Functions satisfying $f(x^2) = f(x)$

Question

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) $f(x) \ge 0 \forall x \in[0, 1]$

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 $f(x^2) = f(x)$?

If you want to deduce that $f$ has to be constant consider $f(x) = f(x^2) = f(x^4) = \ldots = f(x^{2^n})$. What happens with $x^{2^n}$ when $n\to\infty$ when $x\in[0,1)$? — Winther, Jun 26 '19 at 20:17
Also: https://math.stackexchange.com/q/1719910/42969, https://math.stackexchange.com/q/2010453/42969 — Martin R, Jun 26 '19 at 20:20
For any $x\in[0,1]$ and $n$ an integer, we have $f\left(x^{2^n}\right)=f(x).$ When $x\in(0,1)$, then $\lim_{n\to 0} x^{2^n}\to 0.$ — Thomas Andrews, Jun 26 '19 at 20:21

score 1 · Answer 1 · edited Jun 27 '19 at 18:55

1

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.

edited Jun 27 '19 at 18:55

Thomas Andrews

177,126

answered Jun 26 '19 at 20:20

wjmccann

3,085

It should be $f\left(x^{2^n}\right)=f\left(x^{2^{-n}}\right)=k.$ – Thomas Andrews Jun 26 '19 at 20:23
@Winther well that certainly explains why this felt much to easy! This is why I should read more carefully when I do problems :/ – wjmccann Jun 26 '19 at 20:29
@Winther I fixed my answer, hopefully it should be good now! – wjmccann Jun 26 '19 at 20:41
Thank u so much.. I got it. – user684646 Jun 28 '19 at 17:54
@user684646 If this answered your question try marking it as the answer so anyone who finds this in the future can see that it worked – wjmccann Jul 01 '19 at 18:53

score 0 · Answer 2 · answered Jun 26 '19 at 20:18

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?

Functions satisfying $f(x^2) = f(x)$

2 Answers2