Theorem 1.16 of "An Introduction to Difference Equations" by Saber Elaydi

Question

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.

Answer 1

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).

Answer 2

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.