# Moreau's decomposition theorem

(Difference between revisions)
 Revision as of 01:04, 11 July 2009 (edit) (→Moreau's theorem)← Previous diff Revision as of 01:05, 11 July 2009 (edit) (undo) (→Characterization of the projection)Next diff → Line 1: Line 1: == Characterization of the projection == == Characterization of the projection == - 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 $v\in\mathcal C$. Denote by $P_{\mathcal C}$ the projection mapping onto $\mathcal C$. 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 ==

## 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 $LaTeX: v\in\mathcal C$. Denote by $LaTeX: P_{\mathcal C}$ the projection mapping onto $LaTeX: \mathcal C$. 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.

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. 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.