# Moreau's decomposition theorem

(Difference between revisions)
 Revision as of 11:11, 17 July 2009 (edit)← Previous diff Revision as of 11:39, 17 July 2009 (edit) (undo)Next diff → Line 1: Line 1: + Sándor Zoltán Németh + == Projection on closed convex sets == == Projection on closed convex sets == === Projection mapping === === Projection mapping === - Let $(\mathbb H,\langle\cdot,\cdot\rangle)$ be a Hilbert space and $\mathcal C$ a closed convex set in $\mathbb H.$ The '''projection mapping''' $P_{\mathcal C}$ onto $\mathcal C$ is the mapping $P_{\mathcal C}:\mathbb H\to\mathbb H$ defined by $P_{\mathcal C}(x)\in\mathcal C$ and Let $(\mathbb H,\langle\cdot,\cdot\rangle)$ be a Hilbert space and $\mathcal C$ a closed convex set in $\mathbb H.$ The '''projection mapping''' $P_{\mathcal C}$ onto $\mathcal C$ is the mapping $P_{\mathcal C}:\mathbb H\to\mathbb H$ defined by $P_{\mathcal C}(x)\in\mathcal C$ and Line 10: Line 11: === Characterization of the projection === === Characterization of the projection === - - - Let $(\mathbb H,\langle\cdot,\cdot\rangle)$ be a Hilbert space, $\mathcal C$ a closed convex set in $\mathbb H,\,u\in\mathbb H$ and $v\in\mathcal C.$ Then, $v=P_{\mathcal C}(u)$ if and only if $\langle u-v,w-v\rangle\leq0$ for all $w\in\mathcal C.$ Let $(\mathbb H,\langle\cdot,\cdot\rangle)$ be a Hilbert space, $\mathcal C$ a closed convex set in $\mathbb H,\,u\in\mathbb H$ and $v\in\mathcal C.$ Then, $v=P_{\mathcal C}(u)$ if and only if $\langle u-v,w-v\rangle\leq0$ for all $w\in\mathcal C.$ === Proof === === Proof === - Suppose that $v=P_{\mathcal C}u.$ Let $w\in\mathcal C$ and $t\in (0,1)$ be arbitrary. By using the convexity of $\mathcal C,$ it follows that $(1-t)v+tw\in\mathcal C.$ Then, by using the definition of the projection, we have Suppose that $v=P_{\mathcal C}u.$ Let $w\in\mathcal C$ and $t\in (0,1)$ be arbitrary. By using the convexity of $\mathcal C,$ it follows that $(1-t)v+tw\in\mathcal C.$ Then, by using the definition of the projection, we have Line 44: Line 41: == Moreau's theorem == == Moreau's theorem == - Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces. Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces.

## Revision as of 11:39, 17 July 2009

Sándor Zoltán Németh

## 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.