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Let $f:R\rightarrow R$ be a continuous function. There exists a $p>0$, s.t. $f(px)=f(x),p\neq1$ for all $x\in R$.

Can we prove that $f(x)$ is a constant function?

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    $\begingroup$ What about $p=1$? $\endgroup$ – John Douma Jul 28 '16 at 14:27
  • $\begingroup$ @JohnDouma Oh, I mean $p\neq 1$. Thank you for pointing out. $\endgroup$ – 1024 Jul 28 '16 at 14:31
  • $\begingroup$ Sorry I missed that it has to be contiuous $\endgroup$ – Gregory Grant Jul 28 '16 at 14:35
  • $\begingroup$ Wlog p<1. Otherwise $f(x)=f(\frac{1}{p}x)$. I think you can then establish that it's constant on $[0, \infty]$ . There's a lemma in real analysis which says that if $a_n \to a$ and $f$ is continuous then $f(a_n) \to f(a)$. So, you can consider the sequence $p,p^2,...$. But now repeating this for the same sequence scaled by some arbitrary factor (i.e $ ap, ap^2,ap^3,...$) and since limits are unique, $f(1)=f(a)$ and so on. $\endgroup$ – daruma Jul 28 '16 at 14:40
  • $\begingroup$ And repeat for [-\infty, 0] $\endgroup$ – daruma Jul 28 '16 at 14:41
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Hint: Consider separately the cases $p<1$ and $p>1$. For the case that $p<1$, note that $$f(p^nx)=f(x).$$ What happens if you pass to the limit as $n \rightarrow \infty$? Can you repeat this idea for $p>1$? Hope this helps!

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  • $\begingroup$ Great - glad I could help. $\endgroup$ – Zestylemonzi Jul 28 '16 at 14:39
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Yes if $|p| \not = 1$.

Suppose $|p|\not = 1$. If $p=0$, we're done. If $p\not= 0$, without loss of generality (because we can replace $p$ by $\frac{1}{p}$), we can assume that $0<|p|<1$. Take $x\in \mathbb R$. $p^nx \to 0$ when $n\to +\infty$ so since $f$ is continuous $f(p^nx)\to f(0)$ when $n\to +\infty$. But by our hypothesis, the sequence $f(p^nx)$ is constant (because $f(p^nx)=f(p^{n-1}x)=\dots=f(px)=f(x)$) so that $f(p^nx)=f(0)$ for any $n$ and in particular $f(x)=f(0)$.

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