Added the Mean Value Theorem.
This commit is contained in:
Bokuan Li
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\input{./src/dg/derivative/derivative.tex}
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\input{./src/dg/derivative/sets.tex}
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\input{./src/dg/derivative/mvt.tex}
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src/dg/derivative/mvt.tex
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107
src/dg/derivative/mvt.tex
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\section{The Mean Value Theorem}
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\label{section:mean-value-theorem}
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\begin{definition}[Right-Differentiable]
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\label{definition:right-differentiable-mvt}
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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
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\[
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D^+f(x) = \lim_{t \downto 0} \frac{f(x + t) - f(x)}{t}
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\]
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exists.
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\end{definition}
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\begin{lemma}[{{\cite[Lemma 4.6.2]{Bogachev}}}]
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\label{lemma:right-differentiable-inequality}
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Let $-\infty < a < b < \infty$, $f, g \in C([a,b]; \real)$, and $N \subset [a, b]$ be at most countable such that:
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\begin{enumerate}
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\item[(a)] $f, g$ are right-differentiable on $[a, b] \setminus N$.
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\item[(b)] For every $x \in [a, b] \setminus N$, $D^+f(x) \le D^+g(x)$.
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\end{enumerate}
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then for any $x \in [a, b]$, $f(x) - f(a) \le g(x) - g(a)$.
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\end{lemma}
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\begin{proof}
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First assume that for every $x \in [a, b] \setminus N$, $D^+f(x) < D^+g(x)$.
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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
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\[
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S = \bracs{x \in [a, b] \bigg | f(y) - f(a) \le g(y) - g(a) + \eps\sum_{n \in N(x)}2^{-n} \quad \forall y \in [a, x]}
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\]
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then by continuity of $f$ and $g$, $S$ is closed.
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Let $x \in S$ and suppose that $s < b$. If $x \in [a, b] \setminus N$, then since
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\[
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\lim_{t \downto 0}\frac{f(x + t) - f(x)}{t} < \lim_{t \downto 0}\frac{g(x + t) - f(x)}{t}
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\]
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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, $[x, x + \delta) \subset S$.
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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.
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Therefore $S = [a, b]$. As this holds for all $\eps > 0$, $f(b) - f(a) \le g(b) - g(a)$.
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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 $\eps > 0$, $f(x) - f(a) \le g(x) - g(a)$.
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\end{proof}
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\begin{theorem}[{{\cite[Theorem 4.6.1]{Bogachev}}}]
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\label{theorem:right-differentiable-convex-form}
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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:
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\begin{enumerate}
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\item $f, g$ are right-differentiable on $[a, b] \setminus N$.
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\item For each $x \in [a, b] \setminus N$, $D^+f(x) \in D^+g(x)B$.
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\item $g$ is non-decreasing.
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\end{enumerate}
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then
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\[
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f(b) - f(a) \in [g(b) - g(a)]B
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\]
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\end{theorem}
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\begin{proof}
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Let $\phi \in E^*$ and $x \in [a, b] \setminus N$, then since $g$ is non-decreasing,
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\begin{align*}
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\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} \\
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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} \\
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D^+(\phi \circ f)(x) &\le D^+g(x) \cdot \sup_{z \in B}\phi(z)
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\end{align*}
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By \ref{lemma:right-differentiable-inequality},
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\[
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\phi(f(b) - f(a)) \le (g(b) - g(a)) \cdot \sup_{z \in B}\phi(z)
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\]
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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
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\[
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\phi(f(b) - f(a)) > \sup_{x \in B}\phi[(g(b) - g(a))b] = \phi(g(b) - g(a)) \cdot \sup_{x \in B}\phi(x)
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\]
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which is impossible. Therefore $f(b) - f(a) \in [g(b) - g(a)]B$.
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\end{proof}
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\begin{theorem}[Mean Value Theorem]
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\label{theorem:mean-value-theorem}
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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
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\[
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f(b) - f(a) \in \overline{\text{Conv}\bracs{Df(x)(b - a)| x \in (a, b) \setminus N}}
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\]
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\end{theorem}
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\begin{proof}
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By \ref{proposition:derivative-sets-real}, $f$ is right-differentiable on $(a, b) \setminus N$ with
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\[
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D^+f(x) = Df(x) \in \bracs{Df(x)|x \in (a, b) \setminus N}
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\]
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for all $x \in (a, b)$. Let $g(x) = x$, then by \ref{theorem:right-differentiable-convex-form},
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\[
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f(b) - f(a) \in \overline{(b - a)\text{Conv}\bracs{Df(x)|x \in (a, b) \setminus N}}
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\]
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\end{proof}
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\begin{proposition}
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\label{proposition:zero-derivative-constant}
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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.
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\end{proposition}
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\begin{proof}
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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}), $f(y) - f(x) = 0$.
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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.
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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 $y \in W$.
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Since $W$ is both open and closed, $W$ must coincide with $V$.
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\end{proof}
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@@ -39,6 +39,12 @@
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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, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
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\end{definition}
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\begin{definition}]
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\label{definition:derivative-garden}
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Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b \subset 2^E$ be the collection of all finite, compact, and bounded subsets, respectively, then differentiability with respect to $\sigma^E_{\text{Fin}}, \sigma^E_{c}, \sigma^E_b$ correspond to \textbf{Gateaux}, \textbf{Hadamard}, and \textbf{Fréchet} differentiability.
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\end{definition}
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\begin{proposition}[{{\cite[Proposition 4.5.2]{Bogachev}}}]
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\label{proposition:chain-rule-sets}
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Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be upward-directed families that contain all finite sets. If:
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@@ -130,3 +136,32 @@
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\]
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is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
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\end{definition}
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\begin{proposition}
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\label{proposition:derivative-sets-real}
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Let $E$ be a separated topological vector space and $\sigma \subset B(\real)$ be an upward-directed system that contains finite sets, then
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\begin{enumerate}
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\item $\mathcal{R}_{\sigma}(\real; E) = \mathcal{R}_{B(\real)}(\real; E)$. Hence, all forms of $\sigma$-differentiability on $\real$ are equivalent.
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\item For any $U \subset \real$ open, $f: U \to E$, and $x_0 \in U$, $f$ is ($\sigma$-)differentiable at $x_0$ if and only if
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\[
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\lim_{t \to 0}\frac{f(x + t) - f(x)}{t}
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\]
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exists. In which case, the above limit is identified with the ($\sigma$-)derivative of $f$ at $0$.
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\end{enumerate}
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\end{proposition}
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\begin{proof}
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(1): Let $r \in \mathcal{R}_\sigma(\real; E)$. For any $R > 0$ and $U \in \cn_E(0)$, there exists $\delta > 0$ such that $t^{-1}r(tR), t^{-1}r(-tR) \in U$ for all $t \in (0, \delta)$. Thus $t^{-1}r(tB(0, R)) \subset U$, and $r \in \mathcal{R}_{B(\real)}(\real; E)$.
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(2): Suppose that $f$ is differentiable at $x_0$, then there exists $r \in \mathcal{R}_\sigma$ such that for any $t \in \real$ with $x_0 + t \in U$,
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\begin{align*}
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f(x_0 + t) - f(x_0) &= Df(x_0)(t) + r(t) \\
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\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1) + t^{-1}r(t) \\
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\lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} &= Df(x_0)(1)
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\end{align*}
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Now suppose that $v = \lim_{t \to 0}\frac{f(x + t) - f(x)}{t}$ exists. Let $T: \real \to E$ be defined by $t \mapsto tv$, then
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\[
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\lim_{t \to 0}\frac{f(x_0 + t) - f(x_0) - Tt}{t} = \lim_{t \to 0}\frac{f(x_0 + t) - f(x_0)}{t} - v = 0
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\]
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and $Df(x_0) = T$.
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\end{proof}
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