Let $f: \mathbb{R} \to \mathbb{R}$ be continuously differentiable with $f(0) = 0$ and $|f'(0)| < 1$... Question: Prove that there exists $r > 0$ such that $M:=\max_{[-r,r]}|f'(x)| < 1$.
I have two ideas on how to approach this but not sure if either is leading me in the right direction.
Answer 1: Fix $\epsilon > 0$. Since $f$ is continuosly differentiable, then $f'$ is a continuous function. In particular, $f'$ is continuous at $x = 0$. That is, if $|x - 0| = |x| < \delta$, then $|f'(x) - f(0)| < \epsilon$. Note that, for any $x \in \mathbb{R}$ such that $|x| < \delta$, we have that
$$|f'(x)| = |f'(x) - f'(0) + f'(0)| \leq |f'(x) - f'(0)| + |f'(0)| < \epsilon + 1.$$
So, now I can say that since $f'$ is continuous on $\mathbb{R}$, then $f'$ is continuous on $[-r, r]$ for any $r \in \mathbb{R}$. Since $[-r, r]$ is compact, then there exists $x_0 \in [-r, r]$ such that $f'(x) \leq f'(x_0)$ for all $x \in [-r, r]$.
And I am confused on how to tie all of this information together.
Answer 2: My other approach is to use the mean value theorem for $f'$ on the intervals $(-r, 0)$ and $(0, r)$ but I am not sure that this is helpful.
Any suggestions would be much appreciated.
 A: Let's simplify the notation a bit: let
$$
g(x) = f'(x).
$$
As you said, $g$ is continuous on the entire real line, and $g(0) < 1$.
Let $\epsilon$ be half the distance from $g(0)$ to $1$:
$$
\epsilon := {1 \over 2}(1 - g(0)).
$$
(Here I suggest sketching a picture of the graph of a generic continuous $g(x)$ in a neighborhood of $x=0$.)
As $g$ is continuous at $x = 0$, there exists a $\delta > 0$ such that
$$
|g(0) - g(x)| < \epsilon \quad \mbox{for all $x \in [-\delta, \delta]$}.
$$
Now, recall: we chose $\epsilon$ so that $g(0) + \epsilon$ is "half-way upward from $g(0)$ to $1$" and is, accordingly, "half-way" downward from $1$ to $g(0)$.
Therefore, for all $x \in [-\delta, \delta]$,
$$
g(x) \leq g(0) + \epsilon = 1 - \epsilon < 0.
$$
This $\delta$ is your desired $r$.
A: Since $\bigl\lvert f'(0)\bigr\rvert<1$ and since $f'$ is continuous, there is a $\varepsilon>0$ such that$$\bigl(\forall x\in(-\varepsilon,\varepsilon)\bigr):\bigl\lvert f'(x)\bigr\rvert<1.$$So, take $r=\frac\varepsilon2$.
