二階錐跡函數的自協調性

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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：張毓麟博士. The self-concordancy of the trace functions on SOCs.. 研究生：莊國裕. 中華民國一百零四年七月.

(2) The self-concordancy of the trace functions on SOCs.. By Kuo-Yu Chuang. Advisor Yu-Lin Chang. Department of Mathematics, National Taiwan Normal University Taipei, Taiwan. July, 2015.

(3) 誌. 謝. 在這兩年的碩士生涯，對於支持我的家人、指導過我的老師及陪伴著我的朋友們，我十分地感謝他們無私的幫忙。首先，特別感謝我的指導教授張毓麟老師。在我剛進研究所還沒有方向的時候，帶我進入了最佳化這門領域，在我每每遇到困難時總是不吝嗇伸出援手，謝謝老師您的教導。感謝陳界山老師及柯春旭老師擔任我的口試委員，在口試時給予了我很多建議，讓我受益良多。感謝吳荻文同學、張育瑋同學及藍茂愷同學，在我研究過程遇到困難時給予提點，讓我在完成論文的路上順利許多。感謝讀書會的夥伴們長久以來的陪伴，陪我在碩士生涯中解決了許多令人感興趣的問題，也提供了我很多解決問題的想法。僅將這份論文獻給所有關心我的師長、朋友們，以及我的家人！. 莊國裕. 謹誌. 2015 年 7 月.

(4) Contents 1. Introduction…………….......….……….………..………..…………………..1 2. Preliminaries…………………..……….……….....……..…………………….3 3. Main results………………………………………...……..……………………..7 4. In the future…………………………………………………….…..…….…....17 5. References………………………………………….………………………..…..17.

(5) The self-concordancy of the trace functions on SOCs.. July 17, 2015 Abstract. The strongly non-degenerate self-corcordant functions are the key to applying interior-point method. In this paper, we check the self-concordancy and the strongly non-degenerate self-concordancy for some examples, and we want to establish strongly non-degenerate self-concordancy of some functions associated with second-order cone, called SOC trace functions. Key words. Second-order cone, trace function, self-concordant.. 1. Introduction. We let ⟨ , ⟩ denote an arbitrary inner product on Rn . The inner product ⟨ , ⟩ induces a norm on Rn , ∥x∥ := ⟨x, x⟩1/2 . If S is p.d., then S defined a new inner product, namely, ⟨x, y⟩S := ⟨x, Sy⟩. Let ∥ ∥S denote the norm induced by ⟨ , ⟩S . The second-order cone (SOC) in Rn , also called Lorentz cone, is the set defined as { } n n−1 K := (x1 , x2 ) ∈ R × R | x1 ≥ ∥x2 ∥ ,. (1). where ∥·∥ denotes the Euclidean norm. When n = 1, Kn reduces to the set of nonnegative real numbers R+ . As shown in [5], Kn is also a set composed of the squared elements from Jordan algebra (Rn , ◦), where the Jordan product “◦” is a binary operation defined by x ◦ y := (⟨x, y⟩, x1 y2 + y1 x2 ) (2) for any x = (x1 , x2 ), y = (y1 , y2 ) ∈ R×Rn−1 . Here for any x ∈ Rn , we use x1 to denote the first component of x, and x2 to denote the vector consisting of the rest n − 1 components. 1.

(6) From [4, 5], we recall that each x ∈ Rn admits a spectral decomposition associated with Kn of the following form (2) x = λ1 (x)u(1) (3) x + λ2 (x)ux , (i). where λi (x) and ux for i = 1, 2 are the spectral values and the associated spectral vectors of x, respectively, defined by λi (x) = x1 + (−1)i ∥x2 ∥, u(i) x =. ) 1( 1, (−1)i x¯2 , 2. (4). x2 ∥ = 1. with x¯2 = ∥xx22 ∥ if x2 ̸= 0, and otherwise x¯2 being any vector in Rn−1 such that ∥¯ When x2 ̸= 0, the spectral factorization is unique. The determinant and trace of x are defined as det(x) := λ1 (x)λ2 (x) and tr(x) := λ1 (x) + λ2 (x), respectively. With the spectral decomposition above, for any given scalar function ϕ : J ⊆ R → R, we may define a vector-valued function ϕsoc : S ⊆ Rn → Rn by (2) ϕsoc (x) := ϕ(λ1 (x))u(1) x + ϕ(λ2 (x))ux. (5). where J is an interval (finite or infinite, open or closed) of R, and S is the domain of ϕsoc determined by ϕ. Then, we can define the SOC trace function associated with ϕ ϕtr (x) := ϕ(λ1 (x)) + ϕ(λ2 (x)) = tr(ϕsoc (x)) ∀x ∈ S.. (6). Chen, Liao and Pan [2] give the following conclusion: Lemma 1.1. [2, Lemma2.2] For any given ϕ : J ⊆ R → R, let ϕsoc : S → Rn and ϕtr : S → R be given by (5) and (6), respectively. Asumme that J is open. Then, the following results holds. (a) The domain S of ϕsoc and ϕtr is also open. (b) If ϕ is (continuously) differentiable on J, then ϕsoc is (continuously) differentiable ′ on S. Moreover, for any x ∈ S, ∇ϕsoc = ϕ (x1 )I if x2 = 0, and otherwise [ ∇ϕsoc (x) =. ]. xT. b(x). c(x) ∥x22 ∥. x xT. c(x) ∥xx22 ∥ a(x)I + (b(x) − a(x)) ∥x22 ∥22. where. 2. ,. (7).

(7) ϕ′ (λ2 (x)) + ϕ′ (λ1 (x)) , 2 ϕ′ (λ2 (x)) − ϕ′ (λ1 (x)) c(x) = , 2 ϕ′ (λ2 (x)) − ϕ′ (λ1 (x)) a(x) = . λ2 (x) − λ1 (x) b(x) =. (c) If ϕ is (continuously) differentiable, then ϕtr is (continuously) differentiable on S ′ with ∇ϕtr (x) = (ϕ )soc (x); if ϕ is twice (continuously) differentiable, then ϕtr is ′ twice (continuously) differentiable on S with ∇2 ϕtr (x) = ∇(ϕ )soc (x). Chen, Liao and Pan [2] also give the follow conclusion: Theorem 1.2. [2, Theorem 2.3] For any given ϕ : J → R, let ϕsoc : S → Rn and ϕtr : S → R be given by (5) and (6), respectively. Asumme that J is open. If ϕ is twice differentiable on J, then ′′. ′. (a) ϕ (t) ≥ 0 for any t ∈ J ⇐⇒ ∇(ϕ )soc (x) ≽ 0 for any x ∈ S ⇐⇒ ϕtr is convex in S. ′′. ′. (b) ϕ (t) > 0 for any t ∈ J ⇐⇒ ∇(ϕ )soc (x) ≻ 0 for any x ∈ S =⇒ ϕtr is strictly convex in S. Chang and Chen [1] give another proof of Theorem 1.2. By using Schur Complement Theorem, they establish the convexity of SOC trace functions and the compounds of SOC trace functions. Since the strongly non-degenerate self-corcordant functions are the key to applying interior-point method (IPM), naturally, we think that the proof of strongly non-degenerate self-concordancy of SOC trace functions and the compounds of SOC trace functions may be worked.. 2. Preliminaries. This section recalls some results of self-concordant function that will be used in subsequent analysis. More detailed expositions of self-concordant function can be found in the monograph by Renegar[3]. Definition 2.1. Let f be a functional and Df be the domain of f. The functional f is said to be (Frechet) differentiable at x ∈ Df if there exists a vector g(x) satisfying f (x + ∆x) − f (x) − ⟨g(x), ∆x⟩ = 0. ∥∆x∥→0 ∥∆x∥ lim. 3.

(8) The vector g(x) is the gradient of f at x w.r.t. ⟨ , ⟩ Definition 2.2. Let f be a functional and Df be the domain of f. The functional f is said to be twice differentiable at x ∈ Df if f ∈ C 1 and there exists a linear operator H(x) : Rn → Rn satisfying ∥g(x + ∆x) − g(x) − H(x)∆x∥ = 0. ∥∆x∥→0 ∥∆x∥ lim. If it exists, H(x) is said to be the Hessian of f at x w.r.t. ⟨ , ⟩ In this paper, we use Df to denote the domain of the funtctional f, we use ∥ ∥x to denote the norm defined by H(x), and we use Bx (x, 1) to denote the set {y | ∥y−x∥x < 1}. Definition 2.3. A convex real-valued function f is said to be self-concordant if ′′′. ′′. 3. |f (x)| ≤ 2f (x) 2 ∀ x ∈ Df . Definition 2.4. A real-valued function f is said to be strongly non-degenerate selfconcordant if it satisfies the following two conditions: (i) for all x ∈ Df we have Bx (x, 1) ⊆ Df , and (ii) if whenever y ∈ Bx (x, 1) we have 1 − ∥y − x∥x ≤. ∥v∥y 1 ≤ ∥v∥x 1 − ∥y − x∥x. for all v ̸= 0. In this paper, the self-concordant function means the function satisfying Def. 2.3, and the strongly self-concordant function means the function satisfying Def. 2.4. We denote SC be the family of strongly self-concordant functionals. Proposition 2.5. [3, Proposition 2.5.1] Assume f is such that for all x ∈ Df we have Bx (x, 1) ⊆ Df and is such that whenever y ∈ Bx (x, 1) we have 1 ∥v∥y ≤ f or all v ̸= 0. ∥v∥x 1 − ∥y − x∥x Then 1 − ∥y − x∥x ≤. (8). ∥v∥y f or all v ̸= 0. ∥v∥x. From Prop. 2.5, we know that if we want to show that a function is strongly selfconcordant, we only need to show that the function satisfying Def. 2.4(i) and the rightmost inequality in Def. 2.4(ii). From [3], we can obtain the following conclusion: 4.

(9) Theorem 2.6. Assume f is a real-valued function satisfying the following conditionas: (i) for all x ∈ Df we have Bx (x, 1) ⊆ Df , and (ii) if whenever y ∈ Bx (x, 1) we have 1 − ∥y − x∥x ≤. ∥v∥y 1 ≤ f or all v ̸= 0. ∥v∥x 1 − ∥y − x∥x. If f is thrice-differentiable, then ′′′. ′′. 3. |f (x)| ≤ 2f (x) 2 ∀ x ∈ Df . The inverse of this conclusion is not always true, there are some examples in Theorem 3.2 and 3.3. Now, we present how we use the strongly self-concordant function in the interior point method. For a functional f and x ∈ Df , we denote the second-order - or ”quadratic” - approximation at x by 1 qx (y) = f (x) + ⟨g(x), y − x⟩ + ⟨y − x, H(x)(y − x)⟩ 2 The domain of qx is all or Rn . Proposition 2.7. [3, Proposition 1.6.1] The gradient of qx at y is g(x) + H(x)(y − x) and the Hessian is H(x) (regardless of y). Henceforth, assume H(x) is p.d.. Then qx is strictly convex and hence is minimized by the point x+ satisfying g(x) + H(x)(x+ − x) = 0, that is, qx is minimized by the point x+ := x − H(x)−1 g(x) The ”Newton-step at x” is defined to be the difference n(x) := x+ − x = −H(x)−1 g(x). Newton’s method steps x to x + n(x). Proposition 2.8. [3, Proposition 2.2.2] If f ∈ SC, x ∈ Df and y ∈ Bx (x, 1), then |f (y) − qx (y)| ≤. ∥y − x∥3x . 3(1 − ∥y − x∥x ). Theorem 2.9. [3, Theorem2.2.3] Assume f ∈ SC and x ∈ Df . If z minimizes f and z ∈ Bx (x, 1),then x+ := x − H(x)−1 g(x) 5.

(10) satisfies ∥x+ − z∥x ≤. ∥x − z∥2x 1 − ∥x − z∥x. (9). By the definition of strongly self-concordancy, since z ∈ Bx (x, 1) ⊆ Df , thus Bz (z, 1) ⊆ Df . If x ∈ Bz (z, 1), then 1 − ∥x − z∥z ≤. ∥x+ − z∥x ∥x+ − z∥x =⇒ ∥x+ − z∥z ≤ ∥x+ − z∥z 1 − ∥x − z∥z. (10). and ∥x − z∥x 1 ∥x − z∥z ≤ =⇒ ∥x − z∥x ≤ ∥x − z∥z 1 − ∥x − z∥z 1 − ∥x − z∥z. (11). thus 1 − ∥x − z∥x ≥. 1 − 2∥x − z∥z 1 − ∥x − z∥z. (12). If ∥x − z∥z ≤ 21 , applying (9)(10)(11)(12), we have ∥x+ − z∥x ∥x − z∥2x ∥x − z∥2z ≤ ≤ 1 − ∥x − z∥z (1 − ∥x − z∥z )(1 − ∥x − z∥x ) (1 − ∥x − z∥z )3 (1 − ∥x − z∥x ) 2 ∥x − z∥z ≤ (1 − ∥x − z∥z )2 (1 − 2∥x − z∥z ). ∥x+ − z∥z ≤. If ∥x − z∥z ≤ 14 , then ∥x+ − z∥z ≤ and inductively,. 32 ∥x 9. − z∥2z ≤ 4∥x − z∥2z (< 41 , so x+ ∈ Df ),. 1 i ∥xi − z∥z ≤ (4∥x − z∥z )2 4. (13). where x1 = x+ , x2 , . . . is the sequence generated by Newton’s method. The bound (13) makes apparent the rapid convergence of Newton’s method. Theorem 2.10. [3, Theorem 2.2.4] Assume f ∈ SC. If ∥n(x)x ∥ < 1 then ( ∥n(x+ )∥x+ ≤. ∥n(x)∥x 1 − ∥n(x)∥x. )2. A rather unsatisfying, but unavoidable, aspect of general convergence results for Newton’s method is an assumption of x being sufficiently close to a minimizer z, where ”sufficiently close” depend explicitly on z. For general functional, it is impossible to verify 6.

(11) that x is indeed sufficiently close to z without knowing z. For stongly non-degenerate self-concordant functionals, we know that the explicitly depend on z of what consitutes ”sufficiently close” can take a particularly simple form (e.g., we know x is sufficiently close to z if ∥x − z∥z < 41 ), albeit a form which appears still to require knowing z. The next theorem provides means to verify proximity to a minimizer without knowing the minimizer. Theorem 2.11. [3, Theorem 2.2.5] Assume f ∈ SC. If ∥n(x)∥x ≤ then f has a minimizer z and ∥z − x+ ∥x ≤. 1 4. for some x ∈ Df ,. 3∥n(x)∥2x . (1 − ∥n(x)∥x )3. Thus, ∥z − x∥x ≤ ∥n(x)∥x +. 3∥n(x)∥2x . (1 − ∥n(x)∥x )3. From above, we know that strongly self-concordant functions play an important role in interior point methods, thus we want to find ways to contruct more strongly selfconcordant functions. The next two theorems demonstrate other relevant ways for contructing strongly self-concordant functions. Theorem 2.12. [3, Theorem 2.2.6] The set SC is closed under addition; that is, if f1 and f2 are strongly self-concordant functionals satisfying Df1 ∩ Df2 ̸= ϕ, then f1 + f2 : Df1 ∩ Df2 → R is a strongly self-concordant functional. Theorem 2.13. [3, Theorem 2.2.7] If f ∈ SC, Df ⊆ Rm , b ∈ Rm , and A : Rn → Rm is an injective linear operator, then x 7→ f (Ax − b) is a strongly self-concordant functional, assumeing the domain {x : Ax − b ∈ Df } is nonempty.. 3. Main results. In this chapter, we are interesting that whether the proof of strongly self-concordancy of SOC trace functions and the compounds of SOC trace functions can be worked. We prove a half of our aim in the next theorem. Theorem 3.1. For any given scalar funtion ϕ : Dϕ ⊆ R → R, let ϕtr : Dϕtr ⊆ Kn → R be given by (6) where Dϕ = {t | t > 0} and Dϕtr = intKn . Then ϕtr is strongly self-concordant ⇒ ϕ is strongly self-concordant. Proof. Assume ϕtr is strongly self-concordant, we check that ϕ satisfies the two conditions of strongly self-concordancy. 7.

(12) √ (i) Let t ∈ Dϕ . ∀ s√∈ Bt (t, 1), ⟨(s − t), ϕ′′ (t)(s − t)⟩ < 1. This implies ϕ′′ (t)(s − t)2 < 1 Let x = (t, 0)T , and y = (s, 0)T . Then x, y ∈ intKn By applying equation (7), we obtain [ ′′ ] ′′ ϕ (t) 0Tn−1 2 tr ∇ ϕ (x) = = ϕ (t)In . ′′ 0n−1 ϕ (t)In−1 Then, ∥y − x∥x = = =. √ √ √. ⟨(y − x), ϕ′′ (t)(y − x)⟩ ⟨(s − t, 0)T , ϕ′′ (t)I(s − t, 0)T ⟩ ϕ′′ (t)(s − t)2 < 1. That is, y ∈ Bx (x, 1). Since ϕtr is strongly self-concordant, thus y ∈ Bx (x, 1) ⊆ Dϕtr = intKn , and hence s > 0. That is, s ∈ Dϕ , and we obtain Bt (t, 1) ⊆ Dϕ . (ii) Let t ∈ Dϕ . ∀ s ∈ Bt (t, 1), ∀ v ̸= 0. Let x = (t, 0)T , y = (s, 0)T , and vˆ = (v, 0)T . By (i), y ∈ Bx (x, 1). ∥ˆ v ∥y 1 ≤ 1−∥y−x∥ Since ϕtr is strongly self-concordant, ∥ˆ v ∥x x √. ⟨(v, 0)T , ϕ′′ (s)(v, 0)T ⟩ 1 √ ⇒ √ ≤ ′′ ⟨(v, 0)T , ϕ (t)(v, 0)T ⟩ 1 − ⟨(s − t, 0)T , ϕ′′ (t)(s − t, 0)T ⟩ √ ϕ′′ (s)v 2 1 √ ′′ ⇒ ≤ ′′ 2 ϕ (t)v 1 − ϕ (t)(s − t)2 We know that and. ∥v∥s ∥v∥t. √ ′′ √ ′′ 2 ⟨v,ϕ (s)v⟩ = √ ′′ = ϕϕ′′ (s)v , (t)v 2 ⟨v,ϕ (t)v⟩. 1 = = √ ′′ 1− ⟨s−t,ϕ (t)(s−t)⟩ √ ′′ 2 = ϕϕ′′ (s)v ≤ √ ′′1 (t)v 2. 1 1−∥s−t∥t. Thus,. ∥v∥s ∥v∥t. 1−. √. 1−. ϕ (t)(s−t)2. By(i)(ii), ϕ is strongly self-concordant.. 1 ′′. ϕ (t)(s−t)2. =. .. 1 . 1−∥s−t∥t. . Another direction is hard to prove, thus we find some examples to chcek that both SOC trace functions and the compounds of SOC trace functions are strongly self-concordant or not. Theorem 3.2. The following functions are all self-concordant. 8.

(13) (a) ϕ1 (t) = t2 for t > 0. 2 2 n (b) ϕtr 1 (x) = λ1 (x) + λ2 (x) for x ∈ intK .. (c) ϕ2 (t) = t3 for t ≥. √ 3 9 . 6. 3 3 n (d) ϕtr 2 (x) = λ1 (x) + λ2 (x) for x ∈ {x ∈ intK | x1 ≥. √ 3 9 }. 6. (e) ϕ3 (t) = − ln(t) for t > 0. n (f ) ϕtr 3 (x) = − ln(λ1 (x)) − ln(λ2 (x)) for x ∈ intK .. Proof. √ 3 3 ′ ′′ ′′′ ′′′ ′′ 2 2 (a) ϕ1 (t) = 2t, ϕ1 (t) = 2, ϕ√ 1 (t) = 0, |ϕ1 (t)| = 0, 2ϕ1 (t) = 2 · 2 = 4 2 3 ′′′ ′′ Thus, |ϕ1 (t)| = 0 ≤ 4 2 = 2ϕ1 (t) 2 , and hence ϕ1 is self-concordant. 2 2 2 2 2 2 (b) ϕtr 1 (x) = λ1 (x) + λ2 (x) = (x1 − ∥x2 ∥) + (x1 + ∥x2 ∥) [ = 2(x1T+ ∥x ] 2∥ ) 1 0n−1 T T 2 tr ∇ϕtr = 4In , 1 (x) = (4x1 , 4x2 ) = 4(x1 , x2 ) , ∇ ϕ1 (x) = 4 0n−1 In−1 T T 2 ∇2 ϕtr 1 (x)(h, h) = h (4In )h = 4h h = 4∥h∥ , 3 tr T 3 tr ∇ ϕ1 (x)(h, h) = 0n , ∇ ϕ1 (x)(h, h, h) = ⟨0n , h⟩ = 0, 3 tr | ∇3 ϕtr 1 (x) | = max {∇ ϕ1 (x)(h, h, h)} = max {0} = 0 ∥h∥=1. ∥h∥=1. 2 tr 2 | ∇2 ϕtr 1 (x) | = max {∇ ϕ1 (x)(h, h)} = max {4∥h∥ } = 4 ∥h∥=1. 3. ∥h∥=1. 3. 2 = 2 · 4 2 = 2 × 8 = 16. 2 | ∇2 ϕtr 1 (x) | 3 3 tr 2 Thus | ∇ ϕ1 (x) | = 0 ≤ 16 = 2 | ∇2 ϕtr 1 (x) | , and hence ϕtr 1 is self-concordant. ′. ′′. ′′′. (c) ϕ2 (t) = 3t2 , √ϕ2 (t) = 6t, ϕ2 (t) = 6, √ 3 3 3 3 ′′ ′′′ Since t ≥ 69 , 2ϕ2 (t) 2 = 2 · (6t) 2 ≥ 2 · ( 3 9) 2 = 2 · 3 = 6 = |ϕ2 (t)| Thus ϕ2 is self-concordant. 2 3 3 3 3 3 (d) ϕtr 1]− ∥x2 ∥) + (x1 + ∥x2 ∥) = 2(x1 + 3x1 ∥x2 ∥ ), 2 (x) = λ1 (x) [ +2λ2 (x) = (x 6x1 + 6∥x2 ∥2 , (x) = ∇ϕtr 2 12x1 x2. 9.

(14) [ ∇2 ϕtr 2 (x). =. 12x1 12xT2 12x2 12x1 In−1. ∇2 ϕtr 2 (x)(h, h). ]. [ = 12. x1 xT2 x2 x1 In−1. ]. ( [ ]) [ ] x1 xT2 h1 = 12 h1 x2 x1 In−1 h2 [ ] [ ] h1 T T T = 12 h1 x1 + h2 x2 h1 x2 + x1 h2 h2 [. hT2. ]. = 12(h21 x1 + h1 hT2 x2 + h1 xT2 h2 + x1 ∥h2 ∥2 ) [ ] = 12 x1 (h21 + ∥h2 ∥2 ) + 2h1 hT2 x2 ] 12(h21 + ∥h2 ∥2 ) , = 24h1 h2 ] [ ]⟩ ⟨[ 12(h21 + ∥h2 ∥2 ) h1 3 tr ∇ ϕ2 (x)(h, h, h) = , = 12h1 (h21 + ∥h2 ∥2 ) + 24h1 ∥h2 ∥2 24h1 h2 h2 [. ∇3 ϕtr 2 (x)(h, h). | ∇3 ϕtr 2 (x) |. = max {∇3 ϕtr 2 (x)(h, h, h)} ∥h∥=1. = max {12h1 (h21 + ∥h2 ∥2 ) + 24h1 ∥h2 ∥2 } ∥h∥=1. = max {12h1 + 24h1 (1 − h21 )} ∥h∥=1. = max {36h1 − 24h31 } ∥h∥=1 √ = 12 2 We can use the technique in Calculus to prove the last equality. ′ ′′ Let f (h1 ) = 36h1 − 24h31 , then f (h1 ) = 36 − 72h21 , and f (h1 ) = −144h1 ′ We know that the critical happened when f (h1 ) = 0, that is h1 = ± √12 . √ √ √ ′′ ′′ Since f ( √12 ) = −72 2 < 0, f (− √12 ) = 72 2 > 0, thus f ( √12 ) = 12 2 is a local √ maximum, and f (− √12 ) = −12 2 is a local minimum. √ Since ∥h∥ = 1, thus 0 ≤ h1 ≤ 1, and f(0)=0, f(1)=12. Thus f ( √12 ) = 12 2 is a global maximum, and hence the last equality holds.. | ∇2 ϕtr 2 (x) |. = max {∇2 ϕtr 2 (x)(h, h)} ∥h∥=1 ] [ = max {12 x1 (h21 + ∥h2 ∥2 ) + 2h1 hT2 x2 } ∥h∥=1. = max {12(x1 + 2h1 ∥h2 ∥∥x2 ∥)} ∥h∥=1. 1 1 = 12(x1 + 2 · √ · √ · ∥x2 ∥) 2 2 = 12(x1 + ∥x2 ∥) 10.

(15) ( 2 | ∇2 ϕtr 2 (x) |. 3 2. 3 2. 3 2. = 2 · [12(x1 + ∥x2 ∥)] ≥ 2 · (12x1 ) ≥ 2 · √ 3 = 2 · 2 2 · 3 = 12 2 = | ∇3 ϕtr 2 (x) |. ) 23 √ 3 9 12 · 6. Thus ϕtr 2 is self-concordant. 3 tr Remark: If we take the same direction h for ∇2 ϕtr 2 (x)(h, h) and ∇ ϕ2 (x)(h, h, h),then 3 3 tr 2 tr the result |∇ ϕ2 (x)(h,√h, h)|√≤ 2∇ ϕ2 (x)(h, h) 2 may be false. 3 3 For example, if x = ( 69 , − 69 , 0, . . . , 0), h = ( √12 , √12 , 0, . . . , 0),then. [√ ] √ 3 3 9 1 1 9 ∇2 ϕtr · 1 + 2 · √ · √ · (− ) =0 2 (x)(h, h) = 12 6 6 2 2 √ 1 1 1 24 √ √ √ ∇3 ϕtr (x)(h, h, h) = 12 · · 1 + 24 · · = = 12 2 2 2 2 2 2 3. 2 tr 2 thus |∇3 ϕtr 2 (x)(h, h, h)| ̸≤ 2∇ ϕ2 (x)(h, h) ′. ′′. ′′′. (e) ϕ3 (t) = − 1t , ϕ3 (t) = t12 , ϕ3 (t) = − t23 3 3 ′′′ ′′ For t > 0, | ϕ3 (t) | = t23 , 2ϕ3 (t) 2 = 2 · ( t12 ) 2 = 3 ′′′ ′′ Thus, | ϕ3 (t) | ≤ 2ϕ3 (t) 2 , and hence ϕ3 is self-concordant.. 2 t3. 2 2 (f ) ϕtr 3 (x) = −ln(λ1 (x))−ln(λ [ 2 (x))] = −ln(x1 −∥x2 ∥)−ln(x1 +∥x2 ∥) = −ln(x1 −∥x2 ∥ ), 2x1 1 , ∇ϕtr 3 (x) = − x21 −∥x2 ∥2 −2x2. ∇2 ϕtr 3 (x). [ [ ] ] [ ] 1 −1 2x1 2 0T = 2x1 −2x2 + 2 (x21 − ∥x2 ∥2 )2 −2x2 (x1 − ∥x2 ∥2 )2 0 2I [ 2 ] −1 4x1 − 2(x21 − ∥x2 ∥2 ) −4x1 xT2 = 2 −4x1 x2 4x2 xT2 + 2(x21 − ∥x2 ∥2 )I x1 − ∥x2 ∥2. 11.

(16) (. ∇2 ϕtr 3 (x)(h, h). = = =. = =. ∇3 ϕtr 3 (x)(h, h). [ 2 ]) 1 4x1 − 2(x21 − ∥x2 ∥2 ) −4x1 xT2 h h −4x1 x2 4x2 xT2 + 2(x21 − ∥x2 ∥2 )I (x21 − ∥x2 ∥2 )2 [ ] [ ] 1 4x21 h1 − 2(x21 − ∥x2 ∥2 )h1 − 4x1 xT2 h2 h1 h2 −4x1 h1 x2 + 4x2 xT2 h2 + 2(x21 − ∥x2 ∥2 )h2 (x21 − ∥x2 ∥2 )2 1 [4x2 h2 − 2(x21 − ∥x2 ∥2 )h21 − 8x1 h1 xT2 h2 + 4(xT2 h2 )2 2 (x1 − ∥x2 ∥2 )2 1 1 +2(x21 − ∥x2 ∥2 )∥h2 ∥2 ] 1 [4x2 h2 − 8x1 h1 xT2 h2 + 4(xT2 h2 )2 − 2(x21 − ∥x2 ∥2 )(h21 − ∥h2 ∥2 )] 2 (x1 − ∥x2 ∥2 )2 1 1 1 [4(x1 h1 − xT2 h2 )2 − 2(x21 − ∥x2 ∥2 )(h21 − ∥h2 ∥2 )] 2 2 2 (x1 − ∥x2 ∥ ) T. [ ] 2 2x1 T 2 2 2 2 2 = − 2 [4(x1 h1 − x2 h2 ) − 2(x1 − ∥x2 ∥ )(h1 − ∥h2 ∥ )] −2x2 (x1 − ∥x2 ∥2 )3 [ [ ] [ ]] 1 h1 x1 T 2 2 8(x1 h1 − x2 h2 ) + 2 − 2(h1 − ∥h2 ∥ ) −h2 −x2 (x1 − ∥x2 ∥2 )2 [ ] 1 2x1 2 2 2 2 T 2 [2(x1 − ∥x2 ∥ )(h1 − ∥h2 ∥ ) − 8(x1 h1 − x2 h2 ) ] = −2x2 (x21 − ∥x2 ∥2 )3 [ ] 1 h1 8(x1 h1 − xT2 h2 ) + 2 2 2 −h2 (x1 − ∥x2 ∥ ). 3 tr ∇3 ϕtr 3 (x)(h, h, h) = ⟨∇ ϕ3 (x)(h, h), h⟩ 2(x1 h1 − xT2 h2 ) = [2(x21 − ∥x2 ∥2 )(h21 − ∥h2 ∥2 ) − 8(x1 h1 − xT2 h2 )2 ] (x21 − ∥x2 ∥2 )3 8 + 2 (x1 h1 − xT2 h2 )(h21 − ∥h2 ∥2 ) (x1 − ∥x2 ∥2 )2 (x1 h1 − xT2 h2 ) [12(x21 − ∥x2 ∥2 )(h21 − ∥h2 ∥2 ) − 16(x1 h1 − xT2 h2 )2 ] = (x21 − ∥x2 ∥2 )3 3. 2 tr 2 We want to show |∇3 ϕtr 3 (x)| ≤ 2|∇ ϕ3 (x)| , so we may show that 3 2 2 tr 3 tr |∇ ϕ3 (x)(h, h, h)| ≤ 4[∇ ϕ3 (x)(h, h)]. Let α = x1 y1 − xT2 y2 , β = x21 − ∥x2 ∥2 , γ = h21 − ∥h2 ∥2 , then. 12.

(17) 3 3 tr 2 4[∇2 ϕtr 3 (x)(h, h)] − |∇ ϕ3 (x)(h, h, h)| =. 4 α2 2 3 (4α − 2βγ) − (12βγ − 16α2 )2 β6 β6. 1 [4(64α6 − 96α4 βγ + 48α2 β 2 γ 2 − 8β 3 γ 3 ) − α2 (256α4 − 384α2 βγ + 144β 2 γ 2 )] 6 β 1 = [(256α6 − 384α4 βγ + 192α2 β 2 γ 2 − 32β 3 γ 3 ) − (256α6 − 384α4 βγ + 144α2 β 2 γ 2 )] β6 16β 2 γ 2 1 2 2 2 3 3 (48α β γ − 32β γ ) = (3α2 − 2βγ) ≥ 0 = β6 β6. =. We know the last inequality holds from the following discussions: If h21 − ∥h2 ∥2 ≤ 0, then it’s clearly that the inequality holds. If h21 − ∥h2 ∥2 > 0, then |h1 | ≥ ∥h2 ∥, and hence |x1 h1 − xT2 h2 | ≥ |x1 h1 | − |xT2 h2 | ≥ x1 |h1 | − ∥x2 ∥∥h2 ∥ ≥ 0 ⇒ |x1 h1 − xT2 h2 |2 ≥ (x1 |h1 | − ∥x2 ∥∥h2 ∥)2 Thus, α2 − βγ = |x1 h1 − xT2 h2 |2 − (x21 − ∥x2 ∥2 )(h21 − ∥h2 ∥2 ) ≥ (x1 |h1 | − ∥x2 ∥∥h2 ∥)2 − (x21 − ∥x2 ∥2 )(h21 − ∥h2 ∥2 ) = (x21 ∥h2 ∥2 − 2x1 |h1 |∥x2 ∥∥h2 ∥ + ∥x2 ∥2 h21 ) = (x1 ∥h2 ∥ − ∥x2 ∥h1 )2 ≥ 0 Hence, 3α2 − 2βγ = α2 + 2(α2 − βγ) ≥ 0, and the inequality holds. Thus, 2 tr | ∇3 ϕtr 3 (x) | ≤ 2 | ∇ ϕ3 (x) |. 3 2. and hence ϕtr 3 is self-concordant. Theorem 3.3. The following functions are all not strongly self-concordant. (a) ϕ1 (t) = t2 for t > 0. n 2 2 (b) ϕtr 1 (x) = λ1 (x) + λ2 (x) for x ∈ intK .. (c) ϕ2 (t) = t3 for t ≥. √ 3 9 . 6. 13.

(18) 3 3 n (d) ϕtr 2 (x) = λ1 (x) + λ2 (x) for x ∈ {x ∈ intK | x1 ≥. √ 3 9 }. 6. Proof. ′′. ′′. (a) From Theorem3.2(a), ϕ1 (t) = 2, ∥s − t∥2t = ⟨s − t, ϕ1 (t)(s − t)⟩ = 2∥s − t∥2 Take t = 41 > 0, s = − 41 < 0, then ∥s − t∥2t = 2∥s − t∥2 = 2 · ∥(− 41 ) − 14 ∥2 = 2 · 14 = 21 < 1. Thus, there exists t ∈ Dϕ1 , s ∈ / Dϕ1 such that s ∈ Bt (t, 1), and hence ϕ1 is not strongly self-concordant. 2 2 2 (b) From Theorem3.2(b), ∇2 ϕtr 1 (x)(h, h) = 4∥h∥ , ∥y − x∥x = 4∥y − x∥ Take x = (1, 78 , 0, . . . , 0)T ∈ Dϕtr1 , y = (1, 98 , 0, . . . , 0)T ∈ / Dϕtr1 , then 1 ∥y − x∥2x = 4∥y − x∥2 = 4 · [(1 − 1)2 + ( 89 − 78 )2 ] = 4 · 16 <1 Thus, there exists x ∈ Dϕtr1 , y ∈ / Dϕtr1 such that y ∈ Bx (x, 1), tr and hence ϕ1 is not strongly self-concordant. ′′. ′′. (c) From Theorem3.2(c), ϕ2 (t) = 6t, ∥s √− t∥2t = ⟨s − t, ϕ2 (t)(s − t)⟩ = 6t∥s − t∥2 √ √ √ 3 3 3 3 Take t = 69 ≥ 69 , s = 69 − 31 < 69 , √ 2 1 then ∥s − t∥2t = 6t∥s − t∥2 = 3 9 · (− 31 )2 = 9− 3 = √ < 1. 3 81 Thus, there exists t ∈ Dϕ2 , s ∈ / Dϕ2 such that s ∈ Bt (t, 1), and hence ϕ2 is not strongly self-concordant. [ ] 2 2 T (d) From Theorem3.2(d), ∇2 ϕtr 2 (x)(h, h) = 12 x1 (h1 + ∥h2 ∥ ) + 2h1 h2 x2 , 2 T ∥y − x∥2x = √12{x1 [(y1 − x1 )2 + ∥y2 − x√ 2 ∥ ] + 2(y1 − x1 )(y2 − x2 ) x2 } 3 3 / Dϕtr2 , Take x = ( 69 , 0, . . . , 0)T ∈ Dϕtr2 , y = ( 69 − 16 , 0, . . . , 0)T ∈ √ √ √ [ 1 2 ] 3 3 9 12 3 9 2 2 then ∥y − x∥x = 12 · { 6 · (− 6 ) + 0 + 0} = 216 = 189 < 1 Thus, there exists x ∈ Dϕtr2 , y ∈ / Dϕtr2 such that y ∈ Bx (x, 1), tr and hence ϕ2 is not strongly self-concordant. Theorem 3.4. The function ϕ3 (t) = − ln(t) for t > 0 is strongly self-concordant. ′′. Proof. From Theorem 3.2(f), ϕ3 (t) =. 1 , t2. ′′. ∥s − t∥2t = ⟨s − t, ϕ3 (t)(s − t)⟩ =. (i) Let t > 0, ∀s ∈ Bt (t, 1), that is ∥s − t∥2t < 1 ⇒ t12 ∥s − t∥2 < 1 ⇒ (s − t)2 < t2 ⇒ −t < s − t < t ⇒ 0 < s < 2t Thus, s ∈ Dϕ3 , and hence Bt (t, 1) ⊆ Dϕ3. 14. 1 ∥s t2. − t∥2.

(19) (ii) Let t > 0, s ∈ Bt (t, 1), v ̸= 0 Since ∥s − t∥2t = t12 ∥s − t∥2 , thus ∥s − t∥t = 1t |s − t|, and∥v∥t = 1t |v| Since s ∈ Bt (t, 1), thus 1t |s − t| < 1, and |s − t| < t ∥v∥s = ∥v∥t. 1 · |v| s 1 · |v| t ∞ ∑. t s s − t −1 t − s −1 = ( )−1 = (1 + ) = (1 − ) s t t t ∞ t−s k ∑ 1 = ( ) ≤ ∥s − t∥kt = t 1 − ∥s − t∥t k=0 k=0 =. Thus, the rightmost inequality on is proven. By(i)(ii), ϕ3 is strongly self-concordant.. ∥v∥s ∥v∥t. in the definition of strongly self-concordancy . n Theorem 3.5. The function ϕtr 3 (x) = − ln(λ1 (x)) − ln(λ2 (x)) for x ∈ intK is half strongly self-concordant, this means ϕtr 3 satisfying the first condition of strongly selfconcordancy.. Proof. (i) Let x ∈ Dϕ . ∀ y ∈ Bx (x, 1). Let λ1 = x1 − ∥x2 ∥, λ2 = x1 + ∥x2 ∥, µ1 = y1 − ∥y2 ∥, µ2 = y1 + ∥y2 ∥. From Theorem 3.2(f), 1 [4(x1 h1 − xT2 h2 )2 − 2(x21 − ∥x2 ∥2 )(h21 − ∥h2 ∥2 )] − ∥x2 ∥2 )2 1 = [4(x1 h1 − xT2 h2 )2 − 2(x21 − ∥x2 ∥2 )(h21 − ∥h2 ∥2 )] (λ1 λ2 )2. ∇2 ϕtr 3 (x)(h, h) =. (x21. 15.

(20) ∥y − x∥2x = ∇2 ϕtr 3 (x)(y − x, y − x) 4 2 = [x1 (y1 − x1 ) − xT2 (y2 − x2 )]2 − [(y1 − x1 )2 − (∥x2 ∥ − ∥y2 ∥)2 ] 2 (λ1 λ2 ) λ1 λ2 4 = [x1 y1 − xT2 y2 − (x21 − ∥x2 ∥2 )]2 (λ1 λ2 )2 2 − [(y12 − ∥y2 ∥2 ) + (x21 − ∥x2 ∥2 ) − 2(x1 y1 − xT2 y2 )] λ1 λ2 4 2 = [x1 y1 − xT2 y2 − (λ1 λ2 )]2 − [(µ1 µ2 ) + (λ1 λ2 ) − 2(x1 y1 − xT2 y2 )] 2 (λ1 λ2 ) λ1 λ2 1 = [4(x1 y1 )2 + 4(xT2 y2 )2 + 4(λ1 λ2 )2 − 8x1 y1 xT2 y2 − 8x1 y1 λ1 λ2 2 (λ1 λ2 ) +8xT2 y2 λ1 λ2 − 2λ1 λ2 µ1 µ2 − 2(λ1 λ2 )2 + 4x1 y1 λ1 λ2 − 4xT2 y2 λ1 λ2 ] 1 = [2(λ1 λ2 )2 − 4x1 y1 λ1 λ2 + 4xT2 y2 λ1 λ2 − 2λ1 λ2 µ1 µ2 (λ1 λ2 )2 +4(x1 y1 )2 − 8x1 y1 xT2 y2 + 4(xT2 y2 )2 ] 1 = [2(λ1 λ2 )2 − 4λ1 λ2 (x1 y1 − xT2 y2 ) + 4(x1 y1 − xT2 y2 )2 − 2λ1 λ2 µ1 µ2 ] (λ1 λ2 )2 1 = {(λ1 λ2 )2 + [λ1 λ2 − 2(x1 y1 − xT2 y2 )]2 − 2λ1 λ2 µ1 µ2 } 2 (λ1 λ2 ) Since ∥y − x∥x < 1, then ∥y − x∥2x < 1 1 {(λ1 λ2 )2 + [λ1 λ2 − 2(x1 y1 − xT2 y2 )]2 − 2λ1 λ2 µ1 µ2 } < 1 (λ1 λ2 )2 ⇒ [λ1 λ2 − 2(x1 y1 − xT2 y2 )]2 < 2λ1 λ2 µ1 µ2. ⇒. Since [λ1 λ2 − 2(x1 y1 − xT2 y2 )]2 ≥ 0 and λ1 λ2 > 0, thus µ1 µ2 > 0 Assume µ1 < 0 and µ2 < 0, we have λ1 λ2 − 2(x1 y1 − xT2 y2 ) ≥ λ1 λ2 − 2x1 y1 − 2∥x2 ∥∥y2 ∥ λ2 + λ1 µ2 + µ1 λ2 − λ1 µ2 − µ1 −2· = λ1 λ2 − 2 · 2 2 2 2 = λ1 λ2 − λ2 µ2 − λ1 µ1 > 0 Thus,(λ1 λ2 − λ2 µ2 − λ1 µ1 )2 ≤ (λ1 λ2 − 2x1 y1 + xT2 y2 )2 ≤ 2λ2 µ2 λ1 µ1 ⇒ (λ1 λ2 )2 + (λ2µ2 )2 + (λ1 µ1 )2 − 2λ1 λ2 (λ2 µ2 + λ1 µ1 ) + 2λ2 µ2 λ1 µ1 < 2λ2 µ2 λ1 µ1 ⇒ 0 < (λ1 λ2 )2 + (λ2µ2 )2 + (λ1 µ1 )2 < 2λ1 λ2 (λ2 µ2 + λ1 µ1 ) < 0 We get a contradiction, thus µ1 >, µ2 > 0, this means y ∈ Dϕtr3 , and hence Bx (x, 1) ⊆ Dϕtr3 16.

(21) (ii) Need a proof. . 4. In the future. In this paper, we show the followings: n (1) The function ϕtr 3 (x) = − ln(λ1 (x)) − ln(λ2 (x)) for x ∈ intK is half strongly selfconcordant.. (2) For any given scalar funtion ϕ : Dϕ ⊆ R → R, let ϕtr : Dϕtr ⊆ Kn → R be given by (6) where Dϕ = {t | t > 0} and Dϕtr = intKn . Then ϕtr is strongly self-concordant ⇒ ϕ is strongly self-concordant. In the future, we may try to achieve the proofs of the following two guesses: n (1) The function ϕtr 3 (x) = − ln(λ1 (x))−ln(λ2 (x)) for x ∈ intK is strongly self-concordant.. (2) For any given scalar funtion ϕ : Dϕ ⊆ R → R, let ϕtr : Dϕtr ⊆ Kn → R be given by (6) where Dϕ = {t | t > 0} and Dϕtr = intKn . Then ϕ is strongly self-concordant ⇔ ϕtr is strongly self-concordant.. References [1] Y.-L. Chang and J.-S. Chen Convexity of symmetric cone trace functions in Euclidean Jordan Algebra, Journal of Nonlinear and Convex Analysis , vol. 14, no.1, pp. 53-61, 2013. [2] J.-S. Chen, T.-K Liao and S.-H. Pan Using Schur Complement Theorem to prove convexity of some SOC-functions, Journal of Nonlinear and Convex Analysis , vol. 13, no.3, pp. 421-431, 2012. [3] J. Renegar A Mathematical View of Interior-Point Methods in Convex Optimization, MPS-SIAM Series on Optimization, Philadelphia, PA, USA, 2001. [4] M. Fukushima, Z.-Q. Luo and P. Tseng, Smoothing functions for second-order cone complementarity problems, SIAM Journal on Optimization, vol. 12, pp. 436-460, 2002. ´ nyi, Analysis on symmetric Cones, Oxford Mathematical [5] J. Faraut and A. Kora Monographs, Oxford University Press, New York, 1994. 17.

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