Theorem 4

Let $f$ be a locally Lipschitz function on $I \times H$ and $(a, b)$ an arbitrary point of $I \times H$.

1 There exist an interval $K \subset I$, a neighbourhood of $a$ in $I$, and a neighbourhood $V$ of $b$ in $H$ such that, for every point $(t_0, x_0)$ of $K \times V$, there exists an integral (and only one) $u(t, t_0, x_0)$ defined on $K$, with values in $H$ and equal to $x_0$ at the point $t_0$ (in other words, $J(t_0, x_0) \supset K$ for all $(t_0, x_0) \in K \times V$).

2 The map $(t, t_0, x_0) \mapsto u(t, t_0, x_0)$ of $K \times K \times V$ into $H$ is uniformly continuous.

3 There exists a neighbourhood $W \subset V$ of $b$ in $H$ such that, for every point

the equation $x_0 = u(t_0, t, x)$ has a unique solution $x$ on $V$ equal to $u(t, t_0, x_0)$ ("resolution of the integral with respect to the constant of integration").

1 Let $S$ be a ball with centre $b$ and radius $r$ contained in $H$, and $J_0$ an interval contained in $I$, a neighbourhood of $a$ in $I$, such that $f$ is bounded and Lipschitz (with respect to some constant $k$) on $J_0 \times S$; denote by $M$ the supremum of $\|f(t, x)\|$ on $J_0 \times S$. Then there exist (IV, p. 171, th. 1) an interval $J \subset J_0$, a neighbourhood of $a$ in $I$, and an integral $v$ of $x' = f(t, x)$ defined on $J$, with values in $S$ and equal to $b$ at $a$. We shall see that the open ball $V$ with centre $b$ and radius $r/2$, and the intersection $K$ of $J$ with an interval $]a - l, a + l[$, where $l$ is small enough, are as required. Indeed, prop. 7 of IV, p. 174 (applied to the set $J_0 \times S$ and the case where $\alpha = 0$) shows that there exists an integral of $x' = f(t, x)$ defined on $K$, with values in $S$, and equal to $x_0$ at a point $t_0 \in K$, provided that

for every $t \in K$. Now, by the mean value theorem, one has for every $t \in K$; since $\|x_0 - b\| < r/2$ one sees that it suffices to take $l$ such that or the relation (18) to be satisfied for every $(t, t_0, x_0)$ of $K \times K \times V$.

2) By the mean value theorem we have

(20)

for all $t_0, t_1, t_2$ in K and $\mathbf{x}_0$ in V. Now prop. 5 (IV, p. 169) shows that

(21)

for every $t$ and $t_0$ in K, and $\mathbf{x}_1$ and $\mathbf{x}_2$ in V. Finally, if $t_1$ and $t_2$ are any two points in K,

by the mean value theorem, that is to say

since $\mathbf{u}(t, t_2, \mathbf{x}_0)$ is identical to the integral which takes the value $\mathbf{u}(t_1, t_2, \mathbf{x}_0)$ at the point $t_1$, prop. 5 (IV, p. 169) shows that

(22)

for all $t, t_1, t_2$ in K and $\mathbf{x}_0$ in V. The three inequalities (20), (21) and (22) thus prove the uniform continuity of the map $(t, t_0, \mathbf{x}_0) \mapsto \mathbf{u}(t, t_0, \mathbf{x}_0)$ on $K \times K \times V$.

3 By (20), we have $\| \mathbf{u}(t, t_0, \mathbf{x}_0) - \mathbf{x}_0 \| \leq M |t - t_0| \leq 2Ml$ on

If $l$ is taken small enough, so that $2Ml < r/4$, one then sees that if $\mathbf{x}_0$ is any point of the open ball W with centre $\mathbf{b}$ and radius $r/4$, that $\mathbf{u}(t, t_0, \mathbf{x}_0) \in V$ for any $t$ and $t_0$ in K. If $\mathbf{x} = \mathbf{u}(t, t_0, \mathbf{x}_0)$, the function $s \mapsto \mathbf{u}(s, t, \mathbf{x})$ is then defined on K and is equal to the integral of (1) which takes the value $\mathbf{x}$ at the point $t$, that is, to $\mathbf{u}(s, t_0, \mathbf{x}_0)$; in particular

Moreover, if $y \in V$ is such that $\mathbf{x}_0 = \mathbf{u}(t_0, t, y)$, then the integral $s \mapsto \mathbf{u}(s, t, y)$ takes the value $\mathbf{x}_0$ at $t_0$ so is identical to $s \mapsto \mathbf{u}(s, t_0, \mathbf{x}_0)$, which consequently takes the value $\mathbf{x}$ at $t$, which shows that $y = \mathbf{x}$ and concludes the proof.