0
$\begingroup$

Let, $f:(-1,1)\rightarrow\mathbb{R}$ be a function continuous at $x=0$ and given that $f(x)=f(x^2)$ for all $x\in(-1,1)$. Prove that, $f(x)=f(0)$ $\forall x\in(-1,1)$. Ok. First give me some hint.

$\endgroup$
  • 2
    $\begingroup$ Hint first huh? $\endgroup$ – Kaster Dec 19 '13 at 19:51
  • $\begingroup$ what did u say? @Kaster $\endgroup$ – Topology Dec 19 '13 at 19:53
  • 1
    $\begingroup$ Hint: Suppose not. $\endgroup$ – Aeryk Dec 19 '13 at 19:53
  • 6
    $\begingroup$ Hint: $f(0.1) = f(0.01) = f(0.0001) = \dots =f(0)$. $\endgroup$ – Yury Dec 19 '13 at 19:54
  • $\begingroup$ Cancelled roughly twenty misplaced $. $\endgroup$ – Did Jun 19 '14 at 12:23
7
$\begingroup$

$$f(x)=f(x^2)=f(x^4)=\ldots = f(x^{2^n})$$ where $n$ is any natural number

So if we let $n$ be infinitely large then, since $x\in (−1,1)$, $x^{2^n}$ tends to $0$

Hence, $$f(x)=f(0)$$

$\endgroup$
8
$\begingroup$

By induction we have $$ f(x)=f(x^{2^n}) \quad \forall x \in (-1,1), \quad \forall n \in \mathbb{N}. $$ Thus $$ f(x)=\lim_nf(x^{2^n})=f(0) \quad \forall x \in (-1,1). $$

$\endgroup$
2
$\begingroup$

Hint: This can be done by contradiction, with a sequential characterization of continuity.

$\endgroup$
  • $\begingroup$ so in contrary there r two cases. right? first $f(x)$>$f(0)$? like this? $\endgroup$ – Topology Dec 19 '13 at 19:55
  • $\begingroup$ @Mathematics There's no reason to separate into cases. $\endgroup$ – user61527 Dec 19 '13 at 19:57
  • $\begingroup$ So how shall we proceed? @T.Bongers $\endgroup$ – Topology Dec 19 '13 at 19:57
  • $\begingroup$ @Mathematics Think about the hint for a while. $\endgroup$ – user61527 Dec 19 '13 at 19:58
  • $\begingroup$ Ok..Ok @T.Bongers $\endgroup$ – Topology Dec 19 '13 at 19:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.