0
$\begingroup$

I was doing a homework assignment, the problem is the next:

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a $C^{1}$ function such that there is no $x\in{\mathbb{R}}$ with $f(x)=f'(x)=0$. Show that the set $$S=\lbrace{ x \in{[0,1]}: f(x)=0} \rbrace $$ is finite.

I already prove it, but i was trying to find an example of a function $f$ that is no $C^{1}$, I mean at least differentiable in $[0,1]$ but not continuously differentiable and such that $f(x)=f'(x)=0$ but S is not a finite set. Any ideas?

$\endgroup$
1
  • $\begingroup$ Thanks, but is not working because is not continuous in the whole [0,1]. Maybe you missunderstood the problem. $\endgroup$ Sep 28, 2014 at 4:35

1 Answer 1

0
$\begingroup$

Example of differentiable function which is not $C^1$: consider: $$f_1(x) = \begin{cases} x \sin \frac{1}{x}, \mbox{if $x \neq 0$} \\ 0, \mbox{ if $x = 0$}\end{cases}$$ Clearly $f_1$ is continuous at zero. But $f$ is not differentiable at zero, notice that the limit $$\lim_{h \to 0} \frac{h \sin \frac{1}{h}}{h} = \lim_{h \to 0} \sin \frac{1}{h}$$ does not exists. Then, try: $$f_2(x) = \begin{cases} x^2 \sin \frac{1}{x}, \mbox{if $x \neq 0$} \\ 0, \mbox{ if $x = 0$}\end{cases}$$ Again, $f_2$ is continuous at zero. But it is differentiable at $0$ (check!). Is the derivative continuous? If yes, good. If not, try:

$$f_3(x) = \begin{cases} x^3 \sin \frac{1}{x}, \mbox{if $x \neq 0$} \\ 0, \mbox{ if $x = 0$}\end{cases}$$

Go on. What can you say about $f_n$, based on $n$?

$\endgroup$
1
  • $\begingroup$ Thanks, I was thinking just in the first two by separate, but as a sequence of functions it really seems to work. $\endgroup$ Sep 28, 2014 at 4:43

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .