Derivative test for contraction mapping on open sets I have a question on the derivative test for showing that a function is a contraction. The proposition that I know is in this version:

Let $I$ be a closed bounded set of $\mathbb{R}^l$ and $f : I → I$
  a differentiable function with $||f'(x)|| \leq \beta$, 
  $0<\beta<1$, for all $x \in I$. Then f is a contraction with module of
  contraction equal to $\beta$.

Does the following proposition hold too? (I have put in bold the changed parts)
Let $I$ be an open set of $\mathbb{R}^l$ and $f : I \rightarrow \pmb{\mathbb{R}^l}$ a
differentiable function with $||f'(x)|| \leq \beta$,  $0<\beta<1$, for all $x \in I$. Then f is a contraction on $I$ with module of contraction equal to $\beta$. 

The result is true if $I$ is convex, but may fail for a non convex domain.
This example is best described using complex variables. Let
$$
D=\{a<|z|<b\}\setminus(-4,-1),
$$
that is, $D$ is an annulus from which we delete its intersection with the negative real axis. $D$ is open, connected and bounded. Let $f(z)=\sqrt z$. Then $f(D)\subset D$ and $|f'(z)|\le1$ for all $z\in D$. However $f$ is not Lipschitz on $D$ with respect the ususl metric. This can be seen by considerind the points $z_{\pm\delta}=-2\pm\delta\,i$ for $\delta>0$ small.  As $\delta\to0$, the distance between them goes to $0$, but $|f(z_{\delta})-f(z_{-\delta})|\to2\,\sqrt2$
