3 Characterizations of SOC-monotone functions

to appear in Linear Algebra and its Applications, 2012

SOC-monotone and SOC-convex functions v.s. matrix-monotone and matrix-convex functions

Shaohua Pan^†, Yungyen Chiang^‡ and Jein-Shan Chen^§

September 30, 2010

(First revised on January 13, 2012) (Second revised on April 5, 2012)

Abstract. The SOC-monotone function (respectively, SOC-convex function) is a scalar valued function that induces a map to preserve the monotone order (respectively, the convex order), when imposed on the spectral factorization of vectors associated with second- order cones (SOCs) in general Hilbert spaces. In this paper, we provide the sufficient and necessary characterizations for the two classes of functions, and particularly establish that the set of continuous SOC-monotone (respectively, SOC-convex) functions coincides with that of continuous matrix monotone (respectively, matrix convex) functions of order 2.

Keywords: Hilbert space; second-order cone; SOC-monotonicity; SOC-convexity.

AMS subject classifications. 26B05, 26B35, 90C33, 65K05

1 Introduction

Let H be a real Hilbert space of dimension dim(H) ≥ 3 endowed with an inner product h·, ·i and its induced norm k · k. Fix a unit vector e ∈ H and denote by hei^⊥ the orthogonal complementary space of e, i.e., hei^⊥= {x ∈ H | hx, ei = 0} . Then each x can be written as

x = x_e+ x₀e for some x_e ∈ hei^⊥ and x₀ ∈ R.

1This work was supported by National Young Natural Science Foundation (No. 10901058) and the Fundamental Research Funds for the Central Universities.

†Department of Mathematics, South China University of Technology, Wushan Road 381, Tianhe District of Guangzhou 510641, China. Email: shhpan@scut.edu.cn

‡Department of Applied Mathematics, National Sun Yat-sen University, Kaohsiung 80424, Taiwan.

Email: chiangyy@math.nsysu.edu.tw

§Corresponding author. Member of Mathematics Division, National Center for Theoretical Sciences, Taipei Office. The author’s work is supported by National Science Council of Taiwan, Department of Mathematics, National Taiwan Normal University, Taipei, Taiwan 11677. Email: jschen@math.ntnu.edu.tw

The second-order cone (SOC) in H, also called the Lorentz cone, is a set defined by K :=



x ∈ H | hx, ei ≥ 1

√2kxk



=n

x_e+ x₀e ∈ H | x0 ≥ kx_eko .

From [7, Section 2], we know that K is a pointed closed convex self-dual cone. Hence, H becomes a partially ordered space via the relation _K. In the sequel, for any x, y ∈ H, we always write x _K y (respectively, x _K y) when x − y ∈ K (respectively, x − y ∈ intK);

and denote x_e by the vector _kx^xe

ek if x_e 6= 0, and otherwise by any unit vector from hei^⊥. Associated with the second-order cone K, each x = x_e+ x₀e ∈ H can be decomposed as

x = λ₁(x)u₁(x) + λ₂(x)u₂(x), (1) where λi(x) ∈ R and uⁱ(x) ∈ H for i = 1, 2 are the spectral values and the associated spectral vectors of x, defined by

λ_i(x) = x₀+ (−1)ⁱkx_ek, u_i(x) = 1

2(e + (−1)ⁱx_e). (2) Clearly, when x_e 6= 0, the spectral factorization of x is unique by definition.

Let f : J ⊆ R → R be a scalar valued function, where J is an interval (finite or infinite, closed or open) in R. Let S be the set of all x ∈ H whose spectral values λ¹(x) and λ2(x) belong to J . Unless otherwise stated, in this paper S is always taken in this way. By the spectral factorization of x in (1)-(2), it is natural to define f^soc: S ⊆ H → H by

f^soc(x) := f (λ₁(x))u₁(x) + f (λ₂(x))u₂(x), ∀x ∈ S. (3) It is easy to see that the function f^soc is well defined whether x_e= 0 or not. For example, by taking f (t) = t², we have that f^soc(x) = x² = x◦x, where “◦” means the Jordan product and the detailed definition is see in the next section. Note that

(λ1(x) − λ1(y))²+ (λ2(x) − λ2(y))² = 2(kxk²+ kyk²− 2x0y0− 2kxekkyek)

≤ 2 kxk²+ kyk²− 2hx, yi
= 2kx − yk².

We may verify that the domain S of f^soc is open in H if and only if J is open in R. Also, S is always convex since, for any x = x_e+ x₀e, y = y_e+ y₀e ∈ S and β ∈ [0, 1],

λ₁[βx + (1 − β)y] = (βx₀+ (1 − β)y₀) − kβx_e+ (1 − β)y_ek ≥ min{λ₁(x), λ₁(y)}, λ₂[βx + (1 − β)y] = (βx₀+ (1 − β)y₀) + kβx_e+ (1 − β)y_ek ≤ max{λ₂(x), λ₂(y)}, which implies that βx + (1 − β)y ∈ S. Thus, f^soc(βx + (1 − β)y) is well defined.

In this paper we are interested in two classes of special scalar valued functions that induce the maps via (3) to preserve the monotone order and the convex order, respectively.

Definition 1.1 A function f : J → R is said to be SOC-monotone if for any x, y ∈ S, x _K y =⇒ f^soc(x) _K f^soc(y); (4) and f is said to be SOC-convex if, for any x, y ∈ S and any β ∈ [0, 1],

f^soc(βx + (1 − β)y) _K βf^soc(x) + (1 − β)f^soc(y). (5) From Definition 1.1 and equation (3), it is easy to see that the set of SOC-monotone and SOC-convex functions are closed under positive linear combinations and pointwise limits.

The concept of SOC-monotone (respectively, SOC-convex) functions above is a direct extension of those given by [5, 6] to general Hilbert spaces, and is analogous to that of matrix monotone (respectively, matrix convex) functions and more general operator monotone (respectively, operator convex) functions; see, e.g., [17, 15, 14, 2, 11, 23]. Just as the importance of matrix monotone (respectively, matrix convex) functions to the solution of convex semidefinite programming [19, 4], SOC-monotone (respectively, SOC-convex) functions also play a crucial role in the design and analysis of algorithms for convex second- order cone programming [3, 22]. For matrix monotone and matrix convex functions, after the seminal work of L¨owner [17] and Kraus [15], there have been systematic studies and perfect characterizations for them; see [8, 16, 4, 13, 12, 21, 20] and the references therein.

However, the study on SOC-monotone and SOC-convex functions just begins with [5], and the characterizations for them are still imperfect. Particularly, it is not clear what is the relation between the SOC-monotone (respectively, SOC-convex) functions and the matrix monotone (respectively, matrix convex) functions.

In this work, we provide the sufficient and necessary characterizations for SOC-monotone and SOC-convex functions in the setting of Hilbert spaces, and show that the set of con- tinuous SOC-monotone (SOC-convex) functions coincides with that of continuous matrix monotone (matrix convex) functions of order 2. Some of these results generalize those of [5, 6] (see Propositions 3.2 and 4.2), and some are new, which are difficult to achieve by using the techniques of [5, 6] (see, for example, Proposition 4.4). In addition, we also discuss the relations between SOC-monotone functions and SOC-convex functions, verify Conjecture 4.2 in [5] under a little stronger condition (see Proposition 6.2), and present a counterexample to show that Conjecture 4.1 in [5] generally does not hold. It is worthwhile to point out that the analysis in this paper depends only on the inner product of Hilbert spaces, whereas most of the results in [5, 6] are obtained with the help of matrix operations.

Throughout this paper, all differentiability means Fr´echet differentiability. If F : H → H is (twice) differentiable at x ∈ H, we denote by F⁰(x) (F⁰⁰(x)) the first-order F-derivative (the second-order F-derivative) of F at x. In addition, we use Cⁿ(J ) and C^∞(J ) to de- note the set of n times and infinite times continuously differentiable real functions on J , respectively. When f ∈ C¹(J ), we denote by f^[1] the function on J × J defined by

f^[1](λ, µ) :=

( f (λ)−f (µ)

λ−µ if λ 6= µ, f⁰(λ) if λ = µ;

and when f ∈ C²(J ), denote by f^[2] the function on J × J × J defined by f^[2](τ₁, τ₂, τ₃) := f^[1](τ₁, τ₂) − f^[1](τ₁, τ₃)

τ₂− τ₃

if τ₁, τ₂, τ₃ are distinct, and for other values of τ₁, τ₂, τ₃, f^[2] is defined by continuity; e.g., f^[2](τ1, τ1, τ3) = f (τ₃) − f (τ₁) − f⁰(τ₁)(τ₃− τ₁)

(τ₃− τ₁)² , f^[2](τ1, τ1, τ1) = 1

2f⁰⁰(τ1).

For a linear operator L from H into H, we write L ≥ 0 (respectively, L > 0) to mean that L is positive semidefinite (respectively, positive definite), i.e., hh, Lhi ≥ 0 for any h ∈ H (respectively, hh, Lhi > 0 for any 0 6= h ∈ H).

2 Preliminaries

This section recalls some background material and gives several lemmas that will be used in the subsequent sections. We start with the definition of Jordan product [9]. For any x = x_e+ x₀e, y = y_e+ y₀e ∈ H, the Jordan product of x and y is defined as

x ◦ y := (x₀y_e+ y₀x_e) + hx, yie.

A simple computation can verify that for any x, y, z ∈ H and the unit vector e, (i) e ◦ e = e and e ◦ x = x; (ii) x ◦ y = y ◦ x; (iii) x ◦ (x² ◦ y) = x² ◦ (x ◦ y), where x² = x ◦ x; (iv) (x + y) ◦ z = x ◦ z + y ◦ z. For any x ∈ H, define its determinant by

det(x) := λ₁(x)λ₂(x) = x²₀− kx_ek².

Then each x = x_e+ x₀e with det(x) 6= 0 is invertible with respect to the Jordan product, i.e., there is a unique x⁻¹ = (−xe+ x0e)/det(x) such that x ◦ x⁻¹ = e.

We next give several lemmas where Lemma 2.1 is used in Section 3 to characterize SOC- monotonicity, and Lemmas 2.2 and 2.3 are used in Section 4 to characterize SOC-convexity.

Lemma 2.1 Let B := {z ∈ hei^⊥ | kzk ≤ 1}. Then, for any given u ∈ hei^⊥ with kuk = 1 and θ, λ ∈ R, the following results hold.

(a) θ + λhu, zi ≥ 0 for any z ∈ B if and only if θ ≥ |λ|.

(b) θ − kλzk² ≥ (θ − λ²)hu, zi² for any z ∈ B if and only if θ − λ² ≥ 0.

Proof. (a) Suppose that θ + λhu, zi ≥ 0 for any z ∈ B. If λ = 0, then θ ≥ |λ| clearly holds. If λ 6= 0, take z = −sign(λ)u. Since kuk = 1, we have z ∈ B, and consequently, θ+λhu, zi ≥ 0 reduces to θ−|λ| ≥ 0. Conversely, if θ ≥ |λ|, then using the Cauchy-Schwartz inequality yields θ + λhu, zi ≥ 0 for any z ∈ B.

(b) Suppose that θ − kλzk² ≥ (θ − λ²)hu, zi² for any z ∈ B. Then we must have θ − λ² ≥ 0.

If not, for those z ∈ B with kzk = 1 but hu, zi 6= kukkzk, it holds that (θ − λ²)hu, zi² > (θ − λ²)kuk²kzk² = θ − kλzk²,

which contradicts the given assumption. Conversely, if θ − λ² ≥ 0, the Cauchy-Schwartz inequality implies that (θ − λ²)hu, zi² ≤ θ − kλzk² for any z ∈ B. 2

Lemma 2.2 For any given a, b, c ∈ R and x = xe+ x₀e with x_e6= 0, the inequality

akh_ek²− hh_e, x_ei² + b[h₀+ hx_e, h_ei]²+ c[h₀− hx_e, h_ei]² ≥ 0 (6) holds for all h = he+ h0e ∈ H if and only if a ≥ 0, b ≥ 0 and c ≥ 0.

Proof. Suppose that (6) holds for all h = h_e+ h₀e ∈ H. By letting he = x_e, h₀ = 1 and he = −xe, h0 = 1, respectively, we get b ≥ 0 and c ≥ 0 from (6). If a ≥ 0 does not hold, then by taking h_e = q

b+c+1

|a|

ze

kzek with hz_e, x_ei = 0 and h₀ = 1, (6) gives a contradiction

−1 ≥ 0. Conversely, if a ≥ 0, b ≥ 0 and c ≥ 0, then (6) clearly holds for all h ∈ H. 2 Lemma 2.3 Let f ∈ C²(J ) and u_e ∈ hei^⊥ with ku_ek = 1. For any h = h_e+ h₀e ∈ H, define

µ₁(h) := h₀− hu_e, h_ei

√2 , µ₂(h) := h₀+ hu_e, h_ei

√2 , µ(h) :=pkh_ek²− hu_e, h_ei². Then, for any given a, d ∈ R and λ¹, λ2 ∈ J, the following inequality

4f⁰⁰(λ₁)f⁰⁰(λ₂)µ₁(h)²µ₂(h)²+ 2(a − d)f⁰⁰(λ₂)µ₂(h)²µ(h)² +2 (a + d) f⁰⁰(λ₁)µ₁(h)²µ(h)²+ a²− d²
µ(h)⁴

−2 [(a − d) µ₁(h) + (a + d) µ₂(h)]²µ(h)² ≥ 0 (7) holds for all h = h_e+ h₀e ∈ H if and only if

a²− d² ≥ 0, f⁰⁰(λ₂)(a − d) ≥ (a + d)² and f⁰⁰(λ₁)(a + d) ≥ (a − d)². (8) Proof. Suppose that (7) holds for all h = h_e + h₀e ∈ H. Taking h0 = 0 and h_e 6= 0 with hh_e, u_ei = 0, we have µ₁(h) = 0, µ₂(h) = 0 and µ(h) = kh_ek > 0, and then (7) gives a² − d² ≥ 0. Taking h_e 6= 0 such that |hu_e, h_ei| < kh_ek and h₀ = hu_e, h_ei 6= 0, we have µ1(h) = 0, µ2(h) = √

2h0 and µ(h) > 0, and then (7) reduces to the following inequality 4(a − d)f⁰⁰(λ₂) − (a + d)² h²₀+ (a²− d²)(kh_ek²− h²₀) ≥ 0.

This implies that (a − d)f⁰⁰(λ₂) − (a + d)² ≥ 0. If not, by letting h₀ be sufficiently close to khek, the last inequality yields a contradiction. Similarly, taking h with he 6= 0 satisfying

|hu_e, h_ei| < kh_ek and h₀ = −hu_e, h_ei, we get f⁰⁰(λ₁)(a + d) ≥ (a − d)² from (7).

Next, suppose that (8) holds. Then, the inequalities f⁰⁰(λ₂)(a − d) ≥ (a + d)² and f⁰⁰(λ₁)(a + d) ≥ (a − d)² imply that the left-hand side of (7) is greater than

4f⁰⁰(λ₁)f⁰⁰(λ₂)µ₁(h)²µ₂(h)²− 4(a²− d²)µ₁(h)µ₂(h)µ(h)²+ a² − d²
µ(h)⁴,

which is obviously nonnegative if µ₁(h)µ₂(h) ≤ 0. Now assume that µ₁(h)µ₂(h) > 0. If a² − d² = 0, then the last expression is clearly nonnegative, and if a² − d² > 0, then the last two inequalities in (8) imply that f⁰⁰(λ₁)f⁰⁰(λ₂) ≥ (a²− d²) > 0, and therefore,

4f⁰⁰(λ₁)f⁰⁰(λ₂)µ₁(h)²µ₂(h)²− 4(a²− d²)µ₁(h)µ₂(h)µ(h)²+ a²− d²
µ(h)⁴

≥ 4(a²− d²)µ₁(h)²µ₂(h)²− 4(a²− d²)µ₁(h)µ₂(h)µ(h)²+ a² − d²
µ(h)⁴

= (a² − d²)2µ1(h)µ₂(h) − µ(h)²2

≥ 0.

Thus, we prove that inequality (7) holds. The proof is complete. 2

To close this section, we introduce the regularization of a locally integrable real function.

Let ϕ be a real function of class C^∞ with the following properties: ϕ ≥ 0, ϕ is even, the support supp ϕ = [−1, 1], and R

Rϕ = 1. For each ε > 0, let ϕ_ε(t) = ¹_εϕ(_ε^t). Then supp ϕ_ε= [−ε, ε] and ϕ_ε has all the properties of ϕ listed above. If f is a locally integrable real function, we define its regularization of order ε as the function

f_ε(s) :=

Z

f (s − t)ϕ_ε(t)dt = Z

f (s − εt)ϕ(t)dt. (9)

Note that f_ε is a C^∞ function for each ε > 0, and lim_ε→0f_ε(x) = f (x) if f is continuous.

3 Characterizations of SOC-monotone functions

In this section we present some characterizations for SOC-monotone functions, by which the set of continuous SOC-monotone functions is shown to coincide with that of continuous matrix monotone functions of order 2. To this end, we need the following technical lemma.

Lemma 3.1 For any given f : J → R with J open, let f^soc : S → H be defined by (3).

(a) f^soc is continuous on S if and only if f is continuous on J .

(b) f^soc is (continuously) differentiable on S iff f is (continuously) differentiable on J . Also, when f is differentiable on J , for any x = x_e+ x₀e ∈ S and v = v_e+ v₀e ∈ H,

(f^soc)⁰(x)v =







f⁰(x₀)v if x_e = 0;

(b₁(x) − a₀(x))hx_e, v_eix_e+ c₁(x)v₀x_e

+a₀(x)v_e+ b₁(x)v₀e + c₁(x)hx_e, v_eie if x_e 6= 0,

(10)

where a₀(x) = ^{f (λ}_λ^{2(x))−f (λ1(x))}

2(x)−λ1(x) , b₁(x) = ^f0(λ2(x))+f₂ ^0(λ1(x)), c₁(x) = ^{f0(λ2(x))−f}₂ ^0(λ1(x)).

(c) If f is differentiable on J , then for any given x ∈ S and all v ∈ H,

(f^soc)⁰(x)e = (f⁰)^soc(x) and he, (f^soc)⁰(x)vi = hv, (f⁰)^soc(x)i . (d) If f⁰ is nonnegative (respectively, positive) on J , then for each x ∈ S,

(f^soc)⁰(x) ≥ 0 (respectively, (f^soc)⁰(x) > 0).

Proof. (a) Suppose that f^soc is continuous. Let Ω be the set composed of those x = te with t ∈ J . Clearly, Ω ⊆ S, and f^soc is continuous on Ω. Noting that f^soc(x) = f (t)e for any x ∈ Ω, it follows that f is continuous on J . Conversely, if f is continuous on J , then f^soc is continuous at any x = x_e+ x₀e ∈ S with x_e 6= 0 since λ_i(x) and u_i(x) for i = 1, 2 are continuous at such points. Next, let x = x_e + x₀e be an arbitrary element from S with x_e = 0, and we prove that f^soc is continuous at x. Indeed, for any z = z_e+ z₀e ∈ S sufficiently close to x, it is not hard to verify that

kf^soc(z) − f^soc(x)k ≤ |f (λ₂(z)) − f (x₀)|

2 + |f (λ₁(z)) − f (x₀)|

2 +|f (λ₂(z)) − f (λ₁(z))|

2 .

Since f is continuous on J , and λ₁(z), λ₂(z) → x₀ as z → x, it follows that f (λ₁(z)) → f (x₀) and f (λ₂(z)) → f (x₀) as z → x.

The last two equations imply that f^soc is continuous at x.

(b) When f^soc is (continuously) differentiable, using the similar arguments as in part (a) can show that f is (continuously) differentiable. Next assume that f is differentiable. Fix any x = x_e+ x₀e ∈ S. We first consider the case where x_e 6= 0. Since λ_i(x) for i = 1, 2 and _kx^xe

ek are continuously differentiable at such x, it follows that f (λi(x)) and ui(x) are differentiable and continuously differentiable, respectively, at x. Then f^soc is differentiable at such x by the definition of f^soc. Also, an elementary computation shows that

[λi(x)]⁰v = hv, ei + (−1)ⁱhx_e, v − hv, eiei

kx_ek = v0+ (−1)ⁱhx_e, v_ei

kx_ek , (11)

 x_e kxek

⁰

v = v − hv, eie

kxek −hx_e, v − hv, eieix_e

kxek³ = v_e

kxek − hx_e, v_eix_e

kxek³ (12) for any v = v_e+ v₀e ∈ H, and consequently,

[f (λ_i(x))]⁰v = f⁰(λ_i(x))



v₀+ (−1)ⁱhx_e, v_ei kx_ek

 , [u_i(x)]⁰v = 1

2(−1)ⁱ

 v_e

kx_ek − hx_e, v_eix_e kx_ek³

 .

Together with the definition of f^soc, we calculate that (f^soc)⁰(x)v is equal to f⁰(λ₁(x))

2



v₀−hx_e, v_ei kxek

 

e − x_e kxek



− f (λ₁(x)) 2

 v_e

kxek − hx_e, v_eix_e kxek³



+f⁰(λ₂(x)) 2



v₀+ hx_e, v_ei kxek

 

e + x_e kxek



+f (λ₂(x)) 2

 v_e

kxek − hx_e, v_eix_e kxek³



= b₁(x)v₀e + c₁(x) hx_e, v_ei e + c₁(x)v₀x_e+ b₁(x)hx_e, v_eix_e +a₀(x)v_e− a₀(x)hx_e, v_eix_e,

where λ₂(x) − λ₁(x) = 2kx_ek is used for the last equality. Thus, we get (10) for x_e 6= 0.

We next consider the case where x_e = 0. Under this case, for any v = v_e+ v₀e ∈ H, f^soc(x + v) − f^soc(x) = f (x₀+ v₀− kv_ek)

2 (e − v_e) + f (x₀+ v₀ + kv_ek)

2 (e + v_e) − f (x₀)e

= f⁰(x₀)(v₀− kv_ek)

2 e + f⁰(x₀)(v₀+ kv_ek)

2 e

+f⁰(x0)(v0+ kvek)

2 ve− f⁰(x0)(v0− kvek)

2 ve+ o(kvk)

= f⁰(x₀)(v₀e + kv_ekv_e) + o(kvk), where v_e= _kv^ve

ek if v_e 6= 0, and otherwise v_e is an arbitrary unit vector from hei^⊥. Hence, kf^soc(x + v) − f^soc(x) − f⁰(x₀)vk = o(kvk).

This shows that f^soc is differentiable at such x with (f^soc)⁰(x)v = f⁰(x₀)v.

Assume that f is continuously differentiable. From (10), it is easy to see that (f^soc)⁰(x) is continuous at every x with x_e6= 0. We next argue that (f^soc)⁰(x) is continuous at every x with x_e= 0. Fix any x = x₀e with x₀ ∈ J. For any z = z_e+ z₀e with z_e6= 0, we have

k(f^soc)⁰(z)v − (f^soc)⁰(x)vk ≤ |b₁(z) − a₀(z)|kv_ek + |b₁(z) − f⁰(x₀)||v₀|

+|a₀(z) − f⁰(x₀)|kv_ek + |c₁(z)|(|v₀| + kv_ek). (13) Since f is continuously differentiable on J and λ₂(z) → x₀, λ₁(z) → x₀ as z → x, we have

a0(z) → f⁰(x0), b1(z) → f⁰(x0) and c1(z) → 0.

Together with equation (13), we obtain that (f^soc)⁰(z) → (f^soc)⁰(x) as z → x.

(c) The result is direct by the definition of (f⁰)^soc and a simple computation from (10).

(d) Suppose that f⁰(t) ≥ 0 for all t ∈ J . Fix any x = xe+ x0e ∈ S. If xe = 0, the result is direct. It remains to consider the case x_e 6= 0. Since f⁰(t) ≥ 0 for all t ∈ J , we have b₁(x) ≥ 0, b₁(x)−c₁(x) = f⁰(λ₁(x)) ≥ 0, b₁(x)+c₁(x) = f⁰(λ₂(x)) ≥ 0 and a₀(x) ≥ 0. From

part (b) and the definitions of b₁(x) and c₁(x), it follows that for any h = h_e+ h₀e ∈ H, hh, (f^soc)⁰(x)hi = (b₁(x) − a₀(x))hx_e, h_ei²+ 2c₁(x)h₀hx_e, h_ei + b₁(x)h²₀+ a₀(x)kh_ek²

= a₀(x)kh_ek²− hx_e, h_ei² + 1

2(b₁(x) − c₁(x)) [h₀− hx_e, h_ei]² +1

2(b₁(x) + c₁(x)) [h₀+ hx_e, h_ei]² ≥ 0.

This implies that the operator (f^soc)⁰(x) is positive semidefinite. Particularly, if f⁰(t) > 0 for all t ∈ J , we have that hh, (f^soc)⁰(x)hi > 0 for all h 6= 0. The proof is complete. 2

Lemma 3.1(d) shows that the differential operator (f^soc)⁰(x) corresponding to a differ- entiable nondecreasing f is positive semidefinite. So, the differential operator (f^soc)⁰(x) associated with a differentiable SOC-monotone function is also positive semidefinite.

Proposition 3.1 Assume that f ∈ C¹(J ) with J open. Then f is SOC-monotone if and only if (f^soc)⁰(x)h ∈ K for any x ∈ S and h ∈ K.

Proof. If f is SOC-monotone, then for any x ∈ S, h ∈ K and t > 0, we have f^soc(x + th) − f^soc(x) _K 0,

which, by the continuous differentiability of f^soc and the closedness of K, implies that (f^soc)⁰(x)h _K 0.

Conversely, for any x, y ∈ S with x _K y, from the given assumption we have that f^soc(x) − f^soc(y) =

Z 1 0

(f^soc)⁰(x + t(x − y))(x − y)dt ∈ K.

This shows that f^soc(x) _K f^soc(y), i.e., f is SOC-monotone. The proof is complete. 2 Proposition 3.1 shows that the differential operator (f^soc)⁰(x) associated with a differ- entiable SOC-monotone function f leaves K invariant. If, in addition, the linear operator (f^soc)⁰(x) is bijective, then (f^soc)⁰(x) belongs to the automorphism group of K. Such linear operators are important to study the structure of the cone K (see [9]).

Corollary 3.1 Assume that f ∈ C¹(J ) with J open. If f is SOC-monotone, then (a) (f⁰)^soc(x) ∈ K for any x ∈ S;

(b) f^soc is a monotone function, i.e., hf^soc(x) − f^soc(y), x − yi ≥ 0 for any x, y ∈ S.

Proof. Part (a) is direct by using Proposition 3.1 with h = e and Lemma 3.1(c). By part (a), f⁰(τ ) ≥ 0 for all τ ∈ J . Together with Lemma 3.1(d), (f^soc)⁰(x) ≥ 0 for any x ∈ S.

Applying the integral mean-value theorem, it then follows that hf^soc(x) − f^soc(y), x − yi =

Z 1 0

hx − y, (f^soc)⁰(y + t(x − y))(x − y)idt ≥ 0.

This proves the desired result of part (b). The proof is complete. 2

Note that the converse of Corollary 3.1(a) is not correct. For example, for the function f (t) = −t⁻² (t > 0), it is clear that (f⁰)^soc(x) ∈ K for any x ∈ intK, but it is not SOC- monotone by Example 5.1(ii). The following proposition provides another sufficient and necessary characterization for differentiable SOC-monotone functions.

Proposition 3.2 Let f ∈ C¹(J ) with J open. Then f is SOC-monotone if and only if

 f^[1](τ₁, τ₁) f^[1](τ₁, τ₂) f^[1](τ₂, τ₁) f^[1](τ₂, τ₂)



=







f⁰(τ1) f (τ₂) − f (τ₁) τ₂− τ₁ f (τ₁) − f (τ₂)

τ₁− τ₂ f⁰(τ₂)





≥ 0, ∀τ₁, τ₂ ∈ J. (14)

Proof. The equality is direct by the definition of f^[1]. It suffices to prove that f is SOC- monotone if and only if the inequality in (14) holds for any τ₁, τ₂ ∈ J. Assume that f is SOC-monotone. By Proposition 3.1, (f^soc)⁰(x)h ∈ K for any x ∈ S and h ∈ K. Fix any x = x_e+ x₀e ∈ S. It suffices to consider the case where x_e 6= 0. Since (f^soc)⁰(x)h ∈ K for any h ∈ K, we particularly have (f^soc)⁰(x)(z + e) ∈ K for any z ∈ B, where B is the set defined in Lemma 2.1. From Lemma 3.1(b), it follows that

(f^soc)⁰(x)(z + e) = [(b₁(x) − a₀(x)) hx_e, zi + c₁(x)] x_e+ a₀(x)z + [b₁(x) + c₁(x)hx_e, zi] e.

This means that (f^soc)⁰(x)(z + e) ∈ K for any z ∈ B if and only if

b₁(x) + c₁(x)hx_e, zi ≥ 0, (15)

[b₁(x) + c₁(x)hx_e, zi]² ≥ k[ (b₁(x) − a₀(x)) hx_e, zi + c₁(x)]x_e+ a₀(x)zk². (16) By Lemma 2.1(a), we know that (15) holds for any z ∈ B if and only if b¹(x) ≥ |c1(x)|.

Since by a simple computation the inequality in (16) can be simplified as b₁(x)²− c₁(x)²− a₀(x)²kzk² ≥b₁(x)²− c₁(x)²− a₀(x)² hz, x_ei², applying Lemma 2.1(b) yields that (16) holds for any z ∈ B if and only if

b₁(x)²− c₁(x)²− a₀(x)² ≥ 0.

This shows that (f^soc)⁰(x)(z + e) ∈ K for any z ∈ B if and only if

b₁(x) ≥ |c₁(x)| and b₁(x)²− c₁(x)²− a₀(x)² ≥ 0. (17)

The first condition in (17) is equivalent to b₁(x) ≥ 0, b₁(x)−c₁(x) ≥ 0 and b₁(x)+c₁(x) ≥ 0, which, by the expressions of b₁(x) and c₁(x) and the arbitrariness of x, is equivalent to f⁰(τ ) ≥ 0 for all τ ∈ J ; whereas the second condition in (17) is equivalent to

f⁰(τ₁)f⁰(τ₂) − f (τ₂) − f (τ1) τ₂ − τ₁

2

≥ 0, ∀τ₁, τ₂ ∈ J.

The two sides show that the inequality in (14) holds for all τ₁, τ₂ ∈ J.

Conversely, if the inequality in (14) holds for all τ₁, τ₂ ∈ J, then from the arguments above we have (f^soc)⁰(x)(z + e) ∈ K for any x = x_e+ x₀e ∈ S and z ∈ B. This implies that (f^soc)⁰(x)h ∈ K for any x ∈ S and h ∈ K. By Proposition 3.1, f is SOC-monotone. 2

Propositions 3.1 and 3.2 provide the characterizations for continuously differentiable SOC-monotone functions. When f does not belong to C¹(J ), one may check the SOC- monotonicity of f by combining the following proposition with Propositions 3.1 and 3.2.

Proposition 3.3 Let f : J → R be a continuous function on the open interval J, and f^ε be its regularization defined by (9). Then, f is SOC-monotone if and only if f_ε is SOC- monotone on J_ε for every sufficiently small ε > 0, where J_ε:= (a + ε, b − ε) for J = (a, b).

Proof. Throughout the proof, for every sufficiently small ε > 0, we let S_ε be the set of all x ∈ H whose spectral values λ1(x), λ₂(x) belong to J_ε. Assume that f_ε is SOC-monotone on J_ε for every sufficiently small ε > 0. Let x, y be arbitrary vectors from S with x _K y.

Then, for any sufficiently small ε > 0, we have x + εe, y + εe ∈ Sε and x + εe K y + εe.

Using the SOC-monotonicity of f_ε on J_ε yields that f_ε^soc(x + εe) _K f_ε^soc(y + εe). Taking the limit ε → 0 and using the convergence of f_ε^soc(x) → f^soc(x) and the continuity of f^soc on S implied by Lemma 3.1(a), we readily obtain that f^soc(x) _K f^soc(y). This shows that f is SOC-monotone.

Now assume that f is SOC-monotone. Let ε > 0 be an arbitrary sufficiently small real number. Fix any x, y ∈ S_εwith x _K y. Then, for all t ∈ [−1, 1], we have x−tεe, y−tεe ∈ S and x − tεe _K y − tεe. Therefore, f^soc(x − tεe) _K f^soc(y − tεe), which is equivalent to

f (λ₁− tε) + f (λ₂− tε)

2 − f (µ₁ − tε) + f (µ₂ − tε) 2

≥

f (λ₁− tε) − f (λ₂− tε)

2 xe−f (µ₁− tε) − f (µ₂− tε)

2 y_e

. Together with the definition of f_ε, it then follows that

f_ε(λ₁) + f_ε(λ₂)

2 − f_ε(µ₁) + f_ε(µ₂) 2

=

Z  f (λ₁− tε) + f (λ₂− tε)

2 − f (µ₁− tε) + f (µ₂− tε) 2



ϕ(t)dt

≥ Z

f (λ₁− ε) − f (λ₂− ε)

2 x_e− f (µ₁− ε) − f (µ₂− ε)

2 y_e

ϕ(t)dt

≥

Z  f (λ₁− ε) − f (λ₂− ε)

2 x_e−f (µ₁− ε) − f (µ₂− ε)

2 y_e

 ϕ(t)dt

=

f_ε(λ₁) − f_ε(λ₂)

2 x_e− f_ε(µ₁) − f_ε(µ₂)

2 y_e

.

By the definition of f_ε^soc, this shows that f_ε^soc(x) _K f_ε^soc(y), i.e., f_ε is SOC-monotone. 2 From Proposition 3.2 and [2, Theorem V. 3.4], f ∈ C¹(J ) is SOC-monotone if and only if it is matrix monotone of order 2. When the continuous f is not in the class C¹(J ), the result also holds due to Proposition 3.3 and the fact that f is matrix monotone of order n if and only if f_ε is matrix monotone of order n. Thus, we have the following main result.

Theorem 3.1 The set of continuous SOC-monotone functions on the open interval J co- incides with that of continuous matrix monotone functions of order 2 on J .

Remark 3.1 Combining Theorem 3.1 with L¨owner’s theorem [17] shows that if f : J → R is a continuous SOC-monotone function on the open interval J , then f ∈ C¹(J ).

4 Characterizations of SOC-convex functions

This section is devoted itself to the characterizations of SOC-convex functions, and shows that the continuous f is SOC-convex if and only if it is matrix convex of order 2. First, for the first-order differentiable SOC-convex functions, we have the following characterizations.

Proposition 4.1 Assume that f ∈ C¹(J ) with J open. Then, the following results hold.

(a) f is SOC-convex if and only if for any x, y ∈ S,

f^soc(y) − f^soc(x) − (f^soc)⁰(x)(y − x) _K 0. (18) (b) If f is SOC-convex, then (f⁰)^soc is a monotone function on S.

Proof. By following the arguments as in [1, Proposition B.3(a)], the proof of part (a) can be done easily, and we omit the details. From part (a), it follows that for any x, y ∈ S,

f^soc(x) − f^soc(y) − (f^soc)⁰(y)(x − y) _K 0, f^soc(y) − f^soc(x) − (f^soc)⁰(x)(y − x) K 0.

Adding the last two inequalities, we immediately obtain that [(f^soc)⁰(y) − (f^soc)⁰(x)] (y − x) _K 0.

Using the self-duality of K and Lemma 3.1(c) then yields

0 ≤ he, [(f^soc)⁰(y) − (f^soc)⁰(x)] (y − x)i = hy − x, (f⁰)^soc(y) − (f⁰)^soc(x)i . This shows that (f⁰)^soc is monotone. The proof is complete. 2

To provide sufficient and necessary characterizations for twice differentiable SOC-convex functions, we need the following lemma that offers the second-order differential of f^soc. Lemma 4.1 For any given f : J → R with J open, let f^soc: S → H be defined by (3).

(a) f^soc is twice (continuously) differentiable on S if and only if f is twice (continuously) differentiable on J . Furthermore, when f is twice differentiable on J , for any given x = x_e+ x₀e ∈ S and u = u_e+ u₀e, v = v_e+ v₀e ∈ H, we have that

(f^soc)⁰⁰(x)(u, v) = f⁰⁰(x₀)u₀v₀e + f⁰⁰(x₀)(u₀v_e+ v₀u_e) + f⁰⁰(x₀)hu_e, v_eie if x_e = 0; and otherwise

(f^soc)⁰⁰(x)(u, v) = (b₂(x) − a₁(x))u₀hx_e, v_eix_e+ (c₂(x) − 3d(x))hx_e, u_eihx_e, v_eix_e +d(x)[hue, veixe+ hxe, veiue+ hxe, ueive] + c2(x)u0v0xe

+(b₂(x) − a₁(x))hx_e, u_eiv₀x_e+ a₁(x)(v₀u_e+ u₀v_e) +b₂(x)u₀v₀e + c₂(x)[v₀hx_e, u_ei + u₀hx_e, v_ei]e

+a₁(x)hu_e, v_eie + (b₂(x) − a₁(x))hx_e, u_eihx_e, v_eie, (19) where

c₂(x) = f⁰⁰(λ₂(x)) − f⁰⁰(λ₁(x))

2 , b₂(x) = f⁰⁰(λ₂(x)) + f⁰⁰(λ₁(x))

2 ,

a₁(x) = f⁰(λ₂(x)) − f⁰(λ₁(x))

λ2(x) − λ1(x) , d(x) = b₁(x) − a₀(x) kxek . (b) If f is twice differentiable on J , then for any given x ∈ S and u, v ∈ H,

(f^soc)⁰⁰(x)(u, v) = (f^soc)⁰⁰(x)(v, u), hu, (f^soc)⁰⁰(x)(u, v)i = hv, (f^soc)⁰⁰(x)(u, u)i.

Proof. (a) The first part is direct by the given conditions and Lemma 3.1(b), and we only need to derive the differential formula. Fix any u = u_e+ u₀e, v = v_e+ v₀e ∈ H. We first consider the case where x_e = 0. Without loss of generality, assume that u_e 6= 0. For any sufficiently small t > 0, using Lemma 3.1(b) and x + tu = (x₀+ tu₀) + tu_e, we have that

(f^soc)⁰(x + tu)v = [b₁(x + tu) − a₀(x + tu)] hu_e, v_eiu_e+ c₁(x + tu)v₀u_e +a₀(x + tu)v_e+ b₁(x + tu)v₀e + c₁(x + tu)hu_e, v_eie.

In addition, from Lemma 3.1(b), we also have that (f^soc)⁰(x)v = f⁰(x₀)v₀e + f⁰(x₀)v_e. Using the definition of b₁(x) and a₀(x), and the differentiability of f⁰ on J , it follows that

limt→0

b₁(x + tu)v₀e − f⁰(x₀)v₀e

t = f⁰⁰(x₀)u₀v₀e, limt→0

a₀(x + tu)v_e− f⁰(x₀)v_e

t = f⁰⁰(x₀)u₀v_e, limt→0

b₁(x + tu) − a₀(x + tu)

t = 0,

limt→0

c₁(x + tu)

t = f⁰⁰(x0)kuek.

Using the above four limits, it is not hard to obtain that (f^soc)⁰⁰(x)(u, v) = lim

t→0

(f^soc)⁰(x + tu)v − (f^soc)⁰(x)v t

= f⁰⁰(x₀)u₀v₀e + f⁰⁰(x₀)(u₀v_e+ v₀u_e) + f⁰⁰(x₀)hu_e, v_eie.

We next consider the case where xe 6= 0. From Lemma 3.1(b), it follows that (f^soc)⁰(x)v = (b₁(x) − a₀(x)) hx_e, v_ei x_e+ c₁(x)v₀x_e

+a₀(x)v_e+ b₁(x)v₀e + c₁(x) hx_e, v_ei e, which in turn implies that

(f^soc)⁰⁰(x)(u, v) = [(b₁(x) − a₀(x)) hx_e, v_ei x_e]⁰u + [c₁(x)v₀x_e]⁰u

+ [a₀(x)v_e+ b₁(x)v₀e]⁰u + [c₁(x) hx_e, v_ei e]⁰u. (20) By the expressions of a₀(x), b₁(x) and c₁(x) and equations (11)-(12), we calculate that

(b₁(x))⁰u = f⁰⁰(λ₂(x)) [u₀+ hx_e, u_ei]

2 + f⁰⁰(λ₁(x)) [u₀− hx_e, u_ei]

2

= b₂(x)u₀+ c₂(x)hx_e, u_ei, (c₁(x))⁰u = c₂(x)u₀ + b₂(x)hx_e, u_ei, (a₀(x))⁰u = f⁰(λ₂(x)) − f⁰(λ₁(x))

λ₂(x) − λ₁(x) u₀ +b₁(x) − a₀(x)

kx_ek hx_e, u_ei

= a₁(x)u₀+ d(x)hx_e, u_ei, (hx_e, v_ei)⁰u =

 1

kx_eku_e− hx_e, u_ei kx_ek x_e, v_e

 .

Using these equalities and noting that a₁(x) = c₁(x)/kx_ek, we obtain that h

(b₁(x) − a₀(x))hx_e, v_eix_ei⁰ u =h

(b₂(x) − a₁(x))u₀+ (c₂(x) − d(x))hx_e, u_eii

hx_e, v_eix_e +(b₁(x) − a₀(x))

 1

kx_eku_e−hx_e, u_ei kx_ek x_e, v_e

 x_e

+ (b₁(x) − a₀(x)) hx_e, v_ei

 1

kx_eku_e− hx_e, u_ei kx_ek x_e



=h

(b₂(x) − a₁(x))u₀+ (c₂(x) − d(x))hx_e, u_eii

hx_e, v_eix_e +d(x)hu_e, v_eix_e− 2d(x)hx_e, v_eihx_e, u_eix_e+ d(x)hx_e, v_eiu_e; h

a₀(x)v_e+ b₁(x)v₀ei0

u =h

a₁(x)u₀ + d(x)hx_e, u_eii v_e+h

b₂(x)u₀+ c₂(x)hx_e, u_eii v₀e;

h

c₁(x)v₀x_ei0

u =h

c₂(x)u₀+ b₂(x)hx_e, u_eii

v₀x_e+ c₁(x)v₀u_e− hx_e, u_eix_e kx_ek

=h

c₂(x)u₀+ b₂(x)hx_e, u_eii

v₀x_e+ a₁(x)v₀h

u_e− hx_e, u_eix_ei

; and

h

c₁(x)hx_e, v_eiei⁰ u =h

c₂(x)u₀+ b₂(x)hx_e, u_eii

hx_e, v_eie + c₁(x) u_e− hxe, ueixe

kx_ek , v_e

 e

= c₂(x)u₀hx_e, v_eie + (b₂(x) − a₁(x))hx_e, u_ei hx_e, v_ei e + a₁(x)hu_e, v_eie.

Adding the equalities above and using equation (20) yields the formula in (19).

(b) By the formula in part (a), a simple computation yields the desired result. 2 Proposition 4.2 Assume that f ∈ C²(J ) with J open. Then, the following results hold.

(a) f is SOC-convex if and only if for any x ∈ S and h ∈ H, (f^soc)⁰⁰(x)(h, h) ∈ K.

(b) f is SOC-convex if and only if f is convex and for any τ₁, τ₂ ∈ J, f⁰⁰(τ₂)

2

f (τ₂) − f (τ₁) − f⁰(τ₁)(τ₂ − τ₁)

(τ₂ − τ₁)² ≥ f (τ₁) − f (τ₂) − f⁰(τ₂)(τ₁− τ₂) (τ₂− τ₁)²

2

. (21)

(c) f is SOC-convex if and only if f is convex and for any τ₁, τ₂ ∈ J, 1

4f⁰⁰(τ₁)f⁰⁰(τ₂) ≥ f (τ2) − f (τ1) − f⁰(τ1)(τ2− τ1)

(τ₂− τ₁)² · f (τ1) − f (τ2) − f⁰(τ2)(τ1 − τ2) (τ₂− τ₁)² .(22) (d) f is SOC-convex if and only if for any τ₁, τ₂ ∈ J and s = τ₁, τ₂,

 f^[2](τ₂, s, τ₂) f^[2](τ₂, s, τ₁) f^[2](τ₁, s, τ₂) f^[2](τ₁, s, τ₁)



 0.

Proof. (a) Suppose that f is SOC-convex. Since f^soc is twice continuously differentiable by Lemma 4.1(a), we have for any given x ∈ S, h ∈ H and sufficiently small t > 0,

f^soc(x + th) = f^soc(x) + t(f^soc)⁰(x)h + 1

2t²(f^soc)⁰⁰(x)(h, h) + o(t²).