Boundedness of solutions to an ODE and their derivatives The Scenario: Consider the ODE given by
$$
\left\{
\begin{alignat*}{99}
&-g'' + g = F(x,f(x)) \qquad &&x \in [0,L] \\
&g(0) = g(L) = 0
\end{alignat*}
\right.
\tag{1}
$$
where:

*

*$F$ is continuous, i.e. $F(x,f(x)) \in \mathcal{C}([0,L] \times \mathbb{R})$

*$F$ is bounded above, i.e. $|F(x,f(x))| \le M$ for all $(x,f(x)) \in [0,L] \times \mathbb{R}$

*We map $f \mapsto g$ via a "solution operator", in the sense that $Tf = g$ if $g$ is the unique solution to $(1)$

*$f$ is a given function in $\mathcal{C}[0,L]$
From here, I seek to prove that:

*

*$| g(x) | \le M$ for all $x \in [0,L]$

*$\exists N > 0$ a constant, not dependent on $g$ itself, whereby $| g'(x) | \le N$ for each $x \in [0,L]$

The question is just ... how to do so? I've tried looking through the textbooks I'm using for this class (Mingxin Wang's Nonlinear Second Order Parabolic Equations and C.V. Pao's Nonlinear Parabolic & Elliptic Equations) with no luck in seeing something that seems vaguely similar and applicable.
In this class we have been covering the use of Green's functions in solving PDEs and the corresponding ODEs like those above. For instance, identifying $\widetilde{F}(x) := F(x,f(x))$ I can see that $(1)$ has a solution given by
$$g(y) = \int_0^L G(x,y) \cdot \widetilde{F}(x) \, \mathrm{d}x$$
for $x \in [0,L]$ and $G$ the Green's function of the BVP, but I'm not sure how to make any use of this.
We also have been discussed fixed point theory, in particular Schauder's fixed point theorem -- the fact that:

*

*when $X$ is a Banach space

*when $D \subseteq X$ is closed and convex

*when $A : D \to D$ is continuous with $A(D)$ precompact in $X$
then $A$ has a fixed point in $D$. I'm even less sure of how this might be useful, however, for this problem.
I know I haven't actually produced much work, but does anyone have any ideas, or hints or nudges in the right direction?
 A: Regarding your first question: if you assume $g(x_0)>M$ at some point $x_0\in (0,L)$ then  $\max_{[0,L]}g=g(x_1)>M$. Given that that the right hand side $\le M$ you get $g’’(x_1)>0$ which is impossible at a local maximum. You prove similarly that $g$ is never $<-M$ in $(0,L)$.
A: Okay, I think I have an answer, running along with GReyes' answer for the first question. Where the second question is concerned, I'll wait before accepting it, just in case there's something I overlooked. (I struggled with it for a while before getting to an answer that almost seems too simple to be true.)

Part $\#1$: $\newcommand{\nc}{\newcommand}
\nc{\abs}[1]{\left| #1 \right|}
\abs{g} \le M$ on $[0,L]$:
We suppose otherwise, i.e. $\abs{ g(\xi) } > M$ for some $\xi \in [0,L]$. This causes two cases to arise:

*

*$g(\xi) > M > 0$

*$g(\xi) < -M < 0$
(These follow from the absolute value.) We work on case $1$; case $2$ is borderline identical.
By the continuity of $g$ and compactness of $[0,L]$, it must attain its maximum, and so $\exists \eta \in [0,L]$ satisfying
$$g(\eta) = \max_{x \in [0,L]} g(x) \ge g(\xi) > M > 0$$
Using the fact that $g-g'' = F(x,f)$, we have
$$-g''(\eta) + g(\eta) > -g''(\eta) + M$$
whereas
$$-g''(\eta) + g(\eta) = F(\eta,f(\eta)) \le \abs{F(\eta,f(\eta)} \le M$$
Combining the two inequalities we see
$$-g''(\eta) + M < M \implies g''(\eta) > 0$$
Hence, $g$ is concave-up at $\eta$, but as $\eta$ is a maximum, $g$ should be concave-down, a contradiction.

Part $\#2$: $\abs{g'} \le N$ on $[0,L]$ for $N$ not depending on $g$:
Proceeding by contradiction similarly, the fact that $N$ should not depend on $g$ means that, if untrue, then there is a $\nc{\g}{g_\ast} \g \in \mathcal{C}^2[0,L]$ where, for each $r \in \mathbb{R}$, there is a corresponding $\xi_r \in [0,L]$ where $\abs{\g'(\xi_r)} > r$.
Compactness (of $[0,L]$) and continuity (of $\abs{\g'}$) ensure that $\exists \xi \in [0,L]$ for which
$$\abs{ \g'(\xi) } = \max_{x \in [0,L]} \abs{ \g'(x) } \ge \abs{ \g'(\xi_r)} > r > 0$$
In particular,
$$\forall x \in [0,L] \qquad \abs{ \g'(x) } < \abs{ \g'(\xi) } + 1 =: C$$
But since $C \in \mathbb{R}$, the hypothesis must apply to it, and so there is $\xi_C \in [0,L]$ for which
$$\forall x \in [0,L] \qquad \abs{ \g'(\xi_C) } > C > \abs{ \g'(x) }$$
The assignment $x := \xi_C$ gives an obvious contradiction.
