Continuity problem

Given that $f(x)$ is continuous everywhere, and that $f(x) = f\left( x^{2} \right)$ , prove that $f(x)$ is constant.

#Calculus

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

For $0<x<1$ you have $f(x)=f(x^{(2^n)})=f(0)$ by continuity at 0. For $x>1$ you have $f(x)=f(x^{(1/2^n)})=f(1)$ by continuity at 1 . For negative $x$ you have $f(x)=f(x^2)=f(-x)$ . Finally you have $f(0)=f(1-1/n)=f(1)$ by continuity at 1. (Here, $n$ denotes a positive integer.) Note that all we need is continuity at 0 and 1.

Otto Bretscher - 5 years, 8 months ago

If $f(x)$ is continuous , we can simply take $f(x) = kx$ where k is some real constant.
Given , $\Rightarrow f(x) = f(x^2)$ $\Rightarrow kx = kx^2$ $\Rightarrow x(x-1) = 0$ $x = 0 \ \text{or} \ x = 1$
Therefore, $f(x) = 0 \ \text{or} \ f(x) = 1 = \ \text{constant}$ . Hence Proved

Akhil Bansal - 5 years, 8 months ago

How do we know $f(x)$ is a linear function? How do we even know that it is a polynomial? (We don't.)

Jake Lai - 5 years, 7 months ago

If questions states that $f(x)$ is a continuous function, them we can take $f(x)$ as any continuous function, to prove $f(x)$ continuous..
If you choose any continuous function which is not constant, then $f(x) \neq f(x^2)$

Akhil Bansal - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$