Moreau's decomposition theorem

From Wikimization

(Difference between revisions)
Jump to: navigation, search
(Every implicit complementarity problem is equivalent to a fixed point problem)
Line 104: Line 104:
=== References ===
=== 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.
* 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 <math>\mathcal A</math> be a set and <math>F:\mathcal A\to\mathcal A </math> a mapping. The '''fixed point problem''' defined by <math>F\,</math> is the problem
 
- 
-
<center>
 
-
<math>
 
-
Fix(F):\left\{
 
-
\begin{array}{l}
 
-
Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\
 
-
F(x)=x.
 
-
\end{array}
 
-
\right.
 
-
</math>
 
-
</center>
 
- 
-
=== Nonlinear complementarity problems ===
 
- 
-
Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Recall that the dual cone of <math>\mathcal K</math> is the closed convex cone <math>\mathcal K^*=-\mathcal K^\circ,</math> where <math>\mathcal K^\circ</math> is the [[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]] of <math>\mathcal K.</math> The '''nonlinear complementarity problem''' defined by <math>\mathcal K</math> and <math>f\,</math> is the problem
 
- 
-
<center>
 
-
<math>
 
-
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.
 
-
</math>
 
-
</center>
 
- 
-
=== Every nonlinear complementarity problem is equivalent to a fixed point problem ===
 
- 
-
Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then, the nonlinear complementarity problem <math>NCP(f,\mathcal K)</math> is equivalent to the fixed point problem
 
-
<math>Fix(P_{\mathcal K}\circ(I-f)),</math> where <math>I:\mathbb H\to\mathbb H</math> is the identity mapping defined by <math>I(x)=x.\,</math>
 
- 
-
=== Proof ===
 
- 
-
For all <math>x\in\mathbb H</math> denote <math>z=x-f(x)\,</math> and <math>y=-f(x).\,</math> Then, <math>z=x+y.\,</math>
 
-
<br>
 
-
<br>
 
- 
-
Suppose that <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math> Then, <math>z=x+y,\,</math> with <math>x\in\mathcal K,</math> <math>y\in\mathcal K^\circ</math> and <math>\langle x,y\rangle=0.</math> Hence, by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]], we get <math>x=P_{\mathcal K}z.</math> Therefore, <math>x\,</math> is a solution of <math>Fix(P_{\mathcal K}\circ(I-f)).</math>
 
-
<br>
 
-
<br>
 
- 
-
Conversely, suppose that <math>x\,</math> is a solution of <math>Fix(P_{\mathcal K}\circ(I-f)).</math> Then, <math>x\in\mathcal K</math> and by using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]]
 
- 
-
<center>
 
-
<math>z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).</math>
 
-
</center>
 
- 
-
Hence, <math>P_{\mathcal K^\circ}(z)=z-x=y,</math>. Thus, <math>y\in\mathcal K^\circ</math>. [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] also implies that <math>\langle x,y\rangle=0.</math> In conclusion,
 
-
<math>x\in\mathcal K,</math> <math>f(x)=-y\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Therefore, <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math>
 
- 
-
=== An alternative proof without [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] ===
 
- 
-
==== Variational inequalities ====
 
- 
-
Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. The '''variational inequality''' defined by <math>\mathcal C</math> and <math>f\,</math> is the problem
 
- 
-
<center>
 
-
<math>
 
-
VI(f,\mathcal C):\left\{
 
-
\begin{array}{l}
 
-
Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\
 
-
\langle y-x,f(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C.
 
-
\end{array}
 
-
\right.
 
-
</math>
 
-
</center>
 
- 
-
==== Every variational inequality is equivalent to a fixed point problem ====
 
- 
-
Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then the variational inequality <math>VI(f,\mathcal C)</math> is equivalent to the fixed point problem <math>Fix(P_{\mathcal C}\circ(I-f)).</math>
 
- 
-
==== Proof ====
 
- 
-
<math>x\,</math> is a solution of <math>Fix(P_{\mathcal C}\circ(I-f))</math> if and only if
 
-
<math>x=P_{\mathcal C}(x-f(x)).</math> By using the [[Moreau's_decomposition_theorem#Characterization_of_the_projection | characterization of the projection]] the latter equation is equivalent to
 
- 
-
<center>
 
-
<math>\langle x-f(x)-x,y-x\rangle\leq0,</math>
 
-
</center>
 
- 
-
for all <math>y\in\mathcal C.</math> But this holds if and only if <math>x\,</math> is a solution
 
-
of <math>VI(f,\mathcal C).</math>
 
- 
-
===== Remark =====
 
- 
-
The next section shows that the equivalence of variational inequalities and fixed point problems is much stronger than the equivalence of nonlinear complementarity problems and fixed point problems, because each nonlinear complementarity problem is a variational inequality defined on a closed convex cone.
 
- 
-
==== Every variational inequality defined on a closed convex cone is equivalent to a complementarity problem ====
 
- 
-
Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathbb H\to\mathbb H</math> a mapping. Then, the nonlinear complementarity problem <math>NCP(f,\mathcal K)</math> is equivalent to the variational inequality <math>VI(f,\mathcal K).</math>
 
- 
-
==== Proof ====
 
- 
-
Suppose that <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math> Then, <math>x\in\mathcal K,</math> <math>f(x)\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Hence,
 
- 
-
<center>
 
-
<math>\langle y-x,f(x)\rangle\geq 0,</math>
 
-
</center>
 
- 
-
for all <math>y\in\mathcal K.</math> Therefore, <math>x\,</math> is a solution of <math>VI(f,\mathcal K).</math>
 
-
<br>
 
-
<br>
 
- 
-
Conversely, suppose that <math>x\,</math> is a solution of <math>VI(f,\mathcal K).</math> Then,
 
-
<math>x\in\mathcal K</math> and
 
- 
-
<center>
 
-
<math>\langle y-x,f(x)\rangle\geq 0,</math>
 
-
</center>
 
- 
-
for all <math>y\in\mathcal K.</math> Particularly, taking <math>y=0\,</math> and <math>y=2x\,</math>, respectively, we get <math>\langle x,f(x)\rangle=0.</math> Thus, <math>\langle y,f(x)\rangle\geq 0,</math> for all <math>y\in\mathcal K,</math> or equivalently <math>f(x)\in\mathcal K^*.</math> In conclusion, <math>x\in\mathcal K,</math> <math>f(x)\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Therefore, <math>x\,</math> is a solution of <math>NCP(f,\mathcal K).</math>
 
- 
-
==== Concluding the alternative proof ====
 
- 
-
Since <math>\mathcal K</math> is a closed convex cone, the nonlinear complementarity problem <math>NCP(f,\mathcal K)</math> is equivalent to the variational inequality <math>VI(f,\mathcal K),</math> which is equivalent to the fixed point problem <math>Fix(P_{\mathcal K}\circ(I-f)).</math>
 
- 
-
== An application to implicit complementarity problems ==
 
-
=== Implicit complementarity problems ===
 
- 
-
Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f,g:\mathbb H\to\mathbb H</math> two mappings. Recall that the dual cone of <math>\mathcal K</math> is the closed convex cone <math>\mathcal K^*=-\mathcal K^\circ,</math> where <math>\mathcal K^\circ</math> is the
 
-
[[Moreau's_decomposition_theorem#Moreau.27s_theorem |polar]]
 
-
of <math>\mathcal K.</math> The '''implicit complementarity problem''' defined by <math>\mathcal K</math>
 
-
and the ordered pair of mappings <math>(f,g)\,</math> is the problem
 
- 
-
<center>
 
-
<math>
 
-
ICP(f,g,\mathcal K):\left\{
 
-
\begin{array}{l}
 
-
Find\,\,\,u\in\mathbb H\,\,\,such\,\,\,that\\
 
-
g(u)\in\mathcal K,\,\,\,f(u)\in K^*\,\,\,and\,\,\,\langle g(u),f(u)\rangle=0.
 
-
\end{array}
 
-
\right.
 
-
</math>
 
-
</center>
 
- 
-
=== Every implicit complementarity problem is equivalent to a fixed point problem ===
 
- 
-
Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>f,g:\mathbb H\to\mathbb H</math> two mappings. Then, the implicit complementarity problem <math>ICP(f,g,\mathcal K)</math> is equivalent to the [[Moreau's_decomposition_theorem#Fixed_point_problems | fixed point problem]]
 
-
<math>Fix(I-g+P_{\mathcal K}\circ(g-f)),</math> where <math>I:\mathbb H\to\mathbb H</math> is the identity mapping defined by <math>I(x)=x.\,</math>
 
- 
-
=== Proof ===
 
- 
-
For all <math>u\in\mathbb H</math> denote <math>z=g(u)-f(u),\,</math> <math>x=g(u)\,</math> and <math>y=-f(u).\,</math> Then,
 
-
<math>z=x+y.\,</math>
 
-
<br>
 
-
<br>
 
- 
-
Suppose that <math>u\,</math> is a solution of <math>ICP(f,g,\mathcal K).</math> Then, <math>z=x+y,\,</math> with <math>x\in\mathcal K,</math> <math>y\in\mathcal K^\circ</math> and <math>\langle x,y\rangle=0.</math> Hence, by using
 
-
[[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]],
 
-
we get <math>x=P_{\mathcal K}z.</math> Therefore, <math>u\,</math> is a solution of
 
-
<math>Fix(I-g+P_{\mathcal K}\circ(g-f)).</math>
 
-
<br>
 
-
<br>
 
- 
-
Conversely, suppose that <math>u\,</math> is a solution of <math>Fix(I-g+P_{\mathcal K}\circ(g-f)).</math>
 
-
Then, <math>x\in\mathcal K</math> and by using
 
-
[[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]]
 
- 
-
<center>
 
-
<math>z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).</math>
 
-
</center>
 
- 
-
Hence, <math>P_{\mathcal K^\circ}(z)=z-x=y,</math>. Thus, <math>y\in\mathcal K^\circ</math>.
 
-
[[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]]
 
-
also implies that <math>\langle x,y\rangle=0.</math> In conclusion,
 
-
<math>g(u)=x\in\mathcal K,</math> <math>f(u)=-y\in\mathcal K^*</math> and <math>\langle x,f(x)\rangle=0.</math> Therefore, <math>u\,</math> is a solution of <math>ICP(f,g,\mathcal K).</math>
 
- 
-
=== Remark ===
 
- 
-
In particular if <math>g=I,</math> we obtain the result
 
-
[[Moreau%27s_decomposition_theorem#Every_nonlinear_complementarity_problem_is_equivalent_to_a_fixed_point_problem | Every nonlinear complementarity problem is equivalent to a fixed point problem]],
 
-
but the more general result [[Moreau%27s_decomposition_theorem#Every_implicit_complementarity_problem_is_equivalent_to_a_fixed_point_problem | Every implicit complementarity problem is equivalent to a fixed point problem]] has no known connection with variational inequalities. Therefore, using [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] is essential for proving the latter result.
 

Revision as of 11:11, 17 July 2009

Contents

Projection on closed convex sets

Projection mapping

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

Theorem (Moreau). Let LaTeX: \mathcal K be a closed convex cone in the Hilbert space LaTeX: (\mathbb H,\langle\cdot,\cdot\rangle) and LaTeX: \mathcal K^\circ its polar cone; that is, the closed convex cone defined by LaTeX: \mathcal K^\circ=\{a\in\mathbb H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\}.

For LaTeX: x,y,z\in\mathbb 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: 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: u=z-x.\, Then, LaTeX: \langle x,u\rangle=0. It remains to show that LaTeX: u=y.\, First, we prove that LaTeX: u\in\mathcal K^\circ. For this we have to show that LaTeX: \langle u,p\rangle\leq0, for all LaTeX: p\in\mathcal K. By using the characterization of the projection, we have

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

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

    LaTeX: 
\langle z-u,q-u\rangle=\langle x,q-u\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: u=y.\,

notes

For definition of convex cone see Convex cone, Wikipedia; in finite dimension see Convex cones, Wikimization.

For definition of polar cone in finite dimension, see more at Dual cone and polar cone.

LaTeX: \mathcal K^{\circ\circ}=K Extended Farkas' lemma.

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.
Personal tools