# First Eigenvalue of the Laplacian of a Geodesic Ball and Area-Based
Symmetrization of its Metric Tensor

Vicent Gimeno  Departament de Matemàtiques- IMAC, Universitat Jaume I,
Castelló, Spain<EMAIL_ADDRESS>and  Erik Sarrion-Pedralva**  Departament de
Matemàtiques-IMAC, Universitat Jaume I, Castelló, Spain.
<EMAIL_ADDRESS>
###### Abstract.

Given a Riemmanian manifold, we provide a new method to compute a sharp upper
bound for the first eigenvalue of the Laplacian for the Dirichlet problem on a
geodesic ball of radius less than the injectivity radius of the manifold. This
upper bound is obtained by transforming the metric tensor into a rotationally
symmetric metric tensor that preserves the area of the geodesic spheres. The
provided upper bound can be computed using only the area function of the
geodesic spheres contained in the geodesic ball and it is sharp in the sense
that the first eigenvalue of geodesic ball coincides with our upper bound if
and only if the mean curvature pointed inward of each geodesic sphere is a
radial function.

###### Key words and phrases:

First Dirichlet Eigenvalue, Symmetrization, Laplace Operator, Geodesic Ball

###### 2020 Mathematics Subject Classification:

Primary 53C20; Secondary 58C40

* Work partially supported by the Research Program of University Jaume I Project UJI-B2018-35, and DGI -MINECO grant (FEDER) MTM2017-84851-C2-2-P

** Work partially supported by the Research Program of University Jaume I
Project UJI-B2018-35, and DGI -MINECO grant (FEDER) MTM2017-84851-C2-2-P, and
grant for predoctoral research GVA-ESF ACIF/2019/096.

##  1\. Introduction

Let $(M,g)$ be a $n$-dimensional Riemannian manifold, and let $\Omega\subset
M$ be a precompact domain with smooth boundary $\partial\Omega$. The first
eigenvalue $\lambda_{1,g}(\Omega)$ of the Laplacian for the Dirichlet problem
on $\Omega$ is the smallest $\lambda\in\mathbb{R}$ such that there exists a
non trivial associated eigenfunction $\phi$ satisfying

$\left\\{\begin{array}[]{rcll}\Delta_{g}\phi+\lambda\phi&=0,&\text{ on
}&\Omega,\\\ \phi&=0,&\text{ on }&\partial\Omega,\end{array}\right.$

where $\Delta_{g}$ is the Laplacian operator with respect to the metric tensor
$g$, _i.e._ ,

$\Delta_{g}=\frac{1}{\sqrt{\det g}}\sum_{i,j=1}^{n}\frac{\partial}{\partial
x^{i}}\left(\sqrt{\det g\,}g^{ij}\frac{\partial}{\partial x^{j}}\right)$

in local coordinates $(x^{1},\cdots,x^{n})$. Upper and lower bounds for the
first eigenvalue have been widely studied in terms of geometric invariants of
the domain $\Omega$.

In [8] and [7] Cheng obtained upper and lower bounds for the first eigenvalue
of the Laplacian for the Dirichlet problem on a geodesic ball as follows: If
the Ricci curvatures ${\rm Ric}_{g}$ of $M$ are bounded from below by the
Ricci curvatures of a $n$-dimensional simply connected real space form
$(\mathbb{M}_{\kappa}^{n},g_{\kappa})$ of constant sectional curvature
$\kappa$, _i.e._ ,

${\rm Ric}_{g}\geq(n-1)\kappa,$

then the first eigenvalue $\lambda_{1,g}({\rm B}_{R}(p))$ of the Laplacian for
the Dirichlet problem of a geodesic ball ${\rm B}_{R}(p)$ of radius $R$
centered at $p\in M$ is bounded from above by

(1.1) $\lambda_{1,g}({\rm B}_{R}(p))\leq\lambda_{1}(R,\kappa),$

where $\lambda_{1}(R,\kappa)$ is the first Dirichlet eigenvalue of a geodesic
ball of radius $R$ in $\mathbb{M}_{\kappa}^{n}$. Conversely, if the sectional
curvatures of $M$ are bounded from above by the sectional curvature of
$(\mathbb{M}_{\kappa}^{n},g_{\kappa})$, _i.e._ ,

${\rm sec}_{g}\leq\kappa,$

then the first eigenvalue $\lambda_{1,g}({\rm B}_{R}(p))$ of the Laplacian for
the Dirichlet problem of a geodesic ball ${\rm B}_{R}(p)$ of radius
$R<\min\\{{\rm inj}_{g}(p),\pi/\sqrt{\kappa}\\}$111where ${\rm inj}_{g}(p)$ is
the injectivity radius of $p$ and $\pi/\sqrt{\kappa}$ is replaced by $+\infty$
if $\kappa\leq 0$. is bounded from below by

(1.2) $\lambda_{1}({\rm B}_{R}(p))\geq\lambda_{1}(R,\kappa).$

Moreover, inequalities (1.1) and (1.2) are sharp because equality is attained
in both inequalities if and only if ${\rm B}_{R}(p)$ is isometric to the
geodesic ball of radius $R$ in $\mathbb{M}_{\kappa}^{n}$.

In [2], Bessa and Montenegro, have obtained the same upper and lower bounds
for the first eigenvalue of the geodesic ball $B_{R}(p)$ but assuming that the
mean curvature of the geodesic spheres of $(M,g)$ centered at $p\in M$ are
bounded by the mean curvature of the geodesic spheres in
$\mathbb{M}_{\kappa}^{n}$. In particular, if the mean curvature pointed inward
${\rm H}_{{\rm S}_{t}(p)}$ of the geodesic spheres ${\rm S}_{t}(p)$ of $M$ are
bounded from above (resp. from below) by the mean curvature pointed inward
${\rm H}(t,\kappa)$ of the geodesic sphere with the same radius of
$\mathbb{M}_{\kappa}^{n}$, _i.e._ ,

${\rm H}_{{\rm S}_{t}(p)}\leq{\rm
H}(t,\kappa)\quad\left(\vphantom{\dfrac{1}{1}}{\rm resp.}\quad{\rm H}_{{\rm
S}_{t}(p)}\geq{\rm H}(t,\kappa)\right),$

for all point in ${\rm S}_{t}(p)$ and for all $t\in(0,R)$, then

(1.3) $\lambda_{1,g}({\rm
B}_{R}(p))\leq\lambda_{1}(R,\kappa)\quad\left(\vphantom{\dfrac{1}{1}}{\rm
resp.}\quad\lambda_{1,g}({\rm B}_{R}(p))\geq\lambda_{1}(R,\kappa)\right).$

In this case, like in the case where is assumed lower bounds for the Ricci or
upper bounds for the sectional curvatures , inequality (1.3) is sharp. But
now, instead of an isometry between geodesic balls, equality is attained in
(1.3) if and only if ${\rm H}_{{\rm S}_{t}(p)}={\rm H}(t,\kappa)$ for all
$t\in(0,R)$. Observe that the conclusion of an equality between the mean
curvatures pointed inward of geodesic spheres is a weaker than the conclusion
of an isometry between geodesic balls. Indeed, in example 5.3 of [3] is shown
a $4$-dimensional geodesic ball non-isometric to the geodesic ball of
$\mathbb{M}_{\kappa}^{4}$, but with ${\rm H}_{{\rm S}_{t}(p)}={\rm
H}(t,\kappa)$ for all $t\in(0,R)$ .

To bound the sectional or Ricci curvatures, or to bound the mean curvature of
the geodesic spheres, imply to control the behavior of the isoperimetric
quotients. An alternative way to obtain bounds for the first eigenvalue of the
Laplacian makes use of the _isoperimetric inequalities_ and the so-called
_symmetrizations_. One of the most classical symmetrizations is the _Schwarz
symmetrization_ (see [1, 19, 15, 6]). The Schwarz symmetrization of a compact
open domain $D\subseteq M$ is the unique geodesic ball ${\rm
B}_{R(D)}^{\kappa}$ of $\mathbb{M}_{\kappa}^{n}$ satisfying that ${\rm
vol}\left(D\right)={\rm vol}\left({\rm B}_{R(D)}^{\kappa}\right)$.

In [9, 13], Faber and Krahn showed that if for all open set $D\subseteq M$,
consisting in a disjoint union of regular domains, the volume of the perimeter
of $D$ is greater than the volume of the perimeter of its Schwarz
symmetrization ${\rm B}_{R(D)}^{\kappa}$, _i.e._ , if

${\rm vol}\left(\partial D\right)\geq{\rm vol}\left(\partial{\rm
B}_{R(D)}^{\kappa}\right),$

then, for all precompact domain $\Omega\subset M$,

(1.4) $\lambda_{1}(\Omega)\geq\lambda_{1}\left(R(\Omega),\kappa\right).$

Like in the results of Cheng, inequality (1.4) is sharp in the sense that
equality is attained if and only if $\Omega$ is isometric to ${\rm
B}_{R(\Omega)}^{\kappa}$.

Instead of bounding the first eigenvalue assuming certain geometric hypothesis
on the underlying Riemannian manifold, the first eigenvalue can be computed
directly using the so-called Poisson hierarchy (see [15, 16, 12]). Let $(M,g)$
be a Riemannian manifold, the tower of moments of $\Omega\subset M$ is the
family of functions $\displaystyle\\{u_{k}\\}_{k=0}^{\infty}$, defined
inductively as the following sequence of solutions to a hierarchy of boundary
value problems in $\Omega\subset M$:

Let us define

$u_{0}=1\quad\text{ on }\quad\Omega,$

and for $k\geq 1$,

$\left\\{\begin{array}[]{rcll}\Delta_{g}u_{k}+k\,u_{k-1}&=0,&\text{ on
}&\Omega,\\\ u_{k}&=0,&\text{ on }&\partial\Omega.\end{array}\right.$

Moreover, the _moment spectrum of $\Omega$_ is defined as the family of
integrals

$\mathcal{A}_{k}(\Omega)=\int_{\Omega}u_{k}\,d{\rm V}_{g}.$

In [17], McDonald and Meyers proved that the moment spectrum of $\Omega$ can
be used to compute the first eigenvalue of the Laplacian for the Dirichlet
problem on $\Omega$ by

$\lambda_{1,g}(\Omega)=\sup\left\\{\eta\geq
0\,:\,\lim_{k\to\infty}\sup\left(\frac{\eta}{2}\right)^{n}\frac{\mathcal{A}_{k}(\Omega)}{\Gamma(n+1)}<\infty\right\\}.$

More recently, in [3], Bessa, Jorge and the first author have proved that

(1.5)
$\lambda_{1,g}(\Omega)=\lim_{k\to\infty}(k+1)\dfrac{\lVert{u}_{k}\rVert_{2}}{\lVert{u}_{k+1}\rVert_{2}},$

where $\lVert\cdot\rVert_{2}$ denotes the $L_{2}$-norm on $\Omega$. In the
particular case when the domain is a geodesic ball $B_{R}^{w}$ of radius $R$
in a Riemannian model $\mathbb{M}_{w}^{n}$, see section 2.2 for the definition
of a Riemannian model, in [12] is proved that

(1.6)
$\lambda_{1,g_{w}}(B_{R}^{w})=\lim_{k\to\infty}\frac{ku_{k-1}(0)}{u_{k}(0)}=\lim_{k\to\infty}\frac{k\mathcal{A}_{k-1}(B_{R}^{w})}{\mathcal{A}_{k}(B_{R}^{w})}.$

In this paper we prove that an upper bound for the first eigenvalue of the
Laplacian for the Dirichlet problem of a geodesic ball $B_{R}(p)$ can be
computed using only the _area function of the geodesic spheres_ ,

${\rm A}_{g}(t):={\rm vol}_{g}({\rm S}_{t}(p)),$

and the following family of functions $\\{T_{k}\\}_{k=0}^{\infty}$,
constructed recursively from the area function, where

$T_{0}(t)=1$

and for $k\geq 1$,

(1.7) $T_{k}(t)=\int_{t}^{R}\frac{\int_{0}^{\sigma}T_{k-1}(s){\rm
A}_{g}(s)\,ds}{{\rm A}_{g}(\sigma)}\,d\sigma.$

Our upper bound for the first eigenvalue of ${\rm B}_{R}(p)$, obtained using
the above family of functions $\\{T_{k}\\}$, is stated in the following

######  Theorem 1.

Let $(M,g)$ be a Riemannian manifold, let $p\in M$ be a point of $M$ with
injectivity radius ${\rm inj}_{g}(p)$, and let ${\rm B}_{R}(p)$ be the
geodesic ball of radius $R$ centered at $p$. Suppose that $R<{\rm
inj}_{g}(p)$, then the first eigenvalue $\lambda_{1,g}({\rm B}_{R}(p))$ of the
Laplacian for the Dirichlet problem on ${\rm B}_{R}(p)$ is bounded by

(1.8) $\lambda_{1,g}({\rm
B}_{R}(p))\leq\lim_{k\to\infty}\left(\frac{\int_{0}^{R}T_{k}^{2}(t){\rm
A}_{g}(t)\,dt}{\int_{0}^{R}T_{k+1}^{2}(t){\rm A}_{g}(t)\,dt}\right)^{1/2}.$

Furthermore, equality is attained in (1.8) if and only if, for any
$t\in(0,R)$, the mean curvature pointed inward ${\rm H}_{{\rm S}_{t}(p)}$ of
the geodesic sphere ${\rm S}_{t}(p)$ of radius $t$ centered at $p$ is a radial
function. Namely, equality in (1.8) is attained if and only if there exists a
smooth function $h(t)$ such that

${\rm H}_{{\rm S}_{t}(p)}=h(t)\quad{\rm for\;any}\quad 0<t<R.$

In this work instead of use the Schwarz symmetrization (as Faber and Krahn in
[9, 13]), we will construct a rotationally symmetric metric tensor
$\widetilde{g}$ for the geodesic ball ${\rm B}_{R}(p)$ with respect to the
metric tensor $g$, such that the volume of the geodesic spheres of $({\rm
B}_{R}(p),\widetilde{g})$ will coincide with the volume of the geodesic
spheres with the same radius of $({\rm B}_{R}(p),g)$. Namely,

${\rm A}_{g}(t)={\rm A}_{\widetilde{g}}(t).$

This rotationally symmetric metric tensor allows us to find a comparison for
the first eigenvalue of the Laplacian for the Dirichlet problem but without
using an isoperimetric inequality as hypothesis as in Faber and Krahn Theorem.

The family of functions (1.7) are related with the moment functions of the
Poisson hierarchy $\\{\widetilde{u}_{k}\\}$ associated to $\widetilde{g}$. In
fact, in [3], is proved that since $\widetilde{g}$ is a rotationally symmetric
metric tensor

${\widetilde{u}_{k}(q)=k!T_{k}({\rm
dist}_{\widetilde{g}}(p,q))},\quad\|\widetilde{u}_{k}\|_{2}=k!\left(\int_{0}^{R}T^{2}_{k}(t)A_{g}(t)\,dt\right)^{1/2}.$

Hence by (1.5),

$\lambda_{1,\widetilde{g}}({\rm
B}_{R}(p))=\lim_{k\to\infty}\left(\frac{\int_{0}^{R}T_{k}^{2}(t){\rm
A}_{g}(t)\,dt}{\int_{0}^{R}T_{k+1}^{2}(t){\rm A}_{g}(t)\,dt}\right)^{{1/2}}.$

Then, inequality (1.8) can be rewritten as

$\lambda_{1,g}({\rm B}_{R}(p))\leq\lambda_{1,\widetilde{g}}({\rm B}_{R}(p)).$

Notice that in the statement of the Theorem 1 there are no conditions on the
Ricci or sectional curvatures (as in the hypothesis of Cheng in [8] and [7]),
neither on the mean curvature of the geodesic spheres (as in the hypothesis of
Bessa and Montenegro in [2]). But the geodesic spheres, in the critical metric
tensor where equality in (1.8) is attained, have radial mean curvature as in
the result of Bessa and Montenegro.

Since $(B_{R},\widetilde{g})$ is a Riemannian model of radius $R$ we can apply
(1.6) and hence

(1.9) $\displaystyle\lambda_{1,g}({\rm
B}_{R}(p))\leq\lambda_{1,\widetilde{g}}({\rm B}_{R}(p))=$
$\displaystyle\lim_{k\to\infty}\left(\frac{\int_{0}^{R}T_{k}^{2}(t){\rm
A}_{g}(t)\,dt}{\int_{0}^{R}T_{k+1}^{2}(t){\rm A}_{g}(t)\,dt}\right)^{{1/2}}$
$\displaystyle=$
$\displaystyle\lim_{k\to\infty}\frac{T_{k-1}(0)}{T_{k}(0)}=\lim_{k\to\infty}\frac{\int_{0}^{R}T_{k-1}(t){\rm
A}_{g}(t)\,dt}{\int_{0}^{R}T_{k}(t){\rm A}_{g}(t)\,dt}\cdot$

We must remark here that in the classical symmetrization results, the
symmetrizated domain minimizes the first eigenvalue but in our result the ball
with the rotationally symmetric metric tensor maximizes the first eigenvalue.
Moreover, in the following example we show that our upper bounds can not be
obtained from the classical comparison with the Ricci curvature neither from
the comparison with the mean curvature.

######  Example 1.1.

Let $(r,\theta)$ be polar coordinates in $\mathbb{R}^{2}$ around
$\vec{0}\in\mathbb{R}^{2}$. Let us endow $\mathbb{R}^{2}$ with the following
metric

$g=dr\otimes dr+\left(r+\varphi(r)\cos(\theta)\right)^{2}d\theta\otimes
d\theta,$

with

$\varphi:[0,\infty)\to[0,\infty),\quad
t\mapsto\varphi(t):=\left\\{\begin{array}[]{lcl}0,&{\rm if}&t\leq 2,\\\
e^{-\frac{1}{(t-2)^{2}}},&{\rm if}&t>2.\end{array}\right.$

Hence,

${\rm
A}_{g}(t)=\int_{0}^{2\pi}\left(t+\varphi(t)\cos(\theta)\right)d\theta=2\pi t.$

By using the Theorem 1 we conclude that, for the geodesic disc of radius $R$,

(1.10) $\lambda_{1,g}({\rm B}_{R}(\vec{0}))\leq\lambda_{1,g_{\rm can}}({\rm
B}_{R}(\vec{0}))=\frac{j_{0}^{2}}{R^{2}}\approx\frac{5,78319}{R^{2}},$

where $g_{\rm can}$ is the canonical metric tensor in $\mathbb{R}^{2}$,
$g_{\rm can}=dr\otimes dr+r^{2}\,d\theta\otimes d\theta$, and $j_{0}$ is the
first zero of the Bessel function $J_{0}$.

On the other hand, since the Ricci curvature is given by

$K(g)=\left\\{\begin{array}[]{lcl}0,&{\rm for}&r\leq 2,\\\
\frac{2\,(3(r-4)r+10)\cos(\theta)}{(r-2)^{6}\left(\cos(\theta)+e^{\frac{1}{(r-2)^{2}}}r\right)},&{\rm
for}&r>2,\end{array}\right.$

for $r>2$, there are regions where $K(g)<0$. Hence, for a geodesic ball of
radius $R>2$, the bound (1.10) can not be obtained by using the comparison of
Cheng with the Ricci curvature (where it is needed $K(g)\geq 0$). Moreover,
since the mean curvature ${\rm H}(t,\theta)$ pointing inward of the geodesic
sphere ${\rm S}_{t}(\vec{0})$ is given by

${\rm H}(t,\theta)=\left\\{\begin{array}[]{lcl}1/t,&{\rm for}&t\leq 2,\\\
1/t-\frac{(t-4)((t-2)t+2)\cos(\theta)}{(t-2)^{3}t\left(\cos(\theta)+e^{\frac{1}{(t-2)^{2}}}t\right)},&{\rm
for}&t>2,\end{array}\right.$

for $t>2$ there are points in the sphere ${\rm S}_{t}(\vec{0})$ where ${\rm
H}(t,\theta)>1/t$. Hence, for a geodesic ball of radius $R>2$, the bound
(1.10) can not be obtained by using the comparison of Bessa and Montenegro
with the mean curvature of the geodesic spheres (where it is needed ${\rm
H}(t,\theta)\leq 1/t$). Notice that, since for $R>2$ the mean curvature
pointed inward of the geodesic spheres ${\rm S}_{t}(\vec{0})$ is not a radial
function for any $t\in(2,R)$, then equality in (1.10) can not be obtained,
namely

(1.11) $\lambda_{1,g}({\rm B}_{R}(\vec{0}))<\lambda_{1,g_{\rm can}}({\rm
B}_{R}(\vec{0}))\quad{\rm for}\quad R>2.$

The above inequality allows us to state that in $(\mathbb{R}^{2},g)$ there
exists a domain $\Omega$ with symmetrizated radius $R(\Omega)>0$ with respect
to the Euclidean space $(\mathbb{R}^{2},g_{\rm can})$, _i.e._ , with

${\rm vol}_{g}(\Omega)={\rm vol}_{g_{\rm can}}(B_{R(\Omega)}(\vec{0}))=\pi
R^{2}(\Omega)$

such that

${\rm vol}_{g}(\partial\Omega)<{\rm vol}_{g_{\rm can}}(\partial
B_{R(\Omega)}(\vec{0}))=2\pi R(\Omega).$

Because otherwise, if for any domain $\Omega$, ${\rm
vol}_{g}(\partial\Omega)\geq 2\pi{R(\Omega)}$, then by Faber-Krahn Theorem
$\lambda_{1,g}({\rm B}_{R}(\vec{0}))$ should be greater or equal to
$\lambda_{1,g_{\rm can}}({\rm B}_{R}(\vec{0}))$ but this is a contradiction
with inequality (1.11).

From Theorem 1 we can compare the first eigenvalue of a geodesic ball of
$(M,g)$ with the first eigenvalue of a geodesic ball of a real space form
$\mathbb{M}^{n}_{\kappa}$ assuming certain behavior of the area function ${\rm
A}_{g}(t)$ as follows

######  Theorem 2.

Let $(M,g)$ be a Riemannian manifold, let $p\in M$ be a point of $M$ with
injectivity radius ${\rm inj}_{g}(p)$. Let ${\rm A}_{\kappa}(t)$ be the area
function of the geodesic sphere of radius $t$ in $\mathbb{M}^{n}_{\kappa}$.
Suppose that $R<{\rm inj}_{g}(p)$ and that for any $t<R$ the function

$t\mapsto\frac{{\rm A}_{g}(t)}{{\rm A}_{\kappa}(t)}$

is a decreasing function. Then, the first eigenvalue $\lambda_{1,g}({\rm
B}_{R}(p))$ of the Laplacian for the Dirichlet problem on the geodesic ball
${\rm B}_{R}(p)$ of radius $R$ centered at $p$ is bounded by

(1.12) $\lambda_{1,g}({\rm B}_{R}(p))\leq\lambda_{1}(R,\kappa),$

with equality in (1.12) if and only if, for any $t\in(0,R)$, the mean
curvature pointed inward ${\rm H}_{{\rm S}_{t}(p)}$ of the geodesic sphere
${\rm S}_{t}(p)$ is equal to the mean curvature pointed inward of the geodesic
sphere of radius $t$ in $\mathbb{M}_{\kappa}^{n}$, namely

${\rm H}_{{\rm
S}_{t}(p)}=(n-1)\frac{S_{\kappa}^{\prime}(t)}{S_{\kappa}(t)},\quad{\rm
for\;all}\quad 0<t<R,$

where

$S_{\kappa}(t):=\left\\{\begin{array}[]{rcl}\frac{\sin(\sqrt{\kappa}t)}{\sqrt{\kappa}},&{\rm
if}&\kappa>0,\\\ t,&{\rm if}&\kappa=0,\\\
\frac{\sinh(\sqrt{-\kappa}t)}{\sqrt{-\kappa}},&{\rm
if}&\kappa<0.\end{array}\right.$

Observe that when the Ricci curvatures of $M$ are bounded from below by the
Ricci curvatures of $\mathbb{M}_{\kappa}^{n}$, _i.e._ , ${\rm
Ric}_{g}\geq(n-1)\kappa$ (as the hypothesis of Cheng in [8]) the function

$t\mapsto\frac{{\rm A}_{g}(t)}{{\rm A}_{\kappa}(t)}$

is a decreasing function for any $t<{\rm inj}_{g}(p)$ (for more details see
[5]). Hence, our hypothesis about the behavior of the above quotient is weaker
than the original hypothesis of Cheng but only up to the injectivity radius
${\rm inj}_{g}(p)$. Moreover, to characterize the equality, Cheng shows that
the equality is attained if and only if the ball ${\rm B}_{R}(p)$ of $M$ is
isometric to the ball with the same radius of $\mathbb{M}^{n}_{\kappa}$. But,
with our hypothesis, equality is attained if and only if the mean curvature of
the geodesic sphere ${\rm S}_{t}(p)$ is
$(n-1)\frac{S_{\kappa}^{\prime}(t)}{S_{\kappa}(t)}$ for all $t<R$.

Theorem 2 is a particular case of Theorem 4.1 of Section 4 obtained by using
our Theorem 1. Indeed, the general case is obtained when

$t\mapsto\frac{{\rm A}_{g}(t)}{{\rm A}_{W}(t)}$

is decreasing where ${\rm A}_{W}(t)$ is the volume of the geodesic sphere of
radius $t$ in some Riemannian model.

### Acknowledgments

The authors are grateful to professor Vicente Palmer for his valuable help and
for his useful comments and suggestions during the preparation of the present
paper.

##  2\. Volume-based rotational symmetrization of the metric tensor

The main tool of this paper consists in the transformation of the metric
tensor $g$ of a Riemannian manifold $(M,g)$ to a metric tensor $\widetilde{g}$
such that $({\rm B}_{R}(p),\widetilde{g})$ is rotationally symmetric and the
volume of the geodesic spheres of $({\rm B}_{R}(p),\widetilde{g})$ coincides
with the volume of geodesic spheres of $({\rm B}_{R}(p),g)$. In this section
we show in Theorem 2.2 that this new metric tensor $\widetilde{g}$ is smooth,
in Proposition 2.3 we compare the behavior of the distance function with
respect $g$ and $\widetilde{g}$, as well as the area function for geodesic
spheres and the expression of the Laplacian with respect to the new metric
tensor $\widetilde{g}$. Finally, in Corollary 2.6 we characterize the first
eigenfunction and first eigenvalue of $({\rm B}_{R}(p),\widetilde{g})$.

###  2.1. Definition and smoothness of the rotationally symmetric metric
tensor of comparison

Let $(M,g)$ be a Riemannian manifold, let $p\in M$ be a point in $M$, let
${\rm B}_{R}(p)$ be the geodesic ball of $M$ centered at $p$ of radius $R<{\rm
inj}_{g}(p)$, and let us denote by

$\mathcal{B}_{R}(\vec{0}):=\left\\{v\in T_{p}M\,:\,g(v,v)<R^{2}\right\\}.$

The exponential map $\exp_{p}:T_{p}M\to M$ induces a diffeomorphism from
$\mathcal{B}_{R}(\vec{0})$ to ${\rm B}_{R}(p)$. Let $\\{E^{i}\\}_{i=1}^{n}$ be
an orthonormal basis of $T_{p}M$ and let us denote by $x^{1},\ldots,x^{n}$ the
_normal coordinate functions_ associated to the orthonormal basis
$\\{E^{i}\\}_{i=1}^{n}$ with respect to $g$ given by
$x^{i}(q):=g(E^{i},\exp_{p}^{-1}(q))$. Then the map

(2.1) $\zeta:{\rm B}_{R}(p)\to\mathbb{D}_{R}(\vec{0}),\quad
q\mapsto\zeta(q):=\left(x^{1}(q),\ldots,x^{n}(q)\right)\quad$

is a diffeomorphism from ${\rm B}_{R}(p)$ to the open Euclidean ball
$\mathbb{D}_{R}(\vec{0})$,

$\mathbb{D}_{R}(\vec{0}):=\left\\{x\in\mathbb{R}^{n}\,:\,(x^{1})^{2}+\cdots+(x^{n})^{2}<R^{2}\right\\}.$

Since $\zeta=(x^{1},\ldots,x^{n}):{\rm B}_{R}(p)\to\mathbb{D}_{R}(\vec{0})$ is
a coordinate system (or chart) in ${\rm B}_{R}(p)$ then, for any $q\in{\rm
B}_{R}(p)$, the coordinate vectors

$\left\\{\left.\frac{\partial}{\partial
x^{1}}\right|_{q},\ldots,\left.\frac{\partial}{\partial
x^{n}}\right|_{q}\right\\}$

form a basis for the tangent space $T_{q}({\rm B}_{R}(p))$ (see for instance
Theorem 12 of [18]). For each $1\leq i\leq n$ the vector field
$\frac{\partial}{\partial x^{i}}$ on ${\rm B}_{R}(p)$ sending each $q$ to
$\left.\frac{\partial}{\partial x^{i}}\right|_{q}$ is called the _$i$ th
coordinate vector field_ and the one-forms

$\left\\{dx^{1},\ldots,dx^{n}\right\\}\quad{\rm given}\;{\rm by}\quad
dx^{i}\left(\frac{\partial}{\partial x^{j}}\right)={\delta_{i,j}}$

are called _coordinate one-forms_. Moreover, for any smooth function $f:{\rm
B}_{R}(p)\to\mathbb{R}$, the one-form $df$ can be obtained as

$df=\sum_{i=1}^{n}\frac{\partial f}{\partial x^{i}}dx^{i}.$

Let us introduce the following functions $r$ and $\pi$ associated $\zeta$

(2.2) $\begin{array}[]{ll}r:{\rm B}_{R}(p)\to\mathbb{R},&q\mapsto
r(q):=\sqrt{(x^{1}(q))^{2}+\cdots+(x^{n}(q))^{2}},\vphantom{\dfrac{1}{1}}\\\
\pi:{\rm
B}_{R}(p)-\\{p\\}\to\mathbb{S}^{n-1}_{1},&q\mapsto\pi(q):=\dfrac{\zeta(q)}{r(q)}.\end{array}$

In the following definition we introduce the rotationally symmetric metric
tensor of comparison by using the above functions.

######  Definition 2.1 (Rotationally symmetric metric tensor of comparison).

Let $(M,g)$ be a $n$-dimensional Riemannian manifold. Let ${\rm B}_{R}(p)$ be
the geodesic ball of radius $R$ centered at $p{\in M}$. Suppose that $R<{\rm
inj}_{g}(p)$. Let $\\{E^{i}\\}_{i=1}^{n}$ be an orthonormal basis of $T_{p}M$
and let $\zeta=(x^{1},\ldots,x^{n}):{\rm B}_{R}(p)\to\mathbb{D}_{R}(\vec{0})$
be the normal coordinate functions associated to $\\{E^{i}\\}_{i=1}^{n}$. The
_rotationally symmetric metric tensor of comparison_ $\widetilde{g}$
associated to $g$ is the metric tensor given by

(2.3) $\widetilde{g}=\left\\{\begin{array}[]{lcl}dr\otimes
dr+(\omega_{g}^{2}\circ r)\pi^{*}g_{\mathbb{S}^{n-1}_{1}},&{\rm on}&{\rm
B}_{R}(p)-\\{p\\},\\\ \displaystyle\sum_{i=1}^{n}dx^{i}\otimes dx^{i},&{\rm
on}&p,\end{array}\right.$

where $r,\pi$ are given by (2.2), $\pi^{*}g_{\mathbb{S}^{n-1}_{1}}$ is the
pullback by $\pi$ of the canonical metric tensor $g_{\mathbb{S}^{n-1}_{1}}$ of
$\mathbb{S}^{n-1}_{1}$, 222Recall that the canonical metric tensor
$g_{\mathbb{S}^{n-1}_{1}}$ of $\mathbb{S}^{n-1}_{1}$ is the metric tensor that
inherits
$\mathbb{S}^{n-1}_{1}:=\\{x\in\mathbb{R}^{n}\,:\,\sum_{i=1}^{n}(x^{i})^{2}=1\\}$
when is considered as a submanifold of $\mathbb{R}^{n}$ with the canonical
metric tensor $\sum_{i=1}^{n}dx^{i}\otimes dx^{i}$. and
$\omega_{g}:[0,R)\to\mathbb{R}_{+}$ is the positive function given by

(2.4) $t\mapsto\omega_{g}(t):=\left(\frac{{\rm A}_{g}(t)}{{\rm
vol}\left({\mathbb{S}_{1}^{n-1}}\right)}\right)^{\frac{1}{n-1}}$

where $A_{g}(t)$ is the volume of the sphere $S_{t}(p)$ of radius $t$ centered
at $p$, _i.e._ ,

${\rm A}_{g}(t):={\rm vol}_{g}({\rm S}_{t}(p)).$

The main result of this section is to show that the metric tensor
$\widetilde{g}$ is well defined. Indeed, in the following Theorem we will
prove the smoothness of this new metric tensor.

######  Theorem 2.2.

Let $(M,g)$ be a $n$-dimensional Riemannian manifold. Let ${\rm B}_{R}(p)$ be
the geodesic ball of radius $R$ centered at $p{\in M}$. Suppose that $R<{\rm
inj}_{g}(p)$. Then, the rotationally symmetric metric tensor of comparison
$\widetilde{g}$ associated to $g$ is smooth in ${\rm B}_{R}(p)$.

###### Proof.

First, we are proving that $\omega_{g}(t)$ can be rewritten as

$\omega_{g}(t)=t\left(1+t^{2}\varphi(t^{2})\right)$

with some positive smooth function $\varphi$. The area function ${\rm
A}_{g}(t)$ is a smooth function up to its injectivy radius (see [5]) and it
has Taylor expansion about $t=0$ given by (see Theorem 3.1 of [10])

${\rm A}_{g}(t)=a_{0}t^{n-1}+a_{2}t^{n+1}+a_{4}t^{n+3}+\cdots$

for some constants $a_{2k}\in\mathbb{R}$, $k\in\mathbb{N}$, with $a_{0}={\rm
vol}\left(\mathbb{S}^{n-1}\right)$. In particular, ${\rm A}_{g}(0)=0$, the
derivatives ${\rm A}_{g}^{(k)}(0)=0$ for $k=1,\dots,n-2$, and the derivatives
${\rm A}_{g}^{(n+2k)}(0)=0$ for $k\in\mathbb{Z}$. Since every derivative of
${\rm A}_{g}(t)$ vanishes up to $n-1$ order and since ${\rm A}_{g}(t)$ is a
smooth function up to $t={\rm inj}_{g}(p)$, then we can use for all
$t\in[0,{\rm inj}_{g}(p)]$ the Taylor expansion with integral form of the
remainder (see [20] for instance) and we can rewrite ${\rm A}_{g}(t)$ as

${\rm A}_{g}(t)=\dfrac{1}{(n-2)!}\int_{0}^{t}(t-x)^{n-2}{\rm
A}_{g}^{(n-1)}(x)\,dx.$

By using the change of variable $x=st$ in the above expression, we can express

${\rm A}_{g}(t)=a_{0}\,t^{n-1}f(t),\quad
f(t):=\dfrac{1}{a_{0}(n-2)!}\int_{0}^{1}(1-s)^{n-2}{\rm
A}_{g}^{(n-1)}(st)\,ds.$

The function $f(t)$ is a positive smooth function with

$f^{(k)}(t)=\frac{1}{a_{0}(n-2)!}\int_{0}^{1}(1-s)^{n-2}s^{k}{\rm
A}_{g}^{(n-1+k)}(st)\,ds.$

In particular, $f(0)=1$ and, since ${\rm A}_{g}^{(n+2k)}(0)=0$ for
$k\in\mathbb{Z}$, the odd order derivatives of $f(t)$ vanish at $0$. In fact,

$f^{(2k+1)}(0)=\frac{1}{a_{0}(n-2)!}\int_{0}^{1}(1-s)^{n-2}s^{k}{\rm
A}_{g}^{(n+2k)}(0)\,ds=0$

for all $k\in\mathbb{N}$. Then $f$ can be extended to a smooth even function
$\widetilde{f}(t)$ with $\widetilde{f}(0)=1$, given by

$\widetilde{f}(t):=\left\\{\begin{array}[]{lcr}f(t),&{\rm if}&t\geq 0,\\\
f(-t),&{\rm if}&t<0.\end{array}\right.$

Since $\widetilde{f}$ is a smooth even function, it can be expressed as (see
[21])

$\widetilde{f}(t)=h(t^{2})$

with a positive smooth function $h$. Notice that $h(0)=f(0)=1$. We can
therefore express the area function as

(2.5) ${\rm A}_{g}(t)=a_{0}\,t^{n-1}h(t^{2})={\rm
vol}\left(\mathbb{S}^{n-1}\right)t^{n-1}h(t^{2}).$

Now we define the function $F(t):=\left(h(t)\right)^{\frac{1}{n-1}}$. Then,
since $h(t)>0$ on $0\leq t<{{\rm inj}_{g}(p)}$ and $h(0)=1$, then $F$ is
smooth with $F(0)=1$. Hence, from (2.4), we obtain that

$\omega_{g}(t)=t\,F(t^{2}).$

On the other hand, since $F$ is a positive smooth function with $F(0)=1$ then
we can express $F$ as

$F(t)=1+\int_{0}^{t}F^{\prime}(x)\,dx=1+t\int_{0}^{1}F^{\prime}(st)\,ds$

Thus, we can rewrite $F$ as

$F(t)=1+t\varphi(t),\quad\varphi(t):=\int_{0}^{1}F^{\prime}(st)\,ds.$

This implies that $F(t^{2})=1+t^{2}\varphi(t^{2})$ and

(2.6) $\omega_{g}(t)=t\left(1+t^{2}\varphi(t^{2})\right).$

Finally, to prove the smoothness of the rotationally symmetric metric tensor
of comparison observe that since on ${\rm B}_{R}(p)-\\{p\\}$

$\sum_{i=1}^{n}dx^{i}\otimes dx^{i}=dr\otimes
dr+r^{2}\pi^{*}g_{\mathbb{S}^{n-1}_{1}}\quad{\rm and}\quad
dr=\sum_{i=1}^{n}\frac{x^{i}}{r}dx^{i},$

the metric tensor of comparison $\widetilde{g}$ can be expressed in normal
coordinates with respect to any orthonormal basis $\\{E_{i}\\}_{i=1}^{n}$ as

(2.7)
$\begin{array}[]{lcr}\widetilde{g}&=&\displaystyle\sum_{i,j=1}^{n}\left(\delta_{i\,j}+\dfrac{\omega_{g}^{2}(r)-r^{2}}{r^{4}}\left(r^{2}\,\delta_{i\,j}-x^{i}x^{j}\right)\right)\,dx^{i}\otimes
dx^{j}.\end{array}$

Then, applying equation (2.6),

(2.8)
$\begin{array}[]{lcl}\widetilde{g}&=&\displaystyle\sum_{i,j=1}^{n}\left(\delta_{i\,j}+\dfrac{r^{2}\left(1+r^{2}\varphi(r^{2})\right)^{2}-r^{2}}{r^{4}}\left(r^{2}\,\delta_{i\,j}-x^{i}x^{j}\right)\right)\,dx^{i}\otimes
dx^{j}\\\
&=&\displaystyle\sum_{i,j=1}^{n}\left(\vphantom{\dfrac{1}{1}}\delta_{i\,j}+\left(2\varphi(r^{2})+r^{2}\varphi^{2}(r^{2})\right)\left(r^{2}\,\delta_{i\,j}-x^{i}x^{j}\right)\right)\,dx^{i}\otimes
dx^{j}.\end{array}$

Since $x_{i},r^{2}$ are smooth functions from ${\rm B}_{R}(p)$ to $\mathbb{R}$
and $\varphi$ is a smooth function of $\mathbb{R}_{+}$ the Theorem follows.

∎

Observe from the proof of Theorem that what is proved is that
$\omega_{g}(t)=t\,F(t^{2})$ with $F(t)$ being a smooth function satisfying
$F(0)=1$. Moreover in the proof of the Theorem is proved that any metric
tensor on ${\rm B}_{R}(p)-\\{p\\}$ of the form

(2.9) $dr\otimes dr+(\omega^{2}\circ r)\pi^{*}g_{\mathbb{S}^{n-1}_{1}}$

with positive warping function $\omega(t)$, can be extended to a smooth metric
tensor in ${\rm B}_{R}(p)$ if there exists a smooth function $F(t)$ with
$F(0)=1$ such that $\omega(t)=t\,F(t^{2})$. This is equivalent to the
classical condition on the warping function $\omega$, which is

(2.10)
$\omega(0)=0,\quad\omega^{\prime}(0)=1\quad\text{and}\quad\omega^{(2k)}(0)=0\quad\text{for
all}\quad k\in\mathbb{N}^{*},$

where $\omega^{(2k)}$ denotes all the even derivatives of $\omega$. Indeed,
assuming $\omega(t)=t\,F(t^{2})$ with $F(0)=1$ it is easy to check that
$\omega$ satisfies (2.10). In the other direction, if $\omega$ satisfies
condition (2.10) we can construct an even smooth function
$f:\mathbb{R}\to\mathbb{R}_{+}$ such that $\omega(t)=tf(t)$ and then, using
[21] as in the proof of Theorem 2.2, we obtain $f(t)=F(t^{2})$.

###  2.2. Properties of the rotationally symmetric metric tensor of comparison

In this subsection of the paper we show some properties of the rotationally
symmetric metric tensor of comparison $\widetilde{g}$. First of all, we will
clarify in which sense the metric tensor $\widetilde{g}$ is rotationally
symmetric. The Orthogonal group ${\rm O}(n)=\\{R\in{\rm
GL}(n)\,:\,R^{T}R=RR^{T}=\mathfrak{1}_{n}\\}$ acts on
$\mathbb{D}_{R}(\vec{0})$ by

${\rm
O}(n)\times\mathbb{D}_{R}(\vec{0})\to\mathbb{D}_{R}(\vec{0}),\quad(R,x)\mapsto
R\,x.$

By using the diffeomorphism $\zeta:{\rm B}_{R}(p)\to\mathbb{D}_{R}(\vec{0})$,
defined in (2.1), we can define the action of ${\rm O}(n)$ on ${\rm
B}_{R}(p)$, by

${\rm O}(n)\times{\rm B}_{R}(p)\to{\rm
B}_{R}(p),\quad(R,q)\mapsto\zeta^{-1}(R\,\zeta(q)).$

When ${\rm B}_{R}(p)$ is endowed with the metric tensor $\widetilde{g}$, the
group ${\rm O}(n)$ acts by isometries. Since $\widetilde{g}$ remains invariant
under the action of the Orthogonal group we will say that the metric tensor
$\widetilde{g}$ is _rotationally symmetric_.

In [11] is defined an $n$-dimensional Riemannian manifold $(M,g)$ as a
_Riemannian model_ if the following conditions are satisfied:

  1. (1)

There is a chart of $M$ that covers all $M$, and the image of this chart in
$\mathbb{R}^{n}$ is a ball $\mathbb{D}_{R_{0}}$ of radius
$R_{0}\in(0,+\infty]$.

  2. (2)

The metric tensor $g$ in the polar coordinates $(r,\theta)$ in the above chart
has the form given by (2.9) with $\omega$ a positive function.

The number $R_{0}$ is called the radius of the model $M$. Observe that given a
Riemannian model $(M,g)$ the metric tensor $g$ is rotationally symmetric in
any geodesic ball of radius $R<R_{0}$. On the other hand, given a Riemannian
manifold $(M,g)$, the geodesic ball $({\rm B}_{R}(p),\widetilde{g})$ of radius
$R<{\rm inj}_{g}(p)$ endowed with the rotationally symmetric metric tensor of
comparison $\widetilde{g}$ associated to $g$ is a Riemannian model of radius
$R$.

The expression of the distance function, area function, and Laplacian of
functions with respect to $g$ and $\widetilde{g}$ are given in the following

######  Proposition 2.3.

Let $(M,g)$ be a $n$-dimensional Riemannian manifold. Let ${\rm B}_{R}(p)$ be
the geodesic ball of radius $R<{\rm inj}_{g}(p)$ centered at $p{\in M}$, let
$\exp_{p}:T_{p}M\to M$ be the exponential map associated to $g$, let
$\widetilde{g}$ be the rotationally symmetric metric tensor of comparison
associated to $g$. Then for any $q\in{\rm B}_{R}(p)$,

  1. (1)

$\displaystyle\nabla r(q)=\widetilde{\nabla}r(q)=\partial
r(q)=\sum_{i=1}^{n}\frac{x^{i}(q)}{r(q)}\left.\frac{\partial}{\partial
x^{i}}\right|_{q}$,

  2. (2)

$r(q):=\|\exp^{-1}(q)\|={\rm dist}_{g}(p,q)={\rm dist}_{\widetilde{g}}(p,q)$,

  3. (3)

$g(\nabla r(q),\nabla
r(q))=\widetilde{g}(\widetilde{\nabla}r(q),\widetilde{\nabla}r(q))=1$,

  4. (4)

${\rm A}_{\widetilde{g}}(t)={\rm vol}_{\widetilde{g}}(S_{t}(p))={\rm
vol}\left(\mathbb{S}^{n-1}_{1}\right)\omega_{g}^{n-1}(t)={\rm A}_{g}(t)$,

  5. (5)

For any smooth function $f:{\rm B}_{R}(p)\to\mathbb{R}$,

(2.11)
$\Delta_{\widetilde{g}}f=(n-1)\dfrac{\omega^{\prime}_{g}(r)}{\omega_{g}(r)}\dfrac{\partial
f}{\partial r}+\dfrac{\partial^{2}f}{\partial
r^{2}}+\dfrac{1}{\omega^{2}_{g}(r)}\Delta_{\mathbb{S}_{1}^{n-1}}(f\circ\pi^{-1})\circ\pi,$

where $\nabla$ and $\widetilde{\nabla}$ denote the gradient with respect to
$g$ and $\widetilde{g}$ respectively and $\Delta_{\mathbb{S}_{1}^{n-1}}$
denotes the Laplacian of the $(n-1)$-dimensional usual unit sphere.

Since $({\rm B}_{R}(p),\widetilde{g})$ is a Riemannian model, before to prove
the Theorem 1 we will need the following consideration about first
eigenfunction and the first eigenvalue for the Dirichlet problem in a geodesic
ball of a Riemannian model.

######  Proposition 2.4.

Let $(M,g_{\omega})$ be a $n$-dimensional Riemannian model, let ${\rm
B}_{R}(p)$ be a geodesic ball centered at $p\in M$ with radius $R<{\rm
inj}_{g}(p)$. Suppose that the smooth metric tensor on ${\rm
B}_{R}(p)-\\{p\\}$ is given by

$g_{\omega}=dr\otimes dr+(\omega^{2}\circ r)\pi^{*}g_{\mathbb{S}^{n-1}_{1}}$

with $\omega:[0,R)\to\mathbb{R}_{+}$ a positive function. Then, any positive
first eigenfunction $\phi_{1}$ of the Laplacian $\Delta_{g_{\omega}}$ for the
Dirichlet problem on ${\rm B}_{R}(p)$ is radial, $\phi_{1}(q)=f_{1}(r(q))$
with $f_{1}$ a smooth function such that

$f_{1}^{\prime}(0)=0\;{\rm and}\;f_{1}^{\prime}(t)<0\;{\rm for}\;t\in(0,R].$

###### Proof.

If $\phi_{1}$ is any positive first eigenfunction of the Laplacian
$\Delta_{g_{\omega}}$ for the Dirichlet problem on a rotationally symmetric
geodesic ball ${\rm B}_{R}(p)$ then, since $g_{w}$ is rotationally symmetric ,
$\phi_{1}$ is a radial function, _i.e._ , we can rewrite, for all $q\in{\rm
B}_{R}(p)$, the first eigenfunction as $\phi_{1}(q)=f_{1}(r(q))$ where $f_{1}$
is a positive real valued smooth function. Moreover, an easy computation leads
to

$\Delta_{g_{w}}\phi_{1}(q)=(n-1)\dfrac{\omega^{\prime}(r(q))}{\omega(r(q))}f_{1}^{\prime}(r(q))+f_{1}^{\prime\prime}(r(q))=-\lambda_{1}\left({\rm
B}_{R}(p)\right)f_{1}(r(q)),$

where $\lambda_{1}\left({\rm B}_{R}(p)\right)>0$ is the first eigenvalue of
the Laplacian for the Dirichlet problem in ${\rm B}_{R}(p)$. Hence, for any
$t\in[0,R]$,

(2.12)
$(n-1)\dfrac{\omega^{\prime}(t)}{\omega(t)}f_{1}^{\prime}(t)+f_{1}^{\prime\prime}(t)=-\lambda_{1}\left({\rm
B}_{R}(p)\right)f_{1}(t)$

By using that $\phi_{1}$ is a radial function it is know that
$f_{1}^{\prime}(0)=0$ (see [4] for instance). On the other hand, from (2.12)
we know that all the critical points of $f_{1}$ are relative maximums. Since
for a real valued smooth function between two relative maximums there is at
least one relative minimum, therefore $f_{1}$ can only have one maximum in
$[0,R)$. Thus, $0$ is the only critical point of $f_{1}$ and then, $f_{1}$ is
a decreasing function in $(0,R)$. Namely, $f_{1}^{\prime}(t)<0$ for all
$t\in(0,R)$.∎

######  Theorem 2.5 ([3]).

Let $(M,g_{\omega})$ be a $n$-dimensional Riemannian model, let ${\rm
B}_{R}(p)$ be a geodesic ball centered at $p$ with radius $R<{\rm
inj}_{g_{\omega}}(p)$. Suppose that the smooth metric tensor on ${\rm
B}_{R}(p)-\\{p\\}$ is given by

(2.13) $g_{\omega}=dr\otimes dr+(\omega^{2}\circ
r)\pi^{*}g_{\mathbb{S}^{n-1}_{1}}$

with $\omega:[0,R)\to\mathbb{R}_{+}$ a positive function. Then, the first
eigenvalue $\lambda_{1,g_{\omega}}({\rm B}_{R}(p))$ of the Laplacian
$\Delta_{g_{\omega}}$ for the Dirichlet problem on ${\rm B}_{R}(p)$ is given
by

$\lambda_{1,g_{\omega}}({\rm B}_{R}(p))=\lim_{k\to\infty}\frac{\left\lVert
T_{k}(t)\right\lVert_{{2}}}{\left\lVert T_{k+1}(t)\right\lVert_{{2}}}$

where $\lVert\cdot\lVert_{{2}}$ denotes the $L_{2}$-norm on ${\rm B}_{R}(p)$
and with

$\left\\{\begin{array}[]{rcl}T_{0}(t)&=&1,\\\
T_{k}(t)&=&\int_{t}^{R}\frac{\int_{0}^{\sigma}T_{k-1}(s)\,\omega^{n-1}(s)\,ds}{\omega^{n-1}(\sigma)}\,d\sigma.\end{array}\right.$

Since $({\rm B}_{R}(p),\widetilde{g})$ is a particular case of Riemannian
model with $\omega=\omega_{g}$, from Proposition 2.4 and Theorem 2.5, we can
state the following

######  Corollary 2.6.

Let $(M,g)$ be a $n$-dimensional Riemannian manifold. Let ${\rm B}_{R}(p)$ be
the geodesic ball of radius $R<{\rm inj}_{g}(p)$ centered at $p\in M$, let
$\widetilde{g}$ be the rotationally symmetric metric tensor of comparison
associated to $g$. Then, any positive first eigenfunction $\phi_{1}$ of the
Laplacian $\Delta_{\widetilde{g}}$ for the Dirichlet problem on ${\rm
B}_{R}(p)$ is radial, $\phi_{1}(q)=f_{1}(r(q))$ with $f_{1}$ a smooth function
such that

(2.14) $f_{1}^{\prime}(0)=0\;{\rm and}\;f_{1}^{\prime}(t)<0\;{\rm
for}\;t\in(0,R]$

and moreover, the first eigenvalue is given by

$\lambda_{1,\widetilde{g}}({\rm
B}_{R}(p))=\lim_{k\to\infty}\left(\frac{\int_{0}^{R}T_{k}^{2}(t){\rm
A}_{g}(t)\,dt}{\int_{0}^{R}T_{k+1}^{2}(t){\rm A}_{g}(t)\,dt}\right)^{{1/2}}$

with

$\left\\{\begin{array}[]{rcl}T_{0}(t)&=&1,\\\
T_{k}(t)&=&\int_{t}^{R}\frac{\int_{0}^{\sigma}{T_{k-1}(s){\rm
A}_{g}(s)}\,ds}{{\rm A}_{g}(\sigma)}\,d\sigma.\end{array}\right.$

##  3\. Proof of the Theorem 1

In this section we will prove Theorem 1. Indeed, in this section we will prove
Theorem 3.1 which is nothing else than Theorem 1 but showing in (3.1) the
comparison of the first eigenvalues $\lambda_{1,g}\left({\rm B}_{R}(p)\right)$
and $\lambda_{1,\widetilde{g}}\left({\rm B}_{R}(p)\right)$ and the equality

$\lambda_{1,\widetilde{g}}\left({\rm
B}_{R}(p)\right)=\lim_{k\to\infty}\left(\frac{\int_{0}^{R}T_{k}^{2}(t){\rm
A}_{g}(t)\,dt}{\int_{0}^{R}T_{k+1}^{2}(t){\rm A}_{g}(t)\,dt}\right)^{{1/2}},$

given in Corollary 2.6.

######  Theorem 3.1.

Let $(M,g)$ be a Riemannian manifold, let $p\in M$ be a point of $M$ with
injectivity radius ${\rm inj}_{g}(p)$, and let ${\rm B}_{R}(p)$ be the
geodesic ball of radius $R$ centered at $p$. Suppose $R<{\rm inj}_{g}(p)$,
then the first eigenvalue $\lambda_{1,g}\left({\rm B}_{R}(p)\right)$ of the
Laplacian for the Dirichlet problem on ${\rm B}_{R}(p)$ is bounded by

(3.1) $\lambda_{1,g}\left({\rm
B}_{R}(p)\right)\leq\lambda_{1,\widetilde{g}}\left({\rm
B}_{R}(p)\right)=\lim_{k\to\infty}\left(\frac{\int_{0}^{R}T_{k}^{2}(t){\rm
A}_{g}(t)\,dt}{\int_{0}^{R}T_{k+1}^{2}(t){\rm
A}_{g}(t)\,dt}\right)^{{1/2}}{,}$

where $\widetilde{g}$ is the rotationally symmetric metric tensor of
comparison associated to $g$. Futhermore, equality is attained in (3.1) if and
only if, for any $t\in(0,R)$, the mean curvature pointed inward ${\rm H}_{{\rm
S}_{t}(p)}$ of the geodesic sphere ${\rm S}_{t}(p)$ of radius $t$ centered at
$p$ is a radial function. Namely, equality in (3.1) is attained if and only if
there exists a smooth function $h(t)$ such that

${\rm H}_{{\rm S}_{t}(p)}=h(t)\quad{\rm for\;{any}}\quad 0<t<R.$

###### Proof.

To obtain upper bounds for the first eigenvalue we compute the Rayleigh
quotient (see [4] for more information about the Rayleigh quotient) with
respect to $({\rm B}_{R}(p),g)$ of the first eigenfunction on $({\rm
B}_{R}(p),\widetilde{g})$, where $\widetilde{g}$ is the rotationally symmetric
metric tensor of comparison defined in Section 2.

Since $R<{\rm inj}_{g}(p)$, the exponential map
$\exp_{p}:\mathcal{B}_{R}(\vec{0})\to{\rm B}_{R}(p)$ induces a diffeomorphism
$\zeta:{\rm B}_{R}(p)\to\mathbb{D}_{R}(\vec{0})$ (see Section 2 for the
definition of the the diffeomorphism $\zeta$). The metric tensor $g$ in ${\rm
B}_{R}(p)-\\{p\\}$ can be expressed as

$g=dr\otimes dr+\sum_{i,j=1}^{n-1}G_{i,j}(r,\theta)\,d\theta^{i}\otimes
d\theta^{j},$

for some positive definite matrix $G_{i,j}$, where we have used the coordinate
system

$\begin{array}[]{lcl}q\mapsto(r(q),\theta^{1}(q),\cdots\theta^{n-1}(q)):=(r(q),\widetilde{\theta}^{1}(\pi(q)),\cdots,\widetilde{\theta}^{n-1}(\pi(q)))\end{array}$

with the maps $q\mapsto r(q)=\|\zeta(q)\|$ and
$q\mapsto\pi(q)=\frac{\zeta(q)}{r(q)}$ and
$\\{\widetilde{\theta^{i}}\\}_{i=1}^{n-1}$ being a system of local coordinates
on the sphere $\mathbb{S}^{n-1}_{1}$. Using this coordinates, the Riemannian
volume element can be obtained as

$d{\rm V}_{g}=\sqrt{\det(G(r,\theta))}\,dr\wedge d\theta^{1}\wedge\cdots\wedge
d\theta^{n-1}.$

The area function of the geodesic sphere ${\rm S}_{t}(p)$ of radius $t$
centered at $p$ is

${\rm
A}_{g}(t)=\int_{\mathbb{S}^{n-1}_{1}}\sqrt{\det(G(t,\theta))}\,d\theta^{1}\wedge\cdots\wedge
d\theta^{n-1}$

and the mean curvature vector field of the geodesic sphere ${\rm S}_{t}(p)$ of
radius $t$ centered at $p$ can be expressed as

$\vec{\rm H}_{S_{t}(p)}=-\left.\frac{\partial}{\partial
s}\ln{\sqrt{\det(G(s,\theta))}}\right|_{s=t}\partial r.$

Now, we will make use of the rotationally symmetric metric tensor of
comparison $\widetilde{g}$ associated to $g$ using the warping function
$\omega_{g}(t)=\left(\frac{{\rm A}_{g}(t)}{{\rm
vol}\left({\mathbb{S}_{1}^{n-1}}\right)}\right)^{\frac{1}{n-1}},$ as

(3.2) $\widetilde{g}=\left\\{\begin{array}[]{lcl}dr\otimes
dr+(\omega_{g}^{2}\circ r)\pi^{*}g_{\mathbb{S}^{n-1}_{1}},&{\rm on}&{\rm
B}_{R}(p)-\\{p\\},\\\ \displaystyle\sum_{i=1}^{n}dx^{i}\otimes dx^{i},&{\rm
on}&p.\end{array}\right.$

The first positive eigenfunction $\phi_{1}$ and the first eigenvalue
$\lambda_{1,\widetilde{g}}({\rm B}_{R}(p))$ of the Laplacian
$\Delta_{\widetilde{g}}$ for the Dirichlet problem on ${\rm B}_{R}(p)$ are
related by the Rayleigh quotient with respect to $\widetilde{g}$

(3.3) $\lambda_{1,\widetilde{g}}({\rm B}_{R}(p))=\frac{\int_{{\rm
B}_{R}(p)}\widetilde{g}(\widetilde{\nabla}\phi_{1}(q),\widetilde{\nabla}\phi_{1}(q))\,d{\rm
V}_{\widetilde{g}}(q)}{\int_{{\rm B}_{R}(p)}\phi_{1}^{2}(q)\,d{\rm
V}_{\widetilde{g}}(q)}.$

Our upper bound for $\lambda_{1,g}({\rm B}_{R}(p))$ in (3.1) is obtained by
using the Rayleigh quotient with respect to $g$ of the first eigenfunction
$\phi_{1}$ with respect to $\widetilde{g}$, _i.e._ ,

(3.4) $\lambda_{1,g}({\rm B}_{R}(p))\leq\frac{\int_{{\rm
B}_{R}(p)}g(\nabla\phi_{1}(q),\nabla\phi_{1}(q))\,d{\rm V}_{g}(q)}{\int_{{\rm
B}_{R}(p)}\phi_{1}^{2}(q)\,d{\rm V}_{g}(q)}.$

But, the right hand side of the above inequality can be expressed as

(3.5) $\frac{\int_{{\rm
B}_{R}(p)}g(\nabla\phi_{1}(q),\nabla\phi_{1}(q))\,d{\rm V}_{g}(q)}{\int_{{\rm
B}_{R}(p)}\phi_{1}^{2}(q)\,d{\rm V}_{g}(q)}=\frac{\int_{{\rm
B}_{R}(p)}\widetilde{g}(\widetilde{\nabla}\phi_{1}(q),\widetilde{\nabla}\phi_{1}(q))\,d{\rm
V}_{\widetilde{g}}(q)}{\int_{{\rm B}_{R}(p)}\phi_{1}^{2}(q)\,d{\rm
V}_{\widetilde{g}}(q)}$

because taking into account that $\phi_{1}(q)=f_{1}(r(q))$ is a decreasing
radial function (see Corollary 2.6) and by using the co-area formula (see [4])
and Proposition 2.3

$\displaystyle\int_{{\rm B}_{R}(p)}\phi_{1}^{2}(q)\,d{\rm V}_{g}(q)$
$\displaystyle=\int_{{\rm B}_{R}(p)}\dfrac{f^{2}_{1}(r(q))}{g\left(\nabla
r(q),\nabla r(q)\right)}\,g\left(\nabla r(q),\nabla r(q)\right)\,d{\rm
V}_{g}(q)$ $\displaystyle=\int_{0}^{R}\left(\int_{\left\\{q\in
M\,:\,r(q)=t\right\\}}\dfrac{f_{1}^{2}(r(q))}{g\left(\nabla r(q),\nabla
r(q)\right)}\,d{\rm A}_{g}(q)\right)dt$
$\displaystyle=\int_{0}^{R}f_{1}^{2}(t){\rm
A}_{g}(t)\,dt=\int_{0}^{R}f_{1}^{2}(t){\rm A}_{\widetilde{g}}(t)\,dt$
$\displaystyle=\int_{0}^{R}\left(\int_{\left\\{q\in
M\,:\,r(q)=t\right\\}}\dfrac{f_{1}^{2}(r(q))}{\widetilde{g}\left(\widetilde{\nabla}r(q),\widetilde{\nabla}r(q)\right)}\,d{\rm
A}_{\widetilde{g}}(q)\right)dt$ $\displaystyle=\int_{{\rm
B}_{R}(p)}\phi_{1}^{2}(q)\,d{\rm V}_{\widetilde{g}}(q).$

Similarly, since the nominator
$g(\nabla\phi_{1}(q),\nabla\phi_{1}(q))=(f_{1}^{\prime}(r(q))^{2}$ is also a
radial function,

$\int_{{\rm B}_{R}(p)}g(\nabla\phi_{1}(q),\nabla\phi_{1}(q))\,d{\rm
V}_{g}(q)=\int_{{\rm
B}_{R}(p)}\widetilde{g}(\widetilde{\nabla}\phi_{1}(q),\widetilde{\nabla}\phi_{1}(q))\,d{\rm
V}_{\widetilde{g}}(q)$

Therefore, by equations (3.3) and (3.5), by inequality (3.4) and by Corollary
2.6, we obtain

$\lambda_{1,g}({\rm B}_{R}(p))\leq\lambda_{1,\widetilde{g}}({\rm
B}_{R}(p))={\lim_{k\to\infty}\left(\frac{\int_{0}^{R}T_{k}^{2}(t){\rm
A}_{g}(t)\,dt}{\int_{0}^{R}T_{k+1}^{2}(t){\rm A}_{g}(t)\,dt}\right)^{1/2}}.$

Furthermore, equality in the above inequality implies that $\phi_{1}$ is also
a first positive eigenfunction of $\Delta_{g}$, therefore for any $q\in{\rm
B}_{R}(p)$

$\Delta_{g}\phi_{1}(q)=f_{1}^{\prime\prime}(r(q))+f^{\prime}_{1}(r(q))\frac{\partial}{\partial
r}\left(\ln{\sqrt{\det(G(r,\theta))}}(q)\right)=-\lambda_{1,g}({\rm
B}_{R}(p))f_{1}(r(q)).$

Then, for any point $q\in{\rm S}_{t}(p)$,

(3.6) $\left(\vphantom{\dfrac{1}{1}}f_{1}^{\prime\prime}(t)+\lambda_{1,g}({\rm
B}_{R}(p))f_{1}(t)\right)\partial r(q)=f_{1}^{\prime}(t)\vec{\rm H}_{{\rm
S}_{t}(p)}(q).$

But, since $\phi_{1}$ is a first positive eigenfunction for
$\Delta_{\widetilde{g}}$, by (2.11),

$f_{1}^{\prime\prime}(t)+\lambda_{1,g}({\rm
B}_{R}(p))f_{1}(t)=-(n-1)\frac{w^{\prime}_{g}(t)}{w_{g}(t)}f_{1}^{\prime}(t).$

Hence, from (3.6) and taking into account that $f_{1}^{\prime}(t)<0$ (see
Corollary 2.6), we can obtain the mean curvature vector field $\vec{\rm
H}_{{\rm S}_{t}(p)}(q)$ for any point $q\in{\rm S}_{t}(p)$ as

$\vec{\rm H}_{{\rm
S}_{t}(p)}(q)=-(n-1)\frac{w^{\prime}_{g}(t)}{w_{g}(t)}\partial r.$

Therefore, the mean curvature of the geodesic spheres pointed inward given by

${\rm H}_{{\rm S}_{t}(p)}(q)={g(}\vec{\rm H}_{{\rm S}_{t}(p)}(q),-\partial
r{)}=(n-1)\frac{w^{\prime}_{g}(t)}{w_{g}(t)}$

is a radial function as stated.

On the other hand, if the mean curvature of the geodesic spheres is a radial
function, _i.e._ , $\vec{\rm H}_{{\rm S}_{t}(p)}(q)=-h(t)\partial r$, we are
proving that $\lambda_{1,g}({\rm B}_{R}(p))=\lambda_{1,\widetilde{g}}({\rm
B}_{R}(p))$. Indeed, we can prove that $\phi_{1}$ is a positive eigenfunction
of $\Delta_{g}$ because

(3.7) $\displaystyle\Delta_{g}\phi_{1}(q)=$ $\displaystyle
f_{1}^{\prime\prime}(r(q))+f^{\prime}_{1}(r(q))\frac{\partial}{\partial
r}\left(\ln{\sqrt{\det(G(r,\theta))}}(q)\right)$ $\displaystyle=$
$\displaystyle f_{1}^{\prime\prime}(r(q))+f^{\prime}_{1}(r(q))h(t){.}$

But since

$\displaystyle(n-1)\frac{\omega^{\prime}_{g}(t)}{\omega_{g}(t)}=$
$\displaystyle\frac{d}{dt}\ln\omega^{n-1}(t)=\left.\frac{d}{ds}\ln{\rm
A}_{g}(s)\right\lvert_{s=t}$ $\displaystyle=$
$\displaystyle\frac{1}{A_{g}(t)}\int_{\mathbb{S}^{n-1}_{1}}\left.\frac{\partial}{\partial
s}\sqrt{\det(G(s,\theta))}\right\lvert_{s=t}d\theta^{1}\wedge\cdots\wedge
d\theta^{n-1}$ $\displaystyle=$
$\displaystyle\frac{1}{A_{g}(t)}\int_{\mathbb{S}^{n-1}_{1}}\frac{\left.\frac{\partial}{\partial
s}\sqrt{\det(G(s,\theta))}\right\lvert_{s=t}}{\sqrt{\det(G(t,\theta))}}\sqrt{\det(G(t,\theta))}d\theta^{1}\wedge\cdots\wedge
d\theta^{n-1}$ $\displaystyle=$
$\displaystyle\frac{1}{A_{g}(t)}\int_{\mathbb{S}^{n-1}_{1}}h(t)\sqrt{\det(G(t,\theta))}d\theta^{1}\wedge\cdots\wedge
d\theta^{n-1}$ $\displaystyle=$ $\displaystyle h(t),$

from equation (3.7) and (2.11), we have

$\Delta_{g}\phi_{1}(q)=f_{1}^{\prime\prime}(r(q))+(n-1)\frac{\omega^{\prime}_{g}(t)}{\omega_{g}(t)}f^{\prime}_{1}(r(q))=\Delta_{\widetilde{g}}\phi_{1}=-\lambda_{1,\widetilde{g}}({\rm
B}_{R}(p))\phi_{1}$

Hence, $\phi_{1}$ is a positive eigenfunction of $\Delta_{g}$ and the Theorem
follows. ∎

######  Remark 3.2.

Observe that the function $h(t)$ of the Theorem 1 is

$h(t)=(n-1)\frac{w^{\prime}_{g}(t)}{w_{g}(t)}.$

##  4\. Proof of Theorem 2

In this section we will prove Theorem 2 which is a particular case of the
following Theorem 4.1 when $W(t)=S_{\kappa}(t)$ with

$S_{\kappa}(t):=\left\\{\begin{array}[]{rcl}\frac{\sin(\sqrt{\kappa}t)}{\sqrt{\kappa}},&{\rm
if}&\kappa>0,\\\ t,&{\rm if}&\kappa=0,\\\
\frac{\sinh(\sqrt{-\kappa}t)}{\sqrt{-\kappa}},&{\rm
if}&\kappa<0.\end{array}\right.$

######  Theorem 4.1.

Let $(M,g)$ be a Riemannian manifold, and let $p\in M$ be a point of $M$ with
injectivity radius ${\rm inj}_{g}(p)$. Let ${\rm B}_{R}(p)$ be the geodesic
ball of radius $R$ centered at $p$ with $R<{\rm inj}_{g}(p)$. Let
$W:[0,R]\to\mathbb{R}$ be a non-negative smooth function such that the metric
tensor

$g_{W}=dr\otimes dr+(W^{2}\circ r)\pi^{*}g_{\mathbb{S}^{n-1}_{1}}$

is smooth on ${\rm B}_{R}(p)$. Suppose that $R<{\rm inj}_{g}(p)$ and that for
any $t<{R}$ the function

$t\mapsto\frac{{\rm A}_{g}(t)}{{\rm A}_{g_{W}}(t)}$

is a decreasing function. Then, the first eigenvalue $\lambda_{1,g}(({\rm
B}_{R}(p)))$ of the Laplacian for the Dirichlet problem on the geodesic ball
${\rm B}_{R}(p)$ of radius $R$ centered at $p$ is bounded by

(4.1) $\lambda_{1,g}({{\rm B}_{R}(p)})\leq\lambda_{1,g_{W}}({{\rm
B}_{R}(p)}),$

with equality in (4.1) if and only if, for any $t\in(0,R{)}$, the mean
curvature pointed inward ${\rm H}_{{\rm S}_{t}(p)}$ of the geodesic sphere
${\rm S}_{t}(p)$ is

${\rm H}_{{\rm S}_{t}(p)}=(n-1)\frac{W^{\prime}(t)}{W(t)},\quad{\rm
for\;{all}}\quad 0<t<R.$

###### Proof.

From $({\rm B}_{R}(p),g)$ we will symmetrize the metric tensor to obtain the
rotationally symmetric metric tensor

(4.2) $\widetilde{g}=dr\otimes dr+(\omega^{2}_{g}\circ
r)\pi^{*}g_{\mathbb{S}^{n-1}_{1}}$

with $\omega_{g}:[0,R)\to\mathbb{R}_{+}$ being the positive function given by

$t\mapsto\omega_{g}(t):=\left(\frac{{\rm A}_{g}(t)}{{\rm
vol}\left({\mathbb{S}_{1}^{n-1}}\right)}\right)^{\frac{1}{n-1}},$

Hence, by using the Theorem 1

(4.3) $\lambda_{1,g}(({\rm B}_{R}(p)))\leq\lambda_{1,\widetilde{g}}(({\rm
B}_{R}(p))).$

Moreover, since ${\rm A}_{\widetilde{g}}(t)={\rm A}_{g}(t)$, by hypothesis the
function

$t\mapsto\frac{{\rm A}_{\widetilde{g}}(t)}{{\rm A}_{g_{W}}(t)}$

is assumed to be a decreasing function. Therefore

$\frac{\omega_{g}^{\prime}(t)}{\omega_{g}(t)}\leq\frac{W^{\prime}(t)}{W(t)},\quad\text{
for all }t\in[0,R].$

Let us denote by $\phi_{1,W}(q)=f_{1,W}(r(q))$ the first (radial)
eigenfunction of the Laplacian for the Dirichlet problem with respect to the
metric tensor $g_{W}$. Then for any $q\in{\rm S}_{t}(p)$

$\Delta_{g_{W}}\phi_{1,W}(q)=f^{\prime\prime}_{1,W}(t)+(n-1)\frac{W^{\prime}(t)}{W(t)}f^{\prime}_{1,W}(t)=-\lambda_{1,g_{W}}\left({\rm
B}_{R}(t)\right)f_{1,W}(t)$

Hence, for any $q\in{\rm B}_{R}(p)$ with $r(q)=t$,

$\displaystyle-\frac{\triangle_{\widetilde{g}}\phi_{1,W}(q)}{\phi_{1,W}(q)}=$
$\displaystyle\frac{-f^{\prime\prime}_{1,W}(t)-(n-1)\frac{\omega_{g}^{\prime}(t)}{\omega_{g}(t)}f^{\prime}_{1,W}(t)}{f_{1,W}(t)}$
$\displaystyle\leq$
$\displaystyle\frac{-f^{\prime\prime}_{1,W}(t)-(n-1)\frac{W^{\prime}(t)}{W(t)}f^{\prime}_{1,W}(t)}{f_{1,W}(t)}$
$\displaystyle=$ $\displaystyle\lambda_{1,g_{W}}({\rm B}_{R}(p)).$

Finally the Theorem follows by using Barta’s Lemma (see [4] for instance) and
(4.3). ∎

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