# Excercise 1.13 in Brezis's Functional Analysis

This is the Excercise 1.13 in Brezis's Functional Analysis

Let $E=\mathbb{R}^n$ and let $$P=\{x\in\mathbb{R}^n;x_i\geq 0\ \forall i=1,2,...,n\}$$ Let $M$ be a linear subspace of $E$ such that $M\cap P=\{0\}$. Prove that there is some hyperplane $H$ in $E$ such that $$M\subset H \text{ and } H\cap P=\{0\}.$$ [Hint: Show first that $M^{\perp}\cap \text{Int}P\neq\emptyset$]

I know if $f\in M^{\perp}\cap \text{Int}P$, then $f$ is just the functional to make the hyperplane. But how to prove the hint?

Thanks very much!

• I believe $M^{\perp}$ might be the usual perp. ie. $\{x\in\mathbb{R}^n : \langle x,v\rangle = 0 \forall v\in M\}$ Oct 1 '13 at 2:44
• @PrahladVaidyanathan Yes, it's the usual perp. But by the Riesz representation theorem, it is equivelent to a space of functionals. Oct 1 '13 at 2:48
• Why doesn't the following approach work? One can fix a basis for $M$, and define a functional, $f$, which vanishes identically on $M$. Then $M$ is the hyperplane, the kernel of the map and it intersects trivially with $P$ because of the given assumption. Oct 21 '19 at 21:43
• @PrahladVaidyanathan Oct 21 '19 at 21:44
• @Danielsen ...... Oct 21 '19 at 21:44

Suppose, for the sake of contradiction, that $M^\perp \cap \text{Int}\,P = \emptyset$. Then by the first geometric form of Hahn-Banach, there exists $\beta \in \mathbb{R}^n$ such that $\beta \cdot x < \beta \cdot y$ for every $x \in M^\perp$ and every $y \in \text{Int}\,P$. Since $M^\perp$ is a linear space and $\beta \cdot M^\perp$ is bounded, it follows that $\beta \cdot M^\perp = 0$. Moreover, since $0 < \beta \cdot y$ for all $y \in \text{Int}\,P$, we have $\beta \in P$. Therefore, $0 \neq \beta \in M^{\perp\perp} = M$, which is a contradiction.
Therefore, let $v \in M^\perp \cap \text{Int}\,P$, and let $H = \{v\}^\perp$. Then $M \subseteq H$ and $P \cap H = \{0\}$, since for any $0 \neq x \in P$ and $v \in \text{Int}\,P$, we have $x \cdot v > 0$.
Theorem 1 (Brezis). Let $A \subset E$ and $B \subset E$ be two nonempty convex subsets such that $A \cap B = \emptyset$. Assume that one of them is open. Then there exists a closed hyperplane that separates $A$ and $B$.