Complementarity problem

(Difference between revisions)
Jump to: navigation, search

Ranjelin (Talk | contribs)
(New page: == An application to nonlinear complementarity problems == === Fixed point problems === Let $\mathcal A$ be a set and $F:\mathcal A\to\mathcal A$ a mapping. The ''...)
Next diff →

An application to nonlinear complementarity problems

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 $LaTeX: I(x)=x.\,$

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 $LaTeX: Fix(P_{\mathcal K}\circ(I-f)).$

An application to implicit complementarity problems

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.