Constrained Best Approximation with Nonsmooth Nonconvex Constraints

TL;DR

This paper investigates the best approximation problem under nonsmooth nonconvex constraints, characterizing the perturbation property via strong CHIP and extending classical results with new dual and Lagrange multiplier conditions.

Contribution

It introduces a dual cone characterization and sufficient conditions for strong CHIP in nonsmooth nonconvex constraints, extending existing approximation theory.

Findings

Characterization of perturbation property via strong CHIP.

Dual cone representation of the constraint set.

Lagrange multiplier conditions under non-smooth Abadie's qualification.

Abstract

In this paper, we consider the constraint set $K$ of inequalities with nonsmooth nonconvex constraint functions. We show that under Abadie's constraint qualification the "perturbation property" of the best approximation to any $x$ in $\R^n$ from a convex set $\tK:=C \cap K$ is characterized by the strong conical hull intersection property (strong CHIP) of $C$ and $K,$ where $C$ is a non-empty closed convex subset of $\R^n$ and the set $K$ is represented by $K:=\{x\in \R^n : g_j(x) \le 0, \ \forall \ j=1,2,\ldots,m \}$ with $g_j : \R^n \lrar \R$ $(j=1,2, \cdots,m)$ is a tangentially convex function at a given point $\bar x \in K.$ By using the idea of tangential subdifferential and a non-smooth version of Abadie's constraint qualification, we do this by first proving a dual cone characterization of the constraint set $K.$ Moreover, we present sufficient conditions for which the strong CHIP property holds. In particular, when the set $\tK$ is closed and convex, we show that the Lagrange multiplier characterization of best approximation holds under a non-smooth version of Abadie's constraint qualification. The obtained results extend many corresponding results in the context of constrained best approximation. Several examples are provided to clarify the results.

Equations

$$f^{\prime }(\bar{x},\nu ):=\underset{\alpha ⟶0^{+}}{\lim }\frac{f(\bar{x}+\alpha \nu )-f(\bar{x})}{\alpha },$$

$${\partial }_{T}f(\bar{x}):=\{x^{\ast }\in {\mathbb{R}}^n:⟨x^{\ast },\nu ⟩\le f^{\prime }(\bar{x},\nu ),\forall \nu \in {\mathbb{R}}^n\}.$$

$$f^{\prime }(\bar{x},\nu )=\underset{x^{\ast }\in {\partial }_{T}f(\bar{x})}{\max }⟨x^{\ast },\nu ⟩.$$

$$K:=\{x\in {\mathbb{R}}^n:g_j(x)\le 0,\forall j=1,2,\dots ,m\},$$

$$\tilde{K}:=C\cap K,$$

$$I:=\{1,2,\cdots ,m\}.$$

$$I(\bar{x}):=\{j\in I:g_j(\bar{x})=0\}.$$

$$W^{\circ }:=\{\lambda \in {\mathbb{R}}^n:⟨\lambda ,y⟩\le 0,\forall y\in W\},$$

$$N_H(x):=\{u\in {\mathbb{R}}^n:⟨u,t-x⟩\le 0,\forall t\in H\}.$$

$$N_H(x)=(H-x{)}^{\circ },(x\in {\mathbb{R}}^n).$$

$$T_U(x)$$

$$D(\bar{x}):=\{x^{\ast }\in {\mathbb{R}}^n:⟨x^{\ast },{\eta }_j⟩\le 0,\forall {\eta }_j\in {\partial }_T{g}_j(\bar{x}),\forall j\in I(\bar{x})\},$$

$$g_j(\bar{x}+{\beta }_{k,p}{\alpha }_k{x}_k^{\ast })\le 0,\text{mbox}forallsufficientlylargep\in \mathbb{N},\forall k\ge 1,\forall j\in I.$$

$$g_j^{\prime }(\bar{x},{\alpha }_k{x}_k^{\ast })$$

$$g_j^{\prime }(\bar{x},x_k^{\ast })\le 0,\forall k\ge 1.$$

$$g_j^{\prime }(\bar{x},x^{\ast })\le 0,\forall j\in I(\bar{x}).$$

$$⟨x^{\ast },{\eta }_j⟩\le 0,\forall {\eta }_j\in {\partial }_T{g}_j(\bar{x}),\forall j\in I(\bar{x}),$$

$$(NRCQ)⟹(NACQ).$$

$$g_1(x_1,x_2):=|x_2|-x_1,g_2(x_1,x_2):=1-x_1^2-(x_2-1{)}^2,\text{mbox}and$$

$$g_3(x_1,x_2):=1-x_1^2-(x_2+1{)}^2,\forall (x_1,x_2)\in {\mathbb{R}}^2.$$

$$K$$

$$C:=\{(x_1,x_2)\in {\mathbb{R}}^2:x_1\ge 0\},$$

$$g_1^{\prime }(\bar{x},(t_1,t_2))=|t_2|-t_1,g_2^{\prime }(\bar{x},(t_1,t_2))=2t_2,\text{mbox}and$$

$$g_3^{\prime }(\bar{x},(t_1,t_2))=-2t_2,\forall (t_1,t_2)\in {\mathbb{R}}^2.$$

$${\partial }_T{g}_1(\bar{x})=\text{mbox}\mathrm{co}\{(-1,-1),(-1,1)\},{\partial }_T{g}_2(\bar{x})=\{(0,2)\},\text{mbox}and$$

$${\partial }_T{g}_3(\bar{x})=\{(0,-2)\}.$$

$$T_{\tilde{K}}(\bar{x})=D(\bar{x})=\{(t_1,0)\in {\mathbb{R}}^2:t_1\ge 0\},$$

$$d(x,W):=\underset{w\in W}{\inf }\Vert x-w\Vert .$$

$$P_W(x):=\{w\in W:\Vert x-w\Vert =d(x,W)\}.$$

$$\displaystyle\bigg{(}\bigcap_{j=1}^{m}C_{j}-x\bigg{)}^{\circ}=\sum_{j=1}^{m}\bigg{(}C_{j}-x\bigg{)}^{\circ}.$$

$$0\in \partial f(\bar{x})+N_C(\bar{x}),$$

$$\text{mbox}\mathrm{dom}(f):=\{x\in {\mathbb{R}}^n:f(x)<+\infty \}.$$

Full text

Key words. tangentially convex; constraint qualification; Lagrange multipliers; constrained best approximation; strong conical hull intersection property; perturbation property.

2010 (AMS) Mathematics Subject Classification. 41A29; 41A50; 90C26; 90C46.

1 Introduction

Finding suitable conditions for ”perturbation property” has been of substantial interest in constrained best approximation as it is often easier to compute the best approximation from a closed convex set $C$ than from the constraint set $K.$ The merit and motivation for such characterization $($perturbation property$)$ is inspired from [5, Chapter 10]. ”Characterizing constrained interpolation from a convex set” is one of the applications of the ”perturbation property” $($for more details, see [5]$).$

For many years, a great deal of attention has been focusing on the case where the constraint set $K:=\{x\in\mathbb{R}^{n}:g_{j}(x)\leq 0,\ \forall\ j=1,2,\ldots,m\}$ is a closed convex set and has a convex representation in the sense that $g_{j}$ $(j=1,2,\cdots,m)$ is a convex function [5, 6, 7, 9, 12, 13]. Various characterizations of the perturbation property have been given by using local constraint qualifications such as the strong conical hull intersection property (strong CHIP) of $C$ and $K$ at the best approximation [4, 7, 9, 12].

In this paper, we study the problem of whether the best approximation to any $x\in\mathbb{R}^{n}$ from the closed convex set $\tilde{K}:=C\cap K$ can be characterized by the best approximation to a perturbation $x-x^{*}$ of $x$ from a closed convex set $C\subseteq\mathbb{R}^{n}$ for some $x^{*}$ in a certain cone in $\mathbb{R}^{n},$ where $K:=\{x\in\mathbb{R}^{n}:g_{j}(x)\leq 0,\ \forall\ j=1,2,\ldots,m\}$ with $g_{j}:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ $(j=1,2,\cdots,m)$ is a tangentially convex function at the best approximation. We show that the strong CHIP of $C$ and $K$ at the best approximation continues to completely characterize the perturbation property of the best approximation from the closed convex set $\tilde{K}:=C\cap K$ under Abadie’s constraint qualification.

Indeed, by using the idea of tangential subdifferential and a non-smooth version of Abadie’s constraint qualification (which is the weakest qualification among the other well known constraint qualifications), we prove this by first establishing a dual cone characterization of the constraint set $K.$ In the special case, when $\tilde{K}$ is a closed convex set, we show that the Lagrange multipliers characterization of best approximation holds under a non-smooth version of Abadie’s constraint qualification. Our results recapture the corresponding known results of [4, 6, 7, 9, 10, 11, 12, 13, 14, 16]. Several illustrative examples are presented to clarify our results.

The paper has the following structure. In Section 2, we provide the basic results on tangentially convex functions and a non-smooth version of the constraint qualifications. Dual cone characterizations of the constraint set $K$ and sufficient conditions for which the strong CHIP holds are presented in Section 3. In Section 4, we first show that the strong CHIP completely characterizes the perturbation property of the best approximation. Finally, we show that under a non-smooth version of Abadie’s constraint qualification the Lagrange multipliers characterization of best approximation holds. Also, several examples are presented to illustrate our results.

2 Preliminaries

We start this section by fixing notations and preliminaries that will be used later. Recall [3] that for a function $f:\mathbb{R}^{n}\longrightarrow\mathbb{R},$ the directional derivative of $f$ at a point $\bar{x}\in\mathbb{R}^{n}$ in the direction $\nu\in\mathbb{R}^{n}$ is defined by

[TABLE]

if the limit exists. Recall [15] that a function $f:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ is called tangentially convex at a point $\bar{x}\in\mathbb{R}^{n},$ if $f^{\prime}(\bar{x},\cdot)$ is a real valued convex function.

It should be noted that if the function $f$ is tangentially convex at a point $\bar{x}\in\mathbb{R}^{n},$ then, since $f^{\prime}(\bar{x},\cdot)$ is a positively homogeneous function, we conclude that $f^{\prime}(\bar{x},\cdot)$ is a sublinear function on $\mathbb{R}^{n}.$

The tangential subdifferential of a function $f:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ at a point $\bar{x}\in\mathbb{R}^{n}$ is defined by

[TABLE]

If $f$ is tangentially convex at $\bar{x},$ then, $\partial_{T}f(\bar{x})\neq\emptyset,$ and moreover, $f^{\prime}(\bar{x},\cdot)$ is the support functional of $\partial_{T}f(\bar{x}),$ i.e., for each $\nu\in\mathbb{R}^{n},$ we have

[TABLE]

It should be noted that if $f$ is a convex function, then, $\partial_{T}f(x)=\partial f(x)$ for each $x\in\mathbb{R}^{n},$ where $\partial f(x)$ is the classical convex subdifferential of $f$ at $x.$

Remark 2.1**.**

Note that if the function $f:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ is tangentially convex at a point $\bar{x}\in\mathbb{R}^{n},$ then, $f^{\prime}(\bar{x},\cdot)$ is a real valued convex function on $\mathbb{R}^{n},$ and hence, $f^{\prime}(\bar{x},\cdot)$ is a continuous function on $\mathbb{R}^{n}.$

Now, let $K\subseteq\mathbb{R}^{n}$ be defined by

[TABLE]

where $g_{j}:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ $(j=1,2,\cdots,m)$ is a tangentially convex function at a given point $\bar{x}\in K.$ Let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset,$ and let $S:=\mathbb{R}_{+}^{m}.$ Note that $K$ is not necessarily a closed or a convex set. Let

[TABLE]

and

[TABLE]

For a point $\bar{x}\in K,$ we define

[TABLE]

For a set $W\subseteq\mathbb{R}^{n},$ let

[TABLE]

where we denote $\langle\cdot,\cdot\rangle$ for the inner product of $\mathbb{R}^{n}.$ The normal cone to a set $H\subseteq\mathbb{R}^{n}$ at a point $x\in\mathbb{R}^{n}$ is defined by

[TABLE]

It is clear that

[TABLE]

Let $U$ be a subset of $\mathbb{R}^{n},$ and let $x\in\mathbb{R}^{n}.$ We recall [2, 3] the contingent cone of $U$ at $x$ is defined by

[TABLE]

We now introduce a non-smooth version of the linearized tangential cone:

[TABLE]

where $\bar{x}\in K.$

Note that the non-smooth linearized tangential cone $D(\bar{x})$ reduces to its counterpart in the case of differentiability [2, 3]. Moreover, $D(\bar{x})$ is a convex cone.

We now present the definition of the nearly convexity which has been given in [11]. Let $V$ be a non-empty subset of $\mathbb{R}^{n}$ and $x\in V.$

Definition 2.1**.**

$($Nearly Convex at $x\in V).$* The set $V$ is nearly convex at the point $x\in V$ if for each $y\in V$ there exists a sequence $\{t_{k}\}_{k\geq 1}$ of positive real numbers with $t_{k}\longrightarrow 0^{+}$ such that $x+t_{k}(y-x)\in V$ for all sufficiently large $k\in\mathbb{N}.$

The set $V$ is called nearly convex whenever it is nearly convex at each of its points. It is easy to check that if $V$ is convex, then it is nearly convex at each $x\in V.$ As shown in [8], the nearly convexity may hold at a point for a non-convex set $($for more details and illustrative examples related to the nearly convexity, see [8, 11]$).$*

Lemma 2.1**.**

Let $K$ be closed, given by $(\ref{4.1}),$ and let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset.$ Let $\tilde{K}:=C\cap K,$ and $\bar{x}\in\tilde{K}.$ Assume that $K$ is nearly convex at the point $\bar{x}.$ Then, $T_{\tilde{K}}(\bar{x})\subseteq D(\bar{x}),$ where $T_{\tilde{K}}(\bar{x})$ and $D(\bar{x})$ defined by $(\ref{1.2})$ and $(\ref{1.3}),$ respectively.

*Proof:  *Let $x^{*}\in T_{\tilde{K}}(\bar{x})$ be arbitrary. Then there exist sequences $\{\alpha_{k}\}_{k\geq 1}\subset\mathbb{R}_{++}$ and $\{x_{k}^{*}\}_{k\geq 1}\subset\mathbb{R}^{n}$ such that $\alpha_{k}\longrightarrow 0^{+},$ $x_{k}^{*}\longrightarrow x^{*}$ and $\bar{x}+\alpha_{k}x_{k}^{*}\in\tilde{K}$ for all $k\geq 1.$ Since, by the hypothesis, $K$ is nearly convex at the point $\bar{x}$ and $\bar{x}+\alpha_{k}x_{k}^{*}\in K$ for all $k\geq 1,$ it follows from Definition 2.1 that, for each $k\geq 1,$ there exists a sequence $\{\beta_{k,p}\}_{p\geq 1}\subset\mathbb{R}_{++}$ with $\beta_{k,p}\longrightarrow 0^{+}$ $($as $p\longrightarrow+\infty)$ such that $\bar{x}+\beta_{k,p}(\bar{x}+\alpha_{k}x_{k}^{*}-\bar{x})\in K$ for all sufficiently large $p\in\mathbb{N}.$ This implies that

[TABLE]

Since $g_{j}$ $(j\in I)$ is tangentially convex at $\bar{x},$ it follows, by the definition, that $g_{j}^{\prime}(\bar{x},\cdot)$ is a real valued positively homogeneous and convex function on $\mathbb{R}^{n}.$ Therefore, for each $j\in I(\bar{x}),$ in view of $(\ref{1.4})$ we have

[TABLE]

and hence,

[TABLE]

Since $x_{k}^{*}\longrightarrow x^{*}$ and $g_{j}^{\prime}(\bar{x},\cdot)$ is continuous on $\mathbb{R}^{n}$ $($see Remark 2.1$),$ we conclude from (2.13) that

[TABLE]

This together with $(\ref{4.35})$ implies that

[TABLE]

and so, $x^{*}\in D(\bar{x}),$ which completes the proof.

Now, let us to define the non-smooth versions of Robinson’s constraint qualification and Abadie’s constraint qualification.

Definition 2.2**.**

$($Non-smooth Version of Robinson’s Constraint Qualification $(NRCQ)).$* Let $K=\{x\in\mathbb{R}^{n}:g_{j}(x)\leq 0,\ \forall\ j=1,2,\cdots,m\}$ be as in $($2.4$),$ and let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset.$ Let $\tilde{K}:=C\cap K,$ and $\bar{x}\in\tilde{K}.$ We say that non-smooth Robinson’s constraint qualification holds at $\bar{x}$ if there exists $0\neq\nu\in\mathbb{R}^{n}$ such that for each $j\in I(\bar{x})$ and each $\eta_{j}\in\partial_{T}g_{j}(\bar{x}),$ one has $\langle\eta_{j},\nu\rangle<0,$ where $\partial_{T}g_{j}(\bar{x})$ is the tangential subdifferential of $g_{j}$ at $\bar{x}.$*

Definition 2.3**.**

$($Non-smooth Version of Abadie’s Constraint Qualification $(NACQ)).$* Let $K=\{x\in\mathbb{R}^{n}:g_{j}(x)\leq 0,\ \forall\ j=1,2,\cdots,m\}$ be as in $($2.4$),$ and let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset.$ Let $\tilde{K}:=C\cap K,$ and $\bar{x}\in\tilde{K}.$ We say that non-smooth Abadie’s constraint qualification holds at $\bar{x}$ if $D(\bar{x})\subseteq T_{\tilde{K}}(\bar{x}).$*

Obviously, the above definitions of non-smooth version of constraint qualifications reduce to their counterparts in the case of differentiability [1, 3].

Clearly, in view of Lemma 2.1, Definition 2.2 and Definition 2.3, the following implication holds.

[TABLE]

The following example shows that non-smooth Abadie’s constraint qualification is weaker than non-smooth Robinson’s Constraint Qualification.

Example 2.1**.**

Let $g_{1},g_{2},g_{3}:\mathbb{R}^{2}\longrightarrow\mathbb{R}$ be defined by

[TABLE]

Then, we have

[TABLE]

Let

[TABLE]

and $\tilde{K}:=C\cap K=K.$ Let $\bar{x}:=(0,0)\in\tilde{K}.$ Clearly, $g_{1},g_{2},g_{3}$ are tangentially convex at $\bar{x},$ and $g_{1}(\bar{x})=g_{2}(\bar{x})=g_{3}(\bar{x})=0.$ Moreover, it is easy to check that

[TABLE]

This together with $(\ref{42})$ implies that

[TABLE]

It is clear that non-smooth Robinson’s constraint qualification does not hold at $\bar{x}.$ But, we have

[TABLE]

and hence, non-smooth Abadie’s constraint qualification holds at $\bar{x}.$

Remark 2.2**.**

It is worth noting that in [16] it has been given a characterization of constrained best approximation under smooth Robinson’s constraint qualification with continuously Fréchet differentiable constraints, while in this paper we give characterizations of constrained best approximation under non-smooth Abadie’s constraint qualification with the constraint functions are only tangentially convex at the point of best approximation. Since, in view of $($2.14$)$ and Example 2.1, $(NACQ)$ is weaker than $(NRCQ).$ So, our results are stronger and extend the obtained results in [16] and the corresponding results of [4, 6, 7, 9, 10, 12, 13, 14].

For a non-empty subset $W$ of $\mathbb{R}^{n}$ and an arbitrary point $x\in\mathbb{R}^{n},$ we define

[TABLE]

We say that a point $x_{0}\in W$ is a best approximation $($a projection$)$ of $x\in\mathbb{R}^{n}$ if $\|x-x_{0}\|=d(x,W)$ [17]. The set of all best approximations $($projections$)$ of $x$ in $W$ denoted by $P_{W}(x)$ and is given by:

[TABLE]

The following characterization of best approximation in Hilbert spaces is well known [5].

Lemma 2.2**.**

Let $D$ be a non-empty closed convex subset of a Hilbert space $H,$ $x\in H$ and $x_{0}\in D.$ Then, $x_{0}=P_{D}(x)$ if and only if $x-x_{0}\in(D-x_{0})^{\circ}.$

In the following, we give the notion of strong CHIP. The definition of strong CHIP was first introduced in [7] (see also, [4, 5]).

Definition 2.4**.**

$($Strong CHIP$).$* Let $C_{1},C_{2},\ldots,C_{m}$ be non-empty closed convex sets in $\mathbb{R}^{n},$ and let $x\in\bigcap_{j=1}^{m}C_{j}.$ Then, the collection $\{C_{1},C_{2},\ldots,C_{m}\}$ is said to have the strong CHIP $($canonical hull intersection property$)$ at $x$ if*

[TABLE]

The collection $\{C_{1},C_{2},\ldots,C_{m}\}$ is said to have the strong CHIP if it has the strong CHIP at each $x\in\bigcap_{j=1}^{m}C_{j}.$

We recall [2, 3] the following well known result from the non-smooth analysis.

Theorem 2.1**.**

Let $C\subset\mathbb{R}^{n}$ be a non-empty convex set, and let $f:\mathbb{R}^{n}\longrightarrow\mathbb{R}\cup\{+\infty\}$ be a proper convex function such that ${C\cap\mbox{\rm dom}\,(f)}\neq\emptyset.$ Assume that $\bar{x}\in C$ and $f$ is continuous at $\bar{x}.$ Then, $\bar{x}$ is a global minimizer of the function $f$ over $C$ if and only if

[TABLE]

where domain of the function $f,$ $\mbox{\rm dom}\,(f),$ is defined by

[TABLE]

3 Dual Cone Characterizations of the Constraint Set $K$

In this section, we give dual cone characterizations of the constraint set $K$ at a point $x\in K,$ where $K$ is given by (2.4). Also, we present sufficient conditions for which the strong CHIP property holds.

For each $x\in K,$ put

[TABLE]

where $\lambda:=(\lambda_{1},\lambda_{2},\cdots,\lambda_{m})\in S$ and $S:=\mathbb{R}_{+}^{m}.$

Remark 3.1**.**

Throughout the paper, we assume that the constraint functions $g_{j},$ $j=1,2,\ldots,m,$ are tangentially convex at a given point $\bar{x}\in\tilde{K}:=C\cap K.$

We now give a dual cone characterization of the constraint set $K,$ which has a crucial role for characterizing best approximations by the set $\tilde{K}:=C\cap K.$ Note that $K$ is not necessarily a closed or a convex set.

Theorem 3.1**.**

Let $K$ be closed, given by $(\ref{4.1}),$ and let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset.$ Let $\tilde{K}:=C\cap K,$ $\bar{x}\in\tilde{K}$ and $M(\bar{x})$ be as in $(\ref{4.2}).$ Assume that $K$ is nearly convex at the point $\bar{x}.$ If non-smooth Abadie’s constraint qualification holds at $\bar{x},$ then, $M(\bar{x})=(K-\bar{x})^{\circ}=(\tilde{K}-\bar{x})^{\circ}.$

*Proof:  *It is easy to see that $(K-\bar{x})^{\circ}\subseteq(\tilde{K}-\bar{x})^{\circ}.$ Now, let $u\in(\tilde{K}-\bar{x})^{\circ}$ be arbitrary. Then, $\langle u,y-\bar{x}\rangle\leq 0$ for all $y\in\tilde{K},$ and so,

[TABLE]

Let $h:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ be defined by

[TABLE]

It is clear that $h$ is a continuous convex function on $\mathbb{R}^{n}.$ Now, we show that $h(y)\geq 0$ for all $y\in T_{\tilde{K}}(\bar{x}).$ To this end, let $y\in T_{\tilde{K}}(\bar{x})$ be arbitrary. Then by (2.10) there exist $\{t_{m}\}_{m\geq 1}\subset\mathbb{R}_{++}$ and $\{y_{m}\}_{m\geq 1}\subset\mathbb{R}^{n}$ such that $t_{m}\longrightarrow 0^{+},$ $y_{m}\longrightarrow y$ and $\bar{x}+t_{m}y_{m}\in\tilde{K}$ for all $m\geq 1.$ Thus, in view of (3.16), we conclude that $\langle-u,y_{m}\rangle\geq 0$ for all $m\geq 1.$ This together with the fact that $y_{m}\longrightarrow y$ implies that $\langle-u,y\rangle\geq 0,$ and so,

[TABLE]

Consider the following optimization problem:

[TABLE]

It follows from (3.18) that $y=0\in T_{\tilde{K}}(\bar{x})$ is a global minimizer of the problem (3.19) over $T_{\tilde{K}}(\bar{x}).$ On the other hand, since $K$ is nearly convex at the point $\bar{x},$ in view of Lemma 2.1 and the validity of non-smooth Abadie’s constraint qualification at $\bar{x},$ we have, $D(\bar{x})=T_{\tilde{K}}(\bar{x}).$ Therefore, the problem (3.19) can be represented as the following convex optimization problem:

[TABLE]

Note that $D(\bar{x})$ is a closed convex subset of $\mathbb{R}^{n},$ and $y=0\in D(\bar{x})$ is a global minimizer of the problem (3.20) over $D(\bar{x}).$ In view of (2.3) and (2.11), the problem (3.20) can be represented as the following convex optimization problem:

[TABLE]

Note that $y=0$ $($because $g_{j}^{\prime}(\bar{x},0)=0,\ j\in I(\bar{x}))$ is a global minimizer of the problem (3.21). Let

[TABLE]

and let $H:=\cap_{j\in I(\bar{x})}C_{j}.$ Since $g_{j}^{\prime}(\bar{x},\cdot)$ is convex on $\mathbb{R}^{n}$ $(j\in I),$ it is easy to see that $C_{j}$ is convex for each $j\in I(\bar{x}),$ and hence, $H$ is a convex set. Since, by (3.20), $y=0\in H$ $($note that $g_{j}^{\prime}(\bar{x},0)=0\leq 0,\ j\in I(\bar{x}))$ is a global minimizer of the problem (3.21) over $H,$ it follows from Theorem 2.1 that

[TABLE]

This together with [3, Section 3.3] implies that

[TABLE]

Let

[TABLE]

It is easy to check that $M_{j}(\bar{x})$ is a closed convex cone in $\mathbb{R}^{n}$ for each $j\in I(\bar{x}).$ Now, we claim that

[TABLE]

Assume if possible that there exists $x^{*}\in N_{C_{j}}(0)$ such that $x^{*}\notin M_{j}(\bar{x}).$ Since $M_{j}(\bar{x})$ is a closed convex cone, by using the separation theorem there exists $0\neq\nu\in\mathbb{R}^{n}$ such that

[TABLE]

For simplicity, put $h_{j}(\cdot):=g_{j}^{\prime}(\bar{x},\cdot)$ $(j\in I(\bar{x})).$ Since $g_{j}$ is tangentially convex at $\bar{x},$ it follows that, for each $j\in I(\bar{x}),$ $h_{j}$ is a real valued positively homogeneous and convex function, and so,

[TABLE]

This together with (3.26) implies that $h_{j}^{\prime}(0,\nu)\leq 0$ $(j\in I(\bar{x})).$ On the other hand, since $h_{j}$ $(j\in I(\bar{x}))$ is positively homogeneous, we conclude that $h_{j}^{\prime}(0,\nu)=g_{j}^{\prime}(\bar{x},\nu)$ $(j\in I(\bar{x})).$ So, $g_{j}^{\prime}(\bar{x},\nu)\leq 0$ $(j\in I(\bar{x})),$ and hence, $\nu\in C_{j}$ $(j\in I(\bar{x})).$ But, we have $x^{*}\in N_{C_{j}}(0)$ $(j\in I(\bar{x})).$ Therefore, $\langle\nu,x^{*}\rangle\leq 0,$ which contradicts (3.26). Then, the inclusion (3.25) holds. This together with (3.23) implies that

[TABLE]

So, for each $j\in I(\bar{x}),$ there exists $\lambda_{j}\geq 0$ such that

[TABLE]

It is not difficult to see that $\partial g_{j}^{\prime}(\bar{x},\cdot)(0)=\partial_{T}g_{j}(\bar{x})$ $(j\in I(\bar{x})).$ Thus, it follows from (3.27) that

[TABLE]

But, in view of (3.17), we have $\partial h(0)=\{-u\}.$ Now, for each $j\notin I(\bar{x}),$ put $\lambda_{j}=0.$ Therefore, we obtain from (3.28) that

[TABLE]

and so, by (3.15), $u\in M(\bar{x}).$ Hence, $(\tilde{K}-\bar{x})^{\circ}\subseteq M(\bar{x}).$

Now, we show that $M(\bar{x})\subseteq(K-\bar{x})^{\circ}.$ To this end, let $u\in M(\bar{x})$ be arbitrary. Then, in view of (3.15), there exists $(\lambda_{1},\lambda_{2},\cdots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0$ $(j=1,2,\cdots,m)$ such that

[TABLE]

This implies that, for each $j=1,2,\cdots,m,$ there exists $\eta_{j}\in\partial_{T}g_{j}(\bar{x})$ such that

[TABLE]

Now, let $y\in K$ be arbitrary. Since $\bar{x}\in K$ and $K$ is nearly convex at $\bar{x},$ it follows from Definition 2.1 that there exists a sequence $\{\alpha_{k}\}_{k\geq 1}\subset\mathbb{R}_{++}$ with $\alpha_{k}\longrightarrow 0^{+}$ such that $\bar{x}+\alpha_{k}(y-\bar{x})\in K$ for all sufficiently large $k\in\mathbb{N}.$ So, by (2.4),

[TABLE]

Since $g_{j}$ $(j=1,2,\cdots,m)$ is tangentially convex at $\bar{x},$ it follows from (2.3), (3.29), (3.30) and the fact that $\lambda_{j}=0$ for each $j\notin I(\bar{x})$ (because $\lambda_{j}g_{j}(\bar{x})=0,\ j=1,2,\ldots,m)$ that

[TABLE]

Hence, $u\in(K-\bar{x})^{\circ},$ which completes the proof.

Remark 3.2**.**

It should be noted that in Theorem 3.1, for proving the inclusion $M(\bar{x})\subseteq(K-\bar{x})^{\circ},$ we only require that the set $K$ is nearly convex at the point $\bar{x}$ without the validity of non-smooth Abadie’s constraint qualification at $\bar{x}.$

Corollary 3.1**.**

Let $K$ be closed, given by $(\ref{4.1}),$ and let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset.$ Let $\tilde{K}:=C\cap K,$ and $\bar{x}\in\tilde{K}.$ Assume that $K$ is nearly convex at the point $\bar{x}$ and non-smooth Abadie’s constraint qualification holds at $\bar{x}.$ Then, $\{C,K\}$ has the strong CHIP at $\bar{x}.$

*Proof:  *We first note that one always has

[TABLE]

For the converse inclusion, in view of Theorem 3.1, we conclude that $(\tilde{K}-\bar{x})^{\circ}=(K-\bar{x})^{\circ}.$ Therefore, since $0\in(C-\bar{x})^{\circ},$ we have

[TABLE]

which completes the proof.

The following example shows that the converse statement to Corollary 3.1 is not valid.

Example 3.1**.**

Let $g(x):=|x|-x$ for all $x\in\mathbb{R}.$ Thus, $K=\{x\in\mathbb{R}:g(x)\leq 0\}=[0,+\infty),$ which is closed. Let $C:=(-\infty,0],$ and $\bar{x}=0.$ It is clear that $g$ is tangentially convex at $\bar{x},$ $g^{\prime}(\bar{x},\nu)=|\nu|-\nu$ for all $\nu\in\mathbb{R},$ and $C\cap K=\{0\}.$ Also, we have $g(\bar{x})=0$ and $\partial_{T}g(\bar{x})=[-2,0].$ Note that $K$ is nearly convex at the point $\bar{x}.$ It is easy to check that $(K-\bar{x})^{\circ}=(-\infty,0]$ and $(C-\bar{x})^{\circ}=[0,+\infty).$ Let $\tilde{K}:=C\cap K=\{0\}.$ Then,

[TABLE]

Thus, $\{C,K\}$ has the strong CHIP at $\bar{x}.$ But, on the other hand, we have

[TABLE]

This implies that

[TABLE]

and hence, non-smooth Abadie’s constraint qualification does not hold at $\bar{x}.$

By the following example we show that in Theorem 3.1 the validity of the nearly convexity of $K$ at the point $\bar{x}\in K$ cannot be omitted.

Example 3.2**.**

Let $g_{1},g_{2}:\mathbb{R}\longrightarrow\mathbb{R}$ be defined by

[TABLE]

Then, we have

[TABLE]

which is closed. Let $C:=[2,3],$ $\tilde{K}:=C\cap K=\{2\}$ and $\bar{x}:=2.$ It is easy to see that $g_{1}$ and $g_{2}$ are tangentially convex at $\bar{x},$ $g_{1}(\bar{x})=g_{2}(\bar{x})=0,$ $\partial_{T}g_{1}(\bar{x})=\{-12\}$ and $\partial_{T}g_{2}(\bar{x})=\{2\}.$ Clearly,

[TABLE]

Thus, $D(\bar{x})=T_{\tilde{K}}(\bar{x}),$ and so, non-smooth Abadie’s constraint qualification holds at $\bar{x},$ while it is clear that $K$ is not nearly convex at the point $\bar{x}.$

On the other hand, it is not difficult to check that $M(\bar{x})=\mathbb{R},$ $(\tilde{K}-\bar{x})^{\circ}=\mathbb{R}$ and $(K-\bar{x})^{\circ}=(-\infty,0].$ Hence, Theorem 3.1 does not hold.

The following example shows that non-smooth Abadie’s constraint qualification in Theorem 3.1 cannot be omitted.

Example 3.3**.**

Let $g_{1},g_{2}:\mathbb{R}\longrightarrow\mathbb{R}$ be defined by

[TABLE]

and

[TABLE]

Thus, we have

[TABLE]

which is closed. Let $C:=[0,1],$ $\tilde{K}:=C\cap K=\{0\}$ and $\bar{x}:=0.$ It is easy to check that $g_{1}$ and $g_{2}$ are tangentially convex at $\bar{x},$ $g_{1}(\bar{x})=g_{2}(\bar{x})=0,$ $\partial_{T}g_{1}(\bar{x})=\{0\}$ and $\partial_{T}g_{2}(\bar{x})=\{0\}.$ Also, one can see that

[TABLE]

Therefore, $D(\bar{x})\nsubseteq T_{\tilde{K}}(\bar{x}),$ and hence, non-smooth Abadie’s constraint qualification does not hold at $\bar{x},$ while $K$ is nearly convex at the point $\bar{x}.$

Furthermore, it is easy to see that ${M}(\bar{x})=\{0\},$ $(\tilde{K}-\bar{x})^{\circ}=\mathbb{R}$ and $(K-\bar{x})^{\circ}=[0,+\infty),$ and so, Theorem 3.1 does not hold.

4 Characterizations of Constrained Best Approximation

In this section, we give characterizations of constrained best approximations under non-smooth Abadie’s constraint qualification. Let $K$ be as in (2.4), given by,

[TABLE]

where $g_{j}:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ $(j=1,2,\cdots,m)$ is a tangentially convex function at a given point $\bar{x}\in K.$ Let $S:=\mathbb{R}_{+}^{m},$ and let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset.$ Note that $K$ is not necessarily a closed or a convex set.

The following theorem shows that under non-smooth Abadie’s constraint qualification the ”perturbation property” is characterized by the strong conical hull intersection property (Strong CHIP).

Theorem 4.1**.**

*Let $K$ be closed, given by $(\ref{4.1}),$ and let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset.$ Let $\bar{x}\in\tilde{K}:=C\cap K.$ Assume that $\tilde{K}$ is closed and convex. If $K$ is nearly convex at the point $\bar{x}$ and non-smooth Abadie’s constraint qualification holds at $\bar{x},$ then the following assertions are equivalent.

$(i)$ $\{C,K\}$ has the strong CHIP at $\bar{x},$

$(ii)$ For any $x\in\mathbb{R}^{n},$ $\bar{x}=P_{\tilde{K}}(x)$ if and only if there exist $(\lambda_{1},\lambda_{2},\cdots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ $j=1,2,\cdots,m,$ and*

[TABLE]

Proof:  $[(i)\Longrightarrow(ii)].$ Suppose that $(i)$ holds. Then, by Definition 2.4,

[TABLE]

Also, in view of the hypotheses and Theorem 3.1, we have $M(\bar{x})=(K-\bar{x})^{\circ}.$ So, it follows from (4.34) that

[TABLE]

Now, for any $x\in\mathbb{R}^{n},$ assume that $\bar{x}=P_{\tilde{K}}(x).$ Thus, by Lemma 2.2, one has $x-\bar{x}\in(\tilde{K}-\bar{x})^{\circ}.$ Therefore, in view of (3.15) and (4.35), there exist $\ell\in(C-\bar{x})^{\circ}$ and $(\lambda_{1},\lambda_{2},\cdots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0$ $(j=1,2,\cdots,m)$ such that

[TABLE]

So, for each $j=1,2,\cdots,m,$ there exists $\eta_{j}\in\partial_{T}g_{j}(\bar{x})$ such that

[TABLE]

Then, we conclude that

[TABLE]

for some $(\lambda_{1},\lambda_{2},\cdots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ and some $\eta_{j}\in\partial_{T}g_{j}(\bar{x}),\ j=1,2,\cdots,m.$ Frow now on, by an argument similar to the proof of Theorem 4.1 $($the implication $[(i)\Longrightarrow(ii)])$ in [11] and using Theorem 3.1 the result follows.

$[(ii)\Longrightarrow(i)].$ Let $y\in(\tilde{K}-\bar{x})^{\circ}$ be arbitrary, and let $x:=\bar{x}+y.$ Thus, by Lemma 2.2, $\bar{x}=P_{\tilde{K}}(x).$ Then, in view of the hypothesis $(ii),$ there exist $(\lambda_{1},\lambda_{2},\cdots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ $j=1,2,\cdots,m,$ and

[TABLE]

Again, by Lemma 2.2,

[TABLE]

for some $(\lambda_{1},\lambda_{2},\cdots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ and some $\eta_{j}\in\partial_{T}g_{j}(\bar{x}),\ j=1,2,\cdots,m.$ Now, by an argument similar to the proof of Theorem 4.1 $($the implication $[(ii)\Longrightarrow(i)])$ in [11] and using Theorem 3.1 the proof is completed.

The following examples illustrate Theorem 4.1. Moreover, these examples justify how one can use best approximations to check the strong CHIP without explicitly proving the strong CHIP property.

Example 4.1**.**

Let $g_{1},g_{2}:\mathbb{R}^{2}\longrightarrow\mathbb{R}$ be defined by

[TABLE]

Let $S:=\mathbb{R}_{+}^{2}$ and $C:=\mathbb{R}^{2}.$ It is easy to see that

[TABLE]

which is closed and convex. Let $\tilde{K}:=C\cap K=K$ and $\bar{x}:=(0,0)\in\tilde{K}.$ Note that $\tilde{K}$ is closed and convex. It is clear that $g_{1}$ and $g_{2}$ are tangentially convex at $\bar{x},$ but not convex. Moreover, $g_{1}(\bar{x})=g_{2}(\bar{x})=0,$ and

[TABLE]

Therefore, we have

[TABLE]

So, it is easy to check that

[TABLE]

and hence, Abadie’s constraint qualification holds at $\bar{x}.$ Also, it is clear that $K$ is nearly convex at the point $\bar{x}.$

Now, for any $x:=(0,x_{2})\in\mathbb{R}^{2}$ with $x_{2}\leq 0,$ it is easy to see that

[TABLE]

*where $(\lambda_{1}:=0,\lambda_{2}:=-x_{2})\in S,$ $\lambda_{j}g_{j}(\bar{x})=0$ $(j=1,2),$ and $\eta_{1}:=(-1,1)\in\partial_{T}g_{1}(\bar{x}),$ $\eta_{2}:=(0,-1)\in\partial_{T}g_{2}(\bar{x}).$ Then, in view of Theorem 4.1 *$($the implication $[(ii)\Longrightarrow(i)]),$ we conclude that $\{C,K\}$ has the strong CHIP at $\bar{x}.$ Indeed, one can see

[TABLE]

Example 4.2**.**

Let $g_{1},g_{2}:\mathbb{R}\longrightarrow\mathbb{R}$ be defined by

[TABLE]

Let $S:=\mathbb{R}_{+}^{2}$ and $C:=[1,+\infty).$ Clearly, we have

[TABLE]

which is closed and convex. Let $\tilde{K}:=C\cap K=[1,3]$ and $\bar{x}:=1\in\tilde{K}.$ Thus, $\tilde{K}$ is closed and convex, $g_{1},g_{2}$ are tangentially convex at $\bar{x}$ $($but not convex$),$ $g_{1}(\bar{x})=0,$ $g_{2}(\bar{x})=-4\neq 0,$ and

[TABLE]

This implies that

[TABLE]

Moreover, it is not difficult to show that

[TABLE]

and so, non-smooth Abadie’s constraint qualification holds at $\bar{x}.$ Note that $K$ is nearly convex at the point $\bar{x}.$ It is easy to see that, for any $x\in\mathbb{R}$ with $x\leq 1,$ we have

[TABLE]

where $(\lambda_{1}:=\frac{1-x}{3},\lambda_{2}:=0)\in S,$ $\lambda_{j}g_{j}(\bar{x})=0$ $(j=1,2),$ and $\eta_{1}:=-3\in\partial_{T}g_{1}(\bar{x}),$ $\eta_{2}:=-2\in\partial_{T}g_{2}(\bar{x}).$ Hence, by using Theorem 4.1 $($the implication $[(ii)\Longrightarrow(i)]),$ we conclude that $\{C,K\}$ has the strong CHIP at $\bar{x}.$ Indeed, one can see

[TABLE]

Now, let $x\in\mathbb{R}^{n}$ be fixed, and define the function $h:\mathbb{R}^{n}\longrightarrow[0,+\infty)$ by

[TABLE]

For $\bar{x}\in\mathbb{R}^{n},$ we recall that $\partial h(\bar{x}):=\partial\|\cdot-x\|(\bar{x})$ is given by

[TABLE]

In the following, we give the Lagrange multipliers characterization of constrained best approximation under non-smooth Abadie’s constraint qualification.

Theorem 4.2**.**

*Let $K$ be closed, given by $(\ref{4.1}),$ and let $C$ be a non-empty closed convex subset of $\mathbb{R}^{n}$ such that $C\cap K\neq\emptyset.$ Let $\bar{x}\in\tilde{K}:=C\cap K$ and $x\in\mathbb{R}^{n}.$ Assume that $\tilde{K}$ is closed and convex. If $K$ is nearly convex at the point $\bar{x}$ and non-smooth Abadie’s constraint qualification holds at $\bar{x},$ then the following assertions are equivalent.

$(i)$ $\bar{x}=P_{\tilde{K}}(x).$

$(ii)$ There exist $(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ $j=1,2,\cdots,m,$ and*

[TABLE]

$(iii)$* There exist $(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ $j=1,2,\cdots,m,$ and $\eta_{j}\in\partial_{T}g_{j}(\bar{x})\ (j=1,2,\ldots,m)$ such that*

[TABLE]

where we denote $\partial f(x_{0})$ for the convex subdifferential of a convex function $f:\mathbb{R}^{n}\longrightarrow\mathbb{R}$ at the point $x_{0}\in\mathbb{R}^{n}.$

Proof:  $[(i)\Longleftrightarrow(ii)].$ Since, by the hypothesis, non-smooth Abadie’s constraint qualification holds at $\bar{x}$ and $K$ is nearly convex at the point $\bar{x},$ it follows from Corollary 3.1 that $\{C,K\}$ has the strong CHIP at $\bar{x}.$ Therefore, the implication $[(i)\Longleftrightarrow(ii)]$ follows from Theorem 4.1.

$[(i)\Longrightarrow(iii)].$ We may assume without loss of generality that $x\neq\bar{x}.$ Suppose that $(i)$ holds. Then, we have $\bar{x}=P_{\tilde{K}}(x).$ This together with Lemma 2.2 implies that $x-\bar{x}\in(\tilde{K}-\bar{x})^{\circ}.$ But, in view of Theorem 3.1, one has

[TABLE]

Hence, $x-\bar{x}\in M(\bar{x}).$ Since $M(\bar{x})$ is a cone, we conclude that

[TABLE]

Therefore, it follows from (3.15) that there exist $(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ $j=1,2,\cdots,m,$ and $\eta_{j}\in\partial_{T}g_{j}(\bar{x})\ (j=1,2,\ldots,m)$ such that

[TABLE]

where

[TABLE]

Then, $u\in\mathbb{R}^{n},$ $\|u\|=1$ and

[TABLE]

This together with (4.37) implies that

[TABLE]

Thus, it follows from (4.38) and (4.39) that

[TABLE]

for some $(\lambda_{1},\lambda_{2},\cdots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ and some $\eta_{j}\in\partial_{T}g_{j}(\bar{x}),$ $j=1,2,\cdots,m.$ On the other hand, since $0\in(C-\bar{x})^{\circ},$ we have

[TABLE]

Hence, in view of (4.40) there exist $(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ and $\eta_{j}\in\partial_{T}g_{j}(\bar{x}),$ $j=1,2,\cdots,m,$ such that

[TABLE]

which implies $(iii)$ holds.

$[(iii)\Longrightarrow(i)].$ Suppose that $(iii)$ holds. Then there exist $(\lambda_{1},\lambda_{2},\ldots,\lambda_{m})\in S$ with $\lambda_{j}g_{j}(\bar{x})=0,$ $j=1,2,\cdots,m,$ and $\eta_{j}\in\partial_{T}g_{j}(\bar{x})\ (j=1,2,\ldots,m)$ such that

[TABLE]

Now, let $y\in\tilde{K}$ be arbitrary. So, $y\in K.$ Since $K$ is nearly convex at the point $\bar{x},$ it follows from Definition 2.1 that there exists a sequence $\{\alpha_{k}\}_{k\geq 1}\subset\mathbb{R}_{++}$ with $\alpha_{k}\longrightarrow 0^{+}$ such that $\bar{x}+\alpha_{k}(y-\bar{x})\in K$ for all sufficiently large $k\in\mathbb{N}.$ So, by (2.4),

[TABLE]

Since $g_{j}$ $(j=1,2,\cdots,m)$ is tangentially convex at $\bar{x},$ it follows from (2.3), (4.41) and (4.42) with some $\nu\in(C-\bar{x})^{\circ}$ that

[TABLE]

Note that in the above we used the fact that $\lambda_{j}=0$ for each $j\notin I(\bar{x}),$ because $\lambda_{j}g_{j}(\bar{x})=0$ for all $j=1,2,\ldots,m.$ Therefore, we conclude from (4.43) that $\|\bar{x}-x\|=\inf_{y\in\tilde{K}}\|y-x\|=d(x,\tilde{K}),$ and so, $\bar{x}=P_{\tilde{K}}(x),$ i.e., $(i)$ holds.

Acknowledgements. This research was partially supported by Mahani Mathematical Research Center, grant no. 97/3267.