Characterizations of SOC-monotone and SOC- SOC-convex functions

To close this section, we check with one popular smoothing function, f (t) = 1

2

√

t²+ 4 + t ,

which was proposed by Chen and Harker [38], Kanzow [84], and Smale [138]; and is called the CHKS function. Its corresponding SOC-function is defined by

f^soc(x) = 1 2



(x²+ 4e)¹² + x ,

where e = (1, 0, · · · , 0). The function f (t) is convex and monotone, so we may also wish to know whether it is SOC-convex or SOC-monotone or not. Unfortunately, it is neither SOC-convex nor SOC-monotone for n ≥ 3, though it is both SOC-convex and SOC-monotone for n = 2. The following example demonstrates what we have just said.

Example 2.8. Let f : IR → IR be f (t) =

√t²+ 4 + t

2 . Then, (a) f is not SOC-monotone of order n ≥ 3 on IR;

(b) however, f is SOC-monotone of order 2 on IR;

(c) f is not SOC-convex of order n ≥ 3 on IR;

(d) however, f is SOC-convex of order 2 on IR.

Solution. Again, by Remark 2.1, taking x = (2, 1, −1) and y = (1, 1, 0) gives a coun-terexample for both (a) and (c).

To see (b) and (d), it follows by direct verifications as what we have done before. 

2.2 Characterizations of monotone and

Conjecture 2.1. Let f : (0, ∞) → IR be continuous, convex, and nonincreasing. Then, (a) f is SOC-convex;

(b) −f is SOC-monotone.

Conjecture 2.2. Let f : [0, ∞) → [0, ∞) be continuous. Then,

−f is SOC-convex ⇐⇒ f is SOC-monotone.

Proposition 2.8. Let f : [0, ∞) → [0, ∞) be continuous. If −f is SOC-convex, then f is SOC-monotone.

Proof. Suppose that x _Kn y _Kn 0. For any 0 < λ < 1, we can write λx = λy + (1 − λ) λ

1 − λ(x − y).

Then, using the SOC-convexity of −f yields that f^soc(λx) _Kn λf^soc(y) + (1 − λ)f^soc λ

1 − λ(x − y)

_Kn 0,

where the second inequality is true since f is from [0, ∞) into itself and x − y _Kn 0. This yields f^soc(λx) _Kn λf^soc(y). Now, letting λ → 1, we obtain that f^soc(x) _Kn f^soc(y), which says that f is SOC-monotone. 

The converse of Proposition 2.8 is not true, in general. For counterexample, we consider

f (t) = − cot

−π

2(1 + t)⁻¹+ π

, t ∈ [0, ∞).

Notice that − cot(t) is SOC-monotone on [π/2, π), whereas −^π₂(1+t)⁻¹ is SOC-monotone on [0, ∞). Hence, their compound function f (t) is SOC-monotone on [0, ∞). However,

−f (t) does not satisfy the inequality (2.36) for all t ∈ (0, ∞). For example, when t₁ = 7.7 and t₂ = 7.6, the left hand side of (2.36) equals 0.0080, whereas the right hand side equals 27.8884. This shows that f (t) = − cot(t) is not SOC-concave of order n ≥ 3. In sum-mary, only one direction (“=⇒”) of Conjecture 2.2 holds. Whether Conjecture 2.1 is true or not will be confirmed at the end of Section 2.3.

We notice that if f is not a function from [0, ∞) into itself, then Proposition 2.8 may be false. For instance, f (t) = −t² is SOC-concave, but not SOC-monotone. In other words, the domain of function f is an important factor for such relation. From now on, we will demonstrate various characterizations regarding SOC-convex and SOC-monotone functions.

Proposition 2.9. Let g : J → IR and h : I → J , where J ⊆ IR and I ⊆ IR. Then, the following hold.

(a) If g is SOC-concave and SOC-monotone on J and h is SOC-concave on I, then their composition g ◦ h = g(h(·)) is also SOC-concave on I.

(b) If g is SOC-monotone on J and h is SOC-monotone on I, then g ◦ h = g(h(·)) is SOC-monotone on I.

Proof. (a) For the sake of notation, let g^soc : bS → IRⁿ and h^soc : S → bS be the vector-valued functions associated with g and h, respectively, where S ⊆ IRⁿ and bS ⊆ IRⁿ. Definebg(t) = g(h(t)). Then, for any x ∈ S, it follows from (1.2) and (1.9) that

g^soc(h^soc(x)) = g^soch

h(λ₁(x))u⁽¹⁾_x + h(λ₂(x))u⁽²⁾_x i

= gh(λ₁(x))u⁽¹⁾_x + gh(λ₂(x))u⁽²⁾_x (2.8)

= bg^soc(x).

We next prove that bg(t) is SOC-concave on I. For any x, y ∈ S and 0 ≤ β ≤ 1, from the SOC-concavity of h(t) it follows that

h^soc(βx + (1 − β)y) _Kn βh^soc(x) + (1 − β)h^soc(y).

Using the SOC-monotonicity and SOC-concavity of g, we then obtain that g^soch

h^soc(βx + (1 − β)y)i

_Kn g^soch

βh^soc(x) + (1 − β)h^soc(y)i

_Kn βg^soc[h^soc(x)] + (1 − β)g^soc[h^soc(y)].

This together with (2.8) implies that for any x, y ∈ S and 0 ≤ β ≤ 1, (bg)^soc

βx + (1 − β)y

_Kn β(bg)^soc(x) + (1 − β)(g)b^soc(y).

Consequently, the function bg(t), i.e. g(h(·)) is SOC-concave on I.

(b) If x _Kn y, then h^soc(x) _Kn h^soc(y) due to h being SOC-monotone. Using g being SOC-monotone yields

g^soc(h^soc(x)) _Kn g^soc(h^soc(y)) .

Then, by (2.8), we have (g ◦ h)^soc(x) _Kn (g ◦ h)^soc(y), which is the desired result.  Proposition 2.10. Suppose that f : IR → IR and z ∈ IRⁿ. Let g_z : IRⁿ → IR be defined by g_z(x) := hf^soc(x), zi. Then, f is SOC-convex if and only if g_z is a convex function for all z _Kn 0.

Proof. Suppose f is SOC-convex and let x, y ∈ IRⁿ, λ ∈ [0, 1]. Then, we have f^soc (1 − λ)x + λy
_Kn (1 − λ)f^soc(x) + λf^soc(y),

which implies

g_z((1 − λ)x + λy) = f^soc (1 − λ)x + λy
, z

≤ (1 − λ)f^soc(x) + λf^soc(y), z

= (1 − λ)hf^soc(x) , zi + hf^soc(y), zi

= (1 − λ)g_z(x) + λg_z(y),

where the inequality holds by Property 1.3(d). This says that g_z is a convex function.

For the other direction, from the convexity of g, we obtain

f^soc (1 − λ)x + λy
, z ≤ (1 − λ)f^soc(x) + λf^soc(y), z. Since z _Kn 0, by Property 1.3(d) again, the above yields

f^soc (1 − λ)x + λy
_Kn (1 − λ)f^soc(x) + λf^soc(y), which says f is SOC-convex. 

Proposition 2.11. A differentiable function f : IR → IR is SOC-convex if and only if f^soc(y) _Kn f^soc(x) + ∇f^soc(x)(y − x) for all x, y ∈ IRⁿ.

Proof. From Proposition 1.13, we know that f is differentiable if and only if f^soc is differentiable. Using the gradient formula given therein and following the arguments as in [20, Proposition B.3] or [29, Theorem 2.3.5], the proof can be done easily. We omit the details. 

To discover more characterizations and try to answer the aforementioned conjectures, we develop the second-order Taylor’s expansion for the vector-valued SOC-function f^soc defined as in (1.9), which is crucial to our subsequent analysis. To this end, we assume that f ∈ C⁽²⁾(J ) with J being an open interval in IR and dom(f^soc) is open in IRⁿ (this is true by Proposition 1.4(a)). Given any x ∈ dom(f^soc) and h = (h₁, h₂) ∈ IR × IRⁿ⁻¹, we have x + th ∈ dom(f^soc) for any sufficiently small t > 0. We wish to calculate the Taylor’s expansion of the function f^soc(x + th) at x for any sufficiently small t > 0. In particular, we are interested in finding matrices ∇f^soc(x) and Ai(x) for i = 1, 2, . . . , n such that

f^soc(x + th) = f^soc(x) + t∇f^soc(x)h + 1 2t²











h^TA₁(x)h h^TA₂(x)h

... h^TA_n(x)h











+ o(t²). (2.9)

Again, for convenience, we omit the variable notion x in λ_i(x) and u⁽ⁱ⁾x for i = 1, 2 in the subsequent discussions.

It is known that f^soc is differentiable (respectively, smooth) if and only if f is differ-entiable (respectively, smooth), see Proposition 1.13. Moreover, there holds that

∇f^soc(x) =









b⁽¹⁾ c⁽¹⁾ x^T₂ kx₂k c⁽¹⁾ x₂

kx₂k a⁽⁰⁾I + (b⁽¹⁾− a⁽⁰⁾)x₂x^T₂ kx₂k²









(2.10)

if x2 6= 0; and otherwise

∇f^soc(x) = f⁰(x₁)I, (2.11)

where

a⁽⁰⁾ = f (λ₂) − f (λ₁)

λ₂ − λ₁ , b⁽¹⁾ = f⁰(λ₂) + f⁰(λ₁)

2 , c⁽¹⁾ = f⁰(λ₂) − f⁰(λ₁)

2 .

Therefore, we only need to derive the formula of A_i(x) for i = 1, 2, · · · , n in (2.9).

We first consider the case where x₂ 6= 0 and x₂+ th₂ 6= 0. By the definition (1.9), we see that

f^soc(x + th) = 1

2f (x1+ th1− kx2+ th2k)

"

1

−_kx^x2+th2

2+th2k

#

+1

2f (x₁+ th₁+ kx₂+ th₂k)

"

1

x2+th2

kx₂+th2k

#

=







f (x₁+ th₁− kx₂+ th₂k) + f (x₁+ th₁+ kx₂+ th₂k) f (x₁+ th₁+ kx₂+ th₂k) − f (x₁2+ th₁− kx₂+ th₂k)

2

x₂+ th₂ kx2+ th2k







:=  Ξ₁ Ξ₂

 .

To derive the Taylor’s expansion of f^soc(x + th) at x with x₂ 6= 0, we first write out and expand kx₂+ th₂k. Notice that

kx₂+ th₂k = q

kx₂k²+ 2tx^T₂h₂+ t²kh₂k² = kx₂k s

1 + 2tx^T₂h₂

kx₂k² + t²kh₂k² kx₂k². Therefore, using the fact that

√1 + ε = 1 + 1 2ε − 1

8ε²+ o(ε²),

we may obtain

kx₂+ th₂k = kx₂k



1 + t α kx₂k+ 1

2t² β kx₂k²



+ o(t²), (2.12) where

α = x^T₂h₂

kx₂k, β = kh₂k²− (x^T₂h₂)²

kx₂k² = kh₂k²− α² = h^T₂M_x2h₂, with

M_x2 = I − x₂x^T₂ kx₂k².

Furthermore, from (2.12) and the fact that (1 + ε)⁻¹ = 1 − ε + ε²+ o(ε²), it follows that kx₂+ th₂k⁻¹ = kx₂k⁻¹



1 − t α kx2k +1

2t²

 2 α²

kx2k² − β kx2k²



+ o(t²)



. (2.13) Combining equations (2.12) and (2.13) yields that

x₂+ th₂

kx₂+ th₂k = x₂ kx₂k + t

 h₂

kx₂k − α kx₂k

x₂ kx₂k



+1 2t²



2 α²

kx₂k² − β kx₂k²

 x₂

kx₂k − 2 h₂ kx₂k

α kx₂k



+ o(t²)

= x₂

kx₂k + tM_x2 h₂

kx₂k (2.14)

+1 2t²



3h^T₂x₂x^T₂h₂ kx₂k⁴

x₂

kx₂k− kh₂k² kx₂k²

x₂

kx₂k − 2h₂h^T₂ kx₂k²

x₂ kx₂k



+ o(t²).

In addition, from (2.12), we have the following equalities f (x₁+ th₁− kx₂+ th₂k)

= f



x1+ th1−

 kx2k



1 + t α kx₂k + 1

2t² β kx₂k²



+ o(t²)



= f



λ1 + t(h1 − α) −1 2t² β

kx₂k + o(t²)



(2.15)

= f (λ1) + tf⁰(λ1)(h1− α) +1 2t²



−f⁰(λ1) β

kx₂k + f⁰⁰(λ1)(h1− α)²



+ o(t²) and

f (x₁+ th₁+ kx₂ + th₂k)

= f



λ₂+ t(h₁+ α) + 1 2t² β

kx₂k + o(t²)



(2.16)

= f (λ₂) + tf⁰(λ₂)(h₁+ α) + 1 2t²



f⁰(λ₂) β

kx₂k+ f⁰⁰(λ₂)(h₁+ α)²



+ o(t²).

For i = 0, 1, 2, we define a⁽ⁱ⁾ = f⁽ⁱ⁾(λ2) − f⁽ⁱ⁾(λ1)

λ₂− λ₁ , b⁽ⁱ⁾ = f⁽ⁱ⁾(λ2) + f⁽ⁱ⁾(λ1)

2 , c⁽ⁱ⁾ = f⁽ⁱ⁾(λ2) − f⁽ⁱ⁾(λ1)

2 , (2.17)

where f⁽ⁱ⁾ means the i-th derivative of f and f⁽⁰⁾ is the same as the original f . Then, by the equations (2.15)–(2.17), it can be verified that

Ξ₁ = 1 2



f (x₁+ th₁+ kx₂+ th₂k) + f (x₁+ th₁− kx₂+ th₂k)

= b⁽⁰⁾+ t b⁽¹⁾h₁+ c⁽¹⁾α
+ 1

2t² a⁽¹⁾β + b⁽²⁾(h²₁+ α²) + 2c⁽²⁾h₁α
+ o(t²)

= b⁽⁰⁾+ t



b⁽¹⁾h₁+ c⁽¹⁾h^T₂ x₂ kx2k

 +1

2t²h^TA₁(x)h + o(t²), where

A₁(x) =









b⁽²⁾ c⁽²⁾ x^T₂ kx₂k c⁽²⁾ x₂

kx₂k a⁽¹⁾I + b⁽²⁾− a⁽¹⁾
x2x^T₂ kx₂k²









. (2.18)

Note that in the above expression for Ξ₁, b⁽⁰⁾ is exactly the first component of f^soc(x) and (b⁽¹⁾h₁+ c⁽¹⁾h^T_{2 kx}^x2

2k) is the first component of ∇f^soc(x)h. Using the same techniques again,

1

2 f (x₁+ th₁+ kx₂+ th₂k) − f (x₁+ th₁− kx₂+ th₂k)

= c⁽⁰⁾+ t c⁽¹⁾h₁+ b⁽¹⁾α
+ 1 2t²

 b⁽¹⁾ β

kx2k + c⁽²⁾(h²₁+ α²) + 2b⁽²⁾h₁α



+ o(t²)

= c⁽⁰⁾+ t c⁽¹⁾h₁+ b⁽¹⁾α
+ 1

2t²h^TB(x)h + o(t²), (2.19)

where

B(x) =









c⁽²⁾ b⁽²⁾ x^T₂ kx₂k b⁽²⁾ x₂

kx₂k c⁽²⁾I + b⁽¹⁾

kx₂k − c⁽²⁾

 M_x2









. (2.20)

Using equations (2.19) and (2.14), we obtain that Ξ₂ = 1

2



f (x₁+ th₁+ kx₂+ th₂k) − f (x₁+ th₁− kx₂+ th₂k) x₂+ th₂ kx₂+ th₂k

= c⁽⁰⁾ x₂ kx₂k + t

 x₂

kx₂k(c⁽¹⁾h₁+ b⁽¹⁾α) + c⁽⁰⁾M_x2 h₂ kx₂k

 + 1

2t²W + o(t²), where

W = x₂

kx₂kh^TB(x)h + 2M_x2 h₂

kx₂k c⁽¹⁾h₁+ b⁽¹⁾α
+c⁽⁰⁾



3h^T₂x2x^T₂h2

kx₂k⁴ x2

kx₂k− kh2k² kx₂k²

x2

kx₂k − 2h2h^T₂ kx₂k²

x2

kx₂k

 .

Now we denote

d := b⁽¹⁾− a⁽⁰⁾

kx₂k = 2(b⁽¹⁾− a⁽⁰⁾)

λ₂− λ₁ , U := h^TC(x)h V := 2c⁽¹⁾h₁+ b⁽¹⁾α

kx2k − c⁽⁰⁾2x^T₂h₂

kx2k³ = 2a⁽¹⁾h₁+ 2dx^T₂h₂ kx2k, where

C(x) :=









c⁽²⁾ (b⁽²⁾− a⁽¹⁾) x^T₂ kx₂k (b⁽²⁾− a⁽¹⁾) x₂

kx₂k dI + c⁽²⁾− 3d
x2x^T₂ kx₂k²









. (2.21)

Then U can be further recast as

U = h^TB(x)h + c⁽⁰⁾3h^T₂x₂x^T₂h₂

kx₂k⁴ − c⁽⁰⁾kh₂k²

kx₂k² − 2x^T₂h₂

kx₂k²(c⁽¹⁾h₁+ b⁽¹⁾α).

Consequently,

W = x2

kx₂kU + h₂V.

We next consider the case where x₂ = 0 and x₂+ th₂ 6= 0. By definition (1.9),

f^soc(x + th) = f (x₁+ t(h₁− kh₂k)) 2



 1

− h₂ kh₂k



+ f (x₁+ t(h₁+ kh₂k)) 2



 1 h₂ kh₂k





=







f (x1+ t(h1− kh₂k)) + f (x₁+ t(h1+ kh2k)) f (x₁+ t(h₁+ kh₂k)) − f (x2 ₁+ t(h₁− kh₂k))

2

h₂ kh₂k





.

Using the Taylor expansion of f at x₁, we can obtain that 1

2 h

f (x₁+ t(h₁− kh₂k)) + f (x₁+ t(h₁+ kh₂k))i

= f (x₁) + tf⁽¹⁾(x₁)h₁+ 1

2t²f⁽²⁾(x₁)h^Th + o(t²), 1

2 h

f (x₁+ t(h₁− kh₂k)) − f (x₁+ t(h₁+ kh₂k))i

= tf⁽¹⁾(x₁)h₂+ 1

2t²f⁽²⁾(x₁)2h₁h₂+ o(t²).

Therefore,

f^soc(x + th) = f^soc(x) + tf⁽¹⁾(x₁)h +1

2t²f⁽²⁾(x₁)

 h^Th 2h₁h₂

 .

Thus, under this case, we have that

A₁(x) = f⁽²⁾(x₁)I, A_i(x) = f⁽²⁾(x₁)

 0 ¯e^T_i−1

¯

e_i−1 O



i = 2, · · · , n, (2.22)

where ¯e_j ∈ IRⁿ⁻¹ is the vector whose j-th component is 1 and the others are 0.

Summing up the above discussions gives the following conclusion.

Proposition 2.12. Let f ∈ C⁽²⁾(J ) with J being an open interval in IR (which implies dom(f^soc) is open in IRⁿ). Then, for any x ∈ dom(f^soc), h ∈ IRⁿ and any sufficiently small t > 0, there holds

f^soc(x + th) = f^soc(x) + t∇f^soc(x)h + 1 2t²











h^TA₁(x)h h^TA₂(x)h

... h^TA_n(x)h











+ o(t²),

where ∇f^soc(x) and Ai(x) for i = 1, 2, · · · , n are given by (2.11) and (2.22) if x2 = 0;

and otherwise ∇f^soc(x) and A₁(x) are given by (2.10) and (2.18), respectively, and for i ≥ 2,

A_i(x) = C(x) x_2i

kx₂k+ B_i(x) where

B_i(x) = ve^T_i + e_iv^T, v =



a⁽¹⁾ d x^T₂ kx₂k

T

=



a⁽¹⁾, d kx₂kx₂

 .

From Proposition 2.11 and Proposition 2.12, the following consequence is obtained.

Proposition 2.13. Let f ∈ C⁽²⁾(J ) with J being an open interval in IR (which implies dom(f^soc) is open in IRⁿ). Then, f is SOC-convex if and only if for any x ∈ dom(f^soc) and h ∈ IRⁿ, the vector











h^TA1(x)h h^TA₂(x)h

... h^TAn(x)h











∈ Kⁿ,

where Ai(x) is given as in (2.22).

Now we are ready to show our another main result about the characterization of SOC-monotone functions. Two technical lemmas are needed for the proof. The first one is so-called S-Lemma whose proof can be found in [124].

Lemma 2.3. 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  λA for some λ ≥ 0.

Lemma 2.4. Given θ ∈ IR, a ∈ IRⁿ⁻¹, and a symmetric matrix A ∈ IR^n×n. Let Bⁿ⁻¹:=

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

(a) For any h ∈ Kⁿ, Ah ∈ Kⁿ is equivalent to A 1 z



∈ Kⁿ for any z ∈ Bⁿ⁻¹. (b) For any z ∈ Bⁿ⁻¹, θ + a^Tz ≥ 0 is equivalent to θ ≥ kak.

(c) If A =

 θ a^T

a H



with H being an (n − 1) × (n − 1) symmetric matrix, then for any h ∈ Kⁿ, Ah ∈ Kⁿ is equivalent to θ ≥ kak and there exists λ ≥ 0 such that the matrix

 θ²− kak²− λ θa^T − a^TH θa − H^Ta aa^T − H^TH + λI



 O.

Proof. (a) For any h ∈ Kⁿ, suppose that Ah ∈ Kⁿ. Let h = 1 z



where z ∈ Bⁿ⁻¹. Then h ∈ Kⁿ and the desired result follows. For the other direction, if h = 0, the conclusion is obvious. Now let h := (h₁, h₂) be any nonzero vector in Kⁿ. Then, h₁ > 0 and kh₂k ≤ h₁. Consequently, ^h_h²

1 ∈ Bⁿ⁻¹ and A

 1

h2

h1



∈ Kⁿ. Since Kⁿ is a cone, we have

h₁A

 1

h2

h1



= Ah ∈ Kⁿ.

(b) For z ∈ Bⁿ⁻¹, suppose θ + a^Tz ≥ 0. If a = 0, then the result is clear since θ ≥ 0. If a 6= 0, let z := −_kak^a . Clearly, z ∈ Bⁿ⁻¹ and hence θ +^−a_kak^Ta ≥ 0 which gives θ − kak ≥ 0.

For the other direction, the result follows from the Cauchy Schwarz Inequality:

θ + a^Tz ≥ θ − kak · kzk ≥ θ − kak ≥ 0.

(c) From part(a), Ah ∈ Kⁿ for any h ∈ Kⁿ is equivalent to A 1 z



∈ Kⁿ for any z ∈ Bⁿ⁻¹. Notice that

A 1 z



=

 θ a^T

a H

  1 z



= θ + a^Tz a + Hz

 .

Then, Ah ∈ Kⁿ for any h ∈ Kⁿ is equivalent to the following two things:

θ + a^Tz ≥ 0 for any z ∈ Bⁿ⁻¹ (2.23)

and

(a + Hz)^T(a + Hz) ≤ (θ + a^Tz)², for any z ∈ Bⁿ⁻¹. (2.24) By part(b), (2.23) is equivalent to θ ≥ kak. Now, we write the expression of (2.24) as below:

z^T aa^T − H^TH
z + 2 θa^T − a^TH
z + θ² − a^Ta ≥ 0, for any z ∈ Bⁿ⁻¹, which can be further simplified as

 1 z^T 

 θ²− kak² θa^T − a^TH θa − H^Ta aa^T − H^TH

  1 z



≥ 0, for any z ∈ Bⁿ⁻¹. Observe that z ∈ Bⁿ⁻¹ is the same as

 1 z^T  1 0 0 −I

  1 z



≥ 0.

Thus, by applying the S-Lemma (Lemma 2.3), there exists λ ≥ 0 such that

 θ²− kak² θa^T − a^TH θa − H^Ta aa^T − H^TH



− λ 1 0 0 −I



 O.

This completes the proof of part(c). 

Proposition 2.14. Let f ∈ C⁽¹⁾(J ) with J being an open interval (which implies dom(f^soc) is open in IRⁿ). Then, the following hold.

(a) f is SOC-monotone of order 2 if and only if f⁰(τ ) ≥ 0 for any τ ∈ J . (b) f is SOC-monotone of order n ≥ 3 if and only if the 2 × 2 matrix







f⁽¹⁾(t₁) f (t2) − f (t1) t₂− t₁ f (t₂) − f (t₁)

t2− t1

f⁽¹⁾(t₂)





 O, ∀ t₁, t₂ ∈ J.

Proof. By the definition of SOC-monotonicity, f is SOC-monotone if and only if

f^soc(x + h) − f^soc(x) ∈ Kⁿ (2.25) for any x ∈ dom(f^soc) and h ∈ Kⁿsuch that x + h ∈ dom(f^soc). By the first-order Taylor expansion of f^soc, i.e.,

f^soc(x + h) = f^soc(x) + ∇f^soc(x + th)h for some t ∈ (0, 1),

it is clear that (2.25) is equivalent to ∇f^soc(x + th)h ∈ Kⁿ for any x ∈ dom(f^soc) and h ∈ Kⁿsuch that x + h ∈ dom(f^soc), and some t ∈ (0, 1). Let y := x + th = µ₁v⁽¹⁾+ µ₂v⁽²⁾ for such x, h and t. We next proceed the arguments by the two cases of y₂ 6= 0 and y₂ = 0.

Case (1): y₂ 6= 0. Under this case, we notice that

∇f^soc(y) =

 θ a^T

a H

 , where

θ = ˜b⁽¹⁾, a = ˜c⁽¹⁾ y₂

ky₂k, and H = ˜a⁽⁰⁾I + (˜b⁽¹⁾− ˜a⁽⁰⁾)y₂y^T₂ ky₂k², with

˜

a⁽⁰⁾= f (µ₂) − f (µ₁)

µ₂− µ₁ , ˜b⁽¹⁾ = f⁰(µ₂) + f⁰(µ₁)

2 , ˜c⁽¹⁾ = f⁰(µ₂) − f⁰(µ₁)

2 .

In addition, we also observe that

θ²− kak² = (˜b⁽¹⁾)²− (˜c⁽¹⁾)², θa^T − a^TH = 0 and

aa^T − H^TH = −(˜a⁽⁰⁾)²I +

(˜c⁽¹⁾)²− (˜b⁽¹⁾)²+ (˜a⁽⁰⁾)² y₂y^T₂ ky₂k². Thus, by Lemma 2.4, f is SOC-monotone if and only if

(i) ˜b⁽¹⁾ ≥ |˜c⁽¹⁾|;

(ii) and there exists λ ≥ 0 such that the matrix





(˜b⁽¹⁾)²− (˜c⁽¹⁾)²− λ 0 0 (λ − (˜a⁽⁰⁾)²)I +

(˜c⁽¹⁾)²− (˜b⁽¹⁾)²+ (˜a⁽⁰⁾)² y2y₂^T ky₂k²



 O.

When n = 2, (i) together with (ii) is equivalent to saying that f⁰(µ1) ≥ 0 and f⁰(µ2) ≥ 0.

Then we conclude that f is SOC-monotone if and only if f⁰(τ ) ≥ 0 for any τ ∈ J . When n ≥ 3, (ii) is equivalent to saying that (˜b⁽¹⁾)²− (˜c⁽¹⁾)²− λ ≥ 0 and λ − (˜a⁽⁰⁾)² ≥ 0, i.e., (˜b⁽¹⁾)²− (˜c⁽¹⁾)² ≥ (˜a⁽⁰⁾)². Therefore, (i) together with (ii) is equivalent to







f⁽¹⁾(µ₁) f (µ₂) − f (µ₁) µ₂− µ₁ f (µ₂) − f (µ₁)

µ₂ − µ₁ f⁽¹⁾(µ₂)





 O

for any x ∈ IRⁿ, h ∈ Kⁿ such that x + h ∈ domf^soc, and some t ∈ (0, 1). Thus, we conclude that f is SOC-monotone if and only if







f⁽¹⁾(t₁) f (t₂) − f (t₁) t₂− t₁ f (t2) − f (t1)

t₂− t₁ f⁽¹⁾(t₂)





 O for all t₁, t₂ ∈ J.

Case (2): y₂ = 0. Now we have µ₁ = µ₂ and ∇f^soc(y) = f⁽¹⁾(µ₁)I = f⁽¹⁾(µ₂)I. Hence, f is SOC-monotone is equivalent to f⁽¹⁾(µ₁) ≥ 0, which is also equivalent to







f⁽¹⁾(µ₁) f (µ₂) − f (µ₁) µ₂− µ₁ f (µ₂) − f (µ₁)

µ2 − µ1

f⁽¹⁾(µ₂)





 O

since f⁽¹⁾(µ₁) = f⁽¹⁾(µ₂) and ^{f (µ}_µ^{2)−f (µ1)}

2−µ1 = f⁽¹⁾(µ₁) = f⁽¹⁾(µ₂) by the Taylor formula and µ₁ = µ₂. Thus, similar to Case (1), the conclusion also holds under this case. 

The SOC-convexity and SOC-monotonicity are also connected to their counterparts, matrix-convexity and matrix-monotonicity. Before illustrating their relations, we briefly recall definitions of matrix-convexity and matrix-monotonicity.

Definition 2.2. Let M^san denote n×n self-adjoint complex matrices, σ(A) be the spectrum of a matrix A, and J ⊆ IR be an interval.

(a) A function f : J → IR is called matrix monotone of degree n or n-matrix monotone if, for every A, B ∈ M^san with σ(A) ⊆ J and σ(B) ⊆ J , it holds that

A  B =⇒ f (A)  f (B).

(b) A function f : J → IR is called operator monotone or matrix monotone if it is n-matrix monotone for all n ∈ N.

(c) A function f : J → IR is called matrix convex of degree n or n-matrix convex if, for every A, B ∈ M^san with σ(A) ⊆ J and σ(B) ⊆ J , it holds that

f ((1 − λ)A + λB)  (1 − λ)f (A) + λf (B).

(d) A function f : J → IR is called operator convex or matrix convex if it is n-matrix convex for all n ∈ N.

(e) A function f : J → IR is called matrix concave of degree n or n-matrix concave if

−f is n-matrix convex.

(f ) A function f : J → IR is called operator concave or matrix concave if it is n-matrix concave for all n ∈ N.

In fact, from Proposition 2.14 and [75, Theorem 6.6.36], we immediately have the following consequences.

Proposition 2.15. Let f ∈ C⁽¹⁾(J ) with J being an open interval in IR. Then, the following hold.

(a) f is SOC-monotone of order n ≥ 3 if and only if it is 2-matrix monotone, and f is SOC-monotone of order n ≤ 2 if it is 2-matrix monotone.

(b) Suppose that n ≥ 3 and f is SOC-monotone of order n. Then, f⁰(t0) = 0 for some t₀ ∈ J if and only if f (·) is a constant function on J.

We illustrate a few examples by using either Proposition 2.14 or Proposition 2.15.

Example 2.9. Let f : (0, ∞) → IR be f (t) = ln t. Then, f (t) is SOC-monotone on (0, ∞).

Solution. To see this, it needs to verify that the 2 × 2 matrix







f⁽¹⁾(t1) f (t₂) − f (t₁) t₂− t₁ f (t₂) − f (t₁)

t₂− t₁ f⁽¹⁾(t₂)





=







1 t₁

ln(t₂) − ln(t₁) t₂− t₁ ln(t₂) − ln(t₁)

t₂− t₁

1 t₂







is positive semidefinite for all t₁, t₂ ∈ (0, ∞). 

Example 2.10. (a) For any fixed σ ∈ IR, the function f (t) = _σ−t¹ is SOC-monotone on (σ, ∞).

(b) For any fixed σ ∈ IR, the function f (t) =√

t − σ is SOC-monotone on [σ, ∞).

(c) For any fixed σ ∈ IR, the function f (t) = ln(t − σ) is SOC-monotone on (σ, ∞).

(d) For any fixed σ ≥ 0, the function f (t) = _t+σ^t is SOC-monotone on (−σ, ∞).

Solution. (a) For any t₁, t₂ ∈ (σ, ∞), it is clear to see that







1 (σ − t₁)²

1

(σ − t₂)(σ − t₁) 1

(σ − t2)(σ − t1)

1 (σ − t2)²





 O.

Then, applying Proposition 2.14 yields the desired result.

(b) If x _Kn σe, then (x − σe)^1/2 _Kn 0. Thus, by Proposition 2.14, it suffices to show









1 2√

t₁− σ

√t2− σ −√ t1− σ t₂− t₁

√t₂− σ −√ t₁− σ t₂ − t₁

1 2√

t₂− σ









 O for any t₁, t₂ > 0,

which is equivalent to proving that 1 4√

t₁− σ√

t₂− σ − 1

(√

t₂− σ +√

t₁− σ)² ≥ 0.

This inequality holds by 4√

t₁− σ√

t₂ − σ ≤ (√

t₂− σ +√

t₁− σ)² for any t₁, t₂ ∈ (σ, ∞).

(c) By Proposition 2.14, it suffices to prove that for any t₁, t₂ ∈ (σ, ∞),









1 (t₁− σ)

1

(t₂− t₁)ln t₂− σ t₁− σ

 1

(t₂− t₁)ln t₂− σ t₁− σ

 1

(t₂− σ)









 O,

which is equivalent to showing that 1

(t₁− σ)(t₂− σ) −

 1

(t₂− t₁)ln t₂− σ t₁− σ

2

≥ 0.

Notice that ln t ≤ t − 1 (t > 0), and hence it is easy to verify that

 1

(t₂− t₁)ln t₂− σ t₁− σ

2

≤ 1

(t₁− σ)(t₂− σ). Consequently, the desired result follows.

(d) Since for any fixed σ ≥ 0 and any t₁, t₂ ∈ (−σ, ∞), there holds that







σ (σ + t₁)²

σ

(σ + t₂)(σ + t₁) σ

(σ + t₂)(σ + t₁)

σ (σ + t₂)²





 O, we immediately obtain the desired result from Proposition 2.14. 

We point out that the SOC-monotonicity of order 2 does not imply the 2-matrix monotonicity. For example, f (t) = t² is SOC-monotone of order 2 on (0, ∞) by Exam-ple 2.2(a), but by [75, Theorem 6.6.36] we can verify that it is not 2-matrix monotone.

Proposition 2.15(a) indicates that a continuously differentiable function defined on an open interval must be SOC-monotone if it is 2-matrix monotone.

Next, we exploit Peirce decomposition to derive some characterizations for SOC-convex functions. Let f ∈ C⁽²⁾(J ) with J being an open interval in IR and dom(f^soc) ⊆ IRⁿ. For any x ∈ dom(f^soc) and h ∈ IRⁿ, if x₂ = 0, from Proposition 2.12, we have











h^TA1(x)h h^TA₂(x)h

... h^TAn(x)h











= f⁽²⁾(x₁)

 h^Th 2h₁h₂

 .

Since (h^Th, 2h₁h₂) ∈ Kⁿ, from Proposition 2.13, it follows that f is SOC-convex if and only if f⁽²⁾(x₁) ≥ 0. By the arbitrariness of x₁, f is SOC-convex if and only if f is convex

on J .

For the case of x₂ 6= 0, we let x = λ₁u⁽¹⁾+ λ₂u⁽²⁾, where u⁽¹⁾ and u⁽²⁾ are given by (1.4) with ¯x₂ = _kx^x2

2k. Let u⁽ⁱ⁾ = (0, υ⁽ⁱ⁾₂ ) for i = 3, · · · , n, where υ₂⁽³⁾, · · · , υ⁽ⁿ⁾₂ is any orthonormal set of vectors that span the subspace of IRⁿ⁻² orthogonal to x₂. It is easy to verify that the vectors u⁽¹⁾, u⁽²⁾, u⁽³⁾, · · · , u⁽ⁿ⁾ are linearly independent. Hence, for any given h = (h₁, h₂) ∈ IR × IRⁿ⁻¹, there exists µ_i, i = 1, 2, · · · , n such that

h = µ₁√

2 u⁽¹⁾+ µ₂√

2 u⁽²⁾+

n

X

i=3

µ_iu⁽ⁱ⁾.

From (2.18), we can verify that b⁽²⁾+ c⁽²⁾ and b⁽²⁾− c⁽²⁾ are the eigenvalues of A₁(x) with u⁽²⁾and u⁽¹⁾being the corresponding eigenvectors, and a⁽¹⁾is the eigenvalue of multiplicity n−2 with u⁽ⁱ⁾ = (0, υ₂⁽ⁱ⁾) for i = 3, . . . , n being the corresponding eigenvectors. Therefore,

h^TA₁(x)h = µ²₁(b⁽²⁾− c⁽²⁾) + µ²₂(b⁽²⁾+ c⁽²⁾) + a⁽¹⁾

n

X

i=3

µ²_i

= f⁽²⁾(λ1)µ²₁+ f⁽²⁾(λ2)µ²₂+ a⁽¹⁾µ², (2.26) where

µ² =Pn i=3µ²_i.

Similarly, we can verify that c⁽²⁾+ b⁽²⁾− a⁽¹⁾ and c⁽²⁾− b⁽²⁾+ a⁽¹⁾ are the eigenvalues of









c⁽²⁾ (b⁽²⁾− a⁽¹⁾) x^T₂ kx2k (b⁽²⁾− a⁽¹⁾) x₂

kx₂k dI + c⁽²⁾− d
x2x^T₂ kx₂k²









with u⁽²⁾ and u⁽¹⁾ being the corresponding eigenvectors, and d is the eigenvalue of mul-tiplicity n − 2 with u⁽ⁱ⁾ = (0, υ₂⁽ⁱ⁾) for i = 3, · · · , n being the corresponding eigenvectors.

Notice that C(x) in (2.21) can be decomposed the sum of the above matrix and





0 0

0 −2d x₂x^T₂ kx₂k²



.

Consequently,

h^TC(x)h = µ²₁(c⁽²⁾− b⁽²⁾+ a⁽¹⁾) + µ²₂(c⁽²⁾+ b⁽²⁾− a⁽¹⁾) − d(µ₂− µ₁)²+ dµ². (2.27) In addition, by the definition of B_i(x), it is easy to compute that

h^TB_i(x)h =√

2h_2,i−1(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d)), (2.28)

where h_2i = (h₂₁, . . . , h_2,n−1). From equations (2.26)-(2.28) and the definition of A_i(x) in (2.22), we thus have

n

X

i=2

(h^TA_i(x)h)² = [h^TC(x)h]²+ 2kh₂k²(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d))² +2(µ₂− µ₁)h^TC(x)h(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d))

= [h^TC(x)h]²+ 2(1

2(µ₂− µ₁)²+ µ²)(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d))² +2(µ₂− µ₁)h^TC(x)h(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d))

= [h^TC(x)h + (µ₂− µ₁)(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d))]² +2µ²(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d))²

= [−f⁽²⁾(λ₁)µ²₁+ f⁽²⁾(λ₂)µ²₂+ dµ²]²

+2µ²(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d))². (2.29) On the other hand, by Proposition 2.13, f is SOC-convex if and only if

A₁(x)  O and

n

X

i=2

(h^TA_i(x)h)² ≤ (h^TA₁(x)h)². (2.30) From (2.26) and (2.29)-(2.40), we have that f is SOC-convex if and only if A₁(x)  O and

−f⁽²⁾(λ₁)µ²₁+ f⁽²⁾(λ₂)µ²₂+ dµ²²

+ 2µ²(µ₁(a⁽¹⁾− d) + µ₂(a⁽¹⁾+ d))²

≤f⁽²⁾(λ₁)µ²₁+ f⁽²⁾(λ₂)µ²₂+ a⁽¹⁾µ²²

. (2.31)

When n = 2, it is clear that µ = 0. Then, f is SOC-convex if and only if A₁(x)  O and f⁽²⁾(λ₁)f⁽²⁾(λ₂) ≥ 0.

From the previous discussions, we know that b⁽²⁾− c⁽²⁾ = f⁽²⁾(λ₁), b⁽²⁾ + c⁽²⁾ = f⁽²⁾(λ₂) and a⁽¹⁾ = ^f(1)(λ_λ^{2)−f(1)(λ1)}

2−λ1 are all eigenvalues of A₁(x). Thus, f is SOC-convex if and only if

f⁽²⁾(λ₂) ≥ 0, f⁽²⁾(λ₁) ≥ 0, f⁽¹⁾(λ₂) ≥ f⁽¹⁾(λ₁),

which by the arbitrariness of x is equivalent to saying that f is convex on J .

When n ≥ 3, if µ = 0, then from the discussions above, we know that f is SOC-convex if and only if f is convex. If µ 6= 0, without loss of generality, we assume that µ² = 1.

Then, the inequality (2.41) above is equivalent to 4f⁽²⁾(λ₁)f⁽²⁾(λ₂)µ²₁µ²₂+ (a⁽¹⁾)² − d²

+2f⁽²⁾(λ2)µ²₂(a⁽¹⁾− d) + 2f⁽²⁾(λ1)µ²₁(a⁽¹⁾+ d)

−2

µ²₁(a⁽¹⁾− d)²+ µ²₂(a⁽¹⁾+ d)²+ 2µ₁µ₂((a⁽¹⁾)²− d²)

≥ 0 for any µ₁, µ₂. (2.32)

Now we show that A₁(x)  O and (2.32) holds if and only if f is convex on J and f⁽²⁾(λ₁)(a⁽¹⁾+ d) ≥ (a⁽¹⁾− d)², (2.33) f⁽²⁾(λ2)(a⁽¹⁾− d) ≥ (a⁽¹⁾+ d)². (2.34) Indeed, if f is convex on J , then by the discussions above A₁(x)  O clearly holds. If the inequalities (2.33) and (2.34) hold, then by the convexity of f we have a⁽¹⁾ ≥ |d|. If µ₁µ₂ ≤ 0, then we readily have the inequality (2.32). If µ₁µ₂ > 0, then using a⁽¹⁾ ≥ |d|

yields that

f⁽²⁾(λ₁)f⁽²⁾(λ₂)µ²₁µ²₂ ≥ (a⁽¹⁾)²− d².

Combining with equations (2.33) and (2.34) thus leads to the inequality (2.32). On the other hand, if A₁(x)  O, then f must be convex on J by the discussions above, whereas if the inequality (2.32) holds for any µ₁, µ₂, then by letting µ₁ = µ₂ = 0 yields that

a⁽¹⁾ ≥ |d|. (2.35)

Using the inequality (2.35) and letting µ₁ = 0 in (2.32) then yields (2.33), whereas using (2.35) and letting µ₂ = 0 in (2.32) leads to (2.34). Thus, when n ≥ 3, f is SOC-convex if and only if f is convex on J and (2.33) and (2.34) hold. We notice that (2.33) and (2.34) are equivalent to

1

2f⁽²⁾(λ₁)[f (λ₁) − f (λ₂) + f⁽¹⁾(λ₂)(λ₂ − λ₁)]

(λ₂− λ₁)²

≥ [f (λ₂) − f (λ₁) − f⁽¹⁾(λ₁)(λ₂− λ₁)]² (λ₂− λ₁)⁴

and

1

2f⁽²⁾(λ₂)[f (λ₂) − f (λ₁) − f⁽¹⁾(λ₁)(λ₂− λ₁)]

(λ₂− λ₁)²

≥ [f (λ₁) − f (λ₂) + f⁽¹⁾(λ₂)(λ₂− λ₁)]² (λ₂− λ₁)⁴ . Therefore, f is SOC-convex if and only if f is convex on J , and

1

2f⁽²⁾(t₀)[f (t₀) − f (t) − f⁽¹⁾(t)(t₀− t)]

(t₀− t)²

≥ [f (t) − f (t₀) − f⁽¹⁾(t₀)(t − t₀)]²

(t₀− t)⁴ , ∀ t₀, t ∈ J. (2.36) Summing up the above analysis, we can characterize the SOC-convexity as follows.

Proposition 2.16. Let f ∈ C⁽²⁾(J ) with J being an open interval in IR (which implies dom(f^soc) is open in IRⁿ). Then, the following hold.

(a) f is SOC-convex of order 2 if and only if f is convex.

(b) f is SOC-convex of order n ≥ 3 if and only if f is convex and the inequality (2.36) holds for any t₀, t ∈ J .

By the formulas of divided differences, it is not hard to verify that f is convex on J and (2.36) holds for any t0, t ∈ J if and only if

 4²f (t₀, t₀, t₀) 4²f (t₀, t, t₀) 4²f (t, t₀, t₀) 4²f (t, t, t₀)



 O. (2.37)

This, together with Proposition 2.16 and [75, Theorem 6.6.52], leads to the following results.

Proposition 2.17. Let f ∈ C⁽²⁾(J ) with J being an open interval in IR (which implies dom(f^soc) is open in IRⁿ). Then, the following hold.

(a) f is SOC-convex of order n ≥ 3 if and only if it is 2-matrix convex.

(b) f is SOC-convex of order n ≤ 2 if it is 2-matrix convex.

Proposition 2.17 implies that, if f is a twice continuously differentiable function de-fined on an open interval J and 2-matrix convex, then it must be SOC-convex. Similar to Proposition 2.15(a), when f is SOC-convex of order 2, it may not be 2-matrix convex. For example, f (t) = t³ is SOC-convex of order 2 on (0, +∞) by Example 2.3(c), but it is easy to verify that (2.37) does not hold for this function, and consequently, f is not 2-matrix convex. Using Proposition 2.17, we may prove that the direction “⇐” of Conjecture 2.2 does not hold in general, although the other direction is true due to Proposition 2.8.

Particularly, from Proposition 2.17 and [68, Theorem 2.3], we can establish the following characterizations for SOC-convex functions.

Proposition 2.18. Let f ∈ C⁽⁴⁾(J ) with J being an open interval in IR and dom(f^soc) ⊆ IRⁿ. If f⁽²⁾(t) > 0 for every t ∈ J , then f is SOC-convex of order n with n ≥ 3 if and only if one of the following conditions holds.

(a) For every t ∈ J , the 2 × 2 matrix







f⁽²⁾(t) 2

f⁽³⁾(t) f⁽³⁾(t) 6

6

f⁽⁴⁾(t) 24





 O.

(b) There is a positive concave function c(·) on I such that f⁽²⁾(t) = c(t)⁻³ for every t ∈ J .

(c) There holds that

f (t₀) − f (t) − f⁽¹⁾(t)(t₀− t) (t0− t)²

! 

f (t) − f (t₀) − f⁽¹⁾(t₀)(t − t₀) (t0− t)²

!

≤ 1

4f⁽²⁾(t₀)f⁽²⁾(t). (2.38)

Moreover, f is also SOC-convex of order 2 under one of the above conditions.

Proof. We note that f is convex on J . Therefore, by Proposition 2.17, it suffices to prove the following equivalence:

(2.36) ⇐⇒ assertion (a) ⇐⇒ assertion (b) ⇐⇒ assertion (c).

Case (1). (2.36) ⇒ assertion (a): From the previous discussions, we know that (2.36) is equivalent to (2.33) and (2.34). We expand (2.33) using Taylor’s expansion at λ₁ to the forth order and get

3

4f⁽²⁾(λ1)f⁽⁴⁾(λ1) ≥ (f⁽³⁾(λ1))².

We do the same for the inequality (2.34) at λ₂ and get the inequality 3

4f⁽²⁾(λ₂)f⁽⁴⁾(λ₂) ≥ (f⁽³⁾(λ₂))². The above two inequalities are precisely

3

4f⁽²⁾(t)f⁽⁴⁾(t) ≥ (f⁽³⁾(t))², ∀t ∈ J, (2.39) which is clearly equivalent to saying that the 2 × 2 matrix in (a) is positive semidefinite.

Case (2). assertion (a) ⇒ assertion (b): Take c(t) = [f⁽²⁾(t)]^−1/3 for t ∈ J . Then c is a positive function and f⁽²⁾(t) = c(t)⁻³. By twice differentiation, we obtain

f⁽⁴⁾(t) = 12c(t)⁻⁵[c⁰(t)(t)]²− 3c(t)⁻⁴c⁰⁰(t).

Substituting the last equality into the matrix in (a) then yields that

− 1

16c(t)⁻⁷c⁰⁰(t) ≥ 0,

which, together with c(t) > 0 for every t ∈ J , implies that c is concave.

Case (3). assertion (b) ⇒ assertion (c): We first prove the following fact: if f⁽²⁾(t) is strictly positive for every t ∈ J and the function c(t) = f⁽²⁾(t)−1/3

is concave on J , then

[f (t₀) − f (t) − f⁽¹⁾(t)(t₀− t)]

(t₀− t)² ≤ 1

2f⁽²⁾(t₀)^1/3f⁽²⁾(t)^2/3, ∀ t₀, t ∈ J. (2.40)

Indeed, using the concavity of the function c, it follows that [f (t₀) − f (t) − f⁽¹⁾(t)(t₀− t)]

(t₀− t)² =

Z 1 0

Z u1

0

f⁽²⁾[t + u₂(t₀− t)] du₂du₁

= Z 1

0

Z u1

0

c ((1 − u2)t + u2t0))⁻³du2du1

≤ Z 1

0

Z u1

0

((1 − u₂)c(t) + u₂c(t₀))⁻³du₂du₁. Notice that g(t) = 1/t (t > 0) has the second-order derivative g⁽²⁾(t) = 2/t³. Hence,

[f (t₀) − f (t) − f⁽¹⁾(t)(t₀− t)]

(t₀− t)² ≤ 1

2 Z 1

0

Z u1

0

g⁽²⁾((1 − u₂)c(t) + u₂c(t₀)) du₂du₁

= 1

2

 g(c(t₀)) − g(c(t))

(c(t₀) − c(t))² − g⁽¹⁾(c(t)) c(t₀) − c(t)



= 1

2c(t₀)c(t)c(t)

= 1

2f⁽²⁾(t0)^1/3f⁽²⁾(t)^2/3,

which implies the inequality (2.40). Now exchanging t₀ with t in (2.40), we obtain [f (t) − f (t₀) − f⁽¹⁾(t₀)(t − t₀)]

(t₀− t)² ≤ 1

2f⁽²⁾(t)^1/3f⁽²⁾(t0)^2/3, ∀ t, t0 ∈ J. (2.41) Since f is convex on J by the given assumption, the left hand sides of the inequalities (2.40) and (2.41) are nonnegative, and their product satisfies the inequality of (2.38).

Case (4). assertion (c) ⇒ (2.36): We introduce a function F : J → IR defined by F (t) = 1

2f⁽²⁾(t₀)[f (t₀) − f (t) − f⁽¹⁾(t)(t₀− t)] − [f (t) − f (t₀) − f⁽¹⁾(t₀)(t − t₀)]² (t₀− t)²

if t 6= t₀, and otherwise F (t₀) = 0. We next prove that F is nonnegative on J . It is easy to verify that such F (t) is differentiable on J , and moreover,

F⁰(t) = 1

2f⁽²⁾(t₀)f⁽²⁾(t)(t − t₀)

−2(t − t₀)⁻²[f (t) − f (t₀) − f⁽¹⁾(t₀)(t − t₀)](f⁽¹⁾(t) − f⁽¹⁾(t₀)) +2(t − t₀)⁻³[f (t) − f (t₀) − f⁽¹⁾(t₀)(t − t₀)]²

= 1

2f⁽²⁾(t₀)f⁽²⁾(t)(t − t₀)

−2(t − t₀)⁻³[f (t) − f (t₀) − f⁽¹⁾(t₀)(t − t₀)][f (t₀) − f (t) − f⁽¹⁾(t)(t₀− t)]

= 2(t − t₀) 1

4f⁽²⁾(t₀)f⁽²⁾(t) − (t − t₀)⁻⁴ f (t) − f (t₀) − f⁽¹⁾(t₀)(t − t₀)
f (t₀) − f (t) − f⁽¹⁾(t)(t₀ − t)
i

.

Using the inequality in part(c), we can verify that F (t) has a minimum value 0 at t = t₀, and therefore, F (t) is nonnegative on J . This implies the inequality (2.36). 

We demonstrate a few examples by using either Proposition 2.16, Proposition 2.17, or Proposition 2.18.

Example 2.11. Let f : IR → [0, ∞) be f (t) = e^t. Then, (a) f is SOC-convex of order 2 on IR;

(b) f is not SOC-convex of order n ≥ 3 on IR.

Solution. (a) By applying Proposition 2.16(a), it is clear that f is SOC-convex because exponential function is a convex function on IR.

(b) As below, it is a counterexample which shows f (t) = e^t is not SOC-convex of order n ≥ 3. To see this, we compute that

e[(2,0,−1)+(6,−4,−3)]/2 = e^{(4,−2,−2)}

= e⁴



cosh(2√

2) , sinh(2√

2) · (−2, −2)/(2√ 2)



≈ (463.48, −325.45, −325.45) and

1

2 e^(2,0,−1)+ e^{(6,−4,−3)}

= 1

2e²(cosh(1), 0, − sinh(1)) + e⁶(cosh(5), sinh(5) · (−4, −3)/5)

= (14975, −11974, −8985).

We see that 14975 − 463.48 = 14511.52, but

(−11974, −8985) − (−325.4493, −325.4493)

= 14515 > 14511.52 which is a contradiction. 

Example 2.12. (a) For any fixed σ ∈ IR, the function f (t) = (t − σ)^−r with r ≥ 0 is SOC-convex on (σ, ∞) if and only if 0 ≤ r ≤ 1.

(a) For any fixed σ ∈ IR, the function f (t) = (t − σ)^r with r ≥ 0 is SOC-convex on [σ, ∞) if and only if 1 ≤ r ≤ 2, and f is SOC-concave on [σ, ∞) if and only if 0 ≤ r ≤ 1.

(c) For any fixed σ ∈ IR, the function f (t) = ln(t − σ) is SOC-concave on (σ, ∞).

(d) For any fixed σ ≥ 0, the function f (t) = _t+σ^t is SOC-concave on (−σ, ∞).

Solution. (a) For any fixed σ ∈ IR, by a simple computation, we have that







f⁽²⁾(t) 2

f⁽³⁾(t) f⁽³⁾(t) 6

6

f⁽⁴⁾(t) 24





=







r(r + 1)(t − σ)^−r−2 2

r(r + 1)(−r − 2)(t − σ)^−r−3 r(r + 1)(−r − 2)(t − σ)^−r−3 6

6

r(r + 1)(r + 2)(r + 3)(t − σ)^−r−4 24





. The sufficient and necessary condition for the above matrix being positive semidefinite is

r²(r + 1)²(r + 2)(r + 3)(t − σ)^−2r−6

24 − r²(r + 1)²(r + 2)²(t − σ)^−2r−6

18 ≥ 0, (2.42)

which is equivalent to requiring 0 ≤ r ≤ 1. By Proposition 2.18, it then follows that f is SOC-convex on (σ, +∞) if and only if 0 ≤ r ≤ 1.

(b) For any fixed σ ∈ IR, by a simple computation, we have that







f⁽²⁾(t) 2

f⁽³⁾(t) f⁽³⁾(t) 6

6

f⁽⁴⁾(t) 24





=







r(r − 1)(t − σ)^r−2 2

r(r − 1)(r − 2)(t − σ)^r−3 r(r − 1)(r − 2)(t − σ)^r−3 6

6

r(r − 1)(r − 2)(r − 3)(t − σ)^r−4 24





. The sufficient and necessary condition for the above matrix being positive semidefinite is

r ≥ 1 and r²(r − 1)²(r − 2)(r − 3)(t − σ)^2r−6

24 − r²(r − 1)²(r − 2)²(t − σ)^2r−6

18 ≥ 0,

(2.43) whereas the sufficient and necessary condition for it being negative semidefinite is 0 ≤ r ≤ 1 and r²(r − 1)²(r − 2)(r − 3)(t − σ)t^2r−6

24 −r²(r − 1)²(r − 2)²(t − σ)^2r−6

18 ≥ 0.

(2.44) It is easily shown that (2.43) holds if and only if 1 ≤ r ≤ 2, and (2.44) holds if and only if 0 ≤ r ≤ 1. By Proposition 2.18, this shows that f is SOC-convex on (σ, ∞) if and only if 1 ≤ r ≤ 2, and f is SOC-concave on (σ, ∞) if and only if 0 ≤ r ≤ 1. This together with the definition of SOC-convexity yields the desired result.

(c) Notice that for any t > σ, there always holds that

−







f⁽²⁾(t) 2

f⁽³⁾(t) 6 f⁽³⁾(t)

6

f⁽⁴⁾(t) 24





=







1

2(t − σ)² − 1 3(t − σ)³

− 1

3(t − σ)³

1 4(t − σ)⁴





 O.

Consequently, from Proposition 2.18(a), we conclude that f is SOC-concave on (σ, ∞).

(d) For any t > −σ, it is easy to compute that

−







f⁽²⁾(t) 2

f⁽³⁾(t) 6 f⁽³⁾(t)

6

f⁽⁴⁾(t) 24





=





 1

(t + σ)³ − 1 (t + σ)⁴

− 1

(t + σ)⁴

1 (t + σ)⁵





 O.

By Proposition 2.18 again, we then have that the function f is SOC-concave on (−σ, ∞).



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