Complementarity problem

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− Sándor Zoltán Németh

(In particular, we can have \mathbb H=\mathbb R^n everywhere in this page.)

Contents

Fixed point problems

Let \mathcal A be a set and F:\mathcal A\to\mathcal A a mapping. The fixed point problem defined by F\, is the problem

Fix(F):\left\{ \begin{array}{l} Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\ F(x)=x. \end{array} \right.

Nonlinear complementarity problems

Let \mathcal K be a closed convex cone in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and F:\mathbb H\to\mathbb H a mapping. Recall that the dual cone of \mathcal K is the closed convex cone \mathcal K^*=-\mathcal K^\circ where \mathcal K^\circ is the polar of \mathcal K. The nonlinear complementarity problem defined by \mathcal K and f\, is the problem

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 \mathcal K be a closed convex cone in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and F:\mathbb H\to\mathbb H a mapping. Then the nonlinear complementarity problem NCP(F,\mathcal K) is equivalent to the fixed point problem Fix(P_{\mathcal K}\circ(I-F)) where I:\mathbb H\to\mathbb H is the identity mapping defined by I(x)=x\, and P_{\mathcal K} is the projection onto \mathcal K.

Proof

For all x\in\mathbb H denote z=x-F(x)\, and y=-F(x).\, Then z=x+y.\,

Suppose that x\, is a solution of NCP(F,\mathcal K). Then z=x+y\, with x\in\mathcal K, y\in\mathcal K^\circ, and \langle x,y\rangle=0. Hence, via Moreau's theorem, we get x=P_{\mathcal K}z. Therefore x\, is a solution of Fix(P_{\mathcal K}\circ(I-F)).

Conversely, suppose that x\, is a solution of Fix(P_{\mathcal K}\circ(I-F)). Then x\in\mathcal K and via Moreau's theorem

z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).

Hence P_{\mathcal K^\circ}(z)=z-x=y, thus y\in\mathcal K^\circ. Moreau's theorem also implies that \langle x,y\rangle=0. In conclusion, x\in\mathcal K, F(x)=-y\in\mathcal K^*, and \langle x,F(x)\rangle=0. Therefore x\, is a solution of NCP(F,\mathcal K).

An alternative proof without Moreau's theorem

Variational inequalities

Let \mathcal C be a closed convex set in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and F:\mathbb H\to\mathbb H a mapping. The variational inequality defined by \mathcal C and F\, is the problem

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 \mathcal C be a closed convex set in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and F:\mathbb H\to\mathbb H a mapping. Then the variational inequality VI(F,\mathcal C) is equivalent to the fixed point problem Fix(P_{\mathcal C}\circ(I-F)).

Proof

x\, is a solution of Fix(P_{\mathcal C}\circ(I-F)) if and only if x=P_{\mathcal C}(x-F(x)). Via characterization of the projection, the latter equation is equivalent to

\langle x-F(x)-x,y-x\rangle\leq0

for all y\in\mathcal C. But this holds if and only if x\, is a solution to 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 \mathcal K be a closed convex cone in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and F:\mathbb H\to\mathbb H a mapping. Then the nonlinear complementarity problem NCP(F,\mathcal K) is equivalent to the variational inequality VI(F,\mathcal K).

Proof

Suppose that x\, is a solution of NCP(F,\mathcal K). Then x\in\mathcal K, F(x)\in\mathcal K^*, and \langle x,F(x)\rangle=0. Hence

\langle y-x,F(x)\rangle\geq 0

for all y\in\mathcal K. Therefore x\, is a solution of VI(F,\mathcal K).

Conversely, suppose that x\, is a solution of VI(F,\mathcal K). Then x\in\mathcal K and

\langle y-x,F(x)\rangle\geq 0

for all y\in\mathcal K. Choosing y=0\, and y=2x,\, in particular, we get a system of two inequalities that demands \langle x,F(x)\rangle=0. Thus \langle y,F(x)\rangle\geq 0 for all y\in\mathcal K; equivalently, F(x)\in\mathcal K^*. In conclusion, x\in\mathcal K, F(x)\in\mathcal K^*, and \langle x,F(x)\rangle=0. Therefore x\, is a solution to NCP(F,\mathcal K).

Concluding the alternative proof

Since \mathcal K is a closed convex cone, the nonlinear complementarity problem NCP(F,\mathcal K) is equivalent to the variational inequality VI(F,\mathcal K) which is equivalent to the fixed point problem Fix(P_{\mathcal K}\circ(I-F)).

Implicit complementarity problems

Let \mathcal K be a closed convex cone in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and F,G:\mathbb H\to\mathbb H two mappings. Recall that the dual cone of \mathcal K is the closed convex cone \mathcal K^*=-\mathcal K^\circ where \mathcal K^\circ is the polar of \mathcal K. The implicit complementarity problem defined by \mathcal K and the ordered pair of mappings (F,G)\, is the problem

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^*,\,\,\,\langle G(u),F(u)\rangle=0. \end{array} \right.

Every implicit complementarity problem is equivalent to a fixed point problem

Let \mathcal K be a closed convex cone in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and F,G:\mathbb H\to\mathbb H two mappings. Then the implicit complementarity problem ICP(F,G,\mathcal K) is equivalent to the fixed point problem Fix(I-G+P_{\mathcal K}\circ(G-F)) where I:\mathbb H\to\mathbb H is the identity mapping defined by I(x)=x.\,

Proof

For all u\in\mathbb H denote z=G(u)-G(u),\, x=G(u),\, and y=-F(u).\, Then z=x+y.\,

Suppose that u\, is a solution of ICP(F,G,\mathcal K). Then z=x+y\, with x\in\mathcal K, y\in\mathcal K^\circ, and \langle x,y\rangle=0. Via Moreau's theorem, x=P_{\mathcal K}z. Therefore u\, is a solution of Fix(I-G+P_{\mathcal K}\circ(G-F)).

Conversely, suppose that u\, is a solution of Fix(I-G+P_{\mathcal K}\circ(G-F)). Then x\in\mathcal K and, via Moreau's theorem,

z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).

Hence P_{\mathcal K^\circ}(z)=z-x=y, thus y\in\mathcal K^\circ. Moreau's theorem also implies \langle x,y\rangle=0. In conclusion, G(u)=x\in\mathcal K, F(u)=-y\in\mathcal K^*, and \langle G(u),F(u)\rangle=0. Therefore u\, is a solution of ICP(F,G,\mathcal K).

Remark

If \,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 \mathcal C be a closed convex set in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and f:\mathbb H\to\mathbb R a function. The nonlinear optimization problem defined by \mathcal C and f\, is the problem

NOPT(f,\mathcal C):\left\{ \begin{array}{l} Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ f(x)\leq f(y)\,\,\,for\,\,\,all\,\,\,y\in\mathcal C \end{array} \right. ~\equiv~ \begin{array}{rl} Minimize & f(x)\\ Subject\,\,\,to & x\in\mathcal C \end{array}

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

Let \mathcal C be a closed convex set in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and f:\mathbb H\to\mathbb R a differentiable function. Then any solution of the nonlinear optimization problem NOPT(f,\mathcal C) is a solution of the variational inequality VI(\nabla f,\mathcal C) where \nabla f is the gradient of f.\,

Proof

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

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

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

A convex optimization problem is equivalent to a variational inequality

Let \mathcal C be a closed convex set in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and f:\mathbb H\to\mathbb R a differentiable convex function. Then the nonlinear optimization problem NOPT(f,\mathcal C) is equivalent to the variational inequality VI(\nabla f,\mathcal C) where \nabla f is the gradient of f.\,

Proof

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

Conversely, suppose that x\, is a solution of VI(\nabla f,\mathcal C). By convexity of f\, we have f(y)-f(x)\geq\langle\nabla f(x),y-x\rangle\geq0 for all y\in\mathcal C. Therefore x\, is a solution of 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 \mathcal K be a closed convex cone in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and f:\mathbb H\to\mathbb R a differentiable function. Then any solution of the nonlinear optimization problem NOPT(f,\mathcal K) is a solution of the nonlinear complementarity problem NCP(\nabla f,\mathcal K).

Proof

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

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

Theorem NOPT.   Let \mathcal K be a closed convex cone in the Hilbert space (\mathbb H,\langle\cdot,\cdot\rangle) and f:\mathbb H\to\mathbb R a differentiable convex function. Then the nonlinear optimization problem NOPT(f,\mathcal K) is equivalent to the nonlinear complementarity problem NCP(\nabla f,\mathcal K).

Proof

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

Fat nonlinear programming problem

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

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

\mathcal K=\{x\mid Ax\leq0\}

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

G(x)=\nabla f(x+x_0)


Proof

Let x\in\mathbb R^n be a solution of NP(f,A,b).\, Then it is easy to see that x-x_0\, is a solution of \,NP(g,A,0) where g:\mathbb R^n\to\mathbb R is defined by g(x)=f(x+x_0).\, It follows from Theorem NOPT that x-x_0\, is a solution of NCP(G,\mathcal K) because G(x)=\nabla f(x+x_0)=\nabla g(x).

Remark

If f\, is convex, then the converse of the above results also holds. In other words, NP(f,A,b)\equiv NP(g,A,0)\equiv NOPT(g,\mathcal K)\equiv NCP(G,\mathcal K).

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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