There is a theorem in page 100 of Arnold's Mathematical Methods of Classical Mechanics, which says that:
If $\cfrac{dx}{dt} = f(x) = Ax + R_2(x)$, where $A = \cfrac{\partial f}{\partial x}|_{x = 0}$, $R_2(x) = O(x^2)$, and $\cfrac{dy}{dt} = Ay$, $y(0) = x(0)$, then for any $\vphantom{\cfrac12} T>0$ and for any $\xi > 0$ there exists $\delta > 0$ such that if $|x(0)| < \delta$, then $\vphantom{\cfrac12}|x(t) - y(t)| < \xi \delta$ for all $t$ in the interval $0 < t < T$.
How can I prove this theorem rigorously?
Cross-posted at physics.se here.
Lets assume that $R$ is smooth. Therefore if $|f|,|g| < M \le 1$, then $$ |R(f) - R(g)| \le C_1 M |f-g| .$$
Also define $$ C_2 = \sup_{0 \le t \le T} \|e^{tA}\| .$$
Define $$ B(\epsilon) = \{f(t) \in C([0,T]) : \|f\|_\infty \le \epsilon\} $$ $$ G_{x_0}:B(\epsilon) \to C([0,T]) $$ $$ G_{x_0}f(t) = e^{tA} x_0 + \int_0^t e^{(t-s)A} R(f(s)) \, ds $$ Then $$ \|G_{x_0}(f-g)\|_\infty \le T C_2 C_1\epsilon \|f-g\|_\infty $$ Hence there exists $\delta_1>0$ such that if $\epsilon<\delta_1$, then $G_{x_0}$ is a contraction mapping. Also $$ \|G_{x_0}(f)\|_\infty \le C_2 |x_0| + T C_2 C_1\epsilon^2 $$ Hence there exists a $\delta_2$ such that if $\epsilon, |x_0| < \delta_2$, then the range of $G_{x_0}$ is contained in $B(\epsilon)$.
Therefore if $\epsilon, |x_0| < \min\{\delta_1,\delta_2\}$, then $G$ has a fixed point. That will be the function $x(t)$.
Let $z = x-y$. Then $$ \frac{dz}{dt} = Az + R(x) .$$ $$ z(t) = \int_0^t e^{(t-s)A} R(x(s)) \, ds $$ Therefore $$ \|z\|_\infty \le C_2 C_1 \epsilon^2 .$$ Choose $\delta = \min\{\delta_1,\delta_2,\xi/(C_1 C_2)\}$.