Cleanup

src/dg/derivative/sets.tex

@@ -29,6 +29,7 @@
     \[
     f(x_0 + h) = f(x_0) + Th + r(h)
     \]
+
     for all $h \in V$.
 
     The linear map $T \in L(E; F)$ is the \textbf{$\sigma$-derivative of $f$ at $x_0$}, denoted $D_{\sigma}f(x_0)$.
@@ -56,16 +57,19 @@
     \[
     D_\sigma(g \circ f)(x_0) = D_\tau g(f(x_0)) \circ D_\sigma f(x_0)
     \]
+
 \end{proposition}
 \begin{proof}
     Since $g$ is $\tau$-differentiable at $f(x_0)$, there exists $s \in \mathcal{R}_\tau(F; G)$ such that
     \[
     g(f(x_0) + h) = g \circ f (x_0) + D_\tau g(f(x_0))h + s(h)
     \]
+
     for all $h \in F$ such that $f(x_0) + h \in V$. By differentiability of $f$, there exists $r \in \mathcal{R}_\sigma(E; F)$ such that
     \[
     f(x_0 + h) = f(x_0) + D_\sigma f(x_0)h + r(h)
     \]
+
     for all $h \in E$ such that $x_0 + h \in U$. Therefore for all $h \in E$ with $x_0 + h \in U$,
     \begin{align*}
     g \circ f(x_0 + h) &= g \circ f (x_0) + D_\tau g(f(x_0)) \circ D_\sigma f(x_0)h \\
@@ -87,7 +91,7 @@
         \item For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
         \item For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
     \end{enumerate}
-    and by \ref{proposition:chain-rule-sets}, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
+    and by \autoref{proposition:chain-rule-sets}, $\sigma$-derivatives and $\tau$-derivatives satisfy the Chain rule.
 \end{proposition}
 \begin{proof}
     (1): Let $A \in \sigma$ and $U \in \cn_G(0)$. Since $T$ is continuous, there exists $V \in \cn_F(0)$ such that $T(V) \subset U$. Since $r \in \mathcal{R}_\sigma(E; G)$, there exists $t > 0$ such that $r(sA)/s \in V$ for all $s \in (0, t)$. In which case, $T \circ r(sA)/s \in U$ for all $s \in (0, t)$.
@@ -98,21 +102,24 @@
     \[
     \limv{n} \frac{1}{t_n}s \circ (T + r)(t_na_n) = \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
     \]
+
     Since $\bracs{t_n^{-1}r(t_na_n)|n \in \natp}$ is a convergent sequence, it is contained in a compact set. Thus
     \[
     B = \bracs{Ta_n + \frac{r(t_na_n)}{t_n}\bigg | n \in \natp} \subset T(A) + \bracs{\frac{r(t_na_n)}{t_n}|n \in \natp}
     \]
+
     is contained in a compact set if $A$ is compact, and bounded if $A$ is bounded. Given that $s \in \mathcal{R}_\sigma(E; F)$, $t^{-1}s(tx) \to 0$ as $t \downto 0$ uniformly on $B$. Therefore
     \[
     \limv{n}\frac{1}{t_n}s\braks{t_n\paren{Ta_n + \frac{r(t_na_n)}{t_n}}} = 0
     \]
+
 \end{proof}
 
 
 
 
 \begin{remark}
-    In \ref{definition:differentiation-small}, the system $\sigma$ can be chosen based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain.
+    In \autoref{definition:differentiation-small}, the system $\sigma$ can be chosen based on the bornology of $E$, and the definition of small-ness depends exclusively on $\sigma$. As such, there is an apparent disconnect between differentiation and the topology of the domain.
 
     Consider for example a Hilbert space equipped with its norm and weak topology. The norm itself is differentiable with respect to both topologies, because the bounded sets coincide. Moreover, the data for differentiability needs to only come from a neighbourhood of $0$ in the norm topology. As such, a function may be differentiable even if its domain is too small to have an interior.
 
@@ -130,6 +137,7 @@
         \[
         \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
         \]
+
         exists. In which case, the above limit is identified with the derivative of $f$ at $0$.
         \item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, if $f$ is differentiable at $x_0$, then $f$ is continuous at $x_0$.
     \end{enumerate}
@@ -147,5 +155,6 @@
     \[
     \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
     \]
+
     and $Df(x_0) = T$.
 \end{proof}