Theorem 26.5.3 (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. 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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