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ON YOUNG INEQUALITY UNDER EUCLIDEAN JORDAN ALGEBRA

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ALGEBRA

CHIEN-HAO HUANG, YU-HSUAN HSIAO, YU-LIN CHANG, AND JEIN-SHAN CHEN

Abstract. Recently, some inequalities and trace inequalities associated with second-order cone are established. Most of them are very useful in optimization.

In particular, in our recent work [14], we build up some trace versions of Young inequality in the SOC setting and indicate that the Young inequality does not hold in general. In this paper, we pay attention to Young inequality under Euclidean Jordan algebra. By using spectral decomposition, we extend one trace version of Young inequality to the general setting of symmetric cone. In addition, we provide conditions under which the Young inequality holds in the SOC setting.

Accordingly, one can construct counterexamples in general case.

1. Introduction

The Young inequality states that for any a, b ≥ 0, and p, q are positive real numbers such that 1p +1q = 1, there holds

ap p +bq

q ≥ ab,

where the equality holds if and only if ap = bq. The Arithmetic-Geometric-Mean inequality is a special case of Young inequality:

a2+ b2

2 ≥ ab.

It is very useful in real analysis, as a tool to prove the H¨older’s inequality. In addi- tion, it can be used to estimate the norm of nonlinear terms in PDE theory.

As indicated in [7, 10], the second-order cone (SOC) is often involved in many optimization problems, particularly in the context of applications and solutions methods for the second-order-cone program (SOCP) and second-order-cone comple- mentarity problem (SOCCP) [5, 6, 8, 9, 10]. For designing those solutions methods, spectral decomposition associated with SOC is required. Recently, Chang et al.[4]

defined various means associated with Lorentz cones (also known as second-order cones), which are new concepts and natural extensions of traditional arithmetic mean, harmonic mean, and geometric mean, logarithmic mean. Based on these

2010 Mathematics Subject Classification.

Key words and phrases. Euclidean Jordan algebra, Second-order Cone, Lorentz Cone, Young Inequality.

Jein-Shan Chen. Corresponding author. The author’s work is supported by Ministry of Science and Technology, Taiwan.

1

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means defined on Lorentz cone, some inequalities and trace inequalities are estab- lished, see [10, Chpater 4].

In our recent work [14], we build up some trace versions of Young inequality in the SOC setting and indicate that the Young inequality does not hold in general. In this paper, we pay attention to Young inequality under Euclidean Jordan algebra.

By using spectral decomposition, we extend one trace version of Young inequality to the general setting of symmetric cone. In addition, we provide conditions under which the Young inequality holds in the SOC setting. More specifically, we conclude that the Young inequality associated with SOC holds under one of the following two conditions holds: (i) p = q = 2 or (ii) any two vectors share the same Jordan frame.

Accordingly, one can construct counterexamples in general case.

2. Preliminaries

A Euclidean Jordan algebra [11] is a finite dimensional inner product space (V, h·, ·i) (V for short) over the field of real numbers R equipped with a bilinear map (x, y) 7→ x ◦ y : V × V −→ V, which satisfies the following conditions:

(i) x ◦ y = y ◦ x for all x, y ∈ V;

(ii) x ◦ (x2◦ y) = x2◦ (x ◦ y) for all x, y ∈ V;

(iii) hx ◦ y, zi = hx, y ◦ zi for all x, y, z ∈ V,

where x2 := x ◦ x, and x ◦ y is called the Jordan product of x and y. If a Jordan product only satisfies the conditions (i) and (ii) in the above definition, the algebra V is said to be a Jordan algebra. Moreover, if there is an (unique) element e ∈ V such that x ◦ e = x for all x ∈ V, the element e is called the identity element in V.

Note that a Jordan algebra does not necessarily have an identity element. Through- out this paper, we assume that V is a Euclidean Jordan algebra with an identity element e.

In a given Euclidean Jordan algebra V, the set of squares K := {x2: x ∈ V} is a symmetric cone [11, Theorem III.2.1]. This means that K is a self-dual closed convex cone and, for any two elements x, y ∈ int(K), there exists an invertible linear trans- formation Γ : V −→ V such that Γ(x) = y and Γ(K) = K. Accordingly, there is a natural partial order in V. We write x K y if x − y ∈ K, and x K y if x − y ∈ intK.

For any given x ∈ V, we denote m(x) the degree of the minimal polynomial of x, that is,

m(x) :=

n

k > 0 | {e, x, · · · , xk} is linearly dependento .

Since m(x) ≤ dim(V) where dim(V) is the dimension of V, the rank of V is well- defined by r := max{m(x) | x ∈ V}. In Euclidean Jordan algebra V, an element e(i) ∈ V is an idempotent if (e(i))2 = e(i), and it is a primitive idempotent if it is nonzero and cannot be written as a sum of two nonzero idempotents. The idempo- tents e(i)and e(j)are said to be orthogonal if e(i)◦ e(j)= 0. In addition, we say that

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a finite set {e(1), e(2), · · · , e(r)} of primitive idempotents in V is a Jordan frame if e(i)◦ e(j)= 0 for i 6= j, and

r

X

i=1

e(i) = e.

Note that he(i), e(j)i = he(i)◦ e(j), ei whenever i 6= j. There also exist the so-called spectral decomposition for any element x in V, see below theorem.

Theorem 2.1. [11, Theorem III.1.2] Let V be a Euclidean Jordan algebra. Then there is a number r such that, for every x ∈ V, there exists a Jordan frame {e(1), . . . , e(r)} and real numbers λ1(x), · · · , λr(x) with

x = λ1(x)e(1)+ · · · + λr(x)e(r).

Here, the numbers λi(x) (i = 1, · · · , r) are the spectral values of x, the expression λ1(x)e(1) + · · · + λr(x)e(r) is the spectral decomposition of x. Moreover, tr x :=

Pr

i=1λi(x) is called the trace of x, and det(x) =Qr

i=1λi(x) is call the determinant of x.

Given a Euclidean Jordan algebra V with dim(V) = n > 1, from Proposition III 4.4-4.5 and Theorem V.3.7 in [11], we know that any Euclidean Jordan algebra V and its corresponding symmetric cone K are, in a unique way, a direct sum of simple Euclidean Jordan algebras and the constituent symmetric cones therein, respectively, i.e.,

V = V1× · · · × Vm and K = K1× · · · × Km,

where every Vi is a simple Euclidean Jordan algebra (that cannot be a direct sum of two Euclidean Jordan algebras) with the corresponding symmetric cone Ki for i = 1, · · · , m, and n = Pm

i=1ni (ni is the dimension of Vi). Therefore, for any x = (x1, · · · , xm)T and y = (y1, · · · , ym)T ∈ V with xi, yi∈ Vi, we have

x ◦ y = (x1◦ y1, · · · , xm◦ ym)T ∈ V and hx, yi = hx1, y1i + · · · + hxm, ymi.

For simplicity, we focus on the single symmetric cone K because all the analysis can be carried over to the setting of Cartesian product.

3. Trace version of Young inequality under Euclidean Jordan algebra

In a recent work [14], we established three trace versions of Young inequality in the setting of second-order cone; and also made a conjecture that the eigenvalue version of Young inequality in the SOC setting holds. However, only two trace versions of Young inequality were extended to the setting of symmetric cone (under Euclidean Jordan algebra). In this section, we build up the third trace version of Young inequality based on Gowda’s proof in [13]. To proceed, we first recall the below crucial inequality which was achieved in [1, Theorem 23].

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Theorem 3.1. [1, Theorem 23] Let V be a simple Euclidean Jordan algebra with rank r. For any x, y ∈ K, there holds

tr(x ◦ y) ≤

r

X

i=1

λi(x)λi(y),

where λi(x) and λi(y) are the spectral values of x and y with decreasing order, respectively.

Theorem 3.2. (EJA Young inequality-Type III) Let V be a simple Euclidean Jordan algebra with rank r. For any x, y ∈ V, there holds

tr(|x ◦ y|) ≤ tr |x|p p +|y|q

q



where 1 < p, q < ∞ and 1 p +1

q = 1 and |x| = |λ1(x)|e(1)+ · · · + |λr(x)|e(r).

Proof. For x ◦ y ∈ K ∪ (−K), the desired result follows from [14, Theorem 3.11].

For the remainder case, we suppose x ◦ y is decomposed as x ◦ y =



λ1e(1)+ · · · + λke(k)



−

λk+1e(k+1)+ · · · + λre(r)

 ,

where λi ≥ 0 and k is some positive integer with 1 ≤ k ≤ r. For convenience, we denote

c := e(1)+ · · · + e(k) d := e(1)+ · · · + e(k)−

e(k+1)+ · · · + e(r)

= 2c − e.

Applying the Peirce decomposition [11], we know that V = V(c, 1) ⊕ V(c, 1/2) ⊕ V(c, 0),

where V(c, 1) is a Euclidean Jordan algebra of rank k containing the subspace spanned by {e(1), . . . , e(k)} and V(c, 0) is a Euclidean Jordan algebra of rank r − k containing the subspace spanned by {e(k+1), . . . , e(r)}. Moreover, we write y = u + v + w, where u ∈ V(c, 1), v ∈ V(c, 1/2), and w ∈ V(c, 0). We notice that V(c, 1) ∩ V(c, 0) = {0}, |x ◦ y| = (x ◦ y) ◦ d, and y ◦ d = u − w. On the other hand, suppose that the spectral decomposition of u, w are in the forms of

u := λ1(u)˜e(1)+ · · · + λ1(u)˜e(k) w := λ1(w)˜e(k+1)+ · · · + λr−k(w)˜e(r). Then, we observe that

y ◦ d = u − w = λ1(u)˜e(1)+ · · · + λ1(u)˜e(k)− λ1(w)˜e(k+1)− · · · − λr−k(w)˜e(r). It follow from the proof of Theorem 1.1 in [13] that tr(|y ◦ d|q) ≤ tr(|y|q). Therefore, we obtain

tr(|x ◦ y|) = h|x ◦ y|, ei = h(x ◦ y) ◦ d, ei = hx, y ◦ di.

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In light of Theorem 3.1, we further have tr(|x ◦ y|) ≤

r

X

i=1

λi(x)λi(y ◦ d)

r

X

i=1

 |λi(x)|p

p +|λi(y ◦ d)|q q



= tr(|x|p)

p +tr(|y ◦ d|q) q

≤ tr(|x|p)

p +tr(|y|q) q

= tr |x|p p +|y|q

q

 . Hence, we conclude the desired inequality. 2

Remark 3.3. In the setting of second-order cone, Huang et al. [14] obtained the desired conclusion via establishing the following inequality

tr(|x ◦ y|) ≤ |λ1(x)λ1(y)| + |λ2(x)λ2(y)|.

However, we do not know yet whether the similar inequality in Euclidean Jordan algebra hold or not. In other words, it is a future direction to prove or disprove

tr(|x ◦ y|) ≤

r

X

i=1

i(x)λi(y)|,

for any x, y ∈ V, where λi(x) and λi(y) are the spectral values of x and y with decreasing order, respectively.

4. Counterexample of Young inequality

In this section, we show that the general Young inequality does not hold under Euclidean Jordan algebra. We will show how to construct counterexamples in the SOC setting. According, they serve as counterexamples in the symmetric cone set- ting under Euclidean Jordan algebra. From the construction procedure, we also conclude under what conditions the Young inequality will hold in the SOC setting.

To proceed, we recall some materials regarding the SOC in Rn, an important example of symmetric cones. Officially, the SOC is defined as follows:

Kn:=x = (x0, ¯x) ∈ R × Rn−1| x0≥ k¯xk ,

and the corresponding Jordan product of x and y in Rn with x = (x0, ¯x), y = (y0, ¯y) ∈ R × Rn−1 is given by

x ◦ y :=

 xTy x0y + y¯ 0

 .

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We note that e = (1, 0) ∈ R × Rn−1 acts as the Jordan identity. Later, we need to verify Young inequality in the SOC setting as two vectors belong to Kn with different Jordan frame. To do this, we use vector decomposition and then compare it to the condition with same Jordan frame.

For each x ∈ Rn, it follows from [10, 11, 12] that the spectral decomposition associated with Kn is of the form

(4.1) x = λ1(x)u(1)x + λ2(x)u(2)x .

where λi(x) and ui(x) are called the spectral values and the spectral vectors of x, respectively, which defined by

(4.2) λi(x) = x1+ (−1)i−1kx2k,

(4.3) u(i)x =

 1

2(1, (−1)i−1 xkx2

2k) if x26= 0,

1

2(1, (−1)i−1w) if x2= 0,

for i = 1, 2 with w being any vector in Rn−1 satisfying kwk = 1. When x2 6= 0, the spectral factorization is unique.

Let m be any real number and x ∈ Kn. Then the mth power of x is defined by

(4.4) xm = λ1(x)m

u(1)x + λ2(x)m

u(2)x .

With this definition, we are interested in the properties of Young inequality in the SOC setting. For x Kn 0, y Kn 0, and p, q are positive real numbers such that

1

p +1q = 1, it is clear that the Young inequality holds for p = q = 2 since x2+ y2

2 Kn x ◦ y ⇐⇒ x2+ y2Kn 2x ◦ y

⇐⇒ x2− 2x ◦ y + y2 Kn 0

⇐⇒ (x − y)2 Kn 0.

Now, we first consider x, y ∈ Kn such that x and y shart the same Jordan frame.

Theorem 4.1. Suppose x Kn 0, y Kn 0, and p, q are positive real numbers such that 1p + 1q = 1. Let x, y share the same Jordan frame, that is, x, y have spectral decomposition

x = λ1u1+ λ2u2, y = λ3u3+ λ4u4

with u1 = u3, u2 = u4 or u1 = u4, u2 = u3. Then, we have xpp + yqq Kn x ◦ y.

Moreover, the equality holds if and only if xp = yq.

Proof. Without loss of generality, we assume that u1 = u3, u2 = u4, and the proof can be carried over to the similar case when u1 = u4, u2 = u3. First, we observe that

xp p +yq

q = 1

p(λp1u1+ λp2u2) + 1

q(λq3u1+ λq4u2)

=  λp1 p +λq3

q



u1+ λp2 p +λq4

q

 u2, (4.5)

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and

x ◦ y = (λ1u1+ λ2u2) ◦ (λ3u1+ λ4u2)

= λ1λ3u21+ λ1λ4u1◦ u2+ λ2λ3u2◦ u1+ λ2λ4u22

= λ1λ3u1+ λ2λ4u2. (4.6)

Subtracting (4.5) from (4.6), we see that xp

p +yq

q − x ◦ y =  λp1 p +λq3

q



u1+ λp2 p +λq4

q

 u2



− (λ1λ3u1+ λ2λ4u2)

=  λp1 p +λq3

q − λ1λ3



u1+ λp2 p +λq4

q − λ2λ4

 u2. (4.7)

Since x Kn 0, y Kn 0, we know that x1 ≥ kx2k, y1 ≥ ky2k, and hence λi ≥ 0 for all i = 1, 2, 3, 4. Thus, by the traditional Young inequality for numbers, we have

λp1 p +λq3

q − λ1λ3 ≥ 0 and λp2 p +λq4

q − λ2λ4 ≥ 0.

This means xpp+yqq − x ◦ y ∈ Kn, i.e., xpp+yqq Kn x ◦ y. Moreover, since u1, u2 are linearly independent. From (4.7), we have

xp p +yq

q = x ◦ y ⇐⇒ xp p +yq

q − x ◦ y = 0

⇐⇒  λp1 p + λq3

q − λ1λ3



u1+ λp2 p +λq4

q − λ2λ4

 u2= 0

⇐⇒ λp1 p + λq3

q − λ1λ3 = λp2 p +λq4

q − λ2λ4= 0

⇐⇒ λp1 p + λq3

q = λ1λ3 and λp2 p +λq4

q = λ2λ4

⇐⇒ λp1 = λq3 and λp2 = λq4

⇐⇒ xp = λp1u1+ λp2u2 = λq3u3+ λq4u4 = yq,

where the fifth equivalence holds by Young inequality for real number. In other words, the equality holds if and only if xp = yq. 2

Up to now, we show that the Young inequality holds in the SOC setting under the condition that x, y share the same Jordan frame. However, it is much harder to verify two vectors belong to Kn with different Jordan frame. For notations simplicity, we denote

xp p + yq

q − x ◦ y := (α, β) ∈ R × Rn−1.

Then, it is clear that xpp+yqq K x ◦ y holds if and only if α2− kβk2 ≥ 0. In addition, the equality holds if and only if α2− kβk2 = 0. In order to develop some tools for the case when two vectors belong to Kn with different Jordan frame. As below, we start with calculating α2− kβk2.

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Theorem 4.2. Suppose x Kn 0, y Kn 0, and p, q are positive real numbers such that 1p+1q = 1. Suppose that x = (x1, x2) ∈ R × Rn−1, y = (y1, y2) ∈ R × Rn−1, and

θ = (

arccos kxxT2y2

2kky2k

 ifkx2kky2k 6= 0,

0 otherwise.

that is, θ is the angle between two vectors x2, y2, the spectral decompositions of x, y associated with Kn are

x = λ1u1+ λ2u2, y = λ3u3+ λ4u4. Then, there holds

α2− kβk2= f1+ (1 − cos θ)γ − (1 + cos θ)δ

= f2− (1 + cos θ)γ + (1 − cos θ)δ, (4.8)

where

f1 = λp1λp2

p2q3λq4

q2 + λ1λ2λ3λ4+ 1

pq(λp1λq4+ λp2λq3)

−(λp1 p +λq3

q )λ2λ4+ (λp2 p + λq4

q )λ1λ3, f2 = λp1λp2

p2q3λq4

q2 + λ1λ2λ3λ4+ 1

pq(λp1λq3+ λp2λq4)

−(λp1 p +λq4

q )λ2λ3+ (λp2 p + λq3

q )λ1λ4,

γ = 1

2pq(λp1− λp2)(λq3− λq4) −λ1λ2

2p (λp−11 − λp−12 )(λ3− λ4)

−λ3λ4

2q (λq−13 − λq−14 )(λ1− λ2), δ = (λ1− λ2)23− λ4)2

16 .

Proof. According to the spectral decomposition associated with Kn, we have x = λ1u1+ λ2u2, and y = λ3u3+ λ4u4,

where

λ1 = x1+ kx2k, λ2 = x1− kx2k, u1 = 1 2, x2

2kx2k



, u2= 1 2, −x2

2kx2k

 , λ3 = y1+ ky2k, λ4 = y1− ky2k, u3 = 1

2, y2

2ky2k



, u4 = 1 2, −y2

2ky2k

 .

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Then, we write out

xp

p = λp1

p u1p2

p u2= 1

2p(λp1+ λp2), 1

2p(λp1− λp2) x2

kx2k

 , yq

q = λq3

q u3q4

q u4= 1

2q(λq3+ λq4), 1

2q(λq3− λq4) y2 ky2k

 , x ◦ y = x1y1+ xT2y2, x1y2+ y1x2.

Since we have denoted xpp +yqq − x ◦ y by (α, β), we have

α = 1

2p(λp1+ λp2) + 1

2q(λq3+ λq4) − (x1y1+ xT2y2),

β = 1

2p(λp1− λp2) x2

kx2k + 1

2q(λq3− λq4) y2

ky2k − (x1y2+ y1x2).

Thus, we see that

α2 = 1

4p22p1 + λ2p2 + 2λp1λp2) + 1

4q22q3 + λ2q4 + 2λq3λq4) +(x21y21+ (xT2y2)2+ 2x1y1xT2y2) + 1

2pq(λp1λq3+ λp1λq4+ λp2λq3+ λp2λq4)

−1

p(λp1x1y1+ λp1xT2y2+ λp2x1y1+ λp2xT2y2)

−1

q(λq3x1y1+ λq3xT2y2+ λq4x1y1+ λq4xT2y2), (4.9)

and

kβk2 = 1

4p22p1 + λ2p2 − 2λp1λp2) + 1

4q22q3 + λ2q4 − 2λq3λq4) +(x21ky2k2+ y12kx2k2+ 2x1y1xT2y2) + 1

2pq(λp1λq3− λp1λq4− λp2λq3+ λp2λq4)

−1

p(λp1x1xT2y2

kx2k + λp1y1kx2k − λp2x1xT2y2

kx2k − λp2y1kx2k)

−1

q(λq3x1ky2k + λq3y1xT2y2

ky2k − λq4x1ky2k + λq4y1xT2y2

ky2k).

(4.10)

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Subtracting (4.10) from (4.9) yields

α2− kβk2

= λp1λp2

p2q3λq4

q2 +x21y12+ (xT2y2)2− x21ky2k2− y12kx2k2 + 1

2pq[(λp1λq3+ λp2λq4)(1 − xT2y2

kx2kky2k) + (λp1λq4+ λp2λq3)(1 + xT2y2

kx2kky2k)]

+1

p[λp1x1(xT2y2

kx2k − y1) + λp1kx2k(y1−xT2y2

kx2k)]

−1

p[λp2x1(xT2y2

kx2k + y1) + λp2kx2k(y1+xT2y2 kx2k)]

+1

q[λq3x1(ky2k − y1) + λq3xT2y2( y1

ky2k − 1)]

−1

q[λq4x1(ky2k + y1) + λq4xT2y2( y1

ky2k + 1)]

= λp1λp2

p2q3λq4

q2 + [x21y12+ kx2k2ky2k2cos2θ − x21ky2k2− y21kx2k2] + 1

2pq(λp1λq3+ λp2λq4)(1 − cos θ) + 1

2pq(λp1λq4+ λp2λq3)(1 + cos θ) +1

p[(λp1x1(ky2k cos θ − y1) + λp1kx2k(y1− ky2k cos θ)]

−1

p[(λp2x1(ky2k cos θ + y1) + λp2kx2k(y1+ ky2k cos θ)]

+1

q[λq3x1(ky2k − y1) + λq3kx2k cos θ(y1− ky2k)]

−1

q[λq4x1(ky2k + y1) + λq4kx2k cos θ(y1+ ky2k)]

= λp1λp2

p2q3λq4

q2 +x21(y21− ky2k2) − kx2k2(y21− ky2k2cos2θ) + 1

2pq(λp1λq3+ λp2λq4)(1 − cos θ) + (λp1λq4+ λp2λq3)(1 + cos θ)

−λp1

p λ2(y1− ky2k cos θ) −λp2

p λ1(y1+ ky2k cos θ)

−λq3

q λ4(x1− kx2k cos θ) − λq4

q λ3(x1+ kx2k cos θ).

(4.11)

It is difficult to estimate that the value is positive or negative as the equality contains the term cos θ since −1 ≤ cos θ ≤ 1. For the case that two vectors have the same Jordan frame, we can replace the term cos θ by 1 or −1 since two vectors have the same Jordan frame if and only if | cos θ| = 1. To proceed, we need to discuss two subcases.

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Case 1. Change cos θ to 1, by applying (4.11), we have

α2− kβk2

= λp1λp2

p2q3λq4

q2 +x21(y21− ky2k2) − kx2k2(y21− ky2k2(1 + cos2θ − 1)) + 1

2pq(λp1λq3+ λp2λq4)(1 − 1 + (1 − cos θ)) + 1

2pq(λp1λq4+ λp2λq3)(1 + 1 + (cos θ − 1))

−λp1

p λ2(y1− ky2k(1 + (cos θ − 1))) −λp2

p λ1(y1+ ky2k(1 + (cos θ − 1)))

−λq3

q λ4(x1− kx2k(1 + (cos θ − 1))) −λq4

q λ3(x1+ kx2k(1 + (cos θ − 1)))

=

p1λp2

p2q3λq4

q2 +x21(y21− ky2k2) − kx2k2(y21− ky2k2) i

+ 1

2pq(λp1λq3+ λp2λq4)(1 − 1) + (λp1λq4+ λp2λq3)(1 + 1)

− λ

p 1

p λ2(y1− ky2k) +λp2

p λ1(y1+ ky2k) +λq3

q λ4(x1− kx2k) + λq4

q λ3(x1+ kx2k)o +n

kx2k2ky2k2(cos2θ − 1) + 1

2pq(λp1λq3+ λp2λq4)(1 − cos θ) + (λp1λq4+ λp2λq3)(cos θ − 1)

− − λp1

p λ2ky2k(cos θ − 1) + λp2

p λ1ky2k(cos θ − 1)

−λq3

q λ4kx2k(cos θ − 1) +λq4

q λ3kx2k(cos θ − 1)o

=

p1λp2

p2q3λq4

q2 + λ1λ2λ3λ4+ 1

pq(λp1λq4+ λp2λq3)

−(λp1

p λ2λ4p2

p λ1λ3q3

q λ4λ2q4 q λ3λ1)

o +(1 − cos θ)n

− kx2k2ky2k2(1 + cos θ) + 1

2pq(λp1λq3+ λp2λq4) − (λp1λq4+ λp2λq3)

− λ

p 1

p λ2ky2k −λp2

p λ1ky2k +λq3

q λ4kx2k −λq4

q λ3kx2ko .

(12)

After direct calculation, we further have

α2− kβk2

= λp1λp2

p2q3λq4

q2 + λ1λ2λ3λ4+ 1

pq(λp1λq4+ λp2λq3)

−(λp1 p +λq3

q )λ2λ4+ (λp2 p +λq4

q )λ1λ3

 +(1 − cos θ)n

− kx2k2ky2k2(1 + cos θ) + 1

2pqλp1q3− λq4) + λp2q4− λq3)

−(λp1λ2

p − λ1λp2

p )ky2k + (λq3λ4

q − λ3λq4

q )kx2ko

= λp1λp2

p2q3λq4

q2 + λ1λ2λ3λ4+ 1

pq(λp1λq4+ λp2λq3)

−(λp1 p +λq3

q )λ2λ4+ (λp2 p +λq4

q )λ1λ3

 +(1 − cos θ) − kx2k2ky2k2(1 + cos θ) + 1

2pqλp1q3− λq4) + λp2q4− λq3)

− λ1λ2

p (λp−11 − λp−12 )ky2k +λ3λ4

q (λq−13 − λq−14 )kx2k

= λp1λp2

p2q3λq4

q2 + λ1λ2λ3λ4+ 1

pq(λp1λq4+ λp2λq3)

−(λp1 p +λq3

q )λ2λ4+ (λp2 p +λq4

q )λ1λ3 +(1 − cos θ) − (λ1− λ2

2 )23− λ4

2 )2(1 + cos θ) + 1

2pq(λp1− λp2)(λq3− λq4)

−λ1λ2

p (λp−11 − λp−123− λ4

2 −λ3λ4

q (λq−13 − λq−141− λ2 2

= λp1λp2

p2q3λq4

q2 + λ1λ2λ3λ4+ 1

pq(λp1λq4+ λp2λq3)

−(λp1 p +λq3

q )λ2λ4+ (λp2 p +λq4

q )λ1λ3

 +(1 − cos θ) 1

2pq(λp1− λp2)(λq3− λq4) − λ1λ2

2p (λp−11 − λp−12 )(λ3− λ4)

−λ3λ4

2q (λq−13 − λq−14 )(λ1− λ2) − (1 + cos θ)(λ1− λ2)23− λ4)2 16

:= f1+ (1 − cos θ)γ − (1 + cos θ)δ,

where the third equality holds since kx2k = λ1−λ2 2, ky2k = λ3−λ2 4.

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Case 2. Change cos θ to −1, by applying the same argument in Case 1, we have α2− kβk2

=  λ

p 1λp2

p2q3λq4

q2 + λ1λ2λ3λ4+ 1

pq(λp1λq3+ λp2λq4)

−(λp1 p +λq4

q )λ2λ3+ (λp2 p +λq3

q )λ1λ4



−(1 + cos θ) 1

2pq(λp1− λp2)(λq3− λq4) −λ1λ2

2p (λp−11 − λp−12 )(λ3− λ4)

−λ3λ4

2q (λq−13 − λq−14 )(λ1− λ2) + (1 − cos θ)(λ1− λ2)23− λ4)2 16

:= f2− (1 + cos θ)γ + (1 − cos θ)δ.

Then, the desired result follows.. 2

Notice that f1 and f2 equal to α2− kβk2 while x, y have the same Jordan Frame.

In fact, α2− kβk2 = f1 as cos θ = 1 and α2− kβk2 = f2 as cos θ = −1. We shall establish some properties of them in following lemma.

Lemma 4.3. For any x Kn 0, y Kn 0, suppose the spectral decompositions of x, y are

x = λ1u1+ λ2u2, y = λ3u3+ λ4u4,

and p, q are positive real numbers such that 1p +1q = 1. Then the following hold:

(a): f1 ≥ 0. Moreover, f1= 0 if and only if λp1 = λq3 and λp2 = λq4. (b): f2 ≥ 0. Moreover, f2 = 0 if and only if λi= 0 for i = 1, 2, 3, 4.

Proof. For part(a), we let κ = (κ1, κ2) = (λ12 2,λ1−λ2 2¯e) ∈ R × Rn−1, ω = (ω1, ω2) = (λ32 4,λ3−λ2 4e) ∈ R × R¯ n−1, where ¯e = (1, 0, . . . , 0) ∈ Rn−1, then κ Kn 0, ω Kn 0. Note that κ1 = λ12 2, kκ2k = λ1−λ2 2. By the spectral decomposition (4.1)-(4.3), we have

κ = λ1(κ)u(1)κ + λ2(κ)u(2)κ = λ1u(1)κ + λ2u(2)κ . where

λ1(κ) = λ1+ λ2

2 + λ1− λ2 2 = λ1, λ2(κ) = λ1+ λ2

2 − λ1− λ2 2 = λ2, u(1)κ = 1

2 1,

λ12

2 e

λ12

2

!

= 1 2,e

2

 ,

u(2)κ = 1 2 1, −

λ12

2 e

λ12

2

!

= 1 2, −e

2

 .

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Similarly, the spectral decomposition of ω is

ω = λ1(ω)u(1)ω + λ2(ω)u(2)ω = λ3u(1)κ + λ4u(2)κ . where

λ1(ω) = λ3+ λ4

2 + λ3− λ4 2 = λ3, λ2(ω) = λ3+ λ4

2 − λ3− λ4 2 = λ4, u(1)ω = 1

2 1,

λ34

2 e

λ34

2

!

= 1 2,e

2



= u(1)κ ,

u(2)ω = 1 2 1, −

λ34

2 e

λ34

2

!

= 1 2, −e

2



= u(2)κ .

Since x, κ have the same spectral values λ1, λ2 and y, ω have the same spectral values λ3, λ4, we get via (4.8) that α2− kβk2

κω = f1. Moreover κ, ω share the same Jordan frame u(1)κ , u(2)κ . By Theorem 4.1, it yields

κp p +ωq

q Kn κ ◦ ω and hence f1 = α2− kβk2

κω ≥ 0. Moreover, f1= α2− kβk2

κω = 0 ⇐⇒ κp p +ωq

q = κ ◦ ω

⇐⇒ κp= ωq

⇐⇒ (λp1− λq3)u(1)κ + (λp2− λq4)u(2)κ = 0

⇐⇒ λp1− λq3 = λp2− λq4= 0 (4.12)

⇐⇒ λp1= λq3 and λp2= λq4.

where the equivalence (4.12) holds since u(1)κ , u(2)κ are linearly independent.

For part(b), we let ζ = (ζ1, ζ2) = (λ12 2,λ1−λ2 2e) ∈ R × Rn−1, η = (η1, η2) = (λ42 3, −λ4−λ2 3e) ∈ R × Rn−1, where e = (1, 0, . . . , 0) ∈ Rn−1, then ζ Kn 0, η Kn

0. Similarly,

ζ = λ1(ζ)u(1)ζ + λ2(ζ)u(2)ζ = λ1u(1)ζ + λ2u(2)ζ . η = λ1(η)u(1)η + λ2(η)u(2)η = λ3u(2)ζ + λ4u(1)ζ .

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