Contraction implies bounded derivative

Question

I'm trying to show that if $f:[a,b]\to [a,b]$ is a strict contraction with constant $C\in (0,1)$, then $|f'(x)|<1$ f.a. $x\in[a,b]$. I tried using the the mean value theorem, but this only gives me $$ |f'(c)|=\frac{|f(x)-f(y)|}{|x-y|}\leq C<1 $$ for those $c$. Can I get all $x\in[a,b]$ using those constants? Is there a way to use the fixed point theorem by Banach?

Answer 1

Since $|{ f(x+h) -f(x) \over h} | \le C$ for all $x,h$ such that $x,x+h \in [a,b]$ we have $|f'(x)| \le C$.

Answer 2

If f.a. stands for 'for all' then the statement is false. For example $\frac {|x|} 2$ is a strict contraction and it is not differentiable at 0. However, a strict contraction is absolutely continuous. so it has derivative almost everywhere. At any point in $(a,b)$ where $f$ is differentiable we do get $|f'(x)| <1$ by looking at the definition of derivative. Hence $|f'(x)| <1$ almost everywhere.