Theorem 1.16 of "An Introduction to Difference Equations" by Saber Elaydi Consider the difference equations
$$x(k+1) = f(x(k)) \qquad (1)$$
and
$$y(k+1) = g(y(k)) \qquad (2)$$
where $g = f \circ f$.
In An Introduction to Difference Equations (3e) by Saber Elaydi, in the proof of Theorem 1.16 (on p.32), it is mentioned that for $a$ an equilibrium of (1), if $a$ is an asymptotically stable equilibrium of (2), then it is also an asymptotically stable equilibrium for (1).
How can we prove this?
My progress so far

*

*Since $g(x) = f(f(x))$,

*

*$g'(x) = f'(f(x))f'(x)$

*$g''(x) = f''(f(x)) [f'(x)]^2 + f'(f(x)) f''(x)$

*$g'''(x) = f'''(f(x)) f'(x) [f'(x)]^2 + 2f''(f(x)) f'(x) f''(x) + f''(f(x)) f'(x) f''(x) + f'(f(x))f'''(x)$



*Assume $a$ is an asymptotically stable equilibrium for (2). There are two cases

*

*$|g'(a)| < 1$

*$|g'(a)| = 1$, $g''(a) = 0$, and $g'''(a) < 0$



*Note that $g'(a) = f'(f(a))f'(a) = [f'(a)]^2$

*Hence when $|g'(a)| < 1$, $|f'(a)| < 1$ as well. In this case $a$ is an asymptotically stable equilibrium of (1)

*Now assume $|g'(a)| = 1$,  $g''(a) = 0$, and $g'''(a) < 0$.

*Since $g'(a) = [f'(a)]^2 \geq 0$, $g'(a) = 1$. There are two cases

*

*$f'(a) = 1$.

*$f'(a) = -1$.



*Assume $f'(a) = 1$. In this case

*

*$g''(a) = f''(f(a))[f'(a)]^2 + f'(f(a)) f''(a) = 2f''(a) = 0$. Hence $f''(a) = 0$

*$g'''(a) = f'''(f(a)) f'(a) [f'(a)]^2 + 2f''(f(a)) f'(a) f''(a) + f''(f(a)) f'(a) f''(a) + f'(f(a))f'''(a) = 2f'''(a) < 0$. Hence $f'''(a) < 0$

*It follows that $a$ is an asymptotically stable equilibrium of (2)



*My question is how can we proceed when $f'(a) = -1$

*

*Theorem 1.16 states that if $f'(a) = -1$ and $-f'''(a) - 3/2(f''(a))^2 < 0$, then $a$ is asymptotically stable.

*But since this proposition is used to prove Theorem 1.16, using Theorem 1.16 to prove this proposition would be circular.



 A: Hint: Whether or not a point is an asymptotically stable equilibrium point is a purely topological notion, and in this case using the definitions will suffice (note that if $f$ is continuous, then so is $g=f\circ f$). As a heuristic it might be useful to think of $g$ as a "time change" of $f$; under the action of $g$ everything that happens under the action of $f$ happens twice as fast (and vice versa: under the action of $f$ everything that happens under the action of $g$ happens twice as slow).
A: Here is a proof based on @alp-uzman's hint
Let $a$ be an equilibrium for (1).

*

*Assume $a$ is asymptotically stable w.r.t. (2).

*By definition of stability, for all $\epsilon_1 > 0$, there exists $\delta_1 > 0$ such that $|y_0 - a| < \delta_1$ implies $|g^n(y_0) - a| < \epsilon_1$ for $n > 0$ (or equivalently, $|f^{2n}(y_0) - a| < \epsilon_1$)

*In other words, $\lim_{x\to a} g^n(x) = g^n(a) = a$ for $n > 0$

*Since $f$ is continuous, $\lim_{x\to a} f(x) = f(a)$

*In other words, for all $\epsilon_2 > 0$, there exists $\delta_2 > 0$ such that $|x - a| < \delta_2$ implies $|f(x) - f(a)| < \epsilon_2$

*Now let $\epsilon > 0$ be arbitrary. Pick $\delta = \min(\delta_1(\epsilon), \delta_2(\epsilon), \delta_2(\delta_1(\epsilon)))$.

*It follows that

*

*$|x_0 - a| < \delta \leq \delta_1(\epsilon)$ implies $|f^{2n}(x_0) - a| < \epsilon$ for $n > 0$

*$|x_0 - a| < \delta \leq \delta_2(\epsilon)$ implies $|x_1 - a| < \epsilon$

*$|x_0 - a| < \delta \leq \delta_2(\delta_1(\epsilon))$ implies $|x_1 - a| < \delta_1(\epsilon)$, which again implies that $|f^{2n}(x_1) - a| < \epsilon$ for $n > 0$



*To summarize, $|x_0 - a| < \delta$ implies $|f^n(x_0) - a| < \epsilon$ for $n > 0$. Hence $a$ is a stable equilibrium of (1)

*Proving $a$ is attracting w.r.t. (1) should be straightforward.

