Show that f has a unique fixed point in [a, b]. Full question in comments Suppose f : [a, b] → [a, b] is a differentiable function such that L = sup a≤x≤b |f ′ (x)| satisfies L < 1. Show that f has a unique fixed point in [a, b].  
This is my attempt, wondering if it's correct.
As f is differentiable it is continuous, so by the mean value theorem there exist some c such that $(f(b)-f(a))/(a-b)$ = $f'(c)$
But as  sup a≤x≤b |f ′ (x)| < 1 then $(f(b)-f(a))/(a-b)$ < 1 so $f(b)-f(a)$ < $(a-b)$
Therefore  $f(b)-f(a)$ ≤ α$(a-b)$ for some  0<α<1. Therefore this is a contraction
As [a,b] is closed it is complete so f is a contraction on a complete metric space so f has a unique fixed point.
 A: $f$ is a contraction if for some $\alpha, 0 < \alpha < 1$, it's true for all $x,y \in [a,b]$ that $\lvert f(y) - f(x)\rvert \le \alpha \lvert y - x\rvert$. You have to establish this not just for $a,b$ but for all $x,y$ in the interval. Your proof is easily adapted:
Given $a \le x < y \le b$, by the mean value theorem there is $c \in [x,y]$ such that $(f(y)-f(x))/(y-x) = f'(c)$, so $\lvert f(y) - f(x)\rvert \le L \lvert y - x\rvert$. Thus $f$ is a contraction, so it has a unique fixed point.
A: Since $f:[a,b]\to[a,b]$ it is clear from the intermediate value theorem that $f$ has a fixed point, since if $\phi(x) = f(x)-x$, we must have $\phi(a) \ge 0$ and $\phi(b) \le 0$.
So, you only need to establish uniqueness. Suppose $f(x_k) = x_k$ for $k=1,2$,
then $|x_1-x_2| = |f(x_1)-f(x_2)| \le L |x_1-x_2|$. Since $L<1$, the only
solution is $x_1=x_2$. Hence the fixed point is unique.
Aside: The if $x<y$ then the mean value theorem gives some $\xi \in (x,y)$
such that $f(x)-f(y) = f'(\xi) (x-y)$, and hence $| f(x)-f(y) | \le L |x-y|$.
