Complementarity problem
From Wikimization
m (→Nonlinear complementarity problems) |
m (→Every nonlinear complementarity problem is equivalent to a fixed point problem) |
||
Line 31: | Line 31: | ||
== Every nonlinear complementarity problem is equivalent to a fixed point problem == | == Every nonlinear complementarity problem is equivalent to a fixed point 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{H}</math> a mapping. Then the [[Complementarity_problem#Nonlinear_complementarity_problems | 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{H}</math> a mapping. Then the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] |
- | <math>NCP(F,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] | + | <math>NCP(F,\mathcal{K})</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] |
- | <math>Fix(P_{\mathcal K}\circ(I-F))</math> where <math>I:\,\mathbb{H}\to\mathbb{H}</math> is the identity mapping defined by <math>I(x)=x\,</math> and <math>P_{\mathcal K}</math> is the [[Moreau's_decomposition_theorem#Projection_on_closed_convex_sets | projection onto <math>\mathcal K.</math>]] | + | <math>Fix(P_{\mathcal{K}}\circ(I-F))</math> where <math>I:\,\mathbb{H}\to\mathbb{H}</math> is the identity mapping defined by <math>I(x)=x\,</math> and <math>P_{\mathcal{K}}</math> is the [[Moreau's_decomposition_theorem#Projection_on_closed_convex_sets | projection onto <math>\mathcal{K}.</math>]] |
=== Proof === | === Proof === | ||
Line 40: | Line 40: | ||
<br> | <br> | ||
- | Suppose that <math>x\,</math> is a solution of <math>NCP(F,\mathcal K).</math> Then <math>z=x+y\,</math> with <math>x\in\mathcal K,</math> | + | Suppose that <math>x\,</math> is a solution of <math>NCP(F,\mathcal{K}).</math> Then <math>z=x+y\,</math> with <math>x\in\mathcal{K},</math> |
- | <math>y\in\mathcal K^\circ,</math> and <math>\langle x,y\rangle=0.</math> Hence, via [[Moreau's_decomposition_theorem#Moreau.27s_theorem|Moreau's theorem]], we get <math>x=P_{\mathcal K}z.</math> Therefore <math>x\,</math> is a solution of <math>Fix(P_{\mathcal K}\circ(I-F)).</math> | + | <math>y\in\mathcal{K}^\circ,</math> and <math>\langle x,y\rangle=0.</math> Hence, via [[Moreau's_decomposition_theorem#Moreau.27s_theorem|Moreau's theorem]], we get <math>x=P_{\mathcal{K}}z.</math> Therefore <math>x\,</math> is a solution of <math>Fix(P_{\mathcal{K}}\circ(I-F)).</math> |
<br> | <br> | ||
<br> | <br> | ||
- | Conversely, suppose that <math>x\,</math> is a solution of <math>Fix(P_{\mathcal K}\circ(I-F)).</math> Then <math>x\in\mathcal K</math> and via [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] | + | Conversely, suppose that <math>x\,</math> is a solution of <math>Fix(P_{\mathcal{K}}\circ(I-F)).</math> Then <math>x\in\mathcal{K}</math> and via [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] |
<center> | <center> | ||
- | <math>z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).</math> | + | <math>z=P_{\mathcal{K}}(z)+P_{\mathcal{K}^\circ}(z)=x+P_{\mathcal{K}^\circ}(z).</math> |
</center> | </center> | ||
- | Hence <math>P_{\mathcal K^\circ}(z)=z-x=y,</math> thus <math>y\in\mathcal K^\circ.</math> [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] also implies that <math>\langle x,y\rangle=0.</math> In conclusion, | + | Hence <math>P_{\mathcal{K}^\circ}(z)=z-x=y,</math> thus <math>y\in\mathcal{K}^\circ.</math> [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] also implies that <math>\langle x,y\rangle=0.</math> In conclusion, |
- | <math>x\in\mathcal K,</math> <math>F(x)=-y\in\mathcal K^*,</math> and <math>\langle x,F(x)\rangle=0.</math> Therefore <math>x\,</math> is a solution of <math>NCP(F,\mathcal K).</math> | + | <math>x\in\mathcal{K},</math> <math>F(x)=-y\in\mathcal{K}^*,</math> and <math>\langle x,F(x)\rangle=0.</math> Therefore <math>x\,</math> is a solution of <math>NCP(F,\mathcal{K}).</math> |
=== 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 <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{H}</math> a mapping. The '''variational inequality''' defined by <math>\mathcal C</math> and <math>F\,</math> is the problem | + | 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{H}</math> a mapping. The '''variational inequality''' defined by <math>\mathcal{C}</math> and <math>F\,</math> is the problem |
<center> | <center> | ||
<math> | <math> | ||
- | VI(F,\mathcal C):\,\left\{ | + | VI(F,\mathcal{C}):\,\left\{ |
\begin{array}{l} | \begin{array}{l} | ||
- | Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ | + | Find\,\,\,x\in\mathcal{C}\,\,\,such\,\,\,that\\ |
- | \langle y-x,F(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C. | + | \langle y-x,F(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal{C}. |
\end{array} | \end{array} | ||
\right. | \right. | ||
Line 74: | Line 74: | ||
==== Every fixed point problem defined on closed convex set is equivalent to a variational inequality ==== | ==== Every fixed point problem defined on closed convex set 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>T:\,\mathcal C\to\mathcal C</math> a mapping. Then the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] <math>Fix(T)\,</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(F,\mathcal C),</math> where <math>\,F=I-T.</math> | + | Let <math>\mathcal{C}</math> be a closed convex set in the Hilbert space <math>(\mathbb{H},\langle\cdot,\cdot\rangle)</math> and <math>T:\,\mathcal{C}\to\mathcal{C}</math> a mapping. Then the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] <math>Fix(T)\,</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(F,\mathcal{C}),</math> where <math>\,F=I-T.</math> |
===== Proof ===== | ===== Proof ===== | ||
Suppose that <math>x\,</math> is a solution of <math>Fix(T)\,</math>. Then, | Suppose that <math>x\,</math> is a solution of <math>Fix(T)\,</math>. Then, | ||
- | <math>F(x)=0\,</math> and thus <math>x\,</math> is a solution of <math>VI(F,\mathcal C).</math> | + | <math>F(x)=0\,</math> and thus <math>x\,</math> is a solution of <math>VI(F,\mathcal{C}).</math> |
- | Conversely, suppose that <math>x\,</math> is a solution of <math>VI(F,\mathcal C)</math> and | + | Conversely, suppose that <math>x\,</math> is a solution of <math>VI(F,\mathcal{C})</math> and |
let <math>\,y=T(x).</math> Then, <math>\left\langle y-x,F(x)\right\rangle\geq 0,</math> which is | let <math>\,y=T(x).</math> Then, <math>\left\langle y-x,F(x)\right\rangle\geq 0,</math> which is | ||
equivalent to <math>-\parallel x-T(x)\parallel^2=0.</math> Hence, <math>\,x=T(x)</math>; that is, <math>x\,</math> is a solution of <math>Fix(T)\,</math>. | equivalent to <math>-\parallel x-T(x)\parallel^2=0.</math> Hence, <math>\,x=T(x)</math>; that is, <math>x\,</math> is a solution of <math>Fix(T)\,</math>. | ||
==== Every variational inequality is equivalent to a fixed point problem ==== | ==== Every variational inequality is equivalent to a fixed point problem ==== | ||
- | 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{H}</math> a mapping. Then the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(F,\mathcal C)</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] <math>Fix(P_{\mathcal C}\circ(I-F)).</math> | + | 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{H}</math> a mapping. Then the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(F,\mathcal{C})</math> is equivalent to the [[Complementarity_problem#Fixed_point_problems | fixed point problem]] <math>Fix(P_{\mathcal{C}}\circ(I-F)).</math> |
===== Proof ===== | ===== Proof ===== | ||
- | <math>x\,</math> is a solution of <math>Fix(P_{\mathcal C}\circ(I-F))</math> if and only if | + | <math>x\,</math> is a solution of <math>Fix(P_{\mathcal{C}}\circ(I-F))</math> if and only if |
- | <math>x=P_{\mathcal C}(x-F(x)).</math> Via [[Moreau's_decomposition_theorem#Characterization_of_the_projection|characterization of the projection]], the latter equation is equivalent to | + | <math>x=P_{\mathcal{C}}(x-F(x)).</math> Via [[Moreau's_decomposition_theorem#Characterization_of_the_projection|characterization of the projection]], the latter equation is equivalent to |
<center> | <center> | ||
Line 95: | Line 95: | ||
</center> | </center> | ||
- | for all <math>y\in\mathcal C.</math> But this holds if and only if <math>x\,</math> is a solution | + | for all <math>y\in\mathcal{C}.</math> But this holds if and only if <math>x\,</math> is a solution |
- | to <math>VI(F,\mathcal C).</math> | + | to <math>VI(F,\mathcal{C}).</math> |
===== Remark ===== | ===== Remark ===== | ||
Line 102: | Line 102: | ||
==== 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 <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{H}</math> a mapping. Then the [[Complementarity_problem#Nonlinear_complementarity_problems | 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{H}</math> a mapping. Then the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] |
- | <math>NCP(F,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] | + | <math>NCP(F,\mathcal{K})</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] |
- | <math>VI(F,\mathcal K).</math> | + | <math>VI(F,\mathcal{K}).</math> |
===== Proof ===== | ===== Proof ===== | ||
- | Suppose that <math>x\,</math> is a solution of <math>NCP(F,\mathcal K).</math> Then <math>x\in\mathcal K,</math> <math>F(x)\in\mathcal K^*,</math> and <math>\langle x,F(x)\rangle=0.</math> Hence | + | Suppose that <math>x\,</math> is a solution of <math>NCP(F,\mathcal{K}).</math> Then <math>x\in\mathcal{K},</math> <math>F(x)\in\mathcal{K}^*,</math> and <math>\langle x,F(x)\rangle=0.</math> Hence |
<center> | <center> | ||
Line 113: | Line 113: | ||
</center> | </center> | ||
- | for all <math>y\in\mathcal K.</math> Therefore <math>x\,</math> is a solution of <math>VI(F,\mathcal K).</math> | + | for all <math>y\in\mathcal{K}.</math> Therefore <math>x\,</math> is a solution of <math>VI(F,\mathcal{K}).</math> |
<br> | <br> | ||
<br> | <br> | ||
- | Conversely, suppose that <math>x\,</math> is a solution of <math>VI(F,\mathcal K).</math> Then | + | Conversely, suppose that <math>x\,</math> is a solution of <math>VI(F,\mathcal{K}).</math> Then |
- | <math>x\in\mathcal K</math> and | + | <math>x\in\mathcal{K}</math> and |
<center> | <center> | ||
Line 124: | Line 124: | ||
</center> | </center> | ||
- | for all <math>y\in\mathcal K.</math> Choosing <math>y=0\,</math> and <math>y=2x,\,</math> in particular, we get a system of two inequalities that demands <math>\langle x,F(x)\rangle=0.</math> Thus <math>\langle y,F(x)\rangle\geq 0</math> for all <math>y\in\mathcal K;</math> equivalently, <math>F(x)\in\mathcal K^*.</math> In conclusion, <math>x\in\mathcal K,</math> <math>F(x)\in\mathcal K^*,</math> and <math>\langle x,F(x)\rangle=0.</math> Therefore <math>x\,</math> is a solution to <math>NCP(F,\mathcal K).</math> | + | for all <math>y\in\mathcal{K}.</math> Choosing <math>y=0\,</math> and <math>y=2x,\,</math> in particular, we get a system of two inequalities that demands <math>\langle x,F(x)\rangle=0.</math> Thus <math>\langle y,F(x)\rangle\geq 0</math> for all <math>y\in\mathcal{K};</math> equivalently, <math>F(x)\in\mathcal{K}^*.</math> In conclusion, <math>x\in\mathcal{K},</math> <math>F(x)\in\mathcal{K}^*,</math> and <math>\langle x,F(x)\rangle=0.</math> Therefore <math>x\,</math> is a solution to <math>NCP(F,\mathcal{K}).</math> |
==== Concluding the alternative proof ==== | ==== Concluding the alternative proof ==== | ||
- | Since <math>\mathcal K</math> is a closed convex cone, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(F,\mathcal K)</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(F,\mathcal K)</math> 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]] <math>Fix(P_{\mathcal K}\circ(I-F)).</math> | + | Since <math>\mathcal{K}</math> is a closed convex cone, the [[Complementarity_problem#Nonlinear_complementarity_problems | nonlinear complementarity problem]] <math>NCP(F,\mathcal{K})</math> is equivalent to the [[Complementarity_problem#Variational_inequalities | variational inequality]] <math>VI(F,\mathcal{K})</math> 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]] <math>Fix(P_{\mathcal{K}}\circ(I-F)).</math> |
== Implicit complementarity problems == | == Implicit complementarity problems == |
Revision as of 02:33, 19 November 2011
(In particular, we can have everywhere in this page.)
Fixed point problems
Let be a set and a mapping. The fixed point problem defined by is the problem
Nonlinear complementarity problems
Let be a closed convex cone in the Hilbert space and a mapping. Recall that the dual cone of is the closed convex cone where is the polar of The nonlinear complementarity problem defined by and is the problem
Every nonlinear complementarity problem is equivalent to a fixed point problem
Let be a closed convex cone in the Hilbert space and a mapping. Then the nonlinear complementarity problem is equivalent to the fixed point problem where is the identity mapping defined by and is the projection onto
Proof
For all denote and Then
Suppose that is a solution of Then with
and Hence, via Moreau's theorem, we get Therefore is a solution of
Conversely, suppose that is a solution of Then and via Moreau's theorem
Hence thus Moreau's theorem also implies that In conclusion, and Therefore is a solution of
An alternative proof without Moreau's theorem
Variational inequalities
Let be a closed convex set in the Hilbert space and a mapping. The variational inequality defined by and is the problem
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 be a closed convex set in the Hilbert space and a mapping. Then the fixed point problem is equivalent to the variational inequality where
Proof
Suppose that is a solution of . Then, and thus is a solution of
Conversely, suppose that is a solution of and let Then, which is equivalent to Hence, ; that is, is a solution of .
Every variational inequality is equivalent to a fixed point problem
Let be a closed convex set in the Hilbert space and a mapping. Then the variational inequality is equivalent to the fixed point problem
Proof
is a solution of if and only if Via characterization of the projection, the latter equation is equivalent to
for all But this holds if and only if is a solution to
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 be a closed convex cone in the Hilbert space and a mapping. Then the nonlinear complementarity problem is equivalent to the variational inequality
Proof
Suppose that is a solution of Then and Hence
for all Therefore is a solution of
Conversely, suppose that is a solution of Then and
for all Choosing and in particular, we get a system of two inequalities that demands Thus for all equivalently, In conclusion, and Therefore is a solution to
Concluding the alternative proof
Since is a closed convex cone, the nonlinear complementarity problem is equivalent to the variational inequality which is equivalent to the fixed point problem
Implicit complementarity problems
Let be a closed convex cone in the Hilbert space and two mappings. Recall that the dual cone of is the closed convex cone where is the polar of The implicit complementarity problem defined by and the ordered pair of mappings is the problem
Every implicit complementarity problem is equivalent to a fixed point problem
Let be a closed convex cone in the Hilbert space and two mappings. Then the implicit complementarity problem is equivalent to the fixed point problem where is the identity mapping defined by
Proof
For all denote and Then
Suppose that is a solution of
Then with and
Via Moreau's theorem,
Therefore is a solution of
Conversely, suppose that is a solution of Then and, via Moreau's theorem,
Hence thus . Moreau's theorem also implies In conclusion, and Therefore is a solution of
Remark
If 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 be a closed convex set in the Hilbert space and a function. The nonlinear optimization problem defined by and is the problem
Any solution of a nonlinear optimization problem is a solution of a variational inequality
Let be a closed convex set in the Hilbert space and a differentiable function. Then any solution of the nonlinear optimization problem is a solution of the variational inequality where is the gradient of
Proof
Let be a solution of and an arbitrary point. Then by convexity of we have hence and
Therefore is a solution of
A convex optimization problem is equivalent to a variational inequality
Let be a closed convex set in the Hilbert space and a differentiable convex function. Then the nonlinear optimization problem is equivalent to the variational inequality where is the gradient of
Proof
Any solution of is a solution of
Conversely, suppose that is a solution of By convexity of we have for all Therefore is a solution of
Any solution of a nonlinear optimization problem on a closed convex cone is a solution of a nonlinear complementarity problem
Let be a closed convex cone in the Hilbert space and a differentiable function. Then any solution of the nonlinear optimization problem is a solution of the nonlinear complementarity problem
Proof
Any solution of is a solution of which is equivalent to
A convex optimization problem on a closed convex cone is equivalent to a nonlinear complementarity problem
Theorem NOPT. Let be a closed convex cone in the Hilbert space and a differentiable convex function. Then the nonlinear optimization problem is equivalent to the nonlinear complementarity problem
Proof
is equivalent to which is equivalent to
Fat nonlinear programming problem
Let be a function, and a fat matrix of full rank Then the problem
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 be a differentiable function, and a fat matrix of full rank If is a solution of the fat nonlinear programming problem then is a solution of the nonlinear complementarity problem where is a particular solution of the linear system of equations is the polyhedral cone defined by
and is defined by
Proof
Let be a solution of Then it is easy to see that is a solution of where is defined by It follows from Theorem NOPT that is a solution of because
Remark
If is convex, then the converse of the above results also holds. In other words,
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.