143 lines
7.3 KiB
TeX
143 lines
7.3 KiB
TeX
\section{Higher Derivatives}
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\label{section:higher-derivatives}
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\begin{definition}[$n$-Fold Differentiability]
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\label{definition:n-differentiable-sets}
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
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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
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\begin{enumerate}
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\item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
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\item The derivative $D_\sigma^{n-1}f: U \to B^{n-1}_\sigma(E; F)$ is derivative at $x_0$.
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\end{enumerate}
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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 $x_0$}.
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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 \autoref{proposition:multilinear-identify},
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\[
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D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
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\]
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is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
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\end{definition}
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\begin{theorem}[Symmetry of Higher Derivatives]
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\label{theorem:derivative-symmetric-frechet}
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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 $D^nf(x) \in L^n(E; F)$ is symmetric.
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\end{theorem}
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\begin{proof}[Proof {{\cite[Theorem 5.1.1]{Cartan}}}. ]
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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
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\[
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A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
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\]
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then there exists $r_1 \in \mathcal{R}_{\mathfrak{B}(E)}$ such that
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\begin{align*}
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A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
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&+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
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&- [f(x + k) - f(x) - Df(x)(k)] \\
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&= D^2f(x)(h, k) + r_1(h) \cdot Df(x)(k) \\
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&+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
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&- [f(x + k) - f(x) - Df(x)(k)] \\
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\end{align*}
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Let $B_h: B_E(0, r) \to F$ be defined by
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\begin{align*}
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B_h(k) &= f(x + h + k) - f(x + k) \\
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&- Df(x + h)(k) + Df(x)(k)
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\end{align*}
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then
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\begin{align*}
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B_h(k) - B_h(0) &= f(x + h + k) - f(x + k) \\
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&- Df(x + h)(k) + Df(x)(k) \\
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&-f(x + h) + f(x)
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\end{align*}
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Now, there exists $r_2, r_3 \in \mathcal{R}_{\mathfrak{B}(E)}$ such that for any $k \in B(0, r)$,
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\begin{align*}
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DB_h(k) &= Df(x + h + k) - Df(x + k) - Df(x + h) + Df(x) \\
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&= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) \\
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&- Df(x) - D^2f(x)(k) + r_2(k) + r_3(h) \\
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&=r_2(k) + r_3(h)
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\end{align*}
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By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
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\[
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\norm{B_h(k) - B_h(0)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
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\]
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As the above argument is symmetric,
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\[
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\norm{D^2f(x)(h, k) - D^2f(x)(k, h)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
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\]
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so $D^2f(x)(h, k) - D^2f(x)(k, h) = 0$.
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Now suppose that the proposition holds for $n$. Identify $L^n(E; F) = L^{2}(E; L^{n-2}(E; F))$, then for any $\seqf[n]{x_j} \subset E$,
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\begin{align*}
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Df(x)(x_1, \cdots, x_n) &= Df(x)(x_1, x_2)(x_3, \cdots, x_n) \\
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&= Df(x)(x_2, x_1)(x_3, \cdots, x_n) \\
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&= Df(x)(x_2, x_1, x_3, \cdots, x_n)
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\end{align*}
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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.
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\end{proof}
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\begin{theorem}[Symmetry of Higher Derivatives]
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\label{theorem:derivative-symmetric}
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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 $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
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\end{theorem}
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\begin{proof}[Proof {{\cite[Proposition 4.5.14]{Bogachev}}}. ]
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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
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\[
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D_{\mathfrak{B}(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
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\]
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by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ 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 \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
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\end{proof}
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\begin{proposition}[Power Rule]
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\label{proposition:multilinear-derivative}
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Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
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\[
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T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
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\]
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be symmetric. For any $x \in E$ and $1 \le k \le n$, let $x^{(k)}$ denote the tuple of $x$ repeated $k$ times, then:
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\begin{enumerate}
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\item The mapping $f: E \to F \quad x \mapsto T(x^{(n)})$ is infinitely $\sigma$-differentiable on $E$.
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\item For each $1 \le k \le n$ and $x, h \in E$,
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\[
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Df(x)(h_1, \cdots, h_k) = \frac{n!}{(n-k)!} T(x^{(n-k)}, h_1, \cdots, h_k)
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\]
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In particular, $D^kf = n! \cdot T$.
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\item For each $k > n$ and $x \in E$, $Df(x) = 0$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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Suppose inductively that (2) holds for $0 \le k \le n$. Let $G = B^{k}_\sigma(E; F)$, then $D^k_\sigma f \in B^{n-k}_\sigma(E; G)$ under the identification $B^n_\sigma(E; F) = B^{n-k}_\sigma(E; B^k_\sigma(E; F))$ in \autoref{proposition:multilinear-identify}. By \autoref{theorem:derivative-symmetric}, $D^k_\sigma f$ is also symmetric, so using the Binomial formula,
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\begin{align*}
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D^k_\sigma f(x + h) &= \sum_{r = 0}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)}) \\
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&= f(x) + (n-k)D^k_\sigma f(x^{(n-k-1)}, h) \\
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&+ \underbrace{\sum_{r = 2}^{n-k}{n - k \choose r}D^k_\sigma f(x^{(n-k-r)}, h^{(r)})}_{r(h)}
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\end{align*}
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For each $k \ge 2$, let $A \in \sigma$ and $U \in \cn_F(0)$, then since $D^k_\sigma f \in B^{n-k}_\sigma(E; F)$, there exists $t > 0$ such that
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\[
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\frac{D^k_\sigma f(x^{(n-k)}, (sA)^{(k)})}{t} = s^{k-1}D^k_\sigma f(x^{(n-k)}, A^{(k)}) \subset U
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\]
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for all $s \in (0, t)$. Hence $r \in \mathcal{R}_\sigma(E; G)$, and
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\[
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D^{k+1}_\sigma f(x + h) = f(x) + \frac{n!}{(n-k-1)!}T(x^{(n-k-1)}, h_1, \cdots, h_{k+1}) + r(h)
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\]
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by the inductive hypothesis.
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(3): Since $D^n_\sigma f$ is constant, $D^k_\sigma f = 0$ for all $k > n$.
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\end{proof}
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