# anti-cone

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

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

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

The anti-cone is also called the dual cone.

## Usage

The anti-cone operation is generally applied to subsets of $X$ that are 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 $X^{*}$. The set $\Phi$ often needs to be restricted to a subspace smaller than $X^{*}$, or even the continuous dual space $X^{\prime}$, 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 $t>0$. And if $\phi_{1}(x),\phi_{2}(x)\geq 0$, then clearly $(1-t)\phi_{1}(x)+t\phi_{2}(x)\geq 0$ for $0\leq t\leq 1$. ∎

###### Property 2.

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

 $K^{+}=\{\phi\in\Phi\colon\text{\phi(x) over x\in K is bounded below}\}\,.$
###### 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 $t>0$, the vector $tx$ is in the cone $K$, and the function value $\phi(tx)=t\phi(x)$ would be arbitrarily large negative, and hence unbounded below. ∎

## Topological properties

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

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

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

###### Property 3.

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

###### Proof.

Let $\{\phi_{\alpha}\}\subseteq\Phi$ 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 $\phi\in S^{+}$. ∎

###### 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 $\overline{S}^{+}\supseteq S^{+}$. ∎

## 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 $x$ 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 $K^{+}$ by Property 2. The zero vector is the weak limit of $ty$, as $t\searrow 0$, for any vector $y$. Thus $0\in K$, and we conclude with $\inf_{y\in K}\phi(y)\leq 0$. ∎

###### Property 6.

$K^{++}=\overline{K}$ for any convex cone $K$. (The anti-cone operation on $K^{+}$ 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 $\phi(x)\geq 0$ 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 $\phi\in K^{+}$. ∎

###### Property 7.

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

###### Proof.
 $K^{+}\subseteq L^{+}\implies K=\overline{K}=K^{++}\supseteq L^{++}=\overline{L% }\supseteq L\implies K^{+}\subseteq L^{+}\,.\qed$

## References

• 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.
Title anti-cone Anticone 2013-03-22 17:20:48 2013-03-22 17:20:48 stevecheng (10074) stevecheng (10074) 8 stevecheng (10074) Definition msc 46A03 msc 46A20 anticone dual cone GeneralizedFarkasLemma