\documentclass[reqno]{amsart}
\usepackage{hyperref}

\AtBeginDocument{{\noindent\small
{\em Electronic Journal of Differential Equations},
Vol. 2005(2005), No. 77, pp. 1--10.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu  (login: ftp)}
\thanks{\copyright 2005 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2005/77\hfil A property of Sobolev spaces]
{A property of Sobolev spaces on complete Riemannian manifolds}
\author[O. Milatovic\hfil EJDE-2005/77\hfilneg]
{Ognjen Milatovic}

\address{Ognjen Milatovic \hfill\break
Department of Mathematics and Statistics\\
University of North Florida  \\
Jacksonville, FL 32224, USA}
\email{omilatov@unf.edu}

\date{}
\thanks{Submitted June 25, 2005. Published July 8, 2005.}
\subjclass[2000]{58J05}
\keywords{Complete Riemannian manifold; Sobolev space}

\begin{abstract}
 Let $(M,g)$ be a complete Riemannian manifold with metric $g$
 and the Riemannian volume form $d\nu$. We consider the
 $\mathbb{R}^{k}$-valued functions
 $T\in [W^{-1,2}(M)\cap L_{\rm loc}^{1}(M)]^{k}$ and
 $u\in [W^{1,2}(M)]^{k}$ on $M$, where
 $[W^{1,2}(M)]^{k}$ is a Sobolev space on $M$ and $[W^{-1,2}(M)]^{k}$
 is its dual. We give a sufficient condition for the equality of
 $\langle T, u\rangle$ and the integral of $(T\cdot u)$ over $M$,
 where $\langle\cdot,\cdot\rangle$ is the duality between
 $[W^{-1,2}(M)]^{k}$ and $[W^{1,2}(M)]^{k}$. This is an extension to
 complete Riemannian manifolds of a result of H.~Br\'ezis and
 F.~E.~Browder.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{remark}[theorem]{Remark}

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

\subsection*{The setting}\label{SS:setting}
Let $(M,g)$ be a $C^{\infty}$ Riemannian manifold without
boundary, with metric $g=(g_{jk})$ and $\dim M=n$.  We will assume
that $M$ is connected, oriented, and complete. By $d\nu$ we will
denote the Riemannian volume element of $M$. In any local
coordinates $x^1,\dots,x^n$, we have $d\nu=\sqrt{\det
(g_{jk})}\,dx^1 dx^2\dots dx^n$.

By $L^{2}(M)$ we denote the space of real-valued square integrable
functions on $M$ with the inner product
\[
(u,v)=\int_{M}(uv)\,d\nu.
\]
Unless specified otherwise, in all function spaces below, the
functions are real-valued.

In what follows, $C^{\infty}(M)$ denotes the space of smooth
functions on $M$, $C_{c}^{\infty}(M)$ denotes the space of smooth
compactly supported functions on $M$, $\Omega^{1}(M)$ denotes the
space of smooth 1-forms on $M$, and $L^2(\Lambda^{1}T^*M)$ denotes
the space of square integrable 1-forms on $M$.

By $W^{1,2}(M)$ we denote the completion of $C_{c}^{\infty}(M)$ in
the norm
\[
\|u\|_{W^{1,2}}^{2}=\int_{M}|u|^{2}\,d\nu+ \int_{M}|du|^{2}\,d\nu,
\]
where $d:C^{\infty}(M)\to \Omega^{1}(M)$ is the standard
differential.

\begin{remark}\label{R:sobolev-spaces} \rm
It is well known (see, for example, Chapter 2 in~\cite{Aubin})
that if $(M,g)$ is a complete Riemannian manifold, then
$W^{1,2}(M)=\{u\in L^{2}(M)\colon du\in L^{2}(\Lambda^1 T^*M)\}$.
\end{remark}

By $W^{-1,2}(M)$ we denote the dual space of $W^{1,2}(M)$, and by
$\langle\cdot,\cdot\rangle$ we will denote the duality between
$W^{-1,2}(M)$ and $W^{1,2}(M)$.

In what follows, $[C_{c}^{\infty}(M)]^{k}$, $[L^2(M)]^{k}$,
$[L^{2}(\Lambda^1 T^*M)]^{k}$ and $[W^{1,2}(M)]^{k}$ denote the
space of all ordered $k$-tuples $u=(u_1,u_2,\dots, u_k)$ such that
$u_j\in C_{c}^{\infty}(M)$, $u_j\in L^2(M)$,  $u_j \in
L^{2}(\Lambda^1 T^*M)$, $u_j\in W^{1,2}(M)$, respectively, for all
$1\leq j\leq k$. For $u\in [W^{1,2}(M)]^{k}$, we will  use the
following notation:
\begin{gather}\label{E:du-definition}
du:=(du_1,du_2,\dots, du_k), \\
\label{E:norm-u-definition}
|u|:= (u_{1}^{2}+u_2^{2}+\cdots + u_{k}^{2})^{1/2}, \\
\label{E:norm-du-definition}
|du|:=(|du_1|^2+|du_2|^2+\cdots +
|du_k|^2)^{1/2},
\end{gather}
where $|du_j|$ denotes the length of the cotangent vector $du_j$.

The space $[W^{1,2}(M)]^{k}$ is the completion of
$[C_{c}^{\infty}(M)]^{k}$ in the norm
\[
\|u\|^2_{[W^{1,2}(M)]^{k}}=\int_{M}|u|^2\,d\nu+\int_{M}|du|^2\,d\nu,
\]
where $|u|$ and $|du|$ are as in~(\ref{E:norm-u-definition})
and~(\ref{E:norm-du-definition}) respectively.

\begin{remark} \rm
As in Remark \ref{R:sobolev-spaces}, if $(M,g)$ is a complete
Riemannian manifold, then $[W^{1,2}(M)]^{k}=\{u\in
[L^{2}(M)]^{k}\colon du\in [L^{2}(\Lambda^1 T^*M)]^{k}\}$.
\end{remark}

\subsection*{Assumption (H1)} Assume that
\begin{enumerate}
\item $u=(u_1,u_2,\dots,u_k)\in [W^{1,2}(M)]^{k}$ and \item
$T=(T_1,T_2,\dots, T_k)$, where
$T_1, T_2, \dots, T_k\in W^{-1,2}(M) \cap L^{1}_{\rm loc}(M)$.
\end{enumerate}
Here, the notation $T_j\in W^{-1,2}(M) \cap L^{1}_{\rm loc}(M)$ means
that $T_j$ is a.e.~defined function belonging to $L^{1}_{\rm loc}(M)$
such that
\[
\phi\mapsto \int_{M}T_j\phi\,d\nu, \quad \phi\in
C^{\infty}_{c}(M),
\]
extends continuously to $W^{1,2}(M)$.

 For a.e.~$x\in M$, denote
\begin{gather}\label{E:product-Tu-definition}
(T\cdot u)(x):=\sum_{j=1}^{k} T_j(x)u_j(x),
\\
\label{E:duality-Tu-definition}
\langle T,u \rangle := \sum_{j=1}^{k}\langle T_j,u_j\rangle,
\end{gather}
where $\langle\cdot,\cdot\rangle$ on the right hand side
of~(\ref{E:duality-Tu-definition}) denotes the duality between
$W^{-1,2}(M)$ and $W^{1,2}(M)$.

We now state our main result.

\begin{theorem}\label{T:main}
Assume that $(M,g)$ is a complete Riemannian
manifold. Assume that $u=(u_1,u_2,\dots,u_k)$ and
$T=(T_1,T_2,\dots, T_k)$ satisfy the assumption (H1). Assume that
there exists a function $f\in L^{1}(M)$ such that
\begin{equation}\label{E:assumption-g}
(T\cdot u)(x)\geq f(x), \quad\text{a.e. on }M.
\end{equation}
Then $(T\cdot u)\in L^{1}(M)$ and
\[
\langle T, u\rangle= \int_{M} (T\cdot u)(x)\, d\nu(x).
\]
\end{theorem}

In the following Corollary, by $W^{1,2}(M,\mathbb{C})$,
$W^{-1,2}(M,\mathbb{C})$ and $L_{\rm loc}^{1}(M,\mathbb{C})$ we
denote the complex analogues of spaces $W^{1,2}(M)$, $W^{-1,2}(M)$
and $L_{\rm loc}^{1}(M)$. By $\langle\cdot,\cdot\rangle$ we denote
the Hermitian duality between $W^{-1,2}(M,\mathbb{C})$ and
$W^{1,2}(M,\mathbb{C})$.

\begin{corollary}\label{C:complex-case}
Assume that $(M,g)$ is a complete Riemannian
manifold. Assume that $T\in W^{-1,2}(M,\mathbb{C})\cap L_{\rm
loc}^{1}(M,\mathbb{C})$ and $u\in W^{1,2}(M,\mathbb{C})$. Assume
that there exists a real-valued function $f\in L^1(M)$ such that
\[
\operatorname{Re}(T\bar{u}) \geq f,\quad\text{a.e. on }M.
\]
Then $\operatorname{Re} (T\bar{u})\in L^1(M)$ and
\[
\operatorname{Re} \langle T, u\rangle=\int_{M}
\operatorname{Re}(T\bar{u})\,d\nu.
\]
\end{corollary}

\begin{remark} \rm
Theorem \ref{T:main} and Corollary \ref{C:complex-case} extend the
corresponding results of H.~Br\'ezis and
F.~E.~Browder~\cite{Brezis-Browder-78} from ${\mathbb{R}}^{n}$ to
complete Riemannian manifolds. The results
of~\cite{Brezis-Browder-78} were used, among other applications, in
studying self-adjointness and $m$-accretivity in
$L^2({\mathbb{R}}^{n}, \mathbb{C})$ of Schr\"odinger operators with
singular potentials; see, for example, H.~Br\'ezis and
T.~Kato~\cite{Brezis-Kato79}. Analogously, Theorem~\ref{T:main} and
Corollary~\ref{C:complex-case} can be used in the study of
self-adjoint and $m$-accretive realizations (in the space
$L^2(M,\mathbb{C})$) of Schr\"odinger-type operators with singular
potentials, where $M$ is a complete Riemannian manifold, as well as
in the study of partial differential equations on complete
Riemannian manifolds.
\end{remark}

\section{Proof of~Theorem~\ref{T:main}}

We will adopt the arguments of H.~Br\'ezis and
F.~E.~Browder~\cite{Brezis-Browder-78} to the context of a
complete Riemannian manifold.
In what follows, $F\colon {\mathbb{R}}^{k}\to {\mathbb{R}}^{l}$ is
a $C^{1}$ vector-valued function $F(y)=(F_1(y),F_2(y),\dots,
F_l(y))$. By $dF(y)$ we will denote the derivative of $F$ at
$y=(y_1,y_2,\dots, y_k)$.

\begin{lemma} \label{L:chain-rule-smooth}
Assume that $F\in C^{1}({\mathbb{R}}^{k}, {\mathbb{R}}^{l})$,
$F(0)=0$, and for all $y\in {\mathbb{R}}^{k}$,
\[
|dF(y)|\leq C
\]
where $C\geq 0$ is a constant.

Assume that $u=(u_1,u_2,\dots,u_k)\in [W^{1,2}(M)]^{k}$. Then
$(F\circ u) \in [W^{1,2}(M)]^{l}$, and the following holds:
\begin{equation}\label{E:chain-rule-smooth}
d(F\circ u)= \sum_{j=1}^{k} \frac{\partial F}{\partial u_j}d u_j,
\end{equation}
where
\begin{equation}\label{E:notations}
\frac{\partial F}{\partial u_j}=\Big(\frac{\partial F_1}{\partial
y_j}(u), \frac{\partial F_2}{\partial y_j}(u),\dots, \frac{\partial
F_l}{\partial y_j}(u)\Big).
\end{equation}
(Here the notation $\frac{\partial F_s}{\partial y_j}(u)$, where
$1\leq s\leq l$, denotes the composition of $\frac{\partial
F_s}{\partial y_j}$ and $u$. The notation $d(F\circ u)$ denotes the
ordered $l$-tuple $(d(F_1\circ u), d(F_2\circ u), \dots, d(F_l\circ
u))$, where $d(F_s\circ u)$, $1\leq s\leq l$, is the differential of
the scalar-valued function $F_s\circ u$ on $M$.
\end{lemma}

\begin{proof} Let $u\in [W^{1,2}(M)]^{k}$.
By definition of $[W^{1,2}(M)]^{k}$, the weak derivatives $du_j$,
$1\leq j\leq k$, exist and $du_j\in L^2(M)$. By Lemma 7.5
in~\cite{Gilbarg-Trudinger}, it follows that for all $1\leq s\leq
l$, the following holds:
\[
d(F_{s}\circ u)=\sum_{j=1}^{k} \frac{\partial F_s}{\partial u_j}d
u_j,
\]
where
\[
\frac{\partial F_s}{\partial u_j}= \frac{\partial F_s}{\partial
y_j}(u).
\]
This shows~(\ref{E:chain-rule-smooth}).

Since $dF$ is bounded and since $du_j\in L^2(\Lambda^1T^*M)$, it
follows that $d(F_{s}\circ u)\in L^2(\Lambda^1T^*M)$ for all
$1\leq s\leq l$. Thus $d(F\circ u)\in [L^2(\Lambda^1T^*M)]^{l}$.
Moreover, since $u\in [W^{1,2}(M)]^{k}$ and
\[
|F_s\circ u|=|F_s(u)- F_s(0)|\leq C_1|u|,
\]
where $C_1\geq 0$ is a constant and $|u|$ is as
in~(\ref{E:norm-u-definition}), it follows that $(F_s\circ u)\in
L^2(M)$ for all $1\leq s\leq l$. Thus $(F\circ u) \in [L^2(M)]^{l}$.
Therefore, $(F\circ u)\in [W^{1,2}(M)]^{l}$, and the Lemma is
proven.
\end{proof}

\begin{lemma}\label{L:product-rule}
 Assume that $u$, $v\in
W^{1,2}(M)\cap L^{\infty}(M)$. Then $(uv)\in W^{1,2}(M)$ and
\begin{equation}\label{E:product-rule}
d(uv)=(du)v+u(dv).
\end{equation}
\end{lemma}
\begin{proof} By the remark after the equation (7.18)
in~\cite{Gilbarg-Trudinger}, the equation~(\ref{E:product-rule})
holds if the weak derivatives $du$, $dv$ exist and if $uv\in L_{\rm
loc}^{1}(M)$ and $((du)v+u(dv))\in L_{\rm loc}^{1}(M)$. By the
hypotheses of the Lemma, these conditions are satisfied, and,
hence,~(\ref{E:product-rule}) holds.

Furthermore, since $u$, $v\in W^{1,2}(M)\cap L^{\infty}(M)$, we have
$(uv)\in L^2(M)$. By hypotheses of the Lemma and
by (\ref{E:product-rule}) we have $d(uv)\in L^2(M)$. Thus $(uv)\in
W^{1,2}(M)$, and the Lemma is proven.
\end{proof}

In the next lemma, the statement ``$f\colon \mathbb{R}\to
\mathbb{R}$ is a piecewise smooth function" means that $f$ is
continuous and has piecewise continuous first derivative.

\begin{lemma}\label{L:chain-rule-piecewise-smooth}
Assume that $f\colon \mathbb{R}\to \mathbb{R}$ is a
piecewise smooth function with $f(0)=0$ and $f'\in
L^{\infty}(\mathbb{R})$. Let $S$ denote the set of corner points
of $f$. Assume that $u\in W^{1,2}(M)$. Then $(f\circ u)\in
W^{1,2}(M)$ and
\[
d(f\circ u)=\begin{cases}
  f'(u)\,du & \text{for all $x$ such that } u(x) \notin S\\
  0 & \text{for all $x$  such that }u(x)\in S
\end{cases}
\]
\end{lemma}

\begin{proof}
By the remark in the second paragraph below the equation (7.24)
in~\cite{Gilbarg-Trudinger}, the Lemma follows immediately from
Theorem 7.8 in~\cite{Gilbarg-Trudinger}.
\end{proof}

The following Corollary follows immediately from
Lemma~\ref{L:chain-rule-piecewise-smooth}.

\begin{corollary}\label{C:absolute-value}
Assume that $u\in W^{1,2}(M)$. Then $|u|\in W^{1,2}(M)$ and
\[
d|u|=\begin{cases}
  f'(u)\, du & \text{for all $x$ such that }u(x)\neq 0\\
  0 & \text{for all $x$ such that }u(x)=0\
\end{cases},
\]
where $f(t)=|t|$, $t\in\mathbb{R}$.
\end{corollary}

\begin{remark}\label{R:absolute-value} \rm
 Let $f(t)=|t|$, $t\in\mathbb{R}$.
Let $c$ be a real number. By Lemma 7.7 in~\cite{Gilbarg-Trudinger}
and by Corollary~\ref{C:absolute-value}, we can write
$d|u|=h(u)du$ a.e. on $M$,
where
\[
h(t)= \begin{cases}
  f'(t)& \text{for all }t\neq 0\\
  c & \text{otherwise}.
\end{cases}
\]
\end{remark}

\begin{lemma} \label{L:minimum}
Assume that $u$, $v\in W^{1,2}(M)$ and let
\[
w(x):=\min\{u(x), v(x)\}.
\]
Then $w\in W^{1,2}(M)$ and
\[
|dw|\leq \max\{|du|, |dv|\},\quad\text{a.e. on }M,
\]
where $|du(x)|$ denotes the norm of the cotangent vector $du(x)$.
\end{lemma}

\begin{proof}
We can write
\[
w(x)=\frac{1}{2}(u(x)+v(x)-|u(x)-v(x)|).
\]
Since $u$, $v\in W^{1,2}(M)$, by Corollary~\ref{C:absolute-value} we
have $|u-v|\in W^{1,2}(M)$, and, thus, $w\in W^{1,2}(M)$.
By Remark \ref{R:absolute-value}, we have
\begin{equation}\label{E:w-through-h}
dw(x)=\frac{1}{2}(du(x)+dv(x)-(h(u-v))\cdot(du(x)-dv(x))),
\quad\text{a.e. on }M,
\end{equation}
where $h$ is as in Remark~\ref{R:absolute-value}.

Considering $dw(x)$ on sets $\{x\colon u(x)>v(x)\}$, $\{x\colon
u(x)<v(x)$\} and $\{x\colon u(x)=v(x)\}$, and
using~(\ref{E:w-through-h}), we get
\[
|dw(x)|\leq \max\{|du(x)|, |dv(x)|\}, \quad\text{a.e. on }M.
\]
This concludes the proof of the Lemma.
\end{proof}

\begin{lemma}\label{L:key-lemma}
Let $a>0$. Let $u=(u_1,u_2,\dots,
u_k)$ be in $[W^{1,2}(M)]^{k}$, let $v=(v_1,v_2,\dots, v_k)$ be in
$[W^{1,2}(M)\cap  L^{\infty}(M)]^{k}$, and let
\[
w:=\left((|u|^2+a^2)^{-1/2}\min\{(|u|^2+a^2)^{1/2}-a,
(|v|^2+a^2)^{1/2}-a\}\right)u,
\]
where $|u|$ is as in~(\ref{E:norm-u-definition}).
Then $w\in  [W^{1,2}(M)\cap L^{\infty}(M)]^{k}$ and
\[
|dw|\leq 3\max\{|du|,|dv|\},\quad\text{a.e. on }M,
\]
where $|du|$ is as in~(\ref{E:norm-du-definition}).
\end{lemma}

\begin{proof} Let $\phi=(|u|^2+a^2)^{-1/2}u$. Then
$\phi=F\circ u$, where $F\colon {\mathbb{R}}^{k}\to
{\mathbb{R}}^{k}$ is defined by
\[
F(y)=(|y|^2+a^2)^{-1/2}y, \quad y\in{\mathbb{R}}^k.
\]
Clearly, $F\in C^{1}({\mathbb{R}}^{k},{\mathbb{R}}^{k})$ and
$F(0)=0$. It easily checked that the component functions
\[
F_s(y)=(|y|^2+a^2)^{-1/2}y_s
\]
satisfy
\[
\frac{\partial F_s}{\partial y_j}=\begin{cases}
  -(|y|^2+a^2)^{-3/2}y_{s}y_j & \text{for }s\neq j \\
 (|y|^2+a^2)^{-3/2}(|y|^2-y_{j}^{2}+a^2) & \text{for }s=j.
\end{cases}
\]
Therefore, for all $1\leq s,j\leq k$, we have
\[
\big|\frac{\partial F_s}{\partial y_j}(y)\big|\leq \frac{1}{a},
\]
and, hence, $F$ satisfies the hypotheses of
Lemma~\ref{L:chain-rule-smooth}. Thus,
by Lemma~\ref{L:chain-rule-smooth} we have $(F\circ u)=\phi \in
[W^{1,2}(M)]^{k}$.

We now write the formula for $d\phi=(d\phi_1,d\phi_2,\dots,
d\phi_k)$. We have
\begin{equation}\label{E:d-phi}
d\phi=(|u|^2+a^2)^{-3/2}\Big((|u|^2+a^2)du-\Big(\sum_{j=1}^{k}u_jdu_j
\Big)u\Big),
\end{equation}
where $du$ is as in~(\ref{E:du-definition}).

By (\ref{E:d-phi}), using triangle inequality and Cauchy-Schwarz
inequality, we have
\begin{equation}\label{E:estimate-d-phi}
\begin{aligned}
|d\phi| &\leq (|u|^2+a^2)^{-3/2}\Big((|u|^2+a^2)|du|
+\big|\sum_{j=1}^{k}u_jdu_j\big||u|\Big)\\
&\leq  (|u|^2+a^2)^{-3/2}\left((|u|^2+a^2)|du|+|u||du||u|\right)\\
&\leq  (|u|^2+a^2)^{-3/2}\left((|u|^2+a^2)|du|+(|u|^2+a^2)|du|\right)\\
&=  2 (|u|^2+a^2)^{-1/2}|du|, \quad\text{a.e. on }M,
\end{aligned}
\end{equation}
where $|du_j|$ is the norm of the cotangent vector $du_j$, and $|u|$
and $|du|$ are as in (\ref{E:norm-u-definition})
and (\ref{E:norm-du-definition}) respectively.

Let
\[
\psi:=\min\{(|u|^2+a^2)^{1/2}-a, (|v|^2+a^2)^{1/2}-a\}.
\]
Then
\[
(|u|^2+a^2)^{1/2}-a=G\circ u\quad\text{ and
 }\quad(|v|^2+a^2)^{1/2}-a=G\circ v,
\]
where
\[
G(y)=(|y|^2+a^2)^{1/2}-a,\quad y\in{\mathbb{R}}^{k}.
\]
Clearly, $G\in C^{1}({\mathbb{R}}^{k},{\mathbb{R}})$ and $G(0)=0$,
and
\[
\frac{\partial G}{\partial y_j}=(|y|^2+a^2)^{-1/2}y_j,
\]
It is easily seen that there exists a constant $C_2\geq 0$ such that
$|dG(y)|\leq C_2$ for all $y\in{\mathbb{R}}^{k}$. Hence,
by Lemma~\ref{L:chain-rule-smooth} we have $(G\circ u)\in
W^{1,2}(M)$ and $(G\circ v)\in W^{1,2}(M)$.

Thus, by Lemma \ref{L:minimum} we have $\psi\in W^{1,2}(M)$, and
\[
|d\psi|\leq \max\big\{\big|d((|u|^2+a^2)^{1/2}-a)\big|,
\big|d((|v|^2+a^2)^{1/2}-a)\big|\big\}, \quad\text{a.e. on
}M.
\]
Using triangle inequality and Cauchy-Schwarz inequality, we have
\begin{equation}\label{E:du-norm-estimate}
\begin{aligned}
|d((|u|^2+a^2)^{1/2}-a)|
& = \big|(|u|^2+a^2)^{-1/2}\big(\sum_{j=1}^{k}u_jdu_j\big)\big|\\
&\leq  (|u|^2+a^2)^{-1/2} |u||du| \\
&\leq  |du|,
\end{aligned}
\end{equation}
where $|u|$ and $|du|$ are as in~(\ref{E:norm-u-definition})
and~(\ref{E:norm-du-definition}) respectively.
As in (\ref{E:du-norm-estimate}), we obtain
\[
|d((|v|^2+a^2)^{1/2}-a)| \leq |dv|.
\]
Therefore, we get
\begin{equation}\label{E:norm-estimate-psi}
|d\psi|\leq\max\{|du|,|dv|\}, \quad\text{a.e. on }M,
\end{equation}
where $|d\psi|$ is the norm of the cotangent vector $d\psi$, and
$|du|$ and $|dv|$ are as in~(\ref{E:norm-du-definition}).

By definition of $\phi$ we have $\phi\in [L^{\infty}(M)]^{k}$ and,
by definition of $\psi$ we have
\[
\psi \leq (|v|^2+a^2)^{1/2}-a.
\]
Thus,
\begin{equation}\label{E:bounded-psi}
\psi\leq |v|,
\end{equation}
where $|v|$ is as in~(\ref{E:norm-u-definition}).

Since $v\in [L^{\infty}(M)]^{k}$, we have $\psi\in L^{\infty}(M)$.
We have already shown that $\phi \in [W^{1,2}(M)]^{k}$ and $\psi \in
W^{1,2}(M)$. By Lemma~\ref{L:product-rule} (applied to the
components $\psi\phi_j$, $1\leq j\leq k$, of $\psi\phi$) we have
$w=\psi\phi\in [W^{1,2}(M)]^{k}$ and
\begin{equation}\label{E:product-estimate}
d(\psi\phi)=(d\psi)\phi+\psi(d\phi).
\end{equation}
By (\ref{E:product-estimate}), (\ref{E:estimate-d-phi})
and (\ref{E:norm-estimate-psi}), we have a.e. on $M$:
\begin{align*}
|dw| &=  |(d\psi)\phi+\psi(d\phi)|  \\
&\leq |d\psi||\phi|+|\psi||d\phi| \\
&\leq \left(\max\{|du|,|dv|\}\right) |\phi| + 2
(|u|^2+a^2)^{-1/2}|du||\psi| \\
&\leq  \max\{|du|,|dv|\} + 2|du| \\
&\leq  3\max\{|du|,|dv|\},
\end{align*}
where the third inequality holds since $|\phi|\leq 1$ and
$|\psi|(|u|^2+a^2)^{-1/2}\leq 1$.
This concludes the proof of the Lemma.
\end{proof}


\begin{lemma}\label{L:bounded-compact-supp}
Let $T=(T_1,T_2,\dots, T_k)$ and $u=(u_1,u_2,\dots, u_k)$ be as in
the hypotheses of~Theorem~\ref{T:main}. Additionally, assume that
$u$ has compact support and $u\in [L^{\infty}(M)]^{k}$. Then the
conclusion of Theorem~\ref{T:main} holds.
\end{lemma}

\begin{proof} Since the vector-valued function
$u=(u_1,u_2,\dots,u_k)\in [W^{1,2}(M)]^{k}$ is compactly supported,
it follows that the functions $u_j$ are compactly supported. Thus,
using a partition of unity we can assume that $u_j$ is supported in
a coordinate neighborhood $V_j$. Thus we can use the Friedrichs
mollifiers. Let $\rho_j>0$ and $(u_{j})^{\rho_j}:= J^{\rho_j}u$,
where $J^{\rho_j}$ denotes the Friedrichs mollifying operator as in
Section 5.12 of~\cite{bms}. Then $(u_{j})^{\rho_j}\in
C_{c}^{\infty}(M)$, and, as $\rho_j\to 0+$, we have
$(u_j)^{\rho_j}\to u_j$ in $W^{1,2}(M)$; see, for example, Lemma
5.13 in~\cite{bms}. Thus
\begin{equation}\label{E:weak-convergence-smooth}
\langle T_j,(u_{j})^{\rho_j}\rangle \to \langle
T_j,u_{j}\rangle,\quad\text{as }\rho_j\to 0+,
\end{equation}
where $\langle\cdot,\cdot\rangle$ is as on the right hand side
of~(\ref{E:duality-Tu-definition}).

Since $(u_{j})^{\rho_j}\in C_{c}^{\infty}(M)$ and $T_j\in L_{\rm
loc}^{1}(M)$, we have
\begin{equation}\label{E:equality-mollifiers}
\langle T_j,(u_{j})^{\rho_j}\rangle = \int_{M}(T_j\cdot
(u_j)^{\rho_j})\,d\nu.
\end{equation}
Next, we will show that
\begin{equation}\label{E:limit-u-rho-j}
\lim_{\rho_j\to 0+}\int_{M}(T_j\cdot
(u_j)^{\rho_j})\,d\nu=\int_{M}(T_ju_j)\,d\nu.
\end{equation}
Since $u_j\in L^{\infty}(M)$ is compactly supported, by properties
of Friedrichs mollifiers (see, for example, the proof of Theorem
1.2.1 in \cite{Friedlander-Joshi}) it follows that

\begin{itemize}
\item[(i)] there exists a compact set $K_j$ containing the supports of
$u_j$ and $u_{j}^{\rho_j}$ for all $0<\rho_j<1$, and

\item[(ii)] the following inequality holds for all $\rho_j>0$:
\begin{equation}\label{E:friedrichs-L-infty}
\|u_{j}^{\rho_j}\|_{L^{\infty}}\leq \|u_j\|_{L^{\infty}}.
\end{equation}
\end{itemize}
Since $(u_j)^{\rho_j}\to u_j$ in $L^2(M)$ as $\rho_j\to 0+$, after
passing to a subsequence we have
\begin{equation}\label{E:friedrichs-ae}
(u_{j})^{\rho_j}\to u_j \quad \text{ a.e. on }M,
\quad\text{as }\rho_j\to 0+.
\end{equation}
By (\ref{E:friedrichs-L-infty}) we have
\begin{equation}\label{E:friedrichs-T-u-j-rho}
|T_j(x)(u_j)^{\rho_j}(x)|\leq |T_j(x)|\|u_j\|_{L^{\infty}},\quad
\text{a.e. on }M.
\end{equation}
Since $T_j\in L_{\rm loc}^1(M)$, it follows that $T_j\in
L^1(K_j)$.

By (\ref{E:friedrichs-ae}), (\ref{E:friedrichs-T-u-j-rho}) and
since $T_j\in L^1(K_j)$, using dominated convergence theorem, we
have
\begin{equation}\nonumber
\lim_{\rho_j\to 0+}\int_{M}(T_j\cdot
(u_j)^{\rho_j})\,d\nu=\lim_{\rho_j\to 0+}\int_{K_j}(T_j\cdot
(u_j)^{\rho_j})\,d\nu=\int_{K_j}(T_ju_j)\,d\nu=\int_{M}(T_ju_j)\,d\nu,
\end{equation}
and~(\ref{E:limit-u-rho-j}) is proven.
Now,
using (\ref{E:weak-convergence-smooth}),
(\ref{E:equality-mollifiers}), (\ref{E:limit-u-rho-j})
and the notations (\ref{E:product-Tu-definition})
and (\ref{E:duality-Tu-definition}), we get
\begin{equation}\label{E:smooth-compact}
\begin{aligned}
\langle T, u \rangle
&=  \sum_{j=1}^{k} \langle T_j, u_{j}\rangle \\
&=  \sum_{j=1}^{k} \lim_{\rho_j\to 0+}\langle T_j, (u_{j})^{\rho_j}\rangle \\
&=  \sum_{j=1}^{k} \lim_{\rho_j\to 0+} \int_{M}(T_j\cdot
(u_j)^{\rho_j})\,d\nu \\
&=  \sum_{j=1}^{k} \int_{M}(T_ju_j)\,d\nu \ = \ \int_{M}  (T\cdot  u)\,d\nu.
\end{aligned}
\end{equation}
This concludes the proof of the Lemma.
\end{proof}

\begin{proof}[Proof of Theorem~\ref{T:main}]
Let $u\in [W^{1,2}(M)]^{k}$. By definition of $[W^{1,2}(M)]^{k}$
in Section~\ref{S:main}, there exists a sequence $v^{m}\in
[C_{c}^{\infty}(M)]^{k}$ such that $v^m\to u$ in
$[W^{1,2}(M)]^{k}$, as $m\to +\infty$. In particular, $v^{m}\to u$
in $[L^2(M)]^{k}$, and, hence, we can extract a subsequence, again
denoted by $v^m$, such that $v^m\to u$ a.e. on $M$.

Define a sequence $\lambda^m$ by
\[
\lambda^m:=\Big(|u|^{2}+\frac{1}{m^2}\Big)^{-1/2}
\min\Big\{\Big(|u|^{2}+\frac{1}{m^2}\Big)^{1/2}-\frac{1}{m},
\Big(|v^m|^{2}+\frac{1}{m^2}\Big)^{1/2}-\frac{1}{m}\Big\},
\]
where $v^m$ is the chosen subsequence of $v^m$ such that $v^m\to
u$ a.e. on $M$, as $m\to+\infty$.
Clearly, $0\leq \lambda^m \leq 1$. Define
\begin{equation}\label{E:w-m}
w^m:=\lambda^m u.
\end{equation}
We know that $u\in [W^{1,2}(M)]^{k}$ and $v^m\in
[C_{c}^{\infty}(M)]^{k}$. Thus, by Lemma~\ref{L:key-lemma}, for all
$m=1,2,3,\dots$, we have $w^m\in [W^{1,2}(M)\cap
L^{\infty}(M)]^{k}$, and
\begin{equation}\label{E:derivatives-brezis-browder-78}
|d(w^m)|\leq 3\max\{|du|, |d(v^m)|\},
\end{equation}
where $|du|$ is as in~(\ref{E:norm-u-definition}).
Furthermore, for all $m=1,2,3,\dots$, we have
\begin{equation}\label{E:estimate-w-m}
|w^{m}(x)|\leq |u(x)|,
\end{equation}
where $|\cdot|$ is as in~(\ref{E:norm-u-definition}).

Since $u\in [L^2(M)]^{k}$, by (\ref{E:estimate-w-m}) it follows that
$\{w^m\}$ is a bounded sequence in $[L^2(M)]^{k}$. Since $v^m\to u$
in $[W^{1,2}(M)]^{k}$, it follows that the sequence $\{v^m\}$ is
bounded in $[W^{1,2}(M)]^{k}$. In particular, the sequence
$\{d(v^m)\}$ is bounded in $[L^2(\Lambda^1T^*M)]^{k}$. Hence,
by (\ref{E:derivatives-brezis-browder-78}) it follows that
$\{d(w^m)\}$ is a bounded sequence in $[L^2(\Lambda^1T^*M)]^{k}$.
Therefore, $\{w^m\}$ is a bounded sequence in $[W^{1,2}(M)]^{k}$. By
Lemma V.1.4 in~\cite{Kato66} it follows that there exists a
subsequence of $\{w^m\}$, which we again denote by $\{w^m\}$, such
that $w^m$ converges weakly to some $z\in [W^{1,2}(M)]^{k}$. This
means that for every continuous linear functional $A\in
[W^{-1,2}(M)]^{k}$, we have
\[
A(w_m)\to A(z),\quad\text{as }m\to+\infty.
\]
Since
\[
[W^{1,2}(M)]^{k}\subset [L^2(M)]^{k} \subset [W^{-1,2}(M)]^{k},
\]
it follows that $w^m\to z$ in weakly $[L^2(M)]^{k}$.

We will now show that, as $m\to +\infty$, $w^m\to u$ in
$[L^2(M)]^{k}$. By definition of $w^m$ in~(\ref{E:w-m}) it follows
that $w^m\to u$ a.e. on $M$. Since $u\in [L^2(M)]^{k}$,
using (\ref{E:estimate-w-m}) and dominated convergence theorem we
get $w^m\to u$ in $[L^2(M)]^{k}$, as $m\to+\infty$.

In particular, $w^m\to u$ weakly in $[L^2(M)]^{k}$. Therefore, by
the uniqueness of the weak limit (see, for example, the beginning
of Section III.1.6 in~\cite{Kato66}), we have $z=u$. Therefore,
$w^m\to u$ weakly in $[W^{1,2}(M)]^{k}$.

Thus, since $T\in [W^{-1,2}(M)]^{k}$, we have
\begin{equation}\label{E:t-w-m-weak-convergence}
\langle T, w^m\rangle \to \langle T , u\rangle, \quad\text{as
}m\to+\infty.
\end{equation}
By the definition of $\lambda^m$ and~(\ref{E:w-m}) it follows that
\begin{equation}\label{E:w-m-compact-support}
|w^m(x)|\leq |v^m(x)|.
\end{equation}
Since $v^m\in [C_{c}^{\infty}(M)]^k$,
by (\ref{E:w-m-compact-support}) it follows that the functions $w^m$
have compact support. We have shown earlier that $w^m\in
[W^{1,2}(M)\cap L^{\infty}(M)]^{k}$. Thus,
by Lemma~\ref{L:bounded-compact-supp}, the following equality holds:
\begin{equation}\label{E:t-w-m}
\langle T, w^m \rangle = \int_{M}  (T\cdot
 w^m)\,d\nu.
\end{equation}
Let $f$ be as in the hypotheses of the Theorem. Then
\begin{equation}\label{E:inequality-T}
T\cdot w^m=T\cdot(\lambda^mu)=\lambda^m(T\cdot u)\geq \lambda^m
f\geq -|f|.
\end{equation}
By (\ref{E:inequality-T}) it follows that $T\cdot w^m+|f|\geq 0$.
Consider the sequence $T\cdot w^m+|f|$. Since $f\in L^1(M)$ and
$(T\cdot w^m)\in L^1(M)$, by Fatou's lemma we get
\begin{equation}\label{E:application-fatou}
\int_{M}\liminf_{m\to+\infty}(T\cdot w^m+|f|)\,d\nu \leq
\liminf_{m\to+\infty}\int_{M} (T\cdot w^m+|f|)\,d\nu.
\end{equation}
Since $w^m\to u$ a.e. on $M$ as $m\to+\infty$, we have $T\cdot
w^m\to T\cdot u$ a.e. on $M$ as $m\to+\infty$. Thus,
by (\ref{E:application-fatou}) we have
\[
\int_{M} (T\cdot u +|f|)\,d\nu\leq \int_{M} |f|\,d\nu +
\liminf_{m\to+\infty}\int_{M} (T\cdot w^m)\,d\nu,
\]
and, hence, by (\ref{E:t-w-m}) and~(\ref{E:t-w-m-weak-convergence})
we have
\begin{align*}
\int_{M} (T\cdot u +|f|)\,d\nu
&\leq \int_{M} |f|\,d\nu +
\liminf_{m\to+\infty}\int_{M} (T\cdot w^m)\,d\nu \\
&=  \int_{M}
|f|\,d\nu + \liminf_{m\to+\infty}\langle T, w^m\rangle\\
& = \int_{M} |f|\,d\nu + \langle T ,u\rangle.
\end{align*}
Since $f\in L^1(M)$, we have $(T\cdot u +|f|)\in L^1(M)$, and,
hence, $(T\cdot u)\in L^1(M)$.
We have
\[
|T\cdot w^{m}|=|\lambda^m(T\cdot u)|\leq |T\cdot u|,
\]
and by definition of $w^m$, we get, as $m\to+\infty$,
\[
T\cdot w^{m}\to T\cdot u, \quad\text{a.e. on }M.
\]
Using dominated convergence theorem, we get
\begin{equation}\label{E:limit-integrals}
\lim_{m\to+\infty} \int_{M}  (T\cdot
 w^m)\,d\nu=\int_{M}  (T\cdot u)\,d\nu
\end{equation}
By (\ref{E:limit-integrals}),~(\ref{E:t-w-m})
and~(\ref{E:t-w-m-weak-convergence}), we get
\[
\langle T, u\rangle= \int_{M} (T\cdot u)\, d\nu.
\]
This concludes the proof of the Theorem.
\end{proof}

\begin{proof}[Proof of Corollary \ref{C:complex-case}]
Let $T_1=\operatorname{Re} T$ and $T_2=\operatorname{Im} T$. Let
$u_1=\operatorname{Re} u$ and $u_2=\operatorname{Im} u$. Then
$\operatorname{Re} \langle T, u\rangle=\langle T_1, u_1\rangle +
\langle T_2, u_2\rangle$ and $\operatorname{Re} (T\cdot \bar{u})=
T_1u_1+T_2u_2$. Thus,~Corollary~\ref{C:complex-case} follows from
Theorem~\ref{T:main}.
\end{proof}

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\end{document}