Moreau's decomposition theorem

(Difference between revisions)
 Revision as of 15:43, 10 July 2009 (edit)m ← Previous diff Revision as of 16:00, 10 July 2009 (edit) (undo)Next diff → Line 27: Line 27: and thus $y=P_{\mathcal K^\circ}z$. and thus $y=P_{\mathcal K^\circ}z$. -
• 2$\Rightarrow$1:
• +
• 2$\Rightarrow$1: 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 $y=z-x$. + + We will show that $z-P_{\mathcal K}z=P_{\mathcal K^\circ}z$.
• Indeed, let $q\in K^\circ$ be arbitrary. Then, + +
+ $+ \langle z-(z-P_{\mathcal K}z),q-(z-P_{\mathcal K}z)\rangle +$ +
+

Revision as of 16:00, 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: 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 $LaTeX: y=z-x$.

We will show that $LaTeX: z-P_{\mathcal K}z=P_{\mathcal K^\circ}z$.
• Indeed, let $LaTeX: q\in K^\circ$ be arbitrary. Then,

$LaTeX: \langle z-(z-P_{\mathcal K}z),q-(z-P_{\mathcal K}z)\rangle$