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I'm currently reading through the following theorem.

Theorem: Let $p$ be a hyperbolic fixed point with $|f'(p)|<1$. Then there is an open interval $U$ about $p$ such that if $x\in U$, then $$\lim_{n\to\infty}f^n(x)=p.$$

With a part of the proof (I only typed up to the point that I got stuck):

Proof: Since $f$ be $C^1$, there is $\epsilon<0$ such that $|f'(x)|

Why do we have $|f(x)-f(p)|\leq A|x-p|$ and not $|f(x)-f(p)|< A|x-p|$? Since $|f'(x)|

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    Try $x=p$. $ $ $ $ – 2017-01-24
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    Well if $x=p$, I see that they're equal. But is there a more 'theoretical' reason? – 2017-01-24
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    Well, this simply prevents to ask that $|f(x)-f(p)|$x$ close to $p$ since $0 – 2017-01-24

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