Updated the power rule to the non-symmetric generality.

src/cat/notation/index.tex

@@ -13,6 +13,7 @@
     $\lim_{\longleftarrow} A_i$ & Inverse limit of a downward-directed system. & \autoref{definition:inverse-limit} \\
     $\mathbb{D}_n$, $\mathbb{D}$ & Dyadic rationals of level $n$; all dyadic rationals. & \autoref{definition:dyadic} \\
     $\mathrm{rk}(q)$ & Dyadic rank of $q \in \mathbb{D}$. & \autoref{definition:dyadic-rank} \\
-    $M(x)$ & Unique $M(x) \subset \mathbb{N}^+ \cap [1, \mathrm{rk}(x)]$ such that $x = \sum_{n \in M(x)} 2^{-n}$. & \autoref{proposition:dyadic-subset}
+    $M(x)$ & Unique $M(x) \subset \mathbb{N}^+ \cap [1, \mathrm{rk}(x)]$ such that $x = \sum_{n \in M(x)} 2^{-n}$. & \autoref{proposition:dyadic-subset} \\
+    $[n]$ & $\bracs{1, \cdots, n}$ & N/A
 \end{tabular}
 

src/dg/derivative/higher.tex

@@ -2,30 +2,52 @@
 \label{section:higher-derivatives}
 
 
-\begin{definition}[$n$-Fold Differentiability]
+\begin{definition}[Codomain of Derivatives]
-\label{definition:n-differentiable-sets}
+\label{definition:higher-derivatives-codomain}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $L^{(0)}_\sigma(E; F) = F$. For each $n \in \natp$, inductively define
-
-    Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable at $x_0$} if
-    \begin{enumerate}
-        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
-        \item The derivative $D_\sigma^{n-1}f: U \to B^{n-1}_\sigma(E; F)$ is derivative at $x_0$.
-    \end{enumerate}
-
-    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 $x_0$}.
-
-    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},
     \[
-    D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
+    L^{(n)}_\sigma(E; F) = L(E; L^{(n-1)}_\sigma(E; F)) \subset B_\sigma^n(E; F)
     \]
 
-    is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
+    and equip it with the $\sigma$-uniform topology, then under the identification
+    \[
+    I: L^{(n)}_\sigma(E; F) \to B_\sigma^n(E; F) \quad I\lambda(x_1, \cdots, x_n) = \lambda(x_1)\cdots(x_n)
+    \]
+
+    the space $L^{(n)}_\sigma(E; F)$ is a subspace of $B_\sigma^n(E; F)$. 
+\end{definition}
+\begin{proof}
+    By \autoref{proposition:multilinear-identify}. 
+\end{proof}
+
+
+
+\begin{definition}[$n$-Fold Differentiability]
+\label{definition:n-differentiable-sets}
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
+
+    Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\tilde \sigma$-differentiable at $x_0$} if
+    \begin{enumerate}
+        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold $\tilde \sigma$-differentiable on $V$.
+        \item The derivative $D_\sigma^{n-1}f: U \to B^{(n-1)}_\sigma(E; F)$ is $\tilde \sigma$-differentiable at $x_0$.
+    \end{enumerate}
+
+    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) = D_\sigma^{n}f(x_0)$ is the \textbf{$n$-fold $\tilde \sigma$-derivative of $f$ at $x_0$}.
+
+    If $f: U \to F$ is $n$-fold $\tilde \sigma$-differentiable at every point in $U$, then $f$ is \textbf{$n$-fold $\tilde \sigma$-differentiable on $U$}. Under the \hyperref[identification]{proposition:multilinear-identify} $B_\sigma(E; B_\sigma^{n}(E; F)) = B_\sigma^{(n)}(E; F)$, the mapping
+    \[
+    D_\sigma^{n}f: U \to B^{(n-1)}_\sigma(E; F)
+    \]
+
+    is the \textbf{$n$-fold $\tilde \sigma$-derivative of $f$}.
+
+    If for each $1 \le k \le n$, $D_\sigma^{k}f$ takes value in $L^{(k)}_\sigma(E; F)$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable}, and $D_\sigma^{n}f$ is the \textbf{$n$-fold $\sigma$-derivative of $f$}.
 \end{definition}
 
 
 \begin{definition}[Space of Differentiable Functions]
 \label{definition:differentiable-space}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{S}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $D_\sigma^k(U; F)$ is the \textbf{space of $n$-fold $\sigma$-differentiable functions} from $U$ to $F$. 
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $D_\sigma^k(U; F)$ is the \textbf{space of $n$-fold $\sigma$-differentiable functions} from $U$ to $F$. 
 \end{definition}
 
 \begin{theorem}[Symmetry of Higher Derivatives]
@@ -95,7 +117,7 @@
 
 \begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric}
-    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
+    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in L_\sigma^{(n)}(E; F)$ is symmetric.
 \end{theorem}
 \begin{proof}[Proof {{\cite[Proposition 4.5.14]{Bogachev}}}. ]
     Let $\seqf{h_j} \subset E$, $E_0$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_0 \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^*$, the mapping $\phi \circ g: E_0 \cap U \to K$ is $n$-times Fréchet-differentiable, with
@@ -103,47 +125,69 @@
     D_{\mathfrak{B}(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
     \]
 
-    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^n(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
+    by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^{(n)}(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
 \end{proof}
 
-\begin{proposition}[Power Rule]
+\begin{theorem}[Power Rule]
-\label{proposition:multilinear-derivative}
+\label{theorem:power-rule}
-    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space, $T \in B^n_\sigma(E; F)$.
+
+    For each $1 \le m \le n$, let $\text{Inj}([m]; [n])$ be the set of injective mappings from $[m]$ to $[n]$. For any $\phi \in \text{Inj}([m]; [n])$, denote $\phi^c \in \text{Inj}([n-m]; [n] \setminus \phi([m]))$ as the unique increasing injective map. For each $h \in E^{n-m}$, $k \in E^m$, and $1 \le j \le n$, write
     \[
-    T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
+    (h, k)_\phi = \begin{cases}
+        k_{\phi^{-1}(j)} &j \in \phi([m]) \\
+        h_{(\phi^c)^{-1}(j)} &j \not\in \phi([m])
+    \end{cases}
     \]
 
-    be symmetric. For any $x \in E$ and $1 \le k \le n$, let $x^{(k)}$ denote the tuple of $x$ repeated $k$ times, then:
+    For any $x \in E$, denote $x^{(m)}$ as the tuple of $x$ repeated $m$ times, then the mapping $f: E \to F$ defined by $x \mapsto T(x^{(n)})$ is infinitely $\tilde\sigma$-differentiable on $E$, where
     \begin{enumerate}
-        \item The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
+        \item For each $1 \le m \le n$, $x \in E$, and $h \in E^m$, 
-        \item For each $1 \le k \le n$ and $x, h \in E$,
         \[
-        Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
+        D^{m}_\sigma f(x)(h) = \sum_{\phi \in \text{Inj}([m]; [n])}T((x^{(n-m)}, h)_\phi]
         \]
 
-        In particular, $D^kf = n! \cdot T$.
+        In particular, 
-        \item For each $k > n$ and $x \in E$, $Df(x) = 0$.
+        \[
+        D^{n}_\sigma f(x)(h) = \sum_{\phi \in S_n}T(h_{\phi(1)}, \cdots, h_{\phi(n)})
+        \]
+
+        \item For each $m > n$ and $x \in E$, $D^m_\sigma f(x) = 0$.
     \end{enumerate}
-\end{proposition}
+
+    Notably, if $T \in L^{(n)}(E; F)$ is symmetric or $T \in L^n(E; F)$, then $T$ is infinitely $\sigma$-differentiable on $E$. 
+\end{theorem}
 \begin{proof}
-    Suppose inductively that (2) holds for $0 \le k \le n$. Let $G =  B^{k}_\sigma(E; F)$, then $D^k_\sigma f \in B^{n-k}_\sigma(E; G)$ under the identification $B^n_\sigma(E; F) = B^{n-k}_\sigma(E; B^k_\sigma(E; F))$ in \autoref{proposition:multilinear-identify}. By \autoref{theorem:derivative-symmetric}, $D^k_\sigma f$ is also symmetric, so using the Binomial formula,
+    (1): Let $0 \le m \le n - 1$ and suppose inductively that (1) holds for $m$. For each $x, h \in E$, $S \subset [n-m]$, and $1 \le j \le n - m$,
+    \[
+    [(x, h)_S]_j = \begin{cases}
+        h &j \in S \\
+        x &j \not\in S
+    \end{cases}
+    \]
+    
+    By the Binomial formula, for each $h \in E$ and $k \in E^{m}$,
     \begin{align*}
-        D^k_\sigma f(x + h) &= \sum_{r = 0}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)}) \\
+        D^{m}_\sigma f(x + h)(k) &= \sum_{\phi \in \text{Inj}([m]; [n])}T[((x+h)^{(n-m)}, k)_\phi] \\
-        &= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{S \subset [n-m]}T[((x, h)_S, k)_\phi]
-        &+ \underbrace{\sum_{r = 2}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)})}_{r(h)}
     \end{align*}
 
-    For each $k \ge 2$, let $A \in \sigma$ and $U \in \cn_F(0)$, then since $D^k_\sigma f \in B^{n-k}_\sigma(E; F)$, there exists $t > 0$ such that
+    For $\ell \ge 2$, maps in $B_\sigma^{\ell}(E; F)$ are $\sigma$-small, so
     \[
-    \frac{D^k_\sigma f(x^{(n-k)}, (sA)^{(k)})}{t} = s^{k-1}D^k_\sigma f(x^{(n-k)}, A^{(k)}) \subset U
+    r(h) = \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{\substack{S \subset [n-m] \\ |S| \ge 2}}T[((x, h)_S, k)_\phi]
     \]
 
-    for all $s \in (0, t)$. Hence $r \in \mathcal{R}_\sigma(E; G)$, and
+    is $\sigma$-small. Hence
+    \begin{align*}
+        &D^{(m)}_\sigma f(x + h)(k) - D_\sigma^{(m)}f(x)(k) \\
+        &= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{j = 1}^{n-m}T[((x, h)_{\bracs{j}}, k)_\phi] + r(h) \\
+        &= \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, (h, k))_\phi] + r(h)
+    \end{align*}
+
+    and for any $h \in E^{m+1}$,
     \[
-    D^{k+1}_\sigma f(x + h) = f(x) + \frac{n!}{(n-k-1)!}T(x^{(n-k-1)}, h_1, \cdots, h_{k+1}) + r(h)
+    D_\sigma^{(m+1)}f(x)(h) = \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, h)_\phi]
     \]
 
-    by the inductive hypothesis.
+    (2): By (1), $D^n_\sigma f$ is constant. 
-
-    (3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
 \end{proof}

src/dg/derivative/mvt.tex

@@ -105,7 +105,7 @@
 
 \begin{theorem}[Mean Value Theorem]
 \label{theorem:mean-value-theorem}
-    Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and star shaped at $x \in V$, $f: V \to F$ be Gateaux-differentiable on $V$, then for any $y \in V$,
+    Let $E$ be a topological vector space over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $V \subset E$ be open and star shaped at $x \in V$, and $f: V \to F$ be $\tilde \sigma$-differentiable on $V$, then for any $y \in V$,
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
@@ -113,7 +113,7 @@
     where $[x, y] = \bracs{(1 - t)x + ty|y \in [0, 1]}$.
 \end{theorem}
 \begin{proof}
-    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule \autoref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
+    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$, then $g$ is differentiable with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
 
     By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, $f(y) - f(x) = g(1) - g(0)$ is contained in
     \[
@@ -124,7 +124,7 @@
 
 \begin{proposition}
 \label{proposition:zero-derivative-constant}
-    Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and connected, $f: V \to F$ be Gateaux-differentiable on $V$ such that $Df(x) = 0$ for all $x \in V$, then $f$ is constant.
+    Let $E$ be a topological vector space over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $V \subset E$ be open and connected, and $f: V \to F$ be $\tilde \sigma$-differentiable on $V$ with $Df = 0$, then $f$ is constant.
 \end{proposition}
 \begin{proof}
     Let $x \in V$, then for any $U \in \cn(0)$ circled with $U + x \subset V$ and $y \in U + x$,

src/dg/derivative/sets.tex

@@ -17,34 +17,33 @@
 
 \begin{proposition}
 \label{proposition:differentiation-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, and $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, $(\mathcal{H}, \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
 \end{proposition}
 \begin{proof}
     Let $T \in B_\sigma(E; F)$ and suppose that there exists $V \in \cn_E(0)$ circled and $r \in \mathcal{R}_\sigma(E; F)$ such that $T|_V = r|_V$. For any $x \in V$, $\bracs{x} \in \sigma$, so $T(x) = \lim_{t \downto 0}T(tx)/t = 0$ as $F$ is separated.
 \end{proof}
 
 
-\begin{definition}[Derivative]
+\begin{definition}[$\sigma$-Derivative]
 \label{definition:derivative-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\tilde \sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in B_\sigma(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
     \[
     f(x_0 + h) = f(x_0) + Th + r(h)
     \]
 
     for all $h \in V$.
 
-    The linear map $T \in L(E; F)$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}, denoted $D_{\sigma}f(x_0)$.
+    The linear map $T \in B_\sigma(E; F)$ is the \textbf{$\tilde \sigma$-derivative of $f$ at $x_0$}, denoted $D_{\tilde \sigma}f(x_0)$. If $T \in L(E; F)$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$}, and $T$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}.
 \end{definition}
 
-
 \begin{definition}[Differentiable]
 \label{definition:differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ 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 $f$.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$/$\tilde \sigma$-differentiable on $U$} if it is $\sigma$/$\tilde \sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to B_\sigma(E; F)$ is the \textbf{$\sigma$/$\tilde \sigma$-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.
+    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, precompact, 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}
 
 
@@ -134,11 +133,14 @@
     Let $E$ be a TVS over $K \in \RC$, $F$ be a separated locally convex space over $K$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$. Let $\fF \subset 2^{D_\sigma^n(U; F)}$ be a filter such that:
     \begin{enumerate}[label=(\alph*)]
         \item There exists $f: U \to F$ such that $\fF \to f$ pointwise. 
-        \item There exists $f^{(k)}: U \to B^{k}_\sigma(E; F)$ such that for each $1 \le k \le n$, $x \in U$, and $A \in \sigma$ with $x + [0, 1]A \subset U$, $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on $x + [0, 1]A$. 
+        \item For each $1 \le k \le n$, there exists $f^{(k)}: U \to L^{(k)}_\sigma(E; F)$ such that for all $x \in U$, and $A \in \sigma$ with $x + [0, 1]A \subset U$, $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on $x + [0, 1]A$. 
     \end{enumerate}
 
-    then $f \in D_\sigma^n(U; F)$ and $D^k_\sigma f = f^{(k)}$ for all $1 \le k \le n$. 
+    then $f \in D_\sigma^n(U; F)$ and $D^k_\sigma f = f^{(k)}$ for all $1 \le k \le n$. In particular, if $\sigma$ is saturated, then $(b)$ may be replaced by
-
+    \begin{enumerate}
+        \item[(b)] For each $1 \le k \le n$, there exists $f^{(k)}: U \to L^{(k)}_\sigma(E; F)$ such that $D_\sigma^k(\fF) \to f^{(k)}$ uniformly on every $A \in \sigma$. 
+    \end{enumerate}
+    
 \end{theorem}
 \begin{proof}
     Assume without loss of generality that $n = 1$. For any $\varphi \in D^1_\sigma(U; F)$, $x \in U$, and $h \in E$ such that $x + h \in U$, 

src/dg/derivative/taylor.tex

@@ -62,7 +62,7 @@
 
 \begin{theorem}[Taylor's Formula, Peano Remainder]
 \label{theorem:taylor-peano}
-    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
     \[
     g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
     \]
@@ -74,7 +74,7 @@
     r(h) = g(x_0 + h) - g(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
     \]
 
-    For any $1 \le k \le n$, $D^k_\sigma(x_0) \in B^k_\sigma(E; F)$ is symmetric by \autoref{theorem:derivative-symmetric}. Let $T_k(h) = \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})$, then by \autoref{proposition:multilinear-derivative}, for any $\bracs{t_j}_1^\ell \in E$,
+    For any $1 \le k \le n$, $D^k_\sigma(x_0) \in B^k_\sigma(E; F)$ is symmetric by \autoref{theorem:derivative-symmetric}. Let $T_k(h) = \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})$, then by \autoref{theorem:power-rule}, for any $\bracs{t_j}_1^\ell \in E$,
     \[
     D^\ell_\sigma T_k(h)(t_1, \cdots, t_\ell) = \begin{cases}
         0 &\ell > k \\

src/dg/notation.tex

@@ -12,6 +12,7 @@ Differential geometry is the study of things invariant under change of notation.
     $D_\sigma f(x_0)$ & $\sigma$-derivative of $f$ at $x_0$. & \autoref{definition:derivative-sets} \\
     $D_\sigma^n f$ & $n$-fold $\sigma$-derivative. & \autoref{definition:n-differentiable-sets} \\
     $D_\sigma^n(U; F)$ & $n$-fold $\sigma$-differentiable functions. & \autoref{definition:differentiable-space} \\
-    $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{proposition:multilinear-derivative} \\
+    $L^{(n)}_\sigma(E; F)$ & Codomain of derivatives. $L^{(0)}_\sigma(E; F) = F$, $L^{(n)}_\sigma(E; F) = L(E; L_\sigma^{(n-1)}(E; F))$, equipped with the $\sigma$-uniform topology. & \autoref{definition:higher-derivatives-codomain} \\
-    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt} \\
+    $x^{(k)}$ & Tuple of $x$ repeated $k$ times. & \autoref{theorem:taylor-peano} \\
+    $D^+f(x)$ & Right derivative of $f$ at $x$. & \autoref{definition:right-differentiable-mvt}
 \end{tabular}