Theorem 18.4.2. 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 $D^{n}f(x) \in L^{n}(E; F)$ is symmetric.
Proof. 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
then there exists $r_{1} \in \mathcal{R}_{B(E)}$ such that
Let $B_{h}: B_{E}(0, r) \to F$ be defined by
then
Now, there exists $r_{2}, r_{3} \in \mathcal{R}_{B(E)}$ such that for any $k \in B(0, r)$,
By the Mean Value Theorem,
As the above argument is symmetric,
so $D^{2}f(x)(h, k) - D^{2}f(x)(k, h) = 0$.
Now suppose that the proposition holds for $n$. Identify $L^{n}(E; F) = L^{2}(E; L^{n-2}(E; F))$, then for any $\seqf[n]{x_j}\subset E$,
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.$\square$