169 lines
7.9 KiB
TeX
169 lines
7.9 KiB
TeX
\section{Seminorms}
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\label{section:seminorms}
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\begin{definition}[Convex]
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\label{definition:convex}
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Let $E$ be a vector space over $K \in \RC$, then $A \subset E$ is \textbf{convex} if for any $x, y \in A$, $\bracs{\lambda x + (1 - \lambda) y| \lambda \in [0, 1]} \subset A$.
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\end{definition}
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\begin{lemma}[{{\cite[II.1.1]{SchaeferWolff}}}]
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\label{lemma:convex-interior}
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Let $E$ be a TVS over $K \in \RC$, $A \subset E$ be convex, $x \in A^o$, and $y \in \ol{A}$, then
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\[
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\bracs{tx + (1 - t)y|t \in (0, 1)} \subset A^o
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\]
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\end{lemma}
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\begin{proof}
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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 $\alpha z \in A$.
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Let $\mu = \alpha/(\alpha - 1)$, then since $\alpha < 0$, $\mu \in (0, 1)$ and
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\[
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\mu z + (1 - \mu)\alpha z = \frac{\alpha z}{\alpha - 1} + \frac{(\alpha - 1 - \alpha)\alpha z}{\alpha - 1} = 0
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\]
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By (TVS1) and (TVS2),
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\[
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U = \underbrace{\bracs{\mu w + (1 - \mu)\alpha z|w \in A^o}}_{\subset A} \in \cn^o(0)
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\]
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so $0 \in A^o$.
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\end{proof}
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\begin{lemma}
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\label{lemma:convex-gymnastics}
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Let $E$ be a TVS over $K \in \RC$, $A, B \subset E$ be convex, then the following sets are convex:
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\begin{enumerate}
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\item $A^o$.
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\item $\ol{A}$.
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\item $A + B$.
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\item For any $\lambda \in K$, $\lambda A$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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(1): By \autoref{lemma:convex-interior}.
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(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,
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\[
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\fU = \bracs{tE + (1 - t)F|t \in [0, 1], E \in \fF, F \in \mathfrak{G}} \subset 2^A
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\]
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converges to $tx + (1 - t)y$ by (TVS1) and (TVS2). Hence $tx + (1 - t)y \in \ol{A}$.
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\end{proof}
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\begin{proposition}[{{\cite[II.1.3]{SchaeferWolff}}}]
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\label{proposition:convex-interior-closure}
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Let $E$ be a TVS over $K \in \RC$ and $A \subset E$ be convex. If $A^o \ne \emptyset$, then $\ol{A} = \ol{A^o}$.
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\end{proposition}
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\begin{proof}
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Since $A^o \subset A$, $\ol{A^o} \subset \ol{A}$. Let $x \in A^o$, then for any $y \in \ol{A}$,
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\[
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y \in \ol{\bracs{tx + (1 - t)y|t \in (0, 1)}} \subset \ol{A^o}
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\]
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by \autoref{lemma:convex-interior}.
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\end{proof}
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\begin{definition}[Sublinear Functional]
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\label{definition:sublinear-functional}
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Let $E$ be a vector space over $K \in \RC$, then a \textbf{sublinear functional} is a mapping $\rho: E \to \real$ such that:
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\begin{enumerate}
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\item $\rho(0) = 0$.
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\item For any $x \in E$ and $\lambda \ge 0$, $\rho(\lambda x) = \lambda \rho(x)$.
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\item For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
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\end{enumerate}
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\end{definition}
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\begin{definition}[Seminorm]
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\label{definition:seminorm}
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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:
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\begin{enumerate}
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\item[(SN1)] $\rho(0) = 0$.
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\item[(SN2)] For any $x \in E$ and $\lambda \in K$, $\rho(\lambda x) = \abs{\lambda} \rho(x)$.
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\item[(SN3)] For any $x, y \in E$, $\rho(x + y) \le \rho(x) + \rho(y)$.
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\end{enumerate}
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\end{definition}
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\begin{lemma}
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\label{lemma:continuous-seminorm}
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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:
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\begin{enumerate}
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\item $[\cdot]$ is uniformly continuous.
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\item $[\cdot]$ is continuous.
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\item $[\cdot]$ is continuous at $0$.
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\end{enumerate}
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\end{lemma}
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\begin{proof}
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$(3) \Rightarrow (1)$: 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.
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\end{proof}
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\begin{definition}[Topology Induced by Seminorm]
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\label{definition:seminorm-topology}
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Let $E$ be a vector space over $K \in \RC$ and $\seqi{[\cdot]}$ be seminorms, then:
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\begin{enumerate}
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\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.
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\item The topology induced by $\seqi{d}$ makes $E$ a topological vector space.
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\item For each $i \in I$, $[\cdot]_i: E \to [0, \infty)$ is continuous.
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\end{enumerate}
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The topology induced by $\seqi{d}$ is the \textbf{vector space topology induced by} $\seqi{[\cdot]}$. In addition,
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\begin{enumerate}
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\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 $\seqi{[\cdot]}$.
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\end{enumerate}
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\end{definition}
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\begin{definition}[Gauge/Minkowski Functional]
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\label{definition:gauge}
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Let $E$ be a vector space over $K \in \RC$ and $A \subset E$ be a radial set, then the mapping
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\[
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[\cdot]_A: E \to [0, \infty) \quad x \mapsto \inf\bracsn{\lambda > 0| \lambda^{-1}x \in A}
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\]
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is the \textbf{gauge/Minkowski functional} of $A$, and
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\begin{enumerate}
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\item For any $x \in E$ and $\lambda \ge 0$, $[\lambda x]_A = \lambda [x]_A$.
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\item If $A$ is convex, then for any $x, y \in E$, $[x + y]_A \le [x]_A + [y]_A$.
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\item If $A$ is circled, then for any $x \in E$ and $\lambda \in K$, $[\lambda x]_A = \abs{\lambda}[x]_A$.
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\end{enumerate}
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In particular,
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\begin{enumerate}
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\item[(4)] If $A$ is convex, then $[\cdot]_A$ is a sublinear functional.
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\item[(5)] If $A$ is convex and circled, then $[\cdot]_A$ is a seminorm.
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\end{enumerate}
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\end{definition}
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\begin{proof}
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(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
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\[
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(\lambda + \mu)^{-1}(x + y) = t\lambda^{-1}x + (1 - t)\mu^{-1}y
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\]
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then $(\lambda + \mu)^{-1} \in A$, and $\lambda + \mu \ge [x + y]_A$. Thus $[x + y]_A \le [x]_A + [y]_A$.
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\end{proof}
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\begin{definition}[Locally Convex Space]
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\label{definition:locally-convex}
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Let $E$ be a TVS over $\RC$, then the following are equivalent:
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\begin{enumerate}
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\item There exists a fundamental system of neighborhoods at $0$ consisting of convex sets.
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\item There exists a fundamental system of neighbourhoods at $0$ consisting of convex, circled, and radial sets.
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\item There exists a family of seminorms $\seqi{[\cdot]}$ that induces the topology on $E$.
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\end{enumerate}
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If the above holds, then $E$ is a \textbf{locally convex} space.
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\end{definition}
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\begin{proof}
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$(1) \Rightarrow (2)$: 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.
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$(2) \Rightarrow (3)$: 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.
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$(3) \Rightarrow (1)$: For each $i \in I$ and $r > 0$, $\bracs{x \in E| [x]_i < r}$ is convex.
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\end{proof}
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