# 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.

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If anyone who wants to see as to why $f$ remains constant on $\mathbb{R}$ then can refer this link: mathoverflow.net/questions/31990/… – anonymous Aug 12 '10 at 10:51

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.

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Hey i think you have made the $2x+1$ condition redundant. I think your function will work for even $f(x)=f(3x+1)$ – anonymous Aug 12 '10 at 12:08
How about $x = -1$? – j.p. Aug 12 '10 at 12:58
Jug, if $x=-1$, then $2x+1=-1$ also, so there is no problem at all there. – JDH Aug 12 '10 at 22:48
Upps. Makes me wonder if my brain still works... – j.p. Aug 13 '10 at 9:56

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.

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$f(x)=f(2x+1)$

$f(2^x-1)=f(2(2^x-1)+1)$

$f(2^x-1)=f(2^{x+1}-1)$

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

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