Added the Banach-Steinhaus theorem.

src/fa/tvs/equicontinuous.tex

@@ -0,0 +1,62 @@
+\section{Equicontinuous Families of Linear Maps}
+\label{section:equicontinuous-linear}
+
+\begin{proposition}[{{\cite[IV.4.2]{SchaeferWolff}}}]
+\label{proposition:equicontinuous-linear}
+    Let $E, F$ be TVSs over $K \in RC$ and $\alg \subset \hom(E; F)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $\alg$ is uniformly equicontinuous. 
+        \item $\alg$ is equicontinuous. 
+        \item $\alg$ is equicontinuous at $0$. 
+        \item For each $V \in \cn_F(0)$, there exists $U \in \cn^o(E)$ such that $\bigcup_{T \in \alg}T(U) \subset V$. 
+        \item For each $V \in \cn_F(0)$, $\bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. 
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    (5) $\Rightarrow$ (1): Let $V \in \cn_F(0)$, then $U = \bigcap_{T \in \alg}T^{-1}(V) \in \cn_E(0)$. Thus for any $x, y \in E$ with $x - y \in U$, $Tx - Ty \in V$ for all $T \in \alg$. 
+\end{proof}
+
+\begin{proposition}[{{\cite[IV.4.3]{SchaeferWolff}}}]
+\label{proposition:equicontinuous-linear-closure}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$ be equicontinuous, and $\alg'$ be the closure of $\alg$ in $F^E$ with respect to the product topology, then $\alg'$ is equicontinuous and hence $\alg' \subset L(E; F)$. 
+\end{proposition}
+\begin{proof}
+     By \autoref{proposition:operator-space-completeness}, $\alg' \subset \hom(E; F)$. By \autoref{theorem:arzela-ascoli}, $\alg'$ is equicontinuous.
+\end{proof}
+
+\begin{theorem}[Banach-Steinhaus]
+\label{theorem:banach-steinhaus}
+    Let $E, F$ be TVSs over $K \in \RC$ and $\alg \subset L(E; F)$. Suppose that one of the following holds:
+    \begin{enumerate}
+        \item[(B)] $E$ is a Baire space. 
+        \item[(B')] $E$ and $F$ are locally convex with $E$ being barreled.
+    \end{enumerate}
+
+    and that
+    \begin{enumerate}
+        \item[(E2)] For each $x \in E$, $\alg(x) = \bracs{Tx|T \in \alg}$ is bounded in $F$. 
+    \end{enumerate}
+
+    then
+    \begin{enumerate}
+        \item[(E1)] $\alg$ is equicontinuous.
+        \item[(C1)] The product uniformity and the compact uniformity on $\cf$ coincide. 
+        \item[(C2)] The closure of $\alg$ in $F^E$ is with respect to the product topology is an equicontinuous subset of $L(E; F)$. 
+    \end{enumerate}
+\end{theorem}
+\begin{proof}[Proof, {{\cite[IV.4.2]{SchaeferWolff}}}. ]
+    (B) + (E2) $\Rightarrow$ (E1): Let $V \in \cn_F(0)$ be closed and circled, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is circled and closed. By (E2), $U$ is absorbing, so $E = \bigcup_{n \in \natp}nU$. Since $E$ is Baire, there exists $n \in \natp$, $W \in \cn_E(0)$, and $x \in E$ such that $x + W \subset nU$. As $U$ is circled,
+    \[
+    W \subset nU - nU = nU + nU = 2nU
+    \]
+
+    so $U \in \cn_E(0)$, and $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. 
+
+    (B') + (E2) $\Rightarrow$ (E1): Let $V \in \cn_F(0)$ be convex, circled, and closed, then $U = \bigcap_{T \in \alg}T^{-1}(V)$ is convex, circled, and closed. By (E2), $U$ is absorbing, and hence a barrel in $E$. By (B'), $U \in \cn_E(0)$, $\alg$ is equicontinuous by \autoref{proposition:equicontinuous-linear}. 
+
+    (E1) $\Rightarrow$ (C1) + (C2): By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli} and \autoref{proposition:equicontinuous-linear-closure}.
+\end{proof}
+
+
+
+

src/fa/tvs/index.tex

@@ -13,3 +13,4 @@
 \input{./inductive.tex}
 \input{./vector-function.tex}
 \input{./space-of-linear.tex}
+\input{./equicontinuous.tex}

src/topology/uniform/equicontinuous.tex

@@ -13,7 +13,6 @@
     Let $(X, \fU)$ and $(Y, \fV)$ be uniform spaces, and $\cf \subset UC(X; Y)$, then $\cf$ is \textbf{uniformly equicontinuous} if for every $V \in \fV$, there exists $U \in \fU$ such that $(f \times f)(V) \subset \fU$ for all $f \in \cf$. 
 \end{definition}
 
-
 \begin{theorem}[Arzelà-Ascoli]
 \label{theorem:arzela-ascoli}
     Let $X$ be a topological space, $(Y, \fU)$ be a uniform space, and $\cf \subset C(X; Y)$. If
@@ -24,6 +23,7 @@
     then
     \begin{enumerate}[label=(C\arabic*)]
         \item The product uniformity and the compact uniformity on $\cf$ coincide. 
+        \item The closure of $\cf$ in $Y^X$ with respect to the product topology is equicontinuous. 
     \end{enumerate}
 
     In addition, if $\cf$ satisfies (E1) and
@@ -32,11 +32,11 @@
     \end{enumerate}
 
     then
-    \begin{enumerate}[label=(C\arabic*), start=1]
-        \item $\cf$ is a precompact subset of $Y^X$ with respect to the compact uniformity. 
+    \begin{enumerate}[label=(C\arabic*), start=3]
+        \item $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. 
     \end{enumerate}
 
-    Conversely, if $X$ is a LCH space, then (C2) implies (E1) + (E2).
+    Conversely, if $X$ is a LCH space, then (C3) implies (E1) + (E2).
 \end{theorem}
 \begin{proof}
     (E1) $\Rightarrow$ (C1): By \autoref{proposition:compact-uniform-open}, the compact-open topology coincides with the compact-uniform topology on $C(X; Y)$ and thus $\cf$.
@@ -53,32 +53,18 @@
 
     so the product uniformity and the compact uniformity coincide. 
 
-    (E1) + (E2) $\Rightarrow$ (C2): By \autoref{proposition:totally-bounded-product}, $\prod_{x \in X}\cf(x)$ is totally bounded. Since $\cf \subset \prod_{x \in X}\cf(x)$, $\cf$ is also totally bounded. As the pointwise and compact uniformities coincide, $\cf$ is totally bounded with respect to the compact uniformity. By \autoref{proposition:product-complete} and \autoref{proposition:compact-uniform}, $\prod_{x \in X}\ol{\cf(x)}$ is compact and hence complete. Therefore the closure of $\cf$ with respect to the compact uniformity is compact. 
+    (E1) $\Rightarrow$ (C2): Let $\cf'$ be the closure of $\cf$ in $Y^X$ with respect to the product topology. Let $x \in X$ and entourage $U$ of $Y$. Using \autoref{proposition:goodentourages}, assume without loss of generality that $U$ is closed. Since $\cf$ is equicontinuous, there exists $V \in \cn_X(x)$ such that $(f(x), f(y)) \in U$ for all $f \in \cf$ and $y \in V$. For any element $g \in \cf'$, $(g(x), g(y)) \in \ol U = U$ for all $y \in V$. Therefore $\cf'$ is also equicontinuous. 
 
-    (C2) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$,  
+    (E1) + (E2) $\Rightarrow$ (C3): Using (C2), assume without loss of generality that $\cf$ is closed in $Y^X$ with respect to the product topology. In which case, $\cf$ is a closed subset of $\prod_{x \in X}\ol{\cf(x)}$ with respect to the product topology. By \hyperref[Tychonoff's Theorem]{theorem:tychonoff} and \autoref{proposition:compact-extensions}, $\cf$ is compact in the product topology. By (C1), $\cf$ is also compact in the compact uniform topology.
+
+    (C3) $\Rightarrow$ (E1): Assume that $X$ is a LCH space. Let $x \in X$ and $U \in \fU$ be symmetric, then there exists a compact neighbourhood $V \in \cn_X(x)$. Since $\cf$ is totally bounded, there exists $\seqf{f_j} \subset \cf$ such that for each $g \in \cf$, there exists $1 \le j \le n$ such that $(f_j \times g)(V) \subset U$. For each $1 \le j \le n$, $f_j \in C(X; Y)$, so there exists $V_j \in \cn_X(x)$ with $V_j \subset V$ such that for any $y \in V_j$, $(f_j(x), f_j(y)) \in U$. Let $W = \bigcap_{j = 1}^n V_j$, then for any $g \in \cf$ with $(f_j \times g)(V) \subset U$ and $y \in W$,  
     \[
     (g(x), f_j(x)), (f_j(x), f_j(y)), (f_j(y), g(y)) \in U \circ U \circ U
     \]
 
     Therefore $g(W) \subset (U \circ U \circ U)(g(x))$, and $\cf$ is equicontinuous. 
 
-    (C2) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. 
+    (C3) $\Rightarrow$ (E2): Since the evaluation map is uniformly continuous with respect to the compact uniformity, $\cf(x)$ is totally bounded for all $x \in X$ by \autoref{proposition:totally-bounded-image}. 
 \end{proof}
 
 
-\begin{corollary}
-\label{corollary:arzela-locally-compact}
-    Let $X$ be a LCH space, $Y$ be a uniform space, and $\cf \subset C(X; Y)$ such that:
-    \begin{enumerate}[label=(E\arabic*)]
-        \item $\cf$ is equicontinuous.
-        \item For each $x \in X$, $\cf(x) = \bracs{f(x)|f \in \cf}$ is precompact in $Y$.
-    \end{enumerate}
-
-    then $\cf$ is a precompact subset of $C(X; Y)$ with respect to the compact uniformity. 
-\end{corollary}
-\begin{proof}
-    By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, $\cf$ is a precompact subset of $Y^X$. By \autoref{proposition:lch-compactly-generated}, $C(X; Y)$ is a closed subset of $Y^X$. Therefore $\cf$ is a precompact subset of $C(X; Y)$. 
-\end{proof}
-
-
-

src/topology/uniform/uc.tex

@@ -51,7 +51,7 @@
     \end{enumerate}
 
 
-    known as the \textbf{initial uniformity} on $X$ generated by $\seqi{f}$.
+    The collection $\fU$ is the \textbf{initial uniformity} on $X$ generated by $\seqi{f}$.
 \end{definition}
 \begin{proof}
     (3): Since the diagonal is mapped to the diagonal and $\fB$ is closed under intersections, it is sufficient to verify (UB3) for $\fB$. Let $J \subset I$ be finite and $\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j) \in \fB$, then there exists $\bracs{V_j}_{j \in J}$ such that $V_j \circ V_j \subset U_j$ for each $j \in J$. In which case, for any $(x, y), (y, z) \in f_j^{-1}(V_j)$, $(f(x), f(y)), (f(y), f(z)) \in V_j$ and $(f(x), f(z)) \in U_j$. Thus $(f_j \times f_j)^{-1}(V_j) \circ (f_j \times f_j)^{-1}(V_j) \subset (f_j \times f_j)^{-1}(U_j)$, and
@@ -65,7 +65,7 @@
 
     (U): For any $i \in I$, $\mathfrak{V} \supset (f_i \times f_i)^{-1}(\fU_i)$. By (F2), $\mathfrak{V} \supset \fB$, so $\mathfrak{V} \supset \fU$.
 
-    (4): Let $J \subset I$ finite and $\seqj{U_j}$ such that $U_j \in \fU_j$ for each $j \in J$, then
+    (4): Let $J \subset I$ finite and $\seqj{U}$ such that $U_j \in \fU_j$ for each $j \in J$, then
     \[
     (f \times f)^{-1}\paren{\bigcap_{j \in J}(f_j \times f_j)^{-1}(U_j)} = \bigcap_{j \in J}[(f_j \circ f) \times (f_j \circ f)]^{-1}(U_j)
     \]