If $f:\mathbb{R}^n\to\mathbb{R}^n$ is function of class $C^1$ and $\nabla\cdot f=0$ then the system $\dot{x}=f(x)$ not have a fixed point asymptotically stable.
The Bendixson's criteria says that if the divergence of the function is not zero and does not change sign then the ODE has no closed orbit lying entirely in the domain (simply connected).
I know that the system preserve volumen and that for $n=2$, $\nabla\cdot f=0$ iff the system is Hamiltonian and then the system only may have saddle or center.
How do I prove this statement for any $n$?
A fixed point $x_0$ is asymptotically stable if it is Lyapunov stable and moreover there exists a $\delta_1>0$ such that for all $x\in B(x_0,\delta_1)$
$$\lim_{t\to\infty}F_t(x)=x_0.$$
This means that we have pointwise convergence of a family of integrable (actually continuous) functions on the compact $\overline B(x_0,\delta_1/2)$ to the constant $x_0.$
On the other hand Lyapunov stability implies that there is a $\delta_2>0$ such that $F_t(\overline B(x_0,\delta_2))$ is contained in $B(x_0,1)$ for all $t.$
Let $\delta=\min(\delta_1/2,\delta_2).$ The dominated convergence theorem implies that on $\overline B(x_0,\delta)$ the pointwise convergence is actually $L^1$ convergence:
$$\int_{\overline B(x_0,\delta)}|F_t(x)-x_0|dx\to0.$$
From this it follows that for every $\epsilon>0$
$$\mu\left(\{x\in\overline B(x_0,\delta);|F_t(x)-x_0|>\epsilon\}\right)\to0.$$
where $\mu$ denotes Lebesgue measure.
If we assume volume preservation (as a consequence of vanishing divergence) then the above measure is equal to
$$\mu\left(F_t(\{x\in\overline B(x_0,\delta);|F_t(x)-x_0|>\epsilon\})\right)$$
which means that the total measure of $F_t(\overline B(x_0,\delta))$ becomes arbitrarily small for sufficiently large $t,$ contradicting volume preservation.