Cleanup

src/dg/derivative/mvt.tex

@@ -7,6 +7,7 @@
     \[
     D^+f(x) = \lim_{t \downto 0} \frac{f(x + t) - f(x)}{t}
     \]
+
     exists.
 \end{definition}
 
@@ -26,12 +27,14 @@
     \[
     S = \bracs{x \in [a, b] \bigg | f(y) - f(a) \le g(y) - g(a) + \eps\sum_{n \in N(x)}2^{-n} \quad \forall y \in [a, x]}
     \]
+
     then by continuity of $f$ and $g$, $S$ is closed.
 
     Let $x \in S$ and suppose that $s < b$. If $x \in [a, b] \setminus N$, then since
     \[
     \lim_{t \downto 0}\frac{f(x + t) - f(x)}{t} < \lim_{t \downto 0}\frac{g(x + t) - f(x)}{t}
     \]
+
     there exists $\delta > 0$ such that $f(x + t) - f(x) < g(x + t) < g(x)$ for all $t \in [0, \delta)$. In which case, $[x, x + \delta) \subset S$.
 
     If $x = x_n \in N$, then by continuity of $f$ and $g$, there exists $\delta > 0$ such that $f(x + t) - f(x) \le g(x + t) - g(x) + 2^{-n}$ for all $t \in (0, \delta)$. Hence $[x, x + \delta) \subset S$ as well.
@@ -53,6 +56,7 @@
     \[
     f(b) - f(a) \in [g(b) - g(a)]B
     \]
+
 \end{theorem}
 \begin{proof}
     Let $\phi \in E^*$ and $x \in [a, b] \setminus N$, then since $g$ is non-decreasing,
@@ -61,15 +65,17 @@
         D^+(\phi \circ f)(x) &\le \sup_{z \in B}\phi(z) \cdot \lim_{t \downto 0}\frac{\phi(g(x + t) - g(x))}{t} \\
         D^+(\phi \circ f)(x) &\le  D^+g(x) \cdot \sup_{z \in B}\phi(z)
     \end{align*}
-    By \ref{lemma:right-differentiable-inequality},
+    By \autoref{lemma:right-differentiable-inequality},
     \[
     \phi(f(b) - f(a)) \le (g(b) - g(a)) \cdot \sup_{z \in B}\phi(z)
     \]
 
-    Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the Hahn-Banach theorem (\ref{theorem:hahn-banach-geometric-2}), there exists $\phi \in E^*$ such that
+
+    Suppose that $f(b) - f(a) \not\in [g(b) - g(a)]B$. Given that $B$ is closed and convex, by the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach-geometric-2}, there exists $\phi \in E^*$ such that
     \[
     \phi(f(b) - f(a)) > \sup_{x \in B}\phi[(g(b) - g(a))b] = \phi(g(b) - g(a)) \cdot \sup_{x \in B}\phi(x)
     \]
+
     which is impossible. Therefore $f(b) - f(a) \in [g(b) - g(a)]B$.
 \end{proof}
 
@@ -80,16 +86,19 @@
     \[
     f(b) - f(a) \in \overline{\text{Conv}\bracs{Df(x)(b - a)| x \in (a, b) \setminus N}}
     \]
+
 \end{theorem}
 \begin{proof}
-    By \ref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
+    By \autoref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
     \[
     D^+f(x) = Df(x) \in \bracs{Df(y)|y \in (a, b) \setminus N}
     \]
-    for all $x \in (a, b)$. Let $g(x) = x$, then by \ref{theorem:right-differentiable-convex-form},
+
+    for all $x \in (a, b)$. Let $g(x) = x$, then by \autoref{theorem:right-differentiable-convex-form},
     \[
     f(b) - f(a) \in \overline{(b - a)\text{Conv}\bracs{Df(x)|x \in (a, b) \setminus N}}
     \]
+
 \end{proof}
 
 \begin{theorem}[Mean Value Theorem]
@@ -98,15 +107,17 @@
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
+
     where $[x, y] = \bracs{(1 - t)x + ty|y \in [0, 1]}$.
 \end{theorem}
 \begin{proof}
-    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule \ref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \ref{proposition:derivative-sets-real}.
+    Let $g: [0, 1] \to F$ be defined by $g(t) = f((1 - t)x + ty)$. Since $f$ is Gateaux-differentiable, $g$ is differentiable by the chain rule \autoref{proposition:chain-rule-sets-conditions} with $Dg(t) = Df((1 - t)x + ty)(y - x)$, and continuous by \autoref{proposition:derivative-sets-real}.
 
-    By the Mean Value Theorem (\ref{theorem:mean-value-theorem-line}),
+    By the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line},
     \[
     f(y) - f(x) = g(1) - g(0) \in \overline{\text{Conv}\bracs{Dg(t)|t \in [0, 1]}} = \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}}
     \]
+
 \end{proof}
 
 \begin{proposition}
@@ -118,9 +129,10 @@
     \[
     f(y) - f(x) \in \overline{\text{Conv}\bracs{Df(z)(y - x)|z \in (x, y)}} = \bracs{0}
     \]
-    by the Mean Value Theorem (\ref{theorem:mean-value-theorem-line}).
 
-    Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. For any $y \in W$, by \ref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. Therefore $W \supset W \cup (U + y)$, and $W$ is open by \ref{lemma:openneighbourhood}.
+    by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}.
+
+    Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. For any $y \in W$, by \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. Therefore $W \supset W \cup (U + y)$, and $W$ is open by \autoref{lemma:openneighbourhood}.
 
     For any $y \in \ol W \cap V$, there exists $U \in \cn(0)$ circled such that $U + y \subset V$. As $y \in \ol W \cap V$, $U \cap W \ne \emptyset$. Thus $f(y) = f(x)$, $y \in W$, and $W$ is relatively closed.