Local flows by a contraction argument

By openness choose $R>0$ with $\bar{B}(q_0,2R)\subset U$. Continuity of $X$ and $DX$ at $q_0$ gives finite $M,L\ge 0$ bounding $\mathrm{\parallel }X\mathrm{\parallel }$ on $\bar{B}(q_0,R)$ and $\mathrm{\parallel }DX\mathrm{\parallel }$ on $B(q_0,2R)$. By [F1] $X$ is totally differentiable there, and [F2] applied to the convex ball $B(q_0,2R)$ gives $\mathrm{\parallel }X(u)-X(v)\mathrm{\parallel }\le L\mathrm{\parallel }u-v\mathrm{\parallel }$ for $u,v\in \bar{B}(q_0,R)$.

Choose $\delta >0$ with $M\delta \le R\mathrm{/}2$ and $L\delta <1$, and let $\mathcal{C}:=\{\gamma \in C([-\delta ,\delta ],{\mathbb{R}}^m):\gamma (t)\in \bar{B}(q_0,R)\text{ for every }t\}$ with the supremum norm. It is complete: for a supremum-Cauchy sequence $({\gamma }_k)$, [F5] gives the pointwise limit $\gamma (t)\in {\mathbb{R}}^m$; the supremum-Cauchy estimate, passed to that pointwise limit, gives uniform convergence to $\gamma$. The limit is continuous: given $\epsilon >0$ and $t_0$, choose $k$ with $\mathrm{\parallel }{\gamma }_k-\gamma {\mathrm{\parallel }}_{\mathrm{\infty }}<\epsilon \mathrm{/}3$, then continuity gives $\rho >0$ such that $\mathrm{\parallel }{\gamma }_k(t)-{\gamma }_k(t_0)\mathrm{\parallel }<\epsilon \mathrm{/}3$ whenever $\mathrm{\mid }t-t_0\mathrm{\mid }<\rho$; the triangle inequality yields $\mathrm{\parallel }\gamma (t)-\gamma (t_0)\mathrm{\parallel }<\epsilon$. Also $\gamma (t)\in \bar{B}(q_0,R)$ for every $t$, since this Euclidean ball is closed and each ${\gamma }_k(t)$ lies in it. Thus $\gamma \in \mathcal{C}$. For $p\in \bar{B}(q_0,R\mathrm{/}2)$ define $T_p\gamma (t):=p+{\int }_0^{t}X(\gamma (s))ds$. The integrand is continuous and hence integrable by [F3]. Then $T_p$ maps $\mathcal{C}$ into itself, since $\mathrm{\parallel }{T}_p\gamma (t)-q_0\mathrm{\parallel }\le \mathrm{\parallel }p-q_0\mathrm{\parallel }+M\mathrm{\mid }t\mathrm{\mid }\le R$, and is a contraction with constant at most $L\delta$.

Starting at the constant curve ${\gamma }_0(t)=p$, iterate ${\gamma }_{k+1}=T_p{\gamma }_k$. If $\kappa =L\delta <1$, then $\mathrm{\parallel }{\gamma }_{k+1}-{\gamma }_k{\mathrm{\parallel }}_{\mathrm{\infty }}\le {\kappa }^k\mathrm{\parallel }{\gamma }_1-{\gamma }_0{\mathrm{\parallel }}_{\mathrm{\infty }}$; the geometric tail bound makes $({\gamma }_k)$ uniformly Cauchy. Its limit $\varphi (\cdot ,p)\in \mathcal{C}$ is a fixed point by the contraction bound. Any two fixed points have distance at most $\kappa$ times that distance, so they coincide.

For $g(s)=X(\varphi (s,p))$, continuity and [F3] make every component Riemann integrable. At an interior $t$, the difference quotient of $\varphi (t,p)=p+{\int }_0^{t}g(s)ds$ differs from $g(t)$ by at most ${\sup }_{\mathrm{\mid }s-t\mathrm{\mid }\le \mathrm{\mid }h\mathrm{\mid }}\mathrm{\parallel }g(s)-g(t)\mathrm{\parallel }$, which tends to zero by continuity; thus ${\mathrm{\partial }}_t\varphi =X(\varphi )$. Conversely any $C^1$ solution staying in the ball satisfies the fixed-point equation by [F4], so uniqueness of the fixed point is uniqueness of the IVP.

If $p,p^{\mathrm{\prime }}$ lie in $\bar{B}(q_0,R\mathrm{/}2)$, subtract their fixed-point equations from step 3.1 and take suprema to obtain $\mathrm{\parallel }\varphi (\cdot ,p)-\varphi (\cdot ,p^{\mathrm{\prime }}){\mathrm{\parallel }}_{\mathrm{\infty }}\le \mathrm{\parallel }p-p^{\mathrm{\prime }}\mathrm{\parallel }+L\delta \mathrm{\parallel }\varphi (\cdot ,p)-\varphi (\cdot ,p^{\mathrm{\prime }}){\mathrm{\parallel }}_{\mathrm{\infty }}$. Since $L\delta <1$, division gives the local Lipschitz estimate $(1-L\delta {)}^{-1}\mathrm{\parallel }p-p^{\mathrm{\prime }}\mathrm{\parallel }$.

Whenever the compositions remain in the chosen ball, both $t\mapsto \varphi (t+s,p)$ and $t\mapsto \varphi (t,\varphi (s,p))$ solve the same IVP at $t=0$, with initial value $\varphi (s,p)$. Uniqueness from step 4.1 gives the local flow law; setting $t=-s$ gives the local inverse relation.