# Complementarity problem

### From Wikimization

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Sándor Zoltán Németh | Sándor Zoltán Németh | ||

- | + | == Fixed point problems == | |

Let <math>\mathcal A</math> be a set and <math>F:\mathcal A\to\mathcal A </math> a mapping. The '''fixed point problem''' defined by <math>F\,</math> is the problem | Let <math>\mathcal A</math> be a set and <math>F:\mathcal A\to\mathcal A </math> a mapping. The '''fixed point problem''' defined by <math>F\,</math> is the problem | ||

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</center> | </center> | ||

- | + | == Nonlinear complementarity problems == | |

Let <math>\mathcal K</math> be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Recall that the dual cone of <math>\mathcal K</math> is the closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] <math>\mathcal K^*=-\mathcal K^\circ,</math> where <math>\mathcal K^\circ</math> is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of <math>\mathcal K.</math> The '''nonlinear complementarity problem''' defined by <math>\mathcal K</math> and <math>f\,</math> is the problem | Let <math>\mathcal K</math> be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Recall that the dual cone of <math>\mathcal K</math> is the closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] <math>\mathcal K^*=-\mathcal K^\circ,</math> where <math>\mathcal K^\circ</math> is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of <math>\mathcal K.</math> The '''nonlinear complementarity problem''' defined by <math>\mathcal K</math> and <math>f\,</math> is the problem | ||

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</center> | </center> | ||

- | + | == Every nonlinear complementarity problem is equivalent to a fixed point problem == | |

Let <math>\mathcal K</math> be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(f,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] | Let <math>\mathcal K</math> be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(f,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] | ||

<math>Fix(P_{\mathcal K}\circ(I-f)),</math> where <math>I:\mathbb H\to\mathbb H</math> is the identity mapping defined by <math>I(x)=x\,</math> and <math>P_{\mathcal K}</math> is the [[Moreau's_decomposition_theorem#Projection_on_closed_convex_sets | projection onto <math>\mathcal K.</math>]] | <math>Fix(P_{\mathcal K}\circ(I-f)),</math> where <math>I:\mathbb H\to\mathbb H</math> is the identity mapping defined by <math>I(x)=x\,</math> and <math>P_{\mathcal K}</math> is the [[Moreau's_decomposition_theorem#Projection_on_closed_convex_sets | projection onto <math>\mathcal K.</math>]] | ||

- | + | == Proof == | |

For all <math>x\in\mathbb H</math> denote <math>z=x-f(x)\,</math> and <math>y=-f(x).\,</math> Then, <math>z=x+y.\,</math> | For all <math>x\in\mathbb H</math> denote <math>z=x-f(x)\,</math> and <math>y=-f(x).\,</math> Then, <math>z=x+y.\,</math> | ||

<br> | <br> | ||

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<math>x\in\mathcal K,</math> <math>f(x)=-y\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Therefore, <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math> | <math>x\in\mathcal K,</math> <math>f(x)=-y\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Therefore, <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math> | ||

+ | == An alternative proof without [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] == | ||

- | + | === Variational inequalities === | |

- | + | ||

Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. The '''variational inequality''' defined by <math>\mathcal C</math> and <math>f\,</math> is the problem | Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. The '''variational inequality''' defined by <math>\mathcal C</math> and <math>f\,</math> is the problem | ||

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</center> | </center> | ||

- | + | === Every variational inequality is equivalent to a fixed point problem === | |

Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(f,\mathcal C)</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] <math>Fix(P_{\mathcal C}\circ(I-f)).</math> | Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(f,\mathcal C)</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] <math>Fix(P_{\mathcal C}\circ(I-f)).</math> | ||

- | + | === Proof === | |

<math>x\,</math> is a solution of <math>Fix(P_{\mathcal C}\circ(I-f))</math> if and only if | <math>x\,</math> is a solution of <math>Fix(P_{\mathcal C}\circ(I-f))</math> if and only if | ||

<math>x=P_{\mathcal C}(x-f(x)).</math> By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection | characterization of the projection]] the latter equation is equivalent to | <math>x=P_{\mathcal C}(x-f(x)).</math> By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection | characterization of the projection]] the latter equation is equivalent to | ||

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of <math>VI(f,\mathcal C).</math> | of <math>VI(f,\mathcal C).</math> | ||

- | + | === Remark === | |

The next section shows that the equivalence of [[Complementarity_problem#Variational_inequalities | variational inequalities]] and [[Complementarity_problem#Fixed_point_problems | fixed point problems]] is much stronger than the equivalence of [[Complementarity_problem#Nonlinear_complementarity_problems |nonlinear complementarity problems]] and [[Complementarity_problem#Fixed_point_problems | fixed point problems]], because each [[Complementarity_problem#Nonlinear_complementarity_problems |nonlinear complementarity problem]] is a [[Complementarity_problem#Variational_inequalities | variational inequality]] defined on a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone]. | The next section shows that the equivalence of [[Complementarity_problem#Variational_inequalities | variational inequalities]] and [[Complementarity_problem#Fixed_point_problems | fixed point problems]] is much stronger than the equivalence of [[Complementarity_problem#Nonlinear_complementarity_problems |nonlinear complementarity problems]] and [[Complementarity_problem#Fixed_point_problems | fixed point problems]], because each [[Complementarity_problem#Nonlinear_complementarity_problems |nonlinear complementarity problem]] is a [[Complementarity_problem#Variational_inequalities | variational inequality]] defined on a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone]. | ||

- | + | === Every variational inequality defined on a closed convex cone is equivalent to a complementarity problem === | |

Let <math>\mathcal K</math> be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(f,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(f,\mathcal K).</math> | Let <math>\mathcal K</math> be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(f,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(f,\mathcal K).</math> | ||

- | + | === Proof === | |

Suppose that <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math> Then, <math>x\in\mathcal K,</math> <math>f(x)\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Hence, | Suppose that <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math> Then, <math>x\in\mathcal K,</math> <math>f(x)\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Hence, | ||

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for all <math>y\in\mathcal K.</math> Particularly, taking <math>y=0\,</math> and <math>y=2x\,</math>, respectively, we get <math>\langle x,f(x)\rangle=0.</math> Thus, <math>\langle y,f(x)\rangle\geq 0,</math> for all <math>y\in\mathcal K,</math> or equivalently <math>f(x)\in\mathcal K^*.</math> In conclusion, <math>x\in\mathcal K,</math> <math>f(x)\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Therefore, <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math> | for all <math>y\in\mathcal K.</math> Particularly, taking <math>y=0\,</math> and <math>y=2x\,</math>, respectively, we get <math>\langle x,f(x)\rangle=0.</math> Thus, <math>\langle y,f(x)\rangle\geq 0,</math> for all <math>y\in\mathcal K,</math> or equivalently <math>f(x)\in\mathcal K^*.</math> In conclusion, <math>x\in\mathcal K,</math> <math>f(x)\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Therefore, <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math> | ||

- | + | === Concluding the alternative proof === | |

Since <math>\mathcal K</math> is a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone], the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(f,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(f,\mathcal K),</math> which is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] <math>Fix(P_{\mathcal K}\circ(I-f)).</math> | Since <math>\mathcal K</math> is a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone], the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(f,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(f,\mathcal K),</math> which is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] <math>Fix(P_{\mathcal K}\circ(I-f)).</math> | ||

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</center> | </center> | ||

- | + | == Every implicit complementarity problem is equivalent to a fixed point problem == | |

Let <math>\mathcal K</math> be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f,g:\mathbb H\to\mathbb H</math> two mappings. Then, the [[Complementarity_problem#Implicit_complementarity_problems | implicit complementarity problem]] <math>ICP(f,g,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] | Let <math>\mathcal K</math> be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f,g:\mathbb H\to\mathbb H</math> two mappings. Then, the [[Complementarity_problem#Implicit_complementarity_problems | implicit complementarity problem]] <math>ICP(f,g,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] | ||

<math>Fix(I-g+P_{\mathcal K}\circ(g-f)),</math> where <math>I:\mathbb H\to\mathbb H</math> is the identity mapping defined by <math>I(x)=x.\,</math> | <math>Fix(I-g+P_{\mathcal K}\circ(g-f)),</math> where <math>I:\mathbb H\to\mathbb H</math> is the identity mapping defined by <math>I(x)=x.\,</math> | ||

- | + | == Proof == | |

For all <math>u\in\mathbb H</math> denote <math>z=g(u)-f(u),\,</math> <math>x=g(u)\,</math> and <math>y=-f(u).\,</math> Then, | For all <math>u\in\mathbb H</math> denote <math>z=g(u)-f(u),\,</math> <math>x=g(u)\,</math> and <math>y=-f(u).\,</math> Then, | ||

<math>z=x+y.\,</math> | <math>z=x+y.\,</math> | ||

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=== Remark === | === Remark === | ||

In particular if <math>g=I,</math> we obtain the result | In particular if <math>g=I,</math> we obtain the result | ||

- | [[ | + | [[Complementarity_problem#Every_nonlinear_complementarity_problem_is_equivalent_to_a_fixed_point_problem | ''Every nonlinear complementarity problem is equivalent to a fixed point problem'']], |

- | but the more general result [[ | + | but the more general result [[Complementarity_problem#Every_implicit_complementarity_problem_is_equivalent_to_a_fixed_point_problem | ''Every implicit complementarity problem is equivalent to a fixed point problem'']] has no known connection with [[Complementarity_problem#Variational_inequalities | variational inequalities]]. Therefore, using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] is essential for proving the latter result. |

## Revision as of 08:21, 20 July 2009

Sándor Zoltán Németh

## Fixed point problems

Let be a set and a mapping. The **fixed point problem** defined by is the problem

## Nonlinear complementarity problems

Let be a closed convex cone in the Hilbert space and a mapping. Recall that the dual cone of is the closed convex cone where is the polar of The **nonlinear complementarity problem** defined by and is the problem

## Every nonlinear complementarity problem is equivalent to a fixed point problem

Let be a closed convex cone in the Hilbert space and a mapping. Then, the nonlinear complementarity problem is equivalent to the fixed point problem where is the identity mapping defined by and is the projection onto

## Proof

For all denote and Then,

Suppose that is a solution of Then, with and Hence, by using Moreau's theorem, we get Therefore, is a solution of

Conversely, suppose that is a solution of Then, and by using Moreau's theorem

Hence, . Thus, . Moreau's theorem also implies that In conclusion, and Therefore, is a solution of

## An alternative proof without Moreau's theorem

### Variational inequalities

Let be a closed convex set in the Hilbert space and a mapping. The **variational inequality** defined by and is the problem

### Every variational inequality is equivalent to a fixed point problem

Let be a closed convex set in the Hilbert space and a mapping. Then the variational inequality is equivalent to the fixed point problem

### Proof

is a solution of if and only if By using the characterization of the projection the latter equation is equivalent to

for all But this holds if and only if is a solution of

### Remark

The next section shows that the equivalence of variational inequalities and fixed point problems is much stronger than the equivalence of nonlinear complementarity problems and fixed point problems, because each nonlinear complementarity problem is a variational inequality defined on a closed convex cone.

### Every variational inequality defined on a closed convex cone is equivalent to a complementarity problem

Let be a closed convex cone in the Hilbert space and a mapping. Then, the nonlinear complementarity problem is equivalent to the variational inequality

### Proof

Suppose that is a solution of Then, and Hence,

for all Therefore, is a solution of

Conversely, suppose that is a solution of Then, and

for all Particularly, taking and , respectively, we get Thus, for all or equivalently In conclusion, and Therefore, is a solution of

### Concluding the alternative proof

Since is a closed convex cone, the nonlinear complementarity problem is equivalent to the variational inequality which is equivalent to the fixed point problem

## Implicit complementarity problems

Let be a closed convex cone in the Hilbert space and two mappings. Recall that the dual cone of is the closed convex cone where is the
polar
of The **implicit complementarity problem** defined by
and the ordered pair of mappings is the problem

## Every implicit complementarity problem is equivalent to a fixed point problem

Let be a closed convex cone in the Hilbert space and two mappings. Then, the implicit complementarity problem is equivalent to the fixed point problem where is the identity mapping defined by

## Proof

For all denote and Then,

Suppose that is a solution of Then, with and Hence, by using
Moreau's theorem,
we get Therefore, is a solution of

Conversely, suppose that is a solution of Then, and by using Moreau's theorem

Hence, . Thus, . Moreau's theorem also implies that In conclusion, and Therefore, is a solution of

### Remark

In particular if we obtain the result
*Every nonlinear complementarity problem is equivalent to a fixed point problem*,
but the more general result *Every implicit complementarity problem is equivalent to a fixed point problem* has no known connection with variational inequalities. Therefore, using Moreau's theorem is essential for proving the latter result.