Theorem 2

For every point $(t_0, \mathbf{x}_0)$ of $J \times E$ let $\mathbf{u}(t, t_0, \mathbf{x}_0)$ be the integral of the homogeneous equation (4) defined on $J$ and equal to $\mathbf{x}_0$ at $t_0$. 1. For every point $t \in J$ the map $\mathbf{x}_0 \mapsto \mathbf{u}(t, t_0, \mathbf{x}_0)$ is a bijective bicontinuous linear map $C(t, t_0)$ of $E$ to itself. 2. The map $t \mapsto C(t, t_0)$ of $J$ into $\mathscr{L}(E)$ is identical to the integral of the homogeneous linear differential equation

which takes the value $I$ **(the identity map of* $E$ to itself) at the point $t_0$. 3. For any points $s, t, u$ of $J$

By prop. 1, $\mathbf{u}(t, t_0, \mathbf{x}_1) + \mathbf{u}(t, t_0, \mathbf{x}_2)$ (resp. $\lambda \mathbf{u}(t, t_0, \mathbf{x}_0)$) is an integral of (4) and takes the value $\mathbf{x}_1 + \mathbf{x}_2$ (resp. $\lambda \mathbf{x}_0$) at $t_0$, so, by th. 1 of IV, p. 179, is identical to $\mathbf{u}(t, t_0, \mathbf{x}_1 + \mathbf{x}_2)$ (resp. $\mathbf{u}(t, t_0, \lambda \mathbf{x}_0)$); the map $\mathbf{x}_0 \mapsto \mathbf{u}(t, t_0, \mathbf{x}_0)$ is thus a linear map $C(t, t_0)$ of E into itself, and one can write $\mathbf{u}(t, t_0, \mathbf{x}_0) = C(t, t_0). \mathbf{x}_0$.

Since the map $(X, Y) \mapsto XY$ of $\mathscr{L}(E) \times \mathscr{L}(E)$ into $\mathscr{L}(E)$ is continuous (Gen. Top., X, p. 298, prop. 8), the map $t \mapsto A(t)U$ of J into $\mathscr{L}(E)$ is regulated for all $U \in \mathscr{L}(E)$; further (Gen. Top., X, p. 296)

so one can apply th. I of IV, p. 179 to the homogeneous linear equation (5); let $V(t)$ be the integral of this equation defined on J and equal to $I$ at $t_0$. One has (I, p. 6, prop. 3)

and for $t = t_0$ we have $V(t)\mathbf{x}_0 = I\mathbf{x}_0 = \mathbf{x}_0$; by th. 1 of IV, p. 179 one must have $V(t).\mathbf{x}_0 = C(t, t_0)\mathbf{x}_0$ for all $\mathbf{x}_0 \in E$, that is, $V(t) = C(t, t_0)$; this proves that $C(t, t_0)$ belongs to $\mathscr{L}(E)$, in other words, that $\mathbf{x}_0 \mapsto C(t, t_0).\mathbf{x}_0$ is continuous on E, and that the map $t \mapsto C(t, t_0)$ is the integral of (5) which is equal to $I$ at $t_0$.

Finally, the integral $s \mapsto C(s, u).\mathbf{x}_0$ of (4) is equal to $C(t, u).\mathbf{x}_0$ at the point $t$, so, by definition,

for any $\mathbf{x}_0 \in E$, whence the first relation (6); since $C(s, s) = I$ one has $C(s, t)C(t, s) = I$, for any $s$ and $t$ in J; this proves (Set Theory, II, p. 86, corollary) that $C(t, t_0)$ is a bijective map of E onto itself, with inverse map $C(t_0, t)$. This completes the proof of the theorem.

One says that $C(t, t_0)$ is the resolvent of equation (2) of IV, p. 178.