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I'm trying to solve below exercise, i.e.,

Let $E := \mathbb R^n, P := \{x \in E \mid x_i \ge 0 \quad \forall i = 1, \ldots, n\}$, and $M$ be a linear subspace of $E$ such that $M \cap P = \{0\}$. Then there is some hyperplane $H$ such that $M \subset H$ and $H \cap P = \{0\}$.

Could you please have a check on my attempt?

My attempt: Notice that $M^\perp$ is a closed vector subspace, $(M^\perp)^\perp = \overline M = M$, and $\operatorname{int} P = \{x \in E \mid x_i > 0 \quad \forall i = 1, \ldots, n\}$.

First, we prove that $M^\perp \cap (\operatorname{int} P) \neq \emptyset$. Assume the contrary that $M^\perp \cap (\operatorname{int} P) = \emptyset$. Then by this version of Hahn-Banach theorem, there is $a \in E \setminus \{0\}$ such that $$ \langle a, x \rangle \le \langle a, y \rangle \quad \forall x \in M^\perp, y \in P. $$

It follows that $\langle a, x \rangle =0$ for all $x\in M^\perp$ and thus $a \in M$. Also, $\langle a, y \rangle \ge 0$ for all $y\in P$, so $a \in P$. Hence $a \in P \cap M$, which is a contradiction. Hence there is $a \in M^\perp \cap (\operatorname{int} P)$. Then $a_i >0$ for all $i=1, \ldots, n$.

We define $f \in E^*$ by $f(x) := \langle a, x\rangle$ for all $x\in E$. Then $M \subset \ker f$. Clearly, $f(x)>0$ for all $x \in P \setminus \{0\}$. Then $H := \ker f$ is the required hyperplane.

Analyst
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  • The part 'it follows that $\langle a,x\rangle=0$' is only valid if $a\in P$. Otherwise it's fine. Note that $H$ is just $a^\perp$. – Berci Jun 13 '22 at 21:44
  • @Berci If there is $x_0 \in M^\perp$ such that $\langle a, x_0\rangle \neq 0$, then $\langle a, x\rangle$ can be arbitrarily large or arbitrarily small by picking $x = k x_0$ with $k \in \mathbb Z$. – Analyst Jun 13 '22 at 22:10

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