Theorem 2

Let $I \subset \mathbf{R}$ be an interval (not reducing to a single point) with left-hand endpoint $t_0 \in I$, let $H$ be a nonempty open set in $E$, and $f$ a locally Lipschitz function on $I \times H$.

$1^\circ$ For $x_0 \in H$ there exists a largest interval $J \subset I$, with left-hand endpoint $t_0 \in J$, on which there exists an integral $u$ of the equation $x' = f(t, x)$ taking values in $H$ and equal to $x_0$ at the point $t_0$; this integral is unique.

$2^\circ$ If $J \neq I$ then $J$ is a half-open interval $[t_0, \beta[$ of finite length; further, for every compact subset $K \subset H$ the set $\overline{u}(K)$ is a compact subset of $\mathbf{R}$.

$3^\circ$ If $J$ is bounded, and if $f(t, u(t))$ is bounded on $J$, then $u(t)$ has a left limit $c$ at the right-hand endpoint of $J$; further, if $J \neq I$ then $c$ is a boundary point of $H$ in $E$.

$1^\circ$ Let $\mathfrak{M}$ be the set of intervals $L$ (not reducing to a single point) with left-hand endpoint $t_0 \in L$ which are contained in $I$ and are such that on $L$ there is a solution of (1) (IV, p. 163) with values in $H$ and equal to $x_0$ at $t_0$; by th. 1 (IV, p. 171) the set $\mathfrak{M}$ is not empty. Let $L$ and $L'$ be two intervals belonging to $\mathfrak{M}$, and suppose, for example, that $L \subset L'$; if $u$ and $v$ are two integrals of (1) defined respectively on $L$ and $L'$, with values in $H$, and equal to $x_0$ at $t_0$, we shall see that $v$ is an extension of $u$. Indeed, let $t_1$ be the supremum of the set of $t \in L$ such that $u(s) = v(s)$ for $t_0 \leq s \leq t$; we shall show that $t_1$ is the right-hand endpoint of $L$. If this were not so, we would have $u(t_1) = v(t_1)$ by continuity, and $x_1 = u(t_1)$ would belong to $H$; since $f$ is locally Lipschitz, th. 1 shows that there can exist only one integral of (1) defined on a neighbourhood of $t_1$ with values in $H$ and equal to $x_1$ at $t_1$; it is therefore a contradiction to suppose that $t_1$ is not the right-hand endpoint of $L$. We now see that if $J$ is the union of the intervals $L \in \mathfrak{M}$ there exists one and only one integral $u$ of (1), defined on $J$, with values in $H$ and equal to $x_0$ at $t_0$.

2) Suppose that J ≠ I and let β be the right endpoint of J; if β is the right-hand endpoint of I then β ∈ I (so β is finite) and J = [t_0, β[ by hypothesis. Suppose then that β is not the right-hand endpoint of I; if β ∈ J then u(β) = c belongs to H; by th. 1 there exists an integral of (1) with values in H, defined on an interval

and equal to c at β; then J would not be the largest of the intervals in $\mathfrak{M}$, which is absurd; so J = [t_0, β[.

If K is a compact subset of H then $\bar{u}^{-1}(K)$ is closed in J; we shall see that there exists a γ ∈ J such that $\bar{u}^{-1}(K)$ is contained in [t_0, γ ], which will prove that $\bar{u}^{-1}(K)$ is compact. If not, there would be a point c ∈ K such that (β, c) is a cluster point of the set of points (t, u(t)) such that t < β and u(t) ∈ K. Since β ∈ I and c ∈ H there exists a neighbourhood V of β in I, and an open ball S with centre c and radius r contained in H, such that f is Lipschitz and bounded on V × S; let M be the supremum of $\|f(t, x)\|$ over this set. By hypothesis there exists a t_1 ∈ J such that $\beta - t_1 < r/2M$, $t_1 \in V$ and $\|u(t_1) - c\| \leq r/2$; th. 1 shows that there exists one and only one integral of (1), with values in H, defined on an interval [t_1, t_2] containing β, and equal to u(t_1) at t_1; since this interval coincides with u on the interval [t_1, β[ it follows that J = [t_0, β[ is not the largest interval in $\mathfrak{M}$, which is absurd.

3 Suppose that J is bounded and that $\|f(t, u(t))\| \leq M$ on J; then $\|u'(t)\| \leq M$ on the complement of a countable subset of J; then, by the mean value theorem, $\|u(s) - u(t)\| \leq M |s - t|$ for any s and t in J; by the Cauchy criterion, u has a left limit c at the right endpoint β of J. If J ≠ I then c cannot belong to H, for on extending u by continuity at β, u would be an integral of (1) with values in H, defined on an interval [t_0, β] and equal to x_0 at t_0; then one would have J = [t_0, β], contradicting what we have seen in 2 .