Updated content.

src/fa/tvs/bounded.tex

@@ -0,0 +1,38 @@
+\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}

src/fa/tvs/completion.tex

@@ -0,0 +1,34 @@
+\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}

src/fa/tvs/continuous.tex

@@ -0,0 +1,86 @@
+\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}

src/fa/tvs/definition.tex

@@ -0,0 +1,149 @@
+\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}

src/fa/tvs/index.tex

@@ -0,0 +1,6 @@
+\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}