3 Characterizations of SOC-monotone functions

Full text

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to appear in Linear Algebra and its Applications, 2012

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

1

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 = xe+ x0e for some xe ∈ 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

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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

xe+ x0e ∈ H | x0 ≥ kxeko .

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 xe by the vector kxxe

ek if xe 6= 0, and otherwise by any unit vector from hei. Associated with the second-order cone K, each x = xe+ 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

λi(x) = x0+ (−1)ikxek, ui(x) = 1

2(e + (−1)ixe). (2) Clearly, when xe 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 fsoc: S ⊆ H → H by

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

1(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 fsoc is open in H if and only if J is open in R. Also, S is always convex since, for any x = xe+ x0e, y = ye+ y0e ∈ S and β ∈ [0, 1],

λ1[βx + (1 − β)y] = (βx0+ (1 − β)y0) − kβxe+ (1 − β)yek ≥ min{λ1(x), λ1(y)}, λ2[βx + (1 − β)y] = (βx0+ (1 − β)y0) + kβxe+ (1 − β)yek ≤ max{λ2(x), λ2(y)}, which implies that βx + (1 − β)y ∈ S. Thus, fsoc(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.

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

fsoc(βx + (1 − β)y) K βfsoc(x) + (1 − β)fsoc(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;

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and when f ∈ C2(J ), denote by f[2] the function on J × J × J defined by f[2]1, τ2, τ3) := f[1]1, τ2) − f[1]1, 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]1, τ1, τ3) = f (τ3) − f (τ1) − f01)(τ3− τ1)

3− τ1)2 , f[2]1, τ1, τ1) = 1

2f001).

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 = xe+ x0e, y = ye+ y0e ∈ H, the Jordan product of x and y is defined as

x ◦ y := (x0ye+ y0xe) + 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− kxek2.

Then each x = xe+ 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.

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(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 xe6= 0, the inequality

akhek2− hhe, xei2 + b[h0+ hxe, hei]2+ c[h0− hxe, hei]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 = he+ h0e ∈ H. By letting he = xe, 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 he = q

b+c+1

|a|

ze

kzek with hze, xei = 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 ue ∈ hei with kuek = 1. For any h = he+ h0e ∈ H, define

µ1(h) := h0− hue, hei

√2 , µ2(h) := h0+ hue, hei

√2 , µ(h) :=pkhek2− hue, hei2. Then, for any given a, d ∈ R and λ1, λ2 ∈ J, the following inequality

4f001)f0021(h)2µ2(h)2+ 2(a − d)f0022(h)2µ(h)2 +2 (a + d) f0011(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 = he+ h0e ∈ H if and only if

a2− d2 ≥ 0, f002)(a − d) ≥ (a + d)2 and f001)(a + d) ≥ (a − d)2. (8) Proof. Suppose that (7) holds for all h = he + h0e ∈ H. Taking h0 = 0 and he 6= 0 with hhe, uei = 0, we have µ1(h) = 0, µ2(h) = 0 and µ(h) = khek > 0, and then (7) gives a2 − d2 ≥ 0. Taking he 6= 0 such that |hue, hei| < khek and h0 = hue, hei 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)f002) − (a + d)2 h20+ (a2− d2)(khek2− h20) ≥ 0.

This implies that (a − d)f002) − (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

|hue, hei| < khek and h0 = −hue, hei, we get f001)(a + d) ≥ (a − d)2 from (7).

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Next, suppose that (8) holds. Then, the inequalities f002)(a − d) ≥ (a + d)2 and f001)(a + d) ≥ (a − d)2 imply that the left-hand side of (7) is greater than

4f001)f0021(h)2µ2(h)2− 4(a2− d21(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 f001)f002) ≥ (a2− d2) > 0, and therefore,

4f001)f0021(h)2µ2(h)2− 4(a2− d21(h)µ2(h)µ(h)2+ a2− d2 µ(h)4

≥ 4(a2− d21(h)2µ2(h)2− 4(a2− d21(h)µ2(h)µ(h)2+ a2 − d2 µ(h)4

= (a2 − d2)2µ1(h)µ2(h) − µ(h)22

≥ 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 ϕε(t) = 1εϕ(εt). Then supp ϕε= [−ε, ε] and ϕε 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

fε(s) :=

Z

f (s − t)ϕε(t)dt = Z

f (s − εt)ϕ(t)dt. (9)

Note that fε is a C function for each ε > 0, and limε→0fε(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 fsoc : S → H be defined by (3).

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

(b) fsoc is (continuously) differentiable on S iff f is (continuously) differentiable on J . Also, when f is differentiable on J , for any x = xe+ x0e ∈ S and v = ve+ v0e ∈ H,

(fsoc)0(x)v =

f0(x0)v if xe = 0;

(b1(x) − a0(x))hxe, veixe+ c1(x)v0xe

+a0(x)ve+ b1(x)v0e + c1(x)hxe, veie if xe 6= 0,

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where a0(x) = f (λλ2(x))−f (λ1(x))

2(x)−λ1(x) , b1(x) = f02(x))+f2 01(x)), c1(x) = f02(x))−f2 01(x)).

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

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

(fsoc)0(x) ≥ 0 (respectively, (fsoc)0(x) > 0).

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

kfsoc(z) − fsoc(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 fsoc is continuous at x.

(b) When fsoc 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 = xe+ x0e ∈ S. We first consider the case where xe 6= 0. Since λi(x) for i = 1, 2 and kxxe

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 fsoc is differentiable at such x by the definition of fsoc. Also, an elementary computation shows that

i(x)]0v = hv, ei + (−1)ihxe, v − hv, eiei

kxek = v0+ (−1)ihxe, vei

kxek , (11)

 xe kxek

0

v = v − hv, eie

kxek −hxe, v − hv, eieixe

kxek3 = ve

kxek − hxe, veixe

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

[f (λi(x))]0v = f0i(x))



v0+ (−1)ihxe, vei kxek

 , [ui(x)]0v = 1

2(−1)i

 ve

kxek − hxe, veixe kxek3

 .

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Together with the definition of fsoc, we calculate that (fsoc)0(x)v is equal to f01(x))

2



v0−hxe, vei kxek

 

e − xe kxek



− f (λ1(x)) 2

 ve

kxek − hxe, veixe kxek3



+f02(x)) 2



v0+ hxe, vei kxek

 

e + xe kxek



+f (λ2(x)) 2

 ve

kxek − hxe, veixe kxek3



= b1(x)v0e + c1(x) hxe, vei e + c1(x)v0xe+ b1(x)hxe, veixe +a0(x)ve− a0(x)hxe, veixe,

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

We next consider the case where xe = 0. Under this case, for any v = ve+ v0e ∈ H, fsoc(x + v) − fsoc(x) = f (x0+ v0− kvek)

2 (e − ve) + f (x0+ v0 + kvek)

2 (e + ve) − f (x0)e

= f0(x0)(v0− kvek)

2 e + f0(x0)(v0+ kvek)

2 e

+f0(x0)(v0+ kvek)

2 ve− f0(x0)(v0− kvek)

2 ve+ o(kvk)

= f0(x0)(v0e + kvekve) + o(kvk), where ve= kvve

ek if ve 6= 0, and otherwise ve is an arbitrary unit vector from hei. Hence, kfsoc(x + v) − fsoc(x) − f0(x0)vk = o(kvk).

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

Assume that f is continuously differentiable. From (10), it is easy to see that (fsoc)0(x) is continuous at every x with xe6= 0. We next argue that (fsoc)0(x) is continuous at every x with xe= 0. Fix any x = x0e with x0 ∈ J. For any z = ze+ z0e with ze6= 0, we have

k(fsoc)0(z)v − (fsoc)0(x)vk ≤ |b1(z) − a0(z)|kvek + |b1(z) − f0(x0)||v0|

+|a0(z) − f0(x0)|kvek + |c1(z)|(|v0| + kvek). (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 (fsoc)0(z) → (fsoc)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 xe 6= 0. Since f0(t) ≥ 0 for all t ∈ J , we have b1(x) ≥ 0, b1(x)−c1(x) = f01(x)) ≥ 0, b1(x)+c1(x) = f02(x)) ≥ 0 and a0(x) ≥ 0. From

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part (b) and the definitions of b1(x) and c1(x), it follows that for any h = he+ h0e ∈ H, hh, (fsoc)0(x)hi = (b1(x) − a0(x))hxe, hei2+ 2c1(x)h0hxe, hei + b1(x)h20+ a0(x)khek2

= a0(x)khek2− hxe, hei2 + 1

2(b1(x) − c1(x)) [h0− hxe, hei]2 +1

2(b1(x) + c1(x)) [h0+ hxe, hei]2 ≥ 0.

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

Lemma 3.1(d) shows that the differential operator (fsoc)0(x) corresponding to a differ- entiable nondecreasing f is positive semidefinite. So, the differential operator (fsoc)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 (fsoc)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 fsoc(x + th) − fsoc(x) K 0,

which, by the continuous differentiability of fsoc and the closedness of K, implies that (fsoc)0(x)h K 0.

Conversely, for any x, y ∈ S with x K y, from the given assumption we have that fsoc(x) − fsoc(y) =

Z 1 0

(fsoc)0(x + t(x − y))(x − y)dt ∈ K.

This shows that fsoc(x) K fsoc(y), i.e., f is SOC-monotone. The proof is complete. 2 Proposition 3.1 shows that the differential operator (fsoc)0(x) associated with a differ- entiable SOC-monotone function f leaves K invariant. If, in addition, the linear operator (fsoc)0(x) is bijective, then (fsoc)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) fsoc is a monotone function, i.e., hfsoc(x) − fsoc(y), x − yi ≥ 0 for any x, y ∈ S.

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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), (fsoc)0(x) ≥ 0 for any x ∈ S.

Applying the integral mean-value theorem, it then follows that hfsoc(x) − fsoc(y), x − yi =

Z 1 0

hx − y, (fsoc)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]1, τ1) f[1]1, τ2) f[1]2, τ1) f[1]2, τ2)



=

f01) f (τ2) − f (τ1) τ2− τ1 f (τ1) − f (τ2)

τ1− τ2 f02)

≥ 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, (fsoc)0(x)h ∈ K for any x ∈ S and h ∈ K. Fix any x = xe+ x0e ∈ S. It suffices to consider the case where xe 6= 0. Since (fsoc)0(x)h ∈ K for any h ∈ K, we particularly have (fsoc)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

(fsoc)0(x)(z + e) = [(b1(x) − a0(x)) hxe, zi + c1(x)] xe+ a0(x)z + [b1(x) + c1(x)hxe, zi] e.

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

b1(x) + c1(x)hxe, zi ≥ 0, (15)

[b1(x) + c1(x)hxe, zi]2 ≥ k[ (b1(x) − a0(x)) hxe, zi + c1(x)]xe+ 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, xei2, 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 (fsoc)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)

(11)

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

f01)f02) − 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 (fsoc)0(x)(z + e) ∈ K for any x = xe+ x0e ∈ S and z ∈ B. This implies that (fsoc)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 fε be its regularization defined by (9). Then, f is SOC-monotone if and only if fε is SOC- monotone on Jε for every sufficiently small ε > 0, where Jε:= (a + ε, b − ε) for J = (a, b).

Proof. Throughout the proof, for every sufficiently small ε > 0, we let Sε be the set of all x ∈ H whose spectral values λ1(x), λ2(x) belong to Jε. Assume that fε is SOC-monotone on Jε 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 fε on Jε yields that fεsoc(x + εe) K fεsoc(y + εe). Taking the limit ε → 0 and using the convergence of fεsoc(x) → fsoc(x) and the continuity of fsoc on S implied by Lemma 3.1(a), we readily obtain that fsoc(x) K fsoc(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 ∈ Sεwith x K y. Then, for all t ∈ [−1, 1], we have x−tεe, y−tεe ∈ S and x − tεe K y − tεe. Therefore, fsoc(x − tεe) K fsoc(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 ye

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

fε1) + fε2)

2 − fε1) + fε2) 2

=

Z  f (λ1− tε) + f (λ2− tε)

2 − f (µ1− tε) + f (µ2− tε) 2



ϕ(t)dt

(12)

≥ Z

f (λ1− ε) − f (λ2− ε)

2 xe− f (µ1− ε) − f (µ2− ε)

2 ye

ϕ(t)dt

Z  f (λ1− ε) − f (λ2− ε)

2 xe−f (µ1− ε) − f (µ2− ε)

2 ye

 ϕ(t)dt

=

fε1) − fε2)

2 xe− fε1) − fε2)

2 ye

.

By the definition of fεsoc, this shows that fεsoc(x) K fεsoc(y), i.e., fε 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 fε 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,

fsoc(y) − fsoc(x) − (fsoc)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,

fsoc(x) − fsoc(y) − (fsoc)0(y)(x − y) K 0, fsoc(y) − fsoc(x) − (fsoc)0(x)(y − x) K 0.

Adding the last two inequalities, we immediately obtain that [(fsoc)0(y) − (fsoc)0(x)] (y − x) K 0.

(13)

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

0 ≤ he, [(fsoc)0(y) − (fsoc)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 fsoc. Lemma 4.1 For any given f : J → R with J open, let fsoc: S → H be defined by (3).

(a) fsoc 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 = xe+ x0e ∈ S and u = ue+ u0e, v = ve+ v0e ∈ H, we have that

(fsoc)00(x)(u, v) = f00(x0)u0v0e + f00(x0)(u0ve+ v0ue) + f00(x0)hue, veie if xe = 0; and otherwise

(fsoc)00(x)(u, v) = (b2(x) − a1(x))u0hxe, veixe+ (c2(x) − 3d(x))hxe, ueihxe, veixe +d(x)[hue, veixe+ hxe, veiue+ hxe, ueive] + c2(x)u0v0xe

+(b2(x) − a1(x))hxe, ueiv0xe+ a1(x)(v0ue+ u0ve) +b2(x)u0v0e + c2(x)[v0hxe, uei + u0hxe, vei]e

+a1(x)hue, veie + (b2(x) − a1(x))hxe, ueihxe, veie, (19) where

c2(x) = f002(x)) − f001(x))

2 , b2(x) = f002(x)) + f001(x))

2 ,

a1(x) = f02(x)) − f01(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,

(fsoc)00(x)(u, v) = (fsoc)00(x)(v, u), hu, (fsoc)00(x)(u, v)i = hv, (fsoc)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 = ue+ u0e, v = ve+ v0e ∈ H. We first consider the case where xe = 0. Without loss of generality, assume that ue 6= 0. For any sufficiently small t > 0, using Lemma 3.1(b) and x + tu = (x0+ tu0) + tue, we have that

(fsoc)0(x + tu)v = [b1(x + tu) − a0(x + tu)] hue, veiue+ c1(x + tu)v0ue +a0(x + tu)ve+ b1(x + tu)v0e + c1(x + tu)hue, veie.

(14)

In addition, from Lemma 3.1(b), we also have that (fsoc)0(x)v = f0(x0)v0e + f0(x0)ve. 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)ve− f0(x0)ve

t = f00(x0)u0ve, 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 (fsoc)00(x)(u, v) = lim

t→0

(fsoc)0(x + tu)v − (fsoc)0(x)v t

= f00(x0)u0v0e + f00(x0)(u0ve+ v0ue) + f00(x0)hue, veie.

We next consider the case where xe 6= 0. From Lemma 3.1(b), it follows that (fsoc)0(x)v = (b1(x) − a0(x)) hxe, vei xe+ c1(x)v0xe

+a0(x)ve+ b1(x)v0e + c1(x) hxe, vei e, which in turn implies that

(fsoc)00(x)(u, v) = [(b1(x) − a0(x)) hxe, vei xe]0u + [c1(x)v0xe]0u

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

(b1(x))0u = f002(x)) [u0+ hxe, uei]

2 + f001(x)) [u0− hxe, uei]

2

= b2(x)u0+ c2(x)hxe, uei, (c1(x))0u = c2(x)u0 + b2(x)hxe, uei, (a0(x))0u = f02(x)) − f01(x))

λ2(x) − λ1(x) u0 +b1(x) − a0(x)

kxek hxe, uei

= a1(x)u0+ d(x)hxe, uei, (hxe, vei)0u =

1

kxekue− hxe, uei kxek xe, ve

.

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

(b1(x) − a0(x))hxe, veixei0 u =h

(b2(x) − a1(x))u0+ (c2(x) − d(x))hxe, ueii

hxe, veixe +(b1(x) − a0(x))

1

kxekue−hxe, uei kxek xe, ve

xe

(15)

+ (b1(x) − a0(x)) hxe, vei

 1

kxekue− hxe, uei kxek xe



=h

(b2(x) − a1(x))u0+ (c2(x) − d(x))hxe, ueii

hxe, veixe +d(x)hue, veixe− 2d(x)hxe, veihxe, ueixe+ d(x)hxe, veiue; h

a0(x)ve+ b1(x)v0ei0

u =h

a1(x)u0 + d(x)hxe, ueii ve+h

b2(x)u0+ c2(x)hxe, ueii v0e;

h

c1(x)v0xei0

u =h

c2(x)u0+ b2(x)hxe, ueii

v0xe+ c1(x)v0ue− hxe, ueixe kxek

=h

c2(x)u0+ b2(x)hxe, ueii

v0xe+ a1(x)v0h

ue− hxe, ueixei

; and

h

c1(x)hxe, veiei0 u =h

c2(x)u0+ b2(x)hxe, ueii

hxe, veie + c1(x) ue− hxe, ueixe

kxek , ve

e

= c2(x)u0hxe, veie + (b2(x) − a1(x))hxe, uei hxe, vei e + a1(x)hue, veie.

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, (fsoc)00(x)(h, h) ∈ K.

(b) f is SOC-convex if and only if f is convex and for any τ1, τ2 ∈ J, f002)

2

f (τ2) − f (τ1) − f01)(τ2 − τ1)

2 − τ1)2 ≥ f (τ1) − f (τ2) − f02)(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

4f001)f002) ≥ f (τ2) − f (τ1) − f01)(τ2− τ1)

2− τ1)2 · f (τ1) − f (τ2) − f02)(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]2, s, τ2) f[2]2, s, τ1) f[2]1, s, τ2) f[2]1, s, τ1)



 0.

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

fsoc(x + th) = fsoc(x) + t(fsoc)0(x)h + 1

2t2(fsoc)00(x)(h, h) + o(t2).

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