Consider $f : (0 , 1) \rightarrow [0,1]$ be a continous function . Which of the followings are true ?

  1. $f$ is uniformly continous.

  2. $f$ has a fixed point.

  3. $f$ is a differentiable function.

  4. There is a continous function $\overline f$ on $[0,1]$ such that $\overline f = f $ on $(0,1)$

I have tried :

1) suppose $f(x) =\frac{1}{x}$ is continous but not uniformly continous

3) Suppose $f(x) = |x-\frac{1}{2} |$ which is continous but not differentiable.

4) If $f$ is a contnous function on $ E \subset \mathbb R$, then there exist a continous function $g$ on $\mathbb R$ such that $f = g$ on $E$ if $E$ is closed on $\mathbb R$. I think we can use this result, but i have no counter example.

Please tell me counter example and tell me about (2) option.

Thank you


1 Answer 1


$f(x) = 1/x$ is not a valid counter-example for (1) because it does not map $(0,1)$ into $[0, 1]$. But you can choose $f(x) = \sin(1/x)$ instead. That also serves as counterexample for (4) because $\lim_{x \to 0} f(x)$ does not exist.

$f(x) = x/2$ is a counterexample for (2). (This works only because $f$ is defined on an open interval. A continuous function $F: [0,1] \to [0,1]$ necessarily has a fixed-point, this follows from the intermediate value theorem.)

  • $\begingroup$ Thanks Martin for your prompt reply $\endgroup$
    – Struggler
    May 8, 2015 at 8:23
  • $\begingroup$ @Struggler: You are welcome! Btw., is there a reason why you never have accepted any of the answers given to your questions? If an answer is helpful, you can accept it by clicking on the check mark. That marks the problem as solved, and gives some reputation points to you and to the author of the answer. $\endgroup$
    – Martin R
    May 8, 2015 at 10:33

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