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Let $f : (-1,1) \rightarrow ℝ $ be a function which is continuous at 0. Suppose that $f(x^2)=f(x) $ for all $x \in (-1,1) $.Prove that $f(x) = f(0)$.

I am not sure how to do this proof.

My attempt: when $x = -1, x^2 = 1$ and when $x=1, x^2 = 1$ so for $f(x^2)=f(x)$, $ x$ has to equal $0$

is this correct?

or you can do this by proof by contradiction?

kenny
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  • $x=\pm1$ are not in the domain of $f$, why even consider them? Instead, let $-1<x<1$ and consider the sequence $x,x^2,(x^2)^2,\dots$. What can you say about its limit? And about the values of $f$ at those points? Now recall the sequential criterion for limits of functions. – Alexander Burstein Nov 28 '17 at 21:33

2 Answers2

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Hint $f(x^{2^n})=f(x)$ and $lim_nx^{2^n}=0$.

Tsemo Aristide
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  • I wonder, if there were an alternate proof by contradiction. This wonderful answer came to my mind: https://math.stackexchange.com/a/35057/174539 Do you a similar argument is usable? If yes, how? – Imago Nov 28 '17 at 21:57
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Notice that for $x \in (-1, 1)$, we also have $x^2 \in (-1, 1)$, so $f(x) = f(x^2) = f(x^4) = \dots =f(x^{2^n})$. So, for any $m$, we can write $$f(x) = f(x^{2^m}) = \lim_{n \to \infty} f(x^{2^n}) = f(0)$$ by continuity at $0$. Hence $f(x) = f(0)$ for all $x \in (-1, 1)$.

Your idea isn't particularly useful, since $\pm 1\notin (-1, 1)$, so we can't use the fact that $f(x^2) = f(x)$ on them: $f$ isn't even defined there.

B. Mehta
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  • i dont understand where you get lim n→∞f(x^2^n)=f(0) , could you explain this step please? – kenny Nov 28 '17 at 21:45
  • That follows from continuity of $f$ at $0$ - if you have a sequence $y_n$ of terms getting closer to $0$, continuity gives that $f(y_n) \to f(0)$. – B. Mehta Nov 28 '17 at 21:47