Differentiation and asymptotic equivalents. Following this post on Meta, I am going to (semi) regularly ask questions from competitive mathematics exams, on a variety of topics; and provide a solution a few days later. The goal is not only to list interesting (I hope) exercises for the sake of self-study, but also to obtain (again, hopefully) a variety of techniques to solve them.
Differentiating asymptotic equivalents.

Let $f\colon\mathbb{R}\to\mathbb{R}$ be continuous and non-decreasing. For $x\in\mathbb{R}$, let $F(x)=\int_0^x  f$.

*

*Suppose there exists $\alpha > 0$ such that $F(x)\displaystyle\operatorname*{\sim}_{x\to\infty}\frac{x^\alpha}{\alpha}$. Show that $f(x)\displaystyle\operatorname*{\sim}_{x\to\infty}x^{\alpha-1}$.


*Now, assume that  $F(x)=\frac{x^2}{2}+o(x)$ when $x\to\infty$. Show that $f(x) = x+o(\sqrt{x})$.


*Give counterexamples for 1. (and 2.) when $f$ is not assumed monotone.

Reference: Exercise 4.58 in Exercices de mathématiques: oraux X-ENS (Analyse I), by Francinou, Gianella, and Nicolas (2014) ISBN 978-2842252137.
 A: *

*
Suppose there exists $\alpha > 0$ such that $F(x)\displaystyle\operatorname*{\sim}_{x\to\infty}\frac{x^\alpha}{\alpha}$. Show that $f(x)\displaystyle\operatorname*{\sim}_{x\to\infty}x^{\alpha-1}$.

Let $\epsilon > 0 $. The hypothesis means that there is $N(\epsilon)$ such that for all $x\geq N(\epsilon)$ ;   
$$ (1-\epsilon)\frac{x^\alpha}{\alpha} \leq F(x) \leq (1+\epsilon) \frac{x^\alpha}{\alpha} $$
Now by monotonicity,
$$ hxf(x) \leq \int_{x}^{x+hx} f(t) \,dt =F(x+h) - F(x) \leq (1+\epsilon)\cdot\frac{x^\alpha (1+h)^\alpha}{\alpha} - (1-\epsilon)\cdot\frac{x^\alpha}{\alpha} $$
After some manipulation we get for any $h <1$ ;
$$ \frac{f(x)}{x^{\alpha-1}} \leq (1-\epsilon)\cdot\frac{(1+h)^\alpha-1}{h} + \frac{2\epsilon (1+h)^\alpha}{\alpha h} $$
Taylor expanding we get $(1+h)^\alpha \leq 1 + h + M h^2$, where $M = \max_{y\in[1,2]} \frac{f''(y)}{2!}$. Hence
$$ \frac{f(x)}{x^{\alpha-1}} \leq (1-\epsilon)\cdot (1+Mh) +\frac{2\epsilon}{\alpha}\left( \frac{1}{h} + 1 + Mh \right) $$
We choose $h=\sqrt{\epsilon}$ to get (after rearranging)
$$ \frac{f(x)}{x^{\alpha-1}} \leq 1+\left( M+  \frac{2}{\alpha} \right)\sqrt{\epsilon} + \frac{2}{\alpha}\epsilon + \left(\frac{2M}{\alpha} -M \right)\epsilon \sqrt{\epsilon} = 1 + O \left( \sqrt{\epsilon} \right)$$
for all $x\geq N(\epsilon)$
Hence $\limsup_{x \rightarrow \infty}\frac{f(x)}{x^{\alpha -1}} \leq 1$. Similarly,one can show that $\liminf_{x \rightarrow \infty}\frac{f(x)}{x^{\alpha -1}} \geq 1$, and we are done.



*
Now, assume that  $F(x)=\frac{x^2}{2}+o(x)$ when $x\to\infty$. Show that $f(x) = x+o(\sqrt{x})$.

The same method above works, (where instead of $hx$ we put $h \sqrt{x}$) indeed we have for all $x\geq N(\epsilon)$ ;   
$$ -\epsilon x + \frac{x^2}{2} \leq F(x) \leq \epsilon x + \frac{x^2}{2} $$
So,
$$h \sqrt{x} f(x) \leq \epsilon h\sqrt{x} + xh + \frac{h^2}{2} \implies f(x) \leq \epsilon + x + \frac{\sqrt{x}}{2} + 2 \epsilon \sqrt{x} = x + O(\epsilon) \sqrt{x} $$
Now it is easy to see how to complete the proof.



*
Give counterexamples for 1. (and 2.) when $f$ is not assumed monotone.

For $\alpha > 1 $. We can take $f(x) : = x^{\alpha-1}(1+\sin(x))$. Indeed, one can show that 
$$ \int_{0}^{x} t^{\alpha-1}\sin(t) \, dt = O\left(x^{\alpha-1}\right)$$
as $x \rightarrow \infty$,and hence $F(x) = \frac{x^\alpha}{\alpha} + O\left(x^{\alpha-1}\right) \sim \frac{x^\alpha}{\alpha} $
On the other hand $\frac{f(x)}{x^\alpha}$ is not convergent as $x \rightarrow \infty$
