Problem $2.9.6$ (Perko's ODE): Show $|Y(t)| \le |Y(0)|\exp\left(\int_0^t \|A(s)\|\, ds\right)$ for $Y' = AY$ 
Let $A(t)$ be a continuous real-valued square matrix of size $n\times n$. Show that every solution of the nonautonomous linear system (where $Y(t) \in \mathbb R^n$ for all $t$ in the domain of $Y$) $$Y'(t) = A(t)Y(t)$$
satisfies
$$|Y(t)| \le |Y(0)|\exp\left(\int_0^t \|A(s)\|\, ds\right) \tag{1}$$
Further, show that if $\int_0^\infty \|A(s)\|\, ds < \infty$, then every solution of this
system has a finite limit as $t$ approaches infinity.


My work: Suppose $$|Y(t)| \le |Y(0)|\exp\left(\int_0^t \|A(s)\|\, ds\right)$$ is true; and take $t\to\infty$. Since we don't know if $\lim_{t\to\infty} Y(t)$ exists (yet?), it makes sense to work with $\liminf_{t\to\infty}$ and $\limsup_{t\to\infty}$ instead. So, we have
$$\left|\liminf_{t\to\infty} Y(t)\right| \le |Y(0)|\exp\left(\int_0^\infty \|A(s)\|\, ds\right) \quad\text{and}\quad \left|\limsup_{t\to\infty} Y(t)\right| \le |Y(0)|\exp\left(\int_0^\infty \|A(s)\|\, ds\right)$$
If we can show that $\lim_{t\to\infty} Y(t)$ exists, then we are done (by any of the two inequalities above). How do we do that? Further, we also need to prove $(1)$.
Thanks!
 A: For $t \ge 0$ we have $Y(t)=Y(0)+ \int_0^t A(s)Y(s)ds$, hence
$$
|Y(t)| \le |Y(0)| + \int_0^t \|A(s)\|\cdot|Y(s)|ds.
$$
Now (1) follows from the Grönwall-inequality: https://en.wikipedia.org/wiki/Gr%C3%B6nwall%27s_inequality
If in addition  $\int_0^\infty \|A(s)\| ds < \infty$ we first get from (1) that $Y$ is bounded on $[0, \infty)$:
$$
|Y(t)| \le |Y(0)| \exp(\int_0^\infty \|A(s)\| ds)=:\beta
$$
Let $\varepsilon > 0$ and choose $t_0$ such that $\int_{t_0}^t \|A(s)\| ds < \varepsilon$ $(t \ge t_0)$. For $t>\tau \ge t_0$ we have
$$
|Y(t)-Y(\tau)| \le \int_\tau^t \|A(s)\|\cdot|Y(s)|ds \le \beta \int_{t_0}^t\|A(s)\|ds \le \beta \varepsilon.
$$
Thus, according to the Cauchy-criterion $\lim_{t \to \infty} Y(t)$ exists.
A: $
\def\T{{\rm tr}}
\def\A{\T(A)}
\def\l{\lambda}
\def\qiq{\quad\implies\quad}
$This well known result
$$\eqalign{
&\l = \log|Y| \\
&d\l = \T(Y^{-1}\,dY) \;=\; \T(Y^{-1}AY\,dt) \;=\; \A\;dt \\
&\l-\l_0 = \int_0^t\A\;ds \\
&|Y(t)| = |Y(0)|\cdot\exp\left(\int_0^t\A\;ds\right) \\
}$$
leads to a similar conclusion.
