# Show that derivative less than 1 implies contraction.

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

• Please, could you write down your definition of contractive map? I suspect there are two different ideas of contractivity. You proved that $|f(x)-f(y)|<|x-y|$: is this what you need to prove? – Siminore Jun 13 '13 at 14:25
• I've edited the question – Wooster Jun 13 '13 at 14:34
• The problem is that $f'(c)$ is NOT a constant. Picking different $x,y$ leads to different $c$s. – N. S. Jun 13 '13 at 14:42
• Also, what does the definition of contraction say about domain and codomain? ;) – N. S. Jun 13 '13 at 14:43

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.