# Codimension of the largest subspace of a closed convex cone in a normed vector space

Please, help to prove (or disprove) the following statement.

Let $K \neq V$ be a closed convex cone ($K+K=K$, $\alpha K \subseteq K$ for any $\alpha \geq 0$) in a real normed vector space $V$. If $-K \cup K = V$, then the subspace $-K \cap K$ has codimension 1.

Thanks.

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Do you try any geometric or algebric form of Han-Banach theorem in $V$? –  MathOverview Jan 12 '13 at 11:34

Here's a version of your proof that convinces me:

Set $X := -K \cap K$. We mean to show that $X$ is a subspace of codimension one, i.e. there is a $y_0$ with $X + \mathbb R y_0 = V$.

To that end, choose $y_0 \in K \setminus X$ (this is possible since otherwise $K \subset -K$, so that $-K = -K \cup K = X$) and let $v \in V$ be arbitrary. We mean to find an $\alpha \in \mathbb R$ such that $v + \alpha y_0 \in X$.

Consider the line $v + \mathbb R y_0$. Then by convexity of $K$, the intersection with $K$ is either (1) empty, (2) a point, (3) a line segment, (4) a beam or (5) a full line.

(1): If this is so, then we have $v + \beta y_0 \subset -K \setminus K$ for every $\beta \in \mathbb R$. Since $-K$ is a cone, we also have $1/\beta v + y_0 \subset -K \setminus K$ for every $\beta > 0$ and thus $y_0 \in -K$ since $-K$ is closed; a contradiction to the choice of $y_0$ (namely, $y_0 \in K \setminus (-K \cap K)$). In other words, this case cannot occur.

(2) This point must also lie in $-K$ since the rest of the line lies in $-K$ and $-K$ is convex. Consequently it lies in $X$.

(3) Analogous to (2).

(4) This beam can be of two forms: $$B_+ = \{ v + \beta y_0 \colon \beta \ge \beta_0 \}$$ or $$B_- = \{ v + \beta y_0 \colon \beta \le \beta_0 \}.$$ In the letter case, we get $-1/\beta v - y_0 \in K$ for every $\beta < \min(\beta_0,-1)$ and thus $y_0 \in -K$; a contradiction to the choice of $y_0$.

In the former case, we have $v + \beta v_0 \in K$ exactly if $\beta \ge \beta_0$. Since $-K$ is closed, we also have $v + \beta v_0 \in -K$ for $\beta \le \beta_0$. This implies $v + \beta_0 v_0 \in X$.

(5) Analogous to the case of $B_-$.

That $X$ is a subspace is clear since by construction, it is a closed convex cone with $-X = X$ and $0 \in X$.

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Thanks! Some comments to you post. –  Mikhail Jan 13 '13 at 12:59
1) the union of the first and the third quadrants does not give $\mathbb R^2$. 2) You are certainly right (thanks!): I forgot the option that $v+\mathbb R y_0$ does not intersect $K$. But then $v+\mathbb R y_0$ is a subset of $-K$ and the same arguments as in the proof show that this is impossible. –  Mikhail Jan 13 '13 at 13:29
Indeed, I was thinking of $-K + K$ but wrote $-K \cup K$. –  anonymous Jan 13 '13 at 14:04

An attempt of proof. Please, check.

Let $X:=-K \cap K$. Fix $y_0 \in K\setminus X$. The task is to prove $X+ \mathbb R y_0=V$.

Claim: given $v \in V$, there exist either $\inf \left\{ \beta \in \mathbb R: v+ \beta y_0 \in K \right\}$ or $\sup \left\{ \beta \in \mathbb R: v+ \beta y_0 \in K \right\}$. Indeed, if both do not exist, then $v+ \beta y_0 \in K$ for any $\beta \in \mathbb R$. In particular, $n^{-1}v+y_0 \in K$ and $n^{-1}v-y_0 \in K$ for any natural $n$. Tending $n$ to infinity, this implies a contradiction: $y_0 \in X$ (since $K$ is closed). Let $\alpha$ be the existed (lower or upper) bound. Then $v+ \alpha y_0 \in \partial K$. But $\partial K=X$, since $K$ is closed and $-K \cup K = V$. This completes the proof.

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