Using the Mean Value Theorem, prove the following:
Let $f : [a,b] \rightarrow [a,b]$ be differentiable. Then, $f$ is a contraction mapping if and only if there exists $r \in (0,1)$ such that $\lvert f'(x)\rvert \leq r$ for all $x \in [a,b]$.
I've seen other threads asking about this question, but none of them answer my question: How is the Mean Value Theorem enough to prove this statement?
Here's a proof in one direction:
Suppose $f$ is a contraction mapping.
Thus there exists an $r \in (0,1)$ such that $\lvert f(x)-f(y) \rvert \leq r \lvert x-y \rvert$ for all $x,y \in \mathbb{R}$.
By the Mean Value Theorem, there exists a point $c\in [a,b]$ such that$f(x)-f(y)=f'(c)(x-y)$ so $\lvert f(x)-f(y) \rvert = \lvert f'(c) \rvert \lvert x-y \rvert$
and thus $\lvert f'(c) \rvert \lvert x-y \rvert \leq r \lvert x-y \rvert \implies \lvert f'(c) \rvert \leq r$.
But what I am supposed to obtain is:
$\lvert f'(c)\rvert \leq r$ for all $c \in [a,b]$
How do I prove that this is true for all points, instead of just one point with the Mean Value Theorem? Am I misunderstanding the theorem?
My suggestion and the suggestion of many other commenters is that the mean value theorem is most cleanly applied in the direction where you assume $|f'(x)| \leq r$ for $r\in (0,1)$. For any two $x,y\in [a,b]$ there is $c\in (a,b)$ with $|f(y) - f(x)| = |f'(c)||x-y|$. Now you can use the bound on $f'$ to your advantage.
If $f$ is a contraction mapping, then you've already shown that there is $r\in (0,1)$ such that $$\left|\frac{f(x) - f(y)}{x-y}\right|\leq r$$ for all $x,y\in [a,b]$ with $x\neq y$. What does this mean for the derivative of $f$ in $[a,b]$?