Theorem 3

Let F be a topological space and let f be a map from I × H × F into E such that, for every $\xi \in F$, the function $(t, x) \mapsto f(t, x, \xi)$ is Lipschitz on I × H, and such that, when $\xi$ tends to $\xi_0$, $f(t, x, \xi)$ tends uniformly to $f(t, x, \xi_0)$ on I × H. Let $u_0(t)$ be an integral of $x' = f(t, x, \xi_0)$ defined on an interval $J = [t_0, b[$ contained in I, with values in H, and equal to $x_0$ at $t_0$. For every compact interval $[t_0, t_1]$ contained in J there exists a neighbourhood V of $\xi_0$ in F such that, for every $\xi \in V$, the equation $x' = f(t, x, \xi)$ has an integral (and only one) $u(t, \xi)$ defined on $[t_0, t_1]$, with values in H and equal to $x_0$ at $t_0$; moreover, when $\xi$ tends to $\xi_0$ the solution $u(t, \xi)$ tends uniformly to $u_0(t)$ on $[t_0, t_1]$.

Indeed, let $r > 0$ be such that for $t_0 \leq t \leq t_1$ the closed ball with centre $u_0(t)$ and radius $r$ is contained in $H$; if $f(t, x, \xi)$ is Lipschitz with respect to the constant $k > 0$ on $I \times H$ we take $\alpha$ small enough that $\alpha \frac{e^{k(t_1-t_0)} - 1}{k} < r$; taking $V$ such that, for every $\xi \in V$, one has $\| f(t, x, \xi) - f(t, x, \xi_0) \| \leq \alpha$ on $I \times H$, one can apply prop. 7 of IV, p. 174; moreover,

on $[t_0, t_1]$, which completes the proof of the theorem.