I am looking for an alternative solution (or just some helpful comments) to this problem:
Let $\omega(t)\in\mathbb{R}^q$, $q\in\mathbb{N}$, be a absolutely continous function satisfying: $0<||\omega(t)||\leq \epsilon_{\omega},\, \forall t\geq 0$. Let $A\in\mathbb{R}^{n\times n}$, $q\leq n\in\mathbb{N}$, be a Hurwitz matrix (all its eigenvalues are strictly less than zero); and let $H\in\mathbb{R}^{m\times n}$ , $n\geq m\in\mathbb{N}$ and $B\in\mathbb{R}^{n\times q}$.
Then, find $\epsilon_z(t)>0$ such that, $||z(t)||\leq \epsilon_z(t)$ for all $t\geq 0$, where
\begin{equation} z(t) = \int_0^t He^{A(t-s)}B\,\omega(s)ds. \end{equation}
$z(t)$ is bounded as matrix $A$ is Hurwitz.
My first way to find the value of $\epsilon_z(t)$ was by just taking the norm of $z(t)$ and applying straightforward algebra: \begin{equation} \begin{aligned} ||z(t)|| = \left\vert\left\vert\int_0^t He^{A(t-s)}B\,\omega(s)ds,\right\vert\right\vert \leq \int_0^t \left\vert\left\vert He^{A(t-s)}B\,\omega(s)\right\vert\right\vert ds\leq \mu(t)\epsilon_{\omega}\triangleq \epsilon_z(t) \end{aligned} \end{equation}
where $\mu(t)\triangleq \int_0^t \left\vert\left\vert He^{A(t-s)}B \right\vert\right\vert ds$.
However, I am worried about the conservativeness of this value of $\epsilon_z(t)$. In fact, for a given fixed $t$, I need to find the less conservative bound. I think that this is a more complex solution as it implies, not only to find $\epsilon_z(t)$, but also to assure that there exist some $\omega(t)$ causing that $||z(t)||=\epsilon_z(t)$.
¿Does my analysis guarantee that there exist some $\omega(t)$ that forces the equality?. If not, ¿Could this problem be addressed in other way?.
Thank you in advance!