Moreau's decomposition theorem
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+ | [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 on closed convex sets == | ||
=== Projection mapping === | === 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> | <center> | ||
- | <math>\ | + | <math>\parallel x-P_{\mathcal{C}}(x)\parallel=\min\{\parallel x-y\parallel\,:y\in\mathcal{C}\}.</math> |
</center> | </center> | ||
=== Characterization of the projection === | === 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 | + | |
=== Proof === | === 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> | <center> | ||
<math> | <math> | ||
- | \ | + | \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> | </math> | ||
</center> | </center> | ||
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<center> | <center> | ||
- | <math>\langle u-v,w-v\rangle\leq\frac t2\ | + | <math>\langle u-v,w-v\rangle\leq\frac t2\parallel w-v\parallel^2.</math> |
</center> | </center> | ||
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<br> | <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> | <center> | ||
- | <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> | </center> | ||
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== Moreau's theorem == | == Moreau's theorem == | ||
- | |||
Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces. | Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces. | ||
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under the addition of vectors and multiplication of vectors by positive scalars. | 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> | <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> | </li> | ||
</ol> | </ol> | ||
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=== Proof of Moreau's theorem === | === Proof of Moreau's theorem === | ||
<ul> | <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> | <center> | ||
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</center> | </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> | <center> | ||
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</center> | </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 | + | <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> | <center> | ||
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</center> | </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> | <center> | ||
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</center> | </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> | </ul> | ||
=== notes === | === notes === | ||
- | For definition of ''convex cone'' | + | For definition of ''convex cone'' in finite dimension see [[Convex cones | Convex cones, Wikimization]]. |
- | For definition of ''polar cone'' in finite dimension, see | + | 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 === | === 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 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. |
Current revision
(In particular, we can have everywhere in this page.)
Contents |
Projection on closed convex sets
Projection mapping
Let be a Hilbert space and a closed convex set in The projection mapping onto is the mapping defined by and
Characterization of the projection
Let be a Hilbert space, a closed convex set in and Then if and only if for all
Proof
Suppose that Let and be arbitrary. By using the convexity of it follows that Then, by using the definition of the projection, we have
Hence,
By tending with to we get
Conversely, suppose that for all Then
for all Hence, by using the definition of the projection, we get
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 be a closed convex cone in the Hilbert space and its polar cone; that is, the closed convex cone defined by
For the following statements are equivalent:
- and
- and
Proof of Moreau's theorem
- 12: For all we have
Then, by the characterization of the projection, it follows that Similarly, for all we have
- 21: By using the characterization of the projection, we have for all In particular, if then and if then Thus, Denote Then It remains to show that First, we prove that For this we have to show that for
all By using the characterization of the projection, we have
for all Thus, We also have
for all because By using again the characterization of the projection, it follows that
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.
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.