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 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

Contents

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.

Extended Farkas' lemma

For any closed convex cone LaTeX: \mathcal J in the Hilbert space LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle), denote by LaTeX: \mathcal J^\circ the polar cone of LaTeX: \mathcal J. Let LaTeX: \mathcal K be an arbitrary closed convex cone in LaTeX: \mathcal H. Then, the extended Farkas' lemma asserts that LaTeX: \mathcal K^{\circ\circ}=\mathcal K. Hence, denoting LaTeX: \mathcal L=\mathcal K^\circ, it follows that LaTeX: \mathcal L^\circ=\mathcal K. Therefore, the cones LaTeX: \mathcal K and LaTeX: \mathcal L are called mutually polar pair of cones.

Proof of extended Farkas' lemma

(Sándor Zoltán Németh) Let LaTeX: z\in\mathcal H be arbitrary. Then, by Moreau's theorem we have

LaTeX: 
z=P_{\mathcal K}z+P_{\mathcal K^\circ}z

and

LaTeX: 
z=P_{\mathcal K^\circ}z+P_{\mathcal K^{\circ\circ}}z.

Therefore,

LaTeX: 
P_{\mathcal K^{\circ\circ}}z=P_{\mathcal K}z=z-P_{\mathcal K^\circ}z.

In particular, for any LaTeX: z\in K we have LaTeX: \mathcal K^{\circ\circ}\ni P_{\mathcal K^{\circ\circ}}z=z. Hence, LaTeX: \mathcal \mathcal K^{\circ\circ}\supset K. Similarly, for any LaTeX: z\in K^{\circ\circ} we have LaTeX: z= P_{\mathcal K}z\in\mathcal K. Hence, LaTeX: \mathcal K^{\circ\circ}\subset\mathcal K. Therefore, LaTeX: \mathcal K^{\circ\circ}=\mathcal K.

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