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>
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<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>
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<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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We will show that <math>z-P_{\mathcal K}z=P_{\mathcal K^\circ}z</math>.</li> Indeed, let <math>q\in K^\circ</math> be arbitrary. Then,
 
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Revision as of 16:10, 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.

  • 1LaTeX: \Rightarrow2: 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.
  • 2LaTeX: \Rightarrow1: 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.

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

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