# Moreau's decomposition theorem

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and thus <math>y=P_{\mathcal K^\circ}z</math>.</li> | and thus <math>y=P_{\mathcal K^\circ}z</math>.</li> | ||

<li>2<math>\Rightarrow</math>1: Let <math>x=P_{\mathcal K}z</math>. By the characterization of the projection we have <math>\langle z-x,p-x\rangle\leq0,</math> for all <math>p\in\mathcal K</math>. In particular, if <math>p=0</math>, then <math>\langle z-x,x\rangle\geq0</math> and if <math>p=2x</math>, then <math>\langle z-x,x\rangle\leq0</math>. Thus, <math>\langle z-x,x\rangle=0</math>. Denote <math>y=z-x</math>. Then, <math>\langle x,y\rangle=0</math>. It remained to show that <math>y=P_{\mathcal K^\circ}z</math>. | <li>2<math>\Rightarrow</math>1: Let <math>x=P_{\mathcal K}z</math>. By the characterization of the projection we have <math>\langle z-x,p-x\rangle\leq0,</math> for all <math>p\in\mathcal K</math>. In particular, if <math>p=0</math>, then <math>\langle z-x,x\rangle\geq0</math> and if <math>p=2x</math>, then <math>\langle z-x,x\rangle\leq0</math>. Thus, <math>\langle z-x,x\rangle=0</math>. Denote <math>y=z-x</math>. Then, <math>\langle x,y\rangle=0</math>. It remained to show that <math>y=P_{\mathcal K^\circ}z</math>. | ||

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<center> | <center> | ||

<math> | <math> | ||

- | \langle z-y,q-y\rangle, | + | \langle z-y,q-y\rangle=\langle x,q-y\rangle=\langle x,q\rangle\leq0, |

</math> | </math> | ||

</center> | </center> | ||

+ | |||

+ | for all <math>q\in K^\circ</math>, because <math>x\in K</math> | ||

</ul> | </ul> |

## Revision as of 16:15, 10 July 2009

**Moreau's theorem** is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces.

Let be a closed convex cone in the Hilbert space and its polar. For an arbitrary closed convex set in , denote by the projection onto . For the following two statements are equivalent:

- , and
- and

## Proof

Let be an arbitrary closed convex set in , and . Then, it is well known that if and only if for all . We will call this result the * characterization of the projection*.

- 12: For all we have
.

Then, by the characterization of the projection, it follows that . Similarly, for all we have

- 21: Let . By the characterization of the projection we have for all . In particular, if , then and if , then . Thus, . Denote . Then, . It remained to show that .
for all , because