Proposition 18.2.7. Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset B(E)$ and $\tau \subset B(F)$ correspond to the following families of sets on $E$ and $F$:
Compact sets.
Bounded sets.
then
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 $s \in \mathcal{R}_{\tau}(F; G)$, $s \circ (T + r) \in \mathcal{R}_{\sigma}(E; G)$.
and by Proposition 18.2.6, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
Proof. (1): Let $A \in \sigma$ and $U \in \cn_{G}(0)$. Since $T$ is continuous, there exists $V \in \cn_{F}(0)$ such that $T(V) \subset U$. Since $r \in \mathcal{R}_{\sigma}(E; G)$, there exists $t > 0$ such that $r(sA)/s \in V$ for all $s \in (0, t)$. In which case, $T \circ r(sA)/s \in U$ for all $s \in (0, t)$.
(2): To show that $s \circ (T + r) \in \mathcal{R}_{\sigma}(E; G)$, it is sufficient to show that for every $A \in \sigma$, $\seq{t_n}\subset \real_{> 0}$ with $t_{n} \downto 0$ as $n \to \infty$, and $\seq{a_n}\subset A$,
Since $\bracs{t_n^{-1}r(t_na_n)|n \in \natp}$ is a convergent sequence, it is contained in a compact set. Thus
is contained in a compact set if $A$ is compact, and bounded if $A$ is bounded. Given that $s \in \mathcal{R}_{\sigma}(E; F)$, $t^{-1}s(tx) \to 0$ as $t \downto 0$ uniformly on $B$. Therefore