Complementarity problem

(Difference between revisions)
 Revision as of 12:36, 17 July 2009 (edit)← Previous diff Current revision (14:16, 29 June 2015) (edit) (undo) (→Remark) (166 intermediate revisions not shown.) Line 1: Line 1: - Sándor Zoltán Németh + [http://web.mat.bham.ac.uk/S.Z.Nemeth/ $-$ Sándor Zoltán Németh] - === Fixed point problems === + (In particular, we can have $\mathbb{H}=\mathbb{R}^n$ everywhere in this page.) - 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 + == Fixed point problems == + Let $\mathcal{A}$ be a set and $T:\,\mathcal{A}\to\mathcal{A}$ a mapping. The '''fixed point problem''' defined by $T\,$ is the problem
$[itex] - Fix(F):\left\{ + Fix(T):\,\left\{ \begin{array}{l} \begin{array}{l} - Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\ + \textrm{Find }x\in\mathcal{A}\textrm{ such that}\\\\\\ - F(x)=x. + x=T(x). \end{array} \end{array} \right. \right. Line 15: Line 16: - === Nonlinear complementarity problems === + == Nonlinear complementarity problems == - Let [itex]\mathcal K$ 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 [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of $\mathcal K.$ The '''nonlinear complementarity problem''' defined by $\mathcal K$ and $f\,$ is the 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. Recall that the dual cone of $\mathcal{K}$ is the closed convex cone $\mathcal{K}^*=-\mathcal{K}^\circ$ where $\mathcal{K}^\circ$ is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of $\mathcal{K}.$ The '''nonlinear complementarity problem''' defined by $\mathcal{K}$ and $f\,$ is the problem
$[itex] - NCP(f,\mathcal K):\left\{ + NCP(F,\mathcal{K}):\,\left\{ \begin{array}{l} \begin{array}{l} - Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\ + \textrm{Find }x\in\mathcal{K}\textrm{ such that}\\\\\\ - f(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,f(x)\rangle=0. + F(x)\in\mathcal{K}^*\textrm{ and }\langle x,F(x)\rangle=0. \end{array} \end{array} \right. \right. Line 29: Line 30: - === Every nonlinear complementarity problem is equivalent to a fixed point problem === + == Every nonlinear complementarity problem is equivalent to a fixed point problem == - Let [itex]\mathcal K$ 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 + 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 [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] - $Fix(P_{\mathcal K}\circ(I-f)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x.\,$ + $NCP(F,\mathcal{K})$ is equivalent to the [[Complementarity_problem#Fixed_point_problems | 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 [[Moreau's_decomposition_theorem#Projection_on_closed_convex_sets | projection onto $\mathcal{K}.$]] === Proof === === Proof === - For all $x\in\mathbb H$ denote $z=x-f(x)\,$ and $y=-f(x).\,$ Then, $z=x+y.\,$ + 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, by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], we get $x=P_{\mathcal K}z.$ Therefore, $x\,$ is a solution of $Fix(P_{\mathcal K}\circ(I-f)).$ + 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_decomposition_theorem#Moreau.27s_theorem|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 by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] + Conversely, suppose that $x\,$ is a solution of $Fix(P_{\mathcal{K}}\circ(I-F)).$ Then $x\in\mathcal{K}$ and via [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]]
- $z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).$ + $z=P_{\mathcal{K}}(z)+P_{\mathcal{K}^\circ}(z)=x+P_{\mathcal{K}^\circ}(z).$
- Hence, $P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $y\in\mathcal K^\circ$. [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] also implies that $\langle x,y\rangle=0.$ In conclusion, + Hence $P_{\mathcal{K}^\circ}(z)=z-x=y,$ thus $y\in\mathcal{K}^\circ.$ [[Moreau's_decomposition_theorem#Moreau.27s_theorem | 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 $NCP(f,\mathcal K).$ + $x\in\mathcal{K},$ $F(x)=-y\in\mathcal{K}^*,$ and $\langle x,F(x)\rangle=0.$ Therefore $x\,$ is a solution of $NCP(F,\mathcal{K}).$ + === An alternative proof without [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] === - === An alternative proof without [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] === ==== Variational inequalities ==== ==== 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 + 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
$[itex] - VI(f,\mathcal C):\left\{ + VI(F,\mathcal{C}):\,\left\{ \begin{array}{l} \begin{array}{l} - Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ + \textrm{Find }x\in\mathcal{C}\textrm{ such that}\\\\\\ - \langle y-x,f(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C. + \langle y-x,F(x)\rangle\geq 0,\textrm{ for all }y\in\mathcal{C}. \end{array} \end{array} \right. \right.$ [/itex]
+ + ===== Remark ===== + The next result is not needed for the alternative proof and it can be skipped. However, it is an important property in its own. It was included for the completeness of the ideas. + + ==== Every fixed point problem defined on closed convex set 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 $T:\,\mathcal{C}\to\mathcal{C}$ a mapping. Then the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] $Fix(T)\,$ is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(F,\mathcal{C}),$ where $\,F=I-T.$ + + ===== Proof ===== + Suppose that $x\,$ is a solution of $Fix(T)\,$. Then, + $F(x)=0\,$ and thus $x\,$ is a solution of $VI(F,\mathcal{C}).$ + + Conversely, suppose that $x\,$ is a solution of $VI(F,\mathcal{C})$ and + let $\,y=T(x).$ Then, $\left\langle y-x,F(x)\right\rangle\geq 0,$ which is + equivalent to $-\parallel x-T(x)\parallel^2=0.$ Hence, $\,x=T(x)$; that is, $x\,$ is a solution of $Fix(T)\,$. ==== Every variational inequality is equivalent to a fixed point problem ==== ==== 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 $Fix(P_{\mathcal C}\circ(I-f)).$ + 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 [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(F,\mathcal{C})$ is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] $Fix(P_{\mathcal{C}}\circ(I-F)).$ - ==== Proof ==== + ===== Proof ===== - $x\,$ is a solution of $Fix(P_{\mathcal C}\circ(I-f))$ if and only if + $x\,$ is a solution of $Fix(P_{\mathcal{C}}\circ(I-F))$ if and only if - $x=P_{\mathcal C}(x-f(x)).$ By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection | characterization of the projection]] the latter equation is equivalent to + $x=P_{\mathcal{C}}(x-F(x)).$ Via [[Moreau's_decomposition_theorem#Characterization_of_the_projection|characterization of the projection]], the latter equation is equivalent to
- $\langle x-f(x)-x,y-x\rangle\leq0,$ + $\langle x-F(x)-x,y-x\rangle\leq0$
- for all $y\in\mathcal C.$ But this holds if and only if $x\,$ is a solution + for all $y\in\mathcal{C}.$ But this holds if and only if $x\,$ is a solution - of $VI(f,\mathcal C).$ + to $VI(F,\mathcal{C}).$ ===== Remark ===== ===== 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. + The next section shows that the equivalence of [[Complementarity_problem#Variational_inequalities | variational inequalities]] and [[Complementarity_problem#Fixed_point_problems | fixed point problems]] is much stronger than the equivalence of [[Complementarity_problem#Nonlinear_complementarity_problems |nonlinear complementarity problems]] and [[Complementarity_problem#Fixed_point_problems | fixed point problems]] because each [[Complementarity_problem#Nonlinear_complementarity_problems |nonlinear complementarity problem]] is a [[Complementarity_problem#Variational_inequalities | variational inequality]] defined on a closed convex cone. ==== Every variational inequality defined on a closed convex cone is equivalent to a complementarity problem ==== ==== 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 $VI(f,\mathcal K).$ + 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 [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] + $NCP(F,\mathcal{K})$ is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] + $VI(F,\mathcal{K}).$ - ==== Proof ==== + ===== 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, + 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
- $\langle y-x,f(x)\rangle\geq 0,$ + $\langle y-x,F(x)\rangle\geq 0$
- for all $y\in\mathcal K.$ Therefore, $x\,$ is a solution of $VI(f,\mathcal K).$ + 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, + Conversely, suppose that $x\,$ is a solution of $VI(F,\mathcal{K}).$ Then - $x\in\mathcal K$ and + $x\in\mathcal{K}$ and
- $\langle y-x,f(x)\rangle\geq 0,$ + $\langle y-x,F(x)\rangle\geq 0$
- for all $y\in\mathcal K.$ Particularly, taking $y=0\,$ and $y=2x\,$, respectively, we get $\langle x,f(x)\rangle=0.$ Thus, $\langle y,f(x)\rangle\geq 0,$ for all $y\in\mathcal K,$ or 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 of $NCP(f,\mathcal K).$ + 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 ==== ==== 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 $Fix(P_{\mathcal K}\circ(I-f)).$ + Since $\mathcal{K}$ is a closed convex cone, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] $NCP(F,\mathcal{K})$ is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(F,\mathcal{K})$ which [[Complementarity_problem#Every_variational_inequality_is_equivalent_to_a_fixed_point_problem | is equivalent to]] the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] $Fix(P_{\mathcal{K}}\circ(I-F)).$ + + == Another fixed point problem providing solutions for variational inequalities and complementarity problems == + + Let $\mathcal{C}$ be a closed convex set in the Hilbert space $(\mathbb{H},\langle\cdot,\cdot\rangle),$ + $I:\,\mathbb{H}\to\mathbb{H}$ the identity mapping defined by $I(x)=x$ and $F:\,\mathbb{H}\to\mathbb{H}$ a mapping. Then, for any solution $x\,$ of the fixed point problem $Fix({(I-F)\circ P_{\mathcal{C}}}),$ the projection $P_{\mathcal{C}}(x)$ of $x\,$ onto $\mathcal{C}$is a solution of the variational inequality $VI(F,\mathcal{C}).$ + + === Proof === + Since $x\,$ is a solution of the fixed point problem $Fix({(I-F)\circ P_{\mathcal{C}}})$, we have + $P_{\mathcal{C}}(x)-F(P_{\mathcal{C}}(x))=x.$ Thus, $P_{\mathcal{C}}(P_{\mathcal{C}}(x)-F(P_{\mathcal{C}}(x)))=P_{\mathcal{C}}(x)$. Hence, $P_{\mathcal{C}}(x)$ is a solution of the fixed point problem $Fix(P_{\mathcal{C}}\circ(I-F))$ and [[Complementarity_problem#Every_variational_inequality_is_equivalent_to_a_fixed_point_problem | thus]] a solution of the variational inequality $VI(F,\mathcal{C}).$ In particular, if $\mathcal{C}$ is a closed convex cone, then the variational inequality $VI(F,\mathcal{C})$ [[Complementarity_problem#Every_variational_inequality_defined_on_a_closed_convex_cone_is_equivalent_to_a_complementarity_problem | becomes]] the complementarity problem $NCP(F,\mathcal{C}).$ == Implicit complementarity problems == == 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 + 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 [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of $\mathcal K.$ The '''implicit complementarity problem''' defined by $\mathcal K$ of $\mathcal K.$ The '''implicit complementarity problem''' defined by $\mathcal K$ - and the ordered pair of mappings $(f,g)\,$ is the problem + and the ordered pair of mappings $(F,G)\,$ is the problem
$[itex] - ICP(f,g,\mathcal K):\left\{ + ICP(F,G,\mathcal{K}):\,\left\{ \begin{array}{l} \begin{array}{l} - Find\,\,\,u\in\mathbb H\,\,\,such\,\,\,that\\ + \textrm{Find }u\in\mathbb{H}\textrm{ such that}\\\\\\ - g(u)\in\mathcal K,\,\,\,f(u)\in K^*\,\,\,and\,\,\,\langle g(u),f(u)\rangle=0. + G(u)\in\mathcal{K},\,\,\,F(u)\in\mathcal{K}^*,\,\,\,\langle G(u),F(u)\rangle=0. \end{array} \end{array} \right. \right. Line 127: Line 155: - === Every implicit complementarity problem is equivalent to a fixed point problem === + == Every implicit complementarity problem is equivalent to a fixed point problem == - Let [itex]\mathcal K$ 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 [[Moreau's_decomposition_theorem#Fixed_point_problems | 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 [[Complementarity_problem#Implicit_complementarity_problems | implicit complementarity problem]] $ICP(F,G,\mathcal K)$ is equivalent to the [[Complementarity_problem#Fixed_point_problems | 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 $I(x)=x.\,$ + $Fix(I-G+P_{\mathcal K}\circ(G-F))$ where $I:\,\mathbb{H}\to\mathbb{H}$ is the identity mapping defined by $I(x)=x.\,$ === Proof === === Proof === - For all $u\in\mathbb H$ denote $z=g(u)-f(u),\,$ $x=g(u)\,$ and $y=-f(u).\,$ Then, + For all $u\in\mathbb{H}$ denote $z=G(u)-F(u),\,$ $x=G(u),\,$ and $y=-F(u).\,$ Then $z=x+y.\,$ $z=x+y.\,$

- 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 $\langle x,y\rangle=0.$ Hence, by using + Suppose that $u\,$ is a solution of $ICP(F,G,\mathcal K).$ - [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], + Then $z=x+y\,$ with $x\in\mathcal K,$ $y\in\mathcal K^\circ,$ and $\langle x,y\rangle=0.$ - we get $x=P_{\mathcal K}z.$ Therefore, $u\,$ is a solution of + Via [[Moreau's_decomposition_theorem#Moreau.27s_theorem|Moreau's theorem]], - $Fix(I-g+P_{\mathcal K}\circ(g-f)).$ + $x=P_{\mathcal K}z.$ + Therefore $u\,$ is a solution of + $Fix(I-G+P_{\mathcal K}\circ(G-F)).$

- Conversely, suppose that $u\,$ is a solution of $Fix(I-g+P_{\mathcal K}\circ(g-f)).$ + Conversely, suppose that $u\,$ is a solution of $Fix(I-G+P_{\mathcal K}\circ(G-F)).$ - Then, $x\in\mathcal K$ and by using + Then $x\in\mathcal K$ and, via - [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] + [[Moreau's_decomposition_theorem#Moreau.27s_theorem|Moreau's theorem]],
Line 152: Line 182:
- Hence, $P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $y\in\mathcal K^\circ$. + Hence $P_{\mathcal K^\circ}(z)=z-x=y,$ thus $y\in\mathcal K^\circ$. - [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] + [[Moreau's_decomposition_theorem#Moreau.27s_theorem|Moreau's theorem]] - also implies that $\langle x,y\rangle=0.$ In conclusion, + also implies $\langle x,y\rangle=0.$ In conclusion, - $g(u)=x\in\mathcal K,$ $f(u)=-y\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Therefore, $u\,$ is a solution of $ICP(f,g,\mathcal K).$ + $G(u)=x\in\mathcal K,$ $F(u)=-y\in\mathcal K^*,$ and $\langle G(u),F(u)\rangle=0.$ + Therefore $u\,$ is a solution of $ICP(F,G,\mathcal K).$ === Remark === === Remark === - In particular if $g=I,$ we obtain the result + If $\,G=I,$ in particular, we obtain the result - [[Moreau%27s_decomposition_theorem#Every_nonlinear_complementarity_problem_is_equivalent_to_a_fixed_point_problem | Every nonlinear complementarity problem is equivalent to a fixed point problem]], + [[Complementarity_problem#Every_nonlinear_complementarity_problem_is_equivalent_to_a_fixed_point_problem | ''every nonlinear complementarity problem is equivalent to a fixed point problem'']]. - but the more general result [[Moreau%27s_decomposition_theorem#Every_implicit_complementarity_problem_is_equivalent_to_a_fixed_point_problem | Every implicit complementarity problem is equivalent to a fixed point problem]] has no known connection with variational inequalities. Therefore, using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] is essential for proving the latter result. + But the more general result, [[Complementarity_problem#Every_implicit_complementarity_problem_is_equivalent_to_a_fixed_point_problem | ''every implicit complementarity problem is equivalent to a fixed point problem'']], has no known connection with [[Complementarity_problem#Variational_inequalities | variational inequalities]]. Using [[Moreau's_decomposition_theorem#Moreau.27s_theorem|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 + +
+ $+ NOPT(f,\mathcal C):\,\left\{ + \begin{array}{l} + \textrm{Find }x\in\mathcal C\textrm{ such that}\\\\\\ + f(x)\leq f(y)\textrm{ for all }y\in\mathcal C + \end{array} + \right. + + \,~\equiv~\, + + \begin{array}{ll} + \textrm{Minimize} & f(x)\\ + \textrm{Subject to} & x\in\mathcal C + \end{array} + +$ +
+ + == 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 [[Complementarity_problem#Nonlinear_optimization_problems| nonlinear optimization problem ]] $NOPT(f,\mathcal C)$ is a solution of the [[Complementarity_problem#Variational_inequalities | 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 [[Complementarity_problem#Nonlinear_optimization_problems | nonlinear optimization problem]] $NOPT(f,\mathcal C)$ is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] $VI(\nabla f,\mathcal C)$ where $\nabla f$ is the gradient of $f.\,$ + + === Proof === + Any solution of $NOPT(f,\mathcal C)$ + [[Complementarity_problem#Any_solution_of_a_nonlinear_optimization_problem_is_a_solution_of_a_variational_inequality | 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 [[Complementarity_problem#Nonlinear_optimization_problems | nonlinear optimization problem]] $NOPT(f,\mathcal K)$ is a + solution of the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] $NCP(\nabla f,\mathcal K).$ + + === Proof === + Any solution of $NOPT(f,\mathcal K)$ [[Complementarity_problem#Any_solution_of_a_nonlinear_optimization_problem_is_a_solution_of_a_variational_inequality | is a solution of]] + $VI(\nabla f,\mathcal K)$ which [[Complementarity_problem#Every_variational_inequality_defined_on_a_closed_convex_cone_is_equivalent_to_a_complementarity_problem| 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 [[Complementarity_problem#Nonlinear_optimization_problems | nonlinear optimization problem]] $NOPT(f,\mathcal K)$ is equivalent to the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] + $NCP(\nabla f,\mathcal K).$ + + === Proof === + $NOPT(f,\mathcal K)$ [[Complementarity_problem#A_convex_optimization_problem_is_equivalent_to_a_variational_inequality | is equivalent to]] $VI(\nabla f,\mathcal K)$ which [[Complementarity_problem#Every_variational_inequality_defined_on_a_closed_convex_cone_is_equivalent_to_a_complementarity_problem| is equivalent to]] $NCP(\nabla f,\mathcal K).$ + + == An implicit complementarity problem can be associated to an optimization problem == + + It follows from the definition of an [[Complementarity_problem#Implicit_complementarity_problems | implicit complementarity problem]] that $y$ is a solution of $ICP(F,G,\mathcal{K})$ if and only the minimal value of $\langle G(x),F(x)\rangle$ subject to $G(x)\in\mathcal{K}$ and $F(x)\in\mathcal{K}^*$ is $0$ at $y.$ + + == 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 + +
+ $+ NP(f,A,b):\,\left\{ + \begin{array}{ll} + \textrm{Minimize} & f(x)\\ + \textrm{Subject to} & Ax\leq b + \end{array} + \right. +$ +
+ + 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 [[Complementarity_problem#Fat_nonlinear_programming_problem|fat nonlinear programming problem]] + $NP(f,A,b),\,$ then $x-x_0\in\mathbb{R}^n$ is a solution of the [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear + complementarity problem]] $NCP(G,\mathcal K)$ 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 + $g(x)=f(x+x_0).\,$ + [[Complementarity_problem#A convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem|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, $NP(f,A,b)\equiv NP(g,A,0)\equiv NOPT(g,\mathcal{K})\equiv NCP(G,\mathcal{K}),$ where the first equivalence has a slightly different meaning than the other ones, but it should be self-explanatory. + + Since a very large class of nonlinear programming problems can be reduced to [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problems]], the importance of [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problems]] on polyhedral cones is obvious both from theoretical and practical point of view. + + == Mixed complementarity problems == + + Let $\mathbb{I}$ and $\mathbb{J}$ be two Hilbert spaces and $\mathcal{K}=\mathbb{I}\times K$, where $K$ is an arbitrary closed convex cone in $\mathbb{J}$. Let $G:\,\mathbb{I}\times\mathbb{J}\rightarrow\mathbb{I}$, $H:\,\mathbb{I}\times\mathbb{J}\to\mathbb{J}$ and $F=(G,H):\,\mathbb{I}\times\mathbb{J}\to\mathbb{I}\times\mathbb{J}$. Then, the [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problem]] $NCP(F,\mathcal{K})$ is equivalent to the mixed complementarity problem $MiCP(G,H,K)$ defined by +
+ $+ G(x,y)=0,\textrm{ }K\ni y\perp H(x,y)\in K^*. +$ +
+ + === Proof === + + Indeed, $MiCP(G,H,K)$ is a simple equivalent reformulation of the [[Complementarity_problem#Nonlinear_complementarity_problems|nonlinear complementarity problem]] $NCP(F,\mathcal{K})$, by noting that $\mathcal{K}^*=\{0\}\times K^*$.

Current revision

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

Fixed point problems

Let $LaTeX: \mathcal{A}$ be a set and $LaTeX: T:\,\mathcal{A}\to\mathcal{A}$ a mapping. The fixed point problem defined by $LaTeX: T\,$ is the problem

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

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

$LaTeX: NCP(F,\mathcal{K}):\,\left\{ \begin{array}{l} \textrm{Find }x\in\mathcal{K}\textrm{ such that}\\\\\\ F(x)\in\mathcal{K}^*\textrm{ and }\langle x,F(x)\rangle=0. \end{array} \right.$

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, via 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

$LaTeX: VI(F,\mathcal{C}):\,\left\{ \begin{array}{l} \textrm{Find }x\in\mathcal{C}\textrm{ such that}\\\\\\ \langle y-x,F(x)\rangle\geq 0,\textrm{ for all }y\in\mathcal{C}. \end{array} \right.$

Remark

The next result is not needed for the alternative proof and it can be skipped. However, it is an important property in its own. It was included for the completeness of the ideas.

Every fixed point problem defined on closed convex set 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 $LaTeX: T:\,\mathcal{C}\to\mathcal{C}$ a mapping. Then the fixed point problem $LaTeX: Fix(T)\,$ is equivalent to the variational inequality $LaTeX: VI(F,\mathcal{C}),$ where $LaTeX: \,F=I-T.$

Proof

Suppose that $LaTeX: x\,$ is a solution of $LaTeX: Fix(T)\,$. Then, $LaTeX: F(x)=0\,$ and thus $LaTeX: x\,$ is a solution of $LaTeX: VI(F,\mathcal{C}).$

Conversely, suppose that $LaTeX: x\,$ is a solution of $LaTeX: VI(F,\mathcal{C})$ and let $LaTeX: \,y=T(x).$ Then, $LaTeX: \left\langle y-x,F(x)\right\rangle\geq 0,$ which is equivalent to $LaTeX: -\parallel x-T(x)\parallel^2=0.$ Hence, $LaTeX: \,x=T(x)$; that is, $LaTeX: x\,$ is a solution of $LaTeX: Fix(T)\,$.

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 $LaTeX: x=P_{\mathcal{C}}(x-F(x)).$ Via 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 to $LaTeX: VI(F,\mathcal{C}).$

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 $LaTeX: y\in\mathcal{K}.$ Choosing $LaTeX: y=0\,$ and $LaTeX: y=2x,\,$ in particular, we get a system of two inequalities that demands $LaTeX: \langle x,F(x)\rangle=0.$ Thus $LaTeX: \langle y,F(x)\rangle\geq 0$ for all $LaTeX: y\in\mathcal{K};$ 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 to $LaTeX: NCP(F,\mathcal{K}).$

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

Another fixed point problem providing solutions for variational inequalities and complementarity problems

Let $LaTeX: \mathcal{C}$ be a closed convex set in the Hilbert space $LaTeX: (\mathbb{H},\langle\cdot,\cdot\rangle),$ $LaTeX: I:\,\mathbb{H}\to\mathbb{H}$ the identity mapping defined by $LaTeX: I(x)=x$ and $LaTeX: F:\,\mathbb{H}\to\mathbb{H}$ a mapping. Then, for any solution $LaTeX: x\,$ of the fixed point problem $LaTeX: Fix({(I-F)\circ P_{\mathcal{C}}}),$ the projection $LaTeX: P_{\mathcal{C}}(x)$ of $LaTeX: x\,$ onto $LaTeX: \mathcal{C}$is a solution of the variational inequality $LaTeX: VI(F,\mathcal{C}).$

Proof

Since $LaTeX: x\,$ is a solution of the fixed point problem $LaTeX: Fix({(I-F)\circ P_{\mathcal{C}}})$, we have $LaTeX: P_{\mathcal{C}}(x)-F(P_{\mathcal{C}}(x))=x.$ Thus, $LaTeX: P_{\mathcal{C}}(P_{\mathcal{C}}(x)-F(P_{\mathcal{C}}(x)))=P_{\mathcal{C}}(x)$. Hence, $LaTeX: P_{\mathcal{C}}(x)$ is a solution of the fixed point problem $LaTeX: Fix(P_{\mathcal{C}}\circ(I-F))$ and thus a solution of the variational inequality $LaTeX: VI(F,\mathcal{C}).$ In particular, if $LaTeX: \mathcal{C}$ is a closed convex cone, then the variational inequality $LaTeX: VI(F,\mathcal{C})$ becomes the complementarity problem $LaTeX: NCP(F,\mathcal{C}).$

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

$LaTeX: ICP(F,G,\mathcal{K}):\,\left\{ \begin{array}{l} \textrm{Find }u\in\mathbb{H}\textrm{ such that}\\\\\\ G(u)\in\mathcal{K},\,\,\,F(u)\in\mathcal{K}^*,\,\,\,\langle G(u),F(u)\rangle=0. \end{array} \right.$

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 $LaTeX: \langle x,y\rangle=0.$ Via Moreau's theorem, $LaTeX: x=P_{\mathcal K}z.$ 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 $LaTeX: \langle x,y\rangle=0.$ 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} \textrm{Find }x\in\mathcal C\textrm{ such that}\\\\\\ f(x)\leq f(y)\textrm{ for all }y\in\mathcal C \end{array} \right.

\,~\equiv~\,

\begin{array}{ll} \textrm{Minimize} & f(x)\\ \textrm{Subject to} & x\in\mathcal C \end{array}

$

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

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

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 $LaTeX: VI(\nabla f,\mathcal C).$ By 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

Theorem NOPT.   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 $LaTeX: NCP(\nabla f,\mathcal K).$

An implicit complementarity problem can be associated to an optimization problem

It follows from the definition of an implicit complementarity problem that $LaTeX: y$ is a solution of $LaTeX: ICP(F,G,\mathcal{K})$ if and only the minimal value of $LaTeX: \langle G(x),F(x)\rangle$ subject to $LaTeX: G(x)\in\mathcal{K}$ and $LaTeX: F(x)\in\mathcal{K}^*$ is $LaTeX: 0$ at $LaTeX: y.$

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

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

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 $LaTeX: NCP(G,\mathcal K)$ 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

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

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 from Theorem NOPT that $LaTeX: x-x_0\,$ is a solution of $LaTeX: NCP(G,\mathcal K)$ 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. In other words, $LaTeX: NP(f,A,b)\equiv NP(g,A,0)\equiv NOPT(g,\mathcal{K})\equiv NCP(G,\mathcal{K}),$ where the first equivalence has a slightly different meaning than the other ones, but it should be self-explanatory.

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.

Mixed complementarity problems

Let $LaTeX: \mathbb{I}$ and $LaTeX: \mathbb{J}$ be two Hilbert spaces and $LaTeX: \mathcal{K}=\mathbb{I}\times K$, where $LaTeX: K$ is an arbitrary closed convex cone in $LaTeX: \mathbb{J}$. Let $LaTeX: G:\,\mathbb{I}\times\mathbb{J}\rightarrow\mathbb{I}$, $LaTeX: H:\,\mathbb{I}\times\mathbb{J}\to\mathbb{J}$ and $LaTeX: F=(G,H):\,\mathbb{I}\times\mathbb{J}\to\mathbb{I}\times\mathbb{J}$. Then, the nonlinear complementarity problem $LaTeX: NCP(F,\mathcal{K})$ is equivalent to the mixed complementarity problem $LaTeX: MiCP(G,H,K)$ defined by

$LaTeX: G(x,y)=0,\textrm{ }K\ni y\perp H(x,y)\in K^*.$

Proof

Indeed, $LaTeX: MiCP(G,H,K)$ is a simple equivalent reformulation of the nonlinear complementarity problem $LaTeX: NCP(F,\mathcal{K})$, by noting that $LaTeX: \mathcal{K}^*=\{0\}\times K^*$.