Consider the following initial value problem:
$$ \begin{cases} y'(t) = f(t, y(t)) \\ y(t_0) = y_0 \end{cases} $$
where the flux function $f$ is continuous and locally $y$-Lipschitz on a vertical strip $]a,b[ \times \Bbb R$.
I know that by Cauchy's Existence and Uniqueness Theorem and corollaries there exists a unique maximal solution $y \in C^1(]t_{\min}, t_{\max}[, \Bbb R)$ such that $t_0 \in ]t_{\min}, t_{\max}[ \subseteq ]a,b[$.
Now, my teacher told us that if $$\exists \alpha \in C^0(]a,b[): \forall t \in ]t_{\min}, t_{\max}[, |y(t)| \leq \alpha(t)$$ then $t_{\min} = a$ and $t_{\max}=b$ so that $y$ is also a global solution.
I would like to know if anyone knows this result and its proof or any reference concerning it for I have not been able to find anything on the internet so far.
As always, any comment or answer is welcome and let me know if I can explain myself clearer!
This is a special case of the fact that a nonextendable solution has to leave (to the left and to the right) each compact subset of $(a,b) \times \mathbb{R}$. I will outline the proof of $t_{max} = b$ in your setting: Assume by contradiction that $t_{max} < b$. Consider $y$ restricted to $[t_0, t_{max})$. Then $|y|$ is bounded on $[t_0, t_{max})$ by $\gamma:=\max_{t \in [t_0, t_{max}]} \alpha(t)$. Hence $|y'|$ is bounded on $[t_0, t_{max})$ (since $f$ is bounded on the compact set $[t_0, t_{max}]\times [-\gamma,\gamma]\subseteq (a,b) \times \mathbb{R}$). So $y$ is Lipschitz continuous on $[t_0, t_{max})$, hence $$ y_1:=\lim_{t \to t_{max}-} y(t) $$ exists. From the differential equation you get that $$ \lim_{t \to t_{max}-} y'(t) = f(t_{max},y_1). $$ Now you can extend $y$ to the right by the solution of the IVP $$ z'(t)= f(t,z(t)), \quad z(t_{max})=y_1, $$ in contradiction to the fact that $y$ is nonextendable to the right.