# Moreau's decomposition theorem

(Difference between revisions)
 Revision as of 07:38, 12 July 2009 (edit)m (→Proof of Moreau's theorem)← Previous diff Revision as of 07:39, 12 July 2009 (edit) (undo)m (→Proof of Moreau's theorem)Next diff → Line 72: Line 72: and thus $y=P_{\mathcal K^\circ}z.$ and thus $y=P_{\mathcal K^\circ}z.$ -
• 2$\Rightarrow$1: Let $x=P_{\mathcal K}z.$ By using the characterization of the projection, we have $\langle z-x,p-x\rangle\leq0,$ for all $p\in\mathcal K.$ In particular, if $p=0,\,$ then $\langle z-x,x\rangle\geq0$ and if $p=2x,\,$ then $\langle z-x,x\rangle\leq0.$ Thus, $\langle z-x,x\rangle=0.$ Denote $y=z-x\,$. Then, $\langle x,y\rangle=0.$ It remained to show that $y=P_{\mathcal K^\circ}z.$ First, we prove that $y\in\mathcal K^\circ.$ For this we have to show that $\langle y,p\rangle\leq0$, for +
• 2$\Rightarrow$1: Let $x=P_{\mathcal K}z.$ By using the characterization of the projection, we have $\langle z-x,p-x\rangle\leq0,$ for all $p\in\mathcal K.$ In particular, if $p=0,\,$ then $\langle z-x,x\rangle\geq0$ and if $p=2x,\,$ then $\langle z-x,x\rangle\leq0.$ Thus, $\langle z-x,x\rangle=0.$ Denote $y=z-x.\,$ Then, $\langle x,y\rangle=0.$ It remained to show that $y=P_{\mathcal K^\circ}z.$ First, we prove that $y\in\mathcal K^\circ.$ For this we have to show that $\langle y,p\rangle\leq0$, for all $p\in\mathcal K.$ By using the characterization of the projection, we have all $p\in\mathcal K.$ By using the characterization of the projection, we have

## Projection on closed convex sets

### Projection mapping

Let $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$ be a Hilbert space and $LaTeX: \mathcal C$ a closed convex set in $LaTeX: \mathcal H.$ The projection mapping $LaTeX: P_{\mathcal C}$ onto $LaTeX: \mathcal C$ is the mapping $LaTeX: P_{\mathcal C}:\mathcal H\to\mathcal H$ 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 $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$ be a Hilbert space, $LaTeX: \mathcal C$ a closed convex set in $LaTeX: \mathcal H,\,u\in\mathcal H$ 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 (see more at Convex cone, Wikipedia or for finite dimension at Convex cones, Wikimization).

Theorem (Moreau) Let $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$ and $LaTeX: \mathcal K^\circ$ its polar cone; that is, the closed convex cone defined by $LaTeX: K^\circ=\{a\in\mathcal H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\}$ (for finite dimension see more at Dual cone and polar cone; see also Extended Farkas' lemma). For $LaTeX: x,y,z\in\mathcal H$ 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: Let $LaTeX: x=P_{\mathcal K}z.$ 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 $LaTeX: y=z-x.\,$ Then, $LaTeX: \langle x,y\rangle=0.$ It remained to show that $LaTeX: y=P_{\mathcal K^\circ}z.$ First, we prove that $LaTeX: y\in\mathcal K^\circ.$ For this we have to show that $LaTeX: \langle y,p\rangle\leq0$, for all $LaTeX: p\in\mathcal K.$ By using the characterization of the projection, we have

$LaTeX: \langle y,p\rangle=\langle y,p-x\rangle=\langle z-x,p-x\rangle\leq0,$

for all $LaTeX: p\in\mathcal K$. Thus, $LaTeX: y\in\mathcal K^\circ.$ We also have

$LaTeX: \langle z-y,q-y\rangle=\langle x,q-y\rangle=\langle x,q\rangle\leq0,$

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

### 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 $LaTeX: \mathcal A$ be a set and $LaTeX: F:\mathcal A\to\mathcal A.$ The fixed point problem defined by $LaTeX: F\,$ is the problem

$LaTeX: Fix(F):\left\{ \begin{array}{l} Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\ F(x)=x. \end{array} \right.$

### Nonlinear complementarity problems

Let $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathcal H\to\mathcal H.$ Recall that the dual cone of $LaTeX: \mathcal K$ is the closed convex cone $LaTeX: \mathcal K^*=-\mathcal K^\circ,$ where $LaTeX: \mathcal K^\circ$ is the polar of $LaTeX: \mathcal K.$ The nonlinear complementarity problem defined by $LaTeX: \mathcal K$ and $LaTeX: f\,$ is the problem

$LaTeX: NCP(f,\mathcal K):\left\{ \begin{array}{l} Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\ f(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,f(x)\rangle=0. \end{array} \right.$

### Every nonlinear complementarity problem is equivalent to a fixed point problem

Let $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathcal H\to\mathcal H.$ Then, the nonlinear complementarity problem $LaTeX: NCP(f,\mathcal K)$ is equivalent to the fixed point problem $LaTeX: Fix(P_{\mathcal K}\circ(I-f)),$ where $LaTeX: I:\mathcal H\to\mathcal H$ is the identity mapping defined by $LaTeX: I(x)=x.\,$

### Proof

For all $LaTeX: x\in\mathcal H$ denote $LaTeX: z=x-f(x)\,$ and $LaTeX: y=-f(x).\,$ Then, $LaTeX: z=x+y.\,$

Suppose that $LaTeX: x\,$ is a solution of $LaTeX: NCP(f,\mathcal K).$ Then, $LaTeX: z=x+y,\,$ with $LaTeX: x\in\mathcal K,$ $LaTeX: y\in\mathcal K^\circ$ and $LaTeX: \langle x,y\rangle=0.$ Hence, by using Moreau's theorem, we get $LaTeX: x=P_{\mathcal K}z.$ Therefore, $LaTeX: x\,$ is a solution of $LaTeX: Fix(P_{\mathcal K}\circ(I-f)).$

Conversely, suppose that $LaTeX: x\,$ is a solution of $LaTeX: Fix(P_{\mathcal K}\circ(I-f)).$ Then, $LaTeX: x\in\mathcal K$ and by using Moreau's theorem

$LaTeX: z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).$

Hence, $LaTeX: P_{\mathcal K^\circ}(z)=z-x=y,$. Thus, $LaTeX: y\in\mathcal K^\circ$. Moreau's theorem also implies that $LaTeX: \langle x,y\rangle=0.$ In conclusion, $LaTeX: x\in\mathcal K,$ $LaTeX: f(x)=-y\in\mathcal K^*$ and $LaTeX: \langle x,f(x)\rangle=0.$ Therefore, $LaTeX: x\,$ is a solution of $LaTeX: NCP(f,\mathcal K).$

### An alternative proof without Moreau's theorem

#### Variational inequalities

Let $LaTeX: \mathcal C$ be a closed convex set in the Hilbert space $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathcal H\to\mathcal H.$ The variational inequality defined by $LaTeX: \mathcal C$ and $LaTeX: f\,$ is the problem

$LaTeX: VI(f,\mathcal C):\left\{ \begin{array}{l} Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\ \langle y-x,f(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C. \end{array} \right.$

#### Every variational inequality is equivalent to a fixed point problem

Let $LaTeX: \mathcal C$ be a closed convex set in the Hilbert space $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathcal H\to\mathcal H.$ Then the variational inequality $LaTeX: VI(f,\mathcal C)$ is equivalent to the fixed point problem $LaTeX: Fix(P_{\mathcal C}\circ(I-f)).$

#### Proof

$LaTeX: x\,$ is a solution of $LaTeX: Fix(P_{\mathcal C}\circ(I-f))$ if and only if $LaTeX: x=P_{\mathcal C}(x-f(x)).$ By using the characterization of the projection the latter equation is equivalent to

$LaTeX: \langle x-f(x)-x,y-x\rangle\leq0,$

for all $LaTeX: y\in\mathcal C.$ But this holds if and only if $LaTeX: x\,$ is a solution of $LaTeX: VI(f,\mathcal C).$

##### 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 $LaTeX: \mathcal K$ be a closed convex cone in the Hilbert space $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$ and $LaTeX: f:\mathcal H\to\mathcal H.$ Then, the nonlinear complementarity problem $LaTeX: NCP(f,\mathcal K)$ is equivalent to the variational inequality $LaTeX: VI(f,\mathcal K).$

#### Proof

Suppose that $LaTeX: x\,$ is a solution of $LaTeX: NCP(f,\mathcal K).$ Then, $LaTeX: x\in\mathcal K,$ $LaTeX: f(x)\in\mathcal K^*$ and $LaTeX: \langle x,f(x)\rangle=0.$ Hence,

$LaTeX: \langle y-x,f(x)\rangle\geq 0,$

for all $LaTeX: y\in\mathcal K.$ Therefore, $LaTeX: x\,$ is a solution of $LaTeX: VI(f,\mathcal K).$

Conversely, suppose that $LaTeX: x\,$ is a solution of $LaTeX: VI(f,\mathcal K).$ Then, $LaTeX: x\in\mathcal K$ and

$LaTeX: \langle y-x,f(x)\rangle\geq 0,$

for all $LaTeX: y\in\mathcal K.$ Particularly, taking $LaTeX: y=0\,$ and $LaTeX: y=-x\,$, respectively, we get $LaTeX: \langle x,f(x)\rangle=0.$ Thus, $LaTeX: \langle y,f(x)\rangle\geq 0,$ for all $LaTeX: y\in\mathcal K,$ or equivalently $LaTeX: f(x)\in\mathcal K^*.$ In conclusion, $LaTeX: x\in\mathcal K,$ $LaTeX: f(x)\in\mathcal K^*$ and $LaTeX: \langle x,f(x)\rangle=0.$ Therefore, $LaTeX: x\,$ is a solution of $LaTeX: NCP(f,\mathcal K).$

#### Concluding the alternative proof

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