Neccesary condition for a family of functions to have a period-doubling bifurcation I have the following definition of period-doubling bifurcation for a one-parameter family of functions $\left\{{ 
 F_{\lambda}  }\right\}$ :
Definition. A family of functions $\left\{{ 
  F_{\lambda}  }\right\}$  undergoes a period-doubling bifurcation at the parameter value $\lambda=\lambda_{0}$ if there exist an open interval $I$ and an $\epsilon >0$ such that:

*

*For each $\lambda \in [\lambda_{0}-\epsilon,\lambda_{0}+\epsilon]$, there exist a unique fixed point $p_{\lambda}$ for $F_{\lambda}$ in $I$.


*For $\lambda \in (\lambda_{0}-\epsilon, \lambda_{0})$ , $F_{\lambda}$ has not cycles of period $2$ in $I$ and $p_{\lambda}$ is attracting(resp. repelling).


*For $\lambda \in (\lambda_{0},\lambda_{0}+\epsilon)$, there exist a unique 2-cycle  $q_{\lambda}^1,q_{\lambda}^2$ in $I$ with $F_{\lambda}(q_{\lambda}^1)=q_{\lambda}^2$. This 2-cycle is attracting (resp., repelling). Meanwhile, the fixed point $p_{\lambda}$ is repelling(resp., attracting).


*As $\lambda \rightarrow{} \lambda_{0}$, $q_{\lambda}^i \rightarrow p_{\lambda_{0}}$.
If the above conditions are valid with the symbol > instead of <, we also say that the family have a period-doubling bifurcation. Althought this is certainly a complicated definition, there exist simple examples of this kind of bifurcation. For instance, the family $F_{c}(x)=x^2+c$ has a period-doubling bifurcation at $c=-3/4$, in which case, we can take $\epsilon=1$ and $I=(\infty,-1/2)$.
My question is: why this definition implies that $F_{\lambda_{0}}'(p_{\lambda_{0}})=-1$?. I mean, it's clear for me that there is only two possibilities: $1$ or $-1$. But I still don't see why we can discard the first case. Can anyone explain me please?.
Thank you in advance.
 A: First off, it's pretty easy to see (in the context of the quadratic family) why $F_{\lambda}'(p_{\lambda})=1$ doesn't guarantee period doubling. If you graph
$$F_{\lambda}(x) = x^2 + \lambda$$
for $\lambda$ in a neighborhood of $1/4$, there simply is no period-doubling bifurcation since, for $\lambda>1/4$, there is no fixed point to double:

To be clear, though, there is a bifurcation - just one of a different type. As $\lambda$ increases through $1/4$, two fixed points merge and annihilate each other. This type of bifurcation is called a saddle-node bifurcation. I guess this highlights the true importance of the condition that $|F_{\lambda}'(p)|=1$; that's where some type of bifurcation can occur but there's no immediate guarantee as to which type.

To answer your question as to why period doubling occurs only when the derivative is $-1$, assume that $x_1$ and $x_2$ form a 2-cycle for $F_{\lambda}$ and assume (WLOG) that $x_1<x_2$. Then, over the interval $I=[x_1,x_2]$, the average rate of change of $F_{\lambda}$ is
$$\frac{F_{\lambda}(x_2)-F_{\lambda}(x_1)}{x_2-x_1} = \frac{x_1-x_2}{x_2-x_1} = -1.$$
By the mean value theorem, there is a point $c\in(x_1,x_2)$ such that $F_{\lambda}'(c)=-1$. But, in the context of your definition, you have an $\varepsilon$ decreasing to zero which forces the $x_1$ and $x_2$ to collapse on the fixed point. Thus, $F_{\lambda}(p_\lambda)=-1$.
