64 lines
3.1 KiB
TeX
64 lines
3.1 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, $\sigma \subset B(E)$ be an upward-directed family that contains all finite sets, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
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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 \ref{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{proposition}[{{\cite[Theorem 5.1.1]{Cartan}}}]
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\label{proposition:derivative-symmetric}
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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 $x_0 \in U$, then $D^nf(x_0) \in L^2(E; F)$ is symmetric.
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\end{proposition}
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\begin{proof}
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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
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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 $r_1 \in \mathcal{R}_{B(E)}$ 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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\[
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B_h(k) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k)
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\]
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then
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\[
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B_h(k) - B_h(0) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k) -f(x + h) + f(x)
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\]
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Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ 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) - 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 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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\end{proof}
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