Let $f\in \mathcal{C}^1(\mathbb{R}^n,[0,T])$, and we define the following Backward/Foward ODE : for a small $\epsilon>0$, $$ \begin{cases} \dot x^{\epsilon}(t)=f(x^{\epsilon}(t),t), & t\in [0,T] \\ x^{\epsilon}(\epsilon)=x_0, \end{cases} $$ I want to prove that $x^{\epsilon}$ is well defined on $0$ : Using Picard's theorem we have local existence of a solution $x^{\epsilon}(.)$ in a neighborhood of $\epsilon$ (for example :$]\epsilon-a,\epsilon+a[$ ($a>0$) but how to prove the following statement :
There exists $\epsilon_0>0$, such that $\epsilon<\epsilon_0$ : $x^{\epsilon}$ is well defined on $0$.