Moreau's decomposition theorem

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* J. J. Moreau, Décomposition orthogonale d'un espace hilbertien selon deux cones mutuellement polaires, C. R. Acad. Sci., volume 255, pages 238–240, 1962.
* J. J. Moreau, Décomposition orthogonale d'un espace hilbertien selon deux cones mutuellement polaires, C. R. Acad. Sci., volume 255, pages 238–240, 1962.
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== Extended Farkas' lemma ==
 
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For any closed convex cone <math>\mathcal J</math> in the Hilbert space <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math>, denote by <math>\mathcal J^\circ</math> the polar cone of <math>\mathcal J</math>. Let <math>\mathcal K</math> be an arbitrary closed convex cone in <math>\mathcal H</math>. Then, the extended Farkas' lemma asserts that <math>\mathcal K^{\circ\circ}=\mathcal K.</math> Hence, denoting <math>\mathcal L=\mathcal K^\circ,</math> it follows that <math>\mathcal L^\circ=\mathcal K</math>. Therefore, the cones <math>\mathcal K</math> and <math>\mathcal L</math> are called ''mutually polar pair of cones''.
 
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== Proof of extended Farkas' lemma ==
 
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(S&aacute;ndor Zolt&aacute;n N&eacute;meth) Let <math>z\in\mathcal H</math> be arbitrary. Then, by Moreau's theorem we have
 
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<center>
 
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<math>
 
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z=P_{\mathcal K}z+P_{\mathcal K^\circ}z
 
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</math>
 
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</center>
 
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and
 
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<center>
 
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<math>
 
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z=P_{\mathcal K^\circ}z+P_{\mathcal K^{\circ\circ}}z.
 
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</math>
 
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</center>
 
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Therefore,
 
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<center>
 
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<math>
 
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P_{\mathcal K^{\circ\circ}}z=P_{\mathcal K}z=z-P_{\mathcal K^\circ}z.
 
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</math>
 
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</center>
 
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In particular, for any <math>z\in K</math> we have <math>\mathcal K^{\circ\circ}\ni P_{\mathcal K^{\circ\circ}}z=z</math>. Hence, <math>\mathcal \mathcal K^{\circ\circ}\supset K</math>. Similarly, for any <math>z\in K^{\circ\circ}</math> we have <math>z= P_{\mathcal K}z\in\mathcal K</math>. Hence, <math>\mathcal K^{\circ\circ}\subset\mathcal K</math>. Therefore, <math>\mathcal K^{\circ\circ}=\mathcal K</math>.
 

Revision as of 19:28, 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 statements are equivalent:

  1. LaTeX: z=x+y,\,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 of Moreau's theorem

Let LaTeX: \mathcal C be an arbitrary closed convex set in LaTeX: \mathcal H,\,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. 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. By using the characterization of the projection, we have

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

    for all LaTeX: p\in\mathcal K. Thus, LaTeX: y\in\mathcal K^\circ. We also have

    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. By using again the characterization of the projection, it follows that LaTeX: y=P_{\mathcal K^\circ}z.

References

  • J. J. Moreau, Décomposition orthogonale d'un espace hilbertien selon deux cones mutuellement polaires, C. R. Acad. Sci., volume 255, pages 238–240, 1962.
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