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Cleaned up citation stlye changes.

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Bokuan Li
2026-06-24 14:18:27 -04:00
parent 4764ed2fc4
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8
src/fa/lc/convex.tex
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@@ -29,7 +29,7 @@
\begin{lemma}[{{\cite[II.1.1]{SchaeferWolff}}}] \begin{lemma}
\label{lemma:convex-interior} \label{lemma:convex-interior}
Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^o$, and $y \in \ol{A}$, then Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^o$, and $y \in \ol{A}$, then
\[ \[
@@ -37,7 +37,7 @@
\] \]
\end{lemma} \end{lemma}
\begin{proof} \begin{proof}[Proof, {{\cite[II.1.1]{SchaeferWolff}}}. ]
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 $\alpha z \in A$. 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 $\alpha z \in A$.
Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
@@ -74,11 +74,11 @@
converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$. converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
\end{proof} \end{proof}
\begin{proposition}[{{\cite[II.1.3]{SchaeferWolff}}}] \begin{proposition}
\label{proposition:convex-interior-closure} \label{proposition:convex-interior-closure}
Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^o \ne \emptyset$, then $\ol{A} = \ol{A^o}$. Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^o \ne \emptyset$, then $\ol{A} = \ol{A^o}$.
\end{proposition} \end{proposition}
\begin{proof} \begin{proof}[Proof, {{\cite[II.1.3]{SchaeferWolff}}}. ]
Since $A^o \subset A$, $\ol{A^o} \subset \ol{A}$. Let $x \in A^o$, then for any $y \in \ol{A}$, Since $A^o \subset A$, $\ol{A^o} \subset \ol{A}$. Let $x \in A^o$, then for any $y \in \ol{A}$,
\[ \[
y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o} y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}