# Continuity problem

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

Note by Hobart Pao
2 years, 11 months ago

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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.

- 2 years, 11 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

- 2 years, 11 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.)

- 2 years, 10 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 $$f(x) \neq f(x^2)$$

- 2 years, 10 months ago