# Complementarity problem

(Difference between revisions)
 Revision as of 11:11, 17 July 2009 (edit) (New page: == An application to nonlinear complementarity problems == === Fixed point problems === Let $\mathcal A$ be a set and $F:\mathcal A\to\mathcal A$ a mapping. The ''...)← Previous diff Revision as of 11:17, 17 July 2009 (edit) (undo)Next diff → Line 1: Line 1: - == An application to nonlinear complementarity problems == - === Fixed point problems === === Fixed point problems === - Let $\mathcal A$ be a set and $F:\mathcal A\to\mathcal A$ a mapping. The '''fixed point problem''' defined by $F\,$ is the problem Let $\mathcal A$ be a set and $F:\mathcal A\to\mathcal A$ a mapping. The '''fixed point problem''' defined by $F\,$ is the problem Line 17: Line 14: === Nonlinear complementarity problems === === Nonlinear complementarity problems === - Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Recall that the dual cone of $\mathcal K$ is the closed convex cone $\mathcal K^*=-\mathcal K^\circ,$ where $\mathcal K^\circ$ is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of $\mathcal K.$ The '''nonlinear complementarity problem''' defined by $\mathcal K$ and $f\,$ is the problem Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Recall that the dual cone of $\mathcal K$ is the closed convex cone $\mathcal K^*=-\mathcal K^\circ,$ where $\mathcal K^\circ$ is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of $\mathcal K.$ The '''nonlinear complementarity problem''' defined by $\mathcal K$ and $f\,$ is the problem Line 32: Line 28: === Every nonlinear complementarity problem is equivalent to a fixed point problem === === Every nonlinear complementarity problem is equivalent to a fixed point problem === - Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the fixed point problem Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the fixed point problem $Fix(P_{\mathcal K}\circ(I-f)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x.\,$ $Fix(P_{\mathcal K}\circ(I-f)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x.\,$ === Proof === === Proof === - For all $x\in\mathbb H$ denote $z=x-f(x)\,$ and $y=-f(x).\,$ Then, $z=x+y.\,$ For all $x\in\mathbb H$ denote $z=x-f(x)\,$ and $y=-f(x).\,$ Then, $z=x+y.\,$

Line 54: Line 48: Hence, $P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $y\in\mathcal K^\circ$. [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] also implies that $\langle x,y\rangle=0.$ In conclusion, Hence, $P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $y\in\mathcal K^\circ$. [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] also implies that $\langle x,y\rangle=0.$ In conclusion, $x\in\mathcal K,$ $f(x)=-y\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Therefore, $x\,$ is a solution of $NCP(f,\mathcal K).$ $x\in\mathcal K,$ $f(x)=-y\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Therefore, $x\,$ is a solution of $NCP(f,\mathcal K).$ + === An alternative proof without [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] === === An alternative proof without [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] === - ==== Variational inequalities ==== ==== Variational inequalities ==== - Let $\mathcal C$ be a closed convex set in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. The '''variational inequality''' defined by $\mathcal C$ and $f\,$ is the problem Let $\mathcal C$ be a closed convex set in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. The '''variational inequality''' defined by $\mathcal C$ and $f\,$ is the problem Line 73: Line 66: ==== Every variational inequality is equivalent to a fixed point problem ==== ==== Every variational inequality is equivalent to a fixed point problem ==== - Let $\mathcal C$ be a closed convex set in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then the variational inequality $VI(f,\mathcal C)$ is equivalent to the fixed point problem $Fix(P_{\mathcal C}\circ(I-f)).$ Let $\mathcal C$ be a closed convex set in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then the variational inequality $VI(f,\mathcal C)$ is equivalent to the fixed point problem $Fix(P_{\mathcal C}\circ(I-f)).$ ==== Proof ==== ==== Proof ==== - $x\,$ is a solution of $Fix(P_{\mathcal C}\circ(I-f))$ if and only if $x\,$ is a solution of $Fix(P_{\mathcal C}\circ(I-f))$ if and only if $x=P_{\mathcal C}(x-f(x)).$ By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection | characterization of the projection]] the latter equation is equivalent to $x=P_{\mathcal C}(x-f(x)).$ By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection | characterization of the projection]] the latter equation is equivalent to Line 89: Line 80: ===== Remark ===== ===== 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. 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 ==== ==== Every variational inequality defined on a closed convex cone is equivalent to a complementarity problem ==== - Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the variational inequality $VI(f,\mathcal K).$ Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the variational inequality $VI(f,\mathcal K).$ ==== Proof ==== ==== Proof ==== - Suppose that $x\,$ is a solution of $NCP(f,\mathcal K).$ Then, $x\in\mathcal K,$ $f(x)\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Hence, Suppose that $x\,$ is a solution of $NCP(f,\mathcal K).$ Then, $x\in\mathcal K,$ $f(x)\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Hence, Line 118: Line 106: ==== Concluding the alternative proof ==== ==== Concluding the alternative proof ==== - Since $\mathcal K$ is a closed convex cone, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the variational inequality $VI(f,\mathcal K),$ which is equivalent to the fixed point problem $Fix(P_{\mathcal K}\circ(I-f)).$ Since $\mathcal K$ is a closed convex cone, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the variational inequality $VI(f,\mathcal K),$ which is equivalent to the fixed point problem $Fix(P_{\mathcal K}\circ(I-f)).$ - == An application to implicit complementarity problems == + == Implicit complementarity problems == - === Implicit complementarity problems === + - + Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f,g:\mathbb H\to\mathbb H$ two mappings. Recall that the dual cone of $\mathcal K$ is the closed convex cone $\mathcal K^*=-\mathcal K^\circ,$ where $\mathcal K^\circ$ is the Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f,g:\mathbb H\to\mathbb H$ two mappings. Recall that the dual cone of $\mathcal K$ is the closed convex cone $\mathcal K^*=-\mathcal K^\circ,$ where $\mathcal K^\circ$ is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] Line 141: Line 126: === Every implicit complementarity problem is equivalent to a fixed point problem === === Every implicit complementarity problem is equivalent to a fixed point problem === - Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f,g:\mathbb H\to\mathbb H$ two mappings. Then, the implicit complementarity problem $ICP(f,g,\mathcal K)$ is equivalent to the [[Moreau's_decomposition_theorem#Fixed_point_problems | fixed point problem]] Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f,g:\mathbb H\to\mathbb H$ two mappings. Then, the implicit complementarity problem $ICP(f,g,\mathcal K)$ is equivalent to the [[Moreau's_decomposition_theorem#Fixed_point_problems | fixed point problem]] $Fix(I-g+P_{\mathcal K}\circ(g-f)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x.\,$ $Fix(I-g+P_{\mathcal K}\circ(g-f)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x.\,$ === Proof === === Proof === - For all $u\in\mathbb H$ denote $z=g(u)-f(u),\,$ $x=g(u)\,$ and $y=-f(u).\,$ Then, For all $u\in\mathbb H$ denote $z=g(u)-f(u),\,$ $x=g(u)\,$ and $y=-f(u).\,$ Then, $z=x+y.\,$ $z=x+y.\,$ Line 173: Line 156: === Remark === === Remark === - In particular if $g=I,$ we obtain the result In particular if $g=I,$ we obtain the result [[Moreau%27s_decomposition_theorem#Every_nonlinear_complementarity_problem_is_equivalent_to_a_fixed_point_problem | Every nonlinear complementarity problem is equivalent to a fixed point problem]], [[Moreau%27s_decomposition_theorem#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 [[Moreau%27s_decomposition_theorem#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 variational inequalities. Therefore, using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] is essential for proving the latter result. but the more general result [[Moreau%27s_decomposition_theorem#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 variational inequalities. Therefore, using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] is essential for proving the latter result.

## Contents

### Fixed point problems

Let $LaTeX: \mathcal A$ be a set and $LaTeX: F:\mathcal A\to\mathcal A$ a mapping. The fixed point problem defined by $LaTeX: F\,$ is the problem

$LaTeX: Fix(F):\left\{ \begin{array}{l} Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\ F(x)=x. \end{array} \right.$

### Nonlinear complementarity problems

Let $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathbb H\to\mathbb H$ a mapping. Recall that the dual cone of $LaTeX: \mathcal K$ is the closed convex cone $LaTeX: \mathcal K^*=-\mathcal K^\circ,$ where $LaTeX: \mathcal K^\circ$ is the polar of $LaTeX: \mathcal K.$ The nonlinear complementarity problem defined by $LaTeX: \mathcal K$ and $LaTeX: f\,$ is the problem

$LaTeX: NCP(f,\mathcal K):\left\{ \begin{array}{l} Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\ f(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,f(x)\rangle=0. \end{array} \right.$

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

Let $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathbb H\to\mathbb H$ a mapping. Then, the nonlinear complementarity problem $LaTeX: NCP(f,\mathcal K)$ is equivalent to the fixed point problem $LaTeX: Fix(P_{\mathcal K}\circ(I-f)),$ where $LaTeX: I:\mathbb H\to\mathbb H$ is the identity mapping defined by $LaTeX: I(x)=x.\,$

### Proof

For all $LaTeX: x\in\mathbb H$ denote $LaTeX: z=x-f(x)\,$ and $LaTeX: y=-f(x).\,$ Then, $LaTeX: z=x+y.\,$

Suppose that $LaTeX: x\,$ is a solution of $LaTeX: NCP(f,\mathcal K).$ Then, $LaTeX: z=x+y,\,$ with $LaTeX: x\in\mathcal K,$ $LaTeX: y\in\mathcal K^\circ$ and $LaTeX: \langle x,y\rangle=0.$ Hence, by using Moreau's theorem, we get $LaTeX: x=P_{\mathcal K}z.$ Therefore, $LaTeX: x\,$ is a solution of $LaTeX: Fix(P_{\mathcal K}\circ(I-f)).$

Conversely, suppose that $LaTeX: x\,$ is a solution of $LaTeX: Fix(P_{\mathcal K}\circ(I-f)).$ Then, $LaTeX: x\in\mathcal K$ and by using Moreau's theorem

$LaTeX: z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).$

Hence, $LaTeX: P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $LaTeX: y\in\mathcal K^\circ$. Moreau's theorem also implies that $LaTeX: \langle x,y\rangle=0.$ In conclusion, $LaTeX: x\in\mathcal K,$ $LaTeX: f(x)=-y\in\mathcal K^*$ and $LaTeX: \langle x,f(x)\rangle=0.$ Therefore, $LaTeX: x\,$ is a solution of $LaTeX: NCP(f,\mathcal K).$

### An alternative proof without Moreau's theorem

#### Variational inequalities

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

$LaTeX: VI(f,\mathcal C):\left\{ \begin{array}{l} Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ \langle y-x,f(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C. \end{array} \right.$

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

Let $LaTeX: \mathcal C$ be a closed convex set in the Hilbert space $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathbb H\to\mathbb H$ a mapping. Then the variational inequality $LaTeX: VI(f,\mathcal C)$ is equivalent to the fixed point problem $LaTeX: Fix(P_{\mathcal C}\circ(I-f)).$

#### Proof

$LaTeX: x\,$ is a solution of $LaTeX: Fix(P_{\mathcal C}\circ(I-f))$ if and only if $LaTeX: x=P_{\mathcal C}(x-f(x)).$ By using the characterization of the projection the latter equation is equivalent to

$LaTeX: \langle x-f(x)-x,y-x\rangle\leq0,$

for all $LaTeX: y\in\mathcal C.$ But this holds if and only if $LaTeX: x\,$ is a solution of $LaTeX: VI(f,\mathcal C).$

##### 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 $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathbb H\to\mathbb H$ a mapping. Then, the nonlinear complementarity problem $LaTeX: NCP(f,\mathcal K)$ is equivalent to the variational inequality $LaTeX: VI(f,\mathcal K).$

#### Proof

Suppose that $LaTeX: x\,$ is a solution of $LaTeX: NCP(f,\mathcal K).$ Then, $LaTeX: x\in\mathcal K,$ $LaTeX: f(x)\in\mathcal K^*$ and $LaTeX: \langle x,f(x)\rangle=0.$ Hence,

$LaTeX: \langle y-x,f(x)\rangle\geq 0,$

for all $LaTeX: y\in\mathcal K.$ Therefore, $LaTeX: x\,$ is a solution of $LaTeX: VI(f,\mathcal K).$

Conversely, suppose that $LaTeX: x\,$ is a solution of $LaTeX: VI(f,\mathcal K).$ Then, $LaTeX: x\in\mathcal K$ and

$LaTeX: \langle y-x,f(x)\rangle\geq 0,$

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

#### Concluding the alternative proof

Since $LaTeX: \mathcal K$ is a closed convex cone, the nonlinear complementarity problem $LaTeX: NCP(f,\mathcal K)$ is equivalent to the variational inequality $LaTeX: VI(f,\mathcal K),$ which is equivalent to the fixed point problem $LaTeX: Fix(P_{\mathcal K}\circ(I-f)).$

## Implicit complementarity problems

Let $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f,g:\mathbb H\to\mathbb H$ two mappings. Recall that the dual cone of $LaTeX: \mathcal K$ is the closed convex cone $LaTeX: \mathcal K^*=-\mathcal K^\circ,$ where $LaTeX: \mathcal K^\circ$ is the polar of $LaTeX: \mathcal K.$ The implicit complementarity problem defined by $LaTeX: \mathcal K$ and the ordered pair of mappings $LaTeX: (f,g)\,$ is the problem

$LaTeX: ICP(f,g,\mathcal K):\left\{ \begin{array}{l} Find\,\,\,u\in\mathbb H\,\,\,such\,\,\,that\\ g(u)\in\mathcal K,\,\,\,f(u)\in K^*\,\,\,and\,\,\,\langle g(u),f(u)\rangle=0. \end{array} \right.$

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

Let $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f,g:\mathbb H\to\mathbb H$ two mappings. Then, the implicit complementarity problem $LaTeX: ICP(f,g,\mathcal K)$ is equivalent to the fixed point problem $LaTeX: Fix(I-g+P_{\mathcal K}\circ(g-f)),$ where $LaTeX: I:\mathbb H\to\mathbb H$ is the identity mapping defined by $LaTeX: I(x)=x.\,$

### Proof

For all $LaTeX: u\in\mathbb H$ denote $LaTeX: z=g(u)-f(u),\,$ $LaTeX: x=g(u)\,$ and $LaTeX: y=-f(u).\,$ Then, $LaTeX: z=x+y.\,$

Suppose that $LaTeX: u\,$ is a solution of $LaTeX: ICP(f,g,\mathcal K).$ Then, $LaTeX: z=x+y,\,$ with $LaTeX: x\in\mathcal K,$ $LaTeX: y\in\mathcal K^\circ$ and $LaTeX: \langle x,y\rangle=0.$ Hence, by using Moreau's theorem, we get $LaTeX: x=P_{\mathcal K}z.$ Therefore, $LaTeX: u\,$ is a solution of $LaTeX: Fix(I-g+P_{\mathcal K}\circ(g-f)).$

Conversely, suppose that $LaTeX: u\,$ is a solution of $LaTeX: Fix(I-g+P_{\mathcal K}\circ(g-f)).$ Then, $LaTeX: x\in\mathcal K$ and by using Moreau's theorem

$LaTeX: z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).$

Hence, $LaTeX: P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $LaTeX: y\in\mathcal K^\circ$. Moreau's theorem also implies that $LaTeX: \langle x,y\rangle=0.$ In conclusion, $LaTeX: g(u)=x\in\mathcal K,$ $LaTeX: f(u)=-y\in\mathcal K^*$ and $LaTeX: \langle x,f(x)\rangle=0.$ Therefore, $LaTeX: u\,$ is a solution of $LaTeX: ICP(f,g,\mathcal K).$

### Remark

In particular if $LaTeX: g=I,$ 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.