Function Satisfying $f(x)=f(2x+1)$ If $f: \mathbb{R} \to \mathbb{R}$ is a continuous function and satisfies $f(x)=f(2x+1)$, then its not to hard to show that $f$ is a constant.
My question is suppose $f$ is continuous and it satisfies $f(x)=f(2x+1)$, then can the domain of $f$ be restricted so that $f$ doesn't remain a constant. If yes, then give an example of such a function.
 A: Let $f$ have value $1$ on $[0,\infty)$ and value $0$ on $(-\infty,-1]$. This function is not constant (although it is locally constant), and satisfies $f(x)=f(2x+1)$ whenever $x$ is in its domain. 
A: As in the previous proof of $f$ being constant on $\mathbb{R}$, define $g(x) = f(x-1)$, so that $g(x) = g(2x)$; the domains of $f$ and $g$ are just shifted versions of each other.
Certainly, if the domain of $g$ is small enough, say $[2,3]$, then $g$ can be any continuous function, because the domain contains no $x$ and $2x$ at the same time. A more interesting question is: how large can we make the domain so that $g$ will still not be constant? The answer to this is suggested by JDH's answer: if we remove only the single point 0, making the domain $\mathbb{R} \setminus \\{0\\}$, it is disconnected into two components which can independently have constant values.
How big can a domain be on which $g$ is not even locally constant? Remove an arbitrarily small interval around $0$. Take any non-constant continuous function $h$ which is periodic with unit period, and let $g(x) = h(\log_2 x)$. Then $g(x) = g(2x)$ for all $x$, and is continuous everywhere.
A: $f(x)=f(2x+1)$
$f(2^x-1)=f(2(2^x-1)+1)$
$f(2^x-1)=f(2^{x+1}-1)$
Note that the general solution is $f(x)=\Theta(\log_2(x+1))$ , where $\Theta(x)$ is any periodic function with unit period
So if the domain of $f$ can be restricted to $f:(-1,\infty)\to\mathbb{R}$ , then $f$ can be non-constant and can be $\Theta(\log_2(x+1))$ , where $\Theta(x)$ is any non-constant continuous periodic function with unit period
