Prove that $\boldsymbol{\ f}$ is a constant function

Question

if :$$f:[0,1]→\mathbb{R}$$ and for all $x\in[0,1]$ $$f(x)=f\!\left(x^2\right)$$ and for all $b\in[0,1]$ $$ \lim_{ x \to b }f(x)=f(b)$$

prove : for some $c\in\mathbb{R}$, for all $x\in[0,1]$, $f(x)=c$

Answer 1

Hint: For any $x\in(0,1)$, $$ \lim_{n\to\infty}x^{2^n}=0 $$ Show that for $x\in(0,1)$, $$ \lim_{n\to\infty}\,f\!\left(x^{2^n}\right)=f(0) $$ and $$ \lim_{n\to\infty}\,f\!\left(x^{2^n}\right)=f(x) $$ Show that $$ \lim_{x\to1}f(x)=f(1) $$

Answer 2

Letting x be an arbitrary positive and less than 1, f(x)=f(x^2)=f(x^4)... The arguments converge to 0 so the function value must approach f(0). But this is then f(x) by continuity. But x was arbitrary so f is constant on [0,1). Similarly f(1)=f(x).

Answer 3

The function $x\to x^2$ is a bijection of $[0,1]$ and, for instance, $$\lim_{ x \to \frac 12}f(x)=f\left(\frac 12\right)\text{ and } \lim_{ x \to \frac 14 }f(x)=f\left(\frac 14\right)$$ then $$f\left(\frac 12\right)=f\left(\frac 14\right)$$ Hence for all $n$ $$f\left(\frac 12\right)=f\left(\frac{1}{2^n}\right)$$ Similarly for all $n$$$f\left(\frac 1x\right)=f\left(\frac{1}{x^{2^n}}\right)$$