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\part{Functional Analysis}
\label{part:part-fa}
\input{./src/fa/tvs/index.tex}
\input{./src/fa/lc/index.tex}
\input{./src/fa/rs/index.tex}

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\section{Continuous Linear Maps}
\label{section:tvs-convex-morphism}
\begin{proposition}
\label{proposition:tvs-convex-morphism}
Let $E, F$ be locally convex spaces and $T \in \hom(E; F)$, then the following are equivalent:
\begin{enumerate}
\item $T$ is uniformly continuous.
\item $T$ is continuous.
\item $T$ is continuous at $0$.
\item For every continuous seminorm $[\cdot]_F$ on $F$, there exists a continuous seminorm $[\cdot]_E$ on $E$ such that $[Tx]_F \le [x]_E$ for all $x \in E$.
\end{enumerate}
\end{proposition}
\begin{proof}
$(1) \Leftrightarrow (2) \Leftrightarrow (3)$: By \ref{definition:continuous-linear}.
$(2) \Rightarrow (4)$: $x \mapsto [Tx]_F$ is a continuous seminorm on $E$.
$(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \ref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \ref{definition:continuous-linear}.
\end{proof}
\begin{proposition}
\label{proposition:tvs-convex-multilinear}
Let $\seqf{E_j}$ and $F$ be locally convex spaces, and $T: \prod_{j = 1}^n E_j \to F$ be $n$-linear map, then the following are equivalent:
\begin{enumerate}
\item $T$ is continuous.
\item For every continuous seminorm $[\cdot]_F$ on $F$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$, such that for every $x \in \prod_{j = 1}^n E_j$,
\[
[Tx]_F \le \prod_{j = 1}^n [x_j]_{E_j}
\]
\end{enumerate}
\end{proposition}
\begin{proof}
$(1) \Rightarrow (2)$: By continuity of $T$, there exists continuous seminorms $\seqf{[\cdot]_j}$ on $\seqf{E_j}$ such that for any $x \in \prod_{j = 1}^n E_j$, $\max_{1 \le j \le n}[x_j]_{E_j} < 1$ implies that $[Tx]_F < 1$. In which case, the inequality follows from linearity.
\end{proof}

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\section{Seminorms}
\label{section:seminorms}
\begin{definition}[Convex]
\label{definition:convex}
Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.
\end{definition}
\begin{definition}[Sublinear Functional]
\label{definition:sublinear-functional}
Let $E$ be a vector space over $K \in \RC$, then a \textbf{sublinear functional} is a mapping $\rho: E \to \real$ such that:
\begin{enumerate}
\item $\rho(0) = 0$.
\item For any $x \in E$ and $\lambda \ge 0$, $\rho(\lambda x) = \lambda \rho(x)$.
\item For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
\end{enumerate}
\end{definition}
\begin{definition}[Seminorm]
\label{definition:seminorm}
Let $E$ be a vector space over $K \in \RC$, then a \textbf{seminorm} on $E$ is a mapping $\rho: E \to [0, \infty)$ such that:
\begin{enumerate}
\item $\rho(0) = 0$.
\item For any $x \in E$ and $\lambda \in K$, $\rho(\lambda x) = \abs{\lambda} \rho(x)$.
\item For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
\end{enumerate}
\end{definition}
\begin{lemma}
\label{lemma:continuous-seminorm}
Let $E$ be a TVS over $K \in \RC$ and $[\cdot]: E \times E \to [0, \infty)$ be a seminorm on $E$, then the following are equivalent:
\begin{enumerate}
\item $[\cdot]$ is uniformly continuous.
\item $[\cdot]$ is continuous.
\item $[\cdot]$ is continuous at $0$.
\end{enumerate}
\end{lemma}
\begin{proof}
$(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]} \le [x - y] < \eps$. By \ref{proposition:tvs-uniform}, $[\cdot]$ is uniformly continuous.
\end{proof}
\begin{definition}[Topology Induced by Seminorm]
\label{definition:seminorm-topology}
Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then:
\begin{enumerate}
\item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudo-metric.
\item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
\item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
\end{enumerate}
The topology induced by $\seqi{d}$ is the \textbf{vector space topology induced by} $\seqi{[\cdot]}$. In addition,
\begin{enumerate}
\item[(U)] For any family $\seqj{[\cdot]}$ of seminorms continuous on $E$, the vector space topology induced by $\seqj{[\cdot]}$ is contained in the vector space topology induced by $\seqi{[\cdot]}$.
\end{enumerate}
\end{definition}
\begin{definition}[Gauge/Minkowski Functional]
\label{definition:gauge}
Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be a radial set, then the mapping
\[
[\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
\]
is the \textbf{gauge/Minkowski functional} of $A$, and
\begin{enumerate}
\item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
\item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
\item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
\end{enumerate}
In particular,
\begin{enumerate}
\item If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
\item If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
\end{enumerate}
\end{definition}
\begin{proof}
(2): Let $\lambda, \mu > 0$ such that $\lambda^{-1}x, \mu^{-1}y \in A$. By convexity, $t\lambda^{-1} + (1 - t)\mu^{-1}y \in A$ for all $t \in [0, 1]$. Let $t \in [0, 1]$ such that
\[
(\lambda + \mu)^{-1}(x + y) = t\lambda^{-1}x + (1 - t)\mu^{-1}y
\]
then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.
\end{proof}
\begin{definition}[Locally Convex Space]
\label{definition:locally-convex}
Let $E$ be a TVS over $\RC$, then the following are equivalent:
\begin{enumerate}
\item There exists a fundamental system of neighborhoods at $0$ consisting of convex sets.
\item There exists a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets.
\item There exists a family of seminorms $\seqi{[\cdot]}$ that induces the topology on $E$.
\end{enumerate}
If the above holds, then $E$ is a \textbf{locally convex} space.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By \ref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \bracs{tx + (1 - t)y|x, y \in V}$ be the convex hull of $V$, then $W \subset U$ is convex and circled.
$(2) \Rightarrow (3)$: For each $V \in \cn(0)$ convex, circled, and radial, let $[\cdot]_V: E \to [0, \infty)$ be its gauge, then $[\cdot]_V$ is a seminorm. For each $x, y \in X$ and $r > 0$, $[x - y]_V < r$ if and only if $x - y \in rV$. Thus the uniformity induced by $[\cdot]_V$ corresponds to the uniformity generated by $\bracs{U_{rV}| r > 0}$, where $U_V = \bracs{(x, y) \in E|x - y \in V}$. Since this holds for all $U \in \cn(0)$, the topology of $E$ and the topology induced by $\bracs{[\cdot]_V| V \in \cn(0), \text{ convex, circled, radial}}$ coincide.
$(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.
\end{proof}

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\chapter{Locally Convex Spaces}
\label{chap:lc}
\input{./src/fa/lc/convex.tex}
\input{./src/fa/lc/continuous.tex}

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\section{Functions of Bounded Variation}
\label{section:bv}
\begin{definition}[Total Variation]
\label{definition:total-variation}
Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, $f: [a, b] \to E$, and $P \in \scp([a, b])$ be a partition, then
\[
V_{\rho, p}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
\]
is the \textbf{variation} of $f$ with respect to $\rho$ and $P$. The supremum over all such partitions
\[
[f]_{\var, \rho} = \sup_{P \in \scp([a, b])}V_{\rho, P}(f)
\]
is the \textbf{total variation} of $f$ on $[a, b]$ with respect to $\rho$.
If $E$ is a normed space, then the variation and total variation of $f$ is taken with respect to its norm.
\end{definition}
\begin{definition}[Bounded Variation, {{\cite[Proposition X.1.1]{Lang}}}]
\label{definition:bounded-variation}
Let $E$ be a locally convex space, $\rho$ be a continuous seminorm on $E$, and $f: [a, b] \to E$. If $[f]_{\var, \rho} < \infty$, then $f$ is of \textbf{bounded variation} with respect to $\rho$.
The space $BV([a, b]; E)$ is the set of functions $[a, b] \to E$ of bounded variation with respect to every continuous seminorm on $E$, and
\begin{enumerate}
\item $BV([a, b]; E)$ is a vector space.
\item For each continuous seminorm $\rho$ on $E$, $[\cdot]_{\var, \rho}$ is a seminorm on $BV([a, b]; E)$.
\item Let $\fF$ be a filter on $BV([a, b]; E)$ and $f: [a, b] \to E$. If
\begin{enumerate}
\item $\pi_x(\fF) \to f(x)$ for all $x \in [a, b]$.
\item For every continuous seminorm $\rho$ on $E$, there exists $U \in \fF$ such that $\sup_{g \in U}[g]_{\var, \rho} = M_\rho < \infty$.
\end{enumerate}
then $f \in BV([a, b]; E)$ with $[f]_{\var, \rho} \le M_\rho$.
\item For any $f \in BV([a, b]; E)$ and continuous seminorm $\rho$ on $E$, $\sup_{x \in [a, b]}\rho(f(x)) \le \rho(f(a)) + [f]_{\var, \rho}$.
\end{enumerate}
If $(E, \norm{\cdot}_E)$ is a normed space, then
\begin{enumerate}
\item[(5)] $f$ has at most countably many discontinuities.
\end{enumerate}
\end{definition}
\begin{proof}
(3): Let $\rho$ be a continuous seminorm on $E$ and $P \in \scp([a, b])$, then by assumption (a),
\[
V_{\rho, P}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
= \lim_{g, \fF}\sum_{j = 1}^n \rho(g(x_j) - g(x_{j - 1}))
= \lim_{g \in \fF}V_{\rho, P}(g)
\]
By assumption (b), $[0, M_\rho]$ is in the filter generated by $V_{\rho, P}(\fF)$. Thus $V_{\rho, P}(f) \le M_\rho$. As this holds for all $P \in \scp([a, b])$, $V_{\rho, P}(f) \le M_\rho$, and $f \in BV([a, b]; E)$.
(5): For each $n \in \nat$, let
\[
D_n = \bracs{x \in [a, b]|\forall \eps > 0, \exists y \in (x - \eps, x + \eps): \norm{f(x) - f(y)}_E \ge 1/n}
\]
then $D = \bigcup_{n \in \nat}D_n$ is the set of discontinuity points of $f$. If $D$ is uncountable, then there exists $N \in \nat$ such that $D_n$ is infinite.
Fix $N \in \nat$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then
\begin{enumerate}
\item[(a)] $|E_k| \ge N - k$.
\item[(b)] $E_k \subset I_k^o$.
\end{enumerate}
for $k = 1$.
Let $k \le N$ and suppose inductively that $E_k, I_k$ have been constructed. Let $x_k \in E_k$, then by (b), there exists $\eps > 0$ such that $[x_k - \eps, x_k + \eps] \subset I_k$ and $|E_k \setminus [x_k - \eps, x_k + \eps]| \ge N - k$. Let $y_k \in [x_k - \eps, x_k + \eps]$ such that $\norm{f(x_k) - f(y_k)} \ge 1/n$, $I_{k + 1} = I_k \setminus [x_k - \eps, x_k + \eps]$, and $E_{k+1} = E_k \setminus [x_k - \eps, x_k + \eps]$, then $I_k$ and $E_k$ satisfies (a) and (b).
Therefore there exists pairs $\bracs{(x_k, y_k)|1 \le k \le N}$ such that $\norm{f(x_k) - f(y_k)}_E \ge 1/n$ for all $n$, and the smallest interval containing each $(x_k, y_k)$ are pairwise disjoint. Thus $[f]_{\var} \ge N/n$ for all $N \in \nat$, so $[f]_{\var} = \infty$.
\end{proof}

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\chapter{The Riemann-Stieltjes Integral}
\label{chap:rs-integral}
\input{./src/fa/rs/partition.tex}
\input{./src/fa/rs/bv.tex}
\input{./src/fa/rs/rs.tex}
\input{./src/fa/rs/rs-bv.tex}

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\section{Partitions}
\label{section:partitions}
\begin{definition}[Partition]
\label{definition:partition-interval}
Let $[a, b] \subset \real$, then a \textbf{partition} of $[a, b]$ is a sequence
\[
P = \seqfz{x_j} = [a = x_0 \le \cdots \le x_n = b]
\]
The collection $\scp([a, b])$ is the set of all partitions of $[a, b]$.
\end{definition}
\begin{definition}[Tagged Partition]
\label{definition:tagged-partition}
Let $[a, b] \subset \real$, then a \textbf{tagged partition} of $[a, b]$ is a pair $(P = \seqfz{x_j}, c = \seqf{c_j})$ such that $c_j \in [x_{j - 1}, x_j]$ for each $1 \le j \le n$.
The collection $\scp_t([a, b])$ is the set of all tagged partitions of $[a, b]$.
\end{definition}
\begin{definition}[Mesh]
\label{definition:mesh}
Let $P$ be a partition of $[a, b] \subset \real$, then
\[
\sigma(P) = \max_{1 \le j \le n}(x_j - x_{j - 1})
\]
is the \textbf{mesh} of $P$.
\end{definition}
\begin{definition}[Fine]
\label{definition:partition-refinement}
Let $P = \seqfz[m]{x_j}, Q = \seqfz{y_j} \in \scp([a, b])$, then $Q$ is \textbf{finer} than $P$ if for every $0 \le j \le m$, there exists $0 \le k \le m$ such that $x_j = y_k$. For any $P, Q \in \scp([a, b])$, denote $P \le Q$ if $Q$ is finer than $P$, then
\begin{enumerate}
\item $\scp([a, b])$/$\scp_t([a, b])$ equipped with $\le$ is a upward-directed set.
\item If $P \le Q$, then $\sigma(P) \ge \sigma(Q)$.
\item For any $\eps > 0$, there exists $P \in \scp([a, b])$ with $\sigma(P) < \eps$.
\end{enumerate}
If $(P, c), (Q, d) \in \scp_t([a, b])$, then $(Q, d)$ is finer than $(P, c)$ if $Q$ is finer than $P$.
\end{definition}

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\section{Riemann-Stieltjes Integrals and Functions of Bounded Variationo}
\label{section:rs-bv}
\begin{proposition}
\label{proposition:rs-bound}
Let $[a, b] \subset \real$, $E_1, E_2, H$ be locally convex spaces, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to E_2$.
Let $[\cdot]_H$ be a continuous seminorm on $H$, then there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that for any $f \in RS([a, b], G)$,
\[
\braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_1 \cdot [g]_{\var, 2}
\]
\end{proposition}
\begin{proof}
By \ref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that $[xy]_H \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
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 \le \sum_{j = 1}^n [f(c_j)]_1[G(x_j) - G(x_{j - 1})]_2 \\
&\le \sup_{x \in [a, b]}[f]_1 \cdot V_{2, P}(G) \le \sup_{x \in [a, b]}[f]_1 \cdot [g]_{\var, 2}
\end{align*}
\end{proof}
\begin{proposition}
\label{proposition:rs-complete}
Let $[a, b] \subset \real$, $E_1, E_2, H$ be locally convex spaces, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G \in BV([a, b]; E_2)$.
For each continuous seminorm $\rho$ on $H$ and $f: [a, b] \to E$, define
\[
[f]_{u, \rho} = \sup_{x \in [a, b]}f(\rho)
\]
Let $\net{f} \subset RS([a, b], G)$ such that:
\begin{enumerate}
\item[(a)] $\rho(f_\alpha - f) \to 0$ for all continuous seminorm $\rho$ on $E_1$.
\item[(b)] $\lim_{\alpha \in A}\int_a^b f_\alpha dG$ exists.
\end{enumerate}
then $f \in RS([a, b], G)$ and $\int_a^b f dG = \lim_{\alpha \in A}\int_a^b f_\alpha dG$. In particular,
\begin{enumerate}
\item If $H$ is complete, then condition (a) may be omitted.
\item If $H$ is sequentially complete and $A = \nat$, then condition (b) may be omitted.
\end{enumerate}
\end{proposition}
\begin{proof}
Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
\begin{align*}
\rho\paren{S(P, c, f, G) - \lim_{\alpha \in A}\int_a^b f_\alpha dG} &\le \rho(S(P, c, f - f_\alpha, G)) \\
&+ \rho\paren{\int_a^b f_\alpha dG - \lim_{\alpha \in A}\int_a^b f_\alpha dG} \\
&+ \rho\paren{S(P, c, f_\alpha, G) - \int_a^b f_\alpha dG}
\end{align*}
Let $\rho$ be a continuous seminorm on $H$, and $[\cdot]_1$ and $[\cdot]_2$ be continuous seminorms on $E_1$ and $E_2$ such that $\rho(xy) \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
Let $\eps > 0$, then by assumption (a) and (b), there exists $\alpha \in A$ such that:
\begin{enumerate}
\item $[f - f_\alpha]_1 < \eps/(3[G]_{\var, 2})$.
\item $\rho\paren{\int_a^b f_\alpha dG - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps/3$.
\end{enumerate}
Since $f_\alpha \in RS([a, b], G)$, there exists $P_0 \in \scp([a, b])$ such that if $P \ge P_0$,
\begin{enumerate}
\item[(3)] $\rho\paren{S(P, c, f_\alpha, G) - \int_a^b f_\alpha dG} < \eps/3$.
\end{enumerate}
Thus for any $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$,
\[
\rho\paren{S(P, c, f, G) - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps
\]
\end{proof}
\begin{proposition}
\label{proposition:rs-bv-continuous}
Let $[a, b] \subset \real$, $E_1, E_2$ be locally convex spaces, $H$ be a sequentially complete locally convex space, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
Let $f \in C([a, b]; E_1)$, $G \in BV([a, b]; E_2)$, then
\begin{enumerate}
\item $f \in RS([a, b], G)$.
\item For any $\seq{(P_n, t_n)} \subset \scp_t([a, b])$ with $\sigma(P_n) \to 0$,
\[
\int_a^b fdG = \limv{n}S(P_n, t_n, f, G)
\]
\end{enumerate}
\end{proposition}
\begin{proof}
Let $\rho$ be a continuous seminorm on $H$, and $[\cdot]_1$ and $[\cdot]_2$ be continuous seminorms on $E_1$ and $E_2$ such that $\rho(xy) \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
Let $(P = \seqfz{x_j}, c = \seqf{c_j}), (Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_t([a, b])$ with $Q \ge P$, then
\begin{align*}
\rho(S(P, c, f, G) - S(Q, d, f, G)) &\le \sum_{j = 1}^n \sum_{y_k \in [x_{j - 1}, x_j]}[f(c_j) - f(d_k)]_1[G(y_k) - G(y_{k - 1})]_2 \\
&\le \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \sigma(P) \end{array}}[f(x) - f(y)]_1 \cdot [G]_{\var, 2}
\end{align*}
Therefore for any two $(P, c), (Q, d) \in \scp_t([a, b])$,
\[
\rho(S(P, c, f, G) - S(Q, d, f, G)) \le 2 \cdot \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \max(\sigma(P), \sigma(Q)) \end{array}}[f(x) - f(y)]_1 \cdot [G]_{\var, 2}
\]
by passing through a common refinement. Since $f \in C([a, b]; E_1)$, this bound tends to $0$ as $\max(\sigma(P), \sigma(Q))$ tends to $0$, so $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is a Cauchy net.
In addition, for any $\seq{(P_n, t_n)}$ as in (2), $\limv{n}S(P_n, t_n, f, G)$ exists by sequential completeness. Since $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is Cauchy, the limit $\lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)$ exists as well and is equal to $\limv{n}S(P_n, t_n, f, G)$.
\end{proof}

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\section{Riemann-Stieltjes Sums and Integrals}
\label{section:tvs-rs-integral}
\begin{definition}[Riemann-Stieltjes Sum, {{\cite[Section X.1]{Lang}}}]
\label{definition:rs-sum}
Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, and $G: [a, b] \to E_2$.
Let $f: [a, b] \to E_1$ and $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
\[
S(P, c, f, G) = \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]
\]
is the \textbf{Riemann-Stieltjes sum} of $f$ with respect to $G$ and $(P, c)$.
\end{definition}
\begin{definition}[Riemann-Stieltjes Integral, {{\cite[Section X.1]{Lang}}}]
\label{definition:rs-integral}
Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $G: [a, b] \to E_2$.
Let $f: [a, b] \to E_2$, then $f$ is \textbf{Riemann-Stieltjes integrable} with respect to $G$ if the limit
\[
\int_a^b f dG = \int_a^b f(t)dG(t) = \lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)
\]
exists. In which case, $\int_a^b fdG$ is the \textbf{Riemann-Stieltjes integral} of $G$.
The set $RS([a, b], G)$ is the vector space of all \textbf{Riemann-Stieltjes integrable functions} with respect to $G$.
\end{definition}
\begin{lemma}[Summation by Parts, {{\cite[Proposition 1.4]{Lang}}}]
\label{lemma:sum-by-parts}
Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map, $f: [a, b] \to E_1$, $G: [a, b] \to E_2$, and $(P, c) \in \scp_t([a, b])$, then
\[
S(P, c, f, G) + S(P', c', G, f) = f(b)G(b) - f(a)G(a)
\]
where $P' = \seqfz[n+1]{y_j} = [a, c_1, \cdots, c_n, b]$ and $c' = \seqf[n+1]{d_j} = [x_0, \cdots, x_n]$.
\end{lemma}
\begin{proof}
Denote $c_0 = a$ and $c_{n+1} = b$, then
\begin{align*}
S(P, c, f, G) &= \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]
= \sum_{j = 1}^n f(c_j)G(x_j) - \sum_{j = 1}^n f(c_j)G(x_{j - 1}) \\
&= f(c_n)G(x_n)- f(c_0)G(x_0) + \sum_{j = 1}^n f(c_{j - 1})G(x_{j-1}) - \sum_{j = 1}^n f(c_j)G(x_{j - 1}) \\
&= f(c_n)G(x_n)- f(c_0)G(x_0) - \sum_{j = 1}^n G(x_{j - 1})[f(c_j) - f(c_{j - 1})] \\
&= f(c_{n+1})G(x_n) - f(c_0)G(x_0) - \sum_{j = 1}^{n+1}G(x_{j - 1})[f(c_j) - f(c_{j - 1})] \\
&= f(b)G(b) - f(a)G(a) - S(P', c', G, f)
\end{align*}
\end{proof}
\begin{theorem}[Integration by Parts]
\label{theorem:rs-ibp}
Let $[a, b] \subset \real$, $E_1, E_2, H$ be TVSs over $F \in \RC$, and $E_1 \times E_2 \to H$ with $(x, y) \mapsto xy$ be a continuous bilinear map.
Let $f: [a, b] \to E_1$ and $G: [a, b] \to E_2$, then $f \in RS([a, b], G)$ if and only if $G \in RS([a, b], f)$. In which case,
\[
\int_a^b f dG + \int_a^b G df = f(b)G(b) - f(a)G(a)
\]
\end{theorem}
\begin{proof}
Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_F(0)$, then there exits $P_0 = \seqfz{x_j} \in \scp([a, b])$ such that $S(P, c, f, G) - \int_a^b fdG \in U$ for all $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$. Let
\[
Q_0 = [x_0, x_1, x_1, \cdots, x_n, x_n]
\]
then for any $(Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_t([a, b])$ with $Q \ge Q_0$,
\[
f(b)G(b) - f(a)G(a) - \int_a^b fdG - S(Q, d, G, f) =
\int_a^b fdG - S(Q', d', G, f)
\]
by \ref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $\int_a^b fdG - S(Q', d', G, f) \in U$.
\end{proof}

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\section{Bounded Sets}
\label{section:bounded}
\subsection{Bounded Sets}
\label{subsection:tvs-bounded}
\begin{definition}[Bounded]
\label{definition:bounded}
Let $E$ be a TVS over $K \in \RC$ and $B \subset E$, then $B$ is \textbf{bounded} if for every $U \in \cn(0)$, there exists $\lambda \in K$ such that $\lambda U \supset B$.
\end{definition}
\begin{proposition}[{{\cite[1.5.1]{SchaeferWolff}}}]
\label{proposition:bounded-operations}
Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$ be bounded, then the following sets are bounded:
\begin{enumerate}
\item Any $C \subset B$.
\item The closure $\ol{B}$.
\item $\lambda B$ where $\lambda \in K$.
\item $A \cup B$.
\item $A + B$.
\end{enumerate}
\end{proposition}
\begin{proof}
Let $U \in \cn(0)$.
(2): Using \ref{proposition:uniform-neighbourhoods}, assume without loss of generality that $U$ is closed. Let $0 \ne \lambda \in K$ with $\lambda U \supset B$, then since $\lambda U$ is closed, $\lambda U \supset \ol B$.
(4), (5): By \ref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$, and $\lambda, \lambda' \in K$ such that $\lambda V \supset A$ and $\lambda' V \supset B$.
Let $\mu > \abs{\lambda}, \abs{\lambda'}$, then
\[
\mu U \supset \mu V \supset \lambda V \cup \lambda' V \supset A \cup B
\]
and
\[
\mu U \supset \mu(V + V) \supset \lambda V + \lambda' V \supset A + B
\]
\end{proof}

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\section{The Hausdorff Completion}
\label{section:tvs-complete}
\begin{definition}[Hausdorff Completion of TVS]
\label{definition:tvs-completion}
Let $E$ be a TVS over $K \in \RC$, then there exists $(\wh E, \iota)$ such that:
\begin{enumerate}
\item $\wh E$ is a complete Hausdorff TVS.
\item $\iota \in L(E; \wh E)$.
\item[(U)] For any $(F, T)$ satisfying (1) and (2), there exists a unique $\ol{T} \in L(\wh E; F)$ such that the following diagram commutes:
\end{enumerate}
Moreover,
\begin{enumerate}
\item[(4)] $\iota(E)$ is dense in $\wh E$.
\end{enumerate}
The pair $(\wh E, \iota)$ is the \textbf{Hausdorff completion} of $E$.
\end{definition}
\begin{proof}
All claims of (1), (2), (U), and (4), except the linearity of maps and the fact that $\wh E$ is a TVS is proven via the Hausdorff completion (\ref{definition:hausdorff-completion}).
Using \ref{proposition:initial-completion}, identify $\wh E \times \wh E$ with $\wh{E \times E}$ and $K \times \wh E$ with $\wh{K \times E}$ as uniform spaces. By \ref{proposition:hausdorff-uniform-factor}, there exists operations $\wh E \times \wh E \to \wh E$ and $K \times \wh E \to \wh E$ such that the following diagrams commute
\[
\xymatrix{
\widehat E \times \widehat E \ar@{->}[r] & \widehat E & & K \times \widehat E \ar@{->}[r] & \widehat E \\
E \times E \ar@{->}[u] \ar@{->}[r] & E \ar@{->}[u] & & K \times E \ar@{->}[u] \ar@{->}[r] & E \ar@{->}[u]
}
\]
By continuity and the density of $\iota(E)$ in $E$, $\wh E$ with these operations forms a TVS, and $T$ is linear.
\end{proof}
\begin{remark}[{{\cite[Section 1.1]{SchaeferWolff}}}]
The Hausdorff completion works in general with arbitrary valuated fields. Though the completion yields a TVS over the completion of the field, the field need not to be complete.
\end{remark}

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\section{Continuous Linear Maps}
\label{section:tvs-linear-maps}
\begin{definition}[Continuous Linear Map]
\label{definition:continuous-linear}
Let $E, F$ be TVSs over $K \in \RC$, and $T \in \hom({E, F})$ be a linear map, then the following are equivalent:
\begin{enumerate}
\item $T$ is uniformly continuous.
\item $T$ is continuous.
\item $T$ is continuous at $0$.
\end{enumerate}
If the above holds, then $T$ is a \textbf{continuous linear map}. The set $L(E; F)$ denotes the vector space of all continuous linear maps from $E$ to $F$.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2) \Rightarrow (3)$: By \ref{proposition:uniform-continuous} and \ref{definition:continuity}.
$(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$. Using \ref{proposition:tvs-uniform} and \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $U$ and $V$ are symmetric and translation-invariant.
For any $x, y \in V$, $x - y \in V(0)$, so $Tx - Ty \in U(0)$, $Ty \in U(Tx)$ by symmetry, and $(Tx, Ty) \in U$. Therefore $T$ is uniformly continuous.
\end{proof}
\begin{definition}[Continuous Multilinear Map]
\label{definition:continuous-multilinear}
Let $\seqf{E}$, $F$ be TVSs over $K \in \RC$, then the set $L^n(E_1, \cdots, E_n; F) = L^n(\seqf{E_j}; F)$ is the space of all continuous $n$-linear maps from $\prod_{j = 1}^n E_j$ to $F$.
\end{definition}
\begin{proposition}
\label{proposition:continuous-bounded}
Let $E, F$ be TVSs over $K \in \RC$ and $T \in L(E; F)$, then for any $B \subset E$ bounded, $T(B)$ is also bounded.
\end{proposition}
\begin{proof}
Let $U \in \cn_F(0)$, then $T^{-1}(U) \in \cn_E(0)$, so there exists $\lambda \in K$ such that $\lambda T^{-1}(U) = T^{-1}(\lambda U) \supset B$ and $\lambda U \supset T(B)$.
\end{proof}
\begin{definition}[Initial Uniformity]
\label{definition:tvs-initial}
Let $E$ be a vector space over $K \in \RC$, $\seqi{F}$ be TVSs, and $\seqi{T}$ where $T_i \in \hom(E; F_i)$ for all $i \in I$, then there exists a uniformity $\fU$ on $E$ such that:
\begin{enumerate}
\item For each $i \in I$, $T_i \in L(E; F_i)$.
\item[(U)] If $\mathfrak{V}$ is a uniformity on $E$ satisfying $(1)$, then $\mathfrak{V} \supset \fU$.
\end{enumerate}
Moreover,
\begin{enumerate}
\item[(3)] $\fU$ is translation-invariant.
\item[(4)] $E$ equipped with the topology induced by $\fU$ is a topological vector space.
\end{enumerate}
The uniformity and its induced topology are the \textbf{initial uniformity/topology} induced by $\seqi{T}$.
\end{definition}
\begin{proof}
(1), (U): By \ref{definition:initial-uniformity}.
Let $U \in \fU$, then there exists $J \subset I$ finite and translation-invariant entourages $\seqj{U}$ such that
\[
U \subset V = \bigcap_{j \in J}(T_j \times T_j)^{-1}(U_j)
\]
(3): For each $j \in J$, $(x, y) \in (T_j \times T_j)^{-1}(U_j)$, and $z \in E$,
\[
(T_j \times T_j)(x + z, y + z) = (T_jx + T_jz, T_jy + T_jz) \in U_j
\]
so $(T_j \times T_j)^{-1}(U_j)$ is translation-invariant, and so is $V$.
(4): By (TVS1) and (TVS2), for each $j \in J$, there exists an entourage $V_j$ of $F_j$ and $\eps_j > 0$ such that for any $(x, x'), (y, y') \in V_j$ and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'} < \eps_j$, $(x + y, x' + y'), (\lambda x, \lambda' x') \in U_j$.
Therefore, for any $(x, x'), (y, y') \in \bigcap_{j \in J} T_j^{-1}(V_j)$ and $\lambda, \lambda' \in K$ with $\abs{\lambda - \lambda'} < \min_{j \in J}\eps$, $(x + y, x' + y'), (\lambda x, \lambda' x') \in V$.
\end{proof}
\begin{definition}[Product Topology]
\label{definition:tvs-product}
Let $\seqi{E}$ be TVSs over $K \in \RC$ and $E = \prod_{i \in I}E_i$ be their product as a vector space, and $\fU$ be the initial uniformity generated by the projection maps, then
\begin{enumerate}
\item $E$ equipped with the topology induced by $\fU$ is a topological vector space.
\item[(U)] For any TVS $F$ over $K$ and $\seqi{T}$ where $T_i \in L(F; E_i)$ for each $i \in I$, there exists a unique $U \in L(F; E)$ such that the following diagram commutes
\[
\xymatrix{
F \ar@{->}[rd]^{T_i} \ar@{->}[d]_{T} & \\
\prod_{i \in I}E_i \ar@{->}[r]_{\pi_i} & E_i
}
\]
\end{enumerate}
The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
\end{definition}

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\section{Vector Space Topologies}
\label{section:tvs-topology}
\begin{definition}[Topological Vector Space]
\label{definition:tvs}
Let $E$ be a vector space over $K \in \bracs{\real, \complex}$ and $\topo \subset 2^E$ be a topology. If
\begin{enumerate}
\item[(TVS1)] $E \times E \to E$ with $(x, y) \mapsto x + y$ is continuous.
\item[(TVS2)] $K \times E \to E$ with $(\lambda, x) \mapsto \lambda x$ is continuous.
\end{enumerate}
then the pair $(E, \topo)$ is a \textbf{topological vector space}.
\end{definition}
\begin{definition}[Translation-Invariant Topology]
\label{definition:translation-invariant-topology}
Let $E$ be a vector space and $\topo$ be a topology on $E$, then $\topo$ is \textbf{translation-invariant} if for any $U \in \topo$ and $y \in E$, $U + y \in \topo$.
\end{definition}
\begin{lemma}
\label{lemma:tvs-translation-invariant}
Let $E$ be a TVS over $K \in \RC$, then the topology of $E$ is translation-invariant.
\end{lemma}
\begin{proof}
Let $U \subset E$ open and $y \in E$, then $U + y$ is the preimage of $U$ by the map $x \mapsto x - y$. By (TVS1), $U + y$ is open.
\end{proof}
\begin{definition}[Translation-Invariant Uniformity]
\label{definition:translation-invariant-uniformity}
Let $E$ be a vector space, $\fU$ be a uniformity on $E$, and $U \in \fU$, then $U$ is \textbf{translation-invariant} if for every $z \in E$,
\[
U = \bracs{(x + z, y + z)|(x, y) \in U}
\]
and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages.
\end{definition}
\begin{lemma}
\label{lemma:translation-invariant-symmetric}
Let $E$ be a vector space and $\fU$ be a translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
\end{lemma}
\begin{proof}
Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \ref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
\end{proof}
\begin{proposition}[{{\cite[1.1.4]{SchaeferWolff}}}]
\label{proposition:tvs-uniform}
Let $E$ be a TVS over $K \in \bracs{\real, \complex}$, then:
\begin{enumerate}
\item There exists a unique translation-invariant uniformity $\fU$ on $E$ that induces the topology on $E$.
\item For each neighbourhood $V \in \cn(0)$, let $U_V = \bracs{(x, y) \in E^2| x - y \in V}$, then for any fundamental system of neighbourhoods $\fB_0$ at $0$, $\fB = \bracs{U_V| V \in \fB_0}$ is a fundamental system of entourages for $\fU$.
\end{enumerate}
The space $E$ will always be assumed to be equipped with its translation-invariant uniformity.
\end{proposition}
\begin{proof}
(2): Firstly, for any $V \in \fB_0$, $U_V$ is translation-invariant.
\begin{enumerate}
\item[(FB1)] For any $V, V' \in \fB_0$, there exists $W \in \fB_0$ such that $W \subset V \cap V'$. In which case, $U_V \cap U_{V'} \supset U_W \in \fB$.
\item[(UB1)] For any $V \in \fB_0$, $0 \in V$, so $\Delta \subset U_V$.
\item[(UB2)] Let $V \in \fB_0$, then by (TVS1), there exists $W \in \fB_0$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_W$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_W \circ U_W \subset U_V$.
\end{enumerate}
By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
(1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.
Let $W \in \cn(0)$, then by \ref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
Let $V \in \mathfrak{V}$. Using \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
\end{proof}
\begin{proposition}
\label{proposition:tvs-closure}
Let $E$ be a TVS over $K \in \RC$, $A \subset E$, and $\fB \subset \cn(0)$ be a fundamental system of neighbourhoods, then
\[
\ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB}
\]
\end{proposition}
\begin{proof}
Let $V \in \cn(0)$ be balanced and $U_V = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_V(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_V$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_V(A) = A + V$.
Assume without loss of generality that $\fB$ consists of symmetric entourages. By \ref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \ref{proposition:uniformclosure} implies that
\[
\ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A
\]
\end{proof}
\begin{proposition}[{{\cite[1.1.1]{SchaeferWolff}}}]
\label{proposition:tvs-set-operations}
Let $E$ be a TVS over $K \in \RC$ and $A, B \subset E$, then:
\begin{enumerate}
\item If $A$ is open, then $A + B$ is open.
\item If $A$ is closed and $B$ is compact, then $A + B$ is closed.
\end{enumerate}
\end{proposition}
\begin{proof}
$(1)$: For every $x \in B$, $A + x$ is open by \ref{definition:translation-invariant-topology}, so
\[
A + B = \bigcup_{x \in B}(A + x)
\]
is open.
$(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
\[
y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U}
\]
By \ref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
\[
y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U]
\]
Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
\[
y \in \bigcap_{U \in \fF}[(-B) + U + U] \subset \bigcap_{V \in \cn(0)}[(x-B) + V] = \overline{x - B} = x - B
\]
so $x \in y + B \subset A + B$.
\end{proof}
\begin{definition}[Balanced/Circled]
\label{definition:balanced}
Let $E$ be a vector space over $K \in \RC$ and $A \subset E$, then $A$ is \textbf{balanced/circled} if $\lambda A \subset A$ for all $\lambda \in K$ with $\abs{\lambda} \le 1$.
\end{definition}
\begin{definition}[Absorbing/Radial]
\label{definition:absorbing}
Let $E$ be a vector space over $K \in \RC$ and $A, B \subset E$, then $A$ \textbf{absorbs} $B$ if there exists $\lambda \in K$ such that $\lambda A \supset B$, and $A$ is \textbf{absorbing/radial} if it absorbs every point in $E$.
\end{definition}
\begin{proposition}
\label{proposition:tvs-good-neighbourhood-base}
Let $E$ be a topological vector space over $K \in \RC$, then
\begin{enumerate}
\item $E$ admits a fundamental system of neighbourhoods at $0$ consisting of balanced and absorbing sets.
\item The fundamental system of neighbourhoods in $(1)$ can be taken to be open or closed.
\end{enumerate}
\end{proposition}
\begin{proof}
Firstly, (TVS2) implies that every neighbourhood of $0$ is balanced.
By \ref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
Let $U \in \cn^o(0)$ be open. By (TVS2), there exists $r > 0$ such that $\lambda U \subset U$ for all $\lambda \in K$ with $\abs{\lambda} \le r$. Define
\[
V = \bigcup_{\substack{\lambda \in K \\ \abs{\lambda} \le r}} \lambda U \subset U
\]
then for any $x \in V$, there exists $\lambda \in K$ with $\abs{\lambda} \le r$ and $y \in U$ such that $x = \lambda y$. In which case, for any $\mu \in K$ with $\abs{\mu} \le 1$, $\mu(\lambda y) = (\mu \lambda) y \in \mu\lambda U$. Since $\abs{\mu \lambda} \le r$, $\mu \lambda U \subset V$. Thus $V \subset U$ is balanced.
Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^o(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda} \le 1$, $\lambda \overline{V} = \overline{\lambda V} \subset \overline{V}$ by (TVS2). Therefore $\overline{V} \subset U$ is balanced as well.
\end{proof}

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\chapter{Topological Vector Spaces}
\label{chap:tvs}
\input{./src/fa/tvs/definition.tex}
\input{./src/fa/tvs/bounded.tex}
\input{./src/fa/tvs/continuous.tex}