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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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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
View File

@@ -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
View File

@@ -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}