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Symmetry of the derivative (normed/frechet).

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Bokuan Li
2026-02-03 17:57:07 -05:00
parent 7bbbf75213
commit 173727665b
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src/dg/derivative/derivative.tex
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(\lambda f + g)(x_0+h) - (\lambda f + g)(x_0) = \underbrace{[\lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0)]}_{\in \ch}h + \underbrace{(\lambda r + s)}_{\in \calr}(h)
\]
\end{proof}
\begin{definition}[$o(t)$]
\label{definition:little-o}
Let $U \in \cn(0) \subset \real$ and $r: U \to \real$, then $r \in o(t)$ if
\[
\lim_{t \to 0}\frac{o(t)}{t} = 0
\]
\end{definition}
\begin{definition}[Tangent to $0$]
\label{definition:tangent-to-0}
Let $E, F$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, then $\varphi$ is \textbf{tangent to $0$} if for any $W \in \cn_F(0)$, there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
\[
\varphi(tV) \subset r(t)W
\]
for sufficiently small $t \in \real$.
\end{definition}
\begin{definition}[Linear Order at $0$]
\label{definition:linear-order-at-0}
Let $E, F$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, then $\varphi$ is of \textbf{linear order at $0$} if for any $W \in \cn_F(0)$, there exists $V \in \cn_E(0)$ such that
\[
\varphi(tV) \subset tW
\]
for sufficiently small $t \in \real$.
\end{definition}
\begin{lemma}
\label{lemma:tangent-linear-at-0}
Let $E, F, G$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, $V \in \cn_F(0)$, $\varphi, \psi: U \to F$, and $\rho: V \to G$. If $\varphi, \psi, \rho$ are of linear order at $0$, then
\begin{enumerate}
\item $\rho \circ \varphi$ is of linear order at $0$.
\item $\varphi + \psi$ is of linear order at $0$.
\item If one of $\varphi, \rho$ is tangent to $0$, then $\rho \circ \varphi$ is tangent to $0$.
\end{enumerate}
If $\varphi, \psi$ are tangent to $0$, then
\begin{enumerate}
\item[(4)] $\varphi$ is of linear order at $0$.
\item[(5)] $\varphi + \psi$ is tangent at $0$.
\end{enumerate}
Finally, suppose that $\varphi$ is linear.
\begin{enumerate}
\item[(6)] If $\varphi$ is continuous, then $\varphi$ is of linear order at $0$.
\item[(7)] If $\varphi$ is tangent to $0$ and $E$ is Hausdorff, then $\varphi = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1): Let $W \in \cn_G(0)$, then there exists $V_0 \in \cn_F(0)$ with $\rho(tV_0) \subset tW$ and $U_0 \in \cn_E(0)$ with $\varphi(tU_0) \subset tV_0$ for sufficiently small $t$. In which case, $\rho \circ \varphi(tU_0) \subset tW$.
(2): Let $W \in \cn_F(0)$, then there exists $W_0 \in \cn_F(0)$ with $W_0 + W_0 \subset W$ and $V_0 \in \cn_E(0)$ such that $\varphi(tV_0), \psi(tV_0) \subset tW_0$ for sufficiently small $t$. In which case,
\[
\varphi(tV_0) + \psi(tV_0) \subset tW_0 + tW_0 \subset tW
\]
(3): Let $W \in \cn_G(0)$, then there exists $V_0 \in \cn_F(0)$, $U_0 \in \cn_E(0)$, and $r \in o(t)$ such that one of the following holds for sufficiently small $t$:
\begin{enumerate}
\item[(a)] $\varphi(tU_0) \subset tV_0$ and $\rho(tV_0) \subset o(t)W$.
\item[(b)] $\varphi(tU_0) \subset o(t)V_0$ and $\rho(tV_0) \subset tW$.
\end{enumerate}
In both cases, $\rho \circ \varphi(tU_0) \subset o(t)W$.
(4): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. By assumption, there exists $V_0 \in \cn_E(0)$ and $r \in o(t)$ such that
\[
\varphi(tV_0) \subset o(t)W = \frac{r(t)}{t} \cdot tW
\]
for sufficiently small $t$. Since $r \in o(t)$, $\abs{r(t)/t} \le 1$ for sufficiently small $t$. In which case, $\frac{r(t)}{t} \cdot tW \subset tW$.
(5): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Let $V_0 \in \cn_E(0)$ and $r, r' \in o(t)$ such that $\varphi(tV_0) \subset r(t)W$ and $\psi(tV_0) \subset r'(t)W$ for sufficiently small $t$, then
\[
\varphi(tV_0) + \psi(tV_0) \subset r(t)W + r'(t)W \subset [\abs{r(t)} + \abs{r'(t)}]W
\]
(6): Let $W \in \cn_F(0)$, then there exists $V_0 \in \cn_E(0)$ such that $\varphi(V_0) \subset W$. In which case, $\varphi(tV_0) \subset tW$ for all $t \in \real$.
(7): Let $x \in E$ and $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi$ is tangent to $0$ and linear, then there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
\[
\varphi(tV) \subset r(t)W \quad \varphi(V) \subset \frac{r(t)}{t}W
\]
for sufficiently small $t \in \real$. Since $W \in \cn_F(0)$, there exists $\lambda \in K$ such that $x \in \lambda V$. Thus
\[
\varphi(x) \in \varphi(\lambda V) = \lambda\varphi(V) \subset \frac{\lambda r(t)}{t}W
\]
As $r \in o(t)$, there exists $t \in \real$ such that $\abs{\lambda r(t)/t} \le 1$, so $\varphi(x) \in \frac{\lambda r(t)}{t}W \subset W$. Since this holds for all $W \in \cn_F(0)$ and $F$ is Hausdorff, $\varphi(x) \in \ol{\bracs{0}} = \bracs{0}$ by \ref{proposition:tvs-closure} and \ref{lemma:t1}.
\end{proof}
\begin{lemma}
\label{lemma:tangent-to-0}
Let $E, F, G$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, $\psi: U \to F$ be tangent to $0$, then:
\begin{enumerate}
\item For any $\lambda \in L(F; G)$, $\lambda \circ \varphi$ is tangent to $0$.
\item $\varphi + \psi$ is tangent to $0$.
\item If $\varphi$ is linear and $F$ is Hausdorff, then $\varphi = 0$.
\end{enumerate}
\end{lemma}
\begin{proof}
(1): Let $W \in \cn_G(0)$, then $\lambda^{-1}(W) \in \cn_F(0)$, and there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
\[
\lambda \circ \varphi(tV) \subset \lambda (r(t)\lambda^{-1}W) = r(t) \lambda \circ \lambda^{-1} (W) \subset r(t)W
\]
(2): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi, \psi$ are tangent to $0$, there exists $V_1, V_2 \in \cn_E(0)$ and $r_1, r_2 \in o(t)$ such that
\[
\varphi(tV_1) \subset r_1(t)W \quad \psi(tV_2) \subset r_2(t)W
\]
In which case, since $W$ is circled,
\[
\lambda \varphi(t(V_1 \cap V_2)) + \psi(t(V_1 \cap V_2)) \subset r_1(t) W + r_2(t)W \subset (r_1(t) + r_2(t))W
\]
and $r_1 + r_2 \in o(t)$.
(3): Let $x \in E$ and $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi$ is tangent to $0$ and linear, then there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
\[
\varphi(tV) \subset r(t)W \quad \varphi(V) \subset \frac{r(t)}{t}W
\]
for sufficiently small $t \in \real$. Since $W \in \cn_F(0)$, there exists $\lambda \in K$ such that $x \in \lambda V$. Thus
\[
\varphi(x) \in \varphi(\lambda V) = \lambda\varphi(V) \subset \frac{\lambda r(t)}{t}W
\]
As $r \in o(t)$, there exists $t \in \real$ such that $\abs{\lambda r(t)/t} \le 1$, so $\varphi(x) \in \frac{\lambda r(t)}{t}W \subset W$. Since this holds for all $W \in \cn_F(0)$ and $F$ is Hausdorff, $\varphi(x) \in \ol{\bracs{0}} = \bracs{0}$ by \ref{proposition:tvs-closure} and \ref{lemma:t1}.
\end{proof}
\begin{definition}[(Strong Fréchet) Derivative]
\label{definition:derivative}
Let $E, F$ be TVSs over $K \in \RC$ with $F$ being Hausdorff, $U \subset E$ open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{(strongly Fréchet) differentiable at} $x_0$ if there exists $\lambda \in L(E; F)$, $V \in \cn_E(0)$, and $\varphi: V \to F$ tangent to $0$ such that
\[
f(x_0 + h) = f(x_0) + \lambda h + \varphi(h)
\]
for all $h \in V$. In which case, $Df(x_0) = \lambda$ is the unique continuous linear map satisfying the above, and the \textbf{(strong Fréchet) derivative} of $f$ \textbf{at} $x_0$.
If $f$ is differentiable at every $x_0 \in U$, then $f$ is \textbf{(strongly Fréchet) differentiable}, and $Df: U \to L(E; F)$ is the \textbf{(strong Fréchet) derivative} of $f$.
\end{definition}
\begin{proof}
Let $\lambda, \mu \in L(E; F)$ and $\varphi, \psi: V \to F$ be tangent to $0$ such that
\begin{align*}
f(x_0 + h) &= f(x_0) + \lambda h + \varphi(h) \\
&= f(x_0) + \mu h + \varphi(h)
\end{align*}
then
\[
0 = (\lambda - \mu)h + (\varphi - \psi)(h)
\]
By (7) of \ref{lemma:tangent-to-0}, $(\lambda - \mu)$ is tangent to $0$ and thus equal to $0$.
\end{proof}
\begin{proposition}[Chain Rule]
\label{proposition:chain-rule}
Let $E, F, G$ be TVSs over $K \in \RC$ with $F, G$ Hausdorff, $U \subset E$ and $V \subset F$ be open, $f: U \to V$ be differentiable at $x_0 \in U$, and $g: V \to G$ be differentiable at $f(x_0)$, then
\[
D(g \circ f)(x_0) = Dg(f(x_0)) \circ f(x_0)
\]
\end{proposition}
\begin{proof}
By differentiability of $f$ and $g$, there exists $\varphi, \psi$ tangent to $0$ such that
\begin{align*}
(g \circ f)(x_0 + h) &= (g \circ f)(x_0) + Dg(f(x_0))[f(x_0 + h) - f(x_0)] + \varphi(f(x_0 + h) - f(x_0)) \\
&= (g \circ f)(x_0) + Dg(f(x_0))[Df(x_0)(h) + \psi(h)] + \varphi(Df(x_0)(h) + \psi(h)) \\
&= (g \circ f)(x_0) + [Dg(f(x_0)) \circ Df(x_0)](h) \\
&+ [Dg(f(x_0)) \circ \psi + \varphi \circ (Df(x_0) + \psi)](h)
\end{align*}
Since $Dg(f(x_0)) \in L(F; G)$, $Dg(f(x_0)) \circ \psi$ is tangent to $0$ by (6) and (3) \ref{lemma:tangent-linear-at-0}. Similarly, $Df(x_0) + \psi$ is of linear order at $0$ by (4) and (2) of \ref{lemma:tangent-linear-at-0}, so $\varphi \circ (Df(x_0) + \psi)$ is tangent to $0$ by (3) of \ref{lemma:tangent-linear-at-0}. Thus
\[
Dg(f(x_0)) \circ \psi + \varphi \circ (Df(x_0) + \psi)
\]
is tangent to $0$ by (5) of \ref{lemma:tangent-linear-at-0}.
\end{proof}
src/dg/derivative/higher.tex
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@@ -0,0 +1,63 @@
\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}
src/dg/derivative/index.tex
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@@ -4,3 +4,4 @@
\input{./src/dg/derivative/derivative.tex}
\input{./src/dg/derivative/sets.tex}
\input{./src/dg/derivative/mvt.tex}
\input{./src/dg/derivative/higher.tex}
src/dg/derivative/sets.tex
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@@ -1,4 +1,4 @@
\section{Differentiation With Respect to Set Systems}
\section{Differentiation With Respect to Sets}
\label{section:differentiation-set-system}
\begin{definition}[Small]
@@ -119,23 +119,6 @@
A method of extending this sense of differentiability is to require that \textit{every} extension of the function to some open set, or to the entire space is differentiable. Given that this paves way to confusion for related definitions of differentiability, this definition is not formally included here.
\end{remark}
\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}