J Optim Theory Appl (2011) 150:52–64DOI 10.1007/s10957-011-9809-3

Closedness of a Convex Cone and Application by Meansof the End Set of a Convex Set

Hui Hu · Qing Wang

Published online: 8 February 2011© Springer Science+Business Media, LLC 2011

Abstract This article presents new conditions that ensure the closedness of a convexcone in terms of the end set and the extent of its generator. The results significantlyextend the classic condition. The new closedness conditions are utilized to obtain asimple formula of the least global error bound and a suitable regularity condition ofthe set containment problem for sublinear functions.

Keywords Convex set · Convex cone · The end set of a convex set · The extent of aconvex set · Convex function · Sublinear function

1 Introduction

This article presents new conditions that ensure the closedness of a convex cone interms of the end set and the extent of its generator. The results significantly extend theclassic condition. Moreover, the generators satisfying the new conditions can generateall closed and convex cones, while the generators satisfying the classic conditioncan only generate pointed cones. The new conditions are utilized to obtain a simpleformula of the least global error bound and a suitable regularity condition of the setcontainment problem for sublinear functions.

Throughout this article, the discussion is in Rn. A cone is assumed to contain the

origin and a sublinear function is assumed to satisfy f (0) = 0.

The authors wish to thank the referees for their comments and suggestions.

H. Hu (�)Department of Mathematical Sciences, Northern Illinois University, DeKalb, IL 60115, USAe-mail: hu@math.niu.edu

Q. WangThe World Gates, Inc., 11545 Parkwoods Circle, Suite B-1, Alpharetta, GA 30005, USAe-mail: qingwangtct@yahoo.com

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Let ‖ ·‖ denote the Euclidean norm and 〈·, ·〉 denote the inner product. Let S, intS,bdS, convS, So, and S� denote the closure, interior, boundary, convex hull, polar,and negative polar of S respectively. Let B(x, r) denote the closed ball centered at x

with radius r and d(x,S) denote the Euclidean distance from x to S.For a convex set C, let R+C denote the convex cone generated by C and C∞

denote the recession cone of C when C is closed. For x ∈ C, let TC(x) and NC(x)

denote the tangent cone and normal cone of C at x respectively. Let ιC be the indica-tor function and σC be the support function of C.

For a proper and convex function f , let domf , f ∗, f̄ , epif , ∂f (x), ∂∞f (x),f ′(x; ·) denote the domain of finiteness, conjugate, closure, epigraph, subdifferential,singular subdifferential, and the directional derivative of f respectively.

2 The End Set and the Extent of a Convex Set

The end set and the extent of a general convex set was introduced and studied in [1].

Definition 2.1 The end set of a nonempty and convex set C is defined by

E[C] := {x ∈ [0,1]C : tx /∈ [0,1]C for all t > 1}, (1)

where [0,1]C := {tc : t ∈ [0,1], c ∈ C} = conv({0} ∪ C).An equivalent definition of the end set was given by [2, Lemma 1.1]:

E[C] := {x ∈ C : tx /∈ C for all t > 1}. (2)

Definition 2.2 The extent of a nonempty and convex set C is defined by

e[C] := inf{‖x‖ : x ∈ E[C]} iff E[C] �= ∅; otherwise, e[C] := ∞.

Intuitively speaking, the end set consists of the points of radial ends, and the extentmeasures the distance from the origin to the end set. Let us see some examples.

Example 2.1 Let C := {x ∈ Rn : ‖x‖ ≤ 1}. Then, E[C] = {x ∈ R

n : ‖x‖ = 1} ande[C] = 1 > 0. Note that in this simple case, 0 ∈ C but e[C] > 0, R+C is closed.

Example 2.2 Let C := {(x, y) : −1 ≤ x ≤ 1,0 ≤ y ≤ 1}. Then,

E[C] = {(x,1) : −1 ≤ x ≤ 1} ∪ {(1, y) : 0 ≤ y ≤ 1} ∪ {(−1, y) : 0 ≤ y ≤ 1},and e[C] = 1 > 0. In this case, 0 ∈ C, 0 �∈ intC, but e[C] > 0. Note that R+C isclosed.

Example 2.3 Let C := {(x, y) : x2 + (y − 1)2 ≤ 1}. Then,

E[C] = {(x, y) �= (0,0) : x2 + (y − 1)2 = 1} and e[C] = 0.

Note that R+C = {(x, y) : y > 0} ∪ {(0,0)} is not closed.

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Because the end set and the extent of a convex set play important roles in ouranalysis and subsequent applications, we summarize some frequently used propertiesbelow [1, 2], where C is a nonempty and convex set in R

n.

Fact 2.1 0 /∈ E[C], and E[C] ∩ intC = ∅.

Fact 2.2 E[C] = ∅ if and only if [0,1]C is a cone; if and only if C is a truncatedcone.

Fact 2.3 E[C] = E[C]. If C is closed, then E[C] ⊆ C.

Fact 2.4 Suppose that there exist k > 0 and x ∈ Rn satisfying 0 �= kx ∈ C. If M :=

sup{t ≥ 0 : tx ∈ [0,1]C} < ∞, then Mx ∈ E[C].

Fact 2.5 Suppose that there exist k > 0 and x ∈ Rn satisfying 0 �= kx ∈ C. If M :=

sup{t ≥ 0 : tx ∈ C} < ∞, then Mx ∈ E[C].

Fact 2.6 If C is a compact and convex set, and there exist k > 0 and x ∈ Rn satis-

fying 0 �= kx ∈ C, then E[C] = {z ∈ [0,1]C : tz �∈ [0,1]C,∀t > 1}, E[C] �= ∅, ande[C] < ∞.

Fact 2.7 If ∅ �= K �= {0} is a closed and convex cone and 0 < r < ∞, then

E[K ∩ B(0, r)] = {x ∈ K : ‖x‖ = r} and e[K ∩ B(0, r)] = r.

Fact 2.8 E[C × R+] = E[C] × R+, and e[C × R+] = e[C].

Fact 2.9 If C �= {0} is the convex hull of finitely many points, then 0 < e[C] < ∞.Moreover, there exists a face F of C such that 0 /∈ F and e[C] ≥ e[F ] > 0.

Facts 2.1, 2.2, 2.3, and 2.8 follow directly from the definition of the end set.Facts 2.4, 2.6, and 2.9 were given by [1, Lemmas 2.2, 2.3, and 2.5] respectively.

To see that Fact 2.5 holds, there exist tj → M with tj x ∈ C such thattj x → Mx ∈ C. By the definition of M , for all t > 1 we have tMx �∈ C. Therefore,Mx ∈ E[C] by the equivalent definition of the end set.

To see that Fact 2.7 holds, let x ∈ E[K ∩ B(0, r)]. Since K ∩ B(0, r) is closed, bythe definition of the end set (2), we know that x ∈ K ∩ B(0, r), and tx /∈ K ∩ B(0, r)

for all t > 1. x ∈ K ∩ B(0, r) implies that x ∈ K and ‖x‖ ≤ r . Because K is a cone,tx ∈ K for all t > 1. Thus, tx �∈ K ∩B(0, r) for all t > 1 implies that tx �∈ B(0, r) forall t > 1, i.e., t‖x‖ > r for all t > 1. Letting t → 1+, we have ‖x‖ ≥ r . Therefore,E[K ∩ B(0, r)] ⊆ {x ∈ K : ‖x‖ = r}. Next, let x ∈ {x ∈ K : ‖x‖ = r}. Then, x ∈K ∩ B(0, r). Because ‖tx‖ = tr > r for all t > 1, tx �∈ K ∩ B(0, r) for all t > 1. Bythe definition of the end set, x ∈ E[K ∩ B(0, r)], i.e., {x ∈ K : ‖x‖ = r} ⊆ E[K ∩B(0, r)].

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3 The Closedness of Convex Cones

This section presents conditions, that guarantee the closedness of a convex cone gen-erated by a closed and convex set C �= ∅. The conditions are given in terms of the endset and the extent of C.

The well known classic theorem for the closedness of a convex cone states [3,Corollary 9.6.1]: If C is convex, compact, and 0 �∈ C, then R+C is closed. The sim-plicity of this result, however, is coupled with restriction. The cases illustrated inExamples 2.1, 2.2, and 2.3 are beyond its applicable scope. The condition e[C] > 0is a much weaker condition. We’ll show that e[C] > 0 is sufficient to ensure theclosedness of R+C when C is compact or C is closed and 0 ∈ C. We start with thecompact case.

Theorem 3.1 If C is a compact and convex set and e[C] > 0, then K := R+C isclosed.

Proof If C = {0}, then E[C] = ∅, e[C] = ∞, and K = {0} is closed. If C �= {0}, byFacts 2.3 and 2.6 we know that ∅ �= E[C] ⊆ C and e[C] is finite. We claim that

[0,1]C = [0,1]E[C]. (3)

Indeed, since E[C] ⊆ C, we have [0,1]E[C] ⊆ [0,1]C. To show [0,1]C ⊆[0,1]E[C], let 0 < s ≤ 1 and 0 �= x ∈ C. Because C is compact,

1 ≤ M := sup{t ≥ 0 : tx ∈ [0,1]C} < ∞.

By Fact 2.4, Mx ∈ E[C], which implies that sx = (sM−1)Mx ∈ [0,1]E[C]. Thus,[0,1]C ⊆ [0,1]E[C]. It follows from (3) that

K = R+C = R+([0,1]C) = R+([0,1]E[C]) = R+E[C]. (4)

Let xj ∈ K and xj → x̄, we show that x̄ ∈ K . By (4), xj = tj cj , where 0 ≤ tj < ∞and cj ∈ E[C] ⊆ C. First, we show that {tj } must be bounded. Suppose that, onthe contrary, {tj } be unbounded. By passing to a subsequence we may assume thattj → ∞. Then, cj = xj/tj → 0 and thus e[C] = 0, which contradicts the assumptionthat e[C] is positive. Therefore, {tj } is bounded. Without loss of generality, we mayassume that tj → t̄ , cj → c̄ ∈ C, and xj → t̄ c̄. If t̄ c̄ = 0, then t̄ c̄ ∈ K . Otherwise,t̄ > 0. By the compactness of C and c̄ ∈ C,

1 ≤ M := sup{t ≥ 0 : t c̄ ∈ [0,1]C} < ∞.

By Fact 2.4, Mc̄ ∈ E[C]. It follows from (4) that,

x̄ = t̄ c̄ = (t̄M−1)Mc̄ ∈ R+E[C] = K.

Consequently, K = R+C is closed. �

If C is a compact and convex set and 0 /∈ C, then E[C] ⊆ C and

e[C] = inf{‖x‖ : x ∈ E[C]} ≥ inf{‖x‖ : x ∈ C} > 0.

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Thus, Theorem 3.1 naturally extends the classic theorem. Note that if C is not com-pact, then (3) may not hold. For example, if C := {(x, y) : y ≥ x2}, then [0,1]C �=[0,1]E[C]. The condition e[C] > 0 given by Theorem 3.1 is also necessary in thefollowing sense.

Theorem 3.2 A convex cone K is closed if and only if there exists a compact andconvex set C with e[C] > 0 such that K = R+C.

Proof If K = {0}, let C = {0}. Then, E[C] = ∅, e[C] = ∞, and K = {0} is closed.If K �= {0}, then K ∩ B(0,1) is compact and convex, and K = R+(K ∩ B(0,1)). Itfollows from Fact 2.7 that E[K ∩B(0,1)] = {x ∈ K : ‖x‖ = 1} and e[K ∩B(0,1)] =1 > 0. The sufficiency has been established by Theorem 3.1. �

Remark 3.1(i) Theorem 3.2 indicates that the class of compact and convex generators capable

of generating all closed and convex cones may be characterized by e[C] > 0.(ii) A compact and convex generator C not containing the origin can only generate

a pointed cone, i.e., K �= {0} is a closed pointed cone if and only if K can be generatedby a compact and convex set not containing the origin. Indeed, let K := R+C, whereC is compact and convex, and 0 �∈ C. If K is not pointed, then there exists 0 �=x ∈ K such that −x ∈ K . Hence, there exist y, z ∈ C, s > 0, t > 0 satisfying x =sy and −x = tz. It follows that (s/(s + t))y + (t/(s + t))z = (x − x)/(s + t) =0 ∈ C, which contradicts the assumption. Thus K must be pointed. Conversely, ifK �= {0} is a closed and pointed cone, then K = R+C, where C = conv(K ∩ {x ∈R

n : ‖x‖ = 1}) is compact [3, Theorem 17.2]. We show that 0 �∈ C. Suppose that,on the contrary, 0 ∈ C. Then, there exist xj ∈ K ∩ {x ∈ R

n : ‖x‖ = 1} and tj > 0for j = 1, . . . ,m with

∑mj=1 tj = 1 such that 0 = t1x

1 + · · · + tmxm. Therefore, 0 �=t1x

1 + · · · + tm−1xm−1 = −tmxm ∈ K ∩ (−K), i.e., K is not pointed.

Next, we extend the result to a possibly unbounded generator. The idea is to con-struct a compact and convex set which generates the same convex cone, the closed-ness of the convex cone then follows from Theorem 3.1. The following is a keylemma.

Lemma 3.1 Let C be a closed and convex set and C �= {0}.(i) For any 0 < r < ∞, R+C = R+[[0,1]C ∩ B(0, r)].

(ii) If 0 < r ≤ e[C] and r < ∞, then E[[0,1]C ∩ B(0, r)] = {x ∈ [0,1]C : ‖x‖ = r}and e[[0,1]C ∩ B(0, r)] = r .

Proof (i) Since [0,1]C ∩ B(0, r) ⊆ [0,1]C, we know that

R+[[0,1]C ∩ B(0, r)] ⊆ R+[0,1]C = R+C.

For any 0 �= x ∈ C, choose 0 < t < min{‖x‖, r}. Then, t‖x‖−1x ∈ [0,1]C ∩ B(0, r),and

x = (t−1‖x‖)(t‖x‖−1x) ∈ R+[[0,1]C ∩ B(0, r)],

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i.e., C ⊆ R+[[0,1]C ∩ B(0, r)], which implies that

R+C ⊆ R+[[0,1]C ∩ B(0, r)].(ii) Let x ∈ [0,1]C and ‖x‖ = r . Since [0,1]C ∩ intB(0, r) �= ∅, we know that

[0,1]C and B(0, r) satisfy the closed intersection property [4, Lemma 2.3], i.e.,

[0,1]C ∩ B(0, r) = [0,1]C ∩ B(0, r). (5)

It follows from (5) that x ∈ [0,1]C ∩ B(0, r) and tx �∈ [0,1]C ∩ B(0, r) for all t > 1.By the definition of the end set (2), we have that x ∈ E[[0,1]C ∩ B(0, r)]. Thus,

{x ∈ [0,1]C : ‖x‖ = r} ⊆ E[[0,1]C ∩ B(0, r)].To show E[[0,1]C ∩ B(0, r)] ⊆ {x ∈ [0,1]C : ‖x‖ = r}, we consider two cases:

[0,1]C is a cone and [0,1]C is not a cone. In the first case, because [0,1]C is a closedcone, by Facts 2.3, 2.7 and (5),

E[[0,1]C ∩ B(0, r)] = E[[0,1]C ∩ B(0, r)]]= E[[0,1]C ∩ B(0, r)] = {x ∈ [0,1]C : ‖x‖ = r}.

In the second case, because [0,1]C is not a cone, by Fact 2.2 we know that E[C] �= ∅.Let x ∈ E[[0,1]C ∩ B(0, r)]. By (5) and the definition of the end set,

x ∈ [0,1]C ∩ B(0, r) = [0,1]C ∩ B(0, r),

thus x ∈ [0,1]C, ‖x‖ ≤ r ; and

tx /∈ [0,1]C ∩ B(0, r), ∀t > 1. (6)

It remains to show that ‖x‖ ≥ r . If x ∈ E[C], then ‖x‖ ≥ e[C] ≥ r by the assump-tion of this lemma. If x /∈ E[C], since x ∈ [0,1]C, there exists t̄ > 1 satisfyingt̄x ∈ [0,1]C. Since [0,1]C is convex, we have tx ∈ [0,1]C for all t ∈ ]1, t̄[. By(6), we must have tx �∈ B(0, r) for all t ∈ ]1, t̄[, i.e., t‖x‖ > r for all t ∈ ]1, t̄[. Let-ting t → 1+, we obtain ‖x‖ ≥ r . Therefore, E[[0,1]C ∩ B(0, r)] = {x ∈ [0,1]C :‖x‖ = r}. Then, e[[0,1]C ∩ B(0, r)] = r follows directly from the definition of theextent. �

Using Lemma 3.1, we can derive a sufficient condition that guarantees the closed-ness of R+C when C is closed. Note that the following theorem is valid whene[C] = ∞.

Theorem 3.3 Suppose that C be a closed and convex set and e[C] > 0. Then, R+C

is closed if and only if [0,1]C is closed. In particular, if 0 ∈ C or C is compact, thenR+C is closed.

Proof ⇐. Without loss of generality, we may assume C �= {0}. Let 0 < r ≤ e[C] andr < ∞. By Lemma 3.1(i),

R+C = R+[[0,1]C ∩ B(0, r)]. (7)

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By Lemma 3.1(ii),

e[[0,1]C ∩ B(0, r)] = r > 0. (8)

Since [0,1]C is closed, [0,1]C ∩ B(0, r) is compact. It follows from (7), (8), andTheorem 3.1 that R+C = R+[[0,1]C ∩ B(0, r)] is closed.⇒. Let tj ∈ [0,1], xj ∈ C, and tj xj → y ∈ [0,1]C. We show that y ∈ [0,1]C wheny �= 0. Since R+C is closed, there exist s ∈ ]0,∞[ and 0 �= x ∈ C satisfying y =sx. If s ≤ 1, then y ∈ [0,1]C. If s > 1, let M := sup{t ≥ 0 : tx ∈ C}. We discusstwo cases, M < ∞ and M = ∞. In the first case of M < ∞, by Facts 2.3 and 2.5we know that Mx ∈ E[C] ⊆ C. Because y = sx = (sM−1)(Mx) ∈ [0,1]C, by thedefinition of the end set (1), we must have sM−1 ≤ 1. Thus, y ∈ [0,1]C and [0,1]Cis closed. In the second case of M = ∞, we can find p > s such that px ∈ C. Letλ = (s − 1)/(p − 1) ∈ ]0,1[. We have y = sx = λpx + (1 − λ)x ∈ C ⊆ [0,1]C.Therefore, [0,1]C is closed.

In particular, if 0 ∈ C, then [0,1]C = C is closed; if C is compact, then [0,1]C iscompact. By the above proof, R+C is closed. �

Remark 3.2 As seen in the proof, the necessity is true in general, e[C] > 0 is onlyneeded by the sufficiency.

In next example, we illustrate that when the condition 0 ∈ C is removed, the con-clusion of Theorem 3.3 may not hold.

Example 3.1 Let C := {(x,1) : −∞ < x < ∞}. C is closed, unbounded, and 0 �∈ C.It is easy to see that E[C] = C and e[C] = 1, but R+C is not closed. In this case,R+C + C∞ is closed, and R+C = R+C + C∞.

It is well known that if C is closed, convex, and 0 �∈ C, then R+C = R+C + C∞[3, Theorem 9.6]. We summarize the above results into the following main theorem,which extends [3, Theorem 9.6] by replacing the condition 0 �∈ C by e[C] > 0.

Theorem 3.4 Suppose that C be a closed and convex set, and e[C] > 0. Then,

(i) R+C = R+C + C∞.(ii) The following statements are equivalent.

(a) R+C is closed.(b) [0,1]C is closed.(c) C∞ ⊆ [0,1]C.(d) C∞ ⊆ R+C.

In particular, if 0 ∈ C or C is compact, then R+C is closed.

Proof (i) If 0 ∈ C, then R+C + C∞ = R+C. By Theorem 3.3, R+C = R+C + C∞.If 0 �∈ C, then R+C = R+C + C∞ holds by [3, Theorem 9.6].(ii) (a) ⇒ (b). By Theorem 3.3.(b) ⇒ (c). C ⊆ [0,1]C implies C∞ ⊆ ([0,1]C)∞. 0 ∈ [0,1]C implies ([0,1]C)∞ ⊆[0,1]C. Thus, C∞ ⊆ [0,1]C.(c) ⇒ (d). C∞ ⊆ [0,1]C ⊆ R+C.(d) ⇒ (a). By (i), R+C = R+C + C∞ ⊆ R+C + R+C = R+C. �

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Remark 3.3 The implication (a) ⇒ (b) ⇒ (c) ⇒ (d) is true in general without as-suming e[C] > 0 (see Remark 3.2). Thus e[C] > 0 serves as a qualification conditionunder which these necessary conditions become sufficient.

4 Global Error Bounds for Sublinear Functions

Let f be a proper and convex function, and S := {x ∈ Rn : f (x) ≤ 0} �= ∅. f ≤ 0

is said to have a global error bound if there exists 0 < τ < ∞ such that d(x,S) ≤τf+(x) for all x ∈ R

n, where f+(x) = max{0, f (x)}.Suppose that S be closed. f ≤ 0 is said to satisfy the extended BCQ at x ∈ bd(S)

iff NS(x) = R+∂f (x) + ∂∞f (x) [5]. f ≤ 0 is said to satisfy the weak BCQ at x ∈ S

iff NS(x) ⊆ R+∂f (x) + ∂∞f (x) [2].Let C be a convex set and ∅ �= P ⊆ C. P is said to have the segment extension

property with respect to C (P is an SE subset of C), if for any x ∈ C, there existp ∈ P , c ∈ C, and λ ∈ ]0,1[ such that x = (1 − λ)p + λc [2]. It is easy to see that ifK is a convex cone, then {0} is an SE subset of K .

If f is a sublinear and lower semicontinuous (l.s.c.) function, we can characterizethe existence of a global error bound and find the smallest global error bound in termsof e[∂f (0)]. Note that ∂f (0) �= ∅. Indeed, if f is a sublinear and l.s.c. function, thenf (x) = σ∂f (0)(x) [3, Corollary 13.2.1]. Thus f ∗(x) = ι∂f (0)(x). Because f is proper,f ∗ is proper [3, Theorem 12.2]. Therefore, ∂f (0) = domf ∗ �= ∅.

Theorem 4.1 Suppose that f be a sublinear and l.s.c. function, K := {x : f (x) ≤ 0}and 0 < τ < ∞. Then, d(x,K) ≤ τf+(x) for all x ∈ R

n if and only if e[∂f (0)] ≥ τ−1.Moreover, if e[∂f (0)] < ∞, then (e[∂f (0)])−1 is the smallest global error bound; ife[∂f (0)] = ∞, then every τ ∈]0,∞[ is a global error bound.

Proof Since K is a convex cone, {0} is an SE subset of K . By [2, Theorem 4.3], thefollowing statements are equivalent for a proper and convex function

d(x,K) ≤ τf+(x), ∀x ∈ Rn, (9)

NK(0) ⊆ R+∂f (0) + ∂∞f (0) and

inf{‖x‖ : x ∈ E[∂f (0)] ∩ NK(0)} ≥ τ−1.(10)

Because f is sublinear, we can simplify (10). Indeed, since f (0) = 0, ∂f (0) ⊆NK(0). Since ∂f (0) �= ∅, we know that ∂f (0) = ∂f (0) + λ∂∞f (0) for all λ ∈ R+.Thus,

R+∂f (0) + ∂∞f (0) ⊆ NK(0).

Therefore, the weak BCQ in (10) is equivalent to the extended BCQ. Because ∂f (0)

is closed, by Fact 2.3, E[∂f (0)] ⊆ ∂f (0) ⊆ NK(0). Consequently, (10) is equivalentto

NK(0) = R+∂f (0) + ∂∞f (0) and e[∂f (0)] ≥ τ−1. (11)

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Furthermore, since f is sublinear and l.s.c., f (x) = σ∂f (0)(x). Consequently,

f (x) ≤ 0 ⇔ σ∂f (0)(x) ≤ 0 ⇔ 〈x, y〉 ≤ 0, ∀y ∈ ∂f (0) ⇔ x ∈ (∂f (0))�.

Therefore,

K = (∂f (0))� = (R+∂f (0))�. (12)

Since K is a closed and convex cone, by (12) and the Bipolar theorem,

NK(0) = TK(0)� = K� = (R+∂f (0))�� = R+∂f (0). (13)

If e[∂f (0)] ≥ τ−1, then by (13), Theorem 3.4, and (∂f (0))∞ = ∂∞f (0),

NK(0) = R+∂f (0) = R+∂f (0) + ∂∞f (0).

Namely, e[∂f (0)] ≥ τ−1 implies that the extended BCQ holds at 0. This means that(11) is equivalent to e[∂f (0)] ≥ τ−1. Therefore, d(x,K) ≤ τf+(x) for all x ∈ R

n ifand only if e[∂f (0)] ≥ τ−1. By [2, Corollary 4.1], (e[∂f (0)])−1 is the smallest globalerror bound if 0 < e[∂f (0)] < ∞; and every τ ∈ ]0,∞[ is a global error bound ife[∂f (0)] = ∞. �

Now we address the situation that f is not lower semicontinuous. Let

Kf := {x ∈ Rn : f (x) ≤ 0} and Kf̄ := {x ∈ R

n : f̄ (x) ≤ 0}.

The existence of global error bounds for a proper function f can be given in terms ofits closure function f̄ , Kf̄ , and Kf .

Lemma 4.1 If f is a proper and convex function and 0 < τ < ∞, then d(x,Kf ) ≤τf+(x) for all x ∈ R

n if and only if Kf = Kf̄ and d(x,Kf̄ ) ≤ τ f̄+(x) for all x ∈ Rn.

Moreover, the least global error bound for f ≤ 0 is the same as the least global errorbound forf̄ ≤ 0 [6, Theorem 2.1].

Using Lemma 4.1, we can extend Theorem 4.1 to sublinear functions withoutlower semicontinuity.

Theorem 4.2 Suppose that f be sublinear and 0 < τ < ∞. Then, d(x,Kf ) ≤τf+(x) for all x ∈ R

n if and only if Kf = Kf̄ and e[∂f (0)] ≥ τ−1. Moreover, if

e[∂f (0)] < ∞, then (e[∂f (0)])−1 is the smallest global error bound; if e[∂f (0)] =∞, then every 0 < τ < ∞ is a global error bound.

Proof By Lemma 4.1, d(x,Kf ) ≤ τf+(x) for all x ∈ Rn if and only if Kf = Kf̄

and d(x,Kf̄ ) ≤ τ f̄+(x) for all x ∈ Rn. Since f is a sublinear function, f̄ is a sub-

linear and l.s.c. function. Because f (0) = 0, we know that ∂f (0) = ∂f̄ (0). ApplyingTheorem 4.1 to f̄ ≤ 0, we obtain Theorem 4.2. �

J Optim Theory Appl (2011) 150:52–64 61

An important feature of Theorem 4.2 is that the positivity of e[∂f (x)] is onlyrequired at the origin. By this feature, we are able to localize a well known character-ization of global error bounds [7, Theorem 1]: For all x ∈ f −1(0),

f ′(x;h) ≥ τ−1‖h‖, ∀h ∈ NK(x). (14)

Proposition 4.1 Suppose that f be a sublinear and l.s.c. function and τ, r ∈ ]0,∞[.If f ′(x;h) ≥ τ−1‖h‖ ∀h ∈ NK(x) holds for all x ∈ f −1(0) ∩ B(0,2r), then τ is aglobal error bound for f ≤ 0.

Proof Let x ∈ B(0, r) and x /∈ K . There exists x̄ ∈ K such that

d(x,K) = ‖x − x̄‖ ≤ ‖x − 0‖ ≤ r.

Thus, ‖x̄‖ ≤ ‖x̄ − x‖ + ‖x‖ ≤ 2r , i.e., x̄ ∈ B(0,2r). Let v = x − x̄ ∈ NK(x̄). Bythe assumption, f ′(x̄, v) ≥ τ−1‖v‖ = τ−1d(x,K). On the other hand, since f isconvex and f (x̄) = 0, f ′(x̄;v) ≤ f (x̄ + v) − f (x̄) = f (x) ≤ f+(x). It follows thatd(x,K) ≤ τf+(x) for all x ∈ B(0, r), i.e., f ≤ 0 has a local error bound at 0. By [2,Theorem 3.1 and Remark 3.1(ii)], we have e[∂f (0)] ≥ τ−1. By Theorem 4.1, τ is aglobal error bound for f ≤ 0. �

Note that in Proposition 4.1, the condition of (14) must be satisfied in a neighbor-hood of the origin. We use next example to illustrate that (14) holds at 0 ∈ f −1(0)

does not ensure the existence of a global error bound.

Example 4.1 Let C := {(x, y) : x2 + (y − 1)2 ≤ 1} and f (x, y) := σC(x, y). Then,∂f (0,0) = C and e[∂f (0,0)] = 0. Since K = {(x, y) : f (x, y) ≤ 0} = C� = {(0, y) :y ≤ 0}, NK(0,0) = {(x, y) : y ≥ 0}. Let (u, v) ∈ NK(0,0). By [3, Theorem 23.4],f ′((0,0); (u, v)) = σ∂f (0,0)(u, v) ≥ max{ux +vy : (x, y) ∈ {(−1,1), (1,1)}} = |u|+|v| ≥ ‖(u, v)‖. Namely, the condition (14) holds at (0,0) with τ = 1. By Theo-rem 4.1, we know that f ≤ 0 cannot have a global error bound.

5 Regularity Conditions for the Set Containment Problem

In the study of dual conditions characterizing the containment of {x : f (x) ≤ 0}in a polyhedral set, the closedness of R+(epif ∗) serves as a regularity condition[8, Propositions 6.1 and 6.2]. A sufficient condition to ensure the closedness ofR+(epif ∗) is that f is continuous, there exists x0 such that f (x0) < 0, and f ∗(0) <

∞ [8, Proposition 6.2]. This condition, however, cannot be satisfied by any sublinearand l.s.c. function. Indeed, if f is sublinear and l.s.c., then f (x) = σ∂f (0)(x). There-fore, f ∗(0) = ι∂f (0)(0) only has two possible values, 0 and ∞. If there exists x0 suchthat f (x0) < 0, then f ∗(0) = − inff = ∞.

In next theorem, we’ll show that f ∗(0) < ∞ is a necessary condition, and we’llpropose a suitable condition for the closedness of R+(epif ∗) in terms of the extentof ∂f (0).

62 J Optim Theory Appl (2011) 150:52–64

Theorem 5.1 Let f be a proper and convex function.

(i) If R+(epif ∗) is closed, then f ∗(0) < ∞.(ii) If f is sublinear and there exists x0 such that f (x0) < 0, then R+(epif ∗) is

nonclosed.(iii) If f is sublinear, inff ≥ 0, and e[∂f (0)] > 0, then R+(epif ∗) is closed.

Proof (i) If f ∗(0) = ∞, then 0 �∈ domf ∗, and ({0} × R+) ∩ epif ∗ = ∅. Therefore,we have (0,1) /∈ R+(epif ∗). Because f is proper and convex, f ∗ is proper andconvex. Pick any z ∈ domf ∗. For any 0 < t < 1, we have

(1− t)(0,1)+ t (z, f ∗(z)+1) = (tz,1+ tf ∗(z)) = t (z, f ∗(z)+ (1/t)) ∈ R+(epif ∗).

Letting t → 0+, (tz,1+ tf ∗(z)) → (0,1). Since (0,1) �∈ R+(epif ∗), we have shownthat R+(epif ∗) is not closed.

(ii) If f is sublinear and there exists x0 such that f (x0) < 0, then f ∗(0) =− inff = ∞. By (i), R+(epif ∗) is nonclosed.

(iii) inff ≥ 0 if and only if f ∗(0) ≤ 0. Since f is a sublinear function, f ∗(x) =ι∂f (0)(x) ≥ 0 for all x ∈ R

n. Therefore, f ∗(0) = 0 and (0,0) ∈ epif ∗ = ∂f (0) ×R+.By Fact 2.8,

E[epif ∗] = E[∂f (0) × R+] = E[∂f (0)] × R+,

e[epif ∗] = e[∂f (0) × R+] = e[∂f (0)] > 0.

Since epif ∗ is closed and unbounded, (0,0) ∈ epif ∗, and e[epif ∗] > 0, by Theo-rem 3.3 we conclude that R+(epif ∗) is closed. �

The condition, there exists x0 such that f (x0) < 0 and f ∗(0) < ∞, is equivalentto 0 < f ∗(0) < ∞. For a proper and convex function, we give a sufficient conditionthat ensures the closedness of R+(epif ∗).

Theorem 5.2 Suppose that f be a proper and convex function and f ∗(0) ∈ ]0,∞[.Then, R+(epif ∗) is closed if and only if (epif ∗)∞ ⊆ R+(epif ∗).

Proof f ∗(0) > 0 implies that (0,0) /∈ epif ∗. Because f ∗ is a proper, l.s.c., and con-vex function, epif ∗ is closed. Therefore, e[epif ∗] ≥ inf{‖z‖ : z ∈ epif ∗} > 0. Theequivalence follows from Theorem 3.4. �

Remark 5.1 When f is continuous (which is equivalent to f ∗ being 1-coercive [9,II, X. Remark 1.3.10]), it is easy to verify that (epif ∗)∞ ⊆ {0} × R+. Since f ∗(0) ∈]0,∞[, (epif ∗)∞ ⊆ {0} × R+ ⊆ R+(epif ∗) is easily satisfied. Thus, Theorem 5.2extends [8, Proposition 6.2].

6 Concluding Remarks

This article studies the closedness of R+C. The new results significantly extend theclassic condition, and can be applied to unbounded C (e.g., C = epif ∗ in the set

J Optim Theory Appl (2011) 150:52–64 63

containment problem). The least global error bound for g ≤ 0, where g is sublinear,is equal to (e[∂g(0)])−1.

It is natural to look at how the end set may relate to other concepts in optimization,especially to metric regularity and constraint qualification. In [10, 11], a scheme isset up using the image space concepts and techniques [12], and the connection be-tween metric regularity and constraint qualification is studied. It is shown that metricregularity implies GCQ [11, Theorem 4.2].

A stronger connection can be obtained with the help of the end set when all con-straints are convex. Let gi : R

n → R, i = 1, . . . ,m, g := max{gi : i = 1, . . . ,m},G := (g1, . . . , gm), and C := g−1(−R+) = G−1(−R

m+). Let x̄ ∈ bdC, and ε,L bepositive numbers. In the sense of [10, 11] (X = R

n), G is metrically regular (MR)at x̄ with respect to G−1(−R

m+) iff g is MR at x̄ with respect to g−1(−R+); iffd(x,C) ≤ Lg+(x) for all x ∈ B(x̄, ε), i.e, g has a local error bound at x̄.

When g is convex, it is known that [1, Theorems 3.1, 4.1, Corollary 4.1] g isMR at x̄ is equivalent to NC(x) = R+∂g(x) and e[∂g(x)] ≥ L−1 hold for all x ∈B(x̄, ε) ∩ bdC. This can be viewed as: BCQ regulates the directions in ∂g(x) whilee[∂g(x)] ≥ L−1 regulates the magnitude. In this case, e[∂g(x)] ≥ L−1 characterizesthe difference between MR and BCQ.

If gi , i = 1, . . . ,m, are differentiable and convex, then

e[∂g(x)] ≥ L−1, ∀x ∈ B(x̄, ε) ∩ bdC

always holds [1, Theorem 3.2]. In this case, e[∂g(x)] ≥ L−1 is independent of theexistence of Lagrange multipliers. g is MR at x̄ is equivalent to BCQ holds for allx ∈ B(x̄, ε) ∩ bdC [1, Corollary 3.4]. A weaker result was obtained in [13]: g is MRat every point in bdC is equivalent to BCQ holds at every point in bdC. It was anopen question whether or not g is MR at x̄ is equivalent to BCQ holds at x̄. A counterexample was given by [1, Example 3.1].

References

1. Hu, H.: Characterizations of the strong basic constraint qualifications. Math. Oper. Res. 30, 956–965(2005)

2. Hu, H.: Characterizations of local and global error bounds for convex inequalities in Banach spaces.SIAM J. Optim. 18, 309–321 (2007)

3. Rockafellar, R.: Convex Analysis. Princeton University Press, New Jersey (1970)4. Bakan, A., Deutsch, F., Li, W.: Strong CHIP, normality, and linear regularity of convex sets. Trans.

Am. Math. Soc. 357, 3831–3863 (2005)5. Zheng, X., Ng, K.: Metric regularity and constraint qualifications for convex inequalities on Banach

spaces. SIAM J. Optim. 14, 757–772 (2004)6. Hu, H., Wang, Q.: Local and global error bounds for proper functions. Pac. J. Optim. 6, 177–186

(2010)7. Lewis, A., Pang, J.: Error bounds for convex inequality systems. Non-convex Optim. Appl. 27, 75–100

(1997)8. Jeyakumar, V.: Characterizing set containments involving infinite convex constraints and reverse-

convex constraints. SIAM J. Optim. 13, 947–959 (2003)9. Hiriart-Urruty, J.-B., Lemaréchal, C.: Convex Analysis and Minimization Algorithms I, II. Springer,

Berlin (1993)10. Moldovan, A., Pellegrini, L.: On regularity for constrained extremum problems. Part I: Sufficient

optimality conditions. J. Optim. Theory Appl. 142, 147–163 (2009)

64 J Optim Theory Appl (2011) 150:52–64

11. Moldovan, A., Pellegrini, L.: On regularity for constrained extremum problems. Part II: Necessaryoptimality conditions. J. Optim. Theory Appl. 142, 165–183 (2009)

12. Giannessi, F.: Constrained Optimization and Image Space Analysis, Separation of Sets and OptimalityConditions, vol. 1. Springer, New York (2005)

13. Li, W.: Abadie’s constraint qualification, metric regularity, and error bounds for differentiable convexinequalities. SIAM J. Optim. 7, 966–978 (1997)