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### Monotonicity and circular cone monotonicity associated with circular cones

Jinchuan Zhou 1 Department of Mathematics

School of Science

Shandong University of Technology Zibo 255049, P.R. China E-mail: jinchuanzhou@163.com

Jein-Shan Chen 2 Department of Mathematics National Taiwan Normal University

Taipei 11677, Taiwan E-mail: jschen@math.ntnu.edu.tw

April 26, 2015

(1st revised on March 24, 2016) (2nd revised on April 19, 2016)

(3rd revised on April 29, 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.

AMS subject classifications. 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

1The author’s work is supported by National Natural Science Foundation of China (11101248, 11271233) and Shandong Province Natural Science Foundation (ZR2010AQ026, ZR2012AM016).

2Corresponding author. The author’s work is supported by Ministry of Science and Technology, Taiwan.

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half-aperture angle be θ with θ ∈ (0,π2). Then, the n-dimensional circular cone denoted by Lθ can be expressed as

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

Note that L45 corresponds the well-known second-order cone Kn(SOC, for short), which is given by

Kn:= {x = (x1, x2)T ∈ IR × IRn−1| kx2k ≤ 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, 6, 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 to Kn, given by

λi(x) = x1+ (−1)ikx2k and u(i)x = 1 2

 1

(−1)i2



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for i = 1, 2 with ¯x2 := x2/kx2k if x2 6= 0, and ¯x2 being any vector in IRn−1 satisfying k¯x2k = 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 associ- ated with circular cone case. More specifically, from [31, Theorem 3.1], the spectral factorization associated with Lθ for x is in form of

x = λ1(x) · u(1)x + λ2(x) · u(2)x (4) where

 λ1(x) := x1− kx2kctanθ

λ2(x) := x1+ kx2k 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)

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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− kx2kctanθ)

1 + ctan2θ +f (x1 + kx2k tan θ) 1 + tan2θ



−f (x1− kx2kctanθ)ctanθ

1 + ctan2θ + f (x1+ kx2k 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.

Throughout this paper, we always assume that J is an open interval (finite or infinite) in IR, i.e., J := (t, t0) with t, t0 ∈ IR ∪ {±∞}. Denote S the set of all x ∈ IRn whose spectral values λi(x) for i = 1, 2 belong to J , i.e.,

S := {x ∈ IRn| λi(x) ∈ J, i = 1, 2}.

According to [24], 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

min{λ1(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 between Lθ and Kn (see [31]) 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 hx, yi = Pn

i=1xiyi. Due to Lθ = Lπ

2−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 hx, yiA:= hAx, Ayi. Then

Lθ = {x | hx, yiA ≥ 0, ∀y ∈ Lθ} = {x | hAx, Ayi ≥ 0, ∀y ∈ A−1Kn}

= {x | hAx, yi ≥ 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 | hx, yiA≥ 0, ∀y ∈ Kn} = {x | hAx, Ayi ≥ 0, ∀y ∈ Kn}

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= {x | hA2x, yi ≥ 0, ∀y ∈ Kn} = {x | A2x ∈ Kn} = 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 emphasized that the relation Kn = ALθ does 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 Π. For x ∈ Lθ, we have Ax ∈ Kn which implies

ΠLθ(x) = x = A−1(Ax) = A−1ΠKn(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

ΠLθ(x) = 0 and A−1ΠKn(Ax) =

3

23 8



which says ΠLθ(x) 6= A−1ΠKn(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 example, the works on p-order cone [30], homogeneous cone [15, 29], matrix cone [12];

etc. Unlike the symmetric cone case in which the Euclidean Jordan algebra can unify the analysis [13], 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 cru- cial: (i) spectral factorization associated with circular cones; (ii) smooth and nonsmooth analysis for fLθ given as in (7); (iii) the so-called Lθ-convexity; and (iv) Lθ-monotonicity.

The first three points have been studied in [31], [4, 32], and [33], 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 [18]. 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 .

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To end this section, we say a few words about the notations and present the definitions of monotonicity and Lθ-monotonicity. A matrix M ∈ IRn×n is said to be Lθ-invariant if M h ∈ Lθ for all h ∈ Lθ. We write x Lθ y to mean x − y ∈ Lθ and denote Lθ the polar cone of Lθ, i.e.,

Lθ := {y ∈ IRn| hx, yi ≤ 0, ∀x ∈ Lθ}.

Denote e := (1, 0, . . . , 0)T and use λ(M ), λmin(M ), λmax(M ) for the set of all eigenvalues, the minimum, and the maximum of eigenvalues of M , respectively. Besides, Sn means the space of all symmetric matrices in IRn×n and Sn+ is the cone of positive semidefinite matrices. For a mapping g : IRn → IRm, denote by Dg the set of all differentiable points of g. For convenience, we define 0/0 := 0. Given a real-valued function f : J → IR,

(a) f is said to be Lθ-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 6= y.

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

fLθ(x) − fLθ(y), x − y ≥ µkx − yk2, ∀x, y ∈ S.

### 2 Circular cone monotonicity of f

This section is devoted to the study of Lθ-monotonicity. The main purpose is to provide characterizations of Lθ-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 [25]. 2

Lemma 2.2. Given ζ ∈ IR, u ∈ IRn−1, and a symmetric matrix Ξ ∈ IRn×n. Denote B := {z ∈ IRn−1|kzk ≤ 1}. Then, the following statements hold.

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(a) Ξ being Lθ-invariant is equivalent to Ξctanθ z



∈ Lθ for any z ∈ B.

(b) If Ξ =ζ uT

u H



with H ∈ Sn−1, then Ξ is Lθ-invariant is equivalent to

ζ ≥ kuk tan θ and there exists λ ≥ 0 such that

 ζ2− ctan2θkuk2− λ (ζ tan θ)uT − ctanθuTH (ζ tan θ)u − ctanθHu tan2θuuT − H2+ λI



Sn

+ O.

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

ΞLθ ∈ Lθ ⇐⇒ ΞA−1Kn∈ A−1Kn ⇐⇒ AΞA−1Kn ∈ Kn

⇐⇒ AΞA−11 z



∈ Kn⇐⇒ ΞA−11 z



∈ A−1Kn (9)

⇐⇒ ΞA−11 z



∈ Lθ ⇐⇒ Ξctanθ z



∈ Lθ,

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

(b) From (9), we know that

A Ξ A−11 z



=

 ζ tan θuT ctanθu H

 1 z



= ζ + tan θuTz ctanθu + Hz



∈ Kn, which means

ζ + uTz tan θ ≥ 0, ∀z ∈ B, (10)

and

ζ + uTz tan θ ≥ kctanθu + Hzk, ∀z ∈ B. (11) Note that condition (10) is equivalent to

ζ ≥ tan θ max{−uTz|z ∈ B} = tan θkuk and condition (11) is equivalent to

ζ + tan θuTz2

≥ kctanθu + Hzk2, i.e.,

zT(tan2θuuT − H2)z + 2 ζ tan θuT − ctanθuTHz + ζ2− ctan2θuTu ≥ 0 ∀z ∈ B,

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which can be rewritten as

1 zT Θ1 z



≥ 0 ∀z ∈ B, (12)

with

Θ :=

 ζ2− ctan2θuTu (ζ tan θ)uT − ctanθuTH (ζ tan θ)u − ctanθHu tan2θuuT − H2

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

k vT1 0 0 −I

 k v



≥ 0 =⇒ k vT Θk v



≥ 0, ∀k v



∈ IRn. (13) 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 6= 0, dividing by k2 in the left side of (13) yields h

1 (v

k)Ti1 0 0 −I

" 1 v k

#

≥ 0,

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

1 (v k)Ti

Θ" 1 v k

#

≥ 0.

Hence, the right side of (13) holds.

For k = 0, the left side of (13) is kvk ≤ 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 Θ ensures the existence of λ ≥ 0 such that

 ζ2− ctan2θuTu ζ tan θuT − ctanθuTH ζ tan θu − ctanθHu tan2θuuT − H2



− λ1 0 0 −I



Sn

+ O.

Thus, the proof is complete. 2

Lemma 2.3. For a matrix being in form of H := x1 xT2 x2 αI + β ¯x2T2



, where α, β ∈ IR, then

max{x1+ kx2k, x1− β} + max{0, α − x1+ β}

≥ λmax(H) ≥ λmin(H)

≥ min{x1− kx2k, x1− β} + min{0, α − x1+ β}.

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

x1 xT2 x2 αI + β ¯x2T2



=x1 xT2 x2 x1I



− β0 0

0 I − ¯x2T2



+0 0

0 (α − x1 + β)I



and let

1 :=x1 xT2 x2 x1I



− β0 0

0 I − ¯x2T2



, Ω2 :=0 0

0 (α − x1+ β)I

 .

Then, λ(Ω1) = {x1−kx2k, x1+kx2k, x1−β} by [6, Lemma 1] and λ(Ω2) = {0, α−x1+β}.

Thus, the desired result follows from the following facts:

λmin(Ω1 + Ω2) ≥ λmin(Ω1) + λmin(Ω2) and λmax(Ω1+ Ω2) ≤ λmax(Ω1) + λmax(Ω2).

This completes the proof. 2

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) =

f0(x1)I x2 = 0;

 ξ %¯xT2

%¯x2 τ I + (η − τ )¯x2T2



x2 6= 0, (14)

where

τ := f (λ2(x)) − f (λ1(x))

λ2(x) − λ1(x) , ξ := f01(x))

1 + ctan2θ + f02(x)) 1 + tan2θ,

% := − ctanθ

1 + ctan2θf01(x)) + tan θ

1 + tan2θf02(x)), η := ctan2θ

1 + ctan2θf01(x)) + tan2θ

1 + tan2θf02(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 is Lθ-monotone on J if and only if ∇fLθ(x) is Lθ-invariant for all x ∈ S.

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

fLθ(x + th) − fLθ(x)

t Lθ 0. (15)

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

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“⇐” Suppose that ∇fLθ(x) is Lθ-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 ζ ∈ Lθ, we have

ζ, fLθ(x) − fLθ(y) = Z 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) ∈ Lθ because x + t(x − y) ∈ S (since S is convex) and ∇fLθ is Lθ-invariant over S by hypothesis. Since ζ ∈ Lθ is arbitrary, (16) implies fLθ(x) − fLθ(y) ∈ (Lθ) = Lθ, where the last step is due to the fact that Lθ is a closed convex cone. This means fLθ(x) Lθ fLθ(y). 2

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 is Lθ-monotone on J ;

(b) ∂BfLθ(x) is Lθ-invariant for all x ∈ S;

(c) ∂fLθ(x) is Lθ-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 h ∈ Lθ. Taking the limit yields V h Lθ 0. Since V ∈ ∂BfLθ(x) is arbitrary, this says that ∂BfLθ is Lθ-invariant.

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

iβiVi and P

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

iβiVih ∈ Lθ, since Vi is Lθ-invariant and Lθ is convex. Hence ∂fLθ(x) is Lθ- invariant.

“(c) ⇒ (a)” The proof follows from Theorem 2.1 by replacing (16) with ζ, fLθ(x) − fLθ(y) = hζ, V (x − y)i ≤ 0,

for some V ∈ ∂fLθ(z) with z ∈ [x, y] by the mean-value theorem of Lipschitz functions [10]. 2

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

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

f0(t1) − f0(t2)

≥ 0, (17)

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and

f0(t1) f (t2) − f (t1) t2− t1 f (t2) − f (t1)

t2− t1 f0(t2)

S2

+ O, (18)

then f is Lθ-monotone on J .

Proof. According to Theorem 2.1, it suffices to show that ∇fLθ(x) is Lθ-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 = f0(x1)h ∈ Lθ for all h ∈ Lθ, is equivalent to saying f0(x1) ≥ 0.

Case 2: For x2 6= 0, let

H := τ I + (η − τ )¯x2T2.

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

ξ ≥ k%k tan θ (19)

and there exists λ ≥ 0 such that Υ :=

 ξ2− ctan2θ%2− λ ξ tan θ%¯xT2 − ctanθ%¯xT2H ξ tan θ%¯x2− ctanθ%H ¯x2 tan2θ%22T2 − 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 θf01(x)) − ctanθf02(x)) ≤ − tan θf01(x)) + tan θf02(x))

≤ tan θf01(x)) + ctanθf02(x))

⇐⇒ f02(x)) ≥ 0 and f01(x)) ≥ 1 − ctan2θ

2 f02(x)). (21)

This is ensured by (17) and (18). In fact, if tan θ ≥ ctanθ, then we know from (17) that f01(x)) ≥ f02(x)) ≥ (1 − ctan2θ)/2f02(x)) where the second inequality is due to f02(x)) ≥ 0 by (18). If tan θ ≤ ctanθ, then 1 − ctan2θ ≤ 0, and hence f01(x)) ≥ 0 ≥ (1 − ctan2θ)/2f02(x)) since f0i(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)f01(x))2+ 2(1 + ctan2θ)f01(x))f02(x))



= 1

(tan θ + ctanθ)2



(tan2θ − ctan2θ)f01(x))2+ (ctan2θ − tan2θ)f01(x))f02(x))



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+ 1

(tan θ + ctanθ)2 h

2 + tan2θ + ctan2θ i

f01(x))f02(x))

= µ + f01(x))f02(x)), with

µ

:= 1

(tan θ + ctanθ)2



(tan2θ − ctan2θ)f01(x))2+ (ctan2θ − tan2θ)f01(x))f02(x))



= tan θ − ctanθ

tan θ + ctanθf01(x))h

f01(x)) − f02(x))i

≥ 0,

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

= 1

(tan θ + ctanθ)2



− tan2θ + ctan2θf01(x))2+ tan2θ − ctan2θf01(x))f02(x))



¯ xT2

= −tan θ − ctanθ

tan θ + ctanθf01(x))h

f01(x)) − f02(x))i

¯ xT2

= −µ¯xT2.

In the Υ22-entry, we calculate

tan2θ%22T2 − H2 = −τ2I + τ2 + 1

(tan θ + ctanθ)2 h

(tan2θ − ctan2θ)f01(x))2

−2(1 + tan2θ)f01(x))f02(x)) i

!

¯ x2T2

= −τ2I +



τ2+ µ − f01(x))f02(x))



¯ x2T2. Hence, Υ can be rewritten as

Υ = Υ11 Υ12 Υ21 Υ22



= "µ + f01(x))f02(x)) − λ −µ¯xT2

−µ¯x2 (λ − τ2)I +

τ2+ µ − f01(x))f02(x))

¯ x2T2

# .

Now, applying Lemma 2.3 to Υ, we have

λmin(Υ) ≥ min (

Υ11− |µ|, Υ11−

τ2+ µ − f01(x))f02(x)) )

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+ min (

0, 

λ − τ2

− Υ11+

τ2+ µ − f01(x))f02(x)) )

= minn

f01(x))f02(x)) − λ, 2f01(x))f02(x)) − τ2− λo +2 min

n

0, λ − f01(x))f02(x)) o

. (22)

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

f01(x)) ≥ 0, f02(x)) ≥ 0, and f01(x))f02(x)) ≥  f (λ2(x)) − f (λ1(x)) λ2(x) − λ1(x)

2

, which together with (15) yields

f01(x))f02(x)) ≥ 0 and f01(x))f02(x)) − τ2 ≥ 0.

Thus, we can plug λ := f01(x))f02(x)) ≥ 0 into (22), which gives λmin(Υ) ≥ 0. Hence Υ is positive semi-definite. This completes the proof. 2

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 is Lθ- monotone on J if and only if f0(t) ≥ 0 for all t ∈ J and (tan θ−ctanθ)(f0(t1)−f0(t2)) ≥ 0 for all t1, t2 ∈ J with t1 ≤ t2.

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

f02(x)) ≥ 0, f01(x)) ≥ 1 − ctan2θ

2 f02(x)), (23)

and there exists λ ≥ 0 such that

Υ =µ + f01(x))f02(x)) − λ ±µ

±µ µ − f01(x))f02(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− (f01(x))f02(x)) − λ)2− µ2 ≥ 0, which implies λ = f01(x))f02(x)).

Substituting it into (24) yields

Υ = µ ±µ

±µ µ



= µ 1 ±1

±1 1



S2

+ O,

(13)

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

f02(x)) ≥ 0, f01(x)) ≥ 1 − ctan2θ

2 f02(x)), (tan θ − ctanθ)f01(x))f01(x)) − f02(x)) ≥ 0.

Due to the arbitrariness of λi(x) ∈ J and applying similar arguments following (21), the above conditions give f0(t) ≥ 0 for all t ∈ J and (tan θ − ctanθ)(f0(t1) − f0(t2)) ≥ 0 for all t1, t2 ∈ J with t1 ≤ t2. Thus, the proof is complete. 2

### 3 Monotonicity of f

Lθ

In Section 2, we have shown that the circular cone monotonicity fLθ depends on both the monotonicity of f and the range of the angle θ. Now the following questions arise: how about on the relationship of monotonicity of fLθ and the Lθ-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 [26], we readily have

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

(a) fLθ is 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 fLθ is strictly monotone on S;

(c) fLθ is 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). 2 Lemma 3.3. The following statements hold.

(a) For any h ∈ IRn t1

h1− ctanθ¯xT2h22

+ t2

h1+ tan θ¯xT2h22

+ t3

kh2k2 − (¯xT2h2)2

≥ 0 (25) if and only if ti ≥ 0 for i = 1, 2, 3.

(14)

(b) For any h ∈ IRn/{0}

t1

h1− ctanθ¯xT2h22

+ t2

h1+ tan θ¯xT2h22

+ t3

kh2k2− (¯xT2h2)2

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

Proof. (a) The sufficiency is clear, since |¯xT2h2| ≤ k¯x2kkh2k = kh2k 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 hu, ¯x2i = 0 and kuk = 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 kh2k−¯xT2h2 > 0, then the result holds since t3 > 0. If kh2k−¯xT2h2 = 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. 2

If f is differentiable, it is known that fLθ is differentiable [4, 32]. It then follows from Lemma 3.1 that fLθ is monotone on S if and only if ∇fLθ(x) Sn

+ O for all x ∈ S. Hence, to characterize the monotonicity of fLθ, the first thing is to estimate ∇fLθ(x) Sn+ O via 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 f0i(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) = f0(x1)e, it is clear that ∇fLθ(x) Sn

+ O is equivalent to saying f0(x1) ≥ 0. Then, the desired result follows.

Case 2. For x2 6= 0, denote

b1 := f01(x))

1 + ctan2θ and b2 = f02(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

hh, ∇fLθ(x)hi

= (b1+ b2)h21+ 2%h1T2h2+ τ kh2k2+ (b1ctan2θ + b2tan2θ − τ )(¯xT2h2)2

= (b1+ b2)h21+ 2(−b1ctanθ + b2tan θ)h1T2h2+ (b1ctan2θ + b2tan2θ)(¯xT2h2)2 +τ

kh2k2− (¯xT2h2)2

= b1

h21− 2ctanθh1T2h2+ ctan2θ(¯xT2h2)2 + b2

h21+ 2 tan θh1T2h2+ tan2θ(¯xT2h2)2

(15)

+τ

kh2k2− (¯xT2h2)2

= b1

h1− ctanθ¯xT2h22

+ b2

h1+ tan θ¯xT2h22

+ τ

kh2k2 − (¯xT2h2)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., f01(x)) ≥ 0, f02(x)) ≥ 0, and f (λ2(x)) ≥ f (λ1(x)) due to λ2(x) > λ1(x) in this case. 2

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

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

Then for x2 6= 0, ∇fLθ(x) Sn

+ O if and only if f0i(x)) > 0 for i = 1, 2 and f (λ2(x)) >

f (λ1(x)); for x2 = 0, ∇fLθ(x) Sn+ O if and only if f0(x1) > 0.

When f is non-differentiable, we resort to the subdifferential ∂B(fLθ), whose estimate is given in [32].

Lemma 3.4. Let f : IR → IR be strictly continuous. Then, for any x ∈ IRn, the B- differential ∂B(fLθ)(x) is well defined and nonempty. Moreover,

(i) if x2 6= 0, then

B(fLθ)(x) =









 ξ %xT2/kx2k

%x2/kx2k τ I + (η − τ )x2xT2/kx2k2



τ = f λ2(x) − f λ1(x) λ2(x) − λ1(x) ξ − %ctanθ ∈ ∂Bf (λ1(x)) ξ + % tan θ ∈ ∂Bf (λ2(x)) η = ξ − %(ctanθ − tan θ)









;

(27) (ii) if x2 = 0, then

B(fLθ)(x) ⊂





 ξ %wT

%w τ I + (η − τ )wwT



τ ∈ ∂f (λ1(x)), kwk = 1 ξ − %ctanθ ∈ ∂Bf (λ1(x)) ξ + % tan θ ∈ ∂Bf (λ1(x)) η = ξ − %(ctanθ − tan θ)





. (28)

Lemma 3.4 presents an upper estimation on ∂BfLθ(x) when x2 = 0. Here we give an lower estimation.

(16)

Lemma 3.5. Suppose that f is locally Lipschitz at x with x2 = 0. Then,

Bf (x1)I ⊆ ∂BfLθ(x).

Proof. First, take u ∈ ∂Bf (x1). Since f is locally Lipschitz at x1, there exists t ∈ Df satisfying t → x1. Note that λi(te) = t for i = 1, 2. Then, fLθ is differentiable at te, i.e., te ∈ Df and ∇fLθ(te) = f0(t)I. Hence,

uI = lim

t→x1

f0(t)I = lim

te→x∇fLθ(te) ∈ ∂BfLθ(x).

Since u is an arbitrary element in ∂Bf (x1), the desired result follows. 2

Using Lemmas 3.4 and 3.5, the nonsmooth version of Theorem 3.1 is given below.

Theorem 3.2. Given x ∈ IRn, then ∂BfLθ(x) Sn

+ O if and only if ∂Bf (λi(x)) ≥ 0 for i = 1, 2, and f (λ2(x)) ≥ f (λ1(x)).

Proof. Consider the following two cases.

Case 1. For x2 6= 0, according to (27), we know for V ∈ ∂BfLθ(x), there exists vi

Bf (λi(x)) for i = 1, 2 such that ξ − ρctanθ = v1, ξ + ρ tan θ = v2, and V =

 ξ %¯xT2

%¯x2 τ I + (η − τ )¯x2T2

 . Hence,

ξ = v1tan θ + v2ctanθ

tan θ + ctanθ , ρ = v2− v1

tan θ + ctanθ and η = v1ctanθ + v2tan θ tan θ + ctanθ and

hh, V hi

= v1tan θ + v2ctanθ

tan θ + ctanθ h21 + 2 v2− v1

tan θ + ctanθx¯T2h2h1+ v1ctanθ + v2tan θ

tan θ + ctanθ (¯xT2h2)2 +τh

kh2k2− (¯xT2h2)2i

= tan θ

tan θ + ctanθv1h

h1− (¯xT2h2)ctanθi2

+ ctanθ

tan θ + ctanθv2h

h1+ (¯xT2h2) tan θi2

+τh

kh2k2− (¯xT2h2)2i

. (29)

Applying Lemma 3.3, we have hh, V hi ≥ 0 for all h ∈ IRn if and only if v1, v2 ≥ 0, and τ ≥ 0. In other words, ∂BfLθ(x) Sn

+ O if and only if ∂Bf (λi(x)) ≥ 0 for i = 1, 2 and f (λ2(x)) ≥ f (λ1(x)).

Case 2. For x2 = 0, the sufficiency follows by along the same arguments as above. In fact, the every element of the left set (28) is positive semidefinite. Then, the necessity follows from Lemma 3.5. 2

Likewise, we have a nonsmooth version of Corollary 3.1 which is given below.

(17)

Corollary 3.2. Given x ∈ IRn, then for x2 6= 0, ∂BfLθ(x)  0 if and only if ∂Bf (λi(x)) >

0 for i = 1, 2, and f (λ2(x)) > f (λ1(x)); for x2 = 0, ∂BfLθ(x)  0 if and only if

Bf (x1) > 0.

One point needs to be mentioned here. Because we cannot characterize the strict monotonicity by subgradients (see Lemma 3.1), we resort to the definition of monotonic- ity. Indeed, inspired by [18], we obtain the following result.

Theorem 3.3. Suppose that f is locally Lipschitz continuous on J . Then, (a) fLθ is monotone on S if and only if f is nondecreasing on J ;

(b) fLθ is strictly monotone on S if and only if f is strictly increasing on J ;

(c) fLθ is strongly monotone on S with µ > 0 if and only if f is strongly increasing on S with µ > 0.

Proof. (a) =⇒. Take t1, t2 ∈ J with t1 ≤ t2. Since fLθ is monotone, we have 0 ≤ h(t1− t2)e, fLθ(t1e) − fLθ(t2e)i = (t1 − t2)(f (t1) − f (t2)), which implies f (t1) ≤ f (t2).

⇐=. Take x ∈ S. Since f is monotone on J, we know ∂Bf (t) ≥ 0 for all t ∈ J , and hence ∂f (t) ≥ 0 for t ∈ J . Because λi(x) ∈ J for i = 1, 2, it implies ∂Bf (λi(x)) ≥ 0 for i = 1, 2. In addition, f (λ2(x)) − f (λ1(x)) = γ(λ2(x) − λ1(x)) for some γ ∈ ∂f (t0) and t0 ∈ (λ1(x), λ2(x)) ⊆ J . Hence, f (λ2(x)) ≥ f (λ1(x)) due to γ ∈ ∂f (t0) ≥ 0. Applying Theorem 3.2 yields ∂BfLθ(x) Sn

+ O, which means fLθ is monotone on S by Lemma 3.1.

(b) =⇒. Given t1, t2 ∈ J with t1 > t2. Since fLθ is strictly monotone on S, we have



f (t1) − f (t2)



(t1− t2) = hfLθ(t1e) − fLθ(t2e), (t1− t2)ei > 0,

which together with the fact t1 > t2 yields f (t1) > f (t2). This says f is strictly increasing on J .

⇐=. Let x, y ∈ S with x 6= y. We shall show the strict monotonicity of fLθ by checking definition. To proceed, we consider the following four cases.

Case 1. For x2 = y2 = 0, it is clear that hfLθ(x) − fLθ(y), x − yi =

f (x1) − f (y1)

(x1− y1) > 0,

where the last step is due to the fact that f is strictly monotone and x1 6= y1, since (x1, 0) = x 6= y = (y1, 0) under this case.

Case 2. For x2 = 0 and y2 6= 0, we have hfLθ(y) − fLθ(x), y − xi

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