Monotonicity and Circular Cone Monotonicity Associated with Circular Cones

DOI 10.1007/s11228-016-0374-7

Monotonicity and Circular Cone Monotonicity Associated with Circular Cones

Jinchuan Zhou¹· Jein-Shan Chen²

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

© Springer Science+Business Media Dordrecht 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 f^Lθ 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 θx ≤ 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

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])

f^soc(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

λ_i(x)= x1+ (−1)ⁱx2 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− x2ctanθ

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:

f^Lθ(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 λ_i(x) and u⁽ⁱ⁾_x :

f^Lθ(x)=

⎡

⎢⎣

f (x₁− x2ctanθ)

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

−f (x₁− x2ctanθ)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, λ_i 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, 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 [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 betweenLθandKⁿ(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 x, y =

n

i=1xiyi. Due toL^∗_θ = L^π₂−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, yA:= Ax, Ay. Then

L^∗θ = {x | x, yA≥ 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, yA≥ 0, ∀y ∈Kⁿ} = {x | Ax, Ay ≥ 0, ∀y ∈Kⁿ}

= {x | A²x, y ≥ 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 f^Lθ. It should be empha- sized that the relationKⁿ= ALθdoes not guarantee that there exists a similar close relation between f^Lθ and f^soc. For example, take f (t) to be a simple function max{t, 0}, which corresponds to the projection operator . For x∈Lθ, we have Ax∈Kⁿwhich 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 f^Lθ by simply resorting to f^soc. Hence, it is necessary to study f^Lθ 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 crucial:

(i) spectral factorization associated with circular cones; (ii) smooth and nonsmooth analysis for f^Lθ 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 [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 f^Lθ 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 ∈ IR^n×n is said to beLθ-invariant if Mh∈Lθ for all h∈Lθ. We write xLθ yto mean x− y ∈Lθand denoteL^◦_θthe polar cone ofLθ, i.e.,

L^◦θ:= {y ∈ IRⁿ| x, y ≤ 0, ∀x ∈Lθ}.

Denote e:= (1, 0, . . . , 0)^T and use λ(M), λ_min(M), λ_max(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 IR^n×n andSⁿ₊ is the cone of positive semidefinite matrices. For a mapping g: IRⁿ→ IR^m, 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,

xLθ y =⇒ f^Lθ(x)Lθ f^Lθ(y); (b) f^Lθ is said to be monotone on S if

f^Lθ(x)− f^Lθ(y), x− y

≥ 0, ∀x, y ∈ S.

(c) f^Lθ is said to be strictly monotone on S if

f^Lθ(x)− f^Lθ(y), x− y

>0, ∀x, y ∈ S, x = y.

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

f^Lθ(x)− f^Lθ(y), x− y

≥ μx − y², ∀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 y^TAy > 0 for some y. Then, the implication[z^TAz≥ 0 =⇒ z^TBz≥ 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].

Lemma 2.2 Given ζ ∈ IR, u ∈ IRⁿ⁻¹, and a symmetric matrix ∈ IR^n×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 u^T u H



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

and there exists λ≥ 0 such that

ζ²− ctan²θu²− λ (ζ tan θ)u^T − ctanθu^TH (ζtan θ )u− ctanθH u tan²θ uu^T − H²+ λI



Sⁿ₊O.

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

Lθ ∈Lθ ⇐⇒ A⁻¹Kⁿ∈ A⁻¹Kⁿ⇐⇒ AA⁻¹Kⁿ∈Kⁿ

⇐⇒ AA⁻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 u^T ctanθ u H

 1 z



=

ζ+ tan θu^Tz ctanθ u+ Hz



∈Kⁿ, which means

ζ+ u^Tztan θ ≥ 0, ∀z ∈ B, (10)

and

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

ζ ≥ tan θ max{−u^Tz|z ∈ B} = tan θu

and condition (11) is equivalent to



ζ+ tan θu^Tz2

≥ ctanθu + Hz², i.e.,

z^T(tan²θ uu^T − H²)z+ 2

ζtan θ u^T − ctanθu^TH



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

 1 z^T

1 z



≥ 0 ∀z ∈ B, (12)

with

:=

 ζ²− ctan²θ u^Tu (ζtan θ )u^T− ctanθu^TH (ζtan θ )u− ctanθH u tan²θ uu^T − H²

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

 k v^T

 1 0 0 −I

 k v



≥ 0 =⇒ 

k v^T



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 = 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) isv ≤ 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 u^Tu ζtan θ u^T − ctanθu^TH ζtan θ u− ctanθH u tan²θ uu^T − 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 x₂^T x2 αI+ β ¯x2¯x₂^T



, 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 x₂^T x₂ αI+ β ¯x2¯x2^T



=

x1 x^T₂ x2 x1I



− β

0 0

0 I− ¯x2¯x₂^T

 +

0 0

0 (α− x1+ β)I



and let 1:=

x₁ x₂^T x₂ x₁I



− β

0 0

0 I− ¯x2¯x^T2



, 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 f^Lθ defined as in (7). Recall from [4,32] that f^Lθ is differentiable at x if and only if f is differentiable at λi(x)for i= 1, 2 and

∇f^Lθ(x)=

⎧⎨

⎩

f(x1)I x2= 0;

 ξ ¯x₂^T ¯x2 τ I+ (η − τ) ¯x2¯x2^T



x2 = 0, (14)

where

τ := f (λ₂(x))− f (λ1(x))

λ₂(x)− λ1(x) , ξ:= f(λ₁(x))

1+ ctan²θ + f(λ₂(x)) 1+ tan²θ, := − ctanθ

1+ ctan²θf(λ₁(x))+ tan θ

1+ tan²θf(λ₂(x)), η := ctan²θ

1+ ctan²θf(λ₁(x))+ tan²θ

1+ tan²θf(λ₂(x)).

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

Theorem 2.1 Suppose that f : J → IR is differentiable. Then, f isLθ-monotone on J if and only if∇f^Lθ(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∇f^Lθ(x)h∈Lθ. From theLθ-monotonicity of f , we know f^Lθ(x+ th) Lθ f^Lθ(x) for all t > 0. Note thatLθ is a cone. Hence

f^Lθ(x+ th) − f^Lθ(x)

t Lθ 0. (15)

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

∇f^Lθ(x)h∈Lθ.

“⇐” Suppose that ∇f^Lθ(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 f^Lθ(x)Lθ f^Lθ(y).

For any ζ ∈L^◦_θ, we have

ζ, f^Lθ(x)− f^Lθ(y)

=

 1 0

ζ,∇f^Lθ(x+ t(x − y)) (x − y)

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

Note that f is Lipschitz continuous on J if and only if f^Lθ 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) ∂_Bf^Lθ(x) isLθ-invariant for all x∈ S;

(c) ∂f^Lθ(x) isLθ-invariant for all x∈ S.

Proof “(a)⇒ (b)” Take V ∈ ∂Bf^Lθ(x), then by definition of B-subdifferential there exists {xk} ⊂ D_fLθ such that x_k → x and ∇f^Lθ(x_i) → V . According to (15), we obtain

∇f^Lθ(x_i)hLθ 0 for h∈Lθ. Taking the limit yields V hLθ 0. Since V ∈ ∂Bf^Lθ(x)is arbitrary, this says that ∂_Bf^LθisLθ-invariant.

“(b)⇒ (c)” Take V ∈ ∂f^Lθ(x), then by definition, there exists V_i ∈ ∂Bf^Lθ(x)and β_i ∈ [0, 1] such that V = 

iβiVi and

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



iβiVih∈Lθ, since ViisLθ-invariant andLθis convex. Hence ∂f^Lθ(x)isLθ-invariant.

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

ζ, f^Lθ(x)− f^Lθ(y)

= ζ, V (x − y) ≤ 0,

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

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∇f^Lθ(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 ∇f^Lθ(x) being Lθ-invariant, i.e.,

∇f^Lθ(x)h= f(x1)h∈Lθfor all h∈Lθ, is equivalent to saying f(x1)≥ 0.

Case 2 For x₂ = 0, let

H := τI + (η − τ) ¯x2¯x2^T.

Then, applying Lemma 2.2 to the formula of∇f^Lθ(x)in (14),∇f^Lθ(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 ¯x₂^T − ctanθ ¯x₂^TH ξtan θ ¯x2− ctanθ H ¯x2 tan²θ ²¯x2¯x₂^T − 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 θf(λ1(x))− ctanθf(λ2(x))≤ − tan θf(λ1(x))+ tan θf(λ2(x))

≤ tan θf(λ₁(x))+ ctanθf(λ₂(x))

⇐⇒ f(λ2(x))≥ 0 and f(λ1(x))≥1− ctan²θ

2 f(λ2(x)). (21)

This is ensured by (17) and (18). In fact, if tan θ ≥ ctanθ, then we know from (17) that f(λ₁(x))≥ f(λ₂(x))≥ 

(1− ctan²θ )/2

f(λ₂(x))where the second inequality is due

to f(λ₂(x))≥ 0 by (18). If tan θ ≤ ctanθ, then 1 − ctan²θ ≤ 0, and hence f(λ₁(x))≥ 0≥

(1− ctan²θ )/2

f(λ2(x))since f(λi(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 )f(λ1(x))²+ 2(1 + ctan²θ )f(λ1(x))f(λ2(x))



= 1

(tan θ+ ctanθ)²



(tan²θ− ctan²θ )f(λ₁(x))²+ (ctan²θ− tan²θ )f(λ₁(x))f(λ₂(x))

+ 1

(tan θ+ ctanθ)²



2+ tan²θ+ ctan²θ



f(λ₁(x))f(λ₂(x))

= μ + f(λ1(x))f(λ2(x)), with

μ := 1

(tan θ+ ctanθ)²



(tan²θ− ctan²θ )f(λ₁(x))²+ (ctan²θ− tan²θ )f(λ₁(x))f(λ₂(x))



= tan θ− ctanθ

tan θ+ ctanθf(λ₁(x))

f(λ₁(x))− f(λ₂(x))

≥ 0,

where the last step is due to (17). In the ϒ₁₂-entry and ϒ₂₁-entry, we calculate (ξtan θ ) ¯x^T2 − ctanθ ¯x^T2H

= 1

(tan θ+ ctanθ)²

− tan²θ+ ctan²θ



f(λ1(x))²+

tan²θ− ctan²θ



f(λ1(x))f(λ2(x))

¯x2^T

= −tan θ− ctanθ

tan θ+ ctanθf(λ1(x))

f(λ1(x))− f(λ2(x))

¯x2^T

= −μ ¯x2^T.

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

τ²+ 1

(tan θ+ ctanθ)²



(tan²θ− ctan²θ )f(λ₁(x))²

−2(1 + tan²θ )f(λ₁(x))f(λ₂(x))

¯x2¯x^T₂

= −τ²I+

τ²+ μ − f(λ1(x))f(λ2(x))

¯x2¯x2^T.

Hence, ϒ can be rewritten as ϒ =

ϒ₁₁ ϒ₁₂ ϒ21 ϒ22



=

μ+ f(λ₁(x))f(λ₂(x))− λ −μ ¯x2^T

−μ ¯x2 (λ− τ²)I+

τ²+ μ − f(λ1(x))f(λ2(x))

¯x2¯x₂^T

 . Now, applying Lemma 2.3 to ϒ, we have

λmin(ϒ ) ≥ min

ϒ11− |μ|, ϒ11−

τ²+ μ − f(λ1(x))f(λ2(x))



+ min 0, 

λ− τ²

− ϒ11+

τ²+ μ − f(λ₁(x))f(λ₂(x))

= min

f(λ₁(x))f(λ₂(x))− λ, 2f(λ₁(x))f(λ₂(x))− τ²− λ +2 min!

0, λ− f(λ₁(x))f(λ₂(x))"

. (22)

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

f(λ1(x))≥ 0, f(λ2(x))≥ 0, and f(λ1(x))f(λ2(x))≥

f (λ2(x))− f (λ1(x)) λ2(x)− λ1(x)

2

, which together with (15) yields

f(λ1(x))f(λ2(x))≥ 0 and f(λ1(x))f(λ2(x))− τ²≥ 0.

Thus, we can plug λ:= f(λ₁(x))f(λ₂(x))≥ 0 into (22), which gives λ_min(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,

f(λ2(x))≥ 0, f(λ1(x))≥1− ctan²θ

2 f(λ2(x)), (23)

and there exists λ≥ 0 such that ϒ=

μ+ f(λ₁(x))f(λ₂(x))− λ ±μ

±μ μ− f(λ1(x))f(λ2(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− (f(λ1(x))f(λ2(x))− λ)²− μ² ≥ 0, which implies λ = f(λ1(x))f(λ2(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

⎧⎨

⎩f(λ2(x))≥ 0, f(λ1(x))≥1− ctan²θ

2 f(λ2(x)), (tan θ− ctanθ)f(λ1(x))

f(λ1(x))− f(λ2(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.

3 Monotonicity off^Lθ

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 f^Lθ and theLθ-monotonicity of f ? Whether the monotonicity of f^Lθ 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 (orSⁿ₊O) ⇐⇒ A Sⁿ₊O (orSⁿ₊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) f^Lθis monotone on S if and only if ∂f^Lθ(x)Sⁿ₊ O for all x∈ S;

(b) If ∂f^Lθ(x)Sⁿ₊O for all x∈ S, then f^Lθis strictly monotone on S;

(c) f^Lθis strongly monotone on S if and only if there exists μ > 0 such that ∂f^Lθ(x)Sⁿ₊

μI for all x∈ S.

Lemma 3.2 Suppose that f is Lipschitz continuous. Then,

∂_Bf^Lθ(x)Sⁿ₊O⇐⇒ ∂f^Lθ(x)Sⁿ₊O and ∂Bf^Lθ(x)Sⁿ₊O⇐⇒ ∂f^Lθ(x)Sⁿ₊O.

Proof The result follows immediately from the fact ∂f^Lθ(x)= conv∂Bf^Lθ(x).

Lemma 3.3 The following statements hold.

(a) For any h∈ IRⁿ t1



h1− ctanθ ¯x2^Th2

2

+ t2



h1+ tan θ ¯x^T2h2

2

+ t3

h2²− ( ¯x2^Th2)²

≥ 0 (25)

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

(b) For any h∈ IRⁿ\{0}

t1



h1− ctanθ ¯x2^Th2

2

+ t2



h1+ tan θ ¯x2^Th2

2

+ t3

h2²− ( ¯x2^Th2)²



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

Proof (a) The sufficiency is clear, since| ¯x^T₂h2| ≤  ¯x2h2 = 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 andu = 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. Ifh2− ¯x^T2h₂>0, then the result holds since t₃>0. Ifh2− ¯x2^Th₂= 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.

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

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

Then∇f^Lθ(x)Sⁿ₊O if and only if f(λi(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 ∇f^Lθ(x) = f(x1)e, it is clear that∇f^Lθ(x) Sⁿ₊ O is equivalent to saying f(x1)≥ 0. Then, the desired result follows.

Case 2 For x₂ = 0, denote

b1:= f(λ1(x))

1+ ctan²θ and b2= f(λ2(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, ∇f^Lθ(x)h

= (b1+ b2)h²₁+ 2 h1¯x^T₂h₂+ τh2²+ (b1ctan²θ+ b2tan²θ− τ)( ¯x₂^Th₂)²

= (b1+ b2)h²₁+ 2(−b1ctanθ+ b2tan θ )h₁¯x₂^Th₂+ (b1ctan²θ+ b2tan²θ )(¯x₂^Th₂)² +τ

h2²− ( ¯x₂^Th₂)²

= b1



h²₁− 2ctanθh1¯x2^Th₂+ ctan²θ (¯x2^Th₂)²

+ b2



h²₁+ 2 tan θh1¯x2^Th₂+ tan²θ (¯x2^Th₂)²

 +τ

h2²− ( ¯x2^Th₂)²



= b1



h₁− ctanθ ¯x₂^Th₂

2

+ b2



h₁+ tan θ ¯x₂^Th₂

2

+ τ

h2²− ( ¯x₂^Th₂)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., f(λ1(x))≥ 0, f(λ2(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, ∇f^Lθ(x)Sⁿ₊O if and only if f(λ_i(x)) >0 for i= 1, 2 and f (λ2(x)) >

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

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