# Stable anisotropic minimal hypersurfaces in $\mathbf{R}^{4}$

Otis Chodosh  Department of Mathematics, Stanford University, Building 380,
Stanford, CA 94305, USA<EMAIL_ADDRESS>and  Chao Li  Courant
Institute, New York University, 251 Mercer St, New York, NY 10012, USA
<EMAIL_ADDRESS>
###### Abstract.

We show that a complete, two-sided, stable immersed anisotropic minimal
hypersurface in $\mathbf{R}^{4}$ has intrinsic cubic volume growth, provided
the parametric elliptic integral is $C^{2}$-close to the area functional. We
also obtain an interior volume upper bound for stable anisotropic minimal
hypersurfaces in the unit ball. We can estimate the constants explicitly in
all of our results.

In particular, this paper gives an alternative proof of our recent stable
Bernstein theorem for minimal hypersurfaces in $\mathbf{R}^{4}$. The new proof
is more closely related to techniques from the study of strictly positive
scalar curvature.

##  1\. Introduction

Consider $\Phi:\mathbf{R}^{n+1}\setminus\\{0\\}\to(0,\infty)$ a
$1$-homogeneous $C^{3}_{\textrm{loc}}$ function (i.e., $\Phi(sv)=s\Phi(v)$ for
$s>0$). For $M^{n}\to\mathbf{R}^{n+1}$ a two-sided immersion (with chosen unit
normal field $\nu(x)$), we can define the _anisotropic area functional_

$\mathbf{\Phi}(M)=\int_{M}\Phi(\nu(x))\,d\mu.$

Surfaces minimizing the $\mathbf{\Phi}$-functional arise as the equilibrium
shape of crystalline111We note that in the crystalline setting $\Phi$ is
usually only Lipschitz continuous. materials, as well as scaling limits of
Ising and percolation models (see [7, Chapter 5]).

We say that $M$ is _$\mathbf{\Phi}$ -stationary_ if
$\tfrac{d}{dt}\big{|}_{t=0}\mathbf{\Phi}(M_{t})=0$ for all compactly supported
variations of $M$ (fixing $\partial M$) and that $M$ is _$\mathbf{\Phi}$
-stable_ if in addition
$\tfrac{d^{2}}{dt^{2}}\big{|}_{t=0}\mathbf{\Phi}(M_{t})\geq 0$ for such
variations. Note that if $\Phi(v)=|v|$, $\mathbf{\Phi}$ reduces to the
$n$-dimensional area functional and a $\mathbf{\Phi}$-stable hypersurface is
known as a _stable minimal hypersurface_. We say that $\mathbf{\Phi}$ is
_elliptic_ if the $\Phi$-unit ball $\Phi^{-1}((0,1])\cup\\{0\\}$ is uniformly
convex.

This article is motivated by the following questions:

######  Question 1.

For an anisotropic elliptic functional $\mathbf{\Phi}$, is the flat hyperplane
$\mathbf{R}^{n}\subset\mathbf{R}^{n+1}$ the only complete two-sided
$\mathbf{\Phi}$-stationary and stable immersion in $\mathbf{R}^{n+1}$?

######  Question 2.

If $M^{n}\to\mathbf{R}^{n+1}$ is a complete two-sided
$\mathbf{\Phi}$-stationary and stable immersion (for some anisotropic elliptic
functional $\mathbf{\Phi}$) does $M$ satisfy the intrinsic polynomial volume
growth condition $\operatorname{Vol}(B_{M}(p,\rho))\leq C\rho^{n}$?

By a well-known blowup argument, an affirmative answer to Question 1 yields a
priori interior curvature estimates for $\mathbf{\Phi}$-stable immersions with
boundary, and even for stable immersion with respect to a _parametric elliptic
integrand_ (where $\Phi$ is allowed to also depend on $x$). We also note that
for minimal surfaces one can derive _lower_ polynomial growth bounds (both
intrinsic and extrinsic), but for general $\mathbf{\Phi}$-stationary surfaces
no monotonicity type formula is known, cf. [3, 15, 14]. (On the other hand,
stability can be used to derive a _lower_ volume growth estimate; see
Corollary 17 and [15].)

For the area functional, Question 1 (and thus Question 2) has been completely
resolved in the affirmative when $n=2$ (independently) by Fischer-
Colbrie–Schoen, do Carmo–Peng, and Pogorelov [21, 17, 38] (see also [41]) and
recently when $n=3$ by the authors [9]. In particular, we recall the result of
Pogorelov (yielding a localized volume growth estimate)

######  Theorem 3 ([38], cf. [50, Lemma 34], [35, Theorem 2]).

Suppose that $M^{2}\to\mathbf{R}^{3}$ is a stable minimal immersion so that
the intrinsic ball $B_{M}(p,R)\subset M$ has compact closure in $M$ and is
topologically a disk. Then

$|B_{M}(p,\rho)|\leq\frac{4}{3}\pi\rho^{2}.$

On the other hand, Questions 1 and 2 remains open (even for the area
functional) for $n=4,5,6$. There exist non-flat stable minimal hypersurfaces
(area minimizers) in $\mathbf{R}^{8}$ and beyond [4, 27] (thus answering
Question 1 in the negative), but all known examples satisfy the conclusion of
Question 2. Note that Schoen–Simon–Yau [40] (cf. [46, 42, 51]) have shown that
when $n\leq 5$, a complete two-sided stable minimal immersion satisfying the
volume growth condition in Question 2 must be flat.

For arbitrary elliptic functionals, there are non-flat minimizers for $n\geq
3$ [34, 33], but as in the case of area, all known examples satisfy the
intrinsic volume growth condition in Question 2. When $n=2$, Question 1 is
open for general elliptic functionals but is resolved in the affirmative
assuming quadratic area growth (as shown by White [49]) or assuming the
functional is sufficiently $C^{2}$-close222Throughout this article, _$\Phi$ is
$C^{k,\alpha}$-sufficiently close to area_ will mean that
$\|\Phi-1\|_{C^{k,\alpha}(S^{n})}\leq\varepsilon(n)$ for some fixed
$\varepsilon(n)>0$.  to area (as shown by Lin [29]; see also [28, 45]). Still
for $n=2$, Colding–Minicozzi have given a new proof of Theorem 3 that extends
to show that Question 2 holds in the affimative for functionals sufficiently
$C^{2}$-close to area. When $n\geq 3$, Question 1 is answered in the negative
by considering the non-flat area minimizing solutions constructed by
Mooney–Yang [33] (see also [34, 32]). On the other hand, Winklmann has
resolved Question 1 in the affirmative for $n\leq 5$ under the assumptions
that the functional is sufficiently $C^{4}$-close to area and that the surface
satisfies the growth condition from Question 2.

###  1.1. Main results

In this article, we consider the volume growth problem (Question 2) for
$\mathbf{\Phi}$-stable hypersurfaces $\mathbf{R}^{4}$. In fact, the estimate
we prove here is new even in the case of stable minimal hypersurfaces. As
such, it yields an alternative approach to our recent result [9] (this is
discussed further in Section 2).

We note that all constants in this paper can be given explicitly, see Remark
9.

######  Theorem 4.

Assume that $\Phi$ satisfies

$|v|^{2}\leq D^{2}\Phi(\nu)(v,v)\leq\sqrt{2}|v|^{2},$ (1)

for all $v\in\nu^{\perp}$. Consider $M^{3}\to\mathbf{R}^{4}$ a complete, two-
sided, $\mathbf{\Phi}$-stationary and stable immersion. Suppose $0\in M$, and
$M$ is simply333We note that a standard argument (cf. [21]) shows that if
$M^{3}\to\mathbf{R}^{4}$ is a complete two-sided $\mathbf{\Phi}$-stable
immersion, then so is the immersion from the universal cover. connected. Then
there exist explicit constants $V_{0}=V_{0}(\|\Phi\|_{C^{1}(S^{3})}),Q>0$ such
that

  1. (i)

$|B_{M}(0,\rho)|\leq V_{0}\rho^{3}$, for all $\rho>0$.

  2. (ii)

For each connected component $\Sigma_{0}$ of $\partial B_{M}(0,\rho)$, we have

$\max_{x\in\Sigma_{0}}r(x)\leq Q\min_{x\in\Sigma_{0}}r(x),$

where $r(x):=d_{\mathbf{R}^{4}}(0,x)$.

Note that (1) implies that $\nu\mapsto\Phi(\nu)$ is convex (since
$D^{2}\Phi(\nu)(\nu,\nu)=0$ by $1$-homogeneity). As such, all $\mathbf{\Phi}$
considered in Theorem 4 satisfy the ellipticity condition mentioned
previously.

We note that by combining Theorem 4 with [52], we obtain:

######  Corollary 5.

If $\Phi$ is $C^{4}$-sufficiently close to area, then any two-sided complete
$\mathbf{\Phi}$-stationary and stable immersion is flat.

######  Remark 6.

Although it is not explicitly done in [52], the “sufficiently close”
requirement can be quantified. Alternatively, we note that by combining
Theorem 4 with a contradiction argument in the spirit of [45], Corollary 5
actually holds under the weaker assumption of $C^{2,\alpha}$-closeness (but
with no numerical estimate of the required closeness).

We can also prove a localized version of Theorem 4 more in the spirit of
Pogorelov’s result (cf. Theorem 3). The estimate we prove here is slightly
different, since it considers extrinsic balls, but is an interior444As
observed in [26, §1], the bridge principle for stable minimal surfaces [30]
implies that there cannot be an estimate for the area of a proper stable
minimal immersion $M^{2}\to B_{1}(0)\subset\mathbf{R}^{3}$, even if $M$ is
topologically constrained to be a disk.  estimate. Even for stable minimal
surfaces, we are not aware of such an estimate in $\mathbf{R}^{3}$ with
explicit555Given an area-free curvature estimate (available for minimal
surfaces when $n=2,3$ [41, 9]), one can prove an extrinsic interior Pogorelov
result in the spirit of Theorem 7 by a straightforward contradiction argument
(with no control on the constant). The method used here gives an alternative
proof of this curvature estimate (and extends to certain elliptic integrands)
and yields explicit (and not too large) constants.  constants, cf. Remark 9.

######  Theorem 7.

Suppose that $\Phi$ satisfies (1). Assume $M^{3}\to
B_{1}(0)\subset\mathbf{R}^{4}$ is a proper, two-sided,
$\mathbf{\Phi}$-stationary and stable immersion. Suppose $0\in M$, $M$ is
simply connected, and $\partial M$ is connected. Then there exist explicit
constants $\rho_{0}\in(0,1),V_{1}=V_{1}(\|\Phi\|_{C^{1}(S^{3})})$, such that

$|M_{\rho_{0}}^{*}|\leq V_{1},$

where $M_{\rho_{0}}^{*}$ is the connected component of $M\cap
B_{\mathbf{R}^{4}}(0,\rho_{0})$ that contains $0$.

######  Remark 8.

More generally, we can drop the requirement that $M$ is simply connected and
$\partial M$ is connected. In this case, we have:

$|M_{\rho_{0}}^{*}|\leq V_{1}(b_{1}(M)+E),$

where $E$ is the number of boundary connected components of $M$ and $b_{1}(M)$
is the first Betti number.

######  Remark 9.

One may explicitly compute the constants $V_{0},V_{1}$, $Q$, $\rho_{0}$ as
follows. Let

$\displaystyle c_{0}$ $\displaystyle=\frac{1}{\sqrt{2}-\frac{1}{2}},$
$\displaystyle\lambda$
$\displaystyle=\frac{3}{2}\left(\frac{1}{2}-\frac{3(c_{0}-1)}{8(\frac{1}{\sqrt{2}}-\frac{1}{2})}\right)=\frac{3(5+3\sqrt{2})}{56}\approx
0.495.$

Then we have

$V_{0}=\frac{8\pi
e^{\frac{15\pi}{\lambda}}\|\Phi\|_{C^{1}(S^{3})}}{3\lambda\min_{\nu\in
S^{3}}\Phi(\nu)},\quad Q=e^{\frac{7\pi}{\sqrt{\lambda}}},$

and

$\rho_{0}=e^{-\frac{5\pi}{\sqrt{\lambda}}},\quad
V_{1}=\frac{8\pi\|\Phi\|_{C^{1}(S^{3})}}{3\lambda\min_{\nu\in
S^{3}}\Phi(\nu)}.$

###  1.2. Related work

We recall here some works (beyond those mentioned above) that are related to
this paper. The regularity of hypersurfaces _minimizing_ parametric elliptic
integrands has been studied in several places including [18, 39, 43, 19]. See
also [1, 47] for estimates without the minimizing hypothesis. Existence of
critical points of parametric elliptic integrands has been considered in [48,
49, 13]. Finally, we note that stable solutions for the nonlocal area
functional satisfy an a priori growth estimate (as in Question 2) in all
dimensions [12] (see also [20]).

###  1.3. Notation

We will use the following notation:

  * •

$B_{\mathbf{R}^{n+1}}(0,\rho):=\\{x\in\mathbf{R}^{n+1}:|x|<\rho\\}$.

  * •

$r(x)=\operatorname{dist}_{\mathbf{R}^{n+1}}(0,x)$.

  * •

$M^{n}\to\mathbf{R}^{n+1}$ is an immersion and $g$ the induced Riemannian
metric on $M$.

  * •

$D$ is the connection in $\mathbf{R}^{n+1}$, $\nabla$ is the induced
connection on $M$.

  * •

$\mu$ is the volume form of $g$.

  * •

$B_{M}(0,\rho):=\\{x\in M:\operatorname{dist}_{M,g}(0,x)<\rho\\}$.

  * •

$\nu$ is a choice of unit normal vector field of $M$.

  * •

The shape operator will be written $S=\nabla\nu$, and the second fundamental
form written $A(X,Y)=S(X)\cdot Y$.

  * •

The scalar curvature of $g$ will be denoted by $R$.

  * •

We will use the $\ell^{2}$-norm to define $C^{k}$-norms, i.e.
$\|f\|_{C^{k}}:=(\sum_{j=0}^{k}\|D^{(j)}f\|_{C^{0}}^{2})^{\frac{1}{2}}$.

###  1.4. Organization of the paper

In Section 2 we explain the techniques used in this paper in the special case
of the area functional. The remaining part of the paper contains the details
necessary for the generalization to anisotropic integrands. We begin in
Section 3 with some preliminary results. Section 4 contains a generalization
of the one-ended result for stable minimal hypersurfaces due to Cao–Shen–Zhu
to the case of certain anisotropic integrands. We describe the conformally
changed metric in Section 5 as introduced by Gulliver–Lawson and then combine
these techniques with $\mu$-bubbles to prove the main results in Section 5.
Appendix A contains (well-known) computations of the first and second
variation for elliptic integrands. Appendix B contains an auxiliary result
comparing certain quadratic forms.

###  1.5. Acknowledgements

We are grateful to Fang-Hua Lin and Guido De Philippis for their interest and
for several discussions, as well as Doug Stryker for pointing out a mistake in
an earlier version of the paper, and to the referees for some useful
suggestions. O.C. was supported by an NSF grant (DMS-2016403), a Terman
Fellowship, and a Sloan Fellowship. C.L. was supported by an NSF grant
(DMS-2202343).

##  2\. Volume growth for stable minimal hypersurfaces in $\mathbf{R}^{4}$

In this section, we illustrate how one may use stability to deduce area
estimates for stable minimal immersions $M^{3}\to\mathbf{R}^{4}$. We will
defer certain ancillary results and computation to later sections (where they
were carried out for general $\Phi$-stationary and stable hypersurfaces), and
instead focus on the geometric ideas and consequences.

The main result we will prove here is as follows

######  Theorem 10.

Let $M^{3}\to\mathbf{R}^{4}$ be a complete, two-sided, simply connected,
stable minimal immersion, $0\in M$. Then,

$|B_{M}(0,\rho)|\leq\left(\frac{32\pi}{3}\right)^{\frac{3}{2}}\frac{e^{\frac{30\pi}{\sqrt{3}}}}{6\sqrt{\pi}}\rho^{3}$

for all $\rho\geq 0$.

Combined with the work of Schoen–Simon–Yau [40], this yields a new proof of
our recent result [9]:

######  Corollary 11.

Any complete, two-sided, stable minimal immersion $M^{3}\to\mathbf{R}^{4}$ is
flat.

In fact, we have the following localized volume estimate in the spirit of
Theorem 3.

######  Theorem 12.

Let $M^{3}\to\mathbf{R}^{4}$ be a two-sided, simply connected, stable minimal
immersion, with $0\in M$, $\partial M$ connected, and $M\to
B_{\mathbf{R}^{4}}(0,1)$ proper. Then,

$|M_{\rho_{0}}^{*}|\leq\left(\frac{32\pi}{3}\right)^{\frac{3}{2}}\frac{1}{6\sqrt{\pi}}$

where $M_{\rho_{0}}^{*}$ is the connected component of $M\cap
B_{\mathbf{R}^{4}}(0,r_{0})$ that contains $0$ and
$\rho_{0}=e^{-\frac{10\pi}{\sqrt{3}}}$.

###### Proof of Theorem 10.

The first step is to consider a particular conformal deformation of $(M,g)$.
On $M\setminus\\{0\\}$, consider the conformally deformed metric
$\tilde{g}=r^{-2}g$ (where we recall that $r$ is the _Euclidean_ distance to
the origin and $g$ is the induced metric on $M$). We use
$\tilde{\nabla},\tilde{\mu},\tilde{\Delta}$ to denote the covariant
derivative, the volume form and the Laplacian with respect to $\tilde{g}$,
respectively. This conformal change was first used by Gulliver–Lawson [26] to
study isolated singularities for minimal hypersurfaces in $\mathbf{R}^{n+1}$.

######  Remark 13.

The relevance of the Gulliver–Lawson conformal deformation is a key insight in
our work. Indeed, this allows us to apply tools from the study of _strictly_
positive scalar curvature (cf. Remark 14). Our previous proof of Corollary 11
used tools from _non-negative_ scalar curvature (cf. [36, 37]).666Added in
proof: some time after this paper appeared, Catino–Mastrolia–Roncoroni found a
third proof of Corollary 11, based on a surprising connection between
stability and _non-negative Bakry–Émery Ricci curvature_ [22].

The computations in this part work for minimal immersions
$M^{n}\to\mathbf{R}^{n+1}$ whenever $n\geq 3$. For $\lambda\in\mathbf{R}$,
$\varphi\in C_{0}^{1}(M\setminus\\{0\\})$, consider the quadratic form

$\mathcal{Q}(\varphi):=\int_{M}\left(|\tilde{\nabla}\varphi|_{\tilde{g}}^{2}+(\tfrac{1}{2}\tilde{R}-\lambda)\varphi^{2}\right)d\tilde{\mu},$

where $\tilde{R}$ is the scalar curvature of $\tilde{g}$. One computes (see
Section 5 for details) that

$\displaystyle\mathcal{Q}(r^{\frac{n-2}{2}}\varphi)$
$\displaystyle=\int_{M}\left(r^{2}|\nabla(r^{\frac{n-2}{2}}\varphi)|^{2}+(\tfrac{1}{2}\tilde{R}-\lambda)r^{n-2}\varphi^{2}\right)r^{-n}d\mu$
$\displaystyle=\int_{M}\left(|\nabla\varphi|^{2}+\tfrac{1}{2}R\varphi^{2}+\left(\frac{n}{2}\left(n-\frac{n+2}{2}|\nabla
r|^{2}\right)-\lambda\right)r^{-2}\varphi^{2}\right)d\mu$
$\displaystyle\geq\int_{M}\left(|\nabla\varphi|^{2}+\tfrac{1}{2}R\varphi^{2}+\left(\frac{n(n-2)}{4}-\lambda\right)\right)d\mu.$

By the (traced) Gauss equations, minimality of $M$ implies that
$|A_{M}|^{2}=-R_{g}$. Thus, we can use stability of $M$ to conclude

$\int_{M}\left(|\nabla\varphi|^{2}-|A|^{2}\varphi^{2}\right)d\mu\geq
0\quad\Rightarrow\quad\int_{M}\left(|\nabla\varphi|^{2}+\frac{1}{2}R\varphi^{2}\right)d\mu\geq
0,$

for all $\varphi\in C_{0}^{1}(M)$. Note that we have used the fact that the
scalar curvature of a minimal hypersurface in $\mathbf{R}^{n+1}$ has $R\leq 0$
and that $\tfrac{1}{2}<1$. In particular, choosing $\lambda=\frac{n(n-2)}{4}$
above, we find that $\mathcal{Q}(\varphi)\geq 0$ for any $\varphi\in
C_{0}^{1}(M\setminus\\{0\\})$. Using [21, Theorem 1], there exists $u\in
C^{\infty}(M\setminus\\{0\\})$, $u>0$ in the interior of $M\setminus\\{0\\}$,
such that

$\tilde{\Delta}u\leq-\frac{1}{2}\left(\frac{n(n-2)}{2}-\tilde{R}\right)u.$ (2)

We note that (2) is an integral form of _strictly positive_ scalar curvature.

In the second step, we restrict to the case of $n=3$. We use warped
$\mu$-bubbles to derive geometric inequalities for $3$-manifolds $(N^{3},g)$
admitting a positive function $u$ with (2).

######  Remark 14.

The $\mu$-bubble technique was first used by Gromov [23, Section
5$\tfrac{5}{6}$] (see also [24]). Warped $\mu$-bubbles have previously been
combined with minimal hypersurface techniques to study problems in scalar
curvature and in minimal surfaces, see, e.g. [8, 25, 10, 11, 55, 54].
Precisely, suppose $n=3$ and $\partial N\neq\emptyset$. Then there exists an
open set $\Omega$ containing $\partial N$, $\Omega\subset
B_{\frac{10\pi}{\sqrt{3}}}(\partial N)$, such that each connected component of
$\partial\Omega\setminus\partial N$ is a $2$-sphere with area at most
$\frac{32\pi}{3}$ and intrinsic diameter at most $\frac{4\pi}{\sqrt{3}}$ (see
Lemma 24).

Fix $\rho>0$. By [6], $M\setminus
B_{M}\left(0,e^{\frac{10\pi}{\sqrt{3}}}\rho\right)$ has only one unbounded
component $E$. Denote by $M^{\prime}=M\setminus E$. We apply Remark 14 to
$N=M^{\prime}$, and find $M_{0}\subset M^{\prime}$ with
$\operatorname{dist}_{\tilde{g}}(\partial M_{0},\partial
M^{\prime})\leq\frac{10\pi}{\sqrt{3}}$. The topological assumptions on $M$
force $\partial M_{0}$ to be connected, so $|\partial
M_{0}|_{\tilde{g}}\leq\tfrac{32\pi}{3}$ and $\partial M_{0}$ has intrinsic
diameter $\leq\frac{4\pi}{\sqrt{3}}$. By comparing $g$-distance with
$\tilde{g}$-distance (see (2) in Lemma 25), we find that

$B_{M}(0,\rho)\subset M_{0}\subset B_{M}(0,e^{\frac{10\pi}{\sqrt{3}}}\rho).$

In particular, bounding intrinsic distance by extrinsic distance, we see that
$\sup_{\partial M_{0}}r(x)\leq e^{\frac{10\pi}{\sqrt{3}}}\rho$. Thus, we have

$|B_{M}(0,\rho)|\leq|M_{0}|\leq\frac{1}{6\sqrt{\pi}}|\partial
M_{0}|_{g}^{\frac{3}{2}}\leq\frac{1}{6\sqrt{\pi}}(e^{\frac{10\pi}{\sqrt{3}}}\rho)^{3}|\partial
M_{0}|_{\tilde{g}}^{\frac{3}{2}}\leq\left(\frac{32\pi}{3}\right)^{\frac{3}{2}}\frac{e^{\frac{30\pi}{\sqrt{3}}}}{6\sqrt{\pi}}\rho^{3},$

where in the second step we have used the isoperimetric inequality for minimal
hypersurfaces in Euclidean spaces due to Brendle [5] (cf. [31]). This
completes the proof. ∎

We now consider the requisite changes needed to prove the local result:

###### Proof of Theorem 12.

In the case where $M$ is properly immersed in $B_{1}(0)\subset\mathbf{R}^{4}$,
we proceed similarly as before, and obtain a region $M^{\prime}$ such that
$\operatorname{dist}_{\tilde{g}}(\partial M^{\prime},\partial
B_{1}(0))\leq\frac{10\pi}{\sqrt{3}}$, $|\partial
M^{\prime}|_{\tilde{g}}\leq\frac{32\pi}{3}$, and $\partial M^{\prime}$ is
connected. Again, using Lemma 25, we conclude that

$M_{\rho_{0}}^{*}\subset M^{\prime},$

where $\rho_{0}=e^{-\frac{10\pi}{\sqrt{3}}}$, and $M_{\rho_{0}}^{*}$ is the
connected component of $M\cap B_{\mathbf{R}^{4}}(0,\rho_{0})$ that contains
$0$. Using [5] as above,

$|M_{r_{0}}^{*}|\leq|M^{\prime}|\leq\frac{1}{6\sqrt{\pi}}|\partial
M^{\prime}|_{g}^{\frac{3}{2}}\leq\frac{1}{6\sqrt{\pi}}|\partial
M^{\prime}|_{\tilde{g}}^{\frac{3}{2}}\leq\left(\frac{32\pi}{3}\right)^{\frac{3}{2}}\frac{1}{6\sqrt{\pi}}.$

This completes the proof. ∎

##  3\. Preliminaries on anisotropic integrands

We now consider a general anisotropic elliptic integrand. For
$M^{n}\to\mathbf{R}^{n+1}$ two-sided immersion, we can set

$\mathbf{\Phi}(M)=\int_{M}\Phi(\nu(x))\,d\mu.$

In this section we discuss the first and second variation formulae, as well as
some important consequences to be used later.

###  3.1. First variation

Recall that $M$ is $\mathbf{\Phi}$-stationary means that
$\tfrac{d}{dt}\big{|}_{t=0}\mathbf{\Phi}(M_{t})=0$ for all compactly supported
variations $M_{t}$ fixing $\partial M$. By (21), (22), (23) this is equivalent
to

$\operatorname{div}_{M}(D\Phi(\nu))=\operatorname{tr}_{M}(\Psi(\nu)S_{M})=0,$

which we can interpret as vanishing of the $\mathbf{\Phi}$-mean curvature.
Here, $\Psi(\nu):T\mathbf{R}^{n+1}\to T\mathbf{R}^{n+1}$ is defined by
$\Psi(\nu):X\mapsto D^{2}\Phi(\nu)[X,\cdot]$ and $S_{M}$ is the shape operator
of $M$.

By the calculation in Section A.3, we find that if $M$ is
$\mathbf{\Phi}$-stationary, then for any compactly supported (but not
necessarily normal) vector field $X$ along $\Sigma$, we have

$\int_{M}\Phi(\nu)\operatorname{div}_{M}X+D_{D\Phi(\nu)^{T}}X\cdot\nu=\int_{\partial
M}\Phi(\nu)X\cdot\eta+(X\cdot\nu)D\Phi(\nu)\cdot\eta.$ (3)

By plugging the position vector field into (3), we obtain the following
isoperimetric type inequality.

######  Corollary 15.

Suppose $M^{n}\to\mathbf{R}^{n+1}$ is $\mathbf{\Phi}$-stationary, and the
image of $\partial M$ is contained in $B_{\mathbf{R}^{n+1}}(0,\rho)$ for some
$\rho>0$. Then

$|M|\leq\frac{\rho\|\Phi\|_{C^{1}(S^{n})}}{n\cdot\min_{\nu\in
S^{n}}\Phi(\nu)}|\partial M|.$

###### Proof.

Recall that $r(x)=\operatorname{dist}_{\mathbf{R}^{n+1}}(x,0)$. Plug
$X=\sum_{i=1}^{n+1}x_{i}e_{i}$, the position vector field in
$\mathbf{R}^{n+1}$, into (3). Then $\operatorname{div}_{M}X=n$, and

$D_{D\Phi(\nu)^{T}}X\cdot\nu=\sum_{i}(D_{D\Phi(\nu)^{T}}x_{i})(e_{i}\cdot\nu)=\sum_{i}(D\Phi(\nu)^{T}\cdot
e_{i})(e_{i}\cdot\nu)=D\Phi(\nu)^{T}\cdot\nu=0.$

On the other hand, $|X(x)|\leq r(x)$. Thus, we find (using $\nu,\eta$
orthonormal)

$\int_{M}n\Phi(\nu)\leq\int_{\partial
M}\|\Phi\|_{C^{1}(S^{n})}|X|\leq\rho\|\Phi\|_{C^{1}(S^{n})}|\partial M|.$

This completes the proof. ∎

The next lemma generalizes the traced Gauss equation $R=-|A|^{2}$ (valid for
minimal hypersurfaces) to the case of $\mathbf{\Phi}$-stationary hypersurfaces
in $\mathbf{R}^{4}$, under the assumption that $D^{2}\Phi(\nu)$ is
sufficiently pinched.

######  Lemma 16.

Suppose $\Phi$ satisfies (1) and $M^{3}\to\mathbf{R}^{4}$ is
$\mathbf{\Phi}$-stationary. Then at each point on $M$, the induced scalar
curvature satisfies $R\leq 0$ and

$-R\leq|A|^{2}\leq-c_{0}R,$ (4)

where

$c_{0}=\frac{1}{\sqrt{2}-\frac{1}{2}}\approx 1.09$

###### Proof.

Recall that $\mathbf{\Phi}$-stationarity can be written as
$\operatorname{tr}_{M}(\Psi(\nu)S_{M})=0$. Diagonalizing $A_{M}$ at a given
point, write $k_{i}$ for the principal curvatures of $M$ and $e_{i}$ for
corresponding principal directions. Thus, $\mathbf{\Phi}$-stationarity can be
written as

$0=\sum_{i=1}^{3}a_{i}k_{i}$

where $a_{i}=D^{2}\Phi(\nu)[e_{i},e_{i}]$. Without loss of generality, we can
assume that $a_{1}\leq a_{2}\leq a_{3}$. Note that the pinching assumption (1)
yields

$1\leq a_{1}\leq a_{2}\leq a_{3}\leq\sqrt{2}.$

We have $|A|^{2}=\sum k_{i}^{2}$, $R=2\sum_{i<j}k_{i}k_{j}$. Writing
$k_{3}=-\frac{a_{1}k_{1}+a_{2}k_{2}}{a_{3}}$, we have

$|A|^{2}=Q_{1}(k_{1},k_{2}):=\frac{a_{1}^{2}+a_{3}^{2}}{a_{3}^{2}}k_{1}^{2}+\frac{2a_{1}a_{2}}{a_{3}^{2}}k_{1}k_{2}+\frac{a_{2}^{2}+a_{3}^{2}}{a_{3}^{2}}k_{2}^{2},$
$-R=Q_{2}(k_{1},k_{2}):=\frac{2a_{1}}{a_{3}}k_{1}^{2}+\frac{2(a_{1}+a_{2}-a_{3})}{a_{3}}k_{1}k_{2}+\frac{2a_{2}}{a_{3}}k_{2}^{2}.$

By the Gauss equation, we have $R+|A|^{2}=H^{2}\geq 0$, and hence
$|A|^{2}\geq-R$. Moreover, whenever $(a_{1}+a_{2}-a_{3})^{2}<4a_{1}a_{2}$
(which is guaranteed by, for instance, $a_{3}<4a_{1}$), $Q_{2}$ is a positive
definite quadratic form, and hence $-R$ is nonnegative. Given that
$\frac{a_{3}}{a_{1}},\frac{a_{3}}{a_{2}}\in[1,\sqrt{2}]$, (4) follows from
Appendix B. ∎

###  3.2. Second variation

Suppose now that $M^{n}\to\mathbf{R}^{n+1}$ is $\mathbf{\Phi}$-stationary and
stable. In Section A.2 we derive the following second variation formula.

$\frac{d^{2}}{dt^{2}}\bigg{|}_{t=0}\mathbf{\Phi}(M_{t})=\int_{M}\left\langle\nabla
u,\Psi(\nu)\nabla
u\right\rangle-\operatorname{tr}_{M}\left(\Psi(\nu)S_{M}^{2}\right)u^{2},$ (5)

where $u\nu$ is the variation vector field. Note that stability and (5)
implies that

$\int_{M}|\nabla u|^{2}-\Lambda|A|^{2}u^{2}\geq 0$ (6)

for all $u\in C_{c}^{1}(M\setminus\partial M)$. Here, $\Lambda$ depends on
ellipticity of $\mathbf{\Phi}$. It is important to observe that if $\Phi$
satisfies (1) then $\Lambda\geq\tfrac{1}{\sqrt{2}}$ and in particular

$\int_{M}|\nabla u|^{2}-\frac{1}{\sqrt{2}}|A|^{2}u^{2}\geq 0$ (7)

for all $u\in C_{c}^{1}(M\setminus\partial M)$.

###  3.3. Sobolev inequality and its consequences

In this section, we assume that $n\geq 3$, $M^{n}$ is a two-sided
$\mathbf{\Phi}$-stationary and stable hypersurface immersed in
$\mathbf{R}^{n+1}$, where $\mathbf{\Phi}$ is a general anisotropic elliptic
integral. The Michael-Simon Sobolev inequality [31] implies that for any $f\in
C_{c}^{1}(M)$,

$C_{n}\left(\int_{M}|f|^{\frac{n}{n-1}}\right)^{\frac{n-1}{n}}\leq\int_{M}|\nabla
f|+|fH|.$

(See also [5].)

Replacing $f$ by $f^{\frac{2(n-1)}{n-2}}$, we find:

$C_{n}\left(\int_{M}|f|^{\frac{2n}{n-2}}\right)^{\frac{n-1}{n}}\leq\int_{M}\frac{2(n-1)}{n-2}|f|^{\frac{n}{n-2}}|\nabla
f|+|f|^{\frac{2(n-1)}{n-2}}|H|.$ (8)

By the Hölder inequality,

$\int_{M}|f|^{\frac{2(n-1)}{n-2}}|H|\leq\left(\int_{M}f^{2}H^{2}\right)^{\frac{1}{2}}\left(\int_{M}|f|^{\frac{2n}{n-2}}\right)^{\frac{1}{2}}$

The $\Phi$-stability inequality implies

$\int_{M}f^{2}H^{2}\leq n\int_{M}f^{2}|A|^{2}\leq C(\Phi)\int_{M}|\nabla
f|^{2}.$

Now we use the Hölder inequality on the first term of the right hand of (8)
and conclude the following Sobolev inequality:

$\left(\int_{M}|f|^{\frac{2n}{n-2}}\right)^{\frac{n-2}{n}}\leq
C(n,\Phi)\int_{M}|\nabla f|^{2}.$ (9)

######  Corollary 17.

Suppose $M^{n}\to\mathbf{R}^{n+1}$ is $\mathbf{\Phi}$-stationary and stable.
Assume that $B_{M}(p,\rho)\subset M$ has compact closure. Then,

$|B_{M}(p,\rho/2)|\geq C(n,\Phi)\rho^{n}.$

###### Proof.

For any $u\in C^{1}(M)$ such that $u\geq 0$ and $\Delta u\geq 0$, the Sobolev
inequality (9) and the standard Moser iteration implies that, for any
$\theta\in(0,1)$, $s>0$,

$\sup_{B_{M}(p,\theta\rho)}u\leq
C(n,\theta,\Phi,s)\left(\rho^{-n}\int_{B_{M}(p,\rho)}u^{s}\right)^{1/s}.$

The result follows by taking $u=1$, $s=1$ and $\theta=\tfrac{1}{2}$. ∎

######  Remark 18.

The use of Sobolev inequality for volume lower bound was first used by Allard
[2, 7.5].

######  Corollary 19.

Suppose $M^{n}\to\mathbf{R}^{n+1}$ is two-sided complete,
$\mathbf{\Phi}$-stationary and stable, and $K$ is a compact subset of $M$.
Then each unbounded component of $M\setminus K$ has infinte volume.

###### Proof.

Let $E$ be an unbounded component of $M\setminus K$. Suppose the contrary,
that $|E|<V<\infty$. Choose $\rho$ such that $C(n,\Phi)\rho^{n}>V$. By
completeness, there exists $p\in E$ such that $d_{M}(p,\partial E)>\rho$. Then
we have

$V>|E|>|B_{M}(p,\rho)|>C(n,\Phi)\rho^{n}>V,$

a contradiction. This completes the proof. ∎

Combining (9) and Corollary 19, the same argument as used by Cao–Shen–Zhu [6]
implies the following result:

######  Corollary 20.

If $M^{n}\to\mathbf{R}^{n+1}$ is complete, two-sided
$\mathbf{\Phi}$-stationary and stable immersion with at least two-ends, then
there is a bounded non-constant harmonic function on $M$ with finite Dirichlet
energy.

##  4\. One-endedness

Through this section we assume that $n=3$, $M^{3}\to\mathbf{R}^{4}$ is
$\mathbf{\Phi}$-stationary and stable. By analyzing harmonic functions on $M$,
we will show that $M$ has only one end, if $\mathbf{\Phi}$ satisfies (1)
(following [44, 6]).

######  Lemma 21.

Suppose that $M^{3}$ is a complete, two-sided, $\mathbf{\Phi}$-stationary and
stable immersion in $\mathbf{R}^{4}$, and $u$ is a harmonic function on $M$.
Then

$(\Lambda-\tfrac{1}{\sqrt{2}})\int_{M}\varphi^{2}|A|^{2}|\nabla
u|^{2}+\frac{1}{2}\int_{M}\varphi^{2}|\nabla|\nabla
u||^{2}\leq\int_{M}|\nabla\varphi|^{2}|\nabla u|^{2},$ (10)

for any $\varphi\in C_{0}^{1}(M)$. Here $\Lambda=\Lambda(\Phi)$ is the
constant in (6).

###### Proof.

Fix $p\in M$. Let $k_{i}$ be the principal curvatures, $e_{i}$ be the
corresponding orthonormal principal directions diagonalizing $A_{M}$.

We first show that for any immersed hypersurface $M^{3}$ in $\mathbf{R}^{4}$,
equipped with the induced metric, $p\in M$, and any unit vector $v\in T_{p}M$,
we have

$\operatorname{Ric}(v,v)\geq-\frac{1}{\sqrt{2}}|A|^{2}.$

Write $v=\sum y_{i}e_{i}$. Then $\sum y_{i}^{2}=1$. By the Gauss equation, we
have

$\operatorname{Ric}(e_{i},e_{j})=\sum_{k}\textrm{Rm}(e_{i},e_{k},e_{k},e_{j})=\sum_{k}\left(A(e_{k},e_{k})A(e_{i},e_{j})-A(e_{i},e_{k})A(e_{j},e_{k})\right),$

and thus $\operatorname{Ric}(e_{i},e_{j})=0$ when $i\neq j$, and
$\operatorname{Ric}(e_{i},e_{i})=\sum_{j\neq i}A(e_{i},e_{i})A(e_{j},e_{j})$.
Therefore,

$\operatorname{Ric}(v,v)=\sum_{i}\sum_{j\neq
i}A(e_{j},e_{j})A(e_{i},e_{i})y_{i}^{2}=k_{1}(k_{2}+k_{3})y_{1}^{2}+k_{2}(k_{3}+k_{1})y_{2}^{2}+k_{3}(k_{1}+k_{2})y_{3}^{2}.$

By Cauchy-Schwarz and the AM-GM inequality,

$k_{1}^{2}+k_{2}^{2}+k_{3}^{2}\geq
k_{1}^{2}+\frac{1}{2}(k_{2}+k_{3})^{2}\geq-\sqrt{2}k_{1}(k_{2}+k_{3})\\\
\Rightarrow\quad
k_{1}(k_{2}+k_{3})\geq-\frac{1}{\sqrt{2}}\sum_{i}k_{i}^{2}=-\frac{1}{\sqrt{2}}|A|^{2}.$

Similarly,

$k_{2}(k_{3}+k_{1})\geq-\frac{1}{\sqrt{2}}|A|^{2},\quad
k_{3}(k_{1}+k_{2})\geq-\frac{1}{\sqrt{2}}|A|^{2}.$

Therefore,

$\operatorname{Ric}(v,v)\geq-\frac{1}{\sqrt{2}}|A|^{2}\sum_{i}y_{i}^{2}=-\frac{1}{\sqrt{2}}|A|^{2}.$
(11)

Applying this to $\nabla u$, we conclude that:

$\operatorname{Ric}(\nabla u,\nabla
u)\geq-\frac{1}{\sqrt{2}}|A_{M}|^{2}|\nabla u|^{2}.$

Since $M$ is $\Phi$-stable, (6) yields

$\int_{M}\Lambda|A|^{2}\varphi^{2}\leq\int_{M}|\nabla\varphi|^{2},\quad\forall\varphi\in
C_{0}^{1}(M).$

Replacing $\varphi$ by $|\nabla u|\varphi$, we have:

$\begin{split}\int_{M}\varphi^{2}|\nabla
u|^{2}|A|^{2}&\leq\int_{M}|\nabla\varphi|^{2}|\nabla
u|^{2}+2\int_{M}\left(\varphi|\nabla u|\left\langle\nabla\varphi,\nabla|\nabla
u|\right\rangle+\varphi^{2}|\nabla|\nabla u||^{2}\right)\\\
&=\int_{M}|\nabla\varphi|^{2}|\nabla u|^{2}-\int_{M}\varphi^{2}|\nabla
u|\Delta|\nabla u|.\end{split}$ (12)

By the improved Kato inequality,

$|\nabla^{2}u|^{2}\geq\frac{3}{8}|\nabla u|^{-2}|\nabla|\nabla u|^{2}|^{2}.$

Combined with the Bochner formula and (11), we have:

$\begin{split}\Delta|\nabla u|^{2}&=2\operatorname{Ric}_{M}(\nabla u,\nabla
u)+2|\nabla^{2}u|^{2}\\\ &\geq-\sqrt{2}|A|^{2}|\nabla
u|^{2}+\frac{3}{4}|\nabla u|^{-2}|\nabla|\nabla u|^{2}|^{2}.\end{split}$ (13)

Thus,

$\Delta|\nabla u|\geq-\frac{1}{\sqrt{2}}|A|^{2}|\nabla u|+\frac{1}{2}|\nabla
u|^{-1}|\nabla|\nabla u||^{2}.$ (14)

(10) follows from (12) and (14).

∎

######  Proposition 22.

Suppose $\Phi$ satisfies (1). Then any complete, two-sided, $\Phi$-stable
immersion $M^{3}$ in $\mathbf{R}^{4}$ has only one end.

###### Proof.

Suppose the contrary, that $M$ has at least two ends. Then Corollary 20
implies that $M$ admits a nontrivial harmonic function $u$ with
$\int_{M}|\nabla u|^{2}\leq C<\infty$. For $\rho>0$, take $\varphi\in
C_{c}^{1}(M)$ such that $\varphi|_{B_{M}(0,\rho)}=1$,
$\varphi|_{B_{M}(0,2\rho)}=0$, and $|\nabla\varphi|\leq\tfrac{2}{\rho}$. Then
(10) implies that

$\int_{B_{M}(0,\rho)}(\Lambda-\tfrac{1}{\sqrt{2}})|A|^{2}|\nabla
u|^{2}+\frac{1}{2}|\nabla|\nabla u||^{2}\leq\frac{4}{\rho^{2}}\int_{M}|\nabla
u|^{2}\leq\frac{4C}{\rho^{2}}.$

Here $\Lambda\geq\tfrac{1}{\sqrt{2}}$ by (1). Sending $\rho\to\infty$, we
conclude that

$|\nabla|\nabla u||^{2}\equiv 0.$

In particular, this implies that $|\nabla u|$ is a constant. Since $u$ is
nonconstant, we have that $|\nabla u|>0$. However, this implies that

$\int_{M}1=\frac{1}{|\nabla u|^{2}}\int_{M}|\nabla u|^{2}<\infty,$

contradicting Corollary 19. ∎

##  5\. A conformal deformation of metrics

Take $M^{3}\to\mathbf{R}^{4}$ to be $\mathbf{\Phi}$-stable, where $\Phi$
satisfies (1). In this section we carry out the conformal deformation
technique used by Gulliver-Lawson [26] on $M$.

Consider the function $r(x)=\operatorname{dist}_{\mathbf{R}^{n+1}}(0,x)$ on
$M$, and the position vector field $\vec{X}$. Then $\Delta\vec{X}=\vec{H}$.
Thus, $\Delta(r^{2})=\Delta(\sum x_{i}^{2})=2\vec{X}\cdot\Delta
X+2|\nabla\vec{X}|^{2}=2\vec{X}\cdot\vec{H}+2n$. We find:

$\Delta r=\frac{n}{r}+H(\hat{x}\cdot\nu)-\frac{|\nabla r|^{2}}{r},$

here $\hat{x}=\frac{\vec{X}}{|\vec{X}|}$ is the normalized position vector.

Suppose that $w>0$ is a smooth function on $M^{n}\setminus\\{0\\}$. On
$M\setminus\\{0\\}$, define $\tilde{g}=w^{2}g$. For $\lambda\in\mathbf{R}$,
$\varphi\in C_{c}^{1}(M\setminus\\{0\\})$ consider the quadratic form

$\mathcal{Q}_{w}(\varphi)=\int_{M}\left(|\tilde{\nabla}\varphi|_{\tilde{g}}^{2}+(\tfrac{1}{2}\tilde{R}-\lambda)\varphi^{2}\right)d\tilde{\mu},$

where $\tilde{\nabla}$, $\tilde{R}$, $\tilde{\mu}$ are the gradient, the
scalar curvature and the volume form with respect to $\tilde{g}$,
respectively. One relates the geometric quantities in $g$ and $\tilde{g}$ as
follows:

$|\nabla\varphi|_{g}^{2}=w^{2}|\tilde{\nabla}\varphi|_{\tilde{g}}^{2},\quad
d\mu=w^{-n}d\tilde{\mu}.$

Moreover, we have

$w^{2}\tilde{R}=R-2(n-1)\Delta\log w-(n-1)(n-2)|\nabla\log w|^{2}.$

Denote by
$\tilde{\mathcal{Q}}_{w}(\varphi):=\mathcal{Q}_{w}(w^{\frac{2-n}{2}}\varphi)$.
We compute:

$\begin{split}&\tilde{\mathcal{Q}}_{w}(\varphi)\\\
&=\int_{M}\left(w^{-2}|\nabla(w^{\frac{2-n}{2}}\varphi)|_{g}^{2}+(\tfrac{1}{2}\tilde{R}-\lambda)w^{2-n}\varphi^{2}\right)w^{n}d\mu\\\
&=\int_{M}\left(w^{n-2}|w^{\frac{2-n}{2}}\nabla\varphi-\tfrac{n-2}{2}\varphi
w^{-\frac{n}{2}}\nabla
w|_{g}^{2}+(\tfrac{1}{2}w^{2}\tilde{R}-w^{2}\lambda)\varphi^{2}\right)d\mu\\\
&=\int_{M}\left(|\nabla\varphi-\tfrac{n-2}{2}\varphi\nabla\log
w|_{g}^{2}+(\tfrac{1}{2}w^{2}\tilde{R}-w^{2}\lambda)\varphi^{2}\right)d\mu\\\
&=\int_{M}\left(|\nabla\varphi|_{g}^{2}-\frac{n-2}{2}\left\langle\nabla(\varphi^{2}),\nabla\log
w\right\rangle_{g}+\frac{(n-2)^{2}}{4}|\nabla\log
w|_{g}^{2}\varphi^{2}+(\tfrac{1}{2}w^{2}\tilde{R}-w^{2}\lambda)\varphi^{2}\right)d\mu\\\
&=\int_{M}\left(|\nabla\varphi|_{g}^{2}+\left(\frac{n-2}{2}\Delta\log
w+\frac{(n-2)^{2}}{4}|\nabla\log
w|_{g}^{2}+\tfrac{1}{2}w^{2}\tilde{R}-w^{2}\lambda\right)\varphi^{2}\right)d\mu\\\
&=\int_{M}\left(|\nabla\varphi|_{g}^{2}+\tfrac{1}{2}R\varphi^{2}-\left(\frac{n}{2}\left(\Delta\log
w+\tfrac{(n-2)}{2}|\nabla\log
w|_{g}^{2}\right)+w^{2}\lambda\right)\varphi^{2}\right)d\mu.\end{split}$

We now choose $w=r^{-1}$ on $M\setminus\\{0\\}$. Note that (dropping the $g$
subscript on the norm of the gradient)

$\begin{split}\Delta\log w+\frac{n-2}{2}|\nabla\log w|^{2}&=-\frac{\Delta
r}{r}+\frac{n}{2}\frac{|\nabla r|^{2}}{r^{2}}\\\
&=-\frac{n}{r^{2}}-\frac{H(\hat{x}\cdot\nu)}{r}+\frac{n+2}{2}\frac{|\nabla
r|^{2}}{r^{2}}.\end{split}$

Therefore,

$\begin{split}&\tilde{\mathcal{Q}}_{w}(\varphi)=\int_{M}\left(|\nabla\varphi|^{2}+\tfrac{1}{2}R\varphi^{2}+\left(\frac{n}{2}\left(n+rH(\hat{x}\cdot\nu)-\tfrac{n+2}{2}|\nabla
r|^{2}\right)-\lambda\right)r^{-2}\varphi^{2}\right)d\mu\\\
&\geq\int_{M}\left(|\nabla\varphi|^{2}+\tfrac{1}{2}R\varphi^{2}+\left(\frac{n}{2}\left(n-\tfrac{1}{2}\beta
r^{2}H^{2}-\tfrac{1}{2\beta}-\tfrac{n+2}{2}|\nabla
r|^{2}\right)-\lambda\right)r^{-2}\varphi^{2}\right)d\mu\\\
&=\int_{M}\left(|\nabla\varphi|^{2}+(\tfrac{1}{2}R-\tfrac{n}{4}\beta
H^{2})\varphi^{2}+\left(\frac{n}{2}\left(n-\tfrac{1}{2\beta}-\tfrac{n+2}{2}|\nabla
r|^{2}\right)-\lambda\right)r^{-2}\varphi^{2}\right)d\mu,\end{split}$ (15)

for $\beta>0$ to be chosen later.

By the Gauss equation and Lemma 16,

$H^{2}=|A|^{2}+R\leq(1-c_{0})R.$

Combining with $|\nabla r|\leq 1$, we have

$\tilde{\mathcal{Q}}_{w}(\varphi)\geq\int_{M}\left(|\nabla\varphi|^{2}+(\tfrac{1}{2}+\tfrac{n}{4}\beta(c_{0}-1))R\varphi^{2}+\left(\frac{n}{2}(\tfrac{n-2}{2}-\tfrac{1}{2\beta})-\lambda\right)r^{-2}\varphi^{2}\right)d\mu.$
(16)

On the other hand, (7) and (4) imply that for every $\varphi\in C_{c}^{1}(M)$,

$\int_{M}\left(|\nabla\varphi|^{2}+\frac{1}{\sqrt{2}}R\varphi^{2}\right)d\mu\geq
0.$

Note that $R\leq 0$. Thus, by choosing

$\beta=\frac{4(\frac{1}{\sqrt{2}}-\frac{1}{2})}{n(c_{0}-1)},\quad\lambda=\frac{n}{2}\left(\frac{n-2}{2}-\frac{1}{2\beta}\right)=\frac{n}{2}\left(\frac{n-2}{2}-\frac{n(c_{0}-1)}{8(\frac{1}{\sqrt{2}}-\frac{1}{2})}\right),$

we have that $\tilde{\mathcal{Q}}_{w}(\varphi)\geq 0$ for all $\varphi\in
C_{c}^{1}(M\setminus\\{0\\})$. We summarize these in the following
Proposition.

######  Proposition 23.

Suppose $n\geq 3$, $(M^{n},g)$ is an immersed hypersurface in
$\mathbf{R}^{n+1}$, $\Lambda,c_{0}\in\mathbf{R}$, such that:

$\int_{M}(|\nabla\varphi|^{2}+\Lambda R\varphi^{2})dV_{M}\geq
0,\quad\forall\varphi\in C_{c}^{1}(M),$ $\Lambda>\tfrac{1}{2},\quad c_{0}\geq
1,\quad|A|^{2}\leq-c_{0}R_{M}.$

Then the conformally deformed manifold $(M\setminus\\{0\\},\tilde{g}=r^{-1}g)$
satisfies

$\lambda_{1}(-\tilde{\Delta}+\tfrac{1}{2}\tilde{R})\geq\lambda,$

where
$\lambda=\frac{n}{2}\left(\frac{n-2}{2}-\frac{n(c_{0}-1)}{8(\Lambda-\frac{1}{2})}\right)$.

##  6\. Volume estimates

We first recall a diameter bound for warped $\mu$-bubbles in $3$-manifolds
satisfying $\lambda_{1}(-\Delta+\tfrac{1}{2}R)\geq\lambda>0$.

######  Lemma 24 (Warped $\mu$-bubble area and diameter bound).

Let $(N^{3},g)$ be a $3$-manifold with compact connected boundary satisfying

$\lambda_{1}(-\Delta+\tfrac{1}{2}R)\geq\lambda>0.$ (17)

Suppose there exists $p\in N$ such that $d_{N}(p,\partial
N)\geq\frac{5\pi}{\sqrt{\lambda}}$. Then there exists a connected open set
$\Omega$ containing $\partial N$, $\Omega\subset
B_{\frac{5\pi}{\sqrt{\lambda}}}(\partial N)$, such that each connected
component of $\partial\Omega\setminus\partial N$ is a $2$-sphere with area at
most $\tfrac{8\pi}{\lambda}$ and intrinsic diameter at most
$\frac{2\pi}{\sqrt{\lambda}}$.

###### Proof.

This is an application of estimates for the warped $\mu$-bubbles (see, e.g.
[8, Section 3]). Since $N$ satisfies (17), there exists $u\in C^{\infty}(N)$,
$u>0$ in $\mathring{N}$, such that

$\Delta_{N}u\leq-\tfrac{1}{2}(2\lambda-R_{N})u.$ (18)

Take $\varphi_{0}\in C^{\infty}(M)$ to be a smoothing of $d_{N}(\cdot,\partial
N)$ such that $|\operatorname{Lip}(\varphi_{0})|\leq 2$, and $\varphi_{0}=0$
on $\partial N$. Choose $\varepsilon\in(0,\tfrac{1}{2})$ such that
$\varepsilon,\frac{4}{\sqrt{\lambda}}\pi+2\varepsilon$ are regular values of
$\varphi_{0}$. Define

$\varphi=\frac{\varphi_{0}-\varepsilon}{\frac{4}{\sqrt{\lambda}}+\frac{\varepsilon}{\pi}}-\frac{\pi}{2},$

$\Omega_{1}=\\{x\in N:-\tfrac{\pi}{2}<\varphi<\tfrac{\pi}{2}\\}$, and
$\Omega_{0}=\\{x\in N:-\tfrac{\pi}{2}<\varphi\leq 0\\}$. We have that
$|\operatorname{Lip}(\varphi)|<\tfrac{\sqrt{\lambda}}{2}$. In $\Omega_{1}$,
define $h(x)=-\tfrac{1}{2}\tan(\varphi(x))$. By a direct computation, we have

$\lambda+h^{2}-2|\nabla h|\geq 0.$ (19)

Minimize

$\mathcal{A}(\Omega)=\int_{\partial\Omega}ud\mathcal{H}^{2}-\int_{\Omega_{1}}(\chi_{\Omega}-\chi_{\Omega_{0}})hud\mathcal{H}^{3},$

among Caccioppoli sets $\Omega$ in $\Omega_{1}$ with $\Omega\Delta\Omega_{0}$
is compactly contained in $\Omega_{1}$. By [8, Proposition 12], a minimizer
$\tilde{\Omega}$ exists and has regular boundary. We take $\Omega$ to be the
connected component of $\\{x\in
N:0\leq\varphi_{0}(x)\leq\varepsilon\\}\cup\tilde{\Omega}$ that contains
$\partial N$ (in other words, we disregard any component of $\tilde{\Omega}$
that is disjoint from $\partial N$). We verify that $\Omega$ satisfies the
conclusions of Lemma 24.

Indeed, for any connected component $\Sigma$ of
$\partial\Omega\cap\Omega_{1}$, the stability of $\mathcal{A}$ implies [8,
Lemma 14]:

$\int_{\Sigma}|\nabla\psi|^{2}u-\tfrac{1}{2}(R_{N}-\lambda-2K_{\Sigma})\psi^{2}u+(\Delta_{N}u-\Delta_{\Sigma}u)\psi^{2}\\\
-\tfrac{1}{2}u^{-1}\left\langle\nabla_{N}u,\nu\right\rangle^{2}\psi^{2}-\tfrac{1}{2}(\lambda+h^{2}+2\left\langle\nabla_{N}h,\nu\right\rangle)\psi^{2}u\geq
0,\quad\forall\psi\in C^{1}(\Sigma).$ (20)

Taking $\psi=u^{-\frac{1}{2}}$ and using (18), (19), we conclude that

$\lambda|\Sigma|\leq 2\int_{\Sigma}K_{\Sigma}dA\leq
8\pi\quad\Rightarrow\quad|\Sigma|\leq\frac{8\pi}{\lambda}.$

Note that we have used Gauss–Bonnet, which also implies that $\Sigma$ is a
$2$-sphere. The diameter upper bound follows from [8, Lemma 16 and Lemma 18].
∎

For the next lemma, recall that
$r(x)=\operatorname{dist}_{\mathbf{R}^{m}}(0,x)$.

######  Lemma 25.

Below, $k\geq 2$ and $N^{k}$ is a compact connected manifold, possibly with
boundary.

  1. (1)

Consider an immersion $N^{k}\to\mathbf{R}^{m}\setminus\\{0\\}$. Consider
$p,q\in N$ with $d_{\tilde{g}}(p,q)\leq D$, where $\tilde{g}=r^{-2}g$ and $g$
is the induced metric on $N$. Then $r(p)\leq e^{D}r(q)$.

  2. (2)

Consider an immersion $\varphi:N^{k}\to\mathbf{R}^{m}$ with $0\in\varphi(N)$.
Consider $p,q\in N\setminus\varphi^{-1}(0)$ with $d_{\tilde{g}}(p,q)\leq D$.
Write $g$ for the for the induced metric on $N$ and let
$\bar{r}(x)=d_{g}(\varphi^{-1}(0),x)$ denote the intrinsic distance on $N$.
Then $\bar{r}(p)\leq e^{D}\bar{r}(q)$.

###### Proof.

We first establish (1). Choose a curve $\gamma:[0,L]\to N$, parametrized by
$\tilde{g}$-unit speed, connecting $p$ and $q$, such that $L\leq
D+\varepsilon$. Using $|\nabla r|_{g}\leq 1$, we compute

$\displaystyle\log r(q)-\log r(p)$ $\displaystyle=\int_{0}^{L}\frac{d}{dt}\log
r(\gamma(t))dt$ $\displaystyle=\int_{0}^{L}r(\gamma(t))^{-1}g(\nabla
r,\gamma^{\prime}(t))dt$
$\displaystyle\leq\int_{0}^{L}r(\gamma(t))^{-1}|\nabla
r|_{g}|\gamma^{\prime}(t)|_{g}dt$
$\displaystyle\leq\int_{0}^{L}r(\gamma(t))^{-1}|\gamma^{\prime}(t)|_{g}dt$
$\displaystyle=\int_{0}^{L}|\gamma^{\prime}(t)|_{\tilde{g}}dt=L\leq
D+\varepsilon.$

Thus $r(q)\leq e^{D+\varepsilon}r(p)$. The result follows by sending
$\varepsilon\to 0$.

For (2), we begin by noting that $|\nabla\bar{r}|_{g}=1$ and
$r(x)\leq\bar{r}(x)$ for any $x\in N$. Thus, arguing as above

$\log\bar{r}(q)-\log\bar{r}(p)\leq\int_{0}^{L}\bar{r}(\gamma(t))^{-1}|\gamma^{\prime}(t)|_{g}dt\leq\int_{0}^{L}r(\gamma(t))^{-1}|\gamma^{\prime}(t)|_{g}dt=L.$

The proof is completed as above. ∎

###### Proof of Theorem 4.

Let $r=\operatorname{dist}_{\mathbf{R}^{4}}(\cdot,0)$ and
$\bar{r}=\operatorname{dist}_{M,g}(\cdot,0)$, and consider
$\tilde{g}=r^{-2}g$. Fix $\rho>0$, and consider the geodesic ball
$B_{M}(0,e^{\frac{5\pi}{\sqrt{\lambda}}}\rho)$. By Proposition 22, $M\setminus
B_{M}(0,e^{\frac{5\pi}{\sqrt{\lambda}}}\rho)$ has only one unbounded connected
component $E$. Denote by $M^{\prime}=M\setminus E$. We claim that $\partial
M^{\prime}=\partial E$ is connected. Indeed, since $M^{\prime}$ and $E$ are
both connected, if $\partial M^{\prime}$ has more than one connected
components, then one can find a loop in $M$ intersecting one component of
$\partial M^{\prime}$ exactly once, contradicting that $M$ is simply
connected. Applying Lemma 24 to $(M^{\prime}\setminus\\{0\\},\tilde{g})$, we
find a connected open set $\Omega$ in the $\frac{5\pi}{\sqrt{\lambda}}$
neighborhood of $\partial M^{\prime}$, such that each connected component of
$\partial\Omega\setminus\partial M^{\prime}$ has area bounded by
$\frac{8\pi}{\lambda}$ and diameter bounded by $\frac{2\pi}{\sqrt{\lambda}}$
(we emphasize here that the distance, area and diameter are with respect to
$\tilde{g}$). Let $M_{0}$ be the connected component of
$M^{\prime}\setminus\Omega$ that contains $0$.

We make a few observations about $M_{0}$. First, we claim that $M\setminus
M_{0}$ is connected. To see this, let $M_{1}$ be the union of connected
components of $M^{\prime}\setminus\Omega$ other than $M_{0}$. Then $M\setminus
M_{0}=M_{1}\cup\Omega\cup E$. Note that each connected component of $M_{1}$
share a common boundary with $\Omega$. Since $\Omega$ is connected, so is
$M_{1}\cup\Omega$. Next, we claim that $M_{0}$ has only one boundary
component: otherwise, since both $M_{0}$ and $M\setminus M_{0}$ are connected,
as before we can find a loop in $M$ intersecting a connected component of
$\partial M_{0}$ exactly once, contradicting that $M$ is simply connected.

Denote by $\Sigma=\partial M_{0}$. By (2) in Lemma 25,
$\min_{x\in\Sigma}\bar{r}(x)\geq\rho$. Since $B_{M}(0,\rho)$ is connected,
this implies that $B_{M}(0,\rho)\subset M_{0}$. On the other hand, by
comparing intrinsic to extrinsic distance, we see that
$\max_{x\in\Sigma}r(x)\leq e^{\frac{5\pi}{\sqrt{\lambda}}}\rho$, so

$|\Sigma|_{g}=\int_{\Sigma}d\mu=\int_{\Sigma}r^{2}d\tilde{\mu}\leq
e^{\frac{10\pi}{\sqrt{\lambda}}}\rho^{2}|\Sigma|_{\tilde{g}}\leq\frac{8\pi}{\lambda}e^{\frac{10\pi}{\sqrt{\lambda}}}\rho^{2}.$

Thus, Corollary 15 implies that

$|B_{M}(0,\rho)|_{g}\leq|M_{0}|_{g}\leq\frac{\|\Phi\|_{C^{1}}}{3\min_{\nu\in
S^{3}}\Phi(\nu)}e^{\frac{5\pi}{\sqrt{\lambda}}}\rho|\partial
M_{0}|_{g}\leq\frac{8\pi
e^{\frac{15\pi}{\lambda}}\|\Phi\|_{C^{1}}}{3\lambda\min_{\nu\in
S^{3}}\Phi(\nu)}\rho^{3}.$

This proves the first part of the assertion.

Now consider a connected component $\Sigma_{0}$ of $\partial B_{M}(0,\rho)$,
and let $E$ be the connected component of $M\setminus B_{M}(0,\rho)$ such that
$\partial E$ contains $\Sigma_{0}$. Since $M$ is simply connected, we must
have that $\partial E=\Sigma_{0}$. Apply Lemma 24 to $M\setminus E$, and
obtain a connected surface $\Sigma$ such that
$\operatorname{dist}_{\tilde{g}}(\Sigma_{0},\Sigma)\leq\frac{5\pi}{\sqrt{\lambda}}$,
and $\operatorname{diam}_{\tilde{g}}(\Sigma)\leq\frac{2\pi}{\sqrt{\lambda}}$.
(The proof that $\Sigma$ is connected follows a similar argument as used
above.) By the triangle inequality, we have that
$\operatorname{diam}_{\tilde{g}}(\Sigma_{0})\leq\frac{7\pi}{\sqrt{\lambda}}$.
Thus, Lemma 25 implies that

$\max_{x\in\Sigma_{0}}r(x)\leq
e^{\frac{7\pi}{\sqrt{\lambda}}}\min_{x\in\Sigma_{0}}r(x).$

This proves the assertion. ∎

###### Proof of Theorem 7.

The proof is very similar to that of Theorem 4. We apply Lemma 24 to
$(M\setminus\\{0\\},\tilde{g}=r^{-2}g)$ and find a region $\Omega$ in the
$\tfrac{5\pi}{\sqrt{\lambda}}$ neighborhood of $\partial M$, such that each
connected component of $\Omega\setminus\partial M$ has area bounded by
$\frac{8\pi}{\lambda}$ (again, the distance and area are with respect to
$\tilde{g}$). Let $M^{\prime}$ be the connected component of
$M\setminus\Omega$ that contains $\\{0\\}$. Then $\partial M^{\prime}$ is
connected.

Denote by $\Sigma=\partial M^{\prime}$, and
$\rho_{0}=e^{-\frac{5\pi}{\sqrt{\lambda}}}$. By (1) in Lemma 25,
$\min_{x\in\Omega}r(x)\geq\rho_{0}$. In particular, this implies that
$M_{r_{0}}^{*}\subset M^{\prime}$. We have

$|\Sigma|_{g}=\int_{\Sigma}d\mu=\int_{\Sigma}r^{2}d\tilde{\mu}\leq|\Sigma|_{\tilde{g}}\leq\frac{8\pi}{\lambda}.$

Therefore, Corollary 15 implies that

$|M_{\rho_{0}}^{*}|_{g}\leq|M^{\prime}|\leq\frac{\|\Phi\|_{C^{1}}}{3\min_{\nu\in
S^{3}}\Phi(\nu)}|\Sigma|_{g}\leq\frac{8\pi\|\Phi\|_{C^{1}}}{3\lambda\min_{\nu\in
S^{3}}\Phi(\nu)}.$

This completes the proof. ∎

######  Remark 26.

In the more general case where we do not assume that $M$ is simply connected
or has one end (or boundary component), similar proofs work out. The only
modification here is that $\partial M_{0}$ in the proof of Theorem 4 (or
$\partial M^{\prime}$ in the proof of Theorem 7) has connected components
bounded by $b_{1}(M)+E$, where $E$ is the number of ends if $M$ is complete,
and is the number of boundary components if $M\subset B_{1}(0)$. Thus, we have

$|B_{M,R}(0)|\leq V_{0}(b_{1}(M)+E),$

if $M$ is complete, and

$|M_{\rho_{0}}^{*}|\leq V_{1}(b_{1}(M)+E),$

if $M\subset B_{1}(0)$.

##  Appendix A First and second variation

We derive first and second variations of $\mathbf{\Phi}$ with emphasis on our
geometric applications (see also [16, Appendix A] and [53, Section 2]). For
$M^{n}\to\mathbf{R}^{n+1}$ a two-sided immersion, set

$\mathbf{\Phi}(M):=\int_{M}\Phi(\nu)$

for $\Phi:\mathbf{R}^{n+1}\to(0,\infty)$ an elliptic integrand.

###  A.1. First variation

Consider a $1$-parameter family of surfaces $M_{t}$ with normal speed at $t=0$
given by $u\nu$ (with $u\in C^{1}_{c}(M\setminus\partial M)$). Recall that
$\dot{\nu}=-\nabla u$. We find

$\displaystyle\frac{d}{dt}\bigg{|}_{t=0}\mathbf{\Phi}(M_{t})$
$\displaystyle=\int_{M}\left(Hu\Phi(\nu)-D_{\nabla u}\Phi(\nu)\right)$
$\displaystyle=\int_{M}\left(H\Phi(\nu)+\operatorname{div}_{M}(D\Phi(\nu)^{T})\right)u$
$\displaystyle=\int_{M}\left(H\Phi(\nu)+\operatorname{div}_{M}(D\Phi(\nu)-(D_{\nu}\Phi(\nu))\nu)\right)u$
$\displaystyle=\int_{M}\left(H\Phi(\nu)+\operatorname{div}_{M}(D\Phi(\nu))-(D_{\nu}\Phi(\nu))H\right)u.$

Now, we note that we have that $D\Phi(\nu)\cdot\nu=\Phi(\nu)$ by Euler theorem
for homogeneous functions. Thus, we find that

$\frac{d}{dt}\bigg{|}_{t=0}\mathbf{\Phi}(M_{t})=\int_{M}\operatorname{div}_{M}(D\Phi(\nu))u.$
(21)

Thus,

$H_{\Phi}=\operatorname{div}_{M}(D\Phi(\nu)).$ (22)

vanishes if and only if $M$ is a critical point of $\mathbf{\Phi}$. Let us
rewrite this as follows (with $\\{e_{i}\\}_{i=1}^{n}$ a local orthonormal
frame for $M$):

$\displaystyle\operatorname{div}_{M}(D\Phi(\nu))$
$\displaystyle=\sum_{i=1}^{n}(D_{e_{i}}D\Phi(\nu))\cdot e_{i}$
$\displaystyle=\sum_{i=1}^{n}D^{2}\Phi(\nu)[D_{e_{i}}\nu,e_{i}]$
$\displaystyle=\sum_{i=1}^{n}D^{2}\Phi(\nu)[S_{\Sigma}(e_{i}),e_{i}],$

for $S_{M}$ the shape operator of $M$. Let us define
$\Psi(\nu):T\mathbf{R}^{n+1}\to T\mathbf{R}^{n+1}$ by $\Psi(\nu):X\mapsto
D^{2}\Phi(\nu)[X,\cdot]$. (This is just the $(1,1)$-tensor associated to
$D^{2}\Phi(\nu)$ via the Euclidean metric.)

Then, we find

$H_{\Phi}=\operatorname{tr}_{M}(\Psi(\nu)S_{M})$ (23)

Note that for $\Phi(\nu)=|\nu|$, we have

$D\Phi(\nu)=|\nu|^{-1}\nu,\Psi(\nu)=|\nu|^{-1}\operatorname{Id}-|\nu|^{-3}\nu\otimes\nu^{\flat}$

so in particular, when $|\nu|=1$, we find
$\Psi(\nu)|_{T_{p}\Sigma}=\operatorname{Id}_{T_{p}\Sigma}$. Thus, this
recovers the usual mean curvature.

###  A.2. Second variation

Recall the tube formula:

$\dot{S}=-\nabla^{2}u-S^{2}u$

(where we are regarding $\nabla^{2}u$ as a $(1,1)$-tensor via $g_{M}$). Note
also that the trace of a $(1,1)$-tensor is independent of the metric. Thus, we
find

$\dot{H}_{\Phi}=\operatorname{tr}_{M}(-\Psi(\nu)\nabla^{2}u-\Psi(\nu)S_{M}^{2}u+\Psi(\nu)^{\prime}S_{M})$

Note that

$\Psi(\nu)^{\prime}=-(D_{\nabla u}\Psi)(\nu)$

Hence,

$\dot{H}_{\Phi}=\operatorname{tr}_{M}(-\Psi(\nu)\nabla^{2}u-\Psi(\nu)S_{M}^{2}u-(D_{\nabla
u}\Psi)(\nu)S_{M}).$

Integration on $M$ gives

$\frac{d^{2}}{dt^{2}}\bigg{|}_{t=0}\mathbf{\Phi}(M_{t})=\int_{M}\left\langle\nabla
u,\Psi(\nu)\nabla
u\right\rangle-\operatorname{tr}_{M}\left(\Psi(\nu)S_{M}^{2}\right)u^{2}.$
(24)

Thus, stability implies that

$\int_{M}|\nabla u|^{2}-\Lambda|A|^{2}u^{2}\geq 0,\quad\forall u\in
C_{c}^{1}(M\setminus\partial M).$ (25)

Here $\Lambda$ depends on ellipticity of $\Phi$. In particular, if $\Phi$
satisfies (1), then (5) implies that for $\Phi$-stable surfaces $M$, we have

$\int_{M}|\nabla u|^{2}-\frac{1}{\sqrt{2}}|A_{M}|^{2}u^{2}\geq 0,\quad\forall
u\in C_{c}^{1}(M\setminus\partial M).$ (26)

Note that when $\Phi(X)=|X|$, we have seen that
$\Psi(Y)=|Y|^{-1}\operatorname{Id}-|Y|^{-3}Y\otimes Y^{\flat}$. Hence,

$D_{X}\Psi(\nu)=\frac{d}{dt}\Big{|}_{t=0}\Psi(\nu+X)=2(X\cdot\nu)\operatorname{Id}-X\otimes\nu^{\flat}-\nu\otimes
X^{\flat}$

In particular, $D_{\nabla u}\Psi(\nu)|_{T_{p}M}=0$. Thus, we recover the
standard second variation formula in this case.

###  A.3. First variation through vector fields

We also deduce first variation formula of $\mathbf{\Phi}$ through variations
that are not necessarily normal to $M$. We compute as follows:

$\displaystyle\int_{M}\Phi(\nu)\operatorname{div}_{M}X$
$\displaystyle=\int_{M}\Phi(\nu)\operatorname{div}_{M}X^{T}+\Phi(\nu)(X\cdot\nu)H$
$\displaystyle=\int_{M}\operatorname{div}_{M}(\Phi(\nu)X^{T})-\nabla(\Phi(\nu))\cdot
X^{T}+\Phi(\nu)(X\cdot\nu)H$
$\displaystyle=\int_{M}\operatorname{div}_{M}(\Phi(\nu)X^{T})-D_{D\Phi(\nu)^{T}}\nu\cdot
X^{T}+\Phi(\nu)(X\cdot\nu)H$
$\displaystyle=\int_{M}\operatorname{div}_{M}(\Phi(\nu)X^{T})-D_{D\Phi(\nu)^{T}}X^{T}\cdot\nu+\Phi(\nu)(X\cdot\nu)H$
$\displaystyle=\int_{M}\operatorname{div}_{M}(\Phi(\nu)X^{T})+D_{D\Phi(\nu)^{T}}(X\cdot\nu)-D_{D\Phi(\nu)^{T}}X\cdot\nu+\Phi(\nu)(X\cdot\nu)H$
$\displaystyle=\int_{M}\operatorname{div}_{M}(\Phi(\nu)X^{T})+\operatorname{div}((X\cdot\nu)D\Phi(\nu)^{T})-(X\cdot\nu)\operatorname{div}_{\Sigma}D\Phi(\nu)^{T}$
$\displaystyle\qquad-D_{D\Phi(\nu)^{T}}X\cdot\nu+\Phi(\nu)(X\cdot\nu)H$
$\displaystyle=\int_{M}-(X\cdot\nu)\operatorname{div}_{M}D\Phi(\nu)^{T}-D_{D\Phi(\nu)^{T}}X\cdot\nu+\Phi(\nu)(X\cdot\nu)H$
$\displaystyle\qquad+\int_{\partial
M}\Phi(\nu)X\cdot\eta+(X\cdot\nu)D\Phi(\nu)\cdot\eta$
$\displaystyle=\int_{M}-(X\cdot\nu)\operatorname{div}_{M}D\Phi(\nu)-D_{D\Phi(\nu)^{T}}X\cdot\nu+\int_{\partial
M}\Phi(\nu)X\cdot\eta+(X\cdot\nu)D\Phi(\nu)\cdot\eta.$

Thus, we find that if $H_{\Phi}=0$, then

$\int_{M}\Phi(\nu)\operatorname{div}_{M}X+D_{D\Phi(\nu)^{T}}X\cdot\nu=\int_{\partial
M}\Phi(\nu)X\cdot\eta+(X\cdot\nu)D\Phi(\nu)\cdot\eta.$ (27)

##  Appendix B Some computations for quadratic forms

In this section we explicitly compute the constant $c_{0}$ appeared in Lemma
16. The approach is elementary.

######  Lemma 27.

Let $a_{1}\leq a_{2}\leq a_{3}$ be positive constants such that
$\frac{a_{3}}{a_{1}}\leq\sqrt{2}$. Consider quadratic forms

$Q_{1}(k_{1},k_{2})=\frac{a_{1}^{2}+a_{3}^{2}}{a_{3}^{2}}k_{1}^{2}+\frac{2a_{1}a_{2}}{a_{3}^{2}}k_{1}k_{2}+\frac{a_{2}^{2}+a_{3}^{2}}{a_{3}^{2}}k_{2}^{2},$
$Q_{2}(k_{1},k_{2})=\frac{2a_{1}}{a_{3}}k_{1}^{2}+\frac{2(a_{1}+a_{2}-a_{3})}{a_{3}}k_{1}k_{2}+\frac{2a_{2}}{a_{3}}k_{2}^{2}.$

Then we have $Q_{1}\leq c_{0}Q_{2}$, where

$c_{0}=\frac{1}{\sqrt{2}-\frac{1}{2}}\approx 1.09.$

###### Proof.

Write $\alpha=\frac{a_{1}}{a_{3}}$, $\beta=\frac{a_{2}}{a_{3}}$, with
$2^{-\frac{1}{2}}\leq\alpha\leq\beta\leq 1$. Then

$Q_{1}(k_{1},k_{2})=(1+\alpha^{2})k_{1}^{2}+2\alpha\beta
k_{1}k_{2}+(1+\beta^{2})k_{2}^{2}\\\
=(1+\alpha^{2})\left(k_{1}+\frac{\alpha\beta}{1+\alpha^{2}}k_{2}\right)^{2}+\frac{1+\alpha^{2}+\beta^{2}}{1+\alpha^{2}}k_{2}^{2}.$
(28)

Under the substitution $x=k_{1}+\frac{\alpha\beta}{1+\alpha^{2}}k_{2}$,
$y=k_{2}$, we have $k_{1}+k_{2}+(-\alpha k_{1}-\beta
k_{2})=(1-\alpha)x+\frac{1-\beta-\alpha\beta+\alpha^{2}}{1+\alpha^{2}}y$.
Thus, by Cauchy-Schwartz,

$\displaystyle(Q_{1}-Q_{2})(k_{1},k_{2})$ $\displaystyle=(k_{1}+k_{2}-\alpha
k_{1}-\beta k_{2})^{2}$
$\displaystyle=\left((1-\alpha)x+\frac{1-\beta-\alpha\beta+\alpha^{2}}{1+\alpha^{2}}y\right)^{2}$
$\displaystyle\leq
c_{1}\left((1+\alpha^{2})x^{2}+\frac{1+\alpha^{2}+\beta^{2}}{1+\alpha^{2}}y^{2}\right)=c_{1}Q_{1}(k_{1},k_{2}),$

where
$c_{1}=\frac{(1-\alpha)^{2}}{1+\alpha^{2}}+\left(\frac{1-\beta-\alpha\beta+\alpha^{2}}{1+\alpha^{2}}\right)^{2}\cdot\frac{1+\alpha^{2}}{1+\alpha^{2}+\beta^{2}}$.
This gives $Q_{1}\leq\frac{1}{1-c_{1}}Q_{2}$. Using
$2^{-\frac{1}{2}}\leq\alpha\leq\beta\leq 1$, we have:

$c_{1}\leq\frac{(1-2^{-\frac{1}{2}})^{2}}{1+\frac{1}{2}}+\left(\frac{1-2^{-\frac{1}{2}}}{1+\frac{1}{2}}\right)^{2}\cdot\frac{1+\frac{1}{2}}{2}=\frac{3}{2}-\sqrt{2}.$

The result follows. ∎

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