Exercise 9, § 1

Let I be an interval in $\mathbf{R}$, and $t_0$ a point of I, let S be a ball with radius r and centre $x_0$ in E, let G be the normed space of bounded maps from $I \times S$ into E, the norm of such a map $f$ being the supremum $\|f\|$ of $\|f(t, x)\|$ over $I \times S$. For every $M > 0$ let $G_M$ be the ball $\|f\| \leq M$ in G. Let L be the subset of G formed by the Lipschitz maps of $I \times S$ into E; for every function $f \in L \cap G_M$ let $u = U(f)$ be the integral of $x' = f(t, x)$ such that $u(t_0) = x_0$, with values in S, and defined on the intersection $J_M$ of I and the interval $\left[ t_0 - \frac{r}{M}, t_0 + \frac{r}{M} \right]$ (th. 1).

a) Let $(f_n)$ be a sequence of functions belonging to $L \cap G_M$; if $f_n$ converges uniformly on $I \times S$ to a function $f$, then every cluster point (for the topology of compact convergence) of the sequence of $u_n = U(f_n)$ in the space F of bounded maps from $J_M$ into E, is an integral of $x' = f(t, x)$ taking the value $x_0$ at the point $t_0$. Conversely, every integral v of $x' = f(t, x)$ defined on $J_M$ and such that $v(t_0) = x_0$, is also an integral of an equation $x' = g(t, x)$, where g is Lipschitz and arbitrarily close to $f$ in G (consider the equation

b) Suppose further that E is of finite dimension. Show that if $f \in G_M$ satisfies the hypotheses of lemma 1, then for every $t \in J_M$ the set $H(t)$ of values at the point t of the integrals of $x' = f(t, x)$ which take the value $x_0$ at the point $t_0$ is a compact and connected set (to see that $H(t)$ is closed, use Ascoli’s theorem; to see that it is connected, use a): if $x_1, x_2$ are two points of $H(t)$ and $\varepsilon > 0$ is arbitrary, show that there exists a connected set $\Phi$ of functions g belonging to $L \cap G_M$ such that $\|f - g\| \leq \varepsilon$ for every function $g \in \Phi$, and that the set of values of the functions $U(g)$ at the point t is connected and contains $x_1$ and $x_2$. Conclude by passing to the limit along an ultrafilter finer than the neighbourhood filter of 0 in $\mathbf{R}$.