2 A Generalized Re…nement of Young’s Inequality

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A Generalized Re…nement of Young Inequality And Applications

Mohamed Akkouchi

^y

, Mohamed Amine Ighachane

^z

Received 6 March 2020

Abstract

In this paper, we prove that ifa; b >0and0 1, then for all positive intgerm, we have (a b¹ )^m+r^m0 (a^m² b^m²)²+rm ((ab)^m⁴ b^m²)² _(0;1

2]( ) + ((ab)^m⁴ a^m²)² ₍1 2;1]( )

( a+ (1 )b)^m; (A-I)

wherer0= minf ;1 g; rm= minf2r^m₀ ;(1 r0)^m r^m₀ gand _I( )is the characteristic function of the setI. The inequality (A-I) provides a generalization of an important re…nement of the Young inequality obtained in 2015 by Hirzallah and Kittaneh. The inequality (A-I) extends also another important re-

…nement of the Young inequality obtained in 2015 by J. Zhao and J. Wu which corresponds to the case m= 1.

As applications of the inequality (A-I), we give some re…ned Young type inequalities for the traces, determinants, norms of positive de…nite matrices, and some inequalities concerning the generalized Euclid- ean operator radius.

1 Introduction

The well-known Young’s inequality for scalars asserts that for all positive real numbers a; band0 1;

we have

a b¹ a+ (1 )b: (1)

The inequality (1) implies that form= 1;2;3; :::;

(a b¹ )^m ( a+ (1 )b)^m: (2)

Hirzallah and Kittaneh [3] re…ned Young’s inequality (1) to

(a b¹ )²+r₀²(a b)² ( a+ (1 )b)²; (3) wherer0= minf ;1 g:

Kittaneh and Manasrah [7] re…ned Young’s inequality so that a b¹ +r₀(p

a p

b)² a+ (1 )b; (4)

wherer₀= minf ;1 g:

In 2015, Manasrah and Kittaneh [8] re…nd the inequality (2) by adding the quantityr^m₀ (a^m² b^m²)² to its …rst part, and obtained the following inequality:

(a b¹ )^m+r^m₀(a^m² b^m²)² ( a+ (1 )b)^m; (5)

Mathematics Sub ject Classi…cations: 26D07, 26D15, 26D20; 15A45, 15A60.

yDepartment of Mathematics, Faculy of Sciences-Semlalia, University Cadi Ayyad, Av. Prince My. Abdellah, BP: 2390, Marrakesh (40.000-Marrakech), Morocco

zDepartment of Mathematics, Faculy of Sciences-Semlalia, University Cadi Ayyad, Av. Prince My. Abdellah, BP: 2390, Marrakesh (40.000-Marrakech), Morocco

129

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wherem= 1;2;3; :::.

The inequality (5) is a common generalization of the inequalities (3) and (4).

In 2015, J. Zhao and J. Wu re…ned inequality (2) as follows:

if0< ¹₂, then

a b¹ + (p

a p

b)²+r₁(p⁴ ab p

b)² a+ (1 )b; (6)

and if ¹₂< 1, then

a b¹ + (1 )(p

a p

b)²+r1(p⁴ ab p

a)² a+ (1 )b: (7)

wherer0= minf ;1 g; andr1= minf2r0;1 2r0g:

We can gather the above inequalities (6) and (7) in the following form:

a b¹ +r0(a¹² b¹²)²+r1 ((ab)¹⁴ b¹²)² _(0;¹

2]( ) + ((ab)¹⁴ a¹²)² ₍¹

2;1]( ) a+ (1 )b;

valid for all positive numbersa; b and for all 2[0;1], where for any set I, we denote _I its characteristic function de…ned by

I(x) = 1 ifx2I;

0 ifx =2I:

One of the aims of this paper is to extend the inequality above to the following one:

(a b¹ )^m+r^m₀ (a^m² b^m²)²+rm ((ab)^m⁴ b^m²)² _(0;¹

2]( ) + ((ab)^m⁴ a^m²)² ₍¹

2;1]( )

( a+ (1 )b)^m; (A-I)

wherer0= minf ;1 gandrm= minf2r₀^m;(1 r0)^m r^m₀ g, which will be valid for all positive integerm, for all positive numbersa; band all 2[0;1].

This inequality extends all the previous re…nements of Young’s inequality, this means the inequalities:

(3), (4), (5), (6) and (7).

The inequality (A-I) is proved in the second section.

In sections two and three of this paper, we give applications of the inequality (A-I) to establish new re-

…nements to certain Young type inequalities for the traces, determinants, norms of positive de…nite matrices, and some inequalities concerning the generalized Euclidean operator radius.

2 A Generalized Re…nement of Young’s Inequality

For the proof of our …rst main result, we need to recall the following theorem concerning the celebrated weighted arithmetic-geometric mean (AM-GM) inequality.

Theorem 1 Let n be a positive integer. Fori= 1;2; : : : ; n; letai 0, and let i 0 satisfy Pn

i=1 i = 1:

Then, we have

Yn i=1

a_iⁱ Xn i=1

iai:

By takingn:= 2in the weighted AM-GM inequality, we recapture the classical Young’s inequality.

We need also the following lemma.

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Lemma 2 Let mbe a positive integer and let a real number, such that0 1. Then we have Xm

k=1

m

k k ^k(1 )^{m k} =m ; and

mX1 k=0

m

k (m k) ^k(1 )^{m k}=m(1 ):

where ^m_k is the binomial coe¢ cient.

For a proof of Lemma2, one can see [1].

The …rst main result of this paper reads as follows.

Theorem 3 Letaandbbe two positive numbers and0 1:Then for all positive intgerm, we have the inequality (A-I).

Proof. Suppose that0< ¹₂. We claim that

( a+ (1 )b)^m ^m(a^m² b^m²)² rm((ab)^m⁴ b^m²)² (a b¹ )^m: Indeed, we have, the following identities:

( a+ (1 )b)^m ^m(a^m² b^m²)² r_m((ab)^m⁴ b^m²)²

= Xm k=0

m k

k(1 )^{m k}a^kb^{m k} ^m a^m+b^m 2(ab)^m² rm (ab)^m² +b^m 2(ab)^m⁴b^m²

= (1 )^m ^m r_m b^m+ 2 ^m r_m (ab)^m² +

mX1 k=1

m k

k(1 )^{m k}a^kb^{m k}+ 2r_m(ab)^m⁴b^m²

=

m+1X

k=0 kxk

where x_i is given by

x₀:=b^m with ₀:= (1 )^m ^m r_m ; and for1 k m 1,

x_k:=a^kb^{m k} with _k:= m k

k(1 )^{m k} and

xm:= (ab)^m² with m:= 2 ^m rm ; and

x_m+1:= (ab)^m⁴b^m² with _m:= 2r_m: We have

1. x_k>0 for allk2 f0;1; :::; m+ 1g;

2. k 0for allk2 f0;1; :::; m+ 1g;withPm+1 k=0 i= 1:

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Hence by applying Theorem 1, we get

( a+ (1 )b)^m ^m(a^m² b^m²)² rm((ab)^m⁴ b^m²)²

m+1Y

k=0

x_iⁱ =a ^(m)b ^(m);

where

(m) =

mX1 k=1

m

k k ^k(1 )^{m k}+m

2 2 ^m rm +m 2rm

= Xm k=1

m

k k ^k(1 )^{m k}=m ; (by Lemma 2) and

(m) = m

2 2 ^m r_m +3m 2 r_m +

mX1 k=1

m

k (m k) ^k(1 )^{m k}+m (1 )^m ^m r_m

=

mX1 k=0

m

k (m k) ^k(1 )^{m k} =m(1 ) (by Lemma2):

The case ¹₂ < 1 is established by similar arguments.

This completes the proof.

Remark 1 In Theorem 3above, we have re…ned the inequality (5) by adding to its …rst part the quantity r₀^m(a^m² b^m²)²+rm ((ab)^m⁴ b^m²)² _(0;¹

2]( ) + ((ab)^m⁴ a^m²)² ₍¹

2;1]( )

where r₀ = minf ;1 g andr_m= minf2r^m₀;(1 r₀)^m r^m₀g. So this is a considerable generalization of the re…nements of the Young inequality due to Manasrah and Kittaneh [8].

Theorem 3 also extends the result obtained by J. Zhao and J. Wu in [13] which correspondd to the particular casem= 1.

3 Applications to Young Type Inequalities for the Traces, Deter- minants, and Norms of Positive De…nite Matrices

In this section, we apply Theorem3to provide some improvements to certain re…ned Young type inequalities for the traces, determinants, and norms of positive de…nite matrices obtained by F. Kittaneh and Y. Manasrah in [8].

Let Mn(C)designate the space of all n n complex matrices. A matrixA 2Mn(C)is called positive semide…nite, (denoted as A 0) if x Ax 0 for all x2Cⁿ, and it is called positive de…nite (denoted as A >0) ifx Ax >0for all nonzerox2Cⁿ. The singular values of a matrixA2Mn(C)are the eigenvalues of the positive semide…nite matrixjAj= (A A)¹⁼², denoted bysi(A)fori= 1;2;3::; n:

A norm jjj:jjj on Mn(C) is called unitarily invariant if jjjU AVjjj = jjjAjjj for all A 2 Mn(C) and all unitary matricesU; V 2Mn(C).

The trace norm jj:jjis given byjjAjj¹=trjAj=Pn

k=1sk(A);wheretr is the usual trace for matrices. It is well known that this norm is unitarily invariant.

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A matrix version of Young inequality, proved in [2], asserts that

sj(A B¹ ) sj( A+ (1 )B) forj = 1; : : : ; n: (8) The above singular value inequality entails the following unitarily invariant norm inequality

jjA B¹ jj1 jj A+ (1 )Bjj1:

A determinant version of Young’s inequalities is also known [[4]; p. 467], for positive semide…nite matrices A,B and0 1;

det(A B¹ ) det( A+ (1 )B): (9)

The following inequality, proved in [9], asserts that for all positive semide…nite matrices (and unitarily invariant norms), we have:

jjjA XB¹ jjj jjjAjjj+ (1 )jjjBjjj:

By using the inequality (5), Manasrah and Kittaneh [8] gave re…nements to the following inequalities:

trjA B¹ j) ^m tr( A+ (1 )B) ^m;

det(A B¹ )

m

det( A+ (1 )B)

m

; and

jjjA XB¹ jjj

m

jjjAjjj+ (1 )jjjBjjj

m

: (10)

Here, as application of Theorem3, we give further and enhanced improvements to the above inequalities.

To prove the main results of this section, we need to recall the following two lemmas, the …rst lemma [5]

is a Heinz-Kato type inequality for unitarily invariant norms, and the second lemma (see, e.g., [[4], p. 482]) is the Minkowski inequality for determinants.

Lemma 4 ([5]) Let A; B2Mn(C)be positive semide…nite matrices. Then we have jjjA XB¹ jjj jjjAXjjj jjjXBjjj¹ :

In particular, we have

tr A XB¹ (trA) (trB)¹ : Lemma 5 Let A; B2Mn(C)be positive de…nite matrices. Then we have

det (A+B)¹ⁿ det(A)ⁿ¹ +det(B)ⁿ¹:

The …rst result of this section concerns the determinant of positive de…nite matrices and reads as follows:

Theorem 6 Let A; B2Mn(C)be positive de…nite matrices and 0 1:Then for all positive intger m;

we have

det(A B¹ ) ^m+r^nm₀ det(A)^m² det(B)^m² ²+rnm

h

[det(A) det(B)]^m⁴ det(B)^m² ² _(0;¹

2]( ) + [det(A) det(B)]^m⁴ det(A)^m² ² ₍¹

2;1]( ) i

det ( A+ (1 )B)^m; wherer0= minf ;1 g andrnm= minf2r^nm₀ ;(1 r0)^nm r₀^nmg:

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Proof. We have

det ( A+ (1 )B)^m

= h

det ( A+ (1 )B)¹ⁿ inm

h

det( A)ⁿ¹ + det((1 )B)¹ⁿ inm

(by Lemma5)

= h

det(A)¹ⁿ+ (1 ) det(B)¹ⁿinm

det(A)¹ⁿ det(B)ⁿ¹

1 nm

+r₀^nm det(A)ⁿ¹

mn

2 (det(B))ⁿ¹ ^mn²

2

+rnm [det(A)ⁿ¹ det(B)ⁿ¹]^nm⁴ (det(B)ⁿ¹)^nm² ² _(0;¹

2]( ) + [det(A)ⁿ¹ det(B)ⁿ¹]^nm⁴ (det(A)¹ⁿ)^nm² ² ₍¹

2;1]( ) (by Theorem3)

= det(A B¹ ) ^m+r₀^nm det(A)^m² det(B)^m² ² + [det(A) det(B)]^m⁴ det(B)^m² ² _(0;¹

2]( ) + [det(A) det(B)]^m⁴ det(A)^m² ² ₍¹

2;1]( ):

This ends the proof.

The second result of this section concerns the traces of positive de…nite matrices and reads as follows:

Theorem 7 Let A; B 2M_n(C) be positive de…nite matrices and0 1. Then for all positive integer m, we have

tr(jA B¹ j) ^m+r^m₀ (trA)^m² (trB)^m² ²+rm [(trA)(trB)]^m⁴ (trB)^m² ² _(0;¹

2]( ) + [(trA)(trB)]^m⁴ (trA)^m² ² ₍¹

2;1]( ) [tr( A+ (1 )B)]^m; wherer0= minf ;1 g andrm= minf2r₀^m;(1 r0)^m r^m₀ g:

Proof. We have

tr(jA B¹ j) ^m+r^m₀ (trA)^m² (trB)^m² ²+rnm [(trA)(trB)]^m⁴ (trB)^m² ² _(0;¹

2]( ) + [(trA)(trB)]^m⁴ (trA)^m² ² ₍¹

2;1]( )

(trA) (trB)¹ ^m+r₀^m (trA)^m² (trB)^m² ²+r_nm [(trA)(trB)]^m⁴ (trB)^m² ² _(0;¹

2]( ) + [(trA)(trB)]^m⁴ (trA)^m² ² ₍¹

2;1]( ) (by Lemma4) [tr( A+ (1 )B)]^m (by Theorem3):

This ends the proof.

The third and last result of this section provides an improvement to the inequality (10) and reads as follows:

Theorem 8 Let A; X; B 2Mn(C)be positive semide…nite matrices and 0 1. Then for all positive integerm, we have

jjjA XB¹ jjj^m+r₀^m jjjAXjjj^m² jjjXBjjj^m²

2

+rm

h

(jjjAXjjjjjjXBjjj)^m⁴ jjjXBjjj^m²

2

(0;¹₂]( ) + (jjjAXjjjjjjXBjjj)^m⁴ jjjAXjjj^m² ² (¹₂;1]( )i h

jjjAXjjj+ (1 )jjjXBjjjim

; wherer0= minf ;1 g andrm= minf2r₀^m;(1 r0)^m r^m₀ g:

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Proof. We have

jjjA XB¹ jjj^m+r^m₀ jjjAXjjj^m² jjjXBjjj^m² ²+r_mh

(jjjAXjjjjjjXBjjj)^m⁴ jjjXBjjj^m² ² (0;¹₂]( ) + (jjjAXjjjjjjXBjjj)^m⁴ jjjAXjjj^m² ² (¹₂;1]( )i

h

jjjAXjjj jjjXBjjj¹ im

+r^m₀ jjjAXjjj^m² jjjXBjjj^m² ² +r_mh

(jjjAXjjjjjjXBjjj)^m⁴ jjjXBjjj^m² ² (0;¹₂]( ) + (jjjAXjjjjjjXBjjj)^m⁴ jjjAXjjj^m² ² (¹₂;1]( )i

(by Lemma4) h

jjjAXjjj+ (1 )jjjXBjjj im

(by Theorem3):

This completes the proof.

4 Improvements of Some Inequalities for Generalized Euclidean Operator Radius

Let H be a real or complex Hilbert space. The generalized Euclidean operator radius !p of operators T1; :::; Tn2 His de…ned forp 1as follows (see, [11]):

!_p(T₁; :::; T_n) := sup

jjxjj=1

Xn i=1

jhT_ix; xij^p

1 p:

Concerning the generalized Euclidean operator radius!_p, A. Sheikhhosseini, M. S. Moslehian and K. She- brawi established in [12] the following result:

Theorem 9 Let Ti2B(H) fori= 1;2;3::, andp 2m for somem= 1;2;3; ::Then for 0 1;

!^p_p(T₁; ::; T_n)

Xn i=1

jT_ij^p+ (1 )jT j^pi m

inf

jjxjj=1 (x);

where

(x) := minf ;1 g^m Xn

i=1

hjTij^m^px; xi^m² hjT_ij^m^px; xi^m² ²: The …rst aim of this section is to provide some improvements to Theorem 9.

Before giving our results, we recall the following lemmas. The …rst lemma is known as the generalized mixed Schwarz inequality, this lemma is proved by F. Kittaneh in [6].

Lemma 10 LetT 2 B(H)and 2(0;1). Then

jhT x; yij² hjTj² x; xihjT j²⁽¹ ⁾y; yi; 8x; y2 H:

The second Lemma follows from the spectral theorem for positive operators and Jensen’s inequality, this lemma is proved in [10].

Lemma 11 (McCarthy inequality) Let T 2 B(H) such that T 0 and let x2 H be any unit vector.

Then we have the following assertions:

(a) hT x; xi^r hT^rx; xiforr 1,

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(b) hT^rx; xi hT x; xi^rfor0< r 1:

Now, by using Theorem 3, we give the following re…nement of Theorem9.

Theorem 12 Let Ti 2B(H) fori= 1;2;3::, and p 2mfor some m= 1;2;3; ::Then for all real number satisfying0 1, we have

!^p_p(T1; ::; Tn)

Xn i=1

jTij^m^p + (1 )jT_ij^m^p ^m inf

jjxjj=1 1(x) inf

jjxjj=1 2(x);

where

1(x) :=r^m₀ Xn i=1

hjT_ij^m^px; xi^m² hjT_ij^m^px; xi^m² ²; and

2(x) : =r_m Xn

i=1

h

[hjT_ij^m^px; xihjT_ij^m^px; xi]^m⁴ hjT_ij^m^px; xi^m² ² (0;¹₂]( )

+ [hjTij^m^px; xihjT_ij^m^px; xi]^m⁴ hjTij^m^px; xi^m²

2

(¹₂;1]( ) i

: Proof. For allx2 H, we have the following inequalities:

Xn i=1

jhT_ix; xij^p = Xn

i=1

jhT_ix; xij²

p 2

Xn i=1

hjT_ij² x; xihjT_ij²⁽¹ ⁾x; xi

p

2 (by Lemma 10) Xn

i=1

hjTij^p^mx; xihjT_ij^p(1^m ⁾x; xi ^m(by Lemma11(a)) Xn

i=1

hjTij^m^px; xi hjT_ij^m^px; xi⁽¹ ⁾ ^m(by Lemma 11(b)) Xn

i=1

hjTij^m^px; xi+ (1 )hjT_ij^m^px; xi

m

1(x) ₂(x)(by Theorem3) Xn

i=1

h jT_ij^m^p + (1 )jT_ij^m^p x; xi ^m inf

jjxjj=1 1(x) inf

jjxjj=1 2(x) Xn

i=1

h jT_ij^m^p + (1 )jT_ij^m^p ^mx; xi inf

jjxjj=1 1(x) inf

jjxjj=1 2(x)(by Lemma11(a)) h

Xn i=1

jTij^m^p + (1 )jT_ij^m^p ^mx; xi inf

jjxjj=1 1(x) inf

jjxjj=1 2(x):

By taking the supremum over allx2 Hwithjjxjj= 1, we deduce the result. This completes the proof.

Next, we provide some improvements to the results stated in Theorem 4.1 of [12].

Theorem 13 Let T_i 2 B(H) for i = 1;2;3::, r 1, and p q 1 such that ¹_p + ¹_q = ¹_r. Then for m= 1;2;3:::;

!^mr_p (jT₁j; ::;jT_nj)!^mr_q (jT₁j; ::;jT_nj) 0

@r p

Xn i=1

jT_ij^p +r q

Xn j=1

jT_jj^q 1 A

m

inf

jjxjj=1 1(x) inf

jjxjj=1 2(x);

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where

1(x) := r p

m

0

@ Xn i=1

hjT_ijx; xi^p

m 2

Xn j=1

hjT_jjx; xi^q

m 2

1 A

2

;

and

2(x) : =rm

h Xⁿ

i=1

hjTijx; xi^p Xn i=1

hjT_ijx; xi^q

m 4

Xn i=1

hjT_ijx; xi^q

m 2i2

(2r;+1)(p) +

h Xⁿ

i=1

hjTijx; xi^p Xn i=1

hjT_ijx; xi^q

m 4

Xn i=1

hjTijx; yi^p

m 2i2

(r;2r)(p) ;

andr_m= minf2(_p^r)^m;(1 ^r_p)^m (^r_p)^mg:

Proof. We set := ^r_p, then1 := ^r_q. Therefore, we haver0=^r_p andrm= minf2(^r_p)^m;(1 ^r_p)^m (^r_p)^mg: We puta:=Pn

i=1hjT_ijx; xi^p andb:=Pn

i=1hjT_ijx; xi^q. We apply Theorem3 toaandb.

So, according to Theorem3and Lemma 11(a), we have the following successive inequalities:

2 64

Xn i=1

hjT_ijx; xi^p

!^r_p0

@ Xn j=1

hjT_jjx; xi^q 1 A

r q3

75

m

0

@r p

Xn i=1

hjTijx; xi^p+r q

Xn j=1

hjT_jjx; xi^q 1 A

m

1(x) 2(x) 0

@r p

Xn i=1

hjT_ij^px; xi+r q

Xn j=1

hjT_jj^qx; xi 1 A

m

inf

jjxjj=1 1(x) inf

jjxjj=1 2(x) 0

@h r p

Xn i=1

jTij^p x; xi+h r q

Xn j=1

jT_jj^q x; xi 1 A

m

jjxinfjj=1 1(x) inf

jjxjj=1 2(x);

Taking the supremum overx2 H,jjxjj= 1, we get the result. This completes the proof.

Acknowledgment. The authors thank very much the (anonymous) referee for the report and for the useful comments.

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