Polished A-A and added new lines for broken enumerates.
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@@ -9,7 +9,8 @@
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\begin{enumerate}
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\item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
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\item The derivative $D_\sigma^{n-1}f: U \to B^{n-1}_\sigma(E; F)$ is derivative at $x_0$.
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\end{enumerate}
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\end\{enumerate\}
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In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) \in L(E; B^{n-1}_\sigma(E; F))$ is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
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The mapping $f: U \to F$ is \textbf{$n$-fold $\sigma$-differentiable on $U$} if it is $n$-fold $\sigma$-differentiable at every point in $U$. Under the identification $B_\sigma(E; B^{n-1}_\sigma(E; F)) = B_\sigma^{n}(E; F)$ given by \autoref{proposition:multilinear-identify},
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@@ -17,7 +17,8 @@
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\begin{enumerate}
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\item[(a)] $f, g$ are right-differentiable on $[a, b] \setminus N$.
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\item[(b)] For every $x \in [a, b] \setminus N$, $D^+f(x) \le D^+g(x)$.
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\end{enumerate}
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\end\{enumerate\}
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then for any $x \in [a, b]$, $f(x) - f(a) \le g(x) - g(a)$.
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\end{lemma}
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\begin{proof}
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@@ -51,7 +52,8 @@
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\item $f, g$ are right-differentiable on $[a, b] \setminus N$.
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\item For each $x \in [a, b] \setminus N$, $D^+f(x) \in D^+g(x)B$.
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\item $g$ is non-decreasing.
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\end{enumerate}
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\end\{enumerate\}
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then
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\[
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f(b) - f(a) \in [g(b) - g(a)]B
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@@ -8,7 +8,8 @@
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\item For each $A \in \sigma$, $r(th)/t^n \to 0$ uniformly on $A$.
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\item If $r_t(x) = r(tx)/t^n$, then $r_t \to 0$ as $t \to 0$ with respect to the $\sigma$-uniform topology on $F^E$.
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\item For each $A \in \sigma$, $\seq{a_k} \subset A$, and $\seq{t_k} \subset K \setminus \bracs{0}$ with $t_k \to 0$ as $n \to \infty$, $r(t_ka_k)/t_k^n \to 0$ as $n \to \infty$.
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\end{enumerate}
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\end\{enumerate\}
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If the above holds, then $r$ is \textbf{$\sigma$-small of order $n$}.
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The set $\mathcal{R}_\sigma^n(E; F)$ is the $K$-vector space of all $\sigma$-small functions of order $n$ from $E$ to $F$. For simplicity, $\mathcal{R}_\sigma(E; F)$ denotes $\mathcal{R}_\sigma^1(E; F)$.
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@@ -52,7 +53,8 @@
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\begin{enumerate}
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\item[(a)] For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
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\item[(b)] For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
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\end{enumerate}
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\end\{enumerate\}
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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
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\[
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D_\sigma(g \circ f)(x_0) = D_\tau g(f(x_0)) \circ D_\sigma f(x_0)
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@@ -84,12 +86,14 @@
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\begin{enumerate}
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\item Compact sets.
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\item Bounded sets.
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\end{enumerate}
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\end\{enumerate\}
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then
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\begin{enumerate}
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\item For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
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\item For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
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\end{enumerate}
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\end\{enumerate\}
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and by \autoref{proposition:chain-rule-sets}, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
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\end{proposition}
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\begin{proof}
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