Complementarity problem

(Difference between revisions)
 Revision as of 15:36, 20 July 2009 (edit)← Previous diff Revision as of 15:38, 20 July 2009 (edit) (undo) (→Any solution of a non-thin nonlinear programming problem is a solution of a nonlinear complementarity problem defined by a polyhedral cone)Next diff → Line 258: Line 258: Let $f:\mathbb R^n\to\mathbb R$ be a differentiable function Let $f:\mathbb R^n\to\mathbb R$ be a differentiable function $b\in\mathbb R^m,$ and $b\in\mathbb R^m,$ and - $A\in\mathbb R^{m\times n}$ a matrix of full rank $m,\,$ where $m\leq n.$ If $x\in\mathbb R^n$ is a solution of the non-thin nonlinear programming problem + $A\in\mathbb R^{m\times n}$ a matrix of full rank $m,\,$ where $m\leq n.$ If $x\in\mathbb R^n$ is a solution of the [[Complementarity_problem#Non-thin_nonlinear_programming_problem | non-thin nonlinear programming problem]] - $NP(f,A,b),\,$ then $x-x_0\in\mathbb R^n$ is a solution of the nonlinear + $NP(f,A,b),\,$ then $x-x_0\in\mathbb R^n$ is a solution of the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear - complementarity problem $NCP(G,\mathcal K,0),$ where $x_0\in\mathbb R^n$ is + complementarity problem]] $NCP(G,\mathcal K,0),$ where $x_0\in\mathbb R^n$ is a particular solution of the linear system of equations $Ax=b,\,$ a particular solution of the linear system of equations $Ax=b,\,$ $\mathcal K$ is the polyhedral cone defined by $\mathcal K$ is the polyhedral cone defined by

Revision as of 15:38, 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.

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 $LaTeX: NCP(F,\mathcal K).$

Non-thin nonlinear programming problem

Let $LaTeX: f:\mathbb R^n\to\mathbb R$ be a function $LaTeX: b\in\mathbb R^n,$ and $LaTeX: A\in R^{m\times n}$ 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 non-thin nonlinear programming problem.

Any solution of a non-thin 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 non-thin 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.