
\documentclass[reqno]{amsart}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2003(2003), No. 64, pp. 1--8.\newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu  (login: ftp)}
\thanks{\copyright 2003 Southwest Texas State University.}
\vspace{9mm}}

\begin{document}

\title[\hfilneg EJDE--2003/64\hfil Self-adjointness of Schr\"odinger-type operators]
{Self-adjointness of Schr\"odinger-type operators
with singular potentials on manifolds of bounded geometry}

\author[Ognjen Milatovic\hfil EJDE--2003/64\hfilneg]
{Ognjen Milatovic}  

\address{Ognjen Milatovic \newline
78 Apsley Street, Apt. 1\\
Hudson, MA 01749, USA }
\email{omilatovic@fsc.edu}


\date{}
\thanks{Submitted May 13, 2003. Published June 11, 2003.}
\subjclass[2000]{35P05, 58J50, 47B25, 81Q10}
\keywords{Schr\"odinger operator, self-adjointness, manifold,
bounded geometry, \hfill\break\indent  singular potential}


\begin{abstract}
 We consider the Schr\"odinger type differential expression
   $$ H_V=\nabla^*\nabla+V,  $$
 where $\nabla$ is a $C^{\infty}$-bounded
 Hermitian connection on a Hermitian vector bundle $E$ of bounded
 geometry over a manifold of bounded geometry $(M,g)$ with metric
 $g$ and positive $C^{\infty}$-bounded measure $d\mu$, and
 $V=V_1+V_2$, where $0\leq V_1\in L_{\rm loc}^1(\operatorname{End} E)$
 and $0\geq V_2\in L_{\rm loc}^1(\operatorname{End} E)$ are linear
 self-adjoint bundle endomorphisms. We give a sufficient condition
 for self-adjointness of the operator $S$ in $L^2(E)$ defined by
 $Su=H_Vu$ for all $u\in\operatorname{Dom}(S)=\{u\in
 W^{1,2}(E)\colon \int\langle V_1u,u\rangle\,d\mu<+\infty\text{ and
 }H_Vu\in L^2(E)\}$.
 The proof follows the scheme of T.~Kato, but
 it requires the use of more general version of Kato's inequality
 for Bochner Laplacian operator as well as a result on the
 positivity of $u\in L^2(M)$ satisfying the equation
 $(\Delta_M+b)u=\nu$, where $\Delta_M$ is the scalar Laplacian on
 $M$, $b>0$ is a constant and $\nu\geq 0$ is a positive
 distribution on $M$.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{rem}[theorem]{Remark}

\section{Introduction and  main result}\label{S:main}

Let $(M,g)$ be a $C^{\infty}$ Riemannian manifold without
boundary, with metric $g$, $\dim M=n$.  We will assume that $M$ is
connected. We will also assume that $M$ has bounded geometry.
Moreover, we will assume that we are given a positive
$C^{\infty}$-bounded measure $d\mu$, i.e. in any local coordinates
$x^{1}, x^{2},\dots,x^{n}$ there exists a strictly positive
$C^{\infty}$-bounded density $\rho(x)$ such that
$d\mu=\rho(x)dx^{1}dx^{2}\dots dx^{n}$.

Let $E$ be a Hermitian vector bundle over $M$. We will assume that
$E$ is a bundle of bounded geometry (i.e.~it is supplied by an
additional structure: trivializations of $E$ on every canonical
coordinate neighborhood $U$ such that the corresponding matrix
transition functions $h_{U,U'}$ on all intersections $U\cap U'$
of such neighborhoods are $C^{\infty}$-bounded, i.e.~all
derivatives $\partial_{y}^{\alpha}h_{U,U'}(y)$, where $\alpha$ is
a multiindex, with respect to canonical coordinates are bounded
with bounds $C_{\alpha}$ which do not depend on the chosen pair
$U$, $U'$).

We denote by $L^2(E)$ the Hilbert space of square integrable
sections of $E$ with respect to the scalar product
\begin{equation}\label{E:inner}
       (u,v)  =  \int_{M}\, \langle u(x),v(x)\rangle\, d\mu(x).
\end{equation}
Here $\langle\cdot,\cdot\rangle$ denotes the fiberwise inner
product in $E_x$.

In what follows, $C^{\infty}(E)$ denotes smooth sections of $E$,
and $C_{c}^{\infty}(E)$ denotes smooth compactly supported
sections of $E$.
Let $$\nabla\colon C^{\infty}(E)\to C^{\infty}(T^*M\otimes E)$$ be
a Hermitian connection on $E$ which is $C^{\infty}$-bounded as a
linear differential operator, i.e.~in any canonical coordinate
system $U$ (with the chosen trivializations of $E|_{U}$ and
$(T^*M\otimes E)|_{U}$), $\nabla$ is written in the form
\[
\nabla=\sum_{|\alpha|\leq 1}a_{\alpha}(y)\partial_{y}^{\alpha},
\]
where $\alpha$ is a multiindex, and the coefficients
$a_{\alpha}(y)$ are matrix functions whose derivatives
$\partial_{y}^{\beta}a_{\alpha}(y)$ for any multiindex $\beta$ are
bounded by a constant $C_{\beta}$ which does not depend on the
chosen canonical neighborhood.

We will consider a Schr\"odinger type differential expression of
the form
\[
H_V=\nabla^*\nabla+V,
\]
where $V$ is a linear self-adjoint bundle map $V\in
L_{\rm loc}^{1}(\operatorname{End} E)$.  Here $$\nabla^*\colon
C^{\infty}(T^*M\otimes E)\to C^{\infty}(E)$$ is a differential
operator which is formally adjoint to $\nabla$ with respect to the
scalar product~\eqref{E:inner}.

If we take $\nabla=d$, where $d\colon C^{\infty}(M)\to
\Omega^{1}(M)$ is the standard differential, then $d^*d\colon
C^{\infty}(M)\to C^{\infty}(M)$ is called the scalar Laplacian and
will be denoted by $\Delta_M$.

We make the following assumption on $V$.
\begin{itemize}
\item[(A1)] $V=V_1+V_2$, where $0\leq V_1\in L_{\rm loc}^{1}(\operatorname{End}
E)$ and $0\geq V_2\in L_{\rm loc}^{1}(\operatorname{End} E)$ are
linear self-adjoint bundle maps (here the inequalities are
understood in the sense of operators $E_x\to E_x$).
\end{itemize}
By $W^{1,2}(E)$ we  denote the completion of the space
$C^\infty_c(E)$ with respect to the norm $\|\cdot\|_{1}$ defined
by the scalar product
\[
    (u,v)_{1}  :=  (u,v)  +  (\nabla u,\nabla v)
    \quad u, v\in C^{\infty}_c(E).
\]
By $W^{-1,2}(E)$ we will denote the dual of $W^{1,2}(E)$.


\section{Quadratic forms}

In what follows, all quadratic forms are considered in the Hilbert
space $L^2(E)$.

By $h_0$ we denote the quadratic form
\begin{equation}\label{E:quad-h_0}
h_0(u)=\int |\nabla u|^2 \,d\mu
\end{equation}
with the domain $\text{D}(h_0)=W^{1,2}(E)\subset L^2(E)$. Clearly,
$h_0$ is a non-negative, densely defined and closed form.

By $h_1$ we denote the quadratic form
\begin{equation}\label{E:quad-h_1}
h_1(u)=\int\langle V_1u,u\rangle\,d\mu
\end{equation}
with the domain
\begin{equation}\label{E:d(h_1)}
\text{D}(h_1)=\big\{u\in L^2(E):\int \langle
V_1u,u\rangle\,d\mu<+\infty\big\}.
\end{equation}
Clearly, $h_1$ is a non-negative, densely defined, and closed form.

By $h_2$ we denote the quadratic form
\begin{equation} \label{E:quad-h_2}
h_2(u)=\int\langle V_2u,u\rangle\,d\mu
\end{equation}
with the domain
\begin{equation}\label{E:d(h_2)}
\text{D}(h_2)=\big\{u\in L^2(E):\int \left|\langle
V_2u,u\rangle\right|\,d\mu<+\infty\big\}.
\end{equation}
Clearly, $h_2$ is a densely defined form. Moreover, $h_2$ is
symmetric (but not semi-bounded below).

We make the following assumption on $h_2$.
\begin{itemize}
\item[(A2)] Assume that $h_2$ is $h_0$-bounded with relative
bound $b<1$, i.e.
\begin{enumerate}
\item[(i)] $\text{D}(h_2)\supset\text{D}(h_0)$
\item[(ii)] There exist constants $a\geq 0$ and $0\leq b<1$ such that
\begin{equation}\label{E:h_2-h_0-domination}
|h_2(u)|\leq a\|u\|^2+b|h_0(u)|,\quad\text{for all }u\in\text{D}(h_0),
\end{equation}
where $\|\cdot\|$ denotes the norm in $L^2(E)$.
\end{enumerate}
\end{itemize}

\begin{rem}\label{R:Assumption-B} \rm
With the above assumptions on $(M,g)$, bundle $E$ and connection
$\nabla$, Assumption (A2) holds if $V_2\in L^{p}(\operatorname{End}
E)$, where $p=n/2$ for $n\geq 3$, $p>1$ for $n=2$, and $p=1$ for
$n=1$. The proof is given in the last section of this article.
%Sec.~\ref{S:sufficient-condition}.
\end{rem}

%\subsection*{A realization of $H_V$ in $L^2(E)$}\label{SS:operator-S}
As a realization of $H_V$ in $L^2(E)$, we define the
operator $S$ in $L^2(E)$ by the formula $Su=H_Vu$ on the domain
\begin{equation}\label{E:domain-S}
\operatorname{Dom}(S)=\big\{u\in W^{1,2}(E): \int \langle
V_1u,u\rangle\,d\mu<+\infty\text{ and }H_Vu\in L^2(E)\big\}.
\end{equation}

\begin{rem}\label{R:distrib-interpret} \rm
For all
$u\in\text{D}(h_0)=W^{1,2}(E)$ we have $\nabla^*\nabla u\in
W^{-1,2}(E)$, and from Corollary~\ref{C:diagonal-2} below it follows that for
all $u\in W^{1,2}(E)\bigcap\text{D}(h_1)$, we have $Vu\in
L_{\rm loc}^1(E)$. Thus $H_Vu$ in~\eqref{E:domain-S} is a distributional
section of $E$, and the condition $H_Vu\in L^2(E)$ makes sense.
\end{rem}

We now state the main result.

\begin{theorem}\label{T:main}
Assume that $(M,g)$ is a manifold of bounded geometry with
positive $C^{\infty}$-bounded measure $d\mu$, $E$ is a Hermitian
vector bundle of bounded geometry over $M$, and $\nabla$ is a
$C^{\infty}$-bounded Hermitian connection on $E$. Suppose that
Assumptions (A1) and (A2) hold. Then $S$ is a semi-bounded below
self-adjoint operator.
\end{theorem}

\begin{rem} \rm
Theorem~\ref{T:main} extends a result of T.~Kato, cf.~Theorem
VI.4.6(a) in~\cite{Kato66} (see also remark 5(b) in~\cite{Kato78})
which was proven for the operator $-\Delta+V$, where $\Delta$ is
the standard Laplacian on $\mathbb{R}^n$ with the standard metric
and measure, and $V=V_1+V_2$, where $0\leq V_1 \in L_{\rm
loc}^1({\mathbb{R}}^n)$ and $0\geq V_2 \in L_{\rm
loc}^1({\mathbb{R}}^n)$ are as in Assumption (A1) above, and the
quadratic form $h_2$ corresponding to $V_2$ is as in Assumption
(A2) above.
\end{rem}

\section{Proof of Theorem \ref{T:main}}\label{S:proof-main}

We adopt the arguments from
Sec.~VI.4 in~\cite{Kato66} to our setting with the help of more
general version of Kato's inequality~\eqref{E:kato}.

We begin with the following variant of Kato's inequality
for Bochner Laplacian (for the proof see Theorem 5.7~in~\cite{bms}).
The original version of Kato's inequality was proven
in Kato~\cite{Kato72}.

\begin{lemma}\label{L:kato}
Assume that $(M,g)$ is a Riemannian
manifold. Assume that $E$ is a Hermitian vector bundle over $M$
and $\nabla$ is a Hermitian connection on $E$. Assume that $w\in
L_{\rm loc}^{1}(E)$ and $\nabla^*\nabla w\in L_{\rm loc}^{1}(E)$. Then
\begin{equation}\label{E:kato}
      \Delta_M |w| \leq  \operatorname{Re}\langle
      \nabla^*\nabla w,\operatorname{sign} w\rangle,
\end{equation}
where
\[
      \operatorname{sign} w(x)  =
      \begin{cases}
          \frac{w(x)}{|w(x)|} & \textrm{if } w(x)\neq 0 ,\\
          0 &\text{otherwise.}
      \end{cases}
\]
\end{lemma}

 In what follows, we will use the following Lemma whose
proof is given in Appendix B of~\cite{bms}.

\begin{lemma}\label{L:positivity}
Assume that $(M,g)$ is a manifold of bounded geometry
with a smooth positive measure $d\mu$. Assume that
\[
     \big(\, b+\Delta_M\, \big)\, u \ = \ \nu \ \ge \ 0,
     \quad u\in L^2(M),
\]
where $b>0$, $\Delta_M=d^*d$ is the scalar Laplacian on $M$, and
the inequality $\nu\ge 0$ means that $\nu$ is a positive
distribution on $M$, i.e. $(\nu,\phi)\ge 0$ for any $0\leq\phi\in
C_c^\infty(M)$.
Then $u\ge 0$ (almost  everywhere or, equivalently, as a
distribution).
\end{lemma}

\begin{rem} \rm
It is not known whether Lemma~\ref{L:positivity} holds
if $M$ is an arbitrary complete Riemannian manifold. For more
details about difficulties in the case of arbitrary complete
Riemannian manifolds, see Appendix B of~\cite{bms}.
\end{rem}


\begin{lemma}\label{L-sum-of-forms}
The quadratic form $h:=(h_0+h_1)+h_2$ is densely-defined, semi-bounded
below and closed.
\end{lemma}

\begin{proof} Since $h_0$ and $h_1$ are non-negative and
closed, it follows by Theorem VI.1.31 from~\cite{Kato66} that
$h_0+h_1$ is non-negative and closed. Since $h_1$ is non-negative,
it follows immediately from Assumption (A2) that $h_2$ is
$(h_0+h_1)$-bounded with relative bound $b<1$. Since $h_0+h_1$ is
a closed, non-negative form, by Theorem VI.1.33
from~\cite{Kato66}, it follows that $h=(h_0+h_1)+h_2$ is a closed
semi-bounded below form. Since
$C_{c}^{\infty}(E)\subset\text{D}(h_0)\bigcap \text{D}(h_1)\subset
\text{D}(h_2)$, it follows that $h$ is densely defined.
\end{proof}

In what follows, $h(\cdot,\cdot)$ will denote the corresponding
sesquilinear form obtained from $h$ via polarization identity.

\subsection*{Self-adjoint operator $H$ associated to $h$}\label{SS:operator-H}
Since $h$ is densely defined, closed and semi-bounded below form
in $L^2(E)$, by Theorem VI.2.1 from~\cite{Kato66} there exists a
semi-bounded below self-adjoint operator $H$ in $L^2(E)$ such that
\begin{enumerate}
\item[(i)] $\operatorname{Dom}(H)\subset\text{D}(h)$ and
\[
h(u,v)=(Hu,v)\quad\text{for all }u\in\operatorname{Dom}(H),\text{
and }v\in \text{D}(h).
\]
\item[(ii)] $\operatorname{Dom}(H)$ is a core of $h$.
\item[(iii)] If $u\in\text{D}(h)$, $w\in L^2(E)$ and
$h(u,v)=(w,v)$ holds for every $v$ belonging to a core of $h$, then
$u\in\operatorname{Dom}(H)$ and $Hu=w$. The semi-bounded below
self-adjoint operator $H$ is uniquely determined by the condition
(i).
\end{enumerate}

In what follows we will use the following well-known Lemma.

\begin{lemma}\label{L:diagonal}
Assume  that $0\leq T\in L_{\rm loc}^1(\operatorname{End} E)$ is a
linear self-adjoint bundle map.  Assume also that $u\in Q(T)$,
where $Q(T)=\{u\in L^2(E)\colon \langle T u,u\rangle\in L^1(M)\}$.
Then $Tu\in L_{\rm loc}^1(E)$.
\end{lemma}

\begin{proof}
By adding a constant we can assume that $T\geq 1$
(in the operator sense).
Assume that $u\in Q(T)$.  We choose (in a measurable way) an
orthogonal basis in each fiber $E_x$ and diagonalize $1\leq
T(x)\in\operatorname{End}(E_x)$ to get
$$
T(x)=\mathop{\rm diag}(c_1(x),c_2(x),\dots,c_m(x)),
$$
 where $0<c_j\in
L_{\rm loc}^1(M)$, $j=1,2,\dots,m$ and $m=\dim E_x$.

Let $u_j(x)$ ($j=1,2,\dots,m$) be the components of $u(x)\in E_x$
with respect to the chosen orthogonal basis of $E_x$. Then for all
$x\in M$
\[
\langle Tu,u\rangle=\sum_{j=1}^{m}c_j(x)|u_j(x)|^2.
\]
Since $u\in Q(T)$, we know that $0<\int \langle T
u,u\rangle\,d\mu<+\infty$. Since $c_j>0$, it follows that
$c_j|u_j|^2\in L^1(M)$, for all $j=1,2,\dots,m$.

Now, for all $x\in M$ and $j=1,2,\cdots,m$
\begin{equation}\label{E:this}
2|c_ju_j|=2|c_j||u_j|\leq |c_j|+|c_j||u_j|^2,
\end{equation}
The right hand side of~\eqref{E:this} is clearly in $L_{\rm loc}^1(M)$.
Therefore $c_ju_j\in L_{\rm loc}^1(M)$.
But $(Tu)(x)$ has components $c_j(x)u_j(x)$ ($j=1,2,\dots,m$) with
respect to chosen bases of $E_x$. Therefore $Tu\in L_{\rm loc}^1(E)$,
and the Lemma is proven.
\end{proof}

The following corollary follows immediately from Lemma~\ref{L:diagonal}.
\begin{corollary}\label{C:diagonal-1}
If $u\in \text{D}(h_1)$, then $V_1u\in L_{\rm loc}^1(E)$.
\end{corollary}

\begin{corollary}\label{C:diagonal-2}
If $u\in \text{D}(h)$, then $Vu\in L_{\rm loc}^1(E)$.
\end{corollary}

\begin{proof}
Let $u\in \text{D}(h)=\text{D}(h_0)\bigcap \text{D}(h_1)$. By
Assumption (A1) we have $V=V_1+V_2$, where $0\leq V_1\in L_{\rm
loc}^1(\operatorname{End} E)$ and $0\geq V_2\in L_{\rm
loc}^1(\operatorname{End} E)$. By Corollary~\ref{C:diagonal-1} it
follows that $V_1u\in L_{\rm loc}^1(E)$ and since
$\text{D}(h)\subset \text{D}(h_2)$, by Lemma~\ref{L:diagonal} we
have $-V_2u\in L_{\rm loc}^1(E)$. Thus $Vu\in L_{\rm loc}^1(E)$,
and the corollary is proven.
\end{proof}

\begin{lemma}\label{L:H-subset-S} The following operator relation
holds: $H\subset S$.
\end{lemma}

\begin{proof} We will show that for all $u\in\operatorname{Dom}(H)$,
we have $Hu=H_Vu$.
Let $u\in\operatorname{Dom}(H)$.  By property (i) of operator $H$
we have $u\in\text{D}(h)$, hence
by Corollary~\ref{C:diagonal-2} we get $Vu\in L_{\rm loc}^1(E)$. Then, for any
$v\in C_{c}^{\infty}(E)$, we have
\begin{equation}\label{E:representation}
(Hu,v)=h(u,v)=(\nabla u,\nabla v)+\int\langle Vu,v \rangle\,d\mu,
\end{equation}
where $(\cdot,\cdot)$ denotes the $L^2$-inner product.

The first equality in~\eqref{E:representation} holds by property (i)
of operator $H$, and the second equality holds by
definition of $h$.

Hence, using integration by parts in the first term on the right
hand side of the second equality in~\eqref{E:representation} (see,
for example, Lemma 8.8 from~\cite{bms}), we get
\begin{equation}\label{E:distr}
(u,\nabla^*\nabla v)=\int\langle
Hu-Vu,v\rangle\,d\mu,\quad\text{for all}\quad v\in
C_{c}^{\infty}(E).
\end{equation}
Since $Vu\in L_{\rm loc}^1(E)$ and $Hu\in L^2(E)$, it follows that
$(Hu-Vu)\in L_{\rm loc}^1(E)$, and~\eqref{E:distr} implies
$\nabla^*\nabla u=Hu-Vu$ (as distributional sections of $E$).
Therefore,
\[
\nabla^*\nabla u+Vu=Hu,
\]
and this shows that $Hu=H_Vu$ for all $u\in\operatorname{Dom}(H)$.

Now by definition of $S$ it follows that
$\operatorname{Dom}(H)\subset\operatorname{Dom}(S)$ and $Hu=Su$
for all $u\in\operatorname{Dom}(H)$. Therefore $H\subset S$, and
the Lemma is proven.
\end{proof}

\begin{lemma}\label{L:h0+h1-form-core}
$C_{c}^{\infty}(E)$ is a core of the quadratic form $h_0+h_1$.
\end{lemma}

\begin{proof} We need to show that $C_{c}^{\infty}(E)$ is dense in the
Hilbert space
$\text{D}(h_0+h_1)=\text{D}(h_0)\bigcap\text{D}(h_1)$ with the
inner product
\[
(u,v)_{h_0+h_1}:=h_0(u,v)+h_1(u,v)+(u,v),
\]
where $(\cdot,\cdot)$ is the inner product in $L^2(E)$.

Let $u\in\text{D}(h_0+h_1)$ and $(u,v)_{h_0+h_1}=0$ for all $v\in
C_{c}^{\infty}(E)$. We will show that $u=0$.
We have
\begin{equation}\label{E:preceding-kato}
0  =  h_0(u,v)+h_1(u,v)+(u,v)\\
=  (u,\nabla^*\nabla v)+ \int\langle
V_1u,v \rangle\,d\mu + (u,v).
\end{equation}
Here we used integration by parts in the first term on the right
hand side of the second equality.

By Corollary \ref{C:diagonal-1} it follows that $V_1u\in L_{\rm loc}^1(E)$, and
from~\eqref{E:preceding-kato} we get the following equality of
distributional sections of $E$:
\begin{equation}\label{E:distributional-equality}
\nabla^*\nabla u=(-V_1-1)u.
\end{equation}
 From~\eqref{E:distributional-equality} we have $\nabla^*\nabla u\in
L_{\rm loc}^1(E)$. By Lemma~\ref{L:kato} and
by~\eqref{E:distributional-equality}, we obtain
\begin{equation}\label{E:kato-inequality}
\Delta_M|u| \leq \operatorname{Re}\langle\nabla^*\nabla
u,\operatorname{sign} u\rangle = \langle
-(V_1+1)u,\operatorname{sign} u\rangle \leq -|u|.
\end{equation}
The last inequality in~\eqref{E:kato-inequality} follows since
$V_1\geq 0$ (as an operator $E_x\to E_x$).
Therefore,
\begin{equation}\label{E:pzt}
(\Delta_M+1)|u|\leq 0.
\end{equation}
By Lemma~\ref{L:positivity}, it follows that $|u|\leq 0$. So $u=0$, and
the proof is complete.
\end{proof}

\begin{lemma}\label{L:h-form-core}
$C_{c}^{\infty}(E)$ is a core of the quadratic form
$h=(h_0+h_1)+h_2$.
\end{lemma}

Since the quadratic form $h_2$ is $(h_0+h_1)$-bounded,
the lemma follows immediately from Lemma \ref{L:h0+h1-form-core}.

\begin{proof}[Proof of Theorem~\ref{T:main}]
We will show that $S=H$. By Lemma~\ref{L:H-subset-S} we have $H\subset
S$, so it is enough to show that
$\operatorname{Dom}(S)\subset\operatorname{Dom}(H)$.

Let $u\in\operatorname{Dom}(S)$. By definition of
$\operatorname{Dom}(S)$, we have
$u\in\text{D}(h_0)\subset \text{D}(h_2)$ and $u\in\text{D}(h_1)$.
Hence $u\in\text{D}(h)$.
For all $v\in C_{c}^{\infty}(E)$ we have
\[
h(u,v)=h_0(u,v)+h_1(u,v)+h_2(u,v)=(u,\nabla^*\nabla v)+
\int\langle Vu,v \rangle\,d\mu=(H_Vu,v).
\]
The last equality holds since $H_Vu=Su\in L^2(E)$.
By Lemma~\ref{L:h-form-core} it follows that $C_{c}^{\infty}(E)$ is a
form core of $h$. Now from property (iii) of operator $H$
 we have $u\in\operatorname{Dom}(H)$ with
$Hu=H_Vu$. This concludes the proof of the Theorem.
\end{proof}

\begin{proof}[Proof of Remark \ref{R:Assumption-B}] %\label{S:sufficient-condition}
Let $p$ be as in Remark \ref{R:Assumption-B}. We may assume that
$\|V_2\|_{L^{p}(\operatorname{End} E)}$ is arbitrarily small
because there exists a sequence $V_2^{(k)}\in
L^{\infty}(\operatorname{End} E)\bigcap L^{p}(\operatorname{End}
E)$, $k\in {\mathbb{Z}}_+$, such that
\[
\|V_2^{(k)}-V_2\|_{L^{p}(\operatorname{End} E)}\to 0,\quad\text{as
}k\to\infty,
\]
and $V_2^{(k)}$, $k\in {\mathbb{Z}}_+$, contributes to $h_2$ only
a bounded form.

 For the rest of this article, we will assume that
$\|V_2\|_{L^{p}(\operatorname{End} E)}$ is arbitrarily small.
By Cauchy-Schwartz inequality  and H\"older's inequality we have
\begin{equation}\label{E:sobolev-domination}
\left|\int\langle V_2u,u \rangle\,d\mu\right|\leq \int|\langle
V_2u,u \rangle|\,d\mu \leq \int |V_2||u|^2\,d\mu\leq
\|V_2\|_{L^{p}(\operatorname{End} E)}\|u\|^2_{L^t(E)},
\end{equation}
where $|V_2|$ denotes the norm of the operator $V_2(x)\colon
E_x\to E_x$ and
\begin{equation}\label{E:sob-exp}
\frac{1}{p}+\frac{2}{t}=1.
\end{equation}

With our assumptions on $(M,g)$, $E$ and $\nabla$, the usual
Sobolev embedding theorems for $W^{1,2}({\mathbb{R}}^n)$ also hold
for $W^{1,2}(E)$ (see Sec.~A1.1 in~\cite{Shubin92}).

For $n\geq 3$, we know by hypothesis that $p=n/2$, so
from~\eqref{E:sob-exp} we get $1/t=1/2-1/n$. By the Sobolev embedding
theorem (see, for example, the first part of Theorem 2.10
in~\cite{Aubin}) we have
\[
\|u\|_{L^t(E)}\leq C(\|\nabla u\|_{L^2(T^*M\otimes E)
}+\|u\|_{L^2(E)}),\quad\text{for all }u\in W^{1,2}(E),
\]
where $C>0$ is a positive constant.

For $n=2$, we know by hypothesis that $p>1$, so
from~\eqref{E:sob-exp} we get $2<t<\infty$. In this case, it is
well-known (see, e.g.,~the first part of Theorem 2.10
in~\cite{Aubin}) that
\[
\|u\|_{L^t(E)}\leq C(\|\nabla u\|_{L^2(T^*M\otimes E
)}+\|u\|_{L^2(E)}),\quad\text{for all }u\in W^{1,2}(E),
\]
where $C>0$ is a positive constant.

For $n=1$, we know by hypothesis that $p=1$, so
from~\eqref{E:sob-exp} we get $t=\infty$.  In this case, it is
well-known (see e.g. the second part of Theorem 2.10
in~\cite{Aubin}) that
\[
\|u\|_{L^{\infty}(E)}\leq C(\|\nabla u\|_{L^2(T^*M\otimes E
)}+\|u\|_{L^2(E)}),\quad\text{for all }u\in W^{1,2}(E),
\]
where $C>0$ is a positive constant.

Combining each of the last three inequalities
with~\eqref{E:sobolev-domination}, we get~\eqref{E:h_2-h_0-domination}
(with constant $b<1$ because $\|V_2\|_{L^{p}(\operatorname{End}
E)}$ is arbitrarily small).
\end{proof}

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\end{document}
