How does it follow from $(|f(x)-f(p)|)/|x-p| <a$ that $|f^k(x)-p| \leq a^k|x-p|$

Question

The exercise is asking me to show the latter theorem $|f^k(x)-p| \leq a^k|x-P|$ for $k=2$. But I am confused as to how they even derived that theorem. In the proof of a theorem, that had:

$(1)$ There is a neighborhood $N_{\varepsilon}(p)$ for some $\varepsilon>0$ so that $\frac{|f(x)-f(p)|}{|x-p|} <a$ for $x\in N_{\varepsilon}(p)$ It follows that $(2)$ $|f^k(x)-p| \leq a^k|x-p|$ for $k\geq1$

I do not know how $(1)$ follows. I understand I can multiply over the denominator to get $|f(x)-f(p)| < a|x-p|$ which yields an equation that resembles. to the other inequality but I do not know how they derive it $(2)$. Thank you!

Answer 1

The answer is already contained in geetha290krm’s comment. Anyway…

This is what we know:

There exist $\varepsilon>0$ and $a<1$ such that $|f(x)-p|<a|x-p|$ for every $x \in N_\varepsilon(p)$.

Let $x_0 \in N_\varepsilon(p)$ be fixed.

We can prove that $|f^k(x_0)-p|<a^k|x_0-p|$ for every $k \geq 1$ by induction:

• Base step: The fact that the inequality holds for $k=1$ has been proved already.
• Inductive step: Assume that $|f^{k-1}(x_0)-p|<a^{k-1}|x_0-p|$. In particular, this means that $f^{k-1}(x_0) \in N_\varepsilon(p)$, being $a<1$. Therefore, $|f^k(x_0)-p|=|f(f^{k-1}(x_0))-p|<a|f^{k-1}(x_0)-p|<a a^{k-1}|x_0-p|=a^k|x_0-p|$.