Functions satisfying $f(x^2) = f(x)$ Let $f : [0, 1] \to \Bbb{R}$ be a continuous function such that $f(x^2) = f(x)$ for all $x \in[0, 1]$. 
Which one of the following is not true in general? 
A) $f$ is constant
B) $f$ is uniformly continuous 
C) $f$ is differentiable 
D) $f(x) \ge 0 \forall x \in[0, 1]$


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*If I take $f(x) = -C$ ($C$ is positive constant) then clearly it shows that option D is not true in general. 


But I want some theoretical approach to this question if there is any. 
Like is there anything special about the functions satisfying $f(x^2) = f(x)$? 
 A: Ok I have revised my answer to make sure $f$ is continuous
I believe that $f$ must be constant because we are working on the closed interval $[0,1]$. Start with some point $x\in(0,1)$ and take the sequence of points ${x, x^2, x^4, ...}$. Notice $f(x) = f(x^2) = f(x^4) = ...$
Taking the limit of our sequence $\lim_{n\rightarrow \infty} x_n = 0$. This means that $f(0) = \lim_{n\rightarrow\infty} f(x_n)$. Since $f$ is continuous we know that all possible sequences of points must converge to the same limit at $0$. As such, no matter what our initial $x$ value is, we must always have the same sequence of function values. 
Therefore $f$ must be constant.
A: I would go case-by-case and ask the simple question: "Does this make sense?"
For A), consider that if $f$ is constant then it shouldn't matter what argument it takes - $f(x)=f(y)$ for any $x$ and $y$ in the domain. Suppose that $x^2=y$ - are there any points in the domain where this is true?
For B), if A) or C) is true, then what does this say about the continuity of $f$? Is a constant function uniformly continuous? Is a differentiable function uniformly continuous?
For C), if A) is true or B) is true, then what does this say about the differentiability of $f$? Is a constant function necessarily differentiable? Is a uniformly continuous function differentiable?
