# 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.

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?

• You mean "continuous at 0".
– user14972
Dec 29, 2014 at 10:56

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.

• what does "$f(a\,2^{-n})$" simplify? Dec 29, 2014 at 10:46
• @FirasAliAbdelGhani $a 2^{-n}$ converges to $0$, so $f(a\,2^{-n})$ converges to $f(0)$. The same for $b$. Dec 29, 2014 at 10:46
• $2^{-n}$ simplify the values that can be inserted in the function? Dec 29, 2014 at 10:54
• @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. Dec 29, 2014 at 10:56
• I got it at the end:) thank you. and next time I will write down my thoughts so it will be better coordinated question. Dec 29, 2014 at 11:07

The functional equation implies $f\Bigl(\dfrac{x}{2}\Bigr)=f(x)$, hence $f(x)=f\Bigl(\dfrac{x}{2^n}\Bigr)$ for all $n$.