# Moreau's decomposition theorem

(Difference between revisions)
 Revision as of 18:02, 10 July 2009 (edit)m (→Proof of Moreau's theorem)← Previous diff Revision as of 18:03, 10 July 2009 (edit) (undo)m (→Proof of Moreau's theorem)Next diff → Line 27: Line 27: 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 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 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

• ## Revision as of 18:03, 10 July 2009

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 an arbitrary closed convex set $LaTeX: \mathcal C$ in $LaTeX: \mathcal H$, denote by $LaTeX: P_{\mathcal C}$ the projection onto $LaTeX: \mathcal C$. 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

Let $LaTeX: \mathcal C$ be an arbitrary closed convex set in $LaTeX: \mathcal H,\,u\in\mathcal H$ and $LaTeX: v\in\mathcal C$. Then, it is well known that $LaTeX: v=P_{\mathcal C}u$ if and only if $LaTeX: \langle u-v,w-v\rangle\leq0$ for all $LaTeX: w\in\mathcal C$. We will call this result the characterization of the projection.

• 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.

## Extended Farkas' lemma

For any closed convex cone $LaTeX: \mathcal J$ in the Hilbert space $LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle)$, denote by $LaTeX: \mathcal J^\circ$ the polar cone of $LaTeX: \mathcal J$. Let $LaTeX: \mathcal K$ be an arbitrary closed convex cone in $LaTeX: \mathcal H$. Then, the extended Farkas' lemma asserts that $LaTeX: \mathcal K^{\circ\circ}=\mathcal K.$ Hence, denoting $LaTeX: \mathcal L=\mathcal K^\circ,$ it follows that $LaTeX: \mathcal L^\circ=\mathcal K$. Therefore, the cones $LaTeX: \mathcal K$ and $LaTeX: \mathcal L$ are called mutually polar pair of cones.

## Proof of extended Farkas' lemma

(Sándor Zoltán Németh) Let $LaTeX: z\in\mathcal H$ be arbitrary. Then, by Moreau's theorem we have $LaTeX: z=P_{\mathcal K}z+P_{\mathcal K^\circ}z$

and $LaTeX: z=P_{\mathcal K^\circ}z+P_{\mathcal K^{\circ\circ}}z.$

Therefore, $LaTeX: P_{\mathcal K^{\circ\circ}}z=P_{\mathcal K}z=z-P_{\mathcal K^\circ}z.$

In particular, for any $LaTeX: z\in K$ we have $LaTeX: \mathcal K^{\circ\circ}\ni P_{\mathcal K^{\circ\circ}}z=z$. Hence, $LaTeX: \mathcal \mathcal K^{\circ\circ}\supset K$. Similarly, for any $LaTeX: z\in K^{\circ\circ}$ we have $LaTeX: z= P_{\mathcal K}z\in\mathcal K$. Hence, $LaTeX: \mathcal K^{\circ\circ}\subset\mathcal K$. Therefore, $LaTeX: \mathcal K^{\circ\circ}=\mathcal K$.