Fixed typo.

src/dg/complex/derivative.tex

@@ -34,13 +34,13 @@
 
 \begin{theorem}[Cauchy]
 \label{theorem:cauchy-homotopy}
-    Let $E$ be a separated locally convex space over $\complex$, $U \subset \complex$ be open, $f \in C^1(U; E)$, and $\gamma, \mu \in C([a, b]; \complex)$ be closed, rectifiable paths. If $\gamma$ and $\mu$ are homotopic, then
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $f \in C^1(U; E)$, and $\gamma, \mu \in C([a, b]; U)$ be closed rectifiable paths. If $\gamma$ and $\mu$ are homotopic, then
     \[
     \int_\gamma f = \int_\mu f
     \]
 \end{theorem}
 \begin{proof}[Proof of smooth case. ]
-    Let $\Gamma \in C^\infty([0, 1] \times [a, b]; \complex)$ be a smooth homotopy of loops from $\gamma$ to $\mu$, and
+    Let $\Gamma \in C^\infty([0, 1] \times [a, b]; U)$ be a smooth homotopy of loops from $\gamma$ to $\mu$, and
     \[
     F: [0, 1] \to E \quad t \mapsto \int_{\Gamma (t, \cdot)}f = \int_a^b (f \circ \Gamma)(t, s) \Gamma(t, ds)
     \]
@@ -93,7 +93,7 @@
     \end{enumerate}
     
     Furthermore, by passing through a reparametrisation, assume without loss of generality that:
-    \begin{enumerate}[label=(\alph*)]
+    \begin{enumerate}[label=(\alph*),start=1]
         \item For each $t \in [0, \eps)$, $\Gamma(t, \cdot) = \gamma$.
         \item For each $t \in (1 - \eps, 1]$, $\Gamma(t, \cdot) = \mu$.
         \item For each $t \in [0, 1]$, $\Gamma$ is constant on $\bracs{t} \times ([a, a + \eps] \cup [b - \eps, b])$.
@@ -138,5 +138,146 @@
     \]
 \end{proof}
 
+\begin{definition}
+\label{definition:winding-number-1}
+    Let $U \subset \complex$, $z_0 \in U$, and $r > 0$ such that $\ol{B(z_0, r)} \subset U$, then the path
+    \[
+    \omega_{z_0, r}: [0, 2\pi] \to U \quad \theta \mapsto a + re^{i\theta}
+    \]
+
+    is the \textit{standard path of winding number $1$} at $a$ with radius $r$. 
+\end{definition}
+
+
+\begin{theorem}[Cauchy's Integral Formula]
+\label{theorem:cauchy-formula}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $z_0 \in U$, $r > 0$ such that $\ol{B(z_0, r)} \subset U$, $\gamma \in C([a, b]; \complex)$ be a closed, rectifiable path homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$, and $f \in C^1(U; E)$, then
+    \begin{enumerate}
+        \item $\int_\gamma f = 0$.
+        \item $f(z) = \frac{1}{2\pi i}\int_\gamma \frac{f(z)}{z - z_0}dz$. 
+    \end{enumerate}
+
+    More over, for any $g \in C(U; E)$ that satisfies (2) for all $z_0 \in U$, $r > 0$ with $\ol{B(z_0, r)} \subset U$, closed rectifiable curve $\gamma \in C([a, b]; \complex)$ homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$, 
+    \begin{enumerate}[start=2]
+        \item $g \in C^\infty(U; E)$, where for each $k \in \natz$, 
+        \[
+        D^kg(z_0) = \frac{k!}{2\pi i}\int_{\gamma} \frac{g(z)}{(z - z_0)^{k+1}}dz
+        \]
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    By \autoref{theorem:cauchy-homotopy} and the \hyperref[change of variables formula]{theorem:rs-change-of-variables}, for any $g \in C^1(U \setminus \bracs{z_0}; E)$, 
+    \[
+    \int_\gamma g = \lim_{s \downto 0} \int_{\omega_{z_0, s}} g = \int_0^{2\pi} 
+    = \lim_{s \downto 0}\frac{s}{2\pi} \int_{0}^{2\pi} g \circ \omega_{z_0, s}(\theta) e^{i\theta} d\theta
+    \]
+
+    (1): Since $f \in C(U; E)$, $f$ is bounded on $\ol{B(z_0, r)}$, so for any $s \in (0, r)$,
+    \[
+    \frac{s}{2\pi} \int_{0}^{2\pi} f \circ \omega_{z_0, s}(\theta) e^{i\theta} d\theta \in s\ol{\text{Conv}}(f(\ol{B(z_0, r)}))
+    \]
+
+    As $E$ is locally convex, 
+    \[
+    \int_\gamma g = \lim_{s \downto 0} \int_{\omega_{z_0, s}} g = 0
+    \]
+
+    (2): Since $f \in C(U; E)$, 
+    \begin{align*}
+        \frac{1}{2\pi i}\int_{\gamma} \frac{f(z)}{z - z_0}dz &= \lim_{s \downto 0}\frac{s}{2\pi} \int_{0}^{2\pi} \frac{f \circ \omega_{z_0, s}(\theta)}{\omega_{z_0, s}(\theta) - z_0} e^{i\theta} d\theta \\
+        &= \lim_{s \downto 0}\frac{1}{2\pi}\int_0^{2\pi} f \circ \omega_{z_0, s}(\theta) d\theta = f(z_0)
+    \end{align*}
+
+    (3): Suppose inductively that (3) holds for $k \in \natz$. For sufficiently small $h \in \complex$, 
+    \[
+    \frac{D^kg(z_0 + h) -D^kg(z_0)}{h} = \frac{k!}{2\pi ih} \int_\gamma \frac{g(z)}{(z - z_0-h)^{k+1}} - \frac{g(z)}{(z- z_0)^{k+1}}dz
+    \]
+
+    By \autoref{proposition:difference-quotient-compact}, 
+    \[
+    \lim_{h \to 0}\frac{D^kg(z_0 + h) -D^kg(z_0)}{h} = \frac{(k+1)!}{2\pi i} \int_\gamma \frac{g(z)}{(z - z_0)^{k+2}} dz
+    \]
+
+    Therefore $g \in C^{k+1}(U; E)$ with
+    \[
+    D^{k+1}g(z_0) = \frac{(k+1)!}{2\pi i} \int_\gamma \frac{g(z)}{(z - z_0)^{k+2}} dz
+    \]    
+\end{proof}
+
+\begin{corollary}[Cauchy's Estimate]
+\label{corollary:cauchy-estimate}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, $z_0 \in U$, $r > 0$ such that $\ol{B(z_0, r)} \subset U$, then for any $k \in \natz$ and continuous seminorm $[\cdot]_E: E \to [0, \infty)$, 
+    \[
+    [D^kf(z_0)]_E \le \frac{k!}{r^k} \sup_{z \in \ol{B(z_0, r)}}[f(z)]_E
+    \]
+\end{corollary}
+\begin{proof}
+    By \autoref[Cauchy's Integral Formula]{theorem:cauchy-formula} and \autoref{proposition:rs-bound}, 
+    \begin{align*}
+        D^kf(z_0) &= \frac{k!}{2\pi i}\int_{\omega_{z_0, r}} \frac{f(z)}{(z - z_0)^{k+1}}dz \\
+        [D^kf(z_0)]_E &\le \frac{k!}{2\pi i}\int_0^{2\pi}\frac{[f(z)]_E}{|z - z_0|^{k+1}}dz \\
+        &= \frac{k!}{2\pi i}\int_0^{2\pi}\frac{[f(z)]_E}{r^{k+1}}dz \le \frac{k!}{r^k} \sup_{z \in \ol{B(z_0, r)}}[f(z)]_E
+    \end{align*}
+\end{proof}
+
+
+\begin{definition}[Complex Analytic]
+\label{definition:complex-analytic}
+    Let $E$ be a complete separated locally convex space over $\complex$, $U \subset \complex$ be open, and $f \in C(U; E)$, then the following are equivalent:
+    \begin{enumerate}
+        \item (\textbf{Complex Differentiability}) $f \in C^1(U; E)$. 
+        \item (\textbf{Cauchy-Riemann Equations}) Under the identification of $C = \real^2$, $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \in C(U; E)$ and
+        \[
+        \frac{\partial f}{\partial x} = i\frac{\partial f}{\partial y}
+        \]
+        \item (\textbf{Cauchy's Integral Formula}) For each $z_0 \in U$, $r > 0$ such that $\ol{B(z_0, r)} \subset U$, and closed rectifiable path $\gamma \in C([a, b]; U)$ homotopic to $\omega_{z_0, r}$ on $U \setminus \bracs{z_0}$, 
+        \[
+        f(z_0) = \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z - z_0}dz
+        \]
+        \item (\textbf{Analyticity}) For each $z_0 \in U$ and $r > 0$ such that $\ol{B(z_0, r)} \subset U$, there exists $\seq{a_n} \subset E$ such that $f$ may be expressed as a power series
+        \[
+        f(z) = \sum_{n = 0}^\infty a_n(z - z_0)^n
+        \]
+
+        with radius of convergence at least $r$.
+        \item (\textbf{Weak Holomorphy}) For each $\phi \in E^*$, $\phi \circ f$ satisfies the above. 
+    \end{enumerate}
+
+    If the above holds, then $f$ is \textbf{complex analytic}. 
+\end{definition}
+\begin{proof}
+    (1) $\Leftrightarrow$ (2): \autoref{lemma:complex-analytic}. 
+
+    (1) + (2) $\Rightarrow$ (3): See \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}. 
+
+    (3) $\Rightarrow$ (4): By \hyperref[Cauchy's Integral Formula]{theorem:cauchy-formula}, $f \in C^\infty(U; E)$ where for each $k \in \natz$, 
+    \[
+        D^kf(z_0) = \frac{k!}{2\pi i}\int_{\gamma} \frac{f(z)}{(z - z_0)^{k+1}}dz
+    \]
+    
+    Let
+    \[
+    g(z) = \sum_{k = 0}^\infty \frac{1}{k!} D^kf(z_0)(z - z_0)^n
+    \]
+
+    then by \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate}, for any $k \in \natz$ and continuous seminorm $[\cdot]_E: E \to [0, \infty)$, 
+    \[
+        [D^kf(z_0)]_E \le \frac{k!}{r^k} \sup_{z \in \ol{B(z_0, r)}}[f(z)]_E = \frac{Ck!}{r^k}
+    \]
+
+    Thus $[D^kf(z_0)/k!]_E \le C/r^k$ for all $k \in \natz$, and the radius of convergence of $g$ is at least $r$. 
+    
+    Let $z \in B(z_0, r/2)$, $s = |z - z_0|$, and $n \in \natp$, then by \hyperref[Taylor's Formula]{theorem:taylor-lagrange} and \hyperref[Cauchy's Estimate]{corollary:cauchy-estimate}, 
+    \begin{align*}
+        \braks{f(z) - \sum_{k = 0}^n \frac{1}{k!} D^kf(z_0)(z - z_0)^n}_E &\le s^{n+1} \cdot \sup_{z' \in \ol{B(z_0, s)}} [D^{n+1}f(z')]_E \\
+        &\le \frac{Cs^{n+1}}{(r-s)^{n+1}}
+    \end{align*}
+
+    which tends to $0$ as $n \to \infty$. 
+
+    (4) $\Rightarrow$ (1): By \autoref{theorem:termwise-differentiation}. 
+
+    (5) $\Rightarrow$ (3): By the equivalence of the prior points, for any $\phi \in E^*$, $\phi \circ f$ satisfies (3). By the \hyperref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $f$ also satisfies (3). 
+\end{proof}
 
 

src/dg/derivative/euclid.tex

@@ -34,24 +34,24 @@
 
 \begin{proposition}
 \label{proposition:difference-quotient-compact}
-    Let $E$ be a separated locally convex space, $Y$ be a Hausdorff space, and $f: (a, b) \times Y \to E$. If $f$ is differentiable in the first variable and $\frac{df}{dx} \in C((a, b) \times Y; E)$, then
+    Let $E$ be a separated locally convex space over $K \in \RC$, $U \subset K$ be open, $Y$ be a Hausdorff space, and $f: U \times Y \to E$. If $f$ is differentiable in the first variable and $\frac{df}{dx} \in C(U \times Y; E)$, then
     \[
     \frac{f(x + h, y) - f(x, y)}{h} \to \frac{df}{dx}(x, y)
     \]
 
-    as $h \to 0$, uniformly on compacts. 
+    as $h \to 0$, uniformly on compact sets. 
 \end{proposition}
 \begin{proof}
-    Let $[c, d] \subset (a, b)$ and $K \subset Y$ be compact, then by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, for any $(x, y) \in [c, d] \times K$ and $h \in \real$ with $x + h$, 
+    Let $A \subset U$ and $B \subset Y$ be compact, then by the \hyperref[Mean Value Theorem]{theorem:mean-value-theorem-line}, for any $(x, y) \in A \times B$ and $h \in \real$ with $x + h$, 
     \begin{align*}
         &\frac{f(x + h, y) - f(x, y)}{h} - \frac{df}{dx}(x, y) \\
-        &\in \overline{\text{Conv}}\bracs{\frac{df}{dx}(x + k, y) - \frac{df}{dx}(x, y) \bigg | k \in [-h, h]}
+        &\in \overline{\text{Conv}}\bracs{\frac{df}{dx}(x + k, y) - \frac{df}{dx}(x, y) \bigg | k \in B_K(0, |h|)}
     \end{align*}
 
-    Let $\eps > 0$ such that $[c - \eps, d + \eps] \subset (a, b)$, then since  $\frac{df}{dx} \in C((a, b) \times Y; E)$, $\frac{df}{dx}|_{[c - \eps, d + \eps] \times K}$ is uniformly continuous\footnote{$K$ is a compact Hausdorff space, which comes with a \hyperref[unique uniform structure]{proposition:compact-uniform-structure}. }. Since $E$ is locally convex,
+    Let $\eps > 0$ such that $A + B_K(0, |\eps|) \subset U$, then since  $\frac{df}{dx} \in C(U \times Y; E)$, $\frac{df}{dx}|_{(A + B_K(0, |\eps|)) \times B}$ is uniformly continuous\footnote{$K$ is a compact Hausdorff space, which comes with a \hyperref[unique uniform structure]{proposition:compact-uniform-structure}. }. Since $E$ is locally convex,
     \[
     \frac{f(x + h, y) - f(x, y)}{h} - \frac{df}{dx}(x, y) \to 0
     \]
 
-    uniformly on $[c, d] \times K$.
+    uniformly on $A \times B$.
 \end{proof}

src/dg/derivative/power.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Power Series]
 \label{definition:power-series}
-    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, $\bracsn{T_n}_0^\infty$ with $T_n \in L^n(E; F)$ for each $n \in \natz$, and $a \in E$, then the \textbf{power series} of $\bracsn{T_n}_0^\infty$ about $a$ is the function
+    Let $E, F$ be locally convex spaces $K \in \RC$ with $F$ being complete, $\bracsn{T_n}_0^\infty$ with $T_n \in L^n(E; F)$ for each $n \in \natz$, and $a \in E$, then the \textbf{power series} of $\bracsn{T_n}_0^\infty$ about $a$ is the function
     \[
     f(x) = \sum_{n = 0}^\infty T_n(x - a)^{(n)}
     \]
@@ -13,25 +13,39 @@
 
 \begin{definition}[Radius of Convergence]
 \label{definition:radius-of-convergence}
-    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, and $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$ be a power series about $a \in E$, then $R \in [0, \infty]$ be defined by\footnote{Under the abuse that $1/\infty = 0$ and $1/0 =\infty$.}
+    Let $E$ be a normed space over $K \in \RC$, $F$ be a complete locally convex space over $K$, and $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$ be a power series about $a \in E$, $\rho: F \to [0, \infty)$ be a continuous seminorm on $F$. For each $T \in L^n(E; F)$, let
     \[
-    \frac{1}{R} = \limsup_{n \to \infty}\norm{T_n}_{L^n(E; F)}^{1/n}
+    [T]_{L^n(E; F), \rho} = \sup_{x \in B_E(0, 1)^n}\rho(Tx)
     \]
     
-    is the \textbf{radius of convergence of $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$}. For each $0 < r < R$, the series converges uniformly and absolutely on $B_E(a, r)$. 
+    then $R_\rho \in [0, \infty]$ be defined by\footnote{Under the abuse that $1/\infty = 0$ and $1/0 =\infty$.}
+    \[
+    \frac{1}{R_\rho} = \limsup_{n \to \infty}\norm{T_n}_{L^n(E; F)}^{1/n}
+    \]
+
+    is the \textbf{radius of convergence of $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$} with respect to $\rho$, and
+    \begin{enumerate}
+        \item For each $0 < r < R$, the series converges uniformly and absolutely on $B_E(a, r)$ with respect to $\rho$. 
+        \item Let
+        \[
+        R = \inf\bracs{R_\rho| \rho: F \to [0, \infty) \text{ is a continuous seminorm}}
+        \]
+        
+        the series converges uniformly and absolutely on $B_E(a, R)$, and $R$ is the \textbf{radius of convergence of $\sum_{n = 0}^\infty T_n(x - a)^{(n)}$}.
+    \end{enumerate}
 \end{definition}
 \begin{proof}
     For all $x \in B_E(a, r)$, 
     \[
-        \sum_{n = 0}^\infty \normn{T_n(x - a)^{(n)}}_F \le \sum_{n \in \natz} \norm{T_n}_{L^n(E; F)} \norm{x - a}_E^n \le \sum_{n \in \natz} \norm{T_n}_{L^n(E; F)} r^n
+        \sum_{n = 0}^\infty \rho(T_n(x - a)^{(n)}) \le \sum_{n \in \natz} [T_n]_{L^n(E; F), \rho} \norm{x - a}_E^n \le \sum_{n \in \natz}  r^n[T_n]_{L^n(E; F), \rho}
     \]
 
     For any $s \in (r, R)$, there exists $N \in \natp$ such that $\norm{T_n}_{L^n(E; F)}^{1/n} \le 1/s$ for all $n \ge N$. In which case, 
     \[
-    \sum_{n = 0}^\infty \norm{T_n}_{L^n(E; F)} r^n \le \sum_{n = 0}^N \norm{T_n}_{L^n(E; F)}r^n + \sum_{n \ge N}\frac{r^n}{s^n} < \infty
+    \sum_{n = 0}^\infty r^n[T_n]_{L^n(E; F), \rho} \le \sum_{n = 0}^N r^n[T_n]_{L^n(E; F), \rho} + \sum_{n \ge N}\frac{r^n}{s^n} < \infty
     \]
 
-    As this estimate holds uniformly on $B_E(a, r)$, the series converges uniformly and absolutely on $B_E(a, r)$. 
+    As this estimate holds uniformly on $B_E(a, r)$, the series converges uniformly and absolutely on $B_E(a, r)$ with respect to $\rho$. 
 \end{proof}
 
 \begin{remark}
@@ -42,9 +56,9 @@
 
 \begin{theorem}[Termwise Differentiation]
 \label{theorem:termwise-differentiation}
-    Let $E$ be a normed space over $K \in \RC$, $F$ be a Banach space over $K$, $f(x) = \sum_{n = 0}^\infty T_n(x - a)^{(n)}$ a power series about $a \in E$, and $R$ be its radius of convergence, then 
+    Let $E$ be a normed space over $K \in \RC$, $F$ be a complete locally convex space over $K$, $f(x) = \sum_{n = 0}^\infty T_n(x - a)^{(n)}$ a power series about $a \in E$, and $R$ be its radius of convergence, then 
     \begin{enumerate}
-        \item $f \in C^\infty(B(a, R); F)$.
+        \item $f \in C^\infty(B(a, R); F)$ is infinitely Fréchet differentiable. 
         \item For each $x \in B(a, R)$ and $h \in E$, 
         \[
         Df(x)(h) = \sum_{n = 0}^\infty \sum_{k = 1}^{n+1}T_{n+1}(((x-a)^{(n)}, h)^{\bracs{k}})
@@ -54,14 +68,20 @@
     \end{enumerate}
 \end{theorem}
 \begin{proof}
-    (3): For each $n \in \natz$, let
+    (3): Let $\rho: F \to [0, \infty)$ be a continuous seminorm. For each $n \in \natz$ and $T \in L^n(E; F)$, let
+    \begin{align*}
+    [T]_{L^n(E; F), \rho} &= \sup_{x \in B_E(0, 1)^n}\rho(Tx) \\
+    [T]_{L^n(E; L(E; F)), \rho} &= \sup_{x \in B_E(0, 1)^n}[Tx]_{L(E; F), \rho}
+    \end{align*}
+
+    and
     \[
     S_n(x_1, \cdots, x_{n})(h) = \sum_{k = 1}^{n+1}T_{n+1}(((x_1, \cdots, x_n), h)^{\bracs{k}})
     \]
 
-    then $\norm{S_n}_{L^n(E; L(E; F))} \le (n+1)\norm{T_{n+1}}_{L^{n+1}(E; F)}$. Since $(n+1)^{1/n}$ is convergent and $\{||T_n||_{L^n(E; F)}^{1/n}\}$ is bounded, 
+    then $[S_n]_{L^n(E; L(E; F)), \rho} \le (n+1)[T_{n+1}]_{L^{n+1}(E; F), \rho}$. Since $(n+1)^{1/n}$ is convergent and $\{[T_{n+1}]_{L^{n+1}(E; F), \rho}\}_1^\infty$ is bounded, 
     \[
-    \limsup_{n \to \infty} \norm{S_n}_{L^n(E; L(E; F))}^{1/n} \le \limsup_{n \to \infty}(n+1)^{1/n}\norm{T_{n+1}}_{L^{n+1}(E; F)}^{1/n} = \frac{1}{R}
+    \limsup_{n \to \infty} [S_n]_{L^n(E; L(E; F)), \rho}^{1/n} \le \limsup_{n \to \infty}(n+1)^{1/n}[T_{n+1}]_{L^{n+1}(E; F), \rho}^{1/n} \le \frac{1}{R}
     \]
 
     so the radius of convergence of the proposed series is at least $R$. 

src/fa/notation.tex

@@ -43,8 +43,11 @@
     $T_{f,\rho}(x)$ & Variation function of $f$ with respect to $\rho$. & \autoref{definition:variation-function} \\
     $BV([a,b]; E)$ & Functions of bounded variation. & \autoref{definition:bounded-variation} \\
     $S(P, c, f, G)$ & Riemann-Stieltjes sum $\sum_j f(c_j)[G(x_j)-G(x_{j-1})]$. & \autoref{definition:rs-sum} \\
+    $\int_a^b f dG$, $\int_a^b f(t) G(dt)$ & Riemann-Stieljes integral of $f$ with respect to $G$. & \autoref{definition:rs-integral}  \\
     $RS([a,b], G)$ & Space of RS-integrable functions w.r.t.\ $G$. & \autoref{definition:rs-integral} \\
     $\mathrm{Reg}([a,b], G; E)$ & Regulated functions w.r.t.\ $G$ on $[a,b]$. & \autoref{definition:regulated-function} \\
     $\mu_G$ & Lebesgue-Stieltjes measure associated with $G$. & \autoref{definition:riemann-lebesgue-stieltjes} \\
+    $\int_\gamma f$, $\int_\gamma f(z)dz$ & Path integral of $f$ with respect to $\gamma$. & \autoref{definition:path-integral} \\
+    $PI([a, b], \gamma; E)$ & Space of path integrable functions with respect to $\gamma$. & \autoref{definition:path-integral}
 \end{tabular}
 

src/fa/rs/rs-bv.tex

@@ -3,17 +3,13 @@
 
 \begin{proposition}
 \label{proposition:rs-bound}
-    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to F$.
+    Let $[a, b] \subset \real$, $E, F, H$ be locally convex spaces, $E \times F \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to F$, and $f \in RS([a, b], G)$, then for any continuous seminorms $[\cdot]_E: E \to [0, \infty)$, $[\cdot]_F: F \to [0, \infty)$, and $[\cdot]_H: H \to [0, \infty)$ such that $[xy]_H \le [x]_E[y]_F$ for all $x \in E$ and $y \in F$, 
-
-    Let $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that for any $f \in RS([a, b], G)$,
     \[
     \braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
     \]
 
 \end{proposition}
 \begin{proof}
-    By \autoref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that $[xy]_H \le [x]_E[y]_F$ for all $(x, y) \in E \times F$.
-
     Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
     \begin{align*}
         [S(P, c, f, G)]_H &\le \sum_{j = 1}^n [f(c_j)[G(x_j) - G(x_{j - 1})]]_H \\
@@ -137,7 +133,7 @@
     \limv{n}S(P_n, c_n, f(\cdot, t), \alpha) = \int_a^b f(s, t) \alpha(ds)
     \]
 
-    uniformly for all $t \in [c, d]$. Since $\beta \in BV([c, d]; G)$, 
+    uniformly for all $t \in [c, d]$. Since $f \in C([a, b] \times [c, d]; E)$, $f$ is uniformly continuous by \autoref{proposition:uniform-continuous-compact}, and $\bracs{f(\cdot, t)|t \in [c, d]} \subset C([a, b]; E)$ is uniformly equicontinuous. As $\beta \in BV([c, d]; G)$, 
     \[
     \int_c^d\int_a^b f(s, t) \alpha(ds) \beta(dt) = \limv{n}\int_c^d S(P_n, c_n, f(\cdot, t), \alpha) \beta(dt)
     \]