Moreau's decomposition theorem

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<math>Fix(P_{\mathcal K}\circ(I-f)),</math> where <math>I:\mathcal H\to\mathcal H</math> is the identity mapping defined by <math>I(x)=x.\,</math>
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For all <math>x\in\mathcal H</math> denote
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Revision as of 06:15, 12 July 2009

Contents

Projection on closed convex sets

Projection mapping

Let LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle) be a Hilbert space and LaTeX: \mathcal C a closed convex set in LaTeX: \mathcal H. The projection mapping LaTeX: P_{\mathcal C} onto LaTeX: \mathcal C is the mapping LaTeX: P_{\mathcal C}:\mathcal H\to\mathcal H defined by LaTeX: P_{\mathcal C}(x)\in\mathcal C and

LaTeX: \|x-P_{\mathcal C}(x)\|=\min\{\|x-y\|\mid y\in\mathcal C\}.

Characterization of the projection

Let LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle) be a Hilbert space, LaTeX: \mathcal C a closed convex set in LaTeX: \mathcal H,\,u\in\mathcal H and LaTeX: v\in\mathcal C. Then, 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. Let LaTeX: w\in\mathcal C and LaTeX: t\in (0,1) be arbitrary. By using the convexity of LaTeX: \mathcal C, it follows that LaTeX: (1-t)v+tw\in\mathcal C. 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. Recall that a convex cone in a vector space is a set which is invariant under the addition of vectors and multiplication of vectors by positive scalars (see more at Convex cone, Wikipedia or for finite dimension at Convex cones, Wikimization).

Theorem (Moreau) 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 cone; that is, the closed convex cone defined by LaTeX: K^\circ=\{a\in\mathcal H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\} (for finite dimension see more at Dual cone and polar cone; see also Extended Farkas' lemma). 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 using 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.

An application to nonlinear complementarity problems

Fixed point problems

Let LaTeX: \mathcal A be a set and LaTeX: F:\mathcal A\to\mathcal A. The fixed point problem defined by LaTeX: F\, is the problem

LaTeX: 
Fix(F):\left\{
\begin{array}{l}
Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\
F(x)=x.
\end{array}
\right.

Nonlinear complementarity problems

Let LaTeX: \mathcal K be a closed convex cone in the Hilbert space LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle) and LaTeX: f:\mathcal H\to\mathcal H. Recall that the dual cone of LaTeX: \mathcal K is the closed convex cone LaTeX: \mathcal K^*=-\mathcal K^\circ, where LaTeX: \mathcal K^\circ is the polar of LaTeX: \mathcal K. The nonlinear complementarity problem defined by LaTeX: \mathcal K and LaTeX: f\, is the problem

LaTeX: 
NCP(f,\mathcal K):\left\{
\begin{array}{l} 
Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\ 
f(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,f(x)\rangle=0.
\end{array}
\right.

Every nonlinear complementarity problem is equivalent to a fixed point problem

Let LaTeX: \mathcal K be a closed convex cone in the Hilbert space LaTeX: (\mathcal H,\langle\cdot,\cdot\rangle) and LaTeX: f:\mathcal H\to\mathcal H. Then, the nonlinear complementarity problem LaTeX: NCP(f,\mathcal K) is equivalent to the fixed point problem LaTeX: Fix(P_{\mathcal K}\circ(I-f)), where LaTeX: I:\mathcal H\to\mathcal H is the identity mapping defined by LaTeX: I(x)=x.\,


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