Let $f:\mathbb{R}^2\rightarrow\mathbb{R}^2$ continuous fulfilling $$ \begin{cases} f(t,x)<0, & \text{if $xt>0$}.\\ f(t,x)>0, & \text{if $xt<0$}. \end{cases} $$ Show that, $x(t)\equiv0$ is the only solution to the initial value problem $$\dot{x}=f(t,x)\hspace{1cm} x(0)=0$$
So obviously $x(t)\equiv0$ solves the problem on $\mathbb{R}^2$ in order to show that it's the only solution I would try to use the Picard-Lindelöf theorem. In order to be able to use it I would furthermore have to prove that $f$ is locally Lipschitz in $x$. But how can I do this ? I know of this theorem that says that if $f$ is locally differentiable w.r.t $x$ it's locally Lipschitz. Would that help ?
The continuity of $f$ implies that $f(t, 0) = 0$ for all $t \in \Bbb R$, so that $x(t) \equiv 0$ is a solution.
For any solution $x$ which is defined in an interval $I$ containing $t=0$ $$ \frac{d}{dt}(x^2(t)) = 2 x(t) \dot x(t) = 2 x(t)f(t, x(t)) \begin{cases} \le 0 & \text{ if } t > 0 \\ \ge 0 & \text{ if } t < 0 \end{cases} $$ because if $x(t) \ne 0$ then $x(t)$ and $f(t, x(t))$ have the opposite sign for $t> 0$, and the same sign for $t < 0$.
It follows that $x^2(t) \le x^2(0) = 0$.