# Moreau's decomposition theorem

### From Wikimization

(→Every implicit complementarity problem is equivalent to a fixed point problem) |
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=== References === | === 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. | ||

- | |||

- | == An application to nonlinear complementarity problems == | ||

- | |||

- | === Fixed point problems === | ||

- | |||

- | Let <math>\mathcal A</math> be a set and <math>F:\mathcal A\to\mathcal A </math> a mapping. The '''fixed point problem''' defined by <math>F\,</math> is the problem | ||

- | |||

- | <center> | ||

- | <math> | ||

- | Fix(F):\left\{ | ||

- | \begin{array}{l} | ||

- | Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\ | ||

- | F(x)=x. | ||

- | \end{array} | ||

- | \right. | ||

- | </math> | ||

- | </center> | ||

- | |||

- | === Nonlinear complementarity problems === | ||

- | |||

- | 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. Recall that the dual cone of <math>\mathcal K</math> is the closed convex cone <math>\mathcal K^*=-\mathcal K^\circ,</math> where <math>\mathcal K^\circ</math> is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of <math>\mathcal K.</math> The '''nonlinear complementarity problem''' defined by <math>\mathcal K</math> and <math>f\,</math> is the problem | ||

- | |||

- | <center> | ||

- | <math> | ||

- | NCP(f,\mathcal K):\left\{ | ||

- | \begin{array}{l} | ||

- | Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\ | ||

- | f(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,f(x)\rangle=0. | ||

- | \end{array} | ||

- | \right. | ||

- | </math> | ||

- | </center> | ||

- | |||

- | === 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 nonlinear complementarity problem <math>NCP(f,\mathcal K)</math> is equivalent to the 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> | ||

- | |||

- | === Proof === | ||

- | |||

- | For all <math>x\in\mathbb 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_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> | ||

- | |||

- | 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 by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] | ||

- | |||

- | <center> | ||

- | <math>z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).</math> | ||

- | </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, | ||

- | <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]] === | ||

- | |||

- | ==== 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 | ||

- | |||

- | <center> | ||

- | <math> | ||

- | VI(f,\mathcal C):\left\{ | ||

- | \begin{array}{l} | ||

- | Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ | ||

- | \langle y-x,f(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C. | ||

- | \end{array} | ||

- | \right. | ||

- | </math> | ||

- | </center> | ||

- | |||

- | ==== 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 variational inequality <math>VI(f,\mathcal C)</math> is equivalent to the fixed point problem <math>Fix(P_{\mathcal C}\circ(I-f)).</math> | ||

- | |||

- | ==== Proof ==== | ||

- | |||

- | <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> By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection | characterization of the projection]] the latter equation is equivalent to | ||

- | |||

- | <center> | ||

- | <math>\langle x-f(x)-x,y-x\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 defined on a closed convex cone. | ||

- | |||

- | ==== 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 nonlinear complementarity problem <math>NCP(f,\mathcal K)</math> is equivalent to the variational inequality <math>VI(f,\mathcal K).</math> | ||

- | |||

- | ==== 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, | ||

- | |||

- | <center> | ||

- | <math>\langle y-x,f(x)\rangle\geq 0,</math> | ||

- | </center> | ||

- | |||

- | for all <math>y\in\mathcal K.</math> Therefore, <math>x\,</math> is a solution of <math>VI(f,\mathcal K).</math> | ||

- | <br> | ||

- | <br> | ||

- | |||

- | Conversely, suppose that <math>x\,</math> is a solution of <math>VI(f,\mathcal K).</math> Then, | ||

- | <math>x\in\mathcal K</math> and | ||

- | |||

- | <center> | ||

- | <math>\langle y-x,f(x)\rangle\geq 0,</math> | ||

- | </center> | ||

- | |||

- | for all <math>y\in\mathcal K.</math> Particularly, taking <math>y=0\,</math> and <math>y=2x\,</math>, respectively, we get <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> or 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 of <math>NCP(f,\mathcal K).</math> | ||

- | |||

- | ==== Concluding the alternative proof ==== | ||

- | |||

- | Since <math>\mathcal K</math> is a closed convex cone, the nonlinear complementarity problem <math>NCP(f,\mathcal K)</math> is equivalent to the variational inequality <math>VI(f,\mathcal K),</math> which is equivalent to the fixed point problem <math>Fix(P_{\mathcal K}\circ(I-f)).</math> | ||

- | |||

- | == An application to implicit complementarity problems == | ||

- | === Implicit complementarity problems === | ||

- | |||

- | 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,g:\mathbb H\to\mathbb H</math> two mappings. Recall that the dual cone of <math>\mathcal K</math> is the closed convex cone <math>\mathcal K^*=-\mathcal K^\circ,</math> where <math>\mathcal K^\circ</math> is the | ||

- | [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] | ||

- | of <math>\mathcal K.</math> The '''implicit complementarity problem''' defined by <math>\mathcal K</math> | ||

- | and the ordered pair of mappings <math>(f,g)\,</math> is the problem | ||

- | |||

- | <center> | ||

- | <math> | ||

- | ICP(f,g,\mathcal K):\left\{ | ||

- | \begin{array}{l} | ||

- | Find\,\,\,u\in\mathbb H\,\,\,such\,\,\,that\\ | ||

- | g(u)\in\mathcal K,\,\,\,f(u)\in K^*\,\,\,and\,\,\,\langle g(u),f(u)\rangle=0. | ||

- | \end{array} | ||

- | \right. | ||

- | </math> | ||

- | </center> | ||

- | |||

- | === Every implicit 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,g:\mathbb H\to\mathbb H</math> two mappings. Then, the implicit complementarity problem <math>ICP(f,g,\mathcal K)</math> is equivalent to the [[Moreau's_decomposition_theorem#Fixed_point_problems | fixed point problem]] | ||

- | <math>Fix(I-g+P_{\mathcal K}\circ(g-f)),</math> where <math>I:\mathbb H\to\mathbb H</math> is the identity mapping defined by <math>I(x)=x.\,</math> | ||

- | |||

- | === Proof === | ||

- | |||

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

- | <math>z=x+y.\,</math> | ||

- | <br> | ||

- | <br> | ||

- | |||

- | Suppose that <math>u\,</math> is a solution of <math>ICP(f,g,\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>u\,</math> is a solution of | ||

- | <math>Fix(I-g+P_{\mathcal K}\circ(g-f)).</math> | ||

- | <br> | ||

- | <br> | ||

- | |||

- | Conversely, suppose that <math>u\,</math> is a solution of <math>Fix(I-g+P_{\mathcal K}\circ(g-f)).</math> | ||

- | Then, <math>x\in\mathcal K</math> and by using | ||

- | [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] | ||

- | |||

- | <center> | ||

- | <math>z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).</math> | ||

- | </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, | ||

- | <math>g(u)=x\in\mathcal K,</math> <math>f(u)=-y\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Therefore, <math>u\,</math> is a solution of <math>ICP(f,g,\mathcal K).</math> | ||

- | |||

- | === Remark === | ||

- | |||

- | In particular if <math>g=I,</math> we obtain the result | ||

- | [[Moreau%27s_decomposition_theorem#Every_nonlinear_complementarity_problem_is_equivalent_to_a_fixed_point_problem | Every nonlinear complementarity problem is equivalent to a fixed point problem]], | ||

- | but the more general result [[Moreau%27s_decomposition_theorem#Every_implicit_complementarity_problem_is_equivalent_to_a_fixed_point_problem | Every implicit complementarity problem is equivalent to a fixed point problem]] has no known connection with variational inequalities. Therefore, using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] is essential for proving the latter result. |

## Revision as of 12:11, 17 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

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

For definition of *polar cone* in finite dimension, see more at Dual cone and polar 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.