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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 $F(x)=\int_0^x f$.

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

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

  3. 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.

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  • $\begingroup$ Note, half to myself: I got sidetracked writing my thesis, will come back to this afterwards. In the meantime, setting a bounty to trigger more answers. $\endgroup$ – Clement C. Aug 16 '17 at 14:23
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  1. 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.


  1. 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.


  1. 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 $x \rightarrow \infty$

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  • $\begingroup$ For 2., how exactly can you argue "the same method works"? If I am not mistaken, this would only show $f(x) = x+o(x)$, instead of the stronger statement $f(x)=x+o(\sqrt{x})$. $\endgroup$ – Clement C. Aug 6 '17 at 19:31
  • $\begingroup$ It works exactly the same way if we considered $h\sqrt{x}$ instead of $hx$. That can be easily motivated if we just studied $F(x+k) - F(x)$ [which is motivated by Fundamental theorem of Calculus], where $k$ may depend on $x$ and then find that the best $k$ is $o(\sqrt{x})$ ... I have added details to the post $\endgroup$ – Mahmoud Abdelrazek Aug 7 '17 at 2:37

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