• 沒有找到結果。

# Monotonicity and Circular Cone Monotonicity Associated with Circular Cones

N/A
N/A
Protected

Share "Monotonicity and Circular Cone Monotonicity Associated with Circular Cones"

Copied!
22
0
0

(1)

DOI 10.1007/s11228-016-0374-7

## Monotonicity and Circular Cone MonotonicityAssociated with Circular Cones

Jinchuan Zhou1· Jein-Shan Chen2

Received: 27 April 2015 / Accepted: 2 May 2016 / Published online: 14 May 2016

Abstract The circular cone Lθ is not self-dual under the standard inner product and includes second-order cone as a special case. In this paper, we focus on the monotonicity of fLθ and circular cone monotonicity of f . Their relationship is discussed as well. Our results show that the angle θ plays a different role in these two concepts.

Keywords Circular cone· Monotonicity · Circular cone monotonicity Mathematics Subject Classification (2010) 26A27· 26B35 · 49J52 · 65K10

1 Introduction

The circular cone [11,32] is a pointed closed convex cone having hyperspherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation. Let its half- aperture angle be θ with θ∈ (0, 90). Then, the n-dimensional circular cone denoted byLθ

can be expressed as

Lθ := {x = (x1, x2)T ∈ IR × IRn−1| cos θx ≤ x1}.

The author’s work is supported by National Natural Science Foundation of China (11101248, 11271233) and Shandong Province Natural Science Foundation (ZR2010AQ026, ZR2012AM016).

The author’s work is supported by Ministry of Science and Technology, Taiwan.

 Jein-Shan Chen jschen@math.ntnu.edu.tw

1 Department of Mathematics, School of Science, Shandong University of Technology, Zibo 255049, People’s Republic of China

2 Department of Mathematics, National Taiwan Normal University, Taipei 11677, Taiwan

(2)

Note thatL45 corresponds to the well-known second-order cone Kn (SOC, for short), which is given by

Kn:= {x = (x1, x2)T ∈ IR × IRn−1| x2 ≤ x1}.

There has been much study on SOC, see [5,6,8] and references therein; to the contrast, not much attention has been paid to circular cone at present. For optimization problems involved SOC, for example, second-order cone programming (SOCP) [1,2,17,19,21] and second-order cone complementarity problems (SOCCP) [3,9,14,16,28], the so-called SOC-functions (see [5–7])

fsoc(x)= f (λ1(x))u(1)x + f (λ2(x))u(2)x ∀x = (x1, x2)T ∈ IR × IRn−1 (1) play an essential role on both theory and algorithm aspects. In expression (1), f : J → IR with J ⊆ IR is a real-valued function and x is decomposed as

x= λ1(x)· u(1)x + λ2(x)· u(2)x (2) where λ1(x), λ2(x)and u(1)x , u(2)x are the spectral values and the associated spectral vectors of x with respect toKn, given by

λi(x)= x1+ (−1)ix2 and u(i)x = 1 2

 1

(−1)i¯x2



(3) for i = 1, 2 with ¯x2 := x2/x2 if x2 = 0, and ¯x2 being any vector in IRn−1satisfying

 ¯x2 = 1 if x2 = 0. The decomposition (2) is called the spectral factorization associated with second-order cone for x. Likewise, there is a similar decomposition for x associated with circular cone case. More specifically, from [31, Theorem 3.1], the spectral factorization associated withLθfor x is in form of

x= λ1(x)· u(1)x + λ2(x)· u(2)x (4)

where 

λ1(x) := x1− x2ctanθ

λ2(x) := x1+ x2 tan θ (5)

and ⎧

⎪⎪

⎪⎪

u(1)x := 1 1+ ctan2θ

1 0

0 ctanθ· I

  1

− ¯x2



=

 sin2θ

−(sin θ cos θ) ¯x2



u(2)x := 1 1+ tan2θ

1 0

0 tan θ· I

  1

¯x2



=

 cos2θ (sin θ cos θ )¯x2

 (6)

Analogously, for any given f : IR → IR, we can define the following vector-valued function for the setting of circular cone:

fLθ(x):= f (λ1(x)) u(1)x + f (λ2(x)) u(2)x . (7) For convenience, we sometime write out the explicit expression for (7) by plugging in λi(x) and u(i)x :

fLθ(x)=

⎢⎣

f (x1− x2ctanθ)

1+ ctan2θ +f (x1+ x2 tan θ) 1+ tan2θ

f (x1− x2ctanθ)ctanθ

1+ ctan2θ +f (x1+ x2 tan θ) tan θ 1+ tan2θ

¯x2

⎦ . (8)

Clearly, as θ = 45, the decomposition (4)–(8) reduces to (1)–(3). Since our main target is on circular cone, in the subsequent contexts of the whole paper, λi and u(i)x stands for (5) and (6), respectively.

(3)

Throughout this paper, we always assume that J is an open interval (finite or infinite) in IR, i.e., J := (t, t)with t, t ∈ IR ∪ {±∞}. Denote S the set of all x ∈ IRnwhose spectral values λi(x)for i= 1, 2 belong to J , i.e.,

S:= {x ∈ IRn| λi(x)∈ J, i = 1, 2}.

According to , we know S is open if and only if J is open. In addition, as J is an interval, we know S is convex because

min1(x), λ1(y)} ≤ λ1(βx+ (1 − β)y) ≤ λ2(βx+ (1 − β)y) ≤ max{λ2(x), λ2(y)}.

We point out that there is a close relation betweenLθandKn(see ) as below Kn= ALθ where A:=

tan θ 0 0 I

 .

It is well-known that Kn is a self-dual cone in the standard inner product x, y =

n

i=1xiyi. Due toLθ = Lπ2−θ by [31, Theorem 2.1],Lθ is not a self-dual cone unless θ = 45. In fact, we can construct a new inner product which ensures the circular cone Lθ is self-dual. More precisely, we define an inner product associated with A as x, yA:= Ax, Ay. Then

Lθ = {x | x, yA≥ 0, ∀y ∈Lθ} = {x | Ax, Ay ≥ 0, ∀y ∈ A−1Kn}

= {x | Ax, y ≥ 0, ∀y ∈Kn} = {x | Ax ∈Kn}

= A−1Kn=Lθ.

However, under this new inner product the second-order cone is not self-dual, because (Kn) = {x | x, yA≥ 0, ∀y ∈Kn} = {x | Ax, Ay ≥ 0, ∀y ∈Kn}

= {x | A2x, y ≥ 0, ∀y ∈Kn} = {x | A2xKn} = A−2Kn.

Since we cannot find an inner product such that the circular cone and second-order cone are both self-dual simultaneously, we must choose an inner product from the standard inner product or the new inner product associated with A. In view of the well-known properties regarding second-order cone and second-order cone programming (in which many results are based on the Jordan algebra and second-order cones are considered as self-dual cones), we adopt the standard inner product in this paper.

Our main attention in this paper is on the vector-valued function fLθ. It should be empha- sized that the relationKn= ALθdoes not guarantee that there exists a similar close relation between fLθ and fsoc. For example, take f (t) to be a simple function max{t, 0}, which corresponds to the projection operator . For xLθ, we have AxKnwhich implies

Lθ(x)= x = A−1(Ax)= A−1Kn(Ax).

Unfortunately, the above relation fails to hold when x /Lθ. To see this, we let tan θ = 1/4 and x= (−1, 1)T. Then, it can be verified

Lθ(x)= 0 and A−1Kn(Ax)=

3

23 8



which says Lθ(x) = A−1Kn(Ax). This example undoubtedly indicates that we cannot study fLθ by simply resorting to fsoc. Hence, it is necessary to study fLθ directly, and the results in this paper are neither trivial nor being taken for granted.

Much attention has been paid to symmetric cone optimizations, see [22,23,27] and references therein. Non-symmetric cone optimization research is much more recent; for

(4)

example, the works on p-order cone , homogeneous cone [15,29], matrix cone ;

etc. Unlike the symmetric cone case in which the Euclidean Jordan algebra can unify the analysis , so far no unifying algebra structure has been found for non-symmetric cones.

In other words, we need to study each non-symmetric cones according to their different properties involved. For circular cone, a special non-symmetric cone, and circular cone optimization, like when dealing with SOCP and SOCCP, the following studies are crucial:

(i) spectral factorization associated with circular cones; (ii) smooth and nonsmooth analysis for fLθ given as in (7); (iii) the so-calledLθ-convexity; and (iv) Lθ-monotonicity. The first three points have been studied in [4,31,32], and , respectively. Here, we focus on the fourth item, that is, monotonicity. The SOC-monotonicity of f have been discussed thoroughly in [5,7,24]; and the monotonicity of the spectral operator of symmetric cone has been studied in . The main aim of this paper is studying those monotonicity properties in the framework of circular cone. Our results reveal that the angle θ plays different role in these two concepts. More precisely, the circular cone monotonicity of f depends on f and θ, whereas the monotonicity of fLθ only depends on f .

To end this section, we say a few words about the notations and present the definitions of monotonicity andLθ-monotonicity. A matrix M ∈ IRn×n is said to beLθ-invariant if MhLθ for all hLθ. We write xLθ yto mean x− y ∈Lθand denoteLθthe polar cone ofLθ, i.e.,

Lθ:= {y ∈ IRn| x, y ≤ 0, ∀x ∈Lθ}.

Denote e:= (1, 0, . . . , 0)T and use λ(M), λmin(M), λmax(M)for the set of all eigenval- ues, the minimum, and the maximum of eigenvalues of M, respectively. Besides,Snmeans the space of all symmetric matrices in IRn×n andSn+ is the cone of positive semidefinite matrices. For a mapping g: IRn→ IRm, denote by Dgthe set of all differentiable points of g. For convenience, we define 0/0:= 0. Given a real-valued function f : J → IR, (a) fis said to beLθ-monotone on J if for any x, y∈ S,

xLθ y =⇒ fLθ(x)Lθ fLθ(y); (b) fLθ is said to be monotone on S if

fLθ(x)− fLθ(y), x− y

≥ 0, ∀x, y ∈ S.

(c) fLθ is said to be strictly monotone on S if

fLθ(x)− fLθ(y), x− y

>0, ∀x, y ∈ S, x = y.

(d) fLθ is said to be strongly monotone on S with μ > 0 if

fLθ(x)− fLθ(y), x− y

≥ μx − y2, ∀x, y ∈ S.

2 Circular Cone Monotonicity off

This section is devoted to the study ofLθ-monotonicity. The main purpose is to provide characterizations ofLθ-monotone functions. To this end, we need a few technical lemmas.

Lemma 2.1 Let A, B be symmetric matrices and yTAy > 0 for some y. Then, the implication[zTAz≥ 0 =⇒ zTBz≥ 0] is valid if and only if B Sn+ λA for some λ≥ 0.

Proof This is the well known S-Lemma, see [7, Lemma 3.1] or .

(5)

Lemma 2.2 Given ζ ∈ IR, u ∈ IRn−1, and a symmetric matrix ∈ IRn×n. DenoteB :=

{z ∈ IRn−1|z ≤ 1}. Then, the following statements hold.

(a)  beingLθ-invariant is equivalent to 

ctanθ z



Lθfor any z∈ B.

(b) If =

ζ uT u H



with H ∈ Sn−1, then  isLθ-invariant is equivalent to ζ ≥ u tan θ

and there exists λ≥ 0 such that

ζ2− ctan2θu2− λ (ζ tan θ)uT − ctanθuTH (ζtan θ )u− ctanθH u tan2θ uuT − H2+ λI



Sn+O.

Proof (a) The result follows from the following observation:

LθLθ ⇐⇒ A−1Kn∈ A−1Kn⇐⇒ AA−1KnKn

⇐⇒ AA−1

1 z



Kn⇐⇒ A−1

1 z



∈ A−1Kn

⇐⇒ A−1

1 z



Lθ ⇐⇒ 

ctanθ z



Lθ, (9)

where the third equivalence comes from [7, Lemma 3.2].

(b) From (9), we know that A  A−1

1 z



=

 ζ tan θ uT ctanθ u H

 1 z



=

ζ+ tan θuTz ctanθ u+ Hz



Kn, which means

ζ+ uTztan θ ≥ 0, ∀z ∈ B, (10)

and

ζ+ uTztan θ≥ ctanθu + Hz, ∀z ∈ B. (11) Note that condition (10) is equivalent to

ζ ≥ tan θ max{−uTz|z ∈ B} = tan θu

and condition (11) is equivalent to



ζ+ tan θuTz2

≥ ctanθu + Hz2, i.e.,

zT(tan2θ uuT − H2)z+ 2

ζtan θ uT − ctanθuTH



z+ ζ2− ctan2θ uTu≥ 0 ∀z ∈ B, which can be rewritten as

 1 zT

1 z



≥ 0 ∀z ∈ B, (12)

with

:=

 ζ2− ctan2θ uTu (ζtan θ )uT− ctanθuTH (ζtan θ )u− ctanθH u tan2θ uuT − H2

 . We now claim that (12) is equivalent to the following implication:

 k vT

 1 0 0 −I

 k v



≥ 0 =⇒ 

k vT



k v



≥ 0, ∀

k v



∈ IRn. (13)

(6)

First, we see that (12) corresponds to the case of k= 1 in (13). Hence, it only needs to show how to obtain (13) from (12). We proceed the arguments by considering the following two cases.

For k = 0, dividing by k2in the left side of (13) yields

 1 (v

k)T

 1 0 0 −I

 1 v k



≥ 0, which implies v/k∈ B. Then, it follows from (12) that

 1 (v

k)T

1 v k



≥ 0.

Hence, the right side of (13) holds.

For k= 0, the left side of (13) isv ≤ 0, which says v = 0, i.e., (k, v)T = 0. Therefore, the right side of (13) holds clearly.

Now, applying Lemma 2.1 to ensures the existence of λ≥ 0 such that

 ζ2− ctan2θ uTu ζtan θ uT − ctanθuTH ζtan θ u− ctanθH u tan2θ uuT − H2



− λ

1 0 0 −I



Sn+O.

Thus, the proof is complete.

Lemma 2.3 For a matrix being in form of H :=

x1 x2T x2 αI+ β ¯x2¯x2T



, where α, β ∈ IR, then

max{x1+ x2, x1− β} + max{0, α − x1+ β}

≥ λmax(H )≥ λmin(H )

≥ min{x1− x2, x1− β} + min{0, α − x1+ β}.

Proof First, we split H as sum of three special matrices, i.e.,

x1 x2T x2 αI+ β ¯x2¯x2T



=

x1 xT2 x2 x1I



− β

0 0

0 I− ¯x2¯x2T

 +

0 0

0 (α− x1+ β)I



and let 1:=

x1 x2T x2 x1I



− β

0 0

0 I− ¯x2¯xT2



, 2:=

0 0

0 (α− x1+ β)I

 . Then, λ( 1)= {x1− x2, x1+ x2, x1− β} by [6, Lemma 1] and λ( 2)= {0, α − x1+ β}. Thus, the desired result follows from the following facts:

λmin( 1+ 2)≥ λmin( 1)+ λmin( 2) and λmax( 1+ 2)≤ λmax( 1)+ λmax( 2).

This completes the proof.

Next, we turn our attention to the vector-valued function fLθ defined as in (7). Recall from [4,32] that fLθ is differentiable at x if and only if f is differentiable at λi(x)for i= 1, 2 and

∇fLθ(x)=

⎧⎨

f(x1)I x2= 0;

 ξ ¯x2T ¯x2 τ I+ (η − τ) ¯x2¯x2T



x2 = 0, (14)

(7)

where

τ := f (λ2(x))− f (λ1(x))

λ2(x)− λ1(x) , ξ:= f1(x))

1+ ctan2θ + f2(x)) 1+ tan2θ, := − ctanθ

1+ ctan2θf1(x))+ tan θ

1+ tan2θf2(x)), η := ctan2θ

1+ ctan2θf1(x))+ tan2θ

1+ tan2θf2(x)).

The following result shows that if fLθ is differentiable, then we can characterize the Lθ-monotonicity of f via the gradient∇fLθ.

Theorem 2.1 Suppose that f : J → IR is differentiable. Then, f isLθ-monotone on J if and only if∇fLθ(x) isLθ-invariant for all x∈ S.

Proof “⇒” Suppose that f isLθ-monotone. Take x∈ S and h ∈Lθ, what we want to prove is∇fLθ(x)hLθ. From theLθ-monotonicity of f , we know fLθ(x+ th) Lθ fLθ(x) for all t > 0. Note thatLθ is a cone. Hence

fLθ(x+ th) − fLθ(x)

t Lθ 0. (15)

Since Lθ is closed, taking the limit as t → 0+ yields ∇fLθ(x)h Lθ 0, i.e.,

∇fLθ(x)hLθ.

⇐” Suppose that ∇fLθ(x)isLθ-invariant for all x∈ S. Take x, y ∈ S with x Lθ y(i.e., x−y ∈Lθ). In order to show the desired result, we need to argue that fLθ(x)Lθ fLθ(y).

For any ζLθ, we have

ζ, fLθ(x)− fLθ(y)

=

 1 0

ζ,∇fLθ(x+ t(x − y)) (x − y)

dt≤ 0, (16) where the last step comes from∇fLθ(x+ t(x − y))(x − y) ∈Lθbecause x+ t(x − y) ∈ S (since S is convex) and∇fLθ is Lθ-invariant over S by hypothesis. Since ζLθ is arbitrary, (16) implies fLθ(x)− fLθ(y)∈ (Lθ) = Lθ, where the last step is due to the fact thatLθis a closed convex cone. This means fLθ(x)Lθ fLθ(y).

Note that f is Lipschitz continuous on J if and only if fLθ is Lipschitz continuous on S, see [4,32]. The nonsmooth version of Theorem 2.1 is given below.

Theorem 2.2 Suppose that f : J → IR is Lipschitz continuous on J . Then the following statements are equivalent.

(a) f isLθ-monotone on J ;

(b) BfLθ(x) isLθ-invariant for all x∈ S;

(c) ∂fLθ(x) isLθ-invariant for all x∈ S.

Proof “(a)⇒ (b)” Take V ∈ ∂BfLθ(x), then by definition of B-subdifferential there exists {xk} ⊂ Df such that xk → x and ∇fLθ(xi) → V . According to (15), we obtain

∇fLθ(xi)hLθ 0 for hLθ. Taking the limit yields V hLθ 0. Since V ∈ ∂BfLθ(x)is arbitrary, this says that ∂BfLθisLθ-invariant.

(8)

“(b)⇒ (c)” Take V ∈ ∂fLθ(x), then by definition, there exists Vi ∈ ∂BfLθ(x)and βi[0, 1] such that V = 

iβiVi and

iβi = 1. Thus, for any h ∈ Lθ, we have V h =



iβiVihLθ, since ViisLθ-invariant andLθis convex. Hence ∂fLθ(x)isLθ-invariant.

“(c)⇒ (a)” The proof follows from Theorem 2.1 by replacing (16) with

ζ, fLθ(x)− fLθ(y)

= ζ, V (x − y) ≤ 0,

for some V ∈ ∂fLθ(z)with z∈ [x, y] by the mean-value theorem of Lipschitz functions .

With these preparations, we provide a sufficient condition for theLθ-monotonicity.

Theorem 2.3 Suppose that f : J → IR is differentiable. If for all t1, t2∈ J with t1≤ t2, (tan θ− ctanθ)

f(t1)− f(t2)

≥ 0, (17)

and

⎢⎣ f(t1) f (t2)− f (t1) t2− t1

f (t2)− f (t1) t2− t1

f(t2)

⎥⎦ S2+O, (18)

then f isLθ-monotone on J .

Proof According to Theorem 2.1, it suffices to show that∇fLθ(x)isLθ-invariant for all x∈ S. We proceed by discussing the following two cases.

Case 1 For x2 = 0, in this case it is clear that ∇fLθ(x) being Lθ-invariant, i.e.,

∇fLθ(x)h= f(x1)hLθfor all hLθ, is equivalent to saying f(x1)≥ 0.

Case 2 For x2 = 0, let

H := τI + (η − τ) ¯x2¯x2T.

Then, applying Lemma 2.2 to the formula of∇fLθ(x)in (14),∇fLθ(x)isLθ-invariant if and only if

ξ ≥   tan θ (19)

and there exists λ≥ 0 such that ϒ:=

 ξ2− ctan2θ 2− λ ξtan θ ¯x2T − ctanθ ¯x2TH ξtan θ ¯x2− ctanθ H ¯x2 tan2θ 2¯x2¯x2T − H2+ λI



Sn+O. (20) Hence, to achieve the desired result, it is equivalent to showing that the conditions (17) and (18) can guarantee the validity of the conditions (19) and (20). To check this, we first note that (19) is equivalent to

− tan θf1(x))− ctanθf2(x))≤ − tan θf1(x))+ tan θf2(x))

≤ tan θf1(x))+ ctanθf2(x))

⇐⇒ f2(x))≥ 0 and f1(x))≥1− ctan2θ

2 f2(x)). (21)

This is ensured by (17) and (18). In fact, if tan θ ≥ ctanθ, then we know from (17) that f1(x))≥ f2(x))≥ 

(1− ctan2θ )/2

f2(x))where the second inequality is due

(9)

to f2(x))≥ 0 by (18). If tan θ ≤ ctanθ, then 1 − ctan2θ ≤ 0, and hence f1(x))≥ 0≥

(1− ctan2θ )/2

f2(x))since fi(x))≥ 0 for i = 1, 2 by (18).

Now let us look into the entries of ϒ. In the ϒ11-entry, we calculate ξ2− ctan2θ 2

= 1

(tan θ+ ctanθ)2



(tan2θ− ctan2θ )f1(x))2+ 2(1 + ctan2θ )f1(x))f2(x))



= 1

(tan θ+ ctanθ)2



(tan2θ− ctan2θ )f1(x))2+ (ctan2θ− tan2θ )f1(x))f2(x))

+ 1

(tan θ+ ctanθ)2



2+ tan2θ+ ctan2θ



f1(x))f2(x))

= μ + f1(x))f2(x)), with

μ := 1

(tan θ+ ctanθ)2



(tan2θ− ctan2θ )f1(x))2+ (ctan2θ− tan2θ )f1(x))f2(x))



= tan θ− ctanθ

tan θ+ ctanθf1(x))

f1(x))− f2(x))

≥ 0,

where the last step is due to (17). In the ϒ12-entry and ϒ21-entry, we calculate tan θ ) ¯xT2 − ctanθ ¯xT2H

= 1

(tan θ+ ctanθ)2

− tan2θ+ ctan2θ



f1(x))2+

tan2θ− ctan2θ



f1(x))f2(x))

¯x2T

= −tan θ− ctanθ

tan θ+ ctanθf1(x))

f1(x))− f2(x))

¯x2T

= −μ ¯x2T.

In the ϒ22-entry, we calculate tan2θ 2¯x2¯x2T− H2 = −τ2I+

τ2+ 1

(tan θ+ ctanθ)2



(tan2θ− ctan2θ )f1(x))2

−2(1 + tan2θ )f1(x))f2(x))

¯x2¯xT2

= −τ2I+

τ2+ μ − f1(x))f2(x))

¯x2¯x2T.

Hence, ϒ can be rewritten as ϒ =

ϒ11 ϒ12 ϒ21 ϒ22



=

μ+ f1(x))f2(x))− λ −μ ¯x2T

−μ ¯x2 (λ− τ2)I+

τ2+ μ − f1(x))f2(x))

¯x2¯x2T

 . Now, applying Lemma 2.3 to ϒ, we have

λmin(ϒ ) ≥ min

ϒ11− |μ|, ϒ11−

τ2+ μ − f1(x))f2(x))



+ min 0, 

λ− τ2

− ϒ11+

τ2+ μ − f1(x))f2(x))

= min

f1(x))f2(x))− λ, 2f1(x))f2(x))− τ2− λ +2 min!

0, λ− f1(x))f2(x))"

. (22)

(10)

Using λ1(x)≤ λ2(x)and condition (18) ensures

f1(x))≥ 0, f2(x))≥ 0, and f1(x))f2(x))

f (λ2(x))− f (λ1(x)) λ2(x)− λ1(x)

2

, which together with (15) yields

f1(x))f2(x))≥ 0 and f1(x))f2(x))− τ2≥ 0.

Thus, we can plug λ:= f1(x))f2(x))≥ 0 into (22), which gives λmin(ϒ ) ≥ 0.

Hence ϒ is positive semi-definite. This completes the proof.

Remark 2.1 The condition (17) holds automatically when θ = 45. In other case, some addition requirement needs to be imposed on f . For instance, f is required to be convex as θ ∈ (0, 45)while f is required to be concave as θ ∈ (45,90). This indicates that the angle plays an essential role in the framework of circular cone, i.e., the assumption on f is dependent on the range of the angle.

Based on the above, we can achieve a necessary and sufficient condition for Lθ- monotonicity in the special case of n= 2.

Theorem 2.4 Suppose that f : J → IR is differentiable on J and n = 2. Then, f isLθ- monotone on J if and only if f(t)≥ 0 for all t ∈ J and (tan θ −ctanθ)(f(t1)−f(t2))≥ 0 for all t1, t2∈ J with t1≤ t2.

Proof In light of the proof of Theorem 2.3, we know that f isLθ-monotone if and only if for any x∈ S,

f2(x))≥ 0, f1(x))≥1− ctan2θ

2 f2(x)), (23)

and there exists λ≥ 0 such that ϒ=

μ+ f1(x))f2(x))− λ ±μ

±μ μ− f1(x))f2(x))+ λ



S2+O, (24) where the form of ϒ comes from the fact that ¯x2 = ±1 in this case. It follows from (24) that μ2− (f1(x))f2(x))− λ)2− μ2 ≥ 0, which implies λ = f1(x))f2(x)).

Substituting it into (24) yields ϒ=

 μ ±μ

±μ μ



= μ

 1 ±1

±1 1



S2+O,

which in turn implies μ≥ 0. Hence, the conditions (23) and (24) are equivalent to

⎧⎨

f2(x))≥ 0, f1(x))≥1− ctan2θ

2 f2(x)), (tan θ− ctanθ)f1(x))

f1(x))− f2(x))

≥ 0.

Due to the arbitrariness of λi(x)∈ J and applying similar arguments following (21), the above conditions give f(t)≥ 0 for all t ∈ J and (tan θ − ctanθ)(f(t1)− f(t2))≥ 0 for all t1, t2∈ J with t1≤ t2. Thus, the proof is complete.

(11)

3 Monotonicity offLθ

In Section 2, we have shown that the circular cone monotonicity of f depends on both the monotonicity of f and the range of the angle θ . Now the following questions arise:

how about on the relationship between the monotonicity of fLθ and theLθ-monotonicity of f ? Whether the monotonicity of fLθ also depends on θ ? This is the main motivation of this section. First, for a mapping H : IRn → IRn, let us denote ∂H (x) Sn+ O (or

∂H (x)Sn+ O) to mean that each elements in ∂H (x) is positive semi-definite (or positive definite), i.e.,

∂H (x)Sn+O (orSn+O) ⇐⇒ A Sn+O (orSn+O), ∀A ∈ ∂H(x).

Taking into account of the result in , we readily have

Lemma 3.1 Let f be Lipschitz continuous on J . The following statements hold:

(a) fLθis monotone on S if and only if ∂fLθ(x)Sn+ O for all x∈ S;

(b) If ∂fLθ(x)Sn+O for all x∈ S, then fLθis strictly monotone on S;

(c) fLθis strongly monotone on S if and only if there exists μ > 0 such that ∂fLθ(x)Sn+

μI for all x∈ S.

Lemma 3.2 Suppose that f is Lipschitz continuous. Then,

BfLθ(x)Sn+O⇐⇒ ∂fLθ(x)Sn+O and ∂BfLθ(x)Sn+O⇐⇒ ∂fLθ(x)Sn+O.

Proof The result follows immediately from the fact ∂fLθ(x)= conv∂BfLθ(x).

Lemma 3.3 The following statements hold.

(a) For any h∈ IRn t1



h1− ctanθ ¯x2Th2

2

+ t2



h1+ tan θ ¯xT2h2

2

+ t3

h22− ( ¯x2Th2)2

≥ 0 (25)

if and only if ti ≥ 0 for i = 1, 2, 3.

(b) For any h∈ IRn\{0}

t1



h1− ctanθ ¯x2Th2

2

+ t2



h1+ tan θ ¯x2Th2

2

+ t3

h22− ( ¯x2Th2)2



>0 (26) if and only if ti >0 for i= 1, 2, 3.

Proof (a) The sufficiency is clear, since| ¯xT2h2| ≤  ¯x2h2 = h2 by Cauchy-Schwartz inequality. Now let us show the necessity. Taking h= (1, −ctanθ ¯x2)T, then (25) equal to t1(1+ctan2θ )2≥ 0, which implies t1≥ 0. Similarly, let h = (1, tan θ ¯x2)T, then (25) yields t2(1+ tan2θ )2≥ 0, implying t2≥ 0. Finally, let h = (0, u)T with u satisfying u, ¯x2 = 0 andu = 1, then it follows from (25) that t3≥ 0.

(b) The necessity is the same of the argument as given in part (a). For sufficiency, take a nonzero vector h. Ifh2− ¯xT2h2>0, then the result holds since t3>0. Ifh2− ¯x2Th2= 0, then h2 = β ¯x2. So the left side of (26) takes t1(h1− βctanθ)2+ t2(h1+ β tan θ)2 >0, because h1− βctanθ = h1+ β tan θ = 0 only happened when β = 0 and h1 = 0. This means h2= 0 since h2= β ¯x2, so h= 0.

(12)

If f is differentiable, it is known that fLθ is differentiable [4,32]. It then follows from Lemma 3.1 that fLθis monotone on S if and only if∇fLθ(x)Sn+ Ofor all x∈ S. Hence, to characterize the monotonicity of fLθ, the first thing is to estimate∇fLθ(x)Sn+ Ovia f.

Theorem 3.1 Given x ∈ IRn and suppose that f is differentiable at λi(x) for i = 1, 2.

Then∇fLθ(x)Sn+O if and only if fi(x))≥ 0 for i = 1, 2 and f (λ2(x))≥ f (λ1(x)).

Proof The proof is divided into the following two cases.

Case 1 For x2 = 0, using ∇fLθ(x) = f(x1)e, it is clear that∇fLθ(x) Sn+ O is equivalent to saying f(x1)≥ 0. Then, the desired result follows.

Case 2 For x2 = 0, denote

b1:= f1(x))

1+ ctan2θ and b2= f2(x)) 1+ tan2θ.

Then, ξ= b1+ b2, = −b1ctanθ+ b2tan θ , and η= b1ctan2θ+ b2tan2θ. Hence, for all h= (h1, h2)T ∈ IR × IRn−1, we have

h, ∇fLθ(x)h

= (b1+ b2)h21+ 2 h1¯xT2h2+ τh22+ (b1ctan2θ+ b2tan2θ− τ)( ¯x2Th2)2

= (b1+ b2)h21+ 2(−b1ctanθ+ b2tan θ )h1¯x2Th2+ (b1ctan2θ+ b2tan2θ )(¯x2Th2)2 

h22− ( ¯x2Th2)2

= b1



h21− 2ctanθh1¯x2Th2+ ctan2θ (¯x2Th2)2

+ b2



h21+ 2 tan θh1¯x2Th2+ tan2θ (¯x2Th2)2

 

h22− ( ¯x2Th2)2



= b1



h1− ctanθ ¯x2Th2

2

+ b2



h1+ tan θ ¯x2Th2

2

+ τ

h22− ( ¯x2Th2)2

 .

In light of Lemma 3.3, the desired result is equivalent to b1≥ 0, b2≥ 0, τ = f (λ2(x))− f (λ1(x))

λ2(x)− λ1(x) ≥ 0,

i.e., f1(x))≥ 0, f2(x))≥ 0, and f (λ2(x))≥ f (λ1(x))due to λ2(x) > λ1(x)in this case.

By following almost the same arguments as given in Theorem 3.1, we further obtain the following consequence.

Corollary 3.1 Given x ∈ IRnand suppose that f is differentiable at λi(x) for i = 1, 2.

Then for x2 = 0, ∇fLθ(x)Sn+O if and only if fi(x)) >0 for i= 1, 2 and f (λ2(x)) >

f (λ1(x)); for x2= 0, ∇fLθ(x)Sn+O if and only if f(x1) >0.

When f is non-differentiable, we resort to the subdifferential ∂B(fLθ), whose estimate is given in .

patients with stage I/II disease but not in those with stage III disease.43 A high serum level of VEGF is associated with poor survival among patients with small cell lung

Then we can draw a right triangle with angle θ as in Figure 3 and deduce from the Pythagorean Theorem that the third side has length.. This enables us to read from the

This proves that ρ is the radius of convergence of S, and that parts (i) and (iii) hold.. W EN -C HING L IEN Advanced

A light beam incident on a small circular aperture becomes diffracted and its light intensity pattern after passing through the aperture is a diffraction pattern with circular

This circular recapitulates the prevailing rules and regulations on collection of fees in kindergartens, kindergarten-cum-child care centres and schools with kindergarten

Then, we tested the influence of θ for the rate of convergence of Algorithm 4.1, by using this algorithm with α = 15 and four different θ to solve a test ex- ample generated as

Then, it is easy to see that there are 9 problems for which the iterative numbers of the algorithm using ψ α,θ,p in the case of θ = 1 and p = 3 are less than the one of the

In this paper, we have shown that how to construct complementarity functions for the circular cone complementarity problem, and have proposed four classes of merit func- tions for

According to the authors’ earlier experience on symmetric cone optimization, we believe that spectral decomposition associated with cones, nonsmooth analysis regarding

 F. Goldfarb, Second-order cone programming, Math. Alzalg, M.Pirhaji, Elliptic cone optimization and prime-dual path-following algorithms, Optimization. Terlaky, Notes on duality

We point out that extending the concepts of r-convex and quasi-convex functions to the setting associated with second-order cone, which be- longs to symmetric cones, is not easy

In section 4, based on the cases of circular cone eigenvalue optimization problems, we study the corresponding properties of the solutions for p-order cone eigenvalue

104 As shown in Figure 5, spin-restricted TAO- B3LYP and TAO-B3LYP-D3 (with a θ between 50 and 70 mhartree), TAO-PBE0 (with a θ between 60 and 80 mhartree), and TAO-BHHLYP (with a

• LQCD calculation of the neutron EDM for 2+1 flavors ,→ simulation at various pion masses &amp; lattice volumes. ,→ working with an imaginary θ [th’y assumed to be analytic at θ

Nonlinear Convex Anal. Chen, The variational geometry, projection expression and decomposi- tion associated with ellipsoidal cones, J. Nonlinear Convex Anal. Chen, From symmteric

Chen, Properties of circular cone and spectral factorization associated with circular cone, to appear in Journal of Nonlinear and Convex Analysis, 2013.

It is well-known that, to deal with symmetric cone optimization problems, such as second-order cone optimization problems and positive semi-definite optimization prob- lems, this

Chen, “Alternative proofs for some results of vector- valued functions associated with second-order cone,” Journal of Nonlinear and Convex Analysis, vol.. Chen, “The convex and

In this paper, by using the special structure of circular cone, we mainly establish the B-subdifferential (the approach we considered here is more directly and depended on the

According to the authors’ earlier experience on symmetric cone optimization, we believe that spectral decomposition associated with cones, nonsmooth analysis regarding cone-

For circular cone, a special non-symmetric cone, and circular cone optimization, like when dealing with SOCP and SOCCP, the following studies are cru- cial: (i) spectral

Taking second-order cone optimization and complementarity problems for example, there have proposed many ef- fective solution methods, including the interior point methods [1, 2, 3,

Abstract In this paper, we study the parabolic second-order directional derivative in the Hadamard sense of a vector-valued function associated with circular cone.. The