Prove Fixed Point Theorem using the Mean Value Theorem

Question

Assume $f$ has a finite derivative and $|f'(x)| \leq y < 1$ for all $x \in (a,b)$
$f$ is continuous and $a \leq f(x) \leq b$ for all $x \in [a,b]$. Prove $f$ has a unique fixed point in $[a,b]$.

So far I have for every c in (a,b) |f'(c)| ≤ y => lim x->c |f(x) - f(c)|/|x-c| ≤ y => lim x->c |f(x) - f(c)| ≤ y lim x->c |x-c|
Would that be the definition of a contractive map in R?
Therefore by Banach Fixed Point Theorem, f has a unique fixed point.
Can I prove Banach's theorem using the mean value theorem?

Answer 1

$\lim_{x\to c} |f(x) - f(c)|\leq y\ \lim_{x\to c} |x-c|$

Would that be the definition of a contractive map in R?

No, this is just the statement that $f$ is continuous at $c$, because the right-hand side is $0$. The fixed point theorem will apply, but to show that $f$ is contractive you will want to use the mean value theorem. Suppose that $a\leq z\lt x\leq b$. By the mean value theorem, there is a $c$ in $(z,x)$ such that $f'(c)=\frac{f(x)-f(z)}{x-z}$. Apply absolute values, rearrange, and use the hypothesis on the derivative to conclude that $f$ is contractive.

Answer 2

To show that there is some fixed point, consider $g(x) = f(x)-x$. Then $g(a)\geq 0$ and $g(b) \leq 0$. To show that it is unique, use the mean-value theorem for the two purported fixed points.