$|\phi'(x)<1|$ is sufficient to show that $\phi$ is a contraction, so is $\phi'(x)>1$ sufficient to show $\phi$ isn't a contraction?

Question

By using the boundedness of the derivative, it is easy to show that, if $|\phi(x)<1|$, $\phi$ is a contraction as: $$\|\phi(x)-\phi(y)\|=\left|\int_x^y\phi'(s)ds\right|\le|x-y|\max|\phi(s)|$$ Is there a similar argument for the case that $|\phi'(x)>1|$? It feels like there should be.

Answer 1

If $\phi'(x)>1$ then since $\lim_{h \to 0} {\phi(x+h) -\phi(x) \over h} > 1$ there must be some $\delta>0$ such that if $0<h<\delta$ then $\phi(x+h) -\phi(x) > h$ and so $\phi$ cannot be a contraction.