PlanetMath: anti-cone

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anti-cone

(Definition)

Let be a real vector space, and $\Phi$ be a subspace of linear functionals on .

For any set $S \subseteq X$ , its anti-cone , with respect to $\Phi$ , is the set

$\displaystyle S^+ = \{ \phi \in \Phi \colon \phi(x) \geq 0 \,,$ for all $\displaystyle x \in S \}\,.$

The anti-cone is also called the dual cone.

Usage

The anti-cone operation is generally applied to subsets of that are themselves cones. Recall that a cone in a real vector space generalize the notion of linear inequalities in a finite number of real variables. The dual cone provides a natural way to transfer such inequalities in the original vector space to its dual space. The concept is useful in the theory of duality.

The set $\Phi$ in the definition may be taken to be any subspace of the algebraic dual space . The set $\Phi$ often needs to be restricted to a subspace smaller than , or even the continuous dual space , in order to obtain the nice closure and reflexivity properties below.

Basic properties

Property 1 The anti-cone is a convex cone in $\Phi$ .

Proof. If $\phi(x)$ is non-negative, then so is $t\phi(x)$ for

. And if $\phi_1(x), \phi_2(x) \geq 0$ , then clearly $(1-t)\phi_1(x) + t\phi_2(x) \geq 0$ for $0 \leq t \leq 1$ . $\qedsymbol$

Property 2 If $K \subseteq X$ is a cone, then its anti-cone may be equivalently characterized as:

$\displaystyle K^+ = \{ \phi \in \Phi \colon$ $\phi(x)$ over $x \in K$ is bounded below $\displaystyle \}\,.$

Proof. It suffices to show that if $\inf_{x \in K} \phi(x)$ is bounded below, then it is non-negative. If it were negative, take some $x \in K$ such that $\phi(x) < 0$ . For any

, the vector

is in the cone

, and the function value $\phi(tx) = t\phi(x)$ would be arbitrarily large negative, and hence unbounded below. $\qedsymbol$

Topological properties

Assumptions. Assume that $\Phi$ separates points of . Let have the weak topology generated by $\Phi$ , and let $\Phi$ have the weak-* topology generated by ; this makes and $\Phi$ into Hausdorff topological vector spaces.

Vectors $x \in X$ will be identified with their images $\hat{x}$ under the natural embedding of in its double dual space.

The pairing $(X, \Phi)$ is sometimes called a dual pair; and $(\Phi, X)$ , where is identified with its image in the double dual, is also a dual pair.

Property 3 is weak-* closed.

Proof. Let $\{ \phi_\alpha \} \subseteq \Phi$ be a net converging to $\phi$ in the weak-* topology. By definition, $\hat{x}(\phi_\alpha) = \phi_\alpha(x) \geq 0$ . As the functional $\hat{x}$ is continuous in the weak-* topology, we have $\hat{x}(\phi_\alpha) \to \hat{x}(\phi) \geq 0$ . Hence $\phi \in S^+$ . $\qedsymbol$

Property 4 $\overline{S}^+ = S^+$ .

Proof. The inclusion $\overline{S}^+ \subseteq S^+$ is obvious. And if $\phi(x) \geq 0$ for all $x \in S$ , then by continuity, this holds true for $x \in \overline{S}$ too -- so $\overline{S}^+ \supseteq S^+$ . $\qedsymbol$

Properties involving cone inclusion

Property 5 (Farkas' lemma) Let $K \subseteq X$ be a weakly-closed convex cone. Then $x \in K$ if and only if $\phi(x) \geq 0$ for all $\phi \in K^+$ .

Proof. That $\phi(x) \geq 0$ for $\phi \in K^+$ and $x \in K$ is just the definition. For the converse, we show that if $x \in X \setminus K$ , then there exists $\phi \in K^+$ such that $\phi(x) < 0$ .

If $K=\emptyset$ , then the desired $\phi \in \Phi = K^+$ exists because $\Phi$ can separate the points and 0. If $K \neq \emptyset$ , by the hyperplane separation theorem, there is a $\phi \in \Phi$ such that $\phi(x) < \inf_{y \in K} \phi(y)$ . This $\phi$ will automatically be in by Property 2. The zero vector is the weak limit of , as $t \searrow 0$ , for any vector . Thus $0 \in K$ , and we conclude with $\inf_{y \in K} \phi(y) \leq 0$ . $\qedsymbol$

Property 6 $K^{++} = \overline{K}$ for any convex cone . (The anti-cone operation on is to be taken with respect to .)

Proof. We work with $\overline{K}$ , which is a weakly-closed convex cone. By Property 5, $x \in \overline{K}$ if and only if $\phi(x) \geq 0$ for all $\phi \in \overline{K}^+ = K^+$ . But by definition of the second anti-cone, $\hat{x} \in (K^+)^+$ if and only if $\phi(x) = \hat{x}(\phi) \geq 0$ for all $\phi \in K^+$ . $\qedsymbol$

Property 7 Let and be convex cones in , with weakly closed. Then $K^+ \subseteq L^+$ if and only if $K \supseteq L$ .

Proof.

$\displaystyle K^+ \subseteq L^+ \implies K = \overline{K} = K^{++} \supseteq L^{++} = \overline{L} \supseteq L \implies K^+ \subseteq L^+\,. \qedhere$

$\qedsymbol$

Bibliography

1: B. D. Craven and J. J. Kohila. ``Generalizations of Farkas' Theorem.'' SIAM Journal on Mathematical Analysis. Vol. 8, No. 6, November 1977.
2: David G. Luenberger. Optimization by Vector Space Methods. John Wiley & Sons, 1969.

"anti-cone" is owned by stevecheng.

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