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Bokuan Li
2026-03-06 14:06:15 -05:00
parent 173727665b
commit 5034bc4220
109 changed files with 1184 additions and 410 deletions
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4
src/dg/derivative/derivative.tex
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@@ -15,6 +15,7 @@
\[
f(x_0 + h) = f(x_0) + Th + r(h)
\]
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}
@@ -22,6 +23,7 @@
\[
f(x_0 + h) - f(x_0) = Sh + r(h) = Th + s(h)
\]
for all $h \in V$, then $(S - T)(h) = (s - r)(h)$. By (T), $S - T = 0$. Hence $S = T$.
\end{proof}
@@ -31,6 +33,7 @@
\[
D_{\mathcal{HR}}(\lambda f + g)(x_0) = \lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0)
\]
\end{proposition}
\begin{proof}
Let $V \in \cn_E(0)$ and $r, s \in \calr$ such that
@@ -42,4 +45,5 @@
\[
(\lambda f + g)(x_0+h) - (\lambda f + g)(x_0) = \underbrace{[\lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0)]}_{\in \ch}h + \underbrace{(\lambda r + s)}_{\in \calr}(h)
\]
\end{proof}

82
src/dg/derivative/higher.tex
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@@ -12,22 +12,24 @@
\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},
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},
\[
D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
\]
is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
\end{definition}
\begin{proposition}[{{\cite[Theorem 5.1.1]{Cartan}}}]
\label{proposition:derivative-symmetric}
Let $E, F$ be Banach spaces, $U \subset E$ be open, $n \in \natp$, and $f: U \to F$ be a function $n$-times Fréchet-differentiable at $x_0 \in U$, then $D^nf(x_0) \in L^2(E; F)$ is symmetric.
\end{proposition}
\begin{theorem}[{{\cite[Theorem 5.1.1]{Cartan}}}]
\label{theorem:derivative-symmetric-frechet}
Let $E, F$ be Banach spaces, $U \subset E$ be open, $n \in \natp$, and $f: U \to F$ be a function $n$-times Fréchet-differentiable at $x \in U$, then $D^nf(x) \in L^n(E; F)$ is symmetric.
\end{theorem}
\begin{proof}
First suppose that $n = 2$. Let $r > 0$ such that $B(x_0, 2r) \subset U$, and define $A: B_E(0, r) \times B_E(0, r) \to F$ by
First suppose that $n = 2$. Let $r > 0$ such that $B(x, 2r) \subset U$, and define $A: B_E(0, r) \times B_E(0, r) \to F$ by
\[
A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
\]
then there exists $r_1 \in \mathcal{R}_{B(E)}$ such that
\begin{align*}
A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
@@ -41,23 +43,89 @@
\[
B_h(k) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k)
\]
then
\[
B_h(k) - B_h(0) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k) -f(x + h) + f(x)
\]
Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ such that for any $k \in B(0, r)$,
\begin{align*}
DB_h(k) &= Df(x + h + k) - Df(x + k) - Df(x + h) + Df(x) \\
&= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) - Df(x) - D^2f(x)(k) + r_2(k) + r_3(h) \\
&=r_2(k) + r_3(h)
\end{align*}
By the mean value theorem,
By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
\[
\norm{B_h(k) - B_h(0)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
\]
As the above argument is symmetric,
\[
\norm{D^2f(x)(h, k) - D^2f(x)(k, h)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
\]
so $D^2f(x)(h, k) - D^2f(x)(k, h) = 0$.
Now suppose that the proposition holds for $n$. Identify $L^n(E; F) = L^{2}(E; L^{n-2}(E; F))$, then for any $\seqf[n]{x_j} \subset E$,
\[
Df(x)(x_1, \cdots, x_n) = Df(x)(x_1, x_2)(x_3, \cdots, x_n) = Df(x)(x_2, x_1)(x_3, \cdots, x_n) = Df(x)(x_2, x_1, x_3, \cdots, x_n)
\]
Since any element $\sigma \in S_n$ that does not fix $x_1$ is the composition of the transposition $(12)$ and an element that fixes $x_1$, $Df(x)$ is symmetric.
\end{proof}
\begin{theorem}[{{\cite[Proposition 4.5.14]{Bogachev}}}]
\label{theorem:derivative-symmetric}
Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an upward-directed system 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.
\end{theorem}
\begin{proof}
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
\[
D_{B(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
\]
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}.
\end{proof}
\begin{proposition}[Power Rule]
\label{proposition:multilinear-derivative}
Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a Hausdorff locally convex space, and
\[
T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
\]
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:
\begin{enumerate}
\item The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
\item For each $1 \le k \le n$ and $x, h \in E$,
\[
Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
\]
In particular, $D^kf = n! \cdot T$.
\item For each $k > n$ and $x \in E$, $Df(x) = 0$.
\end{enumerate}
\end{proposition}
\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,
\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)}) \\
&= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
&+ \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
\[
\frac{D^k_\sigma f(x^{(n-k)}, (sA)^{(k)})}{t} = s^{k-1}D^k_\sigma f(x^{(n-k)}, A^{(k)}) \subset U
\]
for all $s \in (0, t)$. Hence $r \in \mathcal{R}_\sigma(E; G)$, and
\[
D^{k+1}_\sigma f(x + h) = f(x) + \frac{n!}{(n-k-1)!}T(x^{(n-k-1)}, h_1, \cdots, h_{k+1}) + r(h)
\]
by the inductive hypothesis.
(3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
\end{proof}

9
src/dg/derivative/index.tex
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@@ -1,7 +1,8 @@
\chapter{Differential Calculus}
\label{chap:diff}
\input{./src/dg/derivative/derivative.tex}
\input{./src/dg/derivative/sets.tex}
\input{./src/dg/derivative/mvt.tex}
\input{./src/dg/derivative/higher.tex}
\input{./derivative.tex}
\input{./sets.tex}
\input{./mvt.tex}
\input{./higher.tex}
\input{./taylor.tex}

28
src/dg/derivative/mvt.tex
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@@ -7,6 +7,7 @@
\[
D^+f(x) = \lim_{t \downto 0} \frac{f(x + t) - f(x)}{t}
\]
exists.
\end{definition}
@@ -26,12 +27,14 @@
\[
S = \bracs{x \in [a, b] \bigg | f(y) - f(a) \le g(y) - g(a) + \eps\sum_{n \in N(x)}2^{-n} \quad \forall y \in [a, x]}
\]
then by continuity of $f$ and $g$, $S$ is closed.
Let $x \in S$ and suppose that $s < b$. If $x \in [a, b] \setminus N$, then since
\[
\lim_{t \downto 0}\frac{f(x + t) - f(x)}{t} < \lim_{t \downto 0}\frac{g(x + t) - f(x)}{t}
\]
there exists $\delta > 0$ such that $f(x + t) - f(x) < g(x + t) < g(x)$ for all $t \in [0, \delta)$. In which case, $[x, x + \delta) \subset S$.
If $x = x_n \in N$, then by continuity of $f$ and $g$, there exists $\delta > 0$ such that $f(x + t) - f(x) \le g(x + t) - g(x) + 2^{-n}$ for all $t \in (0, \delta)$. Hence $[x, x + \delta) \subset S$ as well.
@@ -53,6 +56,7 @@
\[
f(b) - f(a) \in [g(b) - g(a)]B
\]
\end{theorem}
\begin{proof}
Let $\phi \in E^*$ and $x \in [a, b] \setminus N$, then since $g$ is non-decreasing,
@@ -61,15 +65,17 @@
D^+(\phi \circ f)(x) &\le \sup_{z \in B}\phi(z) \cdot \lim_{t \downto 0}\frac{\phi(g(x + t) - g(x))}{t} \\
D^+(\phi \circ f)(x) &\le D^+g(x) \cdot \sup_{z \in B}\phi(z)
\end{align*}
By \ref{lemma:right-differentiable-inequality},
By \autoref{lemma:right-differentiable-inequality},
\[
\phi(f(b) - f(a)) \le (g(b) - g(a)) \cdot \sup_{z \in B}\phi(z)
\]
Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the Hahn-Banach theorem (\ref{theorem:hahn-banach-geometric-2}), there exists $\phi \in E^*$ such that
Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in E^*$ such that
\[
\phi(f(b) - f(a)) > \sup_{x \in B}\phi[(g(b) - g(a))b] = \phi(g(b) - g(a)) \cdot \sup_{x \in B}\phi(x)
\]
which is impossible. Therefore $f(b) - f(a) \in [g(b) - g(a)]B$.
\end{proof}
@@ -80,16 +86,19 @@
\[
f(b) - f(a) \in \overline{\text{Conv}\bracs{Df(x)(b - a)| x \in (a, b) \setminus N}}
\]
\end{theorem}
\begin{proof}
By \ref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
By \autoref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
\[
D^+f(x) = Df(x) \in \bracs{Df(y)|y \in (a, b) \setminus N}
\]
for all $x \in (a, b)$. Let $g(x) = x$, then by \ref{theorem:right-differentiable-convex-form},
for all $x \in (a, b)$. Let $g(x) = x$, then by \autoref{theorem:right-differentiable-convex-form},
\[
f(b) - f(a) \in \overline{(b - a)\text{Conv}\bracs{Df(x)|x \in (a, b) \setminus N}}
\]
\end{proof}
\begin{theorem}[Mean Value Theorem]
@@ -98,15 +107,17 @@
\[
f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
\]
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 \ref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \ref{proposition:derivative-sets-real}.
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}.
By the Mean Value Theorem (\ref{theorem:mean-value-theorem-line}),
By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line},
\[
f(y) - f(x) = g(1) - g(0) \in \overline{\text{Conv}\bracs{Dg(t)|t \in [0, 1]}} = \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
\]
\end{proof}
\begin{proposition}
@@ -118,9 +129,10 @@
\[
f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}} = \bracs{0}
\]
by the Mean Value Theorem (\ref{theorem:mean-value-theorem-line}).
Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. For any $y \in W$, by \ref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. Therefore $W \supset W \cup (U + y)$, and $W$ is open by \ref{lemma:openneighbourhood}.
by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}.
Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. For any $y \in W$, by \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. Therefore $W \supset W \cup (U + y)$, and $W$ is open by \autoref{lemma:openneighbourhood}.
For any $y \in \ol W \cap V$, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. As $y \in \ol W \cap V$, $U \cap W \ne \emptyset$. Thus $f(y) = f(x)$, $y \in W$, and $W$ is relatively closed.

13
src/dg/derivative/sets.tex
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@@ -29,6 +29,7 @@
\[
f(x_0 + h) = f(x_0) + Th + r(h)
\]
for all $h \in V$.
The linear map $T \in L(E; F)$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}, denoted $D_{\sigma}f(x_0)$.
@@ -56,16 +57,19 @@
\[
D_\sigma(g \circ f)(x_0) = D_\tau g(f(x_0)) \circ D_\sigma f(x_0)
\]
\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 \\
@@ -87,7 +91,7 @@
\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.
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 $s \in (0, t)$.
@@ -98,21 +102,24 @@
\[
\limv{n} \frac{1}{t_n}s \circ (T + r)(t_na_n) = \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
\]
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 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.
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.
@@ -130,6 +137,7 @@
\[
\lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
\]
exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
\item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, if $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
\end{enumerate}
@@ -147,5 +155,6 @@
\[
\lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
\]
and $Df(x_0) = T$.
\end{proof}

123
src/dg/derivative/taylor.tex Normal file
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@@ -0,0 +1,123 @@
\section{Taylor's Formula}
\label{section:taylor}
\begin{theorem}[Taylor's Formula, Lagrange Remainder {{\cite[Theorem 4.7.1]{Bogachev}}}]
\label{theorem:taylor-lagrange}
Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $S \subset [a, b]$ be at most countable, $n \in \natp$, and $f \in C^{n}([a, b]; E)$ be $(n+1)$-fold differentiable on $[a, b] \setminus N$, then
\[
f(b) - f(a) - \sum_{k = 1}^n \frac{1}{k!}D^kf(a)(b - a)^k
\]
is contained in the closed convex hull\footnote{It may be possible to sharpen the below claim to include the $1/(n+1)!$ factor. However, I was not able to follow the proof for this.} of
\[
\bracs{D^{n+1}f(s)(t - a)^{n+1} | s \in (a, b) \setminus N, t \in [a, b]}
\]
\end{theorem}
\begin{proof}
If $n = 0$, then the theorem is the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}.
Suppose inductively that the theorem holds for $n$. Let
\[
g: [a, b] \to E \quad t \mapsto f(t) - \sum_{k = 1}^{n+1} \frac{1}{k!}D^kf(a)(t - a)^k
\]
then for any $t \in (0, 1)$,
\begin{align*}
Dg(t) &= Df(t) - \sum_{k = 1}^{n+1} \frac{1}{(k-1)!}D^{k}f(a)(t - a)^{k-1} \\
&= Df(t) - Df(a) - \sum_{k = 1}^{n} \frac{1}{k!}D^{k+1}f(a)(t - a)^{k}
\end{align*}
by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line},
\[
g(b) - g(a) = f(b) - f(a) - \sum_{k = 1}^{n+1} \frac{1}{k!}D^kf(a)(t - a)^k
\]
is contained in the closed convex hull of
\[
\bracs{\braks{Df(t) - Df(a) - \sum_{k = 1}^{n} \frac{1}{k!}D^{k+1}f(a)(t - a)^{k}}(b - a) \bigg | t \in [a, b]}
\]
By the inductive hypothesis applied to $Df$, for any $t \in [a, b]$,
\[
Df(t) - Df(a) - \sum_{k = 1}^n \frac{1}{k!}D^{k+1}f(a)(t - a)^k
\]
is contained in the closed convex hull of
\[
\bracs{D^{n+2}f(s)(t - a)^{n+1} | s \in (a, t) \setminus N}
\]
Therefore
\[
f(b) - f(a) - \sum_{k = 1}^{n+1} \frac{1}{k!}D^kf(a)(b - a)^k
\]
is contained in the convex hull of
\[
\bracs{D^{n+2}f(s)(t - a)^{n+2} | s \in (a, t) \setminus N, t \in [a, b]}
\]
\end{proof}
\begin{theorem}[Taylor's Formula, Peano Remainder {{\cite[Theorem 4.7.3]{Bogachev}}}]
\label{theorem:taylor-peano}
Let $E$ be a topological vector space, $\sigma \subset B(E)$ 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
\[
g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
\]
\end{theorem}
\begin{proof}
Let
\[
r(h) = g(x_0 + h) - g(x) - \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)})
\]
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 $\bracs{t_j}_1^\ell \in E$,
\[
D^\ell_\sigma T_k(h)(t_1, \cdots, t_\ell) = \begin{cases}
0 &\ell > k \\
D^k_\sigma(x_0)(t_1, \cdots, t_\ell) &\ell = k \\
\frac{1}{(k-\ell)!}D^k_\sigma(x_0)(h^{(k - \ell)}, t_1, \cdots, t_\ell) & \ell < k
\end{cases}
\]
so
\[
D^k_\sigma r(0) = D^k_\sigma g(x_0) - D^k_\sigma(x_0) = 0
\]
If $n = 1$, then the theorem holds by definition of the derivative. Now suppose inductively that the theorem holds for $n$. By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
\[
r(h) = r(h) - r(0) \in \overline{\text{Conv}\bracs{D_\sigma r(s)(h)| s \in [0, h]}}
\]
For any $A \in \sigma$ and $t > 0$,
\[
\frac{r(tA)}{t^{n+1}} \subset \overline{\text{Conv}\bracs{t^{-n}D_\sigma r(s)(h)|s \in [0, h], h \in tA}}
\]
Let $U \in \cn_F(0)$ be convex and circled, then by the inductive assumption applied to $D_\sigma r$, there exists $t_0 \in (0, 1)$ such that for any $t \in (0, t_0)$.
\[
\frac{D_\sigma r(tA)}{t^n} \subset \bracs{T \in L(E; F)| T(A) \subset U}
\]
Since $U$ is circled,
\[
\bracs{T \in L(E; F)| T(A) \subset U} = \bracs{T \in L(E; F)| T(tA) \subset U \forall t \in (0, 1)}
\]
so
\[
\bracs{t^{-n}D_\sigma r(s)(h)|s \in [0, h], h \in tA} \subset U
\]
and
\[
\frac{r(tA)}{t^{n+1}} \subset \overline{\text{Conv}\bracs{t^{-n}D_\sigma r(s)(h)|s \in [0, h], h \in tA}} \subset \overline{U}
\]
Therefore $r \in \mathcal{R}_\sigma^{n+1}(E; F)$.
\end{proof}