# Complementarity problem

(Difference between revisions)
 Revision as of 08:21, 20 July 2009 (edit)← Previous diff Revision as of 08:36, 20 July 2009 (edit) (undo)Next diff → Line 16: Line 16: == Nonlinear complementarity problems == == Nonlinear complementarity problems == - Let $\mathcal K$ be a closed [http://en.wikipedia.org/wiki/Convex_cone 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 [http://en.wikipedia.org/wiki/Convex_cone 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 [http://en.wikipedia.org/wiki/Convex_cone 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 [http://en.wikipedia.org/wiki/Convex_cone 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
$[itex] - NCP(f,\mathcal K):\left\{ + NCP(F,\mathcal K):\left\{ \begin{array}{l} \begin{array}{l} Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\ Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\ - f(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,f(x)\rangle=0. + F(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,F(x)\rangle=0. \end{array} \end{array} \right. \right. Line 30: Line 30: == Every nonlinear complementarity problem is equivalent to a fixed point problem == == Every nonlinear complementarity problem is equivalent to a fixed point problem == - Let [itex]\mathcal K$ be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] $NCP(f,\mathcal K)$ is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] + Let $\mathcal K$ be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $F:\mathbb H\to\mathbb H$ a mapping. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] $NCP(F,\mathcal K)$ is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] - $Fix(P_{\mathcal K}\circ(I-f)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x\,$ and $P_{\mathcal K}$ is the [[Moreau's_decomposition_theorem#Projection_on_closed_convex_sets | projection onto $\mathcal K.$]] + $Fix(P_{\mathcal K}\circ(I-F)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x\,$ and $P_{\mathcal K}$ is the [[Moreau's_decomposition_theorem#Projection_on_closed_convex_sets | projection onto $\mathcal K.$]] == 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, $z=x+y.\,$

- Suppose that $x\,$ is a solution of $NCP(f,\mathcal K).$ Then, $z=x+y,\,$ with $x\in\mathcal K,$ $y\in\mathcal K^\circ$ and $\langle x,y\rangle=0.$ Hence, by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], we get $x=P_{\mathcal K}z.$ Therefore, $x\,$ is a solution of $Fix(P_{\mathcal K}\circ(I-f)).$ + Suppose that $x\,$ is a solution of $NCP(F,\mathcal K).$ Then, $z=x+y,\,$ with $x\in\mathcal K,$ $y\in\mathcal K^\circ$ and $\langle x,y\rangle=0.$ Hence, by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], we get $x=P_{\mathcal K}z.$ Therefore, $x\,$ is a solution of $Fix(P_{\mathcal K}\circ(I-F)).$

- Conversely, suppose that $x\,$ is a solution of $Fix(P_{\mathcal K}\circ(I-f)).$ Then, $x\in\mathcal K$ and by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] + Conversely, suppose that $x\,$ is a solution of $Fix(P_{\mathcal K}\circ(I-F)).$ Then, $x\in\mathcal K$ and by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]]
Line 49: Line 49: 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
$[itex] - VI(f,\mathcal C):\left\{ + VI(F,\mathcal C):\left\{ \begin{array}{l} \begin{array}{l} Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ - \langle y-x,f(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C. + \langle y-x,F(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C. \end{array} \end{array} \right. \right. Line 68: Line 68: === Every variational inequality is equivalent to a fixed point problem === === Every variational inequality is equivalent to a fixed point problem === - Let [itex]\mathcal C$ 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 [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(f,\mathcal C)$ is equivalent to the [[Complementarity_problem#Fixed_point_problems | 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 [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(F,\mathcal C)$ is equivalent to the [[Complementarity_problem#Fixed_point_problems | 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
- $\langle x-f(x)-x,y-x\rangle\leq0,$ + $\langle x-F(x)-x,y-x\rangle\leq0,$
for all $y\in\mathcal C.$ But this holds if and only if $x\,$ is a solution for all $y\in\mathcal C.$ But this holds if and only if $x\,$ is a solution - of $VI(f,\mathcal C).$ + of $VI(F,\mathcal C).$ === Remark === === Remark === Line 85: Line 85: === 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 [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] $NCP(f,\mathcal K)$ is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(f,\mathcal K).$ + Let $\mathcal K$ be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $F:\mathbb H\to\mathbb H$ a mapping. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] $NCP(F,\mathcal K)$ is equivalent to the [[Complementarity_problem#Variational_inequalities | 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,
- $\langle y-x,f(x)\rangle\geq 0,$ + $\langle y-x,F(x)\rangle\geq 0,$
- for all $y\in\mathcal K.$ Therefore, $x\,$ is a solution of $VI(f,\mathcal K).$ + for all $y\in\mathcal K.$ Therefore, $x\,$ is a solution of $VI(F,\mathcal K).$

- Conversely, suppose that $x\,$ is a solution of $VI(f,\mathcal K).$ Then, + Conversely, suppose that $x\,$ is a solution of $VI(F,\mathcal K).$ Then, $x\in\mathcal K$ and $x\in\mathcal K$ and
- $\langle y-x,f(x)\rangle\geq 0,$ + $\langle y-x,F(x)\rangle\geq 0,$
- for all $y\in\mathcal K.$ Particularly, taking $y=0\,$ and $y=2x\,$, respectively, we get $\langle x,f(x)\rangle=0.$ Thus, $\langle y,f(x)\rangle\geq 0,$ for all $y\in\mathcal K,$ or equivalently $f(x)\in\mathcal K^*.$ In conclusion, $x\in\mathcal K,$ $f(x)\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Therefore, $x\,$ is a solution of $NCP(f,\mathcal K).$ + for all $y\in\mathcal K.$ Particularly, taking $y=0\,$ and $y=2x\,$, respectively, we get $\langle x,F(x)\rangle=0.$ Thus, $\langle y,F(x)\rangle\geq 0,$ for all $y\in\mathcal K,$ or equivalently $F(x)\in\mathcal K^*.$ In conclusion, $x\in\mathcal K,$ $F(x)\in\mathcal K^*$ and $\langle x,F(x)\rangle=0.$ Therefore, $x\,$ is a solution of $NCP(F,\mathcal K).$ === Concluding the alternative proof === === Concluding the alternative proof === - Since $\mathcal K$ is a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone], the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] $NCP(f,\mathcal K)$ is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(f,\mathcal K),$ which is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] $Fix(P_{\mathcal K}\circ(I-f)).$ + Since $\mathcal K$ is a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone], the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] $NCP(F,\mathcal K)$ is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(F,\mathcal K),$ which is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] $Fix(P_{\mathcal K}\circ(I-F)).$ == Implicit complementarity problems == == Implicit complementarity problems == - Let $\mathcal K$ be a closed [http://en.wikipedia.org/wiki/Convex_cone 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 [http://en.wikipedia.org/wiki/Convex_cone convex cone] $\mathcal K^*=-\mathcal K^\circ,$ where $\mathcal K^\circ$ is the + Let $\mathcal K$ be a closed [http://en.wikipedia.org/wiki/Convex_cone 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 [http://en.wikipedia.org/wiki/Convex_cone 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]] of $\mathcal K.$ The '''implicit complementarity problem''' defined by $\mathcal K$ of $\mathcal K.$ The '''implicit complementarity problem''' defined by $\mathcal K$ - and the ordered pair of mappings $(f,g)\,$ is the problem + and the ordered pair of mappings $(F,G)\,$ is the problem
$[itex] - ICP(f,g,\mathcal K):\left\{ + ICP(F,G,\mathcal K):\left\{ \begin{array}{l} \begin{array}{l} Find\,\,\,u\in\mathbb H\,\,\,such\,\,\,that\\ Find\,\,\,u\in\mathbb H\,\,\,such\,\,\,that\\ - g(u)\in\mathcal K,\,\,\,f(u)\in K^*\,\,\,and\,\,\,\langle g(u),f(u)\rangle=0. + G(u)\in\mathcal K,\,\,\,F(u)\in K^*\,\,\,and\,\,\,\langle G(u),F(u)\rangle=0. \end{array} \end{array} \right. \right. Line 128: Line 128: == Every implicit complementarity problem is equivalent to a fixed point problem == == Every implicit complementarity problem is equivalent to a fixed point problem == - Let [itex]\mathcal K$ be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f,g:\mathbb H\to\mathbb H$ two mappings. Then, the [[Complementarity_problem#Implicit_complementarity_problems | implicit complementarity problem]] $ICP(f,g,\mathcal K)$ is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] + Let $\mathcal K$ be a closed [http://en.wikipedia.org/wiki/Convex_cone convex cone] in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $F,G:\mathbb H\to\mathbb H$ two mappings. Then, the [[Complementarity_problem#Implicit_complementarity_problems | implicit complementarity problem]] $ICP(F,G,\mathcal K)$ is equivalent to the [[Complementarity_problem#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 $z=G(u)-G(u),\,$ $x=G(u)\,$ and $y=-F(u).\,$ Then, $z=x+y.\,$ $z=x+y.\,$

- Suppose that $u\,$ is a solution of $ICP(f,g,\mathcal K).$ Then, $z=x+y,\,$ with $x\in\mathcal K,$ $y\in\mathcal K^\circ$ and $\langle x,y\rangle=0.$ Hence, by using + Suppose that $u\,$ is a solution of $ICP(F,G,\mathcal K).$ Then, $z=x+y,\,$ with $x\in\mathcal K,$ $y\in\mathcal K^\circ$ and $\langle x,y\rangle=0.$ Hence, by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], we get $x=P_{\mathcal K}z.$ Therefore, $u\,$ is a solution of we get $x=P_{\mathcal K}z.$ Therefore, $u\,$ is a solution of - $Fix(I-g+P_{\mathcal K}\circ(g-f)).$ + $Fix(I-G+P_{\mathcal K}\circ(G-F)).$

- Conversely, suppose that $u\,$ is a solution of $Fix(I-g+P_{\mathcal K}\circ(g-f)).$ + Conversely, suppose that $u\,$ is a solution of $Fix(I-G+P_{\mathcal K}\circ(G-F)).$ Then, $x\in\mathcal K$ and by using Then, $x\in\mathcal K$ and by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] Line 155: Line 155: [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] also implies that $\langle x,y\rangle=0.$ In conclusion, also implies that $\langle x,y\rangle=0.$ In conclusion, - $g(u)=x\in\mathcal K,$ $f(u)=-y\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Therefore, $u\,$ is a solution of $ICP(f,g,\mathcal K).$ + $G(u)=x\in\mathcal K,$ $F(u)=-y\in\mathcal K^*$ and $\langle G(u),F(u)\rangle=0.$ Therefore, $u\,$ is a solution of $ICP(F,G,\mathcal K).$ === Remark === === Remark ===

## Revision as of 08:36, 20 July 2009

Sándor Zoltán Németh

## 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 $LaTeX: I(x)=x\,$ and $LaTeX: P_{\mathcal K}$ is the projection onto $LaTeX: \mathcal K.$

## 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 $LaTeX: z=G(u)-G(u),\,$ $LaTeX: x=G(u)\,$ 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 $LaTeX: \langle G(u),F(u)\rangle=0.$ 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.