Solve functional equation $f(x)^2 = f(2x)$ if $f$ is not differentiable

Question

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, ${\displaystyle \lim _{x\to -\infty }f(x)=1, \lim _{x\to +\infty }f(x)=0.}$

Answer 1

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.

Answer 2

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.