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Bokuan Li
2026-03-06 14:06:15 -05:00
parent 173727665b
commit 5034bc4220
109 changed files with 1184 additions and 410 deletions
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10
src/fa/index.tex
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@@ -2,8 +2,8 @@
\label{part:fa}
\input{./src/fa/tvs/index.tex}
\input{./src/fa/lc/index.tex}
\input{./src/fa/norm/index.tex}
\input{./src/fa/rs/index.tex}
\input{./src/fa/lp/index.tex}
\input{./tvs/index.tex}
\input{./lc/index.tex}
\input{./norm/index.tex}
\input{./rs/index.tex}
\input{./lp/index.tex}

5
src/fa/lc/continuous.tex
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@@ -13,11 +13,11 @@
\end{enumerate}
\end{proposition}
\begin{proof}
$(1) \Leftrightarrow (2) \Leftrightarrow (3)$: By \ref{definition:continuous-linear}.
$(1) \Leftrightarrow (2) \Leftrightarrow (3)$: By \autoref{definition:continuous-linear}.
$(2) \Rightarrow (4)$: $x \mapsto [Tx]_F$ is a continuous seminorm on $E$.
$(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \ref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \ref{definition:continuous-linear}.
$(4) \Rightarrow (3)$: Let $U \in \cn_F(0)$ be convex, circled, and radial, then its gauge $[\cdot]_U$ is a continuous seminorm on $F$ by \autoref{definition:locally-convex}. Thus there exists a continuous seminorm $[\cdot]_E$ such that $[Tx]_U \le [x]_E$. In which case, $V = \bracs{x \in E| [x]_E < 1} \in \cn_E(0)$ with $T(V) \subset U$. Therefore $T$ is continuous at $0$, and continuous by \autoref{definition:continuous-linear}.
\end{proof}
\begin{proposition}
@@ -29,6 +29,7 @@
\[
[Tx]_F \le \prod_{j = 1}^n [x_j]_{E_j}
\]
\end{enumerate}
\end{proposition}
\begin{proof}

17
src/fa/lc/convex.tex
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@@ -13,6 +13,7 @@
\[
\bracs{tx + (1 - t)y|t \in (0, 1)} \subset A^o
\]
\end{lemma}
\begin{proof}
Fix $t \in (0, 1)$. Using translation, assume without loss of generality that $tx + (1 - t)y = 0$. In which case, $x = \alpha y$ where $\alpha = (1 - t)/t$. By (TVS2), $\alpha A^o \in \cn^o(y)$. Since $y \in \ol{A}$, $\alpha A^o \cap A \ne \emptyset$, so there exists $z \in A^o$ such that $\alpha z \in A$.
@@ -21,10 +22,12 @@
\[
\mu z + (1 - \mu)\alpha z = \frac{\alpha z}{\alpha - 1} + \frac{(\alpha - 1 - \alpha)\alpha z}{\alpha - 1} = 0
\]
By (TVS1) and (TVS2),
\[
U = \underbrace{\bracs{\mu w + (1 - \mu)\alpha z|w \in A^o}}_{\subset A} \in \cn^o(0)
\]
so $0 \in A^o$.
\end{proof}
@@ -39,12 +42,13 @@
\end{enumerate}
\end{lemma}
\begin{proof}
(1): By \ref{lemma:convex-interior}.
(1): By \autoref{lemma:convex-interior}.
(2): Let $x, y \in \ol{A}$. By \ref{definition:closure}, there exists filters $\fF, \mathfrak{G} \subset 2^A$ such that $\fF$ converges to $x$ and $\mathfrak{G}$ converges to $y$. In which case,
(2): Let $x, y \in \ol{A}$. By \autoref{definition:closure}, there exists filters $\fF, \mathfrak{G} \subset 2^A$ such that $\fF$ converges to $x$ and $\mathfrak{G}$ converges to $y$. In which case,
\[
\fU = \bracs{tE + (1 - t)F|t \in [0, 1], E \in \fF, F \in \mathfrak{G}} \subset 2^A
\]
converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
\end{proof}
@@ -57,7 +61,8 @@
\[
y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}
\]
by \ref{lemma:convex-interior}.
by \autoref{lemma:convex-interior}.
\end{proof}
@@ -97,7 +102,7 @@
\end{enumerate}
\end{lemma}
\begin{proof}
$(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]} \le [x - y] < \eps$. By \ref{proposition:tvs-uniform}, $[\cdot]$ is uniformly continuous.
$(3) \Rightarrow (1)$: Let $\eps > 0$, then there exists $V \in \cn(0)$ such that $[x] < \eps$ for all $x \in V$. In which case, for any $x, y \in E$ with $x - y \in V$, $\abs{[x] - [y]} \le [x - y] < \eps$. By \autoref{proposition:tvs-uniform}, $[\cdot]$ is uniformly continuous.
\end{proof}
@@ -122,6 +127,7 @@
\[
[\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
\]
is the \textbf{gauge/Minkowski functional} of $A$, and
\begin{enumerate}
\item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
@@ -139,6 +145,7 @@
\[
(\lambda + \mu)^{-1}(x + y) = t\lambda^{-1}x + (1 - t)\mu^{-1}y
\]
then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.
\end{proof}
@@ -153,7 +160,7 @@
If the above holds, then $E$ is a \textbf{locally convex} space.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By \ref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \bracs{tx + (1 - t)y|x, y \in V}$ be the convex hull of $V$, then $W \subset U$ is convex and circled.
$(1) \Rightarrow (2)$: Let $U \in \cn(0)$ be convex. By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$. Let $W = \bracs{tx + (1 - t)y|x, y \in V}$ be the convex hull of $V$, then $W \subset U$ is convex and circled.
$(2) \Rightarrow (3)$: For each $V \in \cn(0)$ convex, circled, and radial, let $[\cdot]_V: E \to [0, \infty)$ be its gauge, then $[\cdot]_V$ is a seminorm. For each $x, y \in X$ and $r > 0$, $[x - y]_V < r$ if and only if $x - y \in rV$. Thus the uniformity induced by $[\cdot]_V$ corresponds to the uniformity generated by $\bracs{U_{rV}| r > 0}$, where $U_V = \bracs{(x, y) \in E|x - y \in V}$. Since this holds for all $U \in \cn(0)$, the topology of $E$ and the topology induced by $\bracs{[\cdot]_V| V \in \cn(0), \text{ convex, circled, radial}}$ coincide.

28
src/fa/lc/hahn-banach.tex
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@@ -9,6 +9,7 @@
\[
\phi_{x_0, \lambda}: F + \real x_0 \quad (y + tx_0) \mapsto \phi(y) + \lambda t
\]
then there exists $\lambda \in \real$ such that $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_0$.
\end{lemma}
\begin{proof}
@@ -22,6 +23,7 @@
\[
\lambda \in \braks{\sup_{x \in F}[\phi(x) - \rho(x - x_0)], \inf_{x \in F}[\rho(x + x_0) - \phi(x)]}
\]
then for any $x \in F$ and $t > 0$,
\begin{align*}
\phi(x + tx_0) &= \phi(x) + t\lambda = t\braks{\phi(t^{-1}x) + \lambda} \\
@@ -42,26 +44,29 @@
\end{enumerate}
\end{theorem}
\begin{proof}
(1): Let $x_0 \in E \setminus F$, then by \ref{lemma:hahn-banach}, there exists $\lambda \in \real$ such that if
(1): Let $x_0 \in E \setminus F$, then by \autoref{lemma:hahn-banach}, there exists $\lambda \in \real$ such that if
\[
\phi_{x_0, \lambda}: F + \real x_0 \to \real \quad (x + tx_0) \mapsto x + \lambda t
\]
then $\phi_{x_0, \lambda}(x) \le \rho(x)$ for all $x \in F + \real x_0$. Thus for any $F \subsetneq E$, $\phi$ admits an extension to a subspace that strictly contains $F$.
Let
\[
\mathbf{F} = \bracs{\Phi \in \hom(F'; \real)|F' \supset F, \Phi|_F = \phi, \Phi \le \rho|_{F'}}
\]
For any $\Phi, \Phi' \in \mathbf{F}$, define $\Phi \le \Phi'$ if $\Phi'$ is an extension of $\Phi$. Let $\cf \subset \mathbf{F}$ be a chain. For any $\Phi \in \cf$, let $\cd(\Phi)$ be its domain. Since $\bigcup_{\phi \in \cf}\cd(\Phi)$ is a subspace and $\cf$ is a chain, by \ref{lemma:glue-linear}, there exists $\Phi \in \hom\paren{\bigcup_{\Phi' \in \cf}\cd(\Phi'); \real}$ such that $\Phi|_{\cd(\Phi')} = \Phi'$ for all $\Phi' \in \cf$. Thus $\Phi \in \mathbf{F}$ is an upper bound of $\cf$.
For any $\Phi, \Phi' \in \mathbf{F}$, define $\Phi \le \Phi'$ if $\Phi'$ is an extension of $\Phi$. Let $\cf \subset \mathbf{F}$ be a chain. For any $\Phi \in \cf$, let $\cd(\Phi)$ be its domain. Since $\bigcup_{\phi \in \cf}\cd(\Phi)$ is a subspace and $\cf$ is a chain, by \autoref{lemma:glue-linear}, there exists $\Phi \in \hom\paren{\bigcup_{\Phi' \in \cf}\cd(\Phi'); \real}$ such that $\Phi|_{\cd(\Phi')} = \Phi'$ for all $\Phi' \in \cf$. Thus $\Phi \in \mathbf{F}$ is an upper bound of $\cf$.
By Zorn's lemma, $\mathbf{F}$ admits a maximal element $\Phi$. If $\cd(\Phi) \subsetneq E$, then $\Phi$ is not maximal by the preceding discussion. Therefore $\cd(\Phi) = E$ and $\Phi$ is a desired extension.
(2): Given that $\rho$ is a seminorm, for any $u \in \hom(E; \real)$, $u \le \rho$ if and only if $\abs{u} \le \rho$.
Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \ref{proposition:polarisation-linear}. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U} \le \rho$ and $U|_F = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_F = \phi$ by \ref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
Let $u = \re{\phi}$, then $u \in \hom(E; \real)$ by \autoref{proposition:polarisation-linear}. By (1), there exists $U \in \hom(E; \real)$ such that $\abs{U} \le \rho$ and $U|_F = u$. For each $x \in E$, let $\Phi(x) = U(x) - iU(ix)$, then $\Phi \in \hom(E; \complex)$ and $\Phi|_F = \phi$ by \autoref{proposition:polarisation-linear}. In addition, for any $x \in E$, if $\alpha = \overline{\sgn(\Phi(x))}$, then
\[
|\Phi(x)| = \alpha\Phi(x) = \Phi(\alpha x) = U(\alpha x) \le \rho(\alpha x) = \rho(x)
\]
so $\abs{\Phi} \le \rho$.
\end{proof}
@@ -72,18 +77,20 @@
\begin{proof}
By translation, assume without loss of generality that $0 \in A$. In which case, $A \in \cn^o(0)$ is convex.
Let $[\cdot]_A: E \to [0, \infty)$ be the gauge (\ref{definition:gauge}) of $A$, then $[\cdot]_A$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,
Let $[\cdot]_A: E \to [0, \infty)$ be the \hyperref[gaugeg]{definition:gauge} of $A$, then $[\cdot]_A$ is a sublinear functional on $E$. For any $y, z \in E$ and $t > 0$ with $y, z \in tA$,
\[
\abs{[y]_A - [z]_A} \le [y - z]_A \le t
\]
Hence $[\cdot]_A$ is continuous on $E$.
Let $\phi_0: \real x \to \real$ be defined by $\lambda x \mapsto \lambda$. Since $x \not\in A$, $[x]_A > 1$. Hence $\phi_0|_{Kx} \le [\cdot]_A|_{Kx}$. By the Hahn-Banach Theorem (\ref{theorem:hahn-banach}), there exists $\phi \in \hom(E; \real)$ such that $\phi|_{\real x} = \phi_0$ and $\phi(y) \le [y]_A$ for all $y \in E$.
Let $\phi_0: \real x \to \real$ be defined by $\lambda x \mapsto \lambda$. Since $x \not\in A$, $[x]_A > 1$. Hence $\phi_0|_{Kx} \le [\cdot]_A|_{Kx}$. By the \hyperref[Hahn-Banach Theorem]{theorem:hahn-banach}, there exists $\phi \in \hom(E; \real)$ such that $\phi|_{\real x} = \phi_0$ and $\phi(y) \le [y]_A$ for all $y \in E$.
For any $y \in A \cap (-A)$,
\[
\dpb{y, \phi}{E} \le [y]_A < 1 \quad -\dpb{y, \phi}{E} \le [-y]_A < 1
\]
so $\phi \in E^*$.
Finally, let $y \in A$, then $\dpb{y, \phi}{E} \le [y]_A < 1 = \dpb{x, \phi}{E}$.
@@ -94,9 +101,9 @@
Let $E$ be a TVS over $\real$, $A, B \subset E$ be non-empty convex sets, such that $A \cap B = \emptyset$ and $A$ is open, then there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ with $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
\end{theorem}
\begin{proof}
Let $C = A - B$, then $C \subset E$ is an open convex set by \ref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$.
Let $C = A - B$, then $C \subset E$ is an open convex set by \autoref{lemma:convex-gymnastics}. Since $A \cap B = \emptyset$, $0 \not\in C$.
By \ref{lemma:hahn-banach-separation}, there exists $\phi \in E^*$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E} < 0$ implies that $\dpb{a, \phi}{E} < \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
By \autoref{lemma:hahn-banach-separation}, there exists $\phi \in E^*$ such that $C \subset \bracs{\phi < 0}$. In which case, for any $a \in A$ and $b \in B$, $\dpb{a - b, \phi}{E} < 0$ implies that $\dpb{a, \phi}{E} < \dpb{b, \phi}{E}$. Let $\alpha \in [\sup_{a \in A}\dpb{a, \phi}{E}, \inf_{b \in B}\dpb{b, \phi}{E}]$, then $A \subset \bracs{\phi \le \alpha}$ and $B \subset \bracs{\phi \ge \alpha}$.
\end{proof}
\begin{theorem}[Hahn-Banach, Second Geometric Form {{\cite[Theorem 1.7]{Brezis}}}]
@@ -104,9 +111,9 @@
Let $E$ be a locally convex space over $\real$, $A, B \subset E$ be non-empty convex sets such that $A \cap B = \emptyset$, $A$ is closed, and $B$ is compact, then there exists $\phi \in E^* \setminus \bracs{0}$, $\alpha \in \real$, and $r > 0$ with $A \subset \bracs{\phi \le \alpha - r}$ and $B \subset \bracs{\phi \ge \alpha + r}$.
\end{theorem}
\begin{proof}
Let $C = A - B$, then $C$ is closed by \ref{proposition:tvs-set-operations} with $0 \not\in C$. Since $E$ is locally convex, there exists $U \in \cn^o(0)$ convex such that $U \cap C \ne \emptyset$.
Let $C = A - B$, then $C$ is closed by \autoref{proposition:tvs-set-operations} with $0 \not\in C$. Since $E$ is locally convex, there exists $U \in \cn^o(0)$ convex such that $U \cap C \ne \emptyset$.
By \ref{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ such that $U \subset \bracs{f \le \alpha}$ and $C \subset \bracs{f \ge \alpha}$. In which case, for any $u \in U$, $a \in A$, and $b \in B$,
By \autoref{theorem:hahn-banach-geometric-1}, there exists $\phi \in E^* \setminus \bracs{0}$ and $\alpha \in \real$ such that $U \subset \bracs{f \le \alpha}$ and $C \subset \bracs{f \ge \alpha}$. In which case, for any $u \in U$, $a \in A$, and $b \in B$,
\begin{align*}
\dpb{u, \phi}{E} &\le \dpb{a, \phi}{E} - \dpb{b, \phi}{E} \\
\dpb{b, \phi}{E} + \dpb{u, \phi}{E} &\le \dpb{a, \phi}{E}
@@ -115,6 +122,7 @@
\[
\sup_{b \in B}\dpb{b, \phi}{E} + r \le \inf_{a \in A}\dpb{a, \phi}{E}
\]
\end{proof}
@@ -132,7 +140,7 @@
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \ref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the Hahn-Banach theorem (\ref{theorem:hahn-banach}), there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $\phi \le \rho_M \le \rho$.
(1): Let $\rho_M: E \to [0, \infty)$ be the quotient of $\rho$ by $M$, then $\rho_M \le \rho$ is a continuous seminorm on $E$ by \autoref{definition:quotient-norm}. Let $\phi_0: Kx \to K$ be defined by $\lambda x \mapsto \lambda \rho_M(x)$. By the \hyperref[Hahn-Banach theorem]{theorem:hahn-banach}, there exists $\phi \in \hom{E; K}$ such that $\dpb{x, \phi}{E} = \rho_M(x)$ and $\phi \le \rho_M \le \rho$.
(2): By (1) applied to $M = \bracs{0}$.

14
src/fa/lc/index.tex
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@@ -2,10 +2,10 @@
\label{chap:lc}
\input{./src/fa/lc/convex.tex}
\input{./src/fa/lc/continuous.tex}
\input{./src/fa/lc/quotient.tex}
\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}
\input{./convex.tex}
\input{./continuous.tex}
\input{./quotient.tex}
\input{./projective.tex}
\input{./inductive.tex}
\input{./hahn-banach.tex}
\input{./spaces-of-linear.tex}

22
src/fa/lc/inductive.tex
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@@ -13,12 +13,13 @@
\[
\mathcal{B} = \bracs{U \subset E|U \text{ convex, radial, circled}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
\]
is a fundamental system of neighbourhoods for $E$ at $0$.
\end{enumerate}
The topology $\topo$ is the \textbf{inductive locally convex topology} on $E$ induced by $\seqi{T}$.
\end{definition}
\begin{proof}
(1), (5): To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \ref{proposition:tvs-0-neighbourhood-base}.
(1), (5): To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \autoref{proposition:tvs-0-neighbourhood-base}.
\begin{enumerate}
\item[(TVB1)] Every set in $\mathcal{B}$ is radial and circled by definition.
\item[(TVB2)] For any $U \in \mathcal{B}$, $U$ is circled, so $\frac{1}{2}U + \frac{1}{2}U \subset U$. Since $\frac{1}{2}U$ is also circled and radial, $\frac{1}{2}U \in \mathcal{B}$.
@@ -48,6 +49,7 @@
}
\]
\item[(U)] For any pair $(F, \bracsn{S^i_F}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique $S \in L({E, F})$ such that the following diagram commutes
\[
@@ -57,22 +59,24 @@
}
\]
for all $i \in I$.
\item For any locally convex space $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T^i_E \in L(E_i; F)$ for all $i \in I$.
\item The family
\[
\mathcal{B} = \bracs{U \subset E|U \text{ convex, radial, circled}, (T^i_E)^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
\]
is a fundamental system of neighbourhoods for $E$ at $0$.
\end{enumerate}
The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
\end{definition}
\begin{proof}
Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\ref{proposition:module-direct-limit}). Equip $E$ with the inductive topology (\ref{definition:lc-inductive}) induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\autoref{proposition:module-direct-limit}). Equip $E$ with the \hyperref[inductive topology]{definition:lc-inductive} induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
(U): By (U) of \ref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \ref{definition:lc-inductive}, $S \in L(E; F)$.
(U): By (U) of \autoref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \autoref{definition:lc-inductive}, $S \in L(E; F)$.
(5): By (5) of \ref{definition:lc-inductive}.
(5): By (5) of \autoref{definition:lc-inductive}.
\end{proof}
\begin{remark}
@@ -99,11 +103,13 @@
\[
(1 - \alpha)U + w = (1 - \alpha) U + \alpha x \subset V
\]
so $V \in \cn(0)$.
\item For any $\lambda \in K$ with $\abs{\lambda} \le 1$, $u \in U$, $w \in W$, and $t \in [0, 1]$,
\[
\lambda (1 - t)u + \lambda tw = (1 - t)\lambda u + t \lambda w \in V
\]
as $U$ and $W$ are both circled.
\end{itemize}
so $V \in \cn_E(0)$ is convex and circled.
@@ -128,9 +134,9 @@
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $U \in \cn_{E_n}(0)$. By \ref{lemma:lc-induct-separate}, there exists $\bracs{U_m| m \in \natp, m \ge n} \subset 2^E$ such that $U_n = U$, $U_m \in \cn_{E_m}(0)$ and $U_{m} = U_{m + 1} \cap E_m$ for all $m \in \natp$. Let $V = \bigcup_{m \ge n}U_m$, then $V \cap E_m = U_m$ for all $m \ge n$. In particular, $V \cap E_n = U_n = U$.
(1): Let $U \in \cn_{E_n}(0)$. By \autoref{lemma:lc-induct-separate}, there exists $\bracs{U_m| m \in \natp, m \ge n} \subset 2^E$ such that $U_n = U$, $U_m \in \cn_{E_m}(0)$ and $U_{m} = U_{m + 1} \cap E_m$ for all $m \in \natp$. Let $V = \bigcup_{m \ge n}U_m$, then $V \cap E_m = U_m$ for all $m \ge n$. In particular, $V \cap E_n = U_n = U$.
(2): Let $x \in E \setminus \bracs{0}$, then there exists $n \in \natp$ such that $x \in E_n$. Since $E_n$ is separated, there exists $U \in \cn_{E_n}(0)$ with $x \not\in U$. By \ref{lemma:lc-induct-separate} and (1), there exists $V \in \cn_E(0)$ such that $V \cap E_n = U$, so $x \not\in V$.
(2): Let $x \in E \setminus \bracs{0}$, then there exists $n \in \natp$ such that $x \in E_n$. Since $E_n$ is separated, there exists $U \in \cn_{E_n}(0)$ with $x \not\in U$. By \autoref{lemma:lc-induct-separate} and (1), there exists $V \in \cn_E(0)$ such that $V \cap E_n = U$, so $x \not\in V$.
(3), $\neg (b) \Rightarrow \neg (a)$: If $B \not\subset E_n$ for all $n \in \natp$, then there exists a subsequence $\bracsn{n_k}_0^\infty \subset \natp$ and $\seq{x_k} \subset B$ such that $x_k \in E_{n_{k}} \setminus E_{n_{k - 1}}$ for all $k \in \natp$.
@@ -148,6 +154,7 @@
\[
\fU = \bracs{F + U|F \in \mathcal{F}, U \in \cn_E(0)}
\]
then $\fU$ is also a Cauchy filter, which converges if and only if $\fF$ does.
Since each $E_n$ is complete, it is sufficient to show that there exists $n \in \natp$ such that $F + U \cap E_n \ne \emptyset$ for all $F \in \fF$ and $U \in \cn_E(0)$.
@@ -156,6 +163,7 @@
\[
U = \text{Conv}\paren{\bigcup_{n \in \natp}(U_n \cap E_n)}
\]
then since each $U_n$ is circled, so is $U$. Thus $U \cap E_n \supset U_n \cap E_n \in \cn_{E_n}(0)$, and $U \in \cn_E(0)$.
Now, suppose that $(F_n + U) \cap E_n \ne \emptyset$. Let $y \in (F_n + U) \cap E_n$, then there exists $N \in \natp$, $\bracs{x_k}_1^N \subset E$, $\bracs{\lambda_k}_1^N \subset [0, 1]$, and $z \in F_n$ such that
@@ -168,11 +176,13 @@
\[
\underbracs{y - \sum_{k = 1}^n \lambda_kx_k}_{\in E_n} = \underbrace{z + \sum_{k = n + 1}^N \lambda_kx_k}_{\in F_n + U_n}
\]
which is impossible. Therefore $(F_n + U) \cap E_n = \emptyset$ for all $n \in \natp$.
Finally, since $\fF$ is a Cauchy filter, there exists $F \in \fF$ such that $F - F \subset U$. Let $z \in F$, then there exists $n \in \natp$ such that $z \in E_n$. In which case, for any $y \in F \cap F_n$,
\[
z = y + (z - y) \in y + (F - F) \subset y + U \subset F_n + U
\]
which contradicts the fact that $(F_n + U) \cap E_n = \emptyset$.
\end{proof}

12
src/fa/lc/projective.tex
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@@ -7,10 +7,11 @@
Let $E$ be a vector space over $K \in \RC$, $\seqi{F}$ be locally convex spaces over $K$, and $\seqi{T}$ where $T_i \in \hom(E; F_i)$ for all $i \in I$, then the projective topology on $E$ is locally convex.
\end{proposition}
\begin{proof}
By \ref{definition:tvs-initial},
By \autoref{definition:tvs-initial},
\[
\mathcal{B} = \bracs{\bigcap_{j \in J}T_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \cn_{F_j}(0)}
\]
is a fundamental system of neighbourhoods at $0$. For each $i \in I$, $U_i \in \cn_{F_i}(0)$ convex, $T^{-1}(U_i)$ is also convex. Since each $F_i$ is locally convex, $\mathcal{B}$ contains a fundamental system of neighbourhoods at $0$ consisting of only convex sets.
\end{proof}
@@ -19,7 +20,7 @@
Let $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$ be a downward-directed system of locally convex spaces over $K \in \RC$, then $E = \lim_{\longleftarrow}E_i$ is locally convex.
\end{proposition}
\begin{proof}
By (U) of \ref{definition:tvs-projective-limit} and \ref{definition:tvs-initial}, $E$ is equipped with the projective topology generated by the projection maps $E \to E_i$. By \ref{proposition:lc-projective-topology}, $E$ is locally convex.
By (U) of \autoref{definition:tvs-projective-limit} and \autoref{definition:tvs-initial}, $E$ is equipped with the projective topology generated by the projection maps $E \to E_i$. By \autoref{proposition:lc-projective-topology}, $E$ is locally convex.
\end{proof}
\begin{proposition}[{{\cite[II.5.4]{SchaeferWolff}}}]
@@ -32,6 +33,7 @@
\[
\pi^U_V: E_U \to E_V \quad x + M_U \mapsto x + M_V
\]
then $\pi^U_V \in L(E_U; E_V)$.
\item $(\bracsn{E_U}_{U \in \mathcal{B}}, \bracs{\pi^U_V|U, V \in \mathcal{B}, U \subset V})$ is a downward-directed system of topological vector spaces.
\item The map $\pi \in L(E, \lim_{\longleftarrow}E_U)$ induced by $\bracs{\pi_U}_{U \in \mathcal{B}}$ is a bijection.
@@ -39,10 +41,11 @@
\[
E = \lim_{\longleftarrow}E_U = \lim_{\longleftarrow} \ol E_U
\]
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Since $V \supset U$, $[\cdot]_V \ge [\cdot]_U$, so $M_V \supset M_U$. Thus $\ker(\pi_V) \supset M_U$. By (U) of the quotient (\ref{definition:tvs-quotient}), $\pi_V$ factors through $E_U$ as $\pi^U_V$, so $\pi^U_V \in L(E_U; E_V)$.
(1): Since $V \supset U$, $[\cdot]_V \ge [\cdot]_U$, so $M_V \supset M_U$. Thus $\ker(\pi_V) \supset M_U$. By (U) of the \hyperref[quotient]{definition:tvs-quotient}, $\pi_V$ factors through $E_U$ as $\pi^U_V$, so $\pi^U_V \in L(E_U; E_V)$.
(2): Since $\mathcal{B}$ is a fundamental system of neighbourhoods, it is downward-directed under inclusion. For any $U, V, W \in \mathcal{B}$ with $U \subset V \subset W$, $M_U \supset M_V \supset M_W$. Thus $\pi^U_W = \pi^V_W \circ \pi^U_V$.
@@ -54,11 +57,12 @@
\[
\pi_W(x_U) = \pi_W^U \circ \pi_U(x_U) = \pi_W^U p_U(x)
\]
Thus for any $U' \in \mathcal{B}$ with $U \subset W$, $[x_U - x_{U'}]_W = 0$, and $x_U - x_{U'} \in W$. Therefore $\bracs{x_U}_{U \in \mathcal{B}}$ is a Cauchy net, and converges to $x_0 \in E$ by completeness of $E$.
For any $U \in \mathcal{B}$, $\pi_U(x_0) = \lim_{V \in \mathcal{B}}\pi_U(x_V) = p_U(x)$, so $\pi(x_0) = x$, and $\pi$ is surjective.
(4): Since $\mathcal{B} \subset \cn_E(0)$ is a fundamental system of neighbourhoods, the topology on $E$ is the projective topology generated by $\bracs{\pi_U|U \in \mathcal{B}}$. As $\pi_U \circ \pi^{-1} = p_U \in L(\lim E_U; E_U)$ for all $U \in \mathcal{B}$, $\pi^{-1} \in L(\lim E_U; E)$ by (U) of the projective topology (\ref{definition:tvs-initial}).
(4): Since $\mathcal{B} \subset \cn_E(0)$ is a fundamental system of neighbourhoods, the topology on $E$ is the projective topology generated by $\bracs{\pi_U|U \in \mathcal{B}}$. As $\pi_U \circ \pi^{-1} = p_U \in L(\lim E_U; E_U)$ for all $U \in \mathcal{B}$, $\pi^{-1} \in L(\lim E_U; E)$ by (U) of the \hyperref[projective topology]{definition:tvs-initial}.
Let $x \in \lim\ol{E}_U$ and $V \in \cn(x)$. Since $\mathcal{B}$ is downward-directed and $\lim\ol{E}_U$ is equipped with the projective topology induced by $\bracs{p_U|U \in \mathcal{B}}$, there exists $U \in \mathcal{B}$ and $W \in \cn_{\ol E_U}(x)$ such that $p_U^{-1}(W) \subset V$. As $E_U$ is dense in $\ol E_U$, there exists $y_U \in W$, and $y \in E$ such that $y_U = \pi_U(y)$. Therefore $\pi(y_U) \in p_U^{-1}(W) \subset V$, and $\lim E_U$ is dense in $\lim \ol{E}_U$.

7
src/fa/lc/quotient.tex
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@@ -7,6 +7,7 @@
\[
\rho_M: E/M \to [0, \infty) \quad x + M \mapsto \inf_{y \in x + M}\rho(y)
\]
is the \textbf{quotient} of $\rho$ by $M$.
\end{definition}
\begin{proof}
@@ -19,6 +20,7 @@
\[
\rho_M(x + x') \le \rho(y + y') \le \rho(y) + \rho(y')
\]
As this holds for all $y \in x + M$ and $y' \in x' + M$, $\rho_M(x + x') \le \rho_M(x) + \rho_M(x')$.
\end{proof}
@@ -36,23 +38,26 @@
\widetilde E \ar@{->}[r]_{\tilde f} & F
}
\]
If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$.
\item If $\seqi{\rho}$ is a family of seminorms that induces the topology on $E$, then their quotients by $M$ induces the topology on $\td E$.
\end{enumerate}
The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$.
\end{definition}
\begin{proof}
By \ref{definition:tvs-quotient}, (2), (3), (U) holds, and $\td E$ is a TVS over $K$.
By \autoref{definition:tvs-quotient}, (2), (3), (U) holds, and $\td E$ is a TVS over $K$.
(1): Let $U \subset E$ be convex, then for any $x + M, y + M \in \pi(U)$ and $t \in [0, 1]$,
\[
(tx + M) + ((1 - t)y + M) = (tx + (1-t)y) + M \in U + M = \pi(U)
\]
so $\pi(U)$ is convex. Let $\fB = \bracs{U|U \in \cn_E(0) \text{ convex}}$, then $\bracs{\pi(U)|U \in \fB}$ is a fundamental system of neighbourhoods for the quotient topology on $E/M$. Therefore $E/M$ is locally convex.
(5): By (U), each quotient seminorm is continuous on $\td E$, so the quotient topology contains the topology induced by the quotient seminorms. On the other hand, let $\pi(U) \in \cn_{\td E}(0)$, then there exists $J \subset I$ finite and $r > 0$ such that
\[
\bigcap_{j \in J}B_j(0, r) \subset U
\]
For each $j \in J$, let $\eta_j$ be the quotient of $\rho_j$ by $M$. Let $x + M \in E/M$ with $\eta_j(x) < r$ for all $j \in J$. For each $j \in J$, there exists $y_j \in x + M$ such that $\rho_j(y_j) < r$, so $y_j + M \in \pi(U)$. Therefore $x \in \pi(U)$ as well, and the quotient norms induce the quotient topology on $E/M$.
\end{proof}

4
src/fa/lc/spaces-of-linear.tex
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@@ -7,12 +7,14 @@
\[
[\cdot]_{S, i}: E^T \to [0, \infty) \quad f \mapsto \sup_{x \in S}[f(x)]_{S, i}
\]
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}.
By \autoref{proposition:set-uniform-pseudometric}.
\end{proof}

18
src/fa/lp/definition.tex
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@@ -7,6 +7,7 @@
\[
\norm{f}_{L^p(X; E)} = \norm{f}_{L^p(\mu; E)} = \norm{f}_{L^p(X, \cm, \mu; E)} = \braks{\int \norm{f}_E^p d\mu}^{1/p} < \infty
\]
The set $\mathcal{L}^p(X; E) = \mathcal{L}^p(\mu; E) = \mathcal{L}^p(X, \cm, \mu; E)$ is the space of all $p$-integrable functions on $X$.
\end{definition}
@@ -16,6 +17,7 @@
\[
\norm{f}_{L^\infty(X; E)} = \norm{f}_{L^\infty(\mu; E)} = \norm{f}_{L^\infty(X, \cm, \mu; E)} = \inf\bracs{\alpha \ge 0|\mu(\bracs{f > \alpha}) = 0} < \infty
\]
In which case, $\norm{f}_{L^\infty(X; E)}$ is the \textbf{essential supremum} of $f$.
\end{definition}
@@ -25,6 +27,7 @@
\[
\frac{1}{p} + \frac{1}{q} = 1
\]
\end{definition}
\begin{lemma}
@@ -43,16 +46,19 @@
\[
\int \norm{f}_E \norm{g}_F d\mu \le \norm{f}_{L^p(X; E)}\norm{g}_{L^q(X; F)}
\]
\end{theorem}
\begin{proof}
First suppose that $p = 1$ and $q = \infty$. In this case,
\[
\int \norm{f}_E \norm{g}_F d\mu \le \norm{g}_{L^\infty(X; F)}\int \norm{f}_Ed\mu = \norm{f}_{L^1(X; E)}\norm{g}_{L^\infty(X; F)}
\]
Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)} = \norm{g}_{L^q(X; F)} = 1$. By Young's inequality (\ref{lemma:young-inequality}),
Now suppose that $p, q \in (1, \infty)$ are Hölder conjugates. Assume without loss of generality that $\norm{f}_{L^p(X; E)} = \norm{g}_{L^q(X; F)} = 1$. By \hyperref[Young's inequality]{lemma:young-inequality},
\[
\int \norm{f}_E \norm{g}_F d\mu \le \int \frac{\norm{f}_E^p}{p} + \frac{\norm{g}_F^q}{q} d\mu = \frac{1}{p}\int \norm{f}_E d\mu + \frac{1}{q}\int \norm{g}_F^q d\mu = 1
\]
\end{proof}
\begin{theorem}[Minkowski's Inequality, {{\cite[6.5]{Folland}}}]
@@ -61,6 +67,7 @@
\[
\norm{f + g}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} + \norm{g}_{L^p(X; E)}
\]
\end{theorem}
\begin{proof}
If $p = 1$, then the theorem holds directly.
@@ -70,6 +77,7 @@
\norm{f + g}_{L^\infty(X; E)} \le \norm{f}_{L^\infty(X; E)} + \norm{g}_{L^\infty(X; E)}
\]
Now suppose that $p \in (1, \infty)$, then $p = q(p - 1)$, and
\begin{align*}
\norm{f + g}_E^p &\le (\norm{f}_E + \norm{g}_E)\norm{f + g}_E^{p - 1} \\
@@ -85,10 +93,11 @@
\[
L^p(X, \cm, \mu; E) = \mathcal{L}^p(X, \cm, \mu; E)/\bracs{f|f = 0\text{ a.e.}}
\]
is a normed vector space, known as the $E$-valued \textbf{$L^p$ space} on $(X, \cm, \mu)$.
\end{definition}
\begin{proof}
By Minkowski's Inequality (\ref{theorem:minkowski}).
By \hyperref[Minkowski's Inequality]{theorem:minkowski}.
\end{proof}
\begin{proposition}
@@ -96,11 +105,12 @@
Let $(X, \cm, \mu)$ be a measure space, $E$ be a normed space, and $p \in [1, \infty)$, then $\Sigma(X, \cm; E) \cap L^p(X; E)$ is dense in $L^p(X; E)$.
\end{proposition}
\begin{proof}
Let $f \in L^p(X; E)$. By \ref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$.
Let $f \in L^p(X; E)$. By \autoref{definition:strongly-measurable}, there exists $\seq{f_n} \subset \Sigma(X, \cm; E)$ such that $\norm{f_n}_E \le \norm{f}_E$ and $\norm{f_n - f}_E \to 0$ pointwise as $n \to \infty$.
For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the Dominated Convergence Theorem (\ref{theorem:dct}),
For each $n \in \nat$, $\norm{f_n}_E \le \norm{f}_E$, so $\norm{f_n}_{L^p(X; E)} \le \norm{f}_{L^p(X; E)} < \infty$, and $\norm{f_n - f}_E \le 2\norm{f}_E$. By the \hyperref[Dominated Convergence Theorem]{theorem:dct},
\[
\limv{n}\int \norm{f_n - f}_E^p d\mu = \int \limv{n}\norm{f_n - f}_E^p d\mu = 0
\]
Therefore $\norm{f_n - f}_{L^p(X; E)} \to 0$ as $n \to \infty$.
\end{proof}

2
src/fa/lp/index.tex
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@@ -1,4 +1,4 @@
\chapter{$L^p$ Spaces}
\label{chap:lp}
\input{./src/fa/lp/definition.tex}
\input{./definition.tex}

6
src/fa/norm/index.tex
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@@ -1,6 +1,6 @@
\chapter{Normed Spaces}
\label{chap:normed-spaces}
\input{./src/fa/norm/normed.tex}
\input{./src/fa/norm/linear.tex}
\input{./src/fa/norm/multilinear.tex}
\input{./normed.tex}
\input{./linear.tex}
\input{./multilinear.tex}

3
src/fa/norm/linear.tex
View File

@@ -7,7 +7,8 @@
\[
\norm{\cdot}_{L(E; F)}: L(E; F) \to [0, \infty) \quad T \mapsto \sup_{\substack{x \in E \\ \norm{x}_E = 1}}Tx
\]
\end{proposition}
\begin{proof}
By \ref{proposition:lc-spaces-linear-map}.
By \autoref{proposition:lc-spaces-linear-map}.
\end{proof}

3
src/fa/norm/multilinear.tex
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@@ -16,5 +16,6 @@
\[
\sup_{y \in B_F(0, 1)}\norm{T_yx}_G = \sup_{y \in B_F(0, 1)}\norm{T(x, y)}_G < \infty
\]
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 $y \in F$, $\norm{T(x, y)}_G \le M\norm{x}_E\norm{y}_F$.
by continuity of $y \mapsto T(x, y)$. By the \hyperref[Uniform Boundedness Principle]{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}

11
src/fa/norm/normed.tex
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@@ -21,12 +21,13 @@
\end{enumerate}
\end{proposition}
\begin{proof}
(1) $\Rightarrow$ (2): Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is also circled. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$ and $\abs{\lambda}^{-1}U \subset V$.
(1) $\Rightarrow$ (2): Using \autoref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is also circled. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$ and $\abs{\lambda}^{-1}U \subset V$.
Let $\norm{\cdot}_E: E \to [0, \infty)$ be the gauge (\ref{definition:gauge}) of $U$. For each $r > 0$, let $B_E(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
Let $\norm{\cdot}_E: E \to [0, \infty)$ be the \hyperref[gauge]{definition:gauge} of $U$. For each $r > 0$, let $B_E(0, r) = \bracs{x \in E|\norm{x}_E < r}$, then
\[
\bracs{\lambda U|\lambda > 0} = \bracs{B_E(0, r)|r > 0}
\]
is a fundamental system of neighbourhoods at $0$ for $E$. Therefore $\norm{\cdot}_E$ induces the topology on $E$.
\end{proof}
@@ -51,14 +52,17 @@
\[
\norm{y_n}_F \le \gamma^{n - 1}\norm{y_1}_F = \gamma^{n - 1}\norm{y}_F
\]
Since $\norm{x_n}_E \le C\norm{y_n}_F$,
\[
\sum_{k \in \natp}\norm{x_k}_E \le C\norm{y}_F\sum_{k \in \nat_0}\gamma^k = \frac{C\norm{y}_F}{1 - \gamma}
\]
In addition,
\[
\norm{y - \sum_{k = 1}^n Tx_k}_F = \norm{y_{n+1}}_F \le \gamma^n \norm{y}_F
\]
so $\sum_{n = 1}^\infty Tx_n = y$.
\end{proof}
@@ -72,11 +76,12 @@
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 $\sup_{x \in U}\sup_{T \in \mathcal{T}}\norm{Tx}_{F} < \infty$.
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 \hyperref[Baire Category Theorem]{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}

4
src/fa/rs/bv.tex
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@@ -8,10 +8,12 @@
\[
V_{\rho, p}(f) = \sum_{j = 1}^n \rho(f(x_j) - f(x_{j - 1}))
\]
is the \textbf{variation} of $f$ with respect to $\rho$ and $P$. The supremum over all such partitions
\[
[f]_{\var, \rho} = \sup_{P \in \scp([a, b])}V_{\rho, P}(f)
\]
is the \textbf{total variation} of $f$ on $[a, b]$ with respect to $\rho$.
If $E$ is a normed space, then the variation and total variation of $f$ is taken with respect to its norm.
@@ -45,12 +47,14 @@
= \lim_{g, \fF}\sum_{j = 1}^n \rho(g(x_j) - g(x_{j - 1}))
= \lim_{g \in \fF}V_{\rho, P}(g)
\]
By assumption (b), $[0, M_\rho]$ is in the filter generated by $V_{\rho, P}(\fF)$. Thus $V_{\rho, P}(f) \le M_\rho$. As this holds for all $P \in \scp([a, b])$, $V_{\rho, P}(f) \le M_\rho$, and $f \in BV([a, b]; E)$.
(5): For each $n \in \nat^+$, let
\[
D_n = \bracs{x \in [a, b]|\forall \eps > 0, \exists y \in (x - \eps, x + \eps): \norm{f(x) - f(y)}_E \ge 1/n}
\]
then $D = \bigcup_{n \in \nat^+}D_n$ is the set of discontinuity points of $f$. If $D$ is uncountable, then there exists $N \in \nat^+$ such that $D_n$ is infinite.
Fix $N \in \nat^+$. Let $E_1 = D_n \cap (a, b)$ and $I_1 = (a, b)$, then

8
src/fa/rs/index.tex
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@@ -1,7 +1,7 @@
\chapter{The Riemann-Stieltjes Integral}
\label{chap:rs-integral}
\input{./src/fa/rs/partition.tex}
\input{./src/fa/rs/bv.tex}
\input{./src/fa/rs/rs.tex}
\input{./src/fa/rs/rs-bv.tex}
\input{./partition.tex}
\input{./bv.tex}
\input{./rs.tex}
\input{./rs-bv.tex}

2
src/fa/rs/partition.tex
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@@ -7,6 +7,7 @@
\[
P = \seqfz{x_j} = [a = x_0 \le \cdots \le x_n = b]
\]
The collection $\scp([a, b])$ is the set of all partitions of $[a, b]$.
\end{definition}
@@ -23,6 +24,7 @@
\[
\sigma(P) = \max_{1 \le j \le n}(x_j - x_{j - 1})
\]
is the \textbf{mesh} of $P$.
\end{definition}

7
src/fa/rs/rs-bv.tex
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@@ -9,9 +9,10 @@
\[
\braks{\int_a^bf dG}_H \le \sup_{x \in [a, b]}[f]_1 \cdot [g]_{\var, 2}
\]
\end{proposition}
\begin{proof}
By \ref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that $[xy]_H \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
By \autoref{proposition:tvs-convex-multilinear}, there exists continuous seminorms $[\cdot]_1$ on $E_1$ and $[\cdot]_2$ on $E_2$ such that $[xy]_H \le [x]_1[y]_2$ for all $(x, y) \in E_1 \times E_2$.
Let $(P = \seqfz{x_j}, c = \seqf{c_j}) \in \scp_t([a, b])$, then
\begin{align*}
@@ -28,6 +29,7 @@
\[
[f]_{u, \rho} = \sup_{x \in [a, b]}\rho(f(x))
\]
Let $\net{f} \subset RS([a, b], G)$ such that:
\begin{enumerate}
\item[(a)] For each continuous seminorm $\rho$ on $E_1$, $[f_\alpha - f]_{u, \rho} \to 0$.
@@ -61,6 +63,7 @@
\[
\rho\paren{S(P, c, f, G) - \lim_{\alpha \in A}\int_a^b f_\alpha dG} < \eps
\]
\end{proof}
\begin{proposition}
@@ -74,6 +77,7 @@
\[
\int_a^b fdG = \limv{n}S(P_n, t_n, f, G)
\]
\end{enumerate}
\end{proposition}
\begin{proof}
@@ -88,6 +92,7 @@
\[
\rho(S(P, c, f, G) - S(Q, d, f, G)) \le 2 \cdot \sup_{\begin{array}{c} x, y \in [a, b] \\ |x - y| < \max(\sigma(P), \sigma(Q)) \end{array}}[f(x) - f(y)]_1 \cdot [G]_{\var, 2}
\]
by passing through a common refinement. Since $f \in C([a, b]; E_1)$, this bound tends to $0$ as $\max(\sigma(P), \sigma(Q))$ tends to $0$, so $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is a Cauchy net.
In addition, for any $\seq{(P_n, t_n)}$ as in (2), $\limv{n}S(P_n, t_n, f, G)$ exists by sequential completeness. Since $\angles{S(P, c, f, G)}_{(P, c) \in \scp_t([a, b])}$ is Cauchy, the limit $\lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)$ exists as well and is equal to $\limv{n}S(P_n, t_n, f, G)$.

8
src/fa/rs/rs.tex
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@@ -9,6 +9,7 @@
\[
S(P, c, f, G) = \sum_{j = 1}^n f(c_j)[G(x_j) - G(x_{j - 1})]
\]
is the \textbf{Riemann-Stieltjes sum} of $f$ with respect to $G$ and $(P, c)$.
\end{definition}
@@ -20,6 +21,7 @@
\[
\int_a^b f dG = \int_a^b f(t)G(dt) = \lim_{(P, c) \in \scp_t([a, b])}S(P, c, f, G)
\]
exists. In which case, $\int_a^b fdG$ is the \textbf{Riemann-Stieltjes integral} of $G$.
The set $RS([a, b], G)$ is the vector space of all \textbf{Riemann-Stieltjes integrable functions} with respect to $G$.
@@ -31,6 +33,7 @@
\[
S(P, c, f, G) + S(P', c', G, f) = f(b)G(b) - f(a)G(a)
\]
where $P' = \seqfz[n+1]{y_j} = [a, c_1, \cdots, c_n, b]$ and $c' = \seqf[n+1]{d_j} = [x_0, \cdots, x_n]$.
\end{lemma}
\begin{proof}
@@ -53,16 +56,19 @@
\[
\int_a^b f dG + \int_a^b G df = f(b)G(b) - f(a)G(a)
\]
\end{theorem}
\begin{proof}
Suppose that $f \in RS([a, b], G)$. Let $U \in \cn_F(0)$, then there exits $P_0 = \seqfz{x_j} \in \scp([a, b])$ such that $S(P, c, f, G) - \int_a^b fdG \in U$ for all $(P, c) \in \scp_t([a, b])$ with $P \ge P_0$. Let
\[
Q_0 = [x_0, x_1, x_1, \cdots, x_n, x_n]
\]
then for any $(Q = \seqfz[m]{y_j}, d = \seqf[m]{d_j}) \in \scp_t([a, b])$ with $Q \ge Q_0$,
\[
f(b)G(b) - f(a)G(a) - \int_a^b fdG - S(Q, d, G, f) =
\int_a^b fdG - S(Q', d', G, f)
\]
by \ref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $\int_a^b fdG - S(Q', d', G, f) \in U$.
by \autoref{lemma:sum-by-parts}, where $d$ and $Q'$ contain $\seqfz{x_j}$. Thus $(Q', d') \ge P_0$, and $\int_a^b fdG - S(Q', d', G, f) \in U$.
\end{proof}

6
src/fa/tvs/bounded.tex
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@@ -20,16 +20,18 @@
\begin{proof}
Let $U \in \cn(0)$.
(2): Using \ref{proposition:uniform-neighbourhoods}, assume without loss of generality that $U$ is closed. Let $0 \ne \lambda \in K$ with $\lambda U \supset B$, then since $\lambda U$ is closed, $\lambda U \supset \ol B$.
(2): Using \autoref{proposition:uniform-neighbourhoods}, assume without loss of generality that $U$ is closed. Let $0 \ne \lambda \in K$ with $\lambda U \supset B$, then since $\lambda U$ is closed, $\lambda U \supset \ol B$.
(4), (5): By \ref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$, and $\lambda, \lambda' \in K$ such that $\lambda V \supset A$ and $\lambda' V \supset B$.
(4), (5): By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $V \in \cn(0)$ circled such that $V + V \subset U$, and $\lambda, \lambda' \in K$ such that $\lambda V \supset A$ and $\lambda' V \supset B$.
Let $\mu > \abs{\lambda}, \abs{\lambda'}$, then
\[
\mu U \supset \mu V \supset \lambda V \cup \lambda' V \supset A \cup B
\]
and
\[
\mu U \supset \mu(V + V) \supset \lambda V + \lambda' V \supset A + B
\]
\end{proof}

11
src/fa/tvs/complete-metric.tex
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@@ -31,6 +31,7 @@
\[
T\braks{B_E\paren{0, \frac{Cr}{(1 - \gamma)}}} \supset B_F(0, r)
\]
\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:
@@ -76,6 +77,7 @@
\[
x = \limv{N}x_N = \limv{N}\sum_{n = 1}^N(x_n - x_{n-1})
\]
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}
@@ -92,10 +94,11 @@
\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}),
By the \hyperref[method of successive approximations]{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}
@@ -112,11 +115,13 @@
\[
E = \bigcup_{n \in \natp}nB_E(0, r) \quad \overline{T(E)} = \bigcup_{n \in \natp}\overline{nT(B_E(0, r))}
\]
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,
By the \hyperref[Baire Category Theorem]{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 $E$.
Thus by \autoref{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}

5
src/fa/tvs/completion.tex
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@@ -16,9 +16,9 @@
The pair $(\wh E, \iota)$ is the \textbf{Hausdorff completion} of $E$.
\end{definition}
\begin{proof}
All claims of (1), (2), (U), and (4), except the linearity of maps and the fact that $\wh E$ is a TVS is proven via the Hausdorff completion (\ref{definition:hausdorff-completion}).
All claims of (1), (2), (U), and (4), except the linearity of maps and the fact that $\wh E$ is a TVS is proven via the \hyperref[Hausdorff completion]{definition:hausdorff-completion}.
Using \ref{proposition:initial-completion}, identify $\wh E \times \wh E$ with $\wh{E \times E}$ and $K \times \wh E$ with $\wh{K \times E}$ as uniform spaces. By \ref{proposition:hausdorff-uniform-factor}, there exists operations $\wh E \times \wh E \to \wh E$ and $K \times \wh E \to \wh E$ such that the following diagrams commute
Using \autoref{proposition:initial-completion}, identify $\wh E \times \wh E$ with $\wh{E \times E}$ and $K \times \wh E$ with $\wh{K \times E}$ as uniform spaces. By \autoref{proposition:hausdorff-uniform-factor}, there exists operations $\wh E \times \wh E \to \wh E$ and $K \times \wh E \to \wh E$ such that the following diagrams commute
\[
\xymatrix{
\widehat E \times \widehat E \ar@{->}[r] & \widehat E & & K \times \widehat E \ar@{->}[r] & \widehat E \\
@@ -26,6 +26,7 @@
}
\]
By continuity and the density of $\iota(E)$ in $E$, $\wh E$ with these operations forms a TVS, and $T$ is linear.
\end{proof}

5
src/fa/tvs/continuous.tex
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@@ -13,9 +13,9 @@
If the above holds, then $T$ is a \textbf{continuous linear map}. The set $L(E; F)$ denotes the vector space of all continuous linear maps from $E$ to $F$.
\end{definition}
\begin{proof}
$(1) \Rightarrow (2) \Rightarrow (3)$: By \ref{proposition:uniform-continuous} and \ref{definition:continuity}.
$(1) \Rightarrow (2) \Rightarrow (3)$: By \autoref{proposition:uniform-continuous} and \autoref{definition:continuity}.
$(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$. Using \ref{proposition:tvs-uniform} and \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $U$ and $V$ are symmetric and translation-invariant.
$(3) \Rightarrow (1)$: Let $U$ be an entourage of $F$, there exists an entourage $V$ of $E$ such that $T(V(0)) \subset U(0)$. Using \autoref{proposition:tvs-uniform} and \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $U$ and $V$ are symmetric and translation-invariant.
For any $x, y \in V$, $x - y \in V(0)$, so $Tx - Ty \in U(0)$, $Ty \in U(Tx)$ by symmetry, and $(Tx, Ty) \in U$. Therefore $T$ is uniformly continuous.
\end{proof}
@@ -47,6 +47,7 @@
\prod_{i \in I}E_i \ar@{->}[r]_{\pi_i} & E_i
}
\]
\end{enumerate}
The uniformity $\fU$ and its induced topology are the \textbf{product uniformity/topology}, and $E$ equipped with $\fU$ is the \textbf{product TVS} of $\seqi{E}$.
\end{definition}

34
src/fa/tvs/definition.tex
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@@ -31,6 +31,7 @@
\[
U = \bracs{(x + z, y + z)|(x, y) \in U}
\]
and $\fU$ is \textbf{translation-invariant} if there exists a fundamental system of translation-invariant entourages.
\end{definition}
@@ -39,7 +40,7 @@
Let $E$ be a vector space and $\fU$ be a translation-invariant uniformity, then $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
\end{lemma}
\begin{proof}
Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \ref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
Let $z \in E$, then the map $(x, y) \mapsto (x + z, y + z)$ is a bijection. Thus for any translation-invariant entourages $U, V \in \fU$, $(U \cap V) + z = (U + z) \cap (V + z)$, and $U \cap V$ is translation-invariant. By \autoref{lemma:symmetricfundamentalentourage}, $\fU$ admits a fundamental system of symmetric, translation-invariant entourages.
\end{proof}
@@ -60,13 +61,13 @@
\item[(UB2)] Let $V \in \fB_0$, then by (TVS1), there exists $W \in \fB_0$ such that $W + W \subset V$. For any $(x, y), (y, z) \in U_W$, $(x - y), (y - z) \in W$, so $(x - z) \in V$. Thus $U_W \circ U_W \subset U_V$.
\end{enumerate}
By \ref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
By \autoref{proposition:fundamental-entourage-criterion}, $\fB$ forms a fundamental system of entourages for a translation-invariant uniformity $\fU$ on $E$.
(1): Let $\mathfrak{V}$ be a translation-invariant uniformity on $E$ inducing the topology. For any symmetric, translation-invariant entourage $V \in \mathfrak{V}$, $V(x) = V(0) + x$ for all $x \in E$, and $(x, y) \in V$ if and only if $y - x \in V$, if and only if $x - y \in V$. Thus $V = U_{V(0)}$.
Let $W \in \cn(0)$, then by \ref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
Let $W \in \cn(0)$, then by \autoref{lemma:translation-invariant-symmetric}, there exists a symmetric, translation-invariant entourage $V \in \mathfrak{V}$ such that $V(0) \subset W$, and $V \subset U_W$. Thus $\mathfrak{V} \supset \fU$.
Let $V \in \mathfrak{V}$. Using \ref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
Let $V \in \mathfrak{V}$. Using \autoref{lemma:translation-invariant-symmetric}, assume without loss of generality that $V$ is symmetric and translation-invariant, then there exists $W \in \fB_0$ with $W \subset V(0)$. In which case, $U_W \subset V$, and $\fU \supset \mathfrak{V}$.
\end{proof}
\begin{proposition}
@@ -75,14 +76,16 @@
\[
\ol{A} = \bigcap_{U \in \fB}\bracs{A + U| U \in \fB}
\]
\end{proposition}
\begin{proof}
Let $V \in \cn(0)$ be balanced and $U_V = \bracs{(x, y) \in E \times E| x - y \in V}$, then $y \in U_V(A)$ if and only if there exists $x \in A$ such that $(x, y) \in U_V$. This is equivalent to $x - y \in V$ and $y - x \in V$, so $U_V(A) = A + V$.
Assume without loss of generality that $\fB$ consists of symmetric entourages. By \ref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \ref{proposition:uniformclosure} implies that
Assume without loss of generality that $\fB$ consists of symmetric entourages. By \autoref{proposition:tvs-uniform}, $\bracs{U_V|V \in \fB}$ forms a fundamental system of entourages for $E$, and \autoref{proposition:uniformclosure} implies that
\[
\ol{A} = \bigcap_{V \in \fB}\bracs{U_V(A)| U \in \fB} = \bigcap_{V \in \fB}U + A
\]
\end{proof}
\begin{proposition}[{{\cite[I.1.1]{SchaeferWolff}}}]
@@ -94,10 +97,11 @@
\end{enumerate}
\end{proposition}
\begin{proof}
$(1)$: For every $x \in B$, $A + x$ is open by \ref{definition:translation-invariant-topology}, so
$(1)$: For every $x \in B$, $A + x$ is open by \autoref{definition:translation-invariant-topology}, so
\[
A + B = \bigcup_{x \in B}(A + x)
\]
is open.
$(2)$: Let $x \in \overline{A + B}$, then there exists a filter $\fF \subset 2^{A \cup B}$ converging to $x$. For any $U \in \fF$, $U \cap (A + B) \ne \emptyset$, so $(U - B) \cap A \ne \emptyset$, and $\fB = \bracs{U - B| U \in \fF}$ is a filter base in $A$. By compactness of $A$, there exists $y \in A$ such that
@@ -105,14 +109,17 @@
y \in \bigcap_{U \in \fF}\overline{U - B} = \bigcap_{U \in \fF}\overline{(-B) + U}
\]
By \ref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
By \autoref{proposition:tvs-closure}, $\overline{(-B) + U} \subset (-B) + U + U$, so
\[
y \in \bigcap_{U \in \fF}\overline{(-B) + U} \subset \bigcap_{U \in \fF}[(-B) + U + U]
\]
Since $\fF$ converges to $x$, (TVS1) implies that $\bracs{U + U| U \in \fF}$ contains a neighbourhood base of $x$. Thus
\[
y \in \bigcap_{U \in \fF}[(-B) + U + U] \subset \bigcap_{V \in \cn(0)}[(x-B) + V] = \overline{x - B} = x - B
\]
so $x \in y + B \subset A + B$.
\end{proof}
@@ -137,12 +144,13 @@
\begin{proof}
Firstly, (TVS2) implies that every neighbourhood of $0$ is circled.
By \ref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
By \autoref{proposition:uniform-neighbourhoods}, $E$ admits a fundamental system of neighbourhoods consisting of open sets or closed sets.
Let $U \in \cn^o(0)$ be open. By (TVS2), there exists $r > 0$ such that $\lambda U \subset U$ for all $\lambda \in K$ with $\abs{\lambda} \le r$. Define
\[
V = \bigcup_{\substack{\lambda \in K \\ \abs{\lambda} \le r}} \lambda U \subset U
\]
then for any $x \in V$, there exists $\lambda \in K$ with $\abs{\lambda} \le r$ and $y \in U$ such that $x = \lambda y$. In which case, for any $\mu \in K$ with $\abs{\mu} \le 1$, $\mu(\lambda y) = (\mu \lambda) y \in \mu\lambda U$. Since $\abs{\mu \lambda} \le r$, $\mu \lambda U \subset V$. Thus $V \subset U$ is balanced.
Let $U \in \cn(0)$ be closed, then there exists a balanced neighbourhood $V \in \cn^o(0)$ such that $V \subset U$. In which case, for any $\lambda \in K$ with $0 < \abs{\lambda} \le 1$, $\lambda \overline{V} = \overline{\lambda V} \subset \overline{V}$ by (TVS2). Therefore $\overline{V} \subset U$ is balanced as well.
@@ -166,19 +174,20 @@
\end{enumerate}
\end{proposition}
\begin{proof}
\textbf{Forward:} By \ref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
\textbf{Forward:} By \autoref{proposition:tvs-good-neighbourhood-base}, there exists a fundamental system of neighbourhoods $\fB \subset \cn_E(0)$ consisting of circled and radial sets. By (TVS1), $\fB$ satisfies (TVB1).
\textbf{Converse:} For each $V \in \fB$, let $U_V = \bracs{(x, y) \in E|x - y \in V}$, then $U_V$ is symmetric and translation-invariant by (TVB1). Let
\[
\mathfrak{V} = \bracs{U_V|V \in \fB}
\]
then
\begin{enumerate}
\item[(FB1)] For any $V, V' \in \fB$, there exists $W \in \fB$ with $W \subset V \cap V'$. In which case, $U_{V} \cap U_{V'} \supset U_W \in \mathfrak{V}$.
\item[(UB1)] For any $x \in E$ and $V \in \fB$, $x - x = 0 \in V$, so $\Delta \subset U_V$.
\item[(UB2)] For any $V \in \fB$, by (TVB1), there exists $W \in \fB$ such that $W + W \subset V$. In which case, for any $x, y, z \in E$ with $x - y, y - z \in W$, $x - z \in V$. Therefore $U_W \circ U_W \subset U_V$.
\end{enumerate}
By \ref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.
By \autoref{proposition:fundamental-entourage-criterion}, there exists a unique uniformity $\fU$ on $E$ for which $\mathfrak{V}$ is a fundamental system of entourages.
(1): Since $\mathfrak{V}$ is translation-invariant, so is $\fU$.
@@ -197,10 +206,11 @@
\lambda x - \lambda' x' \in \lambda V + \eps(\abs \mu + 1)V \subset \abs{\lambda} V + \eps(\abs \mu + 1)V
\]
Let $W \in \fB$ and $\eps > 0$ such that $\eps(\abs{\mu}+1) \le 1$, then by repeated application of (TVB1), there exists $V \in \fB$ such that $\abs{\lambda} V + V \subset W$. Therefore scalar multiplication is jointly continuous.
\end{enumerate}
(Uniqueness): Let $\mathcal{S} \subset 2^E$ be a topology on $E$ satisfying (1) and (2), then for each $x \in E$, $\cn_{(E, \mathcal{S})}(x) = \cn_{(E, \mathcal{T})}(x)$. By \ref{proposition:neighbourhoodcharacteristic}, $\mathcal{S} = \mathcal{T}$.
(Uniqueness): Let $\mathcal{S} \subset 2^E$ be a topology on $E$ satisfying (1) and (2), then for each $x \in E$, $\cn_{(E, \mathcal{S})}(x) = \cn_{(E, \mathcal{T})}(x)$. By \autoref{proposition:neighbourhoodcharacteristic}, $\mathcal{S} = \mathcal{T}$.
\end{proof}
@@ -209,5 +219,5 @@
Let $E$ be a TVS over $K \in \RC$, then $E$ is locally connected.
\end{proposition}
\begin{proof}
Let $U \in \cn(0)$ be radial, then for any $y \in U$, the mapping $t \mapsto ty$ is a path from $0$ to $y$ contained in $U$. Thus $U$ is path-connected. By \ref{proposition:tvs-0-neighbourhood-base}, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.
Let $U \in \cn(0)$ be radial, then for any $y \in U$, the mapping $t \mapsto ty$ is a path from $0$ to $y$ contained in $U$. Thus $U$ is path-connected. By \autoref{proposition:tvs-0-neighbourhood-base}, the radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods, so the path-connected neighbourhoods of $0$ forms a fundamental system as well.
\end{proof}

2
src/fa/tvs/dual.tex
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@@ -15,12 +15,14 @@
\[
\im{\dpb{x, \phi}{E}} = \re{-i \dpb{x, \phi}{E}} = -\dpb{ix, u}{E}
\]
so (2) holds.
(2): Since $u \in \hom(E; \real)$, so is $\phi$. On the other hand, for any $x \in E$,
\[
\dpb{ix, \phi}{E} = \dpb{ix, u}{E} - i\dpb{-x, u}{E} = i(\dpb{x, u}{E} - i \dpb{ix, u}{E}) = i \dpb{x, \phi}{E}
\]
\end{proof}
\begin{definition}[Topological Dual]

22
src/fa/tvs/index.tex
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@@ -1,14 +1,14 @@
\chapter{Topological Vector Spaces}
\label{chap:tvs}
\input{./src/fa/tvs/definition.tex}
\input{./src/fa/tvs/metric.tex}
\input{./src/fa/tvs/bounded.tex}
\input{./src/fa/tvs/dual.tex}
\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}
\input{./definition.tex}
\input{./metric.tex}
\input{./bounded.tex}
\input{./dual.tex}
\input{./continuous.tex}
\input{./quotient.tex}
\input{./completion.tex}
\input{./complete-metric.tex}
\input{./projective.tex}
\input{./inductive.tex}
\input{./spaces-of-linear.tex}

11
src/fa/tvs/inductive.tex
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@@ -17,7 +17,8 @@
\[
\mathcal{B} = \bracs{U \subset E|U \text{ radial, circled}, T_i^{-1}(U) \in \cn_{E_i}(0) \forall i \in I}
\]
To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \ref{proposition:tvs-0-neighbourhood-base}.
To see that $\mathcal{B}$ is a fundamental system of neighbourhoods at $0$ for a vector space topology on $E$, it is sufficient to verify the following and apply \autoref{proposition:tvs-0-neighbourhood-base}.
\begin{enumerate}
\item[(TVB1)] Every set in $\mathcal{B}$ is radial and circled by definition.
\item[(TVB2)] For any $U \in \mathcal{B}$, $U$ is circled, so $\frac{1}{2}U + \frac{1}{2}U \subset U$. Since $\frac{1}{2}U$ is also circled and radial, $\frac{1}{2}U \in \mathcal{B}$.
@@ -47,6 +48,7 @@
}
\]
\item[(U)] For any pair $(F, \bracsn{S^i_F}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique $S \in L({E, F})$ such that the following diagram commutes
\[
@@ -56,17 +58,18 @@
}
\]
for all $i \in I$.
\item For any TVS $F$ and $T \in \hom(E; F)$, $T \in L(E; F)$ if and only if $T \circ T^i_E \in L(E_i; F)$ for all $i \in I$.
\end{enumerate}
The pair $(E, \bracsn{T^i_E}_{i \in I})$ is the \textbf{inductive limit} of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$.
\end{definition}
\begin{proof}
Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\ref{proposition:module-direct-limit}). Equip $E$ with the inductive topology (\ref{definition:tvs-inductive}) induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
Let $(E, \bracsn{T^i_E}_{i \in I})$ be the direct limit of $(\seqi{E}, \bracsn{T^i_j| i, j \in I, i \lesssim j})$ as vector spaces over $K$ (\autoref{proposition:module-direct-limit}). Equip $E$ with the \hyperref[inductive topology]{definition:tvs-inductive} induced by $\bracsn{T^i_E}_{i \in I}$, then $(E, \bracsn{T^i_E}_{i \in I})$ satisfies (1), (2), and (3).
(U): By (U) of \ref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \ref{definition:tvs-inductive}, $S \in L(E; F)$.
(U): By (U) of \autoref{proposition:module-direct-limit}, there exists a unique $S \in \hom(E; F)$ such that the given diagram commutes. By (4) of \autoref{definition:tvs-inductive}, $S \in L(E; F)$.
(5): By (5) of \ref{definition:tvs-inductive}.
(5): By (5) of \autoref{definition:tvs-inductive}.
\end{proof}
\begin{definition}[Strict]

25
src/fa/tvs/metric.tex
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@@ -19,6 +19,7 @@
\[
d_i: E \times E \quad (x, y) \mapsto \rho_i(x - y)
\]
then $d_i$ is a pseudometric. The uniform topology on $E$ induced by $\seqi{d}$ is the \textbf{topology induced by $\seqi{\rho}$}, and
\begin{enumerate}
\item The topology induced by $\seqi{\rho}$ is a vector space topology.
@@ -27,11 +28,12 @@
\[
\bracs{\bigcap_{j \in J}B_j(0, r)|J \subset I \text{ finite}, r > 0}
\]
is a fundamental system of neighbourhoods at $0$.
\end{enumerate}
\end{definition}
\begin{proof}
(3): By \ref{definition:pseudometric-uniformity}.
(3): By \autoref{definition:pseudometric-uniformity}.
(2): Each $d_i$ is translation-invariant.
@@ -42,16 +44,19 @@
\[
\lambda x - \lambda' x' = \lambda(x - x') + (\lambda - \lambda')x'
\]
Let $i \in I$ and $\eps > 0$. By (PN5) then there exists $\delta > 0$ such that if $\rho_i(x - x') < \delta$, then $\rho_i(\lambda (x - x')) < \eps$.
On the other hand,
\[
(\lambda - \lambda')x' = (\lambda - \lambda')x + (\lambda - \lambda')(x' - x)
\]
By (PN4), there exists $\delta' \in (0, 1]$ such that if $\abs{\lambda - \lambda'} < \delta'$, then $\rho_i((\lambda - \lambda')x) < \eps$. In which case, since $\delta' \le 1$, (PN2) implies that
\[
\rho_i((\lambda - \lambda')x') < \eps + \rho_i(x' - x) < 2\eps
\]
Therefore $\rho_i(\lambda x - \lambda' x') < 3\eps$.
\end{enumerate}
\end{proof}
@@ -68,7 +73,7 @@
\end{enumerate}
\end{proposition}
\begin{proof}
$(4) \Rightarrow (1)$: By \ref{definition:tvs-pseudonorm-topology}, for each $r > 0$, $\rho^{-1}([0, r)) \in \cn_E(0)$. Thus for any $x, y \in E$, if $x - y \in \rho^{-1}([0, r))$, then $\abs{\rho(x) - \rho(y)} \le r$. Therefore $\rho \in UC(E; [0, \infty))$.
$(4) \Rightarrow (1)$: By \autoref{definition:tvs-pseudonorm-topology}, for each $r > 0$, $\rho^{-1}([0, r)) \in \cn_E(0)$. Thus for any $x, y \in E$, if $x - y \in \rho^{-1}([0, r))$, then $\abs{\rho(x) - \rho(y)} \le r$. Therefore $\rho \in UC(E; [0, \infty))$.
\end{proof}
\begin{lemma}[{{\cite[Theorem I.6.1]{SchaeferWolff}}}]
@@ -82,16 +87,19 @@
\[
U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}
\]
\end{lemma}
\begin{proof}
For each $H \subset \natp$ finite, let
\[
U_H = \sum_{n \in H}V_n \quad \rho_H = \sum_{n \in H}2^{-n}
\]
Define
\[
\rho: E \to [0, 1] \quad x \mapsto \inf\bracs{\rho_H|x \in U_H}
\]
then
\begin{enumerate}
\item[(PN1)] Since $0 \in \bigcap_{H \subset \natp \text{ finite}}U_H$, $\rho(0) = 0$.
@@ -99,10 +107,11 @@
\[
\lambda x \in \sum_{n \in H}\lambda U_n \subset \sum_{n \in H}U_n
\]
so $\rho(\lambda x) \le \rho(x)$.
\item[(PN3)] Let $x, y \in X$ and $M, N \subset \natp$ finite such that $x \in U_M$ and $y \in U_N$. Assume without loss of generality that $\rho_M + \rho_N < 1$, then there exists a unique $P \subset \nat$ finite such that $\rho_P = \rho_M + \rho_N$. In which case, $U_P \supset U_M + U_N$ by assumption (b). Therefore $\rho(x + y) \le \rho(x) + \rho(y)$.
\end{enumerate}
For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1} < 2^n$, so $U_{n+1} \subset \rho^{-1}([0, 2^{-n}))$ by \ref{proposition:dyadic-semigroup-order}. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}} = U_n$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
For any $x \in U_{n+1}$, $\rho(x) \le 2^{-n+1} < 2^n$, so $U_{n+1} \subset \rho^{-1}([0, 2^{-n}))$ by \autoref{proposition:dyadic-semigroup-order}. On the other hand, for any $x \in E$ with $\rho(x) < 2^{-n}$, $x \in U_{2^{-n}} = U_n$. This allows showing the remaining seminorm axioms by considering neighbourhoods of the form $\bracs{U_n|n \in \natp}$.
\begin{enumerate}
\item[(PN4)] Let $x \in X$ and $n \in \natp$. By assumption (a), there exists $\alpha > 0$ such that for any $\lambda \in K$ with $\abs{\lambda} \ge \alpha$, $x \in \lambda U_n$. Therefore for any $\lambda \in K$ with $\abs{\lambda} \le \alpha^{-1}$, $\lambda x \in U_n$, and $\rho(x) \le 2^{-n}$.
\item[(PN5)] Let $\lambda \in K$ and $n \in \natp$. By assumption (b), there exists $m \in \nat$ such that $\lambda U_{n-m} \subset \sum_{j = 1}^m U_{n-m}^j \subset U_n$.
@@ -112,7 +121,7 @@
\begin{remark}
\label{remark:tvs-sequence-pseudonorm}
As discussed in \ref{remark:uniform-sequence-pseudometric} on the proof of \ref{lemma:uniform-sequence-pseudometric}, constructing a pseudometric from entourages by building its level sets is difficult because composing symmetric entourages does not necessarily lead to symmetric entourages. The topological vector space does not have this shortcoming, and as such allows this construction.
As discussed in \autoref{remark:uniform-sequence-pseudometric} on the proof of \autoref{lemma:uniform-sequence-pseudometric}, constructing a pseudometric from entourages by building its level sets is difficult because composing symmetric entourages does not necessarily lead to symmetric entourages. The topological vector space does not have this shortcoming, and as such allows this construction.
\end{remark}
\begin{theorem}[Metrisability of Topological Vector Spaces]
@@ -127,15 +136,15 @@
\end{enumerate}
\end{theorem}
\begin{proof}
(3) $\Rightarrow$ (4): By \ref{theorem:uniform-metrisable}.
(3) $\Rightarrow$ (4): By \autoref{theorem:uniform-metrisable}.
(4) $\Rightarrow$ (1): By \ref{proposition:tvs-good-neighbourhood-base}, there exists $\seq{U_n} \subset \cn_E(0)$ circled and radial such that for each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$. By \ref{lemma:tvs-sequence-pseudonorm}, there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $N \in \natp$, $U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}$. In which case, $\rho$ induces the topology on $E$.
(4) $\Rightarrow$ (1): By \autoref{proposition:tvs-good-neighbourhood-base}, there exists $\seq{U_n} \subset \cn_E(0)$ circled and radial such that for each $n \in \natp$, $U_{n+1} + U_{n+1} \subset U_n$. By \autoref{lemma:tvs-sequence-pseudonorm}, there exists a pseudonorm $\rho: E \to [0, \infty)$ such that for each $N \in \natp$, $U_{n+1} \subset \rho^{-1}([0, 2^{-n})) \subset U_{n}$. In which case, $\rho$ induces the topology on $E$.
\end{proof}
\begin{remark}
\label{remark:tvs-metrisable}
Let $E$ be a TVS over $K \in \RC$. Similar to the case of uniform spaces, there exists a family of pseudonorms $\seqi{\rho}$ that induces the topology on $E$. Using Minkowski functionals (\ref{definition:gauge}), $\seqi{\rho}$ can be taken such that $\rho_i(\lambda x) = \abs{\lambda}\rho_i(x)$ for all $x \in E$ and $\lambda \in K$. However, a single pseudonorm with this property cannot always induce the topology even if the space is metrisable, hence the difference between pseudonorms and seminorms.
Let $E$ be a TVS over $K \in \RC$. Similar to the case of uniform spaces, there exists a family of pseudonorms $\seqi{\rho}$ that induces the topology on $E$. Using \hyperref[Minkowski functionals]{definition:gauge}, $\seqi{\rho}$ can be taken such that $\rho_i(\lambda x) = \abs{\lambda}\rho_i(x)$ for all $x \in E$ and $\lambda \in K$. However, a single pseudonorm with this property cannot always induce the topology even if the space is metrisable, hence the difference between pseudonorms and seminorms.
\end{remark}
\begin{definition}[Locally Bounded]
@@ -148,5 +157,5 @@
Let $E$ be a locally bounded TVS over $K \in \RC$, then there exists a pseudonorm $\rho: E \to [0, \infty)$ that induces the topology on $E$.
\end{proposition}
\begin{proof}
Let $U \in \cn^o(0)$ be bounded. Using \ref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is circled. For each $n \in \natp$, let $U_n = n^{-1}U$. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$, $\abs{\lambda}^{-1}U \subset V$. For any $n \in \natp$ with $n^{-1} < \abs{\lambda}^{-1}$, $U_n \subset V$. Thus $E$ admits a countable fundamental system of neighbourhoods at $0$. By \ref{theorem:tvs-metrisable}, the topology on $E$ is induced by a pseudonorm.
Let $U \in \cn^o(0)$ be bounded. Using \autoref{proposition:tvs-good-neighbourhood-base}, assume without loss of generality that $U$ is circled. For each $n \in \natp$, let $U_n = n^{-1}U$. Let $V \in \cn^o(0)$, then there exists $\lambda \in K$ such that $\lambda V \supset U$, $\abs{\lambda}^{-1}U \subset V$. For any $n \in \natp$ with $n^{-1} < \abs{\lambda}^{-1}$, $U_n \subset V$. Thus $E$ admits a countable fundamental system of neighbourhoods at $0$. By \autoref{theorem:tvs-metrisable}, the topology on $E$ is induced by a pseudonorm.
\end{proof}

17
src/fa/tvs/projective.tex
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@@ -17,31 +17,34 @@
\[
\bracs{\bigcap_{j \in J}T_j^{-1}(U_j) \bigg | J \subset I \text{ finite}, U_j \in \cn_{F_j}(0)}
\]
is a fundamental system of neighbourhoods for $E$ at $0$.
\end{enumerate}
The uniformity $\fU$ and its topology are the \textbf{projective uniformity/topology} induced by $\seqi{T}$.
\end{definition}
\begin{proof}
(1), (U): By \ref{definition:initial-uniformity}.
(1), (U): By \autoref{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}.
(5): By \autoref{definition:continuous-linear} and (4) of \autoref{definition:initial-uniformity}.
(6): By \ref{definition:initial-uniformity}.
(6): By \autoref{definition:initial-uniformity}.
\end{proof}
\begin{definition}[Projective Limit of Topological Vector Spaces]
@@ -57,6 +60,7 @@
E \ar@{->}[u]^{T^E_i} \ar@{->}[ru]_{T^E_j} &
}
\]
\item[(U)] For any pair $(F, \bracsn{S^F_i}_{i \in I})$ satisfying (1), (2), and (3), there exists a unique $S \in L(F; E)$ such that the following diagram commutes
\[
@@ -66,17 +70,18 @@
}
\]
for all $i \in I$.
\item For any TVS $F$ over $K$ and $S \in \hom(F; E)$, $S \in L(F; E)$ if and only if $T^E_i \circ S \in L(F; E_i)$ for all $i \in I$.
\end{enumerate}
The pair $(E, \bracsn{T^E_i}_{i \in I})$ is the \textbf{projective limit} of $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$.
\end{definition}
\begin{proof}
Let $(E, \bracsn{T^E_i}_{i \in I})$ be the inverse limit of $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$ as $K$-vector spaces (\ref{proposition:module-inverse-limit}).
Let $(E, \bracsn{T^E_i}_{i \in I})$ be the inverse limit of $(\seqi{E}, \bracsn{T^i_j|i, j \in I, i \lesssim j})$ as $K$-vector spaces (\autoref{proposition:module-inverse-limit}).
Equip $E$ with the projective topology generated by $\bracsn{T^E_i}_{i \in I}$, then $(E, \bracsn{T^E_i}_{i \in I})$ satisfies (1), (2), and (3).
(5): By (5) of \ref{definition:tvs-initial}.
(5): By (5) of \autoref{definition:tvs-initial}.
(U): By (U) of \ref{proposition:module-inverse-limit}, there exists a unique $S \in \hom(F; E)$ such that the given diagram commutes. By (4), $S \in L(F; E)$.
(U): By (U) of \autoref{proposition:module-inverse-limit}, there exists a unique $S \in \hom(F; E)$ such that the given diagram commutes. By (4), $S \in L(F; E)$.
\end{proof}

13
src/fa/tvs/quotient.tex
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@@ -15,6 +15,7 @@
\widetilde E \ar@{->}[r]_{\tilde f} & F
}
\]
If $F$ is a TVS over $K$ and $f \in L(E; F)$, then $\td f \in L(E; F)$.
\end{enumerate}
The space $\td E = E/M$ is the \textbf{quotient} of $E$ by $M$.
@@ -22,18 +23,19 @@
\begin{proof}
Let $\td E = E/M$ be the algebraic quotient of $E$ by $M$, and equip it with the quotient topology by $\pi$.
(1): By \ref{definition:quotient-topology}, for each $\pi(U) \subset E/M$, $\pi(U)$ is open if and only if $U$ is open. Since the topology on $E$ is translation-invariant, so is the quotient topology on $E/M$. Let
(1): By \autoref{definition:quotient-topology}, for each $\pi(U) \subset E/M$, $\pi(U)$ is open if and only if $U$ is open. Since the topology on $E$ is translation-invariant, so is the quotient topology on $E/M$. Let
\[
\fB = \bracs{\pi(U)| U \in \cn(0) \text{ circled and radial}}
\]
Since the circled and radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods at $0$ by \ref{proposition:tvs-good-neighbourhood-base}, $\fB$ is a fundamental system of neighbourhoods at $0$ for the quotient topology on $E/M$. In addition,
Since the circled and radial neighbourhoods of $0$ forms a fundamental system of neighbourhoods at $0$ by \autoref{proposition:tvs-good-neighbourhood-base}, $\fB$ is a fundamental system of neighbourhoods at $0$ for the quotient topology on $E/M$. In addition,
\begin{enumerate}
\item[(TVB1)] Let $U \in \cn(0)$ be circled and radial. For any $\lambda \in K$ with $\abs{\lambda} \le 1$, $\lambda \pi(U) = \pi(\lambda U) \subset \pi(U)$, so $\pi(U)$ is also circled. For any $x + M \in E/M$, there exists $\lambda \in K$ such that $x \in \lambda U$. In which case, $x \in \lambda U + M = \pi(U)$, so $\pi(U)$ is also radial.
\item[(TVB2)] For any $U \in \cn(0)$ circled and radial, by \ref{proposition:tvs-good-neighbourhood-base}, there exists $W \in \cn(0)$ such that $W + W \subset U$. In which case, $\pi(W) + \pi(W) \subset \pi(U)$.
\item[(TVB2)] For any $U \in \cn(0)$ circled and radial, by \autoref{proposition:tvs-good-neighbourhood-base}, there exists $W \in \cn(0)$ such that $W + W \subset U$. In which case, $\pi(W) + \pi(W) \subset \pi(U)$.
\end{enumerate}
By \ref{proposition:tvs-0-neighbourhood-base}, there exists a unique translation-invariant topology on $E/M$ such that $\fB$ is a fundamental system of neighbourhoods at $0$, which must be the quotient topology on $E/M$. In which case, the quotient topology is a vector space topology by (3) of \ref{proposition:tvs-0-neighbourhood-base}.
By \autoref{proposition:tvs-0-neighbourhood-base}, there exists a unique translation-invariant topology on $E/M$ such that $\fB$ is a fundamental system of neighbourhoods at $0$, which must be the quotient topology on $E/M$. In which case, the quotient topology is a vector space topology by (3) of \autoref{proposition:tvs-0-neighbourhood-base}.
(2), (3), (U): By \ref{definition:quotient-topology}.
(2), (3), (U): By \autoref{definition:quotient-topology}.
\end{proof}
\begin{proposition}
@@ -45,5 +47,6 @@
\[
M = \bigcap_{V \in \cn(0)}M + V
\]
which is equivalent to $E/M$ being Hausdorff.
\end{proof}

16
src/fa/tvs/spaces-of-linear.tex
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@@ -5,11 +5,12 @@
\label{proposition:tvs-set-uniformity}
Let $T$ be a set, $\mathfrak{S} \subset 2^T$ be an upward-directed system of sets, $F$ be a TVS over $K \in \RC$, then
\begin{enumerate}
\item The $\mathfrak{S}$-uniformity on $F^T$ (\ref{definition:set-uniform}) is translation invariant.
\item The $\mathfrak{S}$-uniformity on $F^T$ (\autoref{definition:set-uniform}) 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:
@@ -19,7 +20,7 @@
\end{enumerate}
\end{proposition}
\begin{proof}
(1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. Using \ref{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.
(1): Let $U \subset F \times F$ be an entourage and $S \in \mathfrak{S}$. 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 $f, g, f', g' \in \cf$, $S \in \mathfrak{S}$, 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 $(f + g, f' + g') \in E(S, U) \cap \cf$.
@@ -30,10 +31,11 @@
\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 \ref{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
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}
@@ -50,25 +52,30 @@
\[
I: B_{\mathfrak{S}}^k(E; B_{\mathfrak{S}}(E; F)) \to B^{k+1}_{\mathfrak{S}}(E; F)
\]
defined by
\[
(IT)(x_1, \cdots, x_{k+1}) = T(x_1, \cdots, x_k)(x_{k+1})
\]
is an isomorphism.
\item The map
\[
I: \underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} \to B^k_{\mathfrak{S}}(E; F)
\]
defined by
\[
IT(x_1, \cdots, x_k) = T(x_1)\cdots (x_k)
\]
is an isomorphism.
\end{enumerate}
which allows the identification
\[
\underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k \text{ times}} = B^k_{\mathfrak{S}}(E; F)
\]
under the map $I$ in (2).
\end{proposition}
\begin{proof}
@@ -76,10 +83,12 @@
\[
I^{-1}T: E \to B_{\mathfrak{S}}(E; F) \quad x \mapsto T(x, \cdot)
\]
Let $(x_1, \cdots, x_k) \in E^k$ and $S \in \mathfrak{S}$. Since $\mathfrak{S}$ is upward-directed and contains all singletons, assume without loss of generality that $\bracsn{x_j}_1^k \subset S$. In which case,
\[
T(x_1, \cdots, x_k, S) \subset T(S^{k+1}) \in B(F)
\]
by assumption. Thus $I^{-1}T(x_1, \cdots, x_k) \in B_{\mathfrak{S}}(E; F)$.
In addition, for any $S_1 \in \mathfrak{S}$ and entourage $E(S_2, U)$ of $B_{\mathfrak{S}}(E; F)$ where $S_2 \in \mathfrak{S}$ and $U$ is an entourage of $F$, there exists $S \in \mathfrak{S}$ 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 $I^{-1}T(S^k) \in B(B_{\mathfrak{S}}(E; F))$. Thus $I^{-1}T \in B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F))$.
@@ -92,6 +101,7 @@
\[
\underbrace{B_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; \cdots)))}_{k+1 \text{ times}} = B^k_{\mathfrak{S}}(E; B_{\mathfrak{S}}(E; F)) = B^{k+1}_{\mathfrak{S}}(E; F)
\]
Thus (2) holds for all $k \in \natp$.
\end{proof}