r-凸函數在二階錐和n維實數空間上的一些結果

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(1)國立臺灣師範大學數學系碩士班碩士論文. 指導教授 : 陳界山博士. Some Results on r-convex Functions Associated With Second-Order Cone and Rn. 研究生 : 黃鴻霖. 中華民國一百零二年六月.

(2) Some Results on r-convex Functions Associated With Second-Order Cone and Rn. By Hong-Lin Huang. Advisor Jein-Shan Chen. Departments of Mathematics, National Taiwan Normal University Taipei, Taiwan. June, 2013.

(3) 誌謝感謝家人、主裡的弟兄姊妹對我的照顧與支持。感謝我的指導教授 : 陳界山老師。在他的帶領下我認識了最佳化的領域，並順利完成我的碩士論文。除此之外，我看到老師做研究的熱情，安排好事情後能徹底的執行，我應該向老師學習。感謝口試委員柯春旭老師、張毓麟老師提供的意見。. 黃鴻霖. 謹誌. 2013 年 6 月.

(4) Contents 1 Introduction. 1. 2 Preliminary. 2. 3 SOC-functions. 8. 4 SOC-r-convex Functions. 13. 5 SOC-Quasi-Convex Functions. 18. 6 References. 22.

(5) Some Results on r-convex Functions Associated With Second-Order Cone and Rn June 4, 2013 [摘要] Martos 和 Avriel 獨立的定義一群實數函數，稱為 r-凸函數，而且 Avriel 更進一步的研究它們。擬凸函數包含它們，而且它們包含凸函數。本篇論文給一些 r-凸函數的例子，以及延伸 r-凸函數和擬凸函數的概念到二階錐上。 Abstract. A family of real functions, called r-convex functions, were independently defined by Martos and Avriel and studied by the latter author. This family properly includes the family of convex functions and is included in the family of quasiconvex functions. This paper gives some examples of r-convex functions, extends the r-convexity and quasi-convexity concepts to the second-order cone. Key words. Second-order cone, convex function, monotone function, spectral decomposition, quasiconvex function.. 1. Introduction. The concept of convexity plays a central role in mathematical economics, engineering, management science, and optimization theory. Accordingly, much attention has been paid to its generalization, to the associated generalization of the results previously developed for the classical convexity, and to the discovery of necessary and/or sufficient conditions for a function to have generalized convexities. Some of the known extensions are quasiconvex, r-convex[2, 24], and SOC-convex[8, 7] functions. Further extensions can be found in [19]. It was shown that a single variable continuous midpoint-convex function on R is convex. This result was generalized in [22] by relaxing continuity to lower-semicontinuity and replacing the number 12 with an arbitrary parameter α ∈ (0, 1). An analogous result was obtained in [18, 23] for quasiconvex functions. In this paper, we give some r-convex examples, extend the r-convexity and quasiconvexity concepts to the second-order cone. Applications of r-convexity to optimization theory can be found in [15, 12, 3]. In general, r-convex function is between convex function and quasi-convex function. But the concept of quasi-convex function can’t be extended to the symmetric cone since any two vectors in the Euclidean Jordan algebra can’t be compared under ≼Kn . Hence we consider firstly the case of extending the concept of r-convex function to the symmetric cone. Throughout this paper, Rn denotes the space of n-dimensional real column vectors, C denotes a convex subset of Rn . ⟨· , ·⟩ denotes the Euclidean inner product, ∥ · ∥ the Euclidean norm and r ∈ R. The notation ”:=” means ”define”. For any f : Rn → R, ▽f (x) denotes the gradient of f at x ∈ Rn . T means transpose and C (i) (C) denotes the family of functions which are defined on C ⊆ Rn to R and have the i-th continuous 1.

(6) derivative.. 2. Preliminary. A family of generalized convex functions whose properties will be reviewed here has been independently defined by Martos[17] and Avriel[3] and studied by the latter author. This generalization of the convexity concept relies on the classical definition of convex functions and some well-known results from analysis dealing with weighted means of positive numbers. Let w = (w1 , ..., wm ), q = (q1 , ..., qm ) ∈ Rm be vectors whose components are positive and nonnegative numbers, respectively, such that Σm i=1 qi = 1. Given the vector of weights q, the weighted r-mean of the numbers w1 , ..., wm is defined as in [13]: ( )1/r m  ∑  r  qi (wi ) if r ̸= 0,  Mr (w; q) ≡ Mr (w1 , ..., wm ; q) = mi=1 (1)  ∏  q  if r = 0.  (wi ) i i=1. It is well-known in [13] that for s > r, we have Ms (w1 , ..., wm ; q) ≥ Mr (w1 , ..., wm ; q). (2). for all q1 , ..., qm ≥ 0 with Σm i=1 qi = 1. Definition 2.1. (See [3]) A real function f defined on a convex set C ⊂ Rn is said to be r-convex if for any x, y ∈ C, λ ∈ [0, 1], q2 = λ, q1 = 1 − q2 , q = (q1 , q2 ), we have f (q1 x + q2 y) ≤ ln{Mr (ef (x) , ef (y) ; q)} and by (1) this is equivalent to. { ln[(1 − λ)erf (x) + λerf (y) ]1/r f ((1 − λ)x + λy) ≤ (1 − λ)f (x) + λf (y). if r ̸= 0, if r = 0.. (3). Similarly, f is said to be r-concave on C if the inequality (3) is reversed. It is clear from the above definition that a real function is convex(concave) if and only if it is 0-convex(0concave). Remark 2.1. It follows by Definition 2.1 that: The r-convexity of f on C with r > 0(r < 0) is equivalent to the convexity(concavity) of erf on C. Property 2.1. (See [2] Theorem 3.2) If f : C ⊂ Rn → R is r-convex(r-concave) on C, then f is also s-convex(s-concave) on C for all s > r(s < r). Property 2.2. (See [2] Lemma 1.2) Let the real function f defined on an open convex set C ⊂ Rn be twice continuously differentiable on C. For given (x, r) ∈ C × R with r ̸= 0, define ϕ(x, r) = ∇2 f (x) + r∇f (x)∇f (x)T . Then f is r-convex on C if and only if ϕ is positive semidefinite for all x ∈ C. 2.

(7) Definition 2.2. f : C ⊂ Rn → R is quasiconvex on C if, for all x, y ∈ C, f (λx + (1 − λ)y) ≤ max{f (x), f (y)},. 0 < λ < 1.. Similarly, f is said to be quasiconcave on C if f (λx + (1 − λ)y) ≥ min{f (x), f (y)}. It is well known in [13] that lim Mr (w1 , ..., wm ; q) ≡ M∞ (w1 , ..., wm ) = max{w1 , ..., wm },. r→+∞. lim Mr (w1 , ..., wm ; q) ≡ M−∞ (w1 , ..., wm ) = min{w1 , ..., wm }.. r→−∞. It follows by (2) that M∞ (w1 , ..., wm ) ≥ Mr (w1 , ..., wm ; q) ≥ M−∞ (w1 , ..., wm ) for every real r. Thus, if f is r-convex on C, it is also (+∞)-convex, i.e. f (λx + (1 − λ)y) ≤ max{f (x), f (y)} for every x, y ∈ C, λ ∈ [0, 1]. Similarly, if f is r-concave on C, it is also (−∞)-concave, i.e. f (λx + (1 − λ)y) ≥ min{f (x), f (y)}. Property 2.3. (See [2] Theorem 3.3) Every r-convex(r-concave) function on a convex set C is also quasiconvex(quasiconcave) on C. Property 2.4. (See [2] Theorem 4.2) A real function f is r-convex if and only if (−f ) is (−r)-concave. Property 2.5. (See [2] Theorem 4.3) Let f be r-convex(r-concave), α ∈ R and k > 0. Then f + α is r-convex(r-concave) and kϕ is (r/k)-convex((r/k)-concave). Property 2.6. (See [2] Theorem 4.4) Let ϕ, ψ : C ⊂ Rn → R be r-convex(r-concave), and let α1 , α2 > 0. Then the function θ defined by { ln[α1 erϕ(x) + α2 erψ(x) ]1/r if r ̸= 0, θ(x) = α1 ϕ(x) + α2 ψ(x) if r = 0, is also r-convex(r-concave). Property 2.7. (See [2] Theorem 4.5) Let ϕ : C ⊂ Rn → R be r-convex(r-concave) such that r ≤ 0(r ≥ 0) and let the real function ψ be nondecreasing s-convex(s-concave) on R with s ∈ R. Then, the composite function θ = ψϕ is also s-convex(s-concave). Property 2.8. (See [2] Theorem 4.6) ϕ : C ⊂ Rn → R is r-convex(r-concave) if and only if for every x, y ∈ C the function ψ given by ψ(λ) = ϕ((1 − λ)x + λy) is an r-convex(r-concave) function of λ for 0 ≤ λ ≤ 1. Property 2.9. (See [2] Theorem 6.3) Let ϕ be a twice continuously differentiable real quasiconvex function on an open convex set C ⊂ Rn . If there exists a real number r∗ satisfying −z T ∇2 ϕ(x)z r∗ = sup (4) T 2 x∈C, ∥z∥=1 [z ∇ϕ(x)] whenever z T ∇ϕ(x) ̸= 0, then ϕ is r-convex for every r ≥ r∗ . We get the r-concave analog of the above theorem by replacing supremum in (4) by infimum. 3.

(8) We want to extend the image field of r-convex function to the subset of second-order cone. For any real subset, we can get the associated subset of second-order cone. i.e., for {x| x ≥ 0}, {x| x ≽Kn 0} is the associated subset of second-order cone. The second-order cone(SOC) in Rn , also called the Lorentz cone, is defined by Kn = {x = (x1 , x2 ) ∈ R × Rn−1 | ∥x2 ∥ ≤ x1 } If n = 1, let Kn denote the set of nonnegative real R+ . For any x, y in Rn , we write x ≽Kn y if x − y ∈ Kn and write x ≻Kn y if x − y ∈ int(Kn ). In other words, we have x ≽Kn 0 if and only if x ∈ Kn and x ≻Kn 0 if and only if x ∈ int(Kn ). The relation ≽Kn is a partial ordering but not a linear ordering in Kn , i.e., there exist x, y ∈ Kn such that neither x ≽Kn y nor y ≽Kn x. To see this, for n = 2, let x = (1, 1), y = (1, 0). Then we have x − y = (0, 1) ∈ / Kn , y − x = (0, −1) ∈ / Kn . Recently, the SOC has received much attention in optimization. For SOCP and SOCCP, there needs spectral decomposition associated with SOC. The basic concepts are as below. For any x = (x1 , x2 ) ∈ R × Rn−1 , the vector x can be decomposed as x = λ1 u(1) + λ2 u(2) ,. (5). where λ1 , λ2 and u1 , u2 are the spectral values and the associated spectral vectors of x, respectively, given by λi = x1 + (−1)i ∥x2 ∥, (6) 1 x2  ) if x2 ̸= 0,  (1, (−1)i 2 ∥x2 ∥ (i) u = (7)  1  i (1, (−1) w) if x2 = 0. 2 for i = 1, 2 with w being any vector in Rn−1 satisfying ∥w∥ = 1. If x2 ̸= 0, the decomposition is unique. For any function f : R → R, the following vector-valued function associated with Kn (n ≥ 1) was considered [8,10]: f soc (x) = f (λ1 )u(1) + f (λ2 )u(2) , ∀x = (x1 , x2 ) ∈ R × Rn−1 .. (8). If f is defined only on a subset of R, then f soc is defined on the corresponding subset of Rn . The definition (8) is unambiguous whether x2 ̸= 0 or x2 = 0. The cases of f soc (x) = x1/2 , x2 , exp(x) are discussed in the book of [10]. In fact, the above definition (8) is analogous to one associated with the semidefinite cone S+n [20, 21]. Using the Property 2.9 for r = 1, 2, 3, ..., we want to give the associated r-convex functions. In these examples, all functions are real functions and defined on R. 3t2 are (+∞)-convex. t2 + 1 P roof. f ′ (t) = 3(t2 −1)2 2t, f ′′ (t) = 6(t2 −1)2 +6t·2(t2 −1)2t = 6(t2 −1)((t2 −1)+4t2 ) = 6(t2 − 1)(5t2 − 1). Example 2.1. The function f (t) = (t2 − 1)3 + 1 and g(t) =. 1 − 5t2 −4 −6(t2 − 1)(5t2 − 1) = sup = lim− = ∞. 2 2 4 2 2 3 2 |t|→1 6(t − 1)3 36t (t − 1) t̸=0,±1 t̸=0,±1 6t (t − 1) sup. 4.

(9) 3t2 3 3 · 2t =3− 2 . Then g ′ (t) = 2 . 2 t +1 t +1 (t + 1)2 6(t2 + 1)2 − 6t · 2(t2 + 1)2t 6(t2 + 1 − 4t2 ) 6(1 − 3t2 ) = = . g ′′ (t) = (t2 + 1)4 (t2 + 1)3 (t2 + 1)3 g(t) =. 6(1−3t2 ) −(t2 +1)3 sup 6t 2 t̸=0 [ (t2 +1)2 ]. = sup t̸=0. (3t2 − 1)(t2 + 1) 1 2 1 t2 = sup (3t + ) = ∞. 2 − = lim |t|→∞ 2 6t2 t2 t̸=0 6. . Example 2.2. Let a, b, c ∈ R. The function f (t) = at2 + bt + c is (+∞)-convex for a < 0 and is convex for a > 0. { −2a ∞ a < 0, P roof. f ′ (t) = 2at+b, f ′′ (t) = 2a. sup = 2 0 a > 0. t̸= −b (2at + b) 2a. Proposition 2.1. Let ad > 0, b, c ∈ R. Let D = {t ∈ R|at2 +bt+c > 0} and f : D → R. Then the function f (t) = d · ln(at2 + bt + c) is { (+∞)-convex if d(b2 − 4ac) > 0, 1/(2d)-convex if d(b2 − 4ac) < 0 or b2 − 4ac = 0. P roof.. Let B = 2a(at2 + bt + c), s = B/[−(2at + b)2 ], A = b2 − 4ac. Then. f ′ (t) = sup t̸= −b 2a. d(2at + b) d[2a(at2 + bt + c) − (2at + b)2 ] ′′ , f (t) = . at2 + bt + c (at2 + bt + c)2. d[2a(at2 +bt+c)−(2at+b)2 ] −(at2 +bt+c)2 d(2at+b) 2 [ at 2 +bt+c ]. = sup t̸= −b 2a. 2a(at2 + bt + c) − (2at + b)2 = sup (s + 1)/d. −d(2at + b)2 t̸= −b. (9). 2a. 2a t +2abt+2ac = −s(4a t +4abt+b ), then 2a (2s+1)t +2ab(2s+1)t+(sb2 +2ac) = 0. Let g(t) = 2a2 (2s + 1)t2 + 2ab(2s + 1)t + (sb2 + 2ac) and D = 4a2 b2 (2s + 1)2 − 4 · 2a2 (2s + 1)(sb2 + 2ac). Then g(t) = 0 has a solution if and only if 0 ≤ D = 4a2 (2s + 1)[b2 (2s + 1) − 2(sb2 + 2ac)] = 4a2 (2s + 1)A if and only if s ≥ −1/2 for A > 0 or ≤ −1/2 for A < 0. When A = 0, c = b2 /(4a). Then B = 2a(at2 + bt + b2 /(4a)) = (4a2 t2 + 4abt + b2 )/2 = (2at + b)2 /2 1 (2at + b)2 /2, s = = − , and (9) = 1/(2d). 2 −(2at + b) 2 2 Then f (t) = d · ln(at + bt + c) = d · ln (at2 + bt + b2 /(4a)) = d · ln a (t + b/(2a))2 = d [ln a + 2 ln (t + b/(2a))] . f ′ (t) = 2d/[t + b/(2a)], f ′′ (t) = −2d/[t + b/(2a)]2 . 2 2. 2 2. 2. sup t∈R. 2. 2. 2d/[t + b/(2a)]2 1 = . 2 (2d/[t + b/(2a)]) 2d. When A > 0, let c = b /(8a). Then B = 2a(at2 + bt + b2 /(8a)) = (4a2 t2 + 4abt + b2 − (2at + b)2 /2 − b2 /4 1 b2 2 2 2 b /2)/2 = (2at + b) /2 − b /4, s = =− + , and (9) = ∞ −(2at + b)2 2 4(2at + b)2 for d > 0, or 1/(2d) for d < 0. When A < 0, let c = b2 /(2a). Then B = 2a(at2 +bt+b2 /(2a)) = (4a2 t2 +4abt+2b2 )/2 = (2at + b)2 /2 + b2 /2 1 b2 (2at + b)2 /2 + b2 /2, s = = − − , and (9) = ∞ for d < 0, −(2at + b)2 2 2(2at + b)2 or 1/(2d) for d > 0. 2. 5.

(10) Figure 1. Figure 2: h(x) =. −f ′′ (x) [f ′ (x)]2. Example 2.3. Let u ∈ R, f : (0, ∞) → R, then the function f (t) = tu is (+∞)-convex for u(u − 1) < 0 and is convex for u(u − 1) > 0. P roof. f ′ (t) = utu−1 , f ′′ (t) = u(u − 1)tu−2 { ∞ if u(u − 1) < 0, ′′ ′ 2 −u sup −f (t)/[f (t)] = sup(1 − u)t /u = 0 if u(u − 1) > 0. u̸=0 u̸=0 Proposition 2.2. Let a, d, u ̸= 0, b, c ∈ R, then the function f (t) = d(at2 + bt + c)u is { (+∞)-convex if dau [1 − 1/(2u)] < 0, b2 − 4ac = 0, convex if dau [1 − 1/(2u)] > 0, b2 − 4ac = 0. 6.

(11) [The proof is not completed.] P roof. f ′ (t) = du(at2 + bt + c)u−1 (2at + b), f ′′ (t) = du(u − 1)(at2 + bt + c)u−2 (2at + b)2 + du(at2 + bt + c)u−1 2a = du(at2 + bt + c)u−2 [(u − 1)(2at + b)2 + 2a(at2 + bt + c)]. Let B = (at2 + bt + c)(2at + b), s = 2a(at2 + bt + c)/(2at + b)2 , and A = b2 − 4ac. Then −f ′′ (t) (u − 1)(2at + b)2 + 2a(at2 + bt + c) (at2 + bt + c)−u = sup = sup (u − 1 + s). ′ 2 −du(at2 + bt + c)u (2at + b)2 −du B̸=0 f (t) B̸=0 B̸=0 (10). sup. 2a2 t2 + 2abt + 2ac = s(4a2 t2 + 4abt + b2 ) ⇒ 2a2 (2s − 1)t2 + 2ab(2s − 1)t + (sb2 − 2ac) = 0. Above equation has a solution only when 0 ≤ 4a2 b2 (2s − 1)2 − 4 · 2a2 (2s − 1)(sb2 − 2ac) = 4a2 (2s − 1)[b2 (2s − 1) − 2(sb2 − 2ac)] = −4a2 (2s − 1)A ⇔ s ≥ −1/2 for A < 0 or ≤ −1/2 for A > 0. When A = 0, c = b2 /(4a). Then s = 1/2, f (t) = d[a(t + b/(2a))2 ]u , and ( ) [(2at + b)2 /(4a)]−u 1 (10) = sup u−1+ −du 2 B̸=0   lim = ∞ duau < 0, u > 1/2   2at+b→0     lim =0 duau > 0, u > 1/2   2at+b→±∞     lim = ∞ duau > 0, 0 < u < 1/2 2at+b→0 =  lim =0 duau < 0, 0 < u < 1/2   2at+b→±∞     lim = 0 duau < 0, u < 0   2at+b→0     lim = ∞ duau > 0, u < 0 2at+b→±∞. { { ∞ duau (u − 1/2) < 0 ∞ = = u 0 dua (u − 1/2) > 0 0. dau [1 − 1/(2u)] < 0, dau [1 − 1/(2u)] > 0.. When A < 0, let v > 1 and c = vb2 /(4a). Then s = 1 (v − 1)b2 + . 2 2(2at + b)2. [(2at + b)2 + (v − 1)b2 ]/2 = (2at + b)2. [[(2at + b)2 + (v − 1)b2 ]/(4a)]−u (u − 1 + s) (10) = sup −du A̸=0  2 −u   lim [(v − 1)b ] (u − 1/2 + ∞) = ∞ b ̸= 0, duau < 0, −du(4a)−u = t→ −b 2a   lim −∞(u − 1/2) = ∞ b ̸= 0, duau > 0, u < 0. t→±∞. . Example 2.4. Let f : (−π/2, π/2) → R and g : (0, π/2) → R. Then f (t) = tan(t)/r. 7.

(12) and g(t) = ln(tan(t))/r are |r|-convex for r ̸= 0. P roof.. f ′ (t) = sec2 (t)/r, f ′′ (t) = 2 sec2 (t) tan(t)/r. 2r sin(t) cos2 (t) −f ′′ (t) = sup[−r sin(2t)] = |r|. sup ′ 2 = sup f (t) − cos(t) 1 cos(t) 2 sec2 (t) −4 cos(2t) ′′ g ′ (t) = = · = , g (t) = . r tan(t) r cos2 (t) sin(t) r sin(2t) r sin2 (2t) −g ′′ (t) sup ′ 2 = sup[r cos(2t)] = |r|. g (t). Example 2.5. Let r > 0 and f : ( P roof. f ′ (t) =. . −π π ln(sec(t)) , ) → R. Then f (t) = is (−r)-convex. 2 2 r. tan(t) ′′ sec2 (t) −f ′′ (t) , f (t) = . sup ′ 2 = sup[−r csc2 (t)] = −r for r > 0. r r t̸=0 f (t) t̸=0. 3 SOC-functions For any x = (x1 , x2 ), y = (y1 , y2 ) ∈ R × Rn−1 , we define their Jordan product as x ◦ y = (xT y , y1 x2 + x1 y2 ). We write x2 to mean x ◦ x and write x + y to mean the usual componentwise addition of vectors. Then ◦, +, together with e′ = (1, 0, . . . , 0)T ∈ Rn and ∀x, y, z ∈ Rn have the following basic properties [10, 11] : (1) e′ ◦x = x, (2) x◦y = y◦x, (3) x◦(x2 ◦y) = x2 ◦(x◦y), (4) (x + y) ◦ z = x ◦ z + y ◦ z. Notice that the Jordan product is not associative in general. However, it is power associative, i.e., x ◦ (x ◦ x) = (x ◦ x) ◦ x for all x ∈ Rn . Thus, we may, without fear of ambiguity, write xm for the product of m copies of x and xm+n = xm ◦ xn for all positive integers m and n. We define x0 = e′ . Besides, Kn is not closed under Jordan product. For any x ∈ Kn , it is known that there exists a unique vector in Kn denoted by x1/2 such that (x1/2 )2 = x1/2 ◦ x1/2 = x. Indeed, √ ( ) √ 1 x 2 1/2 2 2 x1 + x1 − ∥x2 ∥ . x = (s, ), where s = 2s 2 In the above formula, the term x2 /s is defined to be the zero vector if x2 = 0 and s = 0, i.e., x = 0. For any x ∈ Rn , we always have x2 ∈ Kn (i.e., x2 ≽Kn 0). Hence, there exists a unique vector (x2 )1/2 ∈ Kn denoted by |x|. It is easy to verify that |x| ≽Kn 0 and x2 = |x|2 for any x ∈ Rn . It is also known that |x| ≽Kn x. For any x ∈ Rn , we define [x]+ to be the nearest point (in Euclidean norm, since Jordan product does not induce a norm) projection of x onto Kn , which is the same definition as in Rn+ . In other words, [x]+ is the optimal solution of the parametric SOCP: [x]+ = argmin{∥x − y∥ |y ∈ Kn }. It is well-known that [x]+ = (x + |x|)/2; see [11, 10]. Property 3.1. ([11] Proposition 3.3). For any x = (x1 , x2 ) ∈ R × Rn−1 , we have 8.

(13) (a) |x| = (x2 )1/2 = |λ1 |u(1) + |λ2 |u(2) . (b) [x]+ = [λ1 ]+ u(1) + [λ2 ]+ u(2) = (x + |x|)/2.. (11). Let CM denote the class of functions g : R → R defined by (8) with gˆ : R → R+ a continuously differentiable convex function satisfying lim (ˆ g (α) − α) = 0 and 0 < α→−∞. gˆ′ (α) < 1 for all α ∈ R.. Property 3.2. ([11] Proposition 4.1) For any g ∈ CM, µ > 0, and x = (x1 , x2 ), y = (y1 , y2 ) ∈ R × Rn−1 , let ϕµ : Rn × Rn → Rn defined by ϕµ (x, y) = x − µg((x − y)/µ). Then the following results hold. (a) ϕµ (x, y) = x − µ[ˆ g (λ1 /µ)u(1) + gˆ(λ2 /µ)u(2) ], where, for i = 1, 2,. u(i). λi = x1 − y1 + (−1)i ∥x2 − y2 ∥, ) { ( 1 i x2 −y2 1, (−1) if x2 ̸= y2 , ∥x2 −y2 ∥ = 2 1 (1, (−1)i w) if x2 = y2 , 2. (12) (13). where w ∈ Rn−1 being an arbitrary vector satisfying ∥w∥ = 1. (b) The pointwise limit ϕ0 = limµ→0+ ϕµ has the formula ϕ0 (x, y) = x − ([λ1 ]+ u(1) + [λ2 ]+ u(2) ) = x − [x − y]+ , where, for i = 1, 2, λi and u(i) are given by (12) and (13), respectively, with w ∈ Rn−1 being an arbitrary vector satisfying ∥w∥ = 1. Next, the concepts of SOC-monotone and SOC-convex functions are introduced in [8]. Let f : R → R, (a) f is said to be SOC −monotone of order n if its corresponding vector-valued function f soc satisfies x ≽Kn y =⇒ f soc (x) ≽Kn f soc (y). (b) f is said to be SOC − convex of order n if its corresponding vector-valued function f soc satisfies f soc ((1 − λ)x + λy) ≼Kn (1 − λ)f soc (x) + λf soc (y) (14) for all x, y ∈ Rn and 0 ≤ λ ≤ 1. The function f is SOC-monotone(respectively, SOC-convex) if f is SOC-monotone of all order n(respectively, SOC-convex of all order n). In particular, we notice that if f is continuous, then the condition(14) can be replace by the more special condition : f soc (. 1 x+y ) ≼Kn (f soc (x) + f soc (y)). 2 2 9.

(14) The concepts of SOC-monotone and SOC-convex functions are analogous to matrix monotone and matrix convex functions [5, 14], and are special cases of operator monotone and operator convex functions [1, 6, 16]. Examples of SOC-monotone and SOC-convex functions are given in [8]. It is clear that the set of SOC-monotone functions and the set of SOC-convex functions are both closed under linear combinations and under pointwise limits. Property 3.3. ([9] Theorem 3.1) domf soc ⊆ Rn . Then,. Let f ∈ C (1) (J) with J being an open interval and. (i) f is SOC-monotone of order 2 if and only if f ′ (τ ) ≥ 0 for any τ ∈ J; (ii) f is SOC-monotone of order n ≥ 3 if and only if the 2 × 2 matrix   f (t2 ) − f (t1 ) (1) f (t1 )   t2 − t1  f (t ) − f (t )  ≽ O for all t1 , t2 ∈ J. 2 1 (1) f (t2 ) t2 − t1 Property 3.4. ([9] Theorem 4.1) Let f ∈ C (2) (J) with J being an open interval in R and domf soc ⊆ Rn . Then f is SOC-convex of order 2 if and only if f is convex; and f is SOC-convex of order n ≥ 3 if and only if f is convex and the inequality 1 (2) [f (t0 ) − f (t) − f (1) (t)(t0 − t)] [f (t) − f (t0 ) − f (1) (t0 )(t − t0 )] f (t0 ) ≥ 2 (t0 − t)2 (t0 − t)4 holds for any t0 , t ∈ J. Corollary 3.1. ([9], Corollary 4.3) Let f ∈ C (4) (J) with J being an open interval in R and domf soc ⊆ Rn . If f (2) (t) > 0 for every t ∈ J, then f is SOC-convex of order n ≥ 3 if and only if for every t ∈ J, the 2 × 2 matrix [ ] f (2) (t)/2 f (3) (t)/6 ≽ O. f (3) (t)/6 f (4) (t)/24 Property 3.5. (Second-order condition) ([4], Theorem 3.3.7) Let S be a nonempty open convex set in Rn , and let f : S → R. Suppose f ∈ C 2 (S), that is, its Hessian or second derivative ▽2 f (x) exists. Then f is convex if and only if ▽2 f (x) ≽ O, ∀x ∈ S. Property 3.6. ([8] Proposition 4.1) Let f : [0, ∞] → [0, ∞] be continuous. If f is SOC-concave. then f is SOC-monotone. Property 3.7. ([8] Proposition 4.2) A function f : R → R is SOC-convex if and only if the real-valued function g(x) := ⟨f soc (x), z⟩ is a convex function ∀z ≽Kn 0.. 10.

(15) Lemma 3.1. (a) f (t) = et is SOC-monotone of order 2 on R. (b) The function f (t) = et is not SOC-monotone of order n ≥ 3 on R. (c) Let x = (x1 , x2 ), y = (y1 , y2 ) ∈ R × Rn−1 with x1 − y1 ≥ ∥x2 ∥ + ∥y2 ∥, then ex ≽Kn ey . P roof. (a) Let x = (x1 , x2 ), y = (y1 , y2 ) ∈ R2 such that x ≽Kn y, then x1 − y1 ≥ |x2 − y2 | ⇒ x1 − y1 ≥ max{x2 − y2 , y2 − x2 }.. (15). Let B = ex1 cosh(x2 ) − ey1 cosh(y2 ), a = ex1 sinh(x2 ), b = ey1 sinh(y2 ). f soc (x) − f soc (y) = ex1 (cosh(x2 ), sinh(x2 )) − ey1 (cosh(y2 ), sinh(y2 )) ≽Kn 0 if and only if B ≥ |a − b|. For a ≥ b. B − a + b = ex1 −x2 − ey1 −y2 ≥ 0 by (15). For a < b. B − b + a = ex1 +x2 − ey1 +y2 ≥ 0 by (15). (b) Let x = (2, 1.2, −1.6), y = (−1, 0, −4), then x − y = (3, 1.2, 2.4) ≽Kn 0. (0, −4) (1.2, −1.6) ) − e−1 (cosh(4), sinh(4) ) 2 4 = [(e4 + 1, .6(e4 − 1), −.8(e4 − 1)) − (e3 + e−5 , 0, −e3 + e−5 )]/2 = (17.7529, 16.0794, −11.3999) Kn 0.. But ex − ey = e2 (cosh(2), sinh(2). since ∥(16.0794, −11.3999)∥ = 19.7105 > 17.7529. We give an alternative proof of (b). ] [ [ ] )2 ( a (b) ea −eb b e f (1) (b) f (a)−f e − eb a−b a−b a+b = det ea −eb ≥0 det f (a)−f (b) =e − a−b f (1) (a) ea a−b a−b. (16). if and only if e(a+b)/2 ≥ (ea − eb )/(a − b) if and only if 1 ≥ (e(a−b)/2 − e(b−a)/2 )/(a − b). Let s = (a − b)/2, then above inequality holds if and only if 1 ≥ sinh(s)/s. By Taylor theorem, sinh(s)/s = 1 + s2 /6 + s4 /120 + ... > 1 for s ̸= 0. Hence (16) does not hold. By Theorem 3.1(ii), f is not SOC-monotone of order n ≥ 3 on R. x2 y2 −ey1 sinh(∥y2 ∥) ∥. ∥x2 ∥ ∥y2 ∥ y2 2 x2 [ex1 cosh(∥x2 ∥) − ey1 cosh(∥y2 ∥)]2 − ∥ex1 sinh(∥x2 ∥) − ey1 sinh(∥y2 ∥) ∥ ∥x2 ∥ ∥y2 ∥ < x2 , y2 > = e2x1 + e2y1 − 2ex1 +y1 [cosh(∥x2 ∥) cosh(∥y2 ∥) − sinh(∥x2 ∥) sinh(∥y2 ∥) ]≥0 ∥x2 ∥∥y2 ∥ ⇐ e2x1 + e2y1 − 2ex1 +y1 cosh(∥x2 ∥ + ∥y2 ∥) ≥ 0 with y2 = −x2 ex1 −y1 + ey1 −x1 e2x1 + e2y1 = ⇔ cosh(∥x2 ∥ + ∥y2 ∥) ≤ = cosh(x1 − y1 ) 2ex1 +y1 2 ⇔ x1 − y1 ≥ ∥x2 ∥ + ∥y2 ∥. (c) ex ≽Kn ey ⇔ ex1 cosh(∥x2 ∥)−ey1 cosh(∥y2 ∥) ≥ ∥ex1 sinh(∥x2 ∥). 11.

(16) Lemma 3.2. Any x = (x1 , x2 ), y = (y1 , y2 ) ∈ R × Rn−1 , and r ̸= 0 such that x ≽Kn y/r, then { rx ≽Kn y r > 0, (17) y ≽Kn rx r < 0. P roof. x ≽Kn y/r imply x1 − y1 /r ≥ ∥x2 − y2 /r∥. For r > 0, we have rx1 − y1 ≥ ∥rx2 − y2 ∥ ⇔ rx ≽Kn y. For r < 0, we have rx1 − y1 ≤ r∥x2 − y2 /r∥ = −|r|∥x2 − y2 /r∥ = −∥rx2 − y2 ∥ ⇔ y1 − rx1 ≥ ∥y2 − rx2 ∥ ⇔ y ≽Kn rx. Let n = 2 and r > 0. By the inequality (17) and Lemma 3.1(a), we have f soc (λx + (1 − λ)y) ≼Kn ln(λerf. soc (x). + (1 − λ)erf. soc (y). rf soc (λx+(1−λ)y). ⇔ rf soc (λx + (1 − λ)y) ≼Kn ln(w) ⇒ e. )/r =: ln(w)/r. ≼Kn w.. Lemma 3.3. ([9], Proposition 5.1) The function f (t) = ln t is SOC-monotone of order n ≥ 2 on (0, ∞). P roof. Since f ′ (t) = 1/t > 0 on t > 0. By Property 3.3(i), f (t) = ln t is SOC-monotone of order 2 on t > 0. For n ≥ 3. By Property 3.3(ii), for all a, b > 0, [ ] [ ] ( )2 (b) ln a−ln b f (1) (b) f (a)−f 1/b ln(a/b) 1 a−b a−b det f (a)−f (b) = det ln a−ln b = − (18) (1) ab a − b 1/a f (a) a−b a−b Let a = ec b for some c ̸= 0, then the equation (18)×a(a − b)2 /b = (ec − 1)2 − c2 ec = e2c − (2 + c2 )ec + 1. Since (2 + c2 )2 − 4 = c4 + 2c2 > 0 for any c ̸= 0, we have the equation (18) > 0. Thus, f (t) = ln t is SOC-monotone of order n ≥ 3 on t > 0. Then 0 ≺Kn erf. soc (λx+(1−λ)y). ≼Kn w ⇒ rf soc (λx+(1−λ)y) ≼Kn ln(w). Note for x2 ̸= 0,. (a) ex ≻Kn 0 ⇔ cosh(∥x2 ∥) ≥ ∥ sinh(∥x2 ∥)∥ ⇔ e−∥x2 ∥ > 0 . ) ( x1 +∥x2 ∥ 2 2 (b) 0 ≺Kn ln(x) ⇔ ln(x1 − ∥x2 ∥ ) > ∥ ln x1 −∥x2 ∥ ∥ = ln(x1 + ∥x2 ∥) − ln(x1 − ∥x2 ∥) ⇔ ln(x1 − ∥x2 ∥) > 0 ⇔ x1 − ∥x2 ∥ > 1. Hence (1, 0) ≺Kn x ⇒ 0 ≺Kn ln(x). (c) ln(1, 0) = (0, 0) and e(0,0) = (1, 0). Example 3.1 (a) The function f (t) = et is SOC-convex of order 2 on R. (b) However, f (t) = et is not SOC-convex of order n ≥ 3 on [0, ∞). P roof.. (a) We want to show that e. x+y 2. ≼Kn (ex + ey )/2.. 12. (19).

(17) For x2 , y2 ̸= 0, let A = ex1 sinh(x2 ) + ey1 sinh(y2 ) − 2e holds if and only if ex1 cosh(x2 ) + ey1 cosh(y2 ) − 2e (20) holds since. x1 +y1 2. x1 +y1 2. sinh((x2 + y2 )/2). Then (19). cosh((x2 + y2 )/2) ≥ |A|.. { x1 +y1 −x2 −y2 2 ≥0 ex1 −x2 + ey1 −y2 − 2e x1 +y1 +x2 +y2 2 ex1 +x2 + ey1 +y2 − 2e ≥0. (20). for A ≥ 0, for A < 0.. The proof in the other cases is similar. (b). √ √ √ 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). (e(2,0,−1) + e(6,−4,−3) )/2. = [e2 (cosh(1), 0, − sinh(1)) + e6 (cosh(5), sinh(5) ∗ (−4, −3)/5)]/2 = (14975, −11974, −8985). 14975 − 463.48 = 14511.52. ∥(−11974, −8985) − (−325.4493, −325.4493)∥ = 14515 > 14511.52.. 4. . SOC-r-convex Functions In this section, C denotes a convex subset of R, f : C → R, r ∈ R, and λ ∈ [0, 1].. Definition 4.1. Given r ∈ R, C a convex subset of R, and a function f : C → R. Let Y ⊂ Rn be corresponding to C. If there is a function f soc : Y → Rn corresponding to f such that for all x, y ∈ Y and λ ∈ [0, 1], we have { ) ( soc soc 1 r ̸= 0, ln λerf (x) + (1 − λ)erf (y) soc r (21) f (λx + (1 − λ)y) ≼Kn soc soc λf (x) + (1 − λ)f (y) r = 0. Then f is said to be SOC-r-convex of order n on C. Similarly, f is said to be SOCr-concave of order n on C if the inequality (21) is reversed. We say f is SOC-rconvex(respectively, SOC-r-concave) on C if f is SOC-r-convex of all order n(respectively, SOC-r-concave of all order n) on C. It is clear from the above definition that a real function is SOC-convex(SOC-concave) if and only if it is SOC-0-convex(SOC-0-concave). And a function f is SOC-r-convex if and only if −f is SOC-(−r)-concave. Property 4.1. ([11] Proposition 3.2) (a) For any x = (x1 , x2 ) ∈ R × Rn−1 , ) { ( x2 x1 if x2 = ̸ 0, e cosh(∥x2 ∥), sinh(∥x2 ∥) ∥x2 ∥ ex = x1 e (1, 0) if x2 = 0, where cosh(α) = (eα + e−α )/2 and sinh(α) = (eα − e−α )/2 for α ∈ R. 13.

(18) (b) For any x = (x1 , x2 ) ∈ int Kn , ln(x) is well defined and ( ) ) { ( x1 +∥x2 ∥ x2 1 2 2 − ln(x ∥x ∥ ), ln if x2 ̸= 0, 2 1 x1 −∥x2 ∥ ∥x2 ∥ ln(x) = 2 ln(x1 )(1, 0) if x2 = 0. Proposition 4.1. Let f : [0, ∞] → [0, ∞] be continuous. If f is SOC-r-concave with r ≥ 0, then f is SOC-monotone. P roof. For any 0 < λ < 1, we can write λx = λy + (1−λ)λ (x − y). If r = 0, then (1−λ) f is SOC-concave and SOC-monotone by Property 3.6. If r > 0, then f soc (λx) ≽Kn λ soc soc 1 ln(λerf (y) + (1 − λ)erf ( 1−λ (x−y)) ) ≽Kn 1r ln(λer(0,0) + (1 − λ)er(0,0) ) ≽Kn 1r ln(λ(1, 0) + r (1 − λ)(1, 0)) ≽Kn 0, where the second inequality is true since f is from [0, ∞) into itself and x − y ≽Kn 0; the third inequality is true by Lemma 3.1(c). Letting λ → 1, we obtain that f soc (x) ≽Kn f soc (y), which says that f is SOC-monotone. Example 4.1. f (t) = t is SOC-r-convex(SOC-r-concave) on R for r > 0(r < 0). [The proof is not completed.] P roof. This proof is directly completed by the definition of SOC-r-convex function. f soc (x) = x, ∀x ∈ Rn , is the vector-valued function corresponding to f . For any x = (x1 , x2 ), y = (y1 , y2 ) ∈ R × Rn−1 and λ ∈ [0, 1], (a) For r > 0, we want to show that 1 soc soc ln(λerf (x) + (1 − λ)erf (y) ). r soc f (λx + (1 − λ)y) = λx + (1 − λ)y = (λx1 + (1 − λ)y1 , λx2 + (1 − λ)y2 ). ( ) { rx2 rx1 e cosh(∥rx ∥), sinh(∥rx ∥) if rx2 ̸= 0, soc 2 2 ∥rx2 ∥ erf (x) = erx = erx1 (1, 0) if rx2 = 0. f soc (λx + (1 − λ)y) ≼Kn. soc. soc. Let (w1 , w2 ) = λerf (x) + (1 − λ)erf (y) . For rx2 = ̸ 0, ry2 ̸= 0, ) ( rx2 rx1 (w1 , w2 ) = λe cosh(∥rx2 ∥), sinh(∥rx2 ∥) ∥rx2 ∥ ( ) ry2 ry1 + (1 − λ)e cosh(∥ry2 ∥), sinh(∥ry2 ∥) . ∥ry2 ∥ w1 = λerx1 cosh(∥rx2 ∥) + (1 − λ)ery1 cosh(∥ry2 ∥), rx2 ry2 w2 = λerx1 sinh(∥rx2 ∥) + (1 − λ)ery1 sinh(∥ry2 ∥) . ∥rx2 ∥ ∥ry2 ∥ w12 = [λerx1 cosh(∥rx2 ∥)]2 + [(1 − λ)ery1 cosh(∥ry2 ∥)]2 + 2λ(1 − λ)erx1 +ry1 cosh(∥rx2 ∥) cosh(∥ry2 ∥). ∥w2 ∥2 = [λerx1 sinh(∥rx2 ∥)]2 + [(1 − λ)ery1 sinh(∥ry2 ∥)]2 ⟨x2 , y2 ⟩ + 2λ(1 − λ)erx1 +ry1 sinh(∥rx2 ∥) sinh(∥ry2 ∥) . ∥x2 ∥∥y2 ∥ 14. (22) (23) (24). (25). (26).

(19) cosh(∥rx2 ∥ + ∥ry2 ∥) = cosh(∥rx2 ∥) cosh(∥ry2 ∥) + sinh(∥rx2 ∥) sinh(∥ry2 ∥) ≥ cosh(∥rx2 ∥) cosh(∥ry2 ∥) − sinh(∥rx2 ∥) sinh(∥ry2 ∥). ⟨x2 , y2 ⟩ ∥x2 ∥∥y2 ∥. ≥ cosh(∥rx2 ∥) cosh(∥ry2 ∥) − sinh(∥rx2 ∥) sinh(∥ry2 ∥) = cosh(∥rx2 ∥ − ∥ry2 ∥) ≥ 1.. (27). Then w12 − ∥w2 ∥2 = [λerx1 ]2 + [(1 − λ)ery1 ]2 + 2λ(1 − λ)erx1 +ry1 [cosh(∥rx2 ∥) cosh(∥ry2 ∥) ⟨x2 , y2 ⟩ ] − sinh(∥rx2 ∥) sinh(∥ry2 ∥) ∥x2 ∥∥y2 ∥ ≥ [λerx1 ]2 + [(1 − λ)ery1 ]2 + 2λ(1 − λ)erx1 +ry1 = [λerx1 + (1 − λ)ery1 ]2 .. (28). We know that g(t) = et is convex. For r > 0, we have er[λx1 +(1−λ)y1 ] = g(r[λx1 + (1 − λ)y1 ]) ≤ λerx1 + (1 − λ)ery1 ⇒ w12 − ∥w2 ∥2 ≥ [λerx1 + (1 − λ)ery1 ]2 ≥ e2r[λx1 +(1−λ)y1 ] ⇒ ln(w12 − ∥w2 ∥2 )/(2r) ≥ λx1 + (1 − λ)y1 . Next, let a = ln(w1 + ∥w2 ∥), b = ln(w1 − ∥w2 ∥), we want to show that w2 1 (a − b) − (λx2 + (1 − λ)y2 )∥ 2r ∥w2 ∥ w2 ⇔ a + b − 2r[λx1 + (1 − λ)y1 ] ≥ ∥(a − b) − 2r(λx2 + (1 − λ)y2 )∥ ∥w2 ∥. (a + b)/(2r) − λx1 + (1 − λ)y1 ≥ ∥. (. )2. (29). w2 − 2r(λx2 + (1 − λ)y2 )∥2 ∥w2 ∥ = (a + b)2 − 4r(a + b)[λx1 + (1 − λ)y1 ] + 4r2 [λx1 + (1 − λ)y1 ]2 ( ) 2 T w2 2 2 − (a − b) − 4r(a − b)[λx2 + (1 − λ)y2 ] + 4r [λx2 + (1 − λ)y2 ] ∥w2 ∥ = 4ab − 4ra[λ(x1 − xT2 w2 /∥w2 ∥) + (1 − λ)(y1 − y2T w2 /∥w2 ∥)] − 4rb[λ(x1 + xT2 w2 /∥w2 ∥) a + b − 2r[λx1 + (1 − λ)y1 ]. − ∥(a − b). + (1 − λ)(y1 + y2T w2 /∥w2 ∥)] + 4r2 [λ2 (x21 − ∥x2 ∥2 ) + (1 − λ)2 (y12 − ∥y2 ∥2 ) + 2λ(1 − λ)(x1 y1 − xT2 y2 )]. (30). Let λ = 1/2, then (30) = 4ab − 2ra[x1 − xT2 w2 /∥w2 ∥ + y1 − y2T w2 /∥w2 ∥] − 2rb[x1 + xT2 w2 /∥w2 ∥ + y1 + y2T w2 /∥w2 ∥] + r2 [x21 − ∥x2 ∥2 + y12 − ∥y2 ∥2 + 2(x1 y1 − xT2 y2 )] = [2a − r(x1 + ∥x2 ∥ + y1 + ∥y2 ∥)][2b − r(x1 − ∥x2 ∥ + y1 − ∥y2 ∥)] + (2br − 2ar)[∥x2 ∥ − xT2 w2 /∥w2 ∥ + ∥y2 ∥ − y2T w2 /∥w2 ∥] + 2r2 [∥x2 ∥∥y2 ∥ − xT2 y2 ] 15. (31).

(20) a ≤ ln([erx1 cosh(∥rx2 ∥) + ery1 cosh(∥ry2 ∥) + erx1 sinh(∥rx2 ∥) + ery1 sinh(∥ry2 ∥)]/2) = ln[(erx1 +∥rx2 ∥ + ery1 +∥ry2 ∥ )/2] Let A = erx1 sinh(∥rx2 ∥) − ery1 sinh(∥ry2 ∥), then a ≥ ln([erx1 cosh(∥rx2 ∥) + ery1 cosh(∥ry2 ∥) + |A|]/2) { ln[(erx1 +∥rx2 ∥ + ery1 −∥ry2 ∥ )/2] if A ≥ 0 = ln[(erx1 −∥rx2 ∥ + ery1 +∥ry2 ∥ )/2] if A < 0 { (rx1 + ∥rx2 ∥ + ry1 − ∥ry2 ∥)/2 if A ≥ 0 ≥ (rx1 − ∥rx2 ∥ + ry1 + ∥ry2 ∥)/2 if A < 0 Similarly, { (rx1 − ∥rx2 ∥ + ry1 + ∥ry2 ∥)/2 b≥ (rx1 + ∥rx2 ∥ + ry1 − ∥ry2 ∥)/2. if A ≥ 0 if A < 0. Let y2 = x2 , then w1 = (erx1 +ery1 ) cosh(∥rx2 ∥)/2, w2 = (erx1 +ery1 ) sinh(∥rx2 ∥) a = ln[(erx1 + ery1 )e∥rx2 ∥ /2], b = ln[(erx1 + ery1 )e−∥rx2 ∥ /2], and (29) holds since w2 − r(x2 + y2 ) = 2rx2 − 2rx2 = 0. (a − b) ∥w2 ∥. rx2 /2, ∥rx2 ∥. rx2 /2, Let y2 = −x2 , then w1 = (erx1 +ery1 ) cosh(∥rx2 ∥)/2, w2 = (erx1 −ery1 ) sinh(∥rx2 ∥) ∥rx2 ∥ { ln[(erx1 −∥rx2 ∥ + ery1 +∥rx2 ∥ )/2] if x1 ≥ y1 , b = b ≥ r(x1 + y1 )/2, and (29) holds since ln[(erx1 +∥rx2 ∥ + ery1 −∥rx2 ∥ )/2] if x1 < y1 , a + b − r(x1 + y1 ) − |a − b| = 2b − r(x1 + y1 ) ≥ 0. (b) For r < 0, ln(w12 − ∥w2 ∥2 )/(2r) ≥ λx1 + (1 − λ)y1 if and only if w12 − ∥w2 ∥2 ≤ e2r[λx1 +(1−λ)y1 ] if and only if sup w12 − ∥w2 ∥2 ≤ e2r[λx1 +(1−λ)y1 ] .. x2 ∈Rn−1. By (27) and (28), we have w12 − ∥w2 ∥2 ≥ [λerx1 ]2 + [(1 − λ)ery1 ]2 + 2λ(1 − λ)erx1 +ry1 cosh(∥rx2 ∥ − ∥ry2 ∥), ⇒ sup w12 − ∥w2 ∥2 = ∞. x2 ∈Rn−1. Hence for r < 0, ln(w12 − ∥w2 ∥2 )/(2r) ≥ λx1 + (1 − λ)y1 does not hold. And, f is not SOC-r-convex on R for r < 0. For r < 0, ln(w12 −∥w2 ∥2 )/(2r) ≤ λx1 +(1−λ)y1 if and only if w12 −∥w2 ∥2 ≥ e2r[λx1 +(1−λ)y1 ] if and only if (32) infn−1 w12 − ∥w2 ∥2 ≥ e2r[λx1 +(1−λ)y1 ] x2 ∈R. 16.

(21) By (27) and (28), (32) holds if and only if [λerx1 ]2 + [(1 − λ)ery1 ]2 + 2λ(1 − λ)erx1 +ry1 ≥ e2r[λx1 +(1−λ)y1 ] if and only if λerx1 + (1 − λ)ery1 ≥ er[λx1 +(1−λ)y1 ] . Left inequality is true since g(t) = et is convex. For x2 · y2 = 0, the proof is similar. By Lemma 3.1 ∼ 3.3, we have the following Lemma. Lemma 4.1. Let z = f soc (λx + (1 − λ)y). Let w = λerf then. soc (x). + (1 − λ)erf. soc (y). ∈ int Kn ,. (i) z ≼Kn ln(w)/r if and only if rz ≼Kn ln(w) if and only if erz ≼Kn w for n = 2 and r > 0. (ii) z ≼Kn ln(w)/r if and only if rz ≽Kn ln(w) if and only if erz ≽Kn w for n = 2 and r < 0. (iii) erz ≼Kn w ⇒ rz ≼Kn ln(w) for n ≥ 2. Hence we have Theorem 4.1. Let f : R → R, then f is SOC-r-convex if erf is SOC-convex(SOCconcave) for n ≥ 2, r > 0(r < 0). For n = 2, we can replace ”if” by ”if and only if”. Example 4.2. Let n = 2, then (a) The function f (t) = t is SOC-r-convex(SOC-r-concave) on R for r > 0(r < 0). (b) The f (t) = t2 is SOC-r-convex(SOC-r-concave) on R ( function ) √ √ R \ (− 1/(−2r), 1/(−2r) ) for r ≥ 0(r < 0). 2 (c) The function f (t) = t3 is SOC-r-convex on R \ (−( )1/3 , 0) for r > 0, and is 3r −2 1/3 SOC-r-concave on R \ (0, ( ) ) for r < 0. 3r (d) The function f (t) = 1/t is SOC-r-convex on [−r/2, ∞) \ {0} for r > 0, and is SOC-r-concave (−∞, −r/2] \ {0} for r < 0. √ (e) The function f (t) = t is SOC-r-convex(SOC-r-concave) on t ≥ 1/r2 (t ≥ 0) for r > 0(r < 0). (f) The function f (t) = ln t is SOC-r-convex(SOC-r-concave) on t > 0 for r > 0(r < 0). rt P roof of (b) Let g(t) = ert , then g ′ (t) = 2rtert , g ′′ (t) = (1 + 2rt2 )2re √ . By Property 2. 2. 2. 1 for r < 0 or 3.5, g is convex if and only if g ′′ (t) ≥ 0 if and only if r ≥ 0 or t ≥ −2r √ 1 t ≤ − −2r for r < 0. Then (b) holds by Property 3.4 and Theorem 4.1.. . The proof of other cases is similar. 17.

(22) 5. SOC-Quasi-Convex Functions In this section, C denotes a convex subset of R, f : C → R, r ∈ R, and λ ∈ [0, 1].. Let t1 , t2 ∈ C. Note lim 1r ln(λerf (t1 ) + (1 − λ)erf (t2 ) ) = max{f (t1 ), f (t2 )}. Hence we r→∞ define the similar concept on the second-order cone as following. Definition 5.1 Let f : C ⊂ R → R, 0 ≤ λ ≤ 1, then f is said to be SOC-quasiconvex of order n on C if, for all x, y ∈ C, we have f soc (λx + (1 − λ)y) ≼Kn f soc (y) + [f soc (x) − f soc (y)]+  soc  if f soc (x) ≽Kn f soc (y), f (x) = f soc (y) if f soc (x) ≺Kn f soc (y),  (f soc (x) + f soc (y) + |f soc (x) − f soc (y)|)/2 if f soc (x) − f soc (y) ∈ / Kn ∪ −Kn . Similarly, f is said to be SOC-quasiconcave of order n if f soc (λx + (1 − λ)y) ≽Kn f soc (x) − [f soc (x) − f soc (y)]+ . Proposition 5.1 Let f be SOC-r-convex on C, then f is also SOC-quasiconvex of order 2 on C. P roof. Let f soc (x) ≽Kn f soc (y). For r = 0, it is easy to see that λf soc (x) + (1 − λ)f soc (y) ≼Kn f soc (x), for 0 ≤ λ ≤ 1. For r ̸= 0, we want to show that ) 1 ( rf soc (x) soc + (1 − λ)erf (y) ≼Kn f soc (x). ln λe r soc. For n = 2, r > 0, by Lemma 4.1, (33) holds if and only if λerf (x) + (1 − λ)erf soc soc soc erf (x) if and only if erf (y) ≼Kn erf (x) if and only if f soc (y) ≼Kn f soc (x). Similarly, for n = 2, r < 0, by Lemma 4.1, (33) holds. From Property 2.1, we naturally have the following question: Let f be SOC-r-convex on C, then f is SOC-s-convex on C for all s ≥ r ? Let f : C ⊂ R → R and λ ∈ [0, 1]. Note (i) z/s ≼Kn z/r for all s ≥ r, sr > 0, z ≽Kn 0. (ii). 1 r. ln(λerf (t1 ) + (1 − λ)erf (t2 ) ) increases as r increases.. 1 r→∞ r. (iii) lim. ln(λerf (t1 ) + (1 − λ)erf (t2 ) ) = max{f (t1 ), f (t2 )}.. Example 5.1. For f (t) = t, n ≥ 2, and p, q > 0, then. 1 soc soc ln(p ∗ erf (x) + q ∗ erf (y) ) = f soc (y) + [f soc (x) − f soc (y)]+ r→∞ r lim. 18. (33) soc (y). ≼K n .

(23) [The proof is not completed.] P roof. Since lim ln(p)/r = 0 for all p > 0, we can take p = q = 1. r→∞. For n=2, any a, b, c, d ∈ R, take x = (a, b), y = (c, d). 1 ln(er(a,b) + er(c,d) ) r→∞ r 1 = lim ln( ear (cosh(br), sinh(br)) + ecr (cosh(dr), sinh(dr)) ) r→∞ r ) ( 1 e(a+b)r + e(c+d)r (a+b)r (c+d)r (a−b)r (c−d)r = lim ln(e +e )(e +e ), ln (a−b)r r→∞ 2r e + e(c−d)r = lim. a + b + c − d a − b + c + d max{a + b, c + d} − max{a − b, c − d}) , }, ). (34) 2 2 2 For (a, b) ≽Kn (c, d), a − c ≥ |b − d| ⇒ a − c ≥ max{b − d, d − b} ⇒ { a+b+c−d a−b+c+d a−b≥c−d a ≥ max{ , }, and . 2 2 a+b≥c+d (max{a, c,. Then the equation (34) = (a, b). For n ≥ 3. Let (w1 , w2 ) = λerx + (1 − λ)ery . For rx2 ̸= 0, ry2 ̸= 0, by (22) ∼ (28), [λerx1 ]2 + [(1 − λ)ery1 ]2 + 2λ(1 − λ)erx1 +ry1 cosh(∥rx2 ∥ − ∥ry2 ∥) ≤ w12 − ∥w2 ∥2 ≤ [λerx1 ]2 + [(1 − λ)ery1 ]2 + 2λ(1 − λ)erx1 +ry1 cosh(∥rx2 ∥ + ∥ry2 ∥). ( ) ( ) w1 + ∥w2 ∥ w2 1 1 2 2 ln(w1 − ∥w2 ∥ ), ln lim ln(w1 , w2 ) = lim r→∞ 2r r→∞ r w1 − ∥w2 ∥ ∥w2 ∥. (35) (36). By (35), max{x1 , y1 , (x1 + y1 + ∥x2 ∥ − ∥y2 ∥)/2, (x1 + y1 − ∥x2 ∥ + ∥y2 ∥)/2} ≤ lim ln(w12 − ∥w2 ∥2 )/(2r). (37). r→∞. ≤ max{x1 , y1 , (x1 + y1 + ∥x2 ∥ + ∥y2 ∥)/2} For x ≽Kn y, we have x1 − y1 ≥ ∥x2 − y2 ∥. Then the inequality (37) becomes x1 ≤ lim ln(w12 − ∥w2 ∥2 )/(2r) ≤ x1 r→∞. (38). The second inequality of (38) is true since w12 − ∥w2 ∥2 = [λerx1 ]2 + [(1 − λ)ery1 ]2 + 2λ(1 − λ)erx1 +ry1 cosh(∥rx2 ∥ + ∥ry2 ∥) only when y2 = −x2 . When y2 = −x2 , x1 − y1 ≥ ∥x2 − y2 ∥ = ∥x2 ∥ + ∥y2 ∥ ⇒ x1 ≥ (x1 + y1 + ∥x2 ∥ + ∥y2 ∥)/2. Hence (38) implies 1 lim 2r ln(w12 − ∥w2 ∥2 ) = x1 . r→∞. 19.

(24) Let a = λerx1 +∥rx2 ∥ , b = λerx1 −∥rx2 ∥ , c = (1 − λ)ery1 +∥ry2 ∥ , d = (1 − λ)ery1 −∥ry2 ∥ , A = ry2 rx2 (a − b − c + d)/2. Then w1 = (a + b + c + d)/2, w2 = a−b + c−d . 2 ∥rx2 ∥ 2 ∥ry2 ∥ { (a + c)/(b + d) if x2 = ky2 , k > 0, w1 + ∥w2 ∥ = w1 − ∥w2 ∥ (w1 + |A|)/(w1 − |A|) if x2 = −ky2 , k > 0 · · · 1. { (a + d)/(b + c) if A ≥ 0, 1. = (b + c)/(a + d) if A < 0. { w2 x2 /∥x2 ∥ if x2 = −ky2 , k > 0, A ≥ 0 or x2 = ky2 , k > 0, = ∥w2 ∥ −x2 /∥x2 ∥ if x2 = −ky2 , k > 0, A < 0. ( ) 1 w1 + ∥w2 ∥ w2 lim ln = r→∞ 2r w1 − ∥w2 ∥ ∥w2 ∥ { x2 (max{x1 + ∥x2 ∥, y1 + ∥y2 ∥} − max{x1 − ∥x2 ∥, y1 − ∥y2 ∥}) 2∥x if x2 = ky2 , k > 0, 2∥ x2 (max{x1 + ∥x2 ∥, y1 − ∥y2 ∥} − max{x1 − ∥x2 ∥, y1 + ∥y2 ∥}) 2∥x2 ∥ if x2 = −ky2 , k > 0. (39) n For x ≽K y, we have x1 − y1 ≥ ∥x2 − y2 ∥. Then (39) = x2 if x2 = ky2 . And 1 1 1 2∥ ln(w1 + ∥w2 ∥) = x1 +∥x . Note x1 = lim 2r ln(w12 − ∥w2 ∥2 ) = lim 2r ln(w1 + lim 2r 2 r→∞. r→∞. 1 ∥w2 ∥) + lim 2r ln(w1 − ∥w2 ∥). r→∞ x2 lim λerx1 +∥rx2 ∥ 2∥x , then (39) = 2∥ r→∞. So. lim 1 r→∞ 2r. ln(w1 − ∥w2 ∥) =. x1 −∥x2 ∥ . 2. x2 for x2 ̸= ky2 , k ∈ R. . r→∞. Note lim w2 = r→∞. n For =) ln(αx1 , αx2 ) ( any x = (x1 , x2 ) ∈ int ) ( K , α > 0, ln(αx) αx 1 αx + ∥αx ∥ 2 1 2 = ln[α2 (x21 − ∥x2 ∥2 )], ln 2( αx1 ( − ∥αx2 ∥ ∥αx ) 2∥ ) x2 1 x + ∥x ∥ 1 2 = 2 ln(α) + ln(x21 − ∥x2 ∥2 ), ln 2 x1 − ∥x2 ∥ ∥x2 ∥ = ln(α)(1, 0) + ln(x).. ln(αesx )/s = (ln(α)(1, 0) + sx) /s by the definition of ln(x) ∀x ∈ int Kn , ∀s ∈ R \ {0}. Question: For any x = (x1 , x2 ), y = (y1 , y2 ) ∈ R × Rn−1 , x, y ∈ int Kn , ln(x ◦ y) = ln(x) + ln(y) ? ln(x ( ◦ y) = ln(x1 y1 + xT2 y2 , x1 y2 + x2 y1 ) ( )) T 1 y + ∥x y + x y ∥ x y + x y x y + x 2 1 2 2 1 1 2 2 1 1 1 2 = ln((x1 y1 + xT2 y2 )2 − ∥x1 y2 + x2 y1 ∥2 ), ln 2 x1 y1 + xT2 y2 − ∥x1 y2 + x2 y1 ∥ ∥x1 y2 + x2 y1 ∥ Let (a, b), (c, d) ∈ int K2 . Then a > |b|, c > |d| ⇒ ac > |bd|. |ac + bd| − |ad + bc| = { ac + bd − ad − bc = (a − b)(c − d) > 0 if ad + bc > 0, ac + bd + ad + bc = (a + b)(c + d) > 0 if ad + bc < 0. ln((a,(b) ◦ (c, d)) = ln(ac + bd, ad + bc) ( ) ) ac + bd + |ad + bc| ad + bc 1 2 2 ln((ac + bd) − (ad + bc) ), ln = 2 ac + bd − |ad + bc| |ad + bc| 20.

(25) ( ( )) 1 (a + b)(c + d) 2 2 2 2 = ln((a − b )(c − d )), ln 2[ (a − b)(c − d) ] 1 a + b c+d 2 2 2 2 ) + (ln(c − d ), ln ) = (ln(a − b ), ln 2( a−b c−d ) ( ) 1 a + |b| b 1 c + |d| d = ln(a2 − b2 ), ln( ) + ln(c2 − d2 ), ln( ) 2 a − |b| |b| 2 c − |d| |d| = ln(a, b) + ln(c, d). √ √ Let (a, b,√ c), (d, i, j) ∈ int K3 . Then a > b2 + c2 , c > i2 + j 2 ⇒ ac > (b2 + c2 )(i2 + j 2 ). |ac { + bd| − |ad + bc| = ac + bd − ad − bc = (a − b)(c − d) > 0 if ad + bc > 0, ac + bd + ad + bc = (a + b)(c + d) > 0 if ad + bc < 0. ln((a, b, c) ◦ (d, i, j)) = ln(ad + bi + cj, ai + db, aj + dc) ( 1 ln((ad + bi + cj)2 − (ai + db)2 − (aj + dc)2 ), = 2 ( ) ) √ ad + bi + cj + (ai + db)2 + (aj + dc)2 (ai + db, aj + dc) √ √ ln ad + bi + cj − (ai + db)2 + (aj + dc)2 (ai + db)2 + (aj + dc)2 √ 1( a + b2 + c2 (b, c) 2 2 2 2 2 2 √ ln(a, b, c) + ln(d, i, j) = ln(a − b − c )(d − i − j ), ln( )√ 2 a − b2 + c2 b2 + c2 √ d + i2 + j 2 (i, j) ) √ + ln( )√ . d − i2 + j 2 i2 + j 2 (a2 − b2 − c2 )(d2 − i2 − j 2 ) − [(ad + bi + cj)2 − (ai + db)2 − (aj + dc)2 ] = −(b2 j 2 + c2 i2 + 2bcij) = −(bj + ci)2 . If ln((a, b, c) ◦ (d, i, j)) = ln(a, b, c) + ln(d, i, j), then bj + ci = 0 by (40). √ √ d + i2 + j 2 a + b 2 + c2 b i √ √ ln( )√ + ln( )√ a − b2 + c2 b2 + c2 d − i2 + j 2 i2 + j 2 ( ) √ ad + bi + cj + (ai + db)2 + (aj + dc)2 ai + db √ √ − ln 2 2 ad + bi + cj − (ai + db) + (aj + dc) (ai + db)2 + (aj + dc)2. (40). (41). Suppose that bj + ci = 0, then any√ k, r ∈ R, let j√= −ki, c = kb, db = rai ̸= 0, s = (1 + r)2 (1 + k 2 ) − 4rk 2 . Then a > 1 + k 2 |b|, c > 1 + k 2 |i|. (41) becomes √ √ a + 1 + k 2 |b| b d + 1 + k 2 |i| i √ √ ln( )√ + ln( )√ a − 1 + k 2 |b| 1 + k 2 |b| d − 1 + k 2 |i| 1 + k 2 |i| 21.

(26) (. ) (1 + r)2 + (1 − r)2 k 2 |a||i| (1 + r)ai √ √ − ln 2 2 2 2 2 2 ra i/b + (1 − k )bi − (1 + r) + (1 − r) k |a||i| (1 + r) + (1 − r)2 k 2 |a||i| √ √ ( 2 √ ) a + b 1 + k 2 rai/b + i 1 + k 2 1 ra + (1 − k 2 )b2 + ab s 1 + r √ √ √ √ = ln( )√ − ln ra2 + (1 − k 2 )b2 − ab s s a − b 1 + k 2 rai/b − i 1 + k 2 1 + k 2 √ √ ( ) √ 1 ra2 + (1 − k 2 )b2 + ab s 1 + r a + b 1 + k 2 ra + b 1 + k 2 √ √ √ √ )√ − ln = ln( ra2 + (1 − k 2 )b2 − ab s s a − b 1 + k 2 ra − b 1 + k 2 1 + k 2 √ ) ( √ ra2 + ab s + 4rk 2 + b2 (1 + k 2 ) 1 ra2 + (1 − k 2 )b2 + ab s 1 + r √ √ √ = ln( )√ − ln ra2 + (1 − k 2 )b2 − ab s s ra2 − ab s + 4rk 2 + b2 (1 + k 2 ) 1 + k 2 (42) √ 2 For a = tb > 0 with t > 1 + k , (42) becomes ( ) √ ( 2 √ ) 1 rt2 + t s + 4rk 2 + (1 + k 2 ) rt + (1 − k 2 ) + t s 1 + r √ √ √ √ − ln ln (43) rt2 + (1 − k 2 ) − t s s rt2 − t s + 4rk 2 + (1 + k 2 ) 1 + k2 √ ( 2 √ )√1+4rk2 /s 2 2 2 2 rt + 1 − k + t s rt + 1 + k + t s + 4rk √ √ (43) = 0 ⇔ = . 2 2 2 rt2 + 1 − k 2 − t s rt + 1 + k − t s + 4rk √ ( 2 √ )√1+4rk2 /s rt + 1 − k 2 + t s rt2 + 1 + k 2 + t s + 4rk 2 √ √ Define g(k) = h(t, r, k) = − rt2 + 1 − k 2 − t s rt2 + 1 + k 2 − t s + 4rk 2 where s = (1 + r)2 (1 + k 2 ) − 4rk 2 . ra2 i/b + (1 − k 2 )bi +. √. Figure 3 For i = 0, (40) = 0 only when b = 0 or j = 0.. References [1] J.S. Aujla and H.L. Vasudeva, Convex and monotone operator functions, Annales Polonici Mathematici, vol. 62, pp. 1-11, 1995. [2] M. Avriel, r-Convex Functions, Math. Program., Vol.2, pp.309-323, 1972. [3] M. Avriel, Solution of Certain Nonlinear Programs Involving r-Convex Functions, J. Optim. Theory Appl., Vol. 11, No. 2, pp.159-174, 1973. 22.

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