We consider the following O.D.E
\begin{align}(1)\;\;\;\begin{cases}x'(t)=f(x(t)) & t\geq 0,\\x(0)=x_0\in \Bbb{R}^n&\end{cases}\end{align} where \begin{align}f: \Bbb{R}^{n}\to \Bbb{R}^{n}\end{align} Assuming that \begin{align}\Vert f(x)-f(y) \Vert\leq k\Vert x-y\Vert,\;\;\forall\;x,y\in\Bbb{R}^n.\end{align} Assuming that $f(x_0)=0.$ Please, do I prove that $x(t,x_0)=x_0,\forall \;t\geq 0,$ where $x(\cdot\,,x_0)$ is the solution of $(1)$. If there are references, I also would appreciate!
$f$ being Lipschitz means, any solution to an initial value problem is unique (see e.g. https://en.wikipedia.org/wiki/Initial_value_problem#Existence_and_uniqueness_of_solutions)
Let us define a function $y(t):=x_0$, then $y'(t)=0$, since it is constant. We also have $y(0)=x_0$. Furthermore $f(y(t))=f(x_0)=0$, hence $y'(t)=0=f(y(t))$. That means $y$ satisfies the differential equation and the initial conditions. Since solutions are unique, we are done, because then $x(t,x_0)=x_0$.