Added the Mean Value Theorem.

src/dg/derivative/mvt.tex

@@ -0,0 +1,107 @@
+\section{The Mean Value Theorem}
+\label{section:mean-value-theorem}
+
+\begin{definition}[Right-Differentiable]
+\label{definition:right-differentiable-mvt}
+    Let $-\infty < a < b < \infty$, $E$ be a separated topological vector space, $f: [a, b] \to E$, and $x \in [a, b)$, then $f$ is \textbf{right-differentiable at $x$} if
+    \[
+    D^+f(x) = \lim_{t \downto 0} \frac{f(x + t) - f(x)}{t}
+    \]
+    exists.
+\end{definition}
+
+\begin{lemma}[{{\cite[Lemma 4.6.2]{Bogachev}}}]
+\label{lemma:right-differentiable-inequality}
+    Let $-\infty < a < b < \infty$, $f, g \in C([a,b]; \real)$, and $N \subset [a, b]$ be at most countable such that:
+    \begin{enumerate}
+        \item[(a)] $f, g$ are right-differentiable on $[a, b] \setminus N$.
+        \item[(b)] For every $x \in [a, b] \setminus N$, $D^+f(x) \le D^+g(x)$.
+    \end{enumerate}
+    then for any $x \in [a, b]$, $f(x) - f(a) \le g(x) - g(a)$.
+\end{lemma}
+\begin{proof}
+    First assume that for every $x \in [a, b] \setminus N$, $D^+f(x) < D^+g(x)$.
+
+    Let $\seq{x_n}$ be an enumeration of $N$. For each $x \in [a, b]$, let $N(x) = \bracs{n \in \natp|x_n \in [a, x)}$. Let $\eps > 0$ and define
+    \[
+    S = \bracs{x \in [a, b] \bigg | f(y) - f(a) \le g(y) - g(a) + \eps\sum_{n \in N(x)}2^{-n} \quad \forall y \in [a, x]}
+    \]
+    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.
+
+    Therefore $S = [a, b]$. As this holds for all $\eps > 0$, $f(b) - f(a) \le g(b) - g(a)$.
+
+    Finally, suppose that $D^+f(x) \le D^+g(x)$ for all $x \in [a, b] \setminus N$. Let $\eps > 0$ and $h(x) = g(x) + \eps(x - a)$. By the preceding case, $f(b) - f(a) \le g(b) - g(a) + \eps(b - a)$. As this holds for all $\eps > 0$, $f(x) - f(a) \le g(x) - g(a)$.
+\end{proof}
+
+\begin{theorem}[{{\cite[Theorem 4.6.1]{Bogachev}}}]
+\label{theorem:right-differentiable-convex-form}
+    Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $f \in C([a, b]; E)$, $g \in C([a, b]; \real)$, $N \subset [a, b]$ be at most countable, and $B \subset E$ be a closed convex set such that:
+    \begin{enumerate}
+        \item $f, g$ are right-differentiable on $[a, b] \setminus N$.
+        \item For each $x \in [a, b] \setminus N$, $D^+f(x) \in D^+g(x)B$.
+        \item $g$ is non-decreasing.
+    \end{enumerate}
+    then
+    \[
+    f(b) - f(a) \in [g(b) - g(a)]B
+    \]
+\end{theorem}
+\begin{proof}
+    Let $\phi \in E^*$ and $x \in [a, b] \setminus N$, then since $g$ is non-decreasing,
+    \begin{align*}
+        \lim_{t \downto 0} \frac{\phi(f(x + t) - f(x))}{t} &\le \lim_{t \downto 0}\sup_{z \in B} \frac{\phi(g(x + t) - g(x)z)}{t} \\
+        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},
+    \[
+    \phi(f(b) - f(a)) \le (g(b) - g(a)) \cdot \sup_{z \in B}\phi(z)
+    \]
+
+    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
+    \[
+    \phi(f(b) - f(a)) > \sup_{x \in B}\phi[(g(b) - g(a))b] = \phi(g(b) - g(a)) \cdot \sup_{x \in B}\phi(x)
+    \]
+    which is impossible. Therefore $f(b) - f(a) \in [g(b) - g(a)]B$.
+\end{proof}
+
+
+\begin{theorem}[Mean Value Theorem]
+\label{theorem:mean-value-theorem}
+    Let $-\infty < a < b < \infty$, $E$ be a separated locally convex space, $S \subset [a, b]$ be at most countable, and $f \in C([a, b]; E)$ be differentiable on $(a, b) \setminus N$, then
+    \[
+    f(b) - f(a) \in \overline{\text{Conv}\bracs{Df(x)(b - a)| x \in (a, b) \setminus N}}
+    \]
+\end{theorem}
+\begin{proof}
+    By \ref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
+    \[
+    D^+f(x) = Df(x) \in \bracs{Df(x)|x \in (a, b) \setminus N}
+    \]
+    for all $x \in (a, b)$. Let $g(x) = x$, then by \ref{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{proposition}
+\label{proposition:zero-derivative-constant}
+    Let $E$ be a topological vector space, $F$ be a separated locally convex space, $V \subset E$ be open and connected, $f: V \to F$ be Gateaux-differentiable on $V$ such that $Df(x) = 0$ for all $x \in V$, then $f$ is constant.
+\end{proposition}
+\begin{proof}
+    Let $x, y \in V$ such that $\bracs{(1 - t)x + ty|t \in [0, 1]} \subset V$. Since $f$ is Gateaux-differentiable, $g: [0, 1] \to F$ defined by $g(t) = f((1 - t)x + ty)$ is differentiable with $Dg(t) = 0$ for all $t \in [0, 1]$. By the Mean Value Theorem (\ref{theorem:mean-value-theorem}), $f(y) - f(x) = 0$.
+
+    Fix $x \in V$ and let $W$ be the connected component of $\bracs{y \in V|f(x) = f(y)}$ containing $x$. Since $E$ is locally path-connected (\ref{proposition:tvs-locally-connected}), $W$ is open.
+
+    Let $y \in \overline{W}$. By \ref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_E(0)$ radial such that $U + y \subset V$. Since any point in $U$ is joined to $y$ via a line segment, $f$ is constant on $U + y$. As $y \in \overline{W}$, $U + y \cap W \ne \emptyset$, so $y \in W$.
+
+    Since $W$ is both open and closed, $W$ must coincide with $V$.
+\end{proof}