How to prove this (corollary of) hyperplane separation theorem?

Question

$X$ is a nonempty convex subset of $\mathbb{R}^n$ whose element is $x=\left(x_1,...,x_n\right)$.

The theorem is as follows.

If for each $x\in X$, there is an $i \in \left\{1,...,n\right\}$ such that $x_i>0$, then there exists $\left(\lambda_1,...,\lambda_n\right)$ where $\lambda_i \geqslant 0$ for all $i$ and $\sum_{i=1}^n \lambda_i=1$, such that $\lambda \cdot x \geqslant 0$ for all $x\in X$ and $\lambda \cdot x>0$, for some $x \in X$.

I was wondering how to prove it. "$\geqslant 0$ for all $x$" should be easy. But I got stuck in "$>0$ for some $x$". I failed to use the proper separation theorem (Theorem 11.3 in Rockafellar (1970)).

Thank you very much!

This was cross-posted at https://math.stackexchange.com/questions/4504475/how-to-prove-this-corollary-of-hyperplane-separation-theorem?noredirect=1#comment9461236_4504475

Answer 1

Let $Y$ be the span of $X$, $C=Y\cap (-\infty,0]^n$. Since $X$ and $C$ are disjoint convex sets in $Y$, there exists a non-zero functional $\eta\in Y^*$ which separates (not strictly) $X$ and $C$: $\eta$ is non-positive on $C$ and non-negative on $X$. Since $X$ is of full dimension in $Y$, we have $\eta(x)>0$ for some $x\in X$.

Now note that the coordinate functionals $x\to x_i$, $i=1,\ldots,n$, and the functional $\eta$ on $Y$ satisfy the property "if $x_i$'s are non-negative, then $\eta$ is non-negative". It yields by duality that $\eta$ is a non-negative linear combination of $x_i$'s (as a functional on $Y$, in particular as a function on $X$.)

Answer 2

$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$Let $\R_-:=(-\infty,0]$ and $\R_+:=[0,\infty)$. The desired statement is equivalent to the following:

Suppose that \begin{equation*} \forall x\in X\ \exists i\in[n]:=\{1,\dots,n\}\ x_i>0. \tag{1}\label{1} \end{equation*} Then there is some $\nu\in\R_+^n$ such that \begin{equation*} \nu\cdot X:=\{\nu\cdot x\colon x\in X\}\subseteq\R_+\quad\text{and}\quad \nu\cdot X\ne\{0\}. \tag{2}\label{2} \end{equation*}

Proof: Condition \eqref{1} means that $X\cap\R_-^n=\emptyset$. So, by Theorem 11.3 in Rockafellar (1970), there is some $\la^1\in\R^n\setminus\{0\}$ such that $\la^1\cdot X\subseteq\R_+$ and $\la^1\cdot\R_-^n\subseteq\R_-$. Also, $\la^1\cdot\R_-^n\subseteq\R_-$ means that $\la^1\in\R_+^n$. Thus, \begin{equation*} \text{$\la^1\in\R_+^n\setminus\{0\}$ and $\la^1\cdot X\subseteq\R_+$.} \tag{3}\label{3} \end{equation*}

So, either \eqref{2} holds with $\la^1$ in place of $\nu$ or $\la^1\cdot X=\{0\}$.
So, without loss of generality (wlog), $\la^1\cdot X=\{0\}$. That is, we have \begin{equation*} X\subseteq V^1:=\{x\in\R^n\colon\la^1\cdot x=0\}. \end{equation*} The condition $\la^1\in\R_+^n\setminus\{0\}$ in \eqref{3} implies that wlog for some $k^1\in[n]$ and all $i\in[n]$ \begin{equation*} \la^1_i>0\text{ if }i\le k^1\quad \text{and}\quad \la^1_i=0\text{ if }i>k^1. \tag{4}\label{4} \end{equation*}

Now condition \eqref{1} implies $X\cap V^1_-=\emptyset$, where $V^1_-:=V^1\cap\R_-^n$. Applying now Theorem 11.3 in Rockafellar (1970) to the Euclidean space $V^1$, we see that there is some $\mu^1\in V^1\setminus\{0\}$ such that \begin{equation*} \mu^1\cdot X\subseteq\R_+ \end{equation*} and $\mu^1\cdot(V^1\cap\R_-^n)\subseteq\R_-$. The latter condition means that for all $x\in\R^n$ we have \begin{equation*} (\la^1\cdot x=0\ \&\ x\in\R_-^n)\implies\mu^1\cdot x\le0 \end{equation*} or, equivalently, \begin{equation*} (\la^1\cdot x=0\ \&\ x\in\R_+^n)\implies\mu^1\cdot x\ge0. \tag{5}\label{5} \end{equation*} Substituting the $i$th standard basis vector $e_i$ for $x$ in \eqref{5}, and recalling \eqref{4}, we get $\mu^1_i=\mu^1\cdot e_i\ge0$ for $i>k$.

So, wlog there is some real $t^1>0$ such that for \begin{equation*} \la^2:=\la^1+t^1\mu^1 \tag{6}\label{6} \end{equation*} and some integer $k^2\in[k^1,n]$ we have \begin{equation*} \la^2_i>0\text{ if }i\le k^2\quad \text{and}\quad \la^2_i=0\text{ if }i>k^2. \tag{4a}\label{4a} \end{equation*} Also, by \eqref{6}, $\la^2\cdot X\subseteq\la^1\cdot X+t^1\mu^1\cdot X\subseteq\R_+$. So, either \eqref{2} holds with $\la^2$ in place of $\nu$ or $\la^2\cdot X=\{0\}$ and hence $\mu^1\cdot X=\{0\}$.
So, wlog, $\la^2\cdot X=\{0\}=\mu^1\cdot X$. So, \begin{equation*} X\subseteq V^2:=\{x\in V^1\colon\la^2\cdot x=0\} =\{x\in V^1\colon\mu^1\cdot x=0\}\subsetneq V^1; \end{equation*} the latter strict inclusion follows because $\mu^1\in V^1\setminus\{0\}$.

Continuing thus, in $n$ similar steps wlog we will get $X\subseteq V^n:=\{0\}$, which contradicts \eqref{1}. $\quad\Box$.

Answer 3

I thought that I had an answer for both part, but for the moment, I succeed only on the first part, which is an application of the first separation theorem given in https://en.wikipedia.org/wiki/Hyperplane_separation_theorem

By assumption, $X$ and $\mathbb{R}_-^n$ are two disjoint non-empty convex subsets of $\mathbb{R}^n$. There exist some non-null vector $c$ such that $X \subset \{x \in \mathbb{R}^n : c \cdot x \ge 0\}$ and $\mathbb{R}_-^n \subset \{x \in \mathbb{R}^n : c \cdot x \le 0\}$.

The last condition applied to opposite of the vectors of the canonical basis of $\mathbb{R}^n$ (containd in $\mathbb{R}_-^n$) forces $c$ to have non-negative coordinates. Of course, one may divide $c$ by the sum of its coordinate (which is strictly positive).