圓錐上的軌跡函數的凸性

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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 Convexity of Circular Cone Trace Functions. 研究生：吳荻文. 中華民國一零四年. 七月.

(2) 謝詞本篇論文的完成感謝張毓麟老師的指導論文與莊國裕同學的幫助，給我許多寶貴的建議與指導，使能順利的完成。以及感謝陳界山老師、林延輯老師等大數學系教導過我的老師們，使我具備了完成本篇論文的能力。然後感謝這個從大學到研究所待了八年的師大分部與當中認識的同學們，為我充實了這八年。.

(3) 中文摘要在Using Schur Complement Theorem to prove convexity of some SOC-functions這篇論文中，證明了在二階錐(Second order cone)上由凸函數(Convex function)所生成的軌跡函數(Trace function)亦為凸函數，我們試著將這個結果擴展到所有圓錐(Circular cone)上。但是，根據圓錐角度的不同，一個凸函數所生成的軌跡函數不一定會是凸函數。在本篇論文中找出一組充份條件，使凸函數所生成的軌跡函數會是凸函數。同時也給出幾個不滿足條件下軌跡函數不是凸函數的情況。這組條件是：圓錐角度大於45 ○且函數同時是遞增函數，或圓錐角度小於45 ○且函數同時是遞減函數。.

(4) INDEX 1.Introduction 2.The Monotone Conditions 3.Counterexamples Out of Conditions Refrerece. 1. 1 2 11 14.

(5) THE CONVEXITY OF CIRCULAR CONE TRACE FUNCTIONS Ti-Wen Wu,Yu-Lin Chang July 5, 2015 Abstract: Uusing the Schur complement theorem[1][2] to find the sufficient conditions which make the Hessian of circular cone trace function positive semi-definite or positive definite, and proved that these monotone conditions can make the trace function convex or strictly convex. After that, some examples are given to explain why we have to satisfy these monotone conditions.. Keywords: Circular cone, Trace, Convexity, Monotone, Schur. 1. Introduction. To find a convex function in the high dimensional subspace, there is a method in Using Schur Complement Theorem to prove convexity of some SOC-functions[2] which make a convex function is second order cone. It proves that in second order cone the trace function of convex function is also convex. Now we want extent this to circular cone, at first we have to know which are the trace functions of circular cone. From[3][4][5][6], the circular cone is defined Lθ := {(x1 , x2 ) ∈ R × Rn−1 | x1 ≥ ||x|| cos θ}. (1.1). := {(x1 , x2 ) ∈ R × Rn−1 | x1 ≥ ||x2 || cot θ} From the spectral factorization[5][4] associated with circular cone,the λ1 , λ2 is: λ1 (x) = x1 − ||x2 || cot θ. (1.2). λ2 (x) = x1 + ||x2 || tan θ Suppose f : R → R is a twice differentiable function , define the trace fonction: f tr (x) := f (λ1 (x)) + f (λ2 (x)) If θ 6=. (1.3). π and f is not constant, then in the set 4 E := {(x1 , 0) ∈ R × Rn−1 } ∈ Lθ. (1.4). f tr is not differentiable. But the gradient of f tr still exist in Lθ \ E, " tr. 0. ∇f (x) = f (λ1 (x)). 1 x2 − cot θ ||x2 ||. 1. #. " 0. + f (λ2 (x)). 1 x2 tan θ ||x2 ||. # (1.5).

(6) also, Hessian is exist in Lθ \ E,  b(x)  ∇2 f tr (x) =  . xT c(x) 2 ||x2 ||. .   x2 x2 xT2  c(x) a(x)I + (d(x) − a(x)) ||x2 || ||x2 ||2 0 0 f (λ2 (x)) tan θ − f (λ1 (x)) cot θ a(x) = ||x2 ||. (1.6). b(x) = f 00 (λ1 (x)) + f 00 (λ2 (x)) c(x) = f 00 (λ2 (x)) tan θ − f 00 (λ1 (x)) cot θ d(x) = f 00 (λ1 (x)) cot2 θ + f 00 (λ2 (x)) tan2 θ. The following proposition and theorem will help us prove the theorems in section 2. Proposition 1.1. Suppose f : R → R is a strictly convex function, • if f is increasing, then f is strictly is increasing. • if f is decreasing, then f is strictly is decreasing. Proof. If f is a is increasing function but not strictly, then there are x, y ∈ R such taht y > x and f (x) = f (y). Suppose 0 < λ < 1, f (x) = λf (x) + (1 − λ)f (y) > f (λx + (1 − λ)y). x < λx + (1 − λ)y and f (x) > f (λx + (1 − λ)y), f is a not is increasing is contradict that f is a is increasing, so f has to strictly is increasing. If f is a is decreasing function but not strictly, then there are x, y ∈ R such taht y > x and f (x) = f (y). Suppose 0 < λ < 1, f (y) = λf (x) + (1 − λ)f (y) > f (λx + (1 − λ)y). λx + (1 − λ)y < y and f (λx + (1 − λ)y) < f (y), f is a not is decreasing is contradict that f is a is decreasing, so f has to strictly is decreasing. A B Theorem 1.1 (Schur Complement Theorem[1][2]). Suppose that a matrix is partitioned as Where BT C A and C are square. This matrix is positive (semi-)definite if and only if A is positive (semi-)definite and C − B T A−1 B positive (semi-)definite.. 2. The Monotone Conditions. In this sectoin, we will find sufficient conditions which imply f tr convex. The first step is: Theorem 2.1. Suppose f : R → R is a twice differentiable convex function, in Lθ \ E, • if b(x) = 0 (Eq.1.6), then ∇2 f tr (x) O ⇐⇒ a(x)(I −. 2. x2 xT2 ) O. ||x2 ||2.

(7) • if b(x) > 0, then (c(x))2 x2 xT2 x2 xT2 − O ||x2 ||2 b(x) ||x2 ||2 x2 xT2 (c(x))2 x2 xT2 ∇2 f tr (x) O ⇐⇒ a(x)I + (d(x) − a(x)) − O ||x2 ||2 b(x) ||x2 ||2. ∇2 f tr (x) O ⇐⇒ a(x)I + (d(x) − a(x)). Proof. f (x) is convex function, so f 00 (x) ≥ 0 ∀x ∈ R. b(x) = f 00 (λ1 (x)) + f 00 (λ2 (x)). If b(x) = 0, then f 00 (λ1 (x)) = f 00 (λ2 (x)) = 0. c(x) = f 00 (λ2 (x)) tan θ − f 00 (λ1 (x)) cot θ, so if b(x) = 0, c(x) = 0. d(x) = f 00 (λ1 (x)) cot2 θ + f 00 (λ2 (x)) tan2 θ, so if b(x) = 0, d(x) = 0. If b(x) = 0, " # 0 0 ∇f tr = . x xT 0 a(x)(I − ||x22 ||22 ) Suppose v = (v1 , v2 ) ∈ R × Rn−1 , v T ∇f tr v = v2T a(x)(I − v T ∇f tr v ≥ 0 if and only if v2T a(x)(I −. x2 xT 2 ||x2 ||2 )v2. x2 xT2 )v2 . ||x2 ||2. ≥ 0, so. ∇2 f tr (x) O ⇐⇒ a(x)(I −. x2 xT2 ) O. ||x2 ||2. x xT. If b(x) > 0, b(x) be positive definite and a(x)I + (d(x) − a(x)) ||x22 ||22 is symmetric. Use the Schur Complement Theorem (Thm. 1.1), we have that (c(x))2 x2 xT2 x2 xT2 − O 2 ||x2 || b(x) ||x2 ||2 (c(x))2 x2 xT2 x2 xT2 − O ∇2 f tr (x) O ⇐⇒ a(x)I + (d(x) − a(x)) 2 ||x2 || b(x) ||x2 ||2. ∇2 f tr (x) O ⇐⇒ a(x)I + (d(x) − a(x)). The second step is: Theorem 2.2. Suppose f : R → R is a twice differentiable convex function, if f (x) is increasing and π π π θ tr is positive semi-definite. 2 > θ ≥ 4 , or if f (x) is decreasing and 4 ≥ θ > 0, then in L \ E, ∇f Proof. If b(x) = 0, use Thm. 2.1, ∇2 f tr (x) O ⇐⇒ a(x)(I − suppose v ∈ Rn−1 v T a(x)(I − ||v|| · ||x2 || ≥< v, x2 >, so ||v||2 −. x2 xT2 ) O. ||x2 ||2. x2 xT2 < v, x2 >2 2 )v = a(x)(||v|| − ). ||x2 ||2 ||x2 ||2. <v,x2 >2 ||x2 ||2. ≥ 0. Then. ∇2 f tr (x) O ⇐⇒ a(x) ≥ 0. 3.

(8) if b(x) > 0, use Thm. 2.1, ∇2 f tr (x) O ⇐⇒ a(x)I + (d(x) − a(x)). x2 xT2 (c(x))2 x2 xT2 − O. ||x2 ||2 b(x) ||x2 ||2. (c(x))2 x2 xT2 x2 xT2 − ||x2 ||2 b(x) ||x2 ||2 x2 xT2 b(x)d(x) − (c(x))2 x2 xT2 = a(x)(I − )+ 2 ||x2 || b(x) ||x2 ||2. a(x)I + (d(x) − a(x)). ∀v ∈ Rn−1 (c(x))2 x2 xT2 x2 xT2 − )v ||x2 ||2 b(x) ||x2 ||2 x2 xT2 b(x)d(x) − (c(x))2 T x2 xT2 = a(x)v T (I − )v + v v ||x2 ||2 b(x) ||x2 ||2. v T (a(x)I + (d(x) − a(x)). if ∇2 f tr (x) O, then ∀v ∈ Rn−1 , a(x)v T (I −. x2 xT2 b(x)d(x) − (c(x))2 T x2 xT2 )v + v v ≥ 0. ||x2 ||2 b(x) ||x2 ||2. Now suppose v ∈ {x ∈ Rn−1 |x = rx2 , r ∈ R} a(x)rxT2 (I −. x2 xT2 b(x)d(x) − (c(x))2 T x2 xT2 b(x)d(x) − (c(x))2 2 )rx + rx rx = r ||x2 ||2 , . 2 2 2 ||x2 ||2 b(x) ||x2 ||2 b(x). and suppose v ∈ {x ∈ Rn−1 | < v · x2 >= 0} a(x)v T (I −. x2 xT2 b(x)d(x) − (c(x))2 T x2 xT2 )v + v v = a(x)||v||2 . 2 ||x2 || b(x) ||x2 ||2. Rn−1 = {x ∈ Rn−1 |x = rx2 , r ∈ R} + {x ∈ Rn−1 | < v · x2 >= 0}, so ∀v ∈ Rn−1 , v = va + vb such that va ∈ {x ∈ Rn−1 |x = rx2 , r ∈ R} and vb ∈ {x ∈ Rn−1 | < v · x2 >= 0}. Then a(x)v T (I −. x2 xT2 b(x)d(x) − (c(x))2 T x2 xT2 b(x)d(x) − (c(x))2 )v + v v = a(x)||va ||2 + ||vb ||2 . 2 2 ||x2 || b(x) ||x2 || b(x). By b(x)d(x) − (c(x))2 f 00 (λ1 (x))f 00 (λ2 (x))(cot θ + tan θ) = ≥ 0. b(x) f 00 (λ1 (x)) + f 00 (λ2 (x) we have. b(x)d(x) − (c(x))2 ||vb ||2 ≥ 0. b(x). Then if a(x) ≥ 0 , ∇2 f tr (x) O. and if ∇2 f tr (x) O, suppose vb = 0, a(x)||va ||2 ≥ 0 ,It is that a(x) ≥ 0. Theat is ∇2 f tr (x) O ⇐⇒ a(x) ≥ 0.. f is convex, so f 00 (i) ≥ 0 ∀ i ∈ R, it means that f 0 is increasing. If x ∈ Lθ \ E, λ2 (x) > λ1 (x), then f 0 (λ2 (x)) ≥ f 0 (λ1 (x)). 4.

(9) So, if f 0 (λ1 (x)) = 0 a(x) = if f 0 (λ1 (x)) 6= 0 a(x) =. f 0 (λ2 (x)) tan θ ≥ 0, . ||x2 ||. f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) ( 0 − cot2 θ) ||x|| f (λ1 (x)). (2.1). Then, if f 0 (λ1 (x)) < 0, f 0 (λ2 (x)) > 0, f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) < 0, 0 ≤ 0. ||x|| f (λ1 (x)) if f 0 (λ1 (x)) < 0, f 0 (λ2 (x)) < 0, ,. f 0 (λ2 (x)) f 0 (λ1 (x)) tan θ < 0, 0 ≤ 1. ||x|| f (λ1 (x)). if f 0 (λ1 (x)) > 0, f 0 (λ2 (x)) > 0. f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) > 0, 0 ≥ 1. ||x|| f (λ1 (x)) Suppose f (x) is increasing and. π 2. >θ≥. π 4,. f 0 (λ1 (x)) ≥ 0 and cot2 θ < 1. then, if f (λ1 (x)) = 0 a(x) ≥ 0. 0. if f (λ1 (x)) > 0 f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) > 0, ( 0 − cot2 θ) ≥ 0 ||x|| f (λ1 (x)) f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) a(x) = > 0( 0 − cot2 θ) ≥ 0 ||x|| f (λ1 (x)) Suppose f (x) is decreasing and. π 4. ≥ θ > 0, f 0 (λ2 (x)) ≥ 0 and cot2 θ > 1.. then, if f (λ1 (x)) = 0, 0 ≥ f (λ2 (x)) ≥ f (λ1 (x)) = 0 a(x) = 0. if f 0 (λ2 (x)) < 0 f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) < 0, ( 0 − cot2 θ) ≤ 0 ||x|| f (λ1 (x)) f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) a(x) = > 0( 0 − cot2 θ) ≥ 0 ||x|| f (λ1 (x)) then, in Lθ \ E, if f (x) is increasing and positive semi-definite.. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥ θ > 0, ∇f tr is. Corollary 2.3. Suppose f : R → R is a twice differentiable function which f 00 (x) > 0 ∀x ∈ R, if f (x) is increasing and π2 > θ ≥ π4 , or if f (x) is decreasing and π4 ≥ θ > 0, then in Lθ \ E, ∇f tr is positive definite. 5.

(10) Proof. If f 00 > 0 then b(x) > 0 and. b(x)d(x) − (c(x))2 f 00 (λ1 (x))f 00 (λ2 (x))(cot θ + tan θ) = > 0. b(x) f 00 (λ1 (x)) + f 00 (λ2 (x) x xT. 2. x xT. 2 2 n−1 ∇2 f tr (x) O ⇐⇒ a(x)I + (d(x) − a(x)) ||x22 ||22 − (c(x)) , v = va + vb such b(x) ||x2 ||2 O. and ∀v ∈ R n−1 n−1 that va ∈ {x ∈ R |x = rx2 , r ∈ R} and vb ∈ {x ∈ R | < v · x2 >= 0}.. a(x)v T (I −. x2 xT2 b(x)d(x) − (c(x))2 T x2 xT2 b(x)d(x) − (c(x))2 )v + v = a(x)||va ||2 + v ||vb ||2 . 2 2 ||x2 || b(x) ||x2 || b(x). If v 6= 0 then at least one of va , vb is not 0. That is ∇2 f tr (x) O ⇐⇒ a(x) > 0. f 00 (x) > 0∀x ∈ R, so f 0 is strictly is increasing, and λ2 > λ1 when x ∈ Lθ \ E, then f 0 (λ2 (x)) > f 0 (λ1 (x)). Suppose f (x) is increasing and so. π 2. >θ≥. π 4,. by Prop.1.1 f (x) is strictly is increasing, so f 0 (x) 6= 0 ∀x ∈ R,. f 0 (λ2 (x)) > f 0 (λ1 (x)) > 0 and cot2 θ < 1. f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) > 0, ( 0 − cot2 θ) > 0 ||x|| f (λ1 (x)) f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) a(x) = ( 0 − cot2 θ) > 0 ||x|| f (λ1 (x)) Suppose f (x) is decreasing and so. π 4. ≥ θ > 0, by Prop.1.1 f (x) is strictly is decreasing, so f 0 (x) 6= 0 ∀x ∈ R,. f 0 (λ1 (x)) < f 0 (λ2 (x)) < 0 and cot2 θ > 1. Then, f 0 (λ2 (x)) f 0 (λ1 (x)) tan θ < 0, ( 0 − cot2 θ) < 0 ||x|| f (λ1 (x)) f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) a(x) = ( 0 − cot2 θ) > 0 ||x|| f (λ1 (x)) Then, in Lθ \ E, if f (x) is increasing and positive definite.. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥ θ > 0, ∇f tr is. ]The third step is: Theorem 2.4. Suppose f : R → R is a twice differentiable convex function, f tr is convex if f (x) is increasing and π2 > θ ≥ π4 , or if f (x) is decreasing and π4 ≥ θ > 0. Proof. To proof f tr is convex, we have to proof by 4 part : Part 1. Suppose [x, y] ⊂ Lθ \ E, 0 ≤ λ ≤ 1, define a function g(λ) = f tr (y + λ(x − y)) g 00 (λ) = (x − y)T ∇2 f tr (y + λ(x − y))(x − y). 6. (2.2).

(11) By the Thm. 2.2, if f (x) is increasing and π2 > θ ≥ π4 , or if f (x) is decreasing and positive definite. So, g 00 (λ) ≥ 0 if 0 ≤ λ ≤ 1. It means that if 0 < λ < 1. π 4. ≥ θ > 0,∇2 f tr is. λf tr (x) + (1 − λ)f tr (y) = λg(1) + (1 − λ)g(0) ≥ g(λ) = f tr (λx + (1 − λ)y). Part 2. Suppose x ∈ Lθ \ E y ∈ E, 1 > t > 0 define the u(t) = y + t(x − y). f tr is continuous, so lim f tr (u(t)) = f tr (y).. t→0. [x, u(t)] ⊂ Lθ \ E, by part 1, if f (x) is increasing and. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥ θ > 0,. λf tr (x) + (1 − λ)f tr (y) = lim λf tr (x) + (1 − λ)f tr (u(t)) t→0. tr. ≥ lim f (λx + (1 − λ)u(t)) = f tr (λx + (1 − λ)y) t→0. Part 3. Suppose x, y ∈ Lθ \ E, and there is a t0 such that 0 < t0 < 1 and t0 x + (1 − t0 )y ∈ E. x = (x1 , x2 ), y = (y1 , y2 ), so t0 x + (1 − t0 )y = (t0 x1 + (1 − t0 )y1 , 0) ,. −t0 t0 x2 = y2 , then ||x2 || = ||y2 ||. 1 − t0 1 − t0. 1>λ>0 λf tr (x) + (1 − λ)f tr (y) = λf (λ1 (x)) + f (λ2 (x)) + (1 − λ)f (λ1 (y)) + f (λ2 (y)) = λf (λ1 (x)) + (1 − λ)f (λ2 (y)) + λf (λ2 (x)) + (1 − λ)f (λ1 (y)) f is convex, so λf (λ1 (x)) + (1 − λ)f (λ2 (y)) + λf (λ2 (x)) + (1 − λ)f (λ1 (y)) ≥ f (λλ1 (x) + (1 − λ)λ2 (y)) + f (λλ2 (x) + (1 − λ)λ1 (y)) f (λλ1 (x) + (1 − λ)λ2 (y)) + f (λλ2 (x) + (1 − λ)λ1 (y) = f (λx1 + (1 − λ)y1 + λ(−||x2 || cot θ) + (1 − λ)||y2 || tan θ) + f (λx1 + (1 − λ)y1 + λ||x2 || tan θ + (1 − λ)(−||y2 || cot θ) If λ = t0 f (t0 λ1 (x) + (1 − t0 )λ2 (y)) + f (t0 λ2 (x) + (1 − t0 )λ1 (y) = f (t0 x1 + (1 − t0 )y1 + t0 (−||x2 || cot θ) + (1 − t0 )||y2 || tan θ) + f (t0 x1 + (1 − t0 )y1 + t0 ||x2 || tan θ + (1 − t0 )(−||y2 || cot θ). = 2f (t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ)) then t0 f tr (x) + (1 − t0 )f tr (y) ≥ 2f (t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ)). 7. (2.3).

(12) Figure 1: λ > t0. Suppose. π 2. >θ≥. Figure 2: λ < t0. π 4.. t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ) ≥ t0 x1 + (1 − t0 )y1 . so if f (x) is increasing, 2f (t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ)) ≥ 2f (t0 x1 + (1 − t0 )y1 ). Suppose. π 4. ≥ θ > 0. t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ) ≤ t0 x1 + (1 − t0 )y1 .. so if f (x) is decreasing, 2f (t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ)) ≥ 2f (t0 x1 + (1 − t0 )y1 ). Then if f (x) is increasing and. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥θ>0. t0 f tr (x) + (1 − t0 )f tr (y) ≥ 2f (t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ)) ≥ 2f (t0 x1 + (1 − t0 )y1 ) = f tr (t0 x + (1 − t0 )y). Let z0 = f tr (t0 x + (1 − t0 )y. If 1 > λ > t0 , λx + (1 − λ)y ∈ [x, z0 ], there is a µ =. λ−t such that 1−t. λx + (1 − λ)y = µx + (1 − µ)z0 . By part 2, if f (x) is increasing and. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥ θ > 0,. λf tr (x) + (1 − λ)f tr (y) = µf tr (x) + (1 − µ)(t0 f tr (x) + (1 − t0 )f tr (y)) ≥ µf tr (x) + (1 − µ)f tr (z0 ) ≥ f tr (λx + (1 − λ)y)(fig:1) 8.

(13) If 0 < λ < t0 , λx + (1 − λ)y ∈ [z0 , y], there is a γ =. λ such that t0. λx + (1 − λ)y = γz0 + (1 − γ)y. By part 2, if f (x) is increasing and. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥ θ > 0,. λf tr (x) + (1 − λ)f tr (y) = γ(t0 f tr (x) + (1 − t0 )f tr (y)) + (1 − γ)f tr (y) ≥ γ(f tr (z0 ) + (1 − γ)f tr (y) ≥ f tr (λx + (1 − λ)y)(fig:2). Part 4. If i ∈ E, i = (i1 , 0), then λ1 (i) = i1 − ||0|| cot θ = i1 , λ2 (i) = i1 + ||0|| tan θ = i1 f tr (x) = f (λ1 (i)) + f (λ2 (i)) = 2f (i1 ). Suppose x, y ∈ E, 0 < λ < 1 λx + (1 − λ)y = (λx1 + (1 − λ)y1 , 0) f tr (x) = 2f (x1 ), f tr (y) = 2f (y1 ), f tr (λx + (1 − λ)y) = 2f (λx1 + (1 − λ)y1 ) f is convex, λf tr (x) + (1 − λ)f tr (y) = λ2f (x1 ) + (1 − λ)2f (y1 ) ≥ 2f (λx1 + (1 − λ)y1 ) = f tr (λx + (1 − λ)y). By these 4 part, f tr is convex in Lθ . Corollary 2.5. Suppose f : R → R is a twice differentiable function which f 00 (x) > 0 ∀x ∈ R, if f (x) is increasing and π2 > θ ≥ π4 , or if f (x) is decreasing and π4 ≥ θ > 0, then f tr is strictly convex. Proof. To proof f tr is strictly convex, we have to proof by 4 part : Part 1. Suppose [x, y] ⊂ Lθ \ E, 0 ≤ λ ≤ 1,Use Eq. 2.2 g 00 (λ) = (x − y)T ∇2 f tr (y + λ(x − y))(x − y). By the Cor. 2.3, if f (x) is increasing and π2 > θ ≥ π4 , or if f (x) is decreasing and positive definite. So, g 00 (λ) > 0 if 0 ≤ λ ≤ 1. It means that if 0 < λ < 1. π 4. ≥ θ > 0,∇2 f tr is. λf tr (x) + (1 − λ)f tr (y) = λg(1) + (1 − λ)g(0) > g(λ) = f tr (λx + (1 − λ)y). Part 2. Suppose x ∈ Lθ \ E y ∈ E, 1 > t > 0, if f 00 (x) > 0 ∀x ∈ R, f is strictly convex. By part 2 of the proof of Thm.2.4, if f (x) is increasing and π2 > θ ≥ π4 , or if f (x) is decreasing and π4 ≥ θ > 0, λf tr (x) + (1 − λ)f tr (y) ≥ f tr (λx + (1 − λ)y). Take the u(t) in part 2 of the proof of Thm.2.4. By part 1, if 1 > t > 0 λf tr (x) + (1 − λ)f tr (u(t)) > f tr (λx + (1 − λ)u(t)). 9.

(14) Now suppose there is a 0 < λt < 1, there is a 0 < λs < λt such that u(λt ) ∈ [x, u(λs )].We know ∃ λs − λt α= that 1 − λt 0 < α < 1 and αx + (1 − α)u(λs ) = u(λt ). αf tr (x) + (1 − α)f tr (u(λs )) > f tr (αx + (1 − α)u(λs )) = f tr (u(λt )). u(λs ) = λs x + (1 − λs )y ∈ [x, y], so λt f tr (x) + (1 − λt )f tr (y) = αf tr (x) + (1 − α)λs f tr (x) + (1 − λs )f tr (y) ≥ λt f tr (x) + (1 − λt )f tr (u(λs )) > f tr (λt x + (1 − λt )y). Part 3. Suppose x, y ∈ Lθ \ E, and there is a t0 such that 0 < t0 < 1 and t0 x + (1 − t0 )y ∈ E. f 00 (x) > 0 ∀x ∈ R, so t0 f tr (x) + (1 − t0 )f tr (y) > 2f (t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ)) (Eq.2.3). Then if f (x) is increasing and. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥ θ > 0,. t0 f tr (x) + (1 − t0 )f tr (y) > 2f (t0 x1 + (1 − t0 )y1 + t||x2 ||(tan θ − cot θ)) ≥ 2f (t0 x1 + (1 − t0 )y1 ) = f tr (t0 x + (1 − t0 )y) If 1 > λ > t0 , λx + (1 − λ)y ∈ [x, z0 ], there is a µ =. λ−t such that 1−t. λx + (1 − λ)y = µx + (1 − µ)z0 . Then, if f (x) is increasing and. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥ θ > 0,. λf tr (x) + (1 − λ)f tr (y) = µf tr (x) + (1 − µ)(t0 f tr (x) + (1 − t0 )f tr (y)) > µf tr (x) + (1 − µ)f tr (z0 ) ≥ f tr (λx + (1 − λ)y) If 0 < λ < t0 , λx + (1 − λ)y ∈ [z0 , y], there is a γ =. λ such that t. λx + (1 − λ)y = γz0 + (1 − γ)y. Then, if f (x) is increasing and. π 2. >θ≥. π 4,. or if f (x) is decreasing and. π 4. ≥ θ > 0,. λf tr (x) + (1 − λ)f tr (y) = γ(t0 f tr (x) + (1 − t0 )f tr (y)) + (1 − γ)f tr (y) > γ(f tr (z0 ) + (1 − γ)f tr (y) ≥ f tr (λx + (1 − λ)y). Part 4. Suppose x, y ∈ E, 0 < λ < 1 f tr (x) = 2f (x1 ), f tr (y) = 2f (y1 ), f tr (λx + (1 − λ)y) = 2f (λx1 + (1 − λ)y1 ) f 00 (x) > 0 ∀x ∈ R,, so f is strictly convex, λf tr (x) + (1 − λ)f tr (y) = λ2f (x1 ) + (1 − λ)2f (y1 ) > 2f (λx1 + (1 − λ)y1 ) = f tr (λx + (1 − λ)y). By these 4 part, f tr is strictly convex in Lθ . 10.

(15) In the other hand, if f tr convex , is f convex? Theorem 2.6. Suppose f : R → R is a twice differentiable function, • if f tr is convex, then f is convex in R+ . • if f tr is strictly convex, then f is strictly convex in R+ . Proof. Suppose x, y ∈ E, 0 < λ < 1 f tr (x) = 2f (x1 ), f tr (y) = 2f (y1 ), f tr (λx + (1 − λ)y) = 2f (λx1 + (1 − λ)y1 ) If f tr is convex, λ2f (x1 ) + (1 − λ)2f (y1 ) = λf tr (x) + (1 − λ)f tr (y) ≥ f tr (λx + (1 − λ)y)) = 2f (λx1 + (1 − λ)y1 , f is convex in R+ . If f tr is strictly convex, λ2f (x1 ) + (1 − λ)2f (y1 ) = λf tr (x) + (1 − λ)f tr (y) > f tr (λx + (1 − λ)y)) = 2f (λx1 + (1 − λ)y1 , f is strictly convex in R+ .. 3. Counterexamples Out of Conditions. In this sectoin, we will give some examples out of conditions to show that trace function from a convex function may be not convex. Example 3.1. f : R → R is ex ,. π 4. ≥θ>0. f 0 (x) = ex , and for x ≥ 0 f 0 (x) > 0, a(x) = = When. π 4. f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) ( 0 − cot2 θ)(Eq.2.1) ||x|| f (λ1 (x)) eλ1 (x) tan θ λ2 (x)−λ1 (x) (e − cot2 θ) ||x||. > θ > 0, cot2 θ > 1. So ln(cot2 θ) > 0. (tan θ + cot θ). And λ2 (x) − λ1 (x) = ||x2 ||(tan θ + cot θ) Then, if ||x2 || <. ln(cot2 θ) (tan θ+cot θ). eλ2 (x)−λ1 (x) − cot2 θ < 0. We have a(x) < 0. ln(2) 3 θ Then in R3 , let x = (10, ln(2) 2 , 0), y = (10, 0, 2 ) ∈ L , cot θ = 2 and λ = 7 . 1. f tr (x) = f tr (y) = (2 4 + 2−1 )e10 1. λf tr (x) + (1 − λ)f tr (y) = (2 4 + 2−1 )e10 −5 5 1 . (2 28 + 2 7 )e10 − (2 4 + 2−1 )e10 = 1146.74977793 > 0 f tr is not convex.. 11.

(16) Example 3.2. f : R → R is e−x ,. π 2. π 4. >θ≥. f 0 (x) = −e−x ,and for x ≥ 0 f 0 (x) < 0, f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) ( 0 − cot2 θ) ||x|| f (λ1 (x)). a(x) =. −e−λ1 (x) tan θ λ1 (x)−λ2 (x) (e − cot2 θ) ||x||. = When. π 2. >θ>. π 4,. cot2 θ < 1. So − ln(cot2 θ) > 0. (tan θ + cot θ). And λ1 (x) − λ2 (x) = −||x2 ||(tan θ + cot θ) Then, if ||x2 || <. − ln(cot2 θ) (tan θ+cot θ). eλ1 (x)−λ2 (x) − cot2 θ > 0. We have a(x) < 0. ln(2) θ Then in R3 , let x = (10, ln(2) 2 , 0), y = (10, 0, 2 ) ∈ L , cot θ =. 1 2. and λ = 37 .. 1. f tr (x) = f tr (y) = (2 4 + 2−1 )e−10 1. λf tr (x) + (1 − λ)f tr (y) = (2 4 + 2−1 )e−10 −5 5 1 . (2 28 + 2 7 )e−10 − (2 4 + 2−1 )e−10 = 0.00000236362 > 0 f tr is not convex.. Example 3.3. f : R → R is − ln(x) ,. π 2. >θ≥. π 4. f 0 (x) = − x1 , ∀x > 0, By the definition Eq.1.1 and Eq.1.2, λ1 ≥ 0 and λ2 ≥ 0. For the x ∈ Lθ with λ1 (x) > 0 and λ2 (x) > 0, f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) ( 0 − cot2 θ) ||x|| f (λ1 (x)) tan θ 1 λ1 (x) =− ( − cot2 θ) ||x|| λ1 (x) λ2 (x). a(x) =. For all x ∈ Lθ with λ1 (x) > 0 and λ2 (x) > 0 λ2 ≥ λ1 , so Then if. λ1 (x) λ2 (x). λ1 (x) ≤ 1. λ2 (x). − cot2 θ > 0 cot θ <. λ1 (x) π π so > θ > . λ2 (x) 2 4. In the other hand, f tr (x) = − ln(λ1 (x)) + − ln(λ2 (x)) = − ln(λ1 (x)λ2 (x)) = − ln(x21 + x1 ||x2 ||(tan θ − cot θ) − ||x2 ||2 ) 12.

(17) In R3 Let x = (10, 1, 0), y = (10, 0, 1) ∈ Lθ , cot θ = 12 , cot θ =. 1 2. and λ = 21 .. f tr (x) = − ln(100 + 15 − 1) = − ln(114) f tr (y) = − ln(100 + 15 − 1) = − ln(114) λf tr (x) + (1 − λ)f tr (y) = − ln(114) 1 1 λx + (1 − λ)y = (10, , ) 2 2 r 225 1 tr f (λx + (1 − λ)y) = − ln(100 + − ) 2 2 (100 +. q. 225 2. − 12 ) < 114, so r tr. f (λx + (1 − λ)y) = − ln(100 +. 225 1 − ) > − ln(114). 2 2. f tr (x) is not convex.. Example 3.4. f : R → R is (x − 1)2 f 0 (x) = 2x − 2,so if x 6= 1 f 0 (x) 6= 0.And if x ∈ / E, λ2 (x) > λ1 (x) By Eq.2.1, if λ1 (x) 6= 1 f 0 (λ1 (x)) tan θ f 0 (λ2 (x)) ( 0 − cot2 θ) ||x|| f (λ1 (x)) (λ1 (x) − 1) tan θ λ2 (x) − 1 ( − cot2 θ) =2 ||x|| λ1 (x) − 1. a(x) =. For 1 > λ2 (x) > λ1 (x) ≥ 0 (λ1 (x) − 1) tan θ λ2 (x) − 1 < 0 and > 1. ||x|| λ1 (x) − 1 So if cot2 θ < 1,We have that a(x) < 0. When π2 > θ > π π 2 > θ > 4. 3 In R , let x = ( 21 , 18 , 0), y = ( 12 , 0, 18 ), cot θ = 12 and λ = 73 .. π 4,. cot2 θ < 1. f tr (x) is not convex when. 1 3 4 λx + (1 − λ)y = ( , , ) 2 56 56 −9 2 −1 2 97 4753 tr tr f (x) = f (y) = ( ) +( ) = = 16 4 256 12544 97 tr tr λf (x) + (1 − λ)f (y) = 256 −9 2 5017 −61 2 tr ) +( ) = f (λx + (1 − λ)y) = ( 112 28 12544 f tr (λx + (1 − λ)y) > λf tr (x) + (1 − λ)f tr (y) f tr (x) is not convex.. 13.

(18) For λ2 (x) > λ1 (x) > 1 (λ1 (x) − 1) tan θ λ2 (x) − 1 > 0 and > 1. ||x|| λ1 (x) − 1 So if cot2 θ > 1, We have that a(x) < 0. f tr (x) is not convex when In R3 , let x = (10, 2, 0), y = (10, 0, 2) ∈ Lθ , cot θ = 2 and λ = 73 .. π 4. > θ > 0.. 6 8 λx + (1 − λ)y = (10, , ) 7 7 f tr (x) = f tr (y) = 52 + 102 = 125 6125 49 43 2 68 2 6473 tr f (λx + (1 − λ)y) = ( ) + ( ) = 7 7 49 f tr (λx + (1 − λ)y) > λf tr (x) + (1 − λ)f tr (y). λf tr (x) + (1 − λ)f tr (y) = 125 =. f tr (x) is not convex To do the same way for each angle θ, we can got that f tr (x) is convex only if θ =. π 4.. References [1] Chrle R.johnson Roger A.Horn. Matrix Analysis. Cambridge, 1985. [2] T-K Liao J-S Chen 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] C-Y Yang Y-L Chang and J-S Chen. Smooth and nonsmooth analyses of vector-valued functions associated with circular cones. Nonlinear Analysis: Theory, Methods and Applications, vol. 85, July, pp. 160-173, 2013. [4] J-S Chen J-C Zhou and H-F Hung. Circular cone convexity and some inequalities associated with circular cones. Journal of Inequalities and Applications, vol. 2013, Article ID 571, 17 pages, 2013. [5] J-C Zhou and J-S Chen. Properties of circular cone and spectral factorization associated with circular cone. Journal of Nonlinear and Convex Analysis, vol. 14, no. 4, pp. 807-816, 2013. [6] Y-L Chang C-Y Yang and J-S Chen. Analysis of nonsmooth vector-valued functions associated with infinite-dimensional second-order cones. Nonlinear Analysis: Theory, Methods and Applications, vol. 74, no. 16, pp. 5766-5783, 2011.. 14.

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