x_{0})$
$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{d^{2}}{dt^{2}}_{|_{t=0}}J(t,\tfrac{d_{0}}{2}+t)=J_{tt}(0,\tfrac{d_{0}}{2})-2J_{t}^{\prime}(\tfrac{d_{0}}{2})-\kappa(y_{0}),$

we derive the boundary conditions

$\displaystyle
J_{tt}(0,\tfrac{-d_{0}}{2})=-2J_{t}^{\prime}(\tfrac{-d_{0}}{2})+\kappa(x_{0}),\quad
J_{tt}(0,\tfrac{d_{0}}{2})=2J_{t}^{\prime}(\tfrac{d_{0}}{2})+\kappa(y_{0}).$

From this, we find

$\displaystyle
J_{tt}(s)=[-2J_{t}^{\prime}(\tfrac{-d_{0}}{2})+\kappa(x_{0})]J^{1,0}(s)+[2J_{t}^{\prime}(\tfrac{d_{0}}{2})+\kappa(y_{0})]J^{0,1}(s).$

After writing $J=J^{1,1}$, we find that

(7.6) $\displaystyle
J_{tt}^{\prime}(\tfrac{-d_{0}}{2})-J_{tt}^{\prime}(\tfrac{d_{0}}{2})=[-2J_{t}^{\prime}(\tfrac{-d_{0}}{2})+\kappa(x_{0})]J^{\prime}(\tfrac{-d_{0}}{2})-[2J_{t}^{\prime}(\tfrac{d_{0}}{2})+\kappa(y_{0})]J^{\prime}(\tfrac{d_{0}}{2}).$

From (7.3), we find the following expression for $J_{t}^{\prime}(\tfrac{\pm
d_{0}}{2}):$

$\displaystyle J^{\prime}_{t}(\tfrac{-d_{0}}{2})$ $\displaystyle=$
$\displaystyle-J^{\prime}(\tfrac{-d_{0}}{2})(J^{1,0})^{\prime}(\tfrac{-d_{0}}{2})+J^{\prime}(\tfrac{d_{0}}{2})(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})$
$\displaystyle J^{\prime}_{t}(\tfrac{d_{0}}{2})$ $\displaystyle=$
$\displaystyle-J^{\prime}(\tfrac{-d_{0}}{2})(J^{1,0})^{\prime}(\tfrac{d_{0}}{2})+J^{\prime}(\tfrac{d_{0}}{2})(J^{0,1})^{\prime}(\tfrac{d_{0}}{2}).$

Combining this and (7.6), we get that

(7.7) $\displaystyle
J_{tt}^{\prime}(0,\tfrac{-d_{0}}{2})-J_{tt}^{\prime}(0,\tfrac{+d_{0}}{2})$
$\displaystyle=$
$\displaystyle[-2\left(-J^{\prime}(\tfrac{-d_{0}}{2})(J^{1,0})^{\prime}(\tfrac{-d_{0}}{2})+J^{\prime}(\tfrac{d_{0}}{2})(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\right)+\kappa(x_{0})]J^{\prime}(\tfrac{-d_{0}}{2})$
$\displaystyle-[2\left(-J^{\prime}(\tfrac{-d_{0}}{2})(J^{1,0})^{\prime}(\tfrac{d_{0}}{2})+J^{\prime}(\tfrac{d_{0}}{2})(J^{0,1})^{\prime}(\tfrac{d_{0}}{2})\right)+\kappa(y_{0})]J^{\prime}(\tfrac{d_{0}}{2}).$

Combining Equations (7.4),(7.5), and (7.7) into (7.1), we find that

$\displaystyle\nabla_{E_{2},E_{2}}^{2}(-\mathcal{C})$
$\displaystyle=\kappa(x_{0})J^{\prime}(\tfrac{-d_{0}}{2})-\kappa(y_{0})J^{\prime}(\tfrac{d_{0}}{2})+\frac{1}{2}\Bigl{[}-\kappa^{\prime}(\tfrac{-d_{0}}{2})+\kappa^{\prime}(\tfrac{d_{0}}{2})\Bigr{]}$
$\displaystyle\quad\quad+|J^{\prime}(\tfrac{-d_{0}}{2})|^{2}(J^{1,0})^{\prime}(\tfrac{-d_{0}}{2})-J^{\prime}(\tfrac{d_{0}}{2})J^{\prime}(\tfrac{-d_{0}}{2})(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})$
$\displaystyle\quad\quad+J^{\prime}(\tfrac{d_{0}}{2})J^{\prime}(\tfrac{-d_{0}}{2})(J^{1,0})^{\prime}(\tfrac{d_{0}}{2})-|J^{\prime}(\tfrac{d_{0}}{2})|^{2}(J^{0,1})^{\prime}(\tfrac{d_{0}}{2}).$

Finally, making use of $J=J^{1,0}+J^{0,1}$ and Lemma 2.9, we obtain the
desired result. ∎

###  7.2.  $E_{1}$ derivatives

We now compute the quantity $\nabla_{E_{1},E_{1}}^{2}(-\mathcal{C})$.

######  Lemma 7.2.

$\displaystyle\nabla^{2}_{E_{1},E_{1}}(-\mathcal{C})=\int\tfrac{J^{2}(s)}{2}\Bigl{[}\kappa_{11}(s)J^{2}(s)+3\kappa_{1}(s){p(s)}+\kappa_{2}(s)q(s)\Bigr{]}\,ds+\mathcal{C}(x,y)\nabla_{e_{2}\oplus(-e_{2})}\mathcal{C}$

where $p(s),q(s)$ are given by

(7.8) $\displaystyle p(s)$
$\displaystyle=-J^{1,0}(s)\int_{\tfrac{-d_{0}}{2}}^{s}\tfrac{J^{0,1}(w)}{(J^{1,0})^{\prime}\left(\tfrac{d_{0}}{2}\right)}J(w)\kappa_{1}(w)\,dw-J^{0,1}(s)\int_{s}^{\tfrac{d_{0}}{2}}\tfrac{J^{1,0}(w)}{(J^{1,0})^{\prime}\left(\tfrac{d_{0}}{2}\right)}J(s)\kappa_{1}(s)\,dw$
(7.9) $\displaystyle q(s)$
$\displaystyle=\tfrac{1}{d_{0}}(\tfrac{d_{0}}{2}-s)J^{\prime}(\tfrac{-d_{0}}{2})+\tfrac{1}{d_{0}}(s+\tfrac{d_{0}}{2})J^{\prime}(\tfrac{d_{0}}{2})-J^{\prime}(s)J(s)$

We make several notes about this computation. We have used $t$ to indicate the
variation instead of $r$, to emphasize that the variations are parametrized
differently. Unlike the previous expression, it involves integral quantities
defined along $\gamma$ so is non-local in nature.

###### Proof.

Consider the geodesics

$\displaystyle\sigma_{x}(t)=\exp_{x}(te_{1}(-\tfrac{d}{2}))\quad\sigma_{y}(t)=\exp_{y}(te_{1}(\tfrac{d}{2}))$

which extend perpendicularly from the endpoints of $\gamma$. For small $t,$
denote $\eta(t,\cdot)$ to be the minimal geodesic between $\sigma_{x}(t)$ and
$\sigma_{y}(t),$ normalized to have unit speed.

Doing so, we find that

$\displaystyle\frac{d^{2}}{dt^{2}}_{|t=0}\mathcal{C}(\sigma_{x}(t),\sigma_{y}(t))=\nabla_{E_{1},E_{1}}^{2}\mathcal{C}(x,y).$

To compute this quantity, we must compute the relevant Jacobi fields and
determine their dependence on $t.$ We find that

$\displaystyle\frac{d^{2}}{dt^{2}}_{|t=0}J^{\prime}(t,-\tfrac{d(\sigma_{x}(t),\sigma_{y}(t))}{2})$
$\displaystyle=$ $\displaystyle
J_{tt}^{\prime}(0,\tfrac{d_{0}}{2})-J^{\prime\prime}(0,\tfrac{d_{0}}{2})\frac{1}{2}\nabla^{2}_{E_{1},E_{1}}d{(x_{0},y_{0})}$
$\displaystyle=$ $\displaystyle
J_{tt}^{\prime}(0,\tfrac{-d_{0}}{2})+\frac{\kappa(x_{0})}{2}\nabla^{2}_{E_{1},E_{1}}d{(x_{0},y_{0})}.$

And similarly,

$\displaystyle\frac{d^{2}}{dt^{2}}_{|t=0}J^{\prime}(t,\tfrac{d(\sigma_{x}(t),\sigma_{y}(t))}{2})=J_{tt}^{\prime}(0,\tfrac{d_{0}}{2})-\frac{\kappa(y_{0})}{2}\nabla^{2}_{E_{1},E_{1}}d{(x_{0},y_{0})}$

From this, it remains to calculate
$J_{tt}^{\prime}(0,\tfrac{-d_{0}}{2})-J_{tt}^{\prime}(0,\tfrac{d_{0}}{2}).$

$J_{tt}$ satisfies the equation

$\displaystyle\begin{cases}J_{tt}^{\prime\prime}+\kappa_{tt}J+2\kappa_{t}J_{t}+\kappa
J_{tt}=0\\\
J_{tt}(0,\tfrac{-d_{0}}{2})=\tfrac{1}{2}J^{\prime}(-\tfrac{d_{0}}{2})\nabla^{2}_{E_{1},E_{1}}d(x,y),\\\
J_{tt}(0,\tfrac{d_{0}}{2})=-\tfrac{1}{2}J^{\prime}(\tfrac{d_{0}}{2})\nabla^{2}_{E_{1},E_{1}}d(x,y).\end{cases}$

Applying Lemma 2.1, we find the following solution for $J_{tt}$:

(7.11) $\displaystyle J_{tt}(s)$
$\displaystyle=\left[\tfrac{1}{2}J^{\prime}(-\tfrac{d_{0}}{2})\nabla^{2}_{E_{1},E_{1}}d\right]J^{1,0}(s)+[-\tfrac{1}{2}J^{\prime}(\tfrac{d_{0}}{2})\nabla^{2}_{E_{1},E_{1}}d]J^{0,1}(s)$
$\displaystyle\quad+J^{1,0}(s)\int_{-\tfrac{d_{0}}{2}}^{s}J^{0,1}(w)\tfrac{m(w)}{(J^{1,0})^{\prime}(\tfrac{d_{0}}{2})}\,dw+J^{0,1}(s)\int_{s}^{\tfrac{d_{0}}{2}}J^{1,0}(w)\tfrac{m(w)}{(J^{1,0})^{\prime}(\tfrac{d_{0}}{2})}\,dw$

where

$m(w)=-\kappa_{tt}J-2\kappa_{t}J_{t}.$

By the same argument, we find that

(7.12) $\displaystyle J_{t}(0,s)=p(s).$

Combining Lemma 2.9 with the identities
$2\mathcal{C}(x,y)=\nabla^{2}_{E_{1}E_{1}}d(x,y),$ and $J=J^{1,0}+J^{0,1},$
Equation (7.11) simplifies to the following:

(7.13) $\displaystyle
J_{tt}^{\prime}(\tfrac{-d_{0}}{2})-J_{tt}^{\prime}(\tfrac{d_{0}}{2})=\mathcal{C}(x,y)\Bigl{[}|J^{\prime}(\tfrac{d_{0}}{2})|^{2}+|J^{\prime}(\tfrac{-d_{0}}{2})|^{2}\Bigr{]}+\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}\underbrace{\kappa_{tt}J^{2}+2\kappa_{t}J_{t}J}_{IV}\,dw,$

Observe that by Lemma 3.21, we get that

$\displaystyle\mathcal{C}(x,y)\nabla_{e_{2}\oplus(-e_{2})}\mathcal{C}(x,y)=\frac{1}{2}\mathcal{C}(x,y)\Bigl{(}|J^{\prime}(\tfrac{d_{0}}{2})|^{2}+|J^{\prime}(\tfrac{-d_{0}}{2})|^{2}+\kappa(x_{0})+\kappa(y_{0})\Bigr{)}.$

It thus only remains to compute the terms $IV$. Since

$\displaystyle\kappa_{t}(s)=\frac{d}{dt}_{|_{t=0}}\kappa(\eta(t,s))=J(s)\langle\nabla\kappa,e_{1}\rangle=\kappa_{1}J(s),$

and

$\displaystyle\kappa_{tt}=\frac{d^{2}}{dt^{2}}_{|_{t=0}}\kappa(\eta(t,s))=\langle\nabla_{\tfrac{\partial\eta}{\partial
t}(t,s)}\nabla\kappa(\eta(t,s)),\tfrac{\partial\eta}{\partial
t}(t,s)\rangle+\langle\nabla\kappa,\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial t}\rangle,$

it remains to compute $\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial t}$ at $t=0.$ For this, we need the following
lemma.

######  Lemma 7.3.

$\displaystyle\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial
t}_{|_{t=0}}=p(s)e_{1}(s)+e_{2}(s)\left(\tfrac{1}{d_{0}}(\tfrac{d_{0}}{2}-s)J^{\prime}(\tfrac{-d_{0}}{2})+\tfrac{1}{d_{0}}(s+\tfrac{d_{0}}{2})J^{\prime}(\tfrac{d_{0}}{2})-J^{\prime}(s)J(s)\right)$

###### Proof of Lemma 7.3.

The derivation is very similar to those in Section 6. Observe that

$\displaystyle\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial t}_{|_{t=0}}$
$\displaystyle=\langle\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial t}_{|_{t=0}},\tfrac{\partial\eta}{\partial
t}\rangle\tfrac{e_{1}(s)}{J(s)}+\langle\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial
t}_{|_{t=0}},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle e_{2}(s),$

so it suffices to compute $\langle\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial t},\tfrac{\partial\eta}{\partial t}\rangle$
and $\langle\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle.$

Note that

$\displaystyle\langle\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle=\frac{d}{dt}_{|_{t=0}}\langle\tfrac{\partial\eta}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle-J^{\prime}(s)J(s),$

where we used the fact that $\partial_{t}\|\eta^{\prime}\|=0$ at $t=0.$ In
order to find $\tfrac{d}{dt}_{|_{t=0}}\langle\tfrac{\partial\eta}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle,$ we use a similar argument
as in the previous section. As before, $\tfrac{\partial\eta}{\partial t}$ is a
Jacobi field which we decompose with an orthonormal frame as
$T=\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}$ and an orthogonal part
$V=V(t,s)\perp\eta^{\prime}.$ We then have that

$\displaystyle J(t,s)=J_{V}(t,s)V+J_{T}(t,s)T.$

Since

$\displaystyle\langle\tfrac{\partial\eta}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle=J_{T}(t,s),$

we have to find $(J_{T})_{t}(0,s).$ With similar arguments as in the previous
section, we find that

$\displaystyle J_{T}^{\prime\prime}(t,s)=0,\quad
J_{T}(t,\tfrac{-d_{0}}{2})=\langle\tfrac{\partial\sigma_{x}}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle(t,\tfrac{-d_{0}}{2}),\quad
J_{T}(t,\tfrac{d_{0}}{2})=\langle\tfrac{\partial\sigma_{y}}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle(t,\tfrac{d_{0}}{2}),$

This gives that

$\displaystyle
J_{T}(t,s)=\tfrac{1}{d_{0}}(\tfrac{d_{0}}{2}-s)\langle\tfrac{\partial\sigma_{x}}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle(t,\tfrac{-d_{0}}{2})+\tfrac{1}{d_{0}}(s+\tfrac{d_{0}}{2})\langle\tfrac{\partial\sigma_{y}}{\partial
t},\tfrac{\eta^{\prime}}{\|\eta^{\prime}\|}\rangle(t,\tfrac{d_{0}}{2}).$

Thus

$\displaystyle(J_{T})_{t}(0,s)=\tfrac{1}{d_{0}}(\tfrac{d_{0}}{2}-s)J^{\prime}(\tfrac{-d_{0}}{2})+\tfrac{1}{d_{0}}(s+\tfrac{d_{0}}{2})J^{\prime}(\tfrac{d_{0}}{2}).$

As $J_{T}(0,s)=0$ we have

$\displaystyle\langle\nabla_{\tfrac{\partial\eta}{\partial
t}}\tfrac{\partial\eta}{\partial t},\tfrac{\partial\eta}{\partial
t}\rangle=\frac{1}{2}\tfrac{d}{dt}_{|_{t=0}}\|J(t,s)\|^{2}=(J_{V})_{t}(0,s)J_{V}(0,s).$

As in the previous section, we find the variation of Jacobi fields as follows.
Note that $J_{V}(t,s)$ solves the equation $J^{\prime\prime}+\kappa J=0$ in
$(\tfrac{-d(\sigma_{x}(t),\sigma_{y}(t))}{2},\tfrac{d(\sigma_{x}(t),\sigma_{y}(t))}{2})$
with boundary conditions
$J(t,\tfrac{-d(\sigma_{x}(t),\sigma_{y}(t))}{2})=J(t,\tfrac{d(\sigma_{x}(t),\sigma_{y}(t))}{2})=1.$
Then taking derivative at $t=0$ gives that $(J_{V})_{t}(0,\cdot)$ solves the
equation

$J^{\prime\prime}+\kappa_{t}J+\kappa
J=0\quad\textup{in}\,(\tfrac{-d_{0}}{2},\tfrac{d_{0}}{2}).$

With boundary condition $J_{t}(0,\tfrac{\pm d_{0}}{2})=0.$ Then by Lemma 2.2,
we get $(J_{V})_{t}(0,s)=p(s)$ as desired. ∎

The proof of Lemma 7.2 follows now from all the previous computations together
with Lemma 7.3. ∎

######  Proposition 7.4.

$\displaystyle\left(\nabla^{2}_{E_{1},E_{1}}+\nabla_{E_{2},E_{2}}^{2}\right)(-\mathcal{C})$
$\displaystyle=\kappa(x)J_{x,y}^{\prime}(\tfrac{-d}{2})-\kappa(y)J_{x,y}^{\prime}(\tfrac{d}{2})+[J_{x,y}^{\prime}(\tfrac{-d}{2})]^{3}-[J_{x,y}^{\prime}(\tfrac{d}{2})]^{3}$
$\displaystyle+\mathcal{C}(x,y)\nabla_{e_{2}\oplus(-e_{2})}\mathcal{C}(x,y)+\mathcal{D}(x,y),$

where $\mathcal{D}(x,y)$ is given as follows

(7.14) $\displaystyle\mathcal{D}(x,y)$
$\displaystyle=-(J_{x,y}^{0,1})^{\prime}(\tfrac{-d}{2})\left(J_{x,y}^{\prime}(\tfrac{d}{2})+J_{x,y}^{\prime}(\tfrac{-d}{2})\right)^{2}$
$\displaystyle\quad+\int\tfrac{J_{x,y}^{2}(s)}{2}\Bigl{[}\Delta\kappa(\gamma_{x,y}(s))J_{x,y}^{2}(s)+3\kappa_{1}(\gamma_{x,y}(s)){p(s)}+\kappa_{2}(\gamma_{x,y}(s))\tilde{q}(s)\Bigr{]}\,ds,$

and $\tilde{q}(s)=q(s)+4J_{x,y}^{\prime}(s)J_{x,y}(s).$

###### Proof.

(7.15)
$\displaystyle\left(\nabla_{E_{1},E_{1}}^{2}+\nabla^{2}_{E_{2},E_{2}}\right)(-\mathcal{C})$
$\displaystyle=$
$\displaystyle\int\tfrac{J^{2}(s)}{2}\Bigl{[}\Delta\kappa(s)J^{2}(s)+3\kappa_{1}(s){p(s)}+\kappa_{2}(s)\left(\tfrac{1}{d_{0}}(\tfrac{d_{0}}{2}-s)J^{\prime}(\tfrac{-d_{0}}{2})+\tfrac{1}{d_{0}}(s+\tfrac{d_{0}}{2})J^{\prime}(\tfrac{d_{0}}{2})+3J^{\prime}(s)J(s)\right)\Bigr{]}\,ds$
$\displaystyle\quad+\kappa(x_{0})J^{\prime}(\tfrac{-d_{0}}{2})-\kappa(y_{0})J^{\prime}(\tfrac{d_{0}}{2})+[J^{\prime}(\tfrac{-d_{0}}{2})]^{3}-[J^{\prime}(\tfrac{d_{0}}{2})]^{3}-(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\left(J^{\prime}(\tfrac{d_{0}}{2})+J^{\prime}(\tfrac{-d_{0}}{2})\right)^{2}$
$\displaystyle\quad+\mathcal{C}(x_{0},y_{0})\nabla_{e_{2}\oplus(-e_{2})}\mathcal{C},$

where we used

$\displaystyle\frac{1}{2}\left(\kappa^{\prime}(\tfrac{d_{0}}{2})-\kappa^{\prime}(\tfrac{-d_{0}}{2})\right)=\frac{1}{2}\int_{\tfrac{-d_{0}}{2}}^{\tfrac{d_{0}}{2}}\kappa^{\prime\prime}J^{4}+4\kappa^{\prime}J^{\prime}J^{3}(w)\,dw.$

∎

######  Remark 7.5.

For a space form $M^{2}_{K}$, it is possible to compute these quantities
directly. Since $C(x,y)=-\textup{tn}_{K}(\tfrac{d(x,y)}{2}),$ one finds that

$\displaystyle\nabla^{2}_{E_{2},E_{2}}(-\mathcal{C})=\frac{d^{2}}{dt^{2}}_{|_{t=0}}\textup{tn}_{K}(\tfrac{d}{2}-t)=2\frac{K\textup{tn}_{K}(\tfrac{d}{2})}{\textup{cs}_{K}^{2}(\tfrac{d}{2})}$

and that

$\displaystyle\nabla^{2}_{E_{1},E_{1}}(-\mathcal{C})=\frac{K}{2\textup{cs}_{K}^{2}(\tfrac{d}{2})}\nabla^{2}_{E_{1},E_{1}}d=-\frac{K\textup{tn}_{K}(\tfrac{d}{2})}{\textup{cs}_{K}^{2}(\tfrac{d}{2})}.$

Moreover, we have that $\mathcal{D}(x,y)=0.$

###  7.3. Proof of Lemma 3.7

In this section we estimate the quantities $\mathcal{D}(x,y)$ and
$\varepsilon(x,y)$. Throughout this proof, we will use the fact that
$J^{1,0},J^{0,1},$ and $J$ are strictly positive in the interval
$[-\tfrac{d}{2},\tfrac{d}{2}].$ As a result, we can apply the comparison
results from Section 2.

We now estimate $\mathcal{D}.$ In view of the maximum principle from Section
3, we are concerned with a lower bound of $\mathcal{D}.$

######  Lemma 7.6.

For any $x,y\in\Omega$ with $d(x,y)\leq\pi/(2\sqrt{\overline{\kappa}})$

$\displaystyle\mathcal{D}(x,y)\geq
2\mathcal{C}(x,y)\left(\frac{-2(\inf\Delta\kappa)_{-}}{\underline{\kappa}}+\frac{3\pi^{2}\sqrt{8}|\nabla\kappa|^{2}_{\infty}}{4\overline{\kappa}\underline{\kappa}}+\frac{4\pi|\nabla\kappa|_{\infty}}{\underline{\kappa}}+2\sqrt{2}\frac{\overline{\kappa}^{2}}{\underline{\kappa}}({\overline{\kappa}-\underline{\kappa}})^{2}\right)$

###### Proof.

Concerning the integral terms, note that

$\displaystyle|p(s)|$ $\displaystyle\leq
J^{1,0}(s)\int_{\tfrac{-d}{2}}^{s}\tfrac{J^{0,1}(w)}{-(J^{1,0})^{\prime}\left(\tfrac{d}{2}\right)}J(w)|\nabla\kappa|(w)\,dw+J^{0,1}(s)\int_{s}^{\tfrac{d}{2}}\tfrac{J^{1,0}(w)}{-(J^{1,0})^{\prime}\left(\tfrac{d}{2}\right)}J(w)|\nabla\kappa|(w)\,dw$
$\displaystyle\leq
J(s)\textup{sn}_{\overline{\kappa}}(d)\Bigl{(}\int_{-\tfrac{d}{2}}^{\tfrac{d}{2}}J^{2}|\nabla\kappa|\,dw\Bigr{)}$
$\displaystyle\leq\frac{\textup{sn}_{\overline{\kappa}}(d)|\nabla\kappa|_{\infty}}{\textup{cs}^{3}_{\overline{\kappa}}(\tfrac{d}{2})}\int_{-\tfrac{d}{2}}^{\tfrac{d}{2}}\textup{cs}^{2}_{\overline{\kappa}}(w)\,dw$
$\displaystyle\leq{d^{2}\sqrt{8}}|\nabla k|_{\infty}$

and that

$\displaystyle|\tilde{q}(s)|=$
$\displaystyle\Bigl{|}\tfrac{1}{d}(\tfrac{d}{2}-s)J^{\prime}(\tfrac{-d}{2})+\tfrac{1}{d}(s+\tfrac{d_{0}}{2})J^{\prime}(\tfrac{d}{2})+3J^{\prime}(s)J(s)\Bigr{|}$
$\displaystyle\leq\left(2+\frac{3}{\textup{cs}_{\overline{\kappa}}(\tfrac{d}{2})}\right)\textup{tn}_{\overline{\kappa}}$
$\displaystyle\leq\tfrac{5\sqrt{\overline{\kappa}}d}{\textup{cs}^{2}_{\overline{\kappa}}(\tfrac{d}{2})}\leq
10\sqrt{\overline{\kappa}}d,$

where we used the fact that $\textup{cs}_{\overline{\kappa}}(\tfrac{d}{2})\geq
1/\sqrt{2}.$

Thus we obtain that

$\displaystyle\int\tfrac{J^{2}(s)}{2}\Bigl{[}\Delta\kappa(s)J^{2}(s)+3\kappa_{1}(s){p(s)}+\kappa_{2}(s)\tilde{q}(s)\Bigr{]}\,ds$
$\displaystyle\geq$ $\displaystyle
2d(\inf\Delta\kappa)_{-}-{3\sqrt{8}|\nabla\kappa|^{2}_{\infty}d^{3}}-10\sqrt{\overline{\kappa}}|\nabla\kappa|_{\infty}d^{2}$
$\displaystyle\geq$ $\displaystyle
2\mathcal{C}(x,y)\Bigl{(}\frac{-2(\inf\Delta\kappa)_{-}}{\underline{\kappa}}+\frac{3\sqrt{8}|\nabla\kappa|^{2}_{\infty}d^{2}}{\underline{\kappa}}+\frac{10\sqrt{\overline{\kappa}}|\nabla\kappa|_{\infty}d}{\underline{\kappa}}\Bigr{)}$

where we again used $\textup{cs}_{\overline{\kappa}}(\tfrac{d}{2})\geq
1/\sqrt{2}.$ Furthermore,

$\displaystyle-(J^{0,1})^{\prime}(\tfrac{-d_{0}}{2})\left(J^{\prime}(\tfrac{d_{0}}{2})+J^{\prime}(\tfrac{-d_{0}}{2})\right)^{2}$
$\displaystyle\geq-\tfrac{1}{\textup{sn}_{\overline{\kappa}}(d)}\textup{tn}_{\underline{\kappa}}^{2}(\tfrac{d}{2})\left(\tfrac{\textup{tn}_{\overline{\kappa}}}{\textup{tn}_{\underline{\kappa}}}-1\right)^{2}$
$\displaystyle\geq\sqrt{2}\overline{\kappa}\mathcal{C}(x,y)\left(\tfrac{\textup{tn}_{\overline{\kappa}}}{\textup{tn}_{\underline{\kappa}}}-1\right)^{2},$

for we estimate

$\displaystyle\frac{\textup{tn}_{\overline{\kappa}}}{\textup{tn}_{\underline{\kappa}}}-1$
$\displaystyle=\frac{1}{\textup{tn}_{\underline{\kappa}}}\int_{\underline{\kappa}}^{\overline{\kappa}}\frac{1}{2\sqrt{\kappa}}\tan(\sqrt{\kappa}\tfrac{d}{2})+\frac{d}{4\textup{cs}_{\kappa}^{2}(\tfrac{d}{2})}\,d\kappa$
$\displaystyle\leq\frac{\overline{\kappa}-\underline{\kappa}}{\textup{tn}_{\underline{\kappa}}}\left(\frac{\textup{tn}_{\overline{\kappa}}(\tfrac{d}{2})}{2\sqrt{\underline{\kappa}}}+\frac{d}{2}\right)$
$\displaystyle\leq\frac{\overline{\kappa}-\underline{\kappa}}{\underline{\kappa}}\left(\frac{\textup{tn}_{\overline{\kappa}}(\tfrac{d}{2})}{d\sqrt{\underline{\kappa}}}+1\right),$

where we used
$\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2})\geq\underline{\kappa}\tfrac{d}{2}$
in the second inequality. Further, since the function $x\mapsto\tan(x)/x$ is
strictly increasing on $(0,\infty)$ (whenever defined), and since
$d\leq\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$

$\displaystyle\frac{\overline{\kappa}-\underline{\kappa}}{\underline{\kappa}}\left(\frac{\textup{tn}_{\overline{\kappa}}(\tfrac{d}{2})}{d\sqrt{\underline{\kappa}}}+1\right)$
$\displaystyle\leq\frac{\overline{\kappa}-\underline{\kappa}}{\underline{\kappa}}\left(\frac{4{\overline{\kappa}}}{\pi\sqrt{\underline{\kappa}}}+1\right).$

Putting it all together, with $D\leq\tfrac{\pi}{2\sqrt{\overline{\kappa}}}$
gives the claim. ∎

In order to obtain estimate (3.7), we proceed with estimating
$\varepsilon(x,y)$ which is defined in (3.12).

######  Lemma 7.7.

For any $x,y\in\Omega$

$\displaystyle-\frac{\varepsilon^{2}(x,y)}{8\textup{tn}_{\underline{\kappa}}(\tfrac{d}{2})}\geq
2\mathcal{C}(x,y)\frac{|\nabla\kappa|^{2}_{\infty}}{\underline{\kappa}^{2}}.$

###### Proof.

Observe that

$\displaystyle\varepsilon(x,y)$
$\displaystyle=\int_{\tfrac{-d}{2}}^{\tfrac{d}{2}}\langle\gamma_{x,y}^{\prime}(s),\nabla\kappa(s)\rangle\,ds\leq
2d(x,y)|\nabla\kappa|_{\infty}$

This gives that