Exercise 3.3

Question

enewcommand{\labelenumi}{$(	extit{\alph{enumi}}\,)$}
\def\eg{	extit{e.g.} }
\def\ie{	extit{i.e.} }
\def\contains{\supset}
\def\Def{	riangleq}
\def\N{\usualSet{N}}
\def\R{\usualSet{R}}
\def\C{\usualSet{C}}
\def\minus{\insmall{-}}

Suppose X is a real vector space (without topology). %
Call a point $x_0\in A \subset X$ an internal point of $A$ if %
$A- x_0$ is an absorbing set. %
%
%enewcommand{\labelenumi}{(\alph{enumi})}
%
  \begin{enumerate}
    \item{%
      Suppose $A$ and $B$ are disjoint convex sets in $X$, %
      and $A$ has an internal point. %
      Prove that there is a nonconstant linear functional $\Lambda$ such that %
      $\Lambda(A)\cap \Lambda(B)$ contains at most one point. %
      (The proof is similar to that of Theorem 3.4) %
    }%
    \item{%
      Show (with $X=\R^2$, for example) that it may not possible to have %
      $\Lambda (A)$ and $\Lambda (B)$ disjoint, under the hypotheses of (a). %
    }
  \end{enumerate}

Cordier · Accepted Answer

enewcommand{\labelenumi}{$( extit{\alph{enumi}}\,)$} \def\eg{ extit{e.g.} } \def\ie{ extit{i.e.} } \def\contains{\supset} \def\Def{ riangleq} \def\N{\usualSet{N}} \def\R{\usualSet{R}} \def\C{\usualSet{C}} \def\minus{\insmall{-}} \begin{proof}% Take $A$ and $B$ as in (a); the trivial case $B=\emptyset$ is discarded. % Since $A-x_0$ is absorbing, so is its convex superset % % $C= A-B - x_0 + b_0$ $(b_0 \in B)$. % % Note that $C$ contains the origin. % Let $p$ be the Minkowski functional of $C$. Since $A$ and $B$ are disjoint, % $b_0-x_0$ is not in $C$, hence $p(b_0-x_0) \geq1$. % We now proceed as in the proof of the Hahn-Banach theorem \citeresultFA{3.4} % to establish the existence of a linear functional % % $\Lambda: X o \R$ such that % % \begin{align} \Lambda \leq p % \end{align} % and % % \begin{align} \Lambda(b_0-x_0) =1. \end{align}% % Then % % \begin{align} \Lambda a -\Lambda b + 1 =\Lambda (a-b+ b_0-x_0) \leq p (a-b+ b_0-x_0) \leq 1\quad (a\in A, b \in B). \end{align} % Hence % % \begin{align}\label{3_3_3} \Lambda a \leq \Lambda b . \end{align} % We now prove that $\Lambda (A) \cap \Lambda (B) $ contains at most one point. % Suppose, to reach a contradiction, that this intersection contains % % $y_1$ and $y_2$. % % There so exists % % $(a_i, b_i)$ %% % in $A imes B$ ($i=1, 2$) such that % % \begin{align}\label{3_3_4} \Lambda a_i= \Lambda b_i = y_i . \end{align} % Assume without loss of generality that $y_1< y_2$. Then, % %\begin{align} %y_1=\Lambda b_1 < \Lambda \left( \frac{1}{2} a_1 ight ) + \Lambda \left( \frac{1}{2} a_2 ight ) = \frac{1}{2} (y_1+y_2) \quad . %\end{align} % \begin{align}\label{3_3_5} 2\cdot y_1= \Lambda b_1+ \Lambda b_1 < \Lambda ( a_1 + a_2) = (y_1+y_2) \quad . \end{align} % Remark that $a_3= \frac{1}{2} (a_1+ a_2)$ lies in the convex set $A$. % This implies % % \begin{align} \Lambda b_1 \overset{( ef{3_3_5})}{<} \Lambda a_3 \overset{( ef{3_3_3})}{\leq} \Lambda b_1\quad ; \end{align} % which is a desired contradiction. (a) is so proved and we now deal with (b). % % ewline ewline oindent % From now on, the space $X$ is $\R^2$. Fetch % % \begin{align} S_1 & \Def \set{(x, y )\in \R^2}{x\leq 0, y \geq 0}, \ S_2 & \Def \set{(x, y )\in \R^2}{x>0, y > 0}, \ A & \Def S_1\cup S_2, \ B & \Def X\setminus A. \end{align} % Pick $(x_i, y_i)$ in $S_i$. % Let $t$ range over the unit interval, and so obtain % % \begin{align} t\cdot \left(\begin{array}{c}x_1 \y_1\end{array} ight)+ (1-t) \cdot \left(\begin{array}{c}x_2 \ y_2\end{array} ight)= \left(\begin{array}{c}t\cdot x_1+(1-t)\cdot x_2 \t\cdot y_1 + (1-t)\cdot y_2\end{array} ight) \in \R imes \R_+\subset A. \end{align} % Thus, every segment that has an extremity in $S_1$ and the other one in $S_2$ % lies in $A$. % Moreover, each $S_i$ is convex. We can now conclude that $A$ is so. % The convexity of $B$ is proved in the same manner. Furthermore, % $A$ hosts a non degenerate triangle, \ie $A^{\circ}$ is nonempty\footnote{% % For a immediate proof of this, remark that a triangle boundary is % compact/closed and apply [1.10] or 2.5 of \cite{BigRudin}. }: % $A$ contains an internal point. % % ewline ewline oindent % Let $L$ be a vector line of $\R^2$. % In other words, $L$ is the null space of a linear functional % % $\Lambda: \R^2 o \R$ % % (to see this, take some nonzero $u$ in $L^\bot$ and set % % $\Lambda x= (x,u)$ % % for all $x$ in $\R^2$). One easily checks that both $A$ and $B$ cut $L$. % Hence % % \begin{align} \Lambda (L)=\{0\}\subset \Lambda (A)\cap \Lambda (B) eq\emptyset\quad. \end{align} % So ends the proof. \end{proof}

Exercise 3.3

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Exercise 3.3

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