Let $D=[0,1]^n$, and $f:D \rightarrow D$ a $C^1$ map.
What is the condtion for the Jacobian of $f$ to be a Contraction map, i.e
$$\exists\, 1>r>0\,, ||f(x)-f(y)||\leq r||x-y||, \, \text{For all } \, x,y\in D$$
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Do you need a necessary condition or a criterion? Can you edit your question as to give us more context please? – Olivier Roche Sep 20 '19 at 07:49
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Ok, I will change the context – BrianTag Sep 20 '19 at 07:53
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@OlivierRoche The criterion is that eigenvalues of the Jacobian should be less strictly than $1$, but I'm stuck proving it – BrianTag Sep 20 '19 at 08:02
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A sufficient condition is that the norm of the Jacobian is bounded by $r$ on $D$. This is an immediate consequence of the Mean value theorem in several variables.
Obviously, you need to consider the matrix norm induced by your $\Vert \cdot \Vert$ norm to apply the Mean value theorem quoted above.
The converse is also true as if $\Vert J_f(x_0)\Vert > r$ at $x_0 \in D$, you'll be able to find a point $x$ "close to" $x_0$ such that $$\Vert f(x) - f(x_0) \Vert > r \Vert x - x_0 \vert$$
mathcounterexamples.net
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Mixing mathcounterexamples.net 's answer and this method to compute an operator norm, one gets following criterion :
If the square root of all eigenvalues of $J_f(x)^T J_f(x)$ is bounded by $r$, then $f$ is $r$-Lipschitz.
Olivier Roche
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