% Aenderungen nach dem Abschicken:
% - fulfil -> fulfill.
% - abstract um Dimensionsangabe erweitert.
% - Regularitaet fuer mehr als eine Richtung eingefuegt.
% - manchmal: minimal <-> nontrivial cone
% - Z. 560 (ca.) homogen(e)ity, e eingefuegt.
% - Z. 1353 (ca.) math(e)matics, e eingefuegt.
\documentclass[12pt]{article}
\usepackage{amssymb}
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\newtheorem{thm}{Theorem}[section]
\newtheorem{df}[thm]{Definition}
\newtheorem{satz}[thm]{Theorem}
\newtheorem{lem}[thm]{Lemma}
\newtheorem{lemdef}[thm]{Lemma and Definition}
\newtheorem{cor}[thm]{Corollary}
\newtheorem{bem}[thm]{Remark}
\newtheorem{rem}[thm]{Remark}
\newtheorem{bsp}[thm]{Example}
\newtheorem{exa}[thm]{Example}
\newtheorem{pro}[thm]{Proposition}
\newtheorem{bez}[thm]{Notation}
\newtheorem{nota}[thm]{Notation}
\def\N{{\mathbb{N}}}
\def\R{{\mathbb{R}}}
\def\Z{{\mathbb{Z}}}
\def\bew{{\bf Proof: }}
\def\bewidee{{\bf Sketch of the Proof: }}
\def\bewvon#1{{\bf Proof of #1: }}
\def\qed{\hspace*{\fill}$\blacksquare$}
\def\r1{{\R^{n+1}}}
\def\reu{{\r1\setminus\{0\}}}
\def\sub{{\rm sub\,}}
\def\div{{\rm div\,}}
\def\dist{{\rm dist\,}}
\def\supp{{\rm supp\,}}
\def\diam{{\rm diam\,}}
\def\oe{{w.~l.~o.~g.\ }}
\def\OE{{W.~l.~o.~g.\ }}
\def\quada{\hspace*{1em}}
\hyphenation{re-pre-sen-table re-gu-la-ri-ty non-tri-vi-al}
\parindent0cm
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\sloppy
\begin{document}
%%%%% 14.11.98 %%%%%
\title{Regularity of directed minimal Cones}
\author{Oliver~C.~Schn\"urer}
\date{January 5, 1999}
\maketitle

%%%%% 27.12.98 %%%%%
\begin{abstract}
This paper deals with directed minimal cones in $\r1$
which have at most one singularity. We show that such cones are
-- besides the trivial cases $\emptyset$, $\r1$ -- half spaces.
Using blow-up techniques we show how this result can be used
to get $C^{1,\lambda}$-regularity for the measure-theoretic
boundary of almost minimal Caccioppoli sets which are
representable as subgraphs in $\R^8$.
\end{abstract}

%%%%% 28.11.98 %%%%%
\section{Introduction}
Let $C\subset\r1$ be a minimal cone with its vertex in the origin,
i.~e.\
$$x\in C,\,\tau>0\quad\Longrightarrow\quad \tau x\in C$$
and
$$\int\limits_\Omega|D\chi_C|\le\int\limits_\Omega|D\chi_F|$$
whenever $\Omega\subset\r1$ is an open bounded set and
$C\vartriangle F\Subset\Omega$. A measurable set
$M\subset\r1$ is called directed with respect to
$w\in\r1\setminus\{0\}$, if the functions
$$\R\ni t\mapsto D_\rho(M,x+tw)\equiv \omega_{n+1}^{-1}\rho^{-(n+1)}
  \int\limits_{B_\rho(x+tw)}\chi_M$$
are monotone decreasing for each $\rho>0$ and $x\in\r1$.
Now let us assume additionally that the cone $C$ is directed, nontrivial
($C\not=\emptyset$ and $C\not=\r1$) and has at most one singularity.
In this case we prove a Regularity Theorem which states that
such a cone is a half space.
\par
This result is well known -- even for non-directed cones --
provided that the dimension is small, i.~e.\ $n\le6$.
Let $M\subset\r1$ be an almost minimal set.
Let $x\in\partial E$, \oe we may assume $x=0$.
Define for $t>0$
$$E_t:=\{x\in\r1:tx\in E\}.$$
For an almost minimal set it is well known that there exists a
sequence $t_i\to0$, $t_i>0$, such that
$$E_{t_i}\to C \mbox{ in } L^1_{loc}(\r1)$$
and $C$ is a nontrivial minimal cone.
The regularity of $\partial E$ at $0$ is connected with the
regularity of $\partial C$ at $0$ by the equivalence
$$0\in\partial^\star E\quad\Longleftrightarrow\quad
  0\in\partial^\star C.$$
Especially if $\partial C$ is a hyperplane through the origin
we get $0\in\partial^\star E$. It is well known that
in the case of small dimensions ($n\le6$) this blow-up
technique yields the equality of the measure-theoretic
boundary $\partial E$ with the reduced boundary $\partial^\star E$
for almost minimal sets $E\subset \r1$, because the only
nontrivial minimal cones in $\r1$ are half spaces. In
$\R^8$ there are counterexamples.
\par
An application of the Regularity Theorem is as follows:
The Regularity Theorem allows us to conclude in the same way
$\partial E=\partial^\star E$ in $\R^8$, when $E$ is a
directed set. This occurs for example when $E$ is representable
as a subgraph.
Let $w\in\R^8\setminus\{0\}$ be the corresponding direction.
Obviously the sets $E_{t_i}$ are again directed with respect to
the same vector $w$. Since the limit in $L^1_{loc}(\r1)$ of sets which
are directed with respect to $w$ is directed, too, the
blow-up technique yields a minimal nontrivial directed
cone $C$. It is well known that the dimension of the singular
set is limited by
$$H^s(\partial E\setminus\partial^\star E)=0\quad\forall s>n-7.$$
If there were another singular point apart from the origin,
the property of $C$ being a cone
would imply $H^1(\partial E\setminus\partial^\star E)=+\infty$,
contradicting $n\le7$. This shows that
there is no singular point besides the origin.
Therefore the Regularity Theorem implies that $C$ is a half space
and consequently we get $\partial E=\partial^\star E$.
This proof is valid for all $n\le7$. The problem in extending
this result to $\r1$ with $n\ge8$ is to show that the singular
set consists of at most one point. In $\R^9$ there is indeed a
counterexample which shows that there exists a nontrivial
minimal directed cone which is different from a half space.
\par
This paper is organized as follows:
In section \ref{defkap} we introduce some notations
and recall some well known facts.
In section \ref{gerkegkap} we consider directed cones and show that
directed cones are weakly star-shaped with respect to certain
points. These properties are essential for the proof
of the Regularity Theorem which is stated and proved in
section \ref{regsatzkap}.\\
The following sections contain applications of the Regularity
Theorem: In section \ref{anwendkap} we apply the Theorem to subgraphs, in
section \ref{vorgegHkap} we show how it can be applied to
getting hypersurfaces
of prescribed mean curvature homeomorphic to $S^n$ for $n\le7$.
Since we additionally use the fact that the cones arising are
directed, we get the results known not only up to $\R^7$
but also in $\R^8$.

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% 20.11.98 %%%%%
\newpage
\section{Preliminaries}\label{defkap}
%%%%% 20.11.98 %%%%%
\begin{bez}{\rm
The characteristic function of a set $M$ will be
denoted by $\chi_M$. If we are concerned with coordinates in $\R^{n+1}$,
$(x,t)$ stands for $(x^1,\ldots,x^n,t)$, $x=(\hat x,x^{n+1})$
also summarizes the first $n$ coordinates to $\hat x$.
Assume in the whole paper $n\ge2$.}
\end{bez}

%%%%% 28.11.98 %%%%%
\begin{df}[almost minimal] %\label{fastminimal}
A measurable set $E \subset \R^{n+1}$ is called almost minimal
in $\Omega\subset\R^{n+1}$, if there exists a $\lambda$ such that
$0<\lambda<1$ and for all sets $A \Subset\Omega$
there are numbers $R$ and $K$ such that
$0 < R < \dist (A, \complement\Omega)$ and $K\ge 0$ which fulfill
the inequality
$$ \int\limits_{B_\rho(x)}|D\chi_E|\le
   \int\limits_{B_\rho(x)}|D\chi_F| + K \rho^{n+2\lambda} $$
for every $x\in A$, $0 < \rho < R$ and
$E \vartriangle F \Subset B_\rho(x)$.\par
$E$ is called  (locally) minimal, if we can choose $K=0$.
\end{df}

%%%%% 01.12.98 %%%%%
In this paper we deal with two kinds of boundaries,
the measure-theoretical boundary and the reduced boundary,
which are defined according to \cite[p.~43]{giusti}.
Obviously these definitions are invariant, if we change
the respective sets by a set of Lebesgue measure zero.
\par
\begin{df}[(measure-theoretical) boundary]
Let $E\subset \R^{n+1}$ be a measurable set.
Define the measure-theoretical boundary
$$\partial E:=\{x\in \R^{n+1}:
  0<|B_r(x)\cap E|<|B_r(x)|\quad\forall r>0\},$$
and the measure-theoretical interior and the
measure-theoretical exterior by
$$E_\mu:=\{x\in\R^{n+1}:\exists r>0: |B_r(x)\cap E|=|B_r(x)|\}$$
and
$$\complement_\mu E:=\{x\in\R^{n+1}:\exists r>0: 0=|B_r(x)\cap E|\},$$
respectively.
\end{df}
\par
According to this definition, $E_\mu$ and $\complement_\mu E$
are open sets. $\R^{n+1}$ is the disjoint union of $E_\mu$,
$\complement_\mu E$ and $\partial E$.
Later-on we will denote $E_\mu$ resp.\ $\complement_\mu E$ simply by
$E$ resp.\ $\complement E$.
If $E$ is an almost minimal set, $E$ and $E_\mu$ lie in the
same equivalence class. This follows immediately, since
the boundary of $E$ is the union of a differentiable
manifold and of a set of $H^s$ measure zero, if $s>n-7$, and is
therefore a set of $H^{n+1}$ measure zero.
So we will always assume \oe that an almost minimal set is open.

%%%%% 20.11.98 %%%%%
\begin{df}[reduced boundary]
Let $E\subset \R^{n+1}$ be a Caccioppoli set. The
reduced boundary $\partial^\star E$ consists of those points
$x\in \R^{n+1}$, for which
$$\int\limits_{B_\rho(x)}|D\chi_E|>0 \quad\forall \rho>0$$
is valid and the limit
$$\lim\limits_{\rho\to 0} \nu_\rho(x)=\nu(x)$$
exists. $\nu_\rho(x)$ is defined by
$$\nu_\rho(x):=\frac{\int\limits_{B_\rho(x)}D\chi_E}
  {\int\limits_{B_\rho(x)}|D\chi_E|}$$
and additionally we demand
$$|\nu(x)|=1.$$
$\nu(x)$ is called the inner (unit) normal at $E$ in $x$.
\end{df}

%%%%% 18.10.98 %%%%%
\begin{df}[singularity]
$x\in\R^{n+1}$ is called a singularity of the measurable
set $M\subset\R^{n+1}$, if $x\in\partial M\setminus\partial^\star M$.
$\partial M\setminus\partial^\star M$ is called the singular
set of $M$ or the singular set of $\partial M$.
\par
A point in the boundary which is not singular is called regular.
\end{df}

%%%%% 26.11.98 %%%%%
\begin{df}[cone]
$C\subset\R^{n+1}$ is called a cone with vertex $x$, if
$$y\in C\quad\Longrightarrow\quad x+\tau(y-x)\in C\quad\forall 0<\tau.$$
If $C\not=\emptyset$ and $C\not=\R^{n+1}$, $C$ is called
a nontrivial cone.
\end{df}

%%%%% 28.11.98 %%%%%
\begin{lemdef}\label{dichtgleichmass}
Let $E\subset \R^{n+1}$ be an almost minimal Caccioppoli set in the
open set $\Omega\subset\R^{n+1}$. Assume $0\in\partial E$.
If $E_t:=\{x\in\r1:tx\in E\}$ converges in $L^1_{loc}(\r1)$ to a
set $C$ for a given sequence $t_i\to0$, $t_i>0$, then $C$ is a cone
which is different from $\r1$ and $\emptyset$.
{\rm Such a cone $C$ is called a blow-up cone of $E$ around $0$.}
\end{lemdef}
\bew
Proceed as in \cite [Theorem~9.3]{giusti} and use
the estimates of \cite[p.~118]{mm} and
\cite[Proposition, p.~137]{mm}.
\qed

%%%%% 28.11.98 %%%%%
\begin{bem}\label{inkleinendimensionen}{\rm {}
In small dimensions, i.~e.\ for $n+1\le8$, the blow-up cone $C$
of an almost minimal set around a point of its boundary
has no singularity apart from the origin:
If there were another singularity, this would imply
that a half-line would be singular, because $C$ is a cone.
But this is a contradiction to the fact, that
$H^s(\partial^\star C\setminus\partial C)=0$ for $s>n-7$.
\par
Recall that the reduced boundary of a minimal set
is an analytic manifold. So, in particular, the
boundary of a minimal cone is analytic apart
from the origin, if $n+1\le8$.
}\end{bem}

%%%%% 26.11.98 %%%%%
\begin{df}[subgraph]
Let $\varphi:A\to[-\infty,+\infty]$ be a measurable function.
Define the subgraph of the function $\varphi$ by
$${\rm sub\,} \varphi :=
\{(x, t) \in A\times \R : t < \varphi(x)\}.$$
\end{df}

%%%%% 28.11.98 %%%%%
\begin{bem}
{\rm If a cone with its vertex in the origin is a subgraph of a
function $u$, then $u$ is positive homogeneous of degree $1$.}
\end{bem}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% 18.11.98 %%%%%
\newpage
\section{Directed Cones}\label{gerkegkap}
%%%%% 29.11.98 %%%%%
\begin{df}[directed set]
Let $E\subset \R^{n+1}$ be a measurable set,
$y\in\R^{n+1}\setminus\{0\}$.
$$D_{\rho}(E, x):=
  \omega_{n+1}^{-1}\rho^{-n-1}\int\limits_{B_{\rho}(x)}\chi_E$$
is called approximative density of $E$ in $x$ with respect
to the radius $\rho$.
\par
A set $E$ is called directed with respect to $y$, if the maps
$$f_{x,\rho}:\R\to [0,1],\quad t\mapsto D_\rho(E,x+ty)$$
are monotone decreasing with respect to $t$ for any
$x\in\R^{n+1}$, $\rho>0$, such $y$ is called direction of the
set.
\end{df}

%%%%% 18.11.98 %%%%%
\begin{bem}
{\rm It is possible that a set has different linearly
independent directions.}
\end{bem}

%%%%% 28.11.98 %%%%%
\begin{lem}\label{untereinanderlemma}
Let $M\subset\R^{n+1}$ be a measurable set which is directed
with respect to $v$. Then for any $x\in\R^{n+1}$ and $t>0$
we have the implications
$$x\in\complement M\quad\Longrightarrow\quad x+tv\in\complement M$$
and
$$x\in M\quad\Longrightarrow\quad x-tv\in M.$$
\end{lem}
\bew
According to the definition of the
measure-theoretical complement of a set
there exists in the case $x\in\complement M$ a $\rho>0$
such that $D_\rho(M,x)=0$.
Since $M$ is directed with respect to $v$, it follows for $t>0$ that
$$D_\rho(M,x+tv)\le D_\rho(M,x)=0.$$
$D_\rho(\cdot,\cdot)$ is non-negative, so $D_\rho(M,x+tv)=0$.
This implies $x+tv\in\complement M$.
\par
In the case $x\in M$ the proof is similar:
$x\in M\Rightarrow\exists\rho>0:D_\rho(M,x)=1$.
$M$ is a directed set. Now $t>0$ implies
$D_\rho(M,x-tv)\ge D_\rho(M,x)=1$ and it follows $x-tv\in M$ as above.
\qed

%%%%% 28.11.98 %%%%%
\begin{cor}
Let $M\subset\R^{n+1}$ be a measurable set which is directed
with respect to $v$.
If $x\in\R^{n+1}$ and $t>0$ are such that $x\in\partial M$ and
$x+tv\in\partial M$, it follows that $x+\tau v\in\partial M$ for
$0\le\tau\le t$.
\end{cor}

%%%%% 28.11.98 %%%%%
\begin{lem}
Let $M\subset\R^{n+1}$ be a measurable set, directed
with respect to $v$. For $x_0\in\partial M$
exactly one of the following possibilities occurs:\\
\quada {\rm (i)} $\exists t\not=0:x_0+tv\in\partial M$,\\
\quada {\rm (ii)} $\forall t>0:x_0+tv\in\complement M$ and $x_0-tv\in M$.
\end{lem}
\bew
Assume $x_0+tv\not\in\partial M$ for $t\not=0$,
i.~e.\ $x_0+tv\in M\cup\complement M$ for $t\not=0$.
Therefore we have to show that the second possibility occurs:
\par
For $t>0$ the possibility $x_0+tv\in M$ is excluded:
Assume $t>0$. Then $x_0+tv\in M$ cannot happen, because
in accordance to Lemma \ref{untereinanderlemma} $x_0+tv\in M$ implies
$x_0=(x_0+tv)-tv\in M$ contradicting $x_0\in\partial M$.
It follows $x_0+tv\in\complement M$ for $t>0$.
$x_0-tv\in\complement M$ does not occur for $t>0$ for a
similar reason. Thus the statement is proved.
\qed

%%%%% 28.11.98 %%%%%
\begin{lem}\label{negkomplementlemma}
Let $M\subset\R^{n+1}$ be a measurable set which is
directed with respect to $v$. Then  $\complement M$ and
$-M\equiv\{x\in\R^{n+1}:-x\in M\}$ are directed
with respect to $-v$. Especially $-\complement M$ is
again directed with respect to $v$.
\end{lem}
\bew
The statement follows at once from the equations
$$D_\rho(M,y)+D_\rho(\complement M,y)=1$$
and
$$D_\rho(-M, -x-\tau v)=D_\rho(M, x+\tau v)$$
which are valid for any $x\in\r1$ and any $\rho>0$.
\qed

%%%%% 28.11.98 %%%%%
\begin{lem}\label{cistgerichteterkegel}
Let $E\subset\R^{n+1}$ be representable as a subgraph.
Let $t_i\to0$, $t_i>0$ be as before such that
$E_{t_i}\equiv\{x\in\R^{n+1}:t_ix\in E\}$ converges in
$L^1_{loc}(\R^{n+1})$ to a cone $C$. Then $C$ is
directed with respect to $e_{n+1}$.
\end{lem}
\bew
As a subgraph, $E$ is directed with respect to $e_{n+1}$
(cf.\ Remark \ref{subgleichger}).
If $C$ were not directed with respect to $e_{n+1}$, there
would be $x\in\R^{n+1}$, $\rho>0$ and $t>0$ such that
$$\int\limits_{B_\rho(x)}\chi_C<\int\limits_{B_\rho(x+te_{n+1})}\chi_C.$$
In the same way as for $E$ we get that $E_{t_i}$ is directed with
respect to $e_{n+1}$. Because of the convergence
$E_{t_i}\to C$ in $L^1_{loc}(\R^{n+1})$ we immediately get
a contradiction to the inequality above.
\qed

%%%%% 18.11.98 %%%%%
\begin{df}[weakly star-shaped]
A set $M\subset\R^{n+1}$ is called weakly star-shaped with
center $x$, if for all $z\in M$ we have
$x+\tau(z-x)\in M$ if $0<\tau\le 1$.
\end{df}

%%%%% 18.11.98 %%%%%
The following Lemma is essential for the proof of the
Regularity Theorem:
\begin{lem}\label{sternlemma}
Let $C$ be an open cone with vertex in the origin which is
directed with respect to $e_{n+1}$.
Then $C$ is weakly star-shaped with center $x=(0,-t)$ for
all $t>0$.
\end{lem}
\bew
Let $y=(\hat y,y^{n+1})\in C$ and $\tau$ with $0<\tau<1$ be arbitrary.
We show that $x+\tau(y-x)=(\tau\hat y,-t+\tau(y^{n+1}+t))$ is an
element of $C$. Being a cone, $C$ contains $(\tau\hat y,\tau y^{n+1})$
because of $y\in C$. $-t(1-\tau)$ is negative and so Lemma
\ref{untereinanderlemma} implies
$(\tau\hat y,\tau y^{n+1})-t(1-\tau)(0,1)=
  (\tau\hat y,-t+\tau(y^{n+1}+t))\in C$.
\qed

%%%%% 28.11.98 %%%%%
\begin{bem}\label{subgleichger}{\rm
For a measurable set $M\subset\R^{n+1}$ such that
$H^{n+1}(\partial M)=0$ the following two statements
are equivalent:\\
\quada {\rm (i)} $M$ is the subgraph of a measurable function $u$,\\
\quada {\rm (ii)} $M$ is directed with respect to $e_{n+1}$.
\par
The measurable function $u$ in (i) is given by
$$u(\hat x):=\sup\{t\in\R:(\hat x,t)\in M\}.$$}
\end{bem}
\bew
``(i)$\Longrightarrow$(ii)'':\\
Let $M={\rm sub\,}u$ be a given set.
It follows $\chi_M(x,t)=1$ for $t<u(x)$ and
$\chi_M(x,t)=0$ for $t>u(x)$, i.~e.\ $\chi_M(x,t)$ is
monotone decreasing with respect to $t$ for any $x\in\R^n$.
By integrating, we get for any $\rho>0$,
$\tau>0$ and $x\in\R^{n+1}$
$$\int\limits_{B_\rho(x)}\chi_M(z)dz
\ge\int\limits_{B_\rho(x)}\chi_M(z+\tau e_{n+1})dz
=\int\limits_{B_\rho(x+\tau e_{n+1})}\chi_M(y)dy.$$
Therefore $M$ is a directed set.\par
``(ii)$\Longrightarrow$(i)'':\\
Define for $\hat x\in\R^n$
$$u(\hat x):=\sup\left\{t\in\R:(\hat x,t)\in M\right\}.$$
We remark that $(\hat x,t)\in M$ is equivalent to the
existence of a $\rho>0$ such that $D_\rho(M,(\hat x, t))=1$.
Since $M$ is a measurable set the function $u$ is measurable.
Define $U:={\rm sub\,}u$.\\
{\bf a)} $M\subset U$:\\
Assume $(\hat x,t)\in M$. Since the supremum in the definition
of $u(\hat x)$ is not assumed, it follows $u(\hat x)>t$ and
therefore $(\hat x,t)\in U$.\par
{\bf b)} $U\subset M$:\\
Assume $(\hat x,t)\in U$, i.~e.\ $u(\hat x)>t$.
The definition of $u(\hat x)$ implies that there exists a
$\tau$ such that $u(\hat x)>\tau>t$ and $(\hat x,\tau)\in M$.
Now Lemma \ref{untereinanderlemma} yields $(\hat x,t)\in M$, since
$t-\tau<0$ and $(\hat x,\tau)+(t-\tau)(0,1)=(\hat x,t)$.
\qed

%%%%% 26.12.98 %%%%%
\def\Dir{{\rm Dir\,}}

%%%%% 26.12.98 %%%%%
\begin{df}[cone of directions]\label{defconeofdirections}
Let $E\subset\r1$ be a measurable set. $\Dir(E)$, the cone
of directions, is defined to be the set of all directions of
$E$ together with the origin.
\end{df}

%%%%% 26.12.98 %%%%%
\begin{rem}
The Definition \ref{defconeofdirections} is equivalent to
\begin{eqnarray*}
\Dir(E)&:=&\{y\in\r1:f_{x,\rho}:\R\to[0,1],
  t\mapsto D_\rho(E,x+ty)\\
  && \mbox{{\rm { is monotone decreasing for any }}}x\in\r1
    \mbox{{\rm { and any }}}\rho>0\}.\\
\end{eqnarray*}
\end{rem}

%%%%% 26.12.98 %%%%%
\begin{lem}\label{kegelistmonoid}
Let $E\subset\r1$ be a measurable set. Then $\Dir(E)$ is a closed
cone which is also closed under addition.
\end{lem}
\bew\\
{\bf (i)} The fact, that $\Dir(E)$ is a cone follows immediately
from the Definition \ref{defconeofdirections}.\\
{\bf (ii)} $\Dir(E)$ is closed under addition:\\
Let $y_1,y_2\in\Dir(E)$ be arbitrary. According to the definition this
is equivalent to
$$D_{\rho_1}(E,x_1+t_1y_1)\le D_{\rho_1}(E,x_1+\tau_1y_1)$$
and
$$D_{\rho_2}(E,x_2+t_2y_2)\le D_{\rho_2}(E,x_2+\tau_2y_2)$$
for any $x_1,x_2\in\r1$, $\rho_1,\rho_2>0$ and
$t_1,t_2,\tau_1,\tau_2\in\R$ such that $t_1\ge\tau_1$ and
$t_2\ge\tau_2$.
We have to show now, that
$$D_\rho(E,x+t(y_1+y_2))\le D_\rho(E,x+\tau(y_1+y_2))$$
for any $x\in\r1$, $\rho>0$ and $t,\tau\in\R$ such that $t\ge\tau$.
We choose now $x_1=x+ty_2$, $t_1=t$, $\rho_1=\rho$, $\tau_1=\tau$,
$x_2=x+\tau y_1$, $t_2=t$, $\rho_2=\rho$, $\tau_2=\tau$
and deduce from the inequalities above
\begin{eqnarray*}
D_\rho(E,(x+ty_2)+ty_1)&\le&D_\rho(E,(x+ty_2)+\tau y_1)\\
&=&D_\rho(E,(x+\tau y_1)+t y_2)\\
&\le&D_\rho(E,(x+\tau y_1)+\tau y_2)\\
\end{eqnarray*}
verifying the claimed inequality.\\
{\bf (iii)} $\Dir(E)$ is a closed set:\\
Let $y_i\in\Dir(E)$ for $i\in\N$ such that $y_i\to y$ as
$i\to\infty$. We have to show $y\in\Dir(E)$.
Assume in contrast $y\not\in\Dir(E)$. Then there exists $x\in\r1$,
$\rho>0$, $t,\tau\in\R$ such that $t\ge\tau$ and
$$D_\rho(E,x+ty)-D_\rho(E,x+\tau y)>0.$$
\OE we can assume $\tau=0$. $y_i\in\Dir(E)$ implies
$$D_\rho(E,x+ty_i)-D_\rho(E,x)\le0.$$
Since $D_\rho(E,x+ty_i)$ converges to $D_\rho(E,x+ty)$
as $i$ tends to infinity we get a contradiction and the statement
follows.
\qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% 18.11.98 %%%%%
\newpage
\section{Regularity Theorem}\label{regsatzkap}

%%%%% 28.11.98 %%%%%
\begin{df}\label{aufeinerseite}
An open set $M\subset\R^{n+1}$ is said to lie on one side
of a hyperplane $T$ if an adjusted rotation and translation of
the coordinate system yields the following situation
$$x=(\hat{x},x^{n+1})\in T\quad\Longleftrightarrow\quad x^{n+1}=0,$$
$$x=(\hat{x},x^{n+1})\in M\quad\Longrightarrow\quad x^{n+1}>0.$$
\end{df}

%%%%% 28.11.98 %%%%%
\begin{lem}\label{c1kegelisthalbraum}
Let $C\subset\R^{n+1}$ be a cone which is representable as a
subgraph of a $C^1$-function $u$ in a neighborhood of the vertex.
Then  $\partial C$
and the tangential hyperplane $T$ to $\partial C$ at
the vertex of $C$ coincide and $C$ is a half space.
\end{lem}
\bew
By a translation we can assume \oe that the vertex of the cone
is the origin. Since $C$ is a cone, $u$ is positive
homogeneous of degree $1$ and has a well-defined extension
(\oe $u$)
of the same homogeneity which is defined on the whole $\R^n$.
The subgraph of $u$ is $C$.
Let $v\in\R^n$ be arbitrary. It follows
$$\langle Du(0),v\rangle=\lim\limits_{t\to0}\frac{u(tv)-u(0)}{t}.$$
If we take into account $u(0)=0$ and use the fact that $u$ is a
positive homogeneous function, we deduce for $t>0$
$$\langle Du(0),v\rangle=
  \lim\limits_{t\to0}\frac{t\cdot u(v)}{t}=u(v).$$
The left-hand side of this equality is linear with respect to $v$.
Hence $u$ is a linear function. $C$ is therefore a subgraph of a
linear function and the statement follows.
\qed

%%%%% 28.11.98 %%%%%
\begin{lem}\label{zweipktebene}
Let $C\subset\R^{n+1}$ be an open minimal cone with its vertex
in the origin. If $C$ is directed with respect to $e_{n+1}$
and there exists a $t>0$ such that $z:=(0,t)\in\complement C$ and
$(0,-t)\in C$, then $\partial C$ is a hyperplane.
\end{lem}
\bew
\OE we assume $n\ge7$.
Choose $r>0$ such that $D_r(C,z)=0$ and $D_r(C,-z)=1$.
Define a cone $K$ by
$$K:=\left\{(\hat x,-\tau):\tau>0, \tau r>|\hat x|t\right\}.$$\par
It follows\\
{\bf (i)} $K\subset C$ and $-K\subset\complement C$:\\
Let $(\hat x,-\tau)\in K$ be arbitrary.
Since $K$ is a cone, we get $\left(\hat x\frac{t}{\tau},
-\tau\frac{t}{\tau}\right)=\left(\hat x\frac{t}{\tau},-t\right)\in K$.
Now $\left|\hat x\frac{t}{\tau}\right|<r$ implies
$\left(\hat x\frac{t}{\tau},-t\right)\in B_r(-z)$.
Choose $s>0$ such that
$B_s\left(\left(\hat x\frac{t}{\tau},-t\right)\right)\subset
B_r(-z)$. The equality $D_r(C,-z)=1$ implies
$D_s\left(C,\left(\hat x\frac{t}{\tau},-t\right)\right)=1$
and therefore we get
$\left(\hat x\frac{t}{\tau},-t\right)\in C$.
$C$ is a cone, so we deduce $(\hat x,-\tau)\in C$.
As $(\hat x,-\tau)\in K$ was arbitrary, so we get $K\subset C$.
\par
In the same way
$-K:=\left\{(\hat x,\tau):\tau>0,\tau r>|\hat x|t\right\}
\subset\complement C$ is proved.\par
{\bf (ii)} Representability of $\partial C$ as a graph:\\
We will show, that $\partial C$ can be represented as a
subgraph over $\R^n\setminus\Sigma$ besides a set which has
$H^{n-5}$-measure zero. $\Sigma$ is a closed set of
$H^{n-6}$-measure zero.
\par
For $s>n-7$ we get $H^s(\partial C\setminus\partial^\star C)=0$
by Theorem \cite[Theorem~11.8, p.~134]{giusti}.
This implies especially $H^{n-6}(\partial C\setminus\partial^\star C)=0$.
Define $\pi:\R^n\times\R\to\R^n$ by $\pi((\hat x,x^{n+1}))=\hat x$
and $\Sigma:=\pi(\partial C\setminus\partial^\star C)$.
Since $\pi$ is Lipschitz continuous we deduce
$H^{n-6}(\Sigma)=0$, hence
$H^{n-5}(\Sigma\times\R)=0$ and
$H^{n-5}((\Sigma\times\R)\cap\partial C)=0$.
$\Sigma$ is a closed set, because $\pi$ is continuous and
in view of (i) $(\partial C\setminus\partial^\star C)\cap
(F\times\R)$ is compact for any compact set $F\subset\R^n$.
\par
Let $\hat x\in\R^n\setminus\Sigma$ be arbitrary. $K\subset C$,
hence $\tau>\frac{|\hat x|t}{r}$ implies
$(\hat x,\tau)\in\complement C$ and
$(\hat x,-\tau)\in C$.
$\{\hat x\}\times\R$ is connected, $C$, $\complement C$ und
$\partial C$ are disjoint and we have
$C\cup\complement C\cup\partial C=\R^n$.
Regarding the fact, that $C$ and $\complement C$
are open sets we deduce
$(\{\hat x\}\times\R)\cap\partial C\not=\emptyset$,
because a connected set is not
the disjoint union of two non-empty sets, in this case
$(\{\hat x\}\times\R)\cap C$ with $(\hat x,-\tau)\in C$ and
$(\{\hat x\}\times\R)\cap\complement C$ with
$(\hat x,\tau)\in\complement C$ for all
$\tau>\frac{|\hat x|t}{r}$. The set
$(\{\hat x\}\times\R)\cap\partial C$ is bounded because
$(\hat x,\tau)\in\complement C$ and
$(\hat x,-\tau)\in C$ for all $\tau>\frac{|\hat x|t}{r}$.
$C$ is directed with respect to $e_{n+1}$ and $\partial C$ is closed,
so Lemma \ref{untereinanderlemma} implies
$(\{\hat x\}\times\R)\cap\partial C=\{\hat x\}\times I$ for a
compact interval $I$. The boundary of $C$ is analytic in the
complement of $\Sigma\times\R$ and so $I$ consists of exactly
one point. Therefore we have a function
$u\in L^1_{loc}(\R^n\setminus\Sigma)$, whose graph
coincides with $\partial C\setminus(\Sigma\times\R)$ and we
have the equality $C\cap((\R^n\setminus\Sigma)\times\R)={\rm sub\,} u$.
Observe, however, that $u$ is not automatically analytic, because
the boundary $\partial C$ in the complement of $\Sigma\times\R$
is only analytic as a manifold.\par
{\bf (iii)} Regularity of  $u$:\\
Let $\Omega\Subset\R^n\setminus\Sigma$ be an open ball.
Since the subgraph of $u|_\Omega$ has finite perimeter in
$\Omega\times\R$, i.~e.\
$\int\limits_{\Omega\times\R}|D\chi_{\sub u}|<\infty$,
and $u\in L^\infty(\Omega)$, we can use
\cite[Theorem~1, p.~317]{gia1} to deduce $u\in BV(\Omega)$.
According to \cite{cg1} we get $u\in C^{0,1}(\Omega)$.
But $u$ is also a weak solution of the minimal surface equation
and this implies $u\in C^2(\Omega)$.
$\Omega\Subset\R^n\setminus\Sigma$ was an arbitrary open ball.
Thus we deduce $u\in C^2(\R^n\setminus\Sigma)$. % giusti, p.234.
We have $H^{n-6}(\Sigma)=0$, so Theorem \cite[Theorem 16.9]{giusti}
can be applied, i.~e.\ $u$ can be extended to a
function in $C^2(\R^n)$ which is a solution of the minimal
surface equation. $\sub u$ is an element of the
$L^1$-equivalence class of $C$ independent of the choice
of the extension, because another extension changes $\sub u$
at most by a subset of $\Sigma\times\R$. As
$H^{n-5}(\Sigma\times\R)=0$, the choice of another
extension does not change the equivalence class.
\par
{\bf (iv)} $\partial C$ is a hyperplane:\\
This follows immediately from Lemma \ref{c1kegelisthalbraum}.
\qed

%%%%% 20.11.98 %%%%%
The following Lemma is - besides the Lemma \ref{sternlemma} -
the essential part of the proof of the Regularity Theorem:
\begin{lem}\label{einseitenlemma}
Let $M\subset\R^{n+1}$ be an open set which is
weakly star-shaped with center $x_0\in\partial M$.
If the boundary of $M$ is of class $C^1$ in a
neighborhood of $x_0$, then $M$ lies on one side
of the tangential hyperplane $T=T_{x_0}\partial M\subset\R^{n+1}$.
\end{lem}
\bew
By a translation of the coordinates we can assume
$x_0=(0,0)\in\R^n\times\R$. Since $\partial M$ is
of class $C^1$ in a neighborhood of $x_0$, we reach the following
situation by a suitable rotation of the coordinates:
There exist $R>0$ and $u\in C^1({\hat{B}}_{2R})\equiv
C^1(\{\hat{x}\in\R^n:|\hat{x}|<2R\})$ such that
$u(0)=0$, $Du(0)=0$ and for any $x=(\hat{x},x^{n+1})\in
{\hat{B}}_R\times(-R,R)$ we have the three equivalences
$$x\in\partial M\quad\Longleftrightarrow\quad x^{n+1}=u(\hat{x}),$$
$$x\in M\quad\Longleftrightarrow\quad x^{n+1}>u(\hat{x})$$ and
$$x\in\complement M\quad\Longleftrightarrow\quad x^{n+1}<u(\hat{x}).$$
By construction we have for any $x\in\R^{n+1}$
$$x\in T\quad\Longleftrightarrow\quad x^{n+1}=0.$$
\par
Assume now, the assertion would be false; more precisely: The chosen
position of the coordinates does not fulfill the conditions of
the Definition \ref{aufeinerseite}. That is, the condition
$$x=(\hat{x},x^{n+1})\in M\quad\Longrightarrow\quad x^{n+1}>0$$
is violated for an $x\in\R^{n+1}$. $M$ is an open set, so
there exists also a point
$y_0=({\hat{y}}_0, y^{n+1}_0)\in M$ with $y^{n+1}_0<0$.
First, we observe that ${\hat{y}}_0=0$ cannot occur,
for $M$ is weakly star-shaped with center $x_0=(0,0)$ and
we have for $y_0\in M$ that $\tau y_0$ is in $M$
for $0<\tau\le1$. This implies for $|\tau y_0|<R$, i.~e.\ for
sufficiently small $\tau>0$, the inequality
$$\tau y_0^{n+1}<0=u(0)=u({\hat{y}}_0)$$
which contradicts the assumption
$$x\in M\quad\Longleftrightarrow\quad x^{n+1}>u(\hat{x})$$
for $x\in{\hat{B}}_R\times(-R,R)$.
Assume therefore  ${\hat{y}}_0\not=0$.
Using again the fact that $M$ is weakly star-shaped, we get
$\tau y_0=(\tau{\hat{y}}_0,\tau y^{n+1}_0)\in M$ for $0<\tau\le1$.
Choose now $\varepsilon>0$ such that
$$0<\tau<\varepsilon\quad\Longrightarrow
  \quad\tau y_0\in{\hat{B}}_R\times(-R,R).$$
In ${\hat{B}}_R\times(-R,R)$ the set $M$ was characterized by
the equivalence
$$x=(\hat{x},x^{n+1})\in M\quad\Longleftrightarrow
  \quad x^{n+1}>u(\hat{x}).$$
This implies $u(\tau{\hat{y}}_0)<\tau y_0^{n+1}$
for $0<\tau<\varepsilon$.
If we take into account $u(0)=0$, $|{\hat{y}}_0|\not=0$
and $y_0^{n+1}<0$, we deduce
$$u(\tau{\hat{y}}_0)<\tau y_0^{n+1}<0,$$
$$|u(\tau{\hat{y}}_0)-u(0)|=-u(\tau{\hat{y}}_0)>-\tau y_0^{n+1}$$
and finally
$$\left|\frac{u(\tau{\hat{y}}_0)-u(0)}{\tau|{\hat{y}}_0|}\right|
  >\frac{-y_0^{n+1}}{|{\hat{y}}_0|}$$
for all $0<\tau<\varepsilon$.
The right-hand side of this inequality is independent of $\tau$
and positive, so we get a contradiction for $\tau\to0$
to the fact that the coordinates are chosen such that
we have $Du(0)=0$.
\qed

%%%%% 30.11.98 %%%%%
\begin{satz}[Regularity Theorem]\label{regsatz}
Let $C\subset\R^{n+1}$ be an open nontrivial minimal cone which
is directed with respect to $e_{n+1}$. Assume that the vertex of
$C$ is the origin and that $C$ has at most one singularity,
i.~e.\ $C$ is at most singular in the origin.
Then $\partial C$ is a hyperplane and so $C$ is a half space.
\end{satz}
For $n\le6$ there is nothing to be proved, the statement is true
according to \cite[Theorem~10.11, p.~127]{giusti} even for non-directed
cones.\par
\bew
$\partial C$ is regular apart from the origin, so
$\partial C\setminus\{0\}$ is analytic.
\par
In view of Lemma \ref{zweipktebene}
and due to the fact that $C$ is directed, it suffices to treat the
case that $x_0=(0,t)\in\partial C$ for a $t\not=0$.\par
\OE we can assume $t<0$, if not, consider
$-\complement C\equiv\{x\in\R^{n+1}:-x\in\complement C\}$
instead of $C$. According to Lemma \ref{negkomplementlemma}
this set is directed with respect to $e_{n+1}$.
\par
Now Lemma \ref{sternlemma} implies that $C$ is weakly
star-shaped with center $x_0$. $\partial C$ is regular
apart from the origin. Therefore we have a well-defined
tangential hyperplane $T$ at $\partial C$. With the help
of Lemma \ref{einseitenlemma} we conclude, that $C$ lies
on one side of $T$. Since $C$ is nontrivial,
we get $0\in\partial C$.
Finally \cite[Theorem~15.5, p.~174]{giusti}
implies that $C$ is a half space and $\partial C=T$
is a hyperplane.
\qed

%%%%% 20.11.98 %%%%%
\begin{cor}\label{regularitaetssatzcorollar}
Let $C\subset\R^{n+1}$ be an open nontrivial minimal cone which is
directed with respect to $e_{n+1}$.
If $n+1\le8$ then $\partial C$ is a hyperplane and $C$
is a half space.
\end{cor}
\bew
Remark \ref{inkleinendimensionen} guarantees that $C$
has at most one singularity which is in the vertex if it occurs at all.
Thus the Regularity Theorem \ref{regsatz} yields the statements.
\qed

%%%%% 20.11.98 %%%%%
\begin{bem}{\rm
The formulation of the Regularity Theorem \ref{regsatz}
contains the assumption that the cone $C$ has the
only possible singularity in the vertex. In the proof, however, we
use only the fact, that there is a $t>0$ which satisfies the
following two conditions:\\
\quada {\rm (i)} $(0,t)\in\partial C$ $\Longrightarrow$
  $\partial C$ is a $C^1$-manifold in a neighborhood of $(0,t)$.\\
\quada {\rm (ii)} $(0,-t)\in\partial C$ $\Longrightarrow$
  $\partial C$ is a $C^1$-manifold in a neighborhood of $(0,-t)$.\\
Therefore we get the following corollary:}
\end{bem}

%%%%% 20.11.98 %%%%%
\begin{cor}\label{minregsatzcorollar}
Let $C\subset\r1$ be an open nontrivial minimal cone which is
directed with respect to $e_{n+1}$ and which satisfies the two
conditions (i) and (ii) stated above for a positive $t>0$
(If one of the points $(0,t)$ or $(0,-t)$ is an element of
$\partial C$, we need in fact only the respective condition.).
Then $C$ is a half space.
\end{cor}

%%%%% 20.11.98 %%%%%
\begin{bem}{\rm
The conditions (i) and (ii) in Corollary \ref{minregsatzcorollar}
can be alternatively replaced by one of the following conditions:\\
\quada {\rm (i)} There is a direction $w$ of $C$ which fulfills
  $w\not\in\Sigma$ and $-w\not\in\Sigma$ or
  $w\in\partial C\setminus\Sigma$ or
  $-w\in\partial C\setminus\Sigma$.\\
\quada {\rm (ii)} There is one direction $w$ of $C$
  which fulfills $w\not\in\langle\Sigma\rangle$.\\
\quada {\rm (iii)} $\dim\langle D\rangle>\dim\langle\Sigma\rangle$.\\
Here $D$ is the set of all directions and $\Sigma$ is the singular
set of $C$. The brackets denote the vectorspace which is
spanned by the respective set.}
\end{bem}
\bew
This follows at once from Corollary \ref{minregsatzcorollar},
if we use the fact, that $\partial C\setminus\Sigma$ is an
open $C^1$-manifold.
\qed

%%%%% 26.12.98 %%%%%
\begin{cor}
Let $C\subset\r1$ be an open nontrivial minimal cone with
its vertex in the origin. Suppose $C$ has $k$ linearely
independent directions, $k\in\N$. Then $C$ is a half-space
provided $n+1-k\le7$ or
$H^{k}(\partial C\setminus\partial^\star C)=0$.
\end{cor}
\bew
In the first case ($n+1-k\le7$) we get
$H^k(\partial C\setminus\partial^\star C)=0$, for $k>n-7$
and $C$ is an almost minimal set, so the second case includes
the first one.
\par
Since $C$ has $k$ linearely independent directions we
deduce according to Lemma \ref{kegelistmonoid} $H^k(\Dir(C))=\infty$.
But we assumed that the $k$-dimensional Hausdorff-measure
of the singular set of $C$ vanishes. So we can find
$x\in\reu$ such that
$x\not\in\partial C\setminus\partial^\star C$,
$-x\not\in\partial C\setminus\partial^\star C$ and
$x\in\Dir(C)$. Now the Corollary follows from
Corollary \ref{minregsatzcorollar}.
\qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% 20.11.98 %%%%%
\newpage
\section{Applications}\label{anwendkap}

%%%%% 30.11.98 %%%%%
\begin{satz}[regularity for subgraphs]\label{subgraphanwend}
Assume that $E\subset\R^{n+1}$, $n\le 7$, is a subgraph and
almost minimal (with constant $\lambda$) in $\Omega$,
$\Omega\subset\R^{n+1}$ open,
then we get $\partial E\cap\Omega=\partial^\star E\cap\Omega$
and $\partial E\cap\Omega$ is a $C^{1,\lambda}$-manifold.
\end{satz}
\bew
According to the definitions we get
$\partial^\star E\subset\partial E$.
Let $x_0\in\partial E\cap\Omega$ be an arbitrary point.
Show $x_0\in\partial^\star E$:
By virtue of a translation we can assume \oe $x_0=0$.
The Proposition \cite[Proposition, p.~137]{mm}
guarantees that there is a sequence $t_i\to0$, $t_i>0$, such that
$E_{t_i}:=\{x\in\R^{n+1}:t_ix\in E\}$
converges for $i\to\infty$ in $L^1_{loc}(\R^{n+1})$
to a minimal cone $C$.
Lemma \ref{dichtgleichmass} states that $C$
is a nontrivial cone. In addition we get, according to the
proposition quoted,
$$0\in\partial^\star C\quad\Longleftrightarrow\quad
  x_0=0\in\partial^\star E.$$
As stated above we can
assume \oe that $C$ is an open set. We know from
\cite[Lemma~16.3, p.~184]{giusti}
that $C$ is representable as a subgraph.
Now Theorem \ref{regsatz} implies that $\partial C$ is a
hyperplane, and therefore we get $\partial C=\partial^\star C$.
Using the above stated equivalence we get
$x_0\in\partial^\star E$. Now \cite{tam2,tamanini}
yields that $\partial E\cap\Omega$ is a
$C^{1,\lambda}$-manifold. Thus the statement is proved.
\qed

%%%%% 29.11.98 %%%%%
\begin{thm}\label{simonthm}
Simon's cone
$$K^{2m}:=\left\{x\in\R^{2m}:\sum\limits_{i=1}^m\left(x^i\right)^2<
\sum\limits_{i=m+1}^{2m}\left(x^i\right)^2\right\}$$
is minimal for $m\ge4$.
\end{thm}
\bew
\cite[Theorem~16.4, p.~185]{giusti}
\qed

%%%%% 10.11.98 %%%%%
\begin{lem}\label{neundimensionalesgegenbeispiel}
Theorem \ref{subgraphanwend} is false in $\R^9$, i.~e.\ if
we replace the assumption $E\subset\r1$, $n\le7$, by
$E\subset\r1$, $n\le8$, we cannot prove
$\partial E\cap\Omega=\partial^\star E\cap\Omega$.
\end{lem}
\bew
Let $K:=K^8$ be the minimal cone defined in Theorem \ref{simonthm}.
$K\times\R\subset\R^9$ is obviously the subgraph of the function $u$
defined by
$$u(\hat x):=\left\{\begin{array}{r@{\quad:\quad}l}
  +\infty & \hat x\in K\\
  -\infty & \hat x\not\in K\\
  \end{array}\right.$$
for all $\hat x\in\R^8$.
\cite[Example~16.2, p.~183]{giusti} shows
that $K\times\R$ is minimal.
In this example the measure-theoretical and the reduced boundary
differ, we have $\partial(K\times\R)=
\partial^\star(K\times\R)\dot{\cup}(\{0\}\times\R)$.
This equality follows,
because the reduced boundary of an almost minimal
set $E\subset\r1$ consists exactly of those points $x$ of the
boundary for which $\partial E$ is a $C^1$-manifold in
a neighborhood of $x$.
\qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% 22.11.98 %%%%%
\def\ds{{d\overline{s}}}
\def\tN{{\tilde{N}}}
\def\gab{{\overline{g}_{\alpha\beta}}}
\def\g{{\overline{g}}}
\def\gh{{\hat g}}

%%%%% 23.11.98 %%%%%
\newpage
\section{Prescribed mean Curvature}\label{vorgegHkap}
%%%%% 22.11.98 %%%%%
The paper \cite{cg} treats the following problem:\\
In a complete locally conformally flat $(n+1)$-dimensional
Riemannian manifold $N$ we look for a closed
hypersurface $M$ which is homeomorphic to $S^n$ and
has prescribed mean curvature $f$, $f\in C^{0,1}(N)$, i.~e.\
the equation $H|_M=f(x)$ shall be solved by a hypersurface
of class $C^{2,\alpha}$. The hypersurface is looked for in
an open connected relatively compact subset
$\Omega$ of $N$, which is also regarded - using a diffeomorphism -
as a subset of $\r1$. The boundary of $\Omega$ consists of
two components $M_1$ and $M_2$, which are given
in Euclidian polar coordinates $(x^\alpha)_{0\le\alpha\le n}$,
$x^0\equiv r$, by the functions $u_i\in C^{2,\alpha}(S^n,(0,\infty))$
as graphs over $S^n$:
$M_i={\rm graph\,}\left.u_i\right|_{S^n}=
  \{(z,u_i(z)):z\in S^n\}$.
$M_1$ and $M_2$ act as barriers, i.~e.\ they satisfy
$H|_{M_1}\le f$ and $H|_{M_2}\ge f$, where the respective
unit normal vector is chosen such that the component
$\alpha=0$ is negative.
\par
In the cited paper it is proven, that under the stated
assumptions such a hypersurface $M$ exists provided $n\le6$.
In this chapter we extend the proof and show that such
a hypersurface $M$ exists up to $n=7$.

%%%%% 23.11.98 %%%%%
\begin{thm}\label{hyperflaechenbisdimensionacht}
The problem
``Find a closed hypersurface $M\subset\overline{\Omega}$
of class $C^{2,\alpha}$ such that $\left.H\right|_M=f$, which
is homeomorphic to $S^n$.'' has a solution
if $n\le7$.
\end{thm}
We need some Lemmata:

%%%%% 24.11.98 %%%%%
\begin{lem}\label{randinvarianzlemma}
The metric product $S^n\times\R$ is locally conformally equivalent
to $\reu$.
\end{lem}
\bew
Let $\tN=S^n\times\R$ be the metric product of $S^n$ and $\R$.
The metric of  $\tN$ is given by
$$\ds_\tN^2=dt^2+\sigma_{ij}dx^idx^j,$$
where $(x^i)_{1\le i\le n}$ are coordinates of $S^n$ and $t\in\R$.
Now we identify the manifold $\tN$ with its image in $\r1$ under the
diffeomorphism
\begin{eqnarray*}
\Psi:S^n\times\R &\to& \reu,\\
(x,t) &\mapsto& (x,e^t)\equiv(x,r).\\
\end{eqnarray*}
$\Psi$ is an isometry, if we equip $\reu$ with the metric
$$\frac{1}{r^2}dr^2+\sigma_{ij}dx^idx^j.$$
$((x^i)_{1\le i\le n}, r)$ are polar coordinates of $\reu$.
Now we assume
$(\tN,\ds_\tN^2)=(\reu,\frac{1}{r^2}dr^2+\sigma_{ij}dx^idx^j)$
and deduce
\begin{eqnarray*}
\ds^2_\tN
&=& \frac{1}{r^2}dr^2+\sigma_{ij}dx^idx^j\\
&=& \frac{1}{r^2}(dr^2+r^2\sigma_{ij}dx^idx^j)\\
&=& \frac{1}{r^2}\ds^2_\r1.\\
\end{eqnarray*}
In the last equality we use the fact that $dr^2+r^2\sigma_{ij}dx^idx^j$
is the representation of the standard metric of $\reu$ in
polar coordinates. This equation states that
$(\tN,\ds_\tN^2)$ is a locally conformally flat Riemannian manifold.
\qed

%%%%% 26.11.98 %%%%%
\begin{lem}\label{alssubgraphdarstellbarlemma}
Let $E\subset\r1$ be an almost minimal set which is representable
as a subgraph over $S^n$, i.~e.\
$$E=\{(\hat x,t)\in S^n\times\R^+\subset\reu:t<u(\hat x)\},$$
is a subgraph in polar coordinates.
Assume $u\ge c>0$.
Let $z_0=(0, z_0^{n+1})\subset\r1$ be an arbitrary point
such that $z_0^{n+1}>0$.
If $E_{t_i}=\{y\in\r1:z_0+t_i(y-z_0)\in E\}$
converges for a given sequence $t_i\to0$, $t_i>0$,
$i\in\N\setminus\{0\}$, in $L^1_{loc}(\r1)$
to a cone $C$, then $C$ is representable as a
subgraph over $\R^n\equiv{\langle z_0\rangle}^\perp$.
\end{lem}
\bew
Identify $\R^n$ with $\langle z_0\rangle^\perp$ and introduce
orthogonal coordinates.
Show that there is no $\hat x\in\R^n$ such that
$(\hat x, t)\in C\cup\partial C$ and $(\hat x, \tau)\in\complement C$
with $t>\tau$ or $(\hat x, t)\in C$ and $(\hat x, \tau)\in\partial C$
with $t>\tau$.
\par
{\bf (i)} Assume \oe $z_0^{n+1}=1$. In order to get an
easier representation in coordinates we translate such that $z_0=0$.
Then we have
$$E_{t_i}=\{y\in\r1:t_iy\in E\}.$$
Now, $E$ is representable as a subgraph over $S^n$ with center
$(0,-1)$ and $E_{t_i}$ is representable as a subgraph
over $S^n$ with center $(0,-\frac{1}{t_i})$.
\par
{\bf (ii)} $(\hat x, t)\in C$ and
$(\hat x,\tau)\in\complement C$ $\Longrightarrow$ $t<\tau$:\\
Assume that there is a $\hat x_0\in\R^n$, such that
$(\hat x_0,t_0)\in C\cup\partial C$,
$(\hat x_0,\tau_0)\in\complement C$ and $t_0>\tau_0$.
Then there exists a $\rho>0$ such that
$D_\rho(C,(\hat x_0,\tau_0))=0$. On the other hand we have
$D_\rho(C,(\hat x_0,t_0))>2\varepsilon>0$.
Since $E_{t_i}\to C$ in $L^1_{loc}(\r1)$, we get for sufficiently
large $i$ the inequality
$D_\rho(E_{t_i},(\hat x_0,t_0))>\varepsilon>0$.
Since each $E_{t_i}$ is representable as a subgraph over $S^n$ with
center $\left(0,-\frac{1}{t_i}\right)$ we deduce that
$$(0,\infty)\ni s\mapsto
  D_{s\rho}\left(E_{t_i},\left(0,-\frac{1}{t_i}\right)+sx\right)$$
is monotone decreasing for any $x\in\r1$,
because
\begin{eqnarray*}
&&\chi_{E_{t_i}}\left(\left(0,-\frac{1}{t_i}\right)+s_0y\right)=0,\,
  s_0>0,\,y\in\r1\\
&\Longrightarrow&
  \chi_{E_{t_i}}\left(\left(0,-\frac{1}{t_i}\right)+sy\right)=0
  \quad\forall s\ge s_0\\
\end{eqnarray*}
implies
$$s_0^{-n-1}\int\limits_{B_{s_0\rho}
  \left(\left(0,-\frac{1}{t_i}\right)+s_0x\right)}\chi_{E_{t_i}}\ge
  s^{-n-1}\int\limits_{B_{s\rho}
  \left(\left(0,-\frac{1}{t_i}\right)+sx\right)}
  \chi_{E_{t_i}}\,\forall s\ge s_0.$$
Now, for large $i$ the number
$s_i:=\frac{\tau_0+\frac{1}{t_i}}{t_0+\frac{1}{t_i}}$
satisfies $0<s_i<1$, and therefore we get for such $i$
\begin{eqnarray*}
0<\varepsilon&<&D_\rho(E_{t_i},(\hat x_0,t_0))\\
&=&D_{1\rho}\left(E_{t_i},\left(0,-\frac{1}{t_i}\right)
  +1\left(\hat x_0,t_0+\frac{1}{t_i}\right)\right)\\
&\le&D_{s_i\rho}\left(E_{t_i},\left(0,-\frac{1}{t_i}\right)+
  s_i\left(\hat x_0,t_0+\frac{1}{t_i}\right)\right)\\
&=&D_{s_i\rho}(E_{t_i},(s_i\hat x_0,\tau_0))\\
\end{eqnarray*}
On the other hand $s_i$ converges to $1$ as $i$ tends
to infinity, so we get
$D_\rho(C,(\hat x_0,\tau_0))\ge\varepsilon>0$;
this inequality contradicts $D_\rho(C,(\hat x_0,\tau_0))=0$.
\par
{\bf (iii)} $(\hat x,t)\in C)$ and $(\hat x,\tau)\in\partial C$
$\Longrightarrow$ $t<\tau$:\\
This statement is proved in the same way as (ii).
\par
{\bf (iv)} So we get for each $\hat x\in\R^n$ numbers
$t_1,t_2\in[-\infty,+\infty]$ with $t_1\le t_2$ such that
$$\{\hat x\}\times(-\infty,t_1)\subset C,$$
$$\{\hat x\}\times([t_1,t_2]\cap\R)\subset\partial C$$
and
$$\{\hat x\}\times(t_2,+\infty)\subset\complement C.$$
$H^n$-almost eyerywhere in $\R^n$ we have
$t_1(\hat x)=t_2(\hat x)$: Show only that $H^n$-almost
everywhere the inequality $t_2(\hat x)-t_1(\hat x)<\frac{1}{k}$
for $k\ge1$ is valid.
Define $A_k:=\{\hat x\in\R^n:t_2(\hat x)-t_1(\hat x)\ge\frac{1}{k}\}$.
We get $0=H^{n+1}(\partial C)\ge\frac{1}{k}H^n(A_k)$ and
therefore $H^n(A_k)=0.$
Define $u(\hat x):=t_1(\hat x)$. We get
$C=\sub u$ in $L^1_{loc}(\r1)$. $u$ is measurable, because
$C$ is measurable.
\qed

%%%%% 07.11.98 %%%%%
\def\ea{{\eta^\alpha}}
\def\eb{{\eta^\beta}}
\def\eat{{\tilde{\eta}^\alpha}}
\def\ebt{{\tilde{\eta}^\beta}}
\def\eae{{\eta^\alpha_\varepsilon}}
\def\ebe{{\eta^\beta_\varepsilon}}
\def\eate{{\tilde{\eta}^\alpha_\varepsilon}}
\def\ebte{{\tilde{\eta}^\beta_\varepsilon}}

%%%%% 26.11.98 %%%%%
\begin{lem}[C.~Gerhardt, according to a Seminar]%
\label{fastminimalbleibterhaltenlemma}
Let $E$ be an almost minimal set in
$\Omega\Subset S^n\times\R=\tN=(\tN,\gab)$.
Regard $\tN$ as a subset of $\r1$.
Then $E$ is also almost minimal in
$\Omega\subset\r1=(\r1,\delta_{\alpha\beta})$.
The constant $\lambda$ in the definition of almost
minimal remains unchanged, if $0<\lambda\le\frac{1}{2}$.
\end{lem}
\bew
{\bf (i)} Almost minimal in a manifold is defined
in the same way as in $\r1$. Now the balls are
geodesic balls and the perimeter is defined
using the divergence in the manifold.
Since geodesic and Euclidian balls are comparable, i.~e.\
in any relatively compact subset $A$ and for any $\rho$
such that $0<\rho<R(A)$ there exists a constant $c=c(A)>0$
such that $B^\tN_{c\rho}(x)\subset B^\r1_\rho(x)
\subset B^\tN_{\frac{1}{c}\rho}(x)$ for any geodesic ball
in the respective manifold with center $x\in A$.
Therefore it suffices to prove the statement for
Euclidian balls.
From now on we assume that the chosen coordinates are
such that the standard metric of $\r1$ is represented
by the metric tensor
$(\delta_{\alpha\beta})_{0\le\alpha,\beta\le n}$.
\par
{\bf (ii)} Let $F$ be an arbitrary Caccioppoli set in $\tN$.
Let $\varepsilon>0$, $x_0\in A$ be arbitrary. Choose
$\eta^\alpha_\varepsilon \in C^1_c(B_\rho(x_0))$,
${0\le\alpha\le n}$ such that
\def\eae{{\eta^\alpha_\varepsilon}}
\def\ebe{{\eta^\beta_\varepsilon}}
$\gab\eae\ebe\le1$ and
$$\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_F|\le\varepsilon+
  \int\limits_{B_\rho(x_0)\subset\tN}\chi_FD_\alpha\eae.$$
It follows
\begin{eqnarray*}
\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_F|
&\le&\varepsilon+\int\limits_{B_\rho(x_0)\subset\tN}\chi_FD_\alpha\eae\\
&=&\varepsilon+\int\limits_{B_\rho(x_0)}\chi_F
 \frac{\partial}{\partial x^\alpha}(\sqrt{\overline{g}(x)}\eae).\\
\end{eqnarray*}
For $x\in B_\rho(x_0)$ we get for sufficiently small $R=R(A)$
and $\rho$ such that $0<\rho<R$
\begin{eqnarray*}
\gab(x_0)\frac{\sqrt{\g(x)}}{\sqrt{\g(x_0)}}\eae(x)
  \frac{\sqrt{\g(x)}}{\sqrt{\g(x_0)}}\ebe(x)
&\le&\sup\limits_{y\in A}\sup\limits_{z\in B_\rho(y)}
  \left(\frac{\sqrt{\g(z)}}{\sqrt{\g(x_0)}}\right)^2\cdot\\
&&\cdot((\gab(x_0)-\gab(x))\eta^\alpha(x)\eta^\beta(x)\\
&&\quad+\gab(x)\eta^\alpha(x)\eta^\beta(x))\\
\end{eqnarray*}
where the right-hand side can be estimated from above by
$$1+\rho c(A).$$
These estimates depend especially on the mean value theorem.
$c(A)$ represents a constant which depends only on $A$ and
may have another value each time. We have especially
$R=R(A)=c(A)$.
If we define $\eate=\frac{\sqrt{\g(x)}}{\sqrt{\g(x_0)}}\eae(x)$,
we deduce
\begin{eqnarray*}
\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_F|
&\le&\varepsilon+\int\limits_{B_\rho(x_0)}\chi_F
  \frac{\partial}{\partial x^\alpha}(\sqrt{\g(x_0)}\eate)\\
&\le&\varepsilon+(1+\rho c(A))\cdot\sup\left\{
  \int\limits_{B_\rho(x_0)}\chi_F
  \frac{\partial}{\partial x^\alpha}
  \left(\sqrt{\g(x_0)}\ea\right):\right.\\
&&\eta^\alpha\in
  C^1_c(B_\rho(x_0)),{0\le\alpha\le n},\\
&&\left.\rule{0em}{4.1ex}
  \gab(x_0)\ea(x)\eb(x)\le1\mbox{ for } x\in B_\rho(x_0)\right\}\\
&\equiv&\varepsilon+(1+\rho c(A))\cdot
  P_{(\r1,\g(x_0))}(F,B_\rho(x_0))\\
\end{eqnarray*}
As $\varepsilon>0$ was an arbitrary number, we get
$$\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_F|
  \le(1+\rho c(A))\cdot P_{(\r1,\g(x_0))}(F,B_\rho(x_0)).$$
\par
{\bf (iv)} Let $F$ be again an arbitrary Caccioppoli set in $\tN$.
Let $0<\rho<R$, $x_0\in A$ and
$\eta^\alpha\in C^1_c(B_\rho(x_0))$ such that
$\gab(x_0)\ea(x)\eb(x)\le1$ be arbitrary.
For $\eat(x)=\frac{\sqrt{\g(x_0)}}{\sqrt{\g(x)}}\ea(x)$ we deduce
as above
$$\gab(x)\eat(x)\ebt(x)\le1+\rho c(A),$$
and we infer
\begin{eqnarray*}
\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_F|
&\ge&\frac{1}{1+\rho c(A)}\int\limits_{B_\rho(x_0)}
  \chi_F\frac{\partial}{\partial x^\alpha}(\sqrt{\g(x)}\eat(x))\\
&\ge&\frac{1}{1+\rho c(A)}\int\limits_{B_\rho(x_0)}
  \chi_F\frac{\partial}{\partial x^\alpha}(\sqrt{\g(x_0)}\ea(x)).\\
\end{eqnarray*}
Now we take the supremum in this inequality over all
$\ea\in C^1_c(B_\rho(x_0))$
such that $\gab(x_0)\ea(x)\eb(x)\le1$.
This yields
$$\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_F|
  \ge\frac{1}{1+\rho c(A)}P_{(\r1,\g(x_0))}(F,B_\rho(x_0))$$
\par
{\bf (v)} The conformal equivalence means
$\gab(x_0)=\vartheta(x_0)\delta_{\alpha\beta}(x_0)$
with $0<\vartheta(x_0)<\infty$. We have
$$P_{(\r1,\g(x_0))}(F,B_\rho(x_0))=
 \varphi(\vartheta(x_0))\int\limits_{B_\rho(x_0)}|D\chi_F|,$$
where $\varphi$ is a positive and continuous function.
\par
{\bf (vi)} Now assume again $E\vartriangle F\Subset B_\rho(x_0)$.
Using the result above, the inequalities of (iii) and (iv) can be
written in the following form:
$$\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_F|\le
  (1+\rho c(A))
  \varphi(\vartheta(x_0))\int\limits_{B_\rho(x_0)}|D\chi_F|$$
and
$$\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_E|\ge
  \frac{1}{1+\rho c(A)}\varphi(\vartheta(x_0))
  \int\limits_{B_\rho(x_0)}|D\chi_E|.$$
These inequalities are valid for any $x_0\in A$.
From (ii) we deduce
$$\int\limits_{B_\rho(x_0)\subset\tN}|D\chi_E|\le
  \int\limits_{B_\rho(x_0)\subset\tN}|D\chi_F|+
  K\rho^{n+2\lambda}.$$
Since $\varphi(\vartheta(x_0))$
is a positive and continuous function of
$x_0$, the combination of the above inequalities yields
$$\int\limits_{B_\rho(x_0)}|D\chi_E|\le
  (1+\rho\tilde c(A))\int\limits_{B_\rho(x_0)}|D\chi_F|
  +(1+\rho c(A))c(A)\rho^{n+2\lambda}.\qquad (\ast)$$
Choose $r$ with $0<r<\rho$ such that
$$\int\limits_{B_\rho(x_0)\setminus B_r(x_0)}|D\chi_E|
  \le\omega_{n+1}\rho^n.$$
Now we get the estimate
\begin{eqnarray*}
\int\limits_{B_\rho(x_0)}|D\chi_{E\cup B_r(x_0)}|
&\le&\int\limits_{B_\rho(x_0)\setminus B_r(x_0)}|D\chi_E|+
  H^n(\partial B_r(x_0))\\
&\le&\omega_{n+1}\rho^n+(n+1)\omega_{n+1}r^n\\
&\le&(n+2)\omega_{n+1}\rho^n=c(A)\rho^n.\\
\end{eqnarray*}
$E\cup B_r(x_0)$ is an allowed choice for $F$ in the
above inequality, so we get
$$\int\limits_{B_\rho(x_0)}|D\chi_E|\le
  c(A)(\rho^n+\rho^{n+2\lambda})\le c(A)\rho^n.$$
We multiply this inequality with $\rho\tilde{c}(A)$
and add it to ($\ast$) and obtain
$$(1+\rho \tilde{c}(A))\int\limits_{B_\rho(x_0)}|D\chi_E|\le
  (1+\rho \tilde{c}(A))\int\limits_{B_\rho(x_0)}|D\chi_F|
  +c(A)(\rho^{n+2\lambda}+\rho^{n+1}).$$
Now replace $\lambda$ by $\min\{\lambda,\frac{1}{2}\}$ and conclude
$$\int\limits_{B_\rho(x_0)}|D\chi_E|
  \le\int\limits_{B_\rho(x_0)}|D\chi_F|
  +c(A)\rho^{n+2\lambda}.$$
This inequality states that $E$ is
almost minimal in the set $\Omega\subset\r1$
equipped with the Euclidian metric.
\qed

%%%%% 26.11.98 %%%%%
Most of the proof of Theorem \ref{hyperflaechenbisdimensionacht}
in \cite{cg} is independent of the dimension,
so the following proof contains mainly the
necessary changes to get a proof which is valid up to
$n=7$:
\par
\bewvon{Theorem \ref{hyperflaechenbisdimensionacht}}
As in the cited paper we get
$u_k\in C^{2,\alpha}(S^n)$, $k\ge3$, such that
$$\left.H\right|_{{\rm graph\,}u_k}=f-\gamma e^{-\mu u_k}[u_k-u_{k-1}]$$
and $u_1\le u_k\le u_{k-1}$.
These functions converge pointwise, since for a fixed $x\in S^n$
we have a monotone decreasing sequence $u_k(x)$ which is
bounded from below. Define $u$ to be the pointwise limit of $u_k$ and
further $\varphi_k=\log u_k$ and $\varphi=\log u$.
As in \cite{cg} we deduce that
$E:={\rm sub\,}\varphi=\{(x,t):t<\varphi(x),\,x\in S^n\}$
is almost minimal in the metric product $S^n\times\R$.
Using Lemma \ref{randinvarianzlemma} and
Lemma \ref{fastminimalbleibterhaltenlemma} we can regard $E$
an an almost minimal subset in $\r1$. Now Lemma
\ref{alssubgraphdarstellbarlemma} yields that the blow-up cone $C$
around a point $x\in\partial E$ is representable as a subgraph,
and therefore $C$ is a directed cone,
so the Regularity Theorem implies that $C$ is a half space.
Therefore we have $x\in\partial^\star E$ for any $x\in\partial E$.
From now on the proof of the Theorem in \cite{cg} is independent
of the dimension and can be used to deduce the final result.
\qed

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%% 24.11.98 %%%%%
\newpage
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\end{thebibliography}

{\sc Oliver C.~Schn\"urer\\
Ruprecht-Karls-Universit\"at\\
Institut f\"ur Angewandte Mathematik\\
Im Neuenheimer Feld 294\\
D-69120 Heidelberg\\
Germany}\\
E-mail address: {\tt schnuere@cip.mathi.uni-heidelberg.de}
\end{document}
