discontinuous right-hand side differential equation signum My question is the following. Let us consider the following differential equation
$\dot{x}=−sign(x)$  
with $sign(0)=1$. I would like to see a formal proof that such an equation does not admit any Carathéodory solution with $x(0)=0$. The intuitive reasoning concerning the vector field jamming the solution sound to me clear, but I would like to prove it.
Thanks.  

Suppose there is some AC solution $x$ defined on $[0,\epsilon)$ for some $\epsilon>0$. Then $\dot{x}(t) = -\operatorname{sgn} x(t)$ for ae. $t \in [0,\epsilon)$.
Let $\phi(t) = {1 \over 2} x^2(t)$, then $\phi$ is also AC ($x$ is bounded since
the derivative is bounded by one), and we have
$\dot{\phi}(t) = x(t) \dot{x}(t) = - |x(t)|$ ae. $t \in [0,\epsilon)$. In particular,
$\phi$ is non increasing, and since $x(0) = 0$, we have $\phi(t) = 0$ for all
$t \in [0,\epsilon)$.
Hence $x(t) = 0$ for all $t \in [0,\epsilon)$ and
so $\dot{x}(t) = 0$ for all $t \in [0,\epsilon)$, but this contradicts
$\dot{x}(t) = -\operatorname{sgn} x(t) = -1$ for ae. $t \in [0,\epsilon)$.
Hence no such $x$ exists.
