I'd like to ask a question about the controllability of a linear system. Any help would be appreciated. $$$$Given a vector $x(t)\in \mathbb{R}^n$ defined over [$0$, $\infty$), we say $x(t)$ converge to zero exponentially as $t→+\infty$, or $\lim \limits_{t \to \infty} x(t) = 0$ exponentially, if there exist two positive constants $\alpha > 0$ and $\gamma > 0$ such that $||x(t)|| \leq\alpha||x(0)||e^{-\gamma t}$
Consider the following state equation: $\dot x = Ax + Bu$, with $A\in \mathbb{R}^{n{\times}n}$ Hurwitz. Show that $\lim \limits_{t \to \infty} x(t) = 0$ exponentially for any $x(0)\in \mathbb{R}^n$ if $\lim \limits_{t \to \infty} u(t) = 0$ exponentially. $$$$ The $x(t)$ can be represented by $x(t) = e^{At}x(0)+ \int_0^t e^{A(t-\tau)}Bu(t)\,d\tau. $ But I cannot reach the mathmetical proof to this question.
Since $A$ is Hurwitz stable, then there exist $\alpha_0,\alpha>0$ such that $||e^{At}||\le\alpha_0 e^{-\alpha t}$ for all $t\ge0$. Moreover, since $u(t)$ converges to zero exponentially fast, then there exist $\beta_0,\beta>0$ such that $||u(t)||\le\beta_0 e^{-\beta t}$ for all $t\ge0$. Let also $M>0$, such that $||B||\le M$.
So, from triangle inequality we have that
$$||x(t)||\le ||e^{At}x(0)||+\left|\left|\int_0^t e^{A(t-\tau)}Bu(\tau)d\tau\right|\right|.$$
We have that $||e^{At}x(0)||\le ||e^{At}||\cdot||x(0)||\le \alpha_0 e^{-\alpha t}||x(0)||\to0$ exponentially.
Similarly, we have that
$\left|\left|\int_0^t e^{A(t-\tau)}Bu(\tau)d\tau\right|\right|\le \int_0^t \left|\left|e^{A(t-\tau)}Bu(\tau)\right|\right|d\tau\le M\alpha_0\beta_0\int_0^te^{-\alpha(t-\tau)-\beta\tau}d\tau=\dfrac{M\alpha_0\beta_0}{\alpha-\beta}(e^{-\beta t}-e^{-\alpha t})\to0$ exponentially.
This shows that $||x(t)||$ converges to zero exponentially fast.