Complementarity problem

From Wikimization

(Difference between revisions)

Revision as of 17:50, 16 August 2009

Sándor Zoltán Németh

1 Fixed point problems
2 Nonlinear complementarity problems
3 Every nonlinear complementarity problem is equivalent to a fixed point problem
- 3.1 Proof
- 3.2 An alternative proof without Moreau's theorem
4 Implicit complementarity problems
5 Every implicit complementarity problem is equivalent to a fixed point problem
- 5.1 Proof
- 5.2 Remark
6 Nonlinear optimization problems
7 Any solution of a nonlinear optimization problem is a solution of a variational inequality
- 7.1 Proof
8 A convex optimization problem is equivalent to a variational inequality
- 8.1 Proof
9 Any solution of a nonlinear optimization problem on a closed convex cone is a solution of a nonlinear complementarity problem
- 9.1 Proof
10 A convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem
- 10.1 Proof
11 Fat nonlinear programming problem
12 Any solution of a fat nonlinear programming problem is a solution of a nonlinear complementarity problem defined by a polyhedral cone
- 12.1 Proof
- 12.2 Remark

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

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.

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 $LaTeX: VI(F,\mathcal C),$ where $LaTeX: F=\nabla f$ 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 the 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

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

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

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 $LaTeX: VI(F,\mathcal C),$ where $LaTeX: F=\nabla f$ is the gradient of $LaTeX: f.\,$

Proof

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

Conversely, suppose that $LaTeX: x\,$ is a solution of $LaTeX: VI(F,\mathcal C).$ Hence, by using the convexity of $LaTeX: f\,$ we have $LaTeX: f(y)-f(x)\geq\langle F(x),y-x\rangle\geq0,$ 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 $LaTeX: NCP(F,\mathcal K),$ where $LaTeX: F=\nabla f$ is the gradient of $LaTeX: f.\,$

Proof

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

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 $LaTeX: NCP(F,\mathcal K),$ where $LaTeX: F=\nabla f$ is the gradient of $LaTeX: f.\,$

Proof

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

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 matrix of full rank $LaTeX: m,\,$ where $LaTeX: m\leq n.$ 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 matrix of full rank $LaTeX: m,\,$ where $LaTeX: m\leq n.$ 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 $LaTeX: NCP(G,\mathcal K,0),$ 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

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

and $LaTeX: G:\mathbb R^n\to\mathbb R^n$ is defined by

$LaTeX: G(x)=\nabla f(x+x_0).$

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 $LaTeX: NP(g,A,0),\,$ where $LaTeX: g:\mathbb R^n\to\mathbb R$ is defined by $LaTeX: g(x)=f(x+x_0).\,$ It follows that $LaTeX: x-x_0\,$ is a solution of $LaTeX: NCP(G,\mathcal K,0),$ because $LaTeX: G(x)=\nabla f(x+x_0)=\nabla g(x).$

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) which 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.

Retrieved from "http://www.convexoptimization.com/wikimization/index.php/Complementarity_problem"

@@ Line 131: / Line 131: @@
 <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 ==
+=== Proof ===
 For all <math>u\in\mathbb H</math> denote <math>z=G(u)-G(u),\,</math> <math>x=G(u)\,</math> and <math>y=-F(u).\,</math> Then,
 <math>z=x+y.\,</math>
@@ Line 157: / Line 157: @@
 <math>G(u)=x\in\mathcal K,</math> <math>F(u)=-y\in\mathcal K^*</math> and <math>\langle G(u),F(u)\rangle=0.</math> Therefore, <math>u\,</math> is a solution of <math>ICP(F,G,\mathcal K).</math>
-== Remark ==
+=== Remark ===
 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'']],
@@ Line 163: / Line 163: @@
 == Nonlinear optimization problems ==
 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 R</math> a function. The
 '''nonlinear optimization problem''' defined by <math>\mathcal C</math> and
@@ Line 180: / Line 179: @@
 == Any solution of a nonlinear optimization problem is a solution of a variational inequality ==
 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 R</math> a  differentiable
 function. Then, any solution of the [[Complementarity_problem#Nonlinear_optimization_problems| nonlinear optimization problem ]] <math>NOPT(f,\mathcal C)</math> is a solution of the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(F,\mathcal C),</math> where <math>F=\nabla f</math> is the gradient of <math>f.\,</math>
-== Proof ==
+=== Proof ===
 Let <math>x\,\in\mathcal C</math> be a solution of <math>NOPT(f,\mathcal C)</math> and
 <math>y\in\mathcal C</math> an arbitrary point. Then, by the convexity of
@@ Line 196: / Line 193: @@
 == A convex optimization problem is equivalent to a variational inequality ==
 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 R</math> a
@@ Line 202: / Line 198: @@
 Then, the [[Complementarity_problem#Nonlinear_optimization_problems | nonlinear optimization problem]] <math>NOPT(f,\mathcal C)</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(F,\mathcal C),</math> where <math>F=\nabla f</math> is the gradient of <math>f.\,</math>
-== Proof ==
+=== Proof ===
 Any solution of <math>NOPT(f,\mathcal C)</math>
 [[Complementarity_problem#Any_solution_of_a_nonlinear_optimization_problem_is_a_solution_of_a_variational_inequality | is a solution of]] <math>VI(F,\mathcal C).</math>
@@ Line 213: / Line 208: @@
 == Any solution of a nonlinear optimization problem on a closed convex cone is a solution of a nonlinear complementarity problem ==
 Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space
 <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb R</math> a differentiable function. Then any solution of the [[Complementarity_problem#Nonlinear_optimization_problems | nonlinear optimization problem]]  <math>NOPT(f,\mathcal K)</math> is a
 solution of the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(F,\mathcal K),</math> where <math>F=\nabla f</math> is the gradient of <math>f.\,</math>
-== Proof ==
+=== Proof ===
 Any solution of <math>NOPT(f,\mathcal K)</math> [[Complementarity_problem#Any_solution_of_a_nonlinear_optimization_problem_is_a_solution_of_a_variational_inequality | is a solution of]]
 <math>VI(F,\mathcal K)</math> which [[Complementarity_problem#Every_variational_inequality_defined_on_a_closed_convex_cone_is_equivalent_to_a_complementarity_problem| is equivalent to]] <math>NCP(F,\mathcal K).</math>
 == A convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem ==
 Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space
 <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb R</math>
@@ Line 230: / Line 222: @@
 <math>NCP(F,\mathcal K),</math> where <math>F=\nabla f</math> is the gradient of <math>f.\,</math>
-== Proof ==
+=== Proof ===
 <math>NOPT(f,\mathcal K)</math> [[Complementarity_problem#Every_variational_inequality_is_equivalent_to_a_fixed_point_problem | is equivalent to]] <math>VI(F,\mathcal K)</math> which [[Complementarity_problem#Every_variational_inequality_defined_on_a_closed_convex_cone_is_equivalent_to_a_complementarity_problem| is equivalent to]] <math>NCP(F,\mathcal K).</math>
 == Fat nonlinear programming problem ==
 Let <math>f:\mathbb R^n\to\mathbb R</math> be a function <math>b\in\mathbb R^n,</math> and
 <math>A\in\mathbb R^{m\times n}</math> a matrix of full rank <math>m,\,</math> where <math>m\leq n.</math>
@@ Line 254: / Line 244: @@
 == Any solution of a fat nonlinear programming problem is a solution of a nonlinear complementarity problem defined by a polyhedral cone ==
 Let <math>f:\mathbb R^n\to\mathbb R</math> be a differentiable function
 <math>b\in\mathbb R^m,</math> and
@@ Line 274: / Line 263: @@
 </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
 solution of <math>NP(g,A,0),\,</math> where <math>g:\mathbb R^n\to\mathbb R</math> is defined by
@@ Line 281: / Line 269: @@
 [[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,0),</math> because <math>G(x)=\nabla f(x+x_0)=\nabla g(x).</math>
-== Remark ==
+=== Remark ===
 If <math>f\,</math> is convex, then the converse of the above results also holds.

Complementarity problem

From Wikimization

Revision as of 17:50, 16 August 2009

Contents

Fixed point problems

Nonlinear complementarity problems

Every nonlinear complementarity problem is equivalent to a fixed point problem

Proof

An alternative proof without Moreau's theorem

Variational inequalities

Every variational inequality is equivalent to a fixed point problem

Proof

Remark

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

Proof

Concluding the alternative proof

Implicit complementarity problems

Every implicit complementarity problem is equivalent to a fixed point problem

Proof

Remark

Nonlinear optimization problems

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

Proof

A convex optimization problem is equivalent to a variational inequality

Proof

Any solution of a nonlinear optimization problem on a closed convex cone is a solution of a nonlinear complementarity problem

Proof

A convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem

Proof

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

Proof

Remark

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