DOI 10.1007/s11590-015-0900-9 O R I G I NA L PA P E R
Characterizations of solution sets of cone-constrained convex programming problems
Xin-He Miao1 · Jein-Shan Chen2
Received: 18 September 2013 / Accepted: 1 May 2015 / Published online: 12 May 2015
© Springer-Verlag Berlin Heidelberg 2015
Abstract In this paper, we consider a type of cone-constrained convex program in finite-dimensional space, and are interested in characterization of the solution set of this convex program with the help of the Lagrange multiplier. We establish necessary conditions for a feasible point being an optimal solution. Moreover, some necessary conditions and sufficient conditions are established which simplifies the corresponding results in Jeyakumar et al. (J Optim Theory Appl 123(1), 83–103,2004). In particular, when the cone reduces to three specific cones, that is, the p-order cone, Lpcone and circular cone, we show that the obtained results can be achieved by easier ways by exploiting the special structure of those three cones.
Keywords Convex programs· Lagrange multipliers · K-Convex mapping · Normal cone· KKT conditions
The author’s work is supported by Ministry of Science and Technology, Taiwan.
X.-H. Miao is supported by National Young Natural Science Foundation (No. 11101302) and National Natural Science Foundation of China (No. 11471241).
B
Jein-Shan Chen [email protected] Xin-He Miao1 Department of Mathematics, School of Science, Tianjin University, Tianjin 300072, People’s Republic of China
2 Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan
1 Introduction
Consider the cone-constrained convex programming problem as follows:
min f(x) s.t. Ax = b,
−g(x) ∈ K, (1)
where A ∈ Rm×n, b ∈ Rm,K is a closed convex cone in Rr, f : Rn → R is a convex function, and g: Rn→ Rr is a continuousK-convex mapping which means for every x, y ∈ Rnand each t∈ [0, 1],
tg(x) + (1 − t)g(y) − g (tx + (1 − t)y) ∈ K.
One important issue for such optimization problem is to characterize the solution set which is also a fundamental topic for many mathematical programming problems. With the help of the characterization of the solution set, we will have a deeper understanding for several important optimization problems including bi-level programming, goal programming and multiple objective programming, and so on. Moreover, it is also essential for understanding the behavior of solution methods for solving mathematical programming problems, see [3,8,13,14]. This is the main motivation to investigate characterizations of the solution set of optimization problems.
In [14], Mangasarian provides a characterization of the solution set of a convex programming problem with differentiable functions. Subsequently, Burke and Ferris [3] present another more specific characterization for the solution set. Recently, char- acterizations of the solution set of problem (1) where g = 0 and f is pseudolinear have been presented in [13]. Jeyakumar et al. [11] describe characterizations of the solution set of a general cone-constrained convex programming problem. Wu and Wu [16] characterize the solution set of a general convex program on a normed vector space.
For the problem (1), the purpose of this paper is to characterize its solution set (see Theorem3.3) which simplifies the conclusions in [11]. Moreover, whenK reduces to p-order cone, Lpcone or circular cone, the obtained characterizations can be reached by other ways via exploiting the special structures of these three specific cones.
Finally, we say a few words about notations which will be used in this paper.
LetR denote the space of real numbers, R+(R++) denote the set consisting of the nonnegative (positive) reals, andRnmean the n-dimensional real vector space. For the setK ⊆ Rn, intK denotes the interior of the set K and ∂K denotes the boundary ofK. Moreover, we write B(x, ε) to mean the open sphere with center x ∈ Rn and radiusε > 0. For the function f : Rn→ R, the convex subdifferential of the function f at x ∈ Rnis denoted by∂ f (x). We denote by x the 2-norm of x which induced by the inner product·.·, i.e., x =√
x, x, where x, y means the inner product of x and y. We usexpto mean the p-norm of x with 1≤ p < ∞ which is defined asxp= (n
i=1|xi|p)1p for any x := (x1, x2, . . . , xn)T ∈ Rn.
2 Preliminaries
In this section, we briefly review some basic concepts and background materials about three specific closed convex cones, which will be extensively used in subsequent analysis. More details can be found in [6,9,10,17].
For problem (1), let F and S be the feasible set and the solution set, respectively, that is,
F :=
x∈ Rn| Ax = b, −g(x) ∈ K and
S := {x ∈ F | f (x) ≤ f (y), ∀y ∈ F}.
The subdifferential of the function f at x is defined as
∂ f (x) :=
ξ ∈ Rn| f (y) − f (x) ≥ ξ, y − x, ∀y ∈ Rn .
If C is a convex set, the normal coneNC(x) of C at x ∈ C is defined by NC(x) :=
ξ ∈ Rn| ξ, y − x ≤ 0, ∀y ∈ C .
It is well known that the subdifferential of the indicator function associated with the convex set C at x ∈ C is the normal cone NC(x). Moreover, if the convex set C is the special convex set C = {x ∈ Rn| Ax = b}, it is easy to verify that, for any x ∈ C, the normal coneNC(x) of C at x is
NC(x) =
ATy| y ∈ Rm .
In other words, the normal coneNC(x) is the range space of AT.
From the convexity of the function f , we know that the function f is continuous.
Since g is a continuousK-convex mapping again, it follows that the problem (1) is a convex optimization problem. If problem (1) satisfies the Slater condition [12], that is, there exists ¯x ∈ Rnsuch that A¯x = b and −g( ¯x) ∈ int K, it is known that a ∈ S if and only if the element a satisfies the KKT conditions, i.e., a∈ F and there exists a Lagrange multiplierλa ∈ Rr such that
0∈ ∂ f (a) + ∂ λTag
(a) +
ATy| y ∈ Rm
, λa ∈ K∗ and λaTg(a) = 0,
whereK∗denotes the dual cone ofK given by K∗=
z∈ Rr| z, x ≥ 0, ∀x ∈ K .
For problem (1), we shall assume throughout that the solution set S is nonempty.
Let a ∈ S. By above analysis, there exists the corresponding Lagrange multiplier λ
such that(a, λa) satisfying the KKT conditions. More specifically, we consider the Lagrange function La(·, λa) : Rn→ R defined by
La(x, λa) := f (x) + λTag(x) for all x ∈ Rn.
From f being convex and g beingK-convex, it follows that for all x, y ∈ Rnand each β ∈ [0, 1],
La(βx + (1 − β)y, λa) = f (βx + (1 − β)y) + λTag(βx + (1 − β)y)
≤ β f (x) + (1 − β) f (y) + βλaTg(x) + (1 − β)λTag(y)
= βLa(x, λa) + (1 − β)La(y, λa).
This demonstrates that the function La(·, λa) is also a convex function.
Next, we review the concepts of three specific closed convex cones and their dual cones.
(1) p-order cone, see [1]. It is a generalization of the second-order cone [4,5,15]
and expressed as follows:
Kp:=
⎧⎨
⎩x ∈ Rn x1≥
n
i=2
|xi|p
1p⎫
⎬
⎭ , (1< p < ∞).
If we write x := (x1, ¯x) ∈ R×R(n−1)with¯x := (x2, . . . , xn)T ∈ R(n−1), the p-order coneKpcan be expressed as
Kp=
x∈ Rn| x1≥ ¯xp
, (1 < p < ∞).
Indeed,Kpis a solid (i.e., intKp= ∅), closed and convex cone, and its dual cone is given by
K∗p=
⎧⎨
⎩y∈ Rn y1≥
n
i=2
|yi|q
1
q
⎫⎬
⎭
or equivalently
K∗p=
y= (y1, ¯y) ∈ R × R(n−1)| y1≥ ¯yq
= Kq,
where q satisfies the condition q > 1 and 1p +1q = 1, and ¯y := (y2, y3, . . . , yn)T ∈ R(n−1). Note that the dual coneK∗pis also a convex cone.
(2) Lpcone, see [10]. Let n∈ N and p := (p1, p2, . . . , pn)T ∈ Rnwith pi > 1.
The Lpcone is defined by
Lp:=
(z, θ, k) ∈ Rn× R+× R+ n
i=1
|zi|pi piθpi−1 ≤ k
,
where|zi|/0 := ∞ if zi = 0; 0 if zi = 0. As shown in [10], we know that Lpis a solid, closed and convex cone, and its dual cone is the switched cone Lqs given by
(w, h, φ) ∈ Lqs ⇐⇒ (w, φ, h) ∈ Lq, where q:= (q1, q2, . . . , qn)T ∈ Rn++such that p1
i +q1i = 1 for each i.
(3) The circular cone Lθ, see [7,18]. The circular cone Lθis defined as follows:
Lθ :=
x= (x1, ¯x) ∈ R × R(n−1)| x cos θ ≤ x1
,
whereθ ∈ (0,π2). Again, as shown in [7,18], we know that Lθ is a solid, closed and convex cone, and its dual cone L∗θis given by
L∗θ =
z= (z1, ¯z) ∈ R × R(n−1)
zcosπ 2 − θ
≤ z1
= Lπ2−θ.
By direct calculation or reference to [18], the circular cone Lθ and its dual cone L∗θ can also be expressed as follows, respectively,
Lθ =
x= (x1, ¯x) ∈ R × R(n−1)| ¯x cot θ ≤ x1
and
L∗θ =
z= (z1, ¯z) ∈ R × R(n−1)| ¯z tan θ ≤ z1
.
Remark 2.1 (a) When p = 2, Kp is exactly the second-order cone which says that p-order cone is a generalization of the second-order cone.
(b) When pi = 2 for all i, we have that the Lp cone is the hyperbolic or rotated second-order cone which is a transformation of the standard second-order cone.
(c) Clearly, the circular cone Lθ includes second-order cone as a special case when the rotation angle is 45◦.
3 Characterizations of solution set with Lagrange multiplier
In this section, we will establish some results which characterizes the solution set of problem (1) in terms of Lagrange multiplier of a solution and subgradients of Lagrange function for the problem (1). we first show a necessary condition for the solution set of problem (1). Then, whenK reduces to the aforementioned specific cones, we show
the same results can be obtained by other ways by exploiting the special structure of those three cones.
Theorem 3.1 For problem (1), let a∈ S. Suppose that the corresponding Lagrange multiplierλa∈ Rr satisfies the conditions:
0∈ ∂ La(a, λa) + {ATy| y ∈ Rm}, λa∈ K∗, and λTag(a) = 0. (2)
Then, the following hold.
(a) Ifλa= 0, then, for each x ∈ S, there exists y ∈ Rm such that
−ATy∈ ∂ f (x).
(b) Ifλa= 0, then, for each x ∈ S and g(x) = 0, we have
−g(x) ∈ ∂K, λa∈ ∂K∗, and λTag(x) = 0.
Proof (a) When λa = 0, since the Lagrange multiplier λa satisfies the condition 0∈ ∂ La(a, λa) + {ATy| y ∈ Rm}, there exists y ∈ Rm such that
−ATy∈ ∂ La(a, λa) = ∂ f (a) + ∂(λTag)(a) = ∂ f (a).
Applying the properties of convex functions, for each x ∈ S and every z ∈ Rn, it follows that
f(z) − f (x) = f (z) − f (a)
≥ −(ATy)T(z − a)
= −(ATy)T(z − x + x − a)
= −(ATy)T(z − x) − (ATy)T(x − a)
= −(ATy)T(z − x),
where the first and last equalities respectively follow from f(x) = f (a) and Ax = Aa= b due to x, a ∈ S, which implies that −ATy∈ ∂ f (x).
(b) Whenλa= 0, from the conditions (2), i.e.,
0∈ ∂ La(a, λa) +
ATy| y ∈ Rm
, λa∈ K∗, and λTag(a) = 0,
we know there exists y ∈ Rmsuch that
−ATy∈ ∂ La(a, λa).
Because f is convex and g isK-convex, the function La(·, λa) is convex as shown earlier in Section 2. Therefore, for every x∈ S, we have
f(x) + λTag(x) = La(x, λa)
≥ La(a, λa) − (ATy)T(x − a)
= La(a, λa)
= f (a) + λTag(a). (3)
This together with f(x) = f (a) and λTag(a) = 0 yield λTag(x) ≥ 0. On the other hand, noting that for every x ∈ S, λa ∈ K∗and−g(x) ∈ K, by the definition of the dual coneK∗, we obtain λTag(x) ≤ 0. Hence, this together with λTag(x) ≥ 0 give λaTg(x) = 0.
Next, we argue thatλa ∈ ∂K∗ and−g(x) ∈ ∂K for every x ∈ S and g(x) = 0.
We prove−g(x) ∈ ∂K only. Similar arguments will apply to the case of λa ∈ ∂K∗. Indeed, we will prove it by contradiction. Suppose that−g(x) ∈ int K. Then, there existsε > 0 such that the open ball B(−g(x), ε) ⊂ K. Thus, for any y ∈ Rr, there existsα > 0 such that −g(x) + αy ∈ B(−g(x), ε) ⊂ K, which gives
−g(x) + αy, λa ≥ 0.
Then, it follows from λa, −g(x) = −λTag(x) = 0 that α y, λa ≥ 0. By the arbitrariness of y inRr, we see thatλa = 0. This contradicts the condition λa = 0.
Thus,−g(x) ∈ ∂K.
Remark 3.1 (i) By Theorem3.1, for every x, y ∈ S, we have
f(x) + λaTg(x) = f (a) + λTag(a) = f (y) + λTag(y).
This explains that Lagrange function La(·, λa) is constant on the solution set S of problem (1).
(ii) By Theorem3.1, for every a, x ∈ S, the Lagrange multiplier λa and the vector
−g(x) solve the complementarity problem [9]:
−g(x) ∈ K, λa ∈ K∗, λaTg(x) = 0.
Now, we show that Theorem3.1(b) can be verified by other ways whenK reduces to the p-order cone, Lpcone or the circular cone. We present the three cases as below.
(1) For the case where K is the p-order cone Kp, let 0 = −g(x) := (h1, ¯h) ∈ Kp⊂ R+× Rr−1and 0= λa := (λ1, ¯λ) ∈ Kq⊂ R+× Rr−1. Note that h1> 0 and λ1> 0. By the definitions of Kpand its dual coneKq, we have
h ≥ ¯h and λ ≥ ¯λ .
Hence, it follows fromλTag(x) = 0 that
0= h1λ1+ ¯hT¯λ
≥ ¯hp¯λq+ ¯hT¯λ
≥ 0,
where the last inequality follows by Hölder’s inequality. This leads to h1= ¯hpand λ1= ¯λqdue to h1λ1> 0, which says λa∈ ∂K∗pand−g(x) ∈ ∂Kp.
(2) For the case whereK is the Lp cone, let 0 = −g(x) := (z, θ, k) ∈ Lp ⊂ Rr−2× R+× R+ and 0 = λa := (w, h, φ) ∈ Lqs ⊂ Rr−2× R+× R+. By the definitions of Lpcone and its dual cone Lqs, we obtain that
r−2
i=1
|zi|pi
piθpi−1 ≤ k and
r−2
i=1
|wi|qi qiφqi−1 ≤ h.
We discuss two subcases.
Case 1: θ = 0 or φ = 0. If θ = 0, then by definition, z = 0 follows. Then λaTg(x) = 0 becomes kφ = 0, and hence φ = 0 because −g(x) = (0, 0, k) = 0.
Also,φ = 0 yields w = 0, so that λa = (0, h, 0). Therefore, −g(x) ∈ ∂ Lp and λa∈ ∂ Lqs.
Case 2:θ > 0 and φ > 0. Then, λTag(x) = 0 that 0= zTw + θh + kφ
≥ zTw + θ
r−2
i=1
|wi|qi qiφqi−1 + φ
r−2
i=1
|zi|pi piθpi−1
= zTw + θφ
r−2
i=1
1 qi|wi
φ|qi + 1 pi|zi
θ|pi
≥ zTw + θφ
r−2
i=1
wi
φ · zi
θ
≥ zTw −
r−2
i=1
wizi
= 0,
where the second inequality follows from Young’s inequality. This implies r−2 i=1
|zi|pi
piθpi −1 = k andr−2
i=1 |wi|qi
qiφqi −1 = h, which says λa∈ ∂ Lqs and−g(x) ∈ ∂ Lp.
(3) For the case whereK is the circular cone Lθ, let 0= −g(x) := (h1, ¯h) ∈ Lθ ⊂ R+× Rr−1and 0= λa := (λ1, ¯λ) ∈ L∗θ ⊂ R+× Rr−1. By the expressions of the circular cone Lθand its dual cone L+θ, we have
¯h cot θ ≤ h1 and ¯λ tan θ ≤ λ1.
Then, fromλaTg(x) = 0, we have
0= h1λ1+ ¯hT¯λ
≥ ¯h cot θ · ¯λ tan θ + ¯hT¯λ
= ¯h¯λ + ¯hT¯λ
≥ 0.
This leads to ¯h cot θ = h1and¯λ tan θ = λ1, which yieldsλa∈ ∂ L∗θand−g(x) ∈
∂ Lθ.
From [16, Theorem 3.1], we have the following theorem which will give the form of the solution set of problem (1) in terms of subgradients.
Theorem 3.2 For problem (1), let a∈ S. Then
S = {x ∈ F | ξ, x − a = 0, ∃ ξ ∈ ∂ f (x) ∩ ∂ f (a)}
= {x ∈ F | ξ, x − a = 0, ∃ ξ ∈ ∂ f (x)}
= {x ∈ F | ξ, x − a ≤ 0, ∃ ξ ∈ ∂ f (x)}.
Proof Let C1, C5and C6be the following sets, respectively,
C1:= {x ∈ F | ξ, x − a = 0, ∃ ξ ∈ ∂ f (x) ∩ ∂ f (a)}, C5:= {x ∈ F | ξ, x − a = 0, ∃ ξ ∈ ∂ f (x)}
and
C6:= {x ∈ F | ξ, x − a ≤ 0, ∃ ξ ∈ ∂ f (x)}.
Then, the sets C1,C5and C6correspond to those in [16, Theorem 3.1], from which
the results follow immediately.
Theorem 3.3 For problem (1), let a∈ S and let λabe the corresponding Lagrange multiplier satisfying the conditions:
0∈ ∂ La(a, λa) +
ATy| y ∈ Rm
, λa∈ K∗, and λTag(a) = 0.
(a) Ifλa= 0, then
S=
x∈ F | ∂ f (a) ∩
−ATy| y ∈ Rm
= ∂ f (x) ∩
−ATy| y ∈ Rm
.
(b) Ifλa= 0, then
S=
x∈ F | λTag(x) = 0, 0 ∈ ∂ La(a, λa) ∩
−ATy| y ∈ Rm
.
Proof (a) For convenience, we denote
¯S =
x ∈ F | ∂ f (a) ∩
−ATy| y ∈ Rm
= ∂ f (x) ∩
−ATy| y ∈ Rm
. Then, we need to argue that S= ¯S as below.
We first verify the direction S ⊂ ¯S. Let C := {x | Ax = b}. By the analysis of section 2, we know that NC(x) = {ATy | y ∈ Rm}. If λa = 0, it follows that La(x, λa) = f (x). Then, we have ∂ La(x, λa) = ∂ f (x). Hence, for any x ∈ S, by proposition 2.1 of [11] again, it is easy to obtain that
∂ f (a) ∩
−ATy| y ∈ Rm
= ∂ f (x) ∩
−ATy| y ∈ Rm . This yields S⊂ ¯S.
Conversely, let x ∈ ¯S. Then, we know that x ∈ F and
∂ f (a) ∩
−ATy| y ∈ Rm
= ∂ f (x) ∩
−ATy| y ∈ Rm .
Since a∈ S and its corresponding Lagrange multiplier λasatisfy the condition 0∈ ∂ La(a, λa) +
ATy| y ∈ Rm ,
we have
−ATy∈ ∂ f (a) ∩
−ATy| y ∈ Rm
= ∂ f (x) ∩
−ATy| y ∈ Rm ,
for some y∈ Rm. Then, it is easy to see that−yTA(x − a) = 0. This together with Theorem3.2implies x∈ S. Hence, the conclusion holds.
(b) Letλa= 0. By the Remark 3.1, we know that the Lagrange function La(·, λa) is constant on the solution set S of the problem (1). Hence, for any x ∈ S and a ∈ S, we have La(x, λa) = La(a, λa) and then for each ξ ∈ ∂ La(x, λa) ∩ {−ATy| y ∈ Rm}, there exists y∈ Rm such that−ATy= ξ. Moreover, we get also that
La(x, λa) − La(a, λa) = 0
= La(a, λa) − La(x, λa)
≥ −(ATy)T(a − x)
= −yTA(a − x)
= 0
= −yTA(x − a),
where the fourth equality holds due to a, x ∈ S ⊂ F. This shows that ξ = −ATy∈
∂ La(a, λa) ∩ {−ATy| y ∈ Rm}. Thus, we have
∂ La(x, λa) ∩
−ATy| y ∈ Rm
⊂ ∂ La(a, λa) ∩
−ATy| y ∈ Rm .
Similarly, with the same arguments, we may verify that
∂ La(a, λa) ∩
−ATy| y ∈ Rm
⊂ ∂ La(x, λa) ∩
−ATy| y ∈ Rm .
Therefore,
∂ La(a, λa) ∩
−ATy| y ∈ Rm
= ∂ La(x, λa) ∩
−ATy| y ∈ Rm . Combining with Corollary 2.6 of [11], this implies
S=
x∈ F
λTag(x) = 0, 0 ∈ ∂ La(a, λa) ∩
−ATy| y ∈ Rm
,
which is the desired result.
Remark 3.2 In the setting of the Banach space andK is a closed convex cone, the corresponding conclusions of Theorem3.3have been obtained, see [11, Corollary 2.5]
and [16, Corollary 3.1]. However, in [11, Corollary 2.5], the expression of the solution set ¯S is more complicated than that given in Theorem3.3. Here, we provide a simplified expression for the solution set S.
Example 3.1 Consider the following nonlinear convex programming problem:
min f(x) =
x12+ x22+ x2
s.t. −g(x) =
−x2
x1
∈ Kp,
where x := (x1, x2)T ∈ R2.
Let F and S be the feasible set and the solution set of the considered problem, respectively. For any x= (x1, x2)T ∈ F, we have
f(x) =
x12+ x22+ x2≥ |x2| + x2≥ 0.
Thus, we know that a= (0, 0)T is a solution of the considered problem, i.e., a ∈ S.
Note that
∂ f (a) = {(0, 1)T} + B,
whereB denotes the closed unit ball of Rn, and
∂ f (x) =
⎧⎪
⎨
⎪⎩
⎛
⎝ x1
x2+ x2, x2
x2+ x2+ 1
⎞
⎠
T⎫
⎪⎬
⎪⎭
for any x = a. For the solution a = (0, 0)T ∈ S, it is easy to see that the corresponding Lagrange multiplierλa = (0, 0)T ∈ Kq. Moreover, we also obtain that(0, 0)T ∈
∂ La(a, λa) = ∂ f (a). Therefore, it follows that
(0, 0)T ∈ ∂ f (x) ⇐⇒ x1= 0, x2≤ 0.
With this, we see that the solution set can be simplified as S = {x = (x1, x2)T ∈ R2| x1= 0, x2≤ 0}.
To close this section, combining Theorem3.3, [16, Corollary 3.1] and the contents of [2, page 267], we immediately obtain the following corollary as a special case.
Corollary 3.1 For problem (1), let a ∈ S. If the K-convex mapping g is an identity mapping, i.e., g(x) = x for all x ∈ Rn, then the following hold.
(a) If the solution a ∈ int K, then
S = {x ∈ F | ∂ f (x) = ∂ f (a)}.
(b) If the solution a∈ ∂K, then
S =
x∈ F
∂f (x) ∩
NC1(x) − {λ | λ ∈ K∗, λTx= 0}
= ∂ f (a) ∩
NC1(a) − {λa| λa∈ ∂K∗, λTax= 0} ,
whereNC1(x) = NC1(a) = {ATy| y ∈ Rm} with C1= {x ∈ Rn| Ax = b}.
Acknowledgments The authors are very grateful to the referees for their constructive comments, which have considerably improved the paper.
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