### 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 L_{θ} 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^{L}^{θ} 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.

half-aperture angle be θ with θ ∈ (0,^{π}_{2}). Then, the n-dimensional circular cone denoted
by Lθ can be expressed as

L_{θ} := {x = (x_{1}, x_{2})^{T} ∈ IR × IR^{n−1}| cos θkxk ≤ x_{1}}.

Note that L_{45}^{◦} corresponds the well-known second-order cone K^{n}(SOC, for short), which
is given by

K^{n}:= {x = (x_{1}, x_{2})^{T} ∈ IR × IR^{n−1}| kx_{2}k ≤ x_{1}}.

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])

f^{soc}(x) = f (λ_{1}(x))u^{(1)}_{x} + f (λ_{2}(x))u^{(2)}_{x} ∀x = (x_{1}, x_{2})^{T} ∈ IR × IR^{n−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 = λ_{1}(x) · u^{(1)}_{x} + λ_{2}(x) · u^{(2)}_{x} (2)
where λ_{1}(x), λ_{2}(x) and u^{(1)}x , u^{(2)}x are the spectral values and the associated spectral vectors
of x with respect to K^{n}, given by

λ_{i}(x) = x_{1}+ (−1)^{i}kx_{2}k and u^{(i)}_{x} = 1
2

1

(−1)^{i}x¯_{2}

(3)

for i = 1, 2 with ¯x2 := x2/kx2k if x2 6= 0, and ¯x2 being any vector in IR^{n−1} satisfying
k¯x_{2}k = 1 if x_{2} = 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 = λ_{1}(x) · u^{(1)}_{x} + λ_{2}(x) · u^{(2)}_{x} (4)
where

λ_{1}(x) := x1− kx2kctanθ

λ_{2}(x) := x_{1}+ kx_{2}k tan θ (5)

and

u^{(1)}x := 1
1 + ctan^{2}θ

1 0

0 ctanθ · I

1

−¯x_{2}

=

sin^{2}θ

−(sin θ cos θ)¯x_{2}

u^{(2)}x := 1
1 + tan^{2}θ

1 0

0 tan θ · I

1

¯
x_{2}

=

cos^{2}θ
(sin θ cos θ)¯x_{2}

(6)

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

f^{L}^{θ}(x) := f (λ_{1}(x)) u^{(1)}_{x} + f (λ_{2}(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^{L}^{θ}(x) =

f (x_{1}− kx_{2}kctanθ)

1 + ctan^{2}θ +f (x_{1} + kx_{2}k tan θ)
1 + tan^{2}θ

−f (x1− kx2kctanθ)ctanθ

1 + ctan^{2}θ + f (x1+ kx2k tan θ) tan θ
1 + tan^{2}θ

¯
x_{2}

. (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^{0}) with t, t^{0} ∈ IR ∪ {±∞}. Denote S the set of all x ∈ IR^{n} whose
spectral values λ_{i}(x) for i = 1, 2 belong to J , i.e.,

S := {x ∈ IR^{n}| λ_{i}(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 L_{θ} and K^{n} (see [31]) as below
K^{n} = ALθ where A :=tan θ 0

0 I

.

It is well-known that K^{n} is a self-dual cone in the standard inner product hx, yi =
Pn

i=1x_{i}y_{i}. Due to L^{∗}_{θ} = L^{π}

2−θ by [31, Theorem 2.1], L_{θ} 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, yi_{A}:= hAx, Ayi. Then

L^{∗}_{θ} = {x | hx, yi_{A} ≥ 0, ∀y ∈ L_{θ}} = {x | hAx, Ayi ≥ 0, ∀y ∈ A^{−1}K^{n}}

= {x | hAx, yi ≥ 0, ∀y ∈ K^{n}} = {x | Ax ∈ K^{n}}

= A^{−1}K^{n}= L_{θ}.

However, under this new inner product the second-order cone is not self-dual, because
(K^{n})^{∗} = {x | hx, yi_{A}≥ 0, ∀y ∈ K^{n}} = {x | hAx, Ayi ≥ 0, ∀y ∈ K^{n}}

= {x | hA^{2}x, yi ≥ 0, ∀y ∈ K^{n}} = {x | A^{2}x ∈ K^{n}} = A^{−2}K^{n}.

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^{L}^{θ}. It should be
emphasized that the relation K^{n} = AL_{θ} does not guarantee that there exists a similar
close relation between f^{L}^{θ} and f^{soc}. For example, take f (t) to be a simple function
max{t, 0}, which corresponds to the projection operator Π. For x ∈ L_{θ}, we have Ax ∈ K^{n}
which implies

ΠL_{θ}(x) = x = A^{−1}(Ax) = A^{−1}ΠK^{n}(Ax).

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

ΠL_{θ}(x) = 0 and A^{−1}ΠK^{n}(Ax) =

_{3}

23 8

which says ΠL_{θ}(x) 6= A^{−1}ΠK^{n}(Ax). This example undoubtedly indicates that we cannot
study f^{L}^{θ} by simply resorting to f^{soc}. Hence, it is necessary to study f^{L}^{θ} 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 f^{L}^{θ} given as in (7); (iii) the so-called L_{θ}-convexity; and (iv) L_{θ}-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 f^{L}^{θ} only depends on
f .

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

L^{◦}_{θ} := {y ∈ IR^{n}| hx, yi ≤ 0, ∀x ∈ L_{θ}}.

Denote e := (1, 0, . . . , 0)^{T} and use λ(M ), λ_{min}(M ), λ_{max}(M ) for the set of all eigenvalues,
the minimum, and the maximum of eigenvalues of M , respectively. Besides, S^{n} means
the space of all symmetric matrices in IR^{n×n} and S^{n}+ is the cone of positive semidefinite
matrices. For a mapping g : IR^{n} → IR^{m}, denote by D_{g} 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 L_{θ} y =⇒ f^{L}^{θ}(x) L_{θ} f^{L}^{θ}(y);

(b) f^{L}^{θ} is said to be monotone on S if

f^{L}^{θ}(x) − f^{L}^{θ}(y), x − y ≥ 0, ∀x, y ∈ S.

(c) f^{L}^{θ} is said to be strictly monotone on S if

f^{L}^{θ}(x) − f^{L}^{θ}(y), x − y > 0, ∀x, y ∈ S, x 6= y.

(d) f^{L}^{θ} is said to be strongly monotone on S with µ > 0 if

f^{L}^{θ}(x) − f^{L}^{θ}(y), x − y ≥ µkx − yk^{2}, ∀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 L_{θ}-monotone functions. To this end, we need a few technical lemmas.

Lemma 2.1. Let A, B be symmetric matrices and y^{T}Ay > 0 for some y. Then, the
implication [z^{T}Az ≥ 0 =⇒ z^{T}Bz ≥ 0] is valid if and only if B _{S}^{n}_{+} 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 ∈ IR^{n−1}, and a symmetric matrix Ξ ∈ IR^{n×n}. Denote
B := {z ∈ IR^{n−1}|kzk ≤ 1}. Then, the following statements hold.

(a) Ξ being L_{θ}-invariant is equivalent to Ξctanθ
z

∈ L_{θ} for any z ∈ B.

(b) If Ξ =ζ u^{T}

u H

with H ∈ S^{n−1}, then Ξ is L_{θ}-invariant is equivalent to

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

ζ^{2}− ctan^{2}θkuk^{2}− λ (ζ tan θ)u^{T} − ctanθu^{T}H
(ζ tan θ)u − ctanθHu tan^{2}θuu^{T} − H^{2}+ λI

_{S}^{n}

+ O.

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

ΞL_{θ} ∈ L_{θ} ⇐⇒ ΞA^{−1}K^{n}∈ A^{−1}K^{n} ⇐⇒ AΞA^{−1}K^{n} ∈ K^{n}

⇐⇒ AΞA^{−1}1
z

∈ K^{n}⇐⇒ ΞA^{−1}1
z

∈ A^{−1}K^{n} (9)

⇐⇒ ΞA^{−1}1
z

∈ L_{θ} ⇐⇒ Ξctanθ
z

∈ L_{θ},

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

(b) From (9), we know that

A Ξ A^{−1}1
z

=

ζ tan θu^{T}
ctanθu H

1 z

= ζ + tan θu^{T}z
ctanθu + Hz

∈ K^{n},
which means

ζ + u^{T}z tan θ ≥ 0, ∀z ∈ B, (10)

and

ζ + u^{T}z tan θ ≥ kctanθu + Hzk, ∀z ∈ B. (11)
Note that condition (10) is equivalent to

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

ζ + tan θu^{T}z2

≥ kctanθu + Hzk^{2},
i.e.,

z^{T}(tan^{2}θuu^{T} − H^{2})z + 2 ζ tan θu^{T} − ctanθu^{T}Hz + ζ^{2}− ctan^{2}θu^{T}u ≥ 0 ∀z ∈ B,

which can be rewritten as

1 z^{T} Θ1
z

≥ 0 ∀z ∈ B, (12)

with

Θ :=

ζ^{2}− ctan^{2}θu^{T}u (ζ tan θ)u^{T} − ctanθu^{T}H
(ζ tan θ)u − ctanθHu tan^{2}θuu^{T} − H^{2}

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

k v^{T}1 0
0 −I

k v

≥ 0 =⇒ k v^{T} Θk
v

≥ 0, ∀k v

∈ IR^{n}. (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 k^{2} in the left side of (13) yields
h

1 (v

k)^{T}i1 0
0 −I

" 1 v k

#

≥ 0,

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

1 (v
k)^{T}i

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

ζ^{2}− ctan^{2}θu^{T}u ζ tan θu^{T} − ctanθu^{T}H
ζ tan θu − ctanθHu tan^{2}θuu^{T} − H^{2}

− λ1 0 0 −I

_{S}^{n}

+ O.

Thus, the proof is complete. 2

Lemma 2.3. For a matrix being in form of H := x_{1} x^{T}_{2}
x_{2} αI + β ¯x_{2}x¯^{T}_{2}

, where α, β ∈ IR, then

max{x_{1}+ kx_{2}k, x_{1}− β} + max{0, α − x_{1}+ β}

≥ λ_{max}(H) ≥ λ_{min}(H)

≥ min{x_{1}− kx_{2}k, x_{1}− β} + min{0, α − x_{1}+ β}.

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

x_{1} x^{T}_{2}
x_{2} αI + β ¯x_{2}x¯^{T}_{2}

=x_{1} x^{T}_{2}
x_{2} x_{1}I

− β0 0

0 I − ¯x_{2}x¯^{T}_{2}

+0 0

0 (α − x_{1} + β)I

and let

Ω_{1} :=x_{1} x^{T}_{2}
x_{2} x_{1}I

− β0 0

0 I − ¯x_{2}x¯^{T}_{2}

, Ω_{2} :=0 0

0 (α − x_{1}+ β)I

.

Then, λ(Ω_{1}) = {x_{1}−kx_{2}k, x_{1}+kx_{2}k, x_{1}−β} by [6, Lemma 1] and λ(Ω_{2}) = {0, α−x_{1}+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 f^{L}^{θ} defined as in (7). Recall
from [4, 32] that f^{L}^{θ} is differentiable at x if and only if f is differentiable at λ_{i}(x) for
i = 1, 2 and

∇f^{L}^{θ}(x) =

f^{0}(x_{1})I x_{2} = 0;

ξ %¯x^{T}_{2}

%¯x_{2} τ I + (η − τ )¯x_{2}x¯^{T}_{2}

x_{2} 6= 0, (14)

where

τ := f (λ_{2}(x)) − f (λ_{1}(x))

λ_{2}(x) − λ_{1}(x) , ξ := f^{0}(λ_{1}(x))

1 + ctan^{2}θ + f^{0}(λ_{2}(x))
1 + tan^{2}θ,

% := − ctanθ

1 + ctan^{2}θf^{0}(λ_{1}(x)) + tan θ

1 + tan^{2}θf^{0}(λ_{2}(x)),
η := ctan^{2}θ

1 + ctan^{2}θf^{0}(λ_{1}(x)) + tan^{2}θ

1 + tan^{2}θf^{0}(λ_{2}(x)).

The following result shows that if f^{L}^{θ} is differentiable, then we can characterize the
L_{θ}-monotonicity of f via the gradient ∇f^{L}^{θ}.

Theorem 2.1. Suppose that f : J → IR is differentiable. Then, f is L_{θ}-monotone on J
if and only if ∇f^{L}^{θ}(x) is L_{θ}-invariant for all x ∈ S.

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

f^{L}^{θ}(x + th) − f^{L}^{θ}(x)

t L_{θ} 0. (15)

Since L_{θ}is closed, taking the limit as t → 0^{+} yields ∇f^{L}^{θ}(x)h L_{θ} 0, i.e., ∇f^{L}^{θ}(x)h ∈ L_{θ}.

“⇐” Suppose that ∇f^{L}^{θ}(x) is L_{θ}-invariant for all x ∈ S. Take x, y ∈ S with x L_{θ} y (i.e.,
x−y ∈ Lθ). In order to show the desired result, we need to argue that f^{L}^{θ}(x) L_{θ} f^{L}^{θ}(y).

For any ζ ∈ L^{◦}_{θ}, we have

ζ, f^{L}^{θ}(x) − f^{L}^{θ}(y) =
Z 1

0

ζ, ∇f^{L}^{θ} x + t(x − y)(x − y) dt ≤ 0, (16)
where the last step comes from ∇f^{L}^{θ}(x + t(x − y))(x − y) ∈ L_{θ} because x + t(x − y) ∈ S
(since S is convex) and ∇f^{L}^{θ} is Lθ-invariant over S by hypothesis. Since ζ ∈ L^{◦}_{θ} is
arbitrary, (16) implies f^{L}^{θ}(x) − f^{L}^{θ}(y) ∈ (L^{◦}_{θ})^{◦} = L_{θ}, where the last step is due to the
fact that L_{θ} is a closed convex cone. This means f^{L}^{θ}(x) L_{θ} f^{L}^{θ}(y). 2

Note that f is Lipschitz continuous on J if and only if f^{L}^{θ} 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 L_{θ}-monotone on J ;

(b) ∂_{B}f^{L}^{θ}(x) is L_{θ}-invariant for all x ∈ S;

(c) ∂f^{L}^{θ}(x) is L_{θ}-invariant for all x ∈ S.

Proof. “(a) ⇒ (b)” Take V ∈ ∂_{B}f^{L}^{θ}(x), then by definition of B-subdifferential there
exists {x_{k}} ⊂ D_{f}_{Lθ} such that x_{k} → x and ∇f^{L}^{θ}(x_{i}) → V . According to (15), we obtain

∇f^{L}^{θ}(x_{i})h L_{θ} 0 for h ∈ L_{θ}. Taking the limit yields V h L_{θ} 0. Since V ∈ ∂_{B}f^{L}^{θ}(x) is
arbitrary, this says that ∂_{B}f^{L}^{θ} is L_{θ}-invariant.

“(b) ⇒ (c)” Take V ∈ ∂f^{L}^{θ}(x), then by definition, there exists V_{i} ∈ ∂_{B}f^{L}^{θ}(x) and
β_{i} ∈ [0, 1] such that V = P

iβ_{i}V_{i} and P

iβ_{i} = 1. Thus, for any h ∈ L_{θ}, we have
V h = P

iβiVih ∈ Lθ, since Vi is Lθ-invariant and Lθ is convex. Hence ∂f^{L}^{θ}(x) is Lθ-
invariant.

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

for some V ∈ ∂f^{L}^{θ}(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 L_{θ}-monotonicity.

Theorem 2.3. Suppose that f : J → IR is differentiable. If for all t_{1}, t_{2} ∈ J with t_{1} ≤ t_{2},
(tan θ − ctanθ)

f^{0}(t_{1}) − f^{0}(t_{2})

≥ 0, (17)

and

f^{0}(t1) f (t_{2}) − f (t_{1})
t_{2}− t_{1}
f (t_{2}) − f (t_{1})

t_{2}− t_{1} f^{0}(t_{2})

_{S}^{2}

+ O, (18)

then f is L_{θ}-monotone on J .

Proof. According to Theorem 2.1, it suffices to show that ∇f^{L}^{θ}(x) is Lθ-invariant for
all x ∈ S. We proceed by discussing the following two cases.

Case 1: For x_{2} = 0, in this case it is clear that ∇f^{L}^{θ}(x) being L_{θ}-invariant, i.e.,

∇f^{L}^{θ}(x)h = f^{0}(x1)h ∈ Lθ for all h ∈ Lθ, is equivalent to saying f^{0}(x1) ≥ 0.

Case 2: For x_{2} 6= 0, let

H := τ I + (η − τ )¯x2x¯^{T}_{2}.

Then, applying Lemma 2.2 to the formula of ∇f^{L}^{θ}(x) in (14), ∇f^{L}^{θ}(x) is L_{θ}-invariant if
and only if

ξ ≥ k%k tan θ (19)

and there exists λ ≥ 0 such that Υ :=

ξ^{2}− ctan^{2}θ%^{2}− λ ξ tan θ%¯x^{T}_{2} − ctanθ%¯x^{T}_{2}H
ξ tan θ%¯x_{2}− ctanθ%H ¯x_{2} tan^{2}θ%^{2}x¯_{2}x¯^{T}_{2} − H^{2}+ λI

_{S}^{n}

+ 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^{0}(λ_{1}(x)) − ctanθf^{0}(λ_{2}(x)) ≤ − tan θf^{0}(λ_{1}(x)) + tan θf^{0}(λ_{2}(x))

≤ tan θf^{0}(λ_{1}(x)) + ctanθf^{0}(λ_{2}(x))

⇐⇒ f^{0}(λ_{2}(x)) ≥ 0 and f^{0}(λ_{1}(x)) ≥ 1 − ctan^{2}θ

2 f^{0}(λ_{2}(x)). (21)

This is ensured by (17) and (18). In fact, if tan θ ≥ ctanθ, then we know from (17)
that f^{0}(λ_{1}(x)) ≥ f^{0}(λ_{2}(x)) ≥ (1 − ctan^{2}θ)/2f^{0}(λ_{2}(x)) where the second inequality
is due to f^{0}(λ2(x)) ≥ 0 by (18). If tan θ ≤ ctanθ, then 1 − ctan^{2}θ ≤ 0, and hence
f^{0}(λ_{1}(x)) ≥ 0 ≥ (1 − ctan^{2}θ)/2f^{0}(λ_{2}(x)) since f^{0}(λ_{i}(x)) ≥ 0 for i = 1, 2 by (18).

Now let us look into the entries of Υ. In the Υ_{11}-entry, we calculate
ξ^{2}− ctan^{2}θ%^{2}

= 1

(tan θ + ctanθ)^{2}

(tan^{2}θ − ctan^{2}θ)f^{0}(λ_{1}(x))^{2}+ 2(1 + ctan^{2}θ)f^{0}(λ_{1}(x))f^{0}(λ_{2}(x))

= 1

(tan θ + ctanθ)^{2}

(tan^{2}θ − ctan^{2}θ)f^{0}(λ_{1}(x))^{2}+ (ctan^{2}θ − tan^{2}θ)f^{0}(λ_{1}(x))f^{0}(λ_{2}(x))

+ 1

(tan θ + ctanθ)^{2}
h

2 + tan^{2}θ + ctan^{2}θ
i

f^{0}(λ1(x))f^{0}(λ2(x))

= µ + f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)),
with

µ

:= 1

(tan θ + ctanθ)^{2}

(tan^{2}θ − ctan^{2}θ)f^{0}(λ1(x))^{2}+ (ctan^{2}θ − tan^{2}θ)f^{0}(λ1(x))f^{0}(λ2(x))

= tan θ − ctanθ

tan θ + ctanθf^{0}(λ_{1}(x))h

f^{0}(λ_{1}(x)) − f^{0}(λ_{2}(x))i

≥ 0,

where the last step is due to (17). In the Υ_{12}-entry and Υ_{21}-entry, we calculate
(ξ tan θ)%¯x^{T}_{2} − ctanθ%¯x^{T}_{2}H

= 1

(tan θ + ctanθ)^{2}

− tan^{2}θ + ctan^{2}θf^{0}(λ1(x))^{2}+ tan^{2}θ − ctan^{2}θf^{0}(λ1(x))f^{0}(λ2(x))

¯
x^{T}_{2}

= −tan θ − ctanθ

tan θ + ctanθf^{0}(λ_{1}(x))h

f^{0}(λ_{1}(x)) − f^{0}(λ_{2}(x))i

¯
x^{T}_{2}

= −µ¯x^{T}_{2}.

In the Υ_{22}-entry, we calculate

tan^{2}θ%^{2}x¯_{2}x¯^{T}_{2} − H^{2} = −τ^{2}I + τ^{2} + 1

(tan θ + ctanθ)^{2}
h

(tan^{2}θ − ctan^{2}θ)f^{0}(λ_{1}(x))^{2}

−2(1 + tan^{2}θ)f^{0}(λ1(x))f^{0}(λ2(x))
i

!

¯
x2x¯^{T}_{2}

= −τ^{2}I +

τ^{2}+ µ − f^{0}(λ1(x))f^{0}(λ2(x))

¯
x2x¯^{T}_{2}.
Hence, Υ can be rewritten as

Υ = Υ_{11} Υ_{12}
Υ_{21} Υ_{22}

= "µ + f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) − λ −µ¯x^{T}_{2}

−µ¯x_{2} (λ − τ^{2})I +

τ^{2}+ µ − f^{0}(λ_{1}(x))f^{0}(λ_{2}(x))

¯
x_{2}x¯^{T}_{2}

# .

Now, applying Lemma 2.3 to Υ, we have

λ_{min}(Υ) ≥ min
(

Υ_{11}− |µ|, Υ_{11}−

τ^{2}+ µ − f^{0}(λ_{1}(x))f^{0}(λ_{2}(x))
)

+ min (

0,

λ − τ^{2}

− Υ_{11}+

τ^{2}+ µ − f^{0}(λ_{1}(x))f^{0}(λ_{2}(x))
)

= minn

f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) − λ, 2f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) − τ^{2}− λo
+2 min

n

0, λ − f^{0}(λ1(x))f^{0}(λ2(x))
o

. (22)

Using λ_{1}(x) ≤ λ_{2}(x) and condition (18) ensures

f^{0}(λ_{1}(x)) ≥ 0, f^{0}(λ_{2}(x)) ≥ 0, and f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) ≥ f (λ_{2}(x)) − f (λ1(x))
λ_{2}(x) − λ_{1}(x)

2

, which together with (15) yields

f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) ≥ 0 and f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) − τ^{2} ≥ 0.

Thus, we can plug λ := f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) ≥ 0 into (22), which gives λ_{min}(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 L_{θ}-
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 L_{θ}-
monotone on J if and only if f^{0}(t) ≥ 0 for all t ∈ J and (tan θ−ctanθ)(f^{0}(t_{1})−f^{0}(t_{2})) ≥ 0
for all t_{1}, t_{2} ∈ J with t_{1} ≤ t_{2}.

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

f^{0}(λ_{2}(x)) ≥ 0, f^{0}(λ_{1}(x)) ≥ 1 − ctan^{2}θ

2 f^{0}(λ_{2}(x)), (23)

and there exists λ ≥ 0 such that

Υ =µ + f^{0}(λ1(x))f^{0}(λ2(x)) − λ ±µ

±µ µ − f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) + λ

_{S}^{2}

+ O, (24)

where the form of Υ comes from the fact that ¯x_{2} = ±1 in this case. It follows from
(24) that µ^{2}− (f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)) − λ)^{2}− µ^{2} ≥ 0, which implies λ = f^{0}(λ_{1}(x))f^{0}(λ_{2}(x)).

Substituting it into (24) yields

Υ = µ ±µ

±µ µ

= µ 1 ±1

±1 1

_{S}^{2}

+ O,

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

f^{0}(λ_{2}(x)) ≥ 0, f^{0}(λ_{1}(x)) ≥ 1 − ctan^{2}θ

2 f^{0}(λ_{2}(x)),
(tan θ − ctanθ)f^{0}(λ_{1}(x))f^{0}(λ_{1}(x)) − f^{0}(λ_{2}(x)) ≥ 0.

Due to the arbitrariness of λ_{i}(x) ∈ J and applying similar arguments following (21), the
above conditions give f^{0}(t) ≥ 0 for all t ∈ J and (tan θ − ctanθ)(f^{0}(t1) − f^{0}(t2)) ≥ 0 for
all t_{1}, t_{2} ∈ J with t_{1} ≤ t_{2}. Thus, the proof is complete. 2

### 3 Monotonicity of f

^{L}

^{θ}

In Section 2, we have shown that the circular cone monotonicity f^{L}^{θ} 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 f^{L}^{θ} and the L_{θ}-monotonicity of f ? Whether
the monotonicity of f^{L}^{θ} also depends on θ? This is the main motivation of this section.

First, for a mapping H : IR^{n} → IR^{n}, let us denote ∂H(x) _{S}^{n}_{+} O (or ∂H(x) _{S}^{n}_{+} O) to
mean that each elements in ∂H(x) is positive semi-definite (or positive definite), i.e.,

∂H(x) _{S}^{n}

+ O (or _{S}^{n}

+ O) ⇐⇒ A _{S}^{n}

+ O (or _{S}^{n}

+ 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^{L}^{θ} is monotone on S if and only if ∂f^{L}^{θ}(x) _{S}^{n}

+ O for all x ∈ S;

(b) If ∂f^{L}^{θ}(x) _{S}^{n}

+ O for all x ∈ S, then f^{L}^{θ} is strictly monotone on S;

(c) f^{L}^{θ} is strongly monotone on S if and only if there exists µ > 0 such that ∂f^{L}^{θ}(x) _{S}^{n}

+

µI for all x ∈ S.

Lemma 3.2. Suppose that f is Lipschitz continuous. Then,

∂_{B}f^{L}^{θ}(x) _{S}^{n}

+ O ⇐⇒ ∂f^{L}^{θ}(x) _{S}^{n}

+ O and ∂_{B}f^{L}^{θ}(x) _{S}^{n}

+ O ⇐⇒ ∂f^{L}^{θ}(x) _{S}^{n}

+ O.

Proof. The result follows immediately from the fact ∂f^{L}^{θ}(x) = conv∂_{B}f^{L}^{θ}(x). 2
Lemma 3.3. The following statements hold.

(a) For any h ∈ IR^{n}
t_{1}

h_{1}− ctanθ¯x^{T}_{2}h_{2}2

+ t_{2}

h_{1}+ tan θ¯x^{T}_{2}h_{2}2

+ t_{3}

kh_{2}k^{2} − (¯x^{T}_{2}h_{2})^{2}

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

(b) For any h ∈ IR^{n}/{0}

t_{1}

h_{1}− ctanθ¯x^{T}_{2}h_{2}2

+ t_{2}

h_{1}+ tan θ¯x^{T}_{2}h_{2}2

+ t_{3}

kh_{2}k^{2}− (¯x^{T}_{2}h_{2})^{2}

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

Proof. (a) The sufficiency is clear, since |¯x^{T}_{2}h_{2}| ≤ k¯x_{2}kkh_{2}k = kh_{2}k by Cauchy-Schwartz
inequality. Now let us show the necessity. Taking h = (1, −ctanθ ¯x_{2})^{T}, then (25) equal
to t1(1 + ctan^{2}θ)^{2} ≥ 0, which implies t1 ≥ 0. Similarly, let h = (1, tan θ ¯x2)^{T}, then (25)
yields t_{2}(1 + tan^{2}θ)^{2} ≥ 0, implying t_{2} ≥ 0. Finally, let h = (0, u)^{T} with u satisfying
hu, ¯x_{2}i = 0 and kuk = 1, then it follows from (25) that t_{3} ≥ 0.

(b) The necessity is the same of the argument as given in part (a). For sufficiency, take a
nonzero vector h. If kh_{2}k−¯x^{T}_{2}h_{2} > 0, then the result holds since t_{3} > 0. If kh_{2}k−¯x^{T}_{2}h_{2} = 0,
then h2 = β ¯x2. So the left side of (26) takes t1(h1 − βctanθ)^{2} + t2(h1 + β tan θ)^{2} > 0,
because h_{1} − βctanθ = h_{1}+ β tan θ = 0 only happened when β = 0 and h_{1} = 0. This
means h_{2} = 0 since h_{2} = β ¯x_{2}, so h = 0. 2

If f is differentiable, it is known that f^{L}^{θ} is differentiable [4, 32]. It then follows from
Lemma 3.1 that f^{L}^{θ} is monotone on S if and only if ∇f^{L}^{θ}(x) _{S}^{n}

+ O for all x ∈ S. Hence,
to characterize the monotonicity of f^{L}^{θ}, the first thing is to estimate ∇f^{L}^{θ}(x) _{S}^{n}_{+} O via
f .

Theorem 3.1. Given x ∈ IR^{n} and suppose that f is differentiable at λ_{i}(x) for i = 1, 2.

Then ∇f^{L}^{θ}(x) _{S}^{n}

+ O if and only if f^{0}(λ_{i}(x)) ≥ 0 for i = 1, 2 and f (λ_{2}(x)) ≥ f (λ_{1}(x)).

Proof. The proof is divided into the following two cases.

Case 1. For x_{2} = 0, using ∇f^{L}^{θ}(x) = f^{0}(x_{1})e, it is clear that ∇f^{L}^{θ}(x) _{S}^{n}

+ O is equivalent
to saying f^{0}(x_{1}) ≥ 0. Then, the desired result follows.

Case 2. For x2 6= 0, denote

b_{1} := f^{0}(λ_{1}(x))

1 + ctan^{2}θ and b_{2} = f^{0}(λ_{2}(x))
1 + tan^{2}θ.

Then, ξ = b_{1}+ b_{2}, % = −b_{1}ctanθ + b_{2}tan θ, and η = b_{1}ctan^{2}θ + b_{2}tan^{2}θ. Hence, for all
h = (h_{1}, h_{2})^{T} ∈ IR × IR^{n−1}, we have

hh, ∇f^{L}^{θ}(x)hi

= (b1+ b2)h^{2}_{1}+ 2%h1x¯^{T}_{2}h2+ τ kh2k^{2}+ (b1ctan^{2}θ + b2tan^{2}θ − τ )(¯x^{T}_{2}h2)^{2}

= (b_{1}+ b_{2})h^{2}_{1}+ 2(−b_{1}ctanθ + b_{2}tan θ)h_{1}x¯^{T}_{2}h_{2}+ (b_{1}ctan^{2}θ + b_{2}tan^{2}θ)(¯x^{T}_{2}h_{2})^{2}
+τ

kh_{2}k^{2}− (¯x^{T}_{2}h_{2})^{2}

= b_{1}

h^{2}_{1}− 2ctanθh_{1}x¯^{T}_{2}h_{2}+ ctan^{2}θ(¯x^{T}_{2}h_{2})^{2}
+ b_{2}

h^{2}_{1}+ 2 tan θh_{1}x¯^{T}_{2}h_{2}+ tan^{2}θ(¯x^{T}_{2}h_{2})^{2}

+τ

kh_{2}k^{2}− (¯x^{T}_{2}h_{2})^{2}

= b_{1}

h_{1}− ctanθ¯x^{T}_{2}h_{2}2

+ b_{2}

h_{1}+ tan θ¯x^{T}_{2}h_{2}2

+ τ

kh_{2}k^{2} − (¯x^{T}_{2}h_{2})^{2}
.
In light of Lemma 3.3, the desired result is equivalent to

b_{1} ≥ 0, b_{2} ≥ 0, τ = f (λ_{2}(x)) − f (λ_{1}(x))
λ_{2}(x) − λ_{1}(x) ≥ 0,

i.e., f^{0}(λ_{1}(x)) ≥ 0, f^{0}(λ_{2}(x)) ≥ 0, and f (λ_{2}(x)) ≥ f (λ_{1}(x)) due to λ_{2}(x) > λ_{1}(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 ∈ IR^{n} and suppose that f is differentiable at λ_{i}(x) for i = 1, 2.

Then for x_{2} 6= 0, ∇f^{L}^{θ}(x) _{S}^{n}

+ O if and only if f^{0}(λ_{i}(x)) > 0 for i = 1, 2 and f (λ_{2}(x)) >

f (λ_{1}(x)); for x_{2} = 0, ∇f^{L}^{θ}(x) _{S}^{n}_{+} O if and only if f^{0}(x_{1}) > 0.

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

Lemma 3.4. Let f : IR → IR be strictly continuous. Then, for any x ∈ IR^{n}, the B-
differential ∂_{B}(f^{L}^{θ})(x) is well defined and nonempty. Moreover,

(i) if x_{2} 6= 0, then

∂B(f^{L}^{θ})(x) =

ξ %x^{T}_{2}/kx_{2}k

%x_{2}/kx_{2}k τ I + (η − τ )x_{2}x^{T}_{2}/kx_{2}k^{2}

τ = f λ2(x) − f λ1(x)
λ_{2}(x) − λ_{1}(x)
ξ − %ctanθ ∈ ∂Bf (λ1(x))
ξ + % tan θ ∈ ∂_{B}f (λ_{2}(x))
η = ξ − %(ctanθ − tan θ)

;

(27)
(ii) if x_{2} = 0, then

∂_{B}(f^{L}^{θ})(x) ⊂

ξ %w^{T}

%w τ I + (η − τ )ww^{T}

τ ∈ ∂f (λ_{1}(x)), kwk = 1
ξ − %ctanθ ∈ ∂_{B}f (λ_{1}(x))
ξ + % tan θ ∈ ∂_{B}f (λ_{1}(x))
η = ξ − %(ctanθ − tan θ)

. (28)

Lemma 3.4 presents an upper estimation on ∂_{B}f^{L}^{θ}(x) when x_{2} = 0. Here we give an
lower estimation.

Lemma 3.5. Suppose that f is locally Lipschitz at x with x_{2} = 0. Then,

∂_{B}f (x_{1})I ⊆ ∂_{B}f^{L}^{θ}(x).

Proof. First, take u ∈ ∂_{B}f (x_{1}). Since f is locally Lipschitz at x_{1}, there exists t ∈ D_{f}
satisfying t → x_{1}. Note that λ_{i}(te) = t for i = 1, 2. Then, f^{L}^{θ} is differentiable at te, i.e.,
te ∈ D_{f}_{Lθ} and ∇f^{L}^{θ}(te) = f^{0}(t)I. Hence,

uI = lim

t→x1

f^{0}(t)I = lim

te→x∇f^{L}^{θ}(te) ∈ ∂_{B}f^{L}^{θ}(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 ∈ IR^{n}, then ∂_{B}f^{L}^{θ}(x) _{S}^{n}

+ O if and only if ∂_{B}f (λ_{i}(x)) ≥ 0 for
i = 1, 2, and f (λ_{2}(x)) ≥ f (λ_{1}(x)).

Proof. Consider the following two cases.

Case 1. For x_{2} 6= 0, according to (27), we know for V ∈ ∂_{B}f^{L}^{θ}(x), there exists v_{i} ∈

∂_{B}f (λ_{i}(x)) for i = 1, 2 such that ξ − ρctanθ = v_{1}, ξ + ρ tan θ = v_{2}, and
V =

ξ %¯x^{T}_{2}

%¯x2 τ I + (η − τ )¯x2x¯^{T}_{2}

. Hence,

ξ = v_{1}tan θ + v_{2}ctanθ

tan θ + ctanθ , ρ = v_{2}− v_{1}

tan θ + ctanθ and η = v_{1}ctanθ + v_{2}tan θ
tan θ + ctanθ
and

hh, V hi

= v_{1}tan θ + v_{2}ctanθ

tan θ + ctanθ h^{2}_{1} + 2 v_{2}− v_{1}

tan θ + ctanθx¯^{T}_{2}h_{2}h_{1}+ v_{1}ctanθ + v_{2}tan θ

tan θ + ctanθ (¯x^{T}_{2}h_{2})^{2}
+τh

kh_{2}k^{2}− (¯x^{T}_{2}h_{2})^{2}i

= tan θ

tan θ + ctanθv_{1}h

h_{1}− (¯x^{T}_{2}h_{2})ctanθi2

+ ctanθ

tan θ + ctanθv_{2}h

h_{1}+ (¯x^{T}_{2}h_{2}) tan θi2

+τh

kh_{2}k^{2}− (¯x^{T}_{2}h_{2})^{2}i

. (29)

Applying Lemma 3.3, we have hh, V hi ≥ 0 for all h ∈ IR^{n} if and only if v1, v2 ≥ 0, and
τ ≥ 0. In other words, ∂_{B}f^{L}^{θ}(x) _{S}^{n}

+ O if and only if ∂_{B}f (λ_{i}(x)) ≥ 0 for i = 1, 2 and
f (λ_{2}(x)) ≥ f (λ_{1}(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.

Corollary 3.2. Given x ∈ IR^{n}, then for x_{2} 6= 0, ∂_{B}f^{L}^{θ}(x) 0 if and only if ∂_{B}f (λ_{i}(x)) >

0 for i = 1, 2, and f (λ2(x)) > f (λ1(x)); for x2 = 0, ∂Bf^{L}^{θ}(x) 0 if and only if

∂_{B}f (x_{1}) > 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) f^{L}^{θ} is monotone on S if and only if f is nondecreasing on J ;

(b) f^{L}^{θ} is strictly monotone on S if and only if f is strictly increasing on J ;

(c) f^{L}^{θ} is strongly monotone on S with µ > 0 if and only if f is strongly increasing on
S with µ > 0.

Proof. (a) =⇒. Take t_{1}, t_{2} ∈ J with t_{1} ≤ t_{2}. Since f^{L}^{θ} is monotone, we have
0 ≤ h(t1− t2)e, f^{L}^{θ}(t1e) − f^{L}^{θ}(t2e)i = (t1 − t2)(f (t1) − f (t2)),
which implies f (t_{1}) ≤ f (t_{2}).

⇐=. Take x ∈ S. Since f is monotone on J, we know ∂_{B}f (t) ≥ 0 for all t ∈ J , and
hence ∂f (t) ≥ 0 for t ∈ J . Because λ_{i}(x) ∈ J for i = 1, 2, it implies ∂_{B}f (λ_{i}(x)) ≥ 0 for
i = 1, 2. In addition, f (λ_{2}(x)) − f (λ_{1}(x)) = γ(λ_{2}(x) − λ_{1}(x)) for some γ ∈ ∂f (t_{0}) and
t_{0} ∈ (λ_{1}(x), λ_{2}(x)) ⊆ J . Hence, f (λ_{2}(x)) ≥ f (λ_{1}(x)) due to γ ∈ ∂f (t_{0}) ≥ 0. Applying
Theorem 3.2 yields ∂_{B}f^{L}^{θ}(x) _{S}^{n}

+ O, which means f^{L}^{θ} is monotone on S by Lemma 3.1.

(b) =⇒. Given t_{1}, t_{2} ∈ J with t_{1} > t_{2}. Since f^{L}^{θ} is strictly monotone on S, we have

f (t1) − f (t2)

(t1− t2) = hf^{L}^{θ}(t1e) − f^{L}^{θ}(t2e), (t1− t2)ei > 0,

which together with the fact t_{1} > t_{2} yields f (t_{1}) > f (t_{2}). This says f is strictly increasing
on J .

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

Case 1. For x_{2} = y_{2} = 0, it is clear that
hf^{L}^{θ}(x) − f^{L}^{θ}(y), x − yi =

f (x_{1}) − f (y_{1})

(x_{1}− y_{1}) > 0,

where the last step is due to the fact that f is strictly monotone and x1 6= y1, since
(x_{1}, 0) = x 6= y = (y_{1}, 0) under this case.

Case 2. For x_{2} = 0 and y_{2} 6= 0, we have
hf^{L}^{θ}(y) − f^{L}^{θ}(x), y − xi