Prove that for any $u_0 \in \mathbb{R}^n,$ there exist $T=T(|u_0|)>0$ and a unique $u \in C([0,T])$ solving $$u(t)=u_0+\int_0^tf(u(s))ds,$$ for all $t \in [0,T].$ Assume that $f$ is locally Lipchitz. Recall that $C([0,T])$ is the Banach space with the norm sup $||u||:=\sup_{s \in [0,T]}|u(s)|.$
This is my proof:
Since f is locally Lipschitz, for any $u_0 \in \mathbb{R}^n,$ there exist $\delta>0$ and $M>0$ such that $$\|f(u)-f(u_0)\| \leq M\|u-u_0\|\,\,,\forall u \in \bar{B}(u_0,\delta).$$ Since $u$ is continuous at $t=0$ then there exist $\delta_0$ such that $$\|u(t)-u(0)\|\leq \delta\,\,,\forall |t| < \delta_0.$$ Let $T <\min\{\delta_0; \dfrac{1}{M}\}$ then we get $$\|u(t)-u_0\|\,\,\,,\forall t \in [0,T].$$ Define $T(u)(t)=u_0+\int_0^tf(u(s))ds$ for all $u \in C([0,T])$ then $T(u) \in C([0,T]).$ Now, I prove $T$ is contractive. Indeed, \begin{align*} ||T(u)-T(v)||&=\sup_{t \in [0,T]}|T(u(t))-T(v(t))|\\ & =\sup_{t \in [0,T]}\left|\int_0^tf(u(s))-f(v(s))ds\right|\\ & \leq \sup_{t \in [0,T]}\int_0^t\left|f(u(s))-f(v(s))\right|ds\\ & \leq \sup_{t \in [0,T]}M\int_0^t\left|u(s)-v(s)\right|ds\\ & \leq \sup_{t \in [0,T]}M \|u-v\|t\\ & \leq MT \|u-v\|. \end{align*} Since $MT <1,$ by the Banach's fixed point, there exists a unique $u \in C([0,T])$ such that $u = T(u)$, which completes the proof.
I wonder my proof is correct? This is because compare to the proof of Picard–Lindelöf theorem, I do not define the mapping T on the space $X:=\{u \in C([0,T] \,\,\,|\,\,\, \|-u_0\| \leq \delta\}$ but using the continuity of $u$.