Inequalities for Relative Operator Entropies and Operator Means

Abstract

The main purpose of this article is to study estimates for the Tsallis relative operator entropy, by using the Hermite-Hadamard inequality. We obtain alternative bounds for the Tsallis relative operator entropy and in the process to derive these bounds, we established the significant relation between the Tsallis relative operator entropy and the generalized relative operator entropy. In addition, we study the properties on monotonicity for the weight of operator means and for the parameter of relative operator entropies.

1 Introduction

In operator theory, there are various characterizations and the relationship between operator monotonicity and operator convexity given, say, by Hansen and Pedersen [10], Chansangiam [1]. In [13], Kubo and Ando have studied the connections between operator monotone functions and operator means. The operator monotone function plays an important role in the theory given by Kubo and Ando. Other information about applications of operator monotone functions to theory of operator mean can be found in [16]. Theory of operator mean plays a central role in operator inequalities, operator equations, network theory, and quantum information theory.

Denoted by B(H), the algebra of bounded linear operators is on a Hilbert space H. We write A > 0 to mean that A is a strictly positive operator, or equivalently, A ≥ 0 and A is invertible. Furuta and Yanagida showed the following inequality with elegant proof [9]:

$$ A{!_{p}}B \le A{\#_{p}}B \le A{\nabla_{p}}B, $$

(1.1)

where we respectively denote p-weighted harmonic operator mean, p-weighted geometric operator mean, and p-weighted arithmetic operator mean by A!_pB ≡ {(1 − p)A^− 1 + p B^− 1}^− 1, A #_pB ≡ A^1/2(A^− 1/2B A^− 1/2)^pA^1/2, and A∇_pB ≡ (1 − p)A + p B for A,B > 0 and p ∈ [0,1].

On the other hand, Tsallis defined the one-parameter extended entropy for the analysis of a physical model in statistical physics in [17]. The properties of the Tsallis relative entropy were studied in [4, 5], by Furuichi, Yanagi, and Kuriyama. The relative operator entropy

$$S\left( {A|B} \right): = {A^{1/2}}\log \left( {{A^{- 1/2}}B{A^{- 1/2}}} \right){A^{1/2}}$$

for two invertible positive operators A and B on a Hilbert space was introduced by Fujii and Kamei in [3]. The parametric extension of the relative operator entropy was introduced by Furuta in [7] as

$$S_{p}\left( {A|B} \right): = {A^{1/2}} \left( A^{-1/2}BA^{-1/2}\right)^{p}\log \left( {{A^{- 1/2}}B{A^{- 1/2}}} \right){A^{1/2}}$$

for $p \in \mathbb {R}$ and two invertible positive operators A and B on a Hilbert space. Note that $S_{0}(A|B) \equiv \lim _{p\to 0}S_{p}(A|B) = S(A|B)$. In [18], Yanagi, Kuriyama, and Furuichi introduced a parametric extension of relative operator entropy by the concept of the Tsallis relative entropy for operators, as

$${T_{p}}\left( {A|B} \right): = \frac{{{A^{1/2}}{{\left( {{A^{- 1/2}}B{A^{- 1/2}}} \right)}^{p}}{A^{1/2}} - A}}{p},\,\,\,\left( { - 1 \le p \le 1,~p \ne 0} \right)$$

where A and B are two strictly positive operators on a Hilbert space H. In [6], we found several results about the Tsallis relative operator entropy. Furuta [8] showed two reverse inequalities involving the Tsallis relative operator entropy T_p(A|B) via generalized Kantorovich constant K(p). The Tsallis relative operator entropy can be rewritten as

$${T_{p}}\left( {A|B} \right) = \frac{{A{\natural_{p}}B - A}}{p},$$

where A♮_pB := A^1/2(A^− 1/2BA^− 1/2)^pA^1/2 for all $p \in \mathbb {R}$. The study of the Tsallis relative operator entropy is often strongly connected to the study of the p-weighted geometric operator mean. It is known that [6]:

$$ A - A{B^{- 1}}A \le {T_{p}}\left( {A|B} \right) \le B - A $$

(1.2)

for strictly positive operators A, B, and p ∈ [− 1,0) ∪(0,1] and $\lim _{p \to 0} {T_{p}}\left ({A|B} \right ) = S\left ({A|B} \right )$.

2 Alternative Estimate of the Tsallis Relative Operator Entropy

We start from the following known properties of the Tsallis relative operator entropy. See [11, Theorem 1] or [12, Theorem 2.5 (ii)] for example.

Proposition 2.1

For any strictly positive operators A and B andp,q ∈ [− 1,0) ∪ (0,1]withp ≤ q, we have

$$T_{p}(A|B) \leq T_{q}(A|B). $$

This proposition can be proven by the monotone increasing of $\frac {x^{p}-1}{p}$ on p ∈ [− 1,0) ∪ (0,1] for any x > 0, and implies the following inequalities (which include the inequalities (1.2)) [18]:

$$A-AB^{-1}A=T_{-1}(A|B) \leq T_{-p}(A|B) \leq S(A|B) \leq T_{p}(A|B) \leq T_{1}(A|B) =B-A $$

for any strictly positive operators A and B and p ∈ (0,1]. The general results were recently established in [15] by the notion of perspective functions. In addition, quite recently, the interesting and significant results for relative operator entropy were given in [2] for the case B ≥ A. In this section, we treat the relations on the Tsallis relative operator entropy under the assumption that strictly positive operators A and B have the ordering A ≤ B or A ≥ B.

In [14], we obtained the estimates on the Tsallis relative operator entropy by the use of the Hermite-Hadamard inequality:

$$f\left( \frac{a+b}{2} \right) \leq \frac{1}{b-a} {{\int}_{a}^{b}} f(t) dt \leq \frac{f(a) +f(b)}{2} $$

for a convex function f(t) defined on the interval [a,b] with a≠b.

Theorem 2.2 ([14])

For any invertible positive operators A and B such thatA ≤ Band− 1 ≤ p ≤ 1withp≠ 0,we have

$$\begin{array}{@{}rcl@{}} && {{A}^{1/2}}{{\left( \frac{{{A}^{-1/2}}B{{A}^{-1/2}}+I}{2} \right)}^{p-1}}\left( {{A}^{-1/2}}B{{A}^{-1/2}}-I \right){{A}^{1/2}} \\ & \le& {{T}_{p}}\left( A|B \right) \le \frac{1}{2}\left( A{{\#}_{p}}B-A{{\natural}_{p-1}}B+B-A \right), \end{array} $$

where I is the identity operator.

The inequalities in Theorem 2.2 are improvements of the inequalities (1.2). In the present paper, we give the alternative bounds for the Tsallis relative operator entropy. The condition A ≤ B in Theorem 2.2 can be modified by uA ≤ B ≤ vA with u ≥ 1 so that we use this style (which is often called a sandwich condition) in the present paper. Note that the condition uA ≤ B ≤ vA with u ≥ 1 includes the condition A ≤ B as a special case, also the condition uA ≤ B ≤ vA with v ≤ 1 includes the condition B ≤ A as a special case.

Theorem 2.3

Let A and B be strictly positive operators such thatuA ≤ B ≤ vAwith u,v > 0and let − 1 ≤ p ≤ 1with p≠ 0. If u ≥ 1, then

$$ S_{p/2}(A|B) \leq T_{p}(A|B) \leq \frac{S(A|B) +S_{p}(A|B)}{2}. $$

(2.1)

If v ≤  1, then the reverse inequalities in (2.1) hold.

Proof

For x ≥ 1 and − 1 ≤ p ≤ 1 with p≠ 0, we define the function f(t) = x^pt log x on 0 ≤ t ≤ 1. Since $\frac {d^{2}f(t)}{dt^{2}} = p^{2}x^{pt} \left (\log x\right )^{3} \geq 0$ for x ≥ 1, the function f(t) is convex on t, for the case x ≥ 1. Thus, we have

$$ x^{p/2} \log x \leq \frac{x^{p} -1 }{p} \leq \left( \frac{x^{p} + 1 }{2} \right) \log x $$

(2.2)

by the Hermite-Hadamard inequality, since ${{\int }_{0}^{1}} f(t) dt = \frac {x^{p} -1}{p}$. Note that I ≤ uI ≤ A^− 1/2BA^− 1/2 ≤ vI from the condition u ≥ 1. By the Kubo-Ando theory [13], it is known that for the representing function f_m(x) = 1mx for operator mean m, the scalar inequality f_m(x) ≤ f_n(x), (x > 0) is equivalent to the operator inequality AmB ≤ AnB for all strictly positive operators A and B. (Hereafter, we omit this description for simplicity in the following proofs.) Thus, we have the inequality

$$\begin{array}{@{}rcl@{}} &&A^{1/2}\left( A^{-1/2}BA^{-1/2} \right)^{p/2} \log \left( A^{-1/2}BA^{-1/2} \right)A^{1/2} \leq \frac{A\#_{p} B-A}{p} \\ &\leq& \frac{A^{1/2}\log \left( A^{-1/2}BA^{-1/2} \right)A^{1/2}+A^{1/2}\left( A^{-1/2}BA^{-1/2} \right)^{p} \log \left( A^{-1/2}BA^{-1/2} \right)A^{1/2}}{2} \end{array} $$

which is the inequality (2.1). The reverse inequalities for the case v ≤ 1 can be similarly shown by the concavity of the function f(t) on t, for the case 0 < x ≤ 1, taking into account the condition 0 < uI ≤ A^− 1/2BA^− 1/2 ≤ vI ≤ I. □

We note that both sides in the inequalities (2.1) and their reverses converge to S(A|B) in the limit p → 0. From the proof of Theorem 2.3, for strictly positive operators A and B, we see the following interesting relation between the Tsallis relative operator entropy T_p(A|B) and the generalized relative operator entropy S_p(A|B),

$${{\int}_{0}^{1}} S_{pt}(A|B) dt =T_{p}(A|B). $$

Remark 2.4

Let A and B be strictly positive operators such that uA ≤ B ≤ vA with u,v > 0 and let − 1 ≤ p ≤ 1 with p≠ 0. For the case 0 < p ≤ 1 and u ≥ 1, we see

$$S(A|B) \leq S_{p/2}(A|B) \leq T_{p}(A|B) \leq \frac{S(A|B) +S_{p}(A|B)}{2} \leq S_{p}(A|B) $$

from the inequalities (2.1) since x^p log x is monotone increasing on 0 < p ≤ 1 and $\left (\frac {x^{p} + 1 }{2}\right ) \log x \leq x^{p} \log x$ for x ≥ 1 and 0 < p ≤ 1. For the case − 1 ≤ p < 0 and v ≤ 1, we also see that the reverse inequalities hold since x^p log x is monotone increasing on − 1 ≤ p < 0 and $\left (\frac {x^{p} + 1 }{2}\right ) \log x \geq x^{p} \log x$ for 0 < x ≤ 1 and − 1 ≤ p < 0.

Remark 2.5

We compare the bounds of $\frac {{{x^{p}} - 1}}{p}$ in the inequalities (2.1) with the result given in [14]:

$${\left( {\frac{{x + 1}}{2}} \right)^{p - 1}}\left( {x - 1} \right) \le \frac{{{x^{p}} - 1}}{p} \le \left( {\frac{{{x^{p - 1}} + 1}}{2}} \right)\left( {x - 1} \right),~~\left( {x \ge 1,~0 < p \le 1} \right).$$

(i) We have no ordering between x^p/2 log x and $\left (\frac {x + 1}{2} \right )^{p-1}\left (x-1\right )$. Indeed, when p = 1/4 and x = 3, $x^{p/2} \log x -\left (\frac {x + 1}{2} \right )^{p-1}\left (x-1\right ) \simeq 0.071123.$ On the other hand, when p = 3/4 and x = 3, $x^{p/2} \log x -\left (\frac {x + 1}{2} \right )^{p-1}\left (x-1\right ) \simeq -0.023104.$ (ii) We have no ordering between $\left (\frac {x^{p}+ 1}{2}\right ) \log x$ and $\left (\frac {x^{p-1}+ 1}{2} \right )\left (x-1\right )$. Indeed, when p = 1/4 and x = 3, $\left (\frac {x^{p-1}+ 1}{2} \right )\left (x-1\right ) - \left (\frac {x^{p}+ 1}{2}\right ) \log x \simeq 0.166458.$ On the other hand, when p = 3/4 and x = 3, $\left (\frac {x^{p-1}+ 1}{2} \right )\left (x-1\right ) - \left (\frac {x^{p}+ 1}{2}\right ) \log x \simeq -0.0416177.$

Therefore, we claim Theorem 2.3 is not a trivial result.

Theorem 2.6

Let A and B be strictly positive operators such thatuA ≤ B ≤ vAwith u ≥ 1and let − 1 ≤ p ≤ 1with p≠ 0. Then, we have

$$\begin{array}{@{}rcl@{}} &&\frac{T_{p}(A|B)-T_{p-1}(A|B)}{2} \leq 4 \left\{ T_{p}\left( A\left| \frac{A+B}{2}\right. \right)-T_{p-1}\left( A\left| \frac{A+B}{2}\right. \right)\right\} \\ &\leq& \frac{T_{p}(A|B)-T_{1}(A|B)}{p-1} \leq \frac{T_{p}(A|B)-T_{p-1}(A|B)}{2} +\frac{A \natural_{2}(B-A)}{4}. \end{array} $$

Proof

It is sufficient to prove the following inequalities for t ≥ 1 and − 1 ≤ p ≤ 1 with p≠ 0,

$$ l_{p}(t) \leq k_{p}(t) \leq c_{p}(t) \leq l_{p}(t) +\frac{(t-1)^{2}}{4}, $$

(2.3)

where

$$\begin{array}{@{}rcl@{}} && l_{p}(t) : = \frac{1}{2}\left( \frac{t^{p}-1}{p} - \frac{t^{p-1}-1}{p-1}\right),\quad c_{p}(t) : = \frac{t^{p} -1}{p(p-1)} -\frac{t-1}{p-1},\\ && k_{p}(t) : = \frac{4}{p}\left\{ \left( \frac{t + 1}{2}\right)^{p} -1\right\} -\frac{4}{p-1}\left\{ \left( \frac{t + 1}{2}\right)^{p-1} -1\right\}. \end{array} $$

Firstly, to prove l_p(t) ≤ k_p(t), we set the function h_p(t) := k_p(t) − l_p(t). Then, we calculate

$$\frac{dh_{p}(t)}{dt} =(t-1) \left\{ \left( \frac{t + 1}{2}\right)^{p-2} -\frac{t^{p-2}}{2}\right\}. $$

We set $g_{p}(t) : = \left (\frac {t + 1}{2}\right )^{p-2} -\frac {t^{p-2}}{2}$. Then, we have

$$\frac{dg_{p}(t)}{dt} =\frac{p-2}{2} \left\{\left( \frac{t + 1}{2}\right)^{p-3} -t^{p-3} \right\} \leq 0,\quad g_{p}(1) = \frac{1}{2}, \quad \lim_{t\to \infty} g_{p}(t) = 0. $$

Thus, we have g_p(t) ≥ 0, that is, $\frac {dh_{p}(t)}{dt} \geq 0$ so that we have h_p(t) ≥ h_p(1) = 0.

Secondly, the inequalities $k_{p}(t) \leq c_{p}(t) \leq l_{p}(t) +\frac {(t-1)^{2}}{4}$ can be proven in the following way. We consider x ≥ 1 and the function $f:[1,x] \to \mathbb {R}$ defined by f(y) = y^p− 2 with p ∈ (0,1]. It follows that f^′(y) = (p − 2)y^p− 3 with f^″(y) = (p − 2)(p − 3)y^p− 4 ≥ 0, so the function f is convex. Therefore, applying the Hermite-Hadamard inequality, we have

$$\left( {x - 1} \right){\left( {\frac{{x + 1}}{2}} \right)^{p - 2}} \le {{\int}_{1}^{x}} {{y^{p - 2}}dy \le \left( {x - 1} \right)\left( \frac{{{x^{p - 2}} + 1}}{2}\right)} $$

which, by integrating, is equivalent to the inequality

$${{\int}_{1}^{t}} {\left( {x - 1} \right){{\left( {\frac{{x + 1}}{2}} \right)}^{p - 2}}dx} \le {{\int}_{1}^{t}} {{{\int}_{1}^{x}} {{y^{p - 2}}dydx} } \le {{\int}_{1}^{t}} {\left( {x - 1} \right)\left( \frac{{{x^{p - 2}} + 1}}{2}\right)} dx.$$

Since we have the computations of the following integrals, for t,x ≥ 1

$$\begin{array}{@{}rcl@{}} &&{\!{\int}_{1}^{t}} {{{\int}_{1}^{x}} {{y^{p - 2}}dydx} } = \frac{{{t^{p}} - 1}}{{p\left( {p - 1} \right)}} - \frac{{\left( {t - 1} \right)}}{{\left( {p - 1} \right)}}, \end{array} $$

(2.4)

$$\begin{array}{@{}rcl@{}} &&{\!{\int}_{1}^{t}} {\left( {x - 1} \right){{\left( {\frac{{x + 1}}{2}} \right)}^{p - 2}}dx} = \frac{2}{{p - 1}}\left( {t - 1} \right){\left( {\frac{{t + 1}}{2}} \right)^{p - 1}} \!- \frac{4}{{p\left( {p - 1} \right)}}\left\{ {{{\left( {\frac{{t + 1}}{2}} \right)}^{p}} \!- 1} \right\} \end{array} $$

and

$$ {{\int}_{1}^{t}} {\left( {x - 1} \right) \left( \frac{{{x^{p - 2}} + 1}}{2}\right)} dx = \frac{1}{2}\left\{ {\frac{{{t^{p}} - 1}}{p} - \frac{{{t^{p - 1}} - 1}}{{p - 1}} + \frac{1}{2}\left( {{t^{2}} - 1} \right) - \left( {t - 1} \right)} \right\}, $$

(2.5)

we obtain the inequality

$$\begin{array}{@{}rcl@{}} &&\frac{2}{{p - 1}}\left( {t - 1} \right){\left( {\frac{{t + 1}}{2}} \right)^{p - 1}} - \frac{4}{{p\left( {p - 1} \right)}}\left\{ {{{\left( {\frac{{t + 1}}{2}} \right)}^{p}} - 1} \right\}\\ & \le& \frac{{{t^{p}} - 1}}{{p\left( {p - 1} \right)}} - \frac{{\left( {t - 1} \right)}}{{\left( {p - 1} \right)}} \le \frac{1}{2}\left\{ {\frac{{{t^{p}} - 1}}{p} - \frac{{{t^{p - 1}} - 1}}{{p - 1}} + \frac{1}{2}\left( {{t^{2}} - 1} \right) - \left( {t - 1} \right)} \right\}. \end{array} $$

By simple calculations, we find the above inequalities are equivalent to the inequalities $k_{p}(t) \leq c_{p}(t) \leq l_{p}(t) +\frac {(t-1)^{2}}{4}$. □

Remark 2.7

We compare Theorem 2.6 and Theorem 2.2 in [14]. The inequalities $k_{p}(t) \leq c_{p}(t) \leq l_{p}(t) +\frac {(t-1)^{2}}{4}$ given in (2.3) are equivalent to the following inequalities

$$ \alpha_{p}(t) \leq \frac{t^{p} -1}{p} \leq \beta_{p}(t), $$

(2.6)

where

$$\begin{array}{@{}rcl@{}} && \alpha_{p}(t) : = \frac{t^{p-1} -1}{p-3} +\frac{p-1}{2(3-p)}(t^{2}-1) +(t-1), \\ && \beta_{p}(t) : = (t-1) + 2 \left( \frac{t + 1}{2}\right)^{p-1} -\frac{4}{p} \left\{ \left( \frac{t + 1}{2}\right)^{p} -1\right\}. \end{array} $$

By the inequalities (2.6) with the Kubo-Ando theory [13], we have

$$\begin{array}{@{}rcl@{}} && B - \frac{3}{2}A - \frac{1 - p}{2\left( 3 - p \right)}BA^{- 1}B - \frac{1}{3 - p}A\natural_{p-1} B \\ & &{}\le T_{p}\left( A|B \right) {}\le{} B - A {}+{} 2\left( BA^{- 1} - I \right) A \natural_{p-1} \left( \frac{A+B}{2} \right)- 4 T_{p} \left( A \left| \frac{A+B}{2} \right.\right), \end{array} $$

(2.7)

which are equivalent to the second and third inequalities given in Theorem 2.6.

We compare both bounds of $\frac {t^{p} -1}{p}$ in (2.6) with the fundamental inequalities

$${\left( {\frac{{t + 1}}{2}} \right)^{p - 1}}\left( {t - 1} \right) \le \frac{{{t^{p}} - 1}}{p} \le \left( {\frac{{{t^{p - 1}} + 1}}{2}} \right)\left( {t - 1} \right),~~\left( {t \ge 1,~0 < p \le 1} \right)$$

to obtain Theorem 2.2. For this purpose, let t ≥ 1 and − 1 ≤ p ≤ 1 with p≠ 0. And we set the functions γ_p(t) and δ_p(t) by

$$\gamma_{p}(t) : = \alpha_{p}(t) - {\left( {\frac{{t + 1}}{2}} \right)^{p - 1}}\left( {t - 1} \right), \quad \delta_{p}(t) : = \left( {\frac{{{t^{p - 1}} + 1}}{2}} \right)\left( {t - 1} \right) - \beta_{p}(t). $$

By numerical computations, we have the following results.

• (i) γ_1/2(3/2) ≃ 0.00118777 and γ_1/2(5/2) ≃− 0.0118756.

• (ii) δ_1/2(3/2) ≃− 0.890458 and δ_1/2(5/2) ≃ 0.795489.

Thus, we conclude that for the Tsallis relative entropy T_p(A|B), there is no ordering between the bounds given in the inequalities (2.7) and the ones given in Theorem 2.2 of [14]. Therefore, we claim Theorem 2.6 is also not a trivial result.

Taking the limit p → 0 in Theorem 2.6, we have the following corollary.

Corollary 2.8

For strictly positive operators A and B such thatuA ≤ B ≤ vAwith u ≥ 1, we have

$$\begin{array}{@{}rcl@{}} &&\frac{S(A|B)-T_{-1}(A|B)}{2} \leq 4 \left\{ S\left( A\left| \frac{A+B}{2}\right. \right)-T_{-1}\left( A\left| \frac{A+B}{2}\right. \right)\right\} \\ & \leq& T_{1}(A|B) - S(A|B) \leq \frac{S(A|B)-T_{-1}(A|B)}{2} +\frac{A \natural_{2}(B-A)}{4}. \end{array} $$

3 Monotonicity on the Parameter of Relative Operator Entropies

In our previous section, we gave interesting relations between the Tsallis relative operator entropy T_p(A|B) [18] and the generalized relative operator entropy S_p(A|B) [7]. In this section, we study the monotonicity on parameter p related to two relative operator entropies T_p(A|B) and S_p(A|B).

Lemma 3.1

For x ≥ 1, we have the inequality 1 − x + x log x ≤ x(log x)². For $\frac {1}{e} \leq x \leq 1$, we also have the same inequality.

Proof

For x ≥ 1, we set the function g(x) ≡ 1 − x + x log x − x(log x)². Then, we calculate g^′(x) = −(1 + log x)log x ≤ 0 so that g(x) ≤ g(1) = 0 for x ≥ 1. We also have g^′(x) ≥ 0 for $\frac {1}{e}\leq x \leq 1$ so that we have g(x) ≤ g(1) = 0. □

Proposition 3.2

Let A and B be strictly positive operators such that uA ≤ B ≤ vAwithu,v > 0and let− 1 ≤ p ≤ 1withp≠ 0. If either(i) u ≥ 1and0 < p ≤ q ≤ 1or(ii) v ≤ 1and − 1 ≤ p ≤ q < 0, then

$$T_{p}(A|B)-S_{p}(A|B) \geq T_{q}(A|B)-S_{q}(A|B). $$

If we assume also either (iii) e^− 1/q ≤ v ≤ 1and 0 < p ≤ q ≤ 1or (iv) 1 ≤ u ≤ e^− 1/pand − 1 ≤ p ≤ q <  0, then the above inequality holds.

Proof

For t > 0 and − 1 ≤ p ≤ 1 with p≠ 0, we set the function $f(p,t) \equiv \frac {t^{p}-1}{p} -t^{p}\log t$. Then, we calculate $\frac {df(p,t)}{dp} =\frac {1}{p^{2}}\left (1-t^{p}+t^{p}\log t^{p}- t^{p} \left (\log t^{p}\right )^{2}\right ) \leq 0$. By the use of Lemma 3.1 with x ≡ t^p ≥ 1 for both cases (i) t ≥ 1 and 0 < p ≤ 1 or (ii) t ≤ 1 and − 1 ≤ p < 0, the desired inequality holds. From Lemma 3.1, we also find that $\frac {df(p,t)}{dp}\leq 0$ for $\frac {1}{e}\leq t^{p} \leq 1$ so that the desired inequality holds for both cases (iii) e^− 1/q ≤ v ≤ 1 and 0 < p ≤ q ≤ 1 or (iv) 1 ≤ u ≤ e^− 1/p and − 1 ≤ p ≤ q < 0. □

Lemma 3.3

Define $g(x) \equiv 1-x+x\log x -\frac {1}{2}x (\log x)^{2}$ forx > 0. If 0 < x ≤ 1, then g(x) ≥ 0. If x ≥ 1, then g(x) ≤ 0.

Proof

It is trivial from $\frac {dg(x)}{dx} = -\frac {1}{2}(\log x)^{2}$. □

Proposition 3.4

Let A and B be strictly positive operators such that uA ≤ B ≤ vAwithu,v > 0and let− 1 ≤ p ≤ 1withp≠ 0. If either(i) u ≥ 1and− 1 ≤ p ≤ q < 0or(ii) v ≤ 1and 0 < p ≤ q ≤ 1, then

$$T_{p}(A|B)-\frac{1}{2}S_{p}(A|B) \leq T_{q}(A|B)-\frac{1}{2}S_{q}(A|B). $$

If we have the condition either (iii) u ≥ 1and 0 < p ≤ q ≤ 1or (iv) v ≤ 1and − 1 ≤ p ≤ q < 0,then

$$T_{p}(A|B)-\frac{1}{2}S_{p}(A|B) \geq T_{q}(A|B)-\frac{1}{2}S_{q}(A|B). $$

Proof

We set the function $f(p,t) \equiv \frac {t^{p}-1}{p}-\frac {1}{2}t^{p}\log t$ for t > 0 and − 1 ≤ p ≤ 1 with p≠ 0. Then, we calculate $\frac {df(p,t)}{dp} = \frac {1}{p^{2}}\left (1-t^{p}+t^{p}\log t^{p}-\frac {1}{2}t^{p}\left (\log t^{p}\right )^{2}\right )$. From Lemma 3.3 with x ≡ t^p, we find $\frac {df(p,t)}{dp} \geq 0$ under the condition either (i) t ≥ 1 and − 1 ≤ p < 0 or (ii) 0 < t ≤ 1 and 0 < p ≤ 1. Similarly, from Lemma 3.3 with x ≡ t^p, we find $\frac {df(p,t)}{dp} \leq 0$ under the condition either (iii) t ≥ 1 and 0 < p ≤ 1 or (iv) 0 < t ≤ 1 and − 1 ≤ p < 0. These imply the conclusion of this proposition by the Kubo-Ando theory. □

Comparing Proposition 3.2 and Proposition 3.4, we show slightly more precise results, in a similar way to these propositions. For this purpose, we need the following lemma.

Lemma 3.5

For x > 0and $c \in \mathbb {R}$, we set the function g(x) ≡ 1 − x + x log x − cx(log x)². Then, we have g(x) ≥ 0under the following three conditions (a) 0 < x ≤ 1and $0<c\leq \frac {1}{2}$,(b) $1 \leq x \leq e^{\frac {1-2c}{c}}$ and$0<c\leq \frac {1}{2}$, or (c) x > 0and c ≤ 0. We also have g(x) ≤ 0under the following two conditions (d) $e^{\frac {1-2c}{c}} \leq x \leq 1$ and$\frac {1}{2} \leq c$ or(e) x ≥ 1and $\frac {1}{2} \leq c$.

Proof

Since we have 1 − x + x log x ≥ 0 for x > 0, we have g(x) ≥ 0 for the case (c). From here, we assume c≠ 0. We calculate g^′(x) = (1 − 2c − c log x)log x. Then, we easily have g(x) ≥ 0 for the case (a), and g(x) ≤ 0 for the case (e). As for the case (b), we find g^′(x) ≥ 0 for $1 \leq x \leq e^{\frac {1-2c}{c}}$ so that g(x) ≥ g(1) = 0 for the case (b). As for the case (d), we also find g^′(x) ≥ 0 for $e^{\frac {1-2c}{c}} \leq x \leq 1$ so that g(x) ≤ g(1) = 0 for the case (d). □

Note that $l(c) \equiv g\left (e^{\frac {1-2c}{c}} \right ) = 1+(1-4c)e^{\frac {1-2c}{c}}$ for c≠ 0, then $l^{\prime }(c)=-\left (\frac {1-2c}{c}\right )^{2}e^{\frac {1-2c}{c}} \leq 0$. Thus, we have $l(c)\geq l\left (\frac {1}{2}\right ) = 0$ for $c \leq \frac {1}{2}$, and $l(c)\leq l\left (\frac {1}{2}\right ) = 0$ for $\frac {1}{2}\leq c$.

Proposition 3.6

Let A and B be strictly positive operators such thatuA ≤ B ≤ vAwith u,v > 0,$c \in \mathbb {R}$ and let − 1 ≤ p ≤ 1with p≠ 0. (A) For $0<c \leq \frac {1}{2}$, we have the inequality

$$ T_{p}(A|B)-c S_{p}(A|B) \leq T_{q}(A|B)-c S_{q}(A|B), $$

(3.1)

under the following conditions (a1), (a2), (b1), or (b2).

• (a1) u ≥ 1and − 1 ≤ p ≤ q < 0.

• (a2) v ≤ 1and 0 < p ≤ q ≤ 1.

• (b1) $1 \leq u \leq v \leq e^{\frac {1-2c}{c q}} \left (\leq e^{\frac {1-2c}{c p}}\right )$ and0 < p ≤ q ≤ 1.

• (b2) $\left (e^{\frac {1-2c}{c q}} \leq \right ) e^{\frac {1-2c}{c p}}\leq u \leq v \leq 1$ and− 1 ≤ p ≤ q < 0.

(B) For $c \geq \frac {1}{2}$, wehave the inequality

$$T_{p}(A|B)-c S_{p}(A|B) \geq T_{q}(A|B)-c S_{q}(A|B),</p><p class="noindent">$$

underthe following conditions (d1), (d2), (e1), or (e2).

(d1):

$\left (e^{\frac {1-2c}{c p}} \leq \right ) e^{\frac {1-2c}{c q}} \leq u \leq v \leq 1$ and0 < p ≤ q ≤ 1.

(d2):

$1 \leq u \leq v \leq e^{\frac {1-2c}{c p}} \left (\leq e^{\frac {1-2c}{c q}} \right )$ and− 1 ≤ p ≤ q < 0.

(e1):

u ≥ 1and 0 < p ≤ q ≤ 1.

(e2):

v ≤ 1and − 1 ≤ p ≤ q < 0.

(C) For c ≤ 0and − 1 ≤ p ≤ q ≤ 1with p≠ 0,q≠ 0, wehave the inequality (3.1).

Proof

We set $f(p,t)\equiv \frac {t^{p}-1}{p} -ct^{p}\log t$ for t > 0 and − 1 ≤ p ≤ 1 with p≠ 0. We calculate $\frac {df(p,t)}{dp} =\frac {1}{p^{2}}\left (1-t^{p}+t^{p}\log t^{p}-c t^{p}\left (\log t^{p}\right )^{2}\right )$. From (a), (b) in Lemma 3.5, we have $\frac {df(p,t)}{dp} \geq 0$ under the conditions (a1), (a2), (b1), (b2), or (c). From (d), (e) in Lemma 3.5, we also have $\frac {df(p,t)}{dp} \leq 0$ under the conditions (d1), (d2), (e1), or (e2). Finally, from (c) in Lemma 3.5, we have $\frac {df(p,t)}{dp} \geq 0$ under the conditions (C). Therefore, we have the inequalities in the present proposition. □

4 Monotonicity on the Weight of Operator Means

In this section, along with the previous section, we study the monotonicity of the weight p in weighted mean, since geometric operator mean is used in the definition of the Tsallis relative operator entropy. We review the following inequalities showing the ordering among three p-weighted means.

$$\left\{ (1-p) + pt^{-1} \right\}^{-1} \leq t^{p} \leq (1-p) + p t,\quad (t>0,\,\, 0\leq p \leq 1). $$

We here give the following propositions.

Proposition 4.1

Let A and B be strictly positive operators, and letp,q ∈ (0,1]. Ifp ≤ q, then

$$\frac{A\nabla_{p} B -A\sharp_{p} B}{p} \geq \frac{A\nabla_{q} B -A\sharp_{q} B}{q}.$$

Proof

Since $h(p) : = \frac {x^{p} - 1}{p} -(x-1)$ is increasing function of p for any x > 0, if p ≤ q, then h(p) ≤ h(q) which is

$$ \frac{x^{p} -1 -p(x-1)}{p} \leq \frac{x^{q} -1 -q(x-1)}{q}. $$

(4.1)

By the Kubo-Ando theory [13], we thus have the desired result. □

We can obtain the following results in relation to Proposition 4.1.

Theorem 4.2

Let A and B be strictly positive operators such thatuA ≤ B ≤ vAwith u,v > 0and let p,q ∈ (0,1)with p ≤ q. If v ≤ 1, then

$$ \frac{A\nabla_{p} B -A\sharp_{p} B}{p(1-p)} \leq \frac{A\nabla_{q} B -A\sharp_{q} B}{q(1-q)}. $$

(4.2)

If u ≥ 1,then the reverse inequality in (4.2) holds.

Proof

Since y^p− 2 ≥ y^q− 2 for y ≤ 1 and 0 < p ≤ q < 1, we have

$$0 \leq {{\int}_{x}^{1}} {{{\int}_{t}^{1}} {\left( {{y^{p - 2}} - {y^{q - 2}}} \right)dydt = } } \frac{{\left( {1 - p} \right) + px - {x^{p}}}}{{p\left( {1 - p} \right)}} - \frac{{\left( {1 - q} \right) + qx - {x^{q}}}}{{q\left( {1 - q} \right)}}. $$

Thus, we have the desired result by the Kubo-Ando theory. Since y^p− 2 ≤ y^q− 2 for y ≥ 1 and 0 < p ≤ q < 1, in addition ${{\int }_{1}^{x}} {{{\int }_{1}^{t}} {{y^{p - 2}}dydt = } } {{\int }_{x}^{1}} {{{\int }_{t}^{1}} {{y^{p - 2}}dydt,} }$ we similarly obtain the following opposite inequality

$$0 \geq {{\int}_{1}^{x}} {{{\int}_{1}^{t}} {\left( {{y^{p - 2}} - {y^{q - 2}}} \right)dydt = } } \frac{{\left( {1 - p} \right) + px - {x^{p}}}}{{p\left( {1 - p} \right)}} - \frac{{\left( {1 - q} \right) + qx - {x^{q}}}}{{q\left( {1 - q} \right)}}. $$

which implies the desired result. □

Proposition 4.3

Let A and B be strictly positive operators such that uA ≤ B ≤ vAwithv ≤ 1and let p,q ∈ (0,1]. Ifp ≤ q, then

$$\frac{A\sharp_{p} B-A!_{p} B}{p} \geq \frac{A\sharp_{q} B-A!_{q} B}{q}. $$

Proof

For 0 < p ≤ 1 and 0 < x ≤ 1, we set f(p,x) = f₁(p,x)f₂(p,x) with $f_{1}(p,x) \equiv \frac {x^{p}}{(1-p) +px^{-1}}$ and $f_{2}(p,x) \equiv \frac {1}{p}\left (1-p+px^{-1}-x^{-p}\right )$. Since (1 − p) + pt − t^p ≥ 0 for t > 0 and 0 ≤ p ≤ 1, f₂(p,x) ≥ 0 for 0 < p ≤ 1 and 0 < x ≤ 1. Putting $t=\frac {1}{x}$ in the inequality (4.1), we find that f₂(p,x) is decreasing. Since it is trivial that f₁(p,x) ≥ 0, we finally show $\frac {df_{1}(p,x)}{dp} \leq 0$. We calculate the first derivative of the function f₁(p,x) by p as

$$\frac{df_{1}(p,x)}{dp} = \frac{x^{p} \left\{ \left( 1+\left( x^{-1} -1\right)p\right)\log x -\left( x^{-1} -1\right) \right\} }{ \left\{ (1-p)+px^{-1} \right\}^{2}}. $$

Putting s ≡ x^− 1 − 1 ≥ 0, we have

$$\left( 1+\left( x^{-1} -1\right)p\right)\log x -\left( x^{-1} -1\right) =-(ps+ 1)\log(s + 1) -s \leq 0 $$

which implies $\frac {df_{1}(p,x)}{dp} \leq 0$. Thus, we find that $\frac {df(p,x)}{dp} \leq 0$ for 0 < p ≤ 1 and 0 < x ≤ 1. Therefore, if p ≤ q, then f(p,x) ≥ f(q,x). □

Lemma 4.4

Let t > 0and 0 ≤ p ≤ 1. If either (i) 0 < t ≤ 1and $0 \leq p \leq \frac {1}{2}$ or(ii) t ≥ 1and $\frac {1}{2} \leq p \leq 1$, then $(1-p)+pt \geq \frac {t-1}{\log t}$.

Proof

Both cases are easily proven from

$$(1-p) + pt \geq \frac{t + 1}{2} \geq \frac{t-1}{\log t}. $$

The first inequality is true by $\left (p-\frac {1}{2}\right )(t-1) \geq 0$ and the second inequality holds for t > 0. □

Lemma 4.5

Let x > 0and 0 ≤ p ≤ 1. If x ≥ 1and $0 \leq p \leq \frac {1}{2}$, then $\log x \geq \frac {x-1}{(1-p)x+p} \geq 0$.If 0 < x ≤ 1and $\frac {1}{2} \leq t \leq 1$, then $\log x \leq \frac {x-1}{(1-p)x+p} \leq 0$.

Proof

Put $x = \frac {1}{t}$ in Lemma 4.4. □

Theorem 4.6

Let A and B be strictly positive operators such that uA ≤ B ≤ vAwith u,v > 0and let p,q ∈ (0,1]. If either (i) u ≥ 1and $0 \leq p \leq q \leq \frac {1}{2}$or (ii) v ≤ 1and $\frac {1}{2} \leq p \leq q \leq 1$, then

$$\begin{array}{@{}rcl@{}} &&\frac{A \sharp_{p} B -A!_{p}B}{p} +pA^{1/2}\left( \log A^{-1/2}BA^{-1/2}\right)^{2}A^{1/2}\\ & \leq& \frac{A \sharp_{q} B -A!_{q}B}{q} +qA^{1/2}\left( \log A^{-1/2}BA^{-1/2}\right)^{2}A^{1/2}. \end{array} $$

Proof

We consider the function

$$f(x,p) \equiv \frac{1}{p}\left( x^{p}-\frac{x}{(1-p)x+p} \right) +p (\log x)^{2}. $$

Then, we calculate

$$\begin{array}{@{}rcl@{}} \frac{df(x,p)}{dp} &=&\frac{d}{dp}\left( \frac{x^{p}-1}{p} -\frac{x-1}{(1-p)x +p}\right) + (\log x)^{2}\\ &=&\frac{d}{dp}\left( \frac{x^{p}-1}{p}\right) +(\log x)^{2} -\left( \frac{x-1}{(1-p)x+p}\right)^{2} \geq 0. \end{array} $$

The last inequality is due to Lemma 4.5 and the fact $\frac {d}{dp}\left (\frac {x^{p}-1}{p}\right ) = \frac {x^{p}}{p^{2}} \left (\log x^{p} -1 +\frac {1}{x^{p}} \right )\geq 0$ by log t ≤ t − 1 for t > 0. Thus, we have f(x,p) ≤ f(x,q) under the condition either (i) x ≥ 1 and $0 \leq p \leq q \leq \frac {1}{2}$ or (ii) 0 < x ≤ 1 and $\frac {1}{2} \leq p \leq q \leq 1$. □