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I am trying to find solutions to the following functional equation:
$$(g(kx))^2=g(x).$$
Here, $x$ is in $mathbb{R}$ and $k$ is a constant. In particular, I'm looking for solutions for $k=2^{-1/4}$. Furthermore, I need that the Fourier inverse of $g$ is a density (i.e., nonnegative and integral over $mathbb{R}$ is 1). The function $g(x)=e^{-x^4}$ satisfies the equation, but does not have a nonnegative Fourier inverse. I have deduced that $g(0)$ must be one, but have little experience with functional equations and am at a loss at how to proceed. (It may very well be the case that there are no other solutions.)

The context is the following. I am tasked with finding (or showing that there exist none) i.i.d. random variables $X$ and $Y$ such that $frac{X+Y}{2^{1/4}}sim X$. If you assume that $X$ and $Y$ have density $f$, then standard Fourier arguments show that $hat{f}=g$ must satisfy the functional equation above. The requirement that the Fourier inverse of $g$ be nonnegative comes from the fact that $f$ is a density.

Any hint, either with the functional equation or the original problem, would be greatly appreciated. In particular, with regards to the original problem, I have been able to deduce that $E[X]$ is either 0 or infinite, and in either case, $E[X^2]$ is infinite, but I am not sure how to proceed to show either existence or non-existence.

Unfinished Attempt: I shall return!

For a fixed $kinmathbb{R}$, I shall find the general solution $g:mathbb{R}tomathbb{R}$ to the functional equation$$big(g(kx)big)^2=g(x)text{ for all }xinmathbb{R},.tag{*}$$
It can be easily seen that $g(0)in{0,1}$. If $k=0$, then it follows immediately that $gequiv 0$ and $gequiv 1$ are the only solutions.

If $k=pm1$, then we get $g(x)in{0,1}$ for all $xinmathbb{R}$ (with the additional requirement that $g$ be an even function in the case $k=-1$). From now on, assume that $knotin{-1,0,+1}$. Clearly, we have $g(x)geq 0$ for every $xinmathbb{R}$.

If $k>1$, then $g$ is completely determined by its behavior on $[+1,+k)$ and $(-k,-1]$ using (*), as well as the exceptional value $epsilon:=g(0)in{0,1}$. Let $g_+:[+1,+k)tomathbb{R}_{geq 0}$ and $g_-:(-k,-1]tomathbb{R}_{geq 0}$ be arbitrary. We have, for each $xinmathbb{R}$,
$$g(x)=begin{cases}sqrt[2^{n(k,x)}]{g_+left(frac{x}{k^{n(k,x)}}right)}&text{if }x>0,,
\epsilon&text{if }x=0,,\
sqrt[2^{n(k,x)}]{g_-left(frac{x}{k^{n(k,x)}}right)}&text{if }x<0,.
end{cases}tag{#}$$
Here, $$n(k,x):=leftlfloorfrac{ln|x|}{ln(k)}rightrfloortext{ for each }xinmathbb{R},.$$

If $0<k<1$, then $g$ is completely determined by its behavior on $(+k,+1]$ and $[-1,-k)$, as well as the value $epsilon:=g(0)in{0,1}$. Let $g_+:(+k,+1]tomathbb{R}_{geq 0}$ and $g_-:[-1,-k)tomathbb{R}_{geq 0}$ be arbitrary. Then, $g$ is also given by (#).

This leaves the case $k<0$ (with $kneq -1$). I still need to look into the Fourier transform condition.