Solve functional equation $f(x)^2 = f(2x)$ if $f$ is not differentiable I'm trying to solve such task:
Given a random variable , =1. It is known that (>2|>)=(>) ∀>0. How to find ?
During this I came to such functional equation:
$f(x)^2 = f(2x)$ where $f(x) = 1 - F_\xi(x)$
I managed to find that if $f$ is differentiable than $f(x) = e^{-x}$ using derivatives and Taylor series, but what if $f$ is not differentiable?
It's is possible to find same answer by solving functional equation, but I can't prove that it's the only answer.
I also got a suggestion to research this functions in $0$ and in positive semi-neighborhood, but I have no idea how to use it.
Can I get any help in proving that $e^{-x}$ is the only answer when $f(x)$ is not differentiable or proving that it's always differentiable?
EDIT: Please consider facts, that: 
$F_\xi(x)$ is non-decreasing,
${\displaystyle \lim _{x\to -\infty }F_{\xi}(x)=0, \lim _{x\to +\infty }F_{\xi}(x)=1.}$
So $f(x)$ will be non-increasing,
${\displaystyle \lim _{x\to -\infty }f(x)=1, \lim _{x\to +\infty }f(x)=0.}$
 A: You can let $f$ take arbitrary non-negative values Ion $[1,2)$ and then use your functional equation to extend it to $(0,\infty)$.  IISimilarly, extend from $(-2,-1]$ to $(-\infty,0)$, and let $f(0)=0$ or $=1$.  One could, for instance,  take $f|_{(-2,1]\cup[1,2)}=\chi_{\mathbb Q}$ on $(-2,1]\cup[1,2)$ and extend according to the functional equation.
A closed-form expression for such an $f$ is (for $x\ne0$):
$$ 
f(x) = {f\left(\operatorname{sgn}(x)2^{\langle\log_2 |x|\rangle}\right)}^{2^{\lfloor\log_2|x|\rfloor}}\tag{*}$$
from which you can read off values of $f(x)$ for arbitrary non-zero $x$, given the values of $f$ on the set $(-1,-1]\cup[1,2)$, if your eyesight is good enough.
Here $\langle t\rangle $ is $t-\lfloor t \rfloor$ is the fractional part of $t$, taking values in $[0,1)$, so $2^{\langle |x|\rangle}\in[1,2)$.
This much from the functional equation.  The question includes the restriction that $f(x)$ be $1$ minus the cdf of a real random variable $\xi$, with $E[\xi]=1$.  Since $f(0)=0$ or $f(0)=1$, we know that $\xi\le0$ with probability $1$ or $\xi\ge0$ with probability $1$; since $E[\xi]=1$ we know the second possibility holds.    Here is a method of producing discontinuous solutions of the functional equation that obey these constraints.  Let $f(x)=1$ for all $x\le 0$, let $f(x)=\alpha$ on $[1,2)$ for some $0\le\alpha1$ and apply the (*) recipe to extend $f$ to all of $\mathbb R$.  The resulting $f$ is the upper-tail cdf for a positive random variable $T$ , whose expectation $E[T]$ might not equal $1$.  Then $\xi=T/E[T]$ is a discrete random variable obeying all the desired conditions, and $P(T>x)=f(\lambda x)$ for some value of $\lambda$ related to $E[T]$.  More generally, all solutions can be produced as follows: let $f(x)=1$ for $x\le1$ $f(x)$ be cadlag and non-increasing  on $[1,2]$ in such a way that $0\le f(2)\le f(1)^2\le1$, and repeat this construction.
A: the function $f(x)=e^{-x}$ is not the only solution to the functional equation
$$
f(x)^2 = f(2x)$.
$$
The function $f(x)=0$ is another solution for this functional equation.
Since you ask about nondifferentiable $f$, the nondifferentiable
function
$$
f(x) = 
\begin{cases}
  1&\quad\text{if}\quad x>0
  \\
  0&\quad\text{if}\quad x\leq0
\end{cases}
$$
also solves the functional equation.
