Boundedness of the input $$
\dot x(t)=Ax(t)+Bu(t),\quad t\in[0,T]\\
x(0)=x_0\\
x(t)\text{ is bounded on }t\in[0,T]\\
A\text{ and }B\text{ are given constants},\quad B\neq 0
$$
My question:

Is $u(t)$ also bounded on $t\in[0,T]$?

Thanks in advance.
 A: Let $x(0)=0$ and $x(t)=t^2\cos (t^{-2})$ for $t\ne 0$.The function $x(t)$ is differentiable at every $t$ but $x'(t)$ is unbounded on $[0,T]$ for any $T>0 $ . (Consider $t=1/\sqrt {2\pi n}$ for $n\in N.$) Note that $x'(t)$ is not continuous at $t=0$. 
A: Ok, if $A,B\in \mathbb R$, then
$x(t)=x_0 e^{tA}+B(e^{\cdot A}*u)(t)$, where $*$ denotes the convolution of two functions. If you assume $u$ to be continuous, then the above function is continuous and hence it attains a maximum on the compact interval $[0,T]$, but even if this is not the case, for fixed $T$ the sought-after boundedness follows from Young's inequality for convolutions.
EDIT: Sorry, as correctly pointed out by @obaarey the non-continuous case is wrong. I was thinking of applying the vector-valued Young estimate $\|e^{\cdot A}*u\|_\infty\le \|e^{\cdot A}\|_\infty \|u\|_1$ (in fact, $e^{\cdot A}*u$ is even an absolutely continuous and hence uniformly continuous vector-valued function), which however goes in the "wrong" direction and shows that an unbounded solution implies an unbounded control.
