Assume $Y \subset \mathbb{R}^{n}$ satisified following properties:
- $Y$ is closed and convex
- $0 \in Y$
- $Y$ has a upper bound, i.e. there exists $y_0 \in \mathbb{R}^{n}$ such that $y \le y_0$ for all $y \in Y$
- for any $y \in Y$, if $x \le y$, then we have $x \in Y$. Here $x \le y$ means that $x_i \le y_i$ for all $i=1, \cdots , n$
I need to prove: for any $x \notin Y$, there exists $p \gg 0$ and $c \ge 0$ such that $p \cdot x > c > p \cdot y$ for all $y \in Y$. Here $p \gg 0$ means that $p_i > 0$ for all $i$.
Here is my try: Since $x \notin Y$ and $Y$ is closed and convex, the hyperplane separation theorem tells us that there exists $p\in\mathbb{R}^{n}, c\in\mathbb{R}$ such that $p\cdot x>c>p\cdot y$ for all $y\in Y$.
Since $0\in Y$, we must have $c>0$ and $\overline{y}\geq0$. In this case, $p\ne0$. We can also have that for all $i=1,2,\cdots,n, p_{i}\geq0$. Assume $p_{i}<0$. Let $\mathbf{1}^{i}$ denote the vector $x\in\mathbb{R}^{n}$ such that $x_{i}=1$ and $x_{j}=0$ for all $j\ne i$. Then given any $\lambda>0$ and any $y\in Y$, $y-\lambda\mathbf{1}^{i}\in Y$. Then $$\lim_{\lambda\rightarrow\infty}p\cdot\left(y-\lambda\mathbf{1}^{i}\right)=p\cdot y-\lambda p_{i}\rightarrow\infty$$ Contradiction with $p\cdot x>c>p\cdot y$ for all $y\in Y$. So $p_{i}\geq0$ for all $i=1,2,\cdots,n$.
However, how should I show that $p \gg 0$? Or can I use this $p$ to construct another $p^{\prime}$ that has this property? I think we need to use $y_0$, but I have no idea how to do it. Could anyone help me? Thanks in advance.