Complementarity problem

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(A convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem)
(Any solution of a fat nonlinear programming problem is a solution of a nonlinear complementarity problem defined by a polyhedral cone)
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Let <math>f:\mathbb R^n\to\mathbb R</math> be a differentiable function,
Let <math>f:\mathbb R^n\to\mathbb R</math> be a differentiable function,
<math>b\in\mathbb R^m,</math> and
<math>b\in\mathbb R^m,</math> and
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<math>A\in\mathbb R^{m\times n}</math> a matrix of full rank <math>m\leq n.</math> If <math>x\in\mathbb R^n</math> is a solution of the [[Complementarity_problem#Fat_nonlinear_programming_problem|fat nonlinear programming problem]]
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<math>A\in\mathbb R^{m\times n}</math> a fat matrix of full rank <math>m\leq n.</math> If <math>x\in\mathbb R^n</math> is a solution of the [[Complementarity_problem#Fat_nonlinear_programming_problem|fat nonlinear programming problem]]
<math>NP(f,A,b),\,</math> then <math>x-x_0\in\mathbb R^n</math> is a solution of the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear
<math>NP(f,A,b),\,</math> then <math>x-x_0\in\mathbb R^n</math> is a solution of the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear
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complementarity problem]] <math>NCP(G,\mathcal K)</math> where <math>x_0\in\mathbb R^n</math> is
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complementarity problem]] <math>NCP(\nabla f(x+x_0),\mathcal K)</math> where <math>x_0\in\mathbb R^n</math> is
a particular solution of the linear system of equations <math>Ax=b,\,</math>
a particular solution of the linear system of equations <math>Ax=b,\,</math>
<math>\mathcal K</math> is the polyhedral cone defined by
<math>\mathcal K</math> is the polyhedral cone defined by
<center>
<center>
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<math>K=\{x\mid Ax\leq0\},</math>
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<math>K=\{x\mid Ax\leq0\}</math>
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</center>
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and
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<math>G:\mathbb R^n\to\mathbb R^n</math> is defined by
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<center>
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<math>G(x)=\nabla f(x+x_0).</math>
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</center>
</center>
=== Proof ===
=== Proof ===
Let <math>x\in\mathbb R^n</math> be a solution of <math>NP(f,A,b).\,</math> Then it is easy to see that <math>x-x_0\,</math> is a
Let <math>x\in\mathbb R^n</math> be a solution of <math>NP(f,A,b).\,</math> Then it is easy to see that <math>x-x_0\,</math> is a
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solution of <math>NP(g,A,0)\,</math> where <math>g:\mathbb R^n\to\mathbb R</math> is defined by
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solution of <math>\,NP\left(f(x+x_0),A,0\right).</math>
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<math>g(x)=f(x+x_0).\,</math>
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[[Complementarity_problem#Any_solution_of_a_nonlinear_optimization_problem_on_a_closed_convex_cone_is_a_solution_of_a_nonlinear_complementarity_problem|It follows that]] <math>x-x_0\,</math> is a solution of <math>NCP(\nabla f(x+x_0),\mathcal K).</math>
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[[Complementarity_problem#Any_solution_of_a_nonlinear_optimization_problem_on_a_closed_convex_cone_is_a_solution_of_a_nonlinear_complementarity_problem|It follows that]] <math>x-x_0\,</math> is a solution of <math>NCP(G,\mathcal K)</math> because <math>G(x)=\nabla f(x+x_0)=\nabla g(x).</math>
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=== Remark ===
=== Remark ===

Revision as of 22:39, 17 August 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 via 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 via 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

If LaTeX: G=I, in particular, 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. Using Moreau's theorem is therefore essential for proving the latter result.

Nonlinear optimization problems

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 R a function. The nonlinear optimization problem defined by LaTeX: \mathcal C and LaTeX: f\, is the problem

LaTeX: 
NOPT(f,\mathcal C):\left\{
\begin{array}{l} 
Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ 
f(x)\leq f(y)\,\,\,for\,\,\,all\,\,\,y\in C.
\end{array}
\right.

Any solution of a nonlinear optimization problem is a solution of a variational inequality

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 R a differentiable function. Then any solution of the nonlinear optimization problem LaTeX: NOPT(f,\mathcal C) is a solution of the variational inequality LaTeX: VI(\nabla f,\mathcal C) where LaTeX: \nabla f is the gradient of LaTeX: f.\,

Proof

Let LaTeX: x\,\in\mathcal C be a solution of LaTeX: NOPT(f,\mathcal C) and LaTeX: y\in\mathcal C an arbitrary point. Then by convexity of LaTeX: \mathcal C we have LaTeX: x+t(y-x)\in\mathcal C. Hence, LaTeX: f(x)\leq f(x+t(y-x)) and therefore

LaTeX: \langle \nabla f(x),y-x\rangle=\displaystyle\lim_{t\searrow 0}\frac{f(x+t(y-x))-f(x)}t\geq0.

Therefore, LaTeX: x\, is a solution of LaTeX: VI(\nabla f,\mathcal C).

A convex optimization problem is equivalent to a variational inequality

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 R a differentiable convex function. Then the nonlinear optimization problem LaTeX: NOPT(f,\mathcal C) is equivalent to the variational inequality LaTeX: VI(\nabla f,\mathcal C) where LaTeX: \nabla f is the gradient of LaTeX: f.\,

Proof

Any solution of LaTeX: NOPT(f,\mathcal C) is a solution of LaTeX: VI(\nabla f,\mathcal C).

Conversely, suppose that LaTeX: x\, is a solution of LaTeX: VI(\nabla f,\mathcal C). Hence, by using the convexity of LaTeX: f\, we have LaTeX: f(y)-f(x)\geq\langle\nabla f(x),y-x\rangle\geq0, for all LaTeX: y\in\mathcal C. Therefore, LaTeX: x\, is a solution of LaTeX: NOPT(f,\mathcal C).

Any solution of a nonlinear optimization problem on a closed convex cone is a solution of a nonlinear 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 R a differentiable function. Then any solution of the nonlinear optimization problem LaTeX: NOPT(f,\mathcal K) is a solution of the nonlinear complementarity problem LaTeX: NCP(\nabla f,\mathcal K).

Proof

Any solution of LaTeX: NOPT(f,\mathcal K) is a solution of LaTeX: VI(\nabla f,\mathcal K) which is equivalent to LaTeX: NCP(\nabla f,\mathcal K).

A convex optimization problem on a closed convex cone is equivalent to a nonlinear 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 R a differentiable convex function. Then the nonlinear optimization problem LaTeX: NOPT(f,\mathcal K) is equivalent to the nonlinear complementarity problem LaTeX: NCP(\nabla f,\mathcal K).

Proof

LaTeX: NOPT(f,\mathcal K) is equivalent to LaTeX: VI(\nabla f,\mathcal K) which is equivalent to LaTeX: NCP(\nabla f,\mathcal K).

Fat nonlinear programming problem

Let LaTeX: f:\mathbb R^n\to\mathbb R be a function, LaTeX: b\in\mathbb R^n, and LaTeX: A\in\mathbb R^{m\times n} a fat matrix of full rank LaTeX: m\leq n. Then the problem

LaTeX: 
NP(f,A,b):\left\{
\begin{array}{rl}
	Minimize & f(x)\\
	Subject\,\,\,to & Ax\leq b
\end{array}
\right.

is called fat nonlinear programming problem.

Any solution of a fat nonlinear programming problem is a solution of a nonlinear complementarity problem defined by a polyhedral cone

Let LaTeX: f:\mathbb R^n\to\mathbb R be a differentiable function, LaTeX: b\in\mathbb R^m, and LaTeX: A\in\mathbb R^{m\times n} a fat matrix of full rank LaTeX: m\leq n. If LaTeX: x\in\mathbb R^n is a solution of the fat nonlinear programming problem LaTeX: NP(f,A,b),\, then LaTeX: x-x_0\in\mathbb R^n is a solution of the nonlinear complementarity problem LaTeX: NCP(\nabla f(x+x_0),\mathcal K) where LaTeX: x_0\in\mathbb R^n is a particular solution of the linear system of equations LaTeX: Ax=b,\, LaTeX: \mathcal K is the polyhedral cone defined by

LaTeX: K=\{x\mid Ax\leq0\}

Proof

Let LaTeX: x\in\mathbb R^n be a solution of LaTeX: NP(f,A,b).\, Then it is easy to see that LaTeX: x-x_0\, is a solution of LaTeX: \,NP\left(f(x+x_0),A,0\right). It follows that LaTeX: x-x_0\, is a solution of LaTeX: NCP(\nabla f(x+x_0),\mathcal K).

Remark

If LaTeX: f\, is convex, then the converse of the above results also holds.

We note that there are also many nonlinear programming problems defined by skinny matrices (i.e., m>n) that can be reduced to complementarity problems.

Since a very large class of nonlinear programming problems can be reduced to nonlinear complementarity problems, the importance of nonlinear complementarity problems on polyhedral cones is obvious both from theoretical and practical point of view.

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