During my integral calculus studies I came across the following function:
Let $\Omega \subseteq \mathbb{R}^n$ a non-empty open set and $f \in C^1(\Omega, \mathbb{R}^n)$ such that $|\det f'(x)| < 1$ and $f(K) \subseteq K$ for some convex compact set $K \subset \Omega$.
Which left me wondering if $\varphi : K \to K : x \mapsto f(x)$ is a contraction. It is reasonable to expect that to be true because of the following evidences
(i) $|\det f'(x)| < 1$ implies that $f$ is not a rigid motion, that is, $f$ is not a translation, nor a rotation, nor a reflection and $|\det f'(x)| = |\prod_\limits{i=1}^n \lambda_i | = \prod_\limits{i=1}^n | \lambda_i | < 1$, where $\lambda_i$s are eigenvalues of $f'(x)$. This means that there exists at least one eigenvalue such that $|\lambda| < 1$ and therefore at least one "locally contracting direction" at least as strong as all "locally expanding directions" together. Furthermore, because $f'(x)$ is invertible then $ 0 < |\det f'(x)|$. And this means that $f$ has a predictable local behaviour.
(ii) $\varphi$ is a kind of "measure-contraction mapping" in the sense that
$$ \lambda(\varphi(K)) = \int_{\varphi(K)} dx = \int_{K} |\det \varphi'(x)| dx \le \max\limits_{y \in K} |\det \varphi'(y)| \int_{K} dx = c\ \lambda(K) $$
Where $ c = \max\limits_{y \in K} |\det \varphi'(y)| < 1$. Hence $ \lim\limits_{n\to \infty} \lambda(\varphi^n(K)) = 0 $. And also $0 < \lambda(\varphi^n(K))$.
(iii) Since $\varphi^{n+1}(K \setminus \varphi^n(K)) \subsetneq \varphi^{n+1}(K)$, it follows that $(\varphi^n(K))_{n \in \mathbb{N}}$ is a decreasing sequence of convex compact sets such. So $\lim\limits_{n\to \infty} \varphi^n(K) = \bigcap\limits_{n=1}^{\infty} \varphi^n(K)$ exists and it is not empty; this is the set of fixed points of $\varphi$.
(iv) It is true in $\mathbb{R}^1$ because $|\varphi'(x)| = |\det \varphi'(x)| \le \max\limits_{y \in K} |\det \varphi'(y)| < 1$.
Now I am on the quest to find $L \in [0, 1[$ such that $\|\varphi(x) - \varphi(y)\| \le L \|x - y \|$ but am not progressing.
Any hint or counterexample will be greatly appreciated.
Found a counterexample. Let $\delta \in \left]0, 1\right[$ and
$$ f : \mathbb{R}^2 \to \mathbb{R}^2 : x \mapsto \begin{bmatrix} \delta & 0 \\ 0 & 1 \end{bmatrix} x $$
So $$ f \in C^1(\mathbb{R}^2, \mathbb{R}^2) $$ $$ |\det f'(x)| = |\det f(x)| = |\delta| < 1 $$
Let $B$ the two dimensional closed unit ball and $x \in B$. Then $$ \|f(x)\|^2 = (\delta x_1)^2 + (x_2)^2 < (x_1)^2 + (x_2)^2 = \|x\|^2 \le 1 $$ Hence $f(B) \subseteq B$.
Now consider $\varphi : B \to B : x \mapsto f(x)$. Thus $$ \lim\limits_{n\to \infty} \varphi^n(B) = \{(0, y) : y \in [-1, 1] \} $$
And $$ \lim\limits_{n\to \infty} \lambda(\varphi^n(B)) = \lambda(\{(0, y) : y \in [-1, 1] \}) = 0 $$
But $$ \|\varphi(0, 1) - \varphi(0, -1)\| = 2 = \|(0, 1) - (0, -1)\| $$
Therefore $\varphi$ is not a contraction.