Some generalization of power inequalities for the numerical radius of sum of products of two operators in a Hilbert space

(1)

Adv. Inequal. Appl. 2017, 2017:4 ISSN: 2050-7461

SOME GENERALIZATION OF POWER INEQUALITIES FOR THE NUMERICAL RADIUS OF SUM OF PRODUCTS OF TWO OPERATORS IN A HILBERT SPACE

ADIYASUREN VANDANJAV1, BATZORIG UNDRAKH2,∗

1_{Department of Mathematics, National University of Mongolia, Ulaanbaatar, Mongolia} 2_{Institute of Mathematics, National University of Mongolia, Ulaanbaatar, Mongolia}

Copyright c2017 Vandanjav and Undrakh. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Abstract. In this paper, we generalize some power inequalities numerical radius for the finite number sum of product of two operators in a Hilbert space. Also we generalized some inequalities for the sum of two products using the generalized Lagrange’s identity.

Keywords:numerical range, numerical radius.

2010 AMS Subject Classification:47A12, 47A63.

1. Introduction

LetH be a complex Hilbert space with inner producth·,·i. The numerical rangeofT is the subset

(1) W(T) ={hT x,xi:x∈H,kxk=1}

of the complex plane, whereh·,·iandk · kdenote the standard inner product and norm onH.

∗_{Corresponding author} Received November 4, 2014

(2)

It is known that the numerical range is always convex (see [4]). The spectrum of an operator is contained in the closure of its numerical range.

On a complex Hilbert space the numerical radiusw(T)of an operatorT is given by (2) w(T) =sup{|λ|,λ ∈W(T)}.

Hence, we have

(3) w(αT) =|α|w(T)

for any realα. It is well known thatw(T)is a norm on the Banach algebraB(H)of all bounded linear operatorsT :H→H. The following inequality is well known

(4) 1

2kTk ≤w(T)≤ kTk

For other properties of numerical range and numerical radius, the reader may consult [4] and for other inequalities. See for example [3], [6] and [8].

2. Some Numerical Radius Inequalities

Theorem 2.1.For A_i,B_i∈B(H),i=1,2, ...,n and r≥1we have

(5) w

n

∑

i=1

B∗_iAi

!

≤2−1r· n

∑

i=1

k(A∗_iAi)r+ (B∗iBi)rk 1 r.

Proof.Using the Schwarz inequality in the Hilbert space(H;h., .i)we have

|h(B∗₁A₁+B∗₂A₂+· · ·B∗_nA_n)x,xi| = |h(B∗₁A₁)x,xi+· · ·+h(B∗_nA_nx,x)i|

= |hA₁x,B₁xi+hA₂x,B₂xi+· · · hA_nx,B_nxi|

≤ |hA1x,B1xi|+|hA2x,B2xi|+· · · |hAnx,Bnxi|

≤ kA1xk · kB1xk+· · ·+kAnxk · kBnxk

= n

∑

i=1

h(A∗_iA_i)x,xi12· h₍_B∗

iBi)x,xi 1 2_,

(6)

(3)

Utilizing the arithmetic-geometric mean inequality and byu the power mean inequality for

r≥1 we have,

n

∑

i=1

h(A∗_iA_i)x,xi12· h₍_B∗

iBi)x,xi 1

2 ≤

n

∑

i=1

h(A∗_iA_i)x,xi+h(B∗_iB_i)x,xi

2

≤ n

∑

i=1

_h_A∗

iAix,xir+hB∗iBix,xir

2

1_r (7)

for anyx∈H.

It is known that ifPis a positive operator then for anyr≥1 andx∈Hwithkxk=1 we have the inequality (see [6])

(8) hPx,xir ≤ hPrx,xi.

Applying this property to the positive operatorsA∗_iAandB∗_iB_i,fori=1, . . .n, we deduce that

n

∑

i=1

_h

A∗_iA_ix,xir₊_h_B∗

iBix,xir

2

1_r

≤ n

∑

i=1

_h

(A∗_iA_i)rx,xi+h(B∗_iB_i)rx,xi

2

1_r

= n

∑

i=1

_h

[(A∗_iA_i)r+ (B∗_iB_i)r]x,xi

2

1_r (9)

for anyx∈H. Now combining the inequalities (6), (7) and (9) then the result follows by taking supremum over all unit vectors inH.

This completes the proof.

Corollary 2.2.For A,B∈B(H)we have,

(10) wr(B∗A)≤1

2k(A

∗_A₎r_{+ (}_B∗_B₎r_k

for any r≥1.

Proof.

In particularn=1, and if we putA₁=A,B₁=Bin (5) then we get (10). Corollary 2.3.For A_i∈B(H),i=1, . . . ,n we have,

(11) w(

n

∑

i=1

A_i)≤2−1r n

∑

i=1

k(A∗_iA_i)r+Ik1r

for any r≥1.

(4)

If we putB_i=I for alli=1,2, . . . ,nin (5) then we get (11). Corollary 2.4.For A_i∈B(H),i=1, . . . ,n we have,

(12) w(

n

∑

i=1

A2_i)≤2−1r n

∑

i=1

k(A∗_iA_i)r+ (A_iA∗_i)rk1r

for any r≥1.

Proof.If we putB∗_i =A_ifor alli=1,2, . . . ,nin (5) then we get (12). Corollary 2.5.For A∈B(H)we have,

(13) wr(A)≤ 1

2k(A

∗_A₎r₊_I_k

for any r≥1.

Proof.If we putBi=IandAi=Afor alli=1,2, . . . ,nin (5) then we get (13).

Theorem 2.6.For Ai,Bi∈B(H),i=1, . . . ,n we have,

(14) w

n

∑

i=1

B∗_iA_i

!

≤ n

1−1 r

2 



n

∑

i=1

kA∗_iA_ikr

!1_r

+ n

∑

i=1

kB∗_iB_ikr

!1_r

,

for r≥1.

Proof.Using the Schwarz’s inequality in the Hilbert space(H;h., .i)we have

|h(B∗₁A₁+B∗₂A₂+· · ·B∗_nA_n)x,xi| = |h(B∗₁A₁)x,xi+· · ·+h(B∗_nA_nx,x)i|

= |hA₁x,B₁xi+hA₂x,B₂xi+· · · hA_nx,B_nxi|

≤ |hA₁x,B₁xi|+|hA₂x,B₂xi|+· · · |hA_nx,B_nxi|

≤ kA1xk · kB1xk+· · ·+kAnxk · kBnxk

= n

∑

i=1

h(A∗_iAi)x,xi 1 2· h(_B∗

iBi)x,xi 1 2,

(15)

(5)

Utilizing the arithmetic-geometric mean inequality and power mean inequalityr≥1 we have successively,

n

∑

i=1

h(A∗_iA_i)x,xi12 · h(_B∗

iBi)x,xi 1

2 ≤

n

∑

i=1

h(A∗_iAi)x,xi+h(B∗_iBi)x,xi

2 = 1 2 " n

∑

i=1

h(A∗_iAi)x,xi+ n

∑

i=1

h(B∗_iBi)x,xi

# ≤ 1 2 " n·

∑n_i₌₁h(A_i∗Ai)x,xir

n

1_r

+n·

∑n_i₌₁h(B_i∗Bi)x,xir

n

1_r#

= n

1−1 r 2   n

∑

i=1

h(A∗_iAi)x,xir

!1_r

+ n

∑

i=1

h(B∗_iBi)x,xir

!1_r

. (16)

It is known that ifPis a positive operator, then for any r≥1 andx∈H withkxk=1 we have the inequality (see for example [6])

Applying this property to the positive operatorsA∗_iAandB∗_iB_i,fori=1, . . .n, we deduce that

n1−1r

2 



n

∑

i=1

h(A∗_iA_i)x,xir

!1_r

+ n

∑

i=1

h(B∗_iB_i)x,xir

!1_r



≤ n

1−1_r

2 



n

∑

i=1

h(A∗_iAi)rx,xi

!1_r

+ n

∑

i=1

h(B∗_iBi)rx,xi

!1_r

 (18)

for anyx∈H.

Now combining the inequalities (15), (16) and (18) then the result follows by taking supre-mum over all unit vectors inH.

Theorem 2.7.For Ai,Bi∈B(H),i=1,2, ...,n and r≥1,α ∈(0,1)we have,

(19) w2(

n

∑

i=1

B∗_iA_i)≤n· n

∑

i=1

kα·(A∗_iAi) r

α + (1−_α)(B∗

(6)

Proof.By the Schwarz’s inequality and arithmetic-geometric mean inequality,we have

|h(B∗₁A₁+B∗₂A₂+· · ·B∗_nA_n)x,xi|2 = |h(B∗₁A₁)x,xi+· · ·+hB∗_nA_ni|2

≤ n

∑

i=1

|h(B∗_iAi)x,xi|

!2

≤ n·

n

∑

i=1

|h(B∗_iA_i)x,xi|2

(20)

≤ n·

n

∑

i=1

h(A∗_iA_i)x,xi · h(B∗_iB_i)x,xi

for anyx∈H. It is well known that (see for example [6]) ifPis positive operator andq∈(0,1], then for anyu∈H,kuk=1, we have

(21) hPqu,ui ≤ hPu,uiq

Applying this property to the positive operators(A∗_iAi) 1

α and(B∗

iBi) 1

1−α,fori=1,2, . . . ,n,(_α∈

(0,1]), we have

n· n

∑

i=1

h[(A∗_iA_i)α1]αx,xi · h[(B∗

iBi) 1

1−α]1−αx,xi

≤ n·

n

∑

i=1

h(A∗_iAi) 1

αx,xiα· h(B∗

iBi) 1

1−αx,xi1−α (22)

for anyx∈H,kxk=1.Now, utilizing the weighted arithmetic mean-geometric mean inequality, i.e.,aα_b1−α _≤_α_a_{+ (}₁₋_α₎_b_,_α _∈₍₀_,₁₎_,_a_,_b_≥₀_,_{we get}

n· n

∑

i=1

h(A∗_iAi) 1

αx,xiα· h(B∗

iBi) 1

1−αx,xi1−α

≤ n·

n

∑

i=1

[αh(A∗_iAi) 1

αx,xi+ (1−_α)h(B∗

iBi) 1 1−αx,xi] (23)

for anyx∈H,kxk=1. Moreover, by the elementary inequality following from the convexity of the function f(t) =tr,r≥,1 namely

αa+ (1−α)b≤(αar+ (1−α)br)

1

(7)

we deduce that

n· n

∑

i=1

[αh(A∗_iAi) 1

iBi) 1 1−αx,xi]

≤ n·

n

∑

i=1

[αh(A∗_iAi) 1

αx,xir+ (1−_α)h(B∗

iBi) 1

1−αx,xir]

1 r

≤ n·

n

∑

i=1

[αh(A∗_iAi) r

iBi) r 1−αx,xi]

1 r

= n·

n

∑

i=1

[h(α(A∗_iAi) r

α+ (1−_α)(B∗

iBi) r

1−α)x,xi]

1 r

(24)

for any x∈H,kxk=1, where, for last inequality we used the inequality (8) for the positive operators(A∗_iA_i)α1 and(B∗

iBi) 1

1−α,i=1,2, . . . ,n.

Now combining the inequalities (20), (22),(23) and (24) then result follows by taking supre-mum over all unit vectors inH.

Corollary 2.8.For A,B∈B(H)we have,

(25) w2r(B∗A)≤ kα(A∗A)

r

α + (1−_α)(B∗B)

r 1−αk

for any r≥1andα ∈(0,1).

Proof.If we putB_i=IandA_i=Afor alli=1,2, . . . ,nin (5) then we get (25). Corollary 2.9.If we put in (19) Bi=I for all i=1,2, . . . ,n then we have,

(26) w2(

n

∑

i=1

Ai)≤n· n

∑

i=1

kα(A∗_iAi) r

α+ (1−_α)Ik

for any r≥1andα ∈(0,1). In particular forα= 1₂ we have,

(27) w2(

n

∑

i=1

A_i)≤ n

2·

n

∑

i=1

k(A∗_iA_i)2r+Ik

3. Inequalities for the sum of two products

Theorem 3.1.For A,B∈B(H)and r≥1we have

(28) kB∗Akr≤ 1

2(k(A

(8)

Proof.Using the Schwarz inequality in the Hilbert space(H;h., .i)we have

|h(B∗A)x,yi| = |hAx,Byi| ≤ kAxk · kByk

= h(A∗A)x,xi12· h₍_B∗_B₎_y_,_yi 1 2_,

(29)

for anyx,y∈H.

Using the arithmetic-geometric mean inequality and power means inequalityr≥1 we have,

h(A∗A)x,xi21· h(_B∗_B)_y,_yi 1

2 ≤ h(A

∗_A₎_x_,_x_i₊_h₍_B∗_B₎_y_,_y_i

2

≤

_h

A∗Ax,xir₊_h_B∗_By_,_y_ir

2

1_r (30)

for anyx∈H.

It is known that ifPis a positive operator then for anyr≥1 andx∈Hwithkxk=1 we have the inequality (see [6])

Applying this property to the positive operatorsA∗AandB∗B,fori=1, . . .n, we deduce that

(32)

hA∗Ax,xir₊_h_B∗_By_,_y_ir

2

1_r

≤

h(A∗A)rx,xi+h(B∗B)ry,yi

2

1_r

for anyx,y∈H. Now combining the inequalities (30) and (32) then the result follows by taking supremum over all unit vectors inH.

The Theorem 3.1 may be generalized as follows, since the proof is similar , we will omit the proof.

Theorem 3.2.For A_i,B_i∈B(H),i=1,2, ...,n and r≥1we have

(33) k

n

∑

i=1

B∗_iAik ≤2− 1 r ·

n

∑

i=1

k(A∗_iAi)rk+ n

∑

i=1

k(B∗_iBi)rk

!1_r

.

Theorem 3.3.For A,B∈B(H)and r≥1,α ∈(0,1)we have, (34) kB∗Ak2r ≤α· k(A∗A)

r

αk+ (1−_α)k(B∗B)

(9)

Proof.By the Schwarz’s inequality and arithmetic-geometric mean inequality,we have

|h(B∗Ax,yi|2 = |h(Ax,Byi|2≤ kAxk2kByk2

= h(A∗A)x,xi · h(B∗B)y,yi

(35)

for anyx∈H. It is well known that (see for example [6]) ifPis positive operator andq∈(0,1], then for anyu∈H,kuk=1, we have

(36) hPqu,ui ≤ hPu,uiq

Applying this property to the positive operators(A∗_iA_i)α1 and(B∗

iBi) 1

1−α,fori=1,2, . . . ,n,(_α∈

(0,1]), we have

h[(A∗A)α1]αx,xi · h[(B∗B)

1

1−α]1−αy,yi ≤ h(A∗A)

1

αx,xiα· h(B∗B)

1

1−αy,yi1−α

for any x,y∈H,kxk=kyk=1. Now, utilizing the weighted arithmetic-geometric mean in-equality, i.e.,aα_b1−α _≤_α_a_{+ (}₁₋_α₎_b_,_α _∈₍₀_,₁₎_,_a_,_b_≥₀_,_{we get}

h(A∗A)α1x,xiα· h(B∗B_i)

1

1−αy,yi1−α ≤_αh(A∗A)

1

αx,xi+ (1−_α)h(B∗B)

1 1−αy,yi

for anyx,y∈H,kxk=kyk=1.

Moreover, by the elementary inequality following from the convexity of the function f(t) =

tr,r≥1, namely

αa+ (1−α)b≤(αar+ (1−α)br)

1

r_,_α ∈₍₀_,₁₎_,_a_,_b≥₀_,

we deduce that

αh(A∗A)

1

1−αy,yi ≤ h

αh(A∗A)

1

αx,xir+ (1−_α)h(B∗B)

1 1−αy,yir

i1_r

≤ [αh(A∗A)

r

r 1−αy,yi]

1 r

(37)

for anyx,y∈H,kxk=kyk=1, where, for last inequality we used the inequality (36) for the positive operators(A∗A)α1 and(B∗B)

1 1−α.

(10)

The Theorem 3.3 may be generalized as follows, since the proof is similar , so we will omit the proof.

Theorem 3.4. For Ai,Bi∈B(H),i=1,2, ...,n and r≥1,α ∈(0,1)we have,

(38) k

n

∑

i=1

B∗_iA_ik2≤n· n

∑

i=1

α· k(A∗_iAi) r

αk+ (1−_α)k(B∗

iBi) r 1−αk

1/r

Theorem 3.5. For A,B,C,D∈B(H), k∈_Rand r,s≥1we have,

B∗A+kD∗C

2

≤

(A∗A)r+|k|r₍_C∗_C₎r

2

1 r

(B∗B)s+|k|s₍_D∗_D₎s

2

1 s

− |k|pkA∗Ak kD∗Dk −pkB∗Bk kC∗Ck2

(39)

Proof.

By the Schwarz’s inequality in the Hilbert space(H;h·,·i)we have:

|h(B∗A+kD∗C)x,yi|2 = |hB∗Ax,yi+khD∗Cx,yi|2

≤ [|hB∗Ax,yi+|k|hD∗Cx,yi|]2

≤ hhA∗Ax,xi12h_B∗_By_,_yi 1

2 ₊|_k|h_C∗_Cx_,_xi 1

2h_D∗_Dy_,_yi 1 2

i2

,

(40)

for anyx,y∈H.

Now, using the Lagrange’s identity

(ab+kcd)2= (a2+kc2)(b2+kd2)−k(ad−bc)2, a,b,c,d,k∈_R

we then conclude that: h

hA∗Ax,xi12h_B∗_By_,_yi 1

2 ₊|_k|h_C∗_Cx_,_xi 1

2h_D∗_Dy_,_yi 1 2

i2

= (hA∗Ax,xi+|k|hC∗Cx,xi) (hB∗By,yi+|k|hD∗Dy,yi)

(41)

− |k|hA∗Ax,xi12h_D∗_Dy,_yi 1

2− h_B∗_By,_yi 1

2h_C∗_Cx,_xi 1 2

2

for anyx,y∈H.

(11)

r,s≥1 that

(hA∗Ax,xi+|k|hC∗Cx,xi) (hB∗By,yi+|k|hD∗Dy,yi)

≤ 4

2

x,x

1_r

·

(B∗B)s+|k|s₍_D∗_D₎s

2

x,x

1_s (42)

Consequently, by (40), (41) and (42) we have

|h(B∗A+kD∗C)x,yi|2 ≤ 4 _h

[(A∗A)r+|k|r₍_C∗_C₎r_]_x_,_x_i

2

1_r_h

[(B∗B)s+|k|s₍_D∗_D₎s_]_y_,_y_i

2

1_s

− |k|phA∗Ax,xihD∗Dy,yi −phB∗By,yihC∗Cx,xi2

Taking the supremum overx,y∈H,kxk=kyk=1 we deduce the desired inequality (39). Corollary 3.6.For A,B,C,D∈B(H)we have,

B∗A+kD∗C

2 2 ≤

2 1 r ·

(B∗B)r+|k|r₍_D∗_D₎r

2 1 r

− |k|pkA∗Ak kD∗Dk −pkB∗Bk kC∗Ck2

(43)

for any r≥1.

Proof.If we choose in (39) such thatr=sthen we get (43). Corollary 3.7.For A,B,C,D∈B(H)we have,

B∗A+D∗C

2 2 ≤

(A∗A)r+ (C∗C)r

2 1 r ·

(B∗B)s+ (D∗D)s

2 1 s

− phA∗Ax,xihD∗Dy,yi −phB∗By,yihC∗Cx,xi2.

(44)

for any r,s≥1.

Proof.Puttingk=1 in the proof Theorem 3.5 , then we get (44).

(12)

B∗₁A₁+B∗₂A₂· · ·+ +B∗_nA_n n

2

≤

(A∗₁A₁)r+· · ·+ (A∗_nA_n)r

n

1 r ·

(B∗₁B₁)s+· · ·+ (B∗_nB_n)s

n

1 s

(45)

Proof.By the Schwartz inequality in the Hilbert space(H;h·,·i)we have,

|h(B∗₁A₁+B∗₂A₂· · ·+ +B∗_nA_n)x,yi|2

= |h(B∗₁A₁)x,yi+h(B∗₂A₂)x,yi+· · ·+h(B∗_nA_n)x,yi|2

≤ [|h(B∗₁A1)x,yi|+|h(B∗2A2)x,yi|+· · ·+|h(B∗nAn)x,yi|]2

≤ hhA∗₁A1x,xi

1 2h_B∗

1B1y,yi

1

2+· · ·+h_A∗ nAnx,xi

1 2h_B∗

nBny,yi 1 2

i2 (46)

for anyx,y∈H.

Now we are goning to use the following classical Cauchy-Bunyakowski- Schwartz inequality:

(47) (a₁b₁+a₂b₂+· · ·+a_nb_n)2≤(a2₁+a2₂+· · ·+a2_n)(b₁2+b2₂+· · ·+b2_n)

fora₁,a₂, . . . ,a_n,b₁,b₂, . . . ,b_n∈_R. Then we have,

h

hA∗₁A1x,xi

1 2h_B∗

1B1y,yi

1

2+· · ·+h_A∗ nAnx,xi

1 2h_B∗

nBny,yi 1 2

i2

≤(hA∗₁A1x,xi+· · ·+hA∗nAnx,xi) (hB∗1B1y,yi+· · ·+hB∗nBny,yi),

(48)

for anyx,y∈H. Since the function f(t) =tr is convex forr≥1, we have:

(hA∗₁A₁x,xi+· · ·+hA∗_nA_nx,xi) (hB∗₁B₁y,yi+· · ·+hB∗_nB_ny,yi)

(49)

≤ n2

(A∗₁A1)r+· · ·+ (A∗nAn)r

n

x,x

1_r

(B∗₁B1)s+· · ·+ (B∗nBn)s

n

y,y

1_s

(13)

Consequently, by (46), (49)we have,

|h

_B∗

1A1+B∗2A2· · ·+ +B∗nAn

n

x,yi|2

≤

(A∗₁A₁)r+· · ·+ (A∗_nA_n)r

n

x,x

1_r

(B∗₁B₁)s+· · ·+ (B∗_nB_n)s

n

y,y

1_s

for anyx,y∈H,kxk=kyk=1. Taking supremum overx,y∈H,kxk=kyk=1 we deduce the desired inequality (45).

Corollary 3.9.For A1, . . . ,An∈B(H);B1, . . . ,Bn∈B(H)and r,s≥1we have,

(50)

B∗₁A1+· · ·+B∗nAn

n

2

≤

(A∗₁A1)r+· · ·+ (A∗nAn)r

n

1 r ·

(B∗₁B1)s+· · ·+ (B∗nBn)s

n

1 s

.

Proof.From the proof of the Theorem 3.8 we get the inequality (50).

Corollary 3.10.For A₁, . . . ,A_n∈B(H);B₁, . . . ,B_n∈B(H)and r,s≥1we have,

(51)

A₁+A₂+· · ·+A_n n

2r ≤

(A∗₁A₁)r+· · ·+ (A∗_nA_n)r

n

.

Proof. If we choose in the Theorem 3.8 such that B_i=I,i=1, . . . ,n , whereI is the identity operator then we get the inequality (51).

Conflict of Interests

The authors declare that there is no conflict of interests. Acknowledgements

(14)

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