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How do I prove that if $f: \Bbb{R}\rightarrow\Bbb{R}$ is continuous at $0$ and $$f(x)=f(2x)$$ for each $x\in\Bbb{R}$, then $f$ is constant?

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  • $\begingroup$ You mean "continuous at 0". $\endgroup$ – user14972 Dec 29 '14 at 10:56
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If $f(a)\neq f(b)$, then $f(a\,2^{-n})=f(a)$ and $f(b\,2^{-n})=f(b)$ for each $n$. By continuity, $f(0)=f(a)=f(b)$ which is a contradiction.

It is usually appreciated if you also write down your ideas or trials.

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  • $\begingroup$ what does "$f(a\,2^{-n})$" simplify? $\endgroup$ – Firas Ali Abdel Ghani Dec 29 '14 at 10:46
  • $\begingroup$ @FirasAliAbdelGhani $a 2^{-n}$ converges to $0$, so $f(a\,2^{-n})$ converges to $f(0)$. The same for $b$. $\endgroup$ – Peter Franek Dec 29 '14 at 10:46
  • $\begingroup$ $2^{-n}$ simplify the values that can be inserted in the function? $\endgroup$ – Firas Ali Abdel Ghani Dec 29 '14 at 10:54
  • $\begingroup$ @FirasAliAbdelGhani Unfortunately, I don't understand what you don't understand. If you know what is a limit, continuity and $2^{-n}$, then you should probably try to specify more in your question, where is your problem and what are your ideas, then the community could help you more. $\endgroup$ – Peter Franek Dec 29 '14 at 10:56
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    $\begingroup$ I got it at the end:) thank you. and next time I will write down my thoughts so it will be better coordinated question. $\endgroup$ – Firas Ali Abdel Ghani Dec 29 '14 at 11:07
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The functional equation implies $f\Bigl(\dfrac{x}{2}\Bigr)=f(x)$, hence $f(x)=f\Bigl(\dfrac{x}{2^n}\Bigr)$ for all $n$.

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