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 off^Lθ and circular cone monotonicity off. Their relationship is discussed as well. Our results show that the angleθplays 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 beθwithθ∈(0,90^◦). Then, then-dimensional circular cone denoted byLθ

can be expressed as

Lθ := {x=(x1, x2)^T ∈IR×IRⁿ⁻¹|cosθx ≤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 withJ ⊆IR is a real-valued function andxis decomposed as

x=λ1(x)·u⁽¹⁾_x +λ2(x)·u⁽²⁾_x (2) whereλ1(x),λ2(x)andu⁽¹⁾x ,u⁽²⁾x are thespectral valuesand the associatedspectral vectors ofxwith respect toKⁿ, given by

λ_i(x)=x₁+(−1)ⁱx₂ and u⁽ⁱ⁾_x = 1 2

1 (−1)ⁱx¯2

(3) fori = 1,2 withx¯2 := x2/x2ifx2 = 0, andx¯2 being any vector in IRⁿ⁻¹ satisfying ¯x₂ = 1 ifx₂ = 0. The decomposition (2) is called the spectral factorization associated with second-order cone forx. Likewise, there is a similar decomposition forxassociated with circular cone case. More specifically, from [31, Theorem 3.1], the spectral factorization associated withLθforxis in form of

x=λ1(x)·u⁽¹⁾_x +λ2(x)·u⁽²⁾_x (4)

where

λ1(x) := x1− x2ctanθ

λ₂(x) := x₁+ x₂tanθ (5)

and ⎧

⎪⎪

⎨

⎪⎪

⎩

u⁽¹⁾x := 1 1+ctan²θ

1 0 0 ctanθ·I

1

− ¯x₂

=

sin²θ

−(sinθcosθ )x¯2

u⁽²⁾_x := 1

1+tan²θ

1 0 0 tanθ·I

1

¯ x₂

=

cos²θ (sinθcosθ )x¯2

(6)

Analogously, for any givenf :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 inλ_i(x) andu⁽ⁱ⁾x :

f^Lθ(x)=

⎡

⎢⎣

f (x₁− x₂ctanθ )

1+ctan²θ +f (x₁+ x₂tanθ ) 1+tan²θ

−f (x₁− x₂ctanθ )ctanθ

1+ctan²θ +f (x₁+ x₂tanθ )tanθ 1+tan²θ

¯ x₂

⎤

⎥⎦. (8)

Clearly, asθ =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 andu⁽ⁱ⁾_x stands for (5) and (6), respectively.

Throughout this paper, we always assume thatJ is an open interval (finite or infinite) in IR, i.e.,J :=(t, t)witht, t ∈IR∪ {±∞}. DenoteSthe set of allx ∈IRⁿwhose spectral valuesλi(x)fori=1,2 belong toJ, i.e.,

S:= {x∈IRⁿ|λ_i(x)∈J, i =1,2}.

According to [24], we knowSis open if and only ifJ is open. In addition, asJ is an interval, we knowSis 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:=

tanθ 0

0 I

.

It is well-known that Kⁿ is a self-dual cone in the standard inner productx, 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 withA. 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 functionf^Lθ. It should be empha- sized that the relationKⁿ=ALθdoes not guarantee that there exists a similar close relation between f^Lθ andf^soc. For example, takef (t)to be a simple function max{t,0}, which corresponds to the projection operator. Forx∈Lθ, we haveAx∈Kⁿwhich implies

_Lθ(x)=x=A⁻¹(Ax)=A⁻¹_Kⁿ(Ax).

Unfortunately, the above relation fails to hold whenx /∈Lθ. To see this, we let tanθ = 1/4 andx=(−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 studyf^Lθ by simply resorting tof^soc. Hence, it is necessary to studyf^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 onp-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 forf^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 off 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 angleθplays different role in these two concepts. More precisely, the circular cone monotonicity off depends onf and θ, whereas the monotonicity off^Lθ only depends onf.

To end this section, we say a few words about the notations and present the definitions of monotonicity andLθ-monotonicity. A matrixM ∈ IR^n×n is said to beLθ-invariant if Mh∈Lθ for allh∈Lθ. We writexLθ yto meanx−y ∈Lθand denoteL^◦_θthe polar cone ofLθ, i.e.,

L^◦θ:= {y∈IRⁿ| x, y ≤0, ∀x∈Lθ}.

Denotee:=(1,0, . . . ,0)^T and useλ(M),λ_min(M),λ_max(M)for the set of all eigenval- ues, the minimum, and the maximum of eigenvalues ofM, respectively. Besides,Sⁿmeans the space of all symmetric matrices in IR^n×nandSⁿ₊ is the cone of positive semidefinite matrices. For a mappingg:IRⁿ→IR^m, denote byDgthe set of all differentiable points of g. For convenience, we define 0/0:=0. Given a real-valued functionf :J →IR, (a) fis said to beLθ-monotone onJ if for anyx, y∈S,

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

f^Lθ(x)−f^Lθ(y), x−y

≥0, ∀x, y∈S.

(c) f^Lθ is said to be strictly monotone onSif f^Lθ(x)−f^Lθ(y), x−y

>0, ∀x, y∈S, x =y.

(d) f^Lθ is said to be strongly monotone onSwithμ >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 ifBSⁿ₊ λAfor someλ≥0.

Proof This is the well known S-Lemma, see [7, Lemma 3.1] or [25].

Lemma 2.2 Givenζ ∈ 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 anyz∈B. (b) If=

ζ u^T u H

withH ∈Sⁿ⁻¹, thenisLθ-invariant is equivalent to ζ ≥ utanθ

and there existsλ≥0such that

ζ²−ctan²θu²−λ (ζtanθ )u^T −ctanθ u^TH (ζtanθ )u−ctanθ H 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 tanθ u^T ctanθ u H

1 z

=

ζ+tanθ u^Tz ctanθ u+H z

∈Kⁿ, which means

ζ+u^Tztanθ ≥0, ∀z∈B, (10)

and

ζ+u^Tztanθ≥ ctanθ u+H z, ∀z∈B. (11) Note that condition (10) is equivalent to

ζ ≥tanθmax{−u^Tz|z∈B} =tanθu and condition (11) is equivalent to

ζ+tanθ u^Tz2

≥ ctanθ u+H z², i.e.,

z^T(tan²θ uu^T −H²)z+2

ζtanθ u^T −ctanθ u^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 (ζtanθ )u^T−ctanθ u^TH (ζtanθ )u−ctanθ H 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 ofk=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.

Fork =0, dividing byk²in the left side of (13) yields

1 (v

k)^T 1 0 0 −I

1v k

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

1 (v

k)^T 1v k

≥0.

Hence, the right side of (13) holds.

Fork=0, the left side of (13) isv ≤0, which saysv=0, i.e.,(k, v)^T =0. Therefore, the right side of (13) holds clearly.

Now, applying Lemma 2.1 to ensures the existence ofλ≥0 such that ζ²−ctan²θ u^Tu ζtanθ u^T −ctanθ u^TH

ζtanθ u−ctanθ H 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 ofH :=

x1 x₂^T x2 αI+βx¯2x¯₂^T

,whereα, 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 splitHas sum of three special matrices, i.e., x1 x₂^T

x₂ αI+βx¯₂x¯₂^T

= x1 x^T₂

x2 x1I

−β

0 0 0 I− ¯x2x¯₂^T

+

0 0 0 (α−x1+β)I

and let 1:=

x₁ x₂^T x₂ x₁I

−β

0 0 0 I− ¯x₂x¯^T₂

, 2:=

0 0 0 (α−x₁+β)I

. Then,λ(1)= {x1− x2, x1+ x2, x1−β}by [6, Lemma 1] andλ(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 functionf^Lθ defined as in (7). Recall from [4,32] thatf^Lθ is differentiable at x if and only iff is differentiable atλi(x)for i=1,2 and

∇f^Lθ(x)=

⎧⎨

⎩

f(x1)I x2=0;

ξ x¯₂^T x¯2 τ I+(η−τ )x¯2x¯₂^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))+ tanθ

1+tan²θf(λ₂(x)), η := ctan²θ

1+ctan²θf(λ₁(x))+ tan²θ

1+tan²θf(λ₂(x)).

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

Theorem 2.1 Suppose thatf :J →IRis differentiable. Then,f isLθ-monotone onJ if and only if∇f^Lθ(x)isLθ-invariant for allx∈S.

Proof “⇒” Suppose thatfisLθ-monotone. Takex∈Sandh∈Lθ, what we want to prove is∇f^Lθ(x)h∈Lθ. From theLθ-monotonicity off, we knowf^Lθ(x+th)Lθ f^Lθ(x) for allt >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 allx∈S. Takex, y∈SwithxLθ y(i.e., x−y∈Lθ). In order to show the desired result, we need to argue thatf^Lθ(x)Lθ f^Lθ(y).

For anyζ∈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)∈Lθbecausex+t (x−y)∈ S (sinceS is convex) and∇f^Lθ is Lθ-invariant overS by hypothesis. Sinceζ ∈ L^◦_θ is arbitrary, (16) impliesf^Lθ(x)−f^Lθ(y)∈(L^◦_θ)^◦ = Lθ, where the last step is due to the fact thatLθis a closed convex cone. This meansf^Lθ(x)Lθ f^Lθ(y).

Note thatf is Lipschitz continuous onJ if and only iff^Lθ is Lipschitz continuous on S, see [4,32]. The nonsmooth version of Theorem 2.1 is given below.

Theorem 2.2 Suppose thatf :J →IRis Lipschitz continuous onJ. Then the following statements are equivalent.

(a) f isLθ-monotone onJ;

(b) ∂_Bf^Lθ(x)isLθ-invariant for allx∈S;

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

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

∇f^Lθ(x_i)hLθ 0 forh∈Lθ. Taking the limit yieldsV hLθ 0. SinceV ∈∂_Bf^Lθ(x)is arbitrary, this says that∂_Bf^LθisLθ-invariant.

“(b)⇒(c)” TakeV ∈∂f^Lθ(x), then by definition, there existsV_i ∈∂_Bf^Lθ(x)andβ_i ∈ [0,1]such thatV =

iβiVi and

iβi = 1. Thus, for anyh ∈ Lθ, we haveV h =

iβiVih∈Lθ, sinceViisLθ-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 someV ∈∂f^Lθ(z)withz ∈ [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 thatf :J →IRis differentiable. If for allt1, t2∈J witht1≤t2, (tanθ−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)

thenf isLθ-monotone onJ.

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 allh∈Lθ, is equivalent to sayingf(x1)≥0.

Case 2 Forx₂ =0, let

H :=τ I+(η−τ )x¯₂x¯₂^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 ≥ tanθ (19)

and there existsλ≥0 such that ϒ:=

ξ²−ctan²θ ²−λ ξtanθ x¯₂^T −ctanθ x¯₂^TH ξtanθ x¯2−ctanθ Hx¯2 tan²θ ²x¯2x¯₂^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

−tanθf(λ1(x))−ctanθf(λ2(x))≤ −tanθf(λ1(x))+tanθf(λ2(x))

≤tanθf(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 tanθ ≥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

tof(λ₂(x))≥0 by (18). If tanθ ≤ctanθ, then 1−ctan²θ ≤0, and hencef(λ₁(x))≥ 0≥

(1−ctan²θ )/2

f(λ2(x))sincef(λi(x))≥0 fori=1,2 by (18).

Now let us look into the entries ofϒ. In theϒ11-entry, we calculate ξ²−ctan²θ ²

= 1

(tanθ+ctanθ )²

(tan²θ−ctan²θ )f(λ1(x))²+2(1+ctan²θ )f(λ1(x))f(λ2(x))

= 1

(tanθ+ctanθ )²

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

+ 1

(tanθ+ctanθ )²

2+tan²θ+ctan²θ

f(λ₁(x))f(λ₂(x))

= μ+f(λ1(x))f(λ2(x)), with

μ := 1

(tanθ+ctanθ )²

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

= tanθ−ctanθ

tanθ+ctanθf(λ₁(x))

f(λ₁(x))−f(λ₂(x))

≥ 0,

where the last step is due to (17). In theϒ₁₂-entry andϒ₂₁-entry, we calculate (ξtanθ )x¯^T₂ −ctanθ x¯^T₂H

= 1

(tanθ+ctanθ )²

−tan²θ+ctan²θ

f(λ1(x))²+

tan²θ−ctan²θ

f(λ1(x))f(λ2(x)) x¯₂^T

= −tanθ−ctanθ

tanθ+ctanθf(λ1(x))

f(λ1(x))−f(λ2(x))

¯ x₂^T

= −μx¯₂^T.

In theϒ₂₂-entry, we calculate tan²θ ²x¯₂x¯₂^T−H² = −τ²I+

τ²+ 1

(tanθ+ctanθ )²

(tan²θ−ctan²θ )f(λ₁(x))²

−2(1+tan²θ )f(λ₁(x))f(λ₂(x))

¯ x₂x¯^T₂

= −τ²I+

τ²+μ−f(λ1(x))f(λ2(x)) x¯2x¯₂^T. Hence,ϒcan be rewritten as

ϒ =

ϒ₁₁ ϒ₁₂ ϒ21 ϒ22

=

μ+f(λ₁(x))f(λ₂(x))−λ −μx¯₂^T

−μx¯2 (λ−τ²)I+

τ²+μ−f(λ1(x))f(λ2(x))

¯ x2x¯₂^T

. Now, applying Lemma 2.3 toϒ, 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)

Usingλ₁(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 plugλ:=f(λ₁(x))f(λ₂(x))≥0 into (22), which givesλ_min(ϒ ) ≥0.

Henceϒis positive semi-definite. This completes the proof.

Remark 2.1 The condition (17) holds automatically whenθ = 45^◦. In other case, some addition requirement needs to be imposed onf. For instance,fis required to be convex as θ ∈ (0,45^◦)whilef is required to be concave asθ ∈ (45^◦,90^◦). This indicates that the angle plays an essential role in the framework of circular cone, i.e., the assumption onf 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 ofn=2.

Theorem 2.4 Suppose thatf :J →IRis differentiable onJ andn=2. Then,f isLθ- monotone onJ if and only iff(t)≥0for allt∈J and(tanθ−ctanθ )(f(t1)−f(t2))≥0 for allt1, t2∈J witht1≤t2.

Proof In light of the proof of Theorem 2.3, we know thatf isLθ-monotone if and only if for anyx∈S,

f(λ2(x))≥0, f(λ1(x))≥1−ctan²θ

2 f(λ2(x)), (23)

and there existsλ≥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 ofϒ comes from the fact thatx¯2 = ±1 in this case. It follows from (24) thatμ²−(f(λ1(x))f(λ2(x))−λ)²−μ² ≥0, which impliesλ=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 impliesμ≥0. Hence, the conditions (23) and (24) are equivalent to

⎧⎨

⎩f(λ2(x))≥0, f(λ1(x))≥1−ctan²θ

2 f(λ2(x)), (tanθ−ctanθ )f(λ1(x))

f(λ1(x))−f(λ2(x))

≥0.

Due to the arbitrariness ofλi(x)∈J and applying similar arguments following (21), the above conditions givef(t)≥0 for allt∈J and(tanθ−ctanθ )(f(t1)−f(t2))≥0 for allt₁, t₂∈J witht₁≤t₂. Thus, the proof is complete.

3 Monotonicity off^Lθ

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

how about on the relationship between the monotonicity off^Lθ and theLθ-monotonicity off? Whether the monotonicity off^Lθ also depends onθ? 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) ⇐⇒ ASⁿ₊O (orSⁿ₊O), ∀A∈∂H (x).

Taking into account of the result in [26], we readily have

Lemma 3.1 Letf be Lipschitz continuous onJ. The following statements hold:

(a) f^Lθis monotone onSif and only if∂f^Lθ(x)Sⁿ₊ Ofor allx∈S; (b) If∂f^Lθ(x)Sⁿ₊Ofor allx∈S, thenf^Lθis strictly monotone onS;

(c) f^Lθis strongly monotone onSif and only if there existsμ >0such that∂f^Lθ(x)Sⁿ₊

μIfor allx∈S.

Lemma 3.2 Suppose thatf 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 anyh∈IRⁿ t1

h1−ctanθx¯₂^Th2

2

+t2

h1+tanθx¯^T₂h2

2

+t3

h2²−(x¯₂^Th2)²

≥0 (25)

if and only ifti ≥0fori=1,2,3.

(b) For anyh∈IRⁿ\{0} t1

h1−ctanθx¯₂^Th2

2

+t2

h1+tanθx¯₂^Th2

2

+t3

h2²−(x¯₂^Th2)²

>0 (26) if and only ift_i >0fori=1,2,3.

Proof (a) The sufficiency is clear, since| ¯x^T₂h2| ≤ ¯x2h2 = h2by Cauchy-Schwartz inequality. Now let us show the necessity. Takingh=(1,−ctanθx¯₂)^T, then (25) equal to t1(1+ctan²θ )²≥0, which impliest1≥0. Similarly, leth=(1,tanθx¯2)^T, then (25) yields t2(1+tan²θ )²≥0, implyingt2≥0. Finally, leth=(0, u)^T withusatisfyingu,x¯2 =0 andu =1, then it follows from (25) thatt3≥0.

(b) The necessity is the same of the argument as given in part (a). For sufficiency, take a nonzero vectorh. Ifh₂− ¯x^T₂h₂>0, then the result holds sincet₃>0. Ifh₂− ¯x₂^Th₂=0, thenh₂ = βx¯₂. So the left side of (26) takest₁(h₁−βctanθ )²+t₂(h₁+βtanθ )² >0, becauseh₁−βctanθ =h₁+βtanθ =0 only happened whenβ = 0 andh₁ = 0. This meansh2=0 sinceh2=βx¯2, soh=0.

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

Theorem 3.1 Givenx ∈ IRⁿ and suppose thatf is differentiable atλi(x)fori = 1,2.

Then∇f^Lθ(x)Sⁿ₊Oif and only iff(λi(x))≥0fori=1,2andf (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 sayingf(x1)≥0. Then, the desired result follows.

Case 2 Forx₂ =0, denote

b1:= f(λ1(x))

1+ctan²θ and b2= f(λ2(x)) 1+tan²θ.

Then,ξ =b1+b2,= −b1ctanθ+b2tanθ, andη=b1ctan²θ+b2tan²θ. Hence, for allh=(h1, h2)^T ∈IR×IRⁿ⁻¹, we have

h,∇f^Lθ(x)h

= (b₁+b₂)h²₁+2h1x¯^T₂h₂+τh₂²+(b₁ctan²θ+b₂tan²θ−τ )(x¯₂^Th₂)²

= (b₁+b₂)h²₁+2(−b₁ctanθ+b₂tanθ )h₁x¯₂^Th₂+(b₁ctan²θ+b₂tan²θ )(x¯₂^Th₂)² +τ

h₂²−(x¯₂^Th₂)²

= b₁

h²₁−2ctanθ h₁x¯₂^Th₂+ctan²θ (x¯₂^Th₂)² +b₂

h²₁+2 tanθ h₁x¯₂^Th₂+tan²θ (x¯₂^Th₂)² +τ

h₂²−(x¯₂^Th₂)²

= b₁

h₁−ctanθx¯₂^Th₂ 2

+b₂

h₁+tanθx¯₂^Th₂ 2

+τ

h₂²−(x¯₂^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, andf (λ2(x))≥f (λ1(x))due toλ2(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 Givenx ∈ IRⁿand suppose thatf is differentiable atλi(x)fori = 1,2.

Then forx₂ =0,∇f^Lθ(x)Sⁿ₊Oif and only iff(λ_i(x)) >0fori =1,2andf (λ₂(x)) >

f (λ₁(x)); forx₂=0,∇f^Lθ(x)Sⁿ₊Oif and only iff(x₁) >0.

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