Theorem 2.1.3.label Let $E, F$ be Banach spaces, $U \subset E$ and $V \subset F$ be open subsets, and $f \in C^{k}(U \times V; L(E; F))$ ($k \ge 2$).
For any $\xi, \eta \in C^{k}(U \times V; E)$, denote
If for every $\xi, \eta \in C^{k}(U \times V; E)$ and $(x, y) \in U \times V$,
then for every $(x_{0}, y_{0}) \in U \times V$, there exists $U_{0} \in \cn_{E}(x_{0})$, $V_{0} \in \cn_{F}(y_{0})$, and a unique $\alpha \in C^{k}(U_{0} \times V_{0}; V)$ such that
- (1)
For every $y \in V_{0}$, $\alpha(x_{0}, y) = y$.
- (2)
For every $(x, y) \in U_{0} \times V_{0}$,
\[(\partial_{x} \alpha)(x, y) = f(x, \alpha(x, y))\]
Proof. Using translation, assume without loss of generality that $x_{0} = 0$ and $y_{0} = 0$. Let $B \in \cn_{E}(0)$ and
then by Lemma 2.1.2, there exists $\delta > 0$, $B_{0} \in \cn_{E}(0)$, and $\beta: (-\delta, \delta) \times B_{0} \times V_{0}$ such that
- (a)
For each $(z, y) \in B_{0} \times V_{0}$, $\beta(0, z, y) = y$.
- (b)
For each $(t, z, y) \in (-\delta, \delta) \times B_{0} \times V_{0}$,
\[\frac{d}{dt}\beta(t, z, y) = f(tz, \beta(t, z, y)) \cdot z\] - (c)
For each fixed $y \in V_{0}$, if $\beta(t, z) = \beta(t, z, y)$, then for each $(t, x) \in (-\delta, \delta) \times B_{0}$ and $h \in E$,
\begin{align*}(\partial_{t}\partial_{z}\beta)(t, z)\cdot h&= t(\partial_{x}f)(tz, \beta(t, z))\cdot h \cdot z + f(tz, \beta(t, z))\cdot h \\&+ (\partial_{y}f)(tz, \beta(t, z)) \circ (\partial_{z} \beta)(t, z)\cdot h \cdot z\end{align*}
Following (c), let $k(t, z) = (\partial_{z}\beta)(t, z)\cdot h - tf(tz, \beta(t, z))\cdot h$, then
where by assumption,
Therefore by (c),
Since $0$ is a solution to the above equation, $k(t) = 0$ by the uniqueness of solutions to ODEs. Hence for every $t \in (-\delta, \delta)$ and $z \in B_{0}$,
By adjusting $\delta$ and $B_{0}$, assume without loss of generality that $\delta > 1$. In which case, if $\alpha(x) = \beta(1, x, y)$, then
$\square$