# Moreau's decomposition theorem

(Difference between revisions)
 Revision as of 06:01, 11 July 2009 (edit)← Previous diff Revision as of 06:05, 11 July 2009 (edit) (undo)Next diff → Line 1: Line 1: - = Projection on closed convex sets = + == Projection on closed convex sets == - == Projection mapping == + === Projection mapping === Let $(\mathcal H,\langle\cdot,\cdot\rangle)$ be a Hilbert space and $\mathcal C$ a closed convex set in $\mathcal H$. The '''projection mapping''' $P_{\mathcal C}$ onto $\mathcal C$ is the mapping $P_{\mathcal C}:\mathcal H\to\mathcal H$ defined by $P_{\mathcal C}(x)\in\mathcal C$ and Let $(\mathcal H,\langle\cdot,\cdot\rangle)$ be a Hilbert space and $\mathcal C$ a closed convex set in $\mathcal H$. The '''projection mapping''' $P_{\mathcal C}$ onto $\mathcal C$ is the mapping $P_{\mathcal C}:\mathcal H\to\mathcal H$ defined by $P_{\mathcal C}(x)\in\mathcal C$ and Line 9: Line 9: - == Characterization of the projection == + === Characterization of the projection === Line 15: Line 15: Let $(\mathcal H,\langle\cdot,\cdot\rangle)$ be a Hilbert space, $\mathcal C$ a closed convex set in $\mathcal H,\,u\in\mathcal H$ 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 $(\mathcal H,\langle\cdot,\cdot\rangle)$ be a Hilbert space, $\mathcal C$ a closed convex set in $\mathcal H,\,u\in\mathcal H$ 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 $v=P_{\mathcal C}u$ and let $w\in\mathcal C$ be arbitrary. By using the convexity of $\mathcal C$, it follows that $(1-t)v+tw\in\mathcal C$, for all $t\in (0,1)$. Then, by using the definition of the projection, we have Suppose that $v=P_{\mathcal C}u$ and let $w\in\mathcal C$ be arbitrary. By using the convexity of $\mathcal C$, it follows that $(1-t)v+tw\in\mathcal C$, for all $t\in (0,1)$. Then, by using the definition of the projection, we have Line 43: Line 43: for all $w\in\mathcal C$. Hence, by using the definition of the projection, we get $v=P_{\mathcal C}u$. for all $w\in\mathcal C$. Hence, by using the definition of the projection, we get $v=P_{\mathcal C}u$. - = Moreau's theorem = + == 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 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 Line 56: Line 56: - == Proof of Moreau's theorem == + === Proof of Moreau's theorem ===
Line 93: Line 93:
- == 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.

## 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 $LaTeX: v=P_{\mathcal C}u$ and let $LaTeX: w\in\mathcal C$ be arbitrary. By using the convexity of $LaTeX: \mathcal C$, it follows that $LaTeX: (1-t)v+tw\in\mathcal C$, for all $LaTeX: t\in (0,1)$. 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 cone, 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 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.