My question is as follows :
Let $f$ and $g$ are positive continuous functions on $\mathbb R$ with $g\leq f$ everywhere. Assume that IVP $$x’=f(x), x(0)=0$$ has a solution defined on all of $\mathbb R$, then prove that the IVP $$x’=g(x), x(0)=0$$ also has a solution defined on all of $\mathbb R$. I am thinking as follows::
As first IVP has a solution which does not blow up in finite time and because $g\leq f$ everywhere, hence second IVP solution does not have possibility to blow up in finite time. Therefore, second IVP also have a solution defined on all of $\mathbb R$. Am I correct? Thank you.
Your idea is correct. You can formalize your idea as follows: Let $f,g \in C(\mathbb{R},(0,\infty))$ with $g \le f$, and let $x:\mathbb{R} \to \mathbb{R}$ and $y:(\omega_-,\omega_+) \to \mathbb{R}$ be the solutions of $x'(t)=f(x(t)), x(0)=0$ and $y'(t)=g(y(t)),y(0)=0$, respectively, with $y$ nonextendable. The functions $x,y$ are increasing, and by the method for ODEs of separate variables the solutions are unique and $$ \int_0^{x(t)} \frac{1}{f(s)} ds = t = \int_0^{y(t)} \frac{1}{g(s)} ds \quad (t \in (\omega_-,\omega_+)). $$ Since $1/f \le 1/g$ we get $0\le y(t) \le x(t)$ $(t \in [0,\omega_+))$ and $x(t) \le y(t) \le 0$ $(t \in (\omega_-,0])$. Thus, there is no blow up of $y$ in finite time, neither to the left nor to the right. So $\omega_-=-\infty$ and $\omega_+=\infty$.