I have been struggling with a problem for a long time. I have boiled it down to finding a fixed point of a mapping in a specific set.
Let $p:\mathbb{R}\to\mathbb{R}_{>0}$ be an analytic function with $\int_{-\infty}^{\infty}p(x)~d x=1$. Let also $\beta\geq0$. The mapping is defined by $$M[f]=\frac{p}{1+\beta~(f'/p)' }$$ on the set $\{f\in C^\omega:1+\beta~(f'/p)'>0\}$, where $C^\omega$ is the set of analytic functions over $\mathbb{R}$ and $>$ is point-wise.
Let $\mathcal{F}$ be the set of analytic functions $f:\mathbb{R}\to\mathbb{R}_{>0}$ with the property that $\lim_{x\to+\infty}f(x)=\lim_{x\to-\infty}f(x)$. I want to find conditions under which $M$ has a fixed point in $\mathcal{F}$.
What I know so far
Clearly, $p\in\mathcal{F}$. So, if $\beta=0$, $p$ is the unique fixed point of $M$ and it is in $\mathcal{F}$. The difficult case is when $\beta>0$.
Trying to find other fixed points, I tried plugging in values. So, say $f=kp$ for some $k>0$. Then, I get the condition that $p$ has to satisfy $$(\log{p})''=\frac{k(k-1)}{\beta}$$ which means that $p$ is Gaussian. So, if $p$ is Gaussian with variance $\sigma^2$ and $\beta<\sigma^2/4$, I can find a $k\in(0,1)$ for which $M$ has a fixed point in $\mathcal{F}$.
I suspect that for ``small enough'' values of $\beta$, $M$ should have a fixed point in $\mathcal{F}$ but I want to derive conditions in the form of some upper bound without requiring a particular form for $p$. For example, I would like to say that if $\beta< \bar\beta$ (where $\bar\beta$ is some function of $p$), then $M$ has a fixed point in $\mathcal{F}$.
I worked towards applying $M$ iteratively and began with $p$ as my initial condition. Clearly, if we require $M[p]\in\mathcal{F}$, it has to be that $$(\log p)''>-\frac{1}{\beta}.$$ I understand that showing that if $M^n[p]\in\mathcal{F}$ then $M^{n+1}[p]\in\mathcal{F}$ would be desirable but I'm not sure it's true.