From patchwork Tue Feb 25 03:31:48 2014 Content-Type: text/plain; charset="utf-8" MIME-Version: 1.0 Content-Transfer-Encoding: 7bit Subject: Updated From: Francisco Figueirido X-Patchwork-Id: 3757 Message-Id: <317d7cb86c8f51156ed5.1393299108@miniclam> To: mercurial-devel@selenic.com Cc: francisco@imagine-sw.com Date: Mon, 24 Feb 2014 22:31:48 -0500 # HG changeset patch # User francisco # 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 non-negative $t$ (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 one-dimensional subspace generated by $x$. Take any $0<\alpha1$ we have that $x/s = x/s + (1-1/s)\,0 \in A$. +\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 one-dimensional subspace generated by $x$. Since $p_A(x)\ge 1$ we can define the non-zero 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 Hahn-Banach 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 one-dimensional 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 $z_1-z_2\in\Kernel(\phi)$ 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:normed-space}) 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{Radon-Nikodym Theorem\\Conditional Expectation} \label{ap:radon-nikodym} 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