# How to prove this function is actually constant? [duplicate]

Let $f$ be function $\:f:\mathbb{R}\rightarrow \mathbb{R}$ which for any $x$, $f\left(x\right)=f\left(\frac{x}{2}\right)$. Also, $f$ continuous at $x=0$.
Show that $f$ is constant function.

So far i saw something here Does $f(x) = f(2x)$ for all real $x$, imply that $f(x)$ is a constant function?, but i don't understand the solution that he was given. I thought about assuming that there exist $a,b$ such that $f\left(a\right)\ne f\left(b\right)$ and get a contradict.

Any ideas about what give me that $f$ is continuous at $0$? and any ideas about a way to prove that? tnx in advance!

• @ADG first, i wrote in my question that i look at this. second, look for "Also" that i wrote and find the difference between the questions. – user2637293 Feb 19 '15 at 15:35
• the only difference is that you can't understand that solution, better comment under the answer your confusion? – RE60K Feb 19 '15 at 15:38
• @ADG there, f is continuous in every point. here just in 0. – user2637293 Feb 19 '15 at 15:39
• that doesn't make a difference, we are only concerned about x=0. – RE60K Feb 19 '15 at 15:40

$f(x)=f(\frac{x}{2})=f(\frac{x}{4})=f(\frac{x}{8})=...=f(\frac{x}{2^n})$.
• tnx @Sloan. i figured this steps you wrote. i just don't understand the conclusion , "so f(x)=limn→∞f(x/2n)=f(0) by continuity.". why it derives from continuity? and it derives from continuity in every point or just in $0$? – user2637293 Feb 19 '15 at 15:48
• As the others have pointed out, $x=0$ is the sole point of interest; $f$ should be constantly $f(0)$ as $f(x)=f(\frac{x}{2^n})$ for arbitrarily large values of $n$ and for any $x$. – Sloan Feb 19 '15 at 15:55