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\markboth{\hfil Fredholm linear operators \hfil EJDE--1999/44}
{EJDE--1999/44\hfil M. Cecchi, M. Furi, M. Marini, \& M. P. Pera \hfil}
\begin{document}
\title{\vspace{-1in}\parbox{\linewidth}{\footnotesize\noindent
{\sc Electronic Journal of Differential Equations},
Vol. {\bf 1999}(1999), No.~44, pp. 1--16. \newline
ISSN: 1072-6691. URL: http://ejde.math.swt.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.swt.edu \quad ftp ejde.math.unt.edu (login: ftp)}
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%
Fredholm linear operators associated with ordinary differential
equations on noncompact intervals
\thanks{ {\em 1991 Mathematics Subject Classifications:} 34B05, 47A53.
\hfil\break\indent
{\em Key words and phrases:} Fredholm operators, noncompact
intervals.
\hfil\break\indent
\copyright 1999 Southwest Texas State University and University of
North Texas. \hfil\break\indent
Submitted March 9, 1999. Published October 31, 1999.} }
\date{}
\author{Mariella Cecchi, Massimo Furi,\\
Mauro Marini, \& Maria Patrizia Pera}
\maketitle
\begin{abstract}
In the noncompact interval $J=[a,\infty )$ we consider a linear
problem of the form $Lx=y,\; x \in S$, where $L$ is a first order
differential operator, $y$ a locally summable function in $J$, and
$S$ a subspace of the Fr\'{e}chet space of the locally absolutely
continuous functions in $J$. In the general case, the restriction of
$L$ to $S$ is not a Fredholm operator. However, we show that, under
suitable assumptions, $S$ and $L(S)$ can be regarded as subspaces of
two quite natural spaces in such a way that $L$ becomes a Fredholm
operator between them. Then, the solvability of the problem will be
reduced to the task of finding linear functionals defined in a
convenient subspace of $L_{loc}^{1}(J,{\mathbb R}^{n})$ whose
``kernel intersection'' coincides with $L(S)$. We will prove that,
for a large class of ``boundary sets'' $S$, such functionals can
be obtained by reducing the analysis to the case when the function
$y$ has compact support. Moreover, by adding a suitable stronger
topological assumption on $S$, the functionals can be represented
in an integral form. Some examples illustrating our results are
given as well.
\end{abstract}
\def\theequation{\arabic{section}.\arabic{equation}}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{example}[theorem]{Example}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{proposition}[theorem]{Proposition}
\section{Introduction}
\setcounter{equation}{0}
In the noncompact interval $J = [a, \infty)$ consider the linear
problem
\begin{eqnarray} \label{e1.1}
&Lx = y &\\
&y \in L_{\rm loc}^1(J,{\mathbb R}^n), \quad x \in S \,,&\nonumber
\end{eqnarray}
where $S$ is a subspace of $AC_{\rm loc}(J,{\mathbb R}^n)$, the Fr\'echet
space of all the locally absolutely continuous functions $x: J \to
{\mathbb R}^n$, $L: AC_{\rm loc}(J,{\mathbb R}^n)\to L_{\rm
loc}^1(J,{\mathbb R}^n)$ is
the first order differential operator defined by $(Lx)(t) =
x'(t)+A(t)x(t)$, with $A(\cdot)$ a $n \times n$ matrix with entries in
the space $L_{\rm loc}^1(J,{\mathbb R})$ of locally summable functions from
$J$ to ${\mathbb R}$. It is known that, in the case when the problem is
considered in a compact interval $I$ and $S$ is a finite codimensional
closed subspace of $AC(I,{\mathbb R}^n)$ (i.e.\ $S$ is the intersection of
the kernels of finitely many linearly independent bounded real
functionals on $AC(I,{\mathbb R}^n)$), the Fredholm index of the
restriction of $L$ to $S$ is well defined and equals $n-r$, where $r$
is the codimension of $S$ in $AC(I,{\mathbb R}^n)$.
However, in the problems we are interested here, the codimension of
$L(S)$ in $L_{\rm loc}^1(J,{\mathbb R}^n)$ may not be finite (see e.g.\
Example
\ref{ex3.1}). Thus, in general, the restriction of $L$ to $S$ is not a
Fredholm operator. Since the theory of Fredholm operators is a useful
tool in the solvability of linear boundary value problems (see, e.g.,
\cite{Rudolf}), it turns out to be of some interest to investigate
whether or not $S$ and $L(S)$ can be regarded as subspaces of some
``natural'' and suitable smaller spaces in such a way that $L$ becomes
a Fredholm operator between these two spaces. To this end, let
$AC_0(J,{\mathbb R}^n)$ denote the subspace of $AC(J,{\mathbb R}^n)$ of those
functions having compact support in $J$ and set $S_{\infty} = S +
AC_0(J,{\mathbb R}^n)$ (here the index $\infty$ is suggested by the fact
that, in order to decide whether or not a function belongs to the
space $S + AC_0(J,{\mathbb R}^n)$, it is enough to consider its behavior in
a neighborhood of $\infty $). Hence, $S$ can be regarded as a subspace
of $S_{\infty}$ and, thus, $L(S)$ of $L(S_{\infty})$. Consequently,
one can restate problem (\ref{e1.1}) in $S_{\infty}$ as follows:
\begin{eqnarray} \label{e1.2}
&Lx = y & \\
&y \in L(S_{\infty}), \quad x \in S \subset S_{\infty}\,.&\nonumber
\end{eqnarray}
By assuming that the codimension of $S$ in $S_{\infty}$ is finite, it
is easy to show (see Section 3) that $L: S \to L(S_{\infty})$ is a
Fredholm operator of index $m-r$, where $m = \dim {\rm Ker}L \cap
{S_{\infty}}$ and $r = \dim S_{\infty}/S$. Thus $S_{\infty}$ and
$L(S_{\infty})$ play the same role as, in a compact interval, $AC$ and
$L^1$ respectively, and $m$ is the analogue of $n$. For this reason,
it turns out convenient to restrict our attention to those $y \in
L_{\rm loc}^1(J,{\mathbb R}^n)$ which actually belong to $L(S_{\infty})$. In
this way, after some suitable sufficient conditions for $y \in
L_{\rm loc}^1(J,{\mathbb R}^n)$ to be in $L(S_{\infty})$ are given, when $L(S)
\neq L(S_{\infty})$, the solvability of (\ref{e1.1}) will be reduced
to the problem of finding functionals which individuate $L(S)$ in
$L(S_{\infty})$, that is, real linear functionals defined in a
convenient subspace of $L_{\rm loc}^1(J, {\mathbb R}^n)$ whose ``kernel
intersection'' coincides with $L(S)$. We will show that, for a large
class of ``boundary sets'' $S$, such functionals can be obtained by
reducing the analysis to the case when the function $y$ in problem
(\ref{e1.2}) has compact support.
The plan of the paper is as follows. In Section 2 we investigate the
solvability of the equation $Lx = y$, with $x$ in an asymptotic space;
that is, a subspace $E$ of $AC_{\rm loc}(J,{\mathbb R}^n)$ containing
$AC_0(J,{\mathbb R}^n)$. In Section 3, the solvability of (\ref{e1.2}) is
preliminary reduced to that of the same problem with $y$ in the
subspace $L_0^1(J,{\mathbb R}^n)$ of $L_{\rm loc}^1(J,{\mathbb R}^n)$ of the
functions with essential compact support. In Section 4 we show that, by
adding a suitable topological assumption on the ``boundary space'' $S$,
the functionals individuating $L(S)\cap L_0^1(J,{\mathbb R}^n)$ in
$L_0^1(J,{\mathbb R}^n)$ can be represented in an integral form. In our main
result (Theorem (\ref{t4.3}) below) we prove, with a stronger topological
assumption on $S$, that the same integral expressions individuate
$L(S)$ in a convenient subspace of $L(S_{\infty})$. Some examples
illustrating our results are given.
There exists a vast literature concerning the solvability of linear
boundary value problems on unbounded intervals. For quite recent
results on this topic we suggest, e.g., \cite{Elias}, \cite{Kig} and
references therein. Let us point out that the solvability of the
linear problem (\ref{e1.1}) arises for instance in the investigation
of nonlinear boundary value problems on noncompact intervals and it
is, in some sense, a useful preliminary when such a study is carried
out by means of topological methods. This nonlinear point of view will
be considered in a forthcoming paper. For a different approach to
nonlinear problems see \cite{Conti}, \cite{O'R1}, \cite{O'R2} and
references therein.
We close the Introduction with some notation.
We denote by
$AC_{\rm loc} := AC_{\rm loc}(J,{\mathbb R}^n)$,
$AC_0 := AC_0(J,{\mathbb R}^n)$,
$L_{\rm loc}^1 := L_{\rm loc}^1(J,{\mathbb R}^n)$,
$L_0^1 := L_0^1(J,{\mathbb R}^n)$,
$L_{\rm loc}^{\infty} := L_{\rm loc}^{\infty}(J,{\mathbb R}^n)$.
We will denote by $\langle \cdot ,\cdot \rangle$ the standard inner
product in ${\mathbb R}^n$ and by $|\cdot |$ the corresponding Euclidean
norm. Moreover, $|A|$ stands for the operator norm of an $n \times n$
matrix $A$.
Using a well-known standard notation, given $f: J \to {\mathbb R}^n$ and
$g: J \to {\mathbb R}_{+}$, we will write $f(t) = O(g(t))$ (as $t \to
\infty $) if there exist $M > 0$ and $b > a$ such that $|f(t)| \leq
Mg(t)$ for all $t > b$.
Given two subspaces $A$ and $B$ of a vector space $V$, by abuse of
notation, we write $\dim B/A$ instead of $\dim B/(A \cap B)$. Moreover,
by codimension of $A$ with respect to $B$ we mean the codimension
of $A \cap B$ in $B$.
To avoid cumbersome notation, the restriction of a linear operator
$\Lambda: A \to B$ to a subspace $C$ of $A$ (as domain) and to a
subspace $D$ of $B$ (as codomain) will be often denoted by
$\Lambda: C \to D$.
\section{Asymptotic spaces}
\setcounter{equation}{0}
A subspace of $AC_{\rm loc}$ containing $AC_0$ (respectively, of
$L_{\rm loc}^1$ containing $L_0^1$) will be called {\it asymptotic}. As
the examples below will illustrate, the term ``asymptotic'' is
justified by the fact that, in order to decide whether or not a
function $x \in AC_{\rm loc}$ belongs to such a subspace, it is enough to
consider the behavior of $x$ for $t$ sufficiently large. Moreover, a
subset $E$ of $AC_{\rm loc}$ will be said {\it strongly asymptotic} if it
contains a space of the type $AC_{\varphi} = \big\{x \in AC_{\rm loc}:x =
O(\varphi) \big\}$ for some continuous function $\varphi :J \to
(0,\infty)$. An analogous definition can be given in $L_{\rm loc}^1$ as
well. Since $AC_0\subset AC_{\varphi}$ for any continuous $\varphi:
J \to (0,\infty)$, then any strongly asymptotic space is obviously
asymptotic. Clearly, the converse is not true since, for instance,
$AC_0$ itself is asymptotic but not strongly asymptotic.
These are some examples of asymptotic spaces:
\begin{itemize}
\item $\big\{x \in AC_{\rm loc}: x(\infty) := lim_{t \to \infty}x(t) = 0
\big\};$
\item $AC_{\rm loc} \cap L^1(J,{\mathbb R}^n);$
\item $\big\{x \in AC_{\rm loc}: x$ is bounded $\big\};$
\item $\big\{x \in AC_{\rm loc}: x'(\infty) = 0\big\}.$
\end{itemize}
We observe that the first three spaces are strongly asymptotic as well.
The following space is clearly not asymptotic:
\begin{itemize}
\item $\big\{x \in AC_{\rm loc}: x(0) = x(\infty) = 0 \big\}.$
\end{itemize}
However, this space has codimension one in the (strongly) asymptotic
space $\big\{x \in AC_{\rm loc}: x(\infty) = 0 \big\}.$
Others interesting examples are given by the so-called Corduneanu
spaces (see \cite {Corduneanu}). Namely, given a nonnegative
continuous function $g: J \to {\mathbb R}$, the space
\[
\big\{x \in AC_{\rm loc}: \exists M>0 \hbox{ such that } |x(t)| \leq
Mg(t), t \in J \big\}
\]
is either asymptotic or not, provided $g$ is strictly
positive or vanishes somewhere in $J$.
In this section we will study the solvability of the problem
\begin{eqnarray} \label{e2.1}
&Lx = y &\\
&y \in L_{\rm loc}^1\,, \quad x \in E \,,& \nonumber
\end{eqnarray}
where $J = [a, \infty)$, $L: AC_{\rm loc}\to L_{\rm loc}^1$ is the first
order differential operator defined by
\[
(Lx)(t) = x'(t)+A(t)x(t),
\]
with $A(\cdot)$ a $n \times n$ matrix with entries in
$L_{\rm loc}^1(J,{\mathbb R})$, and $E$ is an asymptotic subspace of
$AC_{\rm loc}$. Given $y \in L_{\rm loc}^1 $, by a solution of (\ref{e2.1})
we mean a locally absolutely continuous function $x: J \to {\mathbb R}^n$
satisfying $Lx(t) = y(t)$ a.e. in $J$ and belonging to $E$.
Clearly, the operator $L$ maps asymptotic subspaces of $AC_{\rm loc}$ in
asymptotic subspaces of $L_{\rm loc}^1$. The following result shows that
a similar assertion holds true for strongly asymptotic spaces.
\begin{theorem}
\label{t2.1}
Let $E$ be a subspace of $AC_{\rm loc}$ and assume that there
exists a continuous $\varphi: J \rightarrow (0,\infty)$ such that $x =
O(\varphi)$ implies $x \in E$. Then there exist a continuous function
$\psi: J \rightarrow (0,\infty)$ and a linear right inverse $K$ of
$L|_E$ defined on the space
\[
L_{\psi}^1 = \big\{y \in L_{\rm loc}^1: y = O(\psi)\big\}
\]
such that $K(y) = O(\varphi)$, for all $y \in L_{\psi}^1$. In
addition, $K$ is continuous in the subset of $L_{\psi}^1$ given by
\[
D_{M \psi} = \big\{y \in L_{\rm loc}^1:|y(t)|\leq M \psi (t),
\hbox{ for a.a. } t \in J \big\}, \quad M>0.
\]
\end{theorem}
\paragraph{Proof.}
Let $X(t)$ be a fundamental matrix of $Lx = 0$. For any
$y \in L_{\rm loc}^1$ such that $|X^{-1}(\cdot)||y| \in L^1(J,{\mathbb R})$,
the function
\[
K(y)(t) = -X(t)\int_t^{\infty}X^{-1}(s)y(s)ds, \quad t \geq a
\]
belongs to $AC_{\rm loc}$ and solves the differential equation $Lx = y$.
Thus, it is enough to find $\psi >0$ such that, when $y \in
L_{\psi}^1$ one gets $|X^{-1}(\cdot)||y| \in L^1(J,{\mathbb R})$ and $K(y)
= O(\varphi)$. Indeed, the continuity of $K$ in $D_{M \psi}$ is a
consequence of the Lebesgue Convergence Theorem.
Let us show that, given a continuous $\varphi > 0$, there exists a
continuous $\psi > 0$ such that $|X^{-1}(\cdot)| \psi \in
L^1(J,{\mathbb R})$ and
\[
|X(t)|\int_t^{\infty}|X^{-1}(s)|\psi (s)ds \leq \varphi (t), \quad t \geq a.
\]
In fact, take any $C^1$ function $\sigma$ satisfying $0 < \sigma(t)
\leq \varphi (t)/|X(t)|$ for $t \geq a$, $\sigma(\infty) = 0$,
$\sigma'(t) < 0$ for $t \geq a$, $\sigma' \in L^1(J,{\mathbb R})$, and
define $\psi(t) = -\sigma'(t)/|X^{-1}(t)|$.
Consequently,
\[
|K(y)(t)|\leq |X(t)|\int_t^{\infty}|X^{-1}(s)||y(s)|ds \leq \varphi
(t)
\]
as claimed.
\hfill$\diamondsuit$\medskip
We point out that the proof of the above theorem gives an explicit
formula to get a function $\psi$ which ensures the solvability of
(\ref{e2.1}) whenever $y = O(\psi)$. This formula is deduced from the
particular form of the operator $K$. This, in some sense, is an
universal inverse (i.e.\ its expression is independent of the space
$E$) and is a good inverse in the case when the homogeneous problem
$Lx = 0$, $x \in E$, admits only the trivial solution. However, when
this condition is not satisfied, an accurate choice of an inverse
allows us to deduce a more convenient formula for the function $\psi$
(clearly, the bigger is $\psi$, the better).
To illustrate this, consider the following example.
\begin{example}
{\rm
Consider in $J = [0,\infty)$ the system
\begin{eqnarray*}
&x' + x = y&\\
&x \in E\,,&
\end{eqnarray*}
where $E = \big\{x \in AC_{\rm loc}(J,{\mathbb R}):x = O(1) \big\}$ is the
space of bounded (locally absolutely continuous) real functions on $J$.
The problem is solvable for any $y \in L_{\rm loc}^1(J,{\mathbb R})$ such that
$y = O(1)$. In fact, the inverse
\[
K(y)(t) = \int_0^{t}e^{s-t}y(s)ds
\]
of $Lx = x' + x$ is such that when $y = O(1)$ one has $|K(y)(t)|\leq M
\int_0^{t}e^{s-t}ds \leq M$, for some $M>0$. Thus, $K(y) \in E$.
Observe that if we had used the inverse of the proof of Theorem
\ref{t2.1} we would have obtained the following sufficient condition
for the solvability of the system: $e^{t}y(t) \in L^1(J,{\mathbb R})$,
which is more restrictive than $y = O(1)$.}
\end{example}
\section{Algebraic results}
\setcounter{equation}{0}
Let $S$ be a subspace of $AC_{\rm loc}$ and (as in the Introduction)
denote by $S_{\infty}$ the subspace of $AC_{\rm loc}$ given by $S_{\infty}
= S + AC_0$. According to the definition of Section 2, $S_{\infty}$ is
an asymptotic space which clearly coincides with $S$ if and only if
$AC_0 \subset S$.
In this section we are concerned with the solvability of the following
linear problem:
\begin{eqnarray} \label{e3.1}
&Lx = y& \\
&y \in L(S_{\infty}) ,\quad x \in S\,. \nonumber
\end{eqnarray}
Assume that
\noindent $\bf H_1)$ $\dim S_\infty /S$ is finite.
This assumption plays the same role as that of assigning, in the
compact intervals, a finite number of independent linear conditions.
For instance $H_1)$ is not satisfied for
\[
S = \big\{x \in AC_{\rm loc}(J,{\mathbb R}):x(n) = 0 \hbox{ for all } n \in
{\mathbb N}\big\}.
\]
In this case $S_{\infty} = AC_{\rm loc}(J,{\mathbb R})$, so that $\dim
S_{\infty}/S = \infty$.
As the examples below show, $H_1)$ turns out to be satisfied in many
problems where, however, the codimension of $S$ in $AC_{\rm loc}$ is not
finite.
\begin{itemize}
\item[a)] Let $S = \big\{x \in AC_{\rm loc}(J,{\mathbb R}): x(\infty) =
0\big\}$. Then, $S_{\infty} = S$,\\
$\dim AC_{\rm loc}/S = \infty $, $\dim S_{\infty}/S = 0$;
\item[b)] Let $S = \big\{x \in AC_{\rm loc}(J,{\mathbb R}): x(1) = x(\infty) =
0 \big\}$. Then,\\
$S_{\infty} = \big\{x \in AC_{\rm loc}(J,{\mathbb R}): x(\infty) = 0 \big\}$,
$\dim AC_{\rm loc}/S = \infty $, $\dim S_{\infty}/S = 1$;
\item[c)] Let $S = \big\{x \in AC_{\rm loc}(J,{\mathbb R})\cap
L^1(J,{\mathbb R}):
\int_0^{\infty}x(t)dt = 0 \big\}$. Then,\\
$S_{\infty} = AC_{\rm loc}(J,{\mathbb R})\cap L^1(J,{\mathbb R})$,
$\dim AC_{\rm loc}/S = \infty $, $\dim S_{\infty}/S = 1$.
\end{itemize}
The following are the corresponding problems in a compact interval
$I = [a,b]$.
\begin{itemize}
\item[a')] Let $S = \big\{x \in AC(I,{\mathbb R}):x(b) = 0\big\}$. Then,
$\dim AC/S = 1$;
\item[b')] Let $S = \big\{x \in AC(I,{\mathbb R}):x(a) = x(b) = 0\big\}$.
Then, $\dim AC/S = 2$;
\item[c')] Let $S = \big\{x \in AC(I,{\mathbb R}):\int_{a}^{b}x(t)dt =
0\big\}$. Then, $\dim AC/S = 1$.
\end{itemize}
Let us point out that, as the example below illustrates, not only the
codimension of $S$ in $AC_{\rm loc}$ but also that of $L(S)$ in
$L^1_{\rm loc}$ may not be finite. This means that, even assuming $H_1)$,
the theory of Fredholm operators does not apply directly to some of
the problems we are interested in. Recall that a linear operator
between vector spaces is said to be a Fredholm operator if it has
finite dimensional kernel and finite codimensional image. The index of
a Fredholm operator is the difference between the dimension of its
kernel and the codimension of its image.
\begin{example}
\label{ex3.1}
{\rm
Consider in $J = [1,\infty)$ the system
\begin{eqnarray*}
&Lx = y&\\
&y \in L_{\rm loc}^1(J,{\mathbb R})\,\quad x \in S\,,&
\end{eqnarray*}
where
\[
S = \big\{x \in AC_{\rm loc}(J,{\mathbb R}):x\hbox{ is bounded} \big\}.
\]
Clearly, $S = S_{\infty}$ and the functions $y_n(t) = t^{-1/n}, t\geq
1, n\in{\mathbb N}$ are in $L^1_{\rm loc}(J,{\mathbb R})$ but, as one can check,
do not belong to $L(S)$. Thus, the codimension of $L(S)$ in
$L^1_{\rm loc}(J,{\mathbb R})$ is not finite.}
\end{example}
To avoid the difficulty presented before, it turns out to be useful to
restrict our operator $L$ between the spaces $S$ and $L(S_{\infty})$.
Actually, the surjective operator $L|_{S_{\infty}}:S_{\infty}\to
L(S_{\infty})$ is Fredholm of index $m = \dim {\rm Ker}L\cap
S_{\infty}$. Thus, taking into account $H_1)$, one can immediately
show that $L|_S: S \to L(S_{\infty})$ is Fredholm of index $m-r$,
where we denote $r = \dim S_{\infty}/S$. In fact, since the inclusion
$i: S \to S_{\infty}$ is Fredholm of index $-r$, the composition $L|_S
= L|_{S_{\infty}}\circ i$ has index ${\rm ind}L|_S = {\rm ind}
L|_{S_{\infty}} + {\rm ind\,} i = m-r$. In particular, the codimension
of $L(S)$ in $L(S_{\infty})$ is finite.
Summarizing, we have proved the following.
\begin{proposition}
\label{p3.2}
Let $H_1)$ be satisfied. Denote $m = \dim {\rm Ker}L \cap S_{\infty}$,
$p = \dim {\rm Ker} L \cap S$, $q = \dim L(S_{\infty})/L(S)$ and $r =
\dim S_{\infty}/S$.
Then, $L: S \to L(S_{\infty})$ is a Fredholm operator whose index
{\rm ind}$L := p-q$ satisfies the equality
\[
p-q = m-r.
\]
Consequently, $q = p-m+r$.
\end{proposition}
The following well-known elementary lemma is stated here for
completeness, since will be used several times in the sequel.
\begin{lemma}
\label{l3.3}({\rm Quotient's Lemma}). Let $A$, $B$ and $C$ be vector
spaces and $\pi: A \to B$, $\gamma: A \to C$ be linear operators.
Assume that $\pi$ is surjective and ${\rm Ker\,}\pi \subset {\rm
Ker\,}\gamma$. Then there exists a unique linear operator $\sigma:
B \to C$ such that $\gamma = \sigma \circ \pi$. Moreover, $\sigma$ is
injective if and only if ${\rm Ker\,}\pi = {\rm Ker\,} \gamma$, and
$\sigma$ is surjective if and only if so is $\gamma$.
\end{lemma}
Proposition \ref{p3.2} above shows that the computation of $q$ can be
reduced to the easier task of calculating the integers $p,m,r$.
We will prove below that, in order to determine such integers, we can
reduce our attention to consider $L$ as an operator acting between the
spaces $AC_0 \oplus {\rm Ker}L$ and $L_0^1$, which are independent of
$S$ and easier to handle than $S_{\infty}$ and $L(S_{\infty})$
respectively, since $AC_0$ and $L_0^1$ are spaces of functions having
compact support. For instance, as ii) of Lemma \ref{l3.4} will show,
the integer $r$ can be computed by determining the codimension of $S$
in $AC_0$, and so on. We point out that the spaces $AC_0 \oplus {\rm
Ker}L$ and $L_0^1$ are related each other since, as easily seen,
$AC_0 \oplus {\rm Ker}L = L^{-1}(L_0^1)$.
In Lemma \ref{l3.4} and Theorem \ref{t3.5} below an algebraic analysis
of problem (\ref{e1.2}) is carried out. As a consequence of this
analysis, we prove that $L:(AC_0 \oplus {\rm Ker}L) \cap S \to L_0^1$
is still Fredholm and has the same index as $L: S \to L(S_{\infty}).$
Moreover, since these two operators have the same kernel, the
codimensions of $L(S)$ in $L(S_{\infty})$ and $L(S) \cap L_0^1(J,{\bf
R}^n)$ in $L_0^1(J,{\mathbb R}^n)$ turn out to be the same.
\begin{lemma}
\label{l3.4}
Assume $H_1)$ is satisfied. Then, the following equalities hold:
\begin{itemize}
\item[\rm i)] $(AC_0 \oplus {\rm Ker}L) \cap S_{\infty} = ({\rm
Ker}L \cap S_{\infty}) \oplus AC_0$;
\item[\rm ii)] $\dim (AC_0/S) = \dim ((AC_0\oplus {\rm Ker}L)
\cap S_{\infty})/S) = \dim S_{\infty}/S.$
\end{itemize}
\end{lemma}
\paragraph{Proof.}
i) Let us show first that $(AC_0 \oplus {\rm Ker}L) \cap
S_{\infty} \subset ({\rm Ker}L \cap S_{\infty}) \oplus AC_0$. Take $x
\in (AC_0 \oplus {\rm Ker}L) \cap S_{\infty}$. Then, there exist $x_0
\in AC_0$ and $x_1 \in {\rm Ker}L$ such that $x = x_0 + x_1$. Since
$S_{\infty} \supset AC_0$, then $x_1 = x-x_0$ belongs to $S_{\infty}$.
Thus, $x \in ({\rm Ker}L \cap S_{\infty}) \oplus AC_0$. The converse
inclusion is obvious.
ii) Consider the commutative diagram
\[
\begin{array}{ccc}
AC_0 & \stackrel{i}{\longrightarrow} & S_{\infty} \\
\downarrow & & \downarrow \\
AC_0/S & \stackrel{\sigma}{\longrightarrow} & S_{\infty}/S
\end{array}
\]
where the vertical arrows are the canonical projections, $i$ is the
inclusion, and $\sigma$ is well-defined and injective because of
Quotient's Lemma. Since $S_{\infty} = S+AC_0$, one gets immediately
that $\sigma$ is onto; consequently, $AC_0/S$ and $S_{\infty}/S$ have
the same dimension. Consider now the diagram
\[
\begin{array}{ccccc}
AC_0 & \stackrel{i_1}{\longrightarrow} & {(AC_0 \oplus {\rm Ker}L)
\cap S_{\infty}} & \stackrel{i_{2}}{\longrightarrow} & S_{\infty} \\
\downarrow & & \downarrow & & \downarrow \\
AC_0/S & \stackrel{\sigma_1}{\longrightarrow} & {((AC_0 \oplus {\rm
Ker}L)\cap S_{\infty})/S} & \stackrel{\sigma_{2}}{\longrightarrow} &
S_{\infty}/S
\end{array}
\]
where $i_1$ and $i_{2}$ are inclusions, the vertical arrows are
canonical projections and, as above, $\sigma_1$ and $\sigma_{2}$ are
well-defined and injective. The assertion follows immediately from the
fact that $\dim AC_0/S = \dim S_{\infty}/S$.
\hfill$\diamondsuit$\medskip
\begin{theorem}
\label{t3.5} Assume $H_1)$ is satisfied. Then, the operator
$L:(AC_0 \oplus {\rm Ker}L) \cap S \to L_0^1$ is Fredholm and has the
same index as $L: S \to L(S_{\infty})$. In addition, $\dim L_0^1/L(S)
= \dim L(S_{\infty})/L(S)$.
\end{theorem}
\paragraph{Proof.}
The operator $L:(AC_0\oplus {\rm Ker}L)\cap S_{\infty}\to
L_0^1$ is clearly onto and, by $i)$ of Lemma \ref{l3.4}, its kernel
coincides with ${\rm Ker}L \cap S_{\infty}$. Thus, it is Fredholm of
index $m = \dim {\rm Ker}L \cap S_{\infty}$. Now, as previously, the
index of the restriction $L:(AC_0 \oplus {\rm Ker}L)\cap S \to L_0^1$
can be computed by means of the composition
\[
(AC_0\oplus {\rm Ker}L) \cap S \stackrel{i}{\to }(AC_0\oplus {\rm Ker}
L)\cap S_{\infty}\stackrel{L}{\to }L_0^1
\]
where, by $H_1)$ and $ii)$ of Lemma \ref{l3.4}, the inclusion $i$ has
index $-r$ (recall that $r = \dim S_{\infty}/S$). Consequently, the
operator $L$ has index $m-r$ either considered between $S$ and
$L(S_{\infty})$ (as previously shown) or between $(AC_0 \oplus {\rm
Ker}L)\cap S$ and $L_0^1$ (recall that $L_0^1 \subset L(S_{\infty})$).
This proves the first part of the assertion. Now, by Proposition
\ref{p3.2} we have $m-r = p-q$, where $p = \dim {\rm Ker}L \cap S$ and
$q = \dim L(S_{\infty})/L(S)$. Hence, we also obtain {\rm
ind}$L|_{(AC_0 \oplus {\rm Ker}L) \cap S} = p-q$. On the other hand, by
the definition of Fredholm index,
\[
{\rm ind}L|_{(AC_0 \oplus {\rm Ker}L) \cap S} = \dim {\rm Ker}
L|_{(AC_0 \oplus {\rm Ker}L) \cap S}-\dim L_0^1/{L((AC_0 \oplus
{\rm Ker}L) \cap S).}
\]
Observe now that ${\rm Ker}L|_{(AC_0 \oplus {\rm Ker}L) \cap S}$ still
coincides with ${\rm Ker}L \cap S$ and, thus, has dimension $p$.
Moreover, one has $L((AC_0 \oplus {\rm Ker}L)\cap S) = L(S)\cap L_0^1$.
Therefore, $\dim L_0^1/L(S) = q = \dim L(S_{\infty})/{L(S)}$, which
achieves the proof.
\hfill$\diamondsuit$\medskip
\section{The main result}
\setcounter{equation}{0}
In Theorem \ref{t3.5} we have shown that $q := \dim
(L(S_{\infty})/{L(S))}$ equals $\dim (L_0^1/{L(S))}$ and, by
Proposition \ref{p3.2} we have obtained that $q$ can be computed by
the formula $q = p - m + r$, with $ p = \dim ({\rm Ker}L \cap S)$, $m
= \dim ({\rm Ker}L \cap S_{\infty})$, $r = \dim(S_{\infty}/S)$. Thus,
if $q = 0$ the solvability of (\ref{e1.1}) is equivalent to the
solvability of the associated problem
\begin{eqnarray} \label{e4.1}
&Lx = y& \\
&y \in L_{\rm loc}^1\,,\quad x \in S_{\infty}\,.&\nonumber
\end{eqnarray}
That is, given $y \in L_{\rm loc}^1$, (\ref{e1.1}) is solvable if and only
if the same is true for (\ref{e4.1}). Assume now we know how to solve
this associated problem, for example by means of Theorem \ref{t2.1},
and consider the case when $q>0$. We want to find conditions on $y \in
L(S_{\infty})$ ensuring the solvability of the original problem
(\ref{e1.1}). We will show that, when $S$ satisfies some reasonable
conditions, this can be done by imposing $y$ to belong to the kernel
of $q$ linearly independent functionals expressed in a convenient
integral form.
Theorem \ref{t4.1} below shows that such ``integral'' functionals
exist in the special case when $y$ has compact support. Moreover, with
an additional topological assumption on $S$, the same functionals can
be adapted to ensure the solvability of problem (\ref{e1.1}) (see
Theorem \ref{t4.3}).
To this end, given any $b>a$, consider the following closed subspace
of $AC_0$:
\[
AC_{[a,b]} = \big\{x \in AC_0:x(t) = 0\hbox{ for } t \geq b \big\}.
\]
We will say that $S$ is {\it locally closed} in $S_{\infty}$ if $S
\cap AC_{[a,b]}$ is closed in $S_{\infty}$ for any $b>a$.
\begin{theorem}
\label{t4.1}
Let $S$ be a subspace of $AC_{\rm loc}$ such that
\begin{itemize}
\item[\rm H$_1$)] $\dim {S_{\infty}/S}$ is finite;
\item[\rm H$_2$)] $S$ is locally closed in $S_{\infty}$.
\end{itemize}
Suppose the codimension $q$ of $L(S)$ in $L(S_{\infty)}$ is nonzero.
Then there exist $\alpha_1,\dots,\alpha_q \in L_{\rm loc}^{\infty}$ such
that, given $y \in L_0^1$, $y \in L(S)$ if and only if
\[
\int_{a}^{\infty}\langle \alpha_{i}(t), y(t)\rangle dt = 0, \quad
\forall\, i = 1,\dots,q.
\]
\end{theorem}
\paragraph{Proof.}
By Theorem \ref{t3.5}, $q = \dim L_0^1/{L(S)}$. Hence,
there exist $q$ linearly independent functionals on $L_0^1$, say
$\lambda_1,\dots,\lambda_q$, such that $y \in L(S)\cap L_0^1$ if and
only if $\lambda_i(y) = 0$ for all $i = 1,..,q$. Let us show that any
$\lambda_i$ is locally continuous, i.e.\ $\forall b>a$ any $\lambda_i$
is continuous in the Banach subspace
\[
L_{[a,b]}^1 = \big\{y \in L_0^1:y(t) = 0\hbox{ for a.a. }t \geq b \big\}
\]
of $L_0^1$. Since $L(S)\cap L_{[a,b]}^1$ has finite codimension in
$L_{[a,b]}^1$, it is enough to prove that it is closed in $L_{[a,b]}^1$.
Observe first that the subspace $AC_{[a,b]} \oplus {\rm Ker}L$ of
$AC_{\rm loc}$ is a Banach space, being the sum of a Banach space and a
finite dimensional space, and that the restriction
$L: AC_{[a,b]} \oplus {\rm Ker}L \to L_{[a,b]}^1$ is continuous, onto
and has finite dimensional kernel. Moreover, since $AC_{[a,b]}$ is
contained in $S_{\infty}$, by assumption $H_{2})$ it follows that $S
\cap AC_{[a,b]}$ is closed in $AC_{[a,b]}$. Hence, it is not difficult
to prove that, ${\rm Ker}L$ being finite dimensional, the subspace
$S \cap (AC_{[a,b]}\oplus {\rm Ker}L)$ is closed in $AC_{[a,b]}\oplus
{\rm Ker}L$. Thus, recalling that a continuous Fredholm operator
between Banach spaces sends closed subspaces into closed
subspaces, it follows immediately that $L(S \cap (AC_{[a,b]} \oplus
{\rm Ker}L)) = L(S)\cap L_{[a,b]}^1$ is closed in $L_{[a,b]}^1$, as
claimed.
Let $b_j \to \infty $, $b_j > a$, be an increasing sequence of numbers
and set $I_j = [a, b_j]$. Since $L_{[a, b_j]}^{\infty}$ is the dual of
$L_{[a, b_j]}^1$, then, for any $j \in {\mathbb N}$ and $i = 1,\dots,q$,
there exists $\alpha_j^i \in L_{[a, b_j]}^{\infty}$ such that
$\lambda_i(y) = \int_{I_j}\langle \alpha_j^i(t),y(t)\rangle dt$ for
any $y \in L_{[a, b_j]}^1$. Clearly, if $j1\,.
\end{array}
\right.
\]
}
\end{example}
Notice that, given two subspaces $F_1$ and $F_{2}$ of a vector space
$F$, if $F_1$ and $F_{2}$ have the same finite codimension and one is
contained into the other, then they necessarily coincide. Referring to
Theorem \ref{t4.1}, let $F_1 = L(S)\cap L_0^1$ and $F_2 = \big\{y \in
L_0^1: \int_{a}^{\infty} \langle \alpha_i(t), y(t) \rangle dt = 0
\hbox{ for all } i = 1,\dots,q \big\}$. Assume that the functions
$\alpha_i$ are linearly independent. Then, $F_2$ has codimension $q$
in $L_0^1$. Consequently, if in Theorem \ref{t4.1} one of the two
conditions ``if'' or ``only if'' is satisfied, then the other one is
satisfied as well.
Now, we are ready to state our main result which depends on the
preliminary investigation in the compact support context contained in
Theorem \ref{t4.1} above.
Given a positive continuous function $\varphi :J \to {\mathbb R}$, let us
denote by $A_{\varphi}$ the closed subset of $AC_{\rm loc}$ given by
\[
A_{\varphi} = \big\{x \in AC_{\rm loc}:|x(t)|\leq \varphi (t),\;t \in J
\big\}.
\]
We have the following
\begin{theorem}
\label{t4.3}
Let $S$ be a subspace of $AC_{\rm loc}$ such that
\begin{itemize}
\item[\rm H$_1$) ] $\dim {S_{\infty}/S}$ is finite;
\item[\rm H$_{2}'$) ] there exists a continuous real function
$\varphi: J\to (0,\infty)$ such that $A_{\varphi}$ is contained in
$S_{\infty}$ and $S\cap A_{\varphi}$ is a closed subset of $AC_{\rm loc}$.
\end{itemize}
Suppose the codimension $q$ of $L(S)$ in $L(S_{\infty})$ is positive
and let $\alpha_1,\dots,\alpha_q \in L_{\rm loc}^{\infty}$ be as in Theorem
\ref{t4.1}. Let $\psi: J \to (0,\infty)$ be continuous and satisfying
the following properties:
\begin{itemize}
\item[a)] $|\alpha_i|\psi \in L^1(J,{\mathbb R})$, for all $i = 1,\dots,q$;
\item[b)] there exists a linear right inverse $K$ of $L|_{S_\infty}$,
defined on the space
\[
L_{\psi}^1 = \big\{y \in L_{\rm loc}^1:y = O(\psi) \big\}
\]
and continuous in the subset
\[
D_{\psi} = \big\{y \in L_{\rm loc}^1:|y(t)|\leq \psi (t)\hbox{, for a.a. }
t \in J \big\}
\]
of $L_{\psi }^1$, such that $K(y) = O(\varphi)$, for all $y \in
L_{\psi }^1$.
\end{itemize}
Then, given $y = O(\psi)$, the problem
\begin{eqnarray*}
&Lx = y& \\
&y \in L_{\rm loc}^1\,,\quad x \in S&
\end{eqnarray*}
has a solution if and only if
\[
\int_{a}^{\infty}\langle \alpha_i(t),y(t)\rangle dt = 0\hbox{,\quad
for
any }i = 1,\dots ,q.
\]
\end{theorem}
Before giving the proof of Theorem \ref{t4.3}, let us make some
comments about its statement.
\begin{remark}
\rm
Assumption $H_{2}')$ above is stronger than $H_{2})$ of Theorem
\ref{t4.1}. To see this, observe first that the condition that $S \cap
A_{\varphi}$ is closed in $AC_{\rm loc}$ for some $\varphi >0$ clearly
implies that $S \cap A_{M\varphi}$ is closed in $AC_{\rm loc}$ for all
$M>0$. To get $H_{2})$ we need to show that if $\{x_n\}$ is a sequence
in $S \cap AC_{[a,b]}$ converging to $x \in S_\infty$, then $x \in S
\cap AC_{[a,b]}$. Obviously, $x \in AC_{[a,b]}$. Let us show that $x$
belongs to $S$ as well. First, notice that there exists $c>0$ such
that $|x_n(t)| \leq c$ for all $n \in {\mathbb N}$ and $t \in [a,b]$. Let
$M>0$ be such that $c \leq M \varphi (t)$ for all $t \in [a,b]$. This
means that $\{x_n\}$ is a sequence in $S \cap A_{M\varphi}$ which, as
observed above, is closed in $AC_{\rm loc}$. Thus, $x$ belongs to $S$, so
that $S$ is locally closed in $S_\infty$, i.e.\ $H_{2})$ is satisfied.
Finally, let us point out that condition $H_{2}')$ is clearly not
equivalent to $H_{2})$ as one can immediately see by taking
$S = AC_0$.
\end{remark}
\begin{remark}
\rm
The existence of a function $\psi$ satisfying $b)$ of Theorem
\ref{t4.3} is ensured by our Theorem \ref{t2.1} above. Moreover, by
reducing $\psi$, if necessary, we can also assume that $a)$ is
satisfied.
\end{remark}
\noindent
{\it Proof of Theorem \ref{t4.3}.} Define $Q_0:L_0^1\rightarrow {\mathbb
R}^q$ by
\[
Q_0(y) = \Big(\int_{a}^{\infty}\langle \alpha_1(t),y(t)\rangle
dt,\dots,\int_{a}^{\infty}\langle \alpha_q(t),y(t)\rangle dt\Big).
\]
By Theorem \ref{t4.1}, given $y \in L_0^1$, one has $Q_0(y) = 0$ if
and only if $y \in L(S)$. Consequently $Q_0$ is onto, its image being
isomorphic to the $q$\hbox{-}dimensional quotient $L_0^1/L(S)$. Let
\[
L_{\psi}^1 = \big\{y \in L_{\rm loc}^1:y = O(\psi) \big\}
\]
and $Q_{\psi}:L_{\psi}^1\to {\mathbb R}^q$ be the linear extension of
$Q_0$ given by
\[
Q_{\psi}(y) = \Big(\int_{a}^{\infty}\langle \alpha_1(t),y(t)\rangle
dt,\dots,\int_{a}^{\infty}\langle \alpha_q(t),y(t)\rangle dt \Big).
\]
We need to show that $Q_{\psi}(y) = 0$ if and only if $y \in L(S)\cap
L_{\psi }^1$. This will be obtained by proving the existence of a
convenient linear operator onto ${\mathbb R}^q$ coinciding with our
integral operator $Q_{\psi}$ in $L_{\psi}^1$ and whose kernel is
$L(S)$. In order to construct such an operator, as previously let $r$
denote the codimension of $S$ in $S_{\infty}$. Hence, there exists a
surjective linear operator $R: S_{\infty}\to {\mathbb R}^r$ such that $x
\in S$ if and only if $R(x) = 0$.
Observe that the restriction $R|_{AC_0}$ of $R$ to $AC_0$ is still
surjective. Indeed, since $S \cap AC_0 = {\rm Ker}R|_{AC_0}$, the
image of $R|_{AC_0}$ is isomorphic to the quotient $AC_0/S$, which, by
Lemma \ref {l3.4}, has dimension $r$.
Let $K_0:L_0^1\to AC_0$ be the restriction of $K$ to the pair of
spaces $L_0^1$ and $AC_0$. Since $S_{\infty} \supset AC_0$, the
composition $T: L_0^1\to {\mathbb R}^r$ given by $T = R \circ K_0$ is
well-defined. Observe that $T$ is onto, since $K_0$ is an isomorphism
and $R|_{AC_0}$ is onto. Moreover, ${\rm Ker}T$ is contained in ${\rm
Ker} Q_0 = L(S)\cap L_0^1$ (indeed if $T(y) = 0$, then $K_0(y) \in S
\cap AC_0$ so that $y = (L \circ K_0)(y) \in L(S)\cap L_0^1$).
Consequently, because of the Quotient Lemma, there exists a unique
$\rho :{\mathbb R}^r \to {\mathbb R}^q$ such that $Q_0 = \rho \circ T$. Since
$K(L_{\psi}^1)$ is contained in $S_{\infty}$, the composition $Q :=
\rho \circ R \circ K$ is an extension of $Q_0$ to $L_{\psi}^1$.
Let us show that, similarly to the case of $Q_0$, one has ${\rm Ker}Q =
L(S)\cap L_{\psi}^1$. Observe first that, since $Q$ is onto, then
${\rm Ker} Q$ has codimension $q$ in $L_{\psi}^1$. The same is true for
$L(S)$ since it is easy to see that $L_{\psi}^1/{L(S)}$ is isomorphic
to $L_0^1/{L(S)}$ (or, equivalently, to $L(S_{\infty})/{L(S)}$).
Therefore, it is enough to prove that, for instance, ${\rm Ker} Q$ is
contained in $L(S)\cap L_{\psi}^1$. To this end, take $y \in {\rm Ker}Q$,
so that $(R \circ K)(y) \in {\rm Ker} \rho $. Clearly, $(R \circ
K)(L(S)\cap L_{\psi}^1) \supset (R \circ K_0)(L(S) \cap L_0^1)$ and $(R
\circ K_0)(L(S)\cap L_0^1) = {\rm Ker} \rho $. Hence, $(R \circ K)(y)
\in (R \circ K)(L(S)\cap L_{\psi}^1)$. Thus, there exists $\tilde{y} \in
L(S)\cap L_{\psi}^1$ such that $K(y - \tilde{y}) \in {\rm Ker} R = S$.
Consequently, $y-\tilde{y} = (L \circ K)(y - \tilde{y}) \in L(S) \cap
L_{\psi}^1$, which implies that $y$ belongs to $L(S)\cap L_{\psi}^1$
as claimed.
Let us now go back to the map $Q_\psi :L^1_\psi \to {\mathbb R}^q$
introduced previously. By using the Lebesgue Convergence Theorem, it
is immediately seen that $Q_\psi$ is continuous in any subset of
$L^1_\psi$ of the form
\[
D_{M\psi} = \big\{y \in L_{\rm loc}^1:|y(t)| \leq M\psi (t),
\hbox{ for a.a. } t \in J \big\}, \quad M>0.
\]
Let us show that $Q$ is continuous in $D_{M\psi}$ as well. To this
end, consider the (closed) subset
\[
A_{M\varphi} = \big\{x \in AC_{\rm loc}: |x(t)| \leq M\varphi (t), t \in
J \big\},\quad M>0 \] of $AC_{\rm loc}$ and let $AC_{\varphi}$ be the vector
space
\[
AC_{\varphi} = \big\{x \in AC_{\rm loc}:x = O(\varphi) \big\} =
\cup_{M>0}A_{M\varphi}.
\]
Suppose $AC_{\varphi}$ equipped with the finest topology which makes
any inclusion $A_{M\varphi}\to AC_{\varphi}$ continuous. That is, $U
\subset AC_{\varphi}$ is open in $AC_{\varphi}$ if and only if $U \cap
A_{M\varphi} $ is open in $A_{M\varphi}$ for all $M>0$. It is not hard
to show (apply, e.g., Theorems 3.3F and 3.3G of \cite {T}) that, with
this topology $AC_{\varphi}$ becomes a Hausdorff topological vector
space.
By assumption $H{_{2}')}$, $S \cap A_{\varphi}$ is closed, which
clearly means that $S$ is closed in $AC_{\varphi}$. Consequently, the
quotient space $AC_{\varphi}/S$ is a finite dimensional Hausdorff
topological vector space. Hence, there exists an injective map
$\sigma: AC_{\varphi}/S \to {\mathbb R}^r$ such that $R = \sigma \circ \pi
$, where $\pi :AC_{\varphi}\to AC_{\varphi}/S$ is the canonical
projection. Such a map $\sigma$ is clearly continuous, since
$AC_{\varphi}/S$ is Hausdorff and finite dimensional. Therefore, $R:
AC_{\varphi}\to {\mathbb R}^r$ is continuous and, so, any restriction of
$R$ to $A_{M\varphi}$ is still continuous. Observe now that in
$A_{M\varphi}$ the topology induced by the one of $AC_{\varphi}$
clearly coincides with the usual topology of $A_{M\varphi}$ (as a
subspace of $AC_{\rm loc}$). Since $K:D_{M\psi} \to A_{M\varphi}$ is
continuous (see Theorem \ref{t2.1}), $Q$ is continuous in $D_{M\psi}$
as claimed. Consequently, $Q_{\psi}$ and $Q$, which are both defined
in $L_{\psi}^1$, continuous in $D_{M\psi}$, $\forall M>0$, and
coinciding in $L_0^1\cap D_{M\psi}$ (which is dense in $D_{M\psi})$,
must coincide in $D_{M\psi}$ and, thus, in $L_{\psi}^1 = \cup_{M > 0}
D_{M\psi}$. That is, for any $y \in L_{\psi}^1$, one has
\[
Q(y) = \Big(\int_{a}^{\infty}\langle \alpha_1(t),y(t)\rangle
dt,\dots,\int_{a}^{\infty}\langle \alpha_{q}(t),y(t)\rangle dt \Big).
\]
This implies $\int_{a}^{\infty}\langle \alpha_i(t),y(t)\rangle dt = 0$
for any $i = 1,\dots,q$ if and only if $y \in L(S)\cap L_{\psi}^1$,
which is the assertion.
\hfill$\diamondsuit$\medskip
The following example illustrates our main result.
\begin{example}
\rm
Consider in $J = [0,\infty)$ the problem
\begin{eqnarray*}
&x'-x = y& \\
&x \in L^1(J,{\mathbb R})&\\
&\int_0^\infty x(t)dt = 0&
\end{eqnarray*}
Here, $S_{\infty} = L^1(J,{\mathbb R})$, $r = 1$, $m = 0$, $p = 0$, so
that $q = p -m + r = 1$. In order to find the function $\alpha$ of
Theorem \ref{t4.1}, consider in $L_0^1$ the inverse
\[
x(t) = K_0(y)(t) = -e^{t}\int_{t}^{\infty}e^{-s}y(s)ds.
\]
By imposing the condition
\[
\int_0^\infty K_0(y)(t)dt = 0
\]
one obtains
\begin{eqnarray*}
\int_0^\infty \Big(\int_{t}^{\infty}e^{t}e^{-s}y(s)ds \Big)dt
&=& \int_0^\infty \Big(\int_0^s e^{t}e^{-s}y(s)dt \Big)ds \\
&=& \int_0^\infty (1-e^{-s})y(s)ds = 0.
\end{eqnarray*}
Thus, $\alpha (t) = 1-e^{-t}$.
Let $\varphi: J \to (0,\infty)$ be any continuous $L^1$ function.
Observe that, as a consequence of the Lebesgue Convergence Theorem,
the assumption $H'_2$) is satisfied for such a $\varphi$. Let us find
a function $\psi$ as in Theorem \ref{t4.3}. First of all we must have
$|\alpha| \psi \in L^1(J,{\mathbb R})$, which implies $\psi \in L^1(J,{\bf
R})$. In addition, take $\varphi$ be $C^1$, positive and such that the
function $\sigma(t) = \varphi (t)e^{-t}$ satisfies the following
conditions:
\begin{itemize}
\item $\sigma(\infty) = 0$;
\item $\sigma'(t)<0$ for $t \in J$;
\item $\sigma' \in L^1(J,{\mathbb R})$.
\end{itemize}
For example take $\varphi (t) = 1/(1+t^2)$.
As in the proof of Theorem \ref{t2.1}, the function
$\psi (t) = -\sigma'(t)e^t$ is such that $K(D_\psi)$ is contained
in $A_\varphi$, where $K:L^1_\psi \to S_\infty$ is the natural
extension of $K_0$ defined above, and, by the Lebesgue Convergence
Theorem, is clearly continuous in $D_\psi$.
Therefore, all the assumptions of Theorem \ref{t4.3} are satisfied, so
that we may conclude that, if $y = O(\psi)$ and
$\int_0^\infty (1-e^{-t})y(t)dt = 0$, then our problem has a solution.
\end{example}
\begin{remark}
\rm
In many situations, the assumption
\begin{itemize}
\item [\rm H$_{3}$)] $S_\infty$ {\it strongly asymptotic and } $S$
{\it closed in } $S_\infty$,
\end{itemize}
which is clearly stronger than $H'_2$), is satisfied. However, as the
following simple example shows, assumption $H_3$) might not be
satisfied in some cases which turn out to be of some interest in the
applications. Take, for instance,
\[
S = \Big\{x \in L^1(J,{\mathbb R})\cap AC_{\rm loc}(J,{\mathbb R}):
\int_a^\infty
x(t)dt = 0 \Big\}.
\]
Thus, $S$ is not closed in $S_\infty = L^1(J,{\mathbb R}) \cap
AC_{\rm loc}(J,{\mathbb R})$, but, by taking a continuous $\varphi >0$ such
that $\int_a^\infty \varphi(t)dt <\infty $, one obtains immediately
that $A_{\varphi}$ is contained in $S_\infty$ and $S \cap A_{\varphi}$
is closed, i.e.\ $H'_2$) is satisfied.
\end{remark}
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{\sc Mariella Cecchi} \\
Dipartimento di Elettronica,\\
Universit\`a di Firenze,\\
Via S. Marta, 3 - 50139 Firenze, Italy\\
e-mail address: cecchi@diefi.die.unifi.it \smallskip
{\sc Massimo Furi} \\
Dipartimento di Matematica Applicata,\\
Universit\`a di Firenze,\\
Via S. Marta, 3 - 50139 Firenze, Italy\\
e-mail address: furi@dma.unifi.it \smallskip
{\sc Mauro Marini} \\
Dipartimento di Elettronica,\\
Universit\`a di Firenze,\\
Via S. Marta, 3 - 50139 Firenze, Italy\\
e-mail address: marini@ingfi1.ing.unifi.it \smallskip
{\sc Maria Patrizia Pera} \\
Dipartimento di Matematica Applicata,\\
Universit\`a di Firenze,\\
Via S. Marta, 3 - 50139 Firenze, Italy\\
e-mail address: pera@dma.unifi.it
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