Moreau's decomposition theorem

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== Proof ==
== Proof ==
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Suppose that <math>v=P_{\mathcal C}u</math> and let <math>w\in\mathcal C</math> be arbitrary. By
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Suppose that <math>v=P_{\mathcal C}u</math> and let <math>w\in\mathcal C</math> be arbitrary. By using the convexity of <math>\mathcal C</math>, it follows that <math>(1-t)v+tw\in\mathcal C</math>, for all <math>t\in (0,1)</math>. Then, by using the definition of the projection, we have
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the convexity of <math>\mathcal C</math> it follows that <math>(1-t)v+tw\in\mathcal C</math>, for
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all <math>t\in (0,1)</math>. Then, by the definition of the projection we have
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<center>
<center>
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</center>
</center>
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By tending with <math>t</math> to <math>0</math> we get <math>\langle u-v,w-v\rangle\leq0</math>.
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By tending with <math>t</math> to <math>0</math>, we get <math>\langle u-v,w-v\rangle\leq0</math>.
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<br>
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<br>
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Conversely, suppose that <math>\langle u-v,w-v\rangle\leq0,</math> for all <math>w\in\mathcal C</math>. Then,
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<center>
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<math>\|u-w\|^2=\|u-v-(w-v)\|^2=\|u-v\|^2-2\langle u-v,w-v\rangle+\|w-v\|^2\geq \|u-v\|^2,</math>
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</center>
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for all <math>w\in\mathcal C</math>. Hence, by using the definition of the projection, we get <math>v=P_{\mathcal C}u</math>.
== Moreau's theorem ==
== Moreau's theorem ==

Revision as of 01:48, 11 July 2009

Contents

Characterization of the projection

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.

Proof

Suppose that LaTeX: v=P_{\mathcal C}u and let LaTeX: w\in\mathcal C be arbitrary. By using the convexity of LaTeX: \mathcal C, it follows that LaTeX: (1-t)v+tw\in\mathcal C, for all LaTeX: t\in (0,1). Then, by using the definition of the projection, we have

LaTeX: \|u-v\|^2\leq\|u-[(1-t)v+tw]\|^2=\|u-v-t(w-v)\|^2=\|u-v\|^2-2t\langle u-v,w-v\rangle+t^2\|w-v\|^2 .

Hence,

LaTeX: \langle u-v,w-v\rangle\leq\frac t2\|w-v\|^2.

By tending with LaTeX: t to LaTeX: 0, we get LaTeX: \langle u-v,w-v\rangle\leq0.

Conversely, suppose that LaTeX: \langle u-v,w-v\rangle\leq0, for all LaTeX: w\in\mathcal C. Then,

LaTeX: \|u-w\|^2=\|u-v-(w-v)\|^2=\|u-v\|^2-2\langle u-v,w-v\rangle+\|w-v\|^2\geq \|u-v\|^2,

for all LaTeX: w\in\mathcal C. Hence, by using the definition of the projection, we get LaTeX: v=P_{\mathcal C}u.

Moreau's theorem

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

  • 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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