Problem. Let $I,J$ be open intervals, $\,f:I\to \mathbb R$, continuous, $\,\varphi :J\to \mathbb R$, continuously differentiable, with $\varphi[J]\subset I$, and $\varphi$ satisfying
$$
\varphi'(t)=f\big(\varphi(t)\big), \quad \text{for all $t\in J$}.
$$
Show that $\varphi$ is monotonic.
This becomes rather straight-forward if we further assume the $f$ is continuously differentiable, or even locally Lipschitz continuous, as this implies that IVP for $x'=f(x)$ enjoy uniqueness, and hence if $f\big(\varphi(t_0)\big)=0$, for some $t_0\in J$, then $\varphi$ is constant. In fact something stronger holds in such case: $\varphi$ is either strictly monotonic or constant.
This problem is also straight-forward if $f(x)\ne 0$, for all $x\in I$, in which case uniqueness kicks in anew.
The hard part is to show monotonicity of $\varphi$ when $f$ is just continuous, and its values include zero.
Best Answer
We shall prove the
It follows that all solutions of autonomous (first order) ODEs - regardless of the regularity of $f$, not even continuity is required - are monotonic, since for such functions the derivative is a function of the value.
Proof: Since $g$ is not monotonic, there are $a,b \in J$ with $g'(a) > 0$ and $g'(b) < 0$. Without loss of generality, we may assume that $a < b$. Pick a $c\in [a,b]$ such that
$$g(c) = \max \{ g(t) : t \in [a,b]\}.$$
We have $c \in (a,b)$ and $g(c) > \max \{g(a),g(b)\}$ since $g'(a) > 0$ and $g'(b) < 0$.
If $g(b) \leqslant g(a)$, let $M = \{t\in (c,b] : g(t) = g(a)\}$. By the intermediate value theorem, $M\neq \varnothing$, and we can choose $u = a,\,v = \min M > c$. Since $g(t) > g(a)$ for $t\in [c,v)$ by the intermediate value theorem, it follows that $g'(v) \leqslant 0 < g'(u)$.
If $g(b) > g(a)$, we choose $u = \max \{ t\in [a,c) : g(t) = g(b)\}$ and $v = b$ and by the same reasoning find $g'(u) \geqslant 0 > g'(v)$.