Romania NMO 2023 Grade 11 P4

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

w

3 users

TOPICS

No tags match your search

M

No tags match your search

M

Romania NMO 2023 Grade 11 P4

G H J

Reveal topic tags

differentiable function

Continous function

L

G H BBookmark kLocked kLocked NReply

Source: Romania National Olympiad 2023

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

196 posts

#1 h Apr 14, 2023, 2:36 AM

Y by

We consider a function $f:\mathbb{R} \rightarrow \mathbb{R}$ for which there exist a differentiable function $g : \mathbb{R} \rightarrow \mathbb{R}$ and exist a sequence $(a_n)_{n \geq 1}$ of real positive numbers, convergent to $0,$ such that

$g'(x) = \lim_{n \to \infty} \frac{f(x + a_n) - f(x)}{a_n}, \forall x \in \mathbb{R}.$
a) Give an example of such a function f that is not differentiable at any point $x \in \mathbb{R}.$

b) Show that if $f$ is continuous on $\mathbb{R}$ , then $f$ is differentiable on $\mathbb{R}.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

95 posts

#2 h Apr 18, 2023, 7:57 AM • 3 Y

Y by AngleChasingXD, RobertRogo, TurtleKing123

For the $(a)$ part simply take
$f(x)= \left\{ \begin{array}{ll} 0 & x \in \mathbb{R}-\mathbb{Q} \\ 1 & x\in \mathbb{Q}\\ \end{array} \right.$ It is easy to prove that this function is not differentiable in any point $x\in \mathbb{R}$ . To show that it indeed works, take $g\equiv 0$ and the sequence $a_n=\frac{1}{n}, \forall n\ge 1$ If $x$ is not rational, neither is $x+a_n$ and if $x$ is rational, so is $x+a_n$ which means that $f(x)=f(x+a_n)$ for any $n$ . Therefore $\lim_{n \to \infty} \frac{f(x + a_n) - f(x)}{a_n}=0=g'(x)$

For the $(b)$ part I will use the following lemma, which overkills this problem:

Lemma: If $(a_n)_{n \geq 1}$ is a sequence of positive real numbers that converges to $0$ , then the set $A=\{ka_n\mid k\in \mathbb{Z}, n\in \mathbb{N^{\star}}\}$ is dense in $\mathbb{R}$ .

This result is not hard to prove since one can simply take two values $a$ and $b$ in $\mathbb{R}$ and specifically find the convenient $k$ and $n$ such that $a<ka_n<b$ .

Returning to our problem, consider the function $h(x)=g(x)-f(x)$ Clearly, $h$ is continuous. Anyways, the given relation translates to $\lim_{n \to \infty} \frac{h(x + a_n) - h(x)}{a_n}=0$ It is easy to prove inductively that $\lim_{n \to \infty} \frac{h(x + ka_n) - h(x)}{ka_n}=0$ for any $k\in \mathbb{Z}$ .

Fix some random $x,y\in \mathbb{R}$ . According to our lemma, there exists a sequence $(c_n)_{n \geq 1}\subset A$ so that $\lim_{n \to \infty} c_n=y-x$ Plugging this in our previous observation gives us $0=\lim_{n \to \infty} \frac{h(x + c_n) - h(x)}{c_n}=\frac{h(y)-h(x)}{y-x}$ because of $h$ 's continuity.

Since $x$ and $y$ have been chosen arbitrarily, it means that $h$ is constant and the conclusion follows immediately.

This post has been edited 6 times. Last edited by MathsLover04, Apr 19, 2023, 8:58 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

11 posts

#3 h Apr 18, 2023, 7:16 PM

Y by

Congrats !

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

200 posts

#4 h Apr 18, 2023, 7:59 PM

Y by

The solution above is wrong. Formally what you would have is that for any $k\in\mathbb{N}$ and $\varepsilon>0$ there exists $n_{k,\varepsilon}$ s.t. $\left|\frac{h(x+ka_n)-h(x)}{ka_n}\right|<\varepsilon$ for all $n>n_{k,\varepsilon}$ . To get your claim you would want this convergence to be uniform in $k$ , i.e. make $n_{k,\varepsilon}$ not dependent on $k$ . This needs a big justification, and it is the entire reason this problem is nontrivial. Hope the professors didn't fall for it.

This post has been edited 1 time. Last edited by Doru2718, Apr 18, 2023, 8:02 PM
Reason: .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

461 posts

#5 h Apr 18, 2023, 10:27 PM

Y by

DanDumitrescu wrote:

We consider a function $f:\mathbb{R} \rightarrow \mathbb{R}$ for which there exist a differentiable function $g : \mathbb{R} \rightarrow \mathbb{R}$ and exist a sequence $(a_n)_{n \geq 1}$ of real positive numbers, convergent to $0,$ such that

$g'(x) = \lim_{n \to \infty} \frac{f(x + a_n) - f(x)}{a_n}, \forall x \in \mathbb{R}.$
a) Give an example of such a function f that is not differentiable at any point $x \in \mathbb{R}.$

Isn't the Weierstrauss Function an example of such a function f?

b) Show that if $f$ is continuous on $\mathbb{R}$ , then $f$ is differentiable on $\mathbb{R}.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

200 posts

#6 h Apr 19, 2023, 2:57 PM • 1 Y

Y by BlueSyrup

It seems that the solution I gave below doesn't actually work. However I think it's still possible to give a solution based on the density idea.

Wrong solution

This post has been edited 1 time. Last edited by Doru2718, Apr 19, 2023, 5:41 PM
Reason: .

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a