# 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>. First, we prove that <math>y\in\mathcal K^\circ</math>. For this we have to show that <math>\langle y,p\rangle\leq0</math>, for | <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>. First, we prove that <math>y\in\mathcal K^\circ</math>. For this we have to show that <math>\langle y,p\rangle\leq0</math>, for | ||

- | all <math>p\in\mathcal K</math>. | + | all <math>p\in\mathcal K</math>. By using the characterization of the projection, we have |

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- | + | for all <math>p\in\mathcal K</math>. Thus, | |

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## Revision as of 15:30, 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 . First, we prove that . For this we have to show that , for
all . By using the characterization of the projection, we have
for all . Thus,

for all , because