Why is every such function constant: $f(x)=f(x^2)$ for $x \in [0,1]$?

Question

Let $f:[0,1] \to \mathbb{R}$ be a continuous function such that $f(x)=f(x^2)$ for all $x \in [0,1]$.

Any hint/idea for proving that $f$ has to be constant?

Answer 1

By a simple induction we have $$f(x)=f(x^{1/2})=f(x^{1/4})=\cdots=f(x^{1/2^n})$$ and since $f$ is continuous then $$f(x)=\lim_{n\to\infty}f(x^{1/2^n})=f(1)$$ hence $f$ is constant.

Answer 2

Here is the hint. Note that for $a>0$, $\lim_{n\to\infty}a^{1/n}=0$. So for $x>0$ $$ f(x)=f(x^{1/2})=f(x^{1/2^2})=\cdots=f(x^{1/2^n}) $$ and then take the limit as $n\to\infty$ and you can finish the rest.

Answer 3

Hint:

Note that e.g. $$f(1/2)=f(1/4)=f(1/16)=\dots$$ Then make use of the fact that $f$ is continuous. You can do this any element of $[0,1)$.

Answer 4

Note that $f(x) = f(x^2) = f(x^4) = \cdots = f(x^{2^n})$.

If $x \in [0,1)$, we see that $x^{2^n} \to 0$, and by continuity we have $f(x^{2^n}) \to f(0)$, from which we get that $f(x) = f(0)$.

By continuity we also get that $f(1) = \lim_{x \to 1} f(x) = f(0)$.