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Added definition of holomorphic functions.
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Bokuan Li
2026-05-16 21:42:56 -04:00
parent 88d71d6654
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147
src/dg/complex/derivative.tex
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@@ -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 \autoref[Hahn-Banach Theorem]{proposition:hahn-banach-utility}, $f$ also satisfies (3).
\end{proof}

12
src/dg/derivative/euclid.tex
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@@ -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}

44
src/dg/derivative/power.tex
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@@ -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 $R$.

3
src/fa/notation.tex
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@@ -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}

6
src/fa/rs/rs-bv.tex
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@@ -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 $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_E$ on $E$ and $[\cdot]_F$ on $F$ such that for any $f \in RS([a, b], G)$,
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 $y \in F$,
\[
\braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_E \cdot [g]_{\text{var}, F}
\]
\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 \\