Adjusted organisation in the TVS chapter.

.vscode/settings.json

@@ -7,5 +7,5 @@
     ],
     "latex.linting.enabled": false,
     "latex-workshop.latex.autoBuild.run": "never",
-    "latex-workshop.latex.texDirs": ["${workspaceFolder}"]
+    "latex-workshop.latex.search.rootFiles.include": ["document.tex"]
 }

document.tex

@@ -16,5 +16,4 @@ Hello this is all my notes.
 \bibliographystyle{alpha} % We choose the "plain" reference style
 \bibliography{refs.bib} % Entries are in the refs.bib file
 
-
 \end{document}

src/cat/gluing/index.tex

@@ -22,6 +22,7 @@
     Thus $\Gamma$ is the graph of a function $f: X \to Y$ with $f|_{U_i} = f_i$ for all $i \in I$.
 \end{proof}
 
+
 \begin{lemma}[Gluing for Linear Functions]
 \label{lemma:glue-linear}
     Let $E, F$ be vector spaces over a field $K$, $\fF$ be a family of subspaces of $E$, and $\bracs{T_V}_{V \in \fF}$ with $T_V \in \hom(V; F)$ for all $V \in \fF$. If:

src/dg/derivative/higher.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[$n$-Fold Differentiability]
 \label{definition:n-differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $\mathcal{H} \subset B_\sigma(E; F)$ be a subspace, and $\mathcal{R}_\sigma = \mathcal{R}_\sigma(E; F)$.
 
     Let $U \subset E$ be open, $f: U \to F$, $x_0 \in U$, and $n > 1$, then $f$ is \textbf{$n$-fold $\sigma$-differentiable at $x_0$} if
     \begin{enumerate}
@@ -31,7 +31,7 @@
     A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
     \]
 
-    then there exists $r_1 \in \mathcal{R}_{B(E)}$ such that
+    then there exists $r_1 \in \mathcal{R}_{\mathfrak{B}(E)}$ such that
     \begin{align*}
         A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
         &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
@@ -55,7 +55,7 @@
     \end{align*}
     
 
-    Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ such that for any $k \in B(0, r)$,
+    Now, there exists $r_2, r_3 \in \mathcal{R}_{\mathfrak{B}(E)}$ such that for any $k \in B(0, r)$,
     \begin{align*}
     DB_h(k) &= Df(x + h + k) - Df(x + k) - Df(x + h) + Df(x) \\
     &= D^2f(x)(h + k) + Df(x) - D^2f(x)(h) \\
@@ -88,12 +88,12 @@
 
 \begin{theorem}[Symmetry of Higher Derivatives]
 \label{theorem:derivative-symmetric}
-    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
+    Let $E$ be a topological vector space over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a separated locally convex space over $K$, $U \subset E$ be open, and $f: E \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then $D_\sigma^nf(x_0) \in B_\sigma^n(E; F)$ is symmetric.
 \end{theorem}
 \begin{proof}[Proof {{\cite[Proposition 4.5.14]{Bogachev}}}. ]
     Let $\seqf{h_j} \subset E$, $E_0$ be the subspace generated by $\seqf{h_j}$, and $g = f|_{E_0 \cap U}: E_0 \cap U \to F$. Since $\sigma$ includes all bounded sets contained in finite-dimensional spaces, for any $\phi \in F^*$, the mapping $\phi \circ g: E_0 \cap U \to K$ is $n$-times Fréchet-differentiable, with
     \[
-    D_{B(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
+    D_{\mathfrak{B}(E_0)}^n(\phi \circ g)(x_0) = \phi \circ D_\sigma^n g(x_0)
     \]
 
     by the \hyperref[chain rule]{proposition:chain-rule-sets-conditions}. By \autoref{theorem:derivative-symmetric-frechet}, $\phi \circ D_\sigma^n g(x_0) \in L^n(E_0; K)$ is symmetric. As this holds for any $\seqf{h_j} \subset E$ and $\phi \in F^*$, $D_{\sigma}^n g(x_0) \in B_\sigma^n(E; F)$ is symmetric by the \hyperref[Hahn-Banach theorem]{proposition:hahn-banach-utility}.
@@ -101,7 +101,7 @@
 
 \begin{proposition}[Power Rule]
 \label{proposition:multilinear-derivative}
-    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an ideal that includes all bounded sets contained in finite-dimensional spaces, $F$ be a Hausdorff locally convex space, and
     \[
     T \in \underbrace{L(E; L(E; \cdots L(E; F) \cdots ))}_{n \text{ times}} \subset B^n(E; F)
     \]

src/dg/derivative/sets.tex

@@ -3,7 +3,7 @@
 
 \begin{definition}[Small]
 \label{definition:differentiation-small}
-    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset B(E)$ be a covering ideal, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $r: E \to F$, and $n \in \natz$, then the following are equivalent:
     \begin{enumerate}
         \item For each $A \in \sigma$, $r(th)/t^n \to 0$ uniformly on $A$.
         \item If $r_t(x) = r(tx)/t^n$, then $r_t \to 0$ as $t \to 0$ with respect to the $\sigma$-uniform topology on $F^E$.
@@ -17,7 +17,7 @@
 
 \begin{proposition}
 \label{proposition:differentiation-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $B_\sigma(E; F)$ be the space of linear operators bounded on sets in $\sigma$, and $\mathcal{R}_\sigma(E; F)$ be the space of $\sigma$-small functions, then $(B_\sigma(E; F), \mathcal{R}_\sigma(E; F))$ is a system of derivatives and remainders.
 \end{proposition}
 \begin{proof}
     Let $T \in B_\sigma(E; F)$ and suppose that there exists $V \in \cn_E(0)$ circled and $r \in \mathcal{R}_\sigma(E; F)$ such that $T|_V = r|_V$. For any $x \in V$, $\bracs{x} \in \sigma$, so $T(x) = \lim_{t \downto 0}T(tx)/t = 0$ as $F$ is separated.
@@ -26,7 +26,7 @@
 
 \begin{definition}[Derivative]
 \label{definition:derivative-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{$\sigma$-differentiable at $x_0$} if there exists $V \in \cn_E(0)$, $T \in L(E; F)$, and $r \in \mathcal{R}_\sigma(E; F)$ such that
     \[
     f(x_0 + h) = f(x_0) + Th + r(h)
     \]
@@ -38,7 +38,7 @@
 
 \begin{definition}[Differentiable]
 \label{definition:differentiable-sets}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset B(E)$ be a covering ideal, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, $\sigma \subset \mathfrak{B}(E)$ be a covering ideal, $U \subset E$ be open, and $f: U \to F$, then $f$ is \textbf{$\sigma$-differentiable on $U$} if it is $\sigma$-differentiable at every point in $U$. In which case, the map $D_\sigma f: U \to L(E; F)$ is the \textbf{$\sigma$-derivative} of $f$.
 \end{definition}
 
 \begin{definition}
@@ -49,7 +49,7 @@
 
 \begin{proposition}[Chain Rule]
 \label{proposition:chain-rule-sets}
-    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset B(E)$ and $\tau \subset B(F)$ be covering ideals. If:
+    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated, $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ be covering ideals. If:
     \begin{enumerate}
         \item[(a)] For any $r \in \mathcal{R}_\sigma(E; F)$ and $T \in L(F; G)$, $T \circ r \in \mathcal{R}_\sigma(E; G)$.
         \item[(b)] For any $r \in \mathcal{R}_\sigma(E; F)$, $T \in L(E; F)$, and $s \in \mathcal{R}_\tau(F; G)$, $s \circ (T + r) \in \mathcal{R}_\sigma(E; G)$.
@@ -82,7 +82,7 @@
 
 \begin{proposition}[{{\cite[Corollary 4.5.4]{Bogachev}}}]
 \label{proposition:chain-rule-sets-conditions}
-    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset B(E)$ and $\tau \subset B(F)$ correspond to the following families of sets on $E$ and $F$:
+    Let $E$, $F$, $G$, be TVSs over $K \in \RC$ with $F, G$ being separated. If $\sigma \subset \mathfrak{B}(E)$ and $\tau \subset \mathfrak{B}(F)$ correspond to the following families of sets on $E$ and $F$:
     \begin{enumerate}
         \item Compact sets.
         \item Bounded sets.

src/dg/derivative/taylor.tex

@@ -62,7 +62,7 @@
 
 \begin{theorem}[Taylor's Formula, Peano Remainder]
 \label{theorem:taylor-peano}
-    Let $E$ be a topological vector space, $\sigma \subset B(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
+    Let $E$ be a topological vector space, $\sigma \subset \mathfrak{B}(E)$ be an upward-directed family that includes bounded sets contained in finite-dimensional subspaces, $F$ be a separated locally convex space, $U \subset E$ be open, and $f: U \to F$ be $n$-fold $\sigma$-differentiable at $x_0 \in U$, then there exists $r \in \mathcal{R}_\sigma^n(E; F)$ such that
     \[
     g(x_0 + h) = g(x_0) + \sum_{k = 1}^n \frac{1}{k!}D^k_\sigma f(x_0)(h^{(k)}) + r(h)
     \]

src/fa/lc/bornologic.tex

@@ -67,10 +67,10 @@
 \label{definition:associated-bornological}
     Let $(E, \mathcal{T})$ be a separated locally convex space over $K \in \RC$, then there exists a locally convex topology $\mathcal{T}_B \subset 2^E$ such that:
     \begin{enumerate}
-        \item $B(E, \mathcal{T}_B) \supset B(E, \mathcal{T})$. 
+        \item $\mathfrak{B}(E, \mathcal{T}_B) \supset \mathfrak{B}(E, \mathcal{T})$. 
-        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$. 
+        \item[(U)] For every $\mathcal{S}$ satisfying (1), $\mathcal{S} \subset \mathcal{T}_B$. In particular, $\mathcal{T} \subset \mathcal{T}_B$ and $\mathfrak{B}(E, \mathcal{T}_B) = \mathfrak{B}(E, \mathcal{T})$. 
         \item $(E, \mathcal{T}_B)$ is a bornological space. 
-        \item Let $\mathcal{B} \subset B(E, \topo)$ be the collection of all convex, circled, bounded, and closed subsets of $E$, ordered by inclusion. For each $B \in \mathcal{B}$, let $(E_B, \rho_B)$ be the normed space associated with $B$ and $\iota_B: E_B \to E$ be the canonical inclusion, then $\mathcal{T}_B$ is the inductive topology induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. 
+        \item Let $\mathcal{B} \subset \mathfrak{B}(E, \topo)$ be the collection of all convex, circled, bounded, and closed subsets of $E$, ordered by inclusion. For each $B \in \mathcal{B}$, let $(E_B, \rho_B)$ be the normed space associated with $B$ and $\iota_B: E_B \to E$ be the canonical inclusion, then $\mathcal{T}_B$ is the inductive topology induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. 
     \end{enumerate}
 
     The space $(E, \mathcal{T}_B)$ is the \textbf{bornological space associated with} $E$. 
@@ -78,14 +78,14 @@
 \begin{proof}
     Let $\mathscr{T}$ be the collection of all topologies satisfying (1), then $\mathcal{T} \in \mathscr{T}$, so $\mathscr{T} \ne \emptyset$. Let $\mathcal{T}_B$ be the projective topology induced by $\mathscr{T}$. 
 
-    (1): Since $\mathcal{T}_B \supset \mathcal{T}$, $B(E, \mathcal{T}) \supset B(E, \mathcal{T}_B)$. Let $B \in (E, \mathcal{T})$ and $U \in \cn_{\mathcal{T}_B}(0)$, then there exists $\mathscr{S} \subset \mathscr{T}$ finite and $\bracsn{U_{\mathcal{S}}}_{\mathcal{S} \in \mathscr{S}}$ such that $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}} \subset U$. In which case, since $B(E, \mathcal{T}) = B(E, \mathcal{S})$ for all $\mathcal{S} \in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well. 
+    (1): Since $\mathcal{T}_B \supset \mathcal{T}$, $\mathfrak{B}(E, \mathcal{T}) \supset \mathfrak{B}(E, \mathcal{T}_B)$. Let $B \in (E, \mathcal{T})$ and $U \in \cn_{\mathcal{T}_B}(0)$, then there exists $\mathscr{S} \subset \mathscr{T}$ finite and $\bracsn{U_{\mathcal{S}}}_{\mathcal{S} \in \mathscr{S}}$ such that $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}} \subset U$. In which case, since $\mathfrak{B}(E, \mathcal{T}) = \mathfrak{B}(E, \mathcal{S})$ for all $\mathcal{S} \in \mathscr{S}$, $\bigcap_{\mathcal{S} \in \mathscr{S}}U_{\mathcal{S}}$ absorbs $B$, so $U$ absorbs $B$ as well. 
     
     (U): By (U) of the \hyperref[projective topology]{definition:tvs-initial}.
 
     (3): Let $\rho: E \to [0, \infty)$ be a seminorm that is bounded on bounded sets, then the topology induced by $\rho$ satisfies (1). By (U), $\rho$ is continuous with respect to $\mathcal{T}_B$, so $(E, \mathcal{T}_B)$ is a bornological space. 
 
-    (4): Let $\mathcal{S}$ be the inductive topology on $E$ induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. Let $B \in B(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, and $\rho: E \to [0, \infty)$ be the gauge of $U$. Since $\iota_{\ol B} \in L(E_{\ol B}; E, \mathcal{S})$, there exists $\lambda > 0$ such that $\rho \le \lambda \iota_{\ol B} \circ \rho_{\ol B}$. Thus $2\lambda U \supset B$ and $(E, \mathcal{S})$ satisfies (1). By (U), $\mathcal{T}_B \supset \mathcal{S}$. 
+    (4): Let $\mathcal{S}$ be the inductive topology on $E$ induced by $\bracsn{\iota_B}_{B \in \mathcal{B}}$. Let $B \in \mathfrak{B}(E, \mathcal{T})$, $U \in \cn_{\mathcal{S}}(0)$, and $\rho: E \to [0, \infty)$ be the gauge of $U$. Since $\iota_{\ol B} \in L(E_{\ol B}; E, \mathcal{S})$, there exists $\lambda > 0$ such that $\rho \le \lambda \iota_{\ol B} \circ \rho_{\ol B}$. Thus $2\lambda U \supset B$ and $(E, \mathcal{S})$ satisfies (1). By (U), $\mathcal{T}_B \supset \mathcal{S}$. 
 
-    On the other hand, since $B(E, \mathcal{T}_B) = B(E, \mathcal{T})$, for each $B \in \mathcal{B}$, $\iota_B: E_B \to (E, \mathcal{T}_B)$ is continuous. Therefore by (U) of the \hyperref[inductive topology]{definition:lc-inductive}, $\mathcal{S} \supset \mathcal{T}_B$. 
+    On the other hand, since $\mathfrak{B}(E, \mathcal{T}_B) = \mathfrak{B}(E, \mathcal{T})$, for each $B \in \mathcal{B}$, $\iota_B: E_B \to (E, \mathcal{T}_B)$ is continuous. Therefore by (U) of the \hyperref[inductive topology]{definition:lc-inductive}, $\mathcal{S} \supset \mathcal{T}_B$. 
 \end{proof}
 

src/fa/notation.tex

@@ -8,7 +8,7 @@
     $E_A$ & Normed space associated with $A \subset E$. & \autoref{definition:lc-associated-normed-space} \\
     $L(E; F)$ & Continuous linear maps $E \to F$. & \autoref{definition:continuous-linear} \\
     $L^n(E_1,\ldots,E_n; F)$ & Continuous $n$-linear maps $\prod E_j \to F$. & \autoref{definition:continuous-multilinear} \\
-    $B(E)$ & Bounded subsets of TVS $E$. & \autoref{definition:bounded} \\
+    $\mathfrak{B}(E)$ & Bounded subsets of TVS $E$. & \autoref{definition:bounded} \\
     $B(T; E)$ & Bounded functions $T \to E$ with uniform topology. & \autoref{definition:bounded-function-space} \\
     $B_\mathfrak{S}^k(E; F)$, $B(E; F)$ & $\mathfrak{S}$-bounded $k$-linear maps; bounded linear maps. & \autoref{definition:bounded-linear-map-space} \\
     $E^*$ & Topological dual of TVS $E$. & \autoref{definition:topological-dual} \\

src/fa/tvs/bounded.tex

@@ -9,7 +9,7 @@
         \item For every $\seq{x_n} \subset B$ and $\seq{\lambda_n} \subset K$ such that $\lambda_n \to 0$, $\lambda_n x_n \to 0$ as $n \to \infty$.
     \end{enumerate}
 
-    If the above holds, then $B$ is \textbf{bounded}. The collection $B(E) = B(E, \topo)$ is the set of all bounded sets of $E$.
+    If the above holds, then $B$ is \textbf{bounded}. The collection $\mathfrak{B}(E) = \mathfrak{B}(E, \topo)$ is the set of all bounded sets of $E$.
 \end{definition}
 \begin{proof}
     (1) $\Rightarrow$ (2): Let $U \in \cn_E(0)$ be circled, then there exists $k \in \natp$ such that $kU \supset B$. Since $\lambda_n \to 0$ as $n \to \infty$, there exists $N \in \natp$ such that $|\lambda_n| \le 1/k$ for all $n \ge N$. In which case, $\lambda_n x_n \in \lambda_n B \subset U$ for all $n \ge N$. 
@@ -19,7 +19,7 @@
 
 \begin{proposition}[{{\cite[I.5.1]{SchaeferWolff}}}]
 \label{proposition:bounded-operations}
-    Let $E$ be a TVS over $K \in \RC$ and $A, B \in B(E)$, then the following sets are bounded:
+    Let $E$ be a TVS over $K \in \RC$ and $A, B \in \mathfrak{B}(E)$, then the following sets are bounded:
     \begin{enumerate}
         \item Any $C \subset B$.
         \item The closure $\ol{B}$.

src/fa/tvs/index.tex

@@ -11,4 +11,5 @@
 \input{./complete-metric.tex}
 \input{./projective.tex}
 \input{./inductive.tex}
-\input{./spaces-of-linear.tex}
+\input{./vector-function.tex}
+\input{./space-of-linear.tex}

src/fa/tvs/space-of-linear.tex

@@ -1,228 +1,156 @@
-\section{Vector-Valued Function Spaces}
+\section{Spaces of Linear Maps}
-\label{section:spaces-linear-map}
+\label{section:space-linear-map-new}
-
+
-\begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
+\begin{definition}[Space of Bounded Linear Maps]
-\label{proposition:tvs-set-uniformity}
+\label{definition:bounded-linear-map-space}
-    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $F$ be a TVS over $K \in \RC$, then
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in \mathfrak{B}(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
-    \begin{enumerate}
+
-        \item The \hyperref[$\sigma$-uniformity]{definition:set-uniform} on $F^T$ is translation invariant.
+    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
-        \item The composition defined by
+\end{definition}
-        \[
+
-        T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
+\begin{proposition}[{{\cite[III.3.3]{SchaeferWolff}}}]
-        \]
+\label{proposition:bounded-linear-map-space-bounded}
-
+    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $A \subset B_\sigma(E; F)$, then the following are equivalent:
-        is continuous.
+    \begin{enumerate}
-    \end{enumerate}
+        \item $A \subset B_\sigma(E; F)$ is bounded with respect to the $\sigma$-uniform topology. 
-
+        \item For each $V \in \cn_F(0)$, $\bigcap_{T \in A}T^{-1}(V)$ absorbs every $S \in \sigma$.
-    For any vector subspace $\cf \subset F^T$, the following are equivalent:
+        \item For every $S \in \sigma$, $\bigcup_{T \in A}T(A)$ is bounded in $F$. 
-    \begin{enumerate}
+    \end{enumerate}
-        \item[(T)] The $\sigma$-uniform topology on $\cf \subset F^E$ is a vector space topology.
+    
-        \item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded.
+\end{proposition}
-    \end{enumerate}
+% Proof omitted because it is obvious. 
-\end{proposition}
+
-\begin{proof}
+
-    (1): Let $U \subset F \times F$ be an entourage and $S \in \sigma$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
+\begin{proposition}
-
+\label{proposition:multilinear-identify}
-    (2): Let $f, g, f', g' \in \cf$, $S \in \sigma$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
+	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
-
+	\begin{enumerate}
-    (T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
+        \item The map
-
+        \[
-    (T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
+        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
-    \begin{align*}
+        \]
-    \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
+
-    &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
+        defined by
-    \end{align*}
+        \[
-    Let $U_0 \in \cn_F(0)$. By (TVS1) and \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then
+        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
-    \[
+        \]
-    \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
+
-    \]
+        is an isomorphism.
-
+        \item The map
-    for all $x \in S$.
+        \[
-\end{proof}
+        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
-
+        \]
-
+
-\begin{definition}[Space of Bounded Functions]
+        defined by
-\label{definition:bounded-function-space}
+        \[
-    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
+        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
-    \begin{enumerate}
+        \]
-        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+
-        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
+        is an isomorphism.
-    \end{enumerate}
+    \end{enumerate}
-\end{definition}
+
-\begin{proof}
+    which allows the identification
-    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
+    \[
-    
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
-    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
+    \]
-
+
-    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
+    under the map $I$ in (2).
-    \[
+\end{proposition}
-    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
+\begin{proof}
-    \]
+    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
-    so $f \in B(T; E)$.
+    \[
-
+    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
-    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
+    \]
-\end{proof}
+
-
+    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
-
+    \[
-\begin{definition}[Space of Bounded Continuous Functions]
+    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in \mathfrak{B}(F)
-\label{definition:bounded-continuous-function-space}
+    \]
-    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
+
-    \begin{enumerate}
+    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
-        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+
-        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
+    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in \mathfrak{B}(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
-    \end{enumerate}
+
-\end{definition}
+    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
-\begin{proof}
+
-    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
+    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
-
+
-    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
+    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
-    
+    \[
-    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
+    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
-\end{proof}
+    \]
-
+
-
+    Thus (2) holds for all $k \in \natp$.
-
+\end{proof}
-\begin{definition}[Space of Bounded Linear Maps]
+
-\label{definition:bounded-linear-map-space}
+\begin{definition}[Strong Operator Topology]
-    Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be an ideal, and $k \in \nat$. The space $B_{\sigma}^k(E; F)$ is the set of all $k$-linear maps $T: E^k \to F$ with $T(S^k) \in B(F)$ for all $S \in \sigma$, equipped with the $\bracsn{S^k| S \in \sigma}$-uniform topology.
+\label{definition:strong-operator-topology}
-
+    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.
-    Let $\fB \subset 2^E$ be the collection of all bounded subsets of $E$, then $B_{\sigma}(E; F) = B(E; F)$ is the \textbf{space of bounded linear maps} from $E$ to $F$. 
+
-\end{definition}
+    The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
-
+\end{definition}
-\begin{proposition}
+
-\label{proposition:multilinear-identify}
+
-	Let $E, F$ be TVSs over $K \in \RC$, $\sigma \subset 2^E$ be a covering ideal, and $k \in \natp$, then
+\begin{proposition}
-	\begin{enumerate}
+\label{proposition:strong-operator-dense}
-        \item The map
+    Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If
-        \[
+    \begin{enumerate}
-        I: B_{\sigma}^k(E; B_{\sigma}(E; F)) \to B^{k+1}_{\sigma}(E; F)
+        \item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$.
-        \]
+        \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. 
-
+    \end{enumerate}
-        defined by
+
-        \[
+    
-        (IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
+    then $T_\alpha \to T$ in $L_s(E; F)$.
-        \]
+\end{proposition}
-
+\begin{proof}
-        is an isomorphism.
+    Let $x \in E$, $U \in \cn_F(Tx)$, and $V \in \cn_F(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_E(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_\alpha(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_0 \in A$ such that for all $\alpha \ge \alpha_0$, $T_\alpha y - Ty \in V$. In which case, for any $\alpha \ge \alpha_0$,
-        \item The map
+    \[
-        \[
+    T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U
-        I: \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} \to B^k_{\sigma}(E; F)
+    \]
-        \]
+\end{proof}
-
+
-        defined by
+
-        \[
+
-        IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
+\begin{definition}[Weak Operator Topology]
-        \]
+\label{definition:weak-operator-topology}
-
+    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F_w^E$ is the \textbf{weak operator topology}.
-        is an isomorphism.
+
-    \end{enumerate}
+    The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
-
+\end{definition}
-    which allows the identification
+
-    \[
+\begin{definition}[Bounded Convergence Topology]
-    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k \text{ times}} = B^k_{\sigma}(E; F)
+\label{definition:bounded-convergence-topology}
-    \]
+    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}.
-
+
-    under the map $I$ in (2).
+    The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
-\end{proposition}
+\end{definition}
-\begin{proof}
+
-    (1): To see that $I$ is surjective, let $T \in B_{\sigma}^{k+1}(E; F)$ and
+\begin{proposition}
-    \[
+\label{proposition:operator-space-completeness}
-    I^{-1}T: E \to B_{\sigma}(E; F) \quad x \mapsto T(x, \cdot)
+    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
-    \]
+    \begin{enumerate}
-
+        \item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
-    Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \sigma$. Since $\sigma$ is a covering ideal, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
+        \item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
-    \[
+    \end{enumerate}
-    T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
+\end{proposition}
-    \]
+\begin{proof}
-
+    (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
-    by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\sigma}(E; F)$.
+    \[
-
+    \phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
-    In addition, for any $S_1 \in \sigma$ and entourage $E(S_2, U)$ of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$, there exists $S \in \sigma$ with $S \supset S_1 \cup S_2$. Given that $T(S^{k+1}) \in B(F)$, there exists $\lambda > 0$ such that $T(S^{k+1}) \subset \lambda U(0)$. In which case, $I^{-1}T(S^k) \subset \lambda E(S, U)(0)$ and $I^{-1}T(S^k) \in B(B_{\sigma}(E; F))$. Thus $I^{-1}T \in B^k_{\sigma}(E; B_{\sigma}(E; F))$.
+    \]
-
+
-    It remains to show that $I$ and $I^{-1}$ is continuous. To this end, let $S \in \sigma$ and $U$ be an entourage of $F$, then for any $T \in E(S^k, E(S, U))(0)$, $IT \in E(S^{k+1}, U)(0)$, so $I$ is continuous.
+    and
-
+    \[
-    On the other hand, let $S_1 \in \sigma$ and $E(S_2, U)$ be an entourage of $B_{\sigma}(E; F)$ where $S_2 \in \sigma$ and $U$ is an entourage of $F$. Let $S \in \sigma$ with $S \supset S_1 \cup S_2$, then for any $T \in E(S^{k+1}, U)(0)$, $I^{-1}T \in E(S^{k}, E(S, U))(0)$. Thus $I^{-1}$ is continuous as well.
+    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - \lambda Tx
-
+    \]
-    (2): The case for $k = 2$ is given by (1). If the proposition holds for $k \in \natp$, then
+
-    \[
+    are continuous with respect to the product topology. Since
-    \underbrace{B_{\sigma}(E; B_{\sigma}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\sigma}(E; B_{\sigma}(E; F)) = B^{k+1}_{\sigma}(E; F)
+    \[
-    \]
+    \hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
-
+    \]
-    Thus (2) holds for all $k \in \natp$.
+
-\end{proof}
+    and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$. 
-
+
-
+    (2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
-
+
-\begin{definition}[Strong Operator Topology]
+\end{proof}
-\label{definition:strong-operator-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F^E$ is the \textbf{strong operator topology}.
-
-    The space $L_s(E; F)$ denotes $L(E; F)$ equipped with the strong operator topology.
-\end{definition}
-
-
-\begin{proposition}
-\label{proposition:strong-operator-dense}
-    Let $E, F$ be TVSs over $K \in \RC$ and $\net{T} \subset L(E; F)$ and $T \in L_s(E; F)$. If
-    \begin{enumerate}
-        \item[(a)] There exists a dense subset $S \subset E$ such that $T_\alpha x \to Tx$ strongly for all $x \in S$.
-        \item[(b)] $\bracs{T_\alpha|\alpha \in A}$ is uniformly equicontinuous. 
-    \end{enumerate}
-
-    
-    then $T_\alpha \to T$ in $L_s(E; F)$.
-\end{proposition}
-\begin{proof}
-    Let $x \in E$, $U \in \cn_F(Tx)$, and $V \in \cn_F(Tx)$ be balanced such that $V + V + V \subset U$. By (b), there exists a balanced neighbourhood $W \in \cn_E(0)$ such that $T(W) \cup \bigcup_{\alpha \in A}T_\alpha(W) \subset V$. By (a), there exists $y \in S \cap (x + W)$ and $\alpha_0 \in A$ such that for all $\alpha \ge \alpha_0$, $T_\alpha y - Ty \in V$. In which case, for any $\alpha \ge \alpha_0$,
-    \[
-    T_\alpha x - Tx = \underbrace{T_\alpha x - T_\alpha y}_{\in V} + \underbrace{T_\alpha y - Ty}_{\in V} + \underbrace{Ty - Tx}_{\in V} \in U
-    \]
-\end{proof}
-
-
-
-\begin{definition}[Weak Operator Topology]
-\label{definition:weak-operator-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fF \subset 2^E$ be the collection of finite subsets of $E$, then the $\fF$-uniform topology on $F_w^E$ is the \textbf{weak operator topology}.
-
-    The space $L_w(E; F) = L_s(E; F_w)$ denotes $L(E; F)$ equipped with the weak operator topology.
-\end{definition}
-
-\begin{definition}[Bounded Convergence Topology]
-\label{definition:bounded-convergence-topology}
-    Let $E, F$ be TVSs over $K \in \RC$, $\fB \subset 2^E$ be the collection of bounded subsets of $E$, then the $\fB$-uniform topology on $L(E; F)$ is the \textbf{topology of bounded convergence}.
-
-    The space $L_b(E; F)$ denotes $L(E; F)$ equipped with the topology of bounded convergence.
-\end{definition}
-
-\begin{proposition}
-\label{proposition:operator-space-completeness}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being separated, then:
-    \begin{enumerate}
-        \item $\hom(E; F)$ is a closed subspace of $F^E$ with respect to the product topology.
-        \item $B(E; F)$ is a closed subspace of $F^E$ with respect to the topology of bounded convergence. In particular, if $F$ is complete, then so is $B(E; F)$.
-    \end{enumerate}
-\end{proposition}
-\begin{proof}
-    (1): For each $x, y \in E$ and $\lambda \in K$, the mappings
-    \[
-    \phi_{x, y}: F^E \to F \quad T \mapsto Tx + Ty - T(x + y)
-    \]
-
-    and
-    \[
-    \psi_{x, \lambda}: F^E \to F \quad T \mapsto T(\lambda x) - \lambda Tx
-    \]
-
-    are continuous with respect to the product topology. Since
-    \[
-    \hom(E; F) = \bigcap_{x, y \in E}\bracsn{\phi_{x, y} = 0} \cap \bigcap_{\substack{x \in E \\ \lambda \in K}}\bracsn{\psi_{x, \lambda} = 0}
-    \]
-
-    and $\bracs{0}$ is closed in $F$, $\hom(E; F)$ is a closed subspace of $F^E$. 
-
-    (2): By \autoref{definition:bounded-function-space} and (1), the space of bounded functions and the space of linear functions from $E$ to $F$ are closed subspaces of $F^E$ with respect to the topology of bounded convergence. Therefore $B(E; F)$ is also a closed subspace.
-
-\end{proof}
-
-

src/fa/tvs/vector-function.tex

@@ -0,0 +1,85 @@
+\section{Vector-Valued Function Spaces}
+\label{section:spaces-linear-map}
+
+\begin{proposition}[{{\cite[III.3.1]{SchaeferWolff}}}]
+\label{proposition:tvs-set-uniformity}
+    Let $T$ be a set, $\sigma \subset 2^T$ be an ideal, and $F$ be a TVS over $K \in \RC$, then
+    \begin{enumerate}
+        \item The \hyperref[$\sigma$-uniformity]{definition:set-uniform} on $F^T$ is translation invariant.
+        \item The composition defined by
+        \[
+        T^F \times T^F \to T^F \quad (f + g)(x) = f(x) + g(x)
+        \]
+
+        is continuous.
+    \end{enumerate}
+
+    For any vector subspace $\cf \subset F^T$, the following are equivalent:
+    \begin{enumerate}
+        \item[(T)] The $\sigma$-uniform topology on $\cf \subset F^E$ is a vector space topology.
+        \item[(B)] For each $f \in \cf$ and $S \in \sigma$, $f(S) \subset E$ is bounded.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (1): Let $U \subset F \times F$ be an entourage and $S \in \sigma$. Using \autoref{proposition:tvs-uniform}, assume without loss of generality that $U$ is translation-invariant. For any $(f, g) \in E(S, U)$, $h \in F^T$, and $x \in S$, $(f(x) + h(x), g(x) + h(x)) \in U$. Thus $(f + h, g + h) \in E(S, U)$ and $E(S, U)$ is translation-invariant.
+
+    (2): Let $f, g, f', g' \in \cf$, $S \in \sigma$, and $U \subset F \times F$ be an entourage. By (TVS1), there exists an entourage $V \subset F \times F$ such that for any $x, x', y, y' \in F$ with $(x, x'), (y, y') \in V$, $(x + y, x' + y') \in U$. If $(f, g), (f', g') \in E(S, V) \cap \cf$, then for any $x \in S$, $(f(x) + g(x), f'(x) + g'(x)) \in U$, so $(f + g, f' + g') \in E(S, U) \cap \cf$.
+
+    (T) $\Rightarrow$ (B): Let $f \in \cf$, $S \in \sigma$ and $U \subset F \times F$ be a symmetric entourage, then $E(S, U)(0)$ is a neighbourhood of $0$ with respect to the $\sigma$-uniform topology. By (TVS2), there exists $\lambda > 0$ such that $f \in \lambda E(S, U)(0)$. In which case, for any $x \in S$, $\lambda^{-1}f(x) \in U(0)$ and $f(x) \in \lambda U(0)$. Thus $f(S) \subset \lambda U(0)$, and $f(S)$ is bounded.
+
+    (T) $\Rightarrow$ (B): Let $f, g \in \cf$, $\lambda, \lambda' \in K$, and $S \in \sigma$, then for any $x \in X$,
+    \begin{align*}
+    \lambda f(x) - \lambda' g(x) &= \lambda f(x) - \lambda' f(x) + \lambda' f(x) - \lambda' g(x) \\
+    &= (\lambda - \lambda')f(x) + \lambda' (f(x) - g(x))
+    \end{align*}
+    Let $U_0 \in \cn_F(0)$. By (TVS1) and \autoref{proposition:tvs-good-neighbourhood-base}, there exists $U \in \cn_F(0)$ circled such that $U + U \subset U_0$. By (TVS2), there exists $V \in \cn_F(0)$ such that $\lambda' V \subset U$. Since $f(S)$ is bounded, there exists $\eps > 0$ with $\eps f(S) \subset U$. In which case, if $\abs{\lambda - \lambda'} < \eps$ and $f(x) - g(x) \in V$ for all $x \in S$, then
+    \[
+    \lambda f(x) - \lambda' g(x) \in (\lambda - \lambda')f(S) + \lambda' V \subset U + U \subset U_0
+    \]
+
+    for all $x \in S$.
+\end{proof}
+
+
+\begin{definition}[Space of Bounded Functions]
+\label{definition:bounded-function-space}
+    Let $T$ be a set, $E$ be a TVS over $K \in \RC$, and $f: T \to E$ be a function, then $f$ is \textbf{bounded} if $f(T)$ is a bounded subset of $E$. The set $B(T; E) \subset E^T$ is the space of all bounded functions from $T$ to $E$, and:
+    \begin{enumerate}
+        \item $B(T; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $B(T; E)$ is a closed subset of $E^T$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $B(T; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): For any $f, g \in B(T; E)$ and $\lambda \in K$, $(\lambda f + g)(T) \subset (\lambda f)(T) + g(T)$, which is bounded by \autoref{proposition:bounded-operations}, so $B(T; E)$ is closed under addition and scalar multiplication. 
+    
+    Since $f(E)$ is bounded for all $f \in B(T; E)$, $B(T; E)$ forms a TVS over $K$ by \autoref{proposition:tvs-set-uniformity}.
+
+    (2): Let $f \in \ol{B(T; E)}$ and $U \in \cn_E(0)$ be circled, then there exists $g \in B(T; E)$ such that $(f - g)(E) \subset U$. Since $g \in B(T; E)$, there exists $\lambda \ge 0$ such that $g(E) \subset \lambda U$. In which case,
+    \[
+    f(E) \subset g(E) + (f - g)(E) \subset (\lambda + 1)U
+    \]
+    so $f \in B(T; E)$.
+
+    If $E$ is complete, then $E^T$ with the uniform topology is complete by \autoref{proposition:set-uniform-complete}. Thus $B(T; E)$ is also complete by \autoref{proposition:complete-closed}.
+\end{proof}
+
+
+\begin{definition}[Space of Bounded Continuous Functions]
+\label{definition:bounded-continuous-function-space}
+    Let $X$ be a topological space and $E$ be a TVS over $K \in \RC$, then $BC(X; E) = B(X; E) \cap C(X; E)$ is the space of bounded and continuous functions from $X$ to $E$, and
+    \begin{enumerate}
+        \item $BC(X; E)$ equipped with the uniform topology and pointwise operations is a TVS over $K \in \RC$. 
+        \item $BC(X; E)$ is a closed subspace of $C(X; E)$ and $B(X; E)$ with respect to the uniform topology. In particular, if $E$ is complete, then so is $BC(X; E)$.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (1): Since addition and scalar multiplication are continuous, $BC(X; E)$ is a subspace of $B(X; E)$, and hence a TVS over $K \in \RC$ by \autoref{definition:bounded-function-space}.
+
+    (2): Since $B(X; E)$ and $C(X; E)$ are both closed subspaces of $E^X$ by \autoref{definition:bounded-function-space}, $BC(X; E)$ is a closed subspace. 
+    
+    If $E$ is complete, then $B(X; E)$ and $C(X; E)$ are both complete under the uniform topology by \autoref{definition:bounded-function-space}. Therefore $BC(X; E)$ is also complete. 
+\end{proof}
+
+
+
+