# Moreau's decomposition theorem

(Difference between revisions)
 Revision as of 15:36, 10 July 2009 (edit) (→Proof)← Previous diff Revision as of 15:43, 10 July 2009 (edit) (undo)m Next diff → Line 14: Line 14:
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• 1$\Rightarrow$2: For all $p\in K$ we have $\langle z-x,p-x\rangle=\langle y,p-x\rangle=\langle y,p\rangle\leq0$. Then, by the characterization of the projection it follows that $x=P_{\mathcal K}z$. Similarly, for all $q\in K^\circ$\langle z-y,q-y\rangle=\langle x,q-y\rangle=\langle x,q\rangle\leq0[/itex] and thus $y=P_{\mathcal K^\circ}z$.
• +
• 1$\Rightarrow$2: For all $p\in K$ we have -
• 1$\Rightarrow$2
• + +
+ $\langle z-x,p-x\rangle=\langle y,p-x\rangle=\langle y,p\rangle\leq0$. +
+ + Then, by the characterization of the projection, it follows that $x=P_{\mathcal K}z$. Similarly, for all $q\in K^\circ we have + + + \langle z-y,q-y\rangle=\langle x,q-y\rangle=\langle x,q\rangle\leq0$ +
+ + and thus $y=P_{\mathcal K^\circ}z$. +
• 2$\Rightarrow$1:

## Revision as of 15:43, 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 two statements are equivalent:

1. $LaTeX: z=x+y$, $LaTeX: x\in\mathcal K, y\in\mathcal K^\circ$ and $LaTeX: \langle x,y\rangle=0$
2. $LaTeX: x=P_{\mathcal K}z$ and $LaTeX: y=P_{\mathcal K^\circ}z$

## Proof

Let $LaTeX: \mathcal C$ be an arbitrary closed convex set in $LaTeX: \mathcal H$, $LaTeX: u\in\mathcal H$ 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: