Moreau's decomposition theorem

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[http://web.mat.bham.ac.uk/S.Z.Nemeth/ <math>-</math> Sándor Zoltán Németh]
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(In particular, we can have <math>\mathbb{H}=\mathbb{R}^n</math> everywhere in this page.)
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== Projection on closed convex sets ==
== Projection on closed convex sets ==
=== Projection mapping ===
=== Projection mapping ===
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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
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Let <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> be a Hilbert space and <math>\mathcal C</math> a closed convex set in <math>\mathcal H.</math> The '''projection mapping''' <math>P_{\mathcal C}</math> onto <math>\mathcal C</math> is the mapping <math>P_{\mathcal C}:\mathcal H\to\mathcal H</math> defined by <math>P_{\mathcal C}(x)\in\mathcal C</math> and
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<center>
<center>
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<math>\|x-P_{\mathcal C}(x)\|=\min\{\|x-y\|\mid y\in\mathcal C\}.</math>
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<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 ===
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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>
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Let <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> be a Hilbert space, <math>\mathcal C</math> a closed convex set in <math>\mathcal H,\,u\in\mathcal 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>
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=== Proof ===
=== Proof ===
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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
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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
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<center>
<center>
<math>
<math>
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\|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,
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\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>
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<math>\langle u-v,w-v\rangle\leq\frac t2\|w-v\|^2.</math>
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<math>\langle u-v,w-v\rangle\leq\frac t2\parallel w-v\parallel^2.</math>
</center>
</center>
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<br>
<br>
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Conversely, suppose that <math>\langle u-v,w-v\rangle\leq0,</math> for all <math>w\in\mathcal C.</math> Then,
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Conversely, suppose that <math>\langle u-v,w-v\rangle\leq0,</math> for all <math>w\in\mathcal C.</math> Then
<center>
<center>
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<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>
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<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 ==
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Moreau's theorem is a fundamental result characterizing projections onto closed convex cones in Hilbert spaces.
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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
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Recall that a '''convex cone''' in a vector space is a set which is invariant
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under the addition of vectors and multiplication of vectors by positive scalars (see more at [http://en.wikipedia.org/wiki/Convex_cone Convex cone, Wikipedia] or for finite dimension at [[Convex cones | Convex cones, Wikimization]]).
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under the addition of vectors and multiplication of vectors by positive scalars.
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'''Theorem (Moreau)''' Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> and <math>\mathcal K^\circ</math> its '''polar cone'''; that is, the closed convex cone defined by <math>K^\circ=\{a\in\mathcal H\mid\langle a,b\rangle\leq0,\,\forall b\in\mathcal K\}</math> (for finite dimension see more at [http://en.wikipedia.org/wiki/Dual_cone_and_polar_cone Dual cone and polar cone]; see also [[Farkas%27_lemma#Extended_Farkas.27_lemma | Extended Farkas' lemma]]). For <math>x,y,z\in\mathcal H</math> the following statements are equivalent:
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'''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>
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For <math>x,y,z\in\mathbb{H}</math> the following statements are equivalent:
<ol>
<ol>
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<li><math>z=x+y,\,x\in\mathcal K,\,y\in\mathcal K^\circ</math> and <math>\langle x,y\rangle=0,</math></li>
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<li><math>z=x+y,\,x\in\mathcal{K},\,y\in\mathcal{K}^\circ</math> and <math>\langle x,y\rangle=0,</math></li>
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<li><math>x=P_{\mathcal K}z</math> and <math>y=P_{\mathcal K^\circ}z.</math>
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<li><math>x=P_{\mathcal{K}}z</math> and <math>y=P_{\mathcal{K}^\circ}z.</math>
</li>
</li>
</ol>
</ol>
=== Proof of Moreau's theorem ===
=== Proof of Moreau's theorem ===
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<ul>
<ul>
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<li>1<math>\Rightarrow</math>2: For all <math>p\in K</math> we have
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<li>1<math>\Rightarrow</math>2: For all <math>p\in\mathcal{K}</math> we have
<center>
<center>
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</center>
</center>
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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
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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>
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and thus <math>y=P_{\mathcal K^\circ}z.</math></li>
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and thus <math>y=P_{\mathcal{K}^\circ}z.</math></li>
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<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 remained 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
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<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
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all <math>p\in\mathcal K.</math> By using the characterization of the projection, we have
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all <math>p\in\mathcal{K}.</math> By using the characterization of the projection, we have
<center>
<center>
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</center>
</center>
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for all <math>p\in\mathcal K.</math> Thus, <math>u\in\mathcal K^\circ.</math> We also have
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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>
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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>
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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>
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</ul>
</ul>
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=== References ===
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=== notes ===
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For definition of ''convex cone'' in finite dimension see [[Convex cones | Convex cones, Wikimization]].
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* 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.
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For definition of ''polar cone'' in finite dimension, see [http://meboo.convexoptimization.com/Meboo.html Convex Optimization & Euclidean Distance Geometry].
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== An application to nonlinear complementarity problems ==
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<math>\mathcal{K}^{\circ\circ}=\mathcal{K}</math> see [[Farkas%27_lemma#Extended_Farkas.27_lemma|Extended Farkas' lemma]].
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=== Fixed point problems ===
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== Applications ==
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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]],
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and [[Projection_on_Polyhedral_Cone#Projection_on_isotone_projection_cones|Projection on isotone projection cone]].
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Let <math>\mathcal A</math> be a set and <math>F:\mathcal A\to\mathcal A.</math> The '''fixed point problem''' defined by <math>F\,</math> is the problem
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=== References ===
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* 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.
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<center>
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<math>
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Fix(F):\left\{
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\begin{array}{l}
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Find\,\,\,x\in\mathcal A\,\,\,such\,\,\,that\\
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F(x)=x.
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\end{array}
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\right.
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</math>
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</center>
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=== Nonlinear complementarity problems ===
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Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathcal H\to\mathcal H.</math> 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
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<center>
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<math>
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NCP(f,\mathcal K):\left\{
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\begin{array}{l}
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Find\,\,\,x\in\mathcal K\,\,\,such\,\,\,that\\
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f(x)\in\mathcal K^*\,\,\,and\,\,\,\langle x,f(x)\rangle=0.
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\end{array}
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\right.
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</math>
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</center>
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=== Every nonlinear complementarity problem is equivalent to a fixed point problem ===
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Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathcal H\to\mathcal H.</math> Then, the nonlinear complementarity problem <math>NCP(f,\mathcal K)</math> is equivalent to the fixed point problem
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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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=== Proof ===
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For all <math>x\in\mathcal H</math> denote <math>z=x-f(x)\,</math> and <math>y=-f(x).\,</math> Then, <math>z=x+y.\,</math>
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<br>
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<br>
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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>
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<br>
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<br>
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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]]
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<center>
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<math>z=P_{\mathcal K}(z)+P_{\mathcal K^\circ}(z)=x+P_{\mathcal K^\circ}(z).</math>
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</center>
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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,
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<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>
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=== An alternative proof without [[Moreau's_decomposition_theorem#Moreau.27s_theorem | Moreau's theorem]] ===
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==== Variational inequalities ====
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Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathcal H\to\mathcal H.</math> The '''variational inequality''' defined by <math>\mathcal C</math> and <math>f\,</math> is the problem
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<center>
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<math>
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VI(f,\mathcal C):\left\{
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\begin{array}{l}
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Find\,\,\,x\in\mathcal C\,\,\,such\,\,\,that\\
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\langle y-x,f(x)\rangle\geq 0,\,\,\,for\,\,\,all\,\,\,y\in\mathcal C.
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\end{array}
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\right.
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</math>
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</center>
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==== Every variational inequality is equivalent to a fixed point problem ====
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Let <math>\mathcal C</math> be a closed convex set in the Hilbert space <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathcal H\to\mathcal 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>
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==== Proof ====
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<math>x\,</math> is a solution of <math>Fix(P_{\mathcal C}\circ(I-f))</math> if and only if
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<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
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<center>
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<math>\langle x-f(x)-x,y-x\rangle\leq0,</math>
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</center>
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for all <math>y\in\mathcal C.</math> But this holds if and only if <math>x\,</math> is a solution
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of <math>VI(f,\mathcal C).</math>
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===== Remark =====
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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.
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==== Every variational inequality defined on a closed convex cone is equivalent to a complementarity problem ====
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Let <math>\mathcal K</math> be a closed convex cone in the Hilbert space <math>(\mathcal H,\langle\cdot,\cdot\rangle)</math> and <math>f:\mathcal H\to\mathcal 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>
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==== Proof ====
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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,
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<center>
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<math>\langle y-x,f(x)\rangle\geq 0,</math>
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</center>
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for all <math>y\in\mathcal K.</math> Therefore, <math>x\,</math> is a solution of <math>VI(f,\mathcal K).</math>
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<br>
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<br>
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Conversely, suppose that <math>x\,</math> is a solution of <math>VI(f,\mathcal K).</math> Then,
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<math>x\in\mathcal K</math> and
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<center>
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<math>\langle y-x,f(x)\rangle\geq 0,</math>
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</center>
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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>
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==== Concluding the alternative proof ====
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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>
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Current revision

LaTeX: - Sándor Zoltán Németh

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

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: \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:

  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\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.
  • 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\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.
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