[PATCH] Updated
Francisco Figueirido
francisco at giantclam.net
Mon Feb 24 21:31:48 CST 2014
# HG changeset patch
# User francisco <francisco at giantclam.net>
# Date 1393296985 18000
# Node ID 317d7cb86c8f51156ed5cf966c4fc0a7e28b323f
# Parent 129b7f5b99664b38e251dd2d64a822f8e9c2f8db
Updated
diff r 129b7f5b9966 r 317d7cb86c8f rw2.tex
 a/rw2.tex Mon Feb 24 00:22:13 2014 0500
+++ b/rw2.tex Mon Feb 24 21:56:25 2014 0500
@@ 9089,13 +9089,53 @@
\]
for any nonnegative $t$ (the case where $t=0$ is trivially true).
Notice that if $x\in A$ then $p_A(x)\le 1$.

Now consider a point $x\not\in A$ and let $\YY\subset\XX$ be the onedimensional subspace generated by $x$. Take any $0<\alpha<p_A(x)$ and define the linear functional
\[
\phi_A( t\,x ) = t\,\alpha \le \abs{t}\,\alpha < p_A( t\, x ).
\]
The HahnBanach Theorem then implies that $\phi_A$ can be extended to the whole of $\XX$ maintaining the bound.
+Notice that:
+\begin{itemize}
+\item If $x\in A$ then $p_A(x)\le 1$.
+\item If $0\in A$ for any $x\in A$ and any $s>1$ we have that $x/s = x/s + (11/s)\,0 \in A$.
+\item If $0\in A$ and $x\not\in A$ then $x/s\in A$ implies that $s>1$ for if $x/s\in A$ for some $s<1$ we would then have, by the above and choosing $t=1/s>1$, that $x=(x/s)/t\in A$ which is a contradiction. This implies that $p_A(x)\ge 1$ for all $x\not\in A$.
+\item If $\XX$ is a topological vector space, $0\in A$ and $x\not\in\closure{A}$ then there is some $\epsilon>0$ such that $x/(1+h)\not\in A$ for any $0\le h\le\epsilon$ and therefore $p_A(x)\ge 1+\epsilon>1$. Conversely, if $p_A(x)=1$ then there is some sequence $s_n$ converging from above to $1$ such that $x/s_n\in A$ and therefore its limit, $x$, is in $\closure{A}$. We thus conclude that $\closure{A}=\{ x : p_A(x) \le 1 \}$.
+\end{itemize}
+Assume $0\in A$ and consider a point $x\not\in A$; let $\YY\subset\XX$ be the onedimensional subspace generated by $x$. Since $p_A(x)\ge 1$ we can define the nonzero linear functional
+\[
+\phi_A( t\,x ) = t\,p_A(x) .
+\]
+If $t<0$ we clearly have $\phi_A(t x) < 0 \le p_A( t x )$; and when $t\ge 0$ we have $\phi(t x) = t p_A(x) = p_A(t x)$, so the HahnBanach Theorem allows us to extend $\phi_A$ to a linear functional defined on the whole $\XX$ space, call it $\phi$, that satisfies the bound $\phi(u)\le p_A(u)$ for any $u\in \XX$. For this extension we then have
+\begin{itemize}
+\item $x\in \{ y : \phi(y)\ge 1\}$.
+\item If $x\not\in\closure{A}$ then $x\in \{ y : \phi(y) > 1\}$.
+\item $A \subset \{ y : \phi(y) \le 1 \}$
+\end{itemize}
+so that the (convex) space defined by the condition $\{ y : \phi(y) = 1\}$ separates $A$ from $x$ (although $x$ could be in the intersection).
+
+\noindent For any linear functional like $\phi$ one defines the kernel of $\phi$ as
+\[
+\Kernel(\phi) = \bigl\{ z : \phi(z) = 0 \bigr\}.
+\]
+Notice that the kernel of $\phi$ has codimension 1: take any point $x$ not in the kernel and consider any other point $y\in\XX$. Then we have $\phi(x)\ne 0$ and therefore
+\[
+u = y  \dfrac{\phi(y)}{\phi(x)} x \in \XX\quad\text{and}\quad \phi(u) = 0
+\]
+so that $y = u + (\phi(y)/\phi(x)) x$ which proves that any point in $\XX$ can be written as the sum of a point $u$ in the kernel of $\phi$ plus a point in the onedimensional linear space generated by $x$.
+Moreover, clearly for any two points $z_1$ and $z_2$ such that $\phi(z_1)=\phi(z_2)=1$ the difference $z_1z_2\in\Kernel(\phi)$ and therefore if we choose some arbitrary point $z_1$ with $\phi(z_1)=1$ we can write
+\[
+\bigl\{ z : \phi(z) = 1 \bigr\} = \bigl\{ w + z_1, \, w\in\Kernel(\phi) \bigr\} = T_{z_1}\left( \Kernel(\phi) \right)
+\]
+where $T_u$ is the translation operator $T_u(y) = y + u$, which is essentially the definition of a hyperplane.
+
+Suppose now that $\XX$ is a normed space (see Appendix~\ref{ap:normedspace}) with norm $x\rightarrow\norm{x}$ and $x\rightarrow p_A(x)$ is the Minkowski functional of a convex set $A$ that contains $0$. Since
+\[
+p_A\left( \dfrac{x}{\norm{x}} \right) = \dfrac{1}{\norm{x}}\, p_A(x) \Rightarrow p_A(x) = \norm{x}\, p_A\left( \dfrac{x}{\norm{x}} \right)
+\]
+we can conclude that the extension of a bounded functional will be continuous if
+\[
+\sup_{u : \norm{u}=1} p_A(u) < \infty.
+\]
+Indeed, in this case we have, for any $x\in\XX$,
+\begin{align*}
+\phi(x) \ge 0 & \Rightarrow \abs{\phi(x)} = \phi(x) \le p_A(x) \le \left( \sup_{u : \norm{u}=1} p_A(u) \right)\,\norm{x} \\
+\phi(x) < 0 & \Rightarrow \abs{\phi(x)} = \phi(x) = \phi(x) \le p_A(x) \le \left( \sup_{u : \norm{u}=1} p_A(u) \right)\,\norm{x}
+\end{align*}
\chapter{RadonNikodym Theorem\\Conditional Expectation}
\label{ap:radonnikodym}
diff r 129b7f5b9966 r 317d7cb86c8f rw2mac.tex
 a/rw2mac.tex Mon Feb 24 00:22:13 2014 0500
+++ b/rw2mac.tex Mon Feb 24 21:56:25 2014 0500
@@ 213,6 +213,8 @@
%% Bold symbol macro for standard LaTeX users
\providecommand{\boldsymbol}[1]{\mbox{\boldmath $#1$}}
+\def\Kernel(#1){\textsf{Kern}\left({#1}\right)}
+
\bibliographystyle{alpha}
\makeatother
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