Show that if $\lim \inf_{t\rightarrow \infty} \int_{t_0}^t \operatorname{tr}\left(A(s)\right)ds= \infty $ then the linear first-order system $x'(t)=A(t)x(t)$ where $A \in C\left(I, \mathbb{R}^{n\times n}\right), x(t) \in \mathbb{R}^n$ has an unbounded solution.
Hey,
We define last section the Wronskian determinant $W(t)=\det(\phi_1(t),..,\phi_n(t))$ with taking $n$ solutions $\phi_1,..,\phi_2$ of the linear first order system.From the Liouville´s formula $W(t)=W(t_0) \exp\left(\int_{t_0}^t \operatorname{tr}(A(s))ds\right)$ it follows that the Wronskian determinant goes to infinity for $t$ going to infinity.
The Matrix $\pi(t,t_0)$ is called principal solution (at $ t_0$) with $ \frac{d}{dt} \pi(t,t_0)=A(t) \pi(t,t_0), \pi(t_0,t_0)=I $.
$\det(\pi(t,t_0))=\exp\left(\int_{t_0}^t\operatorname{tr}(A(s))ds\right)$
$$\lim \inf_{t\rightarrow \infty} \int_{t_0}^t \operatorname{tr}(A(s))ds = \lim \inf_{t\rightarrow \infty} \ln(\det(\pi(t,t_0)))= \infty$$
$\lim \inf_{t\rightarrow \infty} \int_{t_0}^t \operatorname{tr}(A(s))ds = \lim_{t\rightarrow \infty} \inf\{\int_{t_0}^\tau \operatorname{tr}(A(s))ds| \tau \ge t\}$
$\forall K \exists T: \forall t \ge T: \inf\{\int_{t_0}^\tau \operatorname{tr}(A(s))ds| \tau \ge t\}>K$