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/Part 5: Calculus/Chapter 27: Differential Calculus

27.4 Higher Derivatives

Definition 27.4.1 (Codomain of Derivatives).label Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, and $L^{(0)}_{\sigma}(E; F) = F$. For each $n \in \natp$, inductively define

\[L^{(n)}_{\sigma}(E; F) = L(E; L^{(n-1)}_{\sigma}(E; F)) \subset B_{\sigma}^{n}(E; F)\]

and equip it with the $\sigma$-uniform topology, then under the identification

\[I: L^{(n)}_{\sigma}(E; F) \to B_{\sigma}^{n}(E; F) \quad I\lambda(x_{1}, \cdots, x_{n}) = \lambda(x_{1})\cdots(x_{n})\]

the space $L^{(n)}_{\sigma}(E; F)$ is a subspace of $B_{\sigma}^{n}(E; F)$.

Proof. By Proposition 10.13.3.$\square$

Definition 27.4.2 ($n$-Fold Differentiability).label Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, 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 $n$-fold $\tilde \sigma$-differentiable at $x_{0}$ if

  1. (1)

    There exists $V \in \cn_{E}(x_{0})$ such that $f$ is $(n-1)$-fold $\tilde \sigma$-differentiable on $V$.

  2. (2)

    The derivative $D_{\sigma}^{n-1}f: U \to B^{(n-1)}_{\sigma}(E; F)$ is $\tilde \sigma$-differentiable at $x_{0}$.

In which case, $D_{\sigma}(D_{\sigma}^{n-1}f)(x_{0}) = D_{\sigma}^{n}f(x_{0})$ is the $n$-fold $\tilde \sigma$-derivative of $f$ at $x_{0}$.

If $f: U \to F$ is $n$-fold $\tilde \sigma$-differentiable at every point in $U$, then $f$ is $n$-fold $\tilde \sigma$-differentiable on $U$. Under the identification $B_{\sigma}(E; B_{\sigma}^{n}(E; F)) = B_{\sigma}^{(n)}(E; F)$, the mapping

\[D_{\sigma}^{n}f: U \to B^{(n-1)}_{\sigma}(E; F)\]

is the $n$-fold $\tilde \sigma$-derivative of $f$.

If for each $1 \le k \le n$, $D_{\sigma}^{k}f$ takes value in $L^{(k)}_{\sigma}(E; F)$, then $f$ is $n$-fold $\sigma$-differentiable, and $D_{\sigma}^{n}f$ is the $n$-fold $\sigma$-derivative of $f$.

Definition 27.4.3 (Space of Differentiable Functions).label Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $D_{\sigma}^{k}(U; F)$/$\tilde D_{\sigma}^{k}(U; F)$ is the space of $n$-fold $\sigma$/$\tilde \sigma$-differentiable functions from $U$ to $F$.

Definition 27.4.4 (Space of Continuously Differentiable Functions).label Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $n \in \natp$, then $C_{\sigma}^{k}(U; F)$/$\tilde C_{\sigma}^{k}(U; F)$ is the space of $n$-fold continuously $\sigma$/$\tilde \sigma$-differentiable functions from $U$ to $F$.

Theorem 27.4.5 (Symmetry of Higher Derivatives).label 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 [Theorem 5.1.1, Car83]. 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

\[A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)\]

then there exists $r_{1} \in \mathcal{R}_{\mathfrak{B}(E)}$ 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^{2}f(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

\begin{align*}B_{h}(k)&= f(x + h + k) - f(x + k) \\&- Df(x + h)(k) + Df(x)(k)\end{align*}

then

\begin{align*}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)\end{align*}

Now, there exists $r_{2}, r_{3} \in \mathcal{R}_{\mathfrak{B}(E)}$ 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^{2}f(x)(h + k) + Df(x) - D^{2}f(x)(h) \\&- Df(x) - D^{2}f(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 $D^{2}f(x)(h, k) - D^{2}f(x)(k, h) = 0$.

Now suppose that the proposition holds for $n$. Identify $L^{n}(E; F) = L^{2}(E; L^{n-2}(E; F))$, then for any $\seqf[n]{x_j}\subset E$,

\begin{align*}Df(x)(x_{1}, \cdots, x_{n})&= Df(x)(x_{1}, x_{2})(x_{3}, \cdots, x_{n}) \\&= Df(x)(x_{2}, x_{1})(x_{3}, \cdots, x_{n}) \\&= Df(x)(x_{2}, x_{1}, x_{3}, \cdots, x_{n})\end{align*}

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.$\square$

Theorem 27.4.6 (Symmetry of Higher Derivatives).label Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\tilde \sigma$-differentiable at $x_{0} \in U$, then $D_{\sigma}^{n}f(x_{0}) \in L_{\sigma}^{(n)}(E; F)$ is symmetric.

Proof [Proposition 4.5.14, BS17]. Let $\seqf{h_j}\subset E$, $E_{0}$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_{0} \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^{*}$, the mapping $\phi \circ g: E_{0} \cap U \to K$ is $n$-times Fréchet-differentiable, with

\[D_{\mathfrak{B}(E_0)}^{n}(\phi \circ g)(x_{0}) = \phi \circ D_{\sigma}^{n} g(x_{0})\]

by the chain rule. By Theorem 27.4.5, $\phi \circ D_{\sigma}^{n} g(x_{0}) \in L^{n}(E_{0}; K)$ is symmetric. As this holds for any $\seqf{h_j}\subset E$ and $\phi \in F^{*}$, $D_{\sigma}^{n} g(x_{0}) \in B_{\sigma}^{(n)}(E; F)$ is symmetric by the Hahn-Banach theorem.$\square$

Theorem 27.4.7 (Power Rule).label Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $F$ be a separated locally convex space, $T \in B^{n}_{\sigma}(E; F)$.

For each $1 \le m \le n$, let $\text{Inj}([m]; [n])$ be the set of injective mappings from $[m]$ to $[n]$. For any $\phi \in \text{Inj}([m]; [n])$, denote $\phi^{c} \in \text{Inj}([n-m]; [n] \setminus \phi([m]))$ as the unique increasing injective map. For each $h \in E^{n-m}$, $k \in E^{m}$, and $1 \le j \le n$, write

\[(h, k)_{\phi} = \begin{cases}k_{\phi^{-1}(j)}&j \in \phi([m]) \\ h_{(\phi^c)^{-1}(j)}&j \not\in \phi([m])\end{cases}\]

For any $x \in E$, denote $x^{(m)}$ as the tuple of $x$ repeated $m$ times, then the mapping $f: E \to F$ defined by $x \mapsto T(x^{(n)})$ is infinitely $\tilde\sigma$-differentiable on $E$, where

  1. (1)

    For each $1 \le m \le n$, $x \in E$, and $h \in E^{m}$,

    \[D^{m}_{\sigma} f(x)(h) = \sum_{\phi \in \text{Inj}([m]; [n])}T((x^{(n-m)}, h)_{\phi}]\]

    In particular,

    \[D^{n}_{\sigma} f(x)(h) = \sum_{\phi \in S_n}T(h_{\phi(1)}, \cdots, h_{\phi(n)})\]

  2. (2)

    For each $m > n$ and $x \in E$, $D^{m}_{\sigma} f(x) = 0$.

Notably, if $T \in L^{(n)}(E; F)$ is symmetric or $T \in L^{n}(E; F)$, then $T$ is infinitely $\sigma$-differentiable on $E$.

Proof. (1): Let $0 \le m \le n - 1$ and suppose inductively that (1) holds for $m$. For each $x, h \in E$, $S \subset [n-m]$, and $1 \le j \le n - m$,

\[[(x, h)_{S}]_{j} = \begin{cases}h &j \in S \\ x &j \not\in S\end{cases}\]

By the Binomial formula, for each $h \in E$ and $k \in E^{m}$,

\begin{align*}D^{m}_{\sigma} f(x + h)(k)&= \sum_{\phi \in \text{Inj}([m]; [n])}T[((x+h)^{(n-m)}, k)_{\phi}] \\&= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{S \subset [n-m]}T[((x, h)_{S}, k)_{\phi}]\end{align*}

For $\ell \ge 2$, maps in $B_{\sigma}^{\ell}(E; F)$ are $\sigma$-small, so

\[r(h) = \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{\substack{S \subset [n-m] \\ |S| \ge 2}}T[((x, h)_{S}, k)_{\phi}]\]

is $\sigma$-small. Hence

\begin{align*}&D^{(m)}_{\sigma} f(x + h)(k) - D_{\sigma}^{(m)}f(x)(k) \\&= \sum_{\phi \in \text{Inj}([m]; [n])}\sum_{j = 1}^{n-m}T[((x, h)_{\bracs{j}}, k)_{\phi}] + r(h) \\&= \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, (h, k))_{\phi}] + r(h)\end{align*}

and for any $h \in E^{m+1}$,

\[D_{\sigma}^{(m+1)}f(x)(h) = \sum_{\phi \in \text{Inj}([m+1]; [n])}T[(x^{(n-m-1)}, h)_{\phi}]\]

(2): By (1), $D^{n}_{\sigma} f$ is constant.$\square$

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