Using Mean Value Theorem to prove $f$ is a contraction

Question

Using the Mean Value Theorem prove the following:

Let $f:[a,b]\to[a,b]$ be differentiable. Then $f$ is a contraction if and only if there exists a $r\in(0,1)$ such that $|f'(x)|\leqslant r$, for all $x\in[a,b]$.

If I assume $|f'(x)|\leqslant r$.

Using the MVT then $f(b)-f(a)=f'(x)(b-a)$ for some $x\in (a,b)$

Then $|f(b)-f(a)|=|f'(x)(b-a)|\leqslant r|b-a| $then f is a contraction.

However the reverse it has been difficult to prove. Assuming $f$ is a contraction. $f(a)-f(b)\leqslant M(a-b)$ Now I need to prove $f'(x)=M$.

Question:

Could someone provide me a hint?

Thanks in advance!

Answer 1

$$|f'(x)| = \left|\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\right| = \lim_{h \to 0} \frac{|f(x+h) - f(x)|}{|(x+h) - x|} \le M.$$

Answer 2

I am late to the party, are you familiar with the mean value inequality please?

If $f \in C[a, b]$ and differentiable on $(a, b)$, then

$$ \displaystyle \sup_{x \in (a, b)} |f'(x)| \geq |\frac{f(a) - f(b)}{a - b}| \\ $$

$$ \iff |f(a) - f(b)| \leq \sup_{x \in (a, b)} |f'(x)| \cdot |a - b| $$

By assumption, we have $\displaystyle \sup_{x \in (a, b)} |f'(x)| \leq r$, with $r \in (0, 1)$, hence $f$ is a contraction.