Moreau's decomposition theorem
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(→Moreau's theorem) 

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== Moreau's theorem ==  == Moreau's theorem ==  
  Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces  +  Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces. 
  +  
  '''Theorem (Moreau)''' Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>\mathcal K^\circ</math> its '''polar cone'''; that is, the closed convex cone defined by <math>K^\circ=\{a\in\mathbb H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\}</math>  +  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.  
+  
+  '''Theorem (Moreau).''' Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>\mathcal K^\circ</math> its '''polar cone'''; that is, the closed convex cone defined by <math>K^\circ=\{a\in\mathbb H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\}.</math>  
+  
+  For <math>x,y,z\in\mathbb H</math> the following statements are equivalent:  
<ol>  <ol>  
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=== Proof of Moreau's theorem ===  === Proof of Moreau's theorem ===  
  
<ul>  <ul>  
<li>1<math>\Rightarrow</math>2: For all <math>p\in K</math> we have  <li>1<math>\Rightarrow</math>2: For all <math>p\in K</math> we have  
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and thus <math>y=P_{\mathcal K^\circ}z.</math></li>  and thus <math>y=P_{\mathcal K^\circ}z.</math></li>  
  <li>2<math>\Rightarrow</math>1: By using the characterization of the projection, we have <math>\langle zx,px\rangle\leq0,</math> for all <math>p\in\mathcal K.</math> In particular, if <math>p=0,\,</math> then <math>\langle zx,x\rangle\geq0</math> and if <math>p=2x,\,</math> then <math>\langle zx,x\rangle\leq0.</math> Thus, <math>\langle zx,x\rangle=0.</math> Denote <math>u=zx.\,</math> Then, <math>\langle x,u\rangle=0.</math> It  +  <li>2<math>\Rightarrow</math>1: By using the characterization of the projection, we have <math>\langle zx,px\rangle\leq0,</math> for all <math>p\in\mathcal K.</math> In particular, if <math>p=0,\,</math> then <math>\langle zx,x\rangle\geq0</math> and if <math>p=2x,\,</math> then <math>\langle zx,x\rangle\leq0.</math> Thus, <math>\langle zx,x\rangle=0.</math> Denote <math>u=zx.\,</math> Then, <math>\langle x,u\rangle=0.</math> It remains to show that <math>u=y.\,</math> First, we prove that <math>u\in\mathcal K^\circ.</math> For this we have to show that <math>\langle u,p\rangle\leq0,</math> for 
all <math>p\in\mathcal K.</math> By using the characterization of the projection, we have  all <math>p\in\mathcal K.</math> By using the characterization of the projection, we have  
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for all <math>q\in K^\circ,</math> because <math>x\in K.</math> By using again the characterization of the projection, it follows that <math>u=y.\,</math>  for all <math>q\in K^\circ,</math> because <math>x\in K.</math> By using again the characterization of the projection, it follows that <math>u=y.\,</math>  
  
</ul>  </ul>  
  ===  +  === notes === 
+  See more at [http://en.wikipedia.org/wiki/Convex_cone Convex cone, Wikipedia] or for finite dimension at [[Convex cones  Convex cones, Wikimization]].  
+  For finite dimension, see more at [http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone Dual cone and polar cone];  
+  see also [[Farkas%27_lemma#Extended_Farkas.27_lemmaExtended Farkas' lemma]].  
+  
+  === 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.  * J. J. Moreau, Décomposition orthogonale d'un espace hilbertien selon deux cones mutuellement polaires, C. R. Acad. Sci., volume 255, pages 238–240, 1962.  
Revision as of 15:13, 13 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.
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 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: By using the characterization of the projection, we have for all In particular, if then and if then Thus, Denote Then, It remains 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
notes
See more at Convex cone, Wikipedia or for finite dimension at Convex cones, Wikimization.
For finite dimension, see more at Dual cone and polar cone; see also Extended Farkas' lemma.
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 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 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 Particularly, taking and , respectively, we get Thus, for all or equivalently In conclusion, and Therefore, is a solution of
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