Complementarity problem

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== Implicit complementarity problems ==
== Implicit complementarity problems ==
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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. Recall that the dual cone of <math>\mathcal K</math> is the closed convex cone <math>\mathcal K^*=-\mathcal K^\circ,</math> where <math>\mathcal K^\circ</math> is the
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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. 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]]
[[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]]
of <math>\mathcal K.</math> The '''implicit complementarity problem''' defined by <math>\mathcal K</math>
of <math>\mathcal K.</math> The '''implicit complementarity problem''' defined by <math>\mathcal K</math>

Revision as of 16:03, 17 July 2009

Sándor Zoltán Németh

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\, 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)-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.

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