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 $\mathbb{R}.$
Problem
Source: Romania National Olympiad 2023
Tags: real analysis, differentiable function, Continous function
MathsLover04
18.04.2023 10:57
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.
Cristisav
18.04.2023 22:16
Congrats !
Doru2718
18.04.2023 22:59
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.
jayevvms123
19.04.2023 01:27
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 $\mathbb{R}.$
Doru2718
19.04.2023 17:57
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.
Replacing $f$ with $f-g$ we can assume wlog $g=0$. Denote $f_n(x)=\frac{f(x+a_n)-f(x)}{a_n}$. Fix an interval $[a,b]$, let $\varepsilon>0$ and consider the sets $A_N=\{x\in [a,b]: |f_n(x)|<\varepsilon\text{ for all }n\geq N\}$. As $f_N$ are continuous, $A_N$ are open and by the hypothesis $\bigcup_{n}A_n$ covers $[a,b]$. By Heine-Borel, there exists a finite subcover. Hence there exists $N\in\mathbb{N}$ s.t. for all $n\geq N$ and $x\in[a,b]$ we have $|f(x+a_n)-f(x)|<\varepsilon a_n$. Discarting of the first terms of $(a_n)_n$ allows us to assume $N=0$. Summing up $m$ such inequalities we have $|f(x+ma_n)-f(x)|<\varepsilon ma_n$. Using the fact that $(ma_n)_{m,n}$ is dense in $[0,\infty)$ we get $|f(y)-f(x)|<\varepsilon|y-x|$ for all $x,y\in[a,b]$. We are done.