In a previous class, we used the following theorem without proof:
Theorem: Let $f:(\mathbb{R}^n,|\cdot|)\to(\mathbb{R}^n,|\cdot|)$ be differentiable, where $|\cdot|$ denotes the Euclidean norm. Suppose that the spectral radius of the Jacobian of $f$ is less than one on some subset $U$ of $\mathbb{R}^n$. Then $f$ is a contraction mapping on $U$.
With a bit more analysis background than I had when this theorem was originally presented to me, I have been trying to prove this, but have gotten stuck. I have proved what I believe to be a necessary step (reminiscent of that used in the proof for the case $n=1$), where if $\Vert Df\Vert\leq K$ on an open convex subset of $\mathbb{R}^n$, then $f$ is Lipschitz on that subset with Lipschitz constant $K$. However, I am stuck now on showing that $f$ is a contraction mapping.
I have seen similar propositions where if $Df$ is symmetric on $U$, then $f$ is a contraction mapping - the proof quickly follows in this case, as then the operator norm $\Vert Df(x)\Vert$ is equal to the spectral radius of $Df(x)$. However, in general this is not the case, and I haven't been able to figure out where to go in the general case - or if the theorem is even true without that assumption.
My question - are the conditions that I have placed on $Df$ and $U$ in the above theorem sufficient? Or are there more necessary conditions necessary? (Likely, we need $U$ to be convex in order to apply the lemma I have proved.) If $Df$ is not required to be symmetric / if there are no more conditions required on $Df$ for $f$ to be a contraction mapping, what might be the first step to take in the proof?
The correct statement should be : $f$ if Lipschitz with constant $K$ (for some norms on $\mathbb{R}^m$, $\mathbb{R}^n$) if and only if the norm of the Jacobian matrix is $\le K$. Considering a map from a space with a norm to itself, the induced norm of the Jacobian ( or of any square matrix) is $\ge $ spectral radius. And it can be $>$. Hence, using the spectral radius is not good enough in one sense, unless you have some special cases, say the matrix being selfadjoint, and the norm being the one induced from the euclidian metric.