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On further refinements for Young inequalities

Shigeru Furuichi
• Corresponding author
• Department of Information Science, College of Humanities and Sciences, Nihon University, 3-25-40, Sakurajyousui, Setagaya-ku, Tokyo, 156-8550, Japan
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/ Hamid Reza Moradi
Published Online: 2018-12-27 | DOI: https://doi.org/10.1515/math-2018-0115

Abstract

In this paper sharp results on operator Young’s inequality are obtained. We first obtain sharp multiplicative refinements and reverses for the operator Young’s inequality. Secondly, we give an additive result, which improves a well-known inequality due to Tominaga. We also provide some estimates for the difference A1/2(A−1/2BA−1/2)vA1/2-{(1-v)A + vB} for v∉[ 0,1].

MSC 2010: 47A63; 26D07; 47A60

1 Introduction

This note lies in the scope of operator inequalities. We assume that the reader is familiar with the continuous functional calculus and the Kubo-Ando theory [1].

It is to be understood throughout the paper that the capital letters present bounded linear operators acting on a Hilbert space 𝓗. A is positive (written A ≥ 0) in case 〈Ax,x〉 ≥ 0 for all x ∈ 𝓗 also an operator A is said to be strictly positive (denoted by A > 0) if A is positive and invertible. If A and B are self-adjoint, we write BA in case BA ≥ 0. As usual, by I we denote the identity operator.

The weighted arithmetic mean ∇v, geometric mean ♯v, and harmonic mean !v, for v ∈ [0, 1] and a,b > 0, are defined as follows:

$a∇vb=1−va+vb,a♯vb=a1−vbv,a!vb=(1−v)a−1+vb−1−1.$

If v = $\begin{array}{}\frac{1}{2}\end{array}$, we denote the arithmetic, geometric, and harmonic means, respectively, by ∇, ♯ and !, for the simplicity. Like the scalar cases, the operator arithmetic mean, the operator geometric mean, and the operator harmonic mean for A,B > 0 can be stated in the following form:

$A∇vB=1−vA+vB,A♯vB=A12A−12BA−12vA12,A!vB=(1−v)A−1+vB−1−1.$

The celebrated arithmetic-geometric-harmonic-mean inequalities for scalars assert that if a,b > 0, then

$a!vb≤a♯vb≤a∇vb.$(1)

Generalization of the inequalities (1) to operators can be seen as follows: If A,B > 0, then

$A!vB≤A♯vB≤A∇vB.$

The last inequality above is called the operator Young inequality. During the past years, several refinements and reverses were given for Young’s inequality, see for example [2,3,4].

Zuo et al. showed in [5, Theorem 7] that the following inequality holds:

$Kh,2rA♯vB≤A∇vB, r=minv,1−v, Kh,2=h+124h, h=Mm$(2)

whenever 0 < mIBmI < MIAMI or 0 < mIAmI < MIBMI. As the authors mentioned in [5], the inequality (2) improves the following refinement of Young’s inequality involving Specht $\begin{array}{}S\left(t\right)=\frac{{t}^{\frac{1}{t-1}}}{e\mathrm{log}{t}^{\frac{1}{t-1}}}\left(t>0,t\ne 1\right)\end{array}$ (see [6, Theorem 2]),

$ShrA♯vB≤A∇vB.$

Under the above assumptions, Dragomir proved in [7, Corollary 1] that

$A∇vB≤expv1−v2h−12A♯vB.$

We remark that there is no relationship between two constants K(h,2)r and $\begin{array}{}\mathrm{exp}\left[\frac{v\left(1-v\right)}{2}{\left(h-1\right)}^{2}\right]\end{array}$ in general.

In [3, 8] we proved some sharp multiplicative reverses of Young’s inequality. In this brief note, as the continuation of our previous works, we establish sharp bounds for the arithmetic, geometric and harmonic mean inequalities. Moreover, we shall show some additive-type refinements and reverses of Young’s inequality. We will formulate our new results in a more general setting, namely the sandwich assumption sABtA (0 < st). Additionally, we present some Young type inequalities for the wider range of v; i.e., v ∉ [0, 1].

2 Main results

In our previous work [8], we gave new sharp inequalities for reverse Young inequalities. In this section we firstly give new sharp inequalities for Young inequalities, as limited cases in the first inequalities both (i) and (ii) of the following theorem.

Theorem 2.1

Let A,B > 0 such that sABtA for some scalars 0 < st and let fv(x) $\begin{array}{}\equiv \frac{\left(1-v\right)+vx}{{x}^{v}}for\phantom{\rule{thinmathspace}{0ex}}x>0,\end{array}$ and v ∈ [0, 1].

1. If t ≤ 1, then fv(t) Av BAvBfv(s) AvB.

2. If s ≥ 1, then fv(s) AvBAvBfv(t)AvB.

Proof

Since $\begin{array}{}{f}_{v}^{\prime }\left(x\right)\end{array}$ = v(1–v)(x–1)xv-1, fv(x) is monotone decreasing for 0 < x ≤ 1 and monotone increasing for x ≥ 1.

1. For the case 0 < sxt ≤ 1, we have fv(t) ≤ fv(x) ≤ fv(s), which implies fv(t) AvBAv Bfv(s) Av B by the standard functional calculus.□

2. For the case 1 ≤ sxt, we have fv(s) ≤ fv(x) ≤ fv(t) which implies fv(s) Av BAv Bfv(t)Av B by the standard functional calculus.

Remark 2.2

It is worth emphasizing that each assertion in Theorem 2.1 implies the other one. For instance, assume that the assertion (ii) holds, i.e.,

$fvs≤fvx≤fvt, 1≤s≤x≤t.$(3)

Let t ≤ 1, then $\begin{array}{}1\le \frac{1}{t}\le \frac{1}{x}\le \frac{1}{s}.\end{array}$ Hence (3) ensures that

$fv1t≤fv1x≤fv1s.$

So

$1−vt+vt1−v≤1−vx+vx1−v≤1−vs+vs1−v.$

Now, by replacing v by 1–v we get

$1−v+vttv≤1−v+vxxv≤1−v+vssv$

which means

$fvt≤fvx≤fvs, 0

In the same spirit, we can derive (ii) from (i).

Corollary 2.3

Let A, B > 0, m, m′, M, M′ > 0, and v ∈ [0, 1].

1. If 0 < mIAmI < MI ≤ B ≤ MI, then

$m∇vMm♯vMA♯vB≤A∇vB≤m′∇vM′m′♯vM′A♯vB.$(4)

2. If 0 < mIBmI < MIAMI, then

$M∇vmM♯vmA♯vB≤A∇vB≤M′∇vm′M′♯vm′A♯vB.$(5)

Proof

We use again the function $\begin{array}{}{f}_{v}\left(x\right)=\frac{\left(1-v\right)+vx}{{x}^{v}}\end{array}$in this proof.

The condition (i) is equivalent to $\begin{array}{}I\le \frac{M}{m}I\le {A}^{-\frac{1}{2}}B{A}^{-\frac{1}{2}}\le \frac{{M}^{\prime }}{{m}^{\prime }}I,\end{array}$ so that we get $\begin{array}{}{f}_{v}\left(\frac{M}{m}\right)A{\mathrm{♯}}_{v}B\le A{\mathrm{\nabla }}_{v}B\le {f}_{v}\left(\frac{{M}^{\prime }}{{m}^{\prime }}\right)A{\mathrm{♯}}_{v}\end{array}$ B by putting $\begin{array}{}s=\frac{M}{m}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}t=\frac{{M}^{\prime }}{{m}^{\prime }}\end{array}$ in (ii) of Theorem 2.1.

Similarly, the condition (ii) is equivalent to $\begin{array}{}\frac{{m}^{\prime }}{{M}^{\prime }}I\le {A}^{-\frac{1}{2}}B{A}^{-\frac{1}{2}}\le \frac{m}{M}I\le I,\end{array}$ so that we get $\begin{array}{}{f}_{v}\left(\frac{m}{M}\right)A{\mathrm{♯}}_{v}B\le A{\mathrm{\nabla }}_{v}B\le {f}_{v}\left(\frac{{m}^{\prime }}{{M}^{\prime }}\right)A{\mathrm{♯}}_{v}\end{array}$ B by putting $\begin{array}{}s=\frac{{m}^{\prime }}{{M}^{\prime }}\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}t=\frac{m}{M}\end{array}$ of Theorem 2.1.□

Note that the second inequalities in both (i) and (ii) of Theorem 2.1 and Corollary 2.3 are special cases of [8, Theorem A].

Remark 2.4

It is remarkable that the inequalities fv(t) ≤ fv(x) ≤ fv(s) (0 < sxt ≤ 1) given in the proof of Theorem 2.1 are sharp, since the function fv(x) for sxt is continuous. So, all results given from Theorem 2.1 are similarly sharp. As a matter of fact, let A = MI and B = mI, then from LHS of (5), we infer

$A∇vB=(M∇vm)I and A♯vB=(M♯vm)I.$

Consequently,

$M∇vmM♯vmA♯vB=A∇vB.$

To see that the constant $\begin{array}{}\frac{m{\mathrm{\nabla }}_{v}M}{m{\mathrm{♯}}_{v}M}\end{array}$ in the LHS of (4) can not be improved, we consider A = mI and B = MI, then

$m∇vMm♯vMA♯vB=A∇vB.$

By replacing A, B by A–1, B–1, respectively, the refinement and reverse of non-commutative geometric-harmonic mean inequality can be obtained as follows:

Corollary 2.5

Let A, B > 0, m, m′, M, M′ > 0, and v ∈ [0, 1].

1. If 0 < mIAmI < MIBMI, then

$m′!vM′m′♯vM′A♯vB≤A!vB≤m!vMm♯vMA♯vB.$

2. If 0 < mIBmI < MIAMI, then

$M′!vm′M′♯vm′A♯vB≤A!vB≤M!vmM♯vmA♯vB.$

Now, we give a new sharp reverse inequality for Young’s inequality as an additive-type in the following.

Theorem 2.6

Let A,B > 0 such that sABtA for some scalars 0 < st, and v ∈ [0, 1]. Then

$A∇vB−A♯vB≤maxgvs,gvtA$(6)

where gv(x )≡ (1–v) + vxxv for sxt.

Proof

Straightforward differentiation shows that $\begin{array}{}{g}_{v}^{″}\end{array}$ (x) = v(1–v)xv–2 ≥ 0 and gv(x) is continuous on the interval [s,t], so

$gvx≤maxgvs,gvt.$

Therefore, by applying similar arguments as in the proof of Theorem 2.1, we reach the desired inequality (6). This completes the proof of theorem.□

Corollary 2.7

Let A,B > 0 such that mIA,BMI for some scalars 0 < m < M. Then for v ∈ [0, 1],

$A∇vB−A♯vB≤ξA$

where $\begin{array}{}\xi =max\left\{\frac{1}{M}\left(M{\mathrm{\nabla }}_{v}m-M{\mathrm{♯}}_{v}m\right),\frac{1}{m}\left(m{\mathrm{\nabla }}_{v}M-m{\mathrm{♯}}_{v}M\right)\right\}.\end{array}$

Remark 2.8

We claim that if A,B > 0 such that mIA,BMI for some scalars 0 < m < M with h = $\begin{array}{}\frac{M}{m}\end{array}$ and v ∈ [0, 1], then

$A∇vB−A♯vB≤maxgvh,gv1hA≤L1,hlog⁡ShA$

holds, where $\begin{array}{}L\left(x,y\right)=\frac{y-x}{\mathrm{log}y-\mathrm{log}x}\left(x\ne y\right)\end{array}$ is the logarithmic mean and the term S(h) refers to the Specht’s ratio. Indeed, we have the inequalities

$(1−v)+vh−hv≤L(1,h)log⁡S(h),(1−v)+v1h−h−v≤L(1,h)log⁡S(h),$

which were originally proved in [9, Lemma 3.2], thanks to $\begin{array}{}S\left(h\right)=S\left(\frac{1}{h}\right)\phantom{\rule{thinmathspace}{0ex}}and\phantom{\rule{thinmathspace}{0ex}}L\left(1,h\right)=L\left(1,\frac{1}{h}\right).\end{array}$ Therefore, our result, Theorem 2.6, improves the well-known result by Tominaga [9, Theorem 3.1],

$A∇vB−A♯vB≤L1,hlog⁡ShA.$

Since gv(x) is convex, we can not obtain a general result on the lower bound for AvBAvB. However, if we impose the conditions, we can obtain new sharp inequalities for Young inequalities as an additive-type in the first inequalities both (i) and (ii) in the following proposition. (At the same time, of course, we also obtain the upper bounds straightforwardly.)

Proposition 2.9

Let A,B > 0 such that sABtA for some scalars 0 < st, v ∈ [0, 1], and gv is defined as in Theorem 2.6.

1. If t ≤ 1, then gv(t)AAv BAvBgv(s)A.

2. If s ≥ 1, then gv(s)AAvBAvBgv(t)A.

Proof

It follows from the fact that gv(x) is monotone decreasing for 0 < x ≤ 1 and monotone increasing for x ≥ 1.□

Corollary 2.10

Let A, B > 0, m, m′, M, M′ > 0, and v ∈ [0, 1].

1. If 0 < mIAmI < MIBMI, then

$1mm∇vM−m♯vMA≤A∇vB−A♯vB≤1m′m′∇vM′−m′♯vM′A.$

2. If 0 < mIBmI < MIAMI, then

$1MM∇vm−M♯vmA≤A∇vB−A♯vB≤1M′M′∇vm′−M′♯vm′A.$

In the following we use the notations ▽v and ♮v to distinguish from the operator means ∇v and ♯v:

$A▽vB=1−vA+vB,A♮vB=A12A−12BA−12vA12$

for v ∉ [0, 1]. Notice that, since A,B > 0, the expressions AvB and AvB are also well-defined.

Remark 2.11

It is known (and easy to show) that for any A,B > 0,

$A▽vB≤A♮vB,forv∉0,1.$

Assume gv(x) is defined as in Theorem 2.6. By an elementary computation we have

$gv′x>0forv∉0,1and01.$

Now, in the same way as above we have also for any v ∉ [0, 1]:

1. If 0 < mIAmIMIBMI, then

$1m′m′♮vM′−m′▽vM′A≤A♮vB−A▽vB≤1mm♮vM−m▽vMA.$

On account of assumptions, we also infer

$(m′♮vM′−m′▽vM′)I≤A♮vB−A▽vB≤(m♮vM−m▽vM)I.$

2. If 0 < mIBmIMIAMI, then

$1MM♮vm−M▽vmA≤A♮vB−A▽vB≤1M′M′♮vm′−M′▽vm′A.$

On account of assumptions, we also infer

$(M♮vm−M▽vm)I≤A♮vB−A▽vB≤(M′♮vm′−M′▽vm′)I.$

In addition, with the same assumption to Theorem 2.6 except for v ∉ [0, 1], we have

$min{gv(s),gv(t)}A≤A▽vB−A♮vB,$

since we have min{gv(s),gv(t)} ≤ gv(x) by $\begin{array}{}{g}_{v}^{″}\end{array}$ (x) ≤ 0, for v ∉[0, 1].

Acknowledgement

The authors thank anonymous referees for giving valuable comments and suggestions to improve our manuscript. The author (S.F.) was partially supported by JSPS KAKENHI Grant Number 16K05257.

References

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Furuichi S., Moradi H.R., Sababheh M., New sharp inequalities for operator means, Linear Multilinear Algebra., https://doi.org/10.1080/03081087.2018.1461189

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Accepted: 2018-10-18

Published Online: 2018-12-27

Citation Information: Open Mathematics, Volume 16, Issue 1, Pages 1478–1482, ISSN (Online) 2391-5455,

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