Theorem 27.5.1 (Taylor’s Formula, Lagrange Remainder).label 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
\[f(b) - f(a) - \sum_{k = 1}^{n} \frac{1}{k!}D^{k}f(a)(b - a)^{k}\]
is contained in the closed convex hull of
\[\bracs{D^{n+1}f(s)(t - a)^{n+1} | s \in (a, b) \setminus N, t \in [a, b]}\]
Proof [Theorem 4.7.1, BS17]. If $n = 0$, then the theorem is the Mean Value Theorem.
Suppose inductively that the theorem holds for $n$. Let
\[g: [a, b] \to E \quad t \mapsto f(t) - \sum_{k = 1}^{n+1}\frac{1}{k!}D^{k}f(a)(t - a)^{k}\]
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 Mean Value Theorem,
\[g(b) - g(a) = f(b) - f(a) - \sum_{k = 1}^{n+1}\frac{1}{k!}D^{k}f(a)(t - a)^{k}\]
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
\[f(b) - f(a) - \sum_{k = 1}^{n+1}\frac{1}{k!}D^{k}f(a)(b - a)^{k}\]
is contained in the convex hull of
\[\bracs{D^{n+2}f(s)(t - a)^{n+2} | s \in (a, t) \setminus N, t \in [a, b]}\]
$\square$
Theorem 27.5.2 (Taylor’s Formula, Peano Remainder).label Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal containing bounded subsets 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
\[f(x_{0} + h) = f(x_{0}) + \sum_{k = 1}^{n} \frac{1}{k!}D^{k}_{\sigma} f(x_{0})(h^{(k)}) + r(h)\]
for sufficiently small $h$.
Proof [Theorem 4.7.3, BS17]. Let
\[r(h) = f(x_{0} + h) - f(x) - \sum_{k = 1}^{n} \frac{1}{k!}D^{k}_{\sigma} f(x_{0})(h^{(k)})\]
For any $1 \le k \le n$, $D^{k}_{\sigma}(x_{0}) \in B^{k}_{\sigma}(E; F)$ is symmetric by Theorem 27.4.6. Let $T_{k}(h) = \frac{1}{k!}D^{k}_{\sigma} f(x_{0})(h^{(k)})$, then by Theorem 27.4.7, 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
\[D^{k}_{\sigma} r(0) = D^{k}_{\sigma} g(x_{0}) - D^{k}_{\sigma} f(x_{0}) = 0\]
If $n = 1$, then the theorem holds by definition of the derivative. Now suppose inductively that the theorem holds for $n$. By the 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 $r \in \mathcal{R}_{\sigma}^{n+1}(E; F)$.$\square$
Theorem 27.5.3 (Taylor’s Formula, Integral Remainder).label Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $U \subset E$ be open, and $f \in C^{n+1}_{\sigma}(E; F)$, then for any $x_{0} \in U$ and $h \in E$ such that $[x_{0}, x_{0} + h] \subset U$, then
\[f(x_{0} + h) = f(x_{0}) + \sum_{k = 1}^{n} \frac{1}{k!}D^{k}_{\sigma} f(x_{0})(h^{(k)}) + r(h)\]
where
\[r(h) = \int_{0}^{1} \frac{(1 - t)^{n}}{n!}D^{n+1}_{\sigma} f(x_{0} + th)(h^{(n+1)}) dt\]
In particular, for any continuous seminorm $[\cdot]_{F}: F \to [0, \infty)$,
\[[r(h)]_{F} \le \frac{1}{(n+1)!}\cdot \sup_{t \in [0, 1]}[D^{n+1}_{\sigma} f(x_{0} + th)(h^{n+1})]\]
Proof, [Section XIII.6, Lan93]. Firstly, if $n = 0$, then by the Fundamental Theorem of Calculus,
\[f(x_{0} + h) - f(x_{0}) = \int_{0}^{1} D_{\sigma} f(x_{0} + th)(h) dt\]
Assume inductively that the theorem holds for $n \in \natz$. Let
\[u(t) = D^{n+1}_{\sigma} f(x + ty)(h^{(n+1)}) \quad v(t) = -\frac{(1 - t)^{n+1}}{(n+1)!}\]
then $Dv(t) = (1 - t)^{n}/n!$, and using the change of variables formula and integration by parts,
\begin{align*}r(h)&= \frac{1}{n!}\int_{0}^{1} (1 - t)^{n}D^{n+1}_{\sigma} f(x_{0} + th)(h^{(n+1)}) dt \\&= \int_{0}^{1} udv = u(1)v(1) - u(0)v(0) - \int_{0}^{1} vdu \\&= D_{\sigma}^{n+1}(x_{0})(h^{(n+1)}) \\&+ \int_{0}^{1} \frac{(1 - t)^{n+1}}{(n+1)!}D^{n+2}_{\sigma} f(x_{0} + th)(h^{(n+2)}) dt\end{align*}
$\square$
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