Numerical Functional Analysis and Optimization, vol. 32, pp. 507-523, 2011
On the Lorentz cone complementarity problems in infinite-dimensional real Hilbert space
Xin-He Miao1
Department of Mathematics School of Science Tianjin University Tianjin 300072, P.R. China
Jein-Shan Chen 2 Department of Mathematics National Taiwan Normal University
Taipei 11677, Taiwan
October 26, 2010
(revised on February 14, 2011)
Abstract In this paper, we consider the Lorentz cone complementarity problems in infinite-dimensional real Hilbert space. We establish several results which are standard and important when dealing with complementarity problems. These include proving the same growth of the Fishcher-Burmeister merit function and the natural residual merit function, investigating property of bounded level sets under mild conditions via different merit functions, and providing global error bounds through the proposed merit functions. Such results are helpful for further designing solution methods for the Lorentz cone complementarity problems in Hilbert space.
Key words. Lorentz cone, FB-function, NR-function, merit function, error bound, R02- property.
1also affiliated with Department of Mathematics, National Taiwan Normal University as a postdoc fellow. E-mail: xinhemiao@tju.edu.cn
2Corresponding author. Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is partially supported by National Science Council of Taiwan. E-mail:
jschen@math.ntnu.edu.tw.
1 Motivation and Introduction
Recently there has been much attention on symmetric cone optimization, see [4, 13, 14, 20, 21] and references therein. The symmetric cone K is intimately related to Euclidean Jordan algebra since it provides an essential toolbox for the analysis. In addition, the symmetric cone has special structure in Euclidean Jordan algebra (V, ◦, h·, ·i) [8, 12], namely, K = {x2 = x ◦ x | x ∈ V}. It is natural to ask what will happen if we go further beyond Euclidean Jordan algebra. In fact, it is known that the class of Euclidean Jordan algebras belongs to the class of JB-algebras [29]. More specifically, a finite-dimensional JB-algebra coincides with a Euclidean Jordan algebra. There is a subclass of JB-algebras called JB-algebra of finite rank which attracts our attention because every JB-algebra of finite rank is direct sums of spin factors and Euclidean Jordan algebras. What is a spin factor? Indeed, a spin factor has form of IR ⊕ H where H is a Hilbert space.
In view of this, we realize that Hilbert space is the very basic structure when we go beyond a Euclidean Jordan algebra. This is the main motivation why we consider the complementarity problems in Hilbert space. We will focus on real Hilbert space for the sake of convenience and reality.
Let H be a real Hilbert space endowed with an inner product h·, ·i, and write the norm induced by h·, ·i as k · k. In general, the set of squared elements in H is no longer self-dual. We will define a Lorentz cone denoted by Ω which is self-dual in next section.
Then, given a bounded continuous function F : H → H, we will focus on the Lorentz cone complementarity problem (CP for short) which is to find an element z ∈ H such that
z ∈ Ω, w = F (z) ∈ Ω, and hz, wi = 0. (1)
Such a problem is a natural extension of symmetric cone complementarity problems (SCCPs) in Euclidean Jordan algebras. In the finite-dimensional space, a well-known approach for solving the SCCPs is merit function method, which reformulates the SCCPs as a global minimization over Euclidean Jordan algebras via a certain merit function [1, 2, 5, 6, 9, 13, 19, 25, 26]. For this approach, it aims to find a smooth function Φ : V × V → IR+ such that
Φ(x, y) = 0 ⇐⇒ x ∈ K, y ∈ K and hx, yi = 0,
where K is the symmetric cone in V. Then, the SCCPs can be expressed as an uncon- strained smooth minimization problem:
min
x∈V Ψ(x) := Φ(x, F (x)),
we call such a Ψ a merit function for the SCCPs. It is well-known that the complemen- tarity function associated with the symmetric cone plays a key role in the development
of merit function methods. For the approach to be effective, the choice of the comple- mentarity function is crucial. Recently, merit function method was extended to solve the Lorentz cone complementarity problems in the setting of infinite-dimensional real Hilbert space, see [7, 22, 27].
In finite-dimensional space, two popular symmetric cone complementarity functions are the Fischer-Burmeister (FB) symmetric cone complementarity function φFB and the natural residual (NR) symmetric cone complementarity function φNR. Moreover, some properties of these two complementarity functions were studied. For example, the globally Lipschitzian continuty [23], strongly semismooth property [2] and the global Lipschitz continuous gradients [18], etc.
In real Hilbert space H, the Fischer−Burmeister (FB) function was introduced in [7, 27] and defined as
φFB(z, w) := (z2+ w2)1/2− (z + w) ∀z, w ∈ H, (2) where z2 and z1/2 will be explained in next section. Let z+ denote the metric projection ΠΩ(z) of z ∈ H onto the Lorentz cone Ω. Then, the NR complementarity function in infinite-dimensional real Hilbert space is given as follows
φNR(z, w) := z − (z − w)+ ∀z, w ∈ H.
When H = IR, for these two complementarity functions, Tseng [25] proved the following important inequality:
(2 −√
2)kφNR(a, b)k ≤ kφFB(a, b)k ≤ (2 +√
2)kφNR(a, b)k. (3) Recently, Bi, Pan and Chen [1] extended this important inequality to the setting of sym- metric cones. Along this direction, we generalize inequality (3) to the setting of Hilbert space. Next, we come to merit function approach for solving Lorentz cone complemen- tarity problems in Hilbert space. To this end, we define ΦFB : H × H → IR as
ΦFB(z, y) := 1
2kφFB(z, y)k2.
Then, solving problem (1) is equivalent to solving the following unconstrained smooth minimization problem:
minz∈HΨFB(z) := ΦFB(z, F (z)) = 1
2kφFB(z, F (z))k2, (4) where Ψ is called a merit function associated with Ω in H. In finite-dimensional space, Bi, Pan and Chen [1] have established the global error bound property of the FB merit function for the SCCPs. There is another kind of merit function which was also widely studied ([3, 14, 28]) in the setting of finite-dimensional space. It is a slight modification
of the merit function studied by Yamashita and Fukushima, i.e., Ψα : IRn× IRn → IR defined by
Ψα(x) := Φα(x, F (x)) = α
2k(x ◦ F (x))+k2+ ΦFB(x, F (x)), (5) where α ≥ 0. When α = 0, Ψα reduces to FB merit function. When α = 1, Chen [3]
established the global error bound property of the merit function Ψα for the SOCCPs;
Liu, Zhang and Wang [14] obtained the global error bound property of the merit function Ψα for the SCCPs. In this paper, we also generalize Ψα to the setting of Hilbert space and explore similar results. Property of bounded level sets will be discussed as well. In particular, it only takes F being R02-function to guarantee property of bounded level set for Ψα. However, it needs much more stronger conditions for ΨFB to hold such property.
All the results established in this paper are standard and important when dealing with complementarity problems, in particular, they are helpful for further designing solution methods for problem (1).
2 Preliminaries
In this section, we briefly introduce some basic concepts in real (infinite-dimensional or finite-dimensional) Hilbert space H, and review some basic materials. These concepts and materials play important roles in subsequent analysis. More details and related results can be found in [7, 16, 17, 27].
Let H be a infinite-dimensional real Hilbert space with the inner product h·, ·i, and e be an unit vector in H (i.e. kek =phe, ei = 1). In [7], Chiang, Pan and Chen considered the following closed convex cone
Ω(e, r) = {z ∈ H | hz, ei ≥ rkzk}, where r ∈ IR and e ∈ H with 0 < r < 1 and kek = 1. Define
hei⊥:= {x ∈ H | hx, ei = 0},
i.e., hei⊥ is the orthogonal complementarity space of e in H. Since H is a Hilbert space, for any element z ∈ H, there are x ∈ hei⊥ and λ ∈ IR such that z = x + λe (in fact, λ = hz, ei). With this, it can be verified that
Ω(e, r) = {z ∈ H | hz, ei ≥ rkzk} =
z = x + λe | λ ≥ r
√1 − r2kxk
. Hence, for the closed convex cone Ω(e, r), when r = √1
2, we can see Ω(e, 1
√2) = Ω∗(e, 1
√2),
where Ω∗(e,√12) is the dual cone of Ω(e,√12), i.e., Ω∗(e,√12) := {z ∈ H |hz, wi ≥ 0, ∀w ∈ Ω(e,√1
2)}. This illustrates that Ω(e,√1
2) is a self-dual closed convex cone. In particular, if H = IRn and e = (1, 0) ∈ IR × IR(n−1), the set Ω(e,√1
2) coincides with the Lorentz cone (also called second-order cone) Kn in IRn. In view of this, Ω(e,√12) is called the Lorentz cone (or second-order cone) in H. Throughout this paper, for the sake of simplicity, we denote Ω := Ω(e,√1
2). For any z ∈ H, we write z 0(z 0) when z ∈ Ω(z ∈ int(Ω)), and z 0(z ≺ 0) denote −z ∈ Ω(−z ∈ int(Ω)).
Now, we come to the Jordan product in H associated with the Lorentz cone Ω. For any elements z, w ∈ H with z = x + λe and w = y + µe, where x, y ∈ hei⊥ and λ, µ ∈ IR, the Jordan product of z and w is defined by
z ◦ w = µx + λy + hz, wie.
Therefore, z2 means z ◦ z for any z ∈ H. From the definition of Jordan product and direct computation, it is not hard to prove the following properties.
Lemma 2.1 (a) For any z = x + λe ∈ H with x ∈ hei⊥ and λ ∈ IR, there have z2 = 2λx + kzk2e ∈ Ω, and hz, zi = hz2, ei = kzk2.
(b) z ◦ w = w ◦ z and z ◦ e = z for any z, w ∈ H.
(c) (z + w) ◦ v = z ◦ v + w ◦ v for all z, w, v ∈ H.
(d) hz, w ◦ vi = hw, z ◦ vi = hv, z ◦ wi for all z, w, v ∈ H.
(e) If z = x+λe ∈ Ω, there exists a unique element z1/2∈ Ω such that z = (z1/2)◦(z1/2) = (z1/2)2. Here
z1/2 =
0 if z = 0, x
2τ + τ e otherwise, where τ = s
λ +pλ2− kxk2
2 .
(f ) For any z = x + λe ∈ H, if λ2− kxk2 6= 0, then z is invertible with respect to the Jordan product, i.e., there is a unique element z−1∈ H such that z ◦ z−1 = e, where
z−1 = 1
λ2− kxk2(−x + λe).
Moreover, z ∈ int(Ω) if and only if z−1 ∈ int(Ω).
For any z ∈ H, z can be expressed as z = x + λe where x ∈ hei⊥ and λ ∈ IR. It is also easy to verify that z can be decomposed as
z = x + λe = (λ + kxk)e1(z) + (λ − kxk)e2(z) := λ1(z)e1(z) + λ2(z)e2(z),
where ei(z) = 12(e + (−1)i+1 xkxk) for i = 1, 2 when x 6= 0, and ei(z) = 12(e + (−1)i+1x) for˜ i = 1, 2 when x = 0 (˜x is any element in hei⊥ satisfying k˜xk = 1 ). Here λ1(z), λ2(z) and e1(z), e2(z) are called the spectral values and the associated spectral vectors of z, respectively. In addition, if z ∈ Ω, we have
z1/2 =p
λ1(z)e1(z) +p
λ2(z)e2(z).
Clearly, when x 6= 0, the factorizations of z and z1/2 are unique. Note that {e1(z), e2(z)}
is called a Jordan frame in the real Hilbert space H.
Associated with every z ∈ H, we define a linear transformation Lz : H → H as follows Lz(w) := z ◦ w for any w ∈ H.
Lz is called the Lyapunov transformation from H to H. It can be seen that Lz ∈ L(H), where L(H) denotes the Banach space of all continuous linear transformation from H to H.
For a convex cone Ω in H, let ΠΩ denote the metric projection onto Ω [10]. For an z ∈ H, z+ := ΠΩ(z) if and only if z+ ∈ Ω and kz − z+k ≤ kz − wk for all w ∈ Ω. This is also equivalent to hz − z+, w − z+i ≤ 0 for any w ∈ Ω. Since Ω is convex cone, we have z+ is unique. Similarly, z− means ΠΩ(−z). Then, we have the following results.
Lemma 2.2 Let z = x + λe ∈ H. Then the following results hold.
(a) If z 0, z+ = z and z−= 0.
(b) If z 0, z+= 0 and z− = z.
(c) If z /∈ Ω and −z /∈ Ω, there have z+ = kxk + λ
2kxk x +kxk + λ
2 e = max{λ + kxk, 0}e1(z) and
z− = λ − kxk
2kxk x +kxk − λ
2 e = max{kxk − λ, 0}e2(z).
(d) For any z ∈ H, we have z = z+− z− and kzk2 = kz+k2+ kz−k2.
(e) For any z ∈ H and w ∈ Ω, we have hz, wi ≤ hz+, wi and k(z + w)+k ≥ kz+k.
(f ) For any z ∈ Ω and w ∈ H with z2− w2 ∈ Ω, we have z − w ∈ Ω.
Proof. These are well-known results for projection and convex cones. 2
Lemma 2.3 [22, Lemma 3.3] Let H be a real Hilbert space and φFB, ΨFB be given as in (2) and (4), respectively. Then, for any z, w ∈ H, we have
4ΦFB(z, w) ≥ 2kφFB(z, w)+k2 ≥ k(−z)+k2+ k(−w)+k2.
To close this section, we talk about other concepts which play important roles in analysis of boundedness of level sets.
Definition 2.1 Let H be a real Hilbert space. For a bounded continuous mapping F : H → H,
(a) F has uniform P∗-property if there exists ρ > 0 such that
maxi=1,2h(z − w) ◦ (F (z) − F (w)), ei(w)i ≥ ρkz − wk2 ∀z, w ∈ H where ei(w) for i = 1, 2 are the spectral vectors of w;
(b) F is called an R01-function if for any sequence {zk} such that kzkk → ∞, (−zk)+
kzkk → 0 and (−F (zk))+
kzkk → 0 as k → ∞, we have
lim inf
k→∞
hzk, F (zk)i kzkk2 > 0;
(c) F is called an R02-function if for any sequence {zk} such that kzkk → ∞, (−zk)+
kzkk → 0 and (−F (zk))+
kzkk → 0 as k → ∞, we have
lim inf
k→∞
λ1(zk◦ F (zk)) kzkk2 > 0.
The above concepts are taken from [3, 14] in the setting of finite-dimensional space.
Every R01-function is R02-function. In fact, R02-function is equivalent to R0-property defined in [24, Definition 3.2]. Besides, we recall the concept of the strong monotonicity for a bounded continuous function F : H → H. That is, we say that F is strongly monotone with modulus µ > 0 if for any z, w ∈ H, there exist a constant µ > 0 such that
hz − w, F (z) − F (w)i ≥ µkz − wk2.
3 The same growth of FB and NR merit function
In this section, we will give our first main result of this paper. To prove it, the following lemmas will be needed.
Lemma 3.1 Let H be a real Hilbert space. For z ∈ H with z = x + λe, define |z| :=
z++ z−, where z+ and z− are the same as in Lemma 2.2. Then, we have (a) hz+, z−i = 0 and z+◦ z− = 0;
(b) |z|2 = |z2|, k|z|k = kzk and |z| = 2z+− z = z + 2z−.
Proof. The proof is obtained by straightforward calculation. thus it is omitted. 2
Lemma 3.2 Let H be a real Hilbert space. For any z, w, u ∈ H, (a) if z 0, w 0 and z w, we have z1/2 w1/2;
(b) if z 0 and 2z2 = w2+ u2, we have z 12(w + u).
Proof. (a) Let p = z1/2, q = w1/2 and r = p − q = x + λe. To prove z1/2 w1/2, we need to verity that λ ≥ kxk. Note that 0 z − w = p2− (p − r)2 = 2p ◦ r − r2. Let e2 = 12(e − kxkx ) when x 6= 0, and e2 = 12(e − ˜x) (˜x is any element in hei⊥ satisfying k˜xk = 1) when x = 0. Then, it follows that
0 ≤ h2p ◦ r − r2, e2i
= h2p ◦ r, e2i − hr2, e2i
= h2p, r ◦ e2i − (λ − kxk)2he2, e2i
= (λ − kxk)h2p, e2i − (λ − kxk)2
2 ,
which implies (λ−kxk)2 2 ≤ (λ − kxk)h2p, e2i. This together with p 0 and e2 0 yields λ − kxk ≥ 0. The desired result follows.
(b) The arguments are similar to those for [1, Lemma 3.1]. For completeness, we present them as follows. Since w2 + u2− 2w ◦ u = (w − u)2 0, together with 2z2 = w2+ u2, this implies that
z2 = 1
2(w2+ u2) 1
4(w2+ u2) + 1
2w ◦ u = 1
4(w + u)2. From part (a) and z 0, we have z 12|w + u| 12(w + u). 2
Theorem 3.1 Let H be a real Hilbert space. For any z, w ∈ H, there holds (2 −√
2)kφNR(z, w)k ≤ kφFB(z, w)k ≤ (2 +√
2)kφNR(z, w)k. (6) Proof. Fix any z, w ∈ H. If z2+ w2 = 0, we have z = w = 0. Hence, the desired result is obvious. If z2 + w2 6= 0, by Lemma 3.1, we obtain
φNR(z, w) = z − (z − w)+ = 1
2[(z + w) − |z − w|].
Furthermore, we know
φFB(z, w) = z + w − (z2+ w2)1/2
= 2(z − (z − w)+) + 2(z − w)+− (z − w) − (z2+ w2)1/2
= 2φNR(z, w) + |z − w| − (z2+ w2)1/2.
Let p(z, w) = |z − w| − (z2 + w2)1/2. In view of triangle inequality, to prove inequality (6), it suffices to verity that
kp(z, w)k ≤ √
2kφNR(z, w)k. (7)
To see this desired result, applying Lemma 3.2 gives
|z − w| + (z + w) 2(z2+ w2)1/2, (8) Hence, we have
2kp(z, w)k2− 4kφNR(z, w)k2
= 2 kz − wk2+ k(z2+ w2)1/2k2− 2|z − w|, (z2+ w2)1/2
−kz + wk2− k|z − w|k2+ 2h|z − w|, z + wi
= 2kz − wk2− 2h|z − w|, 2(z2+ w2)1/2− (z + w)i
= |z − w|, |z − w| + (z + w) − 2(z2+ w2)1/2
≤ 0,
where the inequality holds is due to that |z − w| 0 and (8). This implies that inequality (7) is true. Then, the proof is complete. 2
4 Error bound of merit functions
Error bound is an important concept that indicates how close an arbitrary point is to the solution set of Lorentz cone complementarity problem (1). Thus, an error bound may be used to provide stopping criterion for an iterative method. As below, for FB merit
function ΨFB and merit function Ψα, we draw conclusions about the error bounds for the solution of infinite-dimensional Lorentz cone complementarity problem (1), respectively.
For FB merit function ΨFB, we obtain a result about the error bounds for the solution of problem (1), which is an extension of [13, Theorem 6.3] in SCCP case.
Proposition 4.1 Suppose that F is strongly monotone with modulus µ > 0 and is Lip- schitz continuous with constant l. Then,
1 (√
2 + 1)(2 + l)
pΨFB(z) ≤ kz − z∗k ≤ l + 1 µ(√
2 − 1)
pΨFB(z),
Proof. Fix any z ∈ H, let N (z) = φNR(z, F (z)). Applying Theorem 3.1, we have (3 − 2√
2)kN (z)k2 ≤ ΨFB(z) ≤ (3 + 2√
2)kN (z)k2. (9)
Since N (z) = φNR(z, F (z)) = z −(z −F (z))+, it follows that F (z)−N (z) = (z −F (z))− 0, z − N (z) = (z − F (z))+ 0 and hF (z) − N (z), z − N (z)i = 0. Because z∗ is the unique solution of problem (1), we have
0 ≥ hF (z) − N (z), z − N (z)i − hF (z) − N (z), z∗− N (z∗)i − hF (z∗) − N (z∗), z − N (z)i
= hF (z) − F (z∗) − (N (z) − N (z∗)), z − z∗− (N (z) − N (z∗))i
≥ hF (z) − F (z∗), z − z∗i − hF (z) − F (z∗), N (z) − N (z∗)i − hz − z∗, N (z) − N (z∗)i
≥ µkz − z∗k2− lkz − z∗kkN (z)k − kz − z∗kkN (z)k,
where the last inequality is due to the strong monotonicity and the Lipschitz continuity of F . Thus, we obtain
kz − z∗k ≤ l + 1
µ kN (z)k. (10)
On the other hand,
kN (z)k = kz − (z − F (z))+− (z∗− (z∗− F (z∗))+)k
≤ kz − z∗k + kz − z∗− (F (z) − F (z∗))k
≤ 2kz − z∗k + kF (z) − F (z∗)k (11)
≤ (2 + l)kz − z∗k.
Combining (9), (10) and (11) leads to 1
(√
2 + 1)(2 + l)
pΨFB(z) ≤ kz − z∗k ≤ l + 1 µ(√
2 − 1)
pΨFB(z) which is the desired result. 2
From Proposition 4.1, we see that if we use FB merit function ΨFB to provide er- ror bound for the solution of problem (1), we need the conditions of F being strongly
monotone and Lipschitz continuous. However, to provide error bound for the solution of (1) via Ψα merit function, we may weaken the aforementioned conditions to uniform P∗-property. We will prove this in Proposition 4.2. For this purpose, we present two technical lemmas first.
Lemma 4.1 [15, Lemma 2.2] Let z = x + λe, w = y + µe ∈ H with x, y ∈ hei⊥ and λ, µ ∈ IR. The following two conditions are equivalent:
(a) z 0, w 0, and hz, wi = 0;
(b) z 0, w 0, and z ◦ w = 0.
In each case, z and w operator commute.
Lemma 4.2 [15, Lemma 2.4] For any z, w ∈ Ω, if z and w operator commute, then z ◦ w ∈ Ω.
Proposition 4.2 Let Ψα be defined as in (5) and α > 0. Suppose that F has uniform P∗-property and the CP (1) has a solution z∗. Then, there exists a scalar τ > 0 such that
τ kz − z∗k2 ≤ k(F (z) ◦ z)+k + k(−F (z))+k + k(−z)+k ∀z ∈ H. (12) Moreover,
kz − z∗k ≤ θΨα(z)14, ∀z ∈ H, where θ is a positive constant.
Proof. Since F has the uniform P∗-property, there exists ρ > 0 such that ρkz − z∗k2 ≤ max
i=1,2h(z − z∗) ◦ (F (z) − F (z∗)), ei(z∗)i, (13) where {ei(z∗)|i = 1, 2} is the Jordan frame about z∗ in H. From z∗ being a solution of (1), it follows that
(z − z∗) ◦ (F (z) − F (z∗)) = z ◦ F (z) − z ◦ F (z∗) − z∗ ◦ F (z).
Note that F (z∗) ∈ Ω, z∗ ∈ Ω and ei(z∗) ∈ Ω. By Lemma 4.1, Lemma 4.2 and Lemma 2.2(e), we have
h−z, F (z∗) ◦ ei(z∗)i ≤ h(−z)+, F (z∗) ◦ ei(z∗)i and
h−F (z), z∗◦ ei(z∗)i ≤ h(−F (z))+, z∗◦ ei(z∗)i.
Moreover, from Lemma 4.1, we have that z∗ and F (z∗) share the same Jordan frame.
Let
z∗ = λ1(z∗)e1(z∗) + λ2(z∗)e2(z∗) and
F (z∗) = λ1(F (z∗))e1(z∗) + λ2(F (z∗))e2(z∗).
Then, it follows that
h(z − z∗) ◦ (F (z) − F (z∗)), ei(z∗)i
= hz ◦ F (z), ei(z∗)i + h−z ◦ F (z∗), ei(z∗)i + h−z∗◦ F (z), ei(z∗)i
= hz ◦ F (z), ei(z∗)i + h−z, F (z∗) ◦ ei(z∗)i + h−F (z), z∗◦ ei(z∗)i
≤ h(z ◦ F (z))+, ei(z∗)i + h(−z)+, F (z∗) ◦ ei(z∗)i + h(−F (z))+, z∗◦ ei(z∗)i
≤ k(z ◦ F (z))+kkei(z∗)k + k(−z)+kkF (z∗) ◦ ei(z∗)k + k(−F (z))+kkz∗◦ ei(z∗)k
≤ 1
√2[k(z ◦ F (z))+k + λi(F (z∗))k(−z)+k + λi(z∗)k(−F (z))+k]
≤ max
1
√2,λi(F (z∗))
√2 ,λi(z∗)
√2
[k(z ◦ F (z))+k + k(−z)+k + k(−F (z))+k]
≤ max
1
√2,λ2(F (z∗))
√2 ,λ1(z∗)
√2
[k(z ◦ F (z))+k + k(−z)+k + k(−F (z))+k] , where the first inequality is from Lemma 2.2(e), and the last inequality is from the facts that λi(F (z∗)) ≥ 0, λ1(z∗) ≥ λ2(z∗) ≥ 0 and z∗◦ F (z∗) = 0, i.e., λi(F (z∗)) · λi(z∗) = 0 for i = 1, 2. Define
τ := ρ
maxn
√1
2,λ2(F (z√ ∗))
2 ,λ1√(z∗)
2
o . This together (12) and (13) yields
τ kz − z∗k2 ≤ k(F (z) ◦ z)+k + k(−F (z))+k + k(−z)+k ∀z ∈ H.
Now, we come to the second part of the proposition. We know that Ψα(z) = α
2k(z ◦ F (z))+k2+ ΨFB(z), which says k(z ◦ F (z))+k ≤q
2
αΨα(z)1/2. Moreover, from Lemma 2.3, it follows that k(−F (z))+k + k(−z)+k ≤ √
2(k(−F (z))+k2+ k(−z)+k2)12
≤ 2√
2ΨFB(z)1/2
≤ 2√
2Ψα(z)1/2. Therefore,
k(F (z) ◦ z)+k + k(−F (z))+k + k(−z)+k ≤ r2
α + 2√ 2
!
Ψα(z)1/2.
Letting θ = q√
2(2√ α+1) τ√
α shows the second part result. 2
The condition α > 0 is necessary in Proposition 4.2. From the proof, we see that θ → ∞ as α → 0. Thus, α 6= 0 is necessary. In fact, when α = 0, we have Ψα = ΨFB. This explains that for the FB merit function ΨFB, we can not reach our desired result under the same conditions in Proposition 4.2.
5 Boundedness of level sets
The boundedness of level sets of a merit function is also important since it is a necessary condition to ensures that the sequence generated by a descent method has at least one accumulation point. In this section, we will show the boundedness of level sets of ΨFB and Ψα under different conditions. The relation between such conditions will be discussed as well.
Lemma 5.1 For any zk, wk ∈ H, let λ2(zk) ≤ λ1(zk) and µ2(wk) ≤ µ1(wk) denote the spectral values of zk and wk, respectively. Then,
(a) if λ2(zk) → −∞ or µ2(wk) → −∞, we have Ψi → ∞ for i = 1, 2, where Ψ1(z) := 1
2(k(−zk)+k2+ k(−wk)+k2) and Ψ2(z) := 1
2kφFB(zk, wk)+k2. (b) if {λ2(zk)} and {µ2(wk)} are bounded below, λ1(zk) → ∞, µ1(wk) → ∞ and kzzkkk ◦
wk
kwkk 6→ 0 as k → ∞, then ΦFB(zk, wk) → ∞.
Proof. (a) See [22, Lemma 4.4].
(b) Suppose that {φFB(zk, wk)} is bounded. Define uk := ((zk)2 + (wk)2)1/2 for each k.
From the definition of φFB, we have zk+ wk= uk− φFB(zk, wk) for each k. Squaring two sides of the above equality leads to
2zk◦ wk = −2uk◦ φFB(zk, wk) + φ2
FB(zk, wk).
Since kzkk ≥ √1
2λ1(zk) and kwkk ≥ √1
2µ1(wk), by the conditions of this lemma, we have lim
k→∞
uk
kzkkkwkk = lim
k→∞
(zk)2
kzkk2kwkk2 + (wk)2 kzkk2kwkk2
1/2
= 0.
This together with the boundedness of {φFB(zk, wk)} implies
k→∞lim
2zk◦ wk
kzkkkwkk = lim
k→∞
−2uk◦ φFB(zk, wk) + φ2FB(zk, wk) kzkkkwkk = 0.
Thus, limk→∞ kzzkkk ◦ kwwkkk = 0, which contradicts the given assumption. Hence, the conclusion is proved. 2
Remark: The condition of Lemma 5.1(b) was discussed in Lemma 4.2 of paper [11] and Lemma 4.1 of paper [3] in finite-dimensional space as well.
Condition A[18]. For any sequence {zk} ⊆ H satisfying kzkk → ∞, if {λ2(zk)} and λ2(F (zk)) are bounded below, and λ1(zk) → ∞, λ1(F (zk)) → ∞ as k → ∞. Then
lim sup
k→∞
zk
kzkk, F (zk) kF (zk)k
> 0.
Using Lemma 5.1 and similar arguments as in [3, Proposition 4.2], we may obtain the following propositions, which are the boundedness of level sets for ΨFB and Ψα, respectively.
Proposition 5.1 Let ΨFB be given as in (4). Assume that F is a strongly monotone function with modulus µ > 0 and satisfies Condition A. Then, the level set
L (γ) := {z ∈ H| ΨFB(z) ≤ γ}
is bounded for all γ ≥ 0.
Proof. We prove this result by contradiction. Suppose there exists an unbounded sequence {zk} ⊆ L (γ) for some γ ≥ 0. We claim that the sequence of the smallest spectral values of zk and F (zk) are bounded below. If not, by Lemma 5.1, we have Ψi(zk) → ∞ for i = 1, 2, which implies ΨFB(zk) → ∞. This contradicts {zk} ⊆ L (γ).
On the other hand, by the strongly monotone property of F , there exists a constant µ > 0 such that
µkzkk2 ≤ hzk, F (zk) − F (0)i
≤ kzkkkF (zk) − F (0)k
≤ kzkk(kF (zk)k + kF (0)k).
This implies that µkzkk ≤ kF (zk)k + kF (0)k. It follows from unboundedness of {zk} and boundedness of F that {F (zk)} is unbounded, which says that λ1(zk) → ∞ and λ1(F (zk)) → ∞ as k → ∞. By Condition A, it gives
lim sup
k→∞
zk
kzkk, F (zk) kF (zk)k
> 0, which implies
lim sup
k→∞
λ1
zk
kzkk ◦ F (zk) kF (zk)k
> 0.
From this, we have kzzkkk◦kF (zF (zkk))k 6→ 0. Together with Lemma 5.1(b), we obtain ΨFB(zk) = ΦFB(zk, F (zk)) → ∞. Hence, this is a contradiction to zk⊆L (γ). The proof is complete.
2
In fact, the condition A and the strong monotonicity of F in Proposition 5.1 can be replaced by the Lipschitz continuity of F and R01-function. This can be verified as below. Since the sequence of the smaller spectral values of zk and F (zk) are bounded below, we have (−zk)+and (−F (zk))+are bounded above. For any sequence zksatisfying kzkk → ∞, by the definition of R01-function, we have
lim inf
k→∞
hzk, F (zk)i
kzkk2 > 0. (14)
This implies that lim sup
k→∞
zk
kzkk, F (zk) kF (zk)k
≥ lim inf
k→∞
zk
kzkk, F (zk) kF (zk)k
= lim inf
k→∞
hzk, F (zk)i kzkk2
kzkk kF (zk)k
> 0,
where the last inequality is due to (14) and the Lipschitz continuity of F .
Next, we show the bounded level sets for Ψα. As will be seen, it requires F being R02-function to guarantee this property. In view of the above remark and the fact that every R01-function is an R02-function, we see this condition is weaker that in Proposition 5.1 though their proofs are similar.
Proposition 5.2 Let Ψα be given as in (5). Suppose that F is an R02-function. Then, the level sets
L (γ) := {z ∈ H| Ψα(z) ≤ γ}.
is bounded for all γ ≥ 0.
Proof. We prove this result by contradiction again. Suppose there exists an unbounded sequence {zk} ⊆ L (γ) for some γ ≥ 0. We claim that the sequence of the smaller spectral values of zk and F (zk) are bounded below. In fact, if not, by Lemma 5.1, we have Ψi(zk) → ∞ for i = 1, 2, which says Ψα(zk) → ∞. This contradicts {zk} ⊆L (γ).
Therefore {(−zk)+} and {(−F (zk))+} are bounded above. Thus, for any sequence {zk} satisfying kzkk → ∞, we have
(−zk)+
kzkk → 0 and (−F (zk))+ kzkk → 0.
By the definition of R02-function, we have lim inf
k→∞
λ1(zk◦ F (zk)) kzkk2 > 0.
This implies that λ1(zk◦ F (zk)) → ∞. Hence, k(zk◦ F (zk))+k → ∞. This together with definition of Ψα, which leads to Ψα(zk) → ∞. This contradicts {zk} ⊆L (γ). Therefore, the desired result is proved. 2
6 Concluding Remarks
In this paper, we have studied the Lorentz cone complementarity problems in real Hilbert space for which we prove the same growth of FB and NR merit functions, provide a global error bound for the solutions via two kinds of merit functions, and discuss the property of the bounded level sets of these two kinds of merit functions under different conditions.
Such results will be helpful and useful for further designing solution methods for solving problem (1).
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