Theorem 18.5.2: Taylor's Formula, Peano Remainder {{\cite[Theorem 4.7.3]{Bogachev}}}

Theorem 18.5.2 (Taylor’s Formula, Peano Remainder [Theorem 4.7.3, BS17]). 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

\[g(x_{0} + h) = g(x_{0}) + \sum_{k = 1}^{n} \frac{1}{k!}D^{k}_{\sigma} f(x_{0})(h^{(k)}) + r(h)\]

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 Theorem 18.4.3. Let $T_{k}(h) = \frac{1}{k!}D^{k}_{\sigma} f(x_{0})(h^{(k)})$, then by Proposition 18.4.4, for any $\bracs{t_j}_{1}^{\ell} \in E$,

\[D^{\ell}_{\sigma} T_{k}(h)(t_{1}, \cdots, t_{\ell}) = \begin{cases}0&\ell > k \\ D^{k}_{\sigma}(x_{0})(t_{1}, \cdots, t_{\ell})&\ell = k \\ \frac{1}{(k-\ell)!}D^{k}_{\sigma}(x_{0})(h^{(k - \ell)}, t_{1}, \cdots, t_{\ell})&\ell < k\end{cases}\]

\[D^{k}_{\sigma} r(0) = D^{k}_{\sigma} g(x_{0}) - D^{k}_{\sigma}(x_{0}) = 0\]

If $n = 1$, then the theorem holds by definition of the derivative. Now suppose inductively that the theorem holds for $n$. By the 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,

\[\bracs{T \in L(E; F)| T(A) \subset U}= \bracs{T \in L(E; F)| T(tA) \subset U \forall t \in (0, 1)}\]

\[\bracs{t^{-n}D_\sigma r(s)(h)|s \in [0, h], h \in tA}\subset U\]

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 $r \in \mathcal{R}_{\sigma}^{n+1}(E; F)$.$\square$

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