Let f be a $C^2$ function : $\mathbb R \to \mathbb R$ (i.e. the second derivative is continuous). Suppose there is a constant $M > 0$ such that in a neighborhood of $+\infty$, $|f''(x)| < M|f '(x)|$, and that $f$ is equivalent to $\exp$ in $+\infty$. Show that $f'$ is also equivalent to $\exp$ in $+\infty$.
First there exists $A$ such that $f'(x)$ $>0$ for all $x ≥ A$ or $<0$ for all $x ≥ A$, by the intermadiate value theorem.
Suppose $f' < 0$ then $f$ decreases on $[A, \infty)$, it contradicts $f \sim \exp$ in $+ \infty$.
So $f' > 0$ on $[A, \infty)$. Then assume $|f''(x)| < M|f'(x)|$ is true on $[A, \infty)$.
$\forall x ≥A$, $f'(x) = \int_{A}^x f''(t)dt + f'(A)$, so $|f'(x)| ≤ M\int_{A}^x f'(t)dt + f'(A)$ i.e.
|$f'(x)| ≤ M(f(x) - f(A)) + f'(A) = Mf(x) + C$
Then, consider $\exp(-x)f(x)$, we deduce from the results above, $\exp(-x)f(x) \to 1$ and $(\exp(-x)f(x))''$ is bounded in $+\infty$.
It can be shown that implies $(\exp(-x)f(x))' \to 0$ and we conclude $f' \sim f \sim \exp$.
Does anyone have another approach ? What if $<$ was $≤$ ? Any generalization ?
Here is another approach.
Lemma Let $a < b$, suppose that
$$1 > {\varepsilon} \geqslant {\sup }_\limits{x \in \left[a , b\right]} \left|f \left(x\right) {e}^{{-x}}-1\right|$$
then there exists $c \in (a , b)$ such that
$$\left|\frac{{f'} \left(c\right)}{f \left(c\right)}-1\right| \leqslant \frac{1}{b-a} \log \left(\frac{1+{\varepsilon}}{1-{\varepsilon}}\right)$$
Proof We use the mean value theorem to get a $c \in \left(a , b\right)$ such that
$$\left(b-a\right) \frac{{f'} \left(c\right)}{f \left(c\right)} = \log \left(f \left(b\right)\right)-\log \left(f \left(a\right)\right) = \left(b-a\right)+\log \left(\frac{f \left(b\right) {e}^{{-b}}}{f \left(a\right) {e}^{{-a}}}\right)$$
hence
$$\left|\frac{{f'} \left(c\right)}{f \left(c\right)}-1\right| = \frac{1}{b-a} \left|\log \left(\frac{f \left(b\right) {e}^{{-b}}}{f \left(a\right) {e}^{{-a}}}\right)\right| \leqslant \frac{1}{b-a} \log \left(\frac{1+{\varepsilon}}{1-{\varepsilon}}\right)$$ $\blacksquare$
We now choose ${\varepsilon} = {\varepsilon} \left(a\right) = {\sup }_\limits{x \geqslant a} \left|f \left(x\right) {e}^{{-x}}-1\right| \mathop{\longrightarrow}\limits_{a \rightarrow \infty } 0$, we choose $b = a+\sqrt{{\varepsilon}}$, so that $c$ depends on $a$ with $\left|c-a\right| \leqslant \sqrt{\varepsilon}$ and $\frac{{f'} \left(c\right)}{f \left(c\right)} \mathop{\longrightarrow}\limits_{a \rightarrow \infty } 1$. We can write
$${f'} \left(a\right) {e}^{{-a}} = {{{\frac{{f'} \left(c\right)}{f \left(c\right)}}\times{\left(f \left(c\right) {e}^{{-c}}\right)}}\times{{e}^{c-a}}}\times{\exp } \left({-\int_{a}^{c}\frac{{f''} \left(t\right)}{{f'} \left(t\right)} d t}\right)$$
This is the product of four terms that tend to $1$ when $a \rightarrow \infty $. It follows that ${f'} \left(a\right) \sim {e}^{a}$.