(Long-time asymptotics). Suppose
$$\int_0^∞\|A(t)\|\,dt < ∞.$$
Show that every solution $x(t)$ of $x'(t)= A(t)x(t)$ converges to some limit: $\lim_{t→∞} x(t) = x_∞.$
(Hint: First show that all solutions are bounded and then use the corresponding integral equation.)
I was able to solve a first part:
I used that the solutions are limited by $\|\phi(t,t_0)\| \leq e^ {\int_0^∞\|A(t)\|\,dt}$. As $\int_0^∞\|A(t)\|\,dt < ∞$, then $\|\phi(t,t_0)\| \leq K $. The problem now is to show that the above limit always exists. Someone can help
Hint
Once you know that $x$ is bounded by some constant $K$ you can use the integral equation to show that $x(t)$ is Cauchy, for $t_1 \leq t_2$ you have: $$\|x(t_1)-x(t_2)\| =\left\|\int_{t_1}^{t_2} A(s) x(s) ds \right\| \leq K \int_{t_1} ^{t_2} \|A(s)\| ds$$ And $\int_{t_1} ^{t_2} \|A(s)\|ds$ goes to $0$ as $t_1$ goes to $+\infty$.
So using the sequential characterization of convergence and Cauchy sequences you can show that, as for Cauchy sequences, as long as your space is complete the function converges.