$$ \dot x(t)=Ax(t)+Bu(t),\quad t\in[0,T]\\ x(0)=x_0\\ u\in L^\infty([0,T],\mathbb R^m)\text{ and }x\in L^\infty([0,T],\mathbb R^n) $$
It is known that $\lim_{a\to 0}||u_a-u_0||_{L^2}=0$, where $u_0$ is a fixed trajectory, $u$ uniquely determined by $a$. I also know that for each $u$, there exists unique $x^u$ for the above problem.
My question:
what additional condition shall I impose on the above problem to prove $$\lim_{a\to 0}||x^{u_a}-x^{u_0}||_{L^\infty}=0$$
Thanks in advance.
Ok, then the solution is given by the variation-of-constants-formula $$x(t)=e^{tA}x_0 +\int_0^t e^{(t-s)A}Bu_a(s)\ ds$$ so your question amounts to asking whether $$\sup_{0\le t\le T}\left\|\int_0^t e^{(t-s)A}B(u_a-u_0)(s)\ ds \right\|$$ (for any norm $\|\cdot\|$ on $\mathbb R^n$) goes to $0$ as $a$ goes to $0$. Now this is certainly the case, because this term can be estimated by $$ \begin{split} \left\|\int_0^t e^{(t-s)A}B(u_a-u_0)(s)\ ds \right\|&\le \int_0^t \left\|e^{(t-s)A}B(u_a-u_0)(s)\right\|\ ds \\ &\le M\|B\|\int_0^t \|(u_a-u_0)(s)\|ds \\ &\le M\|B\| T \|u_a-u_0\|_\infty \to 0 \end{split}$$ where $M$ is the upper bound of the semigroup generated by $A$ (any semigroup has an upper bound on time intervals of finite length).