Farkas' lemma

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=== notes ===
=== notes ===
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For definition of convex cone see [http://en.wikipedia.org/wiki/Convex_cone Convex cone, Wikipedia],
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For definition of ''convex cone'' see [http://en.wikipedia.org/wiki/Convex_cone Convex cone, Wikipedia];
in finite dimension see [[Convex cones | Convex cones, Wikimization]].
in finite dimension see [[Convex cones | Convex cones, Wikimization]].
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For definition of polar cone see [[Moreau%27s_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]],
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For definition of ''polar cone'' see [[Moreau%27s_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]];
in finite dimension see [http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone Dual cone and polar cone].
in finite dimension see [http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone Dual cone and polar cone].

Revision as of 12:48, 11 July 2009

Farkas' lemma is a result used in the proof of the Karush-Kuhn-Tucker (KKT) theorem from nonlinear programming.

It states that if LaTeX: \,A\, is a matrix and LaTeX: \,b a vector, then exactly one of the following two systems has a solution:

  • LaTeX: A^Ty\succeq0 for some LaTeX: y\, such that LaTeX: b^Ty<0~~

or in the alternative

  • LaTeX: Ax=b\, for some LaTeX: x\succeq0

where the notation LaTeX: x\succeq0 means that all components of the vector LaTeX: x are nonnegative.

The lemma was originally proved by Farkas in 1902. The above formulation is due to Albert W. Tucker in the 1950s.

It is an example of a theorem of the alternative; a theorem stating that of two systems, one or the other has a solution, but not both.

Contents

Proof

(Dattorro) Define a convex cone

  • LaTeX: \mathcal{K}=\{y~|~A^Ty\succeq0\}\quad

whose dual cone is

  • LaTeX: \quad\mathcal{K}^*=\{A_{}x~|~x\succeq0\}

From the definition of dual cone LaTeX: \,\mathcal{K}^*\!=\{b~|~b^{\rm T}y\!\geq\!0~~\forall~y\!\in_{}\!\mathcal{K}\} we get

LaTeX: y\in\mathcal{K}~\Leftrightarrow~b^Ty\geq0~~\forall~b\in\mathcal{K}^*

rather,

LaTeX: A^Ty\succeq0~\Leftrightarrow~b^Ty\geq0~~\forall~b\in\{A_{}x~|~x\succeq0\}

Given some LaTeX: {\displaystyle b} vector and LaTeX: y\!\in\!\mathcal{K}~, then LaTeX: {\displaystyle b^Ty\!<\!0} can only mean LaTeX: b\notin\mathcal{K}^*.

An alternative system is therefore simply LaTeX: b\in\mathcal{K}^* and so the stated result follows.

Geometrical Interpretation

Farkas' lemma simply states that either vector LaTeX: \,b belongs to convex cone LaTeX: \mathcal{K}^* or it does not.

When LaTeX: b\notin\mathcal{K}^*, then there is a vector LaTeX: \,y\!\in\!\mathcal{K} normal to a hyperplane separating point LaTeX: \,b from cone LaTeX: \mathcal{K}^*.

References

  • Gyula Farkas, Über die Theorie der Einfachen Ungleichungen, Journal für die Reine und Angewandte Mathematik, volume 124, pages 1–27, 1902.

http://gdz.sub.uni-goettingen.de/no_cache/dms/load/img/?IDDOC=261361


Extended Farkas' lemma

For any closed convex cone LaTeX: \mathcal J in the Hilbert space LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle), denote by LaTeX: \mathcal J^\circ the polar cone of LaTeX: \mathcal J.

Let LaTeX: \mathcal K be an arbitrary closed convex cone in LaTeX: \mathcal H.

Then, the extended Farkas' lemma asserts that LaTeX: \mathcal K^{\circ\circ}=\mathcal K.

Hence, denoting LaTeX: \mathcal L=\mathcal K^\circ, it follows that LaTeX: \mathcal L^\circ=\mathcal K.

Therefore, the cones LaTeX: \mathcal K and LaTeX: \mathcal L are called mutually polar pair of cones.

notes

For definition of convex cone see Convex cone, Wikipedia; in finite dimension see Convex cones, Wikimization.

For definition of polar cone see Moreau's theorem; in finite dimension see Dual cone and polar cone.

Proof of extended Farkas' lemma

(Sándor Zoltán Németh) Let LaTeX: z\in\mathcal H be arbitrary. Then, by Moreau's theorem we have

LaTeX: z=P_{\mathcal K}z+P_{\mathcal K^\circ}z

and

LaTeX: z=P_{\mathcal K^\circ}z+P_{\mathcal K^{\circ\circ}}z.

Therefore,

LaTeX: P_{\mathcal K^{\circ\circ}}z=P_{\mathcal K}z=z-P_{\mathcal K^\circ}z.

In particular, for any LaTeX: z\in K we have LaTeX: \mathcal K^{\circ\circ}\ni P_{\mathcal K^{\circ\circ}}z=z. Hence, LaTeX: \mathcal \mathcal K^{\circ\circ}\supset K. Similarly, for any LaTeX: z\in K^{\circ\circ} we have LaTeX: z= P_{\mathcal K}z\in\mathcal K. Hence, LaTeX: \mathcal K^{\circ\circ}\subset\mathcal K. Therefore, LaTeX: \mathcal K^{\circ\circ}=\mathcal K.

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