Prove using continuity

Question

$g(x)$ is a function with a fixed point $p$. If $g'(x)$ is continuous on $(p-\delta_0,p+\delta_0)$ for some constant $\delta_0>0$ and $|g^{'}(p)|<1$, then there is a positive number $\delta<\delta_0$ such that $$|g^{'}(x)|\leq k<1 $$ for some constant $k$ and for $x$ in $(p-\delta,p+\delta)$. Based on continuity $$|g^{'}(x)-g^{'}(p)|<\epsilon$$ for $x$ in $(p-\delta,p+\delta)$. How do I get to the expression $$|g^{'}(x)|\leq k<1 $$. I need some hints as I haven't done such proofs in a while.

Answer 1

Since $|g'(p)|<1$ the by continuity of the derivative at $p$ and for $\epsilon_0$ such that $|g'(p)|+\epsilon_0 <1 $ we have that $\exists \delta_1>0$ such that $||g'(p)|-|g'(x)|| \leq |g'(x)-g'(p)| \leq \epsilon_0$ for all $x \in (p-\delta_1,p+\delta_1)$

Thus for $k=|g'(p)|+\epsilon_0$ and $\delta < \frac{\min\{\delta_0,\delta_1\}}{2}$ you have the desired conclusion.