to prove that $f(x)=f(0)$

Question

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.

Hint: $f(0.1) = f(0.01) = f(0.0001) = \dots =f(0)$. – Yury Dec 19 '13 at 19:54 — Yury, Dec 19 '13 at 19:54
Cancelled roughly twenty misplaced $. – Did Jun 19 '14 at 12:23 — Did, Jun 19 '14 at 12:23

HorizonsMaths · Answer 1 · 2013-12-20T00:04:50.163

8

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

edited Dec 20 '13 at 00:04

answered Dec 19 '13 at 19:55

HorizonsMaths

16,526

score 7 · Accepted Answer · edited Jun 19 '14 at 11:09

7

$$f(x)=f(x^2)=f(x^4)=\ldots = f(x^{2^n})$$ 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, $$f(x)=f(0)$$

edited Jun 19 '14 at 11:09

Surb

55,662

answered Dec 19 '13 at 20:08

Mathgrad

747
4
13

score 2 · Answer 3 · answered Dec 19 '13 at 19:54

2

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

answered Dec 19 '13 at 19:54

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

to prove that $f(x)=f(0)$

3 Answers3