1° For every continuous map $u$ of $I$ into $H$ the function $t \mapsto f(t, u(t))$ is regulated on $I$; more precisely, the right (resp. left) limit of this function at a point $t_0 \in I$ is equal to the right (resp. left) limit of the function $t \mapsto f(t, u(t_0))$ at the point $t_0$.

2° If $(u_n)$ is a sequence of maps of $I$ into $H$ which converges uniformly to a continuous function $u$ of $I$ into $H$ on every compact subset of $I$, then the sequence of functions $t \mapsto f(t, u_n(t))$ converges uniformly to $f(t, u(t))$ on every compact subset of $I$.

1 Let $c$ be the right limit of $f(t, u(t_0))$ at the point $t_0$; for every $\varepsilon > 0$ there is a compact neighbourhood $V$ of $t_0$ in $I$ such that $\|f(t, u(t_0)) - c\| \leq \varepsilon$ for $t \in V$ and $t > t_0$. On the other hand, there exists $\delta > 0$ such that the relations

imply $\|f(s, x) - f(s, u(t_0))\| \leq \varepsilon$ for all $s \in V$; if $W \subset V$ is a neighbourhood of $t_0$ in $I$ such that $\|u(t) - u(t_0)\| \leq \delta$ for every $t \in W$, then $\|f(t, u(t)) - c\| \leq 2\varepsilon$ for $t \in W$ and $t > t_0$, which proves that $c$ is the right limit of $f(t, u(t))$ at the point $t_0$.

2 Let $K$ be a compact subset of $I$; since $u$ is continuous on $I$, $u(K)$ is a compact subset of $H$; for every $\varepsilon > 0$ and $x \in u(K)$ there exists a number $\delta_x$ such that, for every $y \in H, \|y - x\| \leq \delta_x$ and every $t \in K$, one has $\|f(t, y) - f(t, x)\| \leq \varepsilon$. There is a finite number of points $x_i \in u(K)$ such that the closed balls with centre $x_i$ and radius $\frac{1}{2}\delta_{x_i}$ form a cover of $u(K)$; let $\delta = \operatorname{Min}(\delta_{x_i})$. By hypothesis there exists an integer $n_0$ such that for $n \geq n_0$ one has $\|u_n(t) - u(t)\| \leq \frac{1}{2}\delta$ for every $t \in K$. Now, for every $t \in K$ there exists an index $i$ such that

consequently one has $\|u_n(t) - x_i\| \leq \delta_x$, whence $\|f(t, u_n(t)) - f(t, u(t))\| \leq 2\varepsilon$ for every $t \in K$ and every $n \geq n_0$

For the rest of this section I will denote an interval contained in $\mathbf{R}$, not reducing to a single point, $H$ an open nonempty set contained in the normed space $E$, and $f$ a map from $I \times H$ into $E$ satisfying the hypotheses of lemma 1.

Indeed, by lemma 1, if $u$ is a solution of (1) on $I$, then $f(t, u(t))$ is regulated, so the right-hand side of (6) is defined and equal to $u(t)$ for every $t \in I$. Conversely, if $u$ is a continuous function that satisfies (6) then $f(t, u(t))$ is regulated, by lemma 1, so $u$ has derivative equal to $f(t, u(t))$ except at the points of a countable subset of $I$.

Indeed, such a solution $u$ is a primitive of the continuous function $f(t, u(t))$ (II, p. 66, prop. 3).

Furthermore, we note that a continuous function $f$ on $I \times H$ satisfies the conditions of lemma 1 (Gen. Top., X, p. 286, cor. 3).

In the sequel we shall choose $t_0 \in I$ and $x_0 \in H$ arbitrarily and investigate whether there exist solutions of (1) on $I$ (or on a neighbourhood of $t_0$ in $I$) taking the value $x_0$ at the point $t_0$ (or, what comes to the same, solutions of (6)).

We suppose that $t_0$ is not the right-hand endpoint of J, and prove the proposition for intervals with left-hand endpoint $t_0$. Let $\mathscr{M}$ be the set of solutions of (1) to within $\varepsilon$, each of which takes values in S, is equal to $x_0$ at $t_0$, and is defined on a half open interval $[t_0, b[$ contained in J (the interval depending on the approximate solution under consideration). First we show that $\mathscr{M}$ is not empty. Let c be right limit of $f(t, x_0)$ at $t_0$; by lemma 1 (IV, p. 165) the function $f(t, x_0 + c(t - t_0))$ has a right limit equal to c at $t_0$, so the restriction of the function $x_0 + c(t - t_0)$ to a sufficiently small half open interval $[t_0, b[$ will belong to $\mathscr{M}$.

We order the set $\mathscr{M}$ by the relation "u is a restriction of v", and show that $\mathscr{M}$ is inductive (Set Theory, III, p. 154). Let $(u_\alpha)$ be a totally ordered subset of $\mathscr{M}$ and $[t_0, b_\alpha[$ the interval where $u_\alpha$ is defined: for $b_\alpha \leq b_\beta$ the function $u_\beta$ is thus an extension of $u_\alpha$. The union of the intervals $[t_0, b_\alpha[$ is an interval $[t_0, b[$ contained in J, and there exists one and only one function u defined on $[t_0, b[$ that coincides with $u_\alpha$ on $[t_0, b_\alpha[$ for each $\alpha$; among the $b_\alpha$ there is an increasing sequence $(b_{\alpha_n})$ tending to b; since u agrees with $u_{\alpha_n}$ on $[t_0, b_{\alpha_n}[$ the function u admits a derivative satisfying (7) at all the points of the complement of a countable subset of $[t_0, b[$, and so is the supremum of the set $(u_\alpha)$ in $\mathscr{M}$.

By Zorn's lemma (Set Theory, III, p. 154, th. 2), $\mathscr{M}$ admits a maximal element $u_0$; we shall show that if $[t_0, t_1[$ is the interval where $u_0$ is defined, then either $t_1$ is the right-hand endpoint of J, or else $t_1 - t_0 \geq r/(M + \varepsilon)$. We argue by contradiction, supposing that neither of these conditions holds; first we show that one can extend $u_0$ by continuity at the point $t_1$; in fact, for any s and t in $[t_0, t_1[$,

by the mean value theorem; Cauchy's criterion shows that $u_0$ admits a left limit $x_1 \in S$ at the point $t_1$. Now let $c_1$ be the right limit at $t_1$ of the function $f(t, x_1)$; one has $\|c_1\| \leq M$; the same argument as that at the beginning of the proof shows that one can extend $u_0$ to a half open interval with left-hand endpoint $t_1$ by the function $x_1 + c_1(t - t_1)$, so that the extended function belongs to $\mathscr{M}$, which is absurd. This proves the proposition.

When f is uniformly continuous on $J \times S$ one can prove prop. 3 without using Zorn's lemma (IV, p. 199, exerc. 1a)).

Indeed, if u is any function in this set and s and t are two points of K, then by the mean value theorem

Indeed, by prop. 3, once $n$ is large enough there is an approximate solution $u_n$ of (1) to within $1/n$, defined on $K$, with values in $S$, and equal to $x_0$ at $t_0$. Further, from some value of $n$ on, $u_n(K)$ is contained in a closed ball with centre $x_0$ and radius $< r$, independent of $n$. The set of $u_n$ is equicontinuous (prop. 4), and since $E$ is finite dimensional the set $S$ is relatively compact in $E$; so for every $t \in K$ the set of $u_n(t)$ is relatively compact in $E$. By Ascoli’s theorem (Gen. Top., X, p. 290, th. 2) the set of $u_n$ is relatively compact in the space $\mathscr{F}(K; E)$ of maps from $K$ into $E$ endowed with the uniform norm. Thus there is a sequence extracted from $(u_{n_k})$ of $(u_n)$ which converges uniformly on $K$ to a continuous function $u$. One has $u(K) \subset S$, so $t \mapsto f(t, u(t))$ is defined on $K$; by lemma 1 (IV, p. 165), $f(t, u_{n_k}(t))$ converges uniformly to $f(t, u(t))$ on $K$; by (IV, p. 4, formula (7)), $u_{n_k}$ is a primitive of a function which tends uniformly to $f(t, u(t))$ on $K$, so (II, p. 52, th. 1) $u$ is a solution of (1) on $K$, and equal to $x_0$ at the point $t_0$.

where

From the relation $\|u'(t) - f(t, u(t))\| \leq \varepsilon_1$, valid on the complement of a countable set, one deduces, from the mean value theorem, that

and similarly

whence

By the Lipschitz condition (8) one has

whence, putting $w(t) = \|u(t) - v(t)\|$,

The proposition is thus a consequence of the following lemma:

(12)

where $\varphi$ is a regulated function $\geq 0$ on $[t_0, t_1]$, then, for $t_0 \leq t \leq t_1$,

(13)

Put $y(t) = \int_{t_0}^t k(s) w(s) \, ds$; the relation (12) implies that

(14)

on the complement of a countable set.

Put $z(t) = y(t) \exp \left( - \int_{t_0}^t k(s) \, ds \right)$; then (14) is equivalent to

On applying the mean value theorem (I, p. 15, th. 2) to this inequality, and noting that $z(t_0) = 0$, we obtain

whence

and since $w(t) \leq \varphi(t) + y(t)$ one thus obtains (13).

(15)

This inequality is in fact an immediate consequence of (9) when $t \geq t_0$; to prove it for $t \leq t_0$ it suffices to apply it to the equation

obtained from (1) by the change of variable $t = -s$.

Suppose further that $\mathbf{g}$ is Lipschitz for the constant $k > 0$ on $I \times H$. In these circumstances, if $\mathbf{u}$ is an approximate solution of $\mathbf{x}' = \mathbf{f}(t, \mathbf{x})$ to within $\varepsilon_1$, defined on $I$, with values in $H$, and $\mathbf{v}$ is an approximate solution of $\mathbf{x}' = \mathbf{g}(t, \mathbf{x})$ to within $\varepsilon_2$, defined on $I$, with values in $H$, then, for all $t \in I$

(17)

Indeed

for all $t$ in the complement in $I$ of a countable subset of $I$; in other words, $\mathbf{u}$ is an approximate solution of $\mathbf{x}' = \mathbf{g}(t, \mathbf{x})$ to within $\alpha + \varepsilon_1$, so the inequality (17) follows on applying prop. 5 of IV, p. 169.

Indeed, for $\varepsilon > 0$ sufficiently small, the set $F_\varepsilon$ of approximate solutions of (1) to within $\varepsilon$, defined on K, with values in S, and equal to $\mathbf{x}_0$ at the point $t_0$, is not empty (IV, p. 166, prop. 3); further, if $\mathbf{u}$ and $\mathbf{v}$ belong to $F_\varepsilon$ then, by (15) (IV. p. 170)

for all $t \in K$, so the sets $F_\varepsilon$ form a filter base $\mathscr{G}$ which converges uniformly on $K$ to a continuous function $w$, equal to $x_0$ at $t_0$; also $w$ takes values in $S$, since, for $\varepsilon$ small enough, the functions $u \in F_\varepsilon$ take their values in a closed ball contained in $S$. Since $f(t, u(t))$ tends uniformly on $K$ to $f(t, w(t))$ along $\mathscr{G}$, $w$ satisfies equation (6) of IV, p. 165, so is a solution of (1). The uniqueness of the solution follows immediately from inequality (15) of IV, p. 170 where one takes $\varepsilon_1 = \varepsilon_2 = 0$ and $u(t_0) = v(t_0)$.

We shall say that a function $f$ defined on $I \times H$ is locally Lipschitz if, for every point $(t, x)$ of $I \times H$, there exists a neighbourhood $V$ of $t$ (with respect to $I$) and a neighbourhood $S$ of $x$ such that $f$ is Lipschitz on $V \times S$ (for a constant $k$ depending on $V$ and $S$). By the Borel-Lebesgue theorem, for every compact interval $J \subset I$ and every point $x_0 \in H$ there exists an open ball $S$ with centre $x_0$, contained in $H$, such that $f$ is Lipschitz on $J \times S$; thus $f$ satisfies the hypotheses of lemma 1 of IV, p. 3. When $f$ is locally Lipschitz on $I \times H$ we shall say that the equation $x' = f(t, x)$ is locally Lipschitz on $I \times H$.

We shall generalize and clarify th. 1 of IV, p. 10 for locally Lipschitz equations; we restrict ourselves to the case where $t_0$ is the left endpoint of the interval $I$; one can pass easily to the case where $t_0$ is an arbitrary point of $I$ (cf. IV, §IV ??, p. 9, corollary).

$1^\circ$ 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 .