Exercise 13, § 1

a) Let I = $[t_0, t_0 + c]$ and let $\omega$ be a continuous real function, $\geq 0$, defined on $I \times \mathbf{R}$. Let S be a ball with centre $x_0$ in a complete normed space E, and let f be a continuous map from $I \times S$ into E such that for any $t \in I$, $x_1 \in S$ and $x_2 \in S$ one has $\|f(t, x_1) - f(t, x_2)\| \leq \omega(t, \|x_1 - x_2\|)$. Let u and v be two integrals of $x' = f(t, x)$ defined on I, with values in S, such that $u(t_0) = x_1,\ v(t_0) = x_2$; let w be the maximal integral (exerc. 10) of $z' = \omega(t, z)$ corresponding to the point $(t_0, \|x_1 - x_2\|)$, assumed to be defined on I; show that, on I, one has $\|u(t) - v(t)\| \leq w(t)$. (Let $w(t, \varepsilon)$ be the maximal integral of $z' = \omega(t, z) + \varepsilon$ corresponding to the point $(t_0, \|x_1 - x_2\|)$, where $\varepsilon > 0$ is sufficiently small. Show that $\|u(t) - v(t)\| \leq w(t, \varepsilon)$ for every $\varepsilon > 0$, arguing by contradiction.)

b) In the statement of the hypotheses of a), replace I by the interval $I' = ]t_0 - c, t_0]$. Show that if w is the minimal integral on this interval of $z' = \omega(t, z)$ corresponding to the point $(t_0, \|x_1 - x_2\|)$, then $\|u(t) - v(t)\| \geq w(t)$ on I' (same method).