TITLE: Real continuous function with attractive fixed point has range subset of domain
QUESTION [5 upvotes]: Let $f:\mathbb{R}\to \mathbb{R}$ be $C^1$ with fixed point $p$, such that $|f'(p)|< 1$. Then there exists an interval $I$ containing $p$ such that $f:I\to I$.
I'm trying to prove this rigorously, but there's a roadblock. My best attempt is as follows:
Let $\varepsilon > 0$, such that $\varepsilon$ is small, and consider $f(p+\varepsilon)=f(p)+f'(p)\varepsilon+\mathcal{O}(\varepsilon^2)$ and $f(p-\varepsilon)=f(p)-f'(p)\varepsilon+\mathcal{O}(\varepsilon^2)$.
Then $$|f(p+\varepsilon)-f(p-\varepsilon)|=|\varepsilon(2f'(p))+O(\varepsilon^2)|$$$$\le 2\varepsilon|f'(p)|+|\mathcal{O}(\varepsilon^2)|<2\varepsilon+|\mathcal{O}(\varepsilon^2)|$$
Everything would be as desired if not for the $|\mathcal{O}(\varepsilon^2)|$ term. How does one get rid of it? I could argue that since $\varepsilon$ is small, we can neglect the term, but that is not very rigorous. Would appreciate your insight.

REPLY [6 votes]: You don't need $C^1$, just differentiable at the fixed point $p$ with $|f'(p)| < 1$.
Choose $\epsilon > 0$ with $|f'(p)| < 1 - \epsilon$.  By definition of derivative, there is $\delta > 0$ so that for all $x$ with $|x - p| < \delta$, 
$$ \left| \frac{f(x)-f(p)}{x - p} - f'(p) \right| < \epsilon$$
Thus since $f(p)=p$, if $|x-p|<\delta$ we have $$f(x) \le p + |f'(p) (x-p)| + \epsilon |x-p| < p + |x-p| < p + \delta $$
and similarly
  $$f(x) \ge p - |f'(p)(x-p)| - \epsilon |x-p| > p - |x-p| > p - \delta$$
Thus $f$ maps the interval $(p-\delta, p+\delta)$ into itself.