Proposition 18.2.9. Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be an upward-directed system that contains finite sets, then
$\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
For any $U \subset \real$ open, $f: U \to E$, and $x_{0} \in U$, $f$ is differentiable at $x_{0}$ if and only if
\[\lim_{t \to 0}\frac{f(x + t) - f(x)}{t}\]exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
For any $U \subset \real$ open, $f: U \to E$, and $x_{0} \in U$, if $f$ is differentiable at $x_{0}$, then $f$ is continuous at $x_{0}$.
Proof. (1): Let $r \in \mathcal{R}_{\sigma}(\real; E)$. For any $R > 0$ and $U \in \cn_{E}(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{B(\real)}(\real; E)$.
(2): Suppose that $f$ is differentiable at $x_{0}$, then there exists $r \in \mathcal{R}_{\sigma}$ such that for any $t \in \real$ with $x_{0} + t \in U$,
Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then
and $Df(x_{0}) = T$.$\square$