Added the Mean Value Theorem.

src/dg/derivative/sets.tex

@@ -39,6 +39,12 @@
     Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
 \end{definition}
 
+\begin{definition}]
+\label{definition:derivative-garden}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b \subset 2^E$ be the collection of all finite, compact, and bounded subsets, respectively, then differentiability with respect to $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b$ correspond to \textbf{Gateaux}, \textbf{Hadamard}, and \textbf{Fréchet} differentiability.
+\end{definition}
+
+
 \begin{proposition}[{{\cite[Proposition 4.5.2]{Bogachev}}}]
 \label{proposition:chain-rule-sets}
     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:
@@ -130,3 +136,32 @@
     \]
     is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
 \end{definition}
+
+
+\begin{proposition}
+\label{proposition:derivative-sets-real}
+    Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be an upward-directed system that contains finite sets, then
+    \begin{enumerate}
+        \item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
+        \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is ($\sigma$-)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 ($\sigma$-)derivative of $f$ at $0$.
+    \end{enumerate}
+\end{proposition}
+\begin{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 $x_0 + t \in U$,
+    \begin{align*}
+        f(x_0 + t) - f(x_0) &= Df(x_0)(t) + r(t) \\
+        \frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1) + t^{-1}r(t) \\
+        \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1)
+    \end{align*}
+    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
+    \[
+    \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
+    \]
+    and $Df(x_0) = T$.
+\end{proof}