Updated content.

src/fa/lc/convex.tex

@@ -0,0 +1,101 @@
+\section{Seminorms}
+\label{section:seminorms}
+
+\begin{definition}[Convex]
+\label{definition:convex}
+    Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.
+\end{definition}
+
+\begin{definition}[Sublinear Functional]
+\label{definition:sublinear-functional}
+    Let $E$ be a vector space over $K \in \RC$, then a \textbf{sublinear functional} is a mapping $\rho: E \to \real$ such that:
+    \begin{enumerate}
+        \item $\rho(0) = 0$.
+        \item For any $x \in E$ and $\lambda \ge 0$, $\rho(\lambda x) = \lambda \rho(x)$.
+        \item For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
+    \end{enumerate}
+\end{definition}
+
+
+\begin{definition}[Seminorm]
+\label{definition:seminorm}
+    Let $E$ be a vector space over $K \in \RC$, then a \textbf{seminorm} on $E$ is a mapping $\rho: E \to [0, \infty)$ such that:
+    \begin{enumerate}
+        \item $\rho(0) = 0$.
+        \item For any $x \in E$ and $\lambda \in K$, $\rho(\lambda x) = \abs{\lambda} \rho(x)$.
+        \item For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
+    \end{enumerate}
+\end{definition}
+
+\begin{lemma}
+\label{lemma:continuous-seminorm}
+    Let $E$ be a TVS over $K \in \RC$ and $[\cdot]: E \times E \to [0, \infty)$ be a seminorm on $E$, then the following are equivalent:
+    \begin{enumerate}
+        \item $[\cdot]$ is uniformly continuous.
+        \item $[\cdot]$ is continuous.
+        \item $[\cdot]$ is continuous at $0$.
+    \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.
+\end{proof}
+
+
+\begin{definition}[Topology Induced by Seminorm]
+\label{definition:seminorm-topology}
+    Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then:
+    \begin{enumerate}
+        \item For each $i \in I$, $d_i: E \times E \to [0, \infty)$ defined by $(x, y) \mapsto [x - y]_i$ is a pseudo-metric.
+        \item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
+        \item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
+    \end{enumerate}
+    The topology induced by $\seqi{d}$ is the \textbf{vector space topology induced by} $\seqi{[\cdot]}$. In addition,
+    \begin{enumerate}
+        \item[(U)] For any family $\seqj{[\cdot]}$ of seminorms continuous on $E$, the vector space topology induced by $\seqj{[\cdot]}$ is contained in the vector space topology induced by $\seqi{[\cdot]}$.
+    \end{enumerate}
+\end{definition}
+
+
+\begin{definition}[Gauge/Minkowski Functional]
+\label{definition:gauge}
+    Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be a radial set, then the mapping
+    \[
+    [\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
+    \]
+    is the \textbf{gauge/Minkowski functional} of $A$, and
+    \begin{enumerate}
+        \item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
+        \item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
+        \item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
+    \end{enumerate}
+    In particular,
+    \begin{enumerate}
+        \item If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
+        \item If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
+    \end{enumerate}
+\end{definition}
+\begin{proof}
+    (2): Let $\lambda, \mu > 0$ such that $\lambda^{-1}x, \mu^{-1}y \in A$. By convexity, $t\lambda^{-1} + (1 - t)\mu^{-1}y \in A$ for all $t \in [0, 1]$. Let $t \in [0, 1]$ such that
+    \[
+    (\lambda + \mu)^{-1}(x + y) = t\lambda^{-1}x + (1 - t)\mu^{-1}y
+    \]
+    then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.
+\end{proof}
+
+\begin{definition}[Locally Convex Space]
+\label{definition:locally-convex}
+    Let $E$ be a TVS over $\RC$, then the following are equivalent:
+    \begin{enumerate}
+        \item There exists a fundamental system of neighborhoods at $0$ consisting of convex sets.
+        \item There exists a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets.
+        \item There exists a family of seminorms $\seqi{[\cdot]}$ that induces the topology on $E$.
+    \end{enumerate}
+    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.
+
+    $(2) \Rightarrow (3)$: 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.
+
+    $(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.
+\end{proof}