# Moreau's decomposition theorem

### From Wikimization

(→An application to nonlinear complementarity problems) |
(→Every nonlinear complementarity problem is equivalent to a fixed point problem) |
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<math>Fix(P_{\mathcal K}\circ(I-f)),</math> where <math>I:\mathcal H\to\mathcal H</math> is the identity mapping defined by <math>I(x)=x.\,</math> | <math>Fix(P_{\mathcal K}\circ(I-f)),</math> where <math>I:\mathcal H\to\mathcal H</math> is the identity mapping defined by <math>I(x)=x.\,</math> | ||

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=== Proof === | === Proof === | ||

- | For all <math>x\in\mathcal H</math> denote | + | For all <math>x\in\mathcal H</math> denote <math>z=x-f(x)\,</math> and <math>y=-f(x).\,</math> Then, <math>z=x+y.\,</math> |

+ | <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> <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(I-f)).</math> | ||

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=== An alternative proof without Moreau's theorem === | === An alternative proof without Moreau's theorem === | ||

## Revision as of 06:28, 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