Symmetry of the derivative (normed/frechet).

src/dg/derivative/derivative.tex

@@ -43,173 +43,3 @@
     (\lambda f + g)(x_0+h) - (\lambda f + g)(x_0) = \underbrace{[\lambda D_{\mathcal{HR}}f(x_0) + D_{\mathcal{HR}}g(x_0)]}_{\in \ch}h + \underbrace{(\lambda r + s)}_{\in \calr}(h)
     \]
 \end{proof}
-
-
-
-
-\begin{definition}[$o(t)$]
-\label{definition:little-o}
-    Let $U \in \cn(0) \subset \real$ and $r: U \to \real$, then $r \in o(t)$ if
-    \[
-    \lim_{t \to 0}\frac{o(t)}{t} = 0
-    \]
-\end{definition}
-
-\begin{definition}[Tangent to $0$]
-\label{definition:tangent-to-0}
-    Let $E, F$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, then $\varphi$ is \textbf{tangent to $0$} if for any $W \in \cn_F(0)$, there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
-    \[
-    \varphi(tV) \subset r(t)W
-    \]
-    for sufficiently small $t \in \real$.
-\end{definition}
-
-\begin{definition}[Linear Order at $0$]
-\label{definition:linear-order-at-0}
-    Let $E, F$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, then $\varphi$ is of \textbf{linear order at $0$} if for any $W \in \cn_F(0)$, there exists $V \in \cn_E(0)$ such that
-    \[
-    \varphi(tV) \subset tW
-    \]
-    for sufficiently small $t \in \real$.
-\end{definition}
-
-\begin{lemma}
-\label{lemma:tangent-linear-at-0}
-    Let $E, F, G$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, $V \in \cn_F(0)$, $\varphi, \psi: U \to F$, and $\rho: V \to G$. If $\varphi, \psi, \rho$ are of linear order at $0$, then
-    \begin{enumerate}
-        \item $\rho \circ \varphi$ is of linear order at $0$.
-        \item $\varphi + \psi$ is of linear order at $0$.
-        \item If one of $\varphi, \rho$ is tangent to $0$, then $\rho \circ \varphi$ is tangent to $0$.
-    \end{enumerate}
-    If $\varphi, \psi$ are tangent to $0$, then
-    \begin{enumerate}
-        \item[(4)] $\varphi$ is of linear order at $0$.
-        \item[(5)] $\varphi + \psi$ is tangent at $0$.
-    \end{enumerate}
-    Finally, suppose that $\varphi$ is linear.
-    \begin{enumerate}
-        \item[(6)] If $\varphi$ is continuous, then $\varphi$ is of linear order at $0$.
-        \item[(7)] If $\varphi$ is tangent to $0$ and $E$ is Hausdorff, then $\varphi = 0$.
-    \end{enumerate}
-\end{lemma}
-\begin{proof}
-    (1): Let $W \in \cn_G(0)$, then there exists $V_0 \in \cn_F(0)$ with $\rho(tV_0) \subset tW$ and $U_0 \in \cn_E(0)$ with $\varphi(tU_0) \subset tV_0$ for sufficiently small $t$. In which case, $\rho \circ \varphi(tU_0) \subset tW$.
-
-    (2): Let $W \in \cn_F(0)$, then there exists $W_0 \in \cn_F(0)$ with $W_0 + W_0 \subset W$ and $V_0 \in \cn_E(0)$ such that $\varphi(tV_0), \psi(tV_0) \subset tW_0$ for sufficiently small $t$. In which case,
-    \[
-    \varphi(tV_0) + \psi(tV_0) \subset tW_0 + tW_0 \subset tW
-    \]
-
-    (3): Let $W \in \cn_G(0)$, then there exists $V_0 \in \cn_F(0)$, $U_0 \in \cn_E(0)$, and $r \in o(t)$ such that one of the following holds for sufficiently small $t$:
-    \begin{enumerate}
-        \item[(a)] $\varphi(tU_0) \subset tV_0$ and $\rho(tV_0) \subset o(t)W$.
-        \item[(b)] $\varphi(tU_0) \subset o(t)V_0$ and $\rho(tV_0) \subset tW$.
-    \end{enumerate}
-    In both cases, $\rho \circ \varphi(tU_0) \subset o(t)W$.
-
-    (4): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. By assumption, there exists $V_0 \in \cn_E(0)$ and $r \in o(t)$ such that
-    \[
-    \varphi(tV_0) \subset o(t)W = \frac{r(t)}{t} \cdot tW
-    \]
-    for sufficiently small $t$. Since $r \in o(t)$, $\abs{r(t)/t} \le 1$ for sufficiently small $t$. In which case, $\frac{r(t)}{t} \cdot tW \subset tW$.
-
-    (5): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Let $V_0 \in \cn_E(0)$ and $r, r' \in o(t)$ such that $\varphi(tV_0) \subset r(t)W$ and $\psi(tV_0) \subset r'(t)W$ for sufficiently small $t$, then
-    \[
-    \varphi(tV_0) + \psi(tV_0) \subset r(t)W + r'(t)W \subset [\abs{r(t)} + \abs{r'(t)}]W
-    \]
-
-    (6): Let $W \in \cn_F(0)$, then there exists $V_0 \in \cn_E(0)$ such that $\varphi(V_0) \subset W$. In which case, $\varphi(tV_0) \subset tW$ for all $t \in \real$.
-
-    (7): Let $x \in E$ and $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi$ is tangent to $0$ and linear, then there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
-    \[
-    \varphi(tV) \subset r(t)W  \quad \varphi(V) \subset \frac{r(t)}{t}W
-    \]
-    for sufficiently small $t \in \real$. Since $W \in \cn_F(0)$, there exists $\lambda \in K$ such that $x \in \lambda V$. Thus
-    \[
-    \varphi(x) \in \varphi(\lambda V) = \lambda\varphi(V) \subset \frac{\lambda r(t)}{t}W
-    \]
-    As $r \in o(t)$, there exists $t \in \real$ such that $\abs{\lambda r(t)/t} \le 1$, so $\varphi(x) \in \frac{\lambda r(t)}{t}W \subset W$. Since this holds for all $W \in \cn_F(0)$ and $F$ is Hausdorff, $\varphi(x) \in \ol{\bracs{0}} = \bracs{0}$ by \ref{proposition:tvs-closure} and \ref{lemma:t1}.
-\end{proof}
-
-
-
-\begin{lemma}
-\label{lemma:tangent-to-0}
-    Let $E, F, G$ be TVSs over $K \in \RC$, $U \in \cn_E(0)$, and $\varphi: U \to F$, $\psi: U \to F$ be tangent to $0$, then:
-    \begin{enumerate}
-        \item For any $\lambda \in L(F; G)$, $\lambda \circ \varphi$ is tangent to $0$.
-        \item $\varphi + \psi$ is tangent to $0$.
-        \item If $\varphi$ is linear and $F$ is Hausdorff, then $\varphi = 0$.
-    \end{enumerate}
-\end{lemma}
-\begin{proof}
-    (1): Let $W \in \cn_G(0)$, then $\lambda^{-1}(W) \in \cn_F(0)$, and there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
-    \[
-    \lambda \circ \varphi(tV) \subset \lambda (r(t)\lambda^{-1}W) = r(t) \lambda \circ \lambda^{-1} (W) \subset r(t)W
-    \]
-
-    (2): Let $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi, \psi$ are tangent to $0$, there exists $V_1, V_2 \in \cn_E(0)$ and $r_1, r_2 \in o(t)$ such that
-    \[
-    \varphi(tV_1) \subset r_1(t)W \quad \psi(tV_2) \subset r_2(t)W
-    \]
-    In which case, since $W$ is circled,
-    \[
-    \lambda \varphi(t(V_1 \cap V_2)) + \psi(t(V_1 \cap V_2)) \subset r_1(t) W + r_2(t)W \subset (r_1(t) + r_2(t))W
-    \]
-    and $r_1 + r_2 \in o(t)$.
-
-    (3): Let $x \in E$ and $W \in \cn_F(0)$. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $W$ is circled. Since $\varphi$ is tangent to $0$ and linear, then there exists $V \in \cn_E(0)$ and $r \in o(t)$ such that
-    \[
-    \varphi(tV) \subset r(t)W  \quad \varphi(V) \subset \frac{r(t)}{t}W
-    \]
-    for sufficiently small $t \in \real$. Since $W \in \cn_F(0)$, there exists $\lambda \in K$ such that $x \in \lambda V$. Thus
-    \[
-    \varphi(x) \in \varphi(\lambda V) = \lambda\varphi(V) \subset \frac{\lambda r(t)}{t}W
-    \]
-    As $r \in o(t)$, there exists $t \in \real$ such that $\abs{\lambda r(t)/t} \le 1$, so $\varphi(x) \in \frac{\lambda r(t)}{t}W \subset W$. Since this holds for all $W \in \cn_F(0)$ and $F$ is Hausdorff, $\varphi(x) \in \ol{\bracs{0}} = \bracs{0}$ by \ref{proposition:tvs-closure} and \ref{lemma:t1}.
-\end{proof}
-
-\begin{definition}[(Strong Fréchet) Derivative]
-\label{definition:derivative}
-    Let $E, F$ be TVSs over $K \in \RC$ with $F$ being Hausdorff, $U \subset E$ open, $f: U \to F$, and $x_0 \in U$, then $f$ is \textbf{(strongly Fréchet) differentiable at} $x_0$ if there exists $\lambda \in L(E; F)$, $V \in \cn_E(0)$, and $\varphi: V \to F$ tangent to $0$ such that
-    \[
-    f(x_0 + h) = f(x_0) + \lambda h + \varphi(h)
-    \]
-    for all $h \in V$. In which case, $Df(x_0) = \lambda$ is the unique continuous linear map satisfying the above, and the \textbf{(strong Fréchet) derivative} of $f$ \textbf{at} $x_0$.
-
-    If $f$ is differentiable at every $x_0 \in U$, then $f$ is \textbf{(strongly Fréchet) differentiable}, and $Df: U \to L(E; F)$ is the \textbf{(strong Fréchet) derivative} of $f$.
-\end{definition}
-\begin{proof}
-    Let $\lambda, \mu \in L(E; F)$ and $\varphi, \psi: V \to F$ be tangent to $0$ such that
-    \begin{align*}
-        f(x_0 + h) &= f(x_0) + \lambda h + \varphi(h) \\
-        &= f(x_0) + \mu h + \varphi(h)
-    \end{align*}
-    then
-    \[
-    0 = (\lambda - \mu)h + (\varphi - \psi)(h)
-    \]
-    By (7) of \ref{lemma:tangent-to-0}, $(\lambda - \mu)$ is tangent to $0$ and thus equal to $0$.
-\end{proof}
-
-\begin{proposition}[Chain Rule]
-\label{proposition:chain-rule}
-    Let $E, F, G$ be TVSs over $K \in \RC$ with $F, G$ Hausdorff, $U \subset E$ and $V \subset F$ be open, $f: U \to V$ be differentiable at $x_0 \in U$, and $g: V \to G$ be differentiable at $f(x_0)$, then
-    \[
-    D(g \circ f)(x_0) = Dg(f(x_0)) \circ f(x_0)
-    \]
-\end{proposition}
-\begin{proof}
-    By differentiability of $f$ and $g$, there exists $\varphi, \psi$ tangent to $0$ such that
-    \begin{align*}
-        (g \circ f)(x_0 + h) &= (g \circ f)(x_0) + Dg(f(x_0))[f(x_0 + h) - f(x_0)] + \varphi(f(x_0 + h) - f(x_0)) \\
-        &= (g \circ f)(x_0) + Dg(f(x_0))[Df(x_0)(h) + \psi(h)] + \varphi(Df(x_0)(h) + \psi(h)) \\
-        &= (g \circ f)(x_0) + [Dg(f(x_0)) \circ Df(x_0)](h) \\
-        &+ [Dg(f(x_0)) \circ \psi + \varphi \circ (Df(x_0) + \psi)](h)
-    \end{align*}
-    Since $Dg(f(x_0)) \in L(F; G)$, $Dg(f(x_0)) \circ \psi$ is tangent to $0$ by (6) and (3) \ref{lemma:tangent-linear-at-0}. Similarly, $Df(x_0) + \psi$ is of linear order at $0$ by (4) and (2) of \ref{lemma:tangent-linear-at-0}, so $\varphi \circ (Df(x_0) + \psi)$ is tangent to $0$ by (3) of \ref{lemma:tangent-linear-at-0}. Thus
-    \[
-    Dg(f(x_0)) \circ \psi + \varphi \circ (Df(x_0) + \psi)
-    \]
-    is tangent to $0$ by (5) of \ref{lemma:tangent-linear-at-0}.
-\end{proof}

src/dg/derivative/higher.tex

@@ -0,0 +1,63 @@
+\section{Higher Derivatives}
+\label{section:higher-derivatives}
+
+\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 an upward-directed family that contains all finite sets, $\mathcal{H} \subset B_\sigma(E; F)$ 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}
+        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
+        \item The derivative $D_\sigma^{n-1}f: U \to B^{n-1}_\sigma(E; F)$ is derivative at $x_0$.
+    \end{enumerate}
+    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) \in L(E; B^{n-1}_\sigma(E; F))$ is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
+
+    The mapping $f: U \to F$ is \textbf{$n$-fold $\sigma$-differentiable on $U$} if it is $n$-fold $\sigma$-differentiable at every point in $U$. Under the identification $B_\sigma(E; B^{n-1}_\sigma(E; F)) = B_\sigma^{n}(E; F)$ given by \ref{proposition:multilinear-identify},
+    \[
+    D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
+    \]
+    is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
+\end{definition}
+
+\begin{proposition}[{{\cite[Theorem 5.1.1]{Cartan}}}]
+\label{proposition:derivative-symmetric}
+    Let $E, F$ be Banach spaces, $U \subset E$ be open, $n \in \natp$, and $f: U \to F$ be a function $n$-times Fréchet-differentiable at $x_0 \in U$, then $D^nf(x_0) \in L^2(E; F)$ is symmetric.
+\end{proposition}
+\begin{proof}
+    First suppose that $n = 2$. Let $r > 0$ such that $B(x_0, 2r) \subset U$, and define $A: B_E(0, r) \times B_E(0, r) \to F$ by
+    \[
+    A(h, k) = f(x + h + k) - f(x + h) - f(x + k) + f(x)
+    \]
+    then there exists $r_1 \in \mathcal{R}_{B(E)}$ such that
+    \begin{align*}
+        A(h, k) &= Df(x + h)(k) + Df(x)(k) \\
+        &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
+        &- [f(x + k) - f(x) - Df(x)(k)] \\
+        &= D^2f(x)(h, k) + r_1(h) \cdot Df(x)(k)
+        &+ [f(x + h + k) - f(x + h) - Df(x + h)(k)] \\
+        &- [f(x + k) - f(x) - Df(x)(k)] \\
+    \end{align*}
+    Let $B_h: B_E(0, r) \to F$ be defined by
+    \[
+    B_h(k) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k)
+    \]
+    then
+    \[
+    B_h(k) - B_h(0) = f(x + h + k) - f(x + k) - Df(x + h)(k) + Df(x)(k) -f(x + h) + f(x)
+    \]
+    Now, there exists $r_2, r_3 \in \mathcal{R}_{B(E)}$ 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) - Df(x) - D^2f(x)(k) + r_2(k) + r_3(h) \\
+    &=r_2(k) + r_3(h)
+    \end{align*}
+    By the mean value theorem,
+    \[
+    \norm{B_h(k) - B_h(0)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
+    \]
+    As the above argument is symmetric,
+    \[
+    \norm{D^2f(x)(h, k) - D^2f(x)(k, h)}_F \le \norm{k}_E \cdot o(\norm{k}_E + \norm{h}_E)
+    \]
+    so $D^2f(x)(h, k) - D^2f(x)(k, h) = 0$.
+\end{proof}

src/dg/derivative/index.tex

@@ -4,3 +4,4 @@
 \input{./src/dg/derivative/derivative.tex}
 \input{./src/dg/derivative/sets.tex}
 \input{./src/dg/derivative/mvt.tex}
+\input{./src/dg/derivative/higher.tex}

src/dg/derivative/sets.tex

@@ -1,4 +1,4 @@
-\section{Differentiation With Respect to Set Systems}
+\section{Differentiation With Respect to Sets}
 \label{section:differentiation-set-system}
 
 \begin{definition}[Small]
@@ -119,23 +119,6 @@
     A method of extending this sense of differentiability is to require that \textit{every} extension of the function to some open set, or to the entire space is differentiable. Given that this paves way to confusion for related definitions of differentiability, this definition is not formally included here.
 \end{remark}
 
-\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 an upward-directed family that contains all finite sets, $\mathcal{H} \subset B_\sigma(E; F)$ 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}
-        \item There exists $V \in \cn_E(x_0)$ such that $f$ is $(n-1)$-fold differentiable on $V$.
-        \item The derivative $D_\sigma^{n-1}f: U \to B^{n-1}_\sigma(E; F)$ is derivative at $x_0$.
-    \end{enumerate}
-    In which case, $D_\sigma(D_\sigma^{n-1}f)(x_0) \in L(E; B^{n-1}_\sigma(E; F))$ is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
-
-    The mapping $f: U \to F$ is \textbf{$n$-fold $\sigma$-differentiable on $U$} if it is $n$-fold $\sigma$-differentiable at every point in $U$. Under the identification $B_\sigma(E; B^{n-1}_\sigma(E; F)) = B_\sigma^{n}(E; F)$ given by \ref{proposition:multilinear-identify},
-    \[
-    D_\sigma^{n}f: U \to B^{n-1}_\sigma(E; F)
-    \]
-    is the \textbf{$n$-fold $\sigma$-derivative of $f$ at $x_0$}.
-\end{definition}
 
 
 \begin{proposition}

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}

src/topology/main/baire.tex

@@ -6,13 +6,38 @@
 \label{definition:baire}
     Let $X$ be a topological space, then the following are equivalent:
     \begin{enumerate}
-        \item For any $\seq{A_n}$ nowhere dense, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
+        \item For any $\seq{A_n} \subset 2^X$ nowhere dense, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
-        \item For any $\seq{A_n}$ closed with empty interior, $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
+        \item For any $\seq{A_n} \subset 2^X$ closed with empty interior, $\bigcup_{n \in \nat^+}A_n$ has empty interior.
-        \item For any $\seq{A_n}$ closed with $\bigcup_{n \in \nat^+}A_n = X$, there exists $N \in \nat^+$ such that $\bigcup_{n \le N}A_n$ has non-empty interior.
+        \item For any $\seq{U_n} \subset 2^X$ open and dense, $\bigcap_{n \in \nat^+}U_n$ is dense.
-        \item For any $\seq{U_n}$ open and dense, $\bigcap_{n \in \nat^+}U_n$ is dense.
     \end{enumerate}
     If the above holds, then $X$ is a \textbf{Baire space}.
 \end{definition}
+\begin{proof}
+    $(1) \Rightarrow (2)$: Let $\seq{A_n} \subset 2^X$ be closed with empty interior, then $\seq{A_n}$ are nowhere dense. Hence $\bigcup_{n \in \nat^+}A_n \subsetneq X$.
+
+    $(2) \Rightarrow (3)$: For each $n \in \natp$, let $A_n = U_n^c$, then $A_n$ is closed. For any $\emptyset U \subset A_n$ open, $U \cap U_n \ne \emptyset$ by density of $U_n$, so $A_n$ has empty interior.
+
+    Suppose that $\bigcap_{n \in \natp}U_n$ is not dense, then there exists $\emptyset \ne V \subset X$ open such that $\bigcup_{n \in \natp}A_n \supset V$, which contradicts the fact that $\bigcup_{n \in \natp}A_n$ has non-empty interior.
+
+    $(3) \Rightarrow (1)$: For each $n \in \natp$, let $U_n = A_n^c$, then $U_n$ is open and dense. Since $\bigcap_{n \in \natp}U_n$ is dense, $\bigcup_{n \in \natp}A_n$ has non-empty interior.
+\end{proof}
+
+\begin{lemma}
+\label{lemma:baire-condition}
+    Let $X$ be a topological space. If for any $U \subset X$ open and $\seq{U_n} \subset 2^X$ open and dense, there exists $\seq{V_j} \subset 2^X$ open such that:
+    \begin{enumerate}
+        \item[(a)] For all $n > 1$, $\ol V_n \subset U_n \cap V_{n - 1} \subset U$.
+        \item[(b)] $\bigcap_{j \in \natp} \ol V_j$ is non-empty.
+    \end{enumerate}
+    then $X$ is a Baire space.
+\end{lemma}
+\begin{proof}
+    By assumption (a),
+    \[
+    U \cap \bigcap_{n \in \natp}\ol V_n \subset U \cap \bigcap_{n \in \natp}U_n
+    \]
+    Since $\bigcap_{n \in \natp}\ol V_n \ne \emptyset$ by assumption (b), $U \cap \bigcap_{n \in \natp}U_n \ne \emptyset$, so $\bigcap_{n \in \natp}U_n$ is dense.
+\end{proof}
 
 \begin{theorem}[Baire Category Theorem]
 \label{theorem:baire}
@@ -22,3 +47,15 @@
         \item $X$ is locally compact.
     \end{enumerate}
 \end{theorem}
+\begin{proof}
+    Let $U \subset X$ be open and $\seq{U_n} \subset 2^X$ be open and dense. Let $V_0 = U$. For each $n \in \natp$, by density of $U_n$, there exists $x \in U_n \cap V_{n - 1}$. Since $X$ is regular (\ref{definition:uniform-separated}/\ref{proposition:compact-hausdorff-normal}), there exists $V_{n} \in \cn^o(x)$ such that $x \in V_{n} \subset \ol U_{n} \subset U_n \cap V_{n-1}$. If $X$ is locally compact, choose $V_n$ to be precompact. If $X$ is completely metrisable, choose $V_n$ such that $\text{diam}(V_n) \le 1/n$.
+
+    Now, if $X$ is locally compact, then by the finite intersection property, $\bigcap_{n \in \natp}\ol{V_n} \ne \emptyset$. If $X$ is completely metrisable, then $\seq{V_n}$ is a Cauchy filter base, and converges to at least one point, so $\bigcap_{n \in \natp}\ol{V_n} \ne \emptyset$.
+
+    By \ref{lemma:baire-condition}, $X$ is a Baire space.
+\end{proof}
+
+\begin{definition}[Meagre]
+\label{definition:meagre}
+    Let $X$ be a topological space, then $X$ is \textbf{meagre} if there exists $\seq{A_n} \subset 2^X$ nowhere dense such that $X = \bigcup_{n \in \natp}A_n$.
+\end{definition}