p$)
this $y(x)$ is the solution of a pure integral equation of Volterra-Stieltjes type along with some discrete parts (as seen in \eqref{iddi}). The special case $h(t)=t$ is clearly included in this discussion. For this choice, \eqref{iddi} takes the form
\begin{equation} \label{iddie}
y'(x) = y'(0) - \sum_{n=0}^{p}F(n, y(n))b_n -\int_{p}^{x}F(t,y(t))\,dt,
\end{equation}
``almost" a second order differential equation except for the interface
conditions at a prescribed set of points in $[0,p]$. Under the usual
conditions on $F$ as required by Theorem~\ref{th7}, \eqref{iddie} will
have eventually positive solutions if and only if
$$
\int_{p}^{\infty}t\,F(t, M)\,dt < \infty
$$
holds for some $M>0$ (which is precisely Nehari's necessary and
sufficient criterion for second order nonlinear differential
equations). For this choice of $\sigma$ this result is to be
expected, in some sense, since we are dealing with large $x$
anyhow and so the equation \eqref{iddie} behaves very much like a
differential equation. However, we could \emph{spread} the discrete
part all over the interval $I$ in which case this argument is no
longer tenable, as it is \emph{a priori} conceivable that
oscillations may occur therein (but cannot by Theorem~\ref{th7}).
\section{Proofs}
\begin{proof}[Proof of Theorem~\ref{th}]
We note that $X$ is a closed subset of the Banach space $Y$
above. This is most readily seen by writing the space $X$ as $X =
\{ u \in Y |: 0 \leq \frac{u(t)}{at+b} \leq 1,\text{ for all } t
\geq 0\}$ and applying standard arguments. In addition, it is easy
to see that $X$ is convex. Now we define a map $T$ on $X$ by
setting
\begin{equation} \label{map}
(Tu)(x) = ax+b - \int_{x}^{\infty} (t-x) \, F(t, u(t))\, dt
\end{equation}
for $u \in X$. Note that the right-side of \eqref{map} converges
for each $x \geq 0$, because of \eqref{atk}. Indeed, for $u \in X$,
$x \geq 0$,
\begin{equation} \label{bound}
0 \leq \int_{x}^{\infty}(t-x)F(t, u(t))\, dt \leq \int_{0}^{\infty} t\,
F(t, u(t))\, dt \leq b,
\end{equation}
as $F(t, u(t)) \geq 0$ for such $u$ (which implies that $(Tu)(x) \leq ax+b$)
and the indefinite integral is a non-increasing function of $x$ on
$[0, \infty)$. Since $a \geq 0$, we get that $(Tu)(x) \geq 0$ for any
$x \geq 0$. On the other hand, it is easy to see that for
$u \in X$, $Tu$ is a continuous function on $[0, \infty)$. So,
$TX \subseteq X$.
Next, we prove that $T$ is a continuous map on $X$. For $u, v \in X$,
\begin{align*}
|(Tu)(x) - (Tv)(x) |
& \leq \int_{x}^{\infty} (t-x) | F(t, u(t)) - F(t, v(t)) | \, dt \\
& \leq \int_{x}^{\infty} (t-x) k(t) | u(t) - v(t) |\, dt \\
& \leq \| u - v \|_{Y} \, \int_{0}^{\infty} t\, k(t)\,(at+b)\, dt,
\end{align*}
where we have used \eqref{lip} and the fact that
$\int_{x}^{\infty}(t-x)k(t)(at+b)\, dt$ is a non-increasing function of $x$ for
$x \in [0, \infty)$, since $k(t)(at+b) \geq 0$. It follows that for $x \geq 0$,
\begin{equation}\label{cont}
| \Psi (Tu)(x) - \Psi (Tv)(x) | \leq \frac{1}{b}\, \| u - v
\|_{Y} \, \max\{a,b\}\, \int_{0}^{\infty} t\,(t+1)\, k(t)\, dt,
\end{equation}
from which we conclude that
$$
\| Tu - Tv \|_{Y} \leq \alpha \, \| u - v \|_{Y},
$$
where $\alpha < \infty$ on account of \eqref{teek} and \eqref{tktt}.
It follows that $T$ is continuous on $X$.
Next, we show that $TX$ is compact, that is, $T$ sends bounded
subsets of $X$ onto relatively compact subsets. For $M$ a subset
of $X$ we have that to prove that $TM$ is relatively compact. By
virtue of the isometry $\Psi$, this is equivalent to proving that
$\Psi (T(M))$ is relatively compact. To this end, we use the
measure of noncompactness on $BC(\mathbb{R}^+)$ defined for $A \in
BC(\mathbb{R}^+)$ by
\[
\mu (A)= \lim_{L\to \infty}\left( \lim_{\varepsilon \to 0}
w^L (A,\varepsilon)\right) + \limsup_{t\to\infty}\, {\rm diam}\, A(t),
\]
see [\cite{bg}, Theorem 9.1.1(d), p.46], where
$${\rm diam}\, A(t) = {\rm sup} \{|x(t)-y(t)| : x, y \in A\},$$
and
$$
w^L (A,\varepsilon) = {\rm sup} \{ w^L(x,\varepsilon) : x \in A\},
$$
with
$$
w^L(x,\varepsilon) = {\rm sup} \{|x(t)-x(s)| : t, s \in [0,L], |t-s|
\leq \varepsilon\}.
$$
We fix $\varepsilon >0$, $L>0$, $u\in M\subset X$ and
$t_1,t_2\in\mathbb{R}^+$ with $t_2-t_1\leq \varepsilon$ and,
without loss of generality $t_2>t_1$. Then
\begin{align*}
&|\Psi(Tu)(t_2)-\Psi(Tu)(t_1)|\\
&= \Big|\int_{t_2}^{\infty}\frac{ (s-t_2)F(s,u(s))}{at_2 +b}ds-
\ \int_{t_1}^{\infty}\frac{ (s-t_1)F(s,u(s))}{at_1 +b}ds \Big| \\
&\leq \Big|\ \int_{t_2}^{\infty}\left[\frac{(s-t_2)}{at_2+b}-\
\frac{(s-t_1)}{at_1 +b}\right]F(s,u(s))ds -
\int_{t_1}^{t_2} \frac{(s-t_1)}{at_1 +b}F(s,u(s))ds\Big| \\
& \leq \int_{t_2}^{\infty}\frac{(as
+b)(t_2-t_1)}{(at_2+b)(at_1+b)}F(s,u(s))ds +
\int_{t_1}^{t_2}\frac{(s-t_1)}{at_1 +b}F(s,u(s))ds
\\
&\leq \frac{\varepsilon}{b^2}\ \int_{t_2}^{\infty}(as +b)
F(s,u(s))ds +
\frac{\varepsilon}{b}\int_{t_1}^{t_2}F(s,u(s))ds \\
&\leq \frac{a \varepsilon}{b} + \
\frac{\varepsilon}{b}\int_{t_1}^{\infty}F(s,u(s))ds,
\end{align*}
by \eqref{atk}, since $F \geq 0$ for $u \in M$. Combining these
estimates we deduce that
\begin{equation} \label{28}
\begin{aligned}
|\Psi(Tu)(t_2)-\Psi(Tu)(t_1)|
&\leq \frac{a \varepsilon}{b} +
\frac{\varepsilon}{b}\int_{t_1}^{\infty}F(s,u(s))ds, \\
& \leq \frac{a \varepsilon}{b} + \frac{\varepsilon}{b}\left \{ \int_{t_1}^{t_2}F(s,u(s))ds + \int_{t_2}^{\infty}F(s,u(s))ds \right \}, \\
& \equiv \frac{a \varepsilon}{b} +
\frac{\varepsilon}{b}\left \{ I_1 +I_2 \right \}.
\end{aligned}
\end{equation}
We estimate the two integral quantities in \eqref{28} in turn.
This said, use of \eqref{lip} and \eqref{atk} for $u \in M$ gives
\begin{equation} \label{28a} %\label{28b}
\begin{aligned}
I_2&\leq \int_{t_2}^{\infty} |F(s,u(s))-F(s,0)|ds
+ \int_{t_2}^{\infty} F(s,0)ds \\
&\leq \int_{0}^{\infty} k(s)u(s)ds + \int_{0}^{1}F(s,0)ds
+ \int_{1}^{\infty}sF(s,0)ds \\
&\leq \|u\|_Y \int_0^{\infty}k(s)(as +b)ds
+ \sup_{s \in [0,1]}F(s,0) +\int_{0}^{\infty}sF(s,0)ds \\
&\leq C_2 \equiv \int_0^{\infty} k(s)(as +b)ds + \sup_{s \in
[0,1]}F(s,0) + b
\end{aligned}
\end{equation}
(since, for $u\in M$, $\|u\|_Y \leq 1$), where $C_2$ is finite and
independent of $\varepsilon$, because of \eqref{tktt} and \eqref{kt}.
On the other hand, arguing as in \eqref{28a}, we obtain
\begin{equation}
\begin{aligned}
I_1&\leq \int_{t_1}^{t_2} |F(s,u(s))-F(s,0)|ds+ \int_{t_1}^{t_2} F(s,0)ds \\
&\leq \int_{0}^{\infty} k(s)u(s)ds + \int_{0}^{\infty}F(s,0)ds \\
&\leq C_2
\end{aligned}\label{29a}.
\end{equation}
Combining \eqref{29a}, \eqref{28a} and \eqref{28} we get finally,
\begin{equation}
|\Psi(Tu)(t_2)-\Psi(Tu)(t_1)| \leq \ \frac{a \varepsilon}{b} +
\frac{2C_2 \varepsilon}{b}. \label{30}
\end{equation}
Passing to the supremum over $u \in M \subset X$ we find that
\[
\lim_{\varepsilon \to 0} w^{L}(\Psi (TM),\varepsilon)=0
\]
Since $L>0$ is arbitrary we deduce that
\[
\lim_{L\to \infty} \left(\lim_{\varepsilon \to 0}w^L
(\Psi (TM),\varepsilon)\right)=0.
\]
To complete the proof we need to analyze the term related to the diameter.
Taking $u, v\in M$ and $t\in \mathbb{R}^+$ then, proceeding as in
the continuity argument above leading to \eqref{cont}, we see that
\begin{align*}
|\Psi (Tu)(t)-\Psi (Tv)(t)|
&\leq \frac{\|u-v\|_{Y}}{b} \int_t^{\infty} s(as+b)k(s)ds \\
& \leq \frac{\|u-v\|_Y}{b}\left[a \int_t^{\infty} s^2 k(s)ds
+b \int_t^{\infty}sk(s)ds\right] \\
& \leq \frac{2}{b} \left[a \int_t^{\infty} s^2 k(s)ds +b
\int_t^{\infty} sk(s)ds\right],
\end{align*}
since $\|u-v\|_Y \leq 2$ for $u, v \in M$. Consequently
\[
\limsup_{t\to \infty}\mathop{\rm diam} \Psi (TM)(t)
\leq \frac{2}{b}\limsup_{t\to \infty}\left[ a\int_t^{\infty} s^2 k(s)ds
+b\int_t^{\infty} sk(s)ds \right]=0,
\]
on account of \eqref{teek} and Remark~\ref{uno}. Thus, $\mu (\Psi (TM))=0$.
This fact tells us that $\Psi (TM)$ is relatively compact in
$BC(\mathbb{R}^+)$ and, since $\Psi$ is an isometry, $TM$ is relatively
compact in $Y$. Hence, $T$ is compact, and so Schauder's theorem
gives the existence of a fixed point $u \in X$ for $T$. This fixed
point is necessarily a solution of \eqref{non0} asymptotic to the
line $ax+b$ as $x \to \infty$. This completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{th1}]
Let $Y$ be the Banach space defined in \eqref{spaceY} with the norm
\eqref{normY}, where we replace the interval $[1,\infty)$ by $x_0,\infty)$.
Let $X$ be the closed subset defined by
$X =\{u\in Y: {\sup_{x \geq x_0}\{|u(x)|/|f(x)|\}\leq 2} \}$.
Define a map $T$ on $X$ by $u \in X$,
\begin{equation}
\label{tux}
Tu(x) = f(x) - \int_{x}^{\infty}{(s-x)F(s,u(s))\,ds},\; x\geq x_0.
\end{equation}
Clearly, for $u \in C(I)$ we have, because of our assumptions on $F$,
$Tu \in C(I)$. In addition, \eqref{lip01} gives that for
$u \in X$, $|F(s,u(s))| \leq k(s)|u(s)| + |F(s,0)|$.
Combining this with \eqref{tux}, dividing \eqref{tux} throughout
by $f(x)$ a simple estimation gives that for $ x \geq x_0$,
\[
\Big | \frac{Tu(x)}{f(x)}\Big |
\leq 1 + \frac{1}{|f(x)|}\int_{x}^{\infty}(s-x)|k(s)f(s)|
\Big |\frac{u(s)}{f(s)}\Big | \, ds
+ \frac{1}{|f(x)|}\int_{x}^{\infty}(s-x)|F(s,0)|\, ds.
\]
Now, the use of \eqref{ft} shows that
\begin{equation}
\label{Tuf} \Big | \frac{Tu(x)}{f(x)}\Big | \leq 1 + \Big \|
\frac{u}{f}\Big \|
\frac{1}{\delta}\int_{x}^{\infty}(s-x)|k(s)f(s)|\, ds
+ \frac{1}{\delta}\int_{x}^{\infty}(s-x)|F(s,0)|\, ds,
\end{equation}
which, since $u \in X$ and \eqref{t0} is enforced, furnishes the bound
\[
\Big \| \frac{Tu(x)}{f(x)}\Big \| \leq
1 + 2 \frac{1}{\delta}\frac{\delta}{4}\,+ \frac{1}{\delta}\frac{\delta}{4}
= \frac{7}{4} < 2.
\]
Thus $T$ is a self-map on $X$. In order to show that $T$ is a
contraction on $X$, consider the simple estimate derived from
\eqref{tux}, namely, for $x \geq x_0$,
\begin{equation} \eqref{lip01}
\begin{aligned}
\Big | \frac{Tu(x) - Tv(x)}{f(x)} \Big |
&\leq \frac{1}{|f(x)|} \int_{x}^{\infty}{(s-x)|F(s,u(s))-F(s,v(s))|\, ds},\\
&\leq \frac{1}{|f(x)|} \int_{x}^{\infty}{(s-x)k(s)|u(s)-v(s)|\, ds},\quad
({\rm by}\, ) \\
&\leq \frac{1}{\delta}\|u-v\| \int_{x}^{\infty}{(s-x)k(s)|f(s)| \, ds},
\quad ({\rm by}\, \eqref{ft} ) .
\end{aligned}
\end{equation}
Since the last display is valid for every $x \geq x_0$ it follows
from \eqref{t0} that,
$$
\| Tu - Tv\| \leq (1/4)\, \|u-v\|,
$$
so that $T$ is a contraction on $X$. It is easily seen that the subsequent
fixed point, say $u(x)$, obtained by applying the classical fixed point
theorem of Banach, is a solution of \eqref{non0} satisfying the
conclusion (2) stated in the theorem, since $u \in X$. On the other
hand, since our fixed point $u$ satisfies \eqref{tux}, we have
$$
\frac{u(x)}{f(x)} = 1 - \frac{1}{f(x)}\int_{x}^{\infty} (s-x)F(s,u(s))\, ds.
$$
An estimation of this integral similar to the one leading to the
right-side of \eqref{Tuf} gives that
$$
\lim_{x \to \infty} \frac{1}{f(x)}\int_{x}^{\infty} (s-x)F(s,u(s))\, ds = 0,
$$
on account of the finiteness of all the integrals involved.
This shows that $u(x) \sim f(x)$ as $x \to \infty$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{th2}]
Note that $X$ is a closed subset of the Banach space $BC(\mathbb{R}^+)$.
For $u \in X$ we define a map $T$ by setting
$$
Tu(x) = f(x) - \int_{x}^{\infty}(t-x)F(t, u(t))\, dt, \quad x \geq 0.
$$
Then for $u \in X$ it is clear that $Tu \in C(\mathbb{R}^+)$ and
since $F(t,u(t)) \geq 0$ for such $u$ and all $t \geq 0$, we have
$|Tu(x)| \leq \|f\|_{\infty} + b$, for every $x \geq 0$, where we have
used the fact the integral in question is a non-increasing function of $x$
for all $x \geq 0$. Hence $T$ is a self-map on the ball $X$. Finally,
an argument similar to the corresponding one in Theorem~\ref{th}
gives that $T$ is a contraction on $X$ provided there holds \eqref{tkt}.
This completes the proof.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{th4}]
The necessity is simple. If $y$ is such a solution then set $u=v=y$ throughout.
For the sufficiency we appeal, as usual, to a fixed point theorem.
Consider the space $BC(I)$ of (uniformly) bounded continuous functions
on $I$ with the uniform norm. Since $u, v$ are uniformly bounded by
hypothesis, the subset $X$ defined by
$$
X=\{ y \in BC(I) : u(x) \leq y(x) \leq v(x), x \in I\}
$$
with the induced metric, is complete. Define a map $T$ on $X$ by the usual
\[
Ty(x) = f(x) - \int_{x}^{\infty}(t-x)F(t,y(t))\,dt, \quad x \in I
\]
for $y \in X$. Since $y \in X$, then $y \in L^{\infty}(I)$; it follows
from hypotheses (2) and (3) that the integral on the right is finite
for every $x \in I$ and this defines a continuous function that is
uniformly bounded on $I$. Thus, $Ty$ is continuous and uniformly
bounded on $I$, since $f$ is. Thus, $T$ is well-defined. On the other hand,
by hypothesis (2), $F(t,u(t)) \leq F(t,y(t)) \leq F(t,v(t))$ for
$t \in I$; it follows that, for $x \in I$,
\begin{align*}
Ty(x) &\leq f(x) - \int_{x}^{\infty}(t-x)F(t,u(t))\,dt \leq v(x)\\
& \geq f(x) - \int_{x}^{\infty}(t-x)F(t,v(t))\,dt \geq u(x)
\end{align*}
where we have used assumptions (b) and (c) in order to estimate the integrals.
Thus $T$ is self-map on $X$. That $T$ is a contraction on $X$ follows
the usual route. Briefly, for $x \in I$, $y, z \in X$,
\begin{align*}
|Ty(x) - Tz(x)|
&\leq \int_{x}^{\infty}(t-x)|F(t,y(t))-F(t,z(t))|\, dt\\
&\leq \int_{x}^{\infty}(t-x)k(t)|y(t)-z(t)|\, dt\\
& \leq \left (\int_{x_0}^{\infty}tk(t)\,dt\right ) \; \|y-z\|_{\infty}
\end{align*}
and so,
$$
\|Ty - Tz\|_{\infty} \leq \alpha \|y-z\|_{\infty},
$$
where $\alpha < 1$ is the integral in question (cf., assumption (5)).
\end{proof}
\begin{proof}[Proof of Theorem~\ref{cor01}]
This is clear since we can integrate the differential equation \eqref{non0}
twice to obtain \eqref{ie1} and conversely, if we know that its solution
is $L^{\infty}$, we can differentiate \eqref{ie1} twice to recover
\eqref{non0}. The result follows from an application of the theorem.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thx}]
We integrate the inequalities twice over the half line to obtain both
\eqref{uf} and \eqref{vf}. An application of Theorem~\ref{th4} gives
that \eqref{ie1} has a solution $y(x) \sim f(x)$, as $x \to \infty$.
But the right side of \eqref{ie1} is twice differentiable, consequently
so is $y(x)$, that is \eqref{pert} is satisfied.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{atm}]
First, we show that solutions of \eqref{non2} exist on the half-line, $I$.
Introduce the usual energy functional $E(x)$ on solutions of \eqref{non2} by
\begin{equation}\label{ener}
E(x) = \frac{1}{2}{y'}^2 + \int_{0}^{y}\eta G(x, \eta)\, d\eta
\equiv \frac{1}{2}{y'}^2 + \mathcal{I}(x,y),
\end{equation}
where $\mathcal{I}_{x}(x,y) \leq 0$ by hypothesis. A glance at \eqref{non2}
shows that
$$
E' = y'g + \mathcal{I}_{x} \leq y'g \leq g \sqrt{2E}.
$$
So, $E'E^{-1/2} \leq \sqrt{2}g$ whenever $E>0$. It follows that if the
solution $y(x)$ exists for $x \in [a,b]$, $a \geq x_0$ and, at the same time,
$E(x) > 0$ for such $x$, then
\begin{equation}
\label{ener2} \sqrt{E(b)} \leq \sqrt{E(a)}
+ \frac{\sqrt{2}}{2}\int_{a}^{b}g(t)\,dt.
\end{equation}
Of course, \eqref{ener2} is also true for any interval $[a,b]$ in which
the solution exists. For such an interval we have from \eqref{ener2}
\begin{equation}
\label{ener3} |y'(b)| \leq \sqrt{2E(a)} + \int_{a}^{b}g(t)\,dt,
\end{equation}
so, if the solution exists on an interval $[a,b)$ then it can be continued
to $x=b$ and thus to a right-neighborhood of $b$. Thus, we see that for
any $x \geq x_0$ a solution can be continued throughout $I$.
We now claim that for a given solution $y$ of \eqref{non2} there is an $X$
(depending on $y$) such that we cannot have for
\begin{equation}
\label{yayb}b > a \geq X, y(a)=y(b)=0, y(x) > 0, x \in (a,b).
\end{equation}
Note that by \eqref{tgt} and \eqref{ener3} we can suppose that $X$
is such that $X > 0$ and such that for some $K>0$ we have
\begin{equation}
\label{yKx} |y(x)| < Kx, \quad x \geq X.
\end{equation}
This already implies that all solutions are ``sublinear" or cannot
grow faster than a linear function. We fix this $K$ and consider
the differential equation $$z'' + G(x, Kx)z =0, \quad x \geq X.$$
Since this is a linear equation it is well known that (e.g.,
\cite{rb}), assumption \eqref{tgKt} implies that this equation
has a solution $z(x) \to 1$ as $x \to \infty$. We choose $X_0> X$
so that $z(x) > 0$ for all $x \geq X_0$ and let $b>a \geq X_0$.
Now, writing $y = wz$ and making use of the equation for $z$, we
obtain the second order linear differential equation
\begin{equation}
\label{wz} w''z+2w'z' = g+wz\{G(x,Kx)-G(x,y)\}.
\end{equation}
However, \eqref{yayb}, \eqref{yKx} and $G$ non-decreasing in its
second variable, would imply that the right of \eqref{wz} is positive
in $(a,b)$ so that $(w'z^2)' \geq 0$ on $(a,b)$. Indeed, \eqref{yayb}
would also force $w(a)=w(b)=0$ and $w(x)>0$ in $(a,b)$. On the other hand,
this leads to $w'(a) \geq 0$, $w'(b) \leq 0$. Since $w'z^2$ is non-decreasing,
this implies that $w'=0$, i.e., $w=0$ in $(a,b)$ resulting in a contradiction.
Thus, $y(x)$ is either ultimately positive or it is ultimately non-positive.
Now consider the case where $y(x)$ is ultimately positive. We may suppose
(see \eqref{yKx}) that
\begin{equation}
\label{0yKt}
0 < y(x) < Kx, \quad x \geq X.
\end{equation}
We modify the argument following \eqref{yKx} as follows: Consider
the differential equation $$z_1''+G(x,y(x))z_1 =0 , \quad x\geq
X.$$ As before, the integrability condition \eqref{tgKt} gives
that this will have a solution $z_1(x)$ with $z_1(x) \to 1$ as $x
\to \infty$. We can define $w_1$ as before by $y=w_1z_1$ and find,
as before, that $(w_1'{z_1}^2)' =gz_1$. But \eqref{tgt} along with
the fact that $w_1'{z_1}^2$ is non-decreasing implies that
${w_1}'$ tends to a non-negative finite limit at infinity. The
possibility that ${w_1}'(\infty)>K$ is excluded on account of
\eqref{0yKt}. Hence $ 0 \leq {w_1}'(\infty) \leq K$. If $A\equiv
{w_1}'(\infty)>0$ then necessarily $y(x) \sim Ax$ as $x \to
\infty$.
The other possibility is that \eqref{0yKt} holds but that
${w_1}'(\infty)=0$. In this case, the differential equation for $w_1$ yields
$$
{w_1}'(x) = - \{z_1(x)\}^{-2}\,\int_{x}^{\infty}g(t)z_1(t)\,dt,
$$
and since $z_1(x) \to 1$ as $x \to \infty$ we see that
\begin{equation}
\label{w1}
{w_1}'(x) \sim - \int_{x}^{\infty}g(t)\,dt
\end{equation}
as $x \to \infty$. Note that if \eqref{tgt} were false (for $i=1$)
it would follow that (since $g(x) \geq 0$), $w_1(x) \to -\infty$
as $x \to \infty$ and this contradicts the positivity of $y(x)$
for all large $x$. Thus, \eqref{tgt} is actually a necessary
condition. On the other hand, the hypothesis \eqref{tgt} implies
that $B\equiv w_1(\infty)$ is finite and necessarily non-negative,
because of the positivity of $y$, i.e., $y(x) \sim B$ as $x \to
\infty$, where $B \geq 0$. If $y(x)$ is ultimately
non-positive, we can take it that $y(x) \leq 0$ for $x \geq X$,
and that $y(x) < 0$ on some unbounded subset of $x \geq X$. Since
$yG(x,y) \leq 0$ for $y \leq 0$, we get from \eqref{non2} that
$y'' \geq 0$. Applying Lemma 0 in \cite{fva} with $z(x)\equiv
-y(x)$ we see that $y'(x) \leq 0$ for $x \geq X$. This, in
conjunction with the fact that $y(x) < 0$ on some unbounded subset
implies that $y(x) < 0$ for all sufficiently large $x$. So, $ y(x)
< 0$, $y'(x) \leq 0$, $y'' \geq 0$ for all large $x$ which leads
to a counterpart of the positive solutions result.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{atm2}]
As before we write $F(x,y)=yG(x,y)$ and we proceed as in the proof
of Theorem~\ref{atm} up to \eqref{ener3}. The same argument therein
gives that solutions all exist on some half-axis. Indeed, since
$f'(\infty) = \infty$, by assumption, we can use \eqref{ener3}
to derive that $|y'(x)| \leq (1+\varepsilon)f'(x)$ for all sufficiently
large $x$. In addition, another integration gives us a similar bound for
$y$ in the form $|y(x)| \leq (1+\varepsilon)f(x)$ for all sufficiently
large $x$, say $x \geq X_0$. Next, an integration of \eqref{non2} over
$[X_0, x)$ and use of \eqref{tgt2} and \eqref{fr} shows that
$$
y'(x) \sim \int_{x_0}^{x}g(t)\,dt, \quad x\to \infty.
$$
Finally, one last integration of the preceding equation gives the
desired asymptotic estimate.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{th5}]
Consider the solution $y$ whose initial conditions are $y(a)=a$,
$y'(a)=M$ with right-derivatives, where $a$ is to be chosen
later. Since $y'$ is right-continuous on $I$ there is a $b > a$
such that
\begin{equation} \label{m1}
\frac{M}{2} < |y'(x)| < 2M, \quad{x \in [a,b)}
\end{equation}
Since $y$ is absolutely continuous on $[a,b)$ it follows that
\begin{equation*}
|y(x)| \leq |y(a)| + |\int_{a}^{x}y'(t)\, dt | \leq a+ M(x-a)
\end{equation*}
that is,
\begin{equation} \label{m2}
|y(x)| < Mx, \quad{x \in [a,b)}
\end{equation}
since $M > 1$. Since $y$ is continuous, \eqref{m2} also holds at
$x=b$. It follows from \eqref{nlvs1}, \eqref{nlvs2} that
\begin{equation} \label{m3}
y'(b) = y'(a) - \int_{a}^{b}F(t,y(t))\,d\sigma(t)
\end{equation}
i.e.,
\begin{equation} \label{m4}
| y'(b) - M|\leq \int_{a}^{b}F(t,y(t))\,|d\sigma(t)|.
\end{equation}
On the other hand, $F$ is nondecreasing in its second variable by hypothesis,
so \eqref{m2} and assumption (3) together yield
\begin{equation} \label{m5}
| y'(b) - M|\leq \int_{a}^{\infty}F(t,Mt)\,|d\sigma(t)|.
\end{equation}
Now, by assumption (3) again we can choose (and fix) $a$ so large that
$$
\int_{a}^{\infty}F(t,Mt)\,|d\sigma(t)| < M/4.
$$
Then \eqref{m1} holds for all $b > a$ and, as a result, \eqref{m2}
holds for all $x > a$. From this we see that for given $\varepsilon > 0$,
we can choose $X$ so large that for any $c> b > X$ we have
$$
\int_{b}^{c}F(t,Mt)\,|d\sigma(t)| < \varepsilon,
$$
and a double application of \eqref{m3} and the usual estimates, shows that
$$
|y'(c)-y'(b)| <\int_{b}^{c}F(t,Mt)\,|d\sigma(t)| < \varepsilon,
$$
for $c > b > X$. Hence $y'(x)$ tends to a limit $L$, say, as
$x \to \infty$, and $L \neq 0$ on account of \eqref{m1}.
From this it follows that $y(x)/x = L + o(1)$, i.e., $y$ is
asymptotically linear as $x \to \infty$.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{th6}]
It suffices to show that the
integral equation
\begin{equation}\label{eq1}
y(x)=M-\int_{x}^{\infty}(t-x)\,F(t,y(t))\,
d\sigma(t),
\end{equation}
has a fixed point in $X$ (since the
resulting solution will be absolutely continuous, with a
derivative that is locally of bounded variation and satisfying
\eqref{nlvs2}). So, we define the operator $A$ on $X$ by
$$
(Ay)(x)=M-\int_{x}^{\infty}(t-x)F(t,y(t))\,d\sigma(t).
$$
Since $\sigma$ is non-decreasing and $F \geq 0$, for $y\in X$,
the function defined by
$$
\int_{x}^{\infty}(t-x)\,F(t,y(t))\,d\sigma(t)
$$
is nonincreasing, hence
$$
0\leq\int_{x}^{\infty}(t-x)\,F(t,y(t))\,d\sigma(t)\leq
\int_{0}^{\infty}t\,F(t,y(t))\,d\sigma(t)\leq M,
$$
by hypothesis (a). Consequently, for $y\in X$ we have y
\begin{equation}\label{0ay}
0\leq (Ay)(x)\leq M,\quad {for}x\in I.
\end{equation}
In order to show that for $y\in X$ then
$Ay\in\mathcal{C}(I)$, we note by Fubini's theorem that since
$$
\int_{x}^{\infty}(t-x)F(t,y(t))\, d\sigma(t)
= \int_{x}^{\infty}\int_{t}^{\infty}F(s,y(s))\, d\sigma(s)\, dt,
$$
for every $x \in I$ and the integral of a function that is locally
of bounded variation is locally absolutely continuous, it is in
particular continuous and so, for $y\in X$, we have
$Ay\in\mathcal{C}(I)$. This, in combination with \eqref{0ay} shows
that $Ay \in X$. Hence, the operator $A$ applies $X$ into itself.
Now, we prove that $A$ is continuous on $X$. Indeed,
\begin{align*}
|(Ay_{n})(x)-(Ay)(x)|
&= \Big|\int_{x}^{\infty}(t-x)[F(t,y_{n}(t))-F(t,y(t))]\, d\sigma(t)\Big| \\
&\leq \int_{x}^{\infty}(t-x)|F(t,y_{n}(t))-F(t,y(t))|\, d\sigma(t) \\
&\leq \|y_{n}-y\|_{\infty}\int_{x}^{\infty}(t-x)k(t)\, d\sigma(t) \\
&\leq \|y_{n}-y\|_{\infty}\int_{0}^{\infty}t\,k(t)\,d\sigma(t).
\end{align*}
It follows that $A$ is continuous on $X$ on account of assumption
$(c)$.
The proof that $A$ is compact uses ideas from the theory of
measures on non-compactness. First, we introduce some terminology.
Let us fix a nonempty bounded subset $X$ of $C[0,a]$. For
$\varepsilon >0$ and $y \in X$ denote by $w(y,\varepsilon)$ the
\emph{modulus of continuity} of $y$ defined by
\[
w(y,\varepsilon)= \sup \{| y(t)-y(s)| : t,s \in [0,a],\, | t-s|
\leq \varepsilon\}
\]
Further, let us put
\begin{gather*}
w(X,\varepsilon)=\sup \{w(y,\varepsilon) : y\in X\}\\
w_0 (X)= \lim_{\varepsilon \rightarrow 0} w(X,\varepsilon),
\end{gather*}
It can be shown (see \cite{11}) that the function $\mu(X)=w_0 (X)$
is a regular measure of noncompactness in the space $C[0,a]$.
Now, let $x_{1},x_{2}\in [0,\infty)$ be such that
$x_{2}-x_{1}\leq \varepsilon$ and without loss of generality,
$x_{1}