Symmetry of the derivative (normed/frechet).

src/fa/lc/index.tex

@@ -8,3 +8,4 @@
 \input{./src/fa/lc/projective.tex}
 \input{./src/fa/lc/inductive.tex}
 \input{./src/fa/lc/hahn-banach.tex}
+\input{./src/fa/lc/spaces-of-linear.tex}

src/fa/lc/spaces-of-linear.tex

@@ -0,0 +1,18 @@
+\section{Locally Convex Spaces of Linear Maps}
+\label{section:lc-spaces-linear-map}
+
+\begin{proposition}
+\label{proposition:lc-spaces-linear-map}
+    Let $T$ be a set, $E$ be a locally convex space defined by the seminorms $\seqi{[\cdot]}$, and $\mathfrak{S} \subset 2^T$ be an upward-directed family. For each $i \in I$ and $S \in \mathfrak{S}$, let
+    \[
+    [\cdot]_{S, i}: E^T \to [0, \infty) \quad f \mapsto \sup_{x \in S}[f(x)]_{S, i}
+    \]
+    then the $\mathfrak{S}$-uniform topology on $E^T$ is defined by the seminorms
+    \[
+    \bracs{[\cdot]_{S, i}|S \in \mathfrak{S}, i \in I}
+    \]
+    and hence locally convex.
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:set-uniform-pseudometric}.
+\end{proof}

src/fa/norm/index.tex

@@ -2,3 +2,5 @@
 \label{chap:normed-spaces}
 
 \input{./src/fa/norm/normed.tex}
+\input{./src/fa/norm/linear.tex}
+\input{./src/fa/norm/multilinear.tex}

src/fa/norm/linear.tex

@@ -0,0 +1,13 @@
+\section{Linear Maps}
+\label{section:normed-linear-maps}
+
+\begin{proposition}
+\label{proposition:normed-linear-map-space}
+    Let $E, F$ be normed vector spaces, then the topology on $L_b(E; F)$ is induced by the \textbf{operator norm}
+    \[
+    \norm{\cdot}_{L(E; F)}: L(E; F) \to [0, \infty) \quad T \mapsto \sup_{\substack{x \in E \\ \norm{x}_E = 1}}Tx
+    \]
+\end{proposition}
+\begin{proof}
+    By \ref{proposition:lc-spaces-linear-map}.
+\end{proof}

src/fa/norm/multilinear.tex

@@ -0,0 +1,20 @@
+\section{Multilinear Maps}
+\label{section:normed-multilinear}
+
+\begin{proposition}
+\label{proposition:bilinear-separate}
+    Let $E, F, G$ be normed spaces and $T: E \times F \to G$ be a bilinear map. If:
+    \begin{enumerate}
+        \item For each $x \in E$, $y \mapsto T(x, y)$ is a continuous linear map from $F$ to $G$.
+        \item For each $y \in F$, $x \mapsto T(x, y)$ is a continuous linear map from $E$ to $G$.
+        \item $E$ is a Banach space.
+    \end{enumerate}
+    then $T \in L^2(E, F; G)$.
+\end{proposition}
+\begin{proof}
+    For each $y \in F$, let $T_y \in L(E; G)$ be defined by $x \mapsto T(x, y)$. Let $x \in X$, then
+    \[
+    \sup_{y \in B_F(0, 1)}\norm{T_yx}_G = \sup_{y \in B_F(0, 1)}\norm{T(x, y)}_G < \infty
+    \]
+    by continuity of $y \mapsto T(x, y)$. By the Uniform Boundedness Principle (\ref{theorem:uniform-boundedness}), $M = \sup_{y \in B_F(0, 1)}\norm{T_y}_{L(E; G)} < \infty$. Thus for any $x \in E$ and $y \in F$, $\norm{T(x, y)}_G \le M\norm{x}_E\norm{y}_F$.
+\end{proof}

src/fa/norm/normed.tex

@@ -61,3 +61,22 @@
     \]
     so $\sum_{n = 1}^\infty Tx_n = y$.
 \end{proof}
+
+\begin{theorem}[Uniform Boundedness Principle]
+\label{theorem:uniform-boundedness}
+    Let $E, F$ be normed spaces and $\mathcal{T} \subset L(E; F)$. If
+    \begin{enumerate}
+        \item For every $x \in E$, $\sup_{T \in \mathcal{T}}\norm{Tx}_F < \infty$.
+        \item $E$ is a Banach space.
+    \end{enumerate}
+    then $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} < \infty$.
+\end{theorem}
+\begin{proof}
+    For each $n \in \natp$, let $A_n = \bracs{x \in X|\norm{Tx}_F \le n \forall T \in \mathcal{T}}$, then each $A_n$ is closed with $\bigcup_{n \in \natp}A_n = E$. By the Baire Category Theorem (\ref{theorem:baire}), there exists $n \in \natp$ and $U \subset E$ open such that $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
+
+    Let $x \in U$ and $r > 0$ such that $\overline{B(x, r)} \subset U$, then for any $y \in E$ with $\norm{y}_E \le r$ and $T \in \mathcal{T}$,
+    \[
+    \norm{Ty} = \norm{Ty + Tx - Tx}_E = \normn{T\underbrace{(x + y)}_{\in U}}_E + \norm{Tx}_E \le 2n
+    \]
+    so $\sup_{T \in \mathcal{T}}\norm{T}_{L(E; F)} \le 2n/r$.
+\end{proof}

src/fa/tvs/complete-metric.tex

@@ -0,0 +1,122 @@
+\section{Complete Metric TVSs}
+\label{section:tvs-complete-metric}
+
+\begin{proposition}
+\label{proposition:tvs-complete-metric}
+    Let $E$ be a metric TVS with its topology induced by the pseudonorm $\rho: E \to [0, \infty)$, then the following are equivalent:
+    \begin{enumerate}
+        \item $E$ is complete.
+        \item For any $\seq{x_n} \subset X$ with $\sum_{n \in \natp}\rho(x_n) < \infty$, $\limv{N}\sum_{n = 1}^N x_n$ exists in $E$.
+    \end{enumerate}
+\end{proposition}
+\begin{proof}
+    $(2) \Rightarrow (1)$: Let $\seq{x_n} \subset E$ be a Cauchy sequence, then there exists a subsequence $\seq{n_k} \subset \natp$ such that for each $k \in \natp$, $\rho(x_{n_{k+1}} - x_{n_{k}}) < 2^{-k}$.
+
+    Let $x = x_{n_1} + \limv{N}\sum_{k = 1}^N (x_{n_{k+1}} - x_{n_k})$, then $x = \limv{k}x_{n_k} \in E$. Since $\seq{x_n}$ is a Cauchy sequence that admits a convergent subsequence, it is convergent.
+\end{proof}
+
+\begin{theorem}[Successive Approximations {{\cite[Section III.2]{SchaeferWolff}}}]
+\label{theorem:successive-approximations}
+    Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorm $\rho$ and $\eta$, respectively. Let $T \in L(E; F)$, $r > 0$, $\gamma \in (0, 1)$, and $C \ge 0$. Suppose that for every $y \in B_F(0, r)$, there exists $x \in E$ such that:
+    \begin{enumerate}
+        \item[(a)] $\eta(y - Tx) \le \gamma \eta(y)$.
+        \item[(b)] $\rho(x) \le C \eta(y)$.
+    \end{enumerate}
+    then for any $y \in F$, there exists $\seq{x_n} \subset E$ such that:
+    \begin{enumerate}
+        \item $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)/(1 - \gamma)$.
+        \item $y = \limv{N}\sum_{n = 1}^N Tx_n$.
+    \end{enumerate}
+    In particular,
+    \[
+    T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r)
+    \]
+\end{theorem}
+\begin{proof}
+    Let $y_0 = y$ and $x_0 = 0$. Let $N \in \natz$ and suppose inductively that $\seqf[N]{x_n} \subset E$ has been constructed such that:
+    \begin{enumerate}
+        \item[(I)] $\sum_{n = 1}^N\rho(x_n) \le C\eta(y)\sum_{n = 0}^{N-1}\gamma^{n}$.
+        \item[(II)] $\eta\paren{y - \sum_{n = 1}^N Tx_n} \le \eta(y)\gamma^N$.
+    \end{enumerate}
+    By assumption, there exists $x_{N+1} \in E$ such that:
+    \begin{enumerate}
+        \item[(i)] $\eta\paren{y - \sum_{n = 1}^{N+1} Tx_n} \le \gamma \eta\paren{y - \sum_{n = 1}^N Tx_n} \le \gamma^{N+1}$.
+        \item[(ii)] $\rho(x_{N+1}) \le C\eta\paren{y - \sum_{n = 1}^N Tx_n} \le C\eta(y)\gamma^N$.
+    \end{enumerate}
+    Combining (I) and (ii) shows that $\sum_{n = 1}^N \rho(x_n) \le C \eta(y) \sum_{n = 0}^N \gamma^n$. Therefore there exists $\seq{x_n} \subset E$ such that (I) and (II) holds for all $N \in \natp$.
+
+    By (I), $\sum_{n \in \natp}\rho(x_n) \le C\eta(y)\sum_{n \in \natz}\gamma^n = C \eta(y)/(1 - \gamma)$. By (II), $\limv{N}\eta\paren{y - \limv{N}\sum_{n = 1}^N Tx_n} = \limv{N}\eta(y)\gamma^N = 0$.
+\end{proof}
+
+\begin{proposition}
+\label{proposition:successive-approximation-all}
+    Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorms $\rho$ and $\eta$ respectively, and $T \in L(E; F)$. If
+    \begin{enumerate}
+        \item[(a)] For any $r > 0$, there exists $\delta(r) > 0$ such that $\overline{T(B_E(0, r))} \supset B_F(0, \delta(r))$.
+        \item[(b)] $E$ is complete.
+    \end{enumerate}
+    then for every $s > r$, $T(B_E(0, s)) \supset B_F(0, \delta(r))$.
+\end{proposition}
+\begin{proof}
+    Let $s > r$ and $\seq{s_n}, \seq{\delta_n} \subset (0, \infty)$ such that
+    \begin{enumerate}
+        \item[(i)] $s = \sum_{n \in \natp}s_n$.
+        \item[(ii)] $s_1 = r$.
+        \item[(iii)] For all $n \in \natp$, $\overline{T(B_E(0, s_n))} \supset B_F(0, \delta_n)$.
+        \item[(iv)] $\rho_1 = \rho$.
+    \end{enumerate}
+    Let $y_0 \in B(0, r)$ and $x_0 = 0$. Let $N \in \natp$ and suppose inductively that $\bracs{x_n}_1^N \subset E$ has been constructed such that:
+    \begin{enumerate}
+        \item[(I)] For each $0 \le n \le N - 1$, $\rho(x_{n+1} - x_n) < s_n$.
+        \item[(II)] For each $0 \le n \le N$, $\eta(Tx_n - y) \le \rho_{n+1}$.
+    \end{enumerate}
+    By density of $T(x_N + B_E(0, s_N))$ in $Tx_N + B_F(0, \rho_N)$, there exists $x_{N+1} \in T(x_N + B_E(0, s_N))$ such that $\eta(Tx_{N+1} - y) \le \rho_{N+2}$.
+
+    By (I), $\seq{x_N}$ is a Cauchy sequence, so
+    \[
+    x = \limv{N}x_N = \limv{N}\sum_{n = 1}^N(x_n - x_{n-1})
+    \]
+    exists in $E$. In addition, $\rho(x) \le \sum_{n \in \natp} \rho(x_n - x_{n-1}) < \sum_{n \in \natp}s_n = s$, so $x \in B_E(0, s)$. Finally, $\eta\paren{Tx - y} = \limv{N}\rho(Tx_N - y) = 0$ and $Tx = y$.
+\end{proof}
+
+
+\begin{proposition}
+\label{proposition:coercive-closed-range}
+    Let $E, F$ be metric TVSs over $K \in \RC$ with pseudonorms $\rho$ and $\eta$, respectively, and $T \in L(E; F)$. If
+    \begin{enumerate}
+        \item[(a)] For any $r > 0$, there exists $C \ge 0$ such that for any $y \in T(E)$, there exits $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y)$.
+        \item[(b)] $E$ is complete.
+    \end{enumerate}
+    then $T(E)$ is closed.
+\end{proposition}
+\begin{proof}
+    Let $r > 0$ and $\gamma \in (0, 1)$. For any $y_0 \in B_F(0, r) \cap \overline{T(E)}$, there exists $y \in B_F(0, r)$ such that $\eta(y) \le \eta(y_0)$ and $\eta(y - y_0) \le \gamma \eta(y_0)$. By assumption (a), there exists $x \in T^{-1}(y)$ with $\rho(x) \le C\eta(y) \le C\eta(y_0)$.
+
+    By the method of successive approximations (\ref{theorem:successive-approximations}),
+    \[
+    T(E) \supset T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r) \cap \overline{T(E)}
+    \]
+    As this holds for all $r > 0$, $T(E) \supset \overline{T(E)}$.
+\end{proof}
+
+\begin{theorem}[Open Mapping Theorem]
+\label{theorem:open-mapping}
+    Let $E, F$ be complete metric TVSs over $K \in \RC$, $T \in L(E; F)$ with $T(E)$ dense, then exactly one of the following holds:
+    \begin{enumerate}
+        \item $T(E)$ is meagre.
+        \item $T$ is open.
+    \end{enumerate}
+\end{theorem}
+\begin{proof}
+    Suppose that $T(E)$ is not meagre. Let $r_0 > 0$ and $r > 0$ such that $B_E(0, r) + B_E(0, r) \subset B_E(0, r_0)$, then since $B_E(0, r)$ is absorbing,
+    \[
+    E = \bigcup_{n \in \natp}nB_E(0, r) \quad \overline{T(E)} = \bigcup_{n \in \natp}\overline{nT(B_E(0, r))}
+    \]
+    By the Baire Category Theorem (\ref{theorem:baire}), there exists $N \in \natp$, $s > 0$, and $y \in nT(B_E(0, r))$ such that $B_F(y, s) \subset \overline{nT(B_E(0, r))}$. In which case,
+    \[
+    B_F(0, s) = B_F(y, s) - y \subset \overline{nT(B_E(0, r)) + nT(B_E(0, r))} \subset \overline{nT(B_E(0, r_0))}
+    \]
+    By (TVS2), there exists $t > 0$ such that $n^{-1}B_F(0, s) \supset B_F(0, t)$, so $\overline{T(B_E(0, r_0))} \supset B_F(0, t)$.
+
+    Thus by \ref{proposition:successive-approximation-all}, $B_F(0, t) \subset T(B_E(0, r)) \in \cn_F(0)$ for all $r > r_0$. As $r_0 > 0$ is arbitrary, $T(U) \in \cn_F(0)$ for all $U \in \cn_E(0)$. Therefore $T$ is open by translation-invariance of the topology on $E$.
+\end{proof}

src/fa/tvs/continuous.tex

@@ -34,43 +34,6 @@
     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.
-        \item[(5)] For any TVS $F$ over $K$ and linear map $T \in \hom(F; E)$, $T \in L(F; E)$ if and only if $T_i \circ T \in L(F; F_i)$ for all $i \in I$.
-    \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 $\abs{\lambda - \lambda'} < \min_{j \in J}\eps$, $(x + y, x' + y'), (\lambda x, \lambda' x') \in V$.
-
-    (5): By \ref{definition:continuous-linear} and (4) of \ref{definition:initial-uniformity}.
-\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

src/fa/tvs/index.tex

@@ -8,6 +8,7 @@
 \input{./src/fa/tvs/continuous.tex}
 \input{./src/fa/tvs/quotient.tex}
 \input{./src/fa/tvs/completion.tex}
+\input{./src/fa/tvs/complete-metric.tex}
 \input{./src/fa/tvs/projective.tex}
 \input{./src/fa/tvs/inductive.tex}
 \input{./src/fa/tvs/spaces-of-linear.tex}