to appear in Optimization Letters, 2010
The same growth of FB and NR symmetric cone complementarity functions
Shujun Bi
Department of Mathematics South China University of Technology
Guangzhou 510640, China E-mail: [email protected]
Shaohua Pan1
Department of Mathematics South China University of Technology
Guangzhou 510640, China E-mail: [email protected]
Jein-Shan Chen 2 Department of Mathematics National Taiwan Normal University
Taipei, Taiwan 11677 E-mail: [email protected]
June 10, 2010
(revised on October 20, 2010)
Abstract. We establish that the Fischer-Burmeister (FB) complementarity function and the natural residual (NR) complementarity function associated with the symmetric cone have the same growth, in terms of the classification of Euclidean Jordan algebras. This, on the one hand, provides an affirmative answer to the second open question proposed by Tseng [Growth behavior of a class of merit functions for the nonlinear complementarity problems, Journal of Optimization Theory and Applications, vol. 89, pp. 17–37, 1996.]
for the matrix-valued FB and NR complementarity functions, and on the other hand, extends the third important inequality of Lemma 3.1 in the aforementioned paper to the setting of Euclidean Jordan algebras. It is worthwhile to point out that the proof is surprisingly simple.
1The author’s work is supported by National Young Natural Science Foundation (No. 10901058) and Guangdong Natural Science Foundation (No. 9251802902000001) and the Fundamental Research Funds for the Central Universities (SCUT).
2Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is supported by National Science Council of Taiwan.
Key words. Symmetric cone, FB and NR complementarity functions, growth.
1 Introduction
LetA = (V, ◦, ⟨·, ·⟩V) be a Euclidean Jordan algebra (see Sec. 2 for details). LetK be the set of all squares inV. Given the continuously differentiable mappings F, G : V → V, we consider the symmetric cone complementarity problem (SCCP): to find a vector ζ ∈ V such that
F (ζ)∈ K, G(ζ) ∈ K, ⟨F (ζ), G(ζ)⟩V = 0. (1) This class of problems provides a unified framework for the classical nonlinear program- ming and complementarity problem [5] over the nonnegative orthant cone in Rn, the second-order cone optimization and complementarity problem [1], and the semidefinite programming and complementarity problem [16, 21], and becomes one of main research interests in the current optimization field; see, e.g., [4, 6, 10, 13, 17, 12, 18, 19, 22].
Analogous to the three classes of special symmetric cone complementarity problems above, the complementarity function associated with the symmetric cone plays a crucial role in the development of merit function methods and smoothing (nonsmooth) Newton methods for solving the SCCPs. Recall that ϕ : V × V → V is called a symmetric cone complementary function if it satisfies the following equivalence:
ϕ(x, y) = 0 ⇐⇒ x ∈ K, y ∈ K, ⟨x, y⟩V= 0. (2) With such a function, the SCCP can be reformulated as an unconstrained minimization
minζ∈V Ψ(ζ) := 1
2∥ϕ(F (ζ), G(ζ))∥2V, (3)
in the sense that ζ∗ solves (1) if and only if it is a solution of (3) with zero optimal value, where ∥ · ∥V denotes the norm induced by the inner product ⟨·, ·⟩V. If Ψ is continuously differentiable, then the efficient unconstrained minimization methods can be applied for (3) to yield a solution of (1). This method is often known as the merit function approach.
Two most popular choices for ϕ are the NR symmetric cone complementarity function ϕNR and the FB symmetric cone complementarity function ϕFB, respectively, defined as
ϕNR(x, y) := x− (x − y)+ ∀x, y ∈ V (4) and
ϕFB(x, y) := (x + y)− (x2+ y2)1/2 ∀x, y ∈ V, (5) where z+ means the metric projection of z ∈ V onto the symmetric cone K, x2 = x◦ x denotes the Jordan product of x and itself, and x1/2 means the unique square root of
x∈ K, i.e., (x1/2)2 = x. The squared norm of ϕFB induces a smooth merit function with global Lipschitz continuous gradients (see [9, 15]). This implies that finding solutions to (1) is equivalent to seeking solutions of the unconstrained smooth minimization problem
minζ∈V ΨFB(ζ) := 1
2∥ϕFB(F (ζ), G(ζ))∥2V. (6) However, in order to establish the convergence rate of the merit function method for the SCCPs based on (6), the key is to prove that ϕFB and ϕNR has the same order of growth, i.e., to show that there exist constants c1 > 0 and c2 > 0 such that for all x, y∈ V,
c1∥ϕNR(x, y)∥2V ≤ ∥ϕFB(x, y)∥2V≤ c2∥ϕNR(x, y)∥2V. (7) When A is the Euclidean Jordan algebra R equipped with the multiplication of real numbers, Tseng showed in [20, Lemma 3.1] that inequality (7) holds with c1 = 2−√
2 and c2 = 2 +√
2; whenA is the Jordan spin algebra Ln(see Example 2.3 in the next section), Pan, Chen and Li [14] recently established inequality (7) by contradiction. We note that for the case where A is the n × n real symmetric matrix algebra (see Example 2.2 in the next section), in 1998 Tseng [21] proposed an open question “whether the FB func- tion ∥ϕFB(F (ζ), G(ζ))∥2V is bounded above and below by a constant multiple of the NR function∥ϕNR(F (ζ), G(ζ))∥2V”, which is equivalent to asking whether or not inequality (7) holds under this case. To our best knowledge, until now this open question is not resolved.
In this paper, we show that (7) holds with c1 = 2−√
2 and c2 = 2+√
2, which does not only offer an affirmative answer to the open question of [21], but also extends the results of [20, Lemma 3.1] and [14] to the setting of symmetric cones. Particularly, the proof is surprisingly simpler than that of [20, Lemma 3.1] and [14]. As a direct consequence of (7), we also establish the global error bound property of the FB merit function for SCCPs.
2 Preliminaries
This section recalls some results on Euclidean Jordan algebras that will be used in the next section. More detailed expositions of Euclidean Jordan algebras can be found in the monograph by Faraut and Kor´anyi [3] and Koecher’s lecture notes [7].
A Euclidean Jordan algebra is a triple (V, ◦, ⟨·, ·⟩V) where (V, ⟨·, ·⟩V) is a finite dimen- sional inner product space over the real number field R and (x, y) 7→ x ◦ y : V × V → V is a bilinear mapping satisfying the following three conditions:
(i) x◦ y = y ◦ x for all x, y ∈ V;
(ii) x◦ (x2◦ y) = x2◦ (x ◦ y) for all x, y ∈ V, where x2 = x◦ x;
(iii) ⟨x ◦ y, z⟩V =⟨y, x ◦ z⟩V for all x, y, z ∈ V.
Henceforth, we assume that A = (V, ◦, ⟨·, ·⟩V) is a Euclidean Jordan algebra with an element e∈ V (called the unit element) such that x◦e = x for all x ∈ V. By [3, Theorem III. 2.1], the set of squares K := {x2 | x ∈ V} is a symmetric cone. In the following, we present three common examples of Euclidean Jordan algebras.
Example 2.1. ConsiderRn with the (usual) inner product and Jordan product defined respectively as
⟨x, y⟩ =
∑n i=1
xiyi and x◦ y = x ∗ y ∀x, y ∈ Rn
where xi denotes the ith component of x, etc., and x ∗ y denotes the componentwise product of vectors x and y. Then,Rnis a Euclidean Jordan algebra with the nonnegative orthant Rn+ as its cone of squares.
Example 2.2. The algebra Sn of n× n real symmetric matrices. Let Sn×n be the space of all n× n real symmetric matrices with the trace inner product and Jordan product, respectively, defined by
⟨X, Y ⟩T := Tr(XY ) and X◦ Y := 1
2(XY + Y X) ∀X, Y ∈ Sn×n.
Then, (Sn×n,◦, ⟨·, ·⟩T) is a Euclidean Jordan algebra, and we write it asSn. The cone of squares Sn+×n inSn is the set of all positive semidefinite matrices in Sn×n.
Example 2.3. The Jordan spin algebra Ln. Consider Rn (n > 1) with the inner product ⟨·, ·⟩ and Jordan product
x◦ y :=
[ ⟨x, y⟩
x0y + y¯ 0x¯ ]
for any x = (x0; ¯x), y = (y0; ¯y) ∈ R × Rn−1. We denote the Euclidean Jordan algebra (Rn,◦, ⟨·, ·⟩) by Ln. The cone of squares, called the Lorentz cone (or the second-order cone), is given by Ln+:={(x0; ¯x)∈ R × Rn−1 | x0 ≥ ∥¯x∥}.
For x∈ V, let m(x) := min{
k :{e, x, x2,· · · , xk} are linearly dependent}
and define the rank of A by r := max{m(x) : x ∈ V}. Recall that an element c ∈ V is idempotent if c2 = c, and it is a primitive idempotent if it is nonzero and cannot be written as a sum of two nonzero idempotents. One says that a finite set {c1, c2, . . . , ck} of primitive idempotents in V is a Jordan frame if
cj ◦ ci = 0 if j̸= i for all j, i = 1, 2, . . . , k, and ∑k
j=1cj = e.
Now we may state the second version of the spectral decomposition theorem.
Theorem 2.1 [3, Theorem III.1.2] Let A be a Euclidean Jordan algebra with rank r.
Then, for every x ∈ V, there exist a Jordan frame {c1, c2, . . . , cr} and real numbers λ1(x), λ2(x), . . . , λr(x), arranged in the decreasing order λ1(x)≥ · · · ≥ λr(x), such that
x = λ1(x)c1+ λ2(x)c2+· · · + λr(x)cr.
The numbers λj(x) (counting multiplicities), which are uniquely determined by x, are called the eigenvalues of x, and tr(x) =∑r
j=1λj(x) is called the trace of x.
Let ϕ : R → R be a scalar valued function. Then, it is natural to define a vector valued function ϕV :V → V associated with the Euclidean Jordan algebra A [8, 19] by
ϕV(x) := ϕ(λ1(x))c1 + ϕ(λ2(x))c2+· · · + ϕ(λr(x))cr, where x ∈ V has the spectral decomposition x = ∑r
j=1λj(x)cj. This function is also called L¨owner’s operator in recognition of L¨owner’s contribution. When ϕ(t) = t+ :=
max{0, t} for t ∈ R, ϕV(x) becomes the metric projector operator over K:
x+ = (λ1(x))+c1+ (λ1(x))+c2+· · · + (λr(x))+cr ∀x ∈ V;
while ϕ(t) = t− := min{0, t} for t ∈ R, it is the metric projector operator over −K x− = (λ1(x))−c1+ (λ1(x))−c2+· · · + (λr(x))−cr ∀x ∈ V.
In the sequel, we let |x| be L¨owner’s operator induced by ϕ(t) = |t| for t ∈ R. Then,
|x| = x+− x−= 2x+− x = x − 2x− ∀x ∈ V. (8)
Recall that a Euclidean Jordan algebra is said to be simple if it is not the direct sum of two Euclidean Jordan algebras. It is easy to see thatSn and Ln are simple Euclidean Jordan algebras, whereas the Euclidean Jordan algebra in Example 2.1 is not simple. Let Hn×n denote the space of n× n complex Hermitian matrices, Qn×n the space of n× n quaternion Hermitian matrices, andO3×3 the space of 3×3 octonion Hermitian matrices.
Theorem 2.2 [3, Theorem V.3.7] Suppose that A = (V, ◦, ⟨·, ·⟩V) is a simple Euclidean Jordan algebra of rank r ≥ 3. Then, A is isomorphic to one of the following
(i) The algebra Sn of n× n real symmetric matrices given by Example 2.2;
(ii) The algebra Hn of all n× n complex Hermitian matrices with trace inner product
⟨x, y⟩T :=ℜTr(xy∗) and Jordan product x◦ y := 12(xy + yx) for any x, y ∈ Hn×n; (iii) The algebraQnof all n×n quaternionic Hermitian matrices with trace inner product
⟨x, y⟩T :=ℜTr(xy∗) and Jordan product x◦ y := 12(xy + yx) for any x, y ∈ Qn×n;
(iv) The algebra O3 of all 3× 3 octonionic Hermitian matrices with trace inner product
⟨x, y⟩T :=ℜTr(xy∗) and Jordan product x◦ y := 12(xy + yx) for any x, y ∈ O3×3; (v) The Jordan spin algebra Ln given by Example 2.3.
where the notation “∗” means the conjugate transpose, Tr(xy) denotes the trace of xy which is the multiplication of matrices x and y, and ℜa means the real part of a.
Unless otherwise stated, in the rest of this paper, we assume that A = (V, ◦, ⟨·, ·⟩V) is a simple Euclidean Jordan algebra, and denote∥ · ∥V,∥ · ∥ and ∥ · ∥T the norm induced by the inner product⟨·, ·⟩V,⟨·, ·⟩ and ⟨·, ·⟩T, respectively. We also write x ≽Ky (respectively, x≻Ky) to mean x− y ∈ K (respectively, x − y ∈ intK).
3 Main result
To establish the main result of this paper, the following lemma plays an important role.
Lemma 3.1 (a) For any x, y ∈ V, if x ≽K 0, y≽K 0 and x≽Ky, then x1/2 ≽Ky1/2. (b) For any u, v, w ∈ V, if w ≽K 0 and 2w2 = u2+ v2, then there holds that
w≽K 1
2(u + v).
Proof. (a) This is result of [6, Prop. 8], which is also implied by [8].
(b) Since u2+ v2− 2u ◦ v = (u − v) ◦ (u − v) ∈ K, using 2w2 = u2+ v2 yields w2 = 1
2(u2+ v2)≽K 1
4(u2+ v2) + 1
2u◦ v = 1
4(u + v)2. From part(a) and w ≽K0, this implies that w ≽K 1
2|u + v| ≽K 1
2(u + v). 2
Proposition 3.1 Let Ln be the Euclidean Jordan algebra in Example 2.3. Then, (2−√
2)∥ϕNR(x, y)∥ ≤ ∥ϕFB(x, y)∥ ≤ (2 +√
2)∥ϕNR(x, y)∥, ∀x, y ∈ Ln.
Proof. Fix any x, y ∈ V. If ϕNR(x, y) = 0, then we also have ϕFB(x, y) = 0, and the desired result is immediate. Therefore, in the following arguments we assume that ϕNR(x, y)̸= 0. Using equation (8) and the definition of ϕNR, it is not hard to see that
ϕNR(x, y) = 1
2[(x + y)− |x − y|] .
This together with the definition of ϕFB gives
ϕFB(x, y) = 2ϕNR(x, y) +|x − y| − (x2 + y2)1/2
= 2ϕNR(x, y) + z(x, y) (9)
where z(x, y) ≡ |x − y| − (x2 + y2)1/2. By equation (9) and the triangle inequality, it suffices to argue ∥z(x, y)∥ ≤√
2∥ϕNR(x, y)∥, that is,
∥z(x, y)∥2 ≤ 1 2
x + y − |x − y| 2. (10)
Substituting the expression of z(x, y) into (10), we obtain that (10) is equivalent to |x− y| − (x2+ y2)1/2 2 ≤ 1
2∥x + y − |x − y|∥2
⇐⇒ ∥x − y∥2− 2⟨|x − y|, (x2 + y2)1/2⟩ + ∥(x2+ y2)1/2∥2
≤ 1 2
[∥x + y∥2 +∥x − y∥2− 2⟨|x − y|, x + y⟩]
⇐⇒ ∥c(x, y)∥2− ⟨2c(x, y) − (x + y), |x − y|⟩ − 2⟨x, y⟩ ≤ 0 (11) where c(x, y)≡ (x2 + y2)1/2. Thus, to prove the desired result, it suffices to argue that inequality (11) holds. Indeed, since
∥c(x, y)∥2 =⟨c(x, y)2, e⟩ = ⟨x2+ y2, e⟩ = ∥x∥2+∥y∥2, we have
∥c(x, y)∥2− ⟨2c(x, y) − (x + y), |x − y|⟩ − 2⟨x, y⟩
= ∥x − y∥2− ⟨2c(x, y) − (x + y), |x − y|⟩
= ⟨|x − y|, −2c(x, y) + (x + y) + |x − y|⟩. (12) Applying Lemma 3.1 with w = c(x, y), u = (x + y) and v =|x − y|, we know
−2c(x, y) + (x + y) + |x − y| ∈ −Ln+.
This, along with|x − y| ∈ Ln+ and equation (12), shows that inequality (11) holds. 2
Proposition 3.2 Suppose that A = (V, ◦, ⟨·, ·⟩V) is a simple Euclidean Jordan algebra with the rank r≥ 3. Then, it holds that
(2−√
2)∥ϕNR(x, y)∥T ≤ ∥ϕFB(x, y)∥T ≤ (2 +√
2)∥ϕNR(x, y)∥T ∀x, y ∈ V.
Proof. By Theorem 2.2, it suffices to prove that the desired result holds for A = Sn, or Hn, or Qn, or O3. Fix any x, y ∈ V with V = Sn×n, or Hn×n, or Qn×n, or O3×3. If ϕNR(x, y) = 0, the result is direct. Thus, it suffices to consider the case of ϕNR(x, y)̸= 0.
Note that for the simple Euclidean Jordan algebraSn, orHn, orQn, orO3, we still have ϕFB(x, y) = 2ϕNR(x, y) + z(x, y)
with z(x, y)≡ |x − y| − (x2+ y2)1/2. By the triangle inequality, it suffices to prove
∥z(x, y)∥2T≤ 2∥ϕNR(x, y)∥2T = 1 2
x + y − |x − y| 2
T
. (13)
Using the definition of ∥ · ∥2T and noting that ℜTr(uv∗) = ℜTr(uv) = ℜTr(vu) for all u, v ∈ V, an elementary computation yields that (13) is equivalent to
ℜTr[
(x− y)2 + c(x, y)2− 2|x − y|c(x, y)]
≤ 1 2ℜTr[
(x + y)2 + (x− y)2− 2|x − y|(x + y)]
⇐⇒ ℜTr[
−|x − y|(2c(x, y) − (x + y)) + (x − y)2]
≤ 0
⇐⇒ 1
2ℜTr [
|x − y|
(
c(x, y)−(x + y) +|x − y|
2
)]
≥ 0
⇐⇒ 1
2
⟨
|x − y|, c(x, y) − (x + y) +|x − y|
2
⟩
T
≥ 0 (14)
where c(x, y) ≡ (x2+ y2)1/2. Applying Lemma 3.1 with w = c(x, y), u = (x + y) and v =|x−y| yields that c(x, y)−((x+y)+|x−y|)/2 ≽K0. This together with|x−y| ≽K0 implies that inequality (14) holds. Thus, we complete the proof. 2
Combining Prop. 3.1 with Prop. 3.2, we readily obtain the main result of this paper.
Theorem 3.1 Suppose that A = (V, ◦, ⟨·, ·⟩V) is a simple Euclidean Jordan algebra. Let ϕNR and ϕFB be defined by (4) and (5), respectively. Then, it holds that
(2−√
2)∥ϕNR(x, y)∥V≤ ∥ϕFB(x, y)∥V≤ (2 +√
2)∥ϕNR(x, y)∥V ∀x, y ∈ V.
By Theorem 3.1, we may establish the global error bound property for the FB merit function of SCCPs under the jointly uniform Cartesian P -property of F and G. To this end, we next assume that A is a direct product of simple Euclidean Jordan algebras:
A = A1 × A2× · · · × Am,
where eachAi = (Vi,◦, ⟨·, ·⟩Vi) is a simple Euclidean Jordan algebra with∑m
i=1dim(Vi) = dim(V). Then, K = K1× K2× · · · × Km with Ki being a symmetric cone in Vi. For any x, y ∈ V, we write x = (x1, . . . , xm), y = (y1, . . . , ym) with xi, yi ∈ Vi. Then,
x◦ y = (x1◦ y1, . . . , xm◦ ym) and ⟨x, y⟩V=⟨x1, y1⟩V1 +· · · + ⟨xm, ym⟩Vm.
Consequently, the SCCP (1) is equivalent to finding a vector ζ ∈ V such that
Fi(ζ)∈ Ki, Gi(ζ)∈ Ki, ⟨Fi(ζ), Gi(ζ)⟩Vi = 0, i = 1, 2, . . . , m (15) where F = (F1, . . . , Fm) and G = (G1, . . . , Gm) with Fi, Gi:V → Vi.
Definition 3.1 [2] The mappings F and G are said to have the jointly uniform Cartesian P -property if there exists a constant ρ > 0 such that for any ζ, ξ ∈ V, there is an index ν ∈ {1, . . . , m} such that
⟨Fν(ζ)− Fν(ξ), Gν(ζ)− Gν(ξ)⟩Vν ≥ ρ∥ζ − ξ∥2V.
Theorem 3.2 Suppose that F and G have the jointly uniform Cartesian P -property and are globally Lipschitz continuous with constants L1 > 0 and L2 > 0, respectively. If the SCCP (1) has an optimal solution, say ζ∗, then
2−√ 2
(2L1+ L2)2ΨFB(ζ)≤ ∥ζ − ζ∗∥2V≤ (2 +√
2)(L1+ L2)2
ρ2 ΨFB(ζ) ∀ζ ∈ V where the constant ρ is same as in Definition 3.1.
Proof. Fix any ζ ∈ V. Let R(ζ) ≡ (ϕNR(F1(ζ), G1(ζ)), . . . , ϕNR(Fm(ζ), Gm(ζ))) ∈ V.
Then, using Theorem 3.1 and noting that ΨFB(ζ)≡ 12∑m
i=1∥ϕFB(Fi(ζ), Gi(ζ))∥2Vi, we get 2−√
2
2 ∥R(ζ)∥2V≤ ΨFB(ζ)≤ 2 +√ 2
2 ∥R(ζ)∥2V. In addition, using the same arguments as in [9, Theorem 6.3], we have
1
2L1+ L2∥R(ζ)∥V≤ ∥ζ − ζ∗∥V ≤ L1+ L2
ρ ∥R(ζ)∥V.
From the last two inequalities, we immediately obtain the desired result. 2
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