TITLE: Show that derivative less than 1 implies contraction.
QUESTION [5 upvotes]: I am told that f has a continuous derivative and that $a \leq f(x) \leq b$ and $|f'(x)| < 1 \ \forall x \in [a,b]$ and I have to show that $f$ is a contraction.
Now if I take any $x,y \in [a,b]$, the Mean-Value Theorem says that $\exists c \in (a,b)$ such that
$$|f(x) - f(y)| = |f'(c)| |x-y|$$
and so clearly this is a contraction.
However I haven't used the condition that $f$ has a continuous derivative or that $f$ is bounded by $a$ and $b$, why are these conditions necessary?
My definition of contractive is that $|g(x) − g(y)| \leq a|x − y|$
for some real value $0 \leq a < 1$ and for all $x$, $y \in [a,b]$.
Thanks

REPLY [6 votes]: You only proved that $|f(x)-f(y)| < |x-y|$.
But the definition of contraction is that there exists a $C <1$ so that 
$$|f(x)-f(y)| < C|x-y| \,.$$
Hint: If $f'$ is continuous on $[a,b]$, then so is $|f'|$, and hence $|f'|$  attains its maximum at some point $x_0 \in [a,b]$. 
Also, a contraction by definition is a function $f$ from some metric space to itself. Your domain is $[a,b]$ so your codomain has also to be $[a,b]$, that's why you need the other condition.