Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.


share|cite|improve this question
Do you try any geometric or algebric form of Han-Banach theorem in $V$? – MathOverview Jan 12 '13 at 11:34
up vote 1 down vote accepted

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 $0 \in X$.

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.