Moreau's decomposition theorem
From Wikimization
m (→An application to nonlinear complementarity problems) 
(→An application to nonlinear complementarity problems) 

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  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, by using 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(If)).</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, by using [[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(If)).</math> 
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<br>  <br>  
  Conversely, suppose that <math>x\,</math> is a solution of <math>Fix(P_{\mathcal K}\circ(If)).</math> Then, <math>x\in\mathcal K</math> and by using Moreau's theorem  +  Conversely, suppose that <math>x\,</math> is a solution of <math>Fix(P_{\mathcal K}\circ(If)).</math> Then, <math>x\in\mathcal K</math> and by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem  Moreau's theorem]] 
<center>  <center>  
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  Hence, <math>P_{\mathcal K^\circ}(z)=zx=y,</math>. Thus, <math>y\in\mathcal K^\circ</math>. Moreau's theorem also implies that <math>\langle x,y\rangle=0.</math> In conclusion,  +  Hence, <math>P_{\mathcal K^\circ}(z)=zx=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 theorem ===  +  
==== Variational inequalities ====  ==== Variational inequalities ====  
+  
+  Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathcal H\to\mathcal H.</math> The '''variational inequality''' defined by <math>\mathcal C</math> and <math>f\,</math> is the problem  
+  
+  <center>  
+  <math>  
+  VI(f,\mathcal C):\left\{  
+  \begin{array}{l}  
+  Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\  
+  \langle yx,f(x)\rangle=0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C  
+  \end{array}  
+  \right.  
+  </math>  
+  </center>  
==== 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>(\mathcal H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathcal H\to\mathcal H.</math> Then the variational inequality <math>VI(f,\mathcal C)</math> is equivalent to the fixed point problem <math>Fix(P_{\mathcal C}\circ(If)).</math>  
+  
+  ==== Proof ====  
+  
+  <math>x\,</math> is a solution of <math>Fix(P_{\mathcal C}\circ(If))</math> if and only if  
+  <math>x=P_{\mathcal C}(xf(x)).</math> By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection  characterization of the projection]] the latter equation is equivalent to  
+  
+  <center>  
+  <math>\langle xf(x)x,yx\rangle\leq0,</math>  
+  </center>  
+  
+  for all <math>y\in\mathcal C.</math> But this holds if and only if <math>x\,</math> is a solution  
+  of <math>VI(f,\mathcal C).</math>  
+  
+  ===== 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 problem 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 ====  
  > 
Revision as of 07:39, 12 July 2009
Contents

Projection on closed convex sets
Projection mapping
Let be a Hilbert space and a closed convex set in The projection mapping onto is the mapping defined by and
Characterization of the projection
Let be a Hilbert space, a closed convex set in and Then, if and only if for all
Proof
Suppose that Let and be arbitrary. By using the convexity of it follows that Then, by using the definition of the projection, we have
Hence,
By tending with to we get
Conversely, suppose that for all Then,
for all Hence, by using the definition of the projection, we get
Moreau's theorem
Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces. Recall that a convex cone in a vector space is a set which is invariant under the addition of vectors and multiplication of vectors by positive scalars (see more at Convex cone, Wikipedia or for finite dimension at Convex cones, Wikimization).
Theorem (Moreau) Let be a closed convex cone in the Hilbert space and its polar cone; that is, the closed convex cone defined by (for finite dimension see more at Dual cone and polar cone; see also Extended Farkas' lemma). For the following statements are equivalent:
 and
 and
Proof of Moreau's theorem
 12: For all we have
.
Then, by the characterization of the projection, it follows that . Similarly, for all we have
 21: Let . By using the characterization of the projection, we have for all . In particular, if then and if then . Thus, . Denote . Then, . It remained to show that . First, we prove that . For this we have to show that , for
all . By using the characterization of the projection, we have
for all . Thus, . We also have
for all , because . By using again the characterization of the projection, it follows that .
References
 J. J. Moreau, Décomposition orthogonale d'un espace hilbertien selon deux cones mutuellement polaires, C. R. Acad. Sci., volume 255, pages 238–240, 1962.
An application to nonlinear complementarity problems
Fixed point problems
Let be a set and The fixed point problem defined by is the problem
Nonlinear complementarity problems
Let be a closed convex cone in the Hilbert space and 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 Then, the nonlinear 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 Hence, by using Moreau's theorem, we get Therefore, is a solution of
Conversely, suppose that is a solution of Then, and by using 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 The variational inequality defined by and is the problem
Every variational inequality is equivalent to a fixed point problem
Let be a closed convex set in the Hilbert space and Then the variational inequality is equivalent to the fixed point problem
Proof
is a solution of if and only if By using the characterization of the projection the latter equation is equivalent to
for all But this holds if and only if is a solution of
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 problem defined on a closed convex cone.