Proposition 18.2.6. Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be upward-directed families that contain all finite sets. If:
For any $r \in \mathcal{R}_{\sigma}(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_{\sigma}(E; G)$.
For any $r \in \mathcal{R}_{\sigma}(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_{\tau}(F; G)$, $s \circ (T + r) \in \mathcal{R}_{\sigma}(E; G)$.
then for any $U \subset E$ and $V \subset F$ open, $f: U \to V$ $\sigma$-differentiable at $x_{0} \in U$, $g: V \to F$ $\tau$-differentiable at $f(x_{0}) \in V$, $g \circ f: U \to F$ is $\sigma$-differentiable at $x_{0}$ with
Proof. Since $g$ is $\tau$-differentiable at $f(x_{0})$, there exists $s \in \mathcal{R}_{\tau}(F; G)$ such that
for all $h \in F$ such that $f(x_{0}) + h \in V$. By differentiability of $f$, there exists $r \in \mathcal{R}_{\sigma}(E; F)$ such that
for all $h \in E$ such that $x_{0} + h \in U$. Therefore for all $h \in E$ with $x_{0} + h \in U$,
where $D_{\tau} g(f(x_{0})) \circ r \in \mathcal{R}_{\sigma}(E; G)$ by assumption (a), and $d \circ (D_{\sigma} f(x_{0}) + r) \in \mathcal{R}_{\sigma}(E; G)$ by assumption (b).$\square$