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SOC weighted means and trace inequalities

在文檔中 SOC Functions and Their Applications (頁 179-192)

min 𝑎, 𝑏

𝐻 𝑎, 𝑏 𝐺 𝑎, 𝑏 𝐿 𝑎, 𝑏 𝐴 𝑎, 𝑏

m𝑎𝑥 𝑎, 𝑏

𝐻 𝑎, 𝑏 𝐺 𝑎, 𝑏 𝐴 𝑎, 𝑏

𝜆 1

2 𝜆 𝑏

𝑎

𝑀 𝑎, 𝑏

𝜈 1

2 𝜈 1

𝜆 1

Figure 4.1: Relationship between means defined on real number.

G(x, y), we consider their corresponding SOC weighted means as below. For 0 ≤ λ ≤ 1, we let

Aλ(x, y) := (1 − λ)x + λy, (4.12)

Hλ(x, y) := (1 − λ)x−1+ λy−1−1

, (4.13)

Gλ(x, y) := P  x12 

P (x12)yλ

, (4.14)

denote the SOC weighted arithmetic mean, the SOC weighted harmonic mean, and the SOC weighted geometric mean, respectively. According to the definition, it is clear that

A1−λ(x, y) = Aλ(y, x), H1−λ(x, y) = Hλ(y, x), G1−λ(x, y) = Gλ(y, x).

We note that when λ = 1/2, these SOC weighted means coincide with the SOC arithmetic mean A(x, y), the SOC harmonic mean H(x, y), and the SOC geometric mean G(x, y), respectively.

Proposition 4.15. Suppose 0 ≤ λ ≤ 1. Let Aλ(x, y), Hλ(x, y), and Gλ(x, y) be defined as in (4.12), (4.13), and (4.14), respectively. Then, for any x Kn 0 and y Kn 0, there holds

x ∧ y Kn Hλ(x, y) Kn Gλ(x, y) Kn Aλ(x, y) Kn x ∨ y.

Proof. (i) To verify the first inequality, we discuss two cases. For 12(x + y − |x − y|) /∈ Kn, the inequality holds automatically. For 12(x + y − |x − y|) ∈ Kn, we note that

1

2(x + y − |x − y|) Kn x and 12(x + y − |x − y|) Kn y. Then, using the SOC-monotonicity of f (t) = −t−1 shown in Proposition 2.3, we obtain

x−1 Kn  x + y − |x − y|

2

−1

and y−1 Kn  x + y − |x − y|

2

−1

, which imply

(1 − λ)x−1+ λy−1 Kn  x + y − |x − y|

2

−1

.

Next, applying the SOC-monotonicity again to this inequality, we conclude that x + y − |x − y|

2 Kn (1 − λ)x−1+ λy−1−1

.

(ii) For the second and third inequalities, it suffices to verify the third inequality (the second one can be deduced thereafter). Let s = P (x12)y − e, which gives s Kn −e.

Then, applying Lemma 4.1 yields



e + P (x12)y − eλ

Kn e + λh

P (x12)y − ei ,

which is equivalent to

0 Kn (1 − λ)e + λh

P (x12)yi

−

P (x12)yλ

. Since P (x12) is invariant on Kn, we have

0 Kn P (x12)



(1 − λ)e + λh

P (x12)yi

−

P (x12)yλ

= (1 − λ)x + λy − P (x12)



P (x12)y

λ

, and hence

P (x12)

P (x12)yλ

Kn (1 − λ)x + λy. (4.15)

For the second inequality, replacing x and y in (4.15) by x−1 and y−1, respectively, gives P (x12)



P (x12)y−1

λ

Kn (1 − λ)x−1+ λy−1. Using the SOC-monotonicity again, we conclude

(1 − λ)x−1+ λy−1−1

Kn



P (x12)

P (x12)y−1λ−1

= P (x12)

P (x12)yλ

, where the equality holds by Lemma 4.4(b).

(iii) To see the last inequality, we observe that x Kn 12(x + y + |x − y|) and y Kn

1

2(x + y + |x − y|), which imply

(1 − λ)x + λy Kn x + y + |x − y|

2 .

Then, the desired result follows. 

In Section 4.2, we have established three SOC trace versions of Young inequality.

Based on Proposition 4.15, we provide the SOC determinant version of Young inequality.

Proposition 4.16. (Determinant Young inequality) For any x Kn 0 and y Kn 0, there holds

det(x ◦ y) ≤ det xp p +yq

q

 , where 1 < p, q < ∞ and 1

p+ 1 q = 1.

Proof. Since xpp + yqq Kn G1

q(xp, yq) = P (xp2) P (xp2)yq1q

, and hence

det xp p +yq

q



≥ det



P (xp2)

P (xp2)yq1q

= det(x) det(y) ≥ det(x ◦ y)

by [61, Proposition III.4.2] and Proposition 1.2(b).  Now, we consider the family of Heinz means

Mν(a, b) := aνb1−ν + a1−νbν 2

for a, b > 0 and 0 ≤ ν ≤ 1. Following the idea of Kubo-Ando extension in [95], the SOC Heinz mean can be defined as

Mν(x, y) := Gν(x, y) + Gν(y, x)

2 , (4.16)

where x, y Kn 0 and 0 ≤ ν ≤ 1. We point out that an obvious “naive” extension could be

Bν(x, y) := xν ◦ y1−ν + x1−ν ◦ yν

2 . (4.17)

Unfortunately, Bν may not always satisfy the definition of SOC mean. Although it is not an SOC mean, we still are interested in seeking the trace or norm inequality about Bν and other SOC means, and it will be discussed later.

For any positive numbers a, b, it is well-known that

ab ≤ Mν(a, b) ≤ a + b

2 . (4.18)

Together with the proof of Proposition 4.15, we can obtain the following inequality ac-cordingly.

Proposition 4.17. Suppose 0 ≤ ν ≤ 1 and λ = 12. Let A1

2(x, y), G1

2(x, y), and Mν(x, y) be defined as in (4.12), (4.14), and (4.16), respectively. Then, for any x Kn 0 and y Kn 0, there holds

G1

2(x, y) Kn Mν(x, y) Kn A1

2(x, y).

Proof. Consider x Kn 0, y Kn 0 and 0 ≤ ν ≤ 1, from Proposition 4.15, we have

Mν(x, y) = Gν(x, y) + Gν(y, x) 2

Kn Aν(x, y) + Aν(y, x) 2

= A1

2(x, y).

On the other hand, we note that

Mν(x, y) = Gν(x, y) + G1−ν(x, y) 2

=

P (x12)

P (x12)yν

+ P (x12)

P (x12)y1−ν

2

= P (x12)



P (x12)yν

+

P (x12)y1−ν

2

Kn P (x12)



P (x12)yν2

◦

P (x12)y1−ν2 

= G1

2(x, y),

where the inequality holds due to the fact u + v

2 Kn u12 ◦ v12 for any u, v ∈ Kn and the invariant property of P (x12) on Kn. 

Over all, we could have a picture regarding the ordered relationship of these SOC weighted means as depicted in Figure 4.2.

𝑥 ∧ 𝑦

𝐻 𝑥, 𝑦 𝐺 𝑥, 𝑦 𝐿 𝑥, 𝑦 𝐴 𝑥, 𝑦

𝑥 ∨ 𝑦

𝐻 𝑥, 𝑦 𝐺 𝑥, 𝑦 𝐴 𝑥, 𝑦 𝜆

𝑥 𝑦

𝑀 𝑥, 𝑦

𝜈 1 2

𝜈 1

𝜆 1

2

𝜆 1

Figure 4.2: Relationship between means defined on second-order cone.

Up to now, we have extended the weighted harmonic mean, weighted geometric mean, weighted Heinz mean, and weighted arithmetic mean to second-order cone setting. As below, we explore some other inequalities associated with traces of these SOC weighted means. First, by applying Proposition 1.1(b), we immediately obtain the following trace inequalities for SOC weighted means.

Proposition 4.18. Suppose 0 ≤ λ ≤ 1. Let Aλ(x, y), Hλ(x, y), and Gλ(x, y) be defined as in (4.12), (4.13), and (4.14), respectively. For, any x Kn 0 and y Kn 0, there holds

tr(x ∧ y) ≤ tr(Hλ(x, y)) ≤ tr(Gλ(x, y)) ≤ tr(Aλ(x, y)) ≤ tr(x ∨ y).

Proposition 4.19. Suppose 0 ≤ ν ≤ 1 and λ = 12. Let A1

2(x, y), H1

2(x, y), G1

2(x, y), and Mν(x, y) be defined as in (4.12), (4.13), (4.14), and (4.16), respectively. Then, for any x Kn 0 and y Kn 0, there holds

tr(x ∧ y) ≤ tr(H1

2(x, y)) ≤ tr(G1

2(x, y)) ≤ tr(Mν(x, y)) ≤ tr(A1

2(x, y)) ≤ tr(x ∨ y).

As mentioned earlier, there are some well-known means, like Heinz mean Mν(a, b) = aνb1−ν + a1−νbν

2 , for 0 < ν < 1,

which cannot serve as SOC means albeit it is a natural extension. Even though they are not SOC means, it is still possible to derive some trace or norm inequality about these means.

Next, we pay attention to another special inequality. The Powers-Størmers inequality asserts that for s ∈ [0, 1] the following inequality

2Tr AsB1−s ≥ Tr (A + B − |A − B|)

holds for any pair of positive definite matrices A, B. This is a key inequality to prove the upper bound of Chernoff bound, in quantum hypothesis testing theory [4]. In [72, 73], Hoa, Osaka and Tomiyama investigate the generalized Powers-Størmer inequality. More specifically, for any positive matrices A, B and matrix-concave function f , they prove that

Tr(A) + Tr(B) − Tr(|A − B|) ≤ 2Tr

f (A)12g(B)f (A)12 , where g(t) =

( t

f (t), t ∈ (0, ∞)

0, t = 0 . Moreover, Hoa et al. also shows that the Powers-Størmers Inequality characterizes the trace property for a normal linear positive func-tional on a von Neumann algebras and for a linear positive funcfunc-tional on a C-algebra.

Motivated by the above facts, we establish a version of the Powers-Størmers inequality for SOC-monotone function on [0, ∞) in the SOC setting.

Proposition 4.20. For any x, y, z ∈ IRn, there holds tr((x ◦ y) ◦ z) = tr(x ◦ (y ◦ z)).

Proof. From direct computation, we have x ◦ y = (x1y1+ hx2, y2i, x1y2+ y1x2) and tr((x ◦ y) ◦ z) = 2 (x1y1z1 + z1hx2, y2i + x1hy2, z2i + y1hx2, z2i) .

Similarly, we also have y ◦ z = (y1z1+ hy2, z2i, y1z2+ z1y2) and

tr(x ◦ (y ◦ z)) = 2 (x1y1z1 + x1hy2, z2i + y1hx2, z2i + z1hx2, y2i) . Therefore, we conclude the desired result. 

According to the proof in [72, 73], the crucial point is under what conditions of f (t), there holds the SOC-monotonicity of f (t)t . For establishing the SOC version of Powers-Størmers Inequality, it is also a key, which is answered in next proposition.

Proposition 4.21. Let f be a strictly positive, continuous function on [0, ∞). The function g(t) := t

f (t) is SOC-monotone if one of the following conditions holds.

(a) f is matrix-monotone of order 4;

(b) f is matrix-concave of order 3;

(c) For any contraction T : Kn7→ Kn and z ∈ Kn, there holds

fsoc(T z) Kn T fsoc(z). (4.19) Proof. (a) According to [72, Proposition 2.1], the 4-matrix-monotonicity of f would imply the 2-matrix-monotonicity of g, which coincides with the SOC-monotonicity by Proposition 2.23.

(b) From [73, Theorem 2.1], the 3-matrix-concavity of f implies the 2-matrix-monotonicity of g, which coincides with the SOC-monotonicity as well.

(c) Suppose 0 ≺Kn x Kn y, we have P (x12)  P (y12) by SOC-monotonicity of t1/2 and Lemma 4.4, which implies kP (x12)P (y12)k ≤ 1. Hence, P (x12)P (y12) is an contraction.

Then

x = P (x12)(P (y12)y)

=⇒ fsoc(x) = fsoc(P (x12)(P (y12)y))

=⇒ fsoc(x) Kn P (x12)(P (y12)fsoc(y))

⇐⇒ P (x12)fsoc(x) Kn P (y12)fsoc(y)

⇐⇒ x−1◦ fsoc(x) Kn y−1◦ fsoc(y)

⇐⇒ x ◦ (fsoc(x))−1Kn y ◦ (fsoc(y))−1

⇐⇒ gsoc(x) Kn gsoc(y),

where the second implication holds by setting T = P (x12)P (y12) and the first equivalence holds by the invariant property of P (x12) on Kn. 

Remark 4.4. We elaborate more about Proposition 4.21. We notice that the monotonicity and concavity of f are not strong enough to guarantee the SOC-monotonicity of g. Indeed, the SOC-SOC-monotonicity and SOC-concavity only coincides with the 2-matrix-monotonicity and 2-matrix-concavity, respectively. Hence, we need stronger condition on f to assure the SOC-monotonicity of g. Another point to mention is that the condition (4.19) in Proposition 4.21(c) is a similar idea for SOC setting parallel to the following condition:

Cf (A)C  f (CAC) (4.20)

for any positive semidefinite A and a contraction C in the space of matrices. This in-equality (4.20) plays a key role in proving matrix-monotonicity and matrix-convexity.

For more details about this condition, please refer to [72, 73]. To the contrast, it is not clear about how to define (·) associated with SOC. Nonetheless, we figure out that the condition (4.19) may act as a role like (4.20).

Proposition 4.22. Let f : [0, ∞) −→ (0, ∞) be SOC-monotone and satisfy one of the conditions in Proposition 4.21. Then, for any x, y ∈ Kn, there holds

tr(x + y) − tr(|x − y|) ≤ 2tr

fsoc(x)12 ◦ gsoc(y) ◦ fsoc(x)12

, (4.21)

where g(t) = f (t)t if t > 0, and g(0) = 0.

Proof. For any x, y ∈ Kn, it is known that x − y can be expressed as [x − y]+− [x − y]. Let us denote by p := [x − y]+ and q := [x − y]. Then we have

x − y = p − q and |x − y| = p + q and the inequality (4.21) is equivalent to the following

tr(x) − tr

fsoc(x)12 ◦ gsoc(y) ◦ fsoc(x)12

≤ tr(p).

Since y + p Kn y Kn 0 and y + p = x + q Kn x Kn 0, we have gsoc(x) Kn gsoc(y + p) and by Proposition 4.20

tr(x) − tr

fsoc(x)12 ◦ gsoc(y) ◦ fsoc(x)12

= tr

fsoc(x)12 ◦ gsoc(x) ◦ fsoc(x)12

− tr

fsoc(x)12 ◦ gsoc(y) ◦ fsoc(x)12

≤ tr

fsoc(x)12 ◦ gsoc(y + p) ◦ fsoc(x)12

− tr

fsoc(x)12 ◦ gsoc(y) ◦ fsoc(x)12

= tr



fsoc(x)12 ◦ (gsoc(y + p) − gsoc(y)) ◦ fsoc(x)12



≤ tr

fsoc(y + p)12 ◦ (gsoc(y + p) − gsoc(y)) ◦ fsoc(y + p)12

= tr

fsoc(y + p)12 ◦ gsoc(y + p) ◦ fsoc(y + p)12

−tr

fsoc(y + p)12 ◦ gsoc(y) ◦ fsoc(y + p)12

≤ tr(y + p) − tr

fsoc(y)12 ◦ gsoc(y) ◦ fsoc(y)12



= tr(y + p) − tr(y)

= tr(p).

Hence, we prove the assertion. 

As an application we achieve the SOC version of Powers-Størmer’s inequality.

Proposition 4.23. For any x, y ∈ Kn and 0 ≤ λ ≤ 1, there holds

tr (x + y − |x − y|) ≤ 2tr xλ◦ y1−λ ≤ tr (x + y + |x − y|) .

Proof. (i) For the first inequality, taking f (t) = tλ for 0 ≤ λ ≤ 1 and applying Proposition 4.22. It is known that f is matrix-monotone with f ((0, ∞)) ⊆ (0, ∞) and g(t) = f (t)t = t1−λ. Then, the inequality follows from (4.21) in Proposition 4.22.

(ii) For the second inequality, we note that

0 Kn x Kn x + y + |x − y|

2 ,

0 Kn y Kn x + y + |x − y|

2 .

Moreover, for 0 ≤ λ ≤ 1, f (t) = tλ is SOC-monotone on [0, ∞). This implies that 0 Kn xλ Kn  x + y + |x − y|

2

λ

,

0 Kn y1−λ Kn  x + y + |x − y|

2

1−λ

. Then, applying Lemma 4.3 gives

tr xλ◦ y1−λ ≤ tr x + y + |x − y|

2

 , which is the desired result. 

According to the definition of Bλ, we observe that B0(x, y) = B1(x, y) = x + y

2 = A1

2(x, y).

This together with Proposition 4.22 leads to

tr(x ∧ y) ≤ tr (Bλ(x, y)) ≤ tr(x ∨ y).

In fact, we can sharpen the upper bound of tr(Bλ(x, y)) as shown in the following propo-sition, which also shows when the maximum occurs. Moreover, the inequality (4.18) remains true for second-order cone, in the following trace version.

Proposition 4.24. For any x, y ∈ Kn and 0 ≤ λ ≤ 1, there holds 2tr

x12 ◦ y12

≤ tr xλ◦ y1−λ+ x1−λ◦ yλ ≤ tr(x + y), which is equivalent to tr

x12 ◦ y12

≤ tr (Bλ(x, y)) ≤ tr(A1

2(x, y)). In particular, tr x1−λ◦ yλ ≤ tr (Aλ(x, y)) .

Proof. It is clear that the inequalities hold when λ = 0, 1. Suppose that λ 6= 0, 1, we set p = λ1, q = 1−λ1 .

For the first inequality, we write x = ξ1u(1)x + ξ2u(2)x , y = µ1u(1)y + µ2u(2)y by spectral decomposition (1.2)-(1.4). We note that ξi, µj ≥ 0 and u(i)x , u(j)y ∈ Kn for all i, j = 1, 2.

Then

xλ◦ y1−λ+ x1−λ◦ yλ− 2x12 ◦ y12 =

2

X

i,j=1



ξiλµ1−λj + ξ1−λi µλj − 2pξiµj

u(i)x ◦ u(j)y ,

which implies

tr



xλ◦ y1−λ+ x1−λ◦ yλ− 2x12 ◦ y12

= tr

2

X

i,j=1



ξiλµ1−λj + ξi1−λµλj − 2pξiµj

u(i)x ◦ u(j)y

!

=

2

X

i,j=1

tr

ξλiµ1−λj + ξi1−λµλj − 2pξiµj

u(i)x ◦ u(j)y 

=

2

X

i,j=1



ξiλµ1−λj + ξi1−λµλj − 2pξiµj

tr u(i)x ◦ u(j)y 

≥ 0,

where the inequality holds by (4.18) and Property 1.3(d).

For the second inequality, by the trace version of Young inequality in Proposition 4.10, we have

tr xλ ◦ y1−λ

≤ tr (xλ)p

p + (y1−λ)q q



= tr x p +y

q

 , tr x1−λ◦ yλ

≤ tr (x1−λ)q

q + (yλ)p p



= tr x q +y

p

 . Adding up these two inequalities together yields the desired result. 

Chapter 5

Possible Extensions

It is known that the concept of convexity plays a central role in many applications includ-ing mathematical economics, engineerinclud-ing, management science, and optimization theory.

Moreover, much attention has been paid to its generalization, to the associated general-ization of the results previously developed for the classical convexity, and to the discovery of necessary and/or sufficient conditions for a function to have generalized convexities.

Some of the known extensions are quasiconvex functions, r-convex functions [11, 151], and SOC-convex functions as introduced in Chapter 2. Other further extensions can be found in [127, 149]. For a single variable continuous, the midpoint-convex function on IR is also a convex function. This result was generalized in [148] by relaxing continuity to lower-semicontinuity and replacing the number 12 with an arbitrary parameter α ∈ (0, 1).

An analogous consequence was obtained in [111, 149] for quasiconvex functions.

To understand the main idea behind r-convex function, we recall some concepts that were independently defined by Martos [106] and Avriel [12], and has been studied by the latter author. Indeed, this concept relies on the classical definition of convex functions and some well-known results from analysis dealing with weighted means of positive numbers.

Let w = (w1, ..., wm) ∈ IRm, q = (q1, ..., qm) ∈ IRm be vectors whose components are positive and nonnegative numbers, respectively, such that Pm

i=1qi = 1. Given the vector of weights q, the weighted r-mean of the numbers w1, ..., wm is defined as below (see [70]):

Mr(w; q) = Mr(w1, ..., wm; q) :=





m P

i=1

qi(wi)r

1/r

if r 6= 0,

m

Q

i=1

(wi)qi if r = 0.

(5.1)

It is well-known from [70] that for s > r, there holds

Ms(w1, ..., wm; q) ≥ Mr(w1, ..., wm; q) (5.2) for all q1, ..., qm ≥ 0 with Pm

i=1qi = 1. The r-convexity is built based on the aforemen-tioned weighted r-mean. For a convex set S ⊆ IRn, a real-valued function f : S ⊆ IRn

185

IR is said to be r-convex if, for any x, y ∈ S, λ ∈ [0, 1], q1 = λ, q2 = 1 − λ, q = (q1, q2), there has

f (q1x + q2y) ≤ lnMr ef (x), ef (y); q .

From (5.1), it can be verified that the above inequality is equivalent to f (λx + (1 − λ)y) ≤

(

lnλerf (x)+ (1 − λ)erf (y)1/r

if r 6= 0,

λf (x) + (1 − λ)f (y) if r = 0. (5.3) Similarly, f is said to be r-concave on S if the inequality (5.3) is reversed. It is clear from the above definition that a real-valued function is convex (concave) if and only if it is 0-convex (0-concave). Besides, for r < 0 (r > 0), an r-convex (r-concave) function is called superconvex (superconcave); while for r > 0 (r < 0), it is called subconvex (subcon-cave). In addition, it can be verified that the r-convexity of f on C with r > 0 (r < 0) is equivalent to the convexity (concavity) of erf on S.

A function f : S ⊆ IRn→ IR is said to be quasiconvex on S if, for all x, y ∈ S, f (λx + (1 − λ)y) ≤ max {f (x), f (y)} , 0 ≤ λ ≤ 1.

Analogously, f is said to be quasiconcave on S if, for all x, y ∈ S, f (λx + (1 − λ)y) ≥ min {f (x), f (y)} , 0 ≤ λ ≤ 1.

From [70], we know that

r→∞lim Mr(w1, ..., wm; q) ≡ M(w1, ..., wm) = max{w1, ..., wm},

r→−∞lim Mr(w1, · · · , wm; q) ≡ M−∞(w1, ..., wm) = min{w1, · · · , wm}.

Then, it follows from (5.2) that M(w1, ..., wm) ≥ Mr(w1, ..., wm; q) ≥ M−∞(w1, ..., wm) for every real number r. Thus, if f is r-convex on S, it is also (+∞)-convex, that is, f (λx + (1 − λ)y) ≤ max{f (x), f (y)} for every x, y ∈ S and λ ∈ [0, 1]. Similarly, if f is r-concave on S, it is also (−∞)-concave, i.e., f (λx + (1 − λ)y) ≥ min{f (x), f (y)}.

The following review some basic properties regarding r-convex function from [11] that will be used in the subsequent analysis.

Property 5.1. Let f : S ⊆ IRn→ IR. Then, the followings hold.

(a) If f is r-convex (r-concave) on S, then f is also s-convex (s-concave) on S for s > r (s < r).

(b) Suppose that f is twice continuously differentiable on S. For any (x, r) ∈ S × IR, we define

φ(x, r) = ∇2f (x) + r∇f (x)∇f (x)T.

Then, f is r-convex on S if and only if φ is positive semidefinite for all x ∈ S.

(c) Every r-convex (r-concave) function on a convex set S is also quasiconvex (quasi-concave) on S.

(d) f is r-convex if and only if (−f ) is (−r)-concave.

(e) Let f be r-convex (r-concave), α ∈ IR and k > 0. Then f + α is r-convex (r-concave) and k · f is (kr)-convex ((kr)-concave).

(f ) Let φ, ψ : S ⊆ IRn→ IR be r-convex (r-concave) and α1, α2 > 0. Then, the function θ defined by

θ(x) = (

lnα1erφ(x)+ α2erψ(x)1/r

if r 6= 0, α1φ(x) + α2ψ(x) if r = 0, is also r-convex (r-concave).

(g) Let φ : S ⊆ IRn → IR be r-convex (r-concave) such that r ≤ 0 (r ≥ 0) and let the real valued function ψ be nondecreasing s-convex (s-concave) on IR with s ∈ IR.

Then, the composite function θ = ψ ◦ φ is also s-convex (s-concave).

(h) φ : S ⊆ IRn → IR is r-convex (r-concave) if and only if, for every x, y ∈ S, the function ψ given by

ψ(λ) = φ ((1 − λ)x + λy) is an r-convex (r-concave) function of λ for 0 ≤ λ ≤ 1.

(i) Let φ be a twice continuously differentiable real quasiconvex function on an open convex set S ⊆ IRn. If there exists a real number r satisfying

r = sup

x∈S, kzk=1

−zT2φ(x)z

[zT∇φ(x)]2 (5.4)

whenever zT∇φ(x) 6= 0, then φ is r-convex for every r ≥ r. We obtain the r-concave analog of the above theorem by replacing supremum in (5.4) by infimum.

在文檔中 SOC Functions and Their Applications (頁 179-192)