I'm working through "An Introduction to Difference Equations" by Elaydi. The author claims that a certain property of equilibrium points is inherited by the equilibrium points of a related difference equation. The proof is left as an exercise, with which I'm struggling.
We begin with the difference equation: $$ x(n+1) = f(x(n)) $$ where $n$ is some non-negative integer, and $x$ and $f$ are functions to real numbers. We are told that $x^*$ is an equilibrium point of this equation, so that $f(x^*) = x^*$.
A second difference equation is then introduced: $$ y(n+1) = g(y(n)) \textrm{ where } g(y(n)) = f^2(y(n)), $$ where $y$ is a function to the real numbers.
It is then claimed that if $x^*$ is an equilibrium point of the first difference equation, then it is an equilibrium point of the second equation. This is easy to show, as $f(x^*) = x^* \implies g(x^*) = f(f(x^*)) = f(x^*) = x^*$.
However, it is then claimed that if $x^*$ is a stable equilibrium point of the second equation, then it is also a stable equilibrium point of the first equation. An equilibrium point is said to be stable if, for any given $\epsilon > 0$, there exists a $\delta > 0$ so that for all $x$ such that $|x - x^*| < \delta$, then $|f^n(x) - x^*| < \epsilon$ for all $n$. Intuitively, points close enough to a stable equilibrium point stay within a specified radius of that equilibrium point for all time.
The stability of the equilibrium point with respect to the second equation implies that nearby points stay close enough to the equilibrium point for every even number of applications of the update function $f$. However, I fail to see why this should imply that points near to the equilibrium point should stay near when we apply $f$ an odd number of times.
The author is also assuming that $f'(x^*) = -1$, and that $f$ has continuous derivatives existing up to the third order.
Any insight is appreciated.
Edit
I've made some progress. Roughly speaking, if every $x$ is in the image of $f$, then we can write a given $x$ as both $x = f(x_0)$ and as $x = f^0(x)$ for some $x_0$. The orbit of $x$ under $f^2$ is assured to behave in a way consistent with a stable equilibrium point $x^*$, and so is the orbit of $x_0$. The union of these two orbits contains the orbit of $x$ under $f$, which therefore must also be consistent with a stable equilibrium point $x^*$.
To formalize this, we let $S_\epsilon = (x^* - \delta_\epsilon, x^* + \delta_\epsilon)$, where $\delta_\epsilon$ is defined so that if $|x - x^*| < \delta_\epsilon$ then $|g^n(x) - x^*| < \epsilon$ for all $n$. We think of this as an "$\epsilon$-stable region" with respect to $g$. Let $R_\epsilon = (x^* - \epsilon, x^* + \epsilon)$, and we think of this as the "$\epsilon$-range" of points in $S_\epsilon$ with respect to $g$. Points in $S_\epsilon$ will stay in $R_\epsilon$ under repeated application of $g$, as $x^*$ is stable with respect to $g$. If $x$ is some point in $S_\epsilon$ that is the image of some other point $x_0$ also in $S_\epsilon$, then $x = f(x_0)$. By the stability of $x^*$ under $g$, we are assured that $x$ will remain in $R_\epsilon$ upon applying $f$ an even number of times. However, since $x_0$ also remains in $R_\epsilon$ under applications of $f$ an even number of times, this implies that $x$ will remain in $R_\epsilon$ under application of $f$ an odd number of times. So, if every $x \in S_\epsilon$ is the image under $f$ of some point also in $S_\epsilon$, and this holds for all $\epsilon$, then stability of $x^*$ with respect to $g$ implies stability of $x^*$ with respect to $f$.
However, this seems like a pretty strict condition on $f$. I would like a more general result, if possible.
Edit 2
In the appendix, the author states that $x^*$ is stable under $g$ if and only if it is stable under $f$. The reason given is the claim that $|f^{(n)}(x^*)|<1 \iff |g^{(n)}(x^*)| < 1$. For first order derivatives this is true, as $g'(x^*) = f'(x^*)^2$. However, it is not obviously true for second order derivatives, as $g''(x^*) = f''(x^*)f'(x^*)(f'(x^*) + 1)$. Even if this is true, I am (so far) unable to see why this should imply $x^*$ is stable under $g$ if and only if it is stable under $f$.