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Typo fixes.

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This commit is contained in:
Bokuan Li
2026-03-15 12:32:31 -04:00
parent f951ccccdf
commit 62a8e78dfe
3 changed files with 29 additions and 18 deletions
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32
src/dg/derivative/higher.tex
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@@ -35,26 +35,33 @@
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)
&= 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
\[
B_h(k) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k)
\]
\begin{align*}
B_h(k) &= f(x + h + k) - f(x + k) \\
&- Df(x + h)(k) + Df(x)(k)
\end{align*}
then
\[
B_h(k) - B_h(0) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k) -f(x + h) + f(x)
\]
\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 $r_2, r_3 \in \mathcal{R}_{B(E)}$ 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) \\
&= 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 \hyperref[Mean Value Theorem]{theorem:mean-value-theorem},
\[
\norm{B_h(k) - B_h(0)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
@@ -68,9 +75,12 @@
so $D^2f(x)(h, k) - D^2f(x)(k, h) = 0$.
Now suppose that the proposition holds for $n$. Identify $L^n(E; F) = L^{2}(E; L^{n-2}(E; F))$, then for any $\seqf[n]{x_j} \subset E$,
\[
Df(x)(x_1, \cdots, x_n) = Df(x)(x_1, x_2)(x_3, \cdots, x_n) = Df(x)(x_2, x_1)(x_3, \cdots, x_n) = Df(x)(x_2, x_1, x_3, \cdots, x_n)
\]
\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.
\end{proof}