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ON LOG-SUM INEQUALITIES

Supriyo Dutta a and Shigeru Furuichi b
a Department of Mathematics,
National Institute of Technology Agartala,
Barjala, Jirania, Tripura, India - 799046;
bDepartment of Information Science,
College of Humanities and Sciences, Nihon University,
3-25-40, Sakurajyousui, Setagaya-Ku, Tokyo, 156-8550, Japan
CONTACT S. Dutta. Email: [email protected], [email protected] S. Furuichi. Email: [email protected]
Abstract

In information theory, the well-known log-sum inequality is a fundamental tool which indicates the non-negativity for the relative entropy. In this article, we establish a set of inequalities which are similar to the log-sum inequality involving two functions defined on scalars. The parametric extended log-sum inequalities are shown. We extend these inequalities for the commutative matrices. In addition, utilizing the Löwner partial order relation and the Hansen-Pedersen theory for non-commutative positive semi-definite matrices we demonstrate a number of matrix-inequalities analogous to the log-sum inequality.

Keywords: log-sum inequality; deformed logarithm; convex function; Löwner partial order; Hansen-Pedersen theory

1 Introduction

In information theory, the so-called log-sum inequality is a generalization of Shannon inequality (often also called Klein inequality) which is used to prove the non-negativity of relative entropy. The essence of the non-negativity of the relative entropy is the simple inequality lnxx1\ln x\leq x-1 for x>0x>0. Therefore, log-sum inequality is important to study information theory. This is a variant of the Jensen inequality of convex functions, which plays a crucial role for proving the Gibbs’ inequality or the convexity of Kullback-Leibler divergence [1].

Recall that, a function f:Xf:X\rightarrow\mathbb{R} defined on a convex set XX is said to be a convex function if for all x1,x2Xx_{1},x_{2}\in X and for all t[0,1]t\in[0,1] we have tf(x1)+(1t)f(x2)f(tx1+(1t)x2)tf(x_{1})+(1-t)f(x_{2})\geq f(tx_{1}+(1-t)x_{2}). When XX is the set of real numbers, this inequality is mentioned as the Jensen inequality. In general, we represent the Jensen inequality [2], [3] as

i=1ntif(xi)f(i=1ntixi),wherei=1nti=1and0ti1.\sum_{i=1}^{n}t_{i}f(x_{i})\geq f\left(\sum_{i=1}^{n}t_{i}x_{i}\right),~{}\text{where}~{}\sum_{i=1}^{n}t_{i}=1~{}\text{and}~{}0\leq t_{i}\leq 1. (1)

A twice-differentiable real valued function f(x)f(x) on \mathbb{R} is convex if and only if f′′(x)0f^{\prime\prime}(x)\geq 0. Let a1,a2,,ana_{1},a_{2},\cdots,a_{n} and b1,b2,,bnb_{1},b_{2},\cdots,b_{n} be non-negative numbers. As f(x)=xlog(x)f(x)=x\log(x) is a convex function, the Jensen inequality (1) suggests

i=1nbij=1nbjaibilog(aibi)(i=1nbij=1nbjaibi)log(i=1nbij=1nbjaibi),\sum_{i=1}^{n}\frac{b_{i}}{\sum\limits_{j=1}^{n}b_{j}}\frac{a_{i}}{b_{i}}\log\left(\frac{a_{i}}{b_{i}}\right)\geq\left(\sum_{i=1}^{n}\frac{b_{i}}{\sum\limits_{j=1}^{n}b_{j}}\frac{a_{i}}{b_{i}}\right)\log\left(\sum_{i=1}^{n}\frac{b_{i}}{\sum\limits_{j=1}^{n}b_{j}}\frac{a_{i}}{b_{i}}\right),

where ti=bij=1nbjt_{i}=\dfrac{b_{i}}{\sum\limits_{j=1}^{n}b_{j}} and xi=aibix_{i}=\dfrac{a_{i}}{b_{i}}. Now, replacing a=j=1naja=\sum\limits_{j=1}^{n}a_{j} and b=j=1nbjb=\sum\limits_{j=1}^{n}b_{j} we obtain,

i=1nailog(aibi)alog(ab),\sum_{i=1}^{n}a_{i}\log\left(\frac{a_{i}}{b_{i}}\right)\geq a\log\left(\frac{a}{b}\right), (2)

which is the standard log-sum inequality. The inequality is valid for n=n=\infty provided a<a<\infty as well as b<b<\infty. In [4], an analog of log-sum inequality is derived as

i=1nbif(aibi)bf(ab),\sum_{i=1}^{n}b_{i}f\left(\frac{a_{i}}{b_{i}}\right)\geq bf\left(\frac{a}{b}\right),

where ff is strictly convex at ab\dfrac{a}{b}. The equality holds if and only if ai=cbia_{i}=cb_{i}, for i=1,2,ni=1,2,\cdots n, and constant number cc.

Although log sum inequality is important in classical information theory, a detailed study on this topic is not available in the literature, to the best of our knowledge. This article demonstrates a number of inequalities which are analogous to the log-sum inequality. First, we extend the log-sum inequality with two real-valued functions, which is illustrated in Theorem 1. This theorem expands the applicability of the log-sum inequality as it breaks the restriction that aia_{i} and bib_{i} are the real numbers. The convexity of a function is essential for proving log-sum inequality. Considering the concavity of a function, we also find analogous results. Investigating the variants of log-sum inequality for concave functions is crucial as there are classes of functions whose convexity depends on specific values of a parameter, for instance, the qq-deformed logarithm. We also elaborate the matrix analogs of the log-sum inequality. Here we observe that all the inequalities derived for the real functions can be easily extended as trace-form-inequality for commuting self-adjoint matrices. A recent trend in matrix analysis is constructing matrix theoretic counterparts of the known inequalities for real functions, where the matrix inequality is driven by the Löwner partial order relation. Proving the log-sum inequality in this context is non-trivial and difficult. We are able to derive it under several conditions. To the best of our knowledge, this is the first attempt in literature to extend the log-sum inequality for the non-commutative positive semidefinite matrices.

This article is distributed into five sections. Preliminary concepts are discussed before applying them in a mathematical derivation. In section 2, we consider inequalities for real-valued functions. Here, we discuss the log-sum inequality for deformed logarithms, for instance the qq-deformed logarithm. Section 3 is dedicated to the discussion of log-sum inequality as a trace-form-inequality for commuting self-adjoint matrices. It has a number of consequences in quantum information theory. In the next section, we attempt to derive the log-sum inequality for Löwner partial order relation. This section extensively uses the idea of operator monotone, operator convex functions, and the operator Jensen inequality, which we mention at the beginning of section 4. Then we conclude the article.

2 log-sum inequalities for real functions

In this section, we extend the idea of the log-sum inequality for real-valued functions. Both concave and convex functions are considered for the investigation. Interestingly, the qq-deformed logarithm is convex when q<2q<2, which assists us to illustrate two sets of log-sum inequalities. We begin our discussion with the following theorem.

Theorem 1.

Let gg be a real valued function whose domain contains a1,a2,,ana_{1},a_{2},\cdots,a_{n} and b1,b2,,bnb_{1},b_{2},\cdots,b_{n}, such that g(bi)>0g(b_{i})>0 for all i=1,2,ni=1,2,\cdots n. Consider another function f:[mg,Mg]f:\left[m_{g},M_{g}\right]\rightarrow\mathbb{R} for which h(x)=xf(x)h(x)=xf(x) is convex, where mg:=mini{g(ai)g(bi)}m_{g}:=\min\limits_{i}\left\{\dfrac{g(a_{i})}{g(b_{i})}\right\} and Mg:=maxi{g(ai)g(bi)}M_{g}:=\max\limits_{i}\left\{\dfrac{g(a_{i})}{g(b_{i})}\right\}. Then,

i=1ng(ai)f(g(ai)g(bi))(i=1ng(ai))f(i=1ng(ai)i=1ng(bi)).\sum_{i=1}^{n}g(a_{i})f\left(\frac{g(a_{i})}{g(b_{i})}\right)\geq\left(\sum_{i=1}^{n}g(a_{i})\right)f\left(\frac{\sum_{i=1}^{n}g(a_{i})}{\sum_{i=1}^{n}g(b_{i})}\right). (3)
Proof.

Define a=i=1ng(ai)a=\sum\limits_{i=1}^{n}g(a_{i}) and b=i=1ng(bi)>0b=\sum\limits_{i=1}^{n}g(b_{i})>0. Note that, ab=i=1ng(bi)bg(ai)g(bi)\dfrac{a}{b}=\sum\limits_{i=1}^{n}\dfrac{g(b_{i})}{b}\dfrac{g(a_{i})}{g(b_{i})}, which is a convex combination of g(ai)g(bi)\dfrac{g(a_{i})}{g(b_{i})} for i=1,2,ni=1,2,\cdots n. Clearly, the ratio ab\dfrac{a}{b} belongs to the convex set [mg,Mg]\left[m_{g},M_{g}\right]. Now,

i=1ng(ai)f(g(ai)g(bi))=i=1ng(bi)g(ai)g(bi)f(g(ai)g(bi))=i=1nbg(bi)bh(g(ai)g(bi)).\sum_{i=1}^{n}g(a_{i})f\left(\frac{g(a_{i})}{g(b_{i})}\right)=\sum_{i=1}^{n}g(b_{i})\frac{g(a_{i})}{g(b_{i})}f\left(\frac{g(a_{i})}{g(b_{i})}\right)=\sum_{i=1}^{n}b\frac{g(b_{i})}{b}h\left(\frac{g(a_{i})}{g(b_{i})}\right).

As g(bi)>0g(b_{i})>0 for all i={1,2,n}i=\{1,2,\cdots n\} and b=i=1ng(bi)b=\sum\limits_{i=1}^{n}g(b_{i}) we have 0g(bi)b10\leq\dfrac{g(b_{i})}{b}\leq 1 and i=1ng(bi)b=1\sum\limits_{i=1}^{n}\dfrac{g(b_{i})}{b}=1. Now, the Jensen inequality indicates

i=1nbg(bi)bh(g(ai)g(bi))bh(i=1ng(bi)bg(ai)g(bi))=bh(1bi=1ng(ai))=bh(ab).\begin{split}\sum_{i=1}^{n}b\frac{g(b_{i})}{b}h\left(\frac{g(a_{i})}{g(b_{i})}\right)\geq bh\left(\sum_{i=1}^{n}\frac{g(b_{i})}{b}\frac{g(a_{i})}{g(b_{i})}\right)=bh\left(\frac{1}{b}\sum_{i=1}^{n}g(a_{i})\right)=bh\left(\frac{a}{b}\right).\end{split} (4)

Expanding hh we get

bh(ab)=babf(ab)=af(ab)=(i=1ng(ai))f(i=1ng(ai)i=1ng(bi)).bh\left(\frac{a}{b}\right)=b\frac{a}{b}f\left(\frac{a}{b}\right)=af\left(\frac{a}{b}\right)=\left(\sum_{i=1}^{n}g(a_{i})\right)f\left(\frac{\sum\limits_{i=1}^{n}g(a_{i})}{\sum\limits_{i=1}^{n}g(b_{i})}\right).

Combining, we obtain the result. ∎

From now on, we denote mg:=mini{g(ai)g(bi)}m_{g}:=\min\limits_{i}\left\{\dfrac{g(a_{i})}{g(b_{i})}\right\} and Mg:=maxi{g(ai)g(bi)}M_{g}:=\max\limits_{i}\left\{\dfrac{g(a_{i})}{g(b_{i})}\right\} throughout the article, where gg is a real valued function whose domain of definition contains a1,a2,,ana_{1},a_{2},\cdots,a_{n} and b1,b2,,bnb_{1},b_{2},\cdots,b_{n}, such that g(bi)>0g(b_{i})>0 for all i=1,2,ni=1,2,\cdots n.

Defining g(x)=xg(x)=x and f(x)=log(x)f(x)=\log(x) for x>0x\in\mathbb{R}_{>0}, we observe that h(x)=xlog(x)h(x)=x\log(x) is a convex function. Then, Theorem 1 leads us to the log-sum inequality mentioned in equation (2).

Theorem 1 expands the scope of applications of the log-sum inequality. The domain of gg may not be a convex set. We can consider aia_{i} and bib_{i} arbitrarily such that g(bi)>0g(b_{i})>0 for all i=1,2,ni=1,2,\cdots n. In other words, the range of gg must have a non-empty intersection with the set of positive reals.

Let ff be a twice-differentiable function in [mg,Mg]\left[m_{g},M_{g}\right]. The function xf(x)xf(x) is convex if and only if

d2dx2(xf(x))=xf′′(x)+2f(x)0.\frac{d^{2}}{dx^{2}}(xf(x))=xf^{\prime\prime}(x)+2f^{\prime}(x)\geq 0. (5)

If mg0m_{g}\geq 0, then any monotone increasing and convex function ff defined on [mg,Mg]\left[m_{g},M_{g}\right] fulfill equation (5). Now, we are in a position to consider a few special cases of Theorem 1.

Example 1.

Define g(x)=xrg(x)=x^{r} for some real parameter rr, and f(x)=log(x)f(x)=\log(x) for x>0x>0. As h(x)=xlog(x)h(x)=x\log(x) is a convex function applying Theorem 1 we observe

i=1nairlog(airbir)(i=1nair)log(i=1nairi=1nbir).\sum_{i=1}^{n}a_{i}^{r}\log\left(\frac{a_{i}^{r}}{b_{i}^{r}}\right)\geq\left(\sum_{i=1}^{n}a_{i}^{r}\right)\log\left(\frac{\sum\limits_{i=1}^{n}a_{i}^{r}}{\sum\limits_{i=1}^{n}b_{i}^{r}}\right).

Thus we have

i=1nrair{log(ai)log(bi)}(i=1nair){log(i=1nair)log(i=1nbir)}.\sum_{i=1}^{n}ra_{i}^{r}\left\{\log(a_{i})-\log(b_{i})\right\}\geq\left(\sum_{i=1}^{n}a_{i}^{r}\right)\left\{\log\left(\sum_{i=1}^{n}a_{i}^{r}\right)-\log\left(\sum_{i=1}^{n}b_{i}^{r}\right)\right\}. (6)
Example 2.

The qq-deformed logarithm or qq-logarithm [5] is defined by

f(x)=lnq(x)=x1q11q,f(x)=\ln_{q}(x)=\frac{x^{1-q}-1}{1-q}, (7)

for x>0x>0 and q1q\neq 1. Note that xf(x)xf(x) is a convex function for q<2q<2 as d2dx2(xf(x))=2qxq0\dfrac{d^{2}}{dx^{2}}\left(xf(x)\right)=\dfrac{2-q}{x^{q}}\geq 0. The division rule of qq-deformed logarithm can be expressed as

lnq(xy)\displaystyle\ln_{q}\left(\frac{x}{y}\right) =\displaystyle= lnq(x)+lnq(1y)+(1q)lnq(x)lnq(1y)\displaystyle\ln_{q}(x)+\ln_{q}\left(\frac{1}{y}\right)+(1-q)\ln_{q}(x)\ln_{q}\left(\frac{1}{y}\right) (8)
=\displaystyle= lnq(x)lnq(y)1+(1q)lnq(y)=lnq(x)lnq(y)y1q.\displaystyle\frac{\ln_{q}(x)-\ln_{q}(y)}{1+(1-q)\ln_{q}(y)}=\frac{\ln_{q}(x)-\ln_{q}(y)}{y^{1-q}}.

Putting g(x)=xrg(x)=x^{r} for x>0x>0 and real parameter rr as well as f(x)=lnq(x)f(x)=\ln_{q}(x) with q<2q<2 in Theorem 1 we obtain

i=1nairlnq(airbir)(i=1nair)lnq(i=1nairi=1nbir),\sum_{i=1}^{n}a_{i}^{r}\ln_{q}\left(\frac{a_{i}^{r}}{b_{i}^{r}}\right)\geq\left(\sum_{i=1}^{n}a_{i}^{r}\right)\ln_{q}\left(\frac{\sum\limits_{i=1}^{n}a_{i}^{r}}{\sum\limits_{i=1}^{n}b_{i}^{r}}\right),

where a1,a2,ana_{1},a_{2},\cdots a_{n} and b1,b2,bnb_{1},b_{2},\cdots b_{n} are positive real numbers. Now applying equation (8), we find

lnq(i=1nairi=1nbir)=lnq(i=1nair)lnq(i=1nbir)(i=1nbir)1q.\ln_{q}\left(\frac{\sum\limits_{i=1}^{n}a_{i}^{r}}{\sum\limits_{i=1}^{n}b_{i}^{r}}\right)=\frac{\ln_{q}\left(\sum\limits_{i=1}^{n}a_{i}^{r}\right)-\ln_{q}\left(\sum\limits_{i=1}^{n}b_{i}^{r}\right)}{\left(\sum\limits_{i=1}^{n}b_{i}^{r}\right)^{1-q}}.

Combining we get

(i=1nbir)1qi=1nairlnq(airbir)(i=1nair){lnq(i=1nair)lnq(i=1nbir)}.\left(\sum_{i=1}^{n}b_{i}^{r}\right)^{1-q}\sum_{i=1}^{n}a_{i}^{r}\ln_{q}\left(\frac{a_{i}^{r}}{b_{i}^{r}}\right)\geq\left(\sum_{i=1}^{n}a_{i}^{r}\right)\left\{\ln_{q}\left(\sum_{i=1}^{n}a_{i}^{r}\right)-\ln_{q}\left(\sum_{i=1}^{n}b_{i}^{r}\right)\right\}. (9)

For r=1r=1 the above inequality reduces to

(i=1nbi)1qi=1nailnq(aibi)(i=1nai){lnq(i=1nai)lnq(i=1nbi)}.\left(\sum_{i=1}^{n}b_{i}\right)^{1-q}\sum_{i=1}^{n}a_{i}\ln_{q}\left(\frac{a_{i}}{b_{i}}\right)\geq\left(\sum_{i=1}^{n}a_{i}\right)\left\{\ln_{q}\left(\sum_{i=1}^{n}a_{i}\right)-\ln_{q}\left(\sum_{i=1}^{n}b_{i}\right)\right\}. (10)

Instead of g(x)=xrg(x)=x^{r}, one may consider trigonometric, exponential, hyperbolic, or any other function to get new inequalities.

Recall that, a real-valued function ff defined on a convex set XX is said to be concave if f(x)-f(x) is convex. Applying it in equation (1) observe that for a concave function ff we have

i=1ntif(xi)f(i=1ntixi),wherei=1nti=1and0ti1.\sum_{i=1}^{n}t_{i}f(x_{i})\leq f\left(\sum_{i=1}^{n}t_{i}x_{i}\right),~{}\text{where}~{}\sum_{i=1}^{n}t_{i}=1~{}\text{and}~{}0\leq t_{i}\leq 1.

Considering h(x)=xf(x)h(x)=xf(x) as a concave function in equation (4), we obtain

i=1ng(ai)f(g(ai)g(bi))(i=1ng(ai))f(i=1ng(ai)i=1ng(bi)),\sum_{i=1}^{n}g(a_{i})f\left(\frac{g(a_{i})}{g(b_{i})}\right)\leq\left(\sum_{i=1}^{n}g(a_{i})\right)f\left(\frac{\sum\limits_{i=1}^{n}g(a_{i})}{\sum\limits_{i=1}^{n}g(b_{i})}\right),

under the equivalent conditions on aia_{i} and bib_{i} as well as the real valued functions ff and gg as mentioned in Theorem 1.

When q>2q>2 in equation (7) we observe that h(x)=xf(x)h(x)=xf(x) is a concave function. Therefore the inequalities in equation (9) and (10) becomes

(i=1nbir)1qi=1nairlnq(airbir)(i=1nair){lnq(i=1nair)lnq(i=1nbir)}.\left(\sum_{i=1}^{n}b_{i}^{r}\right)^{1-q}\sum_{i=1}^{n}a_{i}^{r}\ln_{q}\left(\frac{a_{i}^{r}}{b_{i}^{r}}\right)\leq\left(\sum_{i=1}^{n}a_{i}^{r}\right)\left\{\ln_{q}\left(\sum_{i=1}^{n}a_{i}^{r}\right)-\ln_{q}\left(\sum_{i=1}^{n}b_{i}^{r}\right)\right\}. (11)

and

(i=1nbi)1qi=1nailnq(aibi)(i=1nai){lnq(i=1nai)lnq(i=1nbi)},\left(\sum_{i=1}^{n}b_{i}\right)^{1-q}\sum_{i=1}^{n}a_{i}\ln_{q}\left(\frac{a_{i}}{b_{i}}\right)\leq\left(\sum_{i=1}^{n}a_{i}\right)\left\{\ln_{q}\left(\sum_{i=1}^{n}a_{i}\right)-\ln_{q}\left(\sum_{i=1}^{n}b_{i}\right)\right\},

respectively, under the similar conditions on aia_{i} and bib_{i}.

If f(x)=log(x)f(x)=\log(x) we find that xf(x)=xf(1x)-xf(x)=xf\left(\dfrac{1}{x}\right) is a concave function. The equation (2) suggests that

i=1nailog(aibi)alog(ab)-\sum_{i=1}^{n}a_{i}\log\left(\frac{a_{i}}{b_{i}}\right)\leq-a\log\left(\frac{a}{b}\right)

which implies

i=1nailog(biai)alog(ba).\sum_{i=1}^{n}a_{i}\log\left(\frac{b_{i}}{a_{i}}\right)\leq a\log\left(\frac{b}{a}\right).

In general, xf(x)=xf(1x)-xf(x)=xf\left(\dfrac{1}{x}\right) does not hold. For instance, consider f(x)=lnq(x)f(x)=\ln_{q}(x), which refers xlnq(x)=xqlnq(1x)-x\ln_{q}(x)=x^{q}\ln_{q}\left(\dfrac{1}{x}\right). This fact leads us to a new inequality which we consider in the following theorem.

Theorem 2.

Let gg be a real-valued function whose domain contains a1,a2,,ana_{1},a_{2},\cdots,a_{n} and b1,b2,,bnb_{1},b_{2},\cdots,b_{n}, such that g(ai)>0g(a_{i})>0 for i=1,2,ni=1,2,\cdots n; and f:[m~g,M~g]f:\left[\widetilde{m}_{g},\widetilde{M}_{g}\right]\rightarrow\mathbb{R} be a function for which h(x)=xf(1x)h(x)=xf\left(\frac{1}{x}\right) is a concave function, where m~g=min1in{g(bi)g(ai)}\widetilde{m}_{g}=\min\limits_{1\leq i\leq n}\left\{\dfrac{g(b_{i})}{g(a_{i})}\right\}, and M~g=max1in{g(bi)g(ai)}\widetilde{M}_{g}=\max\limits_{1\leq i\leq n}\left\{\dfrac{g(b_{i})}{g(a_{i})}\right\}. Then,

i=1ng(ai)f(g(bi)g(ai))(i=1ng(ai))f(i=1ng(bi)i=1ng(ai)).\sum_{i=1}^{n}g(a_{i})f\left(\frac{g(b_{i})}{g(a_{i})}\right)\leq\left(\sum_{i=1}^{n}g(a_{i})\right)f\left(\frac{\sum\limits_{i=1}^{n}g(b_{i})}{\sum\limits_{i=1}^{n}g(a_{i})}\right).
Proof.

It is easy to find that,

i=1ng(ai)f(g(bi)g(ai))=i=1ng(bi)g(ai)g(bi)f(g(bi)g(ai))=i=1nbg(bi)bh(g(ai)g(bi))bh(i=1ng(bi)bg(ai)g(bi))[ash(x)=xf(1x)is concave]=bh(1bi=1ng(ai))=bh(ab)=babf(ba)=af(ba)(i=1ng(ai))f(i=1ng(bi)i=1ng(ai)).\begin{split}&\sum_{i=1}^{n}g(a_{i})f\left(\frac{g(b_{i})}{g(a_{i})}\right)=\sum_{i=1}^{n}g(b_{i})\frac{g(a_{i})}{g(b_{i})}f\left(\frac{g(b_{i})}{g(a_{i})}\right)=\sum_{i=1}^{n}b\frac{g(b_{i})}{b}h\left(\frac{g(a_{i})}{g(b_{i})}\right)\\ \leq&bh\left(\sum_{i=1}^{n}\frac{g(b_{i})}{b}\frac{g(a_{i})}{g(b_{i})}\right)\quad\left[\text{as}~{}h(x)=xf\left(\frac{1}{x}\right)~{}\text{is concave}\right]\\ =&bh\left(\frac{1}{b}\sum_{i=1}^{n}g(a_{i})\right)=bh\left(\frac{a}{b}\right)=b\frac{a}{b}f\left(\frac{b}{a}\right)=af\left(\frac{b}{a}\right)\\ \leq&\left(\sum_{i=1}^{n}g(a_{i})\right)f\left(\frac{\sum\limits_{i=1}^{n}g(b_{i})}{\sum\limits_{i=1}^{n}g(a_{i})}\right).\end{split}

From now on, we use m~g=min1in{g(bi)g(ai)}\widetilde{m}_{g}=\min\limits_{1\leq i\leq n}\left\{\dfrac{g(b_{i})}{g(a_{i})}\right\}, and M~g=max1in{g(bi)g(ai)}\widetilde{M}_{g}=\max\limits_{1\leq i\leq n}\left\{\dfrac{g(b_{i})}{g(a_{i})}\right\} throughout this article, where gg is a real valued function whose domain of definition contains a1,a2,,ana_{1},a_{2},\cdots,a_{n} and b1,b2,,bnb_{1},b_{2},\cdots,b_{n}, such that g(ai)>0g(a_{i})>0 for all i=1,2,ni=1,2,\cdots n.

We know that a twice-differentiable function h(x)h(x) is concave if and only if h′′(x)0h^{\prime\prime}(x)\leq 0, which indicates

d2dx2(xf(1x))=1x3f′′(1x)0.\frac{d^{2}}{dx^{2}}\left(xf\left(\frac{1}{x}\right)\right)=\frac{1}{x^{3}}f^{\prime\prime}\left(\frac{1}{x}\right)\leq 0. (12)

Note that, equation (12) and (5) are not equivalent, in general. For example, consider the function f(x)=xx2+2f(x)=\frac{x}{x^{2}+2}, where x>0x>0. We can calculate

f(x)=2x2(x2+2)2,andf′′(x)=2x(x26)(x2+2)3.f^{\prime}(x)=\frac{2-x^{2}}{\left(x^{2}+2\right)^{2}},~{}\text{and}~{}f^{\prime\prime}(x)=\frac{2x\left(x^{2}-6\right)}{\left(x^{2}+2\right)^{3}}.

Now, equation (12) suggests that xf(1x)xf\left(\dfrac{1}{x}\right) is concave when

1x3f′′(1x)=2x2(16x2)(2x2+1)30that isx16.\frac{1}{x^{3}}f^{\prime\prime}\left(\frac{1}{x}\right)=\frac{2x^{2}(1-6x^{2})}{\left(2x^{2}+1\right)^{3}}\leq 0~{}\text{that is}~{}x\geq\frac{1}{\sqrt{6}}.

But, putting the values of ff^{\prime} and f′′f^{\prime\prime} in equation (5) we observe that xf(x)xf(x) is concave when

xf′′(x)+2f(x)=812x2(x2+2)30whenx23.xf^{\prime\prime}(x)+2f^{\prime}(x)=\frac{8-12x^{2}}{\left(x^{2}+2\right)^{3}}\leq 0~{}\text{when}~{}x\geq\sqrt{\frac{2}{3}}.

Now, replacing f(x)=xx2+2f(x)=\dfrac{x}{x^{2}+2} and g(x)=xg(x)=x in Theorem 2 we observe

i=1nai2bi2ai2+bi2(i=1nai)2(i=1nbi)2(i=1nai)2+(i=1nbi)2,\sum_{i=1}^{n}\frac{a_{i}^{2}b_{i}}{2a_{i}^{2}+b_{i}^{2}}\leq\frac{\left(\sum\limits_{i=1}^{n}a_{i}\right)^{2}\left(\sum\limits_{i=1}^{n}b_{i}\right)}{2\left(\sum\limits_{i=1}^{n}a_{i}\right)^{2}+\left(\sum\limits_{i=1}^{n}b_{i}\right)^{2}},

where aia_{i} and bib_{i} are greater than 16\dfrac{1}{\sqrt{6}}.

3 log-sum inequalities for commuting self-adjoint matrices

We begin this section with a number of basic concepts of matrix analysis. Given any self-adjoint matrix AA, there exists a unitary matrix UU, such that, A=UΛ(A)UA=U\Lambda(A)U^{\dagger}, where Λ(A)=diag{a1,a2,,an}\Lambda(A)=\operatorname{diag}\{a_{1},a_{2},\cdots,a_{n}\} is a diagonal matrix whose diagonal entries, aia_{i} for i=1,2,,ni=1,2,\cdots,n, are the eigenvalues of AA. Throughout this article UU^{\dagger} denotes the conjugate transpose of the matrix UU. If the matrices AA and BB commute, that is AB=BAAB=BA, then there exists a unitary matrix UU, such that, A=UΛ(A)UA=U\Lambda(A)U^{\dagger} and B=UΛ(B)UB=U\Lambda(B)U^{\dagger}, hold simultaneously. We also denote a=trace(A)=i=1naia=\operatorname{trace}(A)=\sum\limits_{i=1}^{n}a_{i}. Let ff be a continuous real valued function defined on an interval JJ and AA be a self-adjoint matrix with eigenvalues in JJ, then

f(A)=Udiag{f(ai):i=1,2,,n}U.f(A)=U\operatorname{diag}\{f(a_{i}):i=1,2,\cdots,n\}U^{\dagger}. (13)

Now, we demonstrate the results derived in Section 2 for the self-adjoint matrices.

Theorem 3.

Let AA and BB be commuting self-adjoint matrices, with the sets of eigenvalues Λ(A)={a1,a2,,an}\Lambda(A)=\{a_{1},a_{2},\cdots,a_{n}\} and Λ(B)={b1,b2,,bn}\Lambda(B)=\{b_{1},b_{2},\cdots,b_{n}\}, respectively. Also, let g:Λ(A)Λ(B)g:\Lambda(A)\cup\Lambda(B)\rightarrow\mathbb{R} be a function, such that, g(bi)>0g(b_{i})>0 for all i=1,2,ni=1,2,\cdots n. In addition, f:[mg,Mg]f:\left[m_{g},M_{g}\right]\rightarrow\mathbb{R} is a function for which xf(x)xf(x) is convex. Then,

trace[g(A)f(g(A)(g(B))1)]trace[g(A)]f(trace(g(A))trace(g(B))).\operatorname{trace}[g(A)f(g(A)(g(B))^{-1})]\geq\operatorname{trace}[g(A)]f\left(\frac{\operatorname{trace}(g(A))}{\operatorname{trace}(g(B))}\right).
Proof.

As AA and BB are commuting self-adjoint matrices there exists an unitary matrix UU, such that,

g(A)=Udiag{g(ai):i=1,2,,n}U,g(B)=Udiag{g(bi):i=1,2,,n}U,and(g(B))1=Udiag{1g(bi):i=1,2,,n}U,\begin{split}&g(A)=U\operatorname{diag}\{g(a_{i}):i=1,2,\cdots,n\}U^{\dagger},\\ &g(B)=U\operatorname{diag}\{g(b_{i}):i=1,2,\cdots,n\}U^{\dagger},\\ \text{and}~{}&(g(B))^{-1}=U\operatorname{diag}\left\{\frac{1}{g(b_{i})}:i=1,2,\cdots,n\right\}U^{\dagger},\end{split}

as g(bi)>0g(b_{i})>0 for all i=1,2,ni=1,2,\cdots n. Therefore,

f(g(A)(g(B))1)=Udiag{f(g(ai)g(bi)):i=1,2,,n}U,f(g(A)(g(B))^{-1})=U\operatorname{diag}\left\{f\left(\frac{g(a_{i})}{g(b_{i})}\right):i=1,2,\cdots,n\right\}U^{\dagger},

which indicates

g(A)f(g(A)(g(B))1)=Udiag{g(ai)f(g(ai)g(bi)):i=1,2,n}U.g(A)f(g(A)(g(B))^{-1})=U\operatorname{diag}\left\{g(a_{i})f\left(\frac{g(a_{i})}{g(b_{i})}\right):i=1,2,\cdots n\right\}U^{\dagger}.

Note that, g(ai)f(g(ai)g(bi))g(a_{i})f\left(\dfrac{g(a_{i})}{g(b_{i})}\right) are the eigenvalues of the matrix g(A)f(g(A)(g(B))1)g(A)f(g(A)(g(B))^{-1}) for i=1,2,ni=1,2,\cdots n. Hence,

trace(g(A)f(g(A)(g(B))1))=i=1ng(ai)f(g(ai)g(bi)).\operatorname{trace}(g(A)f(g(A)(g(B))^{-1}))=\sum_{i=1}^{n}g(a_{i})f\left(\frac{g(a_{i})}{g(b_{i})}\right).

Also, i=1ng(ai)=trace(g(A))\sum\limits_{i=1}^{n}g(a_{i})=\operatorname{trace}(g(A)) and i=1ng(bi)=trace(g(B))\sum\limits_{i=1}^{n}g(b_{i})=\operatorname{trace}(g(B)). Applying Theorem 1 we obtain

i=1ng(ai)f(g(ai)g(bi))=(i=1ng(ai))f(i=1ng(ai)i=1ng(bi))=trace[g(A)]f(trace(g(A))trace(g(B))).\sum_{i=1}^{n}g(a_{i})f\left(\frac{g(a_{i})}{g(b_{i})}\right)=\left(\sum_{i=1}^{n}g(a_{i})\right)f\left(\frac{\sum\limits_{i=1}^{n}g(a_{i})}{\sum\limits_{i=1}^{n}g(b_{i})}\right)=\operatorname{trace}[g(A)]f\left(\frac{\operatorname{trace}(g(A))}{\operatorname{trace}(g(B))}\right).

Combining we get the result. ∎

The following corollary holds trivially from the above theorem.

Corollary 1.

Given positive definite, commuting matrices AA and BB,

trace(exp(Alog(A)))trace(exp(Alog(B)))trace(A)log(trace(A)trace(B)).\operatorname{trace}(\exp(A\log(A)))-\operatorname{trace}(\exp(A\log(B)))\geq\operatorname{trace}(A)\log\left(\frac{\operatorname{trace}(A)}{\operatorname{trace}(B)}\right).
Proof.

Consider f(x)=log(x)f(x)=\log(x) and g(x)=xg(x)=x in Theorem 3. As AA and BB are positive definite we have ai>0a_{i}>0 and bi>0b_{i}>0 for all i=1,2,,ni=1,2,\cdots,n. Note that,

i=1nailogaibi=i=1nlog(aiai)i=1nlog(biai)=trace(exp(Alog(A)))trace(exp(Alog(B))).\sum_{i=1}^{n}a_{i}\log{\frac{a_{i}}{b_{i}}}=\sum_{i=1}^{n}\log\left(a_{i}^{a_{i}}\right)-\sum_{i=1}^{n}\log\left(b_{i}^{a_{i}}\right)=\operatorname{trace}(\exp(A\log(A)))-\operatorname{trace}(\exp(A\log(B))).

Also, applying i=1nai=trace(A)\sum\limits_{i=1}^{n}a_{i}=\operatorname{trace}(A) and i=1nbi=trace(B)\sum\limits_{i=1}^{n}b_{i}=\operatorname{trace}(B) we observe that

trace[g(A)]f(trace(g(A))trace(g(B)))=trace(A)log(trace(A)trace(B)).\operatorname{trace}[g(A)]f\left(\dfrac{\operatorname{trace}(g(A))}{\operatorname{trace}(g(B))}\right)=\operatorname{trace}(A)\log\left(\dfrac{\operatorname{trace}(A)}{\operatorname{trace}(B)}\right).

Combining we get the result. ∎

Theorem 3 is a matrix-theoretic counterpart of Theorem 1, which assists us to establish a number of matrix inequalities, which are derived in Section 2 for real functions. Some of these matrix inequalities have immediate consequences in quantum information theory. Consider the following example.

Example 3.

The matrix counterpart of equation (6) is represented as

trace[Arlog(ArBr)]trace(Ar)[log(trace(Ar))log(trace(Br))],\operatorname{trace}\left[A^{r}\log(A^{r}B^{-r})\right]\geq\operatorname{trace}(A^{r})[\log(\operatorname{trace}(A^{r}))-\log(\operatorname{trace}(B^{r}))], (14)

where AA and BB are self-adjoint commuting matrices as well as BB is positive definite. Putting r=1r=1 in the above inequality we have

trace[Alog(AB1)]trace(A)[log(trace(A))log(trace(B))].\operatorname{trace}\left[A\log(AB^{-1})\right]\geq\operatorname{trace}(A)[\log(\operatorname{trace}(A))-\log(\operatorname{trace}(B))].

Considering AB=BAAB=BA, we can prove that log(AB1)=log(A)log(B)\log(AB^{-1})=\log(A)-\log(B). Replacing it at the left hand side [6], [7, Theorem 3.3]

trace[A(log(A)log(B))]trace(A)[log(trace(A))log(trace(B))].\operatorname{trace}\left[A(\log(A)-\log(B))\right]\geq\operatorname{trace}(A)[\log(\operatorname{trace}(A))-\log(\operatorname{trace}(B))]. (15)

Recall that, in quantum information theory [8] a quantum state is represented by a density matrix ρ\rho which is a positive semidefinite, self-adjoint matrix with trace(ρ)=1\operatorname{trace}(\rho)=1. The von-Neumann entropy is a well-known measure of quantum information which is given by trace(ρlog(ρ))-\operatorname{trace}(\rho\log(\rho)) for a density matrix ρ\rho. Given two density matrices ρ\rho and σ\sigma the quantum relative entropy is determined by D(ρ||σ)=trace[ρ(log(ρ)log(σ))]D(\rho||\sigma)=\operatorname{trace}\left[\rho(\log(\rho)-\log(\sigma))\right].

Considering trace(A)=trace(B)=1\operatorname{trace}(A)=\operatorname{trace}(B)=1 in equation (15), we find that AA and BB are density matrices. Then, the quantum relative entropy given two density matrices AA and BB,

D(A||B)=trace[A(log(A)log(B))]0.D(A||B)=\operatorname{trace}\left[A(\log(A)-\log(B))\right]\geq 0.

Considering B=IB=I, the identity matrix of order nn, in equation (14) we observe that

trace[ArlogAr]trace(Ar)[log(trace(Ar))log(n)].\operatorname{trace}[A^{r}\log A^{r}]\geq\operatorname{trace}(A^{r})[\log(\operatorname{trace}(A^{r}))-\log(n)].

Putting r=1r=1 we have trace[AlogA]trace(A)[log(trace(A))log(n)]\operatorname{trace}[A\log A]\geq\operatorname{trace}(A)[\log(\operatorname{trace}(A))-\log(n)]. In addition, if AA is a density matrix, that is, trace(A)=1\operatorname{trace}(A)=1 then the above inequality indicates trace[AlogA]log(n)\operatorname{trace}[A\log A]\geq-\log(n). The von-Neumann entropy of AA is trace[AlogA]log(n)-\operatorname{trace}[A\log A]\leq\log(n), which is its maximum value.

Example 4.

The matrix counterpart of equation (9) will be given by

(trace(Br))1qtrace[Arlnq(ArBr)](trace(Ar))[lnq(trace(Ar))lnq(trace(Br))],\left(\operatorname{trace}(B^{r})\right)^{1-q}\operatorname{trace}\left[A^{r}\ln_{q}\left(A^{r}B^{-r}\right)\right]\geq\left(\operatorname{trace}(A^{r})\right)\left[\ln_{q}\left(\operatorname{trace}(A^{r})\right)-\ln_{q}\left(\operatorname{trace}(B^{r})\right)\right],

where AA and BB are self-adjoint commuting matrices as well as BB is positive definite. For r=1r=1 we have

(trace(B))1qtrace[Alnq(AB1)](trace(A))[lnq(trace(A))lnq(trace(B))].\left(\operatorname{trace}(B)\right)^{1-q}\operatorname{trace}\left[A\ln_{q}\left(AB^{-1}\right)\right]\geq\left(\operatorname{trace}(A)\right)\left[\ln_{q}\left(\operatorname{trace}(A)\right)-\ln_{q}\left(\operatorname{trace}(B)\right)\right].

Note that in the above two inequalities we consider q<2q<2 to ensure convexity of xlnq(x)x\ln_{q}(x). For q>2q>2 equation (11) suggests

(trace(Br))1qtrace[Arlnq(ArBr)](trace(Ar))[lnq(trace(Ar))lnq(trace(Br))],\left(\operatorname{trace}(B^{r})\right)^{1-q}\operatorname{trace}\left[A^{r}\ln_{q}\left(A^{r}B^{-r}\right)\right]\leq\left(\operatorname{trace}(A^{r})\right)\left[\ln_{q}\left(\operatorname{trace}(A^{r})\right)-\ln_{q}\left(\operatorname{trace}(B^{r})\right)\right],

for self-adjoint commuting matrices AA and BB as well as positive definite matrix BB.

Theorem 2 also indicates a trace inequality, which we mention below without a proof.

Theorem 4.

Let AA and BB be commuting self-adjoint matrices, with the sets of eigenvalues Λ(A)={a1,a2,,an}\Lambda(A)=\{a_{1},a_{2},\cdots,a_{n}\} and Λ(B)={b1,b2,,bn}\Lambda(B)=\{b_{1},b_{2},\cdots,b_{n}\}, respectively. Also, let g:Λ(A)Λ(B)g:\Lambda(A)\cup\Lambda(B)\rightarrow\mathbb{R} be a function, such that, g(bi)>0g(b_{i})>0 for all i=1,2,ni=1,2,\cdots n. In addition, f:[m~g,M~g]f:\left[\widetilde{m}_{g},\widetilde{M}_{g}\right]\rightarrow\mathbb{R} is a function for which h(x)=xf(1x)h(x)=xf\left(\dfrac{1}{x}\right) is a concave. Then,

trace[g(A)f(g(A)(g(B))1)]trace[g(A)]f(trace(g(A))trace(g(B))).\operatorname{trace}[g(A)f(g(A)(g(B))^{-1})]\leq\operatorname{trace}[g(A)]f\left(\frac{\operatorname{trace}(g(A))}{\operatorname{trace}(g(B))}\right).

4 log-sum inequalities with Löwner partial order

Recall that given self-adjoint matrices AA and BB we write ABA\geq B or BAB\leq A if the matrix ABA-B is positive semidefinite. This ordering is called the Löwner partial order. If AA is positive definite we write A>0A>0.

Let Mn(J)M_{n}(J) be the set of all complex matrices of order nn with eigenvalues in JJ\subset\mathbb{R}. Equation (13) indicates that we can redefine a function f:Jf:J\rightarrow\mathbb{R} as a matrix function f:Mn(J)Mn()f:M_{n}(J)\rightarrow M_{n}(\mathbb{C}), the set of all complex matrices of order nn. Let AA and BB be two self-adjoint matrices of order nn with eigenvalues in JJ. Now, a continuous function f:Jf:J\rightarrow\mathbb{R} is said to be matrix monotone of order nn if ABA\leq B implies f(A)f(B)f(A)\leq f(B). Also, ff is said to be a matrix convex function of order nn if

f(λA+(1λ)B)λf(A)+(1λ)f(B),f(\lambda A+(1-\lambda)B)\leq\lambda f(A)+(1-\lambda)f(B),

for all λ[0,1]\lambda\in[0,1] and every pair of matrices AA, and BB of order nn. An operator monotone function is a matrix monotone function for any natural number nn. Similarly, an operator convex function is a matrix convex function for any natural number nn. A function ff is operator concave if f-f is operator convex. Well known operator monotone functions are f1(t)=trf_{1}(t)=t^{r} for 0<r10<r\leq 1 for t[0,)t\in[0,\infty), as well as f2(t)=log(t)f_{2}(t)=\log(t) for t(0,)t\in(0,\infty). In addition, g(t)=trg(t)=t^{r} is operator convex on (0,)(0,\infty) for 1r0-1\leq r\leq 0 and 1r21\leq r\leq 2 [9, 10]. The operator Jensen inequality [11, 12] plays a crucial role for further development, which we mention below:

Lemma 1.

([11]) If ff is a continuous, real function defined on an interval [0,α)[0,\alpha) with α\alpha\leq\infty, the following conditions are equivalent.

  1. 1.

    ff is operator convex and f(0)0f(0)\leq 0.

  2. 2.

    f(AXA)Af(X)Af(A^{\dagger}XA)\leq A^{\dagger}f(X)A for all AA with A1||A||\leq 1 and every self-adjoint XX with spectrum in [0,α)[0,\alpha).

  3. 3.

    f(AXA+BYB)Af(X)A+Bf(Y)Bf(A^{\dagger}XA+B^{\dagger}YB)\leq A^{\dagger}f(X)A+B^{\dagger}f(Y)B for all A,BA,B with AA+BBIA^{\dagger}A+B^{\dagger}B\leq I and all X,YX,Y with spectrum in [0,α)[0,\alpha).

  4. 4.

    f(PXP)Pf(X)Pf(PXP)\leq Pf(X)P for every projection PP and every self-adjoint XX with spectrum in [0,α)[0,\alpha).

We utilize Dirac’s bra-ket notation [13] for simplifying our notations. Recall that |u\ket{u} denotes a column vector of length nn, or equivalently an n×1n\times 1 matrix. The conjugate transpose of |u\ket{u} is given by u|\bra{u}, which is a row vector of length nn. Note that, |uu|\ket{u}\bra{u} is a self-adjoint matrix of order nn. Now the spectral decomposition of any self-adjoint matrix AA can be written as A=i=1nλi|uiui|A=\sum\limits_{i=1}^{n}\lambda_{i}\ket{u_{i}}\bra{u_{i}} where |ui\ket{u_{i}} is a normalized eigenvector corresponding to the eigenvalue λi\lambda_{i}.

The superadditivity of a mean has been shown in [14] and the corresponding advanced results for a mean and a perspective have been studied in [15, 16]. The following proposition is proven by the elementary calculations under a certain condition.

Proposition 1.

Let A1,A2,AmA_{1},A_{2},\cdots A_{m} and B1,B2,BmB_{1},B_{2},\cdots B_{m} be a two sets of positive definite matrices of order nn, and AmIA\geq mI, where i=1mAi=:A\sum\limits_{i=1}^{m}A_{i}=:A. Also, let ff be an operator concave function on an interval [0,α)[0,\alpha) containing the eigenvalues of BiB_{i} and B=i=1mBiB=\sum\limits_{i=1}^{m}B_{i}. Then

i=1mAi12f(Ai12BiAi12)Ai12A12f(A12BA12)A12.\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}f\left(A_{i}^{-\frac{1}{2}}B_{i}A_{i}^{-\frac{1}{2}}\right)A_{i}^{\frac{1}{2}}\leq A^{\frac{1}{2}}f\left(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\right)A^{\frac{1}{2}}.
Proof.

As each BiB_{i} is a self-adjoint operator, the spectral decomposition of BiB_{i} can be written as

Bi=k=1nλk(i)|uk(i)uk(i)|,B_{i}=\sum_{k=1}^{n}\lambda_{k}^{(i)}\ket{u_{k}^{(i)}}\bra{u_{k}^{(i)}},

where |uk(i)\ket{u_{k}^{(i)}} is an eigenvector associated to the eigenvalue λk(i)\lambda_{k}^{(i)} of BiB_{i}. Then we have

Ai1/2BiAi1/2=k=1nλk(i)|wk(i)wk(i)|,A_{i}^{-1/2}B_{i}A_{i}^{-1/2}=\sum_{k=1}^{n}\lambda_{k}^{(i)}|w_{k}^{(i)}\rangle\langle w_{k}^{(i)}|,

where |wk(i):=Ai1/2|uk(i)|w_{k}^{(i)}\rangle:=A_{i}^{-1/2}|u_{k}^{(i)}\rangle. Now for a function ff we have

f(Ai1/2BiAi1/2)=k=1nf(λk(i))|wk(i)wk(i)|.f\left(A_{i}^{-1/2}B_{i}A_{i}^{-1/2}\right)=\sum_{k=1}^{n}f\left(\lambda_{k}^{(i)}\right)|w_{k}^{(i)}\rangle\langle w_{k}^{(i)}|.

Combining we get

Ai1/2f(Ai1/2BiAi1/2)Ai1/2=k=1nf(λk(i))Ai1/2|wk(i)wk(i)|Ai1/2=k=1nf(λk(i))|uk(i)uk(i)|=f(Bi).\begin{split}A_{i}^{1/2}f\left(A_{i}^{-1/2}B_{i}A_{i}^{-1/2}\right)A_{i}^{1/2}=&\sum_{k=1}^{n}f\left(\lambda_{k}^{(i)}\right)A_{i}^{1/2}|w_{k}^{(i)}\rangle\langle w_{k}^{(i)}|A_{i}^{1/2}\\ =&\sum_{k=1}^{n}f\left(\lambda_{k}^{(i)}\right)|u_{k}^{(i)}\rangle\langle u_{k}^{(i)}|=f\left(B_{i}\right).\end{split}

Summing over ii we find

i=1mAi1/2f(Ai1/2BiAi1/2)Ai1/2=i=1mf(Bi).\sum_{i=1}^{m}A_{i}^{1/2}f\left(A_{i}^{-1/2}B_{i}A_{i}^{-1/2}\right)A_{i}^{1/2}=\sum_{i=1}^{m}f(B_{i}). (16)

If AmIA\geq mI, then we have i=1mA12A12=mA1I\sum_{i=1}^{m}A^{-\frac{1}{2}}A^{-\frac{1}{2}}=mA^{-1}\leq I. Now, applying Lemma 1 we find that

A12i=1mf(Bi)A12=i=1mA12f(Bi)A12f(i=1mA12BiA12)=f(A12i=1mBiA12)=f(A12BA12).\begin{split}A^{-\frac{1}{2}}\sum_{i=1}^{m}f(B_{i})A^{-\frac{1}{2}}&=\sum_{i=1}^{m}A^{-\frac{1}{2}}f(B_{i})A^{-\frac{1}{2}}\leq f\left(\sum_{i=1}^{m}A^{-\frac{1}{2}}B_{i}A^{-\frac{1}{2}}\right)\\ &=f\left(A^{-\frac{1}{2}}\sum_{i=1}^{m}B_{i}A^{-\frac{1}{2}}\right)=f\left(A^{-\frac{1}{2}}BA^{-\frac{1}{2}}\right).\end{split}

Multiplying A12A^{\frac{1}{2}} in both side of the above ineuality we get

i=1mf(Bi)A12f((A12)BA12)A12.\sum_{i=1}^{m}f(B_{i})\leq A^{\frac{1}{2}}f\left((A^{-\frac{1}{2}})^{\dagger}BA^{-\frac{1}{2}}\right)A^{\frac{1}{2}}.

Putting the value of f(Bi)f(B_{i}) from equation (16) we find the result. ∎

Remark 1.

If AiA_{i} and BiB_{i} are replaced by positive real numbers aia_{i} and bib_{i} respectively in the above theorem, we get

i=1mai12f(ai12biai12)ai12a12f(a12ba12)a12,\sum_{i=1}^{m}a_{i}^{\frac{1}{2}}f\left(a_{i}^{-\frac{1}{2}}b_{i}a_{i}^{-\frac{1}{2}}\right)a_{i}^{\frac{1}{2}}\leq a^{\frac{1}{2}}f\left(a^{-\frac{1}{2}}ba^{-\frac{1}{2}}\right)a^{\frac{1}{2}},

where a=i=1maia=\sum\limits_{i=1}^{m}a_{i} and b=i=1mbib=\sum\limits_{i=1}^{m}b_{i} as well as ff is a convex function. Simplifying we get

i=1maif(biai)af(ba).\sum_{i=1}^{m}a_{i}f\left(\frac{b_{i}}{a_{i}}\right)\leq af\left(\frac{b}{a}\right). (17)

Considering f(x)=log(x)f(x)=\log(x) we get the usual log-sum inequality.

Corollary 2.

Let A1,A2,,AmA_{1},A_{2},\cdots,A_{m} and B1,B2,,BmB_{1},B_{2},\cdots,B_{m} be two sets of positive definite self-adjoint operators with A=i=1mAiA=\sum\limits_{i=1}^{m}A_{i} and B=i=1mBiB=\sum\limits_{i=1}^{m}B_{i}, such that, A=BA=B. Also, let ff be an operator concave function on an interval [0,α)[0,\alpha) containing the eigenvalues of BiB_{i} and BB as well as f(1)=0f(1)=0. Then

i=1mAi12f(Ai12BiAi12)Ai120.\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}f(A_{i}^{-\frac{1}{2}}B_{i}A_{i}^{-\frac{1}{2}})A_{i}^{\frac{1}{2}}\leq 0.
Proof.

The proof follows trivially from Proposition 1. ∎

Remark 2.

The operator Shannon inequality was given in [17]

i=1mAi12log(Ai12BiAi12)Ai120\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\log\left(A_{i}^{-\frac{1}{2}}B_{i}A_{i}^{-\frac{1}{2}}\right)A_{i}^{\frac{1}{2}}\leq 0

under the assumption i=1mAi=i=1nBi=I\sum\limits_{i=1}^{m}A_{i}=\sum\limits_{i=1}^{n}B_{i}=I. Our condition in Corollary 2 is slightly weaker than this assumption. We also observe that the operator Shannon inequality holds for any operator concave function ff.

If every AiA_{i} is expansive (i.e., AiIA_{i}\geq I), then the condition AmIA\geq mI in Proposition 1 is satisfied. However, we have not obtained a proper result for contractive condition such as AiIA_{i}\leq I. Closing this section, we present a result which does not impose an additional condition on the matrices AiA_{i}. We need the following known facts for proving the next theorem:

Lemma 2.

([18]) Let XX and AA be bounded linear operators on a Hilbert space \mathcal{H}. Suppose that X0X\geq 0 and A1||A||\leq 1. If ff is an operator monotone function defined on [0,)[0,\infty), then Af(X)Af(AXA)A^{\dagger}f(X)A\leq f(A^{\dagger}XA).

Lemma 3.

([9, p.14]) For any square matrix XiX_{i} and positive definite matrices AiA_{i} we have

i=1mXiAi1Xi(i=1mXi)(i=1mAi)1(i=1mXi).\sum_{i=1}^{m}X^{\dagger}_{i}A_{i}^{-1}X_{i}\geq\left(\sum_{i=1}^{m}X_{i}\right)^{\dagger}\left(\sum_{i=1}^{m}A_{i}\right)^{-1}\left(\sum_{i=1}^{m}X_{i}\right).
Theorem 5.

Let A1,A2,,AmA_{1},A_{2},\cdots,A_{m} and B1,B2,,BmB_{1},B_{2},\cdots,B_{m} be two sets of positive definite matrices, as well as A=i=1mAiA=\sum\limits_{i=1}^{m}A_{i} and B=i=1mBiB=\sum\limits_{i=1}^{m}B_{i}. Also, ff is an operator monotone function defined on [0,)[0,\infty). Then we have

1m(i=1mBi12)[i=1mf(Bi12Ai1Bi12)]1(i=1mBi12)B12[f(B12A1B12)]1B12,\frac{1}{m}\left(\sum_{i=1}^{m}B_{i}^{\frac{1}{2}}\right)\left[\sum_{i=1}^{m}f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}\left(\sum_{i=1}^{m}B_{i}^{\frac{1}{2}}\right)\leq B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}, (18)

and

i=1mAi12Bi12[f(Bi12Ai1Bi12)]1Bi12Ai12\displaystyle\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}B_{i}^{\frac{1}{2}}\left[f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}B_{i}^{\frac{1}{2}}A_{i}^{\frac{1}{2}}
1m(i=1mAi12)B12[f(B12A1B12)]1B12(i=1mAi12).\displaystyle\leq\frac{1}{m}\left(\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\right)B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}\left(\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\right). (19)
Proof.

We can write

Bi12Ai1Bi12=Bi12(B12B12)Ai1(B12B12)Bi12=Bi12B12(B12Ai1B12)B12Bi12.B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}=B_{i}^{\frac{1}{2}}\left(B^{-\frac{1}{2}}B^{\frac{1}{2}}\right)A_{i}^{-1}\left(B^{\frac{1}{2}}B^{-\frac{1}{2}}\right)B_{i}^{\frac{1}{2}}=B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}\left(B^{\frac{1}{2}}A_{i}^{-1}B^{\frac{1}{2}}\right)B^{-\frac{1}{2}}B_{i}^{\frac{1}{2}}.

We have AAiA\geq A_{i}, that is A1Ai1A^{-1}\leq A_{i}^{-1}. Also for any complex square matrix XX we have XA1XXAi1XX^{\dagger}A^{-1}X\leq X^{\dagger}A_{i}^{-1}X. Applying these together we obtain

Bi12Ai1Bi12Bi12B12(B12A1B12)B12Bi12.B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\geq B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)B^{-\frac{1}{2}}B_{i}^{\frac{1}{2}}.

As f(t)f(t) is a matrix monotone function, we can write

f(Bi12Ai1Bi12)f(Bi12B12(B12A1B12)B12Bi12).f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\geq f\left(B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)B^{-\frac{1}{2}}B_{i}^{\frac{1}{2}}\right).

From B>BiB>B_{i} for all i=1,2,,ni=1,2,\cdots,n, we can see

I>B12BiB12=(Bi12B12)Bi12B12I>B^{-\frac{1}{2}}B_{i}B^{-\frac{1}{2}}=\left(B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}\right)^{\dagger}B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}

which implies 1>(Bi12B12)Bi12B12=Bi12B1221>||\left(B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}\right)^{\dagger}B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}||=||B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}||^{2}, since A=AA12||A||=||A^{\dagger}A||^{\frac{1}{2}} for every operator AA in general. Thus we have Bi12B12<1||B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}||<1. We also have B12Bi121\|B^{-\frac{1}{2}}B_{i}^{\frac{1}{2}}\|\leq 1 so that we have the following inequality by Lemma 2,

f(Bi12B12(B12A1B12)B12Bi12)(Bi12B12)f(B12A1B12)(B12Bi12).f\left(B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)B^{-\frac{1}{2}}B_{i}^{\frac{1}{2}}\right)\geq\left(B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}\right)f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\left(B^{-\frac{1}{2}}B_{i}^{\frac{1}{2}}\right).

That is,

f(Bi12Ai1Bi12)(Bi12B12)f(B12A1B12)(B12Bi12).f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\geq\left(B_{i}^{\frac{1}{2}}B^{-\frac{1}{2}}\right)f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\left(B^{-\frac{1}{2}}B_{i}^{\frac{1}{2}}\right).

Multiplying Bi12B_{i}^{-\frac{1}{2}} to the both sides, we have

Bi12f(Bi12Ai1Bi12)Bi12B12f(B12A1B12)B12.B_{i}^{-\frac{1}{2}}f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)B_{i}^{-\frac{1}{2}}\geq B^{-\frac{1}{2}}f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)B^{-\frac{1}{2}}.

Taking an inverse of the both sides, we have

Bi12[f(Bi12Ai1Bi12)]1Bi12B12[f(B12A1B12)]1B12.B_{i}^{\frac{1}{2}}\left[f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}B_{i}^{\frac{1}{2}}\leq B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}. (20)

Thus we have

1mi=1mBi12[f(Bi12Ai1Bi12)]1Bi12B12[f(B12A1B12)]1B12.\frac{1}{m}\sum_{i=1}^{m}B_{i}^{\frac{1}{2}}\left[f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}B_{i}^{\frac{1}{2}}\leq B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}. (21)

Applying Lemma 3 to the above, we have

1mi=1mBi12[f(Bi12Ai1Bi12)]1Bi12\displaystyle\frac{1}{m}\sum_{i=1}^{m}B_{i}^{\frac{1}{2}}\left[f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}B_{i}^{\frac{1}{2}}
1m(i=1mBi12)[i=1mf(Bi12Ai1Bi12)]1(i=1mBi12).\displaystyle\geq\frac{1}{m}\left(\sum_{i=1}^{m}B_{i}^{\frac{1}{2}}\right)\left[\sum_{i=1}^{m}f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}\left(\sum_{i=1}^{m}B_{i}^{\frac{1}{2}}\right). (22)

Combining (21) and (22), we get the inequality (18).

To prove the inequality (19), we start from (20). By multiplying Ai12A_{i}^{\frac{1}{2}} to the both sides in (20), we have

Ai12Bi12[f(Bi12Ai1Bi12)]1Bi12Ai12Ai12B12[f(B12A1B12)]1B12Ai12.A_{i}^{\frac{1}{2}}B_{i}^{\frac{1}{2}}\left[f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}B_{i}^{\frac{1}{2}}A_{i}^{\frac{1}{2}}\leq A_{i}^{\frac{1}{2}}B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}A_{i}^{\frac{1}{2}}.

Taking a summation on ii from 11 to mm for the both sides in the above inequality, we obtain

i=1mAi12Bi12[f(Bi12Ai1Bi12)]1Bi12Ai12i=1mAi12B12[f(B12A1B12)]1B12Ai12.\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}B_{i}^{\frac{1}{2}}\left[f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}B_{i}^{\frac{1}{2}}A_{i}^{\frac{1}{2}}\leq\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}A_{i}^{\frac{1}{2}}.

That is, we have

i=1mAi12Bi12[(1)f(Bi12Ai1Bi12)]1Bi12Ai12\displaystyle\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}B_{i}^{\frac{1}{2}}\left[(-1)f\left(B_{i}^{\frac{1}{2}}A_{i}^{-1}B_{i}^{\frac{1}{2}}\right)\right]^{-1}B_{i}^{\frac{1}{2}}A_{i}^{\frac{1}{2}}
i=1mAi12B12[(1)f(B12A1B12)]1B12Ai12.\displaystyle\geq\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}B^{\frac{1}{2}}\left[(-1)f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}A_{i}^{\frac{1}{2}}. (23)

Applying Lemma 3 to the right hand side in (23), we have

i=1mAi12B12[(1)f(B12A1B12)]1B12Ai12(i=1mAi12)((m)B12f(B12A1B12)B12)1(i=1mAi12),\begin{split}&\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}B^{\frac{1}{2}}\left[(-1)f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}A_{i}^{\frac{1}{2}}\\ &\geq\left(\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\right)\left((-m)B^{-\frac{1}{2}}f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)B^{-\frac{1}{2}}\right)^{-1}\left(\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\right),\end{split}

that is,

i=1mAi12B12[f(B12A1B12)]1B12Ai12(1)(i=1mAi12)1mB12[f(B12A1B12)]1B12(i=1mAi12),\begin{split}&-\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}A_{i}^{\frac{1}{2}}\\ &\geq(-1)\left(\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\right)\frac{1}{m}B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}\left(\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\right),\end{split}

which implies

i=1mAi12B12[f(B12A1B12)]1B12Ai12\displaystyle\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}A_{i}^{\frac{1}{2}}
1m(i=1mAi12)B12[f(B12A1B12)]1B12(i=1mAi12).\displaystyle\leq\frac{1}{m}\left(\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\right)B^{\frac{1}{2}}\left[f\left(B^{\frac{1}{2}}A^{-1}B^{\frac{1}{2}}\right)\right]^{-1}B^{\frac{1}{2}}\left(\sum_{i=1}^{m}A_{i}^{\frac{1}{2}}\right). (24)

Combining (23) and (24), we get the inequality (19). ∎

Remark 3.

Considering aia_{i} and bib_{i} are positive real numbers and replacing Ai=:aiA_{i}=:a_{i} and Bi=:biB_{i}=:b_{i} as well as ff is a monotone increasing function we have from equation (18)

1m(i=1mbi12)2[i=1mf(biai)]1b[f(ba)]1.\frac{1}{m}\left(\sum_{i=1}^{m}b_{i}^{\frac{1}{2}}\right)^{2}\left[\sum_{i=1}^{m}f\left(\frac{b_{i}}{a_{i}}\right)\right]^{-1}\leq b\left[f\left(\frac{b}{a}\right)\right]^{-1}. (25)

Also, from equation (19) we can write

i=1maibi[f(biai)]11m(i=1mai12)2b[f(ba)]1.\sum_{i=1}^{m}a_{i}b_{i}\left[f\left(\frac{b_{i}}{a_{i}}\right)\right]^{-1}\leq\frac{1}{m}\left(\sum_{i=1}^{m}a_{i}^{\frac{1}{2}}\right)^{2}b\left[f\left(\frac{b}{a}\right)\right]^{-1}. (26)

5 Conclusion

The log-sum inequality which is mentioned in equation (2) plays a crucial role in classical and quantum information theory. In this article, we present a number of analogous inequalities. Inequality (2) depends on the convexity of the function xlog(x)x\log(x) which was relaxed utilizing the function xf(x)xf(x), in [4]. We illustrate that the concavity of xf(1x)xf(\dfrac{1}{x}) is also efficient for deriving similar inequalities. We discuss these inequalities with two functions instead of the single function ff, which increase their scope of applications. In this context, the log-sum inequality for qq-deformed logarithm is illustrated. We elaborate these results for the commuting self-adjoint matrices as trace-form inequalities. Later we discuss the log-sum inequality for the operator monotone and convex functions in the context of Löwner partial order relation of the positive semi-definite matrices, which is an application of the Hansen operator inequality. The following problem may be attempted in future.

Both inequalities in (25) and (26) do not have same form of log-sum inequality (17). Therefore we have to conclude that the restricted condition (so-called expansivity of matrices AiA_{i}) given in Proposition 1 leads us to obtain the log-sum inequality for non-commutative matrices. However, we could not obtain such an inequality without restricted condition in Theorem 5. In the further studies on this topic, we would like to obtain the log-sum inequality for non-commutative contractive matrices. We hope that the obtained results will be useful to study the parametric information theory in the future.

Funding

SD was a post doctoral fellow at the Indian Institute of Technology Kharagpur during this work. SF was partially supported by JSPS KAKENHI grant number 21K03341.

References

  • [1] Cover TM, Thomas JA. Elements of information theory. John Wiley & Sons; 2012.
  • [2] Jensen JLWV, et al. Sur les fonctions convexes et les inégalités entre les valeurs moyennes. Acta mathematica. 1906;30:175–193.
  • [3] Niculescu C, Persson LE. Convex functions and their applications. Springer; 2006.
  • [4] Csiszár I, Shields PC, et al. Information theory and statistics: A tutorial. Foundations and Trends® in Communications and Information Theory. 2004;1(4):417–528.
  • [5] Kac V, Cheung P. Quantum calculus. Springer Science & Business Media; 2001.
  • [6] Bebiano N, Providência Jr Jd, Lemos R. Matrix inequalities in statistical mechanics. Linear Algebra and its Applications. 2003; 376(1), 265–273.
  • [7] Furuichi S, Yanagi K, Kuriyama K. Fundamental properties of Tsallis relative entropy. Journal of Mathematical Physics. 2004;45(12):4868–4877.
  • [8] Wilde MM. Quantum information theory. Cambridge University Press; 2013.
  • [9] Zhan X. Matrix inequalities. Springer; 2004.
  • [10] Bhatia R. Matrix analysis. Vol. 169. Springer Science & Business Media; 2013.
  • [11] Hansen F, Pedersen GK. Jensen’s inequality for operators and löwner’s theorem. Mathematische Annalen. 1982;258(3):229–241.
  • [12] Hansen F, Pedersen GK. Jensen’s operator inequality. Bulletin of the London Mathematical Society. 2003;35(4):553–564.
  • [13] Dirac PAM. A new notation for quantum mechanics. In: Mathematical Proceedings of the Cambridge Philosophical Society; Vol. 35; Cambridge University Press; 1939. p. 416–418.
  • [14] Kubo F. and Ando T. Means of positive linear operators, Math. Ann., 246(1980), 205–224.
  • [15] Dragomir SS. Operator quasilinearity of some functionals associated with Davis–Choi–Jensen’s inequality for positive maps, Bull. Aust. Math. Soc., 95 (2017), 322-–332.
  • [16] Nikoufar I. Some properties of the Tsallis relative operator φ\varphi–entropy, Journal of Mathematical Physics, Analysis, Geometry, 17(2)(2021), 216-232.
  • [17] Furuta T. Parametric extensions of shannon inequality and its reverse one in hilbert space operators. Linear algebra and its applications. 2004;381:219–235.
  • [18] Hansen F. An operator inequality. Math Ann. 1979/80;246(3):249–250. Available from: https://doi.org/10.1007/BF01371046.