# On the total curvature and extrinsic area growth of surfaces with tamed
second fundamental form

Cristiane M. Brandão  Departamento de Matemática universidade do Ceará-UFC,
60455-760 Fortaleza, CE, Brazil<EMAIL_ADDRESS>and  Vicent Gimeno
Department of Mathematics-INIT-IMAC, Universitat Jaume I, Castelló de la
Plana, Spain<EMAIL_ADDRESS>
###### Abstract.

In this paper we show that a complete and non-compact surface immersed in the
Euclidean space with quadratic extrinsic area growth has finite total
curvature provided the surface has tamed second fundamental form and admits
total curvature. In such a case we obtain as well a generalized Chern-Osserman
inequality. In the particular case of a surface of nonnegative curvature, we
prove that the surface is diffeomorphic to the Euclidean plane if the surface
has tamed second fundamental form, and that the surface is isometric to the
Euclidean plane if the surface has strongly tamed second fundamental form. In
the last part of the paper we characterize the fundamental tone of any
submanifold of tamed second fundamental form immersed in an ambient space with
a pole and quadratic decay of the radial sectional curvatures.

###### Key words and phrases:

Total curvature and Chern-Osserman inequality and Fundamental tone and area
growth

###### 1991 Mathematics Subject Classification:

35P15

Work partially supported by DGI grant MTM2010-21206-C02-02.

##  1\. Introduction

Let $M$ be a complete non-compact surface, the total curvature of $M$ is the
improper integral $\int_{M}K\,dA$ of the Gaussian curvature $K$ with respect
to the volume element $dA$ of $M$. It is said that $M$ admits total curvature
if for any compact exhaustion $\\{\Omega_{i}\\}$ of $M$, the limit

$\int_{M}K\,dA=\lim_{i\to\infty}\int_{\Omega_{i}}K\,dA$

exists. Cohn-Vossen proved in [10] that $\int_{M}K\,dA\leq\chi(M)$, where
$\chi(M)$ is the Euler characteristic of $M$. A well known theorem due to
Huber [18] states that if the negative part of the curvature
$K_{-}=\max\\{-K,0\\}$ has finite integral, namely,

(1.1) $\int_{M}K_{-}\,dA<\infty,$

then, $\int_{M}K\,dA\leq\chi(M)$ and $M$ is conformally equivalent to a
compact Riemann surface with finitely many punctures. Hartman, under the
assumption (1.1) proved in [17] that the area ${\rm A}(B_{r})$ of a geodesic
ball of radius $r$ at a fixed point must grow at most quadratically in $r$.
Reciprocally, Li proved in [23] that if $M$ has at most quadratic area growth,
finite topology and the Gaussian curvature of $M$ is either non-positive or
non-negative, near infinity of each end, then $M$ must have finite total
curvature.

From an extrinsic point of view, in the setting of a minimal surface $M$
immersed in the Euclidean space $\mathbb{R}^{n}$, it is well known, see [9,
20, 25, 26], that if $M$ has finite total curvature then $M$ has finite
topological type and _quadratic extrinsic area growth_ , _i.e._ , there exists
a constant $C$ such that for any $r\in\mathbb{R}_{+}$

(1.2) $\text{Area}(M\cap B_{r}(0))\leq Cr^{2},$

where $B_{r}(0)$ denotes the geodesic ball centered at the origin
$0\in\mathbb{R}^{n}$ of radius $r$.

Conversely, Q. Chen [8], proved that if $M$ is an oriented complete minimal
surface in the Euclidean space $\mathbb{R}^{n}$ with quadratic extrinsic area
growth and finite topological type then $M$ has finite total curvature.

A natural question is whether an equivalent result relating the extrinsic area
growth and the total curvature holds for a boarder class of complete surfaces
in the Euclidean space. The aim of this paper is to provide an answer to this
question under certain control of the second fundamental form of the
immersion. A surface $M$ is said to have tamed second fundamental form if for
a (any) compact exhaustion $\\{\Omega_{i}\\}$ of $M$,

(1.3) $a(M):=\lim_{i\to\infty}\left(\sup_{x\in
M\setminus\Omega_{i}}\left\\{\rho_{M}(x)\|\alpha(x)\|\right\\}\right)<1,$

where $\rho_{M}(x)={\rm dist}_{M}(x_{0},x)$ is the distance function on $M$ to
a fixed point $x_{0}$, and $\|\alpha(x)\|$ is the norm of the second
fundamental form at $\varphi(x)$. The notion of immersion with tamed second
fundamental form was introduced in [5] for submanifolds of $\mathbb{R}^{n}$
and in [4] for submanifolds of Hadamard manifolds. This notion can be
naturally extended to manifolds with a pole and radial sectional curvature
bounded above, see [13]. Under the hypothesis of tamed second fundamental form
and quadratic extrinsic area growth we can state the following result.

######  Theorem 1.1.

Let $M$ be an immersed complete oriented surface of $\mathbb{R}^{n}$ with
curvature function $K$ and tamed second fundamental form. Suppose that $M$
admits total curvature. Then, $M$ has finite total curvature
($\int_{M}KdA>-\infty$), if and only if, $M$ has quadratic extrinsic area
growth, _i.e._ , there exists a constant $C_{1}$ such that,

(1.4) $\text{A}(M\cap B_{r}(0))\leq C_{1}r^{2},$

for any $r$ large enough. Furthermore, if (1.4) holds, then there exists a
constant $C_{0}>0$ such that

(1.5) $\text{A}(M\cap B_{r}(0))\geq C_{0}r^{2},$

for any $r$ large enough.

Observe that the assumption that the surface admits total curvature (finite or
infinite) can be achieved if the surface has semidefinite curvature (either
nonpositive or nonnegative). As observed by Jorge-Meeks [20], any complete
$m$-dimensional submanifold $M$ of $\mathbb{R}^{n}$ homeomorphic to a compact
Riemannian manifold $\overline{M}$ punctured at finite number of points
$\\{p_{1},\ldots,p_{r}\\}$ and having a well defined normal vector at infinity
have $a(M)=0$. In particular, any complete minimal surfaces of
$\mathbb{R}^{n}$ with finite total curvature has tamed second fundamental form
with $a(M)=0$. Anderson [1], showed that a complete $m$-dimensional minimally
immersed submanifold $M$ of $\mathbb{R}^{n}$ has finite total scalar
curvature, $\smallint_{M}\|\alpha\|^{m}dV<\infty$, if and only if $M$ is
$C^{\infty}$\- diffeomorphic to a compact Riemannian manifold $\overline{M}$
punctured at a finite number of points $\\{p_{1},\ldots,p_{r}\\}$ and the
Gauss map $\Phi$ on $M$ extends to a $C^{\infty}$\- map $\overline{\Phi}$ on
$\overline{M}$, where $\|\alpha\|$ is the norm of the second fundamental form
of $M$. In [5], Bessa, Jorge and Montenegro showed that some aspects of
Anderson’s result hold on complete immersed submanifolds of $\mathbb{R}^{n}$
with tamed second fundamental form, i.e. they are properly immersed and have
finite topology, meaning that $M$ is $C^{\infty}$-diffeomorphic to a compact
smooth manifold $\overline{M}$ with boundary. This result was extend by Bessa-
Costa to isometric immersions with tamed second fundamental form into Hadamard
manifolds, [4] and by Gimeno-Palmer in [13] to isometric immersion with tamed
second fundamental form into ambient manifolds with a pole and bounded radial
sectional curvatures. They also have shown that the volume growth and the
number of ends of submanifolds of dimension greater than $2$ are controlled
with an appropriate decay of the extrinsic curvature.

Assuming finite total curvature we can treat the two dimensional case
obtaining the following Chern-Osserman type inequality.

######  Theorem 1.2.

Let $M$ be an oriented surface immersed in $\mathbb{R}^{n}$ with curvature
function $K$ and tamed second fundamental form. Suppose in addition that $M$
has finite total curvature. Then,

(1.6) $\left(1-a(M)^{2}\right)\widetilde{C}_{0}\leq
2\pi\chi(M)-\int_{M}Kd\text{A}\leq C_{1},$

where $\chi(M)$ is the Euler characteristic of $M$ and
$\widetilde{C}_{0},C_{1}$ are positive constants such that

(1.7) $\displaystyle\text{A}(M\cap B_{r}(0))\leq$ $\displaystyle C_{1}r^{2},$
$\displaystyle\text{L}(M\cap S_{r}(0))\geq$ $\displaystyle\widetilde{C}_{0}r,$

for any $r$ large enough.

Cohn-Vossen proved in [10] that any complete and non-compact surface with
nonnegative Gaussian curvature is diffeomorphic to $\mathbb{R}^{2}$, or if
not, it is flat. By using the above theorem 1.2 we can therefore state the
following corollary

######  Corollary 1.3.

Let $M$ be an oriented surface immersed in $\mathbb{R}^{n}$ with nonnegative
curvature function ($K\geq 0$) and tamed second fundamental form. Then, $M$ is
diffeomorphic to $\mathbb{R}^{2}$. Moreover, in the particular case when $M$
is flat, $M$ is isometric to $\mathbb{R}^{2}$.

Petrunin and Tuschmann in [27] , solving a conjecture of Gromov [2] (see also
[11]), proved that if a complete simply connected manifold $M^{n}$ of
dimension greater than $2$ ($n\geq 3$) has nonnegative sectional curvatures
$K\geq 0$ and is _asymptotically flat_ then $M$ is isometric to
$\mathbb{R}^{n}$. Here, asymptotically flat means that

(1.8) $k(t)t^{2}\to 0\quad(t\to\infty),$

where $k(t)$ is the supremum of $|K|$ on $M\setminus B_{t}(o)$ for some fixed
point $o\in M$, ($B_{t}(o)$ being the geodesic ball of $M$ of radius $t$
centered at $o$). We can extend this intrinsic result to dimension $2$ but
using an extrinsic approach. we will say that a surface immersed in
$\mathbb{R}^{n}$ has _strongly_ tamed second fundamental form if for some
$\epsilon$, and for a (any) compact exhaustion $\\{\Omega_{i}\\}$ of $M$,

(1.9) $\lim_{i\to\infty}\sup_{x\in
M\setminus\Omega_{i}}\left\\{\rho_{M}(x)^{1+\epsilon}\|\alpha(x)\|\right\\}<1.$

In such a case we obtain the following corollary

######  Corollary 1.4.

Let $M$ be an oriented surface immersed in $\mathbb{R}^{n}$ with nonnegative
curvature function ($K\geq 0$) and strongly tamed second fundamental form.
Then, $M$ is isometric to $\mathbb{R}^{2}$.

In proposition 3.2 we will show that for a surface $M$ immersed in the
Euclidean space $\mathbb{R}^{n}$ with tamed second fundamental form, a
necessary and sufficient condition to attain quadratic extrinsic area growth
is to have _linear extrinsic perimeter growth_. Namely, there exists a
constant $\widetilde{C}$, such that for any $r$ large enough,

${\rm L}(M\cap S_{r}(0))\leq\widetilde{C}r$

where here $S_{r}(0)$ stands for the geodesic sphere of radius $r$ centered at
$0\in\mathbb{R}^{n}$.

It is also interesting to study the fundamental tone of submanifolds with
tamed second fundamental form. Recall that the fundamental tone
$\lambda^{\ast}(M)$ of a complete and non-compact Riemannian manifold $M$ is
given by

$\lambda^{\ast}(M)=\inf\left\\{\frac{\int_{M}|\nabla
u|^{2}d\mu}{\int_{M}u^{2}d\mu},\,u\in
C^{\infty}_{0}(M)\setminus\\{0\\}\right\\}$

It is well known that complete surfaces with finite total curvature are
parabolic, see [19] and the proof of [24, theorem 12.2]. Taking into account
that surfaces with positive fundamental tone are hyperbolic surfaces, see
[16], one concludes that surfaces with finite total curvature has zero
fundamental tone as well as the surfaces with tamed second fundamental form
with quadratic extrinsic area growth.

Observe that tamed second fundamental form implies certain quadratic decay of
the Gaussian curvature. Actually, for the fundamental tone of submanifolds
with tamed second fundamental form in an ambient space with a pole, we can
state something more general using quadratic decay of the curvature again but
in a completely different approach.

Let $N$ be a Riemannian manifold with a pole $p$ and radial sectional
curvature bounded below

$K^{\rm rad}_{N}(x)\geq B(\rho_{N}(x))$

along the rays issuing from $p$, where $B\in C^{\infty}([0,\infty))$. The
behavior of this comparison function imposes restrictions on the fundamental
tone of tamed immersions as we can show in the following theorem.

######  Theorem 1.5.

Let $\varphi\colon M\hookrightarrow N$ be an isometric immersion of a
$m$-dimensional complete Riemannian manifold into an $n$-dimensional ambient
manifold $N$ which possesses a pole and radial sectional curvatures bounded
from below and above by

(1.10) $B(\rho_{N}(x))\leq K^{\rm rad}_{N}(x)\leq 0.$

With $B\in C^{\infty}[0,\infty)$ such that for any $t>0$,

(1.11) $B(t)\geq\frac{-2}{t^{2}}.$

Suppose moreover that the norm of the second fundamental form of the immersion
is tamed. Namely, inequality (1.3) holds. Then, $M$ has zero fundamental tone
$\lambda^{*}(M)=0$.

Observe that the identity map ${\rm id}:N\to N$ induces an isometric immersion
from $N$ to $N$ with vanishing second fundamental form, so with tamed second
fundamental form. We can therefore state a purely intrinsic counterpart of
theorem 1.5

######  Corollary 1.6.

Let $N$ be an $n$-dimensional Riemannian manifold $N$ which possesses a pole
and radial sectional curvatures bounded from below and above by

(1.12) $B(\rho_{N}(x))\leq K^{\rm rad}_{N}(x)\leq 0.$

With $B\in C^{\infty}[0,\infty)$ such that for any $t>0$,

(1.13) $B(t)\geq\frac{-2}{t^{2}}.$

Then, $N$ has zero fundamental tone, $\lambda^{*}(N)=0$.

Much efford has been made in the understanding of Gap phenomenon for
Riemannian manifolds. The classical results in this field (see for instance
[31, 15, 30, 14] and references therein) state that assuming certain kind of
faster than quadratic decay of the curvature one obtains flatness and isometry
to the Euclidean space. For example, by using theorem 1 of [30], any manifold
$N^{n}$ with a pole, dimension $n>2$, sectional curvatures $K_{N}$ bounded
from below and above by $B(\rho_{N}(x))\leq K_{N}(x)\leq 0$, and with bounding
function $B\in C^{\infty}[0,\infty)$ satisfying

(1.14) $\limsup_{t\to\infty}t^{2}B(t)=0,$

is isometric to $\mathbb{R}^{n}$. Observe, moreover that in the most part of
this paper is assumed only quadratic decay, or even slower than quadratic
decay of the sectional curvature.

##  2\. Preliminaries

Throughout this paper we shall study geometric and analytic properties of
submanifolds immersed in an ambient Riemannian manifold with a pole. Recall
that a Riemannian manifold $N$ is a Riemannian manifold with a pole if there
exists a point $p\in N$ with empty cut locus, ${\rm cut}(p)=\emptyset$. In
such a case the exponential map $\exp_{p}:T_{p}N\to N$ induces a
diffeomorphism between $T_{p}N$ and $N$, and the distance function

$\rho_{N}:N\to\mathbb{R},\quad x\to\rho_{N}(x)={\rm dist}_{N}(p,x),$

is a smooth function in $N\setminus\\{p\\}$. We will suppose moreover that the
radial sectional curvatures of $N$, along the geodesics issuing from $p$, are
bounded from above

(2.1) $K_{N}(x)\leq-G(\rho_{N}(x))$

where $G\colon\mathbb{R}\rightarrow\mathbb{R}$ is a smooth even function. Let
$h$ be the solution of the following Cauchy problem

(2.4) $\displaystyle\left\\{\begin{array}[]{l}h^{\prime\prime}-Gh=0\\\
h(0)=0,h^{\prime}(0)=1\\\ \end{array}\right.$

and let $I=[0,r_{0})\subseteq[0,\infty)$ be the maximal interval where $h$ is
positive. If $G$ satisfies

(2.5) $t\int_{t}^{\infty}G_{-}(s)ds\leq\frac{1}{4},$

since it was shown that in this condition that $h^{\prime}\geq 0$, see [6,
Prop. 1.21], then $I=[0,+\infty)$. The Hessian Comparison Theorem states that

(2.6)
${\rm\textrm{Hess}}\,\rho_{N}(y)\geq\displaystyle\frac{h^{\prime}}{h}(\rho_{N}(y))\\{\langle,\rangle-d\rho_{N}\otimes
d\rho_{N}\\}$

in the sense of quadratic forms. If

$K_{N}(x)\geq-G(\rho_{N}(x))$

then

(2.7)
${\rm\textrm{Hess}}\,\rho_{N}(y)\leq\displaystyle\frac{h^{\prime}}{h}(\rho_{N}(y))\\{\langle,\rangle-d\rho_{N}\otimes
d\rho_{N}\\}$

See [28] and references therein.

Associated to $N$, there are $m$-dimensional model manifolds
$\mathbb{M}_{h}^{m}=[0,\infty)\times\mathbb{S}^{m-1}$ with the metric
$ds^{2}_{h}=dr^{2}+h^{2}\left(r\right)d\theta^{2}$, for every $m\geq 2$, where
$h$ is the solution of (5.8). Observe that these models have, radial sectional
curvatures $-G\left(r\right)$. Let $\varphi\colon M\hookrightarrow N$ be an
isometric immersion of a complete Riemannian $m$-manifold $M$ into $N$. Let
$x_{0}\in M$ and let $\rho_{M}(x)={\rm dist}_{M}(x_{0},x)$ be the distance
function on $M$ to $x_{0}$. Let $\\{K_{i}\\}_{i=0}^{\infty}$ be an exhaustion
sequence of $M$ by nesting compacts sets $K_{i}\subset K_{i+1}$ with $x_{0}\in
K_{0}$. Let $\\{a_{i}(M)\\}\subset[0,\infty]$ be a non-increasing sequence of
numbers defined by

$\begin{array}[]{ccl}a_{i}(M)=\sup\left\\{\displaystyle\frac{h}{h^{\prime}}(\rho_{M}(x))\cdot\|\alpha(x)\|,\,x\in
M\backslash K_{i}\right\\},\end{array}$

where $\|\alpha(x)\|$ is the norm of the second fundamental form at
$\varphi(x)$. It is straightforward to show that the number
$a(M)=\displaystyle\lim a_{i}$ is independent on the sequence $\\{K_{i}\\}$
and on $x_{0}$.

######  Definition 2.1.

The immersion $\varphi$ has tamed second fundamental form if $a(M)<1$.

Consider a smooth function $g:N\to\mathbb{R}$ and the restriction
$f=g\circ\varphi$. Identifying $X$ with $d\varphi(X)$ we have at $q\in M$ and
for every $X\in T_{q}M$ that

(2.8) $\langle\nabla f,X\rangle=df(X)=dg(X)=\langle\nabla g,X\rangle.$

Hence we write

(2.9) $\nabla g=\nabla f+\nabla^{\perp}g,$

where $\nabla^{\perp}g$ is perpendicular to $T_{q}M$. In particular, for the
extrinsic distance function $R=\rho_{N}\circ\varphi$

(2.10) $\nabla\rho_{N}=\nabla R+\nabla^{\perp}\rho_{N}.$

The following result, due to Gimeno-Palmer [13], extends Bessa-Montenegro-
Jorge [5] and Bessa-Costa [4]. We shall present our proof of theorem 2.2 for
the sake of completeness and to clarify the notation used in the paper.

######  Theorem 2.2 (Gimeno-Palmer).

Let $N$ be a Riemannian manifold with a pole and radial sectional curvature
$K_{N}^{\mathop{\rm rad}\nolimits}(x)\leq-G(\rho_{N}(x))$, $G$ satisfying
(2.5). If $\varphi\colon M\hookrightarrow N$ be an isometric immersion of a
complete Riemannian manifold with tamed second fundamental form then

  * i.

$\varphi$ is proper.

  * ii.

$M$ has finite topology.

###### Proof.

Since $a(M)<1$ there exists $c\in(0,1)$ such that $a(M)<c<1$. Thus, there
exists an $i\in\mathbb{N}$ such that $a(M)<a_{i}(M)<c$. This means that there
exists a geodesic ball $B_{M}(r_{0})\subset M$, with $K_{i}\subset
B_{M}(r_{0})$, centered at $x_{0}$ with radius $r_{0}>0$ such that

(2.11)
$\displaystyle\frac{h}{h^{\prime}}(\rho_{M}(x))\cdot\|\alpha(x)\|<c<1,\;\;\;{\rm
for\;\;all}\;\;x\in M\backslash B_{M}(r_{0}).$

To fix the notation, let $x_{0}\in M$, $p=\varphi(x_{0})$ and recall that
$\rho_{M}(x)={\rm dist}_{M}(x_{0},x)$ and $\rho_{N}(y)={\rm dist}_{N}(p,y)$.
Letting $\phi(t)=\int\limits_{0}^{t}h(s)ds$ define $f\colon M\to\mathbb{R}$ by
$f=\phi\circ\rho_{N}\circ\varphi$. It is straightforward to compute that for
$X\in T_{x}M$, (identifying $d\varphi X=X$)

$\displaystyle\textrm{Hess}_{M}f(x)(X,X)$ $\displaystyle=$
$\displaystyle\textrm{Hess}_{N}\phi\circ\rho_{N}(\varphi(x))(X,X)+\langle{\rm\operatorname{\nabla}}\phi\circ\rho_{N},\alpha(X,X)$
$\displaystyle=$ $\displaystyle
h^{\prime}(\rho_{N})\langle{\rm\operatorname{\nabla}}\rho_{{}_{N}},X\rangle^{2}+h(\rho_{N}){\rm\textrm{Hess}}\rho_{N}(d\varphi
X,d\varphi X)$
$\displaystyle+h(\rho_{N})\langle{\rm\operatorname{\nabla}}\rho_{N},\alpha(X,X)\rangle.$

By the Hessian Comparison theorem, we have that,

(2.13)
$\textrm{Hess}\rho_{N}(y)(X,X)\geq\displaystyle\frac{h^{\prime}}{h}\left\\{\|X\|^{2}-\langle\operatorname{\nabla}\rho_{N},X\rangle^{2}\right\\}$

Therefore for every $x\in M\backslash B_{M}(r_{0})$ we have,

$\displaystyle{\rm\textrm{Hess}}f(x)(X,X)$ $\displaystyle\geq$ $\displaystyle
h^{\prime}(\rho_{N})\langle\operatorname{\nabla}\rho_{N},X\rangle^{2}+h(\rho_{N})[\displaystyle\frac{h^{\prime}}{h}(\rho_{N})\|X\|^{2}-\langle\operatorname{\nabla}\rho_{N},X\rangle^{2}]$
$\displaystyle-h(\rho_{N})\|\alpha\|\cdot\|X\|^{2}$ $\displaystyle=$
$\displaystyle
h^{\prime}(\rho_{N})\cdot\|X\|^{2}-h(\rho_{N})\cdot\|\alpha\|\cdot\|X\|^{2}$
$\displaystyle\geq$ $\displaystyle
h^{\prime}(\rho_{N})\cdot(1-c)\cdot\|X\|^{2}.$

Let $\sigma:[0,\rho_{M}(x)]\rightarrow M$ be a minimal geodesic joining
$x_{0}$ to $x$. For all $t>r_{0}$ we have that
$(f\circ\sigma)^{\prime\prime}(t)=\textrm{Hess}f(\sigma(t))(\sigma^{\prime},\sigma^{\prime})\geq
h^{\prime}(t)(1-c)$, where
$h^{\prime}(t)=h^{\prime}(\rho_{N}(\varphi(\sigma(t))))$.

For $t\leq r_{0}$ we have that $(f\circ\sigma)^{\prime\prime}(t)\geq
b=\inf\left\\{{\rm\textrm{Hess}}f(x)(\nu,\nu),x\in
B_{M}(r_{0}),|\nu|=1\right\\}$. Hence

(2.15)
$\begin{array}[]{ccl}(f\circ\sigma)^{\prime}(s)&=&(f\circ\sigma)^{\prime}(0)+\int_{0}^{s}(f\circ\sigma)^{\prime\prime}(\tau)d\tau\\\
\\\
&\geq&(f\circ\sigma)^{\prime}(0)+\int_{0}^{r_{0}}b\,d\tau+\int_{r_{0}}^{s}h^{\prime}(\tau)(1-c)d\tau\\\
\\\ &\geq&(f\circ\sigma)^{\prime}(0)+b\,r_{0}+(1-c)(h(s)-h(r_{0})).\\\
\end{array}$

Now, since $\varphi(x_{0})=p$, $\rho_{N}(\varphi(x_{0}))=0$ then
$(f\circ\sigma)^{\prime}(0)=0$, and $f(x_{0})=0$, therefore

(2.16)
$\begin{array}[]{ccl}f(x)&=&\int_{0}^{\rho_{M}(x)}(f\circ\sigma)^{\prime}(s)ds\\\
\\\
&\geq&\int_{0}^{\rho_{M}(x)}\left\\{b\,r_{0}+(1-c)(h(s)-h(r_{0}))\right\\}ds\\\
\\\
&\geq&b\,r_{0}\,\rho_{M}(x)-(1-c)h(r_{0})\rho_{M}(x)+(1-c)\int_{0}^{\rho_{M}(x)}h(s)ds\\\
\\\
&\geq&(br_{0}-(1-c)h(r_{0}))\rho_{M}(x)+(1-c)\int_{0}^{\rho_{M}(x)}h(s)ds\\\
\end{array}$

Thus

(2.17)
$\phi(\rho_{N}(\varphi(x)))\geq(br_{0}-(1-c)h(r_{0}))\rho_{M}(x)+(1-c)\int_{0}^{\rho_{M}(x)}h(s)ds$

for all $x\in M$. Then we have that $\rho_{N}(\varphi)\to\infty$ when
$\rho_{M}(x)\to\infty$, and $\varphi$ is therefore proper. Now let
$B_{N}(r_{0})$ be the geodesic ball of $N$ centered at $p$ with radius $r_{0}$
and $S_{N}(r_{0})=\partial B_{N}(r_{0})$. Since $\varphi$ is proper and
$a(M)<1$ we can take $r_{0}$ so that

(2.18) $\displaystyle\frac{h}{h^{\prime}}(\rho_{M}(x))\|\alpha(x)\|\leq
c<1,\;\;\;{\rm for\;\;all}\;\;x\in M\backslash\varphi^{-1}(B_{N}(r_{0}))$

and by Sard’s Theorem, see [12, p.79], $r_{0}$ can be chosen so that
$\Gamma_{r_{0}}={\varphi(M)}{\displaystyle\cap}{S_{N}(r_{0})}\not=\emptyset$
is a submanifold of ${\rm dim}\,\Gamma_{r_{0}}=m-1$. For each
$y\in\Gamma_{r_{0}}$, let us denote by $T_{y}\Gamma_{r_{0}}\subset
T_{y}\varphi(M)$ the tangent spaces of $\Gamma_{r_{0}}$ and $\varphi(M)$ at
$y$, respectively. Since ${\rm dim}\,T_{y}\Gamma_{r_{0}}=m-1$ and ${\rm
dim}\,T_{y}\varphi(M)=m$, there exists only one unit vector $\nu(y)\in
T_{y}\varphi(M)$ such that

$T_{y}\varphi(M)=T_{y}\Gamma_{r_{0}}\oplus[[\nu(y)]],$

with $\langle\nu(y),\operatorname{\nabla}\\!\rho_{N}(y)\rangle>0$. This
defines a smooth vector field $\nu$ on a neighborhood $V$ of
$\varphi^{-1}(\Gamma_{r_{0}})$. Here $[[\nu(y)]]$ is the vector space
generated by $\nu(y)$. Consider the function on $\varphi(V)$ defined by

(2.19)
$\begin{array}[]{ccl}\psi(y)=\langle\nu,\operatorname{\nabla}\rho_{N}\rangle(y)=\langle\nu,{\rm\operatorname{\nabla}}R\rangle(y)=\nu(y)(R),\,y=\varphi(x).\end{array}$

Then $\psi(y)=0$ if and only if every $x=\varphi^{-1}(y)\in V$ is a critical
point of the extrinsic distance function $R=\rho_{N}\circ\varphi$. Now for
each $y\in\Gamma_{r_{0}}$ fixed, let us consider the solution $\xi(t,y)$ of
the following Cauchy problem on $\varphi(M)$:

(2.20)
$\left\\{\begin{array}[]{ccl}\xi_{t}(t,y)&=&\displaystyle\frac{1}{\psi}\,\nu(\xi(t,y))\\\
\\\ \xi(0,y)&=&y\\\ \end{array}\right.$

We will prove that along the integral curve $t\mapsto\xi(t,y)$ there are no
critical points for $R$. For this, consider the function $(\psi\circ\xi)(t,y)$
and observe that

(2.21)
$\begin{array}[]{ccl}\psi_{t}&=&\xi_{t}\langle{\rm\operatorname{\nabla}}\rho_{N},\nu\rangle\\\
\\\
&=&\langle\overline{\nabla}_{\xi_{t}}{\rm\operatorname{\nabla}}\rho_{N},\nu\rangle+\langle{\rm\operatorname{\nabla}}\rho_{N},\overline{\nabla}_{\xi_{t}}\nu\rangle\\\
\\\
&=&\displaystyle\frac{1}{\psi}\langle\overline{\nabla}_{\nu}{\rm\operatorname{\nabla}}\rho_{N},\nu\rangle+\displaystyle\frac{1}{\psi}\langle{\rm\operatorname{\nabla}}\rho_{N},\nabla_{\nu}\nu+\alpha(\nu,\nu)\rangle\\\
\\\
&=&\displaystyle\frac{1}{\psi}{\rm\textrm{Hess}}\rho_{N}(\nu,\nu)+\frac{1}{\psi}\left[\langle{\rm\operatorname{\nabla}}\rho_{N},\nabla_{\nu}\nu\rangle+\langle{\rm\operatorname{\nabla}}\rho_{N},\alpha(\nu,\nu)\rangle\right]\\\
\\\
&=&\displaystyle\frac{1}{\psi}\left[{\rm\textrm{Hess}}\rho_{N}(\nu,\nu)+\langle\operatorname{\nabla}\rho_{N},\nabla_{\nu}\nu\rangle+\langle{\rm\operatorname{\nabla}}\rho_{N},\alpha(\nu,\nu)\rangle\right].\end{array}$

Thus

(2.22)
$\begin{array}[]{ccl}\psi_{t}\psi&=&\textrm{Hess}\rho_{N}(\nu,\nu)+\langle\operatorname{\nabla}\rho_{N},\nabla_{\nu}\nu\rangle+\langle\operatorname{\nabla}\rho_{N},\alpha(\nu,\nu)\rangle\end{array}$

Since $\langle\nu,\nu\rangle=1$, we have at once that
$\langle\nabla_{\nu}\nu,\nu\rangle=0$. As $\nabla_{\nu}\nu\in T_{x}M$, we have
that

$\langle{\rm\operatorname{\nabla}}\rho_{N},\nabla_{\nu}\nu\rangle=\langle{\rm\operatorname{\nabla}}R,\nabla_{\nu}\nu\rangle.$

By equation (2.19), we can write
${\rm\operatorname{\nabla}}R(x)=\psi(\varphi(x))\cdot\nu(\varphi(x))$. Since

$\operatorname{\nabla}R(x)\perp T_{\varphi(x)}\Gamma_{\rho_{N}(y)},$

($\Gamma_{\rho_{N}(y)}=\varphi(M)\displaystyle\cap S_{N}(\rho_{N}(y))$). Then

$\langle{\rm\operatorname{\nabla}}\rho_{N},\nabla_{\nu}\nu\rangle=\langle{\rm\operatorname{\nabla}}R,\nabla_{\nu}\nu\rangle=\psi\langle\nu,\nabla_{\nu}\nu\rangle=0.$

Writing

(2.23)
$\nu(y)=\cos\beta(y)\;\operatorname{\nabla}\rho_{N}+\sin\beta(y)\;\omega$

and

(2.24) $\operatorname{\nabla}\rho_{N}(y)=\cos\beta\;\nu(y)+\sin\beta\;\nu^{*}$

where $\langle\omega,\operatorname{\nabla}\rho_{N}\rangle=0$ and
$\langle\nu,\nu^{*}\rangle=0$, the equation (2.22) becomes

(2.25)
$\begin{array}[]{ccl}\psi_{t}\psi=\sin^{2}\beta\;\textrm{Hess}\rho_{N}(\omega,\omega)+\sin\beta\;\langle\nu^{*},\alpha(\nu,\nu)\rangle.\\\
\end{array}$

From (2.23) we have that $\psi(y)=\cos\beta(y)$

(2.26)
$\psi_{t}\psi=\sqrt{1-\psi^{2}}\sqrt{1-\psi^{2}}\textrm{Hess}\rho_{N}(\omega,\omega)+\sqrt{1-\psi^{2}}\langle\nu^{*},\alpha(\nu,\nu)\rangle.\\\
$

Hence

(2.27)
$\begin{array}[]{ccl}\displaystyle\frac{\psi_{t}\psi}{\sqrt{1-\psi^{2}}}&=&\sqrt{1-\psi^{2}}\textrm{Hess}\rho_{N}(\omega,\omega)+\langle\nu^{*},\alpha(\nu,\nu)\rangle.\\\
\end{array}$

Thus we arrive at the following differential equation

(2.28)
$\begin{array}[]{ccl}-(\sqrt{1-\psi^{2}})_{t}&=&\sqrt{1-\psi^{2}}\;\textrm{Hess}\rho_{N}(\omega,\omega)+\langle\nu^{*},\alpha(\nu,\nu)\rangle\\\
\end{array}$

The Hessian Comparison Theorem implies that

(2.29)
$\textrm{Hess}\rho_{N}(\omega,\omega)\geq\displaystyle\frac{h^{\prime}}{h}(\rho_{N}(\xi(t,y))).\\\
$

Substituting it in the equation $(\ref{eqdif1})$ obtain the following
inequality

(2.30)
$\begin{array}[]{ccl}-(\sqrt{1-\psi^{2}})_{t}&\geq&\sqrt{1-\psi^{2}}\;\displaystyle\frac{h^{\prime}}{h}(\rho_{N}(\xi(t,y)))\;+\langle\nu^{*},\alpha(\nu,\nu)\rangle.\\\
\end{array}$

Denoting by $R(t,y)$ the restriction of $R=\rho_{N}\circ\varphi$ to
$\varphi^{-1}(\xi(t,y))$ we have

$R(t,y)=R(\varphi^{-1}(\xi(t,y)))=\rho_{N}(\xi(t,y))$

On the other hand we have that

(2.31)
$R_{t}=\langle{\rm\operatorname{\nabla}}R,\frac{1}{\psi}\nu\rangle=\langle\psi\nu,\frac{1}{\psi}\nu\rangle=1$

then

(2.32) $R(t,y)=t+r_{0}.$

Writing
$\displaystyle\frac{h^{\prime}}{h}(\rho_{N}(\xi(t,y)))=\displaystyle\frac{h^{\prime}}{h}(t+r_{0})$
in $(\ref{eqdif2})$ we have

(2.33)
$\begin{array}[]{ccl}-(\sqrt{1-\psi^{2}})_{t}\geq\sqrt{1-\psi^{2}}\;\displaystyle\frac{h^{\prime}}{h}(t+r_{0})+\langle\nu^{*},\alpha(\nu,\nu)\rangle\\\
\end{array}$

Multiplying $(\ref{eqdif3})$ by $h(t+r_{0})$, obtain

$\displaystyle\begin{array}[]{ccl}-\left[h(t+r_{0})(\sqrt{1-\psi^{2}})_{t}+h^{\prime}(t+r_{0})\sqrt{1-\psi^{2}}\right]&\geq&h(t+r_{0})\langle\nu^{*},\alpha(\nu,\nu)\rangle\\\
\end{array}$

The last inequality can be written as

(2.35)
$\begin{array}[]{ccl}\left[h(t+r_{0})\sqrt{1-\psi^{2}}\right]_{t}&\leq&-h(t+r_{0})\langle\nu^{*},\alpha(\nu,\nu)\rangle\\\
\end{array}$

Integrating $(\ref{eqdif4})$ from $0$ to $t$ the resulting inequality is the
following

$S_{\kappa}(t+r_{0})\sin\beta(\xi(t,y))\leq
S_{\kappa}(r_{0})\sin\beta(y)+\int_{0}^{t}{-S_{k}(s+r_{0})\langle\nu^{*},\alpha(\nu,\nu)\rangle
ds}$

Thus

(2.36)
$\sin\beta(\xi(t,y))\leq\frac{h(r_{0})}{h(t+r_{0})}\sin\beta(y)+\frac{1}{h(t+r_{0})}\int_{0}^{t}{h(s+r_{0})(-\langle\nu^{*},\alpha(\nu,\nu)\rangle)ds}$

Since $a(M)<1$, then

$-\langle\nu^{*},\alpha(\nu,\nu)\rangle(\xi(s,y))\leq\|\alpha(\xi(s,y))\|\leq
c\frac{h^{\prime}}{h}(\rho_{M}(\xi(s,y)))\leq
c\frac{h^{\prime}}{h}(\rho_{N}(\xi(s,y)))$

But
$\displaystyle\frac{h^{\prime}}{h}(\rho_{N}(\xi(s,y)))=\frac{h^{\prime}}{h}(s+r_{0})$
for every $s\geq 0$. Substituting in (2.36), we have

(2.37)
$\begin{array}[]{ccl}\sin\beta(\xi(t,y))&\leq&\displaystyle\frac{h(r_{0})}{h(t+r_{0})}\sin\beta(y)+\frac{c}{h(t+r_{0})}\int_{0}^{t}h^{\prime}(s+r_{0})ds\\\
\\\
&=&\displaystyle\frac{h(r_{0})}{h(t+r_{0})}\sin\beta(y)+\displaystyle\frac{c}{h(t+r_{0})}(h(t+r_{0})-h(r_{0}))\\\
\\\ &=&\displaystyle\frac{h(r_{0})}{h(t+r_{0})}(\sin\beta(y)-c)+c\\\
\end{array}$

We will show that
$\displaystyle\frac{h(r_{0})}{h(t+r_{0})}(\sin\beta(y)-c)+c<1$. Let
$\Upsilon(t)=\displaystyle\frac{h(r_{0})}{h(t+r_{0})}(\sin\beta(y)-c)+c$. We
have that $\Upsilon(0)=\sin\beta<1$ and
$\Upsilon^{\prime}(t)=\displaystyle-\frac{h^{\prime}(t+r_{0})h(r_{0})}{h^{2}(t+r_{0})}(\sin\beta-c)$.
If $\sin\beta\geq c$ then $\Upsilon^{\prime}(t)\leq 0$ and
$\Upsilon(t)\leq\Upsilon(0)$. If $\sin\beta<c$, suppose by contradiction that
there exists a $T>0$ such that $\Upsilon(T)>1$. This implies that
$0>h(r_{0})(\sin\beta-c)>(1-c)h(T+r_{0})>0$. Then

$\displaystyle\sin\beta(\xi(t,y))\leq\displaystyle\frac{h(r_{0})}{h(t+r_{0})}(\sin\beta(y)-c)+c<1$

for all $t\geq 0$. Therefore, along the integral curve $t\mapsto\xi(t,y)$,
there are no critical point for the function $R(x)=\rho_{N}(\varphi(x))$
outside the geodesic ball $B_{N}(r_{0})$. The flow $\xi_{t}$ maps
$S_{N}(r_{0})$ diffeomorphically into to $S_{N}(r_{0}+t)$, for all $t\geq 0$.
The manifold $M$ has therefore finite topology, see also [7]. This concludes
the proof of the theorem 2.2. ∎

Actually, the above theorem is a consequence of the convexity of the extrinsic
distance function on $M\setminus D_{r_{0}}$,
$D_{r_{0}}=\varphi^{-1}(B_{N}(r_{0}))$, see approach given in [13]. In
particular,

######  Theorem 2.3.

[4, 5, 13] Let $\varphi:M\hookrightarrow N$ be an immersion of a complete
Riemannian $m$-manifold $M$ into an $n$-dimensional ambient manifold $N$ with
a pole and radial sectional curvatures$K_{N}$ bounded from above by

$K_{N}\leq\kappa\leq 0.$

Suppose that $\varphi$ has tamed second fundamental form, then:

  1. (1)

$\varphi$ is proper.

  2. (2)

$M$ has finite topology.

  3. (3)

There exists $r_{0}\in M$ such that the extrinsic distance function has no
critical points in $M\setminus D_{r_{0}}$.

  4. (4)

In particular, $M\setminus D_{r_{0}}$ is a disjoint union $\cup_{k}V_{k}$ of
finite number of ends. $M$ has so many ends $\mathcal{E}(M)$ as components
$\partial D_{r_{0}}$ has , and each end $V_{k}$ is diffeomorphic to $\partial
D_{r_{0}}^{k}\times[0,\infty)$, where $\partial D_{r_{0}}^{k}$ denotes the
component of $\partial D_{r_{0}}$ which belongs to $V_{k}$.

We will need the following technical lemma due to Kasue, [21].

######  Lemma 2.4.

[21, see proof of lemma 4]Let $\varphi:M\hookrightarrow N$ be an immersion of
a complete Riemannian $m$-manifold $M$ into an $n$-dimensional ambient
manifold $N$ with a pole and radial sectional curvatures$K_{N}$ bounded from
above by

$K_{N}\leq\kappa\leq 0.$

Suppose that there exists a function $k:\mathbb{R}\to\mathbb{R}$ such that
$\|\alpha\|(x)\leq k(R(x))$, $R(x)=\rho_{N}\circ\varphi(x)$, then for any
$x\in M\setminus D_{r}$ with $r>r_{0}$,

(2.38)
$|\nabla^{\perp}\rho_{\mathbb{M}^{n}(\kappa)}|\leq\delta(R(x))+\frac{1}{S_{\kappa}(R(x))}\int_{r}^{R(x)}S_{\kappa}(s)k(s)ds.$

The function $\delta(t)$ being a decreasing function such that $\delta\to 0$
when $t\to\infty$ and $S_{\kappa}$ being the solution of the following Cauchy
problem

(2.39)
$\left\\{\begin{array}[]{rll}\begin{array}[]{l}S_{\kappa}^{\prime\prime}(t)+\kappa
S_{\kappa}(t)=0,\\\\[2.84544pt]
S_{\kappa}(0)=0,\,\,\,\,\;S_{\kappa}^{\prime}(0)=1.\end{array}\end{array}\right.$

###  2.1. Tamed surfaces and their topology

This paper is concerned with tamed surfaces, hence, by Theorem 2.3, with
surfaces of finite topological type. Recall that a surface $M$ is of finite
topological type if $M$ is homeomorphic to the interior of a compact surface
$\tilde{M}$ with non-empty boundary. A surface of finite topological type has
finitely many ends. Recall also that given a compact subset $D\subset M$ of
$M$, an _end_ $E$ of $M$ with respect to $D$ is a connected unbounded
component of $M\setminus D$

Observe that if $D_{1}\subset D_{2}$ are compact subsets of $M$, then the
number of ends with respect to $D_{1}$ is at most the number of ends with
respect to $D_{2}$ . This monotonicity property allows us to define the number
of ends of a surface.

######  Definition 2.5.

A surface $M$ is said to have _finitely many ends_ if there exists
$0<k<\infty$, such that, for any compact $D\subset M$, the number of ends with
respect to $D$ is at most $k$. In this case, we denote $\mathcal{E}(M)$ to be
the smallest such $k$, and we shall say that $M$ has $\mathcal{E}(M)$ ends.

Obviously, $\mathcal{E}(M)$ must be an integer and, if a surface has finitely
many ends, one readily concludes that there exists $D_{0}\subset M$ such that,
the number of ends with respect to $D_{0}$ is precisely $\mathcal{E}(M)$.

For surfaces of finite topological type one can state the following
proposition

######  Proposition 2.6.

Suppose that $M$ is a surface of finite topological type, then:

  1. (1)

$M$ has finitely many ends, say $\mathcal{E}(M)$ ends.

  2. (2)

$M$ is homeomorphic to a compact surface $\widetilde{M}$ with $\mathcal{E}(M)$
points removed, _i.e._ ,

$M\sim\widetilde{M}\setminus\left\\{p_{1},\cdots,p_{\mathcal{E}(M)}\right\\}$

  3. (3)

There exists a compact domain $\Omega_{0}\subset M$, such that $M$ has
$\mathcal{E}(M)$ ends with respect $\Omega_{0}$, and, every of such ends is
homeomorphic to $\mathbb{R}_{+}\times\mathbb{S}^{1}$ $($every end with respect
to $\Omega_{0}$ is an annular end$)$.

##  3\. Proof of theorem 1.1

Theorem 1.1 will be proved in two steps. In the first step we will prove
theorem 3.1 which is a version of theorem 1.1 but using the linear extrinsic
perimeter growth property instead of the quadratic extrinsic area growth
property. In the second step we will prove proposition 3.2 where the
equivalence between quadratic extrinsic area growth and linear extrinsic
perimeter growth will be stated.

######  Theorem 3.1.

Let $M$ be an immersed complete oriented surface of $\mathbb{R}^{n}$ with
curvature function $K$ and tamed second fundamental form. Suppose that $M$
admits total curvature. Then, $M$ has finite total curvature
($\int_{M}KdA>-\infty$) if and only if $M$ has linear perimeter growth, _i.e._
, there exists a constant $\widetilde{C}_{1}$ such that,

(3.1) $\text{L}(M\cap S_{r}(0))\leq\widetilde{C}_{1}r,$

for $r$ large enough, where ${\rm L}(M\cap S_{r}(0))$ is the perimeter of the
intersection of the $r$-geodesic sphere in $\mathbb{R}^{n}$ centered at
$0\in\mathbb{R}^{n}$ with the surface $M$. Furthermore, if (3.1) holds, then
there exists a constant $\widetilde{C}_{0}>0$ such that

(3.2) $\text{L}(M\cap S_{r}(0))\geq\widetilde{C}_{0}r,$

for $r$ large enough.

###### Proof.

We are going to apply the Gauss-Bonnet theorem to the extrinsic annulus
$A_{r_{0},t}:=D_{t}\setminus D_{r_{0}}$ for $t>r_{0}$. Taking into account
that since the extrinsic distance function
$R=\rho_{\mathbb{R}^{n}}\circ\varphi$ has no critical points on $M\setminus
D_{r_{0}}$, then ${A_{r_{0},t}}$ is a finite union of annuli, and we obtain

(3.3) $\int_{A_{r_{0},t}}KdA+\int_{\partial
A_{r_{0},t}}k_{g}dL=2\pi\chi({A_{r_{0},t}})=0.$

where $K$, $k_{g}$ and $\chi({A_{r_{0},t}})$ denote the Gaussian curvature,
the geodesic curvature and the Euler characteristic respectively. Observe
moreover that $\partial A_{r_{0},t}$ is the union of two level sets

$\partial A_{r_{0},t}=\partial D_{t}\cup\partial D_{r_{0}}.$

Hence,

(3.4) $\int_{A_{r_{0},t}}KdA=\int_{\partial D_{r_{0}}}k_{g}dL-\int_{\partial
D_{t}}k_{g}dL$

But for any $s$, the geodesic curvature $k_{g}^{s}$ of the extrinsic spheres
$\partial D_{s}$ is given by

(3.5) $\displaystyle k_{g}^{s}=$
$\displaystyle-\langle\nabla_{e}e,\frac{\nabla R}{|\nabla
R|}\rangle=\frac{1}{|\nabla R|}\operatorname{Hess}_{M}R(e,e)$ $\displaystyle=$
$\displaystyle\frac{1}{|\nabla
R|}\left(\frac{1}{s}+\langle\nabla^{\perp}{\rho_{\mathbb{R}^{n}}},\alpha(e,e)\rangle\right),$

where $e$ is tangent to $\partial D_{s}$. Then,

(3.6) $\frac{1}{s}\frac{1}{|\nabla
R|}\left(1-s|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|\cdot\|\alpha\|\right)\leq
k_{g}^{s}\leq\frac{1}{s}\frac{1}{|\nabla
R|}\left(1+s|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|\cdot\|\alpha\|\right).$

Since $a(M)<1$, then for any $c\in(a(M),1)$ there exists $t_{c}$ such that

(3.7) $R(x)\|\alpha\|(x)<c,$

for all $R(x)=t>t_{c}$. Using lemma 2.4, we obtain

(3.8)
$|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|\leq\delta(t)+\frac{c(t-t_{c})}{t}\leq\delta(t)+c,$

at any point $x\in M$ with $R(x)=t$ and $t>t_{c}>r_{0}$. In order to simplify
the notation let us denote by

(3.9) $\Lambda_{c}(t):=\delta(t)+c.$

Therefore,

$\begin{array}[]{lllll}\displaystyle\frac{1}{t}\displaystyle\frac{1}{|\nabla
R|}\left(1-c|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|\right)&\leq&k_{g}^{t}&\leq&\displaystyle\frac{1}{t}\displaystyle\frac{1}{|\nabla
R|}\left(1+c|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|\right)\end{array}$

that can be simplified to

$\begin{array}[]{lllll}\displaystyle\frac{1}{t}\displaystyle\frac{1-c|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|}{\left(1-|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|^{2}\right)^{\frac{1}{2}}}&\leq&k_{g}^{t}&\leq&\displaystyle\frac{1}{t}\displaystyle\frac{1+c|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|}{\left(1-|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|^{2}\right)^{\frac{1}{2}}}\end{array}$

and can be rewritten as

(3.10)
$\begin{array}[]{lllll}\displaystyle\frac{1-c\Lambda_{c}(t)}{t}&\leq&k_{g}^{t}&\leq&\displaystyle\frac{1}{t}\displaystyle\frac{1+c\Lambda_{c}(t)}{\left(1-\Lambda^{2}_{c}(t)\right)^{\frac{1}{2}}}\end{array}$

Applying the above inequalities to the extrinsic annulus $A_{t_{1},t_{2}}$ and
using Gauss-Bonnet formula, as in the inequality (3.4) we have,

(3.11) $\begin{array}[]{lll}\displaystyle\frac{{\rm L}(\partial
D_{t_{1}})}{t_{1}}\displaystyle\left(1-c\Lambda_{c}(t_{1})\right)-\displaystyle\frac{{\rm
L}(\partial
D_{t_{2}})}{t_{2}}\frac{1+c\Lambda_{c}(t_{2})}{\left(1-\Lambda^{2}_{c}(t_{2})\right)^{\frac{1}{2}}}&\leq&\int_{A_{t_{1},t_{2}}}KdA\end{array}$

and

(3.12)
$\begin{array}[]{lll}\int_{A_{t_{1},t_{2}}}KdA&\leq&\displaystyle\frac{{\rm
L}(\partial
D_{t_{1}})}{t_{1}}\displaystyle\frac{1+c\Lambda_{c}(t_{1})}{\left(1-\Lambda^{2}_{c}(t_{1})\right)^{\frac{1}{2}}}-\displaystyle\frac{{\rm
L}(\partial
D_{t_{2}})}{t_{2}}\displaystyle\left(1-c\Lambda_{c}(t_{2})\right)\end{array}$

If we suppose that $M$ has linear extrinsic perimeter growth, from inequality
(3.11) and the monotonicity of $\Lambda_{c}$

(3.13) $\begin{array}[]{lll}-\displaystyle
C_{1}\frac{1+c\Lambda_{c}(t_{1})}{\left(1-\Lambda^{2}_{c}(t_{1})\right)^{\frac{1}{2}}}&\leq&\int_{A_{t_{1},t_{2}}}KdA.\end{array}$

Letting $t_{2}\to\infty$, we get the desired $\int_{M}KdA>-\infty$ because the
integral of the curvature is finite on each end of the surface.

On the other hand from inequalities (3.11), (3.12) and the monotonicity of
$\Lambda_{c}$,

(3.14) $\displaystyle\frac{{\rm L}(\partial D_{t_{2}})}{t_{2}}\leq$
$\displaystyle\frac{1}{1-c\Lambda_{c}(t_{1})}\left[\frac{{\rm L}(\partial
D_{t_{1}})}{t_{1}}\frac{1+c\Lambda_{c}(t_{1})}{\left(1-\Lambda^{2}_{c}(t_{1})\right)^{\frac{1}{2}}}-\int_{A_{t_{1},t_{2}}}KdA\right],$
$\displaystyle\frac{{\rm L}(\partial D_{t_{2}})}{t_{2}}\geq$
$\displaystyle\frac{(1-\Lambda_{c}^{2}(t_{1}))^{\frac{1}{2}}}{1+c\Lambda_{c}(t_{1})}\left[\frac{{\rm
L}(\partial
D_{t_{1}})}{t_{1}}(1-c\Lambda_{c}(t_{1}))-\int_{A_{t_{1},t_{2}}}KdA\right].$

If we assume that $M$ admits finite total curvature, for any $\epsilon>0$
there therefore exists $t_{1}$ large enough such that

(3.15) $\left|\int_{A_{t_{1},t_{2}}}KdA\right|<\epsilon.$

Then

(3.16) $\displaystyle\frac{{\rm L}(\partial D_{t_{2}})}{t_{2}}\leq$
$\displaystyle\frac{1}{1-c\Lambda_{c}(t_{1})}\left[\frac{{\rm L}(\partial
D_{t_{1}})}{t_{1}}\frac{1+c\Lambda_{c}(t_{1})}{\left(1-\Lambda^{2}_{c}(t_{1})\right)^{\frac{1}{2}}}+\epsilon\right]:=\widetilde{C}_{1},$
$\displaystyle\frac{{\rm L}(\partial D_{t_{2}})}{t_{2}}\geq$
$\displaystyle\frac{(1-\Lambda_{c}^{2}(t_{1}))^{\frac{1}{2}}}{1+c\Lambda_{c}(t_{1})}\left[\frac{{\rm
L}(\partial
D_{t_{1}})}{t_{1}}(1-c\Lambda_{c}(t_{1}))-\epsilon\right]:=\widetilde{C}_{0}.$

And this finishes the proof of the theorem because for $t$ large enough

(3.17) $\displaystyle\left|\int_{M}KdA\right|<\infty$
$\displaystyle\Longleftrightarrow\quad\exists C_{1}\,\,:\,\,L(\partial
D_{t})\leq C_{1}t,$ $\displaystyle\left|\int_{M}KdA\right|<\infty$
$\displaystyle\Longrightarrow\quad\exists C_{0}\,\,:\,\,L(\partial D_{t})\geq
C_{0}t.$

∎

######  Proposition 3.2.

Let $M$ be an immersed complete oriented surface of $\mathbb{R}^{n}$ with
tamed second fundamental form, then $M$ has quadratic extrinsic area growth,
if and only if, $M$ has linear extrinsic perimeter growth. Namely,

$\text{L}(M\cap
S_{r}(0))\leq\widetilde{C}_{1}r\Longleftrightarrow\text{A}(M\cap B_{r}(0))\leq
C_{1}r^{2},$

for $r$ large enough. Furthermore,

$\text{L}(M\cap S_{r}(0))\geq\widetilde{C}_{0}r\Longrightarrow\text{A}(M\cap
B_{r}(0))\geq C_{0}r^{2},$

###### Proof.

Denote by $D_{t}(o)=M\cap B_{t}(o)$ the extrinsic ball centered at $o\in M$.
Let us observe that by using coarea formula (see for instance [29]) for the
extrinsic distance function $R=\rho_{\mathbb{R}^{n}}\circ\varphi$ on any
extrinsic ball $D_{t}$ with $t>t_{1}>r_{o}$,

(3.18) $\displaystyle{\rm A}(D_{t})=$ $\displaystyle{\rm
A}(D_{t_{1}})+\int_{t_{1}}^{t}\int_{\partial D_{s}(o)}\frac{1}{|\nabla
R|}dLds$

Thus, for any $c\in(a(M),1)$ and $t_{1}$ large enough, taking into account the
monotonocity of the function $\Lambda_{c}$,

(3.19) $\displaystyle{\rm A}(D_{t})=$ $\displaystyle{\rm
A}(D_{t_{1}})+\int_{t_{1}}^{t}\int_{\partial
D_{s}(o)}\frac{1}{\sqrt{1-|\nabla^{\perp}{\rho_{\mathbb{R}^{n}}}|^{2}}}dLds$
$\displaystyle\leq$ $\displaystyle{\rm
A}(D_{t_{1}})+\int_{t_{1}}^{t}\int_{\partial
D_{s}(o)}\frac{1}{\sqrt{1-\Lambda_{c}^{2}(s)}}dLds$ $\displaystyle=$
$\displaystyle{\rm
A}(D_{t_{1}})+\int_{t_{1}}^{t}\frac{1}{\sqrt{1-\Lambda_{c}^{2}(s)}}{\rm
L}(\partial D_{s}(o))ds$ $\displaystyle\leq$ $\displaystyle{\rm
A}(D_{t_{1}})+\frac{1}{\sqrt{1-\Lambda_{c}^{2}(t_{1})}}\int_{t_{1}}^{t}{\rm
L}(\partial D_{s}(o))ds.$

Hence, if we suppose that $M$ has linear extrinsic perimeter growth,

(3.20) $\displaystyle{\rm A}(D_{t})\leq$ $\displaystyle{\rm
A}(D_{t_{1}})+\frac{1}{\sqrt{1-\Lambda_{c}^{2}(t_{1})}}\frac{\widetilde{C}_{1}}{2}(t^{2}-t_{1}^{2})$
$\displaystyle=$ $\displaystyle\left[\frac{{\rm
A}(D_{t_{1}})}{t^{2}}+\frac{1}{\sqrt{1-\Lambda_{c}^{2}(t_{1})}}\frac{\widetilde{C}_{1}}{2}\left(1-\left(\frac{t_{1}}{t}\right)^{2}\right)\right]t^{2}$
$\displaystyle\leq$ $\displaystyle\left[\frac{{\rm
A}(D_{t_{1}})}{{t_{1}}^{2}}+\frac{1}{\sqrt{1-\Lambda_{c}^{2}(t_{1})}}\frac{\widetilde{C}_{1}}{2}\right]t^{2}$

Denoting $C_{1}:=\frac{{\rm
A}(D_{t_{1}})}{{t_{1}}^{2}}+\frac{1}{\sqrt{1-\Lambda_{c}^{2}(t_{1})}}\frac{\widetilde{C}_{1}}{2}$
we conclude that $M$ has quadratic extrinsic area growth.

In order to prove the reverse implication let us consider now the Laplacian of
the extrinsic distance function $R=\rho_{\mathbb{R}^{n}}\circ\varphi$,

(3.21) $\displaystyle\triangle_{M}R^{2}=$ $\displaystyle
4R\left(\frac{1}{R}+\langle\nabla\rho_{\mathbb{R}^{n}},H\rangle\right)\leq
4\left(1+R|H|\right)\leq 4\left(1+R\|\alpha\|\right)$

Applying the divergence theorem in an extrinsic ball $D_{t}$ with $t$ large
enough and $c\in(a(M),1)$ we have

(3.22) $\displaystyle 2t\int_{\partial D_{t}}|\nabla R|dL=$
$\displaystyle\int_{D_{t}}\triangle_{M}R^{2}\,dA=\int_{D_{t_{1}}}\triangle_{M}R^{2}\,dA+\int_{A_{t_{1},t}}\triangle_{M}R^{2}\,dA$
$\displaystyle\leq$
$\displaystyle\int_{D_{t_{1}}}\triangle_{M}R^{2}\,dA+4(1+c){\rm
A}(A_{t_{1},t})$

Then, denoting $A_{1}:=\int_{D_{t_{1}}}\triangle_{M}R^{2}\,dA$, and assuming
that $M$ has quadratic extrinsic area growth

(3.23) $\displaystyle 2t\sqrt{1-\Lambda_{c}^{2}(t)}{\rm L}(\partial
D_{t})\leq$ $\displaystyle A_{1}+4(1+c){\rm A}(A_{t_{1},t})$
$\displaystyle\leq$ $\displaystyle A_{1}+4(1+c){\rm A}(D_{t})\leq
A_{1}+4(1+c)C_{1}t^{2}$ $\displaystyle\leq$
$\displaystyle\left[\frac{A_{1}}{t_{1}^{2}}+4(1+c)C_{1}\right]t^{2}$

Letting
$\widetilde{C}_{1}:=\frac{\frac{A_{1}}{t_{1}^{2}}+4(1+c)C_{1}}{2\sqrt{1-\Lambda_{c}^{2}(t_{1})}}$,
we therefore obtain

(3.24) ${\rm L}(\partial D_{t})\leq\widetilde{C}_{1}t.$

Observe finally that from inequality (3.18) for any $\delta\in(0,1)$ and any
$t\geq\frac{1}{1-\delta}t_{1}$, under the hypothesis of a lower bound for the
extrinsic perimeter growth

(3.25) ${\rm A}(D_{t})\geq A_{1}+\int_{t_{1}}^{t}{\rm L}(\partial
D_{s})ds\geq\frac{\widetilde{C}_{0}}{2}\left(t^{2}-t_{1}^{2}\right)\geq\frac{\widetilde{C}_{0}}{2}t\left(t-t_{1}\right)\geq\frac{\delta\widetilde{C}_{0}}{2}t^{2}.$

Letting $C_{0}$ be $\frac{\delta\widetilde{C}_{0}}{2}$, the proposition
follows. ∎

##  4\. Proof of theorem 1.2 and corollaries 1.3 and 1.4

###### Proof.

Given a surface of finite topological type which admitting total curvature we
can make use of [30, theorem A], for any fixed point $o\in M$

(4.1) $\lim_{t\to\infty}\frac{{\rm A}(t)}{t^{2}/2}=2\pi\chi(M)-\int_{M}KdA$

where ${\rm A}(t)$ is the area of the geodesic ball of radius $t$ centered at
$o\in M$. Denote by $D_{t}(o)=M\cap B_{t}(o)$ the extrinsic ball centered at
$o\in M$. Therefore

(4.2) ${\rm A}(t)\leq{\rm A}(D_{t}(o)).$

Hence,

(4.3) $\displaystyle 2\pi\chi(M)-\int_{M}KdA=$
$\displaystyle\lim_{t\to\infty}\frac{A(t)}{t^{2}/2}$ $\displaystyle\leq$
$\displaystyle\limsup_{t\to\infty}\frac{{\rm A}(D_{t}(o))}{t^{2}/2}$
$\displaystyle\leq$ $\displaystyle C_{1}.$

The upper bound for the inequality of the theorem therefore follows. On the
other hand, using the Gauss-Bonnet theorem for an extrinsic ball of radius $t$
large enough, and inequality (3.10) we obtain

(4.4) $\displaystyle 2\pi\chi(M)-\int_{D_{t}}KdA=$
$\displaystyle\int_{\partial
D_{t}}k_{g}dL\geq\left(1-c\Lambda_{c}(t)\right)\frac{{\rm L}(\partial
D_{t})}{t}$ $\displaystyle\geq$
$\displaystyle\left(1-c\Lambda_{c}(t)\right)C_{0}.$

Letting $t$ tend to infinity and after letting $c$ tend to $a(M)$ the theorem
follows.

In order to prove corollary 1.3, observe that if we assume that $M$ is flat,
by using the above inequality

(4.5) $\chi(M)=2-2g(M)-\mathcal{E}(M)>0$

where $g(M)$ is the genus of $M$ and $\mathcal{E}(M)$ is the number of ends of
$M$. Since $\mathcal{E}(M)\geq 1$, the only option is $\mathcal{E}(M)=1$ and
$g(M)=0$. The surface $M$ is therefore homeomorphic to a sphere with one point
removed. Since the surface is simply connected, metrically complete and with
zero curvature, the surface is isometric to $\mathbb{R}^{2}$ with the
canonical flat metric (see [22, theorem 11.12] for instance).

Moreover, if $M$ has strongly tamed second fundamental form, then $M$ has
tamed fundamental form as well. Hence by applying co-area formula and taking
into account that $\|\alpha(x)\|\leq\frac{c}{t^{1+\epsilon}}$ for any
$c\in(a(M),1)$ and $t=R(x)$ large enough,

(4.6) $\displaystyle\int_{M}\|\alpha\|^{2}dA=$
$\displaystyle\int_{D_{t_{1}}}\|\alpha\|^{2}dA+\int_{A_{t_{1},t}}\|\alpha\|^{2}dA$
$\displaystyle\leq$
$\displaystyle\int_{D_{t_{1}}}\|\alpha\|^{2}dA+\int_{t_{1}}^{t}\int_{\partial
D_{s}}\frac{\|\alpha\|^{2}}{|\nabla R|}dLds$ $\displaystyle\leq$
$\displaystyle\int_{D_{t_{1}}}\|\alpha\|^{2}dA+\int_{t_{1}}^{t}\frac{c^{2}}{s^{2+2\epsilon}\sqrt{1-\Lambda_{c}^{2}(s)}}L(\partial
D_{s})ds$ $\displaystyle\leq$
$\displaystyle\int_{D_{t_{1}}}\|\alpha\|^{2}dA+\frac{c^{2}\,\widetilde{C}_{1}}{\sqrt{1-\Lambda_{c}^{2}(t_{1})}}\int_{t_{1}}^{t}\frac{1}{s^{1+2\epsilon}}ds<\infty.$

By using now theorem 2 of [32], $\int_{M}KdA$ is an integral multiple of
$2\pi$, and using the lower bounds given in the inequality of theorem 1.2 we
conclude that $\int_{M}KdA=0$ because $\chi(M)=1$. Since $M$ is a complete and
flat surface with tamed second fundamental form, $M$ is therefore isometric to
$\mathbb{R}^{2}$ and this finishes the proof of corollary 1.4 ∎

##  5\.  Proof of Theorem 1.5

The first ingredient for the proof of Theorem 1.5 is Barta’s Theorem [3].

######  Theorem 5.1 (Barta).

Let $\Omega$ be a bounded open set with piecewise smooth boundary in a
Riemannian manifold. Let $f\in C^{2}(\Omega)\cap C^{0}(\bar{\Omega})$ with
$f|\Omega>0$ and $f|\partial\Omega=0$. Then the first Dirichlet eigenvalue
$\lambda_{1}(\Omega)$ has the following bounds:

(5.1) $\begin{array}[]{ccl}\displaystyle\sup_{\Omega}(-\frac{\Delta
f}{f})\geq\lambda_{1}(\Omega)\geq\displaystyle\inf_{\Omega}(-\frac{\Delta
f}{f})\\\ \end{array}$

With equality in (5.1) if and only in $f$ is a positive first eigenfunction of
$\Omega$.

We now present the proof of Theorem 1.5.

###### Proof.

Let $\varphi:M\hookrightarrow N$ be an isometric immersion with tamed second
fundamental form of a complete $m$-manifold $M$ into a $n$-manifold $N$ with a
pole $p\in N$ and sectional radial curvatures $B\leq K_{N}\leq 0$. Let
$x_{0}\in M$, $p=\varphi(x_{0})\in N$ and let $\rho_{N}(y)={\rm
dist}_{N}(p,y)$ be the distance function on $N$ and $\rho_{N}\circ\varphi$ the
extrinsic distance on $M$. By the proof of Theorem (2.2) there is an $r_{0}>0$
such that there is no critical points $x\in
M\setminus\varphi^{-1}(B_{N}(r_{0}))$ for $\rho_{N}\circ\varphi$, where
$B_{N}(r_{0})$ is the geodesic ball in $N$ centered at $p$ with radius
$r_{0}$. Let $r>r_{0}$ and let $D_{r}=\varphi^{-1}(B_{N}(r))$ be an extrinsic
ball. Since $\varphi$ is proper we have that $D_{t}$ is precompact with
boundary $\partial D_{t}$ that we may suppose to be smooth for any $t>r_{0}$
by using the regular set theorem. Let $v:B(r)\to\mathbb{R}$ be a positive
first eigenfunction of the geodesic ball $B(r)$ of radius $r$ in the
$l$-dimensional Euclidean space $\mathbb{R}^{l}$, where $l$ is to be
determined. The function $v$ is radial, i.e. $v(x)=v(|x|)$, and satisfies the
following differential equation,

(5.2)
$v^{\prime\prime}(t)+(l-1)\,\frac{v^{\prime}(t)}{t}\,+\lambda_{1}(r)v(t)=0,\,\,\forall\,t\in[0,r].$

With initial data $v(0)=1$, $v^{\prime}(0)=0$. Moreover, $v^{\prime}(t)<0$ for
all $t\in(0,r]$. Where $\lambda_{1}(r)$ is the first Dirichlet eigenvalue of
the geodesic ball $B(r)\subset\mathbb{R}^{l}$ with radius $r$. Define
$\tilde{v}:B_{N}(r)\to\mathbb{R}$ by $\tilde{v}(y)=v\circ\rho_{N}(y)$ and
$f:D_{r}\to\mathbb{R}$ by $f(x)=\tilde{v}\circ\varphi(x)$. By Barta’s Theorem
we have $\lambda_{1}(D_{r})\leq\sup_{D_{r}}(-\triangle f/f)$. The Laplacian
$\triangle f$ at a point $x\in M$ is given by

$\displaystyle\triangle_{M}f(x)$ $\displaystyle=$
$\displaystyle[\displaystyle\sum_{i=1}^{m}\textrm{Hess}\,\tilde{v}(e_{i},e_{i})+\langle\operatorname{\nabla}\tilde{v},\vec{H}\rangle](\varphi(x))$
$\displaystyle=$
$\displaystyle\displaystyle\sum_{i=1}^{m}\left[{v}^{\prime\prime}(\rho_{N})\langle\operatorname{\nabla}\rho_{N},e_{i}\rangle^{2}+v^{\prime}(\rho_{N})\,\textrm{Hess}\,\rho_{N}(e_{i},e_{i})\right]+v^{\prime}(\rho)\langle\operatorname{\nabla}\rho_{N},\vec{H}\rangle$

Where $\textrm{Hess}\,\tilde{v}$ is the Hessian of $\tilde{v}$ in the metric
of $N$ and $\\{e_{i}\\}_{i=1}^{m}$ is an orthonormal basis for $T_{x}M$ where
we made the identification $d\varphi(e_{i})=e_{i}$. We are going to give an
upper bound for $(-\triangle f/f)$ on $\varphi^{-1}(B_{N}(r))$. Let
$x\in\varphi^{-1}(B_{N}(r))$ and choose an orthonormal basis
$\\{e_{1},...,e_{m}\\}$ for $T_{x}M$ such that $\\{e_{2},\ldots,e_{m}\\}$ are
tangent to the distance sphere $\partial B_{N}(R(x))$ of radius
$R(x)=\rho_{N}(\varphi(x))$ and $e_{1}=\frac{\nabla R}{|\nabla R|}$. To
simplify the notation set $t=\rho_{N}(\varphi(x))$, $\triangle_{M}=\triangle$.
Then

$\displaystyle\triangle f(x)$ $\displaystyle=$
$\displaystyle\displaystyle\sum_{i=1}^{m}\left[{v}^{\prime\prime}(t)\langle\operatorname{\nabla}\rho_{N},e_{i}\rangle^{2}+v^{\prime}(t)\,\textrm{Hess}\,\rho_{N}(e_{i},e_{i})\right]+v^{\prime}(t)\langle\operatorname{\nabla}\rho_{N},\vec{H}\rangle$
$\displaystyle=$
$\displaystyle{v}^{\prime\prime}(t)\langle\operatorname{\nabla}\rho_{N},\frac{\nabla
R}{|\nabla R|}\rangle^{2}+v^{\prime}(t)\,\textrm{Hess}\,\rho_{N}(\frac{\nabla
R}{|\nabla R|},\frac{\nabla R}{|\nabla R|})$
$\displaystyle+\displaystyle\sum_{i=2}^{m}v^{\prime}(t)\,\textrm{Hess}\,\rho_{N}(e_{i},e_{i})+v^{\prime}(t)\langle\operatorname{\nabla}\rho_{N},\vec{H}\rangle$

Thus from (5)

$\displaystyle-\frac{\triangle f}{f}(x)$ $\displaystyle=$
$\displaystyle-\frac{v^{\prime\prime}}{v}(t)\langle\operatorname{\nabla}\rho_{N},\frac{\nabla
R}{|\nabla
R|}\rangle^{2}-\frac{v^{\prime}}{v}(t)\,\textrm{Hess}\,\rho_{N}(\frac{\nabla
R}{|\nabla R|},\frac{\nabla R}{|\nabla R|})$
$\displaystyle-\displaystyle\sum_{i=2}^{m}\frac{v^{\prime}}{v}(t)\,\textrm{Hess}\,\rho_{N}(e_{i},e_{i})-\frac{v}{v}^{\prime}(t)\langle\operatorname{\nabla}\rho_{N},\vec{H}\rangle$

The equation (5.2) says that

$-\frac{v^{\prime\prime}}{v}(t)=(l-1)\frac{v^{\prime}}{t\,v}+\lambda_{1}(r)$

By the Hessian Comparison Theorem and the fact $v^{\prime}/v\leq 0$ we have
from equation (5) the following inequality

$\displaystyle-\frac{\triangle f}{f}(x)$ $\displaystyle\leq$
$\displaystyle\lambda_{1}(r)\left(1-|\nabla^{\perp}R|^{2}\right)$
$\displaystyle-\frac{v^{\prime}}{tv}\left[\frac{t\,h^{\prime}}{h}(m-|\nabla^{\perp}R|^{2})-(l-1)|\nabla
R|^{2}+t|\vec{H}|\right]$ $\displaystyle\leq$ $\displaystyle\lambda_{1}(r)$
$\displaystyle-\frac{v^{\prime}}{tv}\left[\frac{t\,h^{\prime}}{h}m-(l-1)|\nabla
R|^{2}+t|\vec{H}|\right]$

where $h$ is the solution of the following problem

(5.8) $\displaystyle\left\\{\begin{array}[]{l}h^{\prime\prime}+Bh=0\\\
h(0)=0,h^{\prime}(0)=1\\\ \end{array}\right.$

Now, to bound $t\frac{h^{\prime}}{h}$ we will make use of the following lemma

######  Lemma 5.2.

Let $h\in C^{\infty}[0,\infty)$ be a positive function with $h(0)=0$ and
$h^{\prime}(0)=1$. Suppose

$\frac{h^{\prime\prime}}{h}(t)\leq\frac{2}{t^{2}},\quad\forall t>0.$

Then

$t\frac{h^{\prime}}{h}(t)\leq 2.$

###### Proof.

Observe that the function $h^{\prime}(t)t^{2}-2h(t)t$ is a decreasing function
on $t$ because

(5.9)
$\frac{d}{dt}\left(h^{\prime}(t)t^{2}-2h(t)t\right)=h^{\prime\prime}(t)t^{2}-2h(t)=h(t)t^{2}\left(\frac{h^{\prime\prime}(t)}{h(t)}-\frac{2}{t^{2}}\right)\leq
0.$

Hence for any $t>0$

(5.10) $h^{\prime}(t)t^{2}-2h(t)t\leq
h^{\prime}(t_{0})t_{0}^{2}-2h(t_{0})t_{0}\leq h^{\prime}(t_{0})t_{0}^{2},$

for any $t_{0}<t$. Then,

(5.11)
$\frac{h^{\prime}(t)t}{h(t)}\leq\frac{h^{\prime}(t_{0})t_{0}^{2}}{h(t)t}+2,$

Now letting $t_{0}$ tend to $0$ we obtain the desired upper bound. ∎

By using the above lemma in inequality (5),

(5.12) $\displaystyle-\frac{\triangle f}{f}(x)$ $\displaystyle\leq$
$\displaystyle\lambda_{1}(r)-\frac{v^{\prime}}{tv}\left[2m-(l-1)|\nabla
R|^{2}+t|\vec{H}|\right]$

Since the immersion is tamed we have that there exists $t_{c}$ such that for
any $c\in(a(M),1)$

(5.13) $R(x)\|\alpha\|(x)\leq c,\quad\forall x\in M\setminus D_{t_{c}}.$

We are going to split the prove in two cases

### Case I

The point $x\in D_{r}$ belongs to $M\setminus D_{t_{c}}$

Since we are assuming that $x\in M\setminus D_{t_{c}}$, then by using
inequality (5.12)

(5.14) $\displaystyle-\frac{\triangle f}{f}(x)\leq$
$\displaystyle\lambda_{1}(r)-\frac{v^{\prime}}{tv}\left[2m-(l-1)(1-\Lambda_{c}(t_{c})^{2})+c\right].$

Because $t|\vec{H}|\leq t|\alpha|\leq c$ and we have used the monotonocity of
the $\Lambda_{c}$ function given in definition (3.9), see also inequality
(3.8). Since the above inequality is true for any $\mathbb{N}\ni l\geq 1$, we
can choose $l$ large enough in such a way that

$2m-(l-1)(1-\Lambda_{c}(t_{c})^{2})+c\leq 0.$

Hence,

(5.15) $-\frac{\triangle f}{f}(x)\leq\lambda_{1}(r),$

for any $x\in M\setminus D_{t_{c}}$.

### Case II

The point $x\in D_{t}$ belongs to $D_{t_{c}}$.

Since $D_{t_{c}}$ is compact, let us set

(5.16) $H_{0}:=\max_{x\in D_{t_{c}}}R(x)|\vec{H}|.$

By using inequality (5.12),

(5.17) $\displaystyle-\frac{\triangle f}{f}(x)$ $\displaystyle\leq$
$\displaystyle\lambda_{1}(r)-\frac{v^{\prime}}{tv}\left[2m+H_{0}\right]$

We need the following technical lemma.

######  Lemma 5.3.

Let $v$ be the function satisfying (5.2). Then,

$-v^{\prime}(t)/t\leq\lambda_{1}(r)$

for all $t\in[0,r]$.

###### Proof.

Consider the function $\gamma:[0,r]\to\mathbb{R}$ given by
$\gamma(t)=\lambda_{1}(r)\cdot t+v^{\prime}(t)$. We know that $v(0)=1$,
$v^{\prime}(0)=0$ and $v^{\prime}(t)\leq 0$ besides $v$ satisfies equation
(5.2). Observe that

$0=v^{\prime\prime}(t)+(l-1)v^{\prime}+\lambda_{1}(r)v\leq
v^{\prime\prime}+\lambda_{1}(r).$

Thus $v^{\prime\prime}\geq-\lambda_{1}(r)$ and
$\gamma^{\prime}(t)=\lambda_{1}(r)+v^{\prime\prime}\geq 0$. Since
$\gamma(0)=0$ we have $\gamma(t)=\lambda_{1}(r)t+v^{\prime}(t)\geq 0$. This
proves the lemma.∎

Since $v$ is a non-increasing positive function we have $v(t)\geq v(t_{c})$.
Applying the inequality (5.17) we obtain

(5.18) $\displaystyle-\frac{\triangle f}{f}(x)$ $\displaystyle\leq$
$\displaystyle\lambda_{1}(r)\left[1+\frac{1}{v(t_{c})}\left(2m+H_{0}\right)\right].$

Thus, finally from case I and Case II, we know that for all
$x\in\varphi^{-1}(B_{N}(r))$

$\displaystyle-(\triangle f/f)(x)$ $\displaystyle\leq$
$\displaystyle\max\left\\{1,1+\frac{1}{v(t_{c})}\left(2m+H_{0}\right)\right\\}\cdot\lambda_{1}(r)$
$\displaystyle=$
$\displaystyle\left[1+\frac{1}{v(t_{c})}\left(2m+H_{0}\right)\right]\cdot\lambda_{1}(r)$

Then by Barta’s Theorem

$\lambda_{1}(D_{r})\leq\left[1+\frac{1}{v(t_{c})}\left(2m+H_{0}\right)\right]\cdot\lambda_{1}(r)$

Observe that $\left[1+\frac{1}{v(t_{c})}\left(2m+H_{0}\right)\right]$ does not
depend on $r$. So letting $r\to\infty$ we have

$\lambda^{\ast}(M)\leq\left[1+\frac{1}{v(t_{c})}\left(2m+H_{0}\right)\right]\cdot\lambda^{\ast}(\mathbb{R}^{l})=0.$

And this finishes the proof of the theorem.∎

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