anti-cone

Let $X$ be a real vector space, and $\mathrm{\Phi }$ be a subspace of linear functionalson $X$.

For any set $S\subseteq X$,its anti-cone $S^{+}$,with respect to $\mathrm{\Phi }$, is the set

$$S^{+}=\{\varphi \in \mathrm{\Phi }:\varphi (x)\ge 0,\text{ for all }x\in S\}.$$

The anti-cone is also called the dual cone.

Usage

The anti-cone operation is generally applied to subsets of $X$that are themselvescones.Recall that a cone in a real vector space generalize the notion oflinear inequalities in a finite number of real variables.The dual cone provides a natural way to transfer suchinequalities in the original vector spaceto its dual space.The concept is useful in the theory ofduality.

The set $\mathrm{\Phi }$ in the definition may be taken to be any subspaceof the algebraic dual space $X^{*}$.The set $\mathrm{\Phi }$ often needs to be restrictedto a subspace smaller than $X^{*}$, or eventhe continuous dual space $X^{\prime }$,in order to obtainthe nice closure and reflexivity properties below.

Basic properties

Property 1.

The anti-cone is a convex cone in $\mathrm{\Phi }$.

Proof.

If $\varphi (x)$ is non-negative, then so is $t\varphi (x)$ for $t>0$.And if ${\varphi }_1(x),{\varphi }_2(x)\ge 0$,then clearly $(1-t){\varphi }_1(x)+t{\varphi }_2(x)\ge 0$ for $0\le t\le 1$.∎

Property 2.

If $K\mathrm{\subseteq }X$ is a cone, then its anti-cone $K^{\mathrm{+}}$ may be equivalentlycharacterized as:

$$K^{+}=\{\varphi \in \mathrm{\Phi }:\varphi (x)\mathit{\text{ over }}x\in K\mathit{\text{ is bounded below}}\}.$$

Proof.

It suffices to show that if ${inf}_{x\in K}\varphi (x)$ is bounded below,then it is non-negative.If it were negative, take some $x\in K$ such that. For any $t>0$, the vector $tx$ is in the cone $K$,and the function value $\varphi (tx)=t\varphi (x)$ would be arbitrarilylarge negative, and hence unbounded below.∎

Topological properties

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

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

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

Property 3.

$S^{+}$ is weak-* closed.

Proof.

Let $\{{\varphi }_{\alpha }\}\subseteq \mathrm{\Phi }$ be a net converging to $\varphi$in the weak-* topology.By definition, $\widehat{x}({\varphi }_{\alpha })={\varphi }_{\alpha }(x)\ge 0$.As the functional $\widehat{x}$ is continuous in the weak-* topology,we have $\widehat{x}({\varphi }_{\alpha })\to \widehat{x}(\varphi )\ge 0$.Hence $\varphi \in S^{+}$.∎

Property 4.

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

Proof.

The inclusion ${\overline{S}}^{+}\subseteq S^{+}$ is obvious.And if $\varphi (x)\ge 0$ for all $x\in S$,then by continuity, this holds true for $x\in \overline{S}$ too —so ${\overline{S}}^{+}\supseteq S^{+}$.∎

Properties involving cone inclusion

Property 5 (Farkas’ lemma).

Let $K\mathrm{\subseteq }X$ be a weakly-closed convex cone.Then $x\mathrm{\in }K$ if and only if $\varphi \mathit{}\mathrm{(}x\mathrm{)}\mathrm{\ge }\mathrm{0}$ for all $\varphi \mathrm{\in }{K}^{\mathrm{+}}$.

Proof.

That $\varphi (x)\ge 0$ for $\varphi \in K^{+}$ and $x\in K$ is just the definition.For the converse, we show that if $x\in X\setminus K$,then there exists $\varphi \in K^{+}$ such that .

If $K=\mathrm{\varnothing }$, then the desired $\varphi \in \mathrm{\Phi }=K^{+}$exists because $\mathrm{\Phi }$ can separate the points $x$ and $0$.If $K\ne \mathrm{\varnothing }$, by the hyperplane separation theorem,there is a $\varphi \in \mathrm{\Phi }$ such that .This $\varphi$ will automatically be in $K^{+}$ by Property 2.The zero vector is the weak limit of $ty$, as $t\searrow 0$,for any vector $y$.Thus $0\in K$, andwe conclude with ${inf}_{y\in K}\varphi (y)\le 0$.∎

Property 6.

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

Proof.

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

Property 7.

Let $K$ and $L$ be convex cones in $X$, with $K$ weakly closed.Then $K^{\mathrm{+}}\mathrm{\subseteq }{L}^{\mathrm{+}}$ if and only if $K\mathrm{\supseteq }L$.

Proof.

$$K^{+}\subseteq L^{+}⟹K=\overline{K}=K^{++}\supseteq L^{++}=\overline{L}\supseteq L⟹K^{+}\subseteq L^{+}.\mathit{∎}$$