the LTV system $\dot x(t)=A(t)x(t)$ is called uniformly stable if $\exists \gamma>0$ such that $\left\| {\Phi \left( {t,{t_0}} \right)} \right\| \leqslant \gamma $ for all $t\ge t_0$ where ${\Phi \left( {t,{t_0}} \right)}$ is the state transition matrix.
my question: say that I know that: $\forall \epsilon>0$ $\exists \delta>0$ such that $\left\| {x\left( 0 \right)} \right\| \leqslant \delta$ $\rightarrow$ $\left\| {x\left( t \right)} \right\| \leqslant \varepsilon $ for all $t\ge t_0$, how could I show system is uniformly stable?
The condition that $||x(t)||\le\dfrac{\epsilon}{\delta}||x(t_0)||$ for all $t\ge t_0$ is equivalent to the definition of uniform stability. Now, we have that
$$||x(t)||=||\Phi(t,t_0)x(t_0)||\le\dfrac{\epsilon}{\delta}||x(t_0)||,\ t\ge t_0.$$
Since, this is true for all $x(t_0)$, then this implies that $||\Phi(t,t_0)||\le\gamma:=\dfrac{\epsilon}{\delta}$ for all $t\ge t_0$.