# Moreau's decomposition theorem

(Difference between revisions)
 Revision as of 04:37, 17 July 2009 (edit) (→Every implicit complementarity problem is equivalent to a fixed point problem)← Previous diff Revision as of 11:11, 17 July 2009 (edit) (undo)Next diff → Line 104: Line 104: === 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 $\mathcal A$ be a set and $F:\mathcal A\to\mathcal A$ a mapping. The '''fixed point problem''' defined by $F\,$ is the problem - -
- $- Fix(F):\left\{ - \begin{array}{l} - Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\ - F(x)=x. - \end{array} - \right. -$ -
- - === Nonlinear complementarity problems === - - Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Recall that the dual cone of $\mathcal K$ is the closed convex cone $\mathcal K^*=-\mathcal K^\circ,$ where $\mathcal K^\circ$ is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of $\mathcal K.$ The '''nonlinear complementarity problem''' defined by $\mathcal K$ and $f\,$ is the problem - -
- $- NCP(f,\mathcal K):\left\{ - \begin{array}{l} - Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\ - f(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,f(x)\rangle=0. - \end{array} - \right. -$ -
- - === Every nonlinear complementarity problem is equivalent to a fixed point problem === - - Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the fixed point problem - $Fix(P_{\mathcal K}\circ(I-f)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x.\,$ - - === Proof === - - For all $x\in\mathbb H$ denote $z=x-f(x)\,$ and $y=-f(x).\,$ Then, $z=x+y.\,$ -
-
- - Suppose that $x\,$ is a solution of $NCP(f,\mathcal K).$ Then, $z=x+y,\,$ with $x\in\mathcal K,$ $y\in\mathcal K^\circ$ and $\langle x,y\rangle=0.$ Hence, by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], we get $x=P_{\mathcal K}z.$ Therefore, $x\,$ is a solution of $Fix(P_{\mathcal K}\circ(I-f)).$ -
-
- - Conversely, suppose that $x\,$ is a solution of $Fix(P_{\mathcal K}\circ(I-f)).$ Then, $x\in\mathcal K$ and by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] - -
- $z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).$ -
- - Hence, $P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $y\in\mathcal K^\circ$. [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] also implies that $\langle x,y\rangle=0.$ In conclusion, - $x\in\mathcal K,$ $f(x)=-y\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Therefore, $x\,$ is a solution of $NCP(f,\mathcal K).$ - - === An alternative proof without [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] === - - ==== Variational inequalities ==== - - Let $\mathcal C$ be a closed convex set in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. The '''variational inequality''' defined by $\mathcal C$ and $f\,$ is the problem - -
- $- VI(f,\mathcal C):\left\{ - \begin{array}{l} - Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ - \langle y-x,f(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C. - \end{array} - \right. -$ -
- - ==== Every variational inequality is equivalent to a fixed point problem ==== - - Let $\mathcal C$ be a closed convex set in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then the variational inequality $VI(f,\mathcal C)$ is equivalent to the fixed point problem $Fix(P_{\mathcal C}\circ(I-f)).$ - - ==== Proof ==== - - $x\,$ is a solution of $Fix(P_{\mathcal C}\circ(I-f))$ if and only if - $x=P_{\mathcal C}(x-f(x)).$ By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection | characterization of the projection]] the latter equation is equivalent to - -
- $\langle x-f(x)-x,y-x\rangle\leq0,$ -
- - for all $y\in\mathcal C.$ But this holds if and only if $x\,$ is a solution - of $VI(f,\mathcal C).$ - - ===== 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 $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f:\mathbb H\to\mathbb H$ a mapping. Then, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the variational inequality $VI(f,\mathcal K).$ - - ==== Proof ==== - - Suppose that $x\,$ is a solution of $NCP(f,\mathcal K).$ Then, $x\in\mathcal K,$ $f(x)\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Hence, - -
- $\langle y-x,f(x)\rangle\geq 0,$ -
- - for all $y\in\mathcal K.$ Therefore, $x\,$ is a solution of $VI(f,\mathcal K).$ -
-
- - Conversely, suppose that $x\,$ is a solution of $VI(f,\mathcal K).$ Then, - $x\in\mathcal K$ and - -
- $\langle y-x,f(x)\rangle\geq 0,$ -
- - for all $y\in\mathcal K.$ Particularly, taking $y=0\,$ and $y=2x\,$, respectively, we get $\langle x,f(x)\rangle=0.$ Thus, $\langle y,f(x)\rangle\geq 0,$ for all $y\in\mathcal K,$ or equivalently $f(x)\in\mathcal K^*.$ In conclusion, $x\in\mathcal K,$ $f(x)\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Therefore, $x\,$ is a solution of $NCP(f,\mathcal K).$ - - ==== Concluding the alternative proof ==== - - Since $\mathcal K$ is a closed convex cone, the nonlinear complementarity problem $NCP(f,\mathcal K)$ is equivalent to the variational inequality $VI(f,\mathcal K),$ which is equivalent to the fixed point problem $Fix(P_{\mathcal K}\circ(I-f)).$ - - == An application to implicit complementarity problems == - === Implicit complementarity problems === - - Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f,g:\mathbb H\to\mathbb H$ two mappings. Recall that the dual cone of $\mathcal K$ is the closed convex cone $\mathcal K^*=-\mathcal K^\circ,$ where $\mathcal K^\circ$ is the - [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] - of $\mathcal K.$ The '''implicit complementarity problem''' defined by $\mathcal K$ - and the ordered pair of mappings $(f,g)\,$ is the problem - -
- $- ICP(f,g,\mathcal K):\left\{ - \begin{array}{l} - Find\,\,\,u\in\mathbb H\,\,\,such\,\,\,that\\ - g(u)\in\mathcal K,\,\,\,f(u)\in K^*\,\,\,and\,\,\,\langle g(u),f(u)\rangle=0. - \end{array} - \right. -$ -
- - === Every implicit complementarity problem is equivalent to a fixed point problem === - - Let $\mathcal K$ be a closed convex cone in the Hilbert space $(\mathbb H,\langle\cdot,\cdot\rangle)$ and $f,g:\mathbb H\to\mathbb H$ two mappings. Then, the implicit complementarity problem $ICP(f,g,\mathcal K)$ is equivalent to the [[Moreau's_decomposition_theorem#Fixed_point_problems | fixed point problem]] - $Fix(I-g+P_{\mathcal K}\circ(g-f)),$ where $I:\mathbb H\to\mathbb H$ is the identity mapping defined by $I(x)=x.\,$ - - === Proof === - - For all $u\in\mathbb H$ denote $z=g(u)-f(u),\,$ $x=g(u)\,$ and $y=-f(u).\,$ Then, - $z=x+y.\,$ -
-
- - Suppose that $u\,$ is a solution of $ICP(f,g,\mathcal K).$ Then, $z=x+y,\,$ with $x\in\mathcal K,$ $y\in\mathcal K^\circ$ and $\langle x,y\rangle=0.$ Hence, by using - [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], - we get $x=P_{\mathcal K}z.$ Therefore, $u\,$ is a solution of - $Fix(I-g+P_{\mathcal K}\circ(g-f)).$ -
-
- - Conversely, suppose that $u\,$ is a solution of $Fix(I-g+P_{\mathcal K}\circ(g-f)).$ - Then, $x\in\mathcal K$ and by using - [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] - -
- $z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).$ -
- - Hence, $P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $y\in\mathcal K^\circ$. - [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] - also implies that $\langle x,y\rangle=0.$ In conclusion, - $g(u)=x\in\mathcal K,$ $f(u)=-y\in\mathcal K^*$ and $\langle x,f(x)\rangle=0.$ Therefore, $u\,$ is a solution of $ICP(f,g,\mathcal K).$ - - === Remark === - - In particular if $g=I,$ 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.

## Projection on closed convex sets

### Projection mapping

Let $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ be a Hilbert space and $LaTeX: \mathcal C$ a closed convex set in $LaTeX: \mathbb H.$ The projection mapping $LaTeX: P_{\mathcal C}$ onto $LaTeX: \mathcal C$ is the mapping $LaTeX: P_{\mathcal C}:\mathbb H\to\mathbb H$ defined by $LaTeX: P_{\mathcal C}(x)\in\mathcal C$ and $LaTeX: \|x-P_{\mathcal C}(x)\|=\min\{\|x-y\|\mid y\in\mathcal C\}.$

### Characterization of the projection

Let $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ be a Hilbert space, $LaTeX: \mathcal C$ a closed convex set in $LaTeX: \mathbb H,\,u\in\mathbb H$ and $LaTeX: v\in\mathcal C.$ Then, $LaTeX: v=P_{\mathcal C}(u)$ if and only if $LaTeX: \langle u-v,w-v\rangle\leq0$ for all $LaTeX: w\in\mathcal C.$

### Proof

Suppose that $LaTeX: v=P_{\mathcal C}u.$ Let $LaTeX: w\in\mathcal C$ and $LaTeX: t\in (0,1)$ be arbitrary. By using the convexity of $LaTeX: \mathcal C,$ it follows that $LaTeX: (1-t)v+tw\in\mathcal C.$ Then, by using the definition of the projection, we have $LaTeX: \|u-v\|^2\leq\|u-((1-t)v+tw)\|^2=\|u-v-t(w-v)\|^2=\|u-v\|^2-2t\langle u-v,w-v\rangle+t^2\|w-v\|^2,$

Hence, $LaTeX: \langle u-v,w-v\rangle\leq\frac t2\|w-v\|^2.$

By tending with $LaTeX: t\,$ to $LaTeX: 0,\,$ we get $LaTeX: \langle u-v,w-v\rangle\leq0.$

Conversely, suppose that $LaTeX: \langle u-v,w-v\rangle\leq0,$ for all $LaTeX: w\in\mathcal C.$ Then, $LaTeX: \|u-w\|^2=\|u-v-(w-v)\|^2=\|u-v\|^2-2\langle u-v,w-v\rangle+\|w-v\|^2\geq \|u-v\|^2,$

for all $LaTeX: w\in\mathcal C.$ Hence, by using the definition of the projection, we get $LaTeX: v=P_{\mathcal C}u.$

## 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 $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle)$ and $LaTeX: \mathcal K^\circ$ its polar cone; that is, the closed convex cone defined by $LaTeX: \mathcal K^\circ=\{a\in\mathbb H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\}.$

For $LaTeX: x,y,z\in\mathbb H$ the following statements are equivalent:

1. $LaTeX: z=x+y,\,x\in\mathcal K,\,y\in\mathcal K^\circ$ and $LaTeX: \langle x,y\rangle=0,$
2. $LaTeX: x=P_{\mathcal K}z$ and $LaTeX: y=P_{\mathcal K^\circ}z.$

### Proof of Moreau's theorem

• 1 $LaTeX: \Rightarrow$2: For all $LaTeX: p\in K$ we have $LaTeX: \langle z-x,p-x\rangle=\langle y,p-x\rangle=\langle y,p\rangle\leq0.$

Then, by the characterization of the projection, it follows that $LaTeX: x=P_{\mathcal K}z.$ Similarly, for all $LaTeX: q\in K^\circ$ we have $LaTeX: \langle z-y,q-y\rangle=\langle x,q-y\rangle=\langle x,q\rangle\leq0$

and thus $LaTeX: y=P_{\mathcal K^\circ}z.$
• 2 $LaTeX: \Rightarrow$1: By using the characterization of the projection, we have $LaTeX: \langle z-x,p-x\rangle\leq0,$ for all $LaTeX: p\in\mathcal K.$ In particular, if $LaTeX: p=0,\,$ then $LaTeX: \langle z-x,x\rangle\geq0$ and if $LaTeX: p=2x,\,$ then $LaTeX: \langle z-x,x\rangle\leq0.$ Thus, $LaTeX: \langle z-x,x\rangle=0.$ Denote $LaTeX: u=z-x.\,$ Then, $LaTeX: \langle x,u\rangle=0.$ It remains to show that $LaTeX: u=y.\,$ First, we prove that $LaTeX: u\in\mathcal K^\circ.$ For this we have to show that $LaTeX: \langle u,p\rangle\leq0,$ for all $LaTeX: p\in\mathcal K.$ By using the characterization of the projection, we have $LaTeX: \langle u,p\rangle=\langle u,p-x\rangle=\langle z-x,p-x\rangle\leq0,$

for all $LaTeX: p\in\mathcal K.$ Thus, $LaTeX: u\in\mathcal K^\circ.$ We also have $LaTeX: \langle z-u,q-u\rangle=\langle x,q-u\rangle=\langle x,q\rangle\leq0,$

for all $LaTeX: q\in K^\circ,$ because $LaTeX: x\in K.$ By using again the characterization of the projection, it follows that $LaTeX: u=y.\,$

### 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. $LaTeX: \mathcal K^{\circ\circ}=K$ 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.