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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 inequalities associated with circular cone. 研 究 生 : 洪浩峰. 中華民國一百零二年六月.

(2) Some inequalities associated with circular cone. By Hao-Feng Hung. Advisor Jein-Shan Chen. Departments of Mathematics, National Taiwan Normal University Taipei, Taiwan. June, 2013.

(3) 誌 謝 感謝我的家人讓我唸研究所,也感謝我的指導教授,陳界 山老師,在他的指導下我認識了最佳化的領域以及 circular cone 的一些性質,並且順利完成我的碩士論文,另外,我看 到老師安排好事情後能徹底執行,讓我覺得我應該向老師學 習,還有也感謝口試委員柯春旭老師、張毓麟老師提供的意 見。. 洪浩峰 謹誌 2013 年 6 月.

(4) Abstract The circular cone is a pointed closed convex cone having hyperspherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation. In this section, we establish some inequalities associated with circular cone, which we believe that they will be useful for further analyzing properties of f Lθ and proving the convergence of interior point methods for optimization problems involved in circular cones. Key words. Second-order cone , circular cone , spectral factorization , determinant , trace..

(5) Contents 1 Introduction and Preliminaries. 1. 2 Main results. 3. 3 References. 8.

(6) 1. Introduction and Preliminaries. The circular cone [1,2] is a pointed closed convex cone having hyperspherical sections orthogonal to its axis of revolution about which the cone is invariant to rotation. Let its half-aperture angle be θ with θ ∈ (0, π/2). Then, the n-dimensional circular cone, denoted by Lθ , can be expressed as Lθ := {x = (x1 , x2 ) ∈ R × Rn−1 | ∥x∥ cos θ ≤ x1 } or is equivalent to Lθ := {x = (x1 , x2 ) ∈ R × Rn−1 | ∥x2 ∥ cot θ ≤ x1 }. Since ⇔ ⇔ ⇔ ⇔. ∥x∥ cos θ ≤ x1 x21 ≥ ∥x∥2 cos2 θ = (x21 + ∥x2 ∥2 ) cos2 θ ∥x2 ∥2 cos2 θ ≤ x21 (1 − cos2 θ) = x21 sin2 θ ∥x2 ∥2 cot2 θ ≤ x21 ∥x2 ∥ cot θ ≤ x1 .. When θ = 45o , the circular cone reduces to the well-known second-order cone (SOC, also called Lorentz cone) given by Kn := {x = (x1 , x2 ) ∈ R × Rn−1 | ∥x2 ∥ ≤ x1 } or is equivalent to Kn := {x = (x1 , x2 ) ∈ R × Rn−1 | ∥x∥ cos 45o ≤ x1 }. With respect to SOC, for any x = (x1 , x2 ) ∈ R × Rn−1 ,we can decompose x as (2) x = λ1 (x)u(1) x + λ2 (x)ux , (1). (1). (2). where λ1 (x), λ2 (x) and ux , ux are the spectral values and the associated spectral vectors of x with respect to Kn , given by λi (x) = x1 + (−1)i ∥x2 ∥,  1 x   (1, (−1)i 2 ), if x2 ̸= 0, 2 ∥x2 ∥ u(i) (2) x = 1   (1, (−1)i 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 (1) is unique.. 1.

(7) 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 (1), for any function f : R → R, the following vector-valued function associated with Kn (n ≥ 1) is considered as (see [6, 7]): (2) f soc (x) = f (λ1 )u(1) x + f (λ2 )ux. (3). where ∀x = (x1 , x2 ) ∈ R × Rn−1 . If f is defined only on a subset of R , then f soc is defined on the corresponding subset of Rn . The definition (3) is unambiguous whether x2 ̸= 0 or x2 = 0. Some inequalities associated with second-order cone For any x ≽Kn 0 and y ≽Kn 0,we have 1.det(x + y) ≥ det(x) + det(y) 2.[det(e + x)1/2 ] ≥ 1 + det(x)1/2 , ∀x ≽Kn 0 3.det(e + x + y) ≤ det(e + x) det(e + y) 4.If x ≽Kn y, then det(x) ≥ det(y) 5.det(αx + (1 − α)y) ≥ α2 det(x) + (1 − α)2 det(y), ∀0 < α < 1 6.If x ≻Kn 0, then the function f (x) = [det(x)]1/n (i) f (x) = [det(x)]1/n = (x21 − ∥x2 ∥2 )1/n for n ≥ 2 is concave on int(Kn ) (ii) If x ≻Kn 0 and y ≻Kn 0, then the inequality det(x + y)1/n ≥ det(x)1/n + det(y)1/n hold for n = 2. Furthermore, det(x + y)1/n ≥ K(det(x)1/n + det(y)1/n ) hold for n ≥ 3, in particular K =. 41/n . 2. Spectral factorization associated with Lθ Theorem 1.1. [2, Theorem 3.1]. For any z = (z1 , z2 ) ∈ R × Rn−1 , one has (2) z = λ1 (z)u(1) z + λ2 (z)uz. 2. (4).

(8) { λ1 (z) = z1 − ∥z2 ∥ cot θ λ2 (z) = z1 + ∥z2 ∥ tan θ. where. and. [   1  u(1)   z = 1 + cot2 θ [  1  u(2)   z = 1 + tan2 θ. with w =. 1. 0. ][. 1. ]. 0 cot θ −w ][ ] 1 0 1 0 tan θ. w. [. (5). sin2 θ. ]. = −(sin θ cos θ)w [ ] cos2 θ = (sin θ cos θ)w. (6). z2 if z2 ̸= 0 , and any vector in Rn−1 satisfying ∥w∥ = 1 if z2 = 0. ∥z2 ∥. Analogous to (3), with the spectral factorization (4), for any function f : R → R, we consider the following vector-valued function associated with Lθ (n ≥ 1): (2) f Lθ (z) = f (λ1 )u(1) z + f (λ2 )uz. (7). where ∀z = (z1 , z2 ) ∈ R × Rn−1 .. 2. Main results. In this section, we establish some inequalities associated with circular cone, which we believe that they will be useful for further analyzing properties of f Lθ and proving the convergence of interior point methods for optimization problems involved in circular cones.. Proposition 2.1. Let x = (x1 , x2 ) ∈ IR × IRn−1 has spectral factorization associated with circular cone given as in (4). Then, the following hold. (a) [det(e + x)]1/2 ≥ 1 + det(x)1/2 for all x ∈ Lθ , where e = (1, 0, · · · , 0) ∈ IRn . (b) det(x + y) ≤ (x1 + y1 )2 sec2 −∥x2 + y2 ∥2 csc2 � for all x + y ∈ Lθ . (c) det(e + x + y) ≤ (x1 + y1 )2 sec2 θ − ∥x2 + y2 ∥2 csc2 θ + 2(tr(x) + tr(y)) + 1 for all x, y ∈ Lθ , where e = (1, 0, · · · , 0) ∈ IRn . (d) det(x+y) ≥ det(x)+det(y)+(∥x2 ∥2 +∥y2 ∥2 ) csc2 θ −(x21 +y12 ) sec2 θ for all x, y ∈ Lθ . (e) det(αx + (1 − α)y) ≥ α2 det(x) + (1 − α)2 det(y) + (α2 ∥x2 ∥2 + (1 − α)2 ∥y2 ∥2 ) csc2 θ −(α2 x21 + (1 − α)2 y12 ) sec2 θ for all x, y ∈ Lθ and ∀0 < α < 1 (f) If x, y ∈ Lθ and x ≽Lθ y, then det(x) ≥ det(y).. 3.

(9) Proof. (a) First, we observe that ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒ ⇐⇒. [det(e + x)]1/2 ≥ 1 + det(x)1/2 det(e + x) ≥ 1 + 2det(x)1/2 + det(x) √ λ1 (e + x)λ2 (e + x) ≥ 1 + 2 λ1 (x)λ2 (x) + λ1 (x)λ2 (x) √ (x1 + 1 − ∥x2 ∥ cot θ)(x1 + 1 + ∥x2 ∥ tan θ) ≥ 1 + 2 λ1 (x)λ2 (x) + λ1 (x)λ2 (x) √ (λ1 (x) + 1)(λ2 (x) + 1) ≥ 1 + 2 λ1 (x)λ2 (x) + λ1 (x)λ2 (x) √ λ1 (x)λ2 (x) + λ1 (x) + λ2 (x) + 1 ≥ 1 + 2 λ1 (x)λ2 (x) + λ1 (x)λ2 (x) √ λ1 (x) + λ2 (x) ≥ 2 λ1 (x)λ2 (x) λ1 (x) + λ2 (x) √ ≥ λ1 (x)λ2 (x). 2. Hence, to prove the desired result, it suffices to show that λ1 (x) + λ2 (x) √ ≥ λ1 (x)λ2 (x) 2 which is clearly true due to A-M inequality and x ∈ Lθ . (b) Since λ1 (x + y) = x1 + y1 − ∥x2 + y2 ∥ cot θ and λ2 (x + y) = x1 + y1 + ∥x2 + y2 ∥ tan θ, we know det(x + y) = λ1 (x + y) · λ2 (x + y) = (x1 + y1 − ∥x2 + y2 ∥ cot θ) (x1 + y1 + ∥x2 + y2 ∥ tan θ) = (x1 + y1 )2 + (x1 + y1 )∥x2 + y2 ∥ tan θ − (x1 + y1 )∥x2 + y2 ∥ cot θ − ∥x2 + y2 ∥2 Using x + y ∈ Lθ gives ∥x2 + y2 ∥ ≤ (x1 + y1 ) tan θ and −(x1 + y1 ) ≤ −∥x2 + y2 ∥ cot θ. Hence, we have. = ≤ = =. det(x + y) (x1 + y1 )2 + (x1 + y1 )∥x2 + y2 ∥ tan θ − (x1 + y1 )∥x2 + y2 ∥ cot θ − ∥x2 + y2 ∥2 (x1 + y1 )2 + (x1 + y1 )2 tan2 θ − ∥x2 + y2 ∥2 cot2 θ − ∥x2 + y2 ∥2 (x1 + y1 )2 (1 + tan2 θ) − ∥x2 + y2 ∥2 (cot2 θ + 1) (x1 + y1 )2 sec2 θ − ∥x2 + y2 ∥2 csc2 θ. which is the desired result. (c) First, we simplify det(e + x + y) as det(e + x + y) = (x1 + y1 + 1 − ∥x2 + y2 ∥ cot θ) (x1 + y1 + 1 + ∥x2 + y2 ∥ tan θ) = (λ1 (x + y) + 1) (λ2 (x + y) + 1) = det(x + y) + tr(x + y) + 1. 4.

(10) Then, using x, y ∈ Lθ (hence x + y ∈ Lθ ) and applying part(b) yields det(e + x + y) ≤ (x1 + y1 )2 sec2 θ − ∥x2 + y2 ∥2 csc2 θ + tr(x + y) + 1. On the other hand, we know tr(x + y) = = ≤ ≤ =. λ1 (x + y) + λ2 (x + y) (x1 + y1 − ∥x2 + y2 ∥ cot θ) + (x1 + y1 + ∥x2 + y2 ∥ tan θ) 2 (λ2 (x) + λ2 (y)) 2 (λ2 (x) + λ2 (y)) + 2 (λ1 (x) + λ1 (y)) 2(tr(x) + tr(y)). From the above, we prove that det(e + x + y) ≤ (x1 + y1 )2 sec2 θ − ∥x2 + y2 ∥2 csc2 θ + 2(tr(x) + tr(y)) + 1 for all x, y ∈ Lθ . (d) First, we compute det(x + y) = λ1 (x + y) · λ2 (x + y) = (x1 + y1 − ∥x2 + y2 ∥ cot θ) (x1 + y1 + ∥x2 + y2 ∥ tan θ) = (x1 + y1 )2 + (x1 + y1 )∥x2 + y2 ∥ tan θ − (x1 + y1 )∥x2 + y2 ∥ cot θ − ∥x2 + y2 ∥2 and. = = = =. det(x) + det(y) λ1 (x)λ2 (x) + λ1 (y)λ2 (y) (x1 − ∥x2 ∥ cot θ)(x1 + ∥x2 ∥ tan θ) + (y1 − ∥y2 ∥ cot θ)(y1 + ∥y2 ∥ tan θ) x21 + x1 ∥x2 ∥ tan θ − x1 ∥x2 ∥ cot θ − ∥x2 ∥2 + y12 + y1 ∥y2 ∥ tan θ − y1 ∥y2 ∥ cot θ − ∥y2 ∥2 x21 + y12 + x1 ∥x2 ∥ tan θ + y1 ∥y2 ∥ tan θ − x1 ∥x2 ∥ cot θ − y1 ∥y2 ∥ cot θ − ∥x2 ∥2 − ∥y2 ∥2 .. Hence, we have det(x + y) − det(x) − det(y) = 2x1 y1 − 2xT2 y2 + (x1 ∥x2 + y2 ∥ + y1 ∥x2 + y2 ∥ − x1 ∥x2 ∥ − y1 ∥y2 ∥) tan θ −(x1 ∥x2 + y2 ∥ + y1 ∥x2 + y2 ∥ − x1 ∥x2 ∥ − y1 ∥y2 ∥) cot θ. Using x, y ∈ Lθ (and hence x + y ∈ Lθ ) gives x1 tan θ ≥ ∥x2 ∥,. −x1 tan θ ≤ −∥x2 ∥,. x1 ≥ ∥x2 ∥ cot θ,. −x1 ≤ −∥x2 ∥ cot θ,. −(x1 + y1 ) ≤ −∥x2 + y2 ∥ cot θ. 5.

(11) Thus, ≥ = ≥ = = =. det(x + y) − det(x) − det(y) 2x1 y1 − 2xT2 y2 + ∥x2 ∥∥x2 + y2 ∥ + ∥y2 ∥∥x2 + y2 ∥ − x21 tan2 θ − y12 tan2 θ −x1 (x1 + y1 ) − y1 (x1 + y1 ) + ∥x2 ∥2 cot2 θ + ∥y2 ∥2 cot2 θ 2x1 y1 − 2xT2 y2 + ∥x2 + y2 ∥(∥x2 ∥ + ∥y2 ∥) − (x21 + y12 ) tan2 θ −(x1 + y1 )2 + (∥x2 ∥2 + ∥y2 ∥2 ) cot2 θ ∥x2 + y2 ∥2 − (x21 + y12 ) tan2 θ − x21 − y12 − 2xT2 y2 + (∥x2 ∥2 + ∥y2 ∥2 ) cot2 θ ∥x2 ∥2 + ∥y2 ∥2 − (x21 + y12 ) tan2 θ − x21 − y12 + (∥x2 ∥2 + ∥y2 ∥2 ) cot2 θ (∥x2 ∥2 + ∥y2 ∥2 )(1 + cot2 θ) − (x21 + y12 )(1 + tan2 θ) (∥x2 ∥2 + ∥y2 ∥2 ) csc2 θ − (x21 + y12 ) sec2 θ. which is the desired result. (e) First, we compute. = = = =. det(αx + (1 − α)y) λ1 (αx + (1 − α)y) · λ2 (αx + (1 − α)y) [αx1 + (1 − α)y1 − ∥αx2 + (1 − α)y2 ∥ cot θ] [αx1 + (1 − α)y1 + ∥αx2 + (1 − α)y2 ∥ tan θ] (αx1 + (1 − α)y1 )2 + (αx1 + (1 − α)y1 )∥αx2 + (1 − α)y2 ∥ tan θ −(αx1 + (1 − α)y1 )∥αx2 + (1 − α)y2 ∥ cot θ − ∥αx2 + (1 − α)y2 ∥2 α2 x21 + 2α(1 − α)x1 y1 + (1 − α)2 y12 +αx1 ∥αx2 + (1 − α)y2 ∥ tan θ + (1 − α)y1 ∥αx2 + (1 − α)y2 ∥ tan θ −αx1 ∥αx2 + (1 − α)y2 ∥ cot θ − (1 − α)y1 ∥αx2 + (1 − α)y2 ∥ cot θ −(α2 ∥x2 ∥2 + 2α(1 − α)xT2 y2 + (1 − α)2 ∥y2 ∥2 ). and. = = = =. α2 det(x) + (1 − α)2 det(y) α2 λ1 (x)λ2 (x) + (1 − α)2 λ1 (y)λ2 (y) α2 (x1 − ∥x2 ∥ cot θ)(x1 + ∥x2 ∥ tan θ) + (1 − α)2 (y1 − ∥y2 ∥ cot θ)(y1 + ∥y2 ∥ tan θ) α2 (x21 + x1 ∥x2 ∥ tan θ − x1 ∥x2 ∥ cot θ − ∥x2 ∥2 ) + (1 − α)2 (y12 + y1 ∥y2 ∥ tan θ − y1 ∥y2 ∥ cot θ − ∥y2 ∥2 ) α2 x21 + (1 − α)2 y12 + α2 x1 ∥x2 ∥ tan θ + (1 − α)2 y1 ∥y2 ∥ tan θ − α2 x1 ∥x2 ∥ cot θ −(1 − α)2 y1 ∥y2 ∥ cot θ − α2 ∥x2 ∥2 − (1 − α)2 ∥y2 ∥2 .. Hence, we have det(αx + (1 − α)y) − α2 det(x) − (1 − α)2 det(y) = 2α(1 − α)x1 y1 − 2α(1 − α)xT2 y2 + (αx1 ∥αx2 + (1 − α)y2 ∥ + (1 − α)y1 ∥αx2 + (1 − α)y2 ∥ −α2 x1 ∥x2 ∥ − (1 − α)2 y1 ∥y2 ∥) tan θ − (αx1 ∥αx2 + (1 − α)y2 ∥ + (1 − α)y1 ∥αx2 + (1 − α)y2 ∥ 6.

(12) −α2 x1 ∥x2 ∥ − (1 − α)2 y1 ∥y2 ∥) cot θ. Using x, y ∈ Lθ (and hence αx + (1 − α)y ∈ Lθ ) gives x1 tan θ ≥ ∥x2 ∥,. −x1 tan θ ≤ −∥x2 ∥,. x1 ≥ ∥x2 ∥ cot θ, −x1 ≤ −∥x2 ∥ cot θ, −(αx1 + (1 − α)y1 ) ≤ −∥αx2 + (1 − α)y2 ∥ cot θ. Thus, ≥. = ≥ = = =. det(αx + (1 − α)y) − α2 det(x) − (1 − α)2 det(y) 2α(1 − α)x1 y1 − 2α(1 − α)xT2 y2 + α∥x2 ∥∥αx2 + (1 − α)y2 ∥ + (1 − α)∥y2 ∥∥αx2 + (1 − α)y2 ∥ −α2 x21 tan2 θ − (1 − α)2 y12 tan2 θ − αx1 (αx1 + (1 − α)y1 ) − (1 − α)y1 (αx1 + (1 − α)y1 ) +α2 ∥x2 ∥2 cot2 θ + (1 − α)2 ∥y2 ∥2 cot2 θ 2α(1 − α)x1 y1 − 2α(1 − α)xT2 y2 + ∥αx2 + (1 − α)y2 ∥(α∥x2 ∥ + (1 − α)∥y2 ∥) −(α2 x21 + (1 − α)2 y12 ) tan2 θ − (αx1 + (1 − α)y1 )2 + (α2 ∥x2 ∥2 + (1 − α)2 ∥y2 ∥2 ) cot2 θ ∥αx2 + (1 − α)y2 ∥2 − (α2 x21 + (1 − α)2 y12 ) tan2 θ − α2 x21 − (1 − α)2 y12 − 2α(1 − α)xT2 y2 +(α2 ∥x2 ∥2 + (1 − α)2 ∥y2 ∥2 ) cot2 θ α2 ∥x2 ∥2 + (1 − α)2 ∥y2 ∥2 − (α2 x21 + (1 − α)2 y12 ) tan2 θ − α2 x21 − (1 − α)2 y12 +(α2 ∥x2 ∥2 + (1 − α)2 ∥y2 ∥2 ) cot2 θ (α2 ∥x2 ∥2 + (1 − α)2 ∥y2 ∥2 )(1 + cot2 θ) − (α2 x21 + (1 − α)2 y12 )(1 + tan2 θ) (α2 ∥x2 ∥2 + (1 − α)2 ∥y2 ∥2 ) csc2 θ − (α2 x21 + (1 − α)2 y12 ) sec2 θ. which is the desired result. (f) Note that x ≽Lθ y implies x − y ∈ Lθ . Hence, x1 − y1 ≥ ∥x2 − y2 ∥ cot θ ≥ 0 ⇐⇒ x1 − y1 − ∥x2 − y2 ∥ cot θ ≥ 0 ⇐⇒ x1 − y1 − (∥x2 ∥ − ∥y2 ∥) cot θ ≥ x1 − y1 − ∥x2 − y2 ∥ cot θ ≥ 0 which says λ1 (x) − λ1 (y) ≥ λ1 (x − y) ≥ 0, that is, λ1 (x) ≥ λ1 (y). Meanwhile, we know λ2 (x) − λ2 (y) = x1 + ∥x2 ∥ tan θ − (y1 + ∥y2 ∥ tan θ) = x1 − y1 + (∥x2 ∥ − ∥y2 ∥) tan θ. Therefore, when ∥x2 ∥ − ∥y2 ∥ ≥ 0 and θ ∈ (0, 90◦ ), we have λ2 (x) ≥ λ2 (y). For ∥x2 ∥ − ∥y2 ∥ ≤ 0, we discuss two cases: (i) if θ ∈ (0, 45◦ ], using x1 −y1 −(∥x2 ∥−∥y2 ∥) cot θ ≥ 0 and cot θ ≥ tan θ for θ ∈ (0, 45◦ ] leads to x1 − y1 + (∥x2 ∥ − ∥y2 ∥) tan θ ≥ 0, i.e, λ2 (x) ≥ λ2 (y); (ii) if θ ∈ [45◦ , 90◦ ), then π2 − θ ∈ (0, 45◦ ]. Applying part(i) yields π x1 − y1 + (∥x2 ∥ − ∥y2 ∥) tan θ = x1 − y1 + (∥x2 ∥ − ∥y2 ∥) cot( − θ) ≥ 0 2 which says λ2 (x) ≥ λ2 (y). From the above, it is clear that det(x) ≥ det(y) when x, y ∈ Lθ 2 and x ≽Lθ y.. 7.

(13) 3. References. [1] J. Dattorro, Convex Optimization and Euclidean Distance Geometry, Meboo Publishing, California, 2005. [2] J-C Zhou and J-S Chen, Properties of circular cone and spectral factorization associated with circular cone, to appear in Journal of Nonlinear and Convex Analysis, 2013. [3] Y-L Chang, C-Y Yang, and J-S Chen, Smooth and nonsmooth analyses of vectorvalued functions associated with circular cones, Nonlinear Analysis: Theory, Methods and Applications, vol. 85, pp. 160-173, July, 2013. [4] 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. [5] M-L Hsu and R-C Tsai, Some Results on Functions Associated With Second-Order Cone, June, 2007. [6] J-S Chen, The convex and monotone functions associated with second-order cone, Optimization, vol. 55, no. 4, pp. 363-385, 2006. [7] J-S Chen, X. Chen, and P. Tseng, Analysis of nonsmooth vector-valued functions associated with second-order cone, Mathematical Programming, vol. 101, no. 1, pp. 95-117, 2004.. 8.

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