Find all functions $f$ satisfying the property that $$ f(x) = f(x/2) $$ for all $x \in \mathbb{R}$
So far I've come up with the following assumptions:
-$f$ is periodic, i.e of form $f(x) = A \cos(g(x))$
-$O(g(x)) = \log_2x$
I've hit a dead end though. Are my assumptions correct? What else should I try?
EDIT: Here's the motivation behind the periodicity:
On
Let $X_n$ denote every interval in the form $[2^n, 2^{n+1})$ (Which means $\cup_{n \in \mathbb{N}} X_n =\mathbb{R}_+^*$ and $X_n \cap X_m = \emptyset$ whenever $n \neq m$). Further, define the equivalence relation $\sim$ by $x \sim y\ \Leftrightarrow \ x=2^n y$, for some $n \in \mathbb{Z}$. Now, take any function $g:X/\sim\ \rightarrow \mathbb{R}$, and define $f_1:\mathbb{R}_+^* \rightarrow \mathbb{R}$ by $$f_1(x):=g([x])$$ ($[x]:=\{2^n x \in \mathbb{R} | n\in \mathbb{Z}\}$). That the above implies the initial condition is trivial (since $x \sim x/2$). If $f(x)=f(x/2)$ for every $x$, then $f(x)=f(2^n x)$, for every $n \in \mathbb{Z}$, which means the value of $f(x)$ depends entirely on $[x]$ i.e. $f(x)=g([x])$, for some $g$.
Define $f_2$ analogously for the negative numbers, choose any number $k \in \mathbb{R}$ and impose $f_3(0)=k$. Then $$f=f_1 \cup f_2 \cup f_3$$ is clearly the most general solution.
As stated in the other answer, continuity implies $f=constant$
Hint $f(x)=f({x\over 2})=f({x\over 2^2})=\dots=f({x\over 2^n})$
If $f$ is continuous then $f(x)=f(0)$