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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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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 $M = \bracs{0}$.