User:Double12/sandbox

Variational Inequality (VI)
Given a subset $$K$$ of the Euclidean $$n$$-dimensional space $$\mathbb{R}^n$$ and a mapping $$F:K\rightarrow\mathbb{R}^n$$, the variational inequality, denoted $$\textrm{VI}(K,F)$$, is to find a vector $$x\in K$$ such that


 * $$ \begin{align}(y-x)^T F(x) &\geq 0, \quad \forall y \in K. \end{align}$$

The set of solutions to this problem is denoted $$\textrm{SOL}(K, F)$$.

Special cases:

 * 1) $$\textrm{VI}(K, q,M)$$: if $$F(x)=q+Mx$$, $$\forall x\in\mathbb{R}^n$$, for some $$q\in\mathbb{R}^n$$ and $$M\in\mathbb{R}^{n\times n}$$.
 * 2) linearly constrained $$\textrm{VI}(K,F)$$: if $$K$$ is a polyhedral set.
 * 3) $$\textrm{AVI}(K, q,M)$$ (affine VI): if $$K$$ is a polyhedral set, and $$F(x)=q+Mx$$.

Relationship between VI and constrained optimization problem
If the set $$K$$ is convex, then any local minimizer $$x$$ of the following problem


 * $$\begin{align}

\min & \quad f(x) \\ s.t. & \quad x\in K,\end{align}$$

must be in the set $$\textrm{SOL}(K,\nabla f)$$.

Reformulations of the VI
Let $$K\subseteq\mathbb{R}^n$$ be closed convex and $$F:K\to \mathbb{R}^n$$ be arbitrary. Then,


 * $$\begin{align}

x\in \textrm{SOL}(K,F)\Leftrightarrow F^{\textrm{nat}}_K(x)=0,\end{align}$$

where $$F^{\textrm{nat}}_K(x)=x-\Pi_K(x-F(x))$$ is called the natural map associated the $$\textrm{VI}(K,F)$$.

Let $$K\subseteq\mathbb{R}^n$$ be closed convex and $$F:K\to \mathbb{R}^n$$ be arbitrary. Then, $$x\in \textrm{SOL}(K,F)$$ if and only if there exists a vector $$z$$ such that $$x=\Pi_K(z)$$ and $$F^{\textrm{nor}}_K(z)=0$$,

where $$F^{\textrm{nor}}_K(z)=F(\Pi_K(z))+z-\Pi_K(z)$$ is called the normal map associated the $$\textrm{VI}(K,F)$$.

Complementarity Problem (CP)
Given a cone $$K$$ and a mapping $$F : K \rightarrow \mathbb{R}^n$$, the complementarity problem, denoted $$\textrm{CP}(K,F)$$, is to find a vector $$x \in \mathbb{R}^n$$ satisfying the following conditions:


 * $$ \begin{align}

K \ni x \perp F(x) \in K^*, \end{align} $$ where the notation $$\perp$$ means "perpendicular" and $$K^*$$ is the dual cone of $$K$$.

Special cases:

 * 1) $$\textrm{NCP}(F)$$ (nonlinear CP): if $$K=\mathbb{R}^n_+$$.
 * 2) $$\textrm{LCP}(q,M)$$ (linear CP): if $$K=\mathbb{R}^n_+$$, and $$F(x)=q+Mx$$.
 * 3) $$\textrm{MiCP}(F)$$ (mixed CP): if $$K=\mathbb{R}^{n_1}\times\mathbb{R}^{n_2}_+$$, with $$n_1+n_2=n$$.

Relationship between VI and CP
Let $$K$$ be a cone in $$\mathbb{R}^n$$. A vector $$x$$ solves the $$\textrm{VI}(K,F)$$ if and only if $$x$$ solves the $$\textrm{CP}(K,F)$$.

Existence and Uniqueness
Let $$K\subseteq\mathbb{R}^n$$ be closed convex and $$F:K\to \mathbb{R}^n$$ be continuous. Consider the following statements:

  There exists a vector $$x^{\textrm{ref}} \in K$$ such that the set $$L_{<}=\{x\in K: F(x)^T (x-x^{\textrm{ref}})<0\}$$ is bounded (possibly empty).   There exist a bounded open set $$\Omega$$ and a vector $$x^{\textrm{ref}} \in K \cap\Omega$$ such that $$ F(x)^T (x-x^{\textrm{ref}})\geq 0, \forall x \in K \cap \textrm{bd}\Omega. $$   The $$\textrm{VI}(K,F)$$ has a solution. 

It holds that $$1\Rightarrow 2\Rightarrow 3$$. Moreover, if the set $$L_{\leq}=\{x\in K: F(x)^T (x-x^{\textrm{ref}})\leq0\}$$ is bounded, then $$\textrm{SOL}(K,F)$$ is nonempty and compact.

Corollaries:

 * 1) Let $$K\subseteq\mathbb{R}^n$$ be compact convex and $$F:K\to \mathbb{R}^n$$ be continuous, then $$\textrm{SOL}(K,F)$$ is nonempty and compact.
 * 2) Let $$K\subseteq\mathbb{R}^n$$ be closed convex and $$F:K\to \mathbb{R}^n$$ be continuous. If there exists a vector $$x^{\textrm{ref}} \in K$$ such that $$F(x)^T (x-x^{\textrm{ref}})\geq0, \forall x \in K$$, then the $$\textrm{VI}(K,F)$$ has a solution.
 * 3) Let $$K\subseteq\mathbb{R}^n$$ be closed convex and $$F:K\to \mathbb{R}^n$$ be continuous. If there exist a vector $$x^{\textrm{ref}} \in K$$ and a scalar $$\xi\geq 0$$ such that $$\liminf_{\|x\|\to\infty} \frac{F(x)^T (x-x^{\textrm{ref}})}{\|x\|^{\xi}}$$ holds, then the $$\textrm{VI}(K,F)$$ has a nonempty compact solution set.

Let $$K\subseteq\mathbb{R}^n$$ be closed convex and $$F:K\to \mathbb{R}^n$$ be continuous. Then,


 * 1) If $$F$$ is strictly monotone on $$K$$, the $$\textrm{VI}(K,F)$$ has at most one solution.
 * 2) If $$F$$ is $$\xi$$-monotone on $$K$$ for some $$\xi>1$$, the $$\textrm{VI}(K,F)$$ has a unique solution $$x^*$$.

Basic Projection Algorithm (BPA)

 * 1) Set $$k = 0$$ and $$\gamma>0$$.
 * 2) If $$x^k=\Pi_K(x^k-F(x^k))$$ stop.
 * 3) Set $$x^{k+1}=\Pi_K(x^k-\gamma F(x^k))$$ and $$k\leftarrow k+1$$; go to Step 2.

Let $$F:K\to \mathbb{R}^n$$, where $$K$$ be a closed convex subset of $$\mathbb{R}^n$$. Suppose $$F$$ is both strongly monotone and Lipschitz continuous on $$K$$ with constants $$\mu$$ and $$L$$, respectively. If $$\gamma<\frac{2\mu}{L^2}$$, every sequence $$\{x^k\}$$ produced by BPA converges to the unique solution of the $$\textrm{VI}(K,F)$$.

Projection Algorithm with Variable Steps (PAVS)

 * 1) Set $$k = 0$$.
 * 2) If $$x^k=\Pi_K(x^k-F(x^k))$$ stop.
 * 3) Choose $$\gamma_k>0$$. Set $$x^{k+1}=\Pi_K(x^k-\gamma_k F(x^k))$$ and $$k\leftarrow k+1$$; go to Step 2.

Let $$K$$ be a closed convex subset of $$\mathbb{R}^n$$ and $$F:K\to \mathbb{R}^n$$ be mapping that is co-coercive on $$K$$ with constant $$c$$. If $$0<\inf_k \gamma_k\leq \sup_k \gamma_k<2c$$, PAVS produces a sequence $$\{x^k\}$$ converging to a solution of the $$\textrm{VI}(K,F)$$.

Extragradient Algorithm (EgA)

 * 1) Set $$k = 0$$ and $$\gamma>0$$.
 * 2) If $$x^k=\Pi_K(x^k-F(x^k))$$ stop.
 * 3) Set $$x^{k+1/2}=\Pi_K(x^k-\gamma F(x^k))$$ and $$x^{k+1}=\Pi_K(x^k-\gamma F(x^{k+1/2}))$$; set $$k\leftarrow k+1$$ and go to Step 2.

Let $$K$$ be a closed convex subset of $$\mathbb{R}^n$$ and $$F:K\to \mathbb{R}^n$$ be mapping that is pseudo monotone on $$K$$ with respect to $$\textrm{SOL}(K,F)$$ and Lipschitz continuous on $$K$$ with constant $$L$$. If $$\gamma<1/L$$, the sequence $$\{x^k\}$$ produced by EgA converges to a solution of the $$\textrm{VI}(K,F)$$.

Tikhonov Regularization Algorithm (TiRA)
Let $$\{\rho_k\}\downarrow 0$$ and $$\{\epsilon_k\}\downarrow 0$$.


 * 1) Set $$k = 0$$.
 * 2) If $$x^k=\Pi_K(x^k-F(x^k))$$ stop.
 * 3) Calculate a point $$x^{k+1}$$ such that $$\|F^{\textrm{nat}}_{\epsilon_k,K}(x^{k+1})\|\leq\rho_k$$, where $$F^{\textrm{nat}}_{\epsilon_k,K}$$ is the natural map of the perturbed $$\textrm{VI}(K,F+\epsilon_k I)$$; set $$k\leftarrow k+1$$ and go to Step 2.

Let $$K$$ be a closed convex subset of $$\mathbb{R}^n$$. Let $$F:K\to \mathbb{R}^n$$ be a continuous $$P_0$$ function on $$K$$ and suppose $$\textrm{VI}(K,F)$$ is nonempty and bounded. Let $$\{x^k\}$$ be a sequence produced by TiRA. The sequence $$\{x^k\}$$ is bounded and each of its limit points belongs to $$\textrm{SOL}(K,F)$$.