Various inequalities between quasi-arithmetic mean and quasi-geometric type means for matrices

TL;DR

This paper investigates inequalities between various quasi-arithmetic and quasi-geometric means of positive semidefinite matrices, establishing conditions on parameters for these inequalities to hold under different matrix orderings.

Contribution

It introduces new inequalities between matrix means and determines necessary and sufficient conditions for these inequalities across different matrix orderings.

Findings

Derived conditions for inequalities between matrix means.

Extended classical inequalities to matrix settings.

Analyzed inequalities under multiple matrix orderings.

Abstract

In this paper, for $0<\alpha<1$, $p>0$ and positive semidefinite matrices $A,B\ge0$, we consider the quasi-extension $\mathcal{A}_{\alpha,p}(A,B):=((1-\alpha)A^p+\alpha B^p)^{1/p}$ of the $\alpha$-weighted arithmetic matrix mean, and the quasi-extensions $\mathcal{M}_{\alpha,p}(A,B):=\mathcal{M}_\alpha(A^p,B^p)^{1/p}$ of several different $\alpha$-weighted geometric-type matrix means $\mathcal{M}_\alpha(A,B)$ such as the $\alpha$-weighted geometric mean in Kubo and Ando's sense and two types of $\alpha$-weighted version of Fiedler and Pt\'ak's spectral geometric mean, as well as the R\'enyi mean and the $\alpha$-weighted Log-Euclidean mean. For these we examine the inequalities $\mathcal{A}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ and $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ of arithmetic-geometric type, where $\triangleleft$ is one of several different matrix orderings varying from the strongest Loewner order to the weakest order determined by trace inequality. For each choice of the above inequalities, our goal is to hopefully obtain the necessary and sufficient condition on $p,q,\alpha$ under which the inequality holds for all $A,B\ge0$.

Tables

	Sufficient cond.	Necessary cond.
${𝒜}_{\alpha ,q}\le {𝒜}_{\alpha ,q}$	$p=q$ or or	$p=q$ or or
$\begin{array}{ccccc}{𝒜}_{\alpha ,p}{\le }_{\mathrm{chao}}{𝒜}_{\alpha ,q}& & & & \\ {𝒜}_{\alpha ,p}{\le }_{\mathrm{near}}{𝒜}_{\alpha ,q}& & & & \\ {𝒜}_{\alpha ,p}{\le }_{\lambda }{𝒜}_{\alpha ,q}& & & & \\ {𝒜}_{\alpha ,p}{\prec }_w{𝒜}_{\alpha ,q}& & & & \\ {𝒜}_{\alpha ,p}{\le }_{\mathrm{Tr}}{𝒜}_{\alpha ,q}& & & & \end{array}$	$p\le q$	$p\le q$

	Sufficient cond.	Necessary cond.
$L{E}_{\alpha }\le {𝒜}_{\alpha ,q}$	none
$\begin{array}{ccccc}L{E}_{\alpha }{\le }_{\mathrm{chao}}{𝒜}_{\alpha ,q}& & & & \\ L{E}_{\alpha }{\le }_{\mathrm{near}}{𝒜}_{\alpha ,q}& & & & \\ L{E}_{\alpha }{\le }_{\lambda }{𝒜}_{\alpha ,q}& & & & \\ L{E}_{\alpha }{\prec }_w{𝒜}_{\alpha ,q}& & & & \\ L{E}_{\alpha }{\le }_{\mathrm{Tr}}{𝒜}_{\alpha ,q}& & & & \end{array}$	arbitrary $q$

	Sufficient cond.	Necessary cond.
$\begin{array}{ccc}{R}_{\alpha ,p}\le {𝒜}_{\alpha ,q}& & \\ R_{\alpha ,p}{\le }_{\mathrm{chao}}{𝒜}_{\alpha ,q}& & \\ R_{\alpha ,p}{\le }_{\mathrm{near}}{𝒜}_{\alpha ,q}& & \end{array}$	none
$\begin{array}{cc}{R}_{\alpha ,p}{\le }_{\lambda }{𝒜}_{\alpha ,q}& \\ R_{\alpha ,p}{\prec }_w{𝒜}_{\alpha ,q}& \end{array}$	$p/2\le q$	$\alpha \left(1-\alpha \right)p\le q$
$R_{\alpha ,p}{\le }_{\mathrm{Tr}}{𝒜}_{\alpha ,q}$	$q\ge 1$ or $p/2\le q$	$q\ge 1$ or $\alpha \left(1-\alpha \right)p\le q$
$R_{1/2,1}{\le }_{\mathrm{Tr}}{𝒜}_{1/2,q}$	$q\ge 1/4$	$q\ge 1/4$

	Sufficient cond.	Necessary cond.
$G_{\alpha ,p}\le {𝒜}_{\alpha ,q}$	$1\le p\le q$	$1\le p\le q$
$\begin{array}{ccc}{G}_{\alpha ,p}{\le }_{\mathrm{chao}}{𝒜}_{\alpha ,q}& & \\ G_{\alpha ,p}{\le }_{\mathrm{near}}{𝒜}_{\alpha ,q}& & \\ G_{\alpha ,p}{\le }_{\lambda }{𝒜}_{\alpha ,q}& & \end{array}$	$p\le q$	$p\le q$
$\begin{array}{cc}{G}_{\alpha ,p}{\prec }_w{𝒜}_{\alpha ,q}& \\ G_{\alpha ,p}{\le }_{\mathrm{Tr}}{𝒜}_{\alpha ,q}& \end{array}$	arbitrary $p,q$

	Sufficient cond.	Necessary cond.
$\begin{array}{cc}S{G}_{\alpha ,p}\le {𝒜}_{\alpha ,q}& \\ S{G}_{\alpha ,p}{\le }_{\mathrm{chao}}{𝒜}_{\alpha ,q}& \end{array}$	none
$S{G}_{\alpha ,p}{\le }_{\mathrm{near}}{𝒜}_{\alpha ,q}$	?	$q/p\ge 1+\max \{\alpha \left(q-1\right),\left(1-\alpha \right)\left(q-1\right)\}$
$S{G}_{\alpha ,p}{\le }_{\lambda }{𝒜}_{\alpha ,q}$	?	$q/p\ge \max \{\alpha ,1-\alpha \}$
$S{G}_{\alpha ,p}{\prec }_w{𝒜}_{\alpha ,q}$	$p/q\le 2\min \{\alpha ,1-\alpha \}$	$q/p\ge \max \{\alpha ,1-\alpha \}$
$S{G}_{\alpha ,p}{\le }_{\mathrm{Tr}}{𝒜}_{\alpha ,q}$	$q\ge 1$ or $p/q\le 2\min \{\alpha ,1-\alpha \}$	$q\ge 1$ or $p\le 2q$

	Sufficient cond.	Necessary cond.
$\begin{array}{cc}\tilde{S}{G}_{\alpha ,p}\le {𝒜}_{\alpha ,q}& \\ \tilde{S}{G}_{\alpha ,p}{\le }_{\mathrm{chao}}{𝒜}_{\alpha ,q}& \end{array}$	none
$\tilde{S}{G}_{\alpha ,p}{\le }_{\mathrm{near}}{𝒜}_{\alpha ,q}$	none for $\alpha \ne 1/2$	?
$\tilde{S}{G}_{\alpha ,p}{\le }_{\lambda }{𝒜}_{\alpha ,q}$	?	$p\le \frac{q}{1-\alpha }$
$\tilde{S}{G}_{\alpha ,p}{\prec }_w{𝒜}_{\alpha ,q}$	$p\le 2\alpha q$	$p\le \frac{q}{1-\alpha }$
$\tilde{S}{G}_{\alpha ,p}{\le }_{\mathrm{Tr}}{𝒜}_{\alpha ,q}$	$q\ge 1$ or $p\le 2\alpha q$	$q\ge 1$ or $p\le \frac{q}{1-\alpha }$

Equations

$$A{▽}_{\alpha }B$$

$$A{!}_{\alpha }B$$

$$A{\#}_{\alpha }B$$

$$A{!}_{\alpha }B\le A{\#}_{\alpha }B\le A{▽}_{\alpha }B.$$

$$L{E}_{\alpha }(A,B):=\exp ((1-\alpha )\log A+\alpha \log B)$$

$$F(A,B):=(A^{-1}\#B{)}^{1/2}A(A^{-1}\#B{)}^{1/2}$$

$$F_{\alpha }(A,B):=(A^{-1}\#B{)}^{\alpha }A(A^{-1}\#B{)}^{\alpha },$$

$${\tilde{F}}_{\alpha }(A,B):=(A^{-1}{\#}_{\alpha }B{)}^{1/2}{A}^{2(1-\alpha )}(A^{-1}{\#}_{\alpha }B{)}^{1/2}$$

$$R_{\alpha ,z}(A,B):=(A^{\frac{1-\alpha }{2z}}{B}^{\frac{\alpha }{z}}{A}^{\frac{1-\alpha }{2z}}{)}^z$$

$${\mathcal{A}}_{\alpha ,p}(A,B):=(A^p{▽}_{\alpha }{B}^p{)}^{1/p}=((1-\alpha )A^p+\alpha B^p{)}^{1/p},$$

$${\mathcal{H}}_{\alpha ,p}(A,B):=(A^p{!}_{\alpha }{B}^p{)}^{1/p}.$$

$$G_{\alpha ,p}(A,B):=(A^p{\#}_{\alpha }{B}^p{)}^{1/p}.$$

$$F(A,B):=(A^{-1}\#B{)}^{1/2}A(A^{-1}\#B{)}^{1/2},$$

$$F_{\alpha }(A,B):=(A^{-1}\#B{)}^{\alpha }A(A^{-1}\#B{)}^{\alpha }.$$

$$S{G}_{\alpha ,p}(A,B):=F_{\alpha }(A^p,B^p{)}^{1/p}.$$

$${\tilde{F}}_{\alpha }(A,B):=(A^{-1}{\#}_{\alpha }B{)}^{1/2}{A}^{2(1-\alpha )}(A^{-1}{\#}_{\alpha }B{)}^{1/2}.$$

$$\tilde{S}{G}_{\alpha ,p}(A,B):={\tilde{F}}_{\alpha }(A^p,B^p{)}^{1/p}.$$

$$R_{\alpha ,p}(A,B):=(A^{\frac{1-\alpha }{2}p}{B}^{\alpha p}{A}^{\frac{1-\alpha }{2}p}{)}^{1/p},$$

$$L{E}_{\alpha }(A,B):=\exp ((1-\alpha )\log A+\alpha \log B).$$

$$L{E}_{\alpha }(A,B):=P_0\exp \{(1-\alpha )P_0(\log A)P_0+\alpha P_0(\log B)P_0\},$$

$$S{G}_{0,p}(A,B)=(s(B)A^{p}s(B){)}^{1/p},S{G}_{1,p}(A,B)=B,$$

$$\tilde{S}{G}_{0,p}(A,B)=A,\tilde{S}{G}_{1,p}(A,B)=(B^{p/2}s(A)B^{p/2}{)}^{1/p}.$$

$${\mathcal{M}}_{\alpha ,p}(A^{-1},B^{-1})={\mathcal{M}}_{\alpha ,p}(A,B{)}^{-1}.$$

$$(A,B)\mapsto A^{1/2}\{(1-\alpha )I+\alpha (A^{-1/2}B{A}^{-1/2}{)}^p{\}}^{1/p}{A}^{1/2},$$

$$\mathcal{M}(A,B)=\underset{\epsilon \searrow 0}{\lim }\mathcal{M}(A+\epsilon I,B+\epsilon I).$$

$$(A+\epsilon I{)}^{-p}\#(B+\epsilon I{)}^p$$

$$\displaystyle\to A^{-p/2}(A^{p/2}B^{p}A^{p/2})^{1/2}A^{-p/2}=A^{-p}\#B^{p}\quad\mbox{as $\varepsilon\searrow 0$},$$

$$\mathcal{M}(A,B)$$

$$\mathcal{M}(A+\epsilon I,B+\epsilon I)$$

$$L{E}_{\alpha }(A,B)=\underset{p\searrow 0}{\lim }{\mathcal{M}}_{\alpha ,p}(A,B).$$

$$S{G}_{\alpha ,p}(A,B){|}_{{\mathcal{H}}_0}$$

$$=\{P_0+p(2\alpha L+P_0(\log A)P_0+o(p)\}{|}_{{\mathcal{H}}_0}$$

$$=\{P_0+p((1-\alpha )P_0(\log A)P_0+\alpha P_0(\log B)P_0)+o(p)\}{|}_{{\mathcal{H}}_0}.$$

$$\frac{1}{p}\log S{G}_{\alpha ,p}(A^p,B^p){|}_{{\mathcal{H}}_0}=\{(1-\alpha )P_0(\log A)P_0+\alpha P_0(\log B)P_0+o(1)\}{|}_{{\mathcal{H}}_0},$$

Full text

2020 Mathematics Subject Classification: 15A45, 47A64

Keywords and phrases: quasi matrix mean, weighted arithmetic mean, weighted harmonic mean, weighted geometric mean, spectral geometric mean, Rényi mean, Log-Euclidean mean, Loewner order, chaotic order, near order, entrywise eigenvalue order, weak majorization, Lie–Trotter–Kato product formula

1. 1 Introduction

2. 2 Preliminaries

3. 2.1 Quasi matrix means

4. 2.2 Matrix order relations

5. 3 Some technical computations for $2\times 2$ matrices

6. 4 Various quasi-arithmetic-geometric inequalities

7. 4.1 $\mathcal{A}_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

8. 4.2 $LE_{\alpha}$ vs. $\mathcal{A}_{\alpha,q}$

9. 4.3 $R_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

10. 4.4 $G_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

11. 4.5 $SG_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

12. 4.6 $\widetilde{S}G_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

13. 5 Concluding remarks

14. A The operator mean version of the Lie–Trotter–Kato product formula for positive semidefinite matrices

1 Introduction

For positive semidefinite matrices (also operators) the most developed two-variable operator/matrix means are Kubo and Ando’s operator means [29], defined corresponding to operator monotone functions $f\geq 0$ on $[0,\infty)$ with $f(1)=1$ as $A\sigma_{f}B:=A^{1/2}f(A^{-1/2}BA^{-1/2})A^{1/2}$ for positive definite matrices (also operators) $A,B>0$ and extended to general positive semidefinite $A,B\geq 0$ as $A\sigma_{f}B:=\lim_{\varepsilon\searrow 0}(A+\varepsilon I)\sigma_{f}(B+\varepsilon I)$. Among them the most typical ones are the $\alpha$-weighted arithmetic, harmonic and geometric means $\triangledown_{\alpha}$, $!_{\alpha}$ and $\#_{\alpha}$ for $0\leq\alpha\leq 1$, whose definitions for $A,B>0$ are

[TABLE]

The so-called arithmetic-geometric-harmonic mean inequality in the Loewner order is

[TABLE]

Geometric-type matrix means not in Kubo and Ando’s sense have recently been in active consideration in matrix analysis, partly motivated by recent development of quantum divergences in quantum information. The most familiar one is the $\alpha$-weighted Log-Euclidean mean

[TABLE]

for $A,B>0$. The spectral geometric mean

[TABLE]

for $A,B>0$ was formerly introduced in [15] and extended in [30] to the $\alpha$-weighted version as

[TABLE]

which has recently been reconsidered in [17, 19, 16] from a new perspective. Another new $\alpha$-weighted version of the spectral geometric mean was also recently introduced in [11] as

[TABLE]

Yet another geometric-type mean discussed in [22, 13] is

[TABLE]

with two parameters $\alpha\in(0,1)$ and $z>0$, which is called the Rényi mean because of its close relation with the $\alpha$-$z$-Rényi divergence [6].

For any matrix mean $\mathcal{M}(A,B)$ we have its quasi-extension $\mathcal{M}(A^{p},B^{p})^{1/p}$ with a parameter $p>0$. For $0<\alpha<1$ and $p>0$ the quasi-extension $(A^{p}\triangledown_{\alpha}B^{p})^{1/p}=((1-\alpha)A^{p}+\alpha B^{p})^{1/p}$ of the $\alpha$-weighted arithmetic mean has been discussed by many authors under the name “operator power mean.” We notice that $LE_{\alpha}(A,B)$ is invariant under taking quasi-extension. In the present paper we denote the quasi-extensions of $A\triangledown_{\alpha}B$, $A\#_{\alpha}B$, $A!_{\alpha}B$, $F(A,B)$ and $\widetilde{F}(A,B)$ respectively by $\mathcal{A}_{\alpha,p}(A,B)$, $G_{\alpha,p}(A,B)$, $\mathcal{H}_{\alpha,p}(A,B)$, $SG_{\alpha}(A,B)$ and $\widetilde{S}G_{\alpha,p}(A,B)$. We also write $R_{\alpha,p}(A,B)$ for $R_{\alpha,z}(A,B)$ with parameter $p=1/z$ instead of $z$. See Section 2.1 for more details on these quasi-extended matrix means. The quasi matrix means $G_{\alpha,p}$, $SG_{\alpha,p}$, $\widetilde{S}G_{\alpha,p}$, $R_{\alpha,p}$ as well as $LE_{\alpha}$ are referred to as “quasi-geometric type matrix means” because all of them are reduced to $A^{1-\alpha}B^{\alpha}$ when $AB=BA$.

Our main aim of this paper is to extend the arithmetic-geometric mean inequality in (1.1) to those for the quasi arithmetic mean and the quasi-geometric type means mentioned above. A special feature of our study is that we consider not only the Loewner order ($X\leq Y$ for $X,Y\geq 0$) but also several of other weaker orderings such as the chaotic order (denoted by $X\leq_{\mathrm{chao}}Y$), the near order recently introduced in [12] (denoted by $X\leq_{\mathrm{near}}Y$), the entrywise eigenvalue order (denoted by $X\leq_{\lambda}Y$), the weak majorization ($X\prec_{w}Y$) and the trace inequality $\mathrm{Tr}\,X\leq\mathrm{Tr}\,Y$ (denoted by $X\leq_{\mathrm{Tr}}Y$). See Section 2.2 for the explicit definitions of these orderings. It may also be stressed that we deal with quasi matrix means for general positive semidefinite matrices though restricted to positive definite matrices in most references. Our goal is to hopefully obtain the necessary and sufficient condition on $p,q,\alpha$ under which the inequality $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A$ and $B$, when $\mathcal{M}_{\alpha,p}$ is one of the quasi-geometric type matrix means and $\triangleleft$ is one of the above matrix orderings.

The structure of the paper is as follows. In Section 2.1, for $0\leq\alpha\leq 1$ and $p>0$ we review the definitions of the above stated quasi matrix means $\mathcal{A}_{\alpha,p}(A,B)$, $G_{\alpha,p}(A,B)$, etc. for general positive semidefinite matrices $A,B\geq 0$ (with the support condition $s(A)\geq s(B)$ for $SG_{\alpha,p}$ and $\widetilde{S}G_{\alpha,p}$). Some general basic facts on these quasi matrix means are summarized. In Section 2.2 we review the above mentioned matrix orderings and summarize some basic properties of them such as the strength relationship between them. In particular, for any quasi matrix mean $\mathcal{M}_{\alpha,p}$ and any matrix ordering $\triangleleft$ among stated above, we show (Theorem 2.3) that if $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B>0$, then the same holds for all $A,B\geq 0$ (with $s(A)\geq s(B)$ for $\mathcal{M}_{\alpha,p}=SG_{\alpha,p}$, $\widetilde{S}G_{\alpha,p}$). Section 3 provides some technical computations for specified $2\times 2$ positive definite matrices as lemmas, which are repeatedly used in Section 4. The main Section 4 are divided into six subsections. When $\mathcal{M}_{\alpha,p}$ is respectively $\mathcal{A}_{\alpha,p}$, $LE_{\alpha}$, $R_{\alpha,p}$, $G_{\alpha,p}$, $SG_{\alpha,p}$ and $\widetilde{S}G_{\alpha,p}$, we examine the inequality $\mathcal{M}_{\alpha,p}\triangleleft\mathcal{A}_{\alpha,q}$ for any choice $\triangleleft$ of the above matrix orderings. At the end of each subsection a table surveying the results of the subsection is attached for the reader’s convenience. Finally in Section 5 several remarks are in order to supplement characteristics and motivation of our study, additional facts, open questions, etc. The paper contains an appendix on the Lie–Trotter–Kato product formula for operator means, which is used in Section 2. This formula for positive semidefinite matrices is expected to be of quite use, while it has been nowhere published in its complete form.

2 Preliminaries

For each $n\in\mathbb{N}$ we write $\mathbb{M}_{n}$ for the $n\times n$ complex matrices. Let $\mathbb{M}_{n}^{+}$ and $\mathbb{M}_{n}^{++}$ be the positive semidefinite $n\times n$ matrices and the positive definite $n\times n$ matrices, respectively. We simply write $A\geq 0$ for $A\in\mathbb{M}_{n}^{+}$ and $A>0$ for $A\in\mathbb{M}_{n}^{++}$. The $n\times n$ identity matrix is denoted by $I_{n}$ or simply $I$. Let $\mathrm{Tr}$ be the usual trace on $\mathbb{M}_{n}$ and $\|X\|_{\infty}$ be the operator norm of $X\in\mathbb{M}_{n}$. We write “for all $A,B\geq 0$” to mean “for all $A,B\in\mathbb{M}_{n}^{+}$ with any $n\in\mathbb{N}$”, and “for all $A,B>0$” to mean “for all $A,B\in\mathbb{M}_{n}^{++}$ with any $n\in\mathbb{N}$”. For $A\geq 0$ there are two options of the convention of $A^{0}$; the one is $A^{0}:=I$ and the other is $A^{0}:=s(A)$, the support projection of $A$. In our discussions below we adopt the latter convention. We write $A^{-1}$ for the generalized inverse of $A$, i.e., the inverse of $A$ under the restriction to the support of $A$. Moreover, for $p<0$ we define $A^{p}:=(A^{-1})^{-p}$.

In this preliminary section we will explain several examples of quasi matrix means and different notions of matrix orders, which provide the basis for our discussions in the main Section 4.

2.1 Quasi matrix means

We first enumerate the definitions of quasi extensions of several binary matrix means for matrices, whose order properties will be discussed in this paper. Let $0\leq\alpha\leq 1$ and $p>0$. Let $A,B\in\mathbb{M}_{n}^{+}$.

(i) The $\alpha$-weighted arithmetic mean is $A\triangledown_{\alpha}B:=(1-\alpha)A+\alpha B$. Define the quasi $\alpha$-weighted arithmetic mean by

[TABLE]

which is also called the ($\alpha$-weighted) matrix power mean.

(ii) For $A,B>0$ the $\alpha$-weighted harmonic mean is $A!_{\alpha}B:=((1-\alpha)A^{-1}+\alpha B^{-1})^{-1}$, extended to general $A,B\geq 0$ as $A!_{\alpha}B:=\lim_{\varepsilon\searrow 0}(A+\varepsilon I)!_{\alpha}(B+\varepsilon I)$. Define the quasi $\alpha$-weighted harmonic mean by

[TABLE]

(iii) For $A,B>0$ the $\alpha$-geometric mean is $A\#_{\alpha}B:=A^{1/2}(A^{-1/2}BA^{-1/2})^{\alpha}A^{1/2}$, extended to $A,B\geq 0$ as $A\#_{\alpha}B:=\lim_{\varepsilon\searrow 0}(A+\varepsilon I)\#_{\alpha}(B+\varepsilon I)$. The geometric mean $\#$ ($=\#_{1/2}$) was first introduced by Pusz and Woronowicz [33], and $\#_{\alpha}$, together with the above $\triangledown_{\alpha}$ and $!_{\alpha}$, is a typical example of Kubo and Ando’s operator means [29]. Define the quasi $\alpha$-weighted geometric mean by

[TABLE]

(iv) For $A,B>0$ the spectral geometric mean due to Fiedler and Pták [15] is

[TABLE]

which was extended to the $\alpha$-weighted version in [30] as

[TABLE]

For recent study of the weighted spectral geometric mean, see, e.g., [28, 17, 19, 16] where $F_{\alpha}(A,B)$ is denoted by $A\natural_{\alpha}B$. We define the quasi $\alpha$-weighted spectral geometric mean by

[TABLE]

Although the definitions of $F_{\alpha}$ and $SG_{\alpha,p}$ are available for any $A,B\geq 0$ with $A^{-p}$ in the generalized sense, they are meaningful as far as $s(A)\geq s(B)$. Indeed, in this case we have $SG_{\alpha,p}(A,B)=\lim_{\varepsilon\searrow 0}SG_{\alpha,p}(A+\varepsilon I,B+\varepsilon I)$ as verified in Proposition 2.2 below. When $s(A)\leq s(B)$, the situation is similar by interchanging $A,B$ since $SG_{\alpha,p}(A,B)=SG_{1-\alpha,p}(B,A)$ for $A,B>0$ (see Remark 2.1(2) below). In this paper we will consider $SG_{\alpha,p}(A,B)$ for $A,B\geq 0$ with $s(A)\geq s(B)$.

(v) For $A,B>0$ another new weighted version of the spectral geometric mean was recently introduced in [11] as

[TABLE]

We define the quasi version of this by

[TABLE]

Note that $\widetilde{F}_{1/2}=F_{1/2}$ ($=F$) and so $\widetilde{S}G_{1/2,p}=SG_{1/2,p}$ for all $p>0$. The definitions of $\widetilde{F}_{\alpha}$ and $\widetilde{S}G_{\alpha,p}$ are meaningful for any $A,B\geq 0$ with $s(A)\geq s(B)$ similarly to $SG_{\alpha,p}$ in (iv); see Proposition 2.2 below. We will use $\widetilde{S}G_{\alpha,p}$ as well as $SG_{\alpha,p}$ in this situation.

(vi) We consider one more quasi matrix mean defined for all $A,B\geq 0$ by

[TABLE]

which is called the Rényi mean in [13]. Note that the trace function $\mathrm{Tr}\,R_{\alpha,1/z}(B,A)$ (for general $\alpha,z>0$) appears as the main component in the definition of the $\alpha$-$z$-Rényi divergence (see [6, 22]) in quantum information, that is the reason for the terminology.

(vii) For $A,B>0$ the Log-Euclidean mean is

[TABLE]

Restricting to $0<\alpha<1$ we extend this to general $A,B\geq 0$ as

[TABLE]

where $P_{0}:=s(A)\wedge s(B)$. This extended definition is justified by Proposition 2.2 and Theorem A.1 in Appendix A. Note that $LE_{\alpha}(A^{p},B^{p})^{1/p}=LE_{\alpha}(A,B)$ holds for all $p>0$, so we have no quasi extension for $LE_{\alpha}$.

In the next remark we collect a few general simple facts on the quasi matrix means defined in (i)–(vii) above.

Remark 2.1**.**

(1) For $\mathcal{M}_{\alpha,p}\in\{\mathcal{A}_{\alpha,p},\mathcal{H}_{\alpha,p},G_{\alpha,p}\}$, it is obvious by definition that $\mathcal{M}_{0,p}(A,B)=A$ and $\mathcal{M}_{1,p}(A,B)=B$ for all $A,B\geq 0$. From our definitions in (iv) and (v) it is easy to verify that for any $A,B\geq 0$ with $s(A)\geq s(B)$ and any $p>0$,

[TABLE]

Also, note that $R_{0,1}(A,B)=(A^{p/2}s(B)A^{p/2})^{1/p}$ and $R_{1,p}(A,B)=(s(A)B^{p}s(A))^{1/p}$ for all $A,B\geq 0$ and $p>0$. Hence the cases $\alpha=0,1$ will be trivial (or quite simple) for our purpose, so we will concentrate our considerations to $0<\alpha<1$ below.

(2) Let $0<\alpha<1$ and $p>0$. It is clear that any $\mathcal{M}_{\alpha,p}$ from $\mathcal{A}_{\alpha,p}$, $\mathcal{H}_{\alpha,p}$, $G_{\alpha,p}$, $LE_{\alpha}$ is symmetric in the sense that $\mathcal{M}_{\alpha,p}(A,B)=\mathcal{M}_{1-\alpha,p}(B,A)$ for all $A,B\geq 0$. This is the case also for $SG_{\alpha,p}$ when restricted to $A,B>0$; see [30]. However, this is not the case for $\widetilde{S}G_{\alpha,p}$.

(3) It is obvious that $\mathcal{A}_{\alpha,p}$ and $\mathcal{H}_{\alpha,p}$ are transformed each other by taking inverse, that is, $\mathcal{A}_{\alpha,p}(A^{-1},B^{-1})=\mathcal{H}_{\alpha,p}(A,B)^{-1}$ for all $A,B>0$ and $p>0$. It is also immediate to see that any $\mathcal{M}_{\alpha,p}$ of the quasi matrix means given in (iii)–(vii) is invariant under inverse, i.e., for any $A,B>0$,

[TABLE]

(4) We note that among quasi matrix means in (i)–(vii), only $\mathcal{A}_{\alpha,1}$, $\mathcal{H}_{\alpha,1}$ and $G_{\alpha,1}$ for $0\leq\alpha\leq 1$ are Kubo and Ando’s operator means. Indeed, although the function $\mathcal{A}_{\alpha,p}(1,x)=(1-\alpha+\alpha x^{p})^{1/p}$ is operator monotone on $(0,\infty)$ for $0\leq\alpha\leq 1$ and $0<p\leq 1$, the corresponding operator mean is

[TABLE]

which is obviously different from $\mathcal{A}_{\alpha,p}(A,B)$ except for $\alpha=0,1$ or $p=1$. The situation for $\mathcal{H}_{\alpha,p}$ is similar. For each $\mathcal{M}$ from $G_{\alpha,p}$, $SG_{\alpha,p}$, $\widetilde{S}G_{\alpha,p}$, $R_{\alpha,p},LE_{\alpha}$, the function $\mathcal{M}(1,x)=x^{\alpha}$ is operator monotone on $(0,\infty)$ for $0\leq\alpha\leq 1$ but the corresponding operator mean is $\#_{\alpha}$, i.e., $G_{\alpha,1}$.

Proposition 2.2**.**

Let $0<\alpha<1$, $p>0$, and $\mathcal{M}$ be any of the quasi matrix means given in (i)–(vii). Let $A,B\geq 0$, with an additional assumption $s(A)\geq s(B)$ when $\mathcal{M}=SG_{\alpha,p}$ or $\widetilde{S}G_{\alpha,p}$. Then we have

[TABLE]

Proof.

When $\mathcal{M}=\mathcal{A}_{\alpha,p}$ or $\mathcal{H}_{\alpha,p}$ or $G_{\alpha,p}$, it is easy to verify the result from the downward continuity of Kubo and Ando’s operator means since $(A+\varepsilon I)^{p}\searrow A^{p}$ and $(B+\varepsilon I)^{p}\searrow B^{p}$ as $\varepsilon\searrow 0$. For $\mathcal{M}=R_{\alpha,p}$ the result is obvious, and for $\mathcal{M}=LE_{\alpha}$ it was verified in [25, Sec. 4]; see also Remark A.2 in Appendix A. For the remaining, we first consider the case $A>0$. Since

[TABLE]

we see that $SG_{\alpha,p}(A+\varepsilon I,B+\varepsilon I)\to SG_{\alpha,p}(A,B)$ as $\varepsilon\searrow 0$. Similarly we have $\widetilde{S}G_{\alpha,p}(A+\varepsilon I,B+\varepsilon I)\to\widetilde{S}G_{\alpha,p}(A,B)$ too. For general $A,B\geq 0$ with $s(A)\geq s(B)$ take the decomposition $\mathbb{C}^{n}=\mathcal{H}_{1}\oplus\mathcal{H}_{2}$ where $\mathcal{H}_{1}$ is the range of $s(A)$ and $\mathcal{H}_{2}:=\mathcal{H}_{1}^{\perp}$. For $\mathcal{M}=SG_{\alpha,p}$ or $\widetilde{S}G_{\alpha,p}$ we can write

[TABLE]

where $I_{k}:=I_{\mathcal{H}_{k}}$, $k=1,2$. From the above shown case it follows that $\mathcal{M}(A_{1}+\varepsilon I_{1},B_{1}+\varepsilon I_{1})\to\mathcal{M}(A_{1},B_{1})$ as $\varepsilon\searrow 0$. Hence the assertion follows. ∎

The next theorem says that the quasi matrix means in (i)–(vi) satisfy the Lie–Trotter–Kato product formula. So we may consider the Log-Euclidean mean $LE_{\alpha}$ as a sort of attractor for those quasi matrix means. Note that when $A,B>0$, the proof is much simpler without use of Theorem A.1 and Remark A.4.

Theorem 2.3**.**

Let $0<\alpha<1$, $p>0$, and $\mathcal{M}_{\alpha,p}$ be any of the quasi matrix means given in (i)–(vi). Let $A,B\geq 0$, with an additional assumption $s(A)\geq s(B)$ when $\mathcal{M}=SG_{\alpha,p}$ or $\widetilde{S}G_{\alpha,p}$. Then we have

[TABLE]

Proof.

The convergences for $\mathcal{A}_{\alpha,p}$, $\mathcal{H}_{\alpha,p}$ and $G_{\alpha,p}$ are special cases of Theorem A.1, and that for $R_{\alpha,p}$ follows from (A.2) in Appendix A. For the remaining, assume that $s(A)\geq s(B)$. Let $P_{0}:=s(A)\wedge s(B)=s(B)$ and $\mathcal{H}_{0}$ be the range of $P_{0}$. Note that $SG_{\alpha,p}(A,B)=\{(A^{-p}\#B^{p})^{\alpha}A^{p}(A^{-p}\#B^{p})\}^{1/p}$ is supported on $\mathcal{H}_{0}$. Set $L:={1\over 2}\{-P_{0}(\log A)P_{0}+P_{0}(\log B)P_{0}\}$, where $\log A$ is defined in the generalized sense as $\log A:=s(A)(\log A)$ and similarly for $\log B$. Then by (A.25) in Remark A.4 one can write $A^{-p}\#B^{p}=P_{0}+pL+o(p)$ as $p\searrow 0$, which implies that $(A^{-p}\#B^{p})^{\alpha}=P_{0}+p\alpha L+o(p)$. Therefore, one has

[TABLE]

Applying this to the Taylor expansion of $\log(1+x)$ gives

[TABLE]

which yields the assertion for $SG_{\alpha,p}$. The proof for $\widetilde{S}G_{\alpha,p}$ is similar, which is omitted and left to the reader. ∎

2.2 Matrix order relations

Here we recall different types of order relations between matrices in $\mathbb{M}_{n}^{+}$. In Section 4 we will examine several of these orderings between quasi matrix means introduced in Section 2.1. Let $X,Y\in\mathbb{M}_{n}^{+}$.

(a) We write $X\leq Y$ as usual to denote the Loewner order, i.e., the positive semidefiniteness order in the sense that $Y-X$ is positive semidefinite.

(b) We write $X\leq_{\mathrm{chao}}Y$ and call it the chaotic order if $s(X)\leq s(Y)$ and $s(X)(\log X)s(X)\leq s(X)(\log Y)s(X)$. When $X,Y>0$, this simply reduces to $\log X\leq\log Y$.

(c) We write $X\leq_{\mathrm{near}}Y$ and call it the near order if $s(X)\leq s(Y)$ and $X\#Y^{-1}\leq I$, where $Y^{-1}$ is the generalized inverse of $Y$. This ordering was introduced in [12] and further discussed in [16] for $X,Y>0$. Note that $\leq_{\mathrm{near}}$ is not transitive (though ‘near’ transitive); see [12, Theorem 2].

(d) Let $\lambda(X)=(\lambda_{1}(X),\dots,\lambda_{n}(X))$ denote the eigenvalues of $X$ in decreasing order with multiplicities. We write $X\leq_{\lambda}Y$ to denote the entrywise eigenvalue order, i.e., $\lambda_{i}(X)\leq\lambda_{i}(Y)$ for each $i=1,\dots,n$.

(e) The weak (sub)majorization $X\prec_{w}Y$ means that

[TABLE]

This is equivalent to that $\|X\|\leq\|Y\|$ for any unitarily invariant norm $\|\cdot\|$; see, e.g., [21, Proposition 4.4.13]. More details on majorization theory are found in [2], [9, Chap. II] and [32].

(f) The weak log-majorization $X\prec_{w\log}Y$ means that

[TABLE]

The log-majorization $X\prec_{\log}Y$ means that $X\prec_{w\log}Y$ and $\prod_{i=1}^{n}\lambda_{i}(X)=\prod_{i=1}^{n}\lambda_{i}(Y)$, i.e., $\det X=\det Y$.

(g) We write $X\leq_{\mathrm{Tr}}Y$ if the trace inequality $\mathrm{Tr}\,X\leq\mathrm{Tr}\,Y$ holds.

In the next proposition we summarize the strength relationship between the matrix orderings defined in (a)–(g) above, together with a few basic properties.

Proposition 2.4**.**

Let $X,Y\geq 0$.

• (1)

$X\leq Y$* holds if and only $\log(tI+X)\leq\log(tI+Y)$ for all $t>0$.*

• (2)

$X\leq_{\mathrm{chao}}Y$* holds if and only if $s(X)\leq s(Y)$ and $X^{p}\#Y^{-p}\leq I$ for all $p>0$ (equivalently, for all $p\in(0,\delta)$ for some $\delta>0$), with $Y^{-p}$ in the generalized sense.*

• (3)

We have

[TABLE]

Proof.

(1) The ‘only if’ is obvious since $\log x$ ($x>0$) is operator monotone. Conversely, assume that $\log(tI+X)\leq\log(tI+Y)$ for all $t>0$. Since

[TABLE]

we have $X\leq Y$.

(2) This equivalence was shown in [1, Theorem 1] when $X,Y>0$. Assuming that $P:=s(X)\leq s(Y)$, we may show that $P(\log X)P\leq P(\log Y)P$ if and only if $X^{p}\#Y^{-p}\leq I$ for all $p>0$ (equivalently, for all $p\in(0,\delta)$). Recall [4, Theorem 2.1] that

[TABLE]

Moreover, by Theorem A.1,

[TABLE]

If $X^{p}\#Y^{-p}\leq I$ for all $p\in(0,\delta)$, then (2.2) gives $P\exp\{P(\log X)P-P(\log Y)P\}\leq I$ so that $P(\log X)P-P(\log Y)P\leq 0$. Conversely, if $P(\log X)P\leq P(\log Y)P$, then it follows from (2.1) and (2.2) that

[TABLE]

so that $X^{p}\#Y^{-p}\leq I$ for all $p>0$.

(3) Assume that $X\leq Y$. Then $s(X)\leq s(Y)$ and $X^{p}\#Y^{-p}\leq Y^{p}\#Y^{-p}=s(Y)\leq I$ for all $p\in(0,1)$. Hence the first implication follows from (2). The second implication is seen from (2) as well. The third was proved in [16, Theorem 2.4], whose proof is valid in the present setting. The remaining implications are obvious or well known; see, e.g., [21, Proposition 4.1.6] for the penultimate implication. ∎

Lemma 2.5**.**

Let $X,Y,X_{n},Y_{n},\geq 0$ for $n\in\mathbb{N}$ be given such that $X_{n}\to X$ and $Y_{n}\to Y$. For each $\triangleleft\in\{\leq,\leq_{\lambda},\prec_{w},\prec_{w\log},\leq_{\mathrm{Tr}}\}$, if $X_{n}\triangleleft Y_{n}$ for all $n\in\mathbb{N}$, then $X\triangleleft Y$. This holds for $\triangleleft=\,\leq_{\mathrm{chao}}$, $\leq_{\mathrm{near}}$ as well under the additional assumption $s(X)\leq s(Y)$.

Proof.

The first statement is immediately seen since $\lambda(X_{n})\to\lambda(X)$ and $\lambda(Y_{n})\to\lambda(Y)$. For the latter, assume that $s(X)\leq s(Y)$ and $X_{n}\leq_{\mathrm{chao}}Y_{n}$ for all $n$. By Proposition 2.4(2) one has $s(X_{n})\leq s(Y_{n})$ and $X_{n}^{p}\#Y_{n}^{-p}\leq I$, $p>0$. For any $p>0$, since

[TABLE]

letting $n\to\infty$ gives $(Y^{p/2}X^{p}Y^{p/2})^{1/2}\leq Y^{p}$ for all $p>0$. Thanks to $s(X)\leq s(Y)$, this in turn implies that $X^{p}\#Y^{-p}\leq s(Y)\leq I$. Hence $X\leq_{\mathrm{chao}}Y$ follows by Proposition 2.4(2) again. The proof for $\leq_{\mathrm{near}}$ is similar. ∎

The next theorem is of quite importance from the viewpoint of the scope of our study.

Theorem 2.6**.**

Let $0<\alpha<1$ and $p,q>0$. Let $\mathcal{M}_{\alpha,p}$ be any of the quasi matrix means in (i)–(vii) of Section 2.1, and $\triangleleft$ be any of $\leq$, $\leq_{\mathrm{chao}}$, $\leq_{\mathrm{near}}$, $\leq_{\lambda}$, $\prec_{w}$, $\prec_{w\log}$, $\leq_{\mathrm{Tr}}$. If $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ (resp., $\mathcal{H}_{\alpha,q}(A,B)\triangleleft\mathcal{M}_{\alpha,p}(A,B)$) holds for all $A,B>0$, then the same holds for all $A,B\geq 0$, where $s(A)\geq s(B)$ is assumed for $\mathcal{M}_{\alpha,p}=SG_{\alpha,p}$, $\widetilde{S}G_{\alpha,p}$.

Proof.

Assume that $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ for all $A,B>0$. Let $A,B\geq 0$ be arbitrary. The assumption implies that $\mathcal{M}_{\alpha,p}(A+\varepsilon I,B+\varepsilon I)\triangleleft\mathcal{A}_{\alpha,q}(A+\varepsilon I,B+\varepsilon I)$. Now $s(A)\geq s(B)$ is assumed when $\mathcal{M}_{\alpha,p}=SG_{\alpha,p}$ or $\widetilde{S}G_{\alpha,p}$. Then by Proposition 2.2 we have $\mathcal{M}_{\alpha,p}(A+\varepsilon I,B+\varepsilon I)\to\mathcal{M}_{\alpha,p}(A,B)$ as well as $\mathcal{A}_{\alpha,q}(A+\varepsilon I,B+\varepsilon I)\to\mathcal{A}_{\alpha,q}(A,B)$ as $\varepsilon\searrow 0$. Moreover, it is clear that $s(\mathcal{M}_{\alpha,p}(A,B))\leq s(A)\vee s(B)=s(\mathcal{A}_{\alpha,q}(A,B))$. Hence we have $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ by Lemma 2.5. The proof is similar for $\mathcal{H}_{\alpha,q}(A,B)\triangleleft\mathcal{M}_{\alpha,p}(A,B)$ in view of $s(\mathcal{M}_{\alpha,p}(A,B))\geq s(A)\wedge s(B)=s(\mathcal{H}_{\alpha,q}(A,B))$. ∎

Concerning the quasi matrix means and matrix orderings mentioned above, some general facts relevant to our study are in order.

Remark 2.7**.**

(1) Let $0<\alpha,\beta<1$ with $\alpha\neq\beta$, and $p,q>0$ be arbitrary. Let

[TABLE]

Let $f_{\alpha}(x):=\mathcal{M}_{\alpha}(1,x)$ and $g_{\beta}(x):=\mathcal{M}_{\beta}(1,x)$ for scalars $A=x>0$ and $B=1$. Since $f_{\alpha}(1)=g_{\beta}(1)=1$, $f^{\prime}_{\alpha}(1)=\alpha$ and $g^{\prime}_{\beta}(1)=\beta$, one has $f_{\alpha}(1)-g_{\beta}(1)=0$ and $f^{\prime}_{\alpha}(1)-g^{\prime}_{\beta}(1)\neq 0$. Therefore, the sign of $f_{\alpha}(x)-g_{\beta}(x)$ is different between the left and the right of $x=1$. This implies that $\mathcal{M}_{\alpha}(A,B)$ and $\mathcal{N}_{\beta}(A,B)$ are not definitively comparable with respect to any ordering. So we may only discuss inequalities between two of the above quasi matrix means under the same $\alpha$.

(2) For any $\alpha\in(0,1)$ it is clear that $G_{\alpha,p}(a,b)=SG_{\alpha,p}(a,b)=\widetilde{S}G(a,b)=R_{\alpha,p}(a,b)=LE_{\alpha}(a,b)=a^{1-\alpha}b^{\alpha}$ for scalars $a,b>0$, independently of $p>0$. For this reason we refer to $G_{\alpha,p},SG_{\alpha,p},\widetilde{S}G,R_{\alpha,p},LE_{\alpha}$ as quasi-geometric type matrix means. Also it is immediate to see that

[TABLE]

for scalars $a,b>0$, and moreover $\mathcal{H}_{\alpha,q}(a,b)\nearrow a^{1-\alpha}b^{\alpha}$ and $\mathcal{A}_{\alpha,p}(a,b)\searrow a^{1-\alpha}b^{\alpha}$ as $p\searrow 0$. Therefore, for each quasi-geometric type mean $\mathcal{M}_{\alpha,p}$ and for each ordering $\triangleleft$ mentioned above, we may consider possible inequalities between $\mathcal{M}_{\alpha,p},\mathcal{H}_{\alpha,q}$ and between $\mathcal{M}_{\alpha,p},\mathcal{A}_{\alpha,q}$ in the respective directions $\mathcal{H}_{\alpha,q}\triangleleft\mathcal{M}_{\alpha,p}$ and $\mathcal{M}_{\alpha,p}\triangleleft\mathcal{A}_{\alpha,q}$ only.

(3) Let $0<\alpha<1$, $p>0$ and $\mathcal{M}_{\alpha,p}$ be any quasi-geometric type mean as in Theorem 2.6. For any $X,Y>0$ the following are straightforward by definitions:

[TABLE]

From these, Remark 2.1(3) and Theorem 2.6 together we see that for any $\triangleleft\in\{\leq,\leq_{\mathrm{chao}},\leq_{\mathrm{near}},\leq_{\lambda}\}$, $\mathcal{H}_{\alpha,q}(A,B)\triangleleft\mathcal{M}_{\alpha,p}(A,B)$ holds for all $A,B>0$ if and only if $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B>0$. So the characterizations of inequalities between $\mathcal{H}_{\alpha,q},\mathcal{M}_{\alpha,p}$ are to large extent reduced to those between $\mathcal{M}_{\alpha,p},\mathcal{A}_{\alpha,q}$.

3 Some technical computations for $2\times 2$ matrices

In this section we perform some technical computations for specified $2\times 2$ matrices, which will be of quite use in the next main section. Below we will often treat $2\times 2$ matrices with parameter $\theta\in\mathbb{R}$, given in the approximate form up to $o(\theta^{2})$ as

[TABLE]

where $a,b,x_{11},x_{22},x_{12}\in\mathbb{R}$, and little-o notation $o(\theta^{2})$ means that ${o(\theta^{2})\over\theta^{2}}\to 0$ as $\theta\to 0$. In the first lemma we explain the way of computing the functional calculus of $X_{\theta}$ based on Taylor’s theorem and Daleckii and Krein’s derivative formulas (see, e.g., [21, Sec. 2.3]).

Lemma 3.1**.**

Let $X_{\theta}$ be given in (3.1) with $a\neq b$ and $f$ be a $C^{2}$-function on an interval containing $a,b$. Then the functional calculus $f(X_{\theta})$ is given in the approximate form as

[TABLE]

where

[TABLE]

In the above, $f^{[1]}(a_{1},a_{2})$ is the first divided difference of $f$ and $f^{[2]}(a_{1},a_{2},a_{3})$ is the second one (see [21, Sec. 2.2]).

Proof.

First, note that since $X_{\theta}\to X_{0}=\begin{bmatrix}a&0\\ 0&b\end{bmatrix}$ as $\theta\to 0$, $f(X_{\theta})$ is well defined for any $\theta$ near [math]. We write $X_{\theta}=X_{0}+\theta H+\theta^{2}K+o(\theta^{2})$ as $\theta\to 0$ where $H:=\begin{bmatrix}0&x_{12}\\ x_{12}&0\end{bmatrix}$ and $K:=\begin{bmatrix}x_{11}&0\\ 0&x_{22}\end{bmatrix}$. By Taylor’s theorem for the functional calculus $f(X)$ at $X_{0}$ (see [21, Theorem 2.3.1]) one has

[TABLE]

where $Df(X_{0})(\cdot)$ and $D^{2}f(X_{0})(\cdot,\cdot)$ are the first and the second Fréchet derivatives of $f(X)$ at $X_{0}$. Thanks to Daleckii and Krein’s derivative formulas (see [21, p. 163]) one can write, with the Schur product $\circ$,

[TABLE]

and

[TABLE]

Hence expression (3.2) holds with (3.3). ∎

Under the weaker assumption that $f$ is a $C^{1}$-function, the reduced version of (3.2) up to $o(\theta)$ (meaning ${o(\theta)\over\theta}\to 0$ as $\theta\to 0$) is given as

[TABLE]

with $y_{12}$ in (3.3).

Example 3.2**.**

The following special cases of Lemma 3.1 will be used below.

• (1)

Let $a=1\neq b>0$ in (3.1) and $f(x)=x^{r}$ on $(0,\infty)$ with $r>0$. Then

[TABLE]

where

[TABLE]

• (2)

Let $a=0\neq b\in\mathbb{R}$ in (3.1) and $f(x)=e^{x}$ on $\mathbb{R}$. Then

[TABLE]

where

[TABLE]

• (3)

Let $a=1$, $b>0$ in (3.1) and $f(x)=\log x$ on $(0,\infty)$. Then

[TABLE]

where

[TABLE]

We will repeatedly utilize the $2\times 2$ positive definite matrices given as follows:

[TABLE]

for $x,y>0$ and $\theta\in\mathbb{R}$. As immediately verified, the approximate form of $B_{\theta}$ up to $o(\theta^{2})$ is

[TABLE]

Lemma 3.3**.**

Let $0<\alpha<1$, $p,q>0$, and $x,y>0$ be such that $(1-\alpha)x^{p}+\alpha y^{p}\neq 1$ and $(1-\alpha)x^{q}+\alpha y^{q}\neq 1$. Let $A_{0}$ and $B_{\theta}$ be given in (3.5). Then we have

[TABLE]

where

[TABLE]

Furthermore, we have

[TABLE]

Proof.

Since $B_{\theta}^{p}$ is written as in (3.6) with $y^{p}$ in place of $y$, it follows that $X_{\theta}:=(1-\alpha)A_{0}^{p}+\alpha B_{\theta}^{p}$ is of the form (3.1) with $a=1$, $b=(1-\alpha)x^{p}+\alpha y^{p}$, $x_{11}=-\alpha(1-y^{p})$, $x_{22}=\alpha(1-y^{p})$ and $x_{12}=\alpha(1-y^{p})$. Since $(1-\alpha)x^{p}+\alpha y^{p}\neq 1$ by assumption, we see by Example 3.2(1) that $\mathcal{A}_{\alpha,p}(A_{0},B_{\theta})=X_{\theta}^{1/p}$ is given in the form (3.7), where

[TABLE]

Hence (3.7) holds with (3.8).

Moreover, since

[TABLE]

it follows that

[TABLE]

where

[TABLE]

Since the term $\alpha^{2}\{\cdots\}$ in (3.10) is equal to

[TABLE]

we have (3.9) as well. ∎

Lemma 3.4**.**

Let $0<\alpha<1$ and $x,y>0$ be such that $x^{1-\alpha}y^{\alpha}\neq 1$. Let $A_{0}$ and $B_{\theta}$ be given in (3.5). Then we have

[TABLE]

where

[TABLE]

Furthermore, when $p>0$ and $(1-\alpha)x^{p}+\alpha y^{p}\neq 1$, we have

[TABLE]

Proof.

We note that

[TABLE]

Since $\log x^{\alpha}y^{1-\alpha}\neq 0$ by assumption, one can apply Example 3.2(2) with $b=\log x^{1-\alpha}y^{\alpha}$, $x_{11}=\alpha\log y$ and $x_{22}=x_{12}=-\alpha\log y$. Hence it follows that $LE_{\alpha}(A_{0},B_{\theta})=e^{X_{\theta}}$ is given in the form (3.12), where

[TABLE]

Hence (3.12) holds with (3.13).

Moreover, since $(1-\alpha)x^{p}+\alpha y^{p}\neq 1$ by assumption, it follows from (3.7) and (3.12) that

[TABLE]

where $w_{0}:=1-x^{1-\alpha}y^{\alpha}$ and $w_{p}$ is given in (3.11). Since $\alpha^{2}\{\cdots\}$ in the last expression is equal to

[TABLE]

we have (3.14). ∎

Lemma 3.5**.**

Let $0<\alpha<1$, $p>0$, and $A_{0},B_{\theta}\in\mathbb{M}_{2}^{++}$ be given in (3.5) with $y=x>0$, $x\neq 1$. Then we have

[TABLE]

where

[TABLE]

Proof.

Since $B_{\theta}^{\alpha p}$ is written as in (3.6) with $x^{\alpha p}$ in place of $y$, it is easy to compute

[TABLE]

Then one can apply Example 3.2(1) to show (3.15) with (3.16) (as in the first paragraph of the proof of Lemma 3.3), whose details are left to the reader. ∎

In the last lemma of this section we explain the way of computing the eigenvalues of $X_{\theta}$ given in (3.1).

Lemma 3.6**.**

Let $X_{\theta}$ be given in (3.1) with $a\neq b$. Then the eigenvalues of $X_{\theta}$ is given in the approximate form as

[TABLE]

Proof.

Since

[TABLE]

the eigenvalues $\lambda_{\pm}(\theta)$ of $X_{\theta}$ are computed up to $o(\theta^{2})$ as

[TABLE]

Hence it is immediate to verify that $\lambda(X_{\theta})=(\lambda_{+}(\theta),\lambda_{-}(\theta))$ has expression (3.17). ∎

4 Various quasi-arithmetic-geometric inequalities

Throughout the section let $0<\alpha<1$ and $p,q>0$. For each quasi matrix mean $\mathcal{M}_{\alpha,p}$ from $\mathcal{A}_{\alpha,p}$, $LE_{\alpha}$, $R_{\alpha,p}$, $G_{\alpha,p}$, $SG_{\alpha,p}$, $\widetilde{S}G_{\alpha,p}$ and for each matrix ordering $\triangleleft$ from $\leq$, $\leq_{\mathrm{chao}}$, $\leq_{\mathrm{near}}$, $\leq_{\lambda}$, $\prec_{w}$, $\leq_{\mathrm{Tr}}$, we will examine the inequality $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$. Our goal is to find the necessary and sufficient condition of $p,q,\alpha$ for the inequality to hold for all $A,B>0$, though we have not succeeded it for all cases. In this paper we do not deal with the inequality $\mathcal{H}_{\alpha,q}\triangledown\mathcal{M}_{\alpha,p}$ (see Remark 2.7(3) and remark (4) of Section 5). Also we do not include the ordering $\prec_{w\log}$ in our considerations, though it would be meaningful to examine the differences between $\leq_{\lambda}$, $\prec_{w\log}$, $\prec_{w}$ (see Proposition 2.4(3)) in quasi-arithmetic-geometric mean inequalities.

The section is divided into six subsections.

4.1 $\mathcal{A}_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

The next theorem characterizes the inequality $\mathcal{A}_{\alpha,p}\leq\mathcal{A}_{\alpha,q}$, which was formerly shown in [7] with restriction to $\alpha=1/2$.

Theorem 4.1**.**

Let $0<\alpha<1$ and $p,q>0$. Then the following conditions are equivalent:

• (i)

$\mathcal{A}_{\alpha,p}(A,B)\leq\mathcal{A}_{\alpha,q}(A,B)$* for all $A,B\geq 0$;*

• (ii)

$\mathcal{A}_{\alpha,p}(A,B)\leq\mathcal{A}_{\alpha,q}(A,B)$* for all $A,B\in\mathbb{M}_{2}^{++}$;*

• (iii)

$p=q$* or $1\leq p<q$ or $1/2\leq p<1\leq q$.*

To prove the theorem, in addition to Lemma 3.3 we need the following lemma.

Lemma 4.2**.**

Define $\tilde{A}_{0},\tilde{B}_{\theta}\in\mathbb{M}_{2}^{+}$ by

[TABLE]

for $\theta\in\mathbb{R}$. Let $0<\alpha<1$ and $0<p,q<1$. Then we have

[TABLE]

Proof.

The proof is a modification of that of [7, Lemma 3.3] where the case $\alpha=1/2$ was treated. It is easy to check that

[TABLE]

where

[TABLE]

Letting $c:=\sqrt{a^{2}+b^{2}}$ ($\leq 1$), as in the proof of [7, Lemma 3.3] one can compute

[TABLE]

As $\theta\to 0$ we estimate

[TABLE]

so that

[TABLE]

Hence we have

[TABLE]

and moreover

[TABLE]

thanks to $0<p<1$. Therefore, we arrive at

[TABLE]

where

[TABLE]

Therefore, we obtain

[TABLE]

as asserted. ∎

Note that although (4.3) is written in the form of (3.1) with $a=\alpha+(1-\alpha)2^{p}$ and $b=0$, we cannot apply Lemma 3.1 to prove Lemma 4.2 because $x^{1/p}$ is not twice differentiable at $x=0$ when $p>1/2$.

Now we are ready to prove Theorem 4.1.

Proof of Theorem 4.1.

(iii)$\implies$(i) is seen from [7, Theorem 2.1]. Indeed, the proof is easy as follows. If $1\leq p<q$ then

[TABLE]

since $x^{p/q}$ is operator concave and $x^{1/p}$ is operator monotone on $[0,\infty)$. If $1/2\leq p<1\leq q$ then

[TABLE]

since $x^{1/p}$ is operator convex and $x^{1/q}$ is operator concave on $[0,\infty)$.

(i)$\implies$(ii) is trivial.

(ii)$\implies$(iii). By considering the inequality for $A=xI_{2}$ and $B=yI_{2}$, $x,y>0$, (ii) implies that $p\leq q$. Hence it suffices to show that (ii) fails to hold when $0<p<1/2$ and $q>p$ and when $0<p<q<1$.

First, assume that $0<p<1/2$ and $q>p$. Consider $A_{0}$ and $B_{\theta}$ in (3.5) with $0<y<1$ and $x=y^{2}$. Since $(1-\alpha)y^{2p}+\alpha y^{p}<1$ and $(1-\alpha)y^{2q}+\alpha y^{q}<1$ clearly, we can apply Lemma 3.3. As $y\searrow 0$ we estimate

[TABLE]

and

[TABLE]

Therefore, the dominant term of the big bracket $[\cdots]$ of the RHS of (3.9) is

[TABLE]

thanks to $2p<1$ when $y>0$ is sufficiently small. For such a $y>0$ the RHS of (3.9) $<0$ if $\theta$ is small enough, which implies that $\mathcal{A}_{\alpha,p}(A_{0},B_{\theta})\not\leq\mathcal{A}_{\alpha,q}(A_{0},B_{\theta})$.

Next, assume that $0<p<q<1$, and apply Lemma 4.2. Let $\tilde{A}_{0}$ and $\tilde{B}_{\theta}$ be given in (4.1). From expression (4.2) it follows that $\det\{\mathcal{A}_{\alpha,q}(\tilde{A}_{0},\tilde{B}_{\theta})-\mathcal{A}_{\alpha,p}(\tilde{A}_{0},\tilde{B}_{\theta})\}<0$ so that $\mathcal{A}_{\alpha,p}(\tilde{A}_{0},\tilde{B}_{\theta})\not\leq\mathcal{A}_{\alpha,q}(\tilde{A}_{0},\tilde{B}_{\theta})$ if $\theta$ is small enough. By continuity there are $A,B\in\mathbb{M}_{2}^{++}$ such that $\mathcal{A}_{\alpha,p}(A,B)\not\leq\mathcal{A}_{\alpha,q}(A,B)$. ∎

The inequalities between $\mathcal{A}_{\alpha,p}$ and $\mathcal{A}_{\alpha,q}$ with respect to other weaker orderings are simple as stated in the next proposition.

Proposition 4.3**.**

Let $0<\alpha<1$ and $p,q>0$. Then the following conditions are equivalent:

• (i)

$\mathcal{A}_{\alpha,p}(A,B)\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$* for all $A,B\geq 0$;*

• (ii)

$\mathcal{A}_{\alpha,p}(a,b)\leq\mathcal{A}_{\alpha,q}(a,b)$* for all scalars $a,b>0$;*

• (iii)

$p\leq q$.

Hence, for any $\triangleleft\in\{\leq_{\mathrm{chao}},\leq_{\mathrm{near}},\leq_{\lambda},\prec_{w},\leq_{\mathrm{Tr}}\}$, $\mathcal{A}_{\alpha,p}\triangleleft\mathcal{A}_{\alpha,q}$ holds for all $A,B\geq 0$ if and only if $p\leq q$.

Proof.

(i)$\implies$(ii) is obvious, and (ii)$\implies$(iii) is easy since (iii) means that $x^{q/p}$ is convex on $(0,\infty)$. Finally, let us show (iii)$\implies$(i). Assume that $p\leq q$. Let $A,B\geq 0$ be arbitrary. Note that

[TABLE]

and similarly $s(\mathcal{A}_{\alpha,q}(A,B))=P$. Since $x^{p/q}$ is operator concave on $[0,\infty)$, one has

[TABLE]

so that

[TABLE]

Hence (i) holds. ∎

The next proposition says that $p>0\mapsto\mathrm{Tr}\,\mathcal{A}_{\alpha,p}(A,B)$ is strictly increasing unless $A=B$.

Proposition 4.4**.**

Let $0<\alpha<1$ and $0<p<q$. Then for every $A,B\geq 0$ the following conditions are equivalent:

• (i)

$\mathrm{Tr}\,\mathcal{A}_{\alpha,p}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$;

• (ii)

$\mathcal{A}_{\alpha,p}(A,B)=\mathcal{A}_{\alpha,q}(A,B)$;

• (iii)

$A=B$.

Proof.

It is obvious that (iii)$\implies$(ii)$\implies$(i). To show (i)$\implies$(iii), assume (i) so that $\mathrm{Tr}\,\mathcal{A}_{\alpha,t}(A,B)$ is constant for $t\in[p,q]$ by Proposition 4.3. Since the function $t\mapsto\mathrm{Tr}\,\mathcal{A}_{\alpha,t}(A,B)$ is real analytic in $t>0$, it follows that $\mathrm{Tr}\,\mathcal{A}_{\alpha,t}(A,B)$ is constant for all $t>0$. In particular, $\mathrm{Tr}\,\mathcal{A}_{\alpha,1/2}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,1}(A,B)$, which gives $\mathrm{Tr}(A^{1/2}-B^{1/2})^{2}=0$ so that $A=B$. ∎

The results of this subsection are summarized as follows:

4.2 $LE_{\alpha}$ vs. $\mathcal{A}_{\alpha,q}$

The order relations between $LE_{\alpha}$ and $\mathcal{A}_{\alpha,p}$ are clarified in the next theorem and proposition. Theorem 4.5 was formerly shown in [7] with restriction to $\alpha=1/2$.

Theorem 4.5**.**

For any $\alpha\in(0,1)$ and any $q>0$ there exist $A,B\in\mathbb{M}_{2}^{++}$ such that $LE_{\alpha}(A,B)\not\leq\mathcal{A}_{\alpha,q}(A,B)$.

Proof.

Let $0<\alpha<1$ and $q>0$. Assume by contradiction that $LE_{\alpha}(A,B)\leq\mathcal{A}_{\alpha,p}(A,B)$ for all $A,B\in\mathbb{M}_{n}^{++}$. Then $LE_{\alpha}(A_{0},B_{\theta})\leq\mathcal{A}_{\alpha,p}(A_{0},B_{\theta})$ holds for $A_{0},B_{\theta}$ in (3.5) with any $x,y>0$ and $\theta\in\mathbb{R}$. Let $0<y<1$ and $x=y^{2}$. Since $x^{1-\alpha}y^{\alpha}\neq 1$ and $(1-\alpha)x^{q}+\alpha y^{q}\neq 1$, by Lemma 3.4 we must have

[TABLE]

As $y\searrow 0$, since we have the approximations

[TABLE]

the LHS of (4.4) can be estimated to

[TABLE]

which is negative when $y>0$ is sufficiently small. This contradicts (4.4). ∎

Proposition 4.6**.**

Let $0<\alpha<1$ and $q>0$. Then for every $A,B\geq 0$ we have $LE_{\alpha}(A,B)\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$. Hence $LE_{\alpha}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ holds for any $\triangleleft\in\{\leq_{\mathrm{chao}},\leq_{\mathrm{near}},\leq_{\lambda},\prec_{w},\leq_{\mathrm{Tr}}\}$ and $A,B\geq 0$.

Proof.

When $0<p<q$, by Proposition 4.3 we have $\mathcal{A}_{\alpha,p}(A,B)\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$ for any $A,B\geq 0$. Letting $p\searrow 0$ gives $LE_{\alpha}(A,B)\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$ by Theorem 2.3 and Lemma 2.5. ∎

In the above proof, by Theorem 2.6 we may assume that $A,B>0$. Thus, the simpler version of Theorem 2.3 for $A,B>0$ is enough to prove the above proposition. However, we cannot completely avoid use of Theorem A.1 because it is necessary in proving Lemma 2.5 via Proposition 2.4(2).

Proposition 4.7**.**

Let $0<\alpha<1$ and $q>0$. Then for every $A,B\geq 0$ the following conditions are equivalent:

• (i)

$\mathrm{Tr}\,LE_{\alpha}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$;

• (ii)

$LE_{\alpha}(A,B)=\mathcal{A}_{\alpha,q}(A,B)$;

• (iii)

$A=B$.

Proof.

It is obvious that (iii)$\implies$(ii)$\implies$(1). If (i) holds, then we have $\mathrm{Tr}\,\mathcal{A}_{\alpha,p}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ for any $p\in(0,q]$ by Propositions 4.3 and 4.6. Hence (iii) follows by Proposition 4.4. ∎

The results of this subsection are summarized as follows:

4.3 $R_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

In this subsection we discuss inequalities between $R_{\alpha,p}$ and $\mathcal{A}_{\alpha,q}$. The next theorem says that the near order inequality $R_{\alpha,p}(A,B)\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}(A,B)$ fails to hold for any $p,q>0$.

Theorem 4.8**.**

For any $\alpha\in(0,1)$ and any $p,q>0$ there exist $A,B\in\mathbb{M}_{2}^{++}$ such that $R_{\alpha,p}(A,B)\not\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}(A,B)$ (hence $R_{\alpha,p}(A,B)\not\leq\mathcal{A}_{\alpha,q}(A,B)$ and $R_{\alpha,p}(A,B)\not\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$).

Proof.

Consider $A_{0},B_{\theta}\in\mathbb{M}_{2}^{++}$ given in (3.5) with $y=x\neq 1$. Let $0<\alpha<1$ and $p,q>0$ be arbitrarily fixed. The estimates of $R_{\alpha,p}(A_{0},B_{\theta})$ and $\mathcal{A}_{\alpha,q}(A_{0},B_{\theta})$ up to $o(\theta)$ are enough to show the theorem. Assume that $R_{\alpha,p}(A_{0},B_{\theta})\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}(A_{0},B_{\theta})$ for any $x>0$ with $x\neq 1$ and any $\theta\in\mathbb{R}$. Now write $X:=R_{\alpha,p}(A_{0},B_{\theta})$ and $Y:=\mathcal{A}_{\alpha,q}(A_{0},B_{\theta})$. Reducing (3.15) to $o(\theta)$ (see (3.4)) one has $X=\begin{bmatrix}1&\theta z_{12}\\ \theta z_{12}&x\end{bmatrix}+o(\theta)$ with $z_{12}$ in (3.16) (without variable $p$ for simplicity). On the other hand, reducing (3.7) and (3.8) with $y=x\neq 1$ one has $Y=\begin{bmatrix}1&\theta\alpha(1-x)\\ \theta\alpha(1-x)&x\end{bmatrix}+o(\theta)$, and applies (3.4) to have $Y^{1/2}=\begin{bmatrix}1&\theta\alpha(1-x^{1/2})\\ \theta\alpha(1-x^{1/2})&x^{1/2}\end{bmatrix}+o(\theta)$ as $\theta\to 0$. Hence we have

[TABLE]

where

[TABLE]

so that by (3.4) again we can further compute

[TABLE]

Then, since $X\leq_{\mathrm{near}}Y$ is equivalent to $(Y^{1/2}XY^{1/2})^{1/2}\leq Y$, we must have

[TABLE]

Indeed, the last equality is seen as, with $v\in\mathbb{R}$,

[TABLE]

Therefore, $\alpha(1-x)={u_{12}\over 1+x}$ must hold, that is, $x^{1/2}z_{12}+\alpha(1-x^{1/2})(1+x^{3/2})=\alpha(1-x^{2})$. Inserting the form of $z_{12}$ in (3.16) gives

[TABLE]

so that $\bigl(x^{{1-\alpha\over 2}p}-x^{{1+\alpha\over 2}p}\bigr)/(1-x^{p})=\alpha$ for all $x>0$, $x\neq 1$. But this is impossible since the LHS goes to [math] ($\neq\alpha$) as $x\searrow 0$. ∎

Proposition 4.9**.**

Let $0<\alpha<1$ and $p,q>0$. If $p/2\leq q$, then we have $R_{\alpha,p}(A,B))\leq_{\lambda}\mathcal{A}_{\alpha,q}(A,B)$ (hence $R_{\alpha,p}(A,B)\prec_{w}\mathcal{A}_{\alpha,q}(A,B)$) for all $A,B\geq 0$.

Proof.

Ando’s matrix Young inequality [3] says that for any $\alpha\in(0,1)$ and $A,B\in\mathbb{M}_{n}^{+}$,

[TABLE]

equivalently, $\lambda^{1/2}(A^{1-\alpha}B^{2\alpha}A^{1-\alpha})\leq\lambda((1-\alpha)A+\alpha B)$. For any $p>0$, by replacing $A,B$ with $A^{p/2},B^{p/2}$ respectively and taking the $2/p$-power, one has

[TABLE]

that is, $R_{\alpha,p}(A,B)\leq_{\lambda}\mathcal{A}_{\alpha,p/2}(A,B)$. Hence the assertion follows from Proposition 4.3. ∎

Theorem 4.10**.**

Let $0<\alpha<1$ and $p,q>0$. If $R_{\alpha,p}(A,B)\leq_{\lambda}\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B\in\mathbb{M}_{2}^{++}$, then $\alpha(1-\alpha)p\leq q$.

Proof.

Let $A_{0}$ and $B_{\theta}$ be given in (3.5) with $y=x>0$ with $x\neq 1$. Then we have expression (3.15) with (3.16). On the other hand, in the present case (where $y=x$), expression (3.7) with (3.8) is simplified as

[TABLE]

where

[TABLE]

By assumption we must have $R_{\alpha,p}(A_{0},B_{\theta})\leq_{\lambda}\mathcal{A}_{\alpha,q}(A_{0},B_{\theta})$ for all $\theta\in\mathbb{R}$, which implies by Lemma 3.6 that

[TABLE]

for all $x>0$, $x\neq 1$. Now, we compute from (3.16)

[TABLE]

and from (4.6)

[TABLE]

Note that (4.7) and (4.8) are equivalent; indeed, dividing both sides of (4.7) by $x$ and then replacing $x$ with $x^{-1}$ transform (4.7) into (4.8). Letting $x\searrow 0$ in (4.7) gives $-{1\over p}\leq-{\alpha(1-\alpha)\over q}$, so that $\alpha(1-\alpha)p\leq q$ follows. ∎

It seems that we can find no further condition other than $\alpha(1-\alpha)p\leq q$ from (4.7) (equivalently, (4.8)) for all $x\neq 1$.

Proposition 4.11**.**

Let $0<\alpha<1$ and $p,q>0$. If $\min\{1,p/2\}\leq q$, then $\mathrm{Tr}\,R_{\alpha,p}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B\geq 0$.

Proof.

When $p/2\leq q$, the asserted trace inequality follows from Proposition 4.9. Assume that $1\leq q$ and $p>0$ is arbitrary. For every $A,B\in\mathbb{M}_{n}^{+}$, by Horn’s log-majorization (see, e.g., [9, 21]) one has

[TABLE]

so that

[TABLE]

where the last inequality is due to Proposition 4.3 since $1\leq q$. ∎

Theorem 4.12**.**

Let $0<\alpha<1$ and $p,q>0$. If $\mathrm{Tr}\,R_{\alpha,p}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B\in\mathbb{M}_{2}^{++}$, then $\min\{1,\alpha(1-\alpha)p\}\leq q$.

Proof.

It suffices to show that we must have $\alpha(1-\alpha)p\leq q$ when $q<1$. Consider again $A_{0},B_{\theta}\in\mathbb{M}_{2}^{++}$ in (3.5) with $y=x>0$, $x\neq 1$. From Lemma 3.5 and (4.5), (4.6) we compute

[TABLE]

Hence, if $\mathrm{Tr}\,R_{\alpha,p}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ for all $A,B\in\mathbb{M}_{2}^{++}$, then we must have

[TABLE]

for all $x>0$, $x\neq 1$. Since $q<1$ by assumption, letting $x\searrow 0$ gives $-1/p\leq-\alpha(1-\alpha)/q$ and hence $\alpha(1-\alpha)p\leq q$ follows. ∎

In particular, when $\alpha=1/2$ and $p=1$, the next example provides the exact characterization for $\mathrm{Tr}\,R_{1/2,1}(A,B)\leq\mathrm{Tr}\mathcal{A}_{1/2,q}(A,B)$.

Example 4.13**.**

Let $\alpha=1/2$ and $p=1$. Then the following conditions are equivalent:

• (i)

$\mathrm{Tr}\,R_{1/2,1}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{1/2,q}(A,B)$ for all $A,B\geq 0$;

• (ii)

$\mathrm{Tr}\,R_{1/2,1}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{1/2,q}(A,B)$ for all $A,B\in\mathbb{M}_{2}^{++}$;

• (iii)

$q\geq 1/4$.

Proof.

(i)$\implies$(ii) is trivial, and (ii)$\implies$(iii) follows from Theorem 4.12. Next let us show that (iii)$\implies$(i) holds. For this, by Proposition 4.3 it suffices to show that $\mathrm{Tr}\,R_{1/2,1}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{1/2,1/4}(A,B)$ for all $A,B\geq 0$. We note that $\mathrm{Tr}\,R_{1/2,1}(A,B)=\mathrm{Tr}\,A^{1/2}B^{1/2}$ and

[TABLE]

Since

[TABLE]

we have

[TABLE]

It follows from (4.10) and (4.11) that $\mathrm{Tr}\,\mathcal{A}_{1/2,1/4}(A,B)\geq{1\over 2}\mathrm{Tr}\bigl(A^{3/4}B^{1/4}+A^{1/4}B^{3/4}\bigr)$. Moreover, by the matrix norm inequality for the Heinz-type means (see [24]) we have

[TABLE]

so that $\mathrm{Tr}\,A^{1/2}B^{1/2}\leq{1\over 2}\mathrm{Tr}(A^{3/4}B^{1/4}+A^{1/4}B^{3/4})$. Hence (i) follows. ∎

Proposition 4.14**.**

Let $0<\alpha<1$ and $p,q>0$ be such that $\min\{1,p/2\}\leq q$. Then for every $A,B\geq 0$ the following conditions are equivalent:

• (i)

$\mathrm{Tr}\,R_{\alpha,p}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$;

• (ii)

$R_{\alpha,p}(A,B)=\mathcal{A}_{\alpha,q}(A,B)$;

• (iii)

$A=B$.

Proof.

It is obvious that (iii)$\implies$(ii)$\implies$(i). Assume (i). If $p/2<q$, then we have $\mathrm{Tr}\,\mathcal{A}_{\alpha,t}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ for any $t\in[p/2,q]$ by Propositions 4.3 and 4.11. Hence (iii) follows from Proposition 4.4. If $p/2=q$, then Proposition 4.9 yields $\lambda(R_{\alpha,p}(A,B))=\lambda(\mathcal{A}_{\alpha,p/2}(A,B))$ and hence

[TABLE]

Hence It follows from [31, Corollary 2.3] that $A^{p/2}=B^{p/2}$ so that $A=B$. If $q\geq 1$, then for any $t\geq p$ we have $\mathrm{Tr}\,R_{\alpha,p}(A,B)\leq\mathrm{Tr}\,R_{\alpha,t}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ by Araki’s log-majorization [5] and Proposition 4.11. Hence $\mathrm{Tr}\,R_{\alpha,t}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ for $t\geq p$, which extends to all $t>0$ thanks to real analyticity of $t\mapsto\mathrm{Tr}\,R_{\alpha,t}(A,B)$ in $t>0$, so we can apply the above case (where $p/2<q$). ∎

The results of this subsection are summarized as follows:

Problem 4.15**.**

There is a gap between the sufficient condition and the necessary condition for $R_{\alpha,p}\leq_{\lambda}\mathcal{A}_{\alpha,q}$, $R_{\alpha,p}\prec_{w}\mathcal{A}_{\alpha,q}$ and $R_{\alpha,p}\leq_{\mathrm{Tr}}\mathcal{A}_{\alpha,q}$. It is also unknown whether $R_{\alpha,p}\leq_{\lambda}\mathcal{A}_{\alpha,q}$ is strictly stronger than $R_{\alpha,p}\prec_{w}\mathcal{A}_{\alpha,q}$, or they are equivalent. Example 4.13 says that the sufficient condition in Proposition 4.11 is not sharp when $\alpha=1/2$ and $p=1$. This suggests us that the complete characterization of $R_{\alpha,p}\leq_{\mathrm{Tr}}\mathcal{A}_{\alpha,q}$ is a complicated problem.

4.4 $G_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

In this subsection we discuss inequalities between $G_{\alpha,p}$ and $\mathcal{A}_{\alpha,q}$. In theory of operator means in Kubo and Ando’s sense [29] it is well known that $A\#_{\alpha}B\leq A\triangledown_{\alpha}B$ for any $\alpha\in(0,1)$ and $A,B\geq 0$. The next theorem characterizes the Loewner inequality $G_{\alpha,p}\leq\mathcal{A}_{\alpha,q}$, extending $\#_{\alpha}\leq\triangledown_{\alpha}$ (for $p=q=1$).

Theorem 4.16**.**

Let $0<\alpha<1$ and $p,q>0$. Then the following conditions are equivalent:

• (i)

$G_{\alpha,p}(A,B)\leq\mathcal{A}_{\alpha,q}(A,B)$* for all $A,B\geq 0$;*

• (ii)

$G_{\alpha,p}(A,B)\leq\mathcal{A}_{\alpha,q}(A,B)$* for all $A,B\in\mathbb{M}_{2}^{++}$;*

• (iii)

$1\leq p\leq q$.

Lemma 4.17**.**

Let $\alpha>0$ and $p,q>0$. Let $A_{0}$ and $B_{\theta}$ be given in (3.5) with $y=x>0$ and $x\neq 1$. Then we have

[TABLE]

where

[TABLE]

Proof.

Similarly to the proof of Lemma 3.5 it is easy to compute

[TABLE]

Then Example 3.2(1) is applied to compute $\bigl(A_{0}^{-p/2}B_{\theta}^{p}A_{0}^{-p/2}\bigr)^{\alpha}$ and hence

[TABLE]

where

[TABLE]

Furthermore, we apply Example 3.2(1) again to show that (4.12) holds with (4.13), whose details are left to the reader. ∎

Proof of Theorem 4.16.

(iii)$\implies$(i). Since $A^{p}\#_{\alpha}B^{p}\leq A^{p}\triangledown_{\alpha}B^{p}$ and $x^{1/p}$ is operator monotone on $[0,\infty)$, we have

[TABLE]

Since $\mathcal{A}_{\alpha,p}(A,B)\leq\mathcal{A}_{\alpha,q}(A,B)$ by Theorem 4.1, the result follows.

(i)$\implies$(ii) is trivial.

(ii)$\implies$(iii). Assume that (ii) holds. Note that this extends to all $A,B\in\mathbb{M}_{2}^{+}$ by continuity. First, let $A_{0},B_{\theta}\in\mathbb{M}_{2}^{++}$ be given as in Lemma 4.17. Then by (4.12) and (4.13) as well as (4.5) and (4.6) we have

[TABLE]

Hence we must have

[TABLE]

(Note that (4.15) is equivalent to (4.14).) It is clear that $p\leq q$ follows from (4.14).

Now assume that $p\leq q$, and let $A_{0}$ be given in (3.5) with $x>0$ satisfying $(1-\alpha)x^{q}\neq 1$, and $\tilde{B}_{\theta}\in\mathbb{M}_{2}^{+}$ be in (4.1). Then by Example 3.2(1) we compute

[TABLE]

where

[TABLE]

On the other hand, since $\tilde{B}_{\theta}^{p}=\tilde{B}_{\theta}=|\phi\rangle\langle\phi|$ where $\phi:=\begin{bmatrix}\cos\theta\\ \sin\theta\end{bmatrix}$, one has

[TABLE]

so that $G_{\alpha}(A_{0}^{p},\tilde{B}_{\theta}^{p})=\|A_{0}^{-p/2}\phi\|^{2(\alpha-1)}|\phi\rangle\langle\phi|$. Hence one writes

[TABLE]

By (4.16) and (4.18) we write

[TABLE]

Therefore, we must have

[TABLE]

We estimate (4.19) and (4.20) as $x\searrow 0$ as follows:

[TABLE]

If $p<1$, then (4.20) $\to-(1-\alpha)^{2}<0$ as $x\searrow 0$, which is impossible since (4.20) $\geq 0$ in the limit $x\searrow 0$. Hence $p\geq 1$ must hold. ∎

The next corollary is a particular case of Theorem 4.16 when $q=1$.

Corollary 4.18**.**

Let $0<\alpha<1$ and $p>0$. Then the following conditions are equivalent:

• (i)

$G_{\alpha,p}(A,B)\leq\alpha A+(1-\alpha)B$* for all $A,B\geq 0$;*

• (ii)

$G_{\alpha,p}(A,B)\leq\alpha A+(1-\alpha)B$* for all $A,B\in\mathbb{M}_{2}^{++}$;*

• (iii)

$p=1$.

Proposition 4.19**.**

Let $0<\alpha<1$. If $0<p\leq q$, then we have $G_{\alpha,p}(A,B)\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$ for all $A,B\geq 0$.

Proof.

Assume that $0<p\leq q$. Since $1\leq q/p$, it follows from Theorem 4.16 that $G_{\alpha,1}(A^{p},B^{p})\leq\mathcal{A}_{\alpha,q/p}(A^{p},B^{p})$, i.e.,

[TABLE]

which implies that

[TABLE]

and

[TABLE]

so that

[TABLE]

that is, $G_{\alpha,p}(A,B)\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$. ∎

Theorem 4.20**.**

Let $0<\alpha<1$ and $p,q>0$. If $G_{\alpha,p}(A,B))\leq_{\lambda}\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B\in\mathbb{M}_{2}^{++}$, then $p\leq q$.

Proof.

We use $A_{0},B_{\theta}\in\mathbb{M}_{2}^{++}$ in (3.5) with $y=x\neq 1$ once again, and argue in the same way as in the proof of Theorem 4.10. Then by assumption we must have the same inequalities as in (4.7) and (4.8) with $z_{11}(p),z_{22}(p),z_{12}(p)$ in (4.13) instead of (3.16). These are specified in the present case as

[TABLE]

for all $x>0$, $x\neq 1$. Hence we have $p\leq q$. ∎

Proposition 4.21**.**

Let $0<\alpha<1$ and $p,q>0$ be arbitrary. Then for every $A,B\geq 0$ we have $G_{\alpha,p}(A,B)\prec_{w\log}\mathcal{A}_{\alpha,q}(A,B)$ and hence $G_{\alpha,p}(A,B)\prec_{w}\mathcal{A}_{\alpha,q}(A,B)$.

Proof.

When $A,B>0$, the result follows from the log-majorization $G_{\alpha,p}(A,B)\prec_{\log}LE_{\alpha}(A,B)$ (see [4, Corollary 2.3]) and Proposition 4.6. For general $A,B\geq 0$ one can take the limit from the result for $A+\varepsilon I$ and $B+\varepsilon I$ as $\varepsilon\searrow 0$. ∎

Proposition 4.22**.**

Let $0<\alpha<1$ and $p,q>0$. Then for every $A,B\geq 0$ the following conditions are equivalent:

• (i)

$\mathrm{Tr}\,G_{\alpha,p}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$;

• (ii)

$G_{\alpha,p}(A,B)=\mathcal{A}_{\alpha,q}(A,B)$;

• (iii)

$A=B$.

Proof.

It suffices to show (i)$\implies$(iii). If (i) holds, then $\mathrm{Tr}\,\mathcal{A}_{\alpha,t}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ for any $t\in(0,q]$ by Propositions 4.3 and 4.21. Hence we have (iii) by Proposition 4.4. ∎

The results of this subsection are summarized as follows:

4.5 $SG_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

In this subsection we discuss inequalities between $SG_{\alpha,p}$ and $\mathcal{A}_{\alpha,q}$. The next theorem says that $SG_{\alpha,p}(A,B)\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$ fails to hold for any $p,q>0$.

Theorem 4.23**.**

For any $\alpha\in(0,1)$ and any $p,q>0$ there exist $A,B\in\mathbb{M}_{2}^{++}$ such that $SG_{\alpha,p}(A,B)\allowbreak\not\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$ (hence $SG_{\alpha,p}(A,B)\not\leq\mathcal{A}_{\alpha,q}(A,B)$).

Proof.

For $A,B>0$ set $Y:=A^{p}$ and $X:=A^{-p}\#B^{p}$; then $X=Y^{-1}\#B^{p}$ and hence $B^{p}=XYX$ by the Riccati lemma, so that $B=(XYX)^{1/p}$. Therefore, $SG_{\alpha,p}(A,B)\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$ is equivalently written as

[TABLE]

Moreover, for any $X,Y>0$, if we set $A:=Y^{1/p}$ and $B:=(XYX)^{1/p}$, then $Y=A^{p}$ and $X=A^{-p}\#B^{p}$. Hence, by replacing $q/p$ with $r$, it suffices to show that for any $r>0$ there exist $X,Y\in\mathbb{M}_{2}^{++}$ such that

[TABLE]

To do this, let $X:=A_{0}$ and $Y:=B_{\theta}$ in (3.5) for $x,y>0$ with $x^{2}y\neq 1$. Since

[TABLE]

one can compute $(XYX)^{r}$ by Example 3.2(1) as

[TABLE]

where

[TABLE]

Hence one can write

[TABLE]

where

[TABLE]

From this, by Example 3.2(3) one can write

[TABLE]

where

[TABLE]

On the other hand, for $x,y>0$ with $x^{2\alpha}y\neq 1$, since

[TABLE]

one can write by Example 3.2(3),

[TABLE]

where

[TABLE]

From (4.25), (4.28) and (4.24) one can write

[TABLE]

Now assume that $r\log(X^{\alpha}YX^{\alpha})\leq\log\{(1-\alpha)Y^{r}+\alpha(XYX)^{r}\}$ for all $X,Y\in\mathbb{M}_{2}^{++}$. Then we must have

[TABLE]

for all $x,y>0$ with $x^{2}y\neq 1$ and $x^{2\alpha}y\neq 1$. Let $x>0$ with $x\neq 1$ be fixed, so that $x^{2}y<1$ and $x^{2\alpha}y<1$ for sufficiently small $y>0$. Since $(x^{2r})^{\alpha}<\alpha x^{2r}+1-\alpha$ so that $\log(\alpha x^{2r}+1-\alpha)-2r\alpha\log x>0$, it follows from (4.30) that $c_{11}-rd_{11}\geq 0$. Let us estimate $c_{11}$ and $d_{11}$ when $y\searrow 0$. As $y\searrow 0$ we have $b\to 0$ and $\log b\approx r\log y$ by (4.24). Since $a_{11}\to-r+(r-1)x^{2}$ and $a_{12}\to x$ by (4.22), we have by (4.24)

[TABLE]

so that by (4.26),

[TABLE]

Moreover, by (4.29),

[TABLE]

Hence it follows that

[TABLE]

which is impossible since $(\alpha x+1-\alpha)^{2}-x^{2\alpha}>0$ and $\log y<0$ as $y\searrow 0$. This implies that (4.21) holds for some $X,Y\in\mathbb{M}_{2}^{++}$. ∎

When $p=q=1$, Theorem 4.23 says that for any $\alpha\in(0,1)$, the Loewner inequality $F_{\alpha}(A,B)\leq A\triangledown_{\alpha}B$ fails to hold. In particular, $F(A,B)\leq{A+B\over 2}$ fails to hold.

Theorem 4.24**.**

Let $0<\alpha<1$ and $p,q>0$. If $SG_{\alpha,p}(A,B)\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}(A,B)$ for all $A,B\in\mathbb{M}_{2}^{++}$, then $q/p\geq 1+\max\{\alpha(q-1),(1-\alpha)(q-1)\}$; hence

[TABLE]

Proof.

As in the first paragraph of the proof of Theorem 4.23 we see that $SG_{\alpha,p}(A,B)\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}(A,B)$ for all $A,B\in\mathbb{M}_{2}^{++}$ if and only if

[TABLE]

where $r:=q/p$. Let $X:=A_{0}$ and $Y:=B_{\theta}$ in (3.5) for $x,y>0$ with $x^{2}y\neq 1$, $x^{2\alpha}y\neq 1$ and $b:=y^{r}(\alpha x^{2r}+1-\alpha)\neq 1$. By (4.23) with (4.24) one has by Example 3.2(1),

[TABLE]

where

[TABLE]

with $b_{11},b_{12}$ in (4.24). Moreover,

[TABLE]

where

[TABLE]

On the other hand, by (4.27) one has by Example 3.2(1) again,

[TABLE]

where

[TABLE]

Moreover,

[TABLE]

where

[TABLE]

Hence one has

[TABLE]

where

[TABLE]

Now assume (4.32), which means that $(M^{1/2}LM^{1/2})^{1/2}\leq M$ so that

[TABLE]

Therefore,

[TABLE]

for all $x,y>0$ with $x^{2}y\neq 1$, $x^{2\alpha}y\neq 1$ and $b\neq 1$. By (4.24) and (4.37) note that

[TABLE]

Furthermore, when $y\searrow 0$ with $x>0$ fixed, we see that

[TABLE]

which implies that

[TABLE]

From (4.39)–(4.41) it follows that when $y\searrow 0$,

[TABLE]

so that

[TABLE]

As $y\searrow 0$, since $b\to 0$, from (4.33), (4.31) and (4.34) we see that

[TABLE]

Moreover, from (4.36) and (4.37),

[TABLE]

Finally, from (4.38),

[TABLE]

Therefore, letting $y\searrow 0$ in (4.42) gives

[TABLE]

for all $x>0$. Letting $x\searrow 0$ in (4.43) gives

[TABLE]

so that $(1-\alpha)(r-1)-(1-\alpha)^{2}(q-1)\geq 0$, implying that $r\geq 1+(1-\alpha)(q-1)$. Examining the coefficient of the maximal order term $x^{{r\over q}+2}$ of (4.43) we also find that

[TABLE]

so that $\alpha(r-1)+\alpha^{2}(q+1)-2q\alpha^{2}\geq 0$, implying that $r\geq 1+\alpha(q-1)$. This is immediate since the assumption of the theorem holds for $1-\alpha$ in place of $\alpha$ from symmetry of $SG_{\alpha,p}$ and $\mathcal{A}_{\alpha,q}$ (see Remark 2.1(2)). Thus the result follows. ∎

Proposition 4.25**.**

Let $0<\alpha<1$ and $p,q>0$.

• (1)

If ${p/q}\leq\min\{\alpha,1-\alpha\}$, then we have $SG_{\alpha,p}(A,B)\prec_{\log}R_{\alpha,q}(A,B)$ for every $A,B\geq 0$ with $s(A)\geq s(B)$.

• (2)

If ${p/q}\leq 2\min\{\alpha,1-\alpha\}$, then we have $SG_{\alpha,p}(A,B)\prec_{w\log}\mathcal{A}_{\alpha,q}(A,B)$ (hence $SG_{\alpha,p}(A,B)\prec_{w}\mathcal{A}_{\alpha,q}(A,B)$) for every $A,B\geq 0$ with $s(A)\geq s(B)$.

Proof.

By continuity (Proposition 2.2) and symmetry mentioned in Remark 2.1(2) we may assume that $A,B>0$ and $0<\alpha\leq 1/2$.

(1) For $0<\alpha\leq 1/2$ and $A,B>0$, by Araki’s log-majorization [5] we have

[TABLE]

Replacing $A,B$ with $A^{p},B^{p}$ gives $SG_{\alpha,p}(A,B)\prec_{\log}R_{\alpha,1/\alpha}(A^{p},B^{p})^{1/p}=R_{\alpha,p/\alpha}(A,B)$. The result follows since $R_{\alpha,p/\alpha}(A,B)\prec_{\log}R_{\alpha,q}(A,B)$ if $p/\alpha\leq q$ by Araki’s log-majorization again.

(2) If $p/q\leq 2\alpha$ and so $p/\alpha\leq 2q$, then we have $R_{\alpha,p/\alpha}(A,B)\leq_{\lambda}\mathcal{A}_{\alpha,q}(A,B)$ by Proposition 4.9. Combining this with (1) gives the result. ∎

Remark 4.26**.**

As for log-majorization in (1) above, for $0<\alpha<1$ and $p,q>0$, it is indeed known that $SG_{\alpha,p}(A,B)\prec_{\log}R_{\alpha,q}(A,B)$ for all $A,B>0$ if and only if $p/q\leq\min\{\alpha,1-\alpha\}$, and that $R_{\alpha,q}(A,B)\prec_{\log}SG_{\alpha,p}(A,B)$ for all $A,B>0$ if and only if $p/q\geq\max\{\alpha,1-\alpha\}$. The details on these facts will be provided in [23], while the ‘if’ part of the latter was indeed shown in [19].

Theorem 4.27**.**

Let $0<\alpha<1$ and $p,q>0$. If $\lambda_{1}(SG_{\alpha,p}(A,B))\leq\lambda_{1}(\mathcal{A}_{\alpha,q}(A,B))$ holds for all $A,B\in\mathbb{M}_{2}^{++}$, then $q/p\geq\max\{\alpha,1-\alpha\}$. Hence, if $SG_{\alpha,p}(A,B)\prec_{w}\mathcal{A}_{\alpha,q}(A,B)$ for all $A,B\in\mathbb{M}_{2}^{++}$, then $q/p\geq\max\{\alpha,1-\alpha\}$.

Proof.

Assume that $\lambda_{1}(SG_{\alpha,p}(A,B))\leq\lambda_{1}(\mathcal{A}_{\alpha,q}(A,B))$ holds for all $A,B\in\mathbb{M}_{2}^{++}$. Then as in the first paragraph of the proof of Theorem 4.23 it follows that

[TABLE]

holds for all $X,Y\in\mathbb{M}_{2}^{++}$, where $r:=q/p$. Now set $X:=A_{0}$ and $Y:=B_{\theta}$ in (3.5) for $x,y>0$ with $x^{2}y\neq 1$. Apply Lemma 3.6 to (4.23) and (4.35) with $q=1$ to find that

[TABLE]

where $b,b_{11},b_{12}$ are in (4.24) via (4.22) and $\hat{s}_{11},\hat{s}_{12}$ are in (4.36) with $q=1$. Hence, for any $x>0$, if $y>0$ is sufficiently small, then we must have

[TABLE]

When $y\searrow 0$, the above inequality becomes

[TABLE]

Letting $x\searrow 0$ in the above gives $-r\leq-\alpha r-(1-\alpha)+(1-\alpha)^{2}$, so that $r\geq\alpha$. Moreover, from symmetry we have $r\geq 1-\alpha$ too, so the result follows. ∎

Note that the sufficient condition in Proposition 4.25 is indeed stricter than the necessary condition in Theorem 4.27, because $2\min\{\alpha,1-\alpha\}<1/\max\{\alpha,1-\alpha\}$.

Proposition 4.28**.**

Let $0<\alpha<1$ and $p,q>0$. If $q\geq 1$ or ${p/q}\leq 2\min\{\alpha,1-\alpha\}$, then $\mathrm{Tr}\,SG_{\alpha,p}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B\geq 0$ with $s(A)\geq s(B)$.

Proof.

If ${p/q}\leq 2\min\{\alpha,1-\alpha\}$, then the result follows from Proposition 4.25(2). If $q\geq 1$, then we have, with $r>0$ satisfying $p/r\leq\min\{\alpha,1-\alpha\}$,

[TABLE]

by Propositions 4.25(1) and 4.11. ∎

Theorem 4.29**.**

Let $0<\alpha<1$ and $p,q>0$. If $\mathrm{Tr}\,SG_{\alpha,p}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B\in\mathbb{M}_{2}^{++}$, then $\min\{1,p/2\}\leq q$.

Proof.

It suffices to show that we must have $p/2\leq q$ when $q<1$. Consider once again $A_{0},B_{\theta}\in\mathbb{M}_{2}^{++}$ in (3.5) with $y=x>0$, $x\neq 1$. Note that

[TABLE]

We compute

[TABLE]

where

[TABLE]

This computation is similar to the previous ones based on Example 3.2(1), so the details are left to the reader. Furthermore, with $H:=\begin{bmatrix}0&\xi_{12}\\ \xi_{12}&0\end{bmatrix}$ and $K:=\begin{bmatrix}\xi_{11}&0\\ 0&\xi_{22}\end{bmatrix}$ we have by the Taylor expansion,

[TABLE]

from which it is easy to compute

[TABLE]

where

[TABLE]

Then by Lemma 3.6 the two eigenvalues of $A_{0}^{p/2}(A_{0}^{-p}\#B_{\theta}^{p})^{2\alpha}A_{0}^{p/2}$ are

[TABLE]

Hence we find that

[TABLE]

On the other hand, we have estimated $\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A_{0},B_{\theta})$ in (4.9). Therefore, we must have

[TABLE]

for all $x>0$, $x\neq 1$. As $x\searrow 0$ note that

[TABLE]

where the last convergence is due to the assumption $q<1$. Therefore, we have $-{2\alpha(1-\alpha)\over p}\leq-{\alpha(1-\alpha)\over q}$, showing that $p/2\leq q$. ∎

Proposition 4.30**.**

Let $0<\alpha<1$ and $p,q>0$ be such that $q\geq 1$ or ${p/q}\leq 2\min\{\alpha,1-\alpha\}$. Then for every $A,B\geq 0$ with $s(A)\geq s(B)$ the following conditions are equivalent:

• (i)

$\mathrm{Tr}\,SG_{\alpha,p}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$;

• (ii)

$SG_{\alpha,p}(A,B)=\mathcal{A}_{\alpha,q}(A,B)$;

• (iii)

$A=B$.

Proof.

It suffices to show (i)$\implies$(iii). Assume (i). Let $r:=\min\{\alpha,1-\alpha\}$; then $\min\{1,(p/r)/2\}\leq q$. By Propositions 4.25(1) and 4.11 we have

[TABLE]

so that $\mathrm{Tr}\,R_{\alpha,p/r}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$. Hence (iii) follows from Proposition 4.14. ∎

The results of this subsection are summarized as follows:

Problem 4.31**.**

At the moment we find no sufficient condition for $SG_{\alpha,p}\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}$ to hold, while a necessary condition is given in Theorem 4.24. It might happen that this inequality never holds for any $p,q>0$. For example, when $\alpha=1/2$, since the necessary condition gives $p\leq{2q\over q+1}$, we notice that $SG_{1/2,p}\leq_{\mathrm{near}}\mathcal{A}_{1/2,q}$ fails to hold for any $q>0$ if $p\geq 2$, and for any $q<1$ if $p=1$. But it is still unknown if $F(A,B)\leq_{\mathrm{near}}{A+B\over 2}$ (the case $p=q=1$) holds for all $A,B>0$. The situation is similar for $SG_{\alpha,p}\leq_{\lambda}\mathcal{A}_{\alpha,q}$ except the case $\alpha\neq 1/2$. When $\alpha=1/2$, since $\lambda(F(A,B))=\lambda\bigl((B^{1/2}AB^{1/2})^{1/2}\bigr)$ (see [15, Theorem 3.2, Item 8]), it is easy to see that if $0<p\leq q$, then $SG_{1/2,p}(A,B)\leq_{\lambda}\mathcal{A}_{1/2,q}(A,B)$ for all $A,B>0$. As for $SG_{\alpha,p}\prec_{w}\mathcal{A}_{\alpha,q}$ and $SG_{\alpha,p}\leq_{\mathrm{Tr}}\mathcal{A}_{\alpha,q}$, there is a rather big gap between the sufficient condition and the necessary condition obtained.

4.6 $\widetilde{S}G_{\alpha,p}$ vs. $\mathcal{A}_{\alpha,q}$

In this subsection we discuss inequalities between $\widetilde{S}G_{\alpha,p}$ and $\mathcal{A}_{\alpha,q}$. The next theorem says that $\widetilde{S}G_{\alpha,p}(A,B)\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}(A,B)$ fails to hold for any $\alpha\in(0,1)\setminus\{1/2\}$ and any $p,q>0$.

Theorem 4.32**.**

For any $\alpha\in(0,1)\setminus\{1/2\}$ and any $p,q>0$ there exist $A,B\in\mathbb{M}_{2}^{++}$ such that $\widetilde{S}G_{\alpha,p}(A,B)\not\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}(A,B)$. Hence, for any $\alpha\in(0,1)$ and any $p,q>0$ there exist $A,B\in\mathbb{M}_{2}^{++}$ such that $\widetilde{S}G_{\alpha,p}(A,B)\not\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}(A,B)$.

Proof.

First, note that the latter assertion follows from the first and Theorem 4.23 since $\widetilde{S}G_{1/2,p}=SG_{1/2,p}$. For $A,B>0$ set $Y:=A^{p}$ and $X:=A^{-p}\#_{\alpha}B^{p}$; then $B^{p}=Y^{-1/2}(Y^{1/2}XY^{1/2})^{1/\alpha}Y^{-1/2}$. We will express this RHS as $Y^{-1}\#_{1/\alpha}X$, though $1/\alpha>1$, in analogy of the geometric mean $\#_{\alpha}$. Then $\widetilde{S}G_{\alpha,p}(A,B)\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}(A,B)$ is equivalently written as

[TABLE]

where $r:=q/p$. Conversely, for any $X,Y>0$, if we set $A:=Y^{1/p}$ and $B:=(Y^{-1}\#_{1/\alpha}X)^{1/p}$, then $Y=A^{p}$ and $X=A^{-p}\#_{\alpha}B^{p}$. Hence it suffices to show that for any $\alpha\in(0,1)\setminus\{1/2\}$ and any $r,q>0$ there exist $X,Y\in\mathbb{M}_{2}^{++}$ for which (4.44) is violated.

Let $X:=B_{\theta}$ and $Y:=A_{0}$ in (3.5) with $y=x^{2\alpha-1}>0$ ($\alpha\neq 1/2$) and $x\neq 1$. Set

[TABLE]

In the following computations we will repeatedly apply the reduced version of Example 3.2(1) up to $o(\theta)$. One computes $X^{1/2}Y^{2(1-\alpha)}X^{1/2}=\begin{bmatrix}1&\theta a_{12}\\ \theta a_{12}&x\end{bmatrix}+o(\theta)$, where

[TABLE]

and hence

[TABLE]

On the other hand, one computes $Y^{-1}\#_{1/\alpha}X=\begin{bmatrix}1&\theta\xi_{12}\\ \theta\xi_{12}&x\end{bmatrix}+o(\theta)$, where

[TABLE]

and hence $(Y^{-1}\#_{1/\alpha}X)^{r}=\begin{bmatrix}1&\theta\zeta_{12}\\ \theta\zeta_{12}&x^{r}\end{bmatrix}+o(\theta)$, where

[TABLE]

Since $(1-\alpha)Y^{r}+\alpha(Y^{-1}\#_{1/\alpha}X)^{r}=\begin{bmatrix}1&\theta\alpha\zeta_{12}\\ \theta\alpha\zeta_{12}&x^{r}\end{bmatrix}+o(\theta)$, one has

[TABLE]

From (4.48) and (4.46) one can further compute

[TABLE]

where

[TABLE]

Furthermore,

[TABLE]

where

[TABLE]

Now assume that $L\leq_{\mathrm{near}}M$ for all $X,Y\in\mathbb{M}_{2}^{++}$. Then we must have $(M^{1/2}LM^{1/2})^{1/2}\leq M$ so that by (4.48) and (4.49),

[TABLE]

Therefore, we must have

[TABLE]

which becomes ${\alpha\zeta_{12}\over 1-x^{r}}={a_{12}\over 1-x}$. By (4.45) and (4.47) this gives

[TABLE]

for all $x>0$ with $x\neq 1$, which fails to hold as $x\searrow 0$ in either case $0<\alpha<1/2$ or $1/2<\alpha<1$. ∎

Proposition 4.33**.**

Let $0<\alpha<1$ and $p,q>0$.

• (1)

If $p\leq\alpha q$, then we have $\widetilde{S}G_{\alpha,p}(A,B)\prec_{\log}R_{\alpha,q}(A,B)$ for every $A,B\geq 0$ with $s(A)\geq s(B)$.

• (2)

If $p\leq 2\alpha q$, then we have $\widetilde{S}G_{\alpha,p}(A,B)\prec_{w\log}\mathcal{A}_{\alpha,q}(A,B)$ (hence $\widetilde{S}G_{\alpha,p}(A,B)\prec_{w}\mathcal{A}_{\alpha,q}(A,B)$) for every $A,B\geq 0$ with $s(A)\geq s(B)$.

Proof.

(1) By continuity (Proposition 2.2) we may assume that $A,B>0$. Similarly to the proof of Proposition 4.25(1) it suffices to show that

[TABLE]

For $0<\alpha\leq 1/2$, by Araki’s log-majorization [5] we have

[TABLE]

The remaining proof is the same as in the proof of Proposition 4.25(1).

(2) follows from (1) and Proposition 4.9 as in the proof of Proposition 4.25(2). ∎

Remark 4.34**.**

Although we have shown the log-majorizations in Propositions 4.25(1) and 4.33(1) from Araki’s log-majorization, they can also be shown by directly applying the familiar antisymmetric tensor power technique (see [4]). Furthermore, it is known that $\widetilde{S}G_{\alpha,p}(A,B)\prec_{\log}R_{\alpha,q}(A,B)$ for all $A,B>0$ if and only if $p/q\leq\alpha$, and that $R_{\alpha,q}(A,B)\prec_{\log}\widetilde{S}G_{\alpha,p}(A,B)$ for all $A,B>0$ if $\alpha\leq 1/2$ and $q\leq p$, and the same holds only if $\alpha\leq 1/2$ and $p/q\geq 1/2$. The details on these facts will be provided in [23].

Theorem 4.35**.**

Let $0<\alpha<1$ and $p,q>0$. If $\lambda_{1}\bigl(\widetilde{S}G_{\alpha,p}(A,B)\bigr)\leq\lambda_{1}(\mathcal{A}_{\alpha,q}(A,B))$ holds for all $A,B\in\mathbb{M}_{2}^{++}$, then $q/p\geq 1-\alpha$. Hence, if $\widetilde{S}G_{\alpha,p}(A,B)\prec_{w}\mathcal{A}_{\alpha,q}(A,B)$ for all $A,B\in\mathbb{M}_{2}^{++}$, then $q/p\geq 1-\alpha$.

Proof.

Assume that $\lambda_{1}\bigl(\widetilde{S}G_{\alpha,p}(A,B)\bigr)\leq\lambda_{1}(\mathcal{A}_{\alpha,q}(A,B))$ holds for all $A,B\in\mathbb{M}_{2}^{++}$, and argue as in the first paragraph of the proof of Theorem 4.32. Then, since $\lambda((X^{1/2}Y^{2(1-\alpha)}X^{1/2})^{r/q})=\lambda((Y^{1-\alpha}XY^{1-\alpha})^{r/q})$, we have

[TABLE]

holds for all $X,Y\in\mathbb{M}_{2}^{++}$, where $r:=q/p$. Now set $X:=B_{\theta}$ and $Y:=A_{0}$ in (3.5) for $x,y>0$ with $xy\neq 1$ and $x^{1-\alpha}y\neq 1$. We compute

[TABLE]

On the other hand, using Example 3.2(1) we compute

[TABLE]

where

[TABLE]

and hence

[TABLE]

where

[TABLE]

Therefore,

[TABLE]

Let $x,y<1$. Note that $x^{2(1-\alpha)}y<1$ and $(1-\alpha)x^{r}+\alpha x^{(1-\alpha)r\over\alpha}y^{r\over\alpha}<1$. Then by Lemma 3.6 we have

[TABLE]

so that

[TABLE]

This implies that

[TABLE]

Letting $y\searrow 0$ gives

[TABLE]

Hence we must have

[TABLE]

Letting $x\searrow 0$ further gives $0\leq\alpha(r-1)+\alpha^{2}$, i.e., $r\geq 1-\alpha$. ∎

Proposition 4.36**.**

Let $0<\alpha<1$ and $p,q>0$. If $\min\{1,{p\over 2\alpha}\}\leq q$, then $\mathrm{Tr}\,\widetilde{S}G_{\alpha,p}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B\geq 0$ with $s(A)\geq s(B)$.

Proof.

If $p\leq 2\alpha q$, then the result follows from Proposition 4.33(2). If $q\geq 1$, then we have, with $r>0$ satisfying $p/r\leq\alpha$,

[TABLE]

by Propositions 4.33(1) and 4.11. ∎

Theorem 4.37**.**

Let $0<\alpha<1$ and $p,q>0$. If $\mathrm{Tr}\,\widetilde{S}G_{\alpha,p}(A,B)\leq\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B\in\mathbb{M}_{2}^{++}$, then $\min\{1,(1-\alpha)p\}\leq q$.

Proof.

It suffices to show that we must have $(1-\alpha)p\leq q$ when $q<1$. Consider $A_{0},B_{\theta}\in\mathbb{M}_{2}^{++}$ in (3.5) with $y=x>0$, $x\neq 1$. Note that

[TABLE]

Using Example 3.2(1) as before we compute

[TABLE]

where

[TABLE]

Hence we find by Lemma 3.6 that

[TABLE]

From this and (4.9) for $\mathrm{Tr}\,\mathcal{A}_{\alpha,q}(A_{0},B_{\theta})$ we must have

[TABLE]

for all $x>0$, $x\neq 1$. As $x\searrow 0$, noting that $\eta_{11}(p)\to-\alpha$, $\eta_{12}(p)\to 0$ and

[TABLE]

we find that the LHS of (4.50) goes to $-{\alpha\over p}$, while the RHS goes to $-{\alpha(1-\alpha)\over q}$ since $q<1$. Therefore, $-{\alpha\over p}\leq-{\alpha(1-\alpha)\over q}$, showing that $p\leq{q\over 1-\alpha}$. ∎

The following is seen similarly to Proposition 4.30.

Proposition 4.38**.**

Let $0<\alpha<1$ and $p,q>0$ be such that $\min\{1,{p\over 2\alpha}\}\leq q$. Then for every $A,B\geq 0$ with $s(A)\geq s(B)$ the following conditions are equivalent:

• (i)

$\mathrm{Tr}\,\widetilde{S}G_{\alpha,p}(A,B)=\mathrm{Tr}\,\mathcal{A}_{\alpha,p}(A,B)$;

• (ii)

$\widetilde{S}G_{\alpha,p}(A,B)=\mathcal{A}_{\alpha,p}(A,B)$;

• (iii)

$A=B$.

The results of this subsection are summarized as follows. Note that the conditions here are not symmetric under interchanging $\alpha$ and $1-\alpha$, unlike those in the previous subsections. This is a reflection of the fact that $\widetilde{S}G_{\alpha,p}$ is not symmetric in the sense of Remark 2.1(2), unlike other quasi matrix means.

Problem 4.39**.**

We find no necessary condition for $\widetilde{S}G_{\alpha,p}\leq_{\mathrm{near}}\mathcal{A}_{\alpha,q}$ and no sufficient condition for $\widetilde{S}G_{\alpha,p}\leq_{\lambda}\mathcal{A}_{\alpha,q}$. The proof of Theorem 4.32 cannot apply to the case $\alpha=1/2$. Indeed, when $\alpha=1/2$, since $\widetilde{S}G_{1/2,p}=SG_{1/2,p}$, the problem has already been pointed out in Problem 4.31. Moreover, as for $\widetilde{S}G_{\alpha,p}\prec_{w}\mathcal{A}_{\alpha,q}$ and $\widetilde{S}G_{\alpha,p}\leq_{\mathrm{Tr}}\mathcal{A}_{\alpha,q}$, there is a big gap between the sufficient condition and the necessary condition, similarly to those for $SG_{\alpha,p}$ and $\mathcal{A}_{\alpha,q}$.

5 Concluding remarks

(1) For each quasi matrix mean $\mathcal{M}_{\alpha,p}\in\{\mathcal{A}_{\alpha,p},LE_{\alpha},R_{\alpha,p},G_{\alpha,p},SG_{\alpha,p},\widetilde{S}G_{\alpha,p}\}$ and for each matrix order $\triangleleft\in\{\leq,\leq_{\mathrm{chao}},\leq_{\mathrm{near}},\leq_{\lambda},\prec_{w},\leq_{\mathrm{Tr}}\}$, we have aimed at finding the necessary and sufficient condition on $p,q,\alpha$ under which the inequality $\mathcal{M}_{\alpha,p}(A,B)\triangleleft\mathcal{A}_{\alpha,q}(A,B)$ holds for all $A,B>0$. When $\mathcal{M}_{\alpha,p}=\mathcal{A}_{\alpha,p},LE_{\alpha},G_{\alpha,p}$, our objective has perfectly been achieved as seen in the tables at the end of Sections 4.1, 4.2 and 4.4. However, when $\mathcal{M}_{\alpha,p}=R_{\alpha,p},SG_{\alpha,p},\widetilde{S}G_{\alpha,p}$, that has not completely be done as seen in the tables of Sections 4.3, 4.5 and 4.6, where there is a gap between the sufficient condition and the necessary condition for some of our target inequalities. Therefore, the problem is still left open for those cases as explained in Problems 4.15, 4.31 and 4.39. We are especially concerned with the question whether $p/2\leq q$ is the necessary and sufficient condition for $R_{\alpha,p}\leq_{\lambda}\mathcal{A}_{\alpha,q}$ to hold or not. This is indeed equivalent to saying whether $|A^{\alpha p}B^{(1-\alpha)p}|^{1/p}\leq_{\lambda}\alpha A+(1-\alpha)B$ holds for all $A,B>0$ only if $p\leq 1$ (in other words, whether Ando’s matrix Young inequality $|A^{\alpha}B^{1-\alpha}|\leq_{\lambda}\alpha A+(1-\alpha)B$ is the best possible case or not).

(2) We have considered an inequality $\mathcal{M}_{\alpha,p}\triangleleft\mathcal{A}_{\alpha,q}$ in the two directions of the sufficiency part (to show the inequality under some condition) and the necessity part (to find a necessary condition for the inequality to hold). The former direction is a more or less easy task by applying well known facts or methods in matrix analysis. Thus we have presented the results in this direction as propositions. On the other hand, the latter direction is computation-oriented, where we provide a counter-example with use of a specific pair of $2\times 2$ matrices, typically the pair $A_{0},B_{\theta}$ given in (3.5). We have prepared in Section 3 some technical computations on $A_{0},B_{\theta}$ based on Taylor’s theorem for matrix functions, which are repeatedly used in the main Section 4. In this way, the results in the latter direction are much involved with plenty of computations though elementary in nature, so we have presented those as theorems. In order to settle the aforementioned open questions by improving the current necessary conditions, we probably need to seek a more sophisticated example of $2\times 2$ matrix pair, or otherwise a matrix pair of higher degrees, though computations with matrix pairs of $3\times 3$ or higher seem difficult to perform by hand. Also, it seems that numerical computations are not so much helpful because the problem is to find the best possible necessary condition.

(3) It is intuitively clear that the implications stated in Proposition 2.4(3) are all strict. This fact has been exemplified in our study of inequalities of quasi matrix means. In fact, the strictness of those implications except for $\leq_{\mathrm{chao}}\ \Rightarrow\ \leq_{\mathrm{near}}$ is manifest in the tables at the end of Sections 4.1–4.4. As for $\leq_{\mathrm{chao}}\ \Rightarrow\ \leq_{\mathrm{near}}$, the following remark is worth noting: Due to Proposition 2.4(2) this implication is equivalent to the so-called Ando–Hiai inequality [4] (i.e., for $X,Y>0$, $X\#Y\leq I\Rightarrow X^{p}\#Y^{p}\leq I$ for $p\geq 1$); see also [13, Proposition 4]. Therefore, its strictness corresponds to $X\#Y\leq I\nRightarrow X^{p}\#Y^{p}\leq I$ for $0<p<1$.

(4) We can apply the method explained in Remark 2.7(3) to have the characterization of $\mathcal{H}_{\alpha,q}\triangleleft\mathcal{H}_{\alpha,p}$ from that of $\mathcal{A}_{\alpha,p}\triangleleft\mathcal{A}_{\alpha,q}$ given in Section 4.1. Furthermore, we have the sufficient condition (or the necessary condition) for $\mathcal{H}_{\alpha,q}\triangleleft\mathcal{M}_{\alpha,p}$ from that for $\mathcal{M}_{\alpha,p}\triangleleft\mathcal{A}_{\alpha,q}$ given in Sections 4.2–4.6 for any $\mathcal{M}_{\alpha,p}\in\{G_{\alpha,p},SG_{\alpha,p},\widetilde{S}G_{\alpha,p},R_{\alpha,p},L_{\alpha}\}$ and $\triangleleft\in\{\leq,\leq_{\mathrm{chao}},\leq_{\mathrm{near}},\leq_{\lambda}\}$. But the inequalities $\mathcal{H}_{\alpha,q}\prec_{w}\mathcal{M}_{\alpha,p}$ and $\mathcal{H}_{\alpha,q}\leq_{\mathrm{Tr}}\mathcal{M}_{\alpha,p}$ are not touched in this paper. As for $\mathcal{H}_{\alpha,p}\triangleleft\mathcal{A}_{\alpha,q}$, we have $\mathcal{H}_{\alpha,p}\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}$ for any $p,q>0$ since Proposition 4.19 yields $\mathcal{H}_{\alpha,p}\leq_{\mathrm{chao}}G_{\alpha,r}\leq_{\mathrm{chao}}\mathcal{A}_{\alpha,q}$ with $r=\min\{p,q\}$. We have also $\mathcal{H}_{\alpha,p}\leq\mathcal{A}_{\alpha,q}$ for any $p,q\geq 1$ since Theorem 4.16 yields $\mathcal{H}_{\alpha,p}\leq G_{\alpha,1}\leq\mathcal{A}_{\alpha,q}$. An interesting problem left open is whether $\mathcal{H}_{\alpha,p}\leq\mathcal{A}_{\alpha,q}$ holds only if $p,q\geq 1$ or not. Here note that for any $q>0$, $\mathcal{H}_{\alpha,p}\leq\mathcal{A}_{\alpha,q}$ fails to hold at least for sufficiently small $p>0$, because otherwise letting $p\searrow 0$ gives $LE_{\alpha}\leq\mathcal{A}_{\alpha,q}$ thanks to Theorem 2.3 and it contradicts Theorem 4.5.

(5) The study of this paper was partly motivated by Milan Mosonyi’s question to the author, asking whether there exists, for $0<\alpha<1$, a ‘reasonable’ $\alpha$-weighted geometric-type mean $\mathcal{M}(A,B)$ ($A,B>0$) other than $\#_{\alpha}$, where ‘reasonable’ is used in the sense that $\mathcal{M}$ satisfies (i) $\mathcal{M}(A,B)=A^{1-\alpha}B^{\alpha}$ if $AB=BA$, (ii) tensor multiplicative $\mathcal{M}(A_{1}\otimes A_{2},B_{1}\otimes B_{2})=\mathcal{M}(A_{1},B_{1})\otimes\mathcal{M}(A_{2},B_{2})$, (iii) block additive $\mathcal{M}(A_{1}\oplus A_{2},B_{1}\oplus B_{2})=\mathcal{M}(A_{1},B_{1})\oplus\mathcal{M}(A_{2},B_{2})$, and (iv) arithmetic-geometric inequality $\mathcal{M}(A,B)\leq(1-\alpha)A+\alpha B$. We examined the possibility of the quasi-geometric type means discussed in this paper to satisfy condition (iv) as it is obvious that they satisfy the other conditions (i)–(iii). But it turned out that none of them other than $\#_{\alpha}$ satisfies (iv); see the tables of Sections 4.2–4.6 and Corollary 4.18 for $G_{\alpha,p}$. Meanwhile, Mosonyi and his coauthors settled the question as follows in [14]: For any $\alpha\in(0,1)$, if an $\alpha$-weighted matrix mean $\mathcal{M}:\mathbb{M}_{n}^{++}\times\mathbb{M}_{n}^{++}\to\mathbb{M}_{n}^{++}$ ($n\in\mathbb{N}$) satisfies (i), (ii*′) (weakly) tensor multiplicative $\mathcal{M}(A^{\otimes n},B^{\otimes n})=\mathcal{M}(A,B)^{\otimes n}$, (ii′′*) scalar tensor multiplicative $\mathcal{M}(aA,bB)=\mathcal{M}(a,b)\mathcal{M}(A,B)$ for $a,b\in(0,\infty)$, (iii) and (iv), then $\mathcal{M}(A,B)=A\#_{\alpha}B$ for all $A,B>0$. This result establishes a remarkable new characterization of the operator mean $\#_{\alpha}$.

(6) In Theorem A.1 of the appendix below, we present the Lie–Trotter–Kato product formula for operator means in the positive semidefinite matrix case, which we have used in Section 2. The author has known Theorem A.1 for long years, without publication though it was briefly explained in [8] without proof. This product formula for operator means in the positive semidefinite case seems unfamiliar even to matrix analysis experts, while that in the positive definite matrix case is rather well known (see, e.g., [25, Lemma 3.3], [20, Sec. 4.3]). So it would be worthwhile for us to take this opportunity to present its complete description.

Acknowledgements

The author thanks Milan Mosonyi for invitation to the workshop at the Erdős Center in July, 2024 and for valuable suggestions which helped to improve this paper.

Appendix A The operator mean version of the Lie–Trotter–Kato product formula for positive semidefinite

matrices

The famous Lie–Trotter–Kato product formula originally established in [35, 27] says that if $H$ and $K$ are lower bounded self-adjoint operators in a Hilbert space $\mathcal{H}$, then $(e^{-H/m}e^{-K/m})^{m}$ converges in strong operator topology to $e^{-(H\dot{+}K)}P_{0}$ as $m\to\infty$, where $H\dot{+}K$ is the form sum (see [26]) and $P_{0}$ is the orthogonal projection onto the closure of the domain of $H\dot{+}K$. According to the proof in [26] (see also [20, Theorem 3.6]) the formula can be modified in the symmetric form with a continuous parameter as

[TABLE]

Furthermore, it is known [27, Sec. 5] that the formula is valid even if $H$ and $K$ have non-dense domains.

For Hermitian matrices $H$ and $K$ the product formula simply becomes the Lie formula

[TABLE]

which has plenty of applications in matrix analysis. The unitary orbital version of this (without limit) is also worth noting [34, 18]. For positive semidefinite (not necessarily positive definite) matrices $A,B$ we consider $H:=-\log A$ and $K:=-\log B$ defined under the restriction to the ranges of the support projections $A^{0}:=s(A)$ and $B^{0}:=s(B)$, respectively. Applying (A.1) (for non-dense domains) to these $H,K$ we have222There seems no literature which provides the proof of (A.2) in a way specialized to the matrix setting.

[TABLE]

where $P_{0}:=A^{0}\wedge B^{0}$.

This appendix is aimed to supply the operator mean version of (A.2) for matrices $A,B\geq 0$. Throughout the appendix let $A,B$ be $n\times n$ positive semidefinite matrices. Define $\log A$ in the generalized sense as $\log A:=(\log A)A^{0}$ restricted on the range of $A^{0}$ (and zero on the range of $A^{0\perp}=I-A^{0}$), and similarly $\log B:=(\log B)B^{0}$. We write $P_{0}:=A^{0}\wedge B^{0}$ as above. Now, let $\sigma$ be a Kubo–Ando’s operator mean with the representing operator monotone function $f$ on $(0,\infty)$, and let $\alpha:=f^{\prime}(1)$. Note that $0\leq\alpha\leq 1$ and if $\alpha=0$ (resp., $\alpha=1$) then $A\sigma B=A$ (resp., $A\sigma B=B)$ so that $(A^{p}\sigma B^{p})^{1/p}=A$ (resp., $(A^{p}\sigma B^{p})^{1/p}=B$) for all $A,B\geq 0$ and $p>0$. So in the rest we assume that $0<\alpha<1$.

Theorem A.1**.**

With the above assumptions, for every $A,B\geq 0$,

[TABLE]

Remark A.2**.**

Note [25, Sect. 4] that the RHS of (A.3) is written as

[TABLE]

so that we may write

[TABLE]

The next lemma is essential to prove the theorem. The proof of the lemma is a slight modification of that of [25, Lemma 4.1].

Lemma A.3**.**

For each $p\in(0,p_{0})$ with some $p_{0}>0$, a Hermitian matrix $Z(p)$ is given in the $2\times 2$ block form as

[TABLE]

where $Z_{0}(p)$ is $m\times m$, $Z_{1}(p)$ is $l\times l$ and $Z_{2}(p)$ is $m\times l$. Assume:

• (a)

$Z_{0}(p)\to Z_{0}$* as $p\searrow 0$,*

• (b)

$\sup\{\|Z_{2}(p)\|_{\infty}:p\in(0,p_{0})\}<\infty$,

• (c)

there is a $\delta>0$ such that $pZ_{1}(p)\leq-\delta I_{l}$ for all $p\in(0,p_{0})$.

Then

[TABLE]

Proof.

We list the eigenvalues of $Z(p)$ in decreasing order (with multiplicities) as

[TABLE]

together with the corresponding orthonormal eigenvectors

[TABLE]

so that we write

[TABLE]

Furthermore, let $\mu_{1}(p)\geq\dots\geq\mu_{m}(p)$ be the eigenvalues of $Z_{0}(p)$ and $\mu_{1}\geq\dots\geq\mu_{m}$ be the eigenvalues of $Z_{0}$ Then $\mu_{i}(p)\to\mu_{i}$ as $p\searrow 0$ thanks to assumption (a). By the majorization result for eigenvalues in [2, Corollary 7.2] we have

[TABLE]

Since

[TABLE]

thanks to assumptions (a)–(c), it follows that for $m<i\leq m+l$, $p\lambda_{i}(p)<-\delta/2$ for any $p>0$ sufficiently small so that

[TABLE]

Hence it suffices to prove that for any sequence $(p_{0}>)\ p_{k}\searrow 0$ there exist a subsequence $\{p_{k}^{\prime}\}$ of $\{p_{k}\}$ and vectors $v_{1},\dots,v_{m}\in\mathbb{C}^{m}$ such that we have for $1\leq i\leq m$

[TABLE]

Indeed, it then follows that $v_{1},\dots,v_{m}$ are orthonormal vectors in $\mathbb{C}^{m}$, so from (A.4) and (A.6) we obtain

[TABLE]

Now, replacing $\{p_{k}\}$ with a subsequence, we may assume that $u_{i}(p_{k})$ itself converges to some $u_{i}\in\mathbb{C}^{m}\oplus\mathbb{C}^{l}$ for $1\leq i\leq m$. Writing $u_{i}(p_{k})=v_{i}^{(k)}\oplus w_{i}^{(k)}$ in $\mathbb{C}^{m}\oplus\mathbb{C}^{l}$, we have

[TABLE]

by assumption (c). For $i=1$, since $\mu_{1}(p_{k})\leq\lambda_{1}(p_{k})$ by (A.5) for $r=1$, it follows from (A.10) that

[TABLE]

so that

[TABLE]

as $k\to\infty$ ($p_{k}\searrow 0$) due to assumptions (a) and (b). Hence we have $w_{1}^{(k)}\to 0$ so that $u_{1}(p_{k})\to u_{1}=v_{1}\oplus 0$ in $\mathbb{C}^{m}\oplus\mathbb{C}^{l}$ (hence $v_{1}^{(k)}\to v_{1}$) for some $v_{1}\in\mathbb{C}^{m}$. From (A.10) again we furthermore have

[TABLE]

Therefore, $\lambda_{1}(p_{k})\to\langle v_{1},Z_{0}v_{1}\rangle=\mu_{1}$ and hence $Z_{0}v_{1}=\mu_{1}v_{1}$ since $\mu_{1}$ is the largest eigenvalue of $Z_{0}$. Next, when $k\geq 2$ and $i=2$, since $\lambda_{2}(p_{k})$ is bounded below by (A.5) for $r=2$, it follows as above that $w_{2}^{(k)}\to 0$ and hence $u_{2}(p_{k})\to u_{2}=v_{2}\oplus 0$ for some $v_{2}\in\mathbb{C}^{m}$. Therefore,

[TABLE]

so that $\lambda_{2}(p_{k})\to\langle v_{2},Z_{0}v_{2}\rangle=\mu_{2}$ and $Z_{0}v_{2}=\mu_{2}v_{2}$ since $\mu_{2}$ is the largest eigenvalue of $Z_{0}$ restricted to $\{v_{1}\}^{\perp}\cap\mathbb{C}^{m}$. Repeating this argument we obtain $v_{1},\dots,v_{m}\in\mathbb{C}^{m}$ for which (A.7)–(A.9) hold for $1\leq i\leq m$. ∎

Proof of Theorem A.1.

Let us divide the proof into two steps. In the proof below we use the $\alpha$-weighted arithmetic mean $\triangledown_{\alpha}$ and the $\alpha$-weighted harmonic mean $!_{\alpha}$. Note that

[TABLE]

Step 1. First, we prove the theorem in the case where $P\sigma Q=P\wedge Q$ for all orthogonal projections $P,Q$ (this is the case, for instance, when $\sigma=\ !_{\alpha}$ or $\#_{\alpha}$ the weighted geometric mean, see [29, Theorem 3.7]). Let $\mathcal{H}_{0}$ be the range of $P_{0}$ ($=A^{0}!_{\alpha}B^{0}=A^{0}\sigma B^{0}$). From the operator monotonicity of $\log x$ ($x>0$) it follows that for every $p>0$,

[TABLE]

For every $\varepsilon>0$ we have

[TABLE]

where $A^{-p}=(A^{-1})^{p}$ and $B^{-p}=(B^{-1})^{p}$ are defined via the generalized inverses. Therefore,

[TABLE]

since the support projection of $A^{0\perp}\triangledown_{\alpha}B^{0\perp}$ is $A^{0\perp}\vee B^{0\perp}=P_{0}^{\perp}$. In the above, $\{\cdots\}^{-1}$ is the generalized inverse (with support $\mathcal{H}_{0}$) and the inequality is due to [10, Corollary 2.3]. Letting $\varepsilon\searrow 0$ in (A.12) gives

[TABLE]

so that

[TABLE]

Combining (A.11) and (A.14) yields

[TABLE]

Since $A^{-p}=e^{-p\log A}A^{0}=A^{0}-p\log A+o(p)$ and similarly $B^{-p}=B^{0}-p\log B+o(p)$ as $p\searrow 0$, we have

[TABLE]

so that

[TABLE]

Therefore,

[TABLE]

Similarly, we have

[TABLE]

so that

[TABLE]

From (A.15), A.17 and (A.19) we obtain

[TABLE]

which yields the required limit formula.

Step 2. For a general operator mean $\sigma$ the integral representation theorem [29, Theorems 3.4, 3.7] says that there are $\theta,\beta\in[0,1]$ and an operator mean $\tau$ such that

[TABLE]

and $P\tau Q=P\wedge Q$ for all orthogonal projections $P,Q$. Moreover, $\tau$ has the representing operator monotone function $g$ on $(0,\infty)$ for which $\gamma:=g^{\prime}(1)\in(0,1)$ and

[TABLE]

We may assume that $0<\theta\leq 1$ since the case $\theta=0$ was shown in Step 1. Moreover, when $\theta=1$, we have $\beta=\alpha\in(0,1)$. At the moment, assume that $0<\theta\leq 1$ and $0<\beta<1$. Let $A,B\geq 0$ be given, and note that $A^{0}\sigma B^{0}=\theta A^{0}\triangledown_{\beta}B^{0}+(1-\theta)(A^{0}\wedge B^{0})$ has the support projection $A^{0}\vee B^{0}$. Let $\mathcal{H}$, $\mathcal{H}_{0}$ and $\mathcal{H}_{1}$ denote the ranges of $A^{0}\vee B^{0}$, $P_{0}=A^{0}\wedge B^{0}$ and $A^{0}\vee B^{0}-P_{0}$, respectively, so that $\mathcal{H}=\mathcal{H}_{0}\oplus\mathcal{H}_{1}$. Note that the support of $A^{p}\sigma B^{p}$ for any $p>0$ is $\mathcal{H}$. We will describe ${1\over p}\log(A^{p}\sigma B^{p})\big|_{\mathcal{H}}$ in the $2\times 2$ block form with respect to the decomposition $\mathcal{H}=\mathcal{H}_{0}\oplus\mathcal{H}_{1}$. Let

[TABLE]

It follows from Step 1 that $\lim_{p\searrow 0}(A^{p}\tau B^{p})^{1/p}=P_{0}e^{Y_{0}}P_{0}$ and hence

[TABLE]

In the above, the fourth equality follows since $\log\bigl(e^{Y_{0}}+o(1)\bigr)=Y_{0}+o(1)$. On the other hand, we have

[TABLE]

Therefore, we have

[TABLE]

Setting

[TABLE]

we write

[TABLE]

where $C$ is a positive definite contraction on $\mathcal{H}$ and $H$ is a Hermitian operator on $\mathcal{H}$. Note that the eigenspace of $C$ for the eigenvalue $1$ is $\mathcal{H}_{0}$. Hence, with a basis consisting of orthonormal eigenvectors for $C$ we may assume that $C$ is diagonal so that $C=\mathrm{diag}(c_{1},\dots,c_{m+l})$ with

[TABLE]

where $m=\dim\mathcal{H}_{0}$ and $m+l=\dim\mathcal{H}$.

Applying Taylor’s theorem (see, e.g., [21, Theorem 2.3.1]) to $\log(C+pH+o(p))$ we have

[TABLE]

where $D(\log x)(C)(\cdot)$ denotes the Fréchet derivative of the functional calculus by $\log x$ at $C$. Daleckii and Krein’s derivative formula (see, e.g., [21, Theorem 2.3.1]) says that

[TABLE]

where $\circ$ is the Schur product and $(\log c_{i}-\log c_{j})/(c_{i}-c_{j})$ is understood as $1/c_{i}$ when $c_{i}=c_{j}$. We write $D(\log x)(C)(H)$ in the $2\times 2$ block form on $\mathcal{H}_{0}\oplus\mathcal{H}_{1}$ as $\begin{bmatrix}Z_{0}&Z_{2}\\ Z_{2}^{*}&Z_{1}\end{bmatrix}$ where $Z_{0}:=P_{0}HP_{0}|_{\mathcal{H}_{0}}$. By (A.21)–(A.23) we can write

[TABLE]

where

[TABLE]

This $2\times 2$ block form of $Z(p):={1\over p}\log(A^{p}\sigma B^{p})\big|_{\mathcal{H}}$ satisfies assumptions (a)–(c) of Lemma A.3 for $p\in(0,p_{0})$ with a sufficiently small $p_{0}>0$. Therefore, the lemma implies that

[TABLE]

on $\mathcal{H}=\mathcal{H}_{0}\oplus\mathcal{H}_{1}$. Finally, we have

[TABLE]

thanks to (A.20). Hence the desired limit formula follows.

For the remaining case where $0<\theta<1$ and $\beta=0$ or $1$, the proof is similar to the above when we take as $\mathcal{H}$ the range of $A^{0}$ (for $\beta=0$) or $B^{0}$ (for $\beta=1$) instead of the range of $A^{0}\vee B^{0}$. ∎

Remark A.4**.**

Assume that the operator mean $\sigma$ satisfies $P\sigma Q=P\wedge Q$ for all orthogonal projections $P,Q$, that is, the representing function $f$ satisfies $f(0^{+})=0$ and $\lim_{x\to\infty}f(x)/x=0$ (see [29, Theorem 3.7]). This is the case when $\sigma\#_{\alpha}$ for instance. Then, from the proof of Step 1 above, for any $A,B\geq 0$ we have a slightly improved form of (A.3) as follows:

[TABLE]

Indeed, set $L:=(1-\alpha)P_{0}(\log A)P_{0}+\alpha P_{0}(\log B)P_{0}$. By (A.13) and (A.16) one has

[TABLE]

Also, by (A.18) one has

[TABLE]

Therefore, (A.24) follows.