Norm of derivative and contractive function

Question

It is a widely know fact (at least for me, I might be wrong) that, given a function $f(x)$, the condition $|f'(x)| < 1$ is enough to see that the function is contractive. However, the problem becomes harder when it comes to several variables. I know that, if we consider a function $f:\mathbb{R}^n \rightarrow \mathbb{R}^n$, if $||Df(x)||_2 < 1 \forall x \in \mathbb{R}^n$, then this function is contractive (here, $||Df(x)||$ means the 2-norm operator of the jacobian matrix at that point).
In many contexts, given the expression, it seems way easier using operator norms like $||\cdot||_1$ or $||\cdot||_{\infty}$. I was wondering if the bounding of the norm of the derivative by $1$ is enough condition for contractiveness for any norm $p$, or $p=2$ is the only case. I know that in $\mathbb{R}^n$ all $p$-norms are equivalent.
Besides and since one can consider the operator norm of non-square matrix, is the same result valid in this case?

Answer 1

Hint: On the real line $\|x||=\frac 1 2|x|$ is also a norm. The condition $\|f'(x)\|=\frac 1 2|f'(x)|<1$ does not gaurantee that $f$ is a contraction for this norm. Can you find a counter-example?