Consider a mapping $T:\mathbb{R}^n\to\mathbb{R}^n$ with the contraction property with regards to the sup norm i.e.
$$\|Tx -Ty\|_\infty<\gamma\|x-y\|_\infty $$ for a $\gamma<1$. If we now consider the pertubed ODE
$$ \dot{x} = ([(Tx)_i - x_i]\mu_i)_{i=1,\dots,n} $$
for a $\sum_i \mu_i = 1$ and $\mu_i> 0$, or alternatively written
$$ \dot{x} = [Tx-x]\mu $$ where the multiplication of $\mu$ with the $Tx-x$ difference should be read pointwise. Then the only point where $\dot{x}=0$ is of course the unique fixed point of $T$. But is it possible to show that this point is an attracting fixed point? I.e. for any starting point $x_0$, $$x(t) \to x^* \quad(t\to\infty)$$
where $x^*$ is the fixed point?
For $\mu_i = 1/n$ this is true since
$$ x(s+t) = x(s) + t\dot{x}(s) + O(t^2) $$
thus $$ \begin{align} \|x^* - x(s+t)\|_\infty &\le \|x^* - x(s) - t\dot{x}(s)\|_\infty + O(t^2)\\ &= \|x^* - x(s) - t[Tx(s)-x(s)]\|_\infty + O(t^2)\\ &\le t\|x^* - Tx(s)\|_\infty + (1-t)\|x^*-x(s)]\|_\infty + O(t^2)\\ &\le \gamma t\|x^* - x(s)\|_\infty + (1-t)\|x^*-x(s)]\|_\infty + O(t^2)\\ &= (1-(1-\gamma) t)\|x^* - x(s)\|_\infty + O(t^2)\\ \end{align} $$
for the contraction parameter $\gamma<1$ thus $1-\gamma>0$ thus $(1-(1-\gamma)t)<1$ for $t>0$.
For $u(s) = \|x^* -x(s)\|_\infty$ which is differentiable almost everywhere we get
$$\begin{align} \dot{u}(s) &= \lim_{t\to0} \frac{\|x^* - x(s+t)\|_\infty - \|x^* - x(s)\|_\infty}{t}\\ &\le \lim_{t\to0} -(1-\gamma)\|x^* - x(s)\|_\infty \end{align}$$
Thus by Gronwall
$$ \|x^* - x(s)\|_\infty= u(s) \le u(0) \exp\left( \int_0^s -(1-\gamma)dt\right) = \|x^* - x(0)\|_\infty \exp\left(- (1-\gamma)s\right) $$
Every time one dimension becomes zero, one hits a discontinuity of the derivative and has to restart, but since there are only a finite amount of dimensions that should be fine.
Update: In my particular case I was able to circumvent this problem by showing that T is also a contraction with regards to the norm
$$\|x\|_\mu = \sum_{i=1}^n \mu_i |x_i|$$
I would stil be curious whether or not this is true in a more general setting though if someone finds an answer.