# Moreau's decomposition theorem

### From Wikimization

(→Proof) |
(→Moreau's theorem) |
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under the addition of vectors and multiplication of vectors by positive scalars. | 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>\mathcal K^\circ=\{a\in\mathbb H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\}.</math> | + | '''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>\mathcal{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: | + | For <math>x,y,z\in\mathbb{H}</math> the following statements are equivalent: |

<ol> | <ol> | ||

- | <li><math>z=x+y,\,x\in\mathcal K,\,y\in\mathcal K^\circ</math> and <math>\langle x,y\rangle=0,</math></li> | + | <li><math>z=x+y,\,x\in\mathcal{K},\,y\in\mathcal{K}^\circ</math> and <math>\langle x,y\rangle=0,</math></li> |

- | <li><math>x=P_{\mathcal K}z</math> and <math>y=P_{\mathcal K^\circ}z.</math> | + | <li><math>x=P_{\mathcal{K}}z</math> and <math>y=P_{\mathcal{K}^\circ}z.</math> |

</li> | </li> | ||

</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\mathcal{K}</math> we have |

<center> | <center> | ||

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</center> | </center> | ||

- | Then, by the characterization of the projection, it follows that <math>x=P_{\mathcal K}z.</math> Similarly, for all <math>q\in K^\circ</math> we have | + | Then, by the characterization of the projection, it follows that <math>x=P_{\mathcal{K}}z.</math> Similarly, for all <math>q\in\mathcal{K}^\circ</math> we have |

<center> | <center> | ||

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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 z-x,p-x\rangle\leq0,</math> for all <math>p\in\mathcal K.</math> In particular, if <math>p=0,\,</math> then <math>\langle z-x,x\rangle\geq0</math> and if <math>p=2x,\,</math> then <math>\langle z-x,x\rangle\leq0.</math> Thus, <math>\langle z-x,x\rangle=0.</math> Denote <math>u=z-x.\,</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 | + | <li>2<math>\Rightarrow</math>1: By using the characterization of the projection, we have <math>\langle z-x,p-x\rangle\leq0,</math> for all <math>p\in\mathcal K.</math> In particular, if <math>p=0,\,</math> then <math>\langle z-x,x\rangle\geq0</math> and if <math>p=2x,\,</math> then <math>\langle z-x,x\rangle\leq0.</math> Thus, <math>\langle z-x,x\rangle=0.</math> Denote <math>u=z-x.\,</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 |

<center> | <center> | ||

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- | for all <math>p\in\mathcal K.</math> Thus, <math>u\in\mathcal K^\circ.</math> We also have | + | for all <math>p\in\mathcal{K}.</math> Thus, <math>u\in\mathcal{K}^\circ.</math> We also have |

<center> | <center> | ||

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</center> | </center> | ||

- | 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\mathcal{K}^\circ,</math> because <math>x\in\mathcal{K}.</math> By using again the characterization of the projection, it follows that <math>u=y.\,</math> |

</ul> | </ul> | ||

## Revision as of 01:59, 19 November 2011

(In particular, we can have everywhere in this page.)

## 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

For definition of *convex cone* in finite dimension see Convex cones, Wikimization.

For definition of *polar cone* in finite dimension, see Convex Optimization & Euclidean Distance Geometry.

## Applications

For applications see Every nonlinear complementarity problem is equivalent to a fixed point problem, Every implicit complementarity problem is equivalent to a fixed point problem, and Projection on isotone projection cone.

### 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.