一個閉凸集上的凸函數之研究

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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. 一個閉凸集上的凸函數之研究. 研究生：黃偉. 中華民國一百零五年六月.

(2) Master’s Thesis Department of Mathematics. The Study on Convex Functions in a Closed Convex Set Author Wei Huang. Supervisor Dr. Yu-Lin Chang. June 30, 2016. National Taiwan Normal University Taiwan.

(3) Acknowledgements. Thanks to my advisor, Dr. Yu-Lin Chang, for giving experience and brand-new aspects that inspire me on diﬀerent areas of mathematics. He taught me how to be a good mathematician. Moreover, he was like a lighthouse of this project and always kept me safe when through the troubled sea. For all of this, I have learned a lot during this two years and deeply appreciate being one of his student. Thanks to the research group, Yu-Hsuan Hsiao, Kuo-Yu Chang and Ti-Wen Wu, for working together and sharing full of courage to face the challenges in daily life. Thanks Dr. Xin-He Miao, Dr. Jein-Shan Chen and Dr. Yu-Lin Chang for being my thesis committee and providing useful suggestions and comments. In addition, they are always patient with our questions. They encouraged us to learn from the questions but not to feel embarrassed about mistakes. Without your help, this paper would not have been possible. Thanks again. Furthermore, for staying six years, I’m glad to thank the Dept. of Mathematics in NTNU for giving us this beautiful environment to meet so many great people and have unique memories in the school. I like the research room M410 in the Mathematics Building as well. This is a good place to thinking and concentrating on our works. I also want to thank Dr. Tamaki Tanaka for kindly arrangement of JSRC program during February 2016 in Niigata University, Japan. That’s the most wonderful experience that I will never forget. It is important to thank my parents. You always encourage me to do whatever I like. There is no limit to measure your timeless love..

(4) Contents 1 Introduction. 1. 2 Preliminaries. 3. 3 Main Results. 10. 4 Conclusions. 18. References. 19.

(5) The Study on Convex Functions in a Closed Convex Set Wei Huang June 30, 2016 Abstract. We extend to a problem of two-variable function from a monotone condition for the convexity of trace function associated with circular cone in this paper. This structure was based on the Schur Complement Theorem. By smoothing the problem, the greatest benefit is spared the solving time and we believe this establishment will useful in many fields. Keywords. Convexity, Optimization, Circular Cone, Trace Function, Positive Semidefiniteness.. 1. Introduction. We want to find out more convex functions in high dimensional spaces. To do this, consider the second-order cone (SOC) in Rn , n o n n 1 K := x = (x1 , x2 ) 2 R ⇥ R : x1 kx2 k .. From [2], Chen, Liao and Pan established the convexity of SOC-trace functions via the Schur Complement Theorem and gave us that SOC-trace functions generated by any convex function is convex. That is, Definition 1.1. [2] For any given scalar function f : J ✓ R ! R, with the spectral factorization associated with second-order cone, define a vector-valued function f soc : S ✓ Rn ! Rn by (2) f soc (x) := f ( 1 (x))u(1) x + f ( 2 (x))ux. (1). and the SOC-trace function f tr (x) :=f ( 1 (x)) + f ( 2 (x)) = tr(f soc (x)) 1. 8x 2 S. (2).

(6) where i (x). 1 i x2 = x1 + ( 1)i kx2 k, u(i) ). x = (1, ( 1) 2 kx2 k. (3). Lemma 1.2. [2.2, 2] For any given f : J ✓ R ! R, let f soc : S ! Rn and f tr : S ! R be given by (1) and (2), respectively. Assume that J is open. Then, the following results hold. (a) The domain S of f soc and f tr is also open. (b) If f is (continuously) di↵erentiable on J, then f soc is (continuously) di↵erentiable on S. Moreover, for any x 2 S, rf soc (x) = f 0 (x1 )I if x2 = 0, and otherwise rf soc (x) = where b(x) =. f 0(. 2 (x))+f. 2. 0(. ". xT. b(x). c(x) kx22 k. c(x) kxx22 k a(x)I + (b(x). 1 (x)). , c(x) =. f 0(. f 0(. 2 (x)). 2. 1 (x)). x xT. a(x)) kx22 k22. , a(x) =. f(. #. ,. (4). 2 (x))+f ( 1 (x)) 2 (x). 1 (x). .. (c) If f is (continuously) di↵erentiable, then f tr is (continuously) di↵erentiable on S with rf tr (x) = (f 0 )soc (x); if f is twice (continuously) di↵erentiable, then f tr is twice (continuously) di↵erentiable on S with r2 f tr (x) = r(f 0 )soc (x). Theorem 1.1. [2.1, 2] For any given f : J ! R, let f soc : S ! Rn and f tr : S ! R be given by (1) and (2), respectively. Assume that J is open. If f is a twice di↵erentiable on J, then (a) f 00 (t). 0 for any t 2 J () r(f 0 )soc (x) ⌫ 0 for any x 2 S () f tr is convex in S.. (b) f 00 (t) > 0 for any t 2 J () r(f 0 )soc (x). 0 8x 2 S =) f tr is strictly convex in S.. Based on this structure, Wu and Chang in [7] use the similar method for circular cones in Rn , n o L✓ := x = (x1 , x2 ) 2 R ⇥ Rn 1 : x1 kx2 kcot✓ , (5) and this convexity depends on di↵erent angles of circular cones. In fact, for any convex function f , there is a sufficient condition of trace functions f tr that implies f tr is convex. That is,. Definition 1.3. [7] For any twice di↵erentiable function f : R ! R, with the spectral factorization associated with circular cone, define the trace function f tr (x) :=f ( 1 (x)) + f ( 2 (x)). 8x = (x1 , x2 ) 2 L✓ 2. (6).

(7) where 1 (x). = x1. kx2 kcot✓,. 2 (x). = x1 + kx2 ktan✓.. (7). Theorem 1.2. [2.4, 7] Suppose that f : R ! R is a twice di↵erentiable convex function. Then f tr (x) is convex if • f (x) is increasing and. ⇡ 4. ✓<. ⇡ 2. or. • f (x) is decreasing and 0 < ✓  ⇡4 . Corollary 1.2.1. [2.5, 7] Suppose that f : R ! R is a twice di↵erentiable function which f 00 (x) > 0 8x 2 R. Then f tr (x) is strictly convex if • f (x) is increasing and. ⇡ 4. ✓<. ⇡ 2. or. • f (x) is decreasing and 0 < ✓  ⇡4 . Theorem 1.3. [2.6, 7] Suppose f : R ! R is a twice di↵erentiable function. If f tr is (strictly) convex, then f is (strictly) convex in R+ . Motivated by this, we try to extend to a closed convex set by adding a new variable p = tan✓ and explore its convexity in this paper. The following structure is also based on the Schur Complement Theorem and it played an important role in proving the positive(semi)definite property on Hessian matrix of the trace function.. 2. Preliminaries. A closed convex set L(p) in Rn+1 , extended from the circular cone in Rn , is defined as n o L(p) := x = (x1 , x2 , p) 2 R ⇥ Rn 1 ⇥ (0, 1) : kx2 k  x1 p (8). where k · k denotes the Euclidean norm. For any twice di↵erentiable function f = f (a, b) : R2 ! R, with the spectral factorization associated with L(p), we can define the trace function f tr (x) :=f ( 1 (x), p) + f ( 2 (x), p). 8x = (x1 , x2 , p) 2 L(p). (9). where 1 (x). = x1. 1 kx2 k , p. Define the set E(p) := {(x1 , 0, p) 2 R ⇥ Rn f tr is di↵erentiable for any x 2 L(p) \ E(p). 3. 2 (x) 1. = x1 + kx2 kp.. (10). ⇥ (0, 1)} 2 L(p), then the trace function.

(8) We can show the gradient 2 1 @f ( 1 (x), p) 6 x2 1 = 4 kx2 k p @a kx2 k 2. p2. of trace function rf tr (x) = rf ( 1 (x), p) + rf ( 2 (x), p) 3 2 3 2 3 1 0 7 @f ( 1 (x), p) 4 5 @f ( 2 (x), p) 4 x2 p 5 0 + 5+ kx2 k @b @a 1 kx2 k. 3 0 @f ( 2 (x), p) 4 5 + 0 @b 1 2 3 2 1 x 1 7 6 = fa ( 1 (x), p) 4 kx22 k p 5 + fa ( 2 (x), p) 4 kx2 k p2. 3. 3 0 x2 5. p 5+4 0 kx2 k fb ( 1 (x), p) + fb ( 2 (x), p) kx2 k (11) 1. 2. It means that @f tr (x) = fa ( 1 (x), p) + fa ( 2 (x), p) @x1 @f tr (x) x2 1 = ( fa ( 1 (x), p) + fa ( 2 (x), p)p) @x2 kx2 k p tr @f (x) kx2 k 1 = (fa ( 1 (x), p) + fa ( 2 (x), p)p) + fb ( 1 (x), p) + fb ( 2 (x), p) @p p p and we can also verify that @ 2 f tr (x) = faa ( 1 (x), p) + faa ( 2 (x), p) := (x) @x21 @ 2 f tr (x) x2 1 x2 = faa ( 1 (x), p)( ) + faa ( 2 (x), p)( p) @x2 @x1 kx2 k p kx2 k x2 1 x2 = ( faa ( 1 (x), p) + faa ( 2 (x), p)p) := (x) kx2 k p kx2 k @ 2 f tr (x) xT 1 xT = 2 ( faa ( 1 (x), p) + faa ( 2 (x), p)p) := 2 (x) @x1 @x2 kx2 k p kx2 k xT. kx2 k x2 kx22 k @ 2 f tr (x) 1 = ( f ( (x), p) + fa ( 2 (x), p)p) a 1 2 @x2 kx2 k2 p x2 xT2 1 1 xT + ( faa ( 1 (x), p)( ) + faa ( 2 (x), p)( 2 p)p) kx2 k kx2 k p p kx2 k 1 T fa ( 1 (x), p) p + fa ( 2 (x), p)p x2 x2 = (I ) kx2 k2 kx2 k T x2 x2 1 x2 xT2 x2 xT2 2 + (f ( (x), p) + f ( (x), p)p ) := (I )↵(x) + (x) aa 1 aa 2 kx2 k2 p2 kx2 k2 kx2 k2 4.

(9) @ 2 f tr (x) 1 = faa ( 1 (x), p)(kx2 k 2 ) + fba ( 1 (x), p) + faa ( 2 (x), p)kx2 k + fba ( 2 (x), p) @p@x1 p kx2 k 1 = (faa ( 1 (x), p) + faa ( 2 (x), p)p) + fba ( 1 (x), p) + fba ( 2 (x), p) p p kx2 k ⇤ := (x) + ⇣(x) p @ 2 f tr (x) 1 = kx2 k(faa ( 1 (x), p) 2 + faa ( 2 (x), p)) + fab ( 1 (x), p) + fab ( 2 (x), p) @x1 @p p kx2 k ⇤ := (x) + ⇣(x) p @ 2 f tr (x) xT h 1 1 1 = 2 ( faa ( 1 (x), p)(kx2 k 2 ) fba ( 1 (x), p)) fa ( 1 (x), p)( 2 ) @p@x2 kx2 k p p p i + (faa ( 2 (x), p)kx2 k + fba ( 2 (x), p))p + fa ( 2 (x), p) xT2 1 ( faa ( 1 (x), p) 2 + faa ( 2 (x), p)p2 ) p p. =. 1 1 xT2 fa ( 1 (x), p) p + fa ( 2 (x), p)p xT2 + ( fba ( 1 (x), p) + fba ( 2 (x), p)p) + kx2 k p p kx2 k T T x x := 2 ( ⇤ (x) + ↵⇤ (x)) + 2 ⌘(x) p kx2 k 2 tr @ f (x) x2 1 = (fa ( 1 (x), p) 2 + fa ( 2 (x), p)) @x2 @p kx2 k p x2 1 1 x2 + kx2 k(faa ( 1 (x), p)( ) 2 + faa ( 2 (x), p)( p)) kx2 k p p kx2 k x2 1 x2 + fab ( 1 (x), p)( ) + fab ( 2 (x), p)( p) kx2 k p kx2 k x2 x2 := ( ⇤ (x) + ↵⇤ (x)) + ⌘(x) p kx2 k h 1 1 2 @ 2 f tr (x) = kx k (faa ( 1 (x), p)(kx2 k 2 ) + fba ( 1 (x), p)) 2 + fa ( 1 (x), p)( 3 ) 2 2 @p p p p i + faa ( 2 (x), p)kx2 k + fba ( 2 (x), p). 1 ) + fbb ( 1 (x), p) + fab ( 2 (x), p)kx2 k + fbb ( 2 (x), p) p2 kx2 k2 1 2kx2 k = (faa ( 1 (x), p) 2 + faa ( 2 (x), p)p2 ) fa ( 1 (x), p) 2 p p p3 kx2 k 1 1 + (fba ( 1 (x), p) + fba ( 2 (x), p)p + fab ( 1 (x), p) + fab ( 2 (x), p)p) p p p + fbb ( 1 (x), p) + fbb ( 2 (x), p) + fab ( 1 (x), p)(kx2 k. :=. kx2 k2 2kx2 k ⇤ (x) + ⌘ (x) + 2 p p. ⇤. (x) 5. 2kx2 k fa ( 1 (x), p). p3.

(10) Then the Hessian matrix of trace function r2 f tr (x) = 2 xT kx2 k ⇤ 2 (x) (x) (x) + ⇣(x) kx2 k p 6 T T x x x x 2 2 x x ⇤ 2 2 6 (x) (I kx2 k22 )↵(x) + kx2 k22 (x) ( (x) + ↵⇤ (x)) + kxx22 k ⌘(x) kx2 k p 4 xT xT kx2 k ⇤ (x) + ⇣(x) p2 ( ⇤ (x) + ↵⇤ (x)) + kx22 k ⌘(x) C p where C = ↵(x) =. kx2 k2 p2. fa (. 1 (x),. (x) +. 2kx2 k ⇤ ⌘ (x) p. p) p1 +fa (. 2 (x),. ⇤. +. (x). p)p. 2kx2 k fa ( 1 (x), p3. ↵⇤ (x) =. kx2 k. ⇤. (x) = faa ( 1 (x), p) + faa ( 2 (x), p) faa ( 1 (x), p) p1 + faa ( 2 (x), p)p. ⇤. (x) = faa ( 1 (x), p) p12 + faa ( 2 (x), p)p2. ⇤. (x) =. fa (. 3. 7 7 (12) 5. p) and. 1 (x),. p) p1 +fa (. 2 (x),. p)p. kx2 k. (x) = fbb ( 1 (x), p) + fbb ( 2 (x), p). (x) = faa ( 1 (x), p) p1 + faa ( 2 (x), p)p. (x) =. faa ( 1 (x), p) p12 + faa ( 2 (x), p)p2. ⇣(x) = fab ( 1 (x), p) + fab ( 2 (x), p) ⌘(x) =. fab ( 1 (x), p) p1 + fab ( 2 (x), p)p. ⌘ ⇤ (x) = fab ( 1 (x), p) p1 + fab ( 2 (x), p)p.. In this section, we focus on establishing convexity of trace function on L(p). To do this, we show that f tr is convex under our assumption r2 f tr is positive semi-definite on L(p) \ E(p). Theorem 2.1. Suppose that f : R2 ! R is a twice di↵erentiable convex function and r2 f tr is positive semi-definite on L(p) \ E(p), then f tr is convex on L(p) if fa < 0. Proof. Case 1. Suppose that x, y,. x + (1. )y 2 L(p) \ E(p), 0 . g( ) := f tr (y + (x.  1 and let. y)).. Then we get g 00 ( ) = (x y)T r2 f tr (y + (x y))(x y). By our assumption, r2 f tr ⌫ 0 on L(p) \ E(p). Thus, g 00 ( ). 0 if 0 .  1.. That is, f tr (x) + (1. )f tr (y) =. g(1) + (1 tr. )g(0). g( ) = f ( x + (1 Hence, f tr is convex when x, y 2 L(p) \ E(p).. 6. )y)..

(11) Case 2. Suppose that x 2 L(p) \ E(p), y 2 E(p) and let u(t) = y + t(x Since f tr is continuous,. y), 0 < t < 1.. lim f tr (u(t)) = f tr (y). t!0. )u(t) 2 L(p) \ E(p), 0   1, by Case 1., if fa  0 and 0 < p < 1, h i f tr (x) + (1 )f tr (y) = lim f tr (x) + (1 )f tr (u(t)). As x + (1. t!0. lim f tr ( x + (1 t!0. )u(t)) = f tr ( x + (1. )y).. Hence, f tr is convex when x 2 L(p) \ E(p), y 2 E(p). Case 3. Suppose that x, y 2 L(p) \ E(p) and there exists a t0 such that 0 < t0 < 1 and t0 x + (1 t0 )y 2 E(p). Let x = (x1 , x2 , px ), y = (y1 , y2 , py ) and t0 x + (1 t0 )y = (t0 x1 + (1 t0 )y1 , 0, t0 px + (1 t0 t0 x2 = y2 ) kx2 k = ky2 k. 1 t0 1 t0 For 0 <. = =. h. t0 )py ). (13) (14). < 1,. f tr (x) + (1 )f tr (y) h i f ( 1 (x), px ) + f ( 2 (x), px ) + (1 f ( 1 (x), px ) + (1. Since f is convex,. h i ) f ( 1 (y), py ) + f ( 2 (y), py ) i h i )f ( 2 (y), py ) + f ( 2 (x), px ) + (1 )f ( 1 (y), py ). f ( ( 1 (x), px ) + (1 )( 2 (y), py )) + f ( ( 2 (x), px ) + (1 )( 1 (y), py )) 1 = f (( x1 kx2 k , px ) + ((1 )y1 + (1 )ky2 kpy , (1 )py )) px 1 +f (( x1 + kx2 kpx , px ) + ((1 )y1 (1 )ky2 k , (1 )py )) py 1 = f ( x1 + (1 )y1 + ( kx2 k ) + (1 )ky2 kpy , px + (1 )py ) px 1 +f ( x1 + (1 )y1 + kx2 kpx + (1 )( ky2 k ), px + (1 )py ). py. 7.

(12) (i) If. = t0 , by applying (14), we have t0 f tr (x) + (1 f (t0 x1 + (1 +f (t0 x1 + (1 = f (t0 x1 + (1 +f (t0 x1 + (1. t0 )f tr (y) t0 )y1 + t0 ( kx2 k. 1 ) + (1 px. t0 )y1 + t0 kx2 kpx + (1. t0 )ky2 kpy , t0 px + (1 t0 )( ky2 k. 1 ), t0 px + (1 py. t0 )py ) t0 )py ). 1 ), t0 px + (1 t0 )py ) px 1 t0 )y1 + t0 kx2 k(px ), t0 px + (1 t0 )py ). py. t0 )y1 + t0 kx2 k(py. Suppose that 0 < px < 1, 0 < py < 1, t0 x1 + (1. t0 )y1 + t0 kx2 k(py. t0 x1 + (1. t0 )y1 + t0 kx2 k(px. 1 )  t0 x1 + (1 px 1 )  t0 x1 + (1 py. t0 )y1. (15). t0 )y1 .. (16). If fa < 0, then t0 f tr (x) + (1. t0 )f tr (y). 1 ), t0 px + (1 t0 )py ) px 1 +f (t0 x1 + (1 t0 )y1 + t0 kx2 k(px ), t0 px + (1 t0 )py ) py 2f (t0 x1 + (1 t0 )y1 , t0 px + (1 t0 )py ) = f tr (t0 x + (1 t0 )y). f (t0 x1 + (1. t0 )y1 + t0 kx2 k(py. (ii) Let z0 = t0 x + (1 t0 )y. If t0 < < 1, x + (1 )y 2 [x, z0 ] and there exists a µ = x + (1. )y = µx + (1. t0 1 t0. µ)z0 .. By Case 2., if fa < 0 and 0 < px < 1, 0 < py < 1, h f tr (x) + (1 )f tr (y) = µf tr (x) + (1 µ) t0 f tr (x) + (1 µf tr (x) + (1 f tr (µx + (1. 8. such that. t0 )f tr (y). µ)f tr (z0 ). µ)z0 ) = f tr ( x + (1. )y).. i.

(13) Figure 1: t0 <. (iii) If 0 <. <1. < t0 , x + (1. Figure 2: 0 <. )y 2 [z0 , y] and there exists a ⌫ = x + (1. )y = ⌫z0 + (1. t0. < t0. such that. ⌫)y.. By Case 2., if fa < 0 and 0 < px < 1, 0 < py < 1, h i f tr (x) + (1 )f tr (y) = ⌫ t0 f tr (x) + (1 t0 )f tr (y) + (1 ⌫f tr (z0 ) + (1 f tr (⌫z0 + (1. ⌫)f tr (y). ⌫)f tr (y). ⌫)y) = f tr ( x + (1. )y).. Hence, f tr is convex when x, y 2 L(p) \ E(p) and there exists a t0 such that 0 < t0 < 1 and t0 x + (1 t0 )y 2 E(p). Case 4. For any x = (x1 , 0, p) 2 E(p), 1 (x) = x1 , 2 (x) = x1 and f tr (x) = 2f (x1 , p). Suppose that x = (x1 , 0, px ), y = (y1 , 0, py ) 2 E(p) and x + (1. )y = ( x1 + (1. )y1 , 0,. px + (1. )py ), 0 .  1.. We have f tr (x) = 2f (x1 , px ) f tr (y) = 2f (y1 , py ) f tr ( x + (1. )y) = 2f ( x1 + (1. 9. )y1 ,. px + (1. )py )..

(14) Since f is convex, f tr (x) + (1. )f tr (y) =. 2f (x1 , px ) + (1 2f ( x1 + (1. )2f (y1 , py ). )y1 ,. tr. = f ( x + (1. px + (1. )py ). )y).. Hence, f tr is convex when x, y 2 E(p). ⇤. Therefore, f tr is convex in L(p).. 3. Main Results. Theorem 3.1 (Schur Complement Theorem). [6] Let A 2 Rm⇥m be a symmetric positive definite matrix, C 2 Rn⇥n be a symmetric matrix and B 2 Rm⇥n . Then,  A B ⌫ 0 () C B T A 1 B ⌫ 0 BT C and. . A B BT C. 0 () C. BT A 1B. 0.. Remark 3.2. Partition the matrix r2 f tr into 4 block matrices. 2 T 6 6 4. (x). x2 kx2 k (x) kx2 k ⇤ (x) + ⇣(x) p. (I. xT 2 p. (. x2 kx2 k. (x). x2 xT x xT 2 )↵(x) + kx22 k22 (x) kx2 k2 T ⇤ (x) + ↵⇤ (x)) + x2 ⌘(x) kx2 k. x2 p (. kx2 k ⇤ (x) + ⇣(x) p ⇤ (x) + ↵⇤ (x)) + x2 ⌘(x) kx2 k. C. 3. 7 7= 5. . A BT. Theorem 3.3. Suppose that f : R2 ! R is a twice di↵erentiable convex function, then the matrix A is positive semi-definite in L(p) \ E(p) if and only if ↵(x) 0. Proof. Let f be a twice di↵erentiable convex function. Then faa. 0. 8(a, b) 2 R2 .. It means that (x) = faa ( 1 (x), p) + faa ( 2 (x), p) Case 1. If (x) = 0, then the matrix " A=. 0 0 (I. 10. 0.. 0 x2 xT 2 )↵(x) kx2 k2. #. .. B C.

(15) For any v = (v1 , v2 ) 2 R ⇥ Rn 1 , we obtain that v T Av = v2T (I. x2 xT2 )↵(x)v2 = (kv2 k2 kx2 k2. < v2 , x 2 > 2 )↵(x). kx2 k2. Since < v2 , x2 > kv2 kkx2 k, < v2 , x 2 > 2 kx2 k2. kv2 k2. kv2 k2 kx2 k2 = 0. kx2 k2. kv2 k2. Thus, the matrix A is positive semi-definite if and only if ↵(x) x xT. 0.. x xT. Case 2. If (x) > 0, (x) is positive definite and (I kx22 k22 )↵(x)+ kx22 k22 (x) is a symmetric matrix. Then, by the Schur Complement Theorem, A is positive semi-definite if and only if x2 xT2 x2 xT2 )↵(x) + (x) kx2 k2 kx2 k2. (I. x2 xT2 ( (x))2 kx2 k2 (x). is positive semi-definite. For any v = va + vb 2 Rn 1 , where va 2 {x 2 Rn {x 2 R. n 1. 1. : x = kx2 , k 2 R}, vb 2 {x 2 Rn. : x = kx2 , k 2 R} + {x 2 R. n 1. 1. :< x, x2 >= 0}. :< x, x2 >= 0} = Rn 1 ,. we obtain that x2 xT2 x2 xT2 x2 xT2 ( (x))2 )↵(x) + (x) ]v kx2 k2 kx2 k2 kx2 k2 (x) T x2 xT2 (x) (x) ( (x))2 T x2 x2 = v T (I )v↵(x) + v v kx2 k2 kx2 k2 (x) T T x2 x2 (x) (x) ( (x))2 T x2 x2 = vaT (I )v ↵(x) + v v a a a kx2 k2 kx2 k2 (x) T T x2 x2 (x) (x) ( (x))2 T x2 x2 +vbT (I )v ↵(x) + v v b b b kx2 k2 kx2 k2 (x) 2 (x) (x) ( (x)) = kva k2 + kvb k2 ↵(x). (x) v T [(I. Since faa ( 1 (x), p)faa ( 2 (x), p)( p1 + p)2 (x) (x) ( (x))2 = (x) faa ( 1 (x), p) + faa ( 2 (x), p) the matrix A is positive semi-definite if and only if ↵(x) 11. 0.. 0,.

(16) ⇤ Corollary 3.3.1. If fa  0, then ↵(x) in L(p) \ E(p) if fa  0.. 0; that is, the matrix A is positive semi-definite. Proof. Let fa  0. Since fa is an increasing function, if. 1 (x). . 2 (x),. fa ( 1 (x), p)  fa ( 2 (x), p)  0. Then, (i) If fa ( 2 (x), p) = 0, ↵(x) =. fa ( 1 (x), p) p1. 0.. kx2 k. (ii) If fa ( 1 (x), p) = 0, 0 = fa ( 1 (x), p)  fa ( 2 (x), p)  0 ) fa ( 2 (x), p) = 0 ) ↵(x) = 0.. (iii) If fa ( 1 (x), p) < 0 and fa ( 2 (x), p) < 0, ↵(x) =. fa ( 1 (x), p) p1 + fa ( 2 (x), p)p. Because ffaa (( Thus, ↵(x). kx2 k 2 (x), 1 (x),. p) p). 0..  1 and. 1 p2. fa ( 1 (x), p)p fa ( 2 (x), p) ( kx2 k fa ( 1 (x), p). =. p) 1, we have ( fa (2 (x), 1 (x), p). 1 ) p2. Therefore, by Theorem 3.3., A is positive semi-definite if fa  0.. 1 ) p2.  0. ⇤. Remark 3.4. Note that if f is twice di↵erentiable convex and fa  0, we can verify that ↵(x) =. fa (. 1 (x),. p) p1 +fa ( kx2 k. 2 (x),. p)p. 0. ⇤. faa ( 1 (x), p) p1 + faa ( 2 (x), p)p. ⇤. (x) = faa ( 1 (x), p) + faa ( 2 (x), p) (x) =. ↵⇤ (x) =. 0. (x) = faa ( 1 (x), p) p12 + faa ( 2 (x), p)p2. 0. ⇤. fa (. 1 (x),. p) p1 +fa ( kx2 k. 2 (x),. p)p. 0. (x) = fbb ( 1 (x), p) + fbb ( 2 (x), p). 0. (x) = faa ( 1 (x), p) p1 + faa ( 2 (x), p)p. (x) =. faa ( 1 (x), p) p12 + faa ( 2 (x), p)p2. ⇣(x) = fab ( 1 (x), p) + fab ( 2 (x), p) ⌘(x) =. fab ( 1 (x), p) p1 + fab ( 2 (x), p)p. ⌘ ⇤ (x) = fab ( 1 (x), p) p1 + fab ( 2 (x), p)p.. Theorem 3.5. Suppose that f : R2 ! R is a twice di↵erentiable convex function. C is positive in L(p) \ E(p) if fa  0. 12. 0.

(17) Proof. For C =. kx2 k2 p2. kx2 k2 faa ( 1 (x), p4 kx2 k f ( (x), p2 ab 1. (x) +. 2kx2 k ⇤ ⌘ (x) p. ⇤. +. (x). 2kx2 k fa ( 1 (x), p3. p), it represented by. kx2 k f ( (x), p2 ab 1. p) p). 2kx2 k fa ( 1 (x), p). p3. p) kx2 k2 faa ( 2 (x), p) kx2 kfab ( 2 (x), p) + kx2 kfab ( 2 (x), p) fbb ( 2 (x), p) fbb ( 1 (x), p). To begin with, we split the first term into three matrices h. kx2 k p2. i  f ( (x), p) f ( (x), p) aa 1 ab 1 1 fab ( 1 (x), p) fbb ( 1 (x), p). ". kx2 k p2. 1. #. .. By the inequality of arithmetic and geometric means, we obtain that kx2 k2 2kx2 k faa ( 1 (x), p) + fbb ( 1 (x), p) + fab ( 1 (x), p) 4 p p2 kx2 k p 2kx2 k 2 2 faa ( 1 (x), p)fbb ( 1 (x), p) + fab ( 1 (x), p) p p2 p It shows that faa ( 1 (x), p)fbb ( 1 (x), p) |fab ( 1 (x), p)| and det. . faa ( 1 (x), p) fab ( 1 (x), p) = faa ( 1 (x), p)fbb ( 1 (x), p) fab ( 1 (x), p) fbb ( 1 (x), p). 0.. (fab ( 1 (x), p))2. 0.(17). The positive semi-definite condition of this symmetric matrix is equivalence to faa ( 1 (x), p) 0, fbb ( 1 (x), p) 0 and determinant of the matrix being also more than or equal to zero. Therefore, the first term is positive. Similarly, splitting the second term, we have   ⇥ ⇤ faa ( 2 (x), p) fab ( 2 (x), p) kx2 k . kx2 k 1 fab ( 2 (x), p) fbb ( 2 (x), p) 1 By the inequality of arithmetic and geometric means, we obtain that. It shows that det. . p. kx2 k2 faa ( 2 (x), p) + fbb ( 2 (x), p) + 2kx2 kfab ( 2 (x), p) p 2kx2 k faa ( 2 (x), p)fbb ( 2 (x), p) + 2kx2 kfab ( 2 (x), p). 0.. |fab ( 2 (x), p)| and. faa ( 2 (x), p)fbb ( 2 (x), p). faa ( 2 (x), p) fab ( 2 (x), p) = faa ( 2 (x), p)fbb ( 2 (x), p) fab ( 2 (x), p) fbb ( 2 (x), p). Therefore, the second term is also positive. 13. (fab ( 2 (x), p))2. 0.(18).

(18) 2k For the last term, it is clear that 2kx fa ( 1 (x), p) p3 Hence, C is positive if fa  0. ⇤. 0 if fa  0.. Now we know that the matrix A is positive semi-definite and C is also positive in L(p) \ E(p) if fa  0. Unfortunately, we want to claim that r2 f tr (x) is positive semidefinite in L(p) \ E(p) but the result is totally di↵erent from what we had expected. The following description will illustrate it. Let f : R2 ! R now be a twice di↵erentiable convex function with fa  0, we consider the following 3 cases. Case 1. When C = 0 and (x) = 0, we obtain that faa ( 1 (x), p) = faa ( 2 (x), p) = 0 ). (x) =. ⇤. (x) = (x) =. ⇤. (x) = 0. and by the inequality (17), (18), (fab ( 1 (x), p))2 (fab ( 2 (x), p)). 0 ) fab ( 1 (x), p) = 0. 2. 0 ) fab ( 2 (x), p) = 0.. It show that ⇣(x) = ⌘(x) = ⌘ ⇤ (x) = 0. Furthermore, fbb ( 1 (x), p) = fbb ( 2 (x), p) = fa ( 1 (x), p) = 0 Since fa is an increasing function, if. 1 (x). . 2 (x),. 0 = fa ( 1 (x), p)  fa ( 2 (x), p)  0 ) fa ( 2 (x), p) = 0. ) ↵(x) = ↵⇤ (x) = 0.. Therefore, the Hessian matrix of trace function 2 3 0 0 0 r2 f tr (x) = 4 0 0 0 5 0 0 0 This result is nonsense for us.. Case 2. When C > 0 and (x) = 0, we obtain that 2. 0 6 r2 f tr (x) = 4 0 (I 0. 0 x2 xT 2 )↵(x) kx2 k2 T x2 ⇤ ↵ (x) p. 3. 0 x2 ⇤ ↵ (x) p ⇤. (x). 2kx2 k fa ( 1 (x), p3. p). 7 5.. Then, since C is positive definite and A is symmetric, by the Schur Complement. 14.

(19) Theorem, we have A. 2. 0 BC 1 B T = 4 0 (I. 0 x2 xT 2 )↵(x) kx2 k2. x2 xT 2 p2. ↵⇤ (x)2 2kx2 k fa ( 1 (x), p3. ⇤ (x). p). For any v = (v1 , v2 ) 2 R ⇥ Rn 1 , v T [A. BC. 1. B T ]v = v2T (I. x2 xT2 )v2 ↵(x) kx2 k2. v2T. 3. 5.. ↵⇤ (x)2. x2 xT2 v2 p2. ⇤ (x). 2kx2 k fa ( 1 (x), p3. p). .. Let v2 = va + vb 2 Rn 1 , where va 2 {x 2 Rn {x 2 R. n 1. 1. : x = kx2 , k 2 R}, vb 2 {x 2 Rn. : x = kx2 , k 2 R} + {x 2 R. n 1. 1. :< x, x2 >= 0}. :< x, x2 >= 0} = Rn 1 ,. we obtain that vaT (I vaT. =. x2 xT2 )va ↵(x) kx2 k2. x2 xT2 va p2. ⇤ (x). vaT. x2 xT2 va p2. ↵⇤ (x)2 ⇤ (x). ↵⇤ (x)2. 2kx2 k fa ( 1 (x), p3. p). 2kx2 k fa ( 1 (x), p3. p). 0. and vbT (I. x2 xT2 )vb ↵(x) kx2 k2. vbT. x2 xT2 vb p2. ↵⇤ (x)2 ⇤ (x). 2kx2 k fa ( 1 (x), p3. p). = kvb k2 ↵(x). 0.. Hence, if fa  0, r2 f tr (x) is negative semi-definite on the direction which is parallel to x2 , but is positive semi-definite on the direction which is orthogonal to x2 . Case 3. When C > 0 and (x) > 0, from (12), we have the Hessian matrix of trace function r2 f tr (x) = 2 3 xT kx2 k ⇤ 2 (x) (x) (x) + ⇣(x) kx2 k p 6 7 x2 xT x2 xT x2 x2 ⇤ x2 ⇤ 2 2 6 7 (x) (I )↵(x) + (x) ( (x) + ↵ (x)) + ⌘(x) kx2 k kx2 k2 kx2 k2 p kx2 k 4 5 xT xT kx2 k ⇤ ⇤ ⇤ 2 2 (x) + ⇣(x) p ( (x) + ↵ (x)) + kx2 k ⌘(x) C p 2. 2k where C = kxp22k (x) + 2kxp2 k ⌘ ⇤ (x) + ⇤ (x) 2kx fa ( 1 (x), p). p3 Since C is positive definite and the matrix A is symmetric, by the Schur Comple-. 15.

(20) ment Theorem, we have A 2 kx2 k ⇤ 2 (. 4. x2 [ kx2 k. (. kx2 k p. BC. 1. BT =. +⌘). p. xT 2 [ kx2 k. C. ⇤ +⇣)( kx2 k ( ⇤ +↵⇤ )+⌘) p. C. ] (I. x2 xT 2 )↵ kx2 k2. (. kx2 k p. ⇤ +⇣)( kx2 k ( ⇤ +↵⇤ )+⌘) p. C. +. (. x2 xT 2 [ kx2 k2. 3. ]. kx2 k ⇤ ( +↵⇤ )+⌘)2 p. C. 5.. ]. Now we need to check whether the upper-left and the lower-right term is positive or not, and it is clear that ( kxp2 k. ⇤. + ⌘)2. C ( kxp2 k (. ⇤. + ↵⇤ ) + ⌘)2 C. kx2 k p. ⇤. + ⌘)2. kx2 k ( p. ⇤. + ↵⇤ ) + ⌘)2. 0. (). C. (. 0. (). C. (. 0 0.. Using the technique of completing the square and the inequality (17), (18), we have C =. (. kx2 k p. ⇤. + ⌘)2. kx2 k2 1 faa ( 1 (x), p)faa ( 2 (x), p)( p)2 2 p p 2kx2 k 1 + (faa ( 2 (x), p)fab ( 1 (x), p) faa ( 1 (x), p)fab ( 2 (x), p))( p p +faa ( 1 (x), p)fbb ( 1 (x), p) + faa ( 1 (x), p)fbb ( 2 (x), p). p). +faa ( 2 (x), p)fbb ( 1 (x), p) + faa ( 2 (x), p)fbb ( 2 (x), p) (fab ( 1 (x), p))2 2fab ( 1 (x), p)fab ( 2 (x), p) (fab ( 2 (x), p))2 2kx2 k fa ( 1 (x), p) (x) p3 = faa ( 1 (x), p)faa ( 2 (x), p) ⇣ kx k 1 faa ( 2 (x), p)fab ( 1 (x), p) faa ( 1 (x), p)fab ( 2 (x), p) ⌘2 2 ( p) + p p faa ( 1 (x), p)faa ( 2 (x), p) nh i 1 2 + faa ( 1 (x), p)faa ( 2 (x), p) + (faa ( 2 (x), p)) faa ( 1 (x), p)faa ( 2 (x), p) h i faa ( 1 (x), p)fbb ( 1 (x), p) (fab ( 1 (x), p))2 + h i faa ( 1 (x), p)faa ( 2 (x), p) + (faa ( 1 (x), p))2 h io 2 faa ( 2 (x), p)fbb ( 2 (x), p) (fab ( 2 (x), p)) 2kx2 k fa ( 1 (x), p)(faa ( 1 (x), p) + faa ( 2 (x), p)) p3. 0.. 16.

(21) and kx2 k ⇤ ( + ↵⇤ ) + ⌘)2 p kx2 k2 n = 4faa ( 1 (x), p)faa ( 2 (x), p) p2 2 1 1 ( faa ( 1 (x), p)fa ( 1 (x), p) 3 faa ( 1 (x), p)fa ( 2 (x), p) kx2 k p p 3 +faa ( 2 (x), p)fa ( 1 (x), p)p + faa ( 2 (x), p)fa ( 2 (x), p)p ) o 1 2 1 2 2 ((fa ( 1 (x), p)) 2 + 2fa ( 1 (x), p)fa ( 2 (x), p) + (fa ( 2 (x), p)) p ) kx2 k2 p n 2kx2 k 1 + 2fab ( 1 (x), p)faa ( 2 (x), p)p + 2fab ( 2 (x), p)faa ( 1 (x), p) p p 1 1 ( fa ( 1 (x), p)fab ( 1 (x), p) 2 + fa ( 1 (x), p)fab ( 2 (x), p) kx2 k p o fa ( 2 (x), p)fab ( 1 (x), p) + fa ( 2 (x), p)fab ( 2 (x), p)p2 ) n 1 1 + faa ( 1 (x), p)fbb ( 1 (x), p) 2 + faa ( 1 (x), p)fbb ( 2 (x), p) 2 p p 2 2 +faa ( 2 (x), p)fbb ( 1 (x), p)p + faa ( 2 (x), p)fbb ( 2 (x), p)p o 1 (fab ( 1 (x), p))2 2 + 2fab ( 1 (x), p)fab ( 2 (x), p) (fab ( 2 (x), p))2 p2 p 2kx2 k 1 fa ( 1 (x), p)(faa ( 1 (x), p) 2 + faa ( 2 (x), p)p2 ). 3 p p C. (. On the other hand, C ( kxp2 k ( ⇤ + ↵⇤ ) + ⌘)2 will be positive, negative or zero as we give various twice di↵erentiable convex functions. For example, let f (a, b) = e a b ,. fa (a, b) = fb (a, b) = e a b  0 faa (a, b) = fab (a, b) = fbb (a, b) = e. a+b. 8(a, b) 2 R2. 0. and det. . faa ( 1 (x), p) fab ( 1 (x), p) = (e fab ( 1 (x), p) fbb ( 1 (x), p). For any x 2 L(p) with. 1 (x). > 0,. 2 (x). 17. a+b. > 0 and. )(e. a+b. 1 (x). ). . (e 2 (x). a+b 2. ) = 0.. from the definition,.

(22) we have C =. (. kx2 k ( p. ⇤. + ↵⇤ ) + ⌘)2. 4kx2 k2 f ( 1 (x), p)f ( 2 (x), p) p2 2 4 4 +( 3 + 2 + + 4)kx2 kf ( 1 (x), p)f ( 2 (x), p) + 2pkx2 k(f ( 2 (x), p))2 p p p 1 2 1 +( 4 )(f ( 1 (x), p))2 + ( 2 + 2 + p2 )f ( 1 (x), p)f ( 2 (x), p) 3 p p p 2 +(2p 1)(f ( 2 (x), p)) .. Then, in Rn+1 , let x = (10, 1, 0.5) 2 L(p) where kx2 k = 1, 1 kx2 k = 8, 2 (x) = x1 + kx2 kp = 10.5, p f ( 1 (x), p)f ( 2 (x), p) = e 19.5 , (f ( 1 (x), p))2 = e 17 , (f ( 2 (x), p))2 = e 1 (x). = x1. 22. and C. (. kx2 k ( p. ⇤. + ↵⇤ ) + ⌘)2 ⇡. 1.2353 ⇥ 10. 6. < 0.. It shows that r2 f tr (x) cannot be ensured whether be positive semi-definite or not. Unfortunately, in the proof of Theorem 2.1., Case 1. and Case 2. were constructed from that assumption. Hence, for any x 2 Rn+1 , f tr is not a convex function in L(p).. 4. Conclusions. In this paper, we pointed out that this trace function has obstacles to develop the convexity associated with a closed convex set L(p) extended by circular cone. However, this may occur from (10) that it cannot be symmetry when we di↵erentiate 1 (x) and 2 (x) with respect to p. According by this, we can modify our trace functions. For example, f tr (x) :=f ( 1 (x)) + f ( 2 (x)) + f (p) h i f tr (x) :=p f ( 1 (x), p) + f ( 2 (x), p). f tr (x) :=pf ( 1 (x), p) + p 1 f ( 2 (x), p). 8x = (x1 , x2 , p) 2 L(p). (19). 8x = (x1 , x2 , p) 2 L(p). (20). 8x = (x1 , x2 , p) 2 L(p).. (21). In the future, we will use the same way to verify that the convexity of trace functions which comes from (19), (20), (21), or others we have not explore now.. 18.

(23) Moreover, there is another way that let ✓ be our new parameter and n o ⇡ ✓⇤ n 1 L := x = (x1 , x2 , ✓) 2 R ⇥ R ⇥ (0, ) : kx2 k  x1 tan✓ 2. (22). kx2 kcot✓,. (23). where. 1 (x). = x1. 2 (x). = x1 + kx2 ktan✓,. then solve the Hessian matrix with respect to ✓.. References [1] Y.-L. Chang, C.-Y. Yang and J.-S. Chen Smooth and nonsmooth analyses of vector-valued functions associated with circular cones, Nonlinear Analysis: Theory, Methods and Applications, vol. 85, pp. 160-173, 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] Jinchuan Zhou, J.-S. Chen and H.-F. Hung Circular cone convexity and some inequalities associated with circular cones, Journal of Inequalities and Applications, 2013:571, 2013. [4] Jinchuan 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. [5] C.-Y. Yang, Y.-L. Chang 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. [6] R. A. Horn and C. R. Johnson, Matrix Analysis, Cambridge University Press, Cambridge, 1986. [7] T.-W. Wu and Y.-L. Chang, The convexity of circular cone trace functions, Master’s thesis, National Taiwan Normal University, Taipei, Taiwan, 2015.. 19.

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