Consider the ODE system $$X'(t) = AX(t)+Bu(t)$$ where $X(t) \in R^n, \; A \in R^{n \times n} \text{ and } B \in R^{n \times m}$. In control theory, we define the set of states reachable as $$A(0,T) = \{ \phi(T,0,0,u(.)) : u \in U \}$$ such that $U=\{\text{all the function integrable in a finite interval}\}, \; u: R \rightarrow R^m$ and $$\phi(T,0,0,u(.)) = \int_0^T e^{(T-s)A} B u(s) ds$$ It is asked to show that if $0<S<T$, then $A(0,S) \subset A(0,T)$
My attempt: In general, I need to show that if we $x$ such that $$X= \int_0^S e^{(S-s)A} B u(s) ds$$ for some $u \in U$, then there might exist $\bar{u} \in U$ such that $$ X = \int_0^T e^{(T-s)A} B \bar{u}(s) ds$$ correct? We have that
$$ X \in A(0,S) \iff X = \int_0^S e^{(S-s)A} B u(s) ds $$ Using the transformation $T-z = S-s$: $$ X \in A(0,S) \iff X = \int_{T-S}^T e^{(T-z)A} B u(z-(T-S)) ds$$
Now, I couldn't find how to fix the limits of integration defining a new function $\bar{u}$ using $u$ such that those limits goes from $0$ to $T$. Any idea? Also, I considered define $\bar{u}$ first, but the change of variables to 'fix' the exponential messed my limits of integration too.
Thanks!
It is well-known that $A(0, S) = \operatorname{Im} W_r(0, S)$ where
$$ W_r(0, S) := \int_0^S e^{(S-\tau) A} B B^T e^{(S-\tau) A^T} d\tau $$
A very nice proof can be found here. Now you just need to prove that $\operatorname{Im} W_r(0, S) \subseteq \operatorname{Im} W_r(0, T)$, which is pretty easy.