Moreau's decomposition theorem

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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:

$LaTeX: z=x+y$ , $LaTeX: x\in\mathcal K, y\in\mathcal K^\circ$ and $LaTeX: \langle x,y\rangle=0$
$LaTeX: x=P_{\mathcal K}z$ 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: 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 $LaTeX: p\in\mathcal K$ . We have
$LaTeX: \langle y,p\rangle=\langle y,p-x\rangle=\langle z-x,p-x\rangle\leq0,$

by the characterization of the projection.

$LaTeX: \langle z-y,q-y\rangle=\langle x,q-y\rangle=\langle x,q\rangle\leq0,$

for all $LaTeX: q\in K^\circ$ , because $LaTeX: x\in K$