Curvature Varifolds with

Orthogonal Boundary

Ernst Kuwert und Marius Müller (University of Freiburg)

###### Abstract

We consider the class $S^{m}_{\perp}(\Omega)$ of $m$-dimensional surfaces in
$\overline{\Omega}\subset{\mathbb{R}}^{n}$ which intersect $S=\partial\Omega$
orthogonally along the boundary. A piece of an affine $m$-plane in
$S^{m}_{\perp}(\Omega)$ is called an orthogonal slice. We prove estimates for
the area by the $L^{p}$-integral of the second fundamental form in three
cases: first when $\Omega$ admits no orthogonal slices, second for $m=p=2$ if
all orthogonal slices are topological disks, and finally for all $\Omega$ if
the surfaces are confined to a neighborhood of $S$. The orthogonality
constraint has a weak formulation for curvature varifolds, we classify those
varifolds of vanishing curvature. As an application, we prove for any $\Omega$
the existence of an orthogonal $2$-varifold which minimizes the $L^{2}$
curvature in the integer rectifiable class.

MSC 2020 Subject Classification. Primary 28A75, Secondary 49Q15, 49Q20.
Keywords. Curvature varifolds, Orthogonal boundary, Second fundamental form,
Area estimate.

##  1 Introduction

For a bounded domain $\Omega\subset{\mathbb{R}}^{n}$ of class $C^{2}$,
consider the class ${\bf S}^{m}_{\perp}(\Omega)$ of compact $C^{1}$ immersed
submanifolds $\Sigma\hookrightarrow\overline{\Omega}$ of dimension $m\leq
n-1$, which are $C^{2}$ in the interior and meet $S=\partial\Omega$
orthogonally along their boundary. Denoting by $\nu^{S}$ the unit normal
interior to $\Omega$, the condition means that $\nu^{S}\in T\Sigma$ along
$\partial\Sigma\hookrightarrow S$. A natural variational problem in the class
${\bf S}^{m}_{\perp}(\Omega)$ is to minimize the $L^{p}$ curvature energy

${\cal E}^{p}(\Sigma)=\int_{\Sigma}|A_{\Sigma}|^{p}\,d\mu_{\Sigma}\quad\mbox{
where }1\leq p<\infty.$

In [1] Alessandroni and the first author constructed critical points of the
Willmore energy subject to the orthogonality constraint, but only in the class
of surfaces with small prescribed area. Here in Section 5 we address the
global problem in the case when the area is not fixed a priori. Area bounds
depending on the curvature energy then play a key role, they are also of
independent interest. It turns out that the area of surfaces in ${\bf
S}^{m}_{\perp}(\Omega)$ has an upper bound in terms of the curvature energy,
unless the domain $\Omega$ is special.

######  Definition 1.1 (Orthogonal Slices)

Let $x_{0}+P\subset{\mathbb{R}}^{n}$ be an affine $m$-plane. Any component
$\Delta\neq\emptyset$ of ${\rm int}(\overline{\Omega}\cap(x_{0}+P))$ will be
called an $m$-slice of $\Omega$. $\Delta$ is an orthogonal $m$-slice if it
meets $S$ orthogonally along its boundary, that is

$\nu^{S}(x)\in P\quad\mbox{ for all }x\in\partial\Delta.$

We have the following estimate.

######  Theorem 1 (Mass Bounds)

Assume that $\Omega\subset{\mathbb{R}}^{n}$ has no orthogonal $m$-slice. Then
there is a constant $C<\infty$ depending on $n$, $m$, $p$ and $\Omega$, such
that

(1) $|\Sigma|+|\partial\Sigma|\leq
C\,\int_{\Sigma}|A_{\Sigma}|^{p}\,d\mu_{\Sigma}\quad\mbox{ for any
}\Sigma\in{\bf S}^{m}_{\perp}(\Omega).$

Moreover for $m=p=2$, the Euler characteristic is also estimated by

(2) $|\chi(\Sigma)|\leq
C\,\int_{\Sigma}|A_{\Sigma}|^{2}\,d\mu_{\Sigma}\quad\mbox{ for any
}\Sigma\in{\bf S}^{2}_{\perp}(\Omega).$

For arbitrary domains $\Omega$ we prove

######  Theorem 2 (Bounds near the Boundary)

Let $\Omega\subset{\mathbb{R}}^{n}$ be a bounded domain of class $C^{2}$, and
denote by $\varrho_{S}>0$ the injectivity radius of $S=\partial\Omega$. Then
for any $\delta\in(0,\varrho_{S})$ there is a constant
$C=C(n,m,p,\Omega,\delta)<\infty$ such that for $\Sigma\in
S^{m}_{\perp}(\Omega)$

(3) $|\Sigma|+|\partial\Sigma|\leq
C\int_{\Sigma}|A_{\Sigma}|^{p}\,d\mu_{\Sigma}\quad\mbox{ whenever
}\Sigma\subset\overline{U_{\delta}(S)}.$

For $m=p=2$ the Euler characteristic $\chi(\Sigma)$ has a corresponding bound.

A variant of Theorem 1 in the case $m=p=2$ is as follows.

######  Theorem 3 (Bounds for Surfaces)

Let $\Omega\subset{\mathbb{R}}^{n}$ be a bounded domain of class $C^{2}$.
Assume that all orthogonal $2$-slices of $\Omega$ are topological disks. Then
there is a constant $C<\infty$ depending on $n$ and $\Omega$ such that

(4) $|\Sigma|+|\partial\Sigma|+|\chi(\Sigma)|\leq
C\,\Big{(}\int_{\Sigma}|A_{\Sigma}|^{2}\,d\mu_{\Sigma}+\chi(\Sigma)^{+}\Big{)}\quad\mbox{
for any }\Sigma\in{\bf S}^{2}_{\perp}(\Omega).$

For $\Sigma$ connected we have $\chi(\Sigma)^{+}=1$ if $\Sigma$ is a disk and
$\chi(\Sigma)^{+}=0$ otherwise. We note that Theorem 3 applies to any convex
$C^{2}$ domain. In Chapter 4 we give an example showing that the assumptions
in both Theorems 1 and 3 cannot be dropped.

There are a number of papers dealing with curvature-mass relations for
confined curves and surfaces. Müller and Röger consider closed $2$-dimensional
surfaces with the type of a sphere in a $3$-dimensional domain, e.g. a ball
[16]. We also refer to the work of Pozzetta [17] and the references therein.
In a different direction, Colding and Minicozzi proved compactness results for
sequences of embedded minimal surfaces in $3$-manifolds, see for instance [7].
A problem which is more closely related to the present paper is studied by
Mondino [15], proving mass bounds for $m$-varifolds in a Riemannian manifold
in terms of their curvature. He applies the concept of curvature varifolds due
to Hutchinson [12]. The mass bounds are obtained under the condition that
there exist no varifolds with zero curvature. In the special case $m=2$, it is
actually sufficient to rule out smooth totally geodesic immersions, see [4].
In our case, we need to introduce a weak formulation of orthogonality along
the boundary for general curvature varifolds. For this we build on the notion
of varifolds with boundary, defined and studied by Mantegazza [14].

######  Definition 1.2 (orthogonal boundary)

Let $V$ be an $m$-varifold in ${\mathbb{R}}^{n}$ with curvature $B\in
L^{1}(V)$ and support in $\overline{\Omega}$, and let $\Gamma$ be an
$(m-1)$-varifold in $S=\partial\Omega$. We say that $V$ is orthogonal to $S$
along $\Gamma$ if for all $\phi\in
C^{1}({\mathbb{R}}^{n}\times{\mathbb{R}}^{n\times n},{\mathbb{R}}^{n})$

$\displaystyle\int_{{\mathbb{R}}^{n}\times G(m,n)}\big{(}D_{P}\phi\cdot
B+\langle{\rm tr\,}B,\phi\rangle+\langle D_{x}\phi,P\rangle\big{)}\,dV(x,P)$
$\displaystyle=$
$\displaystyle-\int_{G_{m-1}(TS)}\langle\nu^{S}(x),\phi(x,\nu^{S}(x)\wedge
Q)\rangle\,d\Gamma(x,Q).$

Here $\nu^{S}(x)\wedge Q$ denotes the $m$-plane spanned by $\nu^{S}(x)$ and
$Q$. The varifold $\Gamma$ in the definition is uniquely determined, and the
class of varifolds $V$ satisfying the definition (for some $\Gamma$) is
denoted by ${\bf CV}^{m}_{\perp}(\Omega)$.

In Section 2 we classify varifolds $V\in{\bf CV}^{m}_{\perp}(\Omega)$ having
curvature $B_{V}=0$; using disintegration we derive a formula to which only
the slices orthogonal to $\Omega$ contribute. In Section 3 we then consider
the infimum

(6) $\varkappa^{m,p}(\Omega)=\inf\big{\\{}\|B_{V}\|_{L^{p}(V)}^{p}:V\in{\bf
CV}^{m}_{\perp}(\Omega),\,{\bf M}(V)=1\big{\\}}.$

We prove the following generalization of Theorem 1.

######  Theorem 4 (Varifold Bounds)

Let $\Omega\subset{\mathbb{R}}^{n}$ be a bounded domain of class $C^{2}$. If
$\Omega$ has no orthogonal $m$-slices, then $\varkappa^{m,p}(\Omega)>0$ for
all $p\in[1,\infty)$. Moreover for any $V\in{\bf CV}^{m}_{\perp}(\Omega)$ with
associated boundary varifold $\Gamma$, we have

(7) $\displaystyle{\bf M}(V)$ $\displaystyle\leq$
$\displaystyle\frac{1}{\varkappa^{m,p}(\Omega)}\,\|B_{V}\|_{L^{p}(V)}^{p},$
(8) $\displaystyle{\bf M}(\Gamma)$ $\displaystyle\leq$ $\displaystyle
C\,\|B_{V}\|_{L^{p}(V)}^{p}\quad\mbox{ where }C=C(n,m,p,\Omega).$

In the final section this result is applied to a variational problem.

######  Theorem 5 (Existence of Minimizers)

For a bounded domain $\Omega\subset{\mathbb{R}}^{n}$ of class $C^{3}$, let
${\bf ICV}^{2}_{\perp}(\Omega)$ be the class of varifolds in ${\bf
CV}^{2}_{\perp}(\Omega)$ which are integer rectifiable. Then the following
infimum is attained:

(9) $\varkappa(\Omega)=\inf\big{\\{}\|B_{V}\|_{L^{2}(V)}^{2}:V\in{\bf
ICV}^{2}_{\perp}(\Omega),\,V\neq 0\big{\\}}.$

The fact that the mass is not prescribed a priori poses a key difficulty. But
since any orthogonal $2$-slice is a minimizer, we may restrict to the case
when such slices do not exist. Then Theorem 4 applies to give an upper mass
bound. The lower bound is proved by contradiction, we show that if a
minimizing sequence $V_{k}$ would converge to zero then

$\liminf_{k\to\infty}\int_{\overline{\Omega}}|B_{V_{k}}|^{2}\,dV_{k}\geq
8\pi=\int_{{\mathbb{S}}^{2}_{+}}|B_{{\mathbb{S}}^{2}}|^{2}\,d\mu_{{\mathbb{S}}^{2}}.$

However, a comparison surface strictly below $8\pi$ is constructed in Lemma
5.1. We note that our proof of Theorem 5 would apply to $m$-varifolds with
curvature in $L^{m}$, once the round $m$-sphere was known to minimize $L^{m}$
curvature among varifolds in ${\mathbb{R}}^{n}$. We expect that for $m=n-1$
the argument involving the Gauß image generalizes to curvature varifolds.

In Section 6 we prove that any compact $C^{2}$ hypersurface
$S\subset{\mathbb{R}}^{n}$ has a $C^{2}$ approximation by smooth surfaces
$S_{i}$ which have no orthogonal $m$-slices. In particular the assumption of
Theorem 1 and Theorem 4 is generically satisfied.

Acknowledgements.  The authors would like to thank Victor Bangert (Freiburg)
and Lei Liu (Wuhan) for helpful discussions.

##  2 Preliminaries

We derive in this section a weak formulation of the orthogonality constraint
which applies to varifolds. For this we recall briefly the definition of
curvature varifolds due to Hutchinson [12], see also Mantegazza [14].

For an open set $U\subset{\mathbb{R}}^{n}$, we consider first the class of
$m$-dimensional, properly embedded submanifolds $\Sigma\subset U$. The
Grassmannian $G(m,n)$ of $m$-dimensional subspaces of ${\mathbb{R}}^{n}$ will
be identified with the set of orthogonal projections
$P\in{\mathbb{R}}^{n\times n}$ of rank $m$. On $\Sigma$ we then have the
associated Gauß map

(10) $P:\Sigma\to G(m,n)\subset{\mathbb{R}}^{n\times n},\,P(x)=T_{x}\Sigma.$

The tangential and normal components of a vector $X\in{\mathbb{R}}^{n}$ are
denoted by $PX=X^{\top}$ and $P^{\perp}X=X^{\perp}$. In the space
$BL({\mathbb{R}}^{n}\times{\mathbb{R}}^{n},{\mathbb{R}}^{n})$ of
${\mathbb{R}}^{n}$-valued bilinear forms, we define

(11) $B(x)(v,w)=(D_{v^{\top}}P)(x)w\quad\mbox{ for }x\in\Sigma,\mbox{ and
}\,v,w\in{\mathbb{R}}^{n}.$

The second fundamental form $A(x)$ is a component of $B(x)$, more precisely

(12) $B(x)(v,w)^{\perp}=D_{v}(Pw)(x)^{\perp}=A(x)(v,w)\quad\mbox{ for }v,w\in
T_{x}\Sigma.$

Reversely, $B(x)$ can be expressed by $A(x)$: for a tangent frame
$\tau_{\alpha}=\tau_{\alpha}(x)$, $1\leq\alpha\leq m$, and fixed $v\in
T_{x}\Sigma$, $w\in{\mathbb{R}}^{n}$ we compute

$\displaystyle D_{v}P\cdot w$ $\displaystyle=$ $\displaystyle D_{v}(Pw)$
$\displaystyle=$ $\displaystyle[D_{v}(Pw)]^{\perp}+\sum_{\alpha=1}^{m}\langle
D_{v}(Pw),\tau_{\alpha}\rangle\,\tau_{\alpha}$ $\displaystyle=$ $\displaystyle
A(v,w^{\top})-\sum_{\alpha=1}^{m}\langle
D_{v}(P^{\perp}w),\tau_{\alpha}\rangle\,\tau_{\alpha}\quad(\mbox{ as
}D_{v}w=0)$ $\displaystyle=$ $\displaystyle
A(v,w^{\top})+\sum_{\alpha=1}^{m}\langle
w^{\perp},D_{v}\tau_{\alpha}\rangle\,\tau_{\alpha}.$

By definition of the second fundamental form, we conclude

(13) $B(v,w)=A(v^{\top},w^{\top})+\sum_{\alpha=1}^{m}\langle
A(v^{\top},\tau_{\alpha}),w^{\perp}\rangle\,\tau_{\alpha}\quad\mbox{ for all
}v,w\in{\mathbb{R}}^{n}.$

In particular
$|B(v,w)|^{2}=|A(v^{\top},w^{\top})|^{2}+\sum_{\alpha=1}^{m}\langle
A(v^{\top},\tau_{\alpha}),w\rangle^{2}$. Chosing an orthonormal basis $v_{k}$
of ${\mathbb{R}}^{n}$ with $v_{\alpha}=\tau_{\alpha}$ for $\alpha=1,\ldots,m$,
we obtain

(14)
$|B|^{2}=\sum_{\alpha,\beta=1}^{m}|A(\tau_{\alpha},\tau_{\beta})|^{2}+\sum_{k=m+1}^{n}\sum_{\alpha,\beta=1}^{m}\langle
A(\tau_{\alpha},\tau_{\beta}),v_{k}\rangle^{2}=2\,|A|^{2}.$

Furthermore, the mean curvature vector is the trace of $B$, i.e.

(15) ${\rm
tr\,}B=\sum_{k=1}^{n}B(v_{k},v_{k})=\sum_{\alpha=1}^{m}A(\tau_{\alpha},\tau_{\alpha})=\vec{H}.$

For completeness we note also the other traces,

$\sum_{k=1}^{n}\langle B(v_{k},w),v_{k}\rangle=\langle H,w\rangle\quad\mbox{
and }\quad\sum_{k=1}^{n}\langle B(v,v_{k}),v_{k}\rangle=0.$

To derive the first variation formula on $\Sigma$, we now compute

(16) ${{\rm div}}_{\Sigma}\,X^{\top}={{\rm div}}_{\Sigma}\,X+\langle
H,X\rangle\quad\mbox{ for any }X\in C^{1}(\Sigma,{\mathbb{R}}^{n}).$

Substituting $X(x)=\phi(x,P(x))$ where $\phi\in
C^{1}(\overline{\Omega}\times{\mathbb{R}}^{n\times p},{\mathbb{R}}^{n})$, we
get

${{\rm
div}}_{\Sigma}\,X=\big{\langle}D_{\tau_{\alpha}}X,\tau_{\alpha}\big{\rangle}=\big{\langle}D_{x}\phi(x,P)\tau_{\alpha},\tau_{\alpha}\big{\rangle}+\big{\langle}D_{P}\phi(x,P)D_{\tau_{\alpha}}P,\tau_{\alpha}\big{\rangle}.$

Using the Hilbert-Schmidt scalar product, we can write

$\big{\langle}D_{x}\phi(x,P)\tau_{\alpha},\tau_{\alpha}\big{\rangle}=\langle
D_{x}\phi(x,P),P\rangle.$

Furthermore, recalling the definiton of $B$ we have

$\displaystyle\big{\langle}D_{P}\phi(x,P)D_{\tau_{\alpha}}P,\tau_{\alpha}\big{\rangle}$
$\displaystyle=$
$\displaystyle\big{\langle}D_{P}\phi(x,P)B(\tau_{\alpha},\cdot),\tau_{\alpha}\big{\rangle}$
$\displaystyle=$
$\displaystyle\big{\langle}D_{P}\phi(x,P)B(e_{i},\cdot),e_{i}\big{\rangle}$
$\displaystyle=$ $\displaystyle\partial_{P^{k}_{j}}\phi^{i}(x,P)B_{ij}^{k}$
$\displaystyle=:$ $\displaystyle D_{P}\phi(x,P)\cdot B.$

Collecting terms we obtain, for $X(x)=\phi(x,P(x))$,

${{\rm div}}_{\Sigma}\,X+\langle H,X\rangle=D_{P}\phi(x,P)\cdot B+\langle{\rm
tr\,}B,\phi(x,P)\rangle+\langle D_{x}\phi(x,P),P\rangle.$

Now associate to $\Sigma$ the $m$-varifold $V=V_{\Sigma}$ in $U$ with weight
measure $\mu_{\Sigma}={\cal H}^{m}\llcorner\Sigma$ and measure
$V^{x}=\delta_{T_{x}\Sigma}$ on $G(m,n)$, for each $x\in\Sigma$. If
$X(x)=\phi(x,P(x))$ has compact support in $\Sigma$, then

$\displaystyle\int_{\Sigma}{{\rm div}}_{\Sigma}X^{\top}\,d\mu_{\Sigma}$
$\displaystyle=$ $\displaystyle\int_{\Sigma}\big{(}{{\rm
div}}_{\Sigma}\,X+\langle H,X\rangle\big{)}\,d\mu_{\Sigma}$ $\displaystyle=$
$\displaystyle\int_{U\times G(m,n)}\\!\\!\\!\\!\big{(}D_{P}\phi(x,P)\cdot
B+\langle{\rm tr\,}B,\phi(x,P)\rangle+\langle
D_{x}\phi(x,P),P\rangle\big{)}\,dV(x,P).$

On the basis of this calculation Hutchinson gave the following definition.

######  Definition 2.1 (varifold curvature)

Let $V$ be a varifold in $U\subset{\mathbb{R}}^{n}$. We say that $V$ has weak
curvature $B\in L^{1}_{{\rm loc}}(V)$, where $B(x,P)\in
BL({\mathbb{R}}^{n}\times{\mathbb{R}}^{n},{\mathbb{R}}^{n})$, if for any
$\phi\in C^{1}(U\times{\mathbb{R}}^{n\times n},{\mathbb{R}}^{n})$ with compact
support in $U$ we have the identity

(17) $\int_{U\times G(m,n)}\big{(}D_{P}\phi(x,P)\cdot B+\langle{\rm
tr\,}B,\phi(x,P)\rangle+\langle D_{x}\phi(x,P),P\rangle\big{)}\,dV(x,P)=0.$

For test functions $\phi=\phi(x)$ the identity simplifies as follows, see
Lemma 38.4 in [18],

$\displaystyle\delta V(\phi)$ $\displaystyle=$ $\displaystyle\int_{U\times
G(m,n)}\langle D\phi(x),P\rangle\,dV(x,P)$ $\displaystyle=$
$\displaystyle-\int_{U\times G(m,n)}\langle{\rm
tr\,}B(x,P),\phi(x)\rangle\,dV(x,P)$ $\displaystyle=$
$\displaystyle-\int_{U}\Big{\langle}\int_{G(m,n)}{\rm
tr\,}B(x,P)\,dV^{x}(P),\,\phi(x)\Big{\rangle}\,d\mu_{V}(x).$

As $|{\rm tr\,}B|\leq\sqrt{m}\,|B|$, we have for $K={\rm spt\,}\phi\times
G(m,n)$

$|\delta V(\phi)|\leq\sqrt{m}\,\|B\|_{L^{1}(V,K)}\,\|\phi\|_{C^{0}(U)}.$

Therefore Definition 2.1 implies that $V$ has weak mean curvature in $U$ given
by

(18) $H_{V}(x)=\int_{G(m,n)}{\rm tr\,}B(x,P)\,dV^{x}(P)\in L^{1}_{{\rm
loc}}(\mu_{V}).$

Now assume that $\Sigma$ is a compact submanifold with boundary
$\partial\Sigma$. Denote by $\eta$ the interior co-normal and by
$\sigma_{\partial\Sigma}$ the induced boundary measure. Then

$\int_{\Sigma}{{\rm
div}}_{\Sigma}X^{\top}\,d\mu_{\Sigma}=-\int_{\partial\Sigma}\langle
X,\eta\rangle\,d\sigma_{\partial\Sigma}.$

Our interest is in the case when $\Sigma\subset\overline{\Omega}$ meets
$S=\partial\Omega$ orthogonally along the boundary. This means
$\eta=\nu^{S}|_{\partial\Sigma}$, and hence

$\int_{\Sigma}{{\rm
div}}_{\Sigma}X^{\top}\,d\mu_{\Sigma}=-\int_{\partial\Sigma}\langle
X,\nu^{S}\rangle\,d\sigma_{\partial\Sigma}.$

We associate to $\partial\Sigma$ the $(m-1)$-varifold $\Gamma$, i.e. $\Gamma$
is a Radon measure on the bundle

$G_{m-1}(TS)=\\{(x,Q):x\in S,\,Q\in G(m-1,n),\,Q\subset T_{x}S\\}.$

For $p:G_{m-1}(TS)\to S$, $p(x,Q)=x$, we have the weight measure

$p_{\ast}\Gamma={\cal H}^{m-1}\llcorner\partial\Sigma=:\sigma_{\Gamma},$

and the vertical measure $\Gamma^{x}=\delta_{T_{x}(\partial\Sigma)}$ for
$x\in\partial\Sigma$. Thus for $\psi\in C^{0}(G_{m-1}(TS))$

$\int_{\partial\Sigma}\psi(x,T_{x}(\partial\Sigma))\,d\sigma_{\partial\Sigma}(x)=\int_{S}\int_{G_{m-1}(T_{x}S)}\psi(x,Q)\,d\Gamma^{x}(Q)\,d\sigma_{\Gamma}(x)=\Gamma(\psi).$

In order to state the orthogonality condition, we adopt the following
notation: for $Q\in G_{m-1}({\mathbb{R}}^{n})$ and $v\in{\mathbb{R}}^{n}$ with
$v\notin Q$, we denote by $v\wedge Q\in G_{m}({\mathbb{R}}^{n})$ the span of
$Q$ and $v$. In particular if $\Sigma$ meets $S$ orthogonally then
$P(x)=\nu^{S}(x)\wedge T_{x}(\partial\Sigma)$, and for any $\phi\in
C^{1}(\overline{\Omega}\times{\mathbb{R}}^{n\times n},{\mathbb{R}}^{n})$ we
can write putting $X(x)=\phi(x,P(x))$

$\displaystyle\int_{\partial\Sigma}\langle\nu^{S},X\rangle\,d\sigma_{\partial\Sigma}$
$\displaystyle=$
$\displaystyle\int_{\partial\Sigma}\big{\langle}\nu^{S}(x),\phi\big{(}x,\nu^{S}(x)\wedge
T_{x}(\partial\Sigma)\big{)}\big{\rangle}\,d\sigma_{\Gamma}(x)$
$\displaystyle=$
$\displaystyle\int_{G_{m-1}(TS)}\langle\nu^{S}(x),\phi(x,\nu^{S}(x)\wedge
Q)\rangle\,d\Gamma(x,Q).$

This suggests the following notion.

######  Definition 2.2 (orthogonal boundary)

Let $V$ be an $m$-varifold in ${\mathbb{R}}^{n}$ with curvature $B\in
L^{1}(V)$ and support in $\overline{\Omega}$, and let $\Gamma$ be an
$(m-1)$-varifold in $S=\partial\Omega$. We say that $V$ is orthogonal to $S$
along $\Gamma$ if for all $\phi\in
C^{1}({\mathbb{R}}^{n}\times{\mathbb{R}}^{n\times n},{\mathbb{R}}^{n})$

$\displaystyle\int_{{\mathbb{R}}^{n}\times G(m,n)}\big{(}D_{P}\phi\cdot
B+\langle{\rm tr\,}B,\phi\rangle+\langle D_{x}\phi,P\rangle\big{)}\,dV(x,P)$
$\displaystyle=$
$\displaystyle-\int_{G_{m-1}(TS)}\langle\nu^{S}(x),\phi(x,\nu^{S}(x)\wedge
Q)\rangle\,d\Gamma(x,Q).$

###### Remark 2.3

A notion of orthogonality at the boundary can be defined also for varifolds
$V$ which only have mean curvature in $L^{1}(\mu_{V})$. For a given Radon
measure $\sigma$ on $S$, one says that $V$ is orthogonal along $\sigma$ if for
all $\phi\in C^{1}_{c}({\mathbb{R}}^{n},{\mathbb{R}}^{n})$

$\delta V(\phi)=\int_{{\mathbb{R}}^{n}\times G(m,n)}\langle
D\phi,P\rangle\,dV(x,P)=-\int_{{\mathbb{R}}^{n}}\langle
H_{V},\phi\rangle\,d\mu_{V}-\int_{S}\langle\nu^{S}(x),\phi(x)\rangle\,d\sigma(x).$

Definition 2.2 implies this property for $H_{V}$ is as in (18) and
$\sigma=\sigma_{\Gamma}$, the weight measure of $\Gamma$. To employ test
functions depending on $P$, Definition 2.2 will be needed.

###### Lemma 2.4

The boundary varifold $\Gamma$ in Definition 2.2 is unique if it exists.

Proof. Subtracting the identities, we get by Radon-Nikodym a signed measure
$\Lambda$ on $G_{m-1}(TS)$ such that for all $\phi\in
C^{1}({\mathbb{R}}^{n}\times{\mathbb{R}}^{n\times n},{\mathbb{R}}^{n})$

$\int_{G_{m-1}(TS)}\langle\nu^{S}(x),\phi(x,\nu^{S}(x)\wedge
Q)\rangle\,d\Lambda(x,Q)=0.$

For given $\varphi\in C^{0}(G_{m-1}(TS))$ we define $\phi(x,P)=\varphi(x,P\cap
T_{x}S)\,\nu^{S}(x)$ whenever $\nu^{S}(x)\in P$. Using a suitable extension we
obtain a function $\phi\in C^{0}({\mathbb{R}}^{n}\times{\mathbb{R}}^{n\times
n},{\mathbb{R}}^{n})$, such that for all $Q\in G_{m-1}(T_{x}S)$

$\langle\nu^{S}(x),\phi(x,\nu^{S}(x)\wedge
Q)\rangle=\varphi(x,(\nu^{S}(x)\wedge Q)\cap T_{x}S)=\varphi(x,Q).$

We conclude

$\int_{G_{m-1}(TS)}\varphi(x,Q)\,d\Lambda(x,Q)=0\quad\mbox{ for all
}\varphi\in C^{0}(G_{m-1}(TS)).$

By density this implies $\Lambda=0$.

###### Remark 2.5

If Definition 2.2 is valid with $V$ and $\Gamma$, then it holds true for
$\lambda V$ and $\lambda\Gamma$ with the same $B$, for any $\lambda>0$.
However, one has the scaling property

(20) $\|B\|_{L^{p}(\lambda V)}=\lambda^{\frac{1}{p}}\,\|B\|_{L^{p}(V)}.$

######  Lemma 2.6 (Comparable Masses)

There exists a constant $C=C(n,\Omega)<\infty$ such that for any $V,\Gamma$ as
in Definition 2.2 one has

(21) $\displaystyle{\bf M}(V)$ $\displaystyle\leq$ $\displaystyle
C\,\big{(}{\bf M}(\Gamma)+\|B\|_{L^{1}(V)}\big{)},$ (22) $\displaystyle{\bf
M}(\Gamma)$ $\displaystyle\leq$ $\displaystyle C\,\big{(}{\bf
M}(V)+\|B\|_{L^{1}(V)}\big{)}.$

Proof. Taking $\phi(x)=x-x_{0}$ for $x_{0}\in\Omega$ in Definition 2.2 we
obtain

$m{\bf M}(V)\leq{\rm diam\,}(\Omega)\,({\bf M}(\Gamma)+\|B\|_{L^{1}(V)}).$

For the reverse inquality, we take $\phi=\varphi\nabla d_{S}$, where $d_{S}$
is the signed distance from $S$ and $0\leq\varphi\leq 1$ is a cutoff function
with $\varphi|_{S}=1$ and support in $U_{\delta}(S)$, for appropriate
$\delta=\delta(\Omega)\leq 1$. This yields

$\displaystyle{\bf M}(\Gamma)$ $\displaystyle=$
$\displaystyle-\int_{G_{m-1}(TS)}\langle\nu^{S},\varphi\nabla
d_{S}\rangle\,d\Gamma$ $\displaystyle=$
$\displaystyle\int_{\overline{\Omega}\times G(m,n)}\big{(}\langle{\rm
tr\,}B,\varphi\nabla d_{S}\rangle+\langle D_{x}(\varphi\nabla
d_{S}),P\rangle\big{)}\,dV$ $\displaystyle\leq$ $\displaystyle
C\big{(}\|B\|_{L^{1}(V)}+{\bf M}(V)\big{)}.$

###### Remark 2.7

By Young’s inequality we have for any $p\in(1,\infty)$

$\|B\|_{L^{1}(V)}\leq\frac{1}{p}\,{\varepsilon}^{-p}\,\|B\|_{L^{p}(V)}^{p}+\frac{p-1}{p}\,{\varepsilon}^{\frac{p}{p-1}}\,{\bf
M}(V).$

Therefore the inequalities (21), (22) hold with $\|B\|_{L^{1}(V)}$ replaced by
$\|B\|_{L^{p}(V)}^{p}$, with a constant that depends additionally on $p$.

######  Lemma 2.8 (compactness)

Let $V_{k}$, $\Gamma_{k}$ be varifolds as in Definition 2.2, such that ${\bf
M}(V_{k})\leq C$ and for some $p\in(1,\infty]$

(23) $\|B_{k}\|_{L^{p}(V_{k})}\leq\Lambda<\infty\quad\mbox{ for all }k.$

After passing to a subsequence, one has $V_{k}\to V$ and $\Gamma_{k}\to\Gamma$
as varifolds in ${\mathbb{R}}^{n}$, and $V$, $\Gamma$ satisfy Definition 2.2
for some $B\in L^{p}(V)$ with $\|B\|_{L^{p}(V)}\leq\Lambda$.

Proof. From Lemma 2.6 we have

${\bf M}(\Gamma_{k})\leq C\,\big{(}{\bf
M}(V_{k})+\|B_{k}\|_{L^{1}(V_{k})}\big{)}\leq C\,\big{(}{\bf M}(V_{k})+{\bf
M}(V_{k})^{1-\frac{1}{p}}\,\|B_{k}\|_{L^{p}(V_{k})}\big{)}\leq C.$

Thus after passing to a subsequence, $V_{k}\to V$ and $\Gamma_{k}\to\Gamma$ as
varifolds in ${\mathbb{R}}^{n}$. Let
$BL=BL({\mathbb{R}}^{n}\times{\mathbb{R}}^{n},{\mathbb{R}}^{n})$, and
introduce the functionals

$L_{k}:C^{0}_{c}({\mathbb{R}}^{n}\times
G(m,n),BL)\to{\mathbb{R}},\,L_{k}(\psi)=\int_{{\mathbb{R}}^{n}\times
G(m,n)}\langle B_{k},\psi\rangle\,dV_{k}.$

By Hölder’s inequality

$\displaystyle L_{k}(\psi)$ $\displaystyle\leq$
$\displaystyle\|B_{k}\|_{L^{p}(V_{k})}\|\psi\|_{L^{\frac{p}{p-1}}(V_{k})}$
$\displaystyle\leq$ $\displaystyle C\|B_{k}\|_{L^{p}(V_{k})}{\bf
M}(V_{k})^{1-\frac{1}{p}}\|\psi\|_{C^{0}({\mathbb{R}}^{n}\times G(m,n))}$
$\displaystyle\leq$ $\displaystyle C\,\|\psi\|_{C^{0}({\mathbb{R}}^{n}\times
G(m,n))}.$

After passing to a subsequence, we have $L_{k}\to L$ in
$C^{0}_{c}(\overline{\Omega}\times G(m,n))^{\prime}$, and

$L(\psi)\leq\Lambda\,\|\psi\|_{L^{\frac{p}{p-1}}(V)}.$

As $L^{\frac{p}{p-1}}(V)^{\prime}=L^{p}(V)$ for any $p\in(1,\infty]$, the
functional $L$ is represented by some $B\in L^{p}(V)$. The lemma is proved.

An immediate consequence of Lemma 2.8 is the following

######  Theorem 6 (Existence of Minimizers)

Let $\Omega\subset{\mathbb{R}}^{n}$ be a bounded $C^{2}$ domain. For any
$m\leq n-1$ and $p\in(1,\infty]$, let

(24) $\varkappa^{m,p}(\Omega)=\inf\\{\|B_{V}\|^{p}_{L^{p}(V)}:V\in{\bf
CV}^{m}_{\perp}(\Omega),{\bf M}(V)=1\\}.$

Then $\varkappa^{m,p}(\Omega)<\infty$, and the infimum is attained.

###### Remark 2.9

In view of Remark 2.5 we have

(25) $\|B_{V}\|^{p}_{L^{p}(V)}\geq\varkappa^{m,p}(\Omega)\,{\bf
M}(V)\quad\mbox{ for all }V\in{\bf CV}^{m}_{\perp}(\Omega).$

###### Remark 2.10

A corresponding existence result holds in the class ${\bf
ICV}^{m}_{\perp}(\Omega)$ of integer rectifiable varifolds, again for
prescribed mass ${\bf M}(V)$. This follows by Allard’s integral compactness
theorem [2], see also Section 5.

##  3 The Case of Zero Curvature

In this section we assume that $V$ is an $m$-varifold in ${\mathbb{R}}^{n}$
with support in $\overline{\Omega}\times G(m,n)$. We can then use test
functions which are continuous on ${\mathbb{R}}^{n}$, since
$\overline{\Omega}$ is compact. We now state from the appendix the following
disintegration procedure for varifolds.

  * •

Let $\nu=(p_{2})_{\ast}V$, where $p_{2}$ is the projection onto $G(m,n)$. For
any $P\in G(m,n)$, up to a set $E\subset G(m,n)$ with $\nu(E)=0$, there exists
a Radon measure $\mu_{P}$ on ${\mathbb{R}}^{n}$ with
$\mu_{P}({\mathbb{R}}^{n})=1$ such that for all $\phi\in
C^{0}({\mathbb{R}}^{n}\times G(m,n))$ one has

$\int_{{\mathbb{R}}^{n}\times
G(m,n)}\phi(x,P)\,dV(x,P)=\int_{G(m,n)}\int_{{\mathbb{R}}^{n}}\phi(x,P)\,d\mu_{P}(x)\,d\nu(P).$

  * •

Put $\mu_{P}^{\perp}=(P^{\perp})_{\ast}\mu_{P}$ for $P\in G(m,n)\backslash E$.
Then for any $z\in P^{\perp}$, up to a set $E_{P}\subset P^{\perp}$ with
$\mu_{P}^{\perp}(E_{P})=0$, there is a Radon measure $\mu_{P,z}$ on $z+P$ with
$\mu_{P,z}(z+P)=1$ such that for all $\varphi\in C^{0}({\mathbb{R}}^{n})$

$\int_{{\mathbb{R}}^{n}}\varphi(x)\,d\mu_{P}(x)=\int_{P^{\perp}}\int_{z+P}\varphi(x)\,d\mu_{P,z}(x)\,d\mu_{P}^{\perp}(z).$

Taking $\phi(x,P)=\chi_{E}(P)$ we see that $V({\mathbb{R}}^{n}\times
E)=\nu(E)=0$. Likewise we compute for the set $A=\\{(x,P):P\notin
E,\,P^{\perp}x\in E_{P}\\}$

$\displaystyle V(A)$ $\displaystyle=$
$\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\int_{z+P}\chi_{A}(x,P)\,d\mu_{P,z}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P)$
$\displaystyle=$
$\displaystyle\int_{G(m,n)}\underbrace{\int_{P^{\perp}}\chi_{E_{P}}(z)\,\mu_{P,z}(z+P)\,d\mu_{P}^{\perp}(z)}_{=0}\,d\nu(P)=0.$

###### Remark 3.1

We can and will assume that the $\mu_{P}$ have support in the compact set
$K_{P}=\\{x\in{\mathbb{R}}^{n}:(x,P)\in{\rm
spt\,}V\\}\subset\overline{\Omega}$. In fact

$\displaystyle\int_{G(m,n)}\int_{{\mathbb{R}}^{n}}\phi(x,P)\chi_{K_{P}}(x)d\mu_{P}(x)\,d\nu(P)$
$\displaystyle=$ $\displaystyle\int_{{\mathbb{R}}^{n}\times
G(m,n)}\phi(x,P)\chi_{{\rm spt\,}V}(x,P)\,dV(x,P)$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}\times G(m,n)}\phi(x,P)\,dV(x,P).$

Thus we can replace the $\mu_{P}$ by $\mu_{P}\llcorner K_{P}$ without change.
Likewise, we may assume that $\mu_{P,z}$, for each $P\in G(m,n)$ and $z\in
P^{\perp}$, is supported in $K_{P,z}={\rm spt\,}\mu_{P}\cap(z+P)$. In
particular, for $x\in z+P$ we then have the implications

$x\in{\rm spt\,}\mu_{P,z}\,\Rightarrow\,x\in{\rm
spt\,}\mu_{P}\,\Rightarrow\,(x,P)\in{\rm spt\,}V.$

###### Lemma 3.2

Let $V$ be an $m$-varifold in ${\mathbb{R}}^{n}$ with support in
$\overline{\Omega}$, which is orthogonal to $S$ along $\Gamma$ and has
curvature $B_{V}=0$. Then for all $P\in G(m,n)$ and all $z\in P^{\perp}$

$\int_{z+P}\langle D\varphi(x),P\rangle\,d\mu_{P,z}(x)=0\quad\mbox{ for all
$\varphi\in C^{1}({\mathbb{R}}^{n},{\mathbb{R}}^{n})$ tangential on $S$.}$

Proof. We use the ansatz $\phi(x,P)=\varphi(x)\psi(P)$ in the orthogonal
boundary condition, see Definition 2.2. This gives, since $\varphi(x)$ is
tangential by assumption

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\int_{{\mathbb{R}}^{n}\times
G(m,n)}\langle D_{x}\phi(x,P),P\rangle\,dV(x,P)$ $\displaystyle=$
$\displaystyle\int_{G(m,n)}\psi(P)\int_{{\mathbb{R}}^{n}}\langle
D\varphi(x),P\rangle\,d\mu_{P}(x)\,d\nu(P).$

Thus for any $P\in G(m,n)$, up to a set $E(\varphi)$ with measure
$\nu(E(\varphi))=0$, we have

$\int_{{\mathbb{R}}^{n}}\langle D\varphi(x),P\rangle\,d\mu_{P}(x)=0.$

Now replace $\varphi(x)$ by $\sigma(P^{\perp}x)\varphi(x)$. We have

$D_{\tau}(\sigma(P^{\perp}x))=D\sigma(P^{\perp}x)P^{\perp}\tau=0\quad\mbox{
for }\tau\in P.$

Thus we get

$0=\int_{P^{\perp}}\sigma(z)\int_{z+P}\langle
D\varphi(x),P\rangle\,d\mu_{P,z}(x)\,d\mu_{P}^{\perp}(z).$

For any $z\in P^{\perp}$, up to a set $E_{P}(\varphi)$ with
$\mu_{P}^{\perp}(E_{P}(\varphi))=0$, we conclude

$0=\int_{z+P}\langle D\varphi(x),P\rangle\,d\mu_{P,z}(x).$

This is the desired result, except that the null sets depend on the test
function $\varphi$. However, the set of $C^{1}$ functions $\varphi$ which are
tangential along $\partial\Omega$ is separable. We can choose fixed null sets
such that the identity holds for $\varphi$ in a countable dense set, and hence
for all $\varphi$ by approximation.

######  Lemma 3.3 (zero curvature)

Let $V$ be a varifold in $\Omega$ with curvature $B=0$. Then for $\nu$-a.e.
$P\in G(m,n)$ and $\mu_{P}^{\perp}$-a.-e. $z\in P^{\perp}$, the following
holds: for any connected component $U$ of $\Omega\cap(z+P)$ one has

(26) $\mu_{P,z}=\theta_{U}\,{\cal L}^{m}\mbox{ on }U,\quad\mbox{ for a
constant }\theta_{U}\geq 0.$

The lemma follows well-known constancy lemma below, using test functions
$\chi(x)=\omega(x)\tau_{\alpha}$ for an orthonormal basis $\tau_{\alpha}$ of
$P$.

######  Lemma 3.4 (constancy)

Let $\mu$ be a Radon measure on an open domain $U\subset{\mathbb{R}}^{m}$, and
assume that for all $\omega\in C^{\infty}_{c}(U)$

$\int_{U}\partial_{\alpha}\omega(x)\,d\mu(x)=0\quad\mbox{ for
}\alpha=1,\ldots,m.$

Then $\mu=\theta\,{\cal L}^{m}$ on $U$ for some $\theta\in[0,\infty)$.

The proof involving mollification is safely omitted.

######  Lemma 3.5 (dichotomy)

Let $V$ be an $m$-varifold in ${\mathbb{R}}^{n}$ with support in
$\overline{\Omega}$, which is orthogonal to $S$ along $\Gamma$ and has
curvature $B_{V}=0$. Then for any $(x_{0},P_{0})\in{\rm spt\,}V$ with
$x_{0}\in S$ one has the alternative

(27) $\nu^{S}(x_{0})\in P_{0}\quad\mbox{ or }\quad P_{0}\subset T_{x_{0}}S.$

In the second case, there exists $\varrho>0$ such that

(28) $(x,P_{0})\in{\rm spt\,}V\quad\mbox{ for all }x\in(x_{0}+P_{0})\cap
B_{\varrho}(x_{0}).$

In particular, $(x_{0}+P_{0})\cap B_{\varrho}(x_{0})\subset\overline{\Omega}$.

Proof.  Let $(x_{0},P_{0})\in{\rm spt\,}V$, and assume

(29) $\nu_{S}(x_{0})\notin P_{0}.$

By continuity, there exists $r_{0}>0$ such that for all $0<\delta,\varrho\leq
r_{0}$ we have

$\nu_{S}(x)\notin P\quad\mbox{ for all }x\in S\cap B_{2\varrho}(x_{0}),\,P\in
B_{\delta}(P_{0}).$

It follows that

$\nu_{S}(x)\wedge Q\notin B_{\delta}(P_{0})\quad\mbox{ for all }x\in S\cap
B_{2\varrho}(x_{0}),\,Q\in G_{m-1}(T_{x}S).$

Thus for all $\varphi\in C^{1}_{c}(B_{2\varrho}(x_{0}),{\mathbb{R}}^{n})$ and
all $\psi\in C^{1}_{c}(B_{\delta}(P_{0}))$, we have

$\displaystyle\int_{S}\langle\nu^{S}(x),\varphi(x)\rangle\int_{G_{m-1}(T_{x}S)}\psi(\nu^{S}(x)\wedge
Q)\,d\Gamma^{x}(Q)\,d\sigma_{\Gamma}(x)$ $\displaystyle=$
$\displaystyle\int_{S\cap
B_{2\varrho}(x_{0})}\langle\nu^{S}(x),\varphi(x)\rangle\int_{G_{m-1}(T_{x}S)}\underbrace{\psi(\nu^{S}(x)\wedge
Q)}_{=0}\,d\Gamma^{x}(Q)\,d\sigma_{\Gamma}(x)=0.$

Definition 2.2, with $B_{V}=0$ and $\phi(x,P)=\varphi(x)\psi(P)$, now implies

$0=\int_{G(m,n)}\psi(P)\int_{{\mathbb{R}}^{n}}\langle
P,D\varphi(x)\rangle\,d\mu_{P}(x)\,d\nu(P).$

Thus for $\nu$-a.-e. $P\in B_{\delta}(P_{0})$ we obtain

$\int_{{\mathbb{R}}^{n}}\langle P,D\varphi(x)\rangle\,d\mu_{P}(x)=0\quad\mbox{
for all }\varphi\in C^{1}_{c}(B_{2\varrho}(x_{0}),{\mathbb{R}}^{n}).$

Fix such a $P\in B_{\delta}(P_{0})$, and consider a test function
$\varphi(x)=\chi(Px)\zeta(P^{\perp}x)$, where $\chi\in C^{1}_{c}(P\cap
B_{\varrho}(Px_{0}),{\mathbb{R}}^{n})$ and $\zeta\in C^{1}_{c}(P^{\perp}\cap
B_{\varrho}(P^{\perp}x_{0}))$. Then

$\int_{P^{\perp}}\zeta(z)\int_{z+P}{{\rm
div}}_{P}\chi(Px)\,d\mu_{P,z}(x)\,d\mu_{P}^{\perp}(z)=0.$

Therefore we conclude, for $\mu_{P}^{\perp}$-a.-e. $z\in P^{\perp}\cap
B_{\varrho}(P^{\perp}x_{0})$,

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\int_{z+P}{{\rm
div}}_{P}\chi(Px)\,d\mu_{P,z}(x)$ $\displaystyle=$
$\displaystyle\int_{z+P}{{\rm div}}_{P}\chi(x-z)\,d\mu_{P,z}(x)\quad\mbox{ for
all }\chi\in C^{1}_{c}(P\cap B_{\varrho}(Px_{0}),{\mathbb{R}}^{n}).$

Thus for $\nu$-a.e. $P\in B_{\delta}(P_{0})$ and $\mu_{P}^{\perp}$-a.-e. $z\in
P^{\perp}\cap B_{\varrho}(P^{\perp}x_{0})$, we have

(30) $\mu_{P,z}=\theta\,{\cal L}^{m}\llcorner(z+P)\quad\mbox{ on }\,z+P\cap
B_{\varrho}(Px_{0})\quad\mbox{ for some }\theta\in[0,\infty).$

We claim that for any $y_{0}\in P_{0}\cap B_{\varrho/2}(0)$, we have
$(x_{0}+y_{0},P_{0})\in{\rm spt\,}V$. Let $B_{\sigma}(x_{0}+y_{0})$ be given,
wlog with $\sigma\leq\frac{\varrho}{4}$. For $x\in B_{\sigma}(x_{0}+y_{0})$ we
have

$|Px-
Px_{0}|\leq|P(x-(x_{0}+y_{0}))|+|(P-P_{0})y_{0}|+|y_{0}|\leq\frac{3\varrho}{4}+\frac{\delta\varrho}{2},$
$|P^{\perp}x-P^{\perp}x_{0}|\leq|P^{\perp}(x-(x_{0}+y_{0}))|+|(P^{\perp}-P^{\perp}_{0})y_{0}|\leq\frac{\varrho}{4}+\frac{\delta\varrho}{2}.$

Thus $|P(x-x_{0})|,\,|P^{\perp}(x-x_{0})|<\varrho$ for all $x\in
B_{\sigma}(x_{0}+y_{0})$, provided $\delta<\frac{1}{4}$. Now we consider a
ball $B_{\varepsilon}(x_{0}+Py_{0})$ and estimate

$|(x_{0}+y_{0})-(x_{0}+Py_{0})|=|P^{\perp}y_{0}|=|(P^{\perp}-P_{0}^{\perp})y_{0}|\leq\frac{\delta\varrho}{2}.$

We choose ${\varepsilon}=\frac{\sigma}{2}$, and conclude

$B_{\varepsilon}(x_{0}+Py_{0})\subset B_{\sigma}(x_{0}+y_{0})\quad\mbox{ for
any }\delta<\frac{\sigma}{\varrho}.$

Now we compute for fixed $P\in B_{\delta}(P_{0})$

$\displaystyle\mu_{P}(B_{\sigma}(x_{0}+y_{0}))$ $\displaystyle\geq$
$\displaystyle\mu_{P}(B_{\varepsilon}(x_{0}+Py_{0}))$ $\displaystyle=$
$\displaystyle\int_{P^{\perp}}\mu_{P,z}\big{(}(z+P)\cap
B_{\varepsilon}(x_{0}+Py_{0})\big{)}\,d\mu_{P}^{\perp}(z)$ $\displaystyle=$
$\displaystyle\int_{P^{\perp}}\theta_{P,z}{\cal L}^{m}\big{(}(z+P)\cap
B_{\varepsilon}(x_{0}+Py_{0})\big{)}\,d\mu_{P}^{\perp}(z)$ $\displaystyle=$
$\displaystyle\int_{P^{\perp}}\theta_{P,z}{\cal L}^{m}\big{(}(z+P)\cap
B_{\varepsilon}(x_{0})\big{)}\,d\mu_{P}^{\perp}(z)$ $\displaystyle=$
$\displaystyle\mu_{P}\big{(}B_{\varepsilon}(x_{0})\big{)}.$

Integrating over all $P\in B_{\delta}(P_{0})$ we conclude

$\displaystyle V\big{(}B_{\sigma}(x_{0}+y_{0})\times B_{\delta}(P_{0})\big{)}$
$\displaystyle=$
$\displaystyle\int_{B_{\delta}(P_{0})}\mu_{P}(B_{\sigma}(x_{0}+y_{0}))\,d\nu(P)$
$\displaystyle\geq$
$\displaystyle\int_{B_{\delta}(P_{0})}\mu_{P}(B_{\varepsilon}(x_{0}))\,d\nu(P)$
$\displaystyle=$ $\displaystyle V\big{(}B_{\varepsilon}(x_{0})\times
B_{\delta}(P_{0})\big{)}>0,$

since $(x_{0},P_{0})\in{\rm spt\,}V$. This proves $(x_{0}+y_{0},P_{0})\in{\rm
spt\,}V$ for any $y_{0}\in P\cap B_{\varrho/2}(0)$. As a direct consequence,
it follows that $(x_{0}+P_{0})\cap
B_{\varrho/2}(x_{0})\subset\overline{\Omega}$.

###### Lemma 3.6

Let $V$ be an $m$-varifold in ${\mathbb{R}}^{n}$ with support in
$\overline{\Omega}$, which is orthogonal to $S$ along $\Gamma$ and has
curvature $B_{V}=0$. Let $x_{0}\in S$, and assume $(x_{0},P_{0})\in{\rm
spt\,}V$ where $P_{0}\subset T_{x_{0}}S$. Then for $z_{0}=P_{0}x_{0}$ we have
near $x_{0}$

$\mu_{P_{0},z_{0}}=\theta\,{\cal L}^{m}\llcorner(z_{0}+P_{0})\quad\mbox{ for
some constant }\theta\geq 0.$

Proof.  For any $\varphi\in C^{1}({\mathbb{R}}^{n},{\mathbb{R}}^{n})$ which is
tangential on $S$, we have by Lemma 3.2

(31) $\int_{z_{0}+P_{0}}\langle
D\varphi(x),P_{0}\rangle\,d\mu_{P_{0},z_{0}}(x)=0.$

Choose a neighborhood $U\times I$ of $x_{0}$ on which $\Omega$ has a subgraph
representation

$\Omega\cap(U\times I)=\\{(y,z)\in U\times I:z<u(y)\\}.$

After rotation we assume $x_{0}=0$, $T_{x_{0}}S={\mathbb{R}}^{n-1}$ and
$P_{0}={\mathbb{R}}^{m}\times\\{0\\}$ where $m\leq n-1$. For any $\sigma\in
C^{1}_{c}(U\times I)$ we consider the test function

$\varphi(y,z)=\sigma(y,z)\,\big{(}e_{\alpha}+\partial_{\alpha}u(y)e_{n}\big{)}\quad\mbox{
where }x=(y,z)\mbox{ and }1\leq\alpha\leq m.$

By construction the vector field $\varphi$ is tangential along
$\partial\Omega$. Furthermore

$\langle
D\varphi(x),P_{0}\rangle=\sum_{\beta=1}^{m}\langle\partial_{\beta}\varphi(x),e_{\beta}\rangle=\partial_{\alpha}\sigma(x).$

Therefore (31) implies

$\int_{{\mathbb{R}}^{m}}\partial_{\alpha}\sigma(x)\,d\mu_{{\mathbb{R}}^{m},0}=0\quad\mbox{
where }(P_{0},z_{0})=({\mathbb{R}}^{m},0).$

Since any function with compact support in ${\mathbb{R}}^{m}\cap U$ can be
extended to a function with compact support in $U\times I$, the claim follows
from Lemma 3.4.

######  Theorem 7 (Orthogonal Varifolds with Curvature Zero)

Let $V$ be an $m$-varifold in ${\mathbb{R}}^{n}$ with support in
$\overline{\Omega}$, which is orthogonal to $S=\partial\Omega$ along an
$(m-1)$-varifold $\Gamma$ and has curvature $B_{V}=0$. Then for all $\phi\in
C^{0}({\mathbb{R}}^{n}\times G(m,n))$ and $\varphi\in C^{0}(G_{m-1}(TS))$,
respectively, the following representations hold:

$\displaystyle V(\phi)$ $\displaystyle=$
$\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\sum_{\Delta\in
S^{m}_{\perp}(\Omega,z+P)}\theta_{\Delta}\int_{\Delta}\phi(x,P)\,d{\cal
L}^{m}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P),$ $\displaystyle\Gamma(\varphi)$
$\displaystyle=$ $\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\sum_{\Delta\in
S^{m}_{\perp}(\Omega,z+P)}\theta_{\Delta}\int_{\partial\Delta}\varphi(x,P\cap
T_{x}S)\,d{\cal H}^{m-1}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P).$

Here $S^{m}_{\perp}(\Omega,z+P)$ is the finite, possibly empty set of
orthogonal $m$-slices of $\Omega$ in the affine plane $z+P$, and the
$\theta_{\Delta}\geq 0$ are constants.

Proof. Let $\Delta$ be any $m$-slice of $\Omega$ in an affine plane $z+P$,
where $P\in G(m,n)$ and $z\in P^{\perp}$. By definition, for any $x\in\Delta$
there exists $\varrho>0$ such that

$(z+P)\cap B_{\varrho}(x)\subset\overline{\Omega}.$

In particular if $x\in\Delta\cap S$ then $P\subset T_{x}S$. Lemma 3.3 and
Lemma 3.6 imply that $\mu_{P,z}=\theta\,{\cal L}^{m}\llcorner(z+P)$ locally
near any $x\in\Delta$. Since $\Delta$ is connected, we conclude

$\mu_{P,z}\llcorner\Delta=\theta_{\Delta}\,{\cal
L}^{m}\llcorner\Delta\quad\mbox{ for some }\theta_{\Delta}\geq 0.$

Actually, Lemma 3.6 requires that $(x,P)\in{\rm spt\,}V$, but otherwise the
statement follows with $\theta_{\Delta}=0$, see Remark 3.1. Now assume that
$\theta_{\Delta}>0$, and consider a boundary point $x\in\partial\Delta\subset
S$. We then have $(x,P)\in{\rm spt\,}V$ again by Remark 3.1. We claim that
$\nu^{S}(x)\in P$. Otherwise Lemma 3.5 yields that $(x+P)\cap
B_{\varrho}(x)\subset\overline{\Omega}$ for some $\varrho>0$. But $\Delta\cap
B_{\varrho}(x)$ is nonempty, thus $(x+P)\cap B_{\varrho}(x)\subset\Delta$ and
$x\in\Delta$, a contradiction. This shows that if $\theta_{\Delta}>0$ then
$\nu^{S}(x)\in P$ for all $x\in\partial\Delta$, in other words $\Delta$ is an
orthogonal $m$-slice. Along $\partial\Delta$ we have

$\nabla({\rm dist}_{S}|_{z+P})(x)=P\,\nabla{\rm dist}_{S}(x)=\nu_{S}(x)\neq
0,$

therefore $\Delta$ is a bounded domain in $z+P$ of class $C^{2}$. We denote by
$\Delta_{P,z}$ the union of the finitely many orthogonal slices $\Delta\subset
z+P$ with $\theta_{\Delta}>0$, and introduce the function

$\theta_{P,z}:z+P\to[0,\infty),\,\theta_{P,z}(x)=\theta_{\Delta}\mbox{ for
}x\in\overline{\Delta}.$

The measure $\tilde{\mu}_{P,z}={\cal L}^{m}\llcorner\theta_{P,z}$ satisfies,
for any $\varphi\in C^{1}(z+P,P)$,

$\displaystyle\int_{z+P}\langle D\varphi(x),P\rangle\,d\tilde{\mu}_{P,z}(x)$
$\displaystyle=$ $\displaystyle\sum_{\Delta\in
S^{m}_{\perp}(\Omega,z+P)}\theta_{\Delta}\int_{\Delta}\langle
D\varphi(x),P\rangle\,d{\cal L}^{m}(x)$ $\displaystyle=$
$\displaystyle-\sum_{\Delta\in
S^{m}_{\perp}(\Omega,z+P)}\theta_{\Delta}\int_{\partial\Delta}\langle\varphi(x),\nu^{S}(x)\rangle\,d{\cal
H}^{m-1}(x).$

In other words we have, in the sense of distributions on $z+P$,

(32) $\nabla\tilde{\mu}_{P,z}=-\theta_{P,z}({\cal
H}^{m-1}\llcorner\partial\Delta_{P,z})\otimes\nu^{S}.$

On functions $\varphi\in C^{0}_{c}({\mathbb{R}}^{n}\times G(m,n))$, we now
consider the $m$-varifold $\tilde{V}$ given by

(33)
$\tilde{V}(\varphi)=\int_{G(m,n)}\int_{P^{\perp}}\int_{z+P}\varphi(x,P)\,d\tilde{\mu}_{P,z}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P).$

For $\varphi\geq 0$ we can estimate

$\displaystyle 0\leq\tilde{V}(\varphi)$ $\displaystyle=$
$\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\int_{\Delta_{P,z}}\varphi(x,P)\,\,d\mu_{P,z}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P)$
$\displaystyle\leq$
$\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\int_{z+P}\varphi(x,P)\,d\mu_{P,z}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P)$
$\displaystyle=$ $\displaystyle V(\varphi).$

In particular $\tilde{V}$ is continuous, in fact we have

$|\tilde{V}(\varphi)|\leq\tilde{V}(|\varphi|)\leq V(|\varphi|)\leq{\bf
M}(V)\,\|\varphi\|_{C^{0}({\mathbb{R}}^{n}\times G(m,n))}.$

It follows that $\Lambda(\varphi):=V(\varphi)-\tilde{V}(\varphi)$ is a Radon
measure on ${\mathbb{R}}^{n}\times G(m,n)$. Moreover
$\tilde{V}(\varphi)=V(\varphi)$ whenever $\varphi$ has compact support in
$\Omega\times G(m,n)$. Therefore $\Lambda$ has support in $S\times G(m,n)$.
Using (32) we further have for $\phi\in C^{1}({\mathbb{R}}^{n}\times
G(m,n),{\mathbb{R}}^{n})$

$\displaystyle\int_{{\mathbb{R}}^{n}\times G(m,n)}\langle
D_{x}\phi(x,P),P\rangle\,d\tilde{V}(x,P)$ $\displaystyle=$
$\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\int_{z+P}\langle
D_{x}\phi(x,P),P\rangle\,d\tilde{\mu}_{P,z}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P)$
$\displaystyle=$
$\displaystyle-\int_{G(m,n)}\int_{P^{\perp}}\int_{\partial\Delta_{P,z}}\theta_{P,z}(x)\langle\nu_{S}(x),\phi(x,P)\rangle\,d{\cal
H}^{m-1}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P).$

On the other hand, $V$ satisfies the orthogonal boundary constraint

$\int_{{\mathbb{R}}^{n}\times G(m,n)}\langle
D_{x}\phi(x,P),P\rangle\,dV(x,P)=-\int_{G_{m-1}(TS)}\langle\nu^{S}(x),\phi(x,\nu^{S}(x)\wedge
Q)\rangle\,\,d\Gamma(x,Q).$

We take a test function of the form $\phi(x,P)=\varphi(x)$ where
$\varphi|_{S}=0$. Then

$\displaystyle 0$ $\displaystyle=$ $\displaystyle\int_{{\mathbb{R}}^{n}\times
G(m,n)}\langle D\varphi(x),P\rangle\,dV(x,P)-\int_{{\mathbb{R}}^{n}\times
G(m,n)}\langle D\varphi(x),P\rangle\,d\tilde{V}(x,P)$ $\displaystyle=$
$\displaystyle\int_{S\times G(m,n)}\langle
D\varphi(x),P\rangle\,d\Lambda(x,P)$ $\displaystyle=$
$\displaystyle\int_{S\times G(m,n)}\langle
D\varphi(x)\nu^{S}(x),\nu^{S}(x)\rangle\,d\Lambda(x,P)$ $\displaystyle=$
$\displaystyle\int_{S}\Lambda^{x}(G(m,n))\,\langle
D\varphi(x)\nu^{S}(x),\nu^{S}(x)\rangle\,d\mu_{\Lambda}(x).$

Taking now $\varphi(x)={\rm dist}_{S}(x)\,\nu^{S}(x)$ yields
$D\varphi(x)\nu^{S}(x)=\nu^{S}(x)$ along $S$. We conclude

$0=\int_{S}\Lambda^{x}(G(m,n))\,d\mu_{\Lambda}(x)=\Lambda({\mathbb{R}}^{n}\times
G(m,n)).$

This shows $\Lambda=0$ and hence $V=\tilde{V}$, which is the desired
representaton for $V$. Finally, we claim that the boundary varifold $\Gamma$
in Definition 2.2 has the representation

$\Gamma(\varphi)=\int_{G(m,n)}\int_{P^{\perp}}\int_{\partial\Delta_{P,z}}\theta_{P,z}(x)\,\varphi(x,P\cap
T_{x}S)\,d{\cal H}^{m-1}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P).$

First note that $P\cap T_{x}S\in G_{m-1}(T_{x}S)$ as $\nu^{S}(x)\in P$ for
$x\in\partial\Delta_{P,z}$. For given $\varphi\in C^{0}(G_{m-1}(TS))$ we
consider as test function $\phi(x,P)=\varphi(x,P\cap T_{x}S)\nu^{S}(x)$,
suitably extended continuously to ${\mathbb{R}}^{n}\times G(m,n)$. From
$V=\tilde{V}$ and $B_{V}=0$ we get

$\displaystyle\int_{G_{m-1}(TS)}\varphi(x,Q)\,d\Gamma(x,Q)$ $\displaystyle=$
$\displaystyle\int_{G_{m-1}(TS)}\langle\nu^{S}(x),\phi(x,\nu^{S}(x)\wedge
Q)\rangle\,d\Gamma(x,Q)$ $\displaystyle=$
$\displaystyle-\int_{{\mathbb{R}}^{n}\times G(m,n)}\langle
D_{x}\phi(x,P),P\rangle\,dV(x,P)$ $\displaystyle=$
$\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\int_{\partial\Delta_{P,z}}\theta_{P,z}(x)\langle\nu_{S}(x),\phi(x,P)\rangle\,d{\cal
H}^{m-1}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P)$ $\displaystyle=$
$\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\int_{\partial\Delta_{P,z}}\theta_{P,z}(x)\varphi(x,P\cap
T_{x}S)\,d{\cal H}^{m-1}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P).$

##  4 Mass and Topology Bounds

In this section Theorem 7 is applied to prove the bounds stated in the
introduction.

Proof of Theorem 4. We first prove that $\varkappa^{m,p}(\Omega)>0$. As
$\varkappa^{m,p}(\Omega)\geq\varkappa^{m,1}(\Omega)$ by definition, it
suffices to consider the case $p=1$. Assume by contradiction that there is a
sequence $V_{k}\in{\bf CV}^{m}_{\perp}(\Omega)$ with ${\bf M}(V_{k})=1$, such
that $\|B_{V_{k}}\|_{L^{1}(V_{k})}\to 0$. For the associated boundary
varifolds $\Gamma_{k}$, we have by Lemma 2.6

${\bf M}(\Gamma_{k})\leq C\,\big{(}{\bf
M}(V_{k})+\|B_{V_{k}}\|_{L^{1}(V_{k})}\big{)}\leq C\quad\mbox{ where
}C=C(n,\Omega).$

Thus we can assume that $V_{k}\to V$ where again ${\bf M}(V)=1$, and also
$\Gamma_{k}\to\Gamma$ as varifolds. By passing to the limit in (2.2) we get

$\int_{{\mathbb{R}}^{n}\times G(m,n)}\langle
D_{x}\phi,P\rangle\,dV(x,P)=-\int_{G_{m-1}(TS)}\langle\nu^{S}(x),\phi(x,\nu^{S}(x)\wedge
Q)\rangle\,d\Gamma(x,Q).$

This means that $V$ is orthogonal to $S$ along $\Gamma$ and has curvature
zero. But $\Omega$ has no orthogonal $m$-slices by assumption, and hence
Theorem 7 implies that $V=0$, a contradiction. For given nonzero $V\in{\bf
CV}^{m}_{\perp}(\Omega)$, let $V^{\prime}=\lambda V$ where $\lambda={\bf
M}(V)^{-1}$, thus ${\bf M}(V^{\prime})=1$. By definition of
$\varkappa^{m,p}(\Omega)$ and (20)

$\|B_{V}\|^{p}_{L^{p}(V)}={\bf
M}(V)\,\|B_{V^{\prime}}\|^{p}_{L^{p}(V^{\prime})}\geq{\bf
M}(V)\,\varkappa^{m,p}(\Omega).$

The estimate for $\Gamma$ follows from Lemma 2.6 and Remark 2.7.

Theorem 1 is just a special case of Theorem 4, except for estimate (2) which
will be proved below when discussing the two-dimensional case.

Proof of Theorem 2. To prove claim (3) it is sufficient to bound $|\Sigma|$,
since then $|\partial\Sigma|$ is estimated by Lemma 2.6. Assume by
contradiction that there is a sequence $\Sigma_{k}\in{\bf
S}^{m}_{\perp}(\Omega)$ with $\Sigma_{k}\subset U_{\delta}(S)$, such that

$\int_{\Sigma_{k}}|A_{\Sigma_{k}}|^{p}\,d\mu_{\Sigma_{k}}<\frac{1}{k}\,|\Sigma_{k}|.$

Passing to $V_{k}=\Sigma_{k}/|\Sigma_{k}|\in{\bf CV}^{m}_{\perp}(\Omega)$ we
have ${\bf M}(V_{k})=1$, and (20) yields

$\|B_{V_{k}}\|_{L^{p}(V_{k})}^{p}=2^{p/2}\|A_{\Sigma_{k}}\|_{L^{p}(\Sigma_{k})}^{p}/|\Sigma_{k}|<2^{p/2}\frac{1}{k}\to
0.$

$V_{k}$ satisfies Definition 2.2 with boundary
$\Gamma_{k}=\partial\Sigma_{k}/|\Sigma_{k}|$. As ${\bf M}(\Gamma_{k})\leq C$
by Lemma 2.6, we may assume that $V_{k}\to V$ and $\Gamma_{k}\to\Gamma$ as
varifolds. It follows that ${\bf M}(V)=1$, $V$ is orthogonal to $S$ and has
curvature zero. Moreover

${\rm spt\,}\mu_{V}\subset\overline{U_{\delta}(S)}.$

Let $\Delta$ be any orthogonal $m$-slice of $\Omega$. Choose $p\in\Delta$ such
that $\varrho={\rm dist}(p,\partial\Delta)$ is maximal. Then there exist at
least two points $x\in\partial\Delta$ such that $p=x+\varrho\nu^{S}(x)$, and
thus the normal injectivity radius of $S$ satisfies $\varrho_{S}\leq\varrho$.
We get for $\Omega_{\delta}=\Omega\backslash\overline{U_{\delta}(S)}$

$|\Delta\cap\Omega_{\delta}|\geq c(m)(\varrho_{S}-\delta)^{m}>0.$

On the other hand, by comparing with a ball containing $\Omega$, we have
$|\Delta|\leq C$ for $C=C(m,\Omega)$. From the representation in Theorem 7 we
now get

$\displaystyle\mu_{V}(\Omega_{\delta})$ $\displaystyle=$
$\displaystyle\int_{G(m,n)}\int_{P^{\perp}}\sum_{\Delta\in
S^{m}_{\perp}(\Omega,z+P)}\theta_{\Delta}\,|\Delta\cap\Omega_{\delta}|\,d\mu_{P}^{\perp}(x)\,d\nu(P)$
$\displaystyle\geq$ $\displaystyle
c(m,\Omega)(\varrho_{S}-\delta)^{m}\int_{G(m,n)}\int_{P^{\perp}}\sum_{\Delta\in
S^{m}_{\perp}(\Omega,z+P)}\theta_{\Delta}\,|\Delta|\,d\mu_{P}^{\perp}(x)\,d\nu(P)$
$\displaystyle=$ $\displaystyle c(m,\Omega)(\varrho_{S}-\delta)^{m}\,M(V)>0.$

This contradicts the above and proves claim (3).

###### Example 4.1

If $\Omega$ is a round ball or an $n$-dimensional ellipsoid, then any
orthogonal slice $\Delta$ contains the origin. Adapting the arguments yields
the following: for any ${\varepsilon}>0$ there is a constant
$C=C(n,m,p,\Omega,{\varepsilon})<\infty$ such that for any $\Sigma\in
S^{m}_{\perp}(\Omega)$

$|\Sigma|+|\partial\Sigma|\leq
C\int_{\Sigma}|A_{\Sigma}|^{p}\,d\mu_{\Sigma}\quad\mbox{ whenever
}\Sigma\subset\Omega\backslash B_{\varepsilon}(0).$

Now turn to the special case when $m=2$ and $p=2$. Let $\Sigma$ be an immersed
surface in $\overline{\Omega}$ with orthogonal boundary condition. For the
geodesic curvature along the boundary with respect to the interior unit
normal, we have

(34) $\varkappa_{g}=h^{S}(\tau,\tau),\quad\mbox{ where }\tau\mbox{ is the unit
tangent along }\partial\Sigma.$

Namely, for a local arclength parametrization $c(s)$ we compute

$\varkappa_{g}=\langle c^{\prime\prime}(s),\nu^{S}(c(s))\rangle=-\langle
c^{\prime}(s),D\nu^{S}(c(s))c^{\prime}(s)\rangle=h^{S}(c^{\prime}(s),c^{\prime}(s)).$

The remaining estimates for the Euler characteristic in Theorem 1 and Theorem
2 now follow easily from the bounds (1) resp. (3) and the Gauß-Bonnet theorem,
using

$2\pi\,|\chi(\Sigma)|\leq\int_{\Sigma}|K_{\Sigma}|\,{\ {\rm
d}}\mu_{\Sigma}+\int_{\partial\Sigma}|\varkappa_{g}|\,ds\leq\frac{1}{2}\int_{\Sigma}|A_{\Sigma}|^{2}\,d\mu_{\Sigma}+\|h^{S}\|_{C^{0}(S)}\,|\partial\Sigma|.$

Proof of Theorem 3. It suffices to prove the bound for $|\Sigma|$. Assume by
contradiction that there are immersed surfaces $\Sigma_{k}\in{\bf
S}^{2}_{\perp}(\Omega)$ such that

$\int_{\Sigma_{k}}|A_{\Sigma_{k}}|^{2}\,d\mu_{\Sigma_{k}}+\chi(\Sigma_{k})^{+}<\frac{1}{k}\,|\Sigma_{k}|.$

The sequence $V_{k}=\Sigma_{k}/|\Sigma_{k}|\in{\bf CV}^{2}_{\perp}(\Omega)$
has norm ${\bf M}(V_{k})=1$ and boundary
$\Gamma_{k}=\partial\Sigma_{k}/|\Sigma_{k}|$. After passing to a subsequence
we have $V_{k}\to V$ and $\Gamma_{k}\to\Gamma$ as varifolds, where Definition
2.2 holds in the limit with curvature $B_{V}=0$. Now the Gauß-Bonnet theorem
implies, denoting by $\tau_{\partial\Sigma_{k}}$ the unit tangent vector of
$\partial\Sigma_{k}$,

$\displaystyle\int_{\partial\Sigma_{k}}h^{S}\big{(}\tau_{\partial\Sigma_{k}},\tau_{\partial\Sigma_{k}}\big{)}\,ds_{\partial\Sigma_{k}}$
$\displaystyle=$
$\displaystyle\int_{\partial\Sigma_{k}}\varkappa_{g}\,ds_{\partial\Sigma_{k}}$
$\displaystyle\leq$ $\displaystyle
2\pi\chi(\Sigma_{k})^{+}-\int_{\Sigma_{k}}K_{\Sigma_{k}}\,d\mu_{\Sigma_{k}}.$

We get a well-defined function $\varphi:G_{1}(TS)\to{\mathbb{R}}$ by putting
$\varphi(x,Q)=h^{S}(x)(\tau,\tau)$ where $\tau\in Q$ with $|\tau|=1$. By the
varifold convergence $\Gamma_{k}\to\Gamma$, we have

$\displaystyle\int_{G_{1}(TS)}\varphi(x,Q)\,d\Gamma(x,Q)$ $\displaystyle=$
$\displaystyle\lim_{k\to\infty}\int_{G_{1}(TS)}\varphi(x,Q)\,d\Gamma_{k}(x,Q)$
$\displaystyle=$
$\displaystyle\lim_{k\to\infty}\frac{1}{|\Sigma_{k}|}\int_{\partial\Sigma_{k}}h^{S}\big{(}\tau_{\partial\Sigma_{k}},\tau_{\partial\Sigma_{k}}\big{)}\,ds_{\partial\Sigma_{k}}$
$\displaystyle\leq$
$\displaystyle\lim_{k\to\infty}\frac{1}{|\Sigma_{k}|}\Big{(}2\pi\chi(\Sigma_{k})^{+}-\int_{\Sigma_{k}}K_{\Sigma_{k}}\,d\mu_{\Sigma_{k}}\Big{)}$
$\displaystyle=$ $\displaystyle 0.$

On the other hand Theorem 7 yields the representation

$\Gamma(\varphi)=\int_{G(2,n)}\int_{P^{\perp}}\sum_{\Delta\in
S^{2}_{\perp}(\Omega,z+P)}\theta_{\Delta}\int_{\partial\Delta}\varphi(x,P\cap
T_{x}S)\,d{\cal H}^{1}(x)\,d\mu_{P}^{\perp}(z)\,d\nu(P).$

Since $\Delta$ is an orthogonal slice, we have by the computation above

$\varphi(x,P\cap
T_{x}S)=h^{S}(x)\big{(}\tau_{\partial\Delta}(x),\tau_{\partial\Delta}(x)\big{)}=\varkappa_{\partial\Delta}(x).$

Here $\Delta$ is a domain in the plane $P$ and $\varkappa_{\partial\Delta}$ is
the Euclidean curvature of its boundary. Now by assumption each $\Delta$ is a
topological disk, therefore by the Hopf Umlaufsatz [11]

$\int_{\partial\Delta}\varphi(x,P\cap T_{x}S)\,d{\cal
H}^{1}(x)=\int_{\partial\Delta}\varkappa_{\partial\Delta}(x)\,d{\cal
H}^{1}(x)=2\pi.$

Inserting we obtain

$\Gamma(\varphi)=2\pi\int_{G(2,n)}\int_{P^{\perp}}\sum_{\Delta\in
S^{2}_{\perp}(\Omega,z+P)}\theta_{\Delta}\,d\mu_{p}^{\perp}(z)\,d\nu(P).$

But this yields, since $|\Delta|\leq C(n,\Omega)$,

$1={\bf M}(V)=\int_{G(2,n)}\int_{P^{\perp}}\sum_{\Delta\in
S^{2}_{\perp}(\Omega,z+P)}\theta_{\Delta}\,|\Delta|\leq
C(n,\Omega)\,\Gamma(\varphi)\leq 0.$

###### Example 4.2

The assumptions in Theorem 1 and Theorem 3 cannot be dropped. Let

$\Omega=\\{(x,y,z)\in{\mathbb{R}}^{3}:1<x^{2}+y^{2}<2,\,-1<z<1\\}.$

If $V_{k}$ is the varifold given by the annulus
$A=\\{(x,y,0):1<x^{2}+y^{2}<2\\}$ with multiplicity $k$, then $V_{k}$ has zero
curvature and orthogonal boundary in $\Omega$, but the mass ${\bf
M}(V_{k})=k\pi$ goes to infinity. By modification one also gets a
counterexample of disk-type surfaces with bounded curvature and orthogonal
boundary in a smooth domain. Namely one can add two copies of the vertical
strip $\\{(x,0,z):1<x<2,\,0<z<1\\}$, thereby obtaining a surface with
orthogonal boundary which can be parametrized on a rectangle. A suitable
approximation yields a smooth example.

##  5 Minimizers with Orthogonal Boundary

Proof of Theorem 5.  If $\Omega$ has an orthogonal slice $\Delta$, then it is
a minimizer with curvature zero and the theorem follows. Otherwise let
$V_{k}\in{\bf CV}^{2}_{\perp}(\Omega)$ be a minimizing sequence. By Theorem 4
we then have

(35) ${\bf M}(V_{k})+{\bf M}(\Gamma_{k})\leq
C\,\|B_{V_{k}}\|_{L^{2}(V_{k})}^{2}\quad\mbox{ where }C=C(n,\Omega).$

Lemma 2.8 yields that after passing to a subsequence, $V_{k}\to V$ and
$\Gamma_{k}\to\Gamma$ as varifolds in ${\mathbb{R}}^{n}$, where $V$, $\Gamma$
satisfy Definition 2.2 and hence $V\in{\bf CV}^{2}_{\perp}(\Omega)$, and
further

$\|B_{V}\|_{L^{2}(V)}\leq\liminf_{k\to\infty}\|B_{V_{k}}\|_{L^{2}(V_{k})}.$

By Remark 2.3, we have for any $\phi\in
C^{1}_{c}({\mathbb{R}}^{n},{\mathbb{R}}^{n})$

$\delta V_{k}(\phi)=-\int_{{\mathbb{R}}^{n}}\langle
H_{V_{k}},\phi\rangle\,d\mu_{V_{k}}-\int_{S}\langle\nu^{S}(x),\phi(x)\rangle\,d\sigma_{\Gamma_{k}}.$

Here $H_{V_{k}}(x)={\rm tr\,}B_{V_{k}}(x,T_{x}\mu_{V_{k}})$ and
$\sigma_{\Gamma_{k}}$ is the weight measure of $\Gamma_{k}$. In particular

$|\delta V_{k}(\phi)|\leq\Big{(}\|H_{V_{k}}\|_{L^{2}(V_{k})}{\bf
M}(V_{k})^{\frac{1}{2}}+{\bf
M}(\Gamma_{k})\Big{)}\|\phi\|_{C^{0}({\mathbb{R}}^{n})}\leq
C\,\|\phi\|_{C^{0}({\mathbb{R}}^{n})}.$

Now Allard’s integral compactness [2] implies $V\in{\bf
ICV}^{2}_{\perp}(\Omega)$. All that remains to show is that $V\neq 0$, i.e. we
need a lower bound

(36) $\liminf_{k\to\infty}{\bf M}(V_{k})>0.$

This is achieved by the following three lemmas. Namely, if we had ${\bf
M}(V_{k})\to 0$, then Lemma 5.2 and Lemma 5.3 would imply

$\liminf_{k\to\infty}\|B_{V_{k}}\|_{L^{2}(V_{k})}^{2}\geq 8\pi.$

But then $V_{k}$ is not a minimizing sequence by Lemma 5.1, a contradiction.

###### Lemma 5.1

For any bounded domain $\Omega\subset{\mathbb{R}}^{n}$ of class $C^{3}$ there
exists a surface $\Sigma\in S^{2}_{\perp}(\Omega)$ with varifold curvature

(37) $\int_{\Sigma}|B_{\Sigma}|^{2}\,d\mu_{\Sigma}<8\pi.$

Proof. Let $S\subset{\mathbb{R}}^{n}$ be given by a graph representation
$y=u(z)$ where $z\in{\mathbb{R}}^{n-1}$ with $|z|\leq 1$, such that

$u(0)=0\quad\mbox{ and }\quad Du(0)=0.$

The upward unit normal of the graph is given by

$\nu(z)=\frac{(-Du(z),1)}{\sqrt{1+|Du(z)|^{2}}}.$

We identify $x\in{\mathbb{R}}^{2}$ with $(x,0)\in{\mathbb{R}}^{n-1}$, and
consider the $2$-dimensional halfsphere

$S_{+}=\\{(x,y):x\in{\mathbb{R}}^{2},\,y\geq
0,\,|x|^{2}+y^{2}=1\\}\subset{\mathbb{R}}^{n}.$

We now define the map

$f:S_{+}\to{\mathbb{R}}^{n},\,f(x,y)=(x,u(x))+y\,\nu(x).$

For $u(x)\equiv 0$ we have $f(x,y)=(x,0)+y(0,1)=(x,y)$. Now consider

$u_{\lambda}(z)=\begin{cases}\frac{1}{\lambda}u(\lambda z)&\mbox{ for
}\lambda\neq 0,\\\ 0&\mbox{ for }\lambda=0.\end{cases}$

Clearly $u_{-\lambda}(z)=-u_{\lambda}(-z)$, and $Du_{\lambda}(z)=Du(\lambda
z)$. Moreover $u_{\lambda}$ depends smoothly on both variables $(\lambda,z)$,
in fact we have

$u_{\lambda}(z)=\int_{0}^{1}\frac{d}{dt}u_{\lambda}(tz)\,dt=\int_{0}^{1}Du(\lambda
tz)\cdot z\,dt.$

The graph of $u_{\lambda}$ has the upward unit normal

$\nu_{\lambda}(z)=\frac{(-Du(\lambda z),1)}{\sqrt{1+|Du(\lambda z)|^{2}}}.$

We obtain the variation

$f_{\lambda}:S_{+}\to{\mathbb{R}}^{n},\,f_{\lambda}(x,y)=(x,u_{\lambda}(x))+y\,\nu_{\lambda}(x).$

Clearly $u_{0}(x)=0$, $\nu_{0}(x)=(0,1)$ and $f_{0}(x,y)=(x,y)$. Now

$\frac{\partial}{\partial\lambda}Du(\lambda z)\cdot
e_{i}|_{\lambda=0}=D^{2}u(0)(z,e_{i})=h^{S}(z,e_{i}).$

Here $h^{S}$ is the second fundamental form of $S$ at $z=0$ with respect to
$(0,1)$. Next

$\frac{\partial}{\partial\lambda}u_{\lambda}(z)|_{\lambda=0}=\int_{0}^{1}D^{2}u(0)(tz,z)\,dt=\frac{1}{2}h^{S}(z,z),$
$\frac{\partial}{\partial\lambda}\nu_{\lambda}(z)|_{\lambda=0}=\frac{\partial}{\partial\lambda}\,\frac{(-Du_{\lambda}(z),1)}{(1+|Du_{\lambda}(z)|^{2})^{\frac{1}{2}}}|_{\lambda=0}=-\sum_{i=1}^{n-1}h^{S}(z,e_{i})e_{i}=D\nu(0)z.$

In summary the velocity field of the variation is

$\phi(x,y):=\frac{\partial}{\partial\lambda}f_{\lambda}(x,y)|_{\lambda=0}=\frac{1}{2}h^{S}(x,x)(0,1)+y\,D\nu(0)x.$

The surface $f(x,y)$ meets the graph of $u(z)$ orthogonally along
$S^{1}=\partial S_{+}\subset{\mathbb{R}}^{2}$. To see this, let $x\in S^{1}$
and choose a unit vector $v\in{\mathbb{R}}^{2}$, $v\perp x$. At the boundary
we have one tangent vector of $f(x,y)$ given by

$\frac{d}{d\alpha}f(\cos\alpha\,x+\sin\alpha\,v,0)|_{\alpha=0}=(v,Du(x)v).$

A second tangent vector is obtained by

$Df(x,0)\cdot(0,1)=\nu(x)=\frac{(-Du(x),1)}{\sqrt{1+|Du(x)|^{2}}}.$

The first vector is tangent to $f|_{S^{1}}$, the second is orthogonal to the
first and is normalized. Hence the second vector is the interior co-normal
$\eta$ along the boundary, and is equal to $\nu(x)$, so that $f(x,y)$
satisfies the orthogonal boundary constraint, and the family
$f_{\lambda}(x,y)$ is admissible for comparison. We have the following first
variation formula for the Willmore functional with boundary terms, see Theorem
1 in [1]:

$\frac{d}{d\lambda}{\cal
W}(f_{\lambda})|_{\lambda=0}=\frac{1}{2}\int_{\Sigma}\langle\vec{W}(f),\phi\rangle\,d\mu_{g}+\frac{1}{2}\int_{\partial\Sigma}\omega(\eta)\,ds_{g}.$

The boundary term is given by

$\omega(\eta)=2\langle\phi,\nabla_{\eta}H\rangle-d\langle\phi,H\rangle(\eta)-\frac{1}{2}|H|^{2}\langle\phi,\eta\rangle.$

In the case of $S_{+}$ we have $\vec{W}=0$, $H(x,y)=-2(x,y)$ and $\nabla H=0$.
Thus

$\langle\phi,H\rangle=-h^{S}(x,x)y-2y\,\langle D\nu(0)x,x\rangle=h^{S}(x,x)y.$

Moreover, $\eta=(0,1)$ along $\partial S_{+}$ and thus
$-d\langle\phi,H\rangle(\eta)=-h^{S}(x,x)$. For the third term on the
boundary, we get directly
$-\frac{1}{2}|H|^{2}\langle\phi,\eta\rangle=-h^{S}(x,x)$. We conclude that

$\frac{d}{d\lambda}{\cal
W}(f_{\lambda})|_{\lambda=0}=-2\int_{S^{1}}h^{S}(x,x)\,ds(x)=-2\pi\,{\rm
tr}_{{\mathbb{R}}^{2}}h^{S}.$

If the sum of the two biggest principal curvatures in direction $(0,1)$ is
positive, then

$\frac{d}{d\lambda}{\cal W}(f_{\lambda})|_{\lambda=0}<0.$

On the other hand, for $\lambda=0$ the integral of $|A^{\circ}|^{2}$ is zero
and hence minimal, therefore its derivative vanishes. Using
$|A|^{2}=|A^{\circ}|^{2}+\frac{1}{2}|H|^{2}$ we conclude

$\frac{d}{d\lambda}\int_{S_{+}}|A_{f_{\lambda}}|^{2}\,d\mu_{f_{\lambda}}|_{\lambda=0}<0.$

Thus for small $\lambda>0$ we have comparison surfaces $f_{\lambda}$ with
curvature energy below the half sphere.

###### Lemma 5.2

Let $V_{k}\in{\bf ICV}(\Omega)$ be a sequence with $\mu_{V_{k}}\to 0$ locally
in $\Omega$. Assume there are points $x_{k}\in{\rm spt\,}\mu_{V_{k}}$ such
that $x_{k}\to x_{0}\in\Omega$. Then

(38) $\liminf_{k\to\infty}\|B_{V_{k}}\|_{L^{2}(V_{k})}^{2}\geq
16\pi=\|B_{{\mathbb{S}}^{2}}\|_{L^{2}({\mathbb{S}}^{2})}^{2}.$

Proof. Let $0<\varrho<{\rm dist}(x,\partial\Omega)$. By the monotonicity
formula, see equations (A.8) and (A.10) in [13], we have for sufficiently
large $k$

$\displaystyle 1$ $\displaystyle\leq$
$\displaystyle\frac{\mu_{V_{k}}(B_{\varrho}(x_{k}))}{\pi\varrho^{2}}+\frac{1}{16\pi}\int_{B_{\varrho}(x_{k})}|H_{V_{k}}|^{2}\,d\mu_{V_{k}}+\frac{1}{2\pi\varrho^{2}}\int_{B_{\varrho}(x_{k})}\langle
H_{V_{k}}(x),x-x_{0}\rangle\,d\mu_{V}$ $\displaystyle\leq$
$\displaystyle\Big{(}\frac{1}{16\pi}+{\varepsilon}\Big{)}\int_{B_{\varrho}(x_{k})}|H_{V_{k}}|^{2}\,d\mu_{V_{k}}+C({\varepsilon})\,\frac{\mu_{V_{k}}(B_{\varrho}(x_{k}))}{\varrho^{2}}.$

Letting $k\to\infty$ and then ${\varepsilon}\searrow 0$ we infer

$\liminf_{k\to\infty}\int_{B_{\varrho}(x_{k})}|H_{V_{k}}|^{2}\,d\mu_{V_{k}}\geq
16\pi.$

From $H_{V}(x)={\rm tr\,}B(x,T_{x}\mu_{V})$ we obviously have
$|B_{V}|^{2}\geq\frac{1}{2}|H_{V}|^{2}$, but this improves by the factor $2$.
Namely, Hutchinson showed the following relations for any
$v\in{\mathbb{R}}^{n}$:

  * •

$B(x,P)(v,P)\subset P^{\perp}$ and $B(x,P)(v,P^{\perp})\subset P$ ([12],
Proposition 5.2.4(iii))

  * •

$\langle B(v,\tau),\nu\rangle=\langle B(v,\nu),\tau\rangle$ for $\tau\in P$,
$\nu\in P^{\perp}$ ([12], Proposition 5.2.4(i)).

Moreover, Mantegazza proved in [14, Theorem 5.4] that
$B(x,P)(P^{\perp}v,\,\cdot\,)\equiv 0$ for $P=T_{x}\mu$, if $V$ is integer
rectifiable. Thus $H_{V}(x)={\rm tr\,}B^{\perp}(x,T_{x}\mu)$ and

(39) $\int_{{\mathbb{R}}^{n}}|H_{V}|^{2}\,d\mu_{V}\leq
2\int_{{\mathbb{R}}^{n}}|B_{V}^{\perp}|^{2}\,d\mu_{V}=\int_{{\mathbb{R}}^{n}}|B_{V}|^{2}\,d\mu_{V}.$

This finishes the proof of the lemma.

###### Lemma 5.3

Let $\Omega\subset{\mathbb{R}}^{n}$ be a bounded domain of class $C^{2}$, and
let $V_{k}\in{\bf ICV}^{2}_{\perp}(\Omega)$ be a sequence with ${\bf
M}(V_{k})\to 0$ and ${\rm spt\,}V_{k}\subset U_{\delta_{k}}(S)$ where
$\delta_{k}\to 0$. Then

(40) $\liminf_{k\to\infty}\int_{\overline{\Omega}}|B_{V_{k}}|^{2}\,dV_{k}\geq
8\pi=\int_{{\mathbb{S}}^{2}_{+}}|B_{{\mathbb{S}}^{2}}|^{2}\,dV_{{\mathbb{S}}^{2}}.$

The rest of this section is devoted to the proof of Lemma 5.3. A monotonicity
formula for $m$-dimensional varifolds with free boundary and mean curvature in
$L^{p}$, $p>m$, was proved by Grüter and Jost in [10], see also De Masi [8].
However, it seems unclear whether the argument of [10] extends to the critical
case $p=m=2$. Our strategy is to reflect the varifold across $S$. This
approach is not possible in the setting of [10], because the estimate for the
mean curvature of the reflected varifold involves the second fundamental form,
see equation (66) below.

In the following, it will be convenient to pullback the geometry by the
reflection map.

###### Definition 5.4

Let $G=(g_{ij})\in C^{1}(U,{\mathbb{R}}^{n\times n})$ be a Riemannian metric
on the open set $U\subset{\mathbb{R}}^{n}$. For an $m$-varifold $V$ on $U$ we
consider the functional

(41) $V^{g}(\phi)=\int_{U\times G(m,n)}\phi(x,P)\,JG(x,P)\,dV(x,P)\quad\mbox{
for }\phi\in C^{0}_{c}(U\times G(m,n)).$

Here $JG(x,P)$ denotes the Riemannian volume element. With respect to a
standard orthonormal basis $\tau_{\alpha}$ of $P$ where $\alpha=1,\ldots,m$,
we have

(42) $JG(x,P)=\big{(}\det
g(x)(\tau_{\alpha},\tau_{\beta})\big{)}^{\frac{1}{2}}=\big{(}\det
PG(x)|_{P}\big{)}^{\frac{1}{2}}.$

The Riemannian mass of $V$ is given by

(43) $M^{g}(V)=\int_{U\times G(m,n)}JG(x,P)\,dV(x,P).$

The first variation of $V^{g}$ with respect to a vector field $X\in
C^{1}_{c}(U,{\mathbb{R}}^{n})$ is defined by

(44) $\delta V^{g}(X)=\frac{d}{dt}M^{g}((\varphi_{t})_{\sharp}V)|_{t=0}.$

######  Lemma 5.5 (Riemannian first variation)

Let $X\in C^{1}_{c}(U,{\mathbb{R}}^{n})$ be a vector field with associated
flow $\varphi_{t}$. Assume that $V$ is an integer rectifiable $m$-varifold in
$U$ with weak second fundamental form $B\in L^{1}(V)$. Then we have the first
variation formula

(45) $\delta V^{g}(X)=-\int_{U}\langle
G(x)H_{g}(x),X(x)\rangle\,JG(x,P(x))\,d\mu_{V}(x),$

where $\mu_{V}$ is the weight measure of $V$ and $P(x)=T_{x}V$ is the
approximate tangent space. The vector $H_{g}$ decomposes as
$H_{g}(x)=\sum_{i=1}^{3}\phi_{i}(x,P(x))$ where

(46) $\displaystyle\phi_{1}(x,P)^{l}$ $\displaystyle=$
$\displaystyle\frac{1}{JG(x,P)}\,g^{lm}(x)\frac{\partial}{\partial
P^{j}_{k}}\big{(}JG\,Q^{i}_{m}\big{)}(x,P)\,B^{k}_{ij}(x,P),$ (47)
$\displaystyle\phi_{2}(x,P)$ $\displaystyle=$
$\displaystyle\frac{1}{JG(x,P)}\,G(x)^{-1}\frac{\partial}{\partial
x^{j}}\big{(}JG\,\,Q\big{)}(x,P)^{\rm T}Pe_{j},$ (48)
$\displaystyle\phi_{3}(x,P)^{i}$ $\displaystyle=$
$\displaystyle-g(x)^{ij}\big{\langle}Q(x,P)G(x)^{-1}\partial_{x^{j}}G(x),P\big{\rangle}.$

Here $Q(x,P)$ is the projection onto $P$ with respect to the inner product
$g(x)$.

Proof. For a diffeomorphism $\varphi\in C^{1}(U,U)$ the pushforward varifold
is defined by

(49) $\varphi_{\sharp}V(\psi)=\int_{U\times
G(m,n)}\psi(\varphi(x),D\varphi(x)P)\,J\varphi(x,P)\,dV(x,P).$

Here by $D\varphi(x)P$ we mean the image subspace, and $J\varphi(x)$ is the
standard Jacobian of $\varphi$ on $P$. The Riemannian mass of
$\varphi_{\sharp}V$ is then

$\displaystyle M^{g}(\varphi_{\sharp}V)$ $\displaystyle=$
$\displaystyle\int_{U\times G(m,n)}JG(y,Q)\,d(\varphi_{\sharp}V)(y,Q)$
$\displaystyle=$ $\displaystyle\int_{U\times
G(m,n)}JG(\varphi(x),D\varphi(x)P)\,J\varphi(x,P)\,dV(x,P).$

Now consider the identity maps $I:(P,g(x))\to P$ and
$J:D\varphi(x)P\to\big{(}D\varphi(x)P,g(\varphi(x))\big{)}$. Then for
$D\varphi(x):P\to D\varphi(x)P$ we have

$\displaystyle J^{g}\varphi(x,P)$ $\displaystyle=$
$\displaystyle\det\big{(}(JD\varphi(x)I)^{\ast}\,JD\varphi(x)I\big{)}^{\frac{1}{2}}$
$\displaystyle=$
$\displaystyle\det(J^{\ast}J)^{\frac{1}{2}}\,\det(D\varphi(x)^{\ast}D\varphi(x))^{\frac{1}{2}}\,\det(I^{\ast}I)^{\frac{1}{2}}$
$\displaystyle=$
$\displaystyle\frac{JG(\varphi(x),D\varphi(x)P)}{JG(x,P)}{J\varphi(x,P)}.$

It follows that

$M^{g}(\varphi_{\sharp}V)=\int_{U\times
G(m,n)}J^{g}\varphi(x,P)\,dV^{g}(x,P)\quad\mbox{ where }dV^{g}=JG\,\,dV.$

Now assume that $\varphi=\varphi_{t}$ is the flow of a vector field $X\in
C^{1}_{c}(U,{\mathbb{R}}^{n})$. Choose an orthonormal basis
$v_{1},\ldots,v_{m}$ of $P$ with respect to $g(x)$, and calculate

$\displaystyle\frac{\partial}{\partial t}J^{g}\varphi_{t}(x,P)|_{t=0}$
$\displaystyle=$ $\displaystyle\frac{\partial}{\partial t}\Big{(}\det
g(\varphi_{t}(x))(D\varphi_{t}(x)v_{\alpha},D\varphi_{t}(x)v_{\beta})\Big{)}^{\frac{1}{2}}|_{t=0}$
$\displaystyle=$ $\displaystyle
g(x)(DX(x)v_{\alpha},v_{\alpha})+\frac{1}{2}D_{X}g(x)(v_{\alpha},v_{\alpha})$
$\displaystyle=$ $\displaystyle{\rm
tr\,}\big{(}Q(x,P)DX(x)|_{P}\big{)}+\frac{1}{2}{\rm
tr\,}\big{(}Q(x,P)G(x)^{-1}D_{X}G(x)|_{P}\big{)}$ $\displaystyle=$
$\displaystyle\langle Q(x,P)\,DX(x),P\rangle+\frac{1}{2}\langle
Q(x,P)\,G(x)^{-1}D_{X}G(x),P\rangle.$

We write

$\displaystyle JG(x,p)\langle Q(x,P)DX(x),P\rangle$ $\displaystyle=$
$\displaystyle\langle D_{x}\big{(}JG(x,P)Q(x,P)X(x)\big{)},P\rangle$
$\displaystyle-\sum_{j=1}^{n}\big{\langle}\frac{\partial}{\partial
x^{j}}\big{(}JG(x,P)Q(x,P)\big{)}\,X(x),Pe_{j}\big{\rangle}.$

We now apply the curvature varifold property with the test vector field
$Y(x,P)=JG(x,P)Q(x,P)X(x)$. Noting that $Y$ is $C^{1}$ and has compact
support, we get

$\displaystyle\int_{U\times G(m,n)}\langle
D_{x}\big{(}JG(x,P)Q(x,P)X(x)\big{)},P\rangle\,dV(x,P)$ $\displaystyle=$
$\displaystyle-\int_{U\times G(m,n)}\big{(}D_{P}Y\cdot B+\langle{\rm
tr\,}B,Y\rangle\big{)}\,dV(x,P).$

Since $V$ is integer rectifiable, it has weak mean curvature

$H^{k}_{V}(x)=\sum_{i=1}^{n}B_{ii}^{k}(x,P(x))\in L^{1}(\mu_{V}),$

where $P(x)$ is the projection onto $T_{x}\mu$ and $\mu_{V}$ is the weight
measure. Now we use $H(x)\perp T_{x}\mu$ $\mu$-almost-everywhere, see Brakke
[5, Chapter 5] or Mantegazza [14, Theorem 5.4]. As $Y(x,P(x))\in T_{x}\mu$,
the last term in (5) vanishes. We conclude

$\displaystyle\delta V^{g}(X)$ $\displaystyle=$
$\displaystyle\frac{d}{dt}M^{g}((\varphi_{t})_{\sharp}V)|_{t=0}$
$\displaystyle=$ $\displaystyle-\int_{U\times G(m,n)}D_{P}Y\cdot B\,dV(x,P)$
$\displaystyle-\int_{U}\sum_{j=1}^{n}\big{\langle}\partial_{x^{j}}(JG\,\,Q)(x,P)X(x),Pe_{j}\big{\rangle}\,dV(x,P)$
$\displaystyle+\frac{1}{2}\int_{U\times G(m,n)}\langle
Q(x,P)G(x)^{-1}D_{X}G(x),P\rangle\,JG(x,P)\,dV(x,P).$

Finally, we rewrite this in Riemannian form, that is

$\delta V^{g}(X)=-\int_{U}\langle
G(x)H_{g}(x),X(x)\rangle\,JG(x,P(x))\,d\mu_{V}(x).$

Now one readily checks this formula for $H_{g}$ as given in the statement of
the Lemma.

######  Lemma 5.6 (local energy bound)

Assume that in the previous lemma, we have

(51) $|G(x)-E_{n}|\leq{\varepsilon}\quad\mbox{ and }\quad|DG(x)|\leq\Lambda.$

Then we can estimate, for $P=P(x)$,

(52)
$\|H_{g}(x)\|^{2}_{g}\,JG(x,P)\leq|H(x)|^{2}+C{\varepsilon}\,|B(x,P)|^{2}+C({\varepsilon},\Lambda).$

Proof.  We consider the sets

$\displaystyle{\cal G}$ $\displaystyle=$
$\displaystyle\\{G\in{\mathbb{R}}^{n\times n}:G^{\rm T}=G,\,|G-{\rm
E}_{n}|\leq\frac{1}{2}\\},$ $\displaystyle{\cal M}$ $\displaystyle=$
$\displaystyle\\{M\in L({\mathbb{R}}^{n},{\mathbb{R}}^{n\times
n}):|M|\leq\Lambda\\}.$

For given $G\in{\cal G}$ and $P\in G(m,n)$, we choose a standard orthonormal
basis $\tau_{1},\ldots,\tau_{m}$ of $P$ and define $g_{\alpha\beta}=\langle
G\tau_{\alpha},\tau_{\beta}\rangle$. The following quantities are well-
defined:

$\displaystyle J(G,P)$ $\displaystyle=$ $\displaystyle\big{(}\det
g_{\alpha\beta}\big{)}^{\frac{1}{2}},$ $\displaystyle Q(G,P)e_{j}$
$\displaystyle=$ $\displaystyle g^{\alpha\beta}\langle
G\tau_{\beta},e_{j}\rangle\,\tau_{\alpha}.$

To see that the functions are smooth, one can choose a smooth orthonormal
basis $\tau_{\alpha}(P)$ for $P$ near a point $P_{0}\in G(m,n)$. Such a basis
can be obtained by taking a basis $\tau_{\alpha}$ for $P_{0}$ and applying the
Gram-Schmidt process to the vectors $P\tau_{\alpha}$. Putting $G=G(x)$ and
$M=DG(x)$, one verifies that the three functions $\phi_{i}(x,P)$ for $P=P(x)$
have the following structure:

$\displaystyle\phi_{1}(x,P)$ $\displaystyle=$ $\displaystyle
f_{1}(G,P)^{ij}_{k}\,B_{ij}^{k},$ $\displaystyle\phi_{2}(x,P)$
$\displaystyle=$ $\displaystyle f_{2}(G,P)\ast M,$
$\displaystyle\phi_{3}(x,P)$ $\displaystyle=$ $\displaystyle f_{3}(G,P)\ast
M.$

For $G={\rm E}_{n}$ and $M=0$ we compute precisely

$\phi_{1}^{l}=\delta^{lm}\frac{\partial P^{i}_{m}}{\partial
P^{j}_{k}}B_{ij}^{k}=B^{l}_{ii}=H^{l}\quad\mbox{ and
}\quad\phi_{2}=\phi_{3}=0.$

It follows that

$\displaystyle|H_{g}(x)-H(x)|$ $\displaystyle\leq$ $\displaystyle
C\,|f_{1}(G,P)-f_{1}({\rm
E}_{n},P)|\,|B|+C\,\big{(}|f_{2}(G,P)|+|f_{3}(G,P)|\big{)}\,|M|$
$\displaystyle\leq$ $\displaystyle C{\varepsilon}\,|B|+C\Lambda.$

We can now estimate further

$\displaystyle\big{|}|H_{g}|^{2}-|H|^{2}\big{|}$ $\displaystyle\leq$
$\displaystyle\big{(}2|H|+C{\varepsilon}\,|B|+C\Lambda\big{)}\cdot\big{(}C{\varepsilon}\,|B|+C\Lambda)$
$\displaystyle\leq$ $\displaystyle
C(|B|+C\Lambda)(C{\varepsilon}|B|+C\Lambda)$ $\displaystyle\leq$
$\displaystyle C{\varepsilon}|B|^{2}+C\Lambda|B|+C\Lambda^{2}$
$\displaystyle\leq$ $\displaystyle
C{\varepsilon}|B|^{2}+C({\varepsilon})\Lambda^{2}.$

Finally we use $\langle Gv,v\rangle\leq(1+C{\varepsilon})|v|^{2}$ and conclude

$\displaystyle\|H_{g}\|_{g}^{2}\,JG$ $\displaystyle\leq$
$\displaystyle(1+C{\varepsilon})|H_{g}|^{2}$ $\displaystyle\leq$
$\displaystyle(1+C{\varepsilon})(|H|^{2}+C{\varepsilon}|B|^{2}+C({\varepsilon})\Lambda^{2})$
$\displaystyle\leq$
$\displaystyle|H|^{2}+C{\varepsilon}|B|^{2}+C({\varepsilon})\Lambda^{2}.$

The lemma is proved.

We now turn to the problem of mapping a varifold by a diffeomorphism.

###### Lemma 5.7

Let $\sigma\in C^{2}(U,\hat{U})$ be a diffeomorphism of open sets
$U,\hat{U}\subset{\mathbb{R}}^{n}$, with pullback metric
$g_{ij}=\langle\partial_{i}\sigma,\partial_{j}\sigma\rangle$. Let $V$ be an
integer rectifiable varifold with compact support in $U$, having second
fundamental form $B\in L^{2}(V)$. Then the pushforward varifold
$\hat{V}=\sigma_{\sharp}V$ has weak mean curvature $\hat{H}\in
L^{2}(\hat{\mu})$, for $\hat{\mu}=\mu_{\hat{V}}$, given by

(53) $\hat{H}\circ\sigma=D\sigma\cdot H_{g}.$

Moreover, we have for $JG(x)=JG(x,T_{x}\mu_{V})$

(54)
$\int_{\sigma(E)}|\hat{H}(y)|^{2}\,d\hat{\mu}(y)=\int_{E}\|H_{g}(x)\|_{g}^{2}\,JG(x)\,d\mu(x).$

Proof. By definition of the varifold pushforward, we have

$\sigma_{\sharp}V(\eta)=\int_{U}\eta(\sigma(x),D\sigma(x)P)J\sigma(x,P)\,dV(x,P).$

For a orthonormal basis $\tau_{\alpha}$ of $P$ we have $\langle
D\sigma(x)\tau_{\alpha},D\sigma(x)\tau_{\beta}\rangle=g(x)(\tau_{\alpha},\tau_{\beta})$,
thus

${\bf
M}(\sigma_{\sharp}V)=\int_{U}J\sigma(x,P)\,dV(x,P)=\int_{U}JG(x,P)\,dV(x,P)=M^{g}(V).$

Now for given $Y\in C^{1}_{c}(\hat{U},{\mathbb{R}}^{n})$ we let $X\in
C^{1}_{c}(U,{\mathbb{R}}^{n})$ be the $\sigma$-related vector field, i.e.
$Y\circ\sigma=D\sigma\cdot X$. Then we have
$\psi_{t}\circ\sigma=\sigma\circ\varphi_{t}$ for the related flows, and

${\bf M}((\psi_{t})_{\sharp}\,\sigma_{\sharp}V)={\bf
M}((\psi_{t}\circ\sigma)_{\sharp}V)={\bf
M}((\sigma\circ\varphi_{t})_{\sharp}V)={\bf
M}(\sigma_{\sharp}\,(\varphi_{t})_{\sharp}V)=M^{g}((\varphi_{t})_{\sharp}V).$

Therefore we obtain

$\displaystyle\delta\hat{V}(Y)$ $\displaystyle=$
$\displaystyle\frac{d}{dt}{\bf
M}((\psi_{t})_{\sharp}(\sigma_{\sharp}V))|_{t=0}$ $\displaystyle=$
$\displaystyle\frac{d}{dt}M^{g}((\varphi_{t})_{\sharp}V)|_{t=0}$
$\displaystyle=$ $\displaystyle-\int_{U}\langle
G(x)H_{g}(x),X(x)\rangle\,JG(x,P(x))\,d\mu(x).$

Transforming back to $\sigma_{\sharp}V$ yields

$\displaystyle\int_{U}\langle G(x)H_{g}(x),X(x)\rangle\,JG(x,P(x))\,d\mu(x)$
$\displaystyle=$ $\displaystyle\int_{U\times G(m,n)}\langle
D\sigma(x)H_{g}(x),D\sigma(x)X(x)\rangle J\sigma(x,P)\,dV(x,P)$
$\displaystyle=$ $\displaystyle\int_{U\times G(m,n)}\langle
D\sigma(\sigma^{-1}(y))H_{g}(\sigma^{-1}(y)),Y(y)\rangle|_{y=\sigma(x)}\,J\sigma(x,P)\,dV(x,P)$
$\displaystyle=$
$\displaystyle\int_{\hat{U}\times{\mathbb{R}}^{n}}\big{\langle}D\sigma(\sigma^{-1}(y))H_{g}(\sigma^{-1}(y)),Y(y)\big{\rangle}\,d(\sigma_{\sharp}V)(x,P).$

We conclude

(55)
$\delta\hat{V}(Y)=-\int_{\hat{U}}\langle\hat{H},Y\rangle\,d\hat{\mu}\quad\mbox{
where }\hat{H}(y)=D\sigma(x)H_{g}(x)|_{x=\sigma^{-1}(y)}.$

Substituting $X(x)=H_{g}(x)$ in the previous calculation, we see finally

$\int_{U}\|H_{g}(x)\|_{g}^{2}JG(x)\,d\mu(x)=\int_{\hat{U}}|\hat{H}(y)|^{2}\,d\hat{\mu}(y).$

We now specialize to the situation where $\Omega\subset{\mathbb{R}}^{n}$ is a
bounded $C^{2}$ domain with boundary $S=\partial\Omega$, and consider the
reflection

(56) $\sigma:U\to U,\,\sigma(x+r\nu^{S}(x))=x-r\nu^{S}(x)\quad\mbox{ where
}(x,r)\in S\times(-\delta,\delta).$

For $U=U_{\delta}(S)$ we assume that $\sigma(U)=U$. We put
$g_{ij}(x)=\langle\partial_{i}\sigma(x),\partial_{j}\sigma(x)\rangle$.

###### Lemma 5.8

For $x\in S$, $v\in{\mathbb{R}}^{n}$ we have, where $P_{S}$ is the projection
onto $TS$,

(57) $g_{ij}(x)=\delta_{ij}\quad\mbox{ and }\quad
Dg_{ij}(x)v=4\langle\nu^{S}(x),v\rangle\,h^{S}(x)(P_{S}e_{i},P_{S}e_{j}).$

Proof. First the $r$-derivative of (56) gives

$D\sigma(x+r\nu^{S}(x))\nu^{S}(x)=-\nu^{S}(x).$

Second let $\gamma(s)$ be a curve in $S$ with $\gamma(0)=x$ and
$\gamma^{\prime}(0)=v\in T_{x}S$. Then we compute, denoting by $W(x)$ the
Weingarten map of $S$,

$\displaystyle({\rm Id}-rW(x))v$ $\displaystyle=$
$\displaystyle\frac{d}{ds}\big{(}\gamma(s)-r\nu(\gamma(s))\big{)}|_{s=0}$
$\displaystyle=$
$\displaystyle\frac{d}{ds}\sigma\big{(}\gamma(s)+r\nu^{S}(\gamma(s))\big{)}|_{s=0}$
$\displaystyle=$ $\displaystyle D\sigma(x+r\nu^{S}(x))\big{(}{\rm
Id}+rW(x)\big{)}v.$

This implies

$D\sigma(x+r\nu^{S}(x))v=({\rm Id}-rW(x))\,({\rm Id}+rW(x))^{-1}v\quad\mbox{
for }v\in T_{x}S.$

For $r=0$ we get as expected

$D\sigma(x)\nu^{S}(x)=-\nu^{S}(x)\quad\mbox{ and }\quad D\sigma(x)v=v.$

This implies $G(x)=D\sigma(x)^{\rm T}D\sigma(x)={\rm E_{n}}$ for $x\in S$.
Moreover it follows that

$DG(x)v=0\quad\mbox{ for }x\in S,\,v\in T_{x}S.$

We compute further

$\frac{\partial}{\partial
r}D\sigma(x+r\nu^{S}(x))\nu^{S}(x)|_{r=0}=\frac{\partial}{\partial
r}(-\nu^{S}(x))=0.$

For the tangential derivative we get, again for $v\in T_{x}S$,

$\displaystyle\frac{\partial}{\partial r}D\sigma(x+r\nu^{S}(x))v|_{r=0}$
$\displaystyle=$ $\displaystyle\frac{\partial}{\partial r}({\rm
Id}-rW(x))\,({\rm Id}+rW(x))^{-1}v|_{r=0}$ $\displaystyle=$
$\displaystyle-2W(x)v.$

Thus we have, again for $v,w\in T_{x}S$.

$\begin{array}[]{lcl}\frac{\partial}{\partial
r}g(x+r\nu^{S}(x))(\nu^{S}(x),\nu^{S}(x))|_{r=0}&=&0,\\\
\frac{\partial}{\partial r}g(x+r\nu^{S}(x))(\nu^{S}(x),v)|_{r=0}&=&0,\\\
\frac{\partial}{\partial
r}g(x+r\nu^{S}(x))(v,w)|_{r=0}&=&4h^{S}(x)(v,w).\end{array}$

The lemma is proved.

We now compute the vector $H_{g}(x)$ as defined in Lemma 5.5 for points $x\in
S$.

###### Lemma 5.9

For $x\in S$ and $P=P(x)$ we have the formula

(58)
$H_{g}(x)=H(x)-2\,\Delta_{P}d_{S}(x)\,P^{\perp}\nu^{S}(x)+4\,P^{\perp}D\nabla
d_{S}(x)\cdot P\nu^{S}.$

Proof. Let $P\in G(m,n)$ be any fixed subspace, with Euclidean orthonormal
basis $\tau_{\alpha}$ for $1\leq\alpha\leq m$. Put
$g_{\alpha\beta}(x)=g(x)(\tau_{\alpha},\tau_{\beta})$. At a point $x\in S$ we
have

$\displaystyle g_{\alpha\beta}(x)$ $\displaystyle=$
$\displaystyle\langle\tau_{\alpha},\tau_{\beta}\rangle=\delta_{\alpha\beta},$
$\displaystyle\partial_{x^{k}}g_{\alpha\beta}(x)$ $\displaystyle=$
$\displaystyle
4\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(x)(\tau_{\alpha},\tau_{\beta}).$

For convenience, let us introduce the notation $\Delta_{P}d_{S}(x)={\rm
tr}_{P}D^{2}d_{S}(x)$. We then have

(59)
$\partial_{x^{k}}JG(x,P)=2\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(\tau_{\alpha},\tau_{\alpha})=2\langle\nu^{S}(x),e_{k}\rangle\,\Delta_{P}d_{S}(x).$

Next we compute for $G(x)=(g_{\alpha\beta}(x))$,

(60)
$\langle\partial_{x^{k}}G,P\rangle=4\,\langle\nu^{S}(x),e_{k}\rangle\,h^{S}(x)(\tau_{\alpha},\tau_{\alpha})=4\,\langle\nu^{S}(x),e_{k}\rangle\,\Delta_{P}d_{S}(x).$

Now we address the $x^{k}$ derivative of $Q(x,P)$. Using again
$g_{\alpha\beta}=g(\tau_{\alpha},\tau_{\beta})$ we have

$Q(x,P)v=g^{\alpha\beta}g(\tau_{\beta},v)\tau_{\alpha}\quad\mbox{ for any
}v\in{\mathbb{R}}^{n}.$

Now we have, again for any $v\in{\mathbb{R}}^{n}$,

$\displaystyle\partial_{x^{k}}g^{\alpha\beta}$ $\displaystyle=$
$\displaystyle-4\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(x)(\tau_{\alpha},\tau_{\beta}),$
$\displaystyle\partial_{x^{k}}g(\tau,v)$ $\displaystyle=$ $\displaystyle
4\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(x)(\tau,v).$

Inserting we have

$\partial_{x^{k}}Q(x,P)v=-4\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(x)(\tau_{\alpha},\tau_{\beta})\langle\tau_{\beta},v\rangle\,\tau_{\alpha}+\,4\,\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(\tau_{\alpha},v)\tau_{\alpha},$

which yields

$\displaystyle\partial_{x^{k}}Q(x,P)v$ $\displaystyle=$ $\displaystyle
4\,\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(x)(\tau_{\alpha},P^{\perp}v)\tau_{\alpha}$
$\displaystyle=$ $\displaystyle 4\,\langle\nu^{S}(x),e_{k}\rangle\,P\,D\nabla
d_{S}(x)P^{\perp}v.$

Now $H_{g}(x)=\sum_{i=1}^{3}\phi_{i}(x,P(x))$, where for
$g_{ij}(x)=\delta_{ij}$ the $\phi_{i}$ are as follows:

$\displaystyle\phi_{1}(x,P)^{l}$ $\displaystyle=$
$\displaystyle\partial_{P^{j}_{k}}(JG\,Q^{i}_{l})(x,P)B_{ij}^{k}(x,P),$
$\displaystyle\phi_{2}(x,P)$ $\displaystyle=$
$\displaystyle(\partial_{x^{k}}JG(x,P))\,Pe_{k}+\partial_{x^{k}}Q(x,P)^{\rm
T}Pe_{k},$ $\displaystyle\phi_{3}(x,P)^{i}$ $\displaystyle=$
$\displaystyle-\langle\partial_{i}G(x),P\rangle.$

For $\phi_{3}$ we have directly

$\phi_{3}(x,P)=-4\langle\nu^{S}(x),e_{k}\rangle\Delta_{P}d_{S}(x)e_{k}=-4\Delta_{P}d_{S}(x)\,\nu^{S}(x).$

For $\phi_{2}(x,P)$ we first calculate

$\displaystyle\langle\partial_{x^{k}}Q(x,P)^{\rm T}v,w\rangle$
$\displaystyle=$ $\displaystyle\langle v,\partial_{x^{k}}Q(x,P)w\rangle$
$\displaystyle=$ $\displaystyle 4\langle\nu^{S}(x),e_{k}\rangle\,\langle
v,D^{2}d_{S}(\tau_{\alpha},P^{\perp}w)\tau_{\alpha}\rangle$ $\displaystyle=$
$\displaystyle 4\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(x)(Pv,P^{\perp}w).$

Inserting we compute further

$\displaystyle\phi_{2}(x,P)$ $\displaystyle=$ $\displaystyle
2\,\langle\nu^{S}(x),e_{k}\rangle\,\Delta_{P}d_{S}(x)\,Pe_{k}$
$\displaystyle+4\,\langle\nu^{S}(x),e_{k}\rangle\,D^{2}d_{S}(x)(Pe_{k},P^{\perp}e_{l})\,e_{l}$
$\displaystyle=$ $\displaystyle
2\,\Delta_{P}d_{S}(x)P\nu^{S}(x)+4\,D^{2}d_{S}(x)(P\nu^{S}(x),P^{\perp}e_{l})e_{l}.$

To simplify further we use

$D^{2}d_{S}(x)(P\nu^{S}(x),P^{\perp}e_{l})=\langle D\nabla
d_{S}(x)P\nu^{S}(x),P^{\perp}e_{l}\rangle e_{l}=P^{\perp}D\nabla
d_{S}(x)P\nu^{S}(x).$

Thus we arrrive at

$\phi_{2}(x,P)=2\,\Delta_{P}d_{S}(x)P\nu^{S}(x)+4\,P^{\perp}D\nabla
d_{S}(x)P\nu^{S}(x).$

It remains to compute $\phi_{1}(x,P)$. This involves only a partial derivative
with respect to the $P$ variable, hence we can assume that $x\in S$ is a fixed
point, in particular we have for all $P$ that $JG(x,P)=1$ and $Q(x,P)=P$. We
then get easily

$\phi_{1}(x,P)^{l}=\frac{\partial P^{i}_{l}}{\partial
P^{j}_{k}}B^{k}_{ij}(x,P)=\delta_{ij}\delta_{kl}B^{k}_{ij}(x,P)=B^{l}_{ii}.$

In other words we just have

$\phi_{1}(x,P(x))=H(x).$

Collecting the three terms we finally have

(62) $H_{g}(x)=H(x)-2\Delta_{P}d_{S}(x)P^{\perp}\nu^{S}(x)+4\,P^{\perp}D\nabla
d_{S}(x)P\nu^{S}(x).$

######  Proposition 5.10 (reflecting the varifold)

Let $\Omega\subset{\mathbb{R}}^{n}$ be a bounded $C^{3}$ domain, and let
$V\in{\bf ICV}^{m}_{\perp}(\Omega)$ be given with ${\rm spt\,}V\subset U$.
Here we assume that $U=U_{\delta}(S)$ is a parallel neighborhood of
$S=\partial\Omega$ such that the reflection $\sigma\in C^{2}(U,U)$ across $S$
is defined. Define

(63) $W(\phi)=V(\phi)+\sigma_{\sharp}V(\phi)\quad\mbox{ for }\phi\in
C^{0}_{c}({\mathbb{R}}^{n}\times G(m,n)).$

Then the integer rectifiable varifold $W$ has the following properties:

(a) $\mu_{W}=2\mu_{V}$ on $S$, and $T_{x}\mu_{W}=T_{x}\mu_{V}\subset T_{x}S$
whenever $T_{x}\mu_{V}$, $x\in S$, exists.

(b) $W$ has weak mean curvature $H_{W}\in L^{2}(\mu_{W})$, given by

(64) $H_{W}(x)=\begin{cases}H_{V}(x)&\mbox{ for }x\in\Omega,\\\ (D\sigma\cdot
H_{g})(\sigma(x))&\mbox{ for
}x\in{\mathbb{R}}^{n}\backslash\overline{\Omega},\\\
\frac{1}{2}\big{(}H_{V}(x)+D\sigma(x)H_{g}(x)\big{)}&\mbox{ for }x\in
S.\end{cases}$

Here $H_{g}$ is as in Lemma 5.5. For $x\in S$ we have precisely

(65) $H_{W}(x)=P^{\top}_{S}H_{V}(x)+\Delta_{P}d_{S}(x)\,\nu^{S}(x).$

(c) Assume $|G-{\rm E_{n}}|\leq{\varepsilon}$ and $|DG|\leq\Lambda$ on $U$.
Then $W$ satisfies

(66) $\int_{{\mathbb{R}}^{n}}|H_{W}|^{2}d\mu_{W}\leq
2\,\int_{\overline{\Omega}}|H_{V}|^{2}d\mu_{V}+C{\varepsilon}\int_{\overline{\Omega}}|B_{V}|^{2}d\mu_{V}+C(\Lambda,{\varepsilon}){\bf
M}(V).$

Proof.  Assuming $V={\bf v}(M,\theta)$ we have $\sigma_{\sharp}V={\bf
v}(\sigma(M),\theta\circ\sigma)$. Thus

$W={\bf v}(M\cup\sigma(M),\theta+\theta\circ\sigma)$

is integer rectifiable and $\mu_{W}=2\mu_{V}$ on $S$. Now if $T_{x}\mu_{V}$
exists for $x\in S$, then $T_{x}\mu_{V}$ is contained in the halfspace
$\langle\nu^{S}(x),v\rangle\geq 0$, and hence $T_{x}\mu_{V}\subset T_{x}S$. In
general $T_{\sigma(x)}\mu_{\sigma_{\sharp}V}=D\sigma(x)T_{x}\mu_{V}$, and
hence $T_{x}\mu_{\sigma_{\sharp}V}=T_{x}V=T_{x}W$ for $x\in S$.

We now turn to (b). In the definition of varifold with orthogonal boundary, we
use a test function $\phi\in C^{1}_{c}({\mathbb{R}}^{n})$ to get the first
variation formula

$\delta V(\phi)=-\int_{\overline{\Omega}}\langle
H(x),\phi(x)\rangle\,d\mu_{V}(x)\\\
-\int_{S}\langle\nu^{S}(x),\phi(x)\rangle\,d\mu_{\Gamma}(x).$

Here $H(x)=\sum_{i=1}^{n}B_{ii}(x,T_{x}\mu_{V})$ and we used that
$\Gamma^{x}(G_{m-1}(T_{x}S))=1$ by definition. We now repeat the computation
of Lemma 5.5. Since now $\phi$ is not compactly supported in $\Omega$, we
obtain additional boundary terms:

$\displaystyle\delta V^{g}(\phi)$ $\displaystyle=$
$\displaystyle-\int_{\overline{\Omega}}\langle
G(x)H_{g}(x),\phi(x)\rangle\,JG(x,P(x))\,d\mu_{V}(x)$
$\displaystyle-\int_{S}\langle\nu^{S}(x),\phi(x)\rangle\,\,d\mu_{\Gamma}(x)-\int_{S}\langle
PH(x),\phi(x)\rangle\,d\mu_{V}(x).$

The last term will eventually be cancelled, however we do not know at present
that the relation $H_{V}\perp T\mu_{V}$ holds on $S$. Replacing $\phi$ by the
$\sigma$-related field $\psi$, i.e. $\phi(\sigma(x))=D\sigma(x)\psi(x)$, we
get

$\displaystyle\delta V^{g}(\psi)$ $\displaystyle=$
$\displaystyle-\int_{\overline{\Omega}}\langle
G(x)H_{g}(x),\psi(x)\rangle\,JG(x,P(x))\,d\mu_{V}(x)$
$\displaystyle+\int_{S}\langle\nu^{S}(x),\phi(x)\rangle\,\,d\mu_{\Gamma}(x)-\int_{S}\langle
PH(x),\phi(x)\rangle\,d\mu_{V}(x)$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{n}\backslash\Omega}\langle
D\sigma(\sigma(y))\cdot
H_{g}(\sigma(y),\phi(y)\rangle\,d\mu_{\sigma_{\sharp}V}$
$\displaystyle+\int_{S}\langle\nu^{S}(x),\phi(x)\rangle\,\,d\mu_{\Gamma}(x)-\int_{S}\langle
PH(x),\phi(x)\rangle\,d\mu_{V}(x).$

Now $\delta(\sigma_{\sharp}V)(\phi)=\delta V^{g}(\psi)$. When adding the
variations of $V$ and $\sigma_{\sharp}V$, the boundary terms involving
$\Gamma$ cancel, and we obtain

$\delta W(\phi)=-\int_{{\mathbb{R}}^{n}}\langle H_{W},\phi\rangle\,d\mu_{W},$

where

$H_{W}(x)=\begin{cases}H_{V}(x)&\mbox{ for }x\in\Omega,\\\
D\sigma(\sigma(x))H_{g}(\sigma(x))&\mbox{ for
}x\in{\mathbb{R}}^{n}\backslash\overline{\Omega}\\\
\frac{1}{2}\big{(}P^{\perp}H_{V}(x)+D\sigma(\sigma(x))H_{g}(\sigma(x))\big{)}&\mbox{
for }x\in S.\end{cases}$

This proves that $W$ has weak mean curvature in $L^{2}(\mu_{W})$. Now
$P\nu^{S}=0$ on $S$, thus Lemma 5.9 yields

$H_{g}(x)=H(x)-2\Delta_{P}d_{S}(x)\nu^{S}(x)\quad\mbox{ on }S.$

But $H_{W}\perp T_{x}\mu_{W}$ by Brakke’s theorem, therefore we get on $S$

$0=P\vec{H}_{W}=\frac{1}{2}PD\sigma(x)H_{V}(x).$

This shows that $H(x)$ is perpendicular to $T_{x}\mu_{V}$ also along $S$, and
claim (b) is proved. Finally claim (c) follows from Lemma 5.6.

Proof of Lemma 5.3.  We apply the reflection method to $V=V_{k}$. For $k$
large, the assumptions of statement (c) in Proposition 5.10 are satisfied.
Using Willmore’s inequality [19], see also [13], we get recalling equation
(39)

$\displaystyle 16\pi$ $\displaystyle\leq$
$\displaystyle\int_{{\mathbb{R}}^{n}}|H_{W_{k}}|^{2}\,d\mu_{W_{k}}$
$\displaystyle\leq$ $\displaystyle
2\int_{\overline{\Omega}}|H_{V_{k}}|^{2}\,d\mu_{V_{k}}+C{\varepsilon}\int_{\overline{\Omega}}|B_{V_{k}}|^{2}\,d\mu_{V_{k}}+C(\Lambda,{\varepsilon}){\bf
M}(V_{k})$ $\displaystyle\leq$
$\displaystyle(2+C{\varepsilon})\int_{\overline{\Omega}}|B_{V_{k}}|^{2}\,d\mu_{V_{k}}+C(\Lambda,{\varepsilon}){\bf
M}(V_{k}).$

The lemma follows by letting $k\to\infty$ and then ${\varepsilon}\searrow 0$.

##  6 Generic Domains

In this section we consider orthogonal $(m-1)$-slices $\Gamma$ of $S$. By this
we mean that $\Gamma$ is a compact, connected, $(m-1)$-dimensional submanifold
of $S$ along which an affine $m$-plane intersects $S$ orthogonally. If
$\Delta$ is an orthogonal $m$-slice of $\Omega$ then each component of
$\partial\Delta$ has this property, but we do not require that $\Gamma$ is
given like this.

######  Theorem 8 (Generic Domains)

Let $\Omega\subset{\mathbb{R}}^{n}$ be a bounded $C^{2}$ domain with boundary
$S=\partial\Omega$. For $m\geq 2$ the following holds:

  * (1)

There exist smooth surfaces $S_{i}$ converging to $S$ in $C^{2}$, which have
no orthogonal $(m-1)$-slices, in particular the domains $\Omega_{i}$ have no
orthogonal $m$-slices.

  * (2)

The set of domains $\Omega$ without orthogonal $m$-slices is open in the
$C^{2}$ topology.

Claim (1) does not extend to $m=1$, in fact for any $x_{0}\in S$ the line
$x_{0}+{\mathbb{R}}\nu^{S}(x_{0})$ intersects $S$ orthogonally. More
interestingly, the segment between points $x_{0},x_{1}\in S$ with
$|x_{0}-x_{1}|={\rm diam}(S)$ is orthogonal to $S$.

The proof starts with a local analysis near a given orthogonal $(m-1)$-slice
$\Gamma_{0}\subset S$ in the plane $z_{0}+P_{0}$, where $P_{0}\in G(m,n)$,
$z_{0}\in{\mathbb{R}}^{n}$ and $P_{0}z_{0}=0$. For any $x\in\Gamma_{0}$ we
have $P_{0}=T_{x}\Gamma_{0}\oplus{\mathbb{R}}\nu^{S}(x)$ and thus
$T_{x}S=T_{x}\Gamma_{0}\oplus P_{0}^{\perp}$. Our first goal is to construct a
parametrization of all nearby $(m-1)$-slices of $S$, whether orthogonal or
not. For suitable ${\varepsilon}_{0}>0$ we chose Fermi type coordinates

$E:\\{(x,v)\in\Gamma_{0}\times P_{0}^{\perp}:|v|<{\varepsilon}_{0}\\}\to
U_{{\varepsilon}_{0}}(\Gamma_{0})\subset S,\,E(x,v)=\pi_{S}(x+v).$

$E$ is a $C^{1}$ diffeomorphism with $DE(x,0)={\rm Id}_{T_{x}S}$. Describing
perturbations of $S$ in direction of the normal $\nu^{S}$ would lead to a loss
of differentiability. Instead we choose a unit vector field $\xi\in
C^{2}(S,{\mathbb{R}}^{n})$ with $\langle\xi,\nu^{S}\rangle\geq\frac{1}{2}$.
For $u\in C^{2}(S)$ we consider the map

$g_{u}\in C^{2}(S,{\mathbb{R}}^{n}),\,g_{u}(x)=x+u(x)\,\xi(x).$

Writing $\nabla$ for the derivative with respect to $x\in S$, we get

$\nabla g_{u}={\rm id}_{S}+du\otimes\xi+u\,\nabla\xi.$

The set of $u\in C^{2}(S)$ with $\|u\|_{C^{2}(S)}<\delta_{0}$ corresponds to a
neighborhood of $S$ in the space of surfaces with $C^{2}$ topology, sending
$u$ to $S_{u}=g_{u}(S)$. The unit normal $\nu_{u}$ along $g_{u}$ is defined by
the equation $N(u,\nu_{u})=0$ where

$N:C^{2}(S)\times C^{1}(S,{\mathbb{R}}^{n})\to C^{1}(TS)\oplus
C^{1}({\mathbb{R}}),\,N(u,\nu)=(\nabla
g_{u})^{\ast}\nu\oplus\frac{1}{2}(|\nu|^{2}-1).$

Now $\nu_{u}=\nu^{S}$ for $u=0$, and we compute

$\displaystyle D_{u}[\nabla g_{u}](0)\,\varphi$ $\displaystyle=$
$\displaystyle\xi\otimes d\varphi+\varphi\,\nabla\xi,$ $\displaystyle
D_{u}N(0,\nu^{S})\,\varphi$ $\displaystyle=$
$\displaystyle\big{(}\langle\nu^{S},\xi\rangle\,{\rm
grad\,}\varphi+\varphi(\nabla\xi)^{\ast}\nu^{S}\big{)}\,\oplus\\{0\\},$
$\displaystyle D_{\nu}N(0,\nu^{S})\,\psi$ $\displaystyle=$ $\displaystyle
P_{TS}\psi\oplus\langle\nu^{S},\psi\rangle.$

Evidently, $D_{\nu}N(0,\nu^{S})$ is an isomorphism, hence $\nu:C^{2}(S)\to
C^{1}(S,{\mathbb{R}}^{n})$, $\nu(u)=\nu_{u}$, is of class $C^{1}$. Moreover at
$u=0$, its Fréchet derivative is

(67) $D\nu(0)\varphi=-\big{(}\langle\nu^{S},\xi\rangle\,{\rm
grad\,}\varphi+\varphi(\nabla\xi)^{\ast}\nu^{S}\big{)}.$

To describe perturbations of the affine plane, we consider the manifold

$M=\\{(z,P)\in{\mathbb{R}}^{n}\times{\mathbb{R}}^{n\times n}:P^{{\rm
T}}=P=P^{2},\,{\rm tr\,}P=m,\,Pz=0\\}.$

For $(z,P,u,v)\in M\times C^{2}(S)\times C^{1}(\Gamma_{0},P_{0}^{\perp})$ with
$\|u\|_{C^{2}(S)}<\delta_{0}$ and
$\|v\|_{C^{0}(\Gamma_{0})}<{\varepsilon}_{0}$ we define $F(z,P,u,v)\in
C^{1}(\Gamma_{0},P_{0}^{\perp})$ by

(68)
$F(z,P,u,v)(x)=P_{0}^{\perp}P^{\perp}\big{(}g_{u}(E(x,v))-z\big{)}|_{v=v(x)}.$

For $|P-P_{0}|^{2}\leq\frac{1}{2}$ we have
$|P_{0}^{\perp}P^{\perp}w|^{2}\geq\frac{1}{2}\,|P^{\perp}w|^{2}$, which then
implies

$F(z,P,u,v)=0\quad\Leftrightarrow\quad g_{u}\big{(}E(x,v(x))\big{)}\in
z+P\mbox{ for all }x\in\Gamma_{0}.$

The function $F$ has derivatives

(69) $D_{u}F(z_{0},P_{0},0,0)\,\varphi=\varphi P_{0}^{\perp}\xi\quad\mbox{ and
}\quad D_{v}F(z_{0},P_{0},0,0)\eta=\eta.$

###### Lemma 6.1

There exist $\delta_{1}\in(0,\delta_{0}]$,
${\varepsilon}_{1}\in(0,{\varepsilon}_{0}]$ and a $C^{1}$ map $\eta:U\to V$
where

$\displaystyle U$ $\displaystyle=$ $\displaystyle\\{(z,P,u)\in M\times
C^{2}(S):|z-z_{0}|,\,|P-P_{0}|,\,\|u\|_{C^{2}(S)}<\delta_{1}\\},$
$\displaystyle V$ $\displaystyle=$ $\displaystyle\\{v\in
C^{1}(\Gamma_{0},P_{0}^{\perp}):\|v\|_{C^{1}(\Gamma_{0})}<{\varepsilon}_{1}\\},$

such that the following statements hold:

  * (1)

for all $(z,P,u)\in U$, $x\in\Gamma_{0}$ and $v\in P_{0}^{\perp}$ with
$|v|<{\varepsilon}_{1}$ we have

(70) $g_{u}(E(x,v))\in z+P\quad\Leftrightarrow\quad v=\eta(z,P,u)(x).$

  * (2)

The map $\gamma(z,P,u):\Gamma_{0}\to{\mathbb{R}}^{n}$,
$\gamma(z,P,u)(x)=g_{u}(E(x,v(x)))$ for $v=\eta(z,P,u)$ is a $C^{1}$
parametrization of an $(m-1)$-slice $\Gamma(z,P,u)$ of $S_{u}$ in the plane
$z+P$.

  * (3)

$\eta(z_{0},P_{0},0)=0$ and $D_{u}\eta(z_{0},P_{0},0)\varphi=-\varphi
P_{0}^{\perp}\xi$.

Proof. Applying the implicit function theorem, we obtain neighborhoods $U,V$
as claimed such that the equation $F(z,P,u,v)=0$ is solved by a function
$v=\eta(z,P,u)$, and hence $g_{u}(E(x,v(x)))\in z+P$ for all $x\in\Gamma_{0}$.
Now consider

$\phi:U\times\Gamma_{0}\times
P_{0}^{\perp},\,\phi(z,P,u,x,v)=P_{0}^{\perp}P^{\perp}\big{(}g_{u}(E(x,v))-z\big{)}.$

Assume there are sequences $(z_{k},P_{k},u_{k})\to(z_{0},P_{0},0)$ and
$x_{k}\in\Gamma_{0}$, such that

$\phi(z_{k},P_{k},u_{k},x_{k},v_{k})=\phi(z_{k},P_{k},u_{k},x_{k},w_{k})=0,$

where $v_{k},w_{k}\in P_{0}^{\perp}$ with $v_{k},w_{k}\to 0$ and $v_{k}\neq
w_{k}$. By compactness, we may assume that $x_{k}\to x_{0}\in\Gamma_{0}$. Then
we obtain by subtracting

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\phi(z_{k},P_{k},u_{k},x_{k},w_{k})-\phi(z_{k},P_{k},u_{k},x_{k},v_{k})$
$\displaystyle=$ $\displaystyle D_{v}\phi(z_{0},P_{0},0,x_{0},0)(w_{k}-v_{k})$
$\displaystyle+\int_{0}^{1}\big{(}D_{v}\phi(z_{k},P_{k},u_{k},x_{k},tw_{k}+(1-t)v_{k})-D_{v}\phi(z_{0},P_{0},0,x_{0},0)\big{)}\cdot(w_{k}-v_{k})\,dt$
$\displaystyle=$ $\displaystyle w_{k}-v_{k}+o(|w_{k}-v_{k}|).$

For $k\to\infty$ we get a contradiction, and the lemma follows.

We now turn to orthogonal slices near $\Gamma_{0}$. For $U$ as in Lemma 6.1 we
define the map $G:U\to C^{1}(\Gamma_{0},P_{0}^{\perp})$ by

(71)
$G(z,P,u)(x)=(P_{0}^{\perp}P^{\perp}\nu_{u})(E(x,v))|_{v=\eta(z,P,u)(x)}.$

We have $G(z,P,u)=0$ if and only if $\gamma(z,P,u)$ parametrizes an orthogonal
$(m-1)$-slice of $S_{u}$. The derivative $D_{u}G(z_{0},P_{0},0)$ is the
operator $L_{0}:C^{2}(S)\to C^{1}(\Gamma_{0},P_{0}^{\perp})$ given by

(72) $L_{0}\varphi=-P_{0}^{\perp}\Big{(}\langle\nu^{S},\xi\rangle\,{\rm
grad\,}\varphi+\varphi\,(\nabla\xi)^{\ast}\nu^{S}+\varphi\,\xi\Big{)}\big{|}_{\Gamma_{0}}.$

In Appendix B we construct a right inverse
$R_{0}:C^{1}(\Gamma_{0},P_{0}^{\perp})\to C^{2}(S)$ to $L_{0}$. The space
$W_{0}={\rm im\,}R_{0}$ is closed in $C^{2}(S)$, and we have the direct sum
decomposition

$C^{2}(S)={\rm ker\,}L_{0}\oplus W_{0},\,\varphi=\varphi-
R_{0}(L_{0}\varphi)\oplus R_{0}(L_{0}\varphi).$

Thus $L_{0}|_{W_{0}}:W_{0}\to C^{1}(\Gamma_{0},P_{0}^{\perp})$ is an
isomorphism.

###### Lemma 6.2

There exist $\delta_{2}\in(0,\delta_{1}]$,
${\varepsilon}_{2}\in(0,{\varepsilon}_{1}]$ and a map $g_{0}\in
C^{1}(A_{0},W_{0})$ where

$A_{0}=\\{(z,P,\chi)\in M\times{\rm
ker\,}L_{0}:|z-z_{0}|,\,|P-P_{0}|,\,\|\chi\|_{C^{2}(S)}<\delta_{2}\\},$

such that for any $u=\chi+w\in{\rm ker\,}L_{0}\oplus W_{0}$ with
$\|u\|_{C^{2}(S)}<{\varepsilon}_{2}$ we have the equivalence

$G(z,P,u)=0\quad\Leftrightarrow\quad\|\chi\|_{C^{2}(S)}<\delta_{2}\mbox{ and
}w=g_{0}(z,P,\chi).$

We define the family of functions

(73)
$C(\Gamma_{0})=\\{u=\chi+g_{0}(z,P,\chi):|z-z_{0}|,|P-P_{0}|\leq\frac{\delta_{2}}{2},\,\|\chi\|_{C^{2}(S)}<\delta_{2}\\}.$

For $\|u\|<{\varepsilon}_{2}$ we have $u\in C(\Gamma_{0})$ if and only if the
slice $\Gamma(z,P,u)$ is orthogonal to $S_{u}$, for some $(z,P)\in M$ with
$|z-z_{0}|,|P-P_{0}|\leq\frac{\delta_{2}}{2}$.

In the following we employ the Hausdorff distance
$d_{H}(\Gamma,\Gamma^{\prime})$ for slices $\Gamma,\Gamma^{\prime}$.

###### Lemma 6.3

There is a constant $\delta=\delta(\Gamma_{0})>0$ such that whenever $\Gamma$
is an orthogonal $(m-1)$-slice of $S_{u}$ in $z+P$ with
$d_{H}(\Gamma,\Gamma_{0}),\,\|u\|_{C^{2}(S)}<\delta$ then $u\in
C(\Gamma_{0})$.

Proof. Let $\Gamma_{k}\subset z_{k}+P_{k}$ be $(m-1)$-slices of $S_{u_{k}}$,
where ${\rm dist}_{H}(\Gamma_{0},\Gamma_{k})\to 0$ and
$\|u_{k}\|_{C^{2}(S)}\to 0$. Then $(z_{k},P_{k})$ (sub-)converges to some
$(z,P)\in M$, where $\Gamma_{0}\subset z+P$ by Hausdorff convergence. We must
have $(z,P)=(z_{0},P_{0})$, otherwise $\Gamma_{0}$ would be contained in an
affine subspace of dimension at most $(m-1)$. For $k$ large we have
$\Gamma_{k}\subset U_{{\varepsilon}_{1}}(\Gamma_{0})$. Lemma 6.1(1) now
implies that $\Gamma_{k}=\Gamma(z_{k},P_{k},u_{k})$. We have
$\eta(z_{k},P_{k},u_{k})\to 0$ in $C^{1}(\Gamma_{0},P_{0}^{\perp})$. Now
decompose $u_{k}=\chi_{k}+w_{k}\in{\rm ker\,}L_{0}\oplus W_{0}$. Then
$|z_{k}-z_{0}|,|P_{k}-P_{0}|\leq\frac{\delta_{2}}{2}$,
$\|u_{k}\|_{C^{2}(S)}<{\varepsilon}_{2}$ and
$\|\chi_{k}\|_{C^{2}(S)}<\delta_{2}$ for $k$ large. As $\Gamma_{k}$ is an
orthogonal slice, we have

$G(z_{k},P_{k},u_{k})(x)=(P_{0}^{\perp}P^{\perp}\nu_{u_{k}})(E(x,v))|_{v=\eta(z_{k},P_{k},u_{k})(x)}=0\quad\mbox{
for all }x\in\Gamma_{0}.$

Therefore Lemma 6.2 and (73) yield $u_{k}\in C(\Gamma_{0})$.

This finishes the local analysis near $\Gamma_{0}$. The following lemma
implies that the set ${\cal O}(S)$ of orthogonal $(m-1)$-slices of $S$,
equipped with Hausdorff distance, becomes a compact metric space.

###### Lemma 6.4

Let $\Gamma_{k}$ be orthogonal $(m-1)$-slices for $S_{u_{k}}$, where
$\|u_{k}\|_{C^{2}(S)}\to 0$. After passing to a subsequence, the $\Gamma_{k}$
converge to an orthogonal $(m-1)$-slice $\Gamma_{0}\neq\emptyset$ of $S$ with
respect to Hausdorff distance.

Proof. We assume that $\Gamma_{k}$ is contained in $z_{k}+P_{k}$ where
$P_{k}z_{k}=0$. Arguing as in the proof of Theorem 2, we get a point $x_{k}\in
z_{k}+P_{k}$ in $\Omega$ with ${\rm dist}(x_{k},S)\geq\varrho_{S}>0$. We may
assume $P_{k}\to P_{0}$ and $x_{k}\to a\in\Omega$, which implies
$z_{k}=P_{k}^{\perp}x_{k}\to P^{\perp}a=:z_{0}$. Now consider a point
$x_{0}=\lim_{k\to\infty}x_{k}\in S$ where $x_{k}\in\Gamma_{k}$. Putting
$S_{k}=S_{u_{k}}$ we have
$\nu^{S}(x_{0})=\lim_{k\to\infty}\nu^{S_{k}}(x_{k})\in z+P$. Let
$A^{\Gamma_{k}}$ be the second fundamental form of $\Gamma_{k}$ as submanifold
of ${\mathbb{R}}^{n}$. As $\Gamma_{k}$ is totally geodesic in $S_{k}$, we have

$A^{\Gamma_{k}}(x)(v,w)=A^{S_{k}}(v,w)\quad\mbox{ for any }v,w\in
T_{x}\Gamma_{k}.$

Hence the norms $\|A^{\Gamma_{k}}\|_{L^{\infty}}$ are bounded. Moreover, Lemma
2.6 implies a bound for the measures $|\Gamma_{k}|$. After passing to a
subsequence, we can assume

${\cal H}^{m-1}\llcorner\Gamma_{k}\to\gamma\quad\mbox{ in
}C^{0}_{c}({\mathbb{R}}^{n})^{\prime}.$

We claim that $\Gamma_{k}$ converges to $\Gamma:={\rm spt\,}\gamma$ in
Hausdorff distance. First assume by contradicition that there exist
$x_{k}\in\Gamma$ with

$\limsup_{k\to\infty}{\rm dist}(x_{k},\Gamma_{k})\geq\delta>0.$

After passing to a subsequence, we get $x_{k}\to x\in\Gamma$ and ${\rm
dist}(x,\Gamma_{k})\geq\delta>0$, a contradiction. Secondly assume, again by
contradiction, that there are $x_{k}\in\Gamma_{k}$ with

$\limsup_{k\to\infty}{\rm dist}(x_{k},\Gamma)\geq\delta>0.$

There is a $\varrho>0$ (the Langer radius) such that for any $k$ and any
$x\in\Gamma_{k}$, the submanifold $\Gamma_{k}$ contains a graphical piece
given by a function

$h_{k}:T_{x}\Gamma_{k}\cap B_{\varrho}(x)\to
T_{x}\Gamma_{k}^{\perp}\quad\mbox{ where }|Dh_{k}|\leq 1,$

see for instance Theorem 2.6 in [6]. In particular $|\Gamma_{k}\cap
B_{\frac{\delta}{2}}(x_{k})|\geq c>0$ uniformly, again a contradiction. It
remains to show that $\Gamma\subset S\cap(z_{0}+P_{0})$ is a
$(m-1)$-dimensional manifold. This follows by representing the slices
$\Gamma_{k}$ locally as graphs, using the implicit function theorem as before.

By Lemma 6.4 the set ${\cal O}(S)$ has a finite covering by the Hausdorff
distance balls

${\cal O}(S)=\bigcup_{i=1}^{\ell}B_{\delta_{i}}(\Gamma_{i})\quad\mbox{ for
}\delta_{i}=\delta(\Gamma_{i})>0\mbox{ as in Lemma }\ref{lemmahausdorff}.$

We claim that if $S_{u}$ has an orthogonal $(m-1)$-slice $\Gamma$ for
$\|u\|_{C^{2}(S)}$ small, then

$u\in\bigcup_{i=1}^{\ell}C_{i}\quad\mbox{ where }C_{i}=C(\Gamma_{i}).$

Namely, let $\Gamma^{\prime}_{k}$ be orthogonal $(m-1)$-slices of $S_{u_{k}}$
where $u_{k}\to 0$ in $C^{2}(S)$. By Lemma 6.4 we can assume that the
$\Gamma^{\prime}_{k}$ converge to an orthogonal $(m-1)$-slice $\Gamma$ of $S$.
Now $\Gamma\in B_{\delta_{i}}(\Gamma_{i})$ for some $i\in\\{1,\ldots,\ell\\}$.
It follows that $d_{H}(\Gamma^{\prime}_{k},\Gamma_{i})<\delta_{i}$ for $k$
large. But then $u_{k}\in C_{i}$ by Lemma 6.3.

It is easy to see that $C_{i}\cap B_{\varrho}(0)$ is closed in
$B_{\varrho}(0)$ for $\varrho>0$ small. For given $u\in B_{\varrho}(0)\cap
C_{i}$ we now construct a perturbation $u^{\prime}\in
U_{i}=B_{\varrho}(0)\backslash C_{i}$. Decompose $u=\chi+w$ and choose a
subspace $X\subset W_{i}$ of dimension $\dim M+1$ with $w\in X$. By Hahn-
Banach there is a continuous projection $Q:W\to X$. Consider the $C^{1}$ map

$\phi:M^{\prime}=\big{\\{}(z,P)\in
M:|z-z_{0}|,|P-P_{0}|\leq\frac{\delta_{2}}{2}\big{\\}}\to
X,\,\phi(z,P)=Q(g_{i}(z,P,\chi)).$

The image of $\phi$ is a Lebesgue null set in $X$. Thus there exists
$w^{\prime}\in X\cap B_{\varepsilon}(w)$ with

$w^{\prime}\,\notin\,Q(g_{i}(M^{\prime},\chi)),\mbox{ in particular
}w^{\prime}\notin g_{i}(M^{\prime},\chi)\cap X.$

It follows that $u^{\prime}=\chi+w^{\prime}\in U_{i}$. Moreover, for those
$j\in\\{1,\ldots,\ell\\}$ for which $u$ is already in $U_{j}$, we can arrange
that also $u^{\prime}\in U_{j}$ by choosing the perturbation sufficiently
small. Repeating this step at most $\ell$ times yields a function
$u_{\varrho}\in\bigcap_{i=1}^{\ell}U_{i}$, which means that $S_{u_{\varrho}}$
has no orthogonal slices. By approximation we even obtain a smooth surface
with that property, applying Lemma 6.4. This finishes the proof of statement
(1) in Theorem 8.

For statement (2) consider a sequence $\Omega_{u_{k}}$ with
$\|u_{k}\|_{C^{2}(S)}\to 0$, where $\Omega_{u_{k}}$ has an orthogonal
$m$-slice $\Delta_{k}$ in $z_{k}+P_{k}$. By Lemma 6.4 we may assume
$(z_{k},P_{k})\to(z,P)$, and $\Gamma_{k}=\partial\Delta_{k}$ converges in
Hausdorff distance to an orthogonal $(m-1)$-slice $\Gamma$ (possibly with
several components) of $S$. Now put

$\Delta=\\{x\in(z+P)\backslash\Gamma:x=\lim_{k\to\infty}x_{k}\mbox{ for some
}x_{k}\in\Delta_{k}\\}.$

By the argument of Theorem 2 which was also used in Lemma 6.4, there are
points $a_{k}\in\Delta_{k}$ with ${\rm dist}(a_{k},S)\geq\varrho_{S}>0$, thus
$\Delta\neq\emptyset$. Now if $x\in\Delta$ and $\varrho<{\rm dist}(x,\Gamma)$,
then $(z_{k}+P_{k})\cap B_{\varrho}(x)\subset\Delta_{k}$ for $k$ large, hence
$(z+P)\cap B_{\varrho}(x)\subset\Delta$. This shows that $\Delta$ is open and
relatively closed in $(z+P)\backslash\Gamma$, and hence $\Delta$ is an
orthogonal $m$-slice of $\Omega$. The proof of Theorem 8 is now complete.

##  7 Appendix

Appendix A. The following disintegration theorem is used in the paper. We
refer to [18], Lemma 38.4, and [3], Theorem 5.3.1, for further information.

######  Theorem 9 (disintegration)

Let $\gamma$ be a Radon measure on $X\times Y$ with compact support, where
$X,Y$ are metric spaces, and denote by $\pi:X\times Y\to Y$ the projection
map. Then $\beta=\pi_{\ast}\gamma$ is a Radon measure, and there is a
$\beta$-a.e. uniquely determined family $(\alpha_{y})_{y\in Y}$ of Radon
probability measures on $X$, such that for any Borel function $\phi:X\times
Y\to[0,\infty]$ one has

(74) $\int_{X\times
Y}\phi\,d\gamma=\int_{Y}\int_{X}\phi(x,y)\,d\alpha_{y}(x)\,d\beta(y).$

Appendix B. In this appendix, we construct the (bounded, linear) right inverse
$R_{0}$ as announced in equation (72). We first construct $R_{0}$ on
${\mathbb{R}}^{n}={\mathbb{R}}^{m}\times{\mathbb{R}}^{p}$.

###### Lemma 7.1

There exists a bounded operator
$R_{0}:C^{0}({\mathbb{R}}^{m},{\mathbb{R}}^{p})\to C^{1}({\mathbb{R}}^{n})$
such that

$R_{0}\psi(x)=0\quad\mbox{ and }\quad{\rm
grad\,}R_{0}\psi(x)=\psi(x)\quad\mbox{ for all }x\in{\mathbb{R}}^{m}.$

Moreover, $\|R_{0}\psi\|_{C^{k+1}({\mathbb{R}}^{n})}\leq
C_{k}\,\|\psi\|_{C^{k}({\mathbb{R}}^{m})}$ and $(R_{0}\psi)(x,z)=0$ for ${\rm
dist}(x,{\rm spt\,}\psi)\geq|z|$.

Proof.  Fix $\eta\in C^{1}_{c}({\mathbb{R}}^{m})$ with support in $B_{1}(0)$
and weight $\int_{{\mathbb{R}}^{m}}\eta(x)\,dx=1$. For $\psi\in
C^{0}({\mathbb{R}}^{m})$ we define $u\in
C^{0}({\mathbb{R}}^{m}\times{\mathbb{R}})$ by

$u(x,z)=z\,(\eta_{z}\ast\psi)(x)\quad\mbox{ where
}\eta_{z}\ast\psi(x)=\int_{{\mathbb{R}}^{m}}\eta(y)\psi(x-zy)\,dy.$

Clearly $u(x,0)=0$. For $z\neq 0$ we have, putting $\sigma_{m,z}=({\rm
sign\,}z)^{m}$,

$(\eta_{z}\ast\psi)(x)=\sigma_{m,z}\int_{{\mathbb{R}}^{m}}z^{-m}\eta\Big{(}\frac{x-y}{z}\Big{)}\,\psi(y)\,dy.$

Differentiating under the integral and transforming back yields

$\displaystyle\partial_{i}u(x,z)$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{m}}\partial_{i}\eta(y)\,\psi(x-zy)\,dy\quad\mbox{
for }i=1,\ldots,m,$ $\displaystyle\partial_{z}u(x,z)$ $\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{m}}\eta(y)\psi(x-zy)\,dy-\int_{{\mathbb{R}}^{m}}\Big{(}m\eta(y)+\sum_{i=1}^{m}y^{i}\partial_{i}\eta(y)\Big{)}\,\psi(x-zy)\,dy.$

Thus $Du\in C^{0}({\mathbb{R}}^{m+1},{\mathbb{R}}^{m+1})$ and
$\|u\|_{C^{1}({\mathbb{R}}^{m+1})}\leq C\,\|\psi\|_{C^{0}({\mathbb{R}}^{m})}$.
Since $\eta$ has weight one while $m\eta+y\cdot D\eta={{\rm
div}}\big{(}\eta(y)y\big{)}$ has weight zero, we conclude
$\partial_{z}u(x,0)=\psi(x)$. Now assume $\psi\in C^{k}({\mathbb{R}}^{m})$ for
some $k\geq 1$. We have for any $\alpha\in{\mathbb{N}}_{0}^{m}$,
$s\in{\mathbb{N}}_{0}$ with $|\alpha|+s=k$

$\displaystyle\partial^{\alpha}(\partial_{z})^{s}(\eta_{z}\ast\psi)(x,z)$
$\displaystyle=$
$\displaystyle\int_{{\mathbb{R}}^{m}}\sum_{\beta\in{\mathbb{N}}_{0}^{m}:|\beta|=s}\eta_{\beta}(y)\,\partial^{\alpha+\beta}\psi(x-zy)\,dy,\quad\mbox{
where }$ $\displaystyle\eta_{\beta}(y)$ $\displaystyle=$
$\displaystyle(-1)^{|\beta|}\frac{s!}{\beta!}\eta(y)y^{\beta}.$

For $s=0$ we have
$\partial^{\alpha}(z\,\eta_{z}\ast\psi)=z\,\partial^{\alpha}(\eta_{z}\ast\psi)=z\,\eta_{z}\ast\partial^{\alpha}\psi$,
while for $s\geq 1$

$\displaystyle\partial^{\alpha}(\partial_{z})^{s}(z\,\eta_{z}\ast\psi)$
$\displaystyle=$ $\displaystyle
z\,\partial^{\alpha}(\partial_{z})^{s}(\eta_{z}\ast\psi)+\partial^{\alpha}(\partial_{z})^{s-1}(\eta_{z}\ast\psi)$
$\displaystyle=$
$\displaystyle\sum_{\beta\in{\mathbb{N}}_{0}^{m}:|\beta|=s}z\,(\eta_{\beta})_{z}\ast\partial^{\alpha+\beta}\psi+\sum_{\beta\in{\mathbb{N}}_{0}^{m}:|\beta|=s-1}(\eta_{\beta})_{z}\ast\partial^{\alpha+\beta}\psi.$

The second sum involves only derivatives of order at most $k-1$. For the first
sum we apply the argument above, with $\eta$ replaced by $\eta_{\beta}$. This
shows $\partial^{\alpha}(\partial_{z})^{s}(z\,\eta_{z}\ast\psi)\in
C^{1}({\mathbb{R}}^{m+1})$ and hence $z\,\eta_{z}\ast\psi\in
C^{k+1}({\mathbb{R}}^{m+1})$. To prove the theorem, we finally put
$u(x,z)=\sum_{j=1}^{p}u_{j}(x,z)$ where
$u_{j}(x,z)=\sum_{j=1}^{p}z_{j}\,\eta_{z_{j}}\ast\psi_{j}(x)$. The claim
follows immediately defining $R_{0}\psi=u$.

Before we proceed to the general case we observe that if ${\rm
spt}(\psi_{j})\subset W$ for all $j=1,...,p$ then for each $\epsilon>0$ one
may also achieve that ${\rm spt}(R_{0}\psi)\subset
B_{2\sqrt{p}\epsilon}(W)\times(-\epsilon,\epsilon)^{p}$. Indeed, let
$\alpha\in C_{0}^{\infty}((-\epsilon,\epsilon)^{p})$ be a fixed function such
that $\alpha\equiv 1$ on $(-\frac{\epsilon}{2},\frac{\epsilon}{2})^{p}$ and
look at $\tilde{R}_{0}:C^{0}({\mathbb{R}}^{m},{\mathbb{R}}^{p})\rightarrow
C^{1}({\mathbb{R}}^{n})$ defined via

(75) $(\tilde{R}_{0}\psi)(x,z)=\alpha(z)(R_{0}\psi)(x,z).$

Clearly, $\tilde{R}_{0}$ has all the properties in the statement of the
previous lemma. Moreover, $(x,z)\in{\rm spt}(\tilde{R}_{0}\psi)$ implies
$z\in(-\epsilon,\epsilon)^{p}$ and $(x,z)\in{\rm spt}(R_{0}\psi)$. One infers
$|z|\leq\sqrt{p}\,\epsilon$ and by the previous lemma this means ${\rm
dist}(x,{\rm spt}(\psi))\leq\sqrt{p}\,\epsilon$.

###### Lemma 7.2

Let $N$ be an $n$-dimensional $C^{l}$-manifold and $M$ be a compact
$m$-dimensional $C^{l}$-submanifold without boundary, $l\geq 1$. Then there
exists a bounded linear operator $R_{0}:C^{0}(M,(TM)^{\perp})\rightarrow
C^{1}(N)$ such that

$R_{0}\psi|_{M}=0\quad{\rm and}\quad{\rm grad\,}R_{0}\psi|_{M}=\psi.$

Further $\|R_{0}\psi\|_{C^{k+1}(N)}\leq C_{k}(M,N)\|\psi\|_{C^{k}(M)}$ for all
$k=0,...,l-1$.

Proof of Lemma 7.2. Let $M$ and $N$ be as in the statement and $\psi\in
C^{k-1}(M,(TM)^{\perp})$ be arbitrary. It is possible to choose a finite
collection of open sets $U_{1},...,U_{r}\subset N$ covering $M$ and maps
$\Phi_{j}\in C^{l}(U_{j};{\mathbb{R}}^{n})$ which are diffeomorphisms onto
their images and satisfy $\Phi_{j}(U_{j}\cap
M)=\Phi_{j}(U_{j})\cap{\mathbb{R}}^{m}$. Note in particular that
$D\Phi_{j}(x)(T_{x}M)={\mathbb{R}}^{m}$. Let now $\beta_{1},...,\beta_{r}$ be
a partition of unity for $M$ such that $\beta_{j}\in C^{l}_{c}(M\cap U_{j})$
for all $j$. By openness of $U_{j}$ it is possible to choose $\epsilon>0$ such
that $B_{2\sqrt{p}\epsilon}(\Phi_{j}({\rm
spt}(\beta_{j})))\times(-\epsilon,\epsilon)^{p}\subset\subset\Phi_{j}(U_{j})$
for all $j=1,...,r$. Next define
$w_{j}:=\beta_{j}[(D\Phi_{j})^{*}]^{-1}\psi\circ\Phi_{j}^{-1}$ and
$u:N\rightarrow{\mathbb{R}}$ to be

$u(x)=\sum_{j=1}^{r}\tilde{R}_{0}w_{j}(\Phi_{j}(x)),$

where $\tilde{R}_{0}$ is chosen as in (75) with $\epsilon>0$ as above, in
particular one has ${\rm
spt}(\tilde{R}_{0}w_{j})\subset\subset\Phi_{j}(U_{j})$. We remark that for the
well-definedness and $C^{k+1}$-smoothness of $\tilde{R}_{0}w_{j}$ we need to
ensure that $w_{j}\in C^{k}({\mathbb{R}}^{m},{\mathbb{R}}^{p})$. This is due
to the fact that $D\Phi_{j}\in C^{l-1}$ and maps $TM$ to ${\mathbb{R}}^{m}$,
which implies that $(D\Phi_{j})^{*}$ maps ${\mathbb{R}}^{p}$ to
$(TM)^{\perp}$. One readily checks that $u|_{M}=0$ and with the chain rule one
obtains for $x\in M$

$\displaystyle{\rm grad}(u)(x)$
$\displaystyle=\sum_{j=1}^{r}D\Phi_{j}(x)^{*}({\rm
grad}(\tilde{R}_{0}w_{j})(\Phi_{j}(x))$
$\displaystyle=\sum_{j=1}^{r}D\Phi_{j}(x)^{*}(\beta_{j}(x)(D\Phi_{j}(x)^{*})^{-1}\psi(x))=\psi(x).$

The claim follows now defining $R_{0}\psi=u$. From the explicit construction
one easily deduces the desired operator norm estimates for $R_{0}$.

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