Moreau's decomposition theorem

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$LaTeX: -$ Sándor Zoltán Németh

(In particular, we can have $LaTeX: \mathbb{H}=\mathbb{R}^n$ everywhere in this page.)

1 Projection on closed convex sets
2 Moreau's theorem
- 2.1 Proof of Moreau's theorem
- 2.2 notes
3 Applications
- 3.1 References

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

$LaTeX: \parallel x-P_{\mathcal{C}}(x)\parallel=\min\{\parallel x-y\parallel\,:y\in\mathcal{C}\}.$

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

$LaTeX: \parallel u-v\parallel^2\leq\parallel u-((1-t)v+tw)\parallel^2=\parallel u-v-t(w-v)\parallel^2=\parallel u-v\parallel^2-2t\langle u-v,w-v\rangle+t^2\parallel w-v\parallel^2,$

Hence,

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

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

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

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:

$LaTeX: z=x+y,\,x\in\mathcal{K},\,y\in\mathcal{K}^\circ$ and $LaTeX: \langle x,y\rangle=0,$
$LaTeX: x=P_{\mathcal{K}}z$ and $LaTeX: y=P_{\mathcal{K}^\circ}z.$

Proof of Moreau's theorem

1 $LaTeX: \Rightarrow$ 2: For all $LaTeX: p\in\mathcal{K}$ 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 $LaTeX: q\in\mathcal{K}^\circ$ we have

$LaTeX: \langle z-y,q-y\rangle=\langle x,q-y\rangle=\langle x,q\rangle\leq0$
and thus $LaTeX: y=P_{\mathcal{K}^\circ}z.$
2 $LaTeX: \Rightarrow$ 1: 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 $LaTeX: q\in\mathcal{K}^\circ,$ because $LaTeX: x\in\mathcal{K}.$ By using again the characterization of the projection, it follows that $LaTeX: u=y.\,$

notes

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

For definition of polar cone in finite dimension, see Convex Optimization & Euclidean Distance Geometry.

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

Applications

For applications see Every nonlinear complementarity problem is equivalent to a fixed point problem, Every implicit complementarity problem is equivalent to a fixed point problem, and Projection on isotone projection cone.

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.

Retrieved from "http://www.convexoptimization.com/wikimization/index.php/Moreau%27s_decomposition_theorem"

Moreau's decomposition theorem

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Contents

Projection on closed convex sets

Projection mapping

Characterization of the projection

Proof

Moreau's theorem

Proof of Moreau's theorem

notes

Applications

References

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@@ Line 1: / Line 1: @@
+[http://web.mat.bham.ac.uk/S.Z.Nemeth/ <math>-</math> Sándor Zoltán Németh]
+(In particular, we can have <math>\mathbb{H}=\mathbb{R}^n</math> everywhere in this page.)
 == Projection on closed convex sets ==
 === Projection mapping ===
+Let <math>(\mathbb{H},\langle\cdot,\cdot\rangle)</math> be a Hilbert space and <math>\mathcal{C}</math> a closed convex set in <math>\mathbb{H}.</math> The '''projection mapping''' <math>P_{\mathcal{C}}</math> onto <math>\mathcal{C}</math> is the mapping <math>P_{\mathcal{C}}:\mathbb{H}\to\mathbb{H}</math> defined by <math>P_{\mathcal{C}}(x)\in\mathcal{C}</math> and
-Let <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> be a Hilbert space and <math>\mathcal C</math> a closed convex set in <math>\mathbb H.</math> The '''projection mapping''' <math>P_{\mathcal C}</math> onto <math>\mathcal C</math> is the mapping <math>P_{\mathcal C}:\mathbb H\to\mathbb H</math> defined by <math>P_{\mathcal C}(x)\in\mathcal C</math> and
 <center>
-<math>\|x-P_{\mathcal C}(x)\|=\min\{\|x-y\|\mid y\in\mathcal C\}.</math>
+<math>\parallel x-P_{\mathcal{C}}(x)\parallel=\min\{\parallel x-y\parallel\,:y\in\mathcal{C}\}.</math>
 </center>
 === Characterization of the projection ===
+Let <math>(\mathbb{H},\langle\cdot,\cdot\rangle)</math> be a Hilbert space, <math>\mathcal{C}</math> a closed convex set in <math>\mathbb{H},\,u\in\mathbb{H}</math> and <math>v\in\mathcal{C}.</math> Then <math>v=P_{\mathcal{C}}(u)</math> if and only if <math>\langle u-v,w-v\rangle\leq0</math> for all <math>w\in\mathcal{C}.</math>
-Let <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> be a Hilbert space, <math>\mathcal C</math> a closed convex set in <math>\mathbb H,\,u\in\mathbb H</math> and <math>v\in\mathcal C.</math> Then, <math>v=P_{\mathcal C}(u)</math> if and only if <math>\langle u-v,w-v\rangle\leq0</math> for all <math>w\in\mathcal C.</math>
 === Proof ===
+Suppose that <math>v=P_{\mathcal{C}}u.</math> Let <math>w\in\mathcal{C}</math> and <math>t\in (0,1)</math> be arbitrary. By using the convexity of <math>\mathcal{C},</math> it follows that <math>(1-t)v+tw\in\mathcal{C}.</math> Then, by using the definition of the projection, we have
-Suppose that <math>v=P_{\mathcal C}u.</math> Let <math>w\in\mathcal C</math> and <math>t\in (0,1)</math> be arbitrary. By using the convexity of <math>\mathcal C,</math> it follows that <math>(1-t)v+tw\in\mathcal C.</math> Then, by using the definition of the projection, we have
 <center>
 <math>
-\|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,
+\parallel u-v\parallel^2\leq\parallel u-((1-t)v+tw)\parallel^2=\parallel u-v-t(w-v)\parallel^2=\parallel u-v\parallel^2-2t\langle u-v,w-v\rangle+t^2\parallel w-v\parallel^2,
 </math>
 </center>
@@ Line 28: / Line 27: @@
 <center>
-<math>\langle u-v,w-v\rangle\leq\frac t2\|w-v\|^2.</math>
+<math>\langle u-v,w-v\rangle\leq\frac t2\parallel w-v\parallel^2.</math>
 </center>
@@ Line 35: / Line 34: @@
 <br>
-Conversely, suppose that <math>\langle u-v,w-v\rangle\leq0,</math> for all <math>w\in\mathcal C.</math> Then,
+Conversely, suppose that <math>\langle u-v,w-v\rangle\leq0,</math> for all <math>w\in\mathcal C.</math> Then
 <center>
-<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>
+<math>\parallel u-w\parallel^2=\parallel u-v-(w-v)\parallel^2=\parallel u-v\parallel^2-2\langle u-v,w-v\rangle+\parallel w-v\parallel^2\geq \parallel u-v\parallel^2,</math>
 </center>
@@ Line 44: / Line 43: @@
 == Moreau's theorem ==
 Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces.
@@ Line 50: / Line 48: @@
 under the addition of vectors and multiplication of vectors by positive scalars.
-'''Theorem (Moreau).''' Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathbb H,\langle\cdot,\cdot\rangle)</math> and <math>\mathcal K^\circ</math> its '''polar cone'''; that is, the closed convex cone defined by <math>\mathcal K^\circ=\{a\in\mathbb H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\}.</math>
+'''Theorem (Moreau).''' Let <math>\mathcal{K}</math> be a closed convex cone in the Hilbert space <math>(\mathbb{H},\langle\cdot,\cdot\rangle)</math> and <math>\mathcal{K}^\circ</math> its '''polar cone'''; that is, the closed convex cone defined by <math>\mathcal{K}^\circ=\{a\in\mathbb{H}\,\mid\,\langle a,b\rangle\leq0,\,\forall b\in\mathcal{K}\}.</math>
-For <math>x,y,z\in\mathbb H</math> the following statements are equivalent:
+For <math>x,y,z\in\mathbb{H}</math> the following statements are equivalent:
 <ol>
-<li><math>z=x+y,\,x\in\mathcal K,\,y\in\mathcal K^\circ</math> and <math>\langle x,y\rangle=0,</math></li>
+<li><math>z=x+y,\,x\in\mathcal{K},\,y\in\mathcal{K}^\circ</math> and <math>\langle x,y\rangle=0,</math></li>
-<li><math>x=P_{\mathcal K}z</math> and <math>y=P_{\mathcal K^\circ}z.</math>
+<li><math>x=P_{\mathcal{K}}z</math> and <math>y=P_{\mathcal{K}^\circ}z.</math>
 </li>
 </ol>
@@ Line 62: / Line 60: @@
 === Proof of Moreau's theorem ===
 <ul>
-<li>1<math>\Rightarrow</math>2: For all <math>p\in K</math> we have
+<li>1<math>\Rightarrow</math>2: For all <math>p\in\mathcal{K}</math> we have
 <center>
@@ Line 68: / Line 66: @@
 </center>
-Then, by the characterization of the projection, it follows that <math>x=P_{\mathcal K}z.</math> Similarly, for all <math>q\in K^\circ</math> we have
+Then, by the characterization of the projection, it follows that <math>x=P_{\mathcal{K}}z.</math> Similarly, for all <math>q\in\mathcal{K}^\circ</math> we have
 <center>
@@ Line 74: / Line 72: @@
 </center>
-and thus <math>y=P_{\mathcal K^\circ}z.</math></li>
+and thus <math>y=P_{\mathcal{K}^\circ}z.</math></li>
-<li>2<math>\Rightarrow</math>1: By using the characterization of the projection, we have <math>\langle z-x,p-x\rangle\leq0,</math> for all <math>p\in\mathcal K.</math> In particular, if <math>p=0,\,</math> then <math>\langle z-x,x\rangle\geq0</math> and if <math>p=2x,\,</math> then <math>\langle z-x,x\rangle\leq0.</math> Thus, <math>\langle z-x,x\rangle=0.</math> Denote <math>u=z-x.\,</math> Then, <math>\langle x,u\rangle=0.</math> It remains to show that <math>u=y.\,</math> First, we prove that <math>u\in\mathcal K^\circ.</math> For this we have to show that <math>\langle u,p\rangle\leq0,</math> for
+<li>2<math>\Rightarrow</math>1: By using the characterization of the projection, we have <math>\langle z-x,p-x\rangle\leq0,</math> for all <math>p\in\mathcal K.</math> In particular, if <math>p=0,\,</math> then <math>\langle z-x,x\rangle\geq0</math> and if <math>p=2x,\,</math> then <math>\langle z-x,x\rangle\leq0.</math> Thus, <math>\langle z-x,x\rangle=0.</math> Denote <math>u=z-x.\,</math> Then <math>\langle x,u\rangle=0.</math> It remains to show that <math>u=y.\,</math> First, we prove that <math>u\in\mathcal{K}^\circ.</math> For this we have to show that <math>\langle u,p\rangle\leq0,</math> for
-all <math>p\in\mathcal K.</math> By using the characterization of the projection, we have
+all <math>p\in\mathcal{K}.</math> By using the characterization of the projection, we have
 <center>
@@ Line 84: / Line 82: @@
 </center>
-for all <math>p\in\mathcal K.</math> Thus, <math>u\in\mathcal K^\circ.</math> We also have
+for all <math>p\in\mathcal{K}.</math> Thus, <math>u\in\mathcal{K}^\circ.</math> We also have
 <center>
@@ Line 92: / Line 90: @@
 </center>
-for all <math>q\in K^\circ,</math> because <math>x\in K.</math> By using again the characterization of the projection, it follows that <math>u=y.\,</math>
+for all <math>q\in\mathcal{K}^\circ,</math> because <math>x\in\mathcal{K}.</math> By using again the characterization of the projection, it follows that <math>u=y.\,</math>
 </ul>
 === notes ===
-For definition of ''convex cone'' see [http://en.wikipedia.org/wiki/Convex_cone Convex cone, Wikipedia]; in finite dimension see [[Convex cones | Convex cones, Wikimization]].
+For definition of ''convex cone'' in finite dimension see [[Convex cones | Convex cones, Wikimization]].
-For definition of ''polar cone'' in finite dimension, see more at [http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone Dual cone and polar cone].
+For definition of ''polar cone'' in finite dimension, see [http://meboo.convexoptimization.com/Meboo.html Convex Optimization & Euclidean Distance Geometry].
-<math>\mathcal K^{\circ\circ}=K</math> [[Farkas%27_lemma#Extended_Farkas.27_lemma|Extended Farkas' lemma]].
+<math>\mathcal{K}^{\circ\circ}=\mathcal{K}</math> see [[Farkas%27_lemma#Extended_Farkas.27_lemma|Extended Farkas' lemma]].
+== Applications ==
+For applications see [[Complementarity_problem#Every_nonlinear_complementarity_problem_is_equivalent_to_a_fixed_point_problem|Every nonlinear complementarity problem is equivalent to a fixed point problem]], [[Complementarity_problem#Every_implicit_complementarity_problem_is_equivalent_to_a_fixed_point_problem|Every implicit complementarity problem is equivalent to a fixed point problem]],
+and [[Projection_on_Polyhedral_Cone#Projection_on_isotone_projection_cones|Projection on isotone projection cone]].
 === References ===
 * J. J. Moreau, D&eacute;composition 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> 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> 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> 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>