\section{Differentiation With Respect to Sets}
\label{section:differentiation-set-system}

\begin{definition}[Small]
\label{definition:differentiation-small}
    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
    \begin{enumerate}
        \item For each $A \in \sigma$, $r(th)/t^n \to 0$ uniformly on $A$.
        \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 $F^E$.
        \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 $n \to \infty$.
    \end{enumerate}

    If the above holds, then $r$ is \textbf{$\sigma$-small of order $n$}.

    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 $\mathcal{R}_\sigma^1(E; F)$.
\end{definition}

\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, 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}[$\sigma$-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{$\tilde \sigma$-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 $T \in B_\sigma(E; F)$ 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{$\sigma$/$\tilde \sigma$-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, relatively 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}[Chain Rule]
\label{proposition:chain-rule-sets}
    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ be covering ideals. If:
    \begin{enumerate}
        \item[(a)] For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
        \item[(b)] For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
    \end{enumerate}

    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
    \[
    D_\sigma(g \circ f)(x_0) = D_\tau g(f(x_0)) \circ D_\sigma f(x_0)
    \]

\end{proposition}
\begin{proof}[Proof {{\cite[Proposition 4.5.2]{Bogachev}}}. ]
    Since $g$ is $\tau$-differentiable at $f(x_0)$, there exists $s \in \mathcal{R}_\tau(F; G)$ such that
    \[
    g(f(x_0) + h) = g \circ f (x_0) + D_\tau g(f(x_0))h + s(h)
    \]

    for all $h \in F$ such that $f(x_0) + h \in V$. By differentiability of $f$, there exists $r \in \mathcal{R}_\sigma(E; F)$ such that
    \[
    f(x_0 + h) = f(x_0) + D_\sigma f(x_0)h + r(h)
    \]

    for all $h \in E$ such that $x_0 + h \in U$. Therefore for all $h \in E$ with $x_0 + h \in U$,
    \begin{align*}
    g \circ f(x_0 + h) &= g \circ f (x_0) + D_\tau g(f(x_0)) \circ D_\sigma f(x_0)h \\
    &+ D_\tau g(f(x_0)) \circ r(h) + s(D_\sigma f(x_0)h + r(h))
    \end{align*}
    where $D_\tau g(f(x_0)) \circ r \in \mathcal{R}_\sigma(E; G)$ by assumption (a), and $d \circ (D_\sigma f(x_0) + r) \in \mathcal{R}_\sigma(E; G)$ by assumption (b).
\end{proof}

\begin{proposition}[{{\cite[Corollary 4.5.4]{Bogachev}}}]
\label{proposition:chain-rule-sets-conditions}
    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ correspond to the following families of sets on $E$ and $F$:
    \begin{enumerate}
        \item Relatively compact sets.
        \item Bounded sets.
    \end{enumerate}

    then
    \begin{enumerate}
        \item For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
        \item For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
    \end{enumerate}

    and by \autoref{proposition:chain-rule-sets}, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
\end{proposition}
\begin{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 $\seq{a_n} \subset A$,
    \[
    \limv{n} \frac{1}{t_n}s \circ (T + r)(t_na_n) = \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
    \]

    Since $\bracs{t_n^{-1}r(t_na_n)|n \in \natp}$ is a convergent sequence, it is contained in a compact set. Thus
    \[
    B = \bracs{Ta_n + \frac{r(t_na_n)}{t_n}\bigg | n \in \natp} \subset T(A) + \bracs{\frac{r(t_na_n)}{t_n}|n \in \natp}
    \]

    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
    \[
    \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
    \]
\end{proof}




\begin{remark}
\label{remark:differentiation-bornology}

    In \autoref{definition:differentiation-small}, the system $\sigma$ can be chosen based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain.

    Consider for example a Hilbert space equipped with its norm and weak topology. The norm itself is differentiable with respect to both topologies, because the bounded sets coincide. Moreover, the data for differentiability needs to only come from a neighbourhood of $0$ in the norm topology. As such, a function may be differentiable even if its domain is too small to have an interior.

    A method of extending this sense of differentiability is to require that \textit{every} extension of the function to some open set, or to the entire space is differentiable. Given that this paves way to confusion for related definitions of differentiability, this definition is not formally included here.
\end{remark}

\begin{theorem}[Interchange of Limits and Derivatives]
\label{theorem:differentiable-uniform-limit}
    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 $\fF \subset 2^{\tilde D_\sigma^n(U; F)}$ be a filter such that:
    \begin{enumerate}[label=(\alph*)]
        \item There exists $f: U \to F$ such that $\fF \to f$ pointwise. 
        \item For each $1 \le k \le n$, there exists $f^{(k)}: U \to B^{k}_\sigma(E; F)$ 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 $f \in \tilde D_\sigma^n(U; F)$ 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 $f^{(k)}: U \to B^{k}_\sigma(E; F)$ 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 $\varphi \in \tilde D^1_\sigma(U; F)$, $x \in U$, and $h \in E$ such that $x + h \in U$, 
    \begin{align*}
        f(x + h) - f(x) - f^{(1)}(x)h &= \underbrace{\varphi(x + h) - \varphi(x) - D_\sigma\varphi(x)h}_{\in \mathcal{R}_\sigma(E; F)} \\
        &+ (f - \varphi)(x + h) - (f - \varphi)(x) \\
        &+ (D_\sigma\varphi - f^{(1)})(x)h 
    \end{align*}

    Since $\fF \to f$ pointwise, for any $S \in \fF$, 
    \[
    (f - \varphi)(x + h) - (f - \varphi)(x) \in \overline{\bracs{(g - \varphi)(x + h) - (g - \varphi)(x)|g \in S}}
    \]
    
    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem}, for any $g \in \tilde D^1_\sigma(U; F)$, 
    \[
    (g - \varphi)(x + h) - (g - \varphi)(x) \in \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + th)h|t \in [0, 1]}
    \]

    Hence
    \[
    (f - \varphi)(x + h) - (f - \varphi)(x) \in  \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + th)h|g \in S, t \in [0, 1]}
    \]

    so for any $t \in (0, 1)$ and $A \in \sigma$, 
    \begin{align*}
    &(f - \varphi)(x + tA) - (f - \varphi)(x) \\
    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)th|g \in S, s \in [0, 1], h \in A} \\
    &= t\ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)h|g \in S, s \in [0, 1], h \in A}
    \end{align*}

    and
    \begin{align*}
    &t^{-1}[(f - \varphi)(x + tA) - (f - \varphi)(x)] \\
    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sth)h|g \in S, s \in [0, 1], h \in A} \\
    &\subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A}
    \end{align*}

    In addition, since $D_\sigma(\fF) \to f^{(1)}$ pointwise, 
    \[
    t^{-1}(f^{(1)} - D_\sigma\varphi)(x)(tA) \subset \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A}
    \]

    as well. 

    Now, let $V \in \cn_F(0)$ be convex and circled. Using assumption (b), let $S \in \fF$ such that for any $\varphi \in S$, 
    \[
    \ol{\conv}\bracs{D_\sigma(g - \varphi)(x + sh)h|g \in S, s \in [0, 1], h \in A} \subset V
    \]
    
    Fix $\varphi \in S$, then as $\varphi$ is differentiable at $x$, there exists $\delta \in (0, 1)$ such that
    \[
    t^{-1}[\varphi(x + tA) - \varphi(x) - D_\sigma\varphi(x)(tA)] \subset V
    \]

    for all $t \in (0, \delta)$. 

    So
    \[
    t^{-1}[f(x + tA) - f(x) - f^{(1)}(x)(tA)] \subset 3V
    \]

    for all $t \in (0, \delta)$. Therefore $f$ is $\tilde \sigma$-differentiable at $x$ with $D_\sigma f(x) = f^{(1)}(x)$. 
    
\end{proof}