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(1)

4 Ongoing work

In this section, we discuss the SOC-monotonicity for the function ln(x) proposed in [7].

For any x = (x 1 , x 2 ) ∈ int(K n ), ln(x) is well defined and

ln(x) =

 

 

 

1 2

Ã

ln ³ x 2 1 − kx 2 k 2 ´ , ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2 kx 2 k

!

if x 2 6= 0,

ln(x 1 )(1, 0) if x 2 = 0.

Proposition 4.1 For f : (0, ∞) → R be f (t) = ln(t) , then f is SOC-monotone on (0, ∞) .

Proof. For any x = (x 1 , x 2 ) Â

Kn

0 and y = (y 1 , y 2 ) Â

Kn

0 , we have the following

inequalities :

 

 

 

 

 

 

x 1 + kx 2 k ≥ x 1 − kx 2 k > 0 y 1 + ky 2 k ≥ y 1 − ky 2 k > 0

|hx 2 , y 2 i| ≤ kx 2 k · ky 2 k ≤ x 1 y 1 . Suppose x º

Kn

y which implies x − y º

Kn

0, we also know

x 1 − y 1 ≥ kx 2 − y 2 k ≥ 0 .

According to the condition of SOC-monotone (12), it suffices to show ln(x) º

Kn

ln(y) . We shall consider the following three cases : (i) x 2 = 0, y 2 = 0 (ii) x 2 6= 0, y 2 = 0 (iii) x 2 6= 0, y 2 6= 0.

Case(i) : x 2 = 0, y 2 = 0

Since ln x = (ln x 1 , 0) and ln y = (ln y 1 , 0), then ln x − ln y = (ln x 1 − ln y 1 , 0) ln x 1 − ln y 1 = ln x 1

y 1 > ln y 1

y 1 = ln 1 = 0.

Therefore ,

ln x − ln y º

Kn

0 ⇒ ln x º

Kn

ln y.

Case(ii) : x 2 6= 0, y 2 = 0 Since ln x = 1

2

Ã

ln ³ x 2 1 − kx 2 k 2 ´ , ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2

kx 2 k

!

and ln y = (ln y 1 , 0),

(2)

then ln x − ln y = 1 2

Ã

ln ³ x 2 1 − kx 2 k 2 ´ − 2 ln y 1 , ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2 kx 2 k

!

. We denote ln x − ln y := 1

2 ( Ξ 1 , Ξ 2 ) ,where

 

 

 

Ξ 1 = ln (x 2 1 − kx 2 k 2 ) − 2 ln y 1 Ξ 2 = ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2 kx 2 k .

To prove f is SOC-monotone, it suffices to verify two thing : Ξ 1 ≥ 0 and kΞ 2 k ≤ Ξ 1 . (1) First, we verify that Ξ 1 ≥ 0.

Ξ 1 = ln ³ x 2 1 − kx 2 k 2 ´ − 2 ln y 1 = ln det(x) − ln det(y) = ln det(x) det(y) .

Based on the Property 2.4(a), if x º

Kn

y, then det(x) ≥ det(y). Hence, we get Ξ 1 ≥ 0.

(2) Second, we remains to show that Ξ 2 1 − kΞ 2 k 2 ≥ 0 . Ξ 2 1 − kΞ 2 k 2

=

·

ln ³ x 2 1 − kx 2 k 2 ´ − 2 ln y 1

¸ 2

−

° °

° °

° ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2 kx 2 k

° °

° °

°

2

=

·

ln ³ x 2 1 − kx 2 k 2 ´ − 2 ln y 1

¸ 2

−

"

ln

à x 1 + kx 2 k x 1 − kx 2 k

! # 2

=

"

ln ³ x 2 1 − kx 2 k 2 ´ − 2 ln y 1 + ln

à x 1 + kx 2 k x 1 − kx 2 k

!#

·

"

ln ³ x 2 1 − kx 2 k 2 ´ − 2 ln y 1 − ln

à x 1 + kx 2 k x 1 − kx 2 k

!#

=

"

ln

à x 2 1 − kx 2 k 2

(ln y 1 ) 2 · x 1 + kx 2 k x 1 − kx 2 k

! #

·

"

ln

à x 2 1 − kx 2 k 2

(ln y 1 ) 2 · x 1 − kx 2 k x 1 + kx 2 k

! #

= 4

"

ln

à x 1 + kx 2 k y 1

! ¸

·

·

ln

à x 1 − kx 2 k y 1

! #

≥ 0.

Based on Property 2.4(b), since x º

Kn

y and y 2 = 0 which imply x 1 + kx 2 k ≥ y 1 and x 1 − kx 2 k ≥ y 1 , then

ln

à x 1 + kx 2 k y 1

!

≥ 1 and ln

à x 1 − kx 2 k y 1

!

≥ 1.

Hence, the last equality is greater than zero. We get Ξ 1 ≥ kΞ 2 k . Therefore,

ln x − ln y º

Kn

0 ⇒ ln x º

Kn

ln y.

(3)

Case(iii) : x 2 6= 0, y 2 6= 0 Since ln x = 1

2

Ã

ln ³ x 2 1 − kx 2 k 2 ´ , ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2 kx 2 k

!

and ln y = 1 2

Ã

ln ³ y 2 1 − ky 2 k 2 ´ , ln

à y 1 + ky 2 k y 1 − ky 2 k

! y 2 ky 2 k

!

then ln x − ln y

= 1 2

Ã

ln ³ x 2 1 − kx 2 k 2 ´ − ln ³ y 1 2 − ky 2 k 2 ´ , ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2

kx 2 k − ln

à y 1 + ky 2 k y 1 − ky 2 k

! y 2

ky 2 k

!

.

We denote ln x − ln y := 1

2 ( Ξ 1 , Ξ 2 ) ,where

 

 

 

Ξ 1 = ln (x 2 1 − kx 2 k 2 ) − ln (y 1 2 − ky 2 k 2 ) Ξ 2 = ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2 kx 2 k − ln

à y 1 + ky 2 k y 1 − ky 2 k

! y 2 ky 2 k .

To prove f is SOC-monotone, it suffices to verify two thing : Ξ 1 ≥ 0 and kΞ 2 k ≤ Ξ 1 . (1) First, we verify that Ξ 1 ≥ 0 .

Ξ 1 = ln det(x) − ln det(y) = ln det(x) det(y) .

Based on the Property 2.4(a), if x º

Kn

y, then det(x) ≥ det(y) . Hence , we get Ξ 1 ≥ 0.

(2) Second, we remains to show that Ξ 2 1 − kΞ 2 k 2 ≥ 0.

Ξ 2 1 − kΞ 2 k 2

=

·

ln ³ x 2 1 − kx 2 k 2 ´ − ln ³ y 1 2 − ky 2 k 2 ´ ¸ 2

−

° °

° °

° ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2

kx 2 k − ln

à y 1 + ky 2 k y 1 − ky 2 k

! y 2

ky 2 k

° °

° °

°

2

=

·

ln ³ x 2 1 − kx 2 k 2 ´ − ln ³ y 1 2 − ky 2 k 2 ´ ¸ 2

−

" Ã

ln x 1 + kx 2 k x 1 − kx 2 k

! 2

− 2 ln x 1 + kx 2 k

x 1 − kx 2 k · ln y 1 + ky 2 k

y 1 − ky 2 k · hx 2 , y 2 i kx 2 kky 2 k +

Ã

ln y 1 + ky 2 k y 1 − ky 2 k

! 2 #

≥

·

ln ³ x 2 1 − kx 2 k 2 ´ − ln ³ y 1 2 − ky 2 k 2 ´ ¸ 2

−

" Ã

ln x 1 + kx 2 k x 1 − kx 2 k

! 2

− 2 ln x 1 + kx 2 k

x 1 − kx 2 k · ln y 1 + ky 2 k

y 1 − ky 2 k · (−1) +

Ã

ln y 1 + ky 2 k y 1 − ky 2 k

! 2 #

=

·

ln ³ x 2 1 − kx 2 k 2 ´ − ln ³ y 1 2 − ky 2 k 2 ´ ¸ 2 −

"

ln λ 2 (x)

λ 1 (x) + ln λ 2 (y) λ 1 (y)

# 2

(4)

=

"

ln λ 1 (x)λ 2 (x)

λ 1 (y)λ 2 (y) − ln λ 2 (x)λ 2 (y) λ 1 (x)λ 1 (y)

#

·

"

ln λ 1 (x)λ 2 (x)

λ 1 (y)λ 2 (y) + ln λ 2 (x)λ 2 (y) λ 1 (x)λ 1 (y)

#

=

Ã

2 ln λ 1 (x) λ 2 (y)

!

·

Ã

2 ln λ 2 (x) λ 1 (y)

!

.

Here we meet some diffculties as below such that we can’t proceed further. We know two things :

(1) λ 2 (x) ≥ λ 1 (x), λ 2 (y) ≥ λ 1 (y),

(2) λ 2 (x) ≥ λ 2 (y), λ 1 (x) ≥ λ 1 (y) if x º

Kn

y.

But we can’t get whether λ 1 (x) > λ 2 (y) or λ 1 (x) < λ 2 (y). We need other skill to complete the proof.

2

Lemma 4.1 For any x = (x 1 , x 2 ) Â

Kn

0 and y = (y 1 , y 2 ) Â

Kn

0 , we have ln (det(αx + (α)y)) ≥ ln (det(x)) α · ln (det(y)) 1−α ∀ 0 < α < 1 .

Proof. The proof has already verified in Proposition 2.4 in [4]. 2

Lemma 4.2 For any x = (x 1 , x 2 ), y = (y 1 , y 2 ) ∈ K n , we have (a) (x 1 + y 1 ) 2 − ky 2 k 2 ≥ 4x 1

q

y 1 2 − ky 2 k 2 ,

(b) ³ x 1 + y 1 − ky 2 k ´ 2 ≥ 4x 1 ³ y 1 − ky 2 k ´ , (c) ³ x 1 + y 1 + ky 2 k ´ 2 ≥ 4x 1 ³ y 1 + ky 2 k ´ ,

(d) x 1 y 1 − hx 2 , y 2 i ≥ q x 2 1 − kx 2 k 2 · q y 1 2 − ky 2 k 2 ,

(e) (x 1 + y 1 ) 2 − kx 2 + y 2 k 2 ≥ 4 q x 2 1 − kx 2 k 2 · q y 1 2 − ky 2 k 2 .

Proof. The proof has already verified in Lemma 3.2 in [4].

2

(5)

Proposition 4.2 For f : (0, ∞) → R be f (t) = ln(t) , then f is SOC-concave on (0, ∞) . Proof. Since f (t) = ln(t) is continuous , then the condition (13) of concave can be replaced by the condition (14). It is clear that we need to show that

f

µ x + y 2

º

Kn

1 2

µ

f (x) + f (y)

.

Let us consider the following three cases : (i) x 2 = 0, y 2 = 0 (ii) x 2 6= 0, y 2 = 0 (iii) x 2 6= 0, y 2 6= 0.

Case(i) : x 2 = 0, y 2 = 0

Since ln x = (ln x 1 , 0) and ln y = (ln y 1 , 0) , then ln

µ x + y 2

=

µ

ln

µ x 1 + y 1 2

, 0

. Thus ,

f

µ x + y 2

− 1 2

µ

f (x) + f (y)

=

µ

ln

µ x 1 + y 1

2

, 0

− 1 2

µ

ln x 1 + ln y 1 , 0

=

µ

ln

µ x 1 + y 1 2

− 1

2 ln x 1 y 1 , 0

=

Ã

ln

à x

1

+y

1

√ 2

x 1 y 1

!

, 0

!

.

As we know x 1 + y 1

2 ≥ √

x 1 y 1 , it implies ln

à x

1

+y

1

√ 2

x 1 y 1

!

≥ 0 . Therefore,

f

µ x + y 2

º

Kn

1 2

µ

f (x) + f (y)

.

Case(ii) : x 2 6= 0, y 2 = 0 Since

ln x = 1 2

Ã

ln ³ x 2 1 − kx 2 k 2 ´ , ln

à x 1 + kx 2 k x 1 − kx 2 k

! x 2 kx 2 k

!

= 1 2

Ã

ln det(x) , ln λ 2 (x) λ 1 (x)

x 2 kx 2 k

!

ln y = ( ln y 1 , 0) = 1 2

³ det(y) , 0 ´

ln x + y

2 = 1

2

 ln det

µ x + y 2

, ln

à λ 2 ( x+y 2 ) λ 1 ( x+y 2 )

! x

2

2 kx

2

k

2

,

(6)

then f

µ x + y 2

− 1 2

µ

f (x) + f (y)

= 1 2

 ln det

µ x + y 2

, ln

à λ 2 ( x+y 2 )

x+y 2

! x

2

2 kx

2

k

2

− 1 2

"

1 2

Ã

ln det(x) + ln det(y) , ln λ 2 (x) λ 1 (x)

x 2 kx 2 k

!#

= 1 2

Ã

ln det

µ x + y 2

− 1

2 ( ln det(x) + ln det(y) ) ,

Ã

ln λ 2 ( x+y 2 ) λ 1 ( x+y 2 ) − 1

2 ln λ 2 (x) λ 1 (x)

!

· x 2 kx 2 k

!

= 1

2 (Ξ 1 , Ξ 2 ),

where

 

 

 

 

Ξ 1 = ln det

µ x + y 2

− 1

2 ( ln det(x) + ln det(y) ) Ξ 2 =

Ã

ln λ 2 ( x+y 2 ) λ 1 ( x+y 2 ) − 1

2 ln λ 2 (x) λ 1 (x)

!

· x 2 kx 2 k .

To prove f is SOC-concave , it suffices to verify two thing : Ξ 1 ≥ 0 and kΞ 2 k ≤ Ξ 1 . (1) First , we verify that Ξ 1 ≥ 0 .

Ξ 1 = ln det

µ x + y 2

− ³ ln det(x) 1/2 + ln det(y) 1/2 ´ .

Based on Lemma 4.1 , ln det

µ x + y 2

≥ ³ ln det(x) 1/2 + ln det(y) 1/2 ´ . Hence , we get Ξ 1 ≥ 0 .

(2) Second , we remains to show that Ξ 2 1 − kΞ 2 k 2 ≥ 0 .

Ξ 2 1 − kΞ 2 k 2

=

"

ln det

µ x + y 2

− 1

2 ( ln det(x) + ln det(y) )

# 2

−

° °

° °

° Ã

ln λ 2 ( x+y 2 ) λ 1 ( x+y 2 ) − 1

2 ln λ 2 (x) λ 1 (x)

!

· x 2

kx 2 k

° °

° °

°

2

=

"

ln det

µ x + y 2

− 1

2 ( ln det(x) + ln det(y) )

# 2

−

"

ln λ 2 ( x+y 2 ) λ 1 ( x+y 2 ) − 1

2 ln λ 2 (x) λ 1 (x)

# 2

=

"

ln det

µ x + y 2

− 1

2 ( ln det(x) + ln det(y) ) +

µ

ln λ 2 ( x+y 2 ) λ 1 ( x+y 2 ) − 1

2 ln λ 2 (x) λ 1 (x)

¶#

·

"

ln det

µ x + y 2

− 1

2 ( ln det(x) + ln det(y) ) −

µ

ln λ 2 ( x+y 2 ) λ 1 ( x+y 2 ) − 1

2 ln λ 2 (x) λ 1 (x)

¶#

(7)

=

"

ln

µ

λ 2

µ x + y 2

¶¶ 2

− 1 2

³ ln (λ 2 (x)) 2 + ln y 1 2 ´

#

·

"

ln

µ

λ 1

µ x + y 2

¶¶ 2

+ 1 2

³ ln (λ 1 (x)) 2 + ln y 2 1 ´

#

=

"

ln

µ x 1 + y 1

2 + ° ° ° x 2

2

° °

°

2

(x 1 + kx 2 k) · y 1

#

·

"

ln

µ x 1 + y 1

2 − ° ° ° x 2

2

° °

°

2

(x 1 − kx 2 k) · y 1

#

=

"

ln (x 1 + y 1 + kx 2 k) 2 4y 1 (x 1 + kx 2 k)

#

·

"

ln (x 1 + y 1 − kx 2 k) 2 4y 1 (x 1 − kx 2 k)

#

≥ 0.

Based on Lemma 4.2 (b),(c) , we have

³ x 1 + y 1 − kx 2 k ´ 2 ≥ 4y 1

³ x 1 − kx 2 k ´ and ³ x 1 + y 1 + kx 2 k ´ 2 ≥ 4y 1

³ x 1 + kx 2 k ´ .

Thus ,

ln (x 1 + y 1 − kx 2 k) 2

4y 1 (x 1 − kx 2 k) ≥ 1 and ln (x 1 + y 1 + kx 2 k) 2 4y 1 (x 1 + kx 2 k) ≥ 1.

Hence , the last equality is greater than zero . We get Ξ 1 ≥ kΞ 2 k . Therefore,

f

µ x + y 2

º

Kn

1 2

µ

f (x) + f (y)

.

Case(iii) : x 2 6= 0, y 2 6= 0

Here we meet the same diffculties as ln(x)-SOC monotone such that we can’t proceed further . We need other skill to complete the proof.

2

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