\section{Higher Derivatives}
\label{section:higher-derivatives}

\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}

\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{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
    \[
    A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
    \]
    then there exists $r_1 \in \mathcal{R}_{B(E)}$ such that
    \begin{align*}
        A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
        &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
        &- [f(x + k) - f(x) - Df(x)(k)] \\
        &= D^2f(x)(h, k) + r_1(h) \cdot Df(x)(k)
        &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
        &- [f(x + k) - f(x) - Df(x)(k)] \\
    \end{align*}
    Let $B_h: B_E(0, r) \to F$ be defined by
    \[
    B_h(k) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k)
    \]
    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,
    \[
    \norm{B_h(k) - B_h(0)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
    \]
    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 $D^2f(x)(h, k) - D^2f(x)(k, h) = 0$.
\end{proof}