From 8a4e6f5ebfde450cf32e5f9ee49e183db58daeb4 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Mon, 2 Feb 2026 17:53:23 -0500 Subject: [PATCH] Defined derivatives. --- src/dg/derivative/derivative.tex | 18 ++--- src/dg/derivative/index.tex | 1 + src/dg/derivative/sets.tex | 119 ++++++++++++++++++++++++++++++- 3 files changed, 127 insertions(+), 11 deletions(-) diff --git a/src/dg/derivative/derivative.tex b/src/dg/derivative/derivative.tex index 63cadf6..a3a868d 100644 --- a/src/dg/derivative/derivative.tex +++ b/src/dg/derivative/derivative.tex @@ -3,19 +3,19 @@ \begin{definition}[Derivatives and Remainders] \label{definition:derivative-system} - Let $E, F$ be TVSs over $K \in \RC$ and $\ch(E; F), \calr(E; F) \subset F^E$ be vector subspaces, then $(\ch, \calr) = (\ch(E; F), \calr(E; F))$ is a pair of \textbf{derivatives} and \textbf{remainders} if + Let $E, F$ be TVSs over $K \in \RC$ and $\ch(E; F), \calr(E; F) \subset F^E$ be vector subspaces, then $\mathcal{HR} = (\ch(E; F), \calr(E; F))$ is a pair of \textbf{derivatives} and \textbf{remainders} if \begin{enumerate} \item[(T)] For any $T \in \ch$, if there exists $V \in \cn_E(0)$ and $r \in \calr$ such that $T|_V = r|_V$, then $T = 0$. \end{enumerate} \end{definition} -\begin{definition}[$(\ch, \calr)$-Differentiability] +\begin{definition}[$\mathcal{HR}$-Differentiability] \label{definition:space-differentiability} - Let $E, F$ be TVSs over $K \in \RC$, $(\ch, \calr)$ be a pair of derivatives and remainders, $U \subset E$ be open, $f: U \to F$ be a function, and $x_0 \in U$, then $f$ is \textbf{$(\ch, \calr)$-differentiable} at $x_0$ if there exists $V \in \cn_E(0)$, $T \in \ch$, and $r \in \calr$ such that + Let $E, F$ be TVSs over $K \in \RC$, $\mathcal{HR}$ be a pair of derivatives and remainders, $U \subset E$ be open, $f: U \to F$ be a function, and $x_0 \in U$, then $f$ is \textbf{$\mathcal{HR}$-differentiable} at $x_0$ if there exists $V \in \cn_E(0)$, $T \in \ch$, and $r \in \calr$ such that \[ f(x_0 + h) = f(x_0) + Th + r(h) \] - for all $h \in V$. In which case, $T = D_{(\ch, \calr)}f(x_0)$ is the unique element of $\ch$ satisfying the above, known as the \textbf{$(\ch, \calr)$-derivative} of $f$ at $x_0$. + for all $h \in V$. In which case, $T = D_{\mathcal{HR}}f(x_0)$ is the unique element of $\ch$ satisfying the above, known as the \textbf{$\mathcal{HR}$-derivative} of $f$ at $x_0$. \end{definition} \begin{proof} Let $S, T \in \ch$, $r, s \in \calr$, and $V \in \cn_E(0)$ such that @@ -27,20 +27,20 @@ \begin{proposition} \label{proposition:derivative-linearity} - Let $E, F$ be TVSs over $K \in \RC$, $(\ch, \calr)$ be a pair of derivatives and remainders, $U \subset E$ be open, $f, g: U \to F$ be functions, and $x_0 \in U$. If $f, g$ are $(\ch, \calr)$-differentiable at $x_0$, then for any $\lambda \in K$, $\lambda f + g$ is $(\ch, \calr)$-differentiable at $x_0$, and + Let $E, F$ be TVSs over $K \in \RC$, $\mathcal{HR}$ be a pair of derivatives and remainders, $U \subset E$ be open, $f, g: U \to F$ be functions, and $x_0 \in U$. If $f, g$ are $\mathcal{HR}$-differentiable at $x_0$, then for any $\lambda \in K$, $\lambda f + g$ is $\mathcal{HR}$-differentiable at $x_0$, and \[ - D_{(\ch, \calr)}(\lambda f + g)(x_0) = \lambda D_{(\ch, \calr)}f(x_0) + D_{(\ch, \calr)}g(x_0) + D_{\mathcal{HR}}(\lambda f + g)(x_0) = \lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0) \] \end{proposition} \begin{proof} Let $V \in \cn_E(0)$ and $r, s \in \calr$ such that \begin{align*} - f(x_0 + h) - f(x_0) &= D_{(\ch, \calr)}f(x_0)h + r(h) \\ - g(x_0 + h) - f(x_0) &= D_{(\ch, \calr)}g(x_0)h + s(h) + f(x_0 + h) - f(x_0) &= D_{\mathcal{HR}}f(x_0)h + r(h) \\ + g(x_0 + h) - f(x_0) &= D_{\mathcal{HR}}g(x_0)h + s(h) \end{align*} then \[ - (\lambda f + g)(x_0+h) - (\lambda f + g)(x_0) = \underbrace{[\lambda D_{(\ch, \calr)}f(x_0) + D_{(\ch, \calr)}g(x_0)]}_{\in \ch}h + \underbrace{(\lambda r + s)}_{\in \calr}(h) + (\lambda f + g)(x_0+h) - (\lambda f + g)(x_0) = \underbrace{[\lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0)]}_{\in \ch}h + \underbrace{(\lambda r + s)}_{\in \calr}(h) \] \end{proof} diff --git a/src/dg/derivative/index.tex b/src/dg/derivative/index.tex index f2848c3..d793c09 100644 --- a/src/dg/derivative/index.tex +++ b/src/dg/derivative/index.tex @@ -2,3 +2,4 @@ \label{chap:diff} \input{./src/dg/derivative/derivative.tex} +\input{./src/dg/derivative/sets.tex} diff --git a/src/dg/derivative/sets.tex b/src/dg/derivative/sets.tex index e930fd6..153773a 100644 --- a/src/dg/derivative/sets.tex +++ b/src/dg/derivative/sets.tex @@ -3,15 +3,130 @@ \begin{definition}[Small] \label{definition:differentiation-small} - Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed family of sets that contains all finite sets, $r: E \to F$, and $n \in \natz$, then the following are equivalent: + Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $r: E \to F$, 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, $\sigma \subset B(E)$ be an upward-directed family contains all finite sets, $B_\sigma(E; F)$ 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. +\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] +\label{definition:derivative-sets} + 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, $f: U \to F$, 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 + \[ + f(x_0 + h) = f(x_0) + Th + r(h) + \] + for all $h \in V$. + + The linear map $T \in L(E; F)$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}, denoted $D_{\sigma}f(x_0)$. +\end{definition} + +\begin{definition}[Differentiable] +\label{definition:differentiable-sets} + 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{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: + \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} + 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 $\sigma \subset B(E)$ and $\tau \subset B(F)$ correspond to the following families of sets on $E$ and $F$: + \begin{enumerate} + \item Finite sets. + \item 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 \ref{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): Firstly, for any $A \subset E$ finite, $(T + r)(A)$ is also finite, so the claim holds directly in the case of finite sets. + + 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} - In \ref{definition:differentiation-small}, the system $\sigma$ can be chosedn 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. + In \ref{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{definition}[$n$-Fold Differentiability] +\label{definition:n-differentiable-sets} + 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, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, 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 $\sigma$-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 \ref{proposition:multilinear-identify}, + \[ + D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F) + \] + is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}. +\end{definition}