SUBADDITIVITY OF AN INTEGRAL TRANSFORM FOR POSITIVE OPERATORS IN HILBERT SPACES

Abstract

For a continuous and positive function $w(\lambda )$ , $\lambda>0$ and $\mu $ a positive measure on $(0,\infty )$ , we consider the integral transform

$$ \begin{align*} \mathcal{D}( w,\mu ) ( T) :=\int_{0}^{\infty }w(\lambda) ( \lambda +T) ^{-1}\,d\mu ( \lambda ) , \end{align*} $$

where the integral is assumed to exist for T a positive operator on a complex Hilbert space H. We show among other things that if B, $A>0,$ then $\mathcal {D}( w,\mu ) $ is operator subadditive on $(0,\infty ) $ , that is,

$$ \begin{align*} \mathcal{D}( w,\mu ) ( A) +\mathcal{D}( w,\mu) ( B) \geq \mathcal{D}( w,\mu )(A+B). \end{align*} $$

From this, we derive that if $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator monotone function on $[0,\infty )$ , then the function $[ f( t) -f( 0) ] t^{-1}$ is operator subadditive on $( 0,\infty ) .$ Also, if $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator convex function on $[0,\infty )$ , then the function $[ f( t) -f( 0) -f_{+}^{\prime }( 0) t ] t^{-2}$ is operator subadditive on $( 0,\infty ) .$

1 Introduction

Consider a complex Hilbert space $( H,\langle \cdot \, ,\cdot \rangle ) $ . An operator T is said to be positive (denoted by $ T\geq 0$ ) if $\langle Tx,x\rangle \geq 0$ for all $x\in H$ and strictly positive (denoted by $T>0$ ) if T is positive and invertible. A real valued continuous function $ f $ on $(0,\infty )$ is said to be operator monotone if $f(A)\geq f(B)$ holds for any $A\geq B>0.$ As usual, by $A\geq B$ , we understand that $A-B\geq 0.$

We have the following representation of operator monotone functions (see for instance [1, pages 144–145]).

Theorem 1.1 (Löwner, [5]).

A function $f:[0,\infty )\rightarrow \mathbb {R}$ is operator monotone in $[0,\infty )$ if and only if it has the representation

(1.1) $$ \begin{align} f( t) =f( 0) +bt+\int_{0}^{\infty }\frac{t\lambda }{t+\lambda }\,d\mu ( \lambda ) , \end{align} $$

where $b\geq 0$ and $\mu $ is a positive measure on $(0,\infty )$ such that

(1.2) $$ \begin{align} \int_{0}^{\infty }\frac{\lambda }{1+\lambda }\,d\mu ( \lambda ) <\infty. \end{align} $$

A real valued continuous function f on an interval I is said to be operator convex (operator concave) on I if

(OC) $$ \begin{align} f( ( 1-\lambda ) A+\lambda B) \leq ( \geq )\, ( 1-\lambda ) f( A) +\lambda f( B) \end{align} $$

in the operator order, for all $\lambda \in [ 0,1] $ and for all selfadjoint operators A and B on a Hilbert space H whose spectra are contained in $I.$ Notice that a function f is operator concave if $-f$ is operator convex.

We have the following representation of operator convex functions [1, page 147].

Theorem 1.2. A function $f:(0,\infty )\rightarrow \mathbb {R}$ is operator convex in $(0,\infty )$ with $f_{+}^{\prime }( 0) \in \mathbb {R}$ if and only if it has the representation

(1.3) $$ \begin{align} f( t) =f( 0) +f_{+}^{\prime }( 0) t+ct^{2}+\int_{0}^{\infty }\frac{t^{2}\lambda }{t+\lambda }\,d\mu (\lambda ) , \end{align} $$

where $c\geq 0$ and $\mu $ is a positive measure on $(0,\infty )$ such that (1.2) holds.

Assume that A, $B\geq 0$ . Moslehian and Najafi [6] showed that $AB+BA$ is positive if and only if the operator subadditivity property holds, that is,

$$ \begin{align*} f( A+B) \leq f( A) +f( B) \end{align*} $$

for all nonnegative operator monotone functions f on $[0,\infty ).$ For some interesting consequences of this result, see [6].

We have the following integral representation for the power function when $ t>0$ , $r\in (0,1]$ (see for instance [1, page 145]),

$$ \begin{align*} t^{r-1}=\frac{\sin ( r\pi ) }{\pi }\int_{0}^{\infty }\frac{\lambda ^{r-1}}{\lambda +t}\,d\lambda. \end{align*} $$

Motivated by these representations, we introduce, for a continuous and positive function $w( \lambda ) $ , $\lambda>0$ , the integral transform

(1.4) $$ \begin{align} \mathcal{D}( w,\mu ) ( t) :=\int_{0}^{\infty }\frac{w( \lambda ) }{\lambda +t}\,d\mu ( \lambda ) , \quad t>0, \end{align} $$

where $\mu $ is a positive measure on $(0,\infty )$ and the integral (1.4) exists for all $t>0.$ For $\mu $ , the usual Lebesgue measure, we put

$$ \begin{align*} \mathcal{D}( w) ( t) :=\int_{0}^{\infty }\frac{w(\lambda ) }{\lambda +t}\,d\lambda , \quad t>0. \end{align*} $$

Now, assume that $T>0$ . By the continuous functional calculus for selfadjoint operators, we can define the positive operator

$$ \begin{align*} \mathcal{D}( w,\mu ) ( T) :=\int_{0}^{\infty }w(\lambda ) ( \lambda +T) ^{-1}\,d\mu ( \lambda ) , \end{align*} $$

where w and $\mu $ are as above. When $\mu $ is the usual Lebesgue measure, for $T>0$ ,

$$ \begin{align*} \mathcal{D}( w) ( T) :=\int_{0}^{\infty }w(\lambda ) ( \lambda +T) ^{-1}\,d\lambda. \end{align*} $$

If we take $\mu $ to be the usual Lebesgue measure and the kernel $w_{r}( \lambda ) =\lambda ^{r-1}$ , $r\in (0,1],$ then

$$ \begin{align*} t^{r-1}=\frac{\sin ( r\pi ) }{\pi }\mathcal{D}( w_{r}) ( t) ,\quad t>0. \end{align*} $$

Motivated by these results, we show among other things that if B, $A>0,$ then

$$ \begin{align*} \mathcal{D}( w,\mu ) ( A) +\mathcal{D}( w,\mu ) ( B) \geq \mathcal{D}( w,\mu ) (A+B) , \end{align*} $$

that is, $\mathcal {D}( w,\mu ) $ is operator subadditive on $( 0,\infty ) .$ From this, if $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator monotone function on $[0,\infty )$ , we show that the function $[ f( t) -f( 0) ] t^{-1}$ is operator subadditive on $( 0,\infty ) .$ Also, if $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator convex function on $[0,\infty )$ , then the function $[ f( t) -f( 0) -f_{+}^{\prime }( 0) t] t^{-2}$ is operator subadditive on $( 0,\infty ) .$ Some examples for integral transforms $\mathcal {D}( \cdot \, ,\cdot ) $ related to the exponential and logarithmic functions are also provided.

2 Subadditivity property

Theorem 2.1. For all A, $B>0$ ,

(2.1) $$ \begin{align} D(w,\mu )( A) +D(w,\mu )( B) \geq D(w,\mu )(A+B) , \end{align} $$

namely, $D(w,\mu )$ is operator subadditive.

Proof. For all A, $B>0,$ by using the representation of $D(w,\mu ),$

(2.2) $$ \begin{align} D(w,\mu )( A) & +D(w,\mu )( B) -D(w,\mu )(A+B) \notag \\ & =\int_{0}^{\infty }w( \lambda ) [ ( A+\lambda )^{-1}+( B+\lambda ) ^{-1}-( A+B+\lambda ) ^{-1}] \,d\mu ( \lambda ). \end{align} $$

For $\lambda \geq 0,$ define the operator

$$ \begin{align*} K_{\lambda }:=( A+\lambda ) ^{-1}+( B+\lambda )^{-1}-( A+B+\lambda ) ^{-1}. \end{align*} $$

If we multiply both sides of $K_{\lambda }$ by $A+B+\lambda ,$ then we obtain successively

(2.3) $$ \begin{align} \hspace{-7pt}( A+B+\lambda ) K_{\lambda }( A+B+\lambda ) & =( A+B+\lambda ) ( A+\lambda ) ^{-1}(A+B+\lambda ) \notag \\ &\quad +( A+B+\lambda ) ( B+\lambda ) ^{-1}(A+B+\lambda ) -A-B-\lambda \notag \\ & =( 1+B( A+\lambda ) ^{-1}) ( A+\lambda +B)\notag \\ &\quad +( A( B+\lambda ) ^{-1}+1) ( A+B+\lambda ) -A-B-\lambda \notag \\ & =A+\lambda +B+B+B( A+\lambda ) ^{-1}B \notag \\ &\quad +A( B+\lambda ) ^{-1}A+A+A+B+\lambda -A-B-\lambda \notag \\ & =B( A+\lambda ) ^{-1}B+A( B+\lambda ) ^{-1}A+2(A+B) +\lambda =:L_{\lambda }.\ \end{align} $$

By multiplying both sides of (2.3) by $( A+B+\lambda ) ^{-1}$ ,

(2.4) $$ \begin{align} K_{\lambda }=( A+B+\lambda ) ^{-1}L_{\lambda }( A+B+\lambda) ^{-1}. \end{align} $$

We then have the representation

$$ \begin{align*} & D(w,\mu )( A) +D(w,\mu )( B) -D(w,\mu )( A+B)\\ & =\int_{0}^{\infty }w( \lambda ) K_{\lambda }\,d\mu ( \lambda) =\int_{0}^{\infty }w( \lambda ) ( A+B+\lambda ) ^{-1}L_{\lambda }( A+B+\lambda ) ^{-1}\,d\mu ( \lambda ) \end{align*} $$

for all A, $B>0.$

Since A, $B>0$ and $\lambda \geq 0,$ we obtain $L_{\lambda }\geq 0$ from (2.3). By (2.4), $K_{\lambda }\geq 0$ , and multiplying by $w( \lambda ) \geq 0$ and integrating over the measure $\mu $ gives (2.1).

Corollary 2.2. Assume that $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator monotone function on $[0,\infty )$ . If A, $B>0$ , then

(2.5) $$ \begin{align} f( A) A^{-1}+f( B) B^{-1}-f( A+B) (A+B) ^{-1} \geq f( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1}] , \end{align} $$

that is, the function $[ f( t) -f( 0) ] t^{-1}$ is operator subadditive on $( 0,\infty ) .$ In particular, if $f( 0) =0,$ then

(2.6) $$ \begin{align} f( A) A^{-1}+f( B) B^{-1}\geq f( A+B) ( A+B) ^{-1}. \end{align} $$

Proof. If $f:[0,\infty )\rightarrow \mathbb {R}$ is operator monotone, then by (1.1),

$$ \begin{align*} \frac{f( t) -f( 0) }{t}-b=\mathcal{D}( \ell ,\mu ) ( t) ,\quad \text{ }t>0 \end{align*} $$

for some positive measure $\mu ,$ where $\ell ( \lambda ) =\lambda $ , $\lambda>0.$ By applying Theorem 2.1 for $\mathcal {D}( \ell ,\mu ) $ and performing the required calculations, we deduce

$$ \begin{align*} 0& \leq D(w,\mu )( A) +D(w,\mu )( B) -D(w,\mu )( A+B) \\ & =[ f( A) -f( 0) ] A^{-1}-b+[ f( B) -f( 0) ] B^{-1}-b -[ f( A+B) -f( 0) ] ( A+B) ^{-1}+b \\ & =f( A) A^{-1}+f( B) B^{-1}-f( A+B) ( A+B) ^{-1} -f( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1}] -b \end{align*} $$

for all A, $B>0,$ which gives

$$ \begin{align*} f( A) A^{-1}+f( B) B^{-1}-f( A+B) ( A+B) ^{-1} & \geq f( 0) ( A^{-1}+B^{-1}-( A+B) ^{-1}) +b \\ & \geq f( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1}] \end{align*} $$

for all A, $B>0$ and (2.5) is obtained.

Remark 2.3. If we take $f( t) =t^{r}$ , $r\in (0,1]$ , in (2.6), then we get the power inequality

$$ \begin{align*} A^{r-1}+B^{r-1}\geq ( A+B) ^{r-1} \end{align*} $$

for all A, $B>0.$

If we take $f( t) =\ln ( t+1) $ in (2.6), then we get the logarithmic inequality

$$ \begin{align*} A^{-1}\ln ( A+1) +B^{-1}\ln ( B+1) \geq (A+B) ^{-1}\ln ( A+B+1). \end{align*} $$

Similar inequalities can be obtained by using the examples of operator monotone functions from [2, 3] and the references therein.

Corollary 2.4. Assume that $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator convex function on $[0,\infty )$ . If A, $B>0,$ then

$$ \begin{align*} &f( A) A^{-2} +f( B) B^{-2}-f( A+B) ( A+B) ^{-2} \\ & \quad\geq f( 0) [ A^{-2}+B^{-2}-( A+B) ^{-2}] +f_{+}^{\prime }( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1}] , \end{align*} $$

that is, the function $[ f( t) -f( 0) -f_{+}^{\prime }( 0) t] t^{-2}$ is operator subadditive on $ ( 0,\infty ) .$

If $f( 0) =0,$ then

(2.7) $$ \begin{align} f( A) A^{-2}+f( B) B^{-2}-f( A+B) ( A+B) ^{-2} \geq f_{+}^{\prime }( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1}]. \end{align} $$

Proof. If $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator convex function on $ [0,\infty ),$ then by (1.3),

$$ \begin{align*} \frac{f( t) -f( 0) -f_{+}^{\prime }( 0) t}{t^{2}}-c=\mathcal{D}( \ell ,\mu ) ( t) \end{align*} $$

for some positive measure $\mu ,$ where $\ell ( \lambda ) =\lambda $ , $\lambda>0.$

By applying Theorem 2.1 for $\mathcal {D}( \ell ,\mu ) $ and performing the required calculations, we deduce

$$ \begin{align*} &0 \leq D(w,\mu )( A) +D(w,\mu )( B) -D(w,\mu )( A+B) \\ &\quad =f( A) A^{-2}-f( 0) A^{-2}-f_{+}^{\prime }( 0) A^{-1}-c +f( B) B^{-2}-f( 0) B^{-2}-f_{+}^{\prime }( 0) B^{-1}-c \\ &\qquad -f( A+B) ( A+B) ^{-2}+f( 0) ( A+B) ^{-2}+f_{+}^{\prime }( 0) ( A+B) ^{-1}+c \\ &\quad =f( A) A^{-2}+f( B) B^{-2}-f( A+B) ( A+B) ^{-2} \\ &\qquad -f( 0) [ A^{-2}+B^{-2}-( A+B) ^{-2}] -f_{+}^{\prime }( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1} ] -c \end{align*} $$

for all A, $B>0.$ From this,

$$ \begin{align*} & f( A) A^{-2} +f( B) B^{-2}-f( A+B) ( A+B) ^{-2} \\ &\quad \geq f( 0) [ A^{-2}+B^{-2}-( A+B) ^{-2}] +f_{+}^{\prime }( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1} ] +c \\ &\quad \geq f( 0) [ A^{-2}+B^{-2}-( A+B) ^{-2}] +f_{+}^{\prime }( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1} ] , \end{align*} $$

which proves (2.7).

Remark 2.5. Let $a>0$ and $p\in \lbrack -1,0)\cup \lbrack 1,2].$ Then for all A, $B>0$ , we have the power inequality

$$ \begin{align*} ( A & +a) ^{p}A^{-2}+( B+a) ^{p}B^{-2}-( A+B+a) ^{p}( A+B) ^{-2} \\ & \geq a^{p}[ A^{-2}+B^{-2}-( A+B) ^{-2}] +pa^{p-1} [ A^{-1}+B^{-1}-( A+B) ^{-1}]. \end{align*} $$

We also have the logarithmic inequality

$$ \begin{align*} ( A & +B) ^{-2}\ln ( A+B+1) -A^{-2}\ln ( A+1) -B^{-2}\ln ( B+1) \\ & \geq ( A+B) ^{-1}-B^{-1}-A^{-1} \end{align*} $$

for all A, $B>0.$

3 Reverse inequalities

We define the difference $\mathcal {D}( w,\mu ) ( \cdot \, ,\cdot ) $ for positive numbers t, s by

$$ \begin{align*} \mathcal{D}( w,\mu ) ( t,s) :=D(w,\mu )( t) +D(w,\mu )( s) -D(w,\mu )( t+s) \geq 0 \end{align*} $$

and the difference for positive operators A, $B,$

$$ \begin{align*} \mathcal{D}( w,\mu ) ( A,B) :=D(w,\mu )( A) +D(w,\mu )( B) -D(w,\mu )( A+B) \geq 0 \end{align*} $$

for a continuous and positive function $w( \lambda ) $ , $\lambda>0$ and $\mu $ a positive measure on $(0,\infty )$ such that the integral (1.4) exists for all $t\geq 0.$ We prove the following reverse inequality.

Theorem 3.1. Assume that there exists positive constants $\alpha $ , $\beta ,$ $\gamma $ and $\delta $ such that

(3.1) $$ \begin{align} 0<\alpha \leq A\leq \beta \quad\text{and}\quad 0<\gamma \leq B\leq \delta. \end{align} $$

Then,

(3.2) $$ \begin{align} 0\leq \mathcal{D}( w,\mu ) ( A,B) \leq \mathcal{D} ( w,\mu ) ( \alpha ,\gamma ) +\bigg( \frac{\beta +\delta }{\alpha +\gamma }-1\bigg) D(w,\mu )(\, \beta +\delta ). \end{align} $$

Proof. Observe that

$$ \begin{align*} ( A+\lambda ) ^{-1}\leq ( \alpha +\lambda ) ^{-1}, \quad ( B+\lambda ) ^{-1}\leq ( \gamma +\lambda ) ^{-1} \end{align*} $$

and

$$ \begin{align*} (\, \beta +\delta +\lambda ) ^{-1}\leq ( A+B+\lambda ) ^{-1}, \end{align*} $$

that is,

$$ \begin{align*} -( A+B+\lambda ) ^{-1}\leq -(\, \beta +\delta +\lambda ) ^{-1}, \end{align*} $$

which gives

$$ \begin{align*} ( A+\lambda ) ^{-1}+( B+\lambda ) ^{-1}-( A+B+\lambda ) ^{-1} \leq ( \alpha +\lambda ) ^{-1}+( \gamma +\lambda ) ^{-1}-(\, \beta +\delta +\lambda ) ^{-1} \end{align*} $$

and so

$$ \begin{align*} ( A & +\lambda ) ^{-1}+( B+\lambda ) ^{-1}-( A+B+\lambda ) ^{-1} \\ & \leq ( \alpha +\lambda ) ^{-1}+( \gamma +\lambda ) ^{-1}-( \alpha +\gamma +\lambda ) ^{-1}+( \alpha +\gamma +\lambda ) ^{-1}-(\, \beta +\delta +\lambda ) ^{-1} \end{align*} $$

for all $\lambda \geq 0.$ If we multiply by $w( \lambda ) \geq 0$ and integrate, then by (2.2),

$$ \begin{align*} & D(w,\mu )( A) +D(w,\mu )( B) -D(w,\mu )(A+B) \\ &\quad \leq D(w,\mu )( \alpha ) +D(w,\mu )( \gamma ) -D(w,\mu )( \alpha +\gamma ) \\ &\qquad +\int_{0}^{\infty }w( \lambda ) [ ( \alpha +\gamma +\lambda ) ^{-1}-(\, \beta +\delta +\lambda ) ^{-1}] \,d\mu ( \lambda ) , \end{align*} $$

which gives

(3.3) $$ \begin{align}&\mathcal{D}(w,\mu)(A,B)\nonumber \\&\quad \le \mathcal{D}(w,\mu)(\alpha,\gamma) + \int_0^\infty w(\lambda)[(\alpha+\gamma+\lambda)^{-1} - (\beta+\delta+\lambda)^{-1}] \,d\mu(\lambda).\end{align} $$

Observe that

$$ \begin{align*} \int_{0}^{\infty } & w( \lambda ) [ ( \alpha +\gamma +\lambda ) ^{-1}-(\, \beta +\delta +\lambda ) ^{-1}] \,d\mu ( \lambda ) \\ & =\int_{0}^{\infty }w( \lambda ) \bigg[ \frac{\beta +\delta -\alpha -\gamma }{( \alpha +\gamma +\lambda ) (\, \beta +\delta +\lambda ) }\bigg] \,d\mu ( \lambda ) \\ & =(\, \beta +\delta -\alpha -\gamma ) \int_{0}^{\infty }\frac{ w( \lambda ) }{( \alpha +\gamma +\lambda ) (\, \beta +\delta +\lambda ) }\,d\mu ( \lambda ) \end{align*} $$

and

$$ \begin{align*} \frac{1}{\alpha +\gamma +\lambda }\leq \frac{1}{\alpha +\gamma }\quad \text{for } \lambda \geq 0, \end{align*} $$

which implies that

$$ \begin{align*} \int_{0}^{\infty }\frac{w( \lambda ) }{( \alpha +\gamma +\lambda ) (\, \beta +\delta +\lambda ) \,}\,d\mu ( \lambda) & \leq \frac{1}{\alpha +\gamma }\int_{0}^{\infty }\frac{w( \lambda ) }{\beta +\delta +\lambda }\,d\mu ( \lambda ) \\ & =\frac{1}{\alpha +\gamma }D(w,\mu )(\, \beta +\delta ). \end{align*} $$

Therefore,

(3.4) $$ \begin{align} \int_{0}^{\infty }w( \lambda ) [ ( \alpha +\gamma +\lambda ) ^{-1}-(\, \beta +\delta +\lambda ) ^{-1}] \,d\mu ( \lambda ) \leq \bigg( \frac{\beta +\delta }{\alpha +\gamma }-1\bigg) D(w,\mu )(\, \beta +\delta ). \end{align} $$

By making use of (3.3) and (3.4), we derive (3.2).

The case of operator monotone functions is the following corollary.

Corollary 3.2. Assume that $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator monotone function on $[0,\infty )$ with $f( 0) =0.$ If $ A$ , $B>0$ satisfy the condition (3.1), then

(3.5) $$ \begin{align} 0& \leq f( A) A^{-1}+f( B) B^{-1}-f( A+B) ( A+B) ^{-1} \notag \\ & \leq f( \alpha ) \alpha ^{-1}+f( \gamma ) \gamma^{-1}-f( \alpha +\gamma ) ( \alpha +\gamma ) ^{-1} +\bigg( \frac{\beta +\delta }{\alpha +\gamma }-1\bigg) f(\, \beta +\delta ) (\, \beta +\delta ) ^{-1}. \end{align} $$

Proof. Write

$$ \begin{align*} \frac{f( t) }{t}-b=\mathcal{D}( \ell ,\mu ) (t) ,\quad t>0 \end{align*} $$

for some positive measure $\mu ,$ where $\ell ( \lambda ) =\lambda $ , $\lambda>0$ . From (3.2),

$$ \begin{align*} 0& \leq f( A) A^{-1}+f( B) B^{-1}-f( A+B) ( A+B) ^{-1} \\ & \leq f( \alpha ) \alpha ^{-1}+f( \gamma ) \gamma ^{-1}-f( \alpha +\gamma ) ( \alpha +\gamma ) ^{-1} +\bigg( \frac{\beta +\delta }{\alpha +\gamma }-1\bigg) \bigg( \frac{ f(\, \beta +\delta ) }{\beta +\delta }-b\bigg) \\ & \leq f( \alpha ) \alpha ^{-1}+f( \gamma ) \gamma ^{-1}-f( \alpha +\gamma ) ( \alpha +\gamma ) ^{-1} +\bigg( \frac{\beta +\delta }{\alpha +\gamma }-1\bigg) \frac{f(\, \beta +\delta ) }{\beta +\delta }, \end{align*} $$

which proves the desired result (3.5).

Remark 3.3. If A, $B>0$ satisfy the condition (3.1) and $r\in (0,1],$ then we have the reverse power inequality

$$ \begin{align*} 0 \leq A^{r-1}+B^{r-1}-( A+B) ^{r-1} \leq \alpha ^{r-1}+\gamma ^{r-1}-( \alpha +\gamma ) ^{r-1}+\bigg( \frac{\beta +\delta }{\alpha +\gamma }-1\bigg) (\, \beta +\delta ) ^{r-1}. \end{align*} $$

The case of operator convex functions is the following corollary.

Corollary 3.4. Assume that $f:[0,\infty )\rightarrow \mathbb {R}$ is an operator convex function on $[0,\infty )$ with $f( 0) =0.$ If A, $B>0$ satisfy the condition (3.1), then

(3.6) $$ \begin{align} 0 & \leq f( A) A^{-2}+f( B) B^{-2}-f( A+B) ( A+B) ^{-2} -f_{+}^{\prime }( 0) [ A^{-1}+B^{-1}-( A+B) ^{-1}] \notag \\ & \leq f( \alpha ) \alpha ^{-2}+f( \gamma ) \gamma ^{-2}-f( \alpha +\gamma ) ( \alpha +\gamma ) ^{-2} -f_{+}^{\prime }( 0) [ \alpha ^{-1}+\gamma ^{-1}-( \alpha +\gamma ) ^{-1}] \notag \\ &\quad +\bigg( \frac{\beta +\delta }{\alpha +\gamma }-1\bigg) \frac{f(\, \beta +\delta ) -f_{+}^{\prime }( 0) (\, \beta +\delta ) }{(\, \beta +\delta ) ^{2}}. \end{align} $$

Proof. The result follows from (3.2) observing, by (1.3), that

$$ \begin{align*} \frac{f( t) -f( 0) -f_{+}^{\prime }( 0) t}{t^{2}}-c=\mathcal{D}( \ell ,\mu ) ( t) \end{align*} $$

for some positive measure $\mu ,$ where $\ell ( \lambda ) =\lambda $ , $\lambda>0.$

Remark 3.5. If A, $B>0$ satisfy the condition (3.1), then by taking $f( t) =-\ln ( t+1)$ in (3.6), we obtain

$$ \begin{align*} 0& \leq ( A+B) ^{-2}\ln ( A+B+1) -A^{-2}\ln ( A+1) -B^{-2}\ln ( B+1) -( A+B) ^{-1}+A^{-1}+B^{-1} \notag \\ & \leq ( \alpha +\gamma ) ^{-2}\ln ( \alpha +\gamma +1) -\alpha ^{-2}\ln ( \alpha +1) -\gamma ^{-2}\ln ( \gamma +1) -( \alpha +\gamma ) ^{-1}+\alpha ^{-1}+\gamma ^{-1} \notag \\ &\quad +\bigg( \frac{\beta +\delta }{\alpha +\gamma }-1\bigg) \frac{(\, \beta +\delta ) -\ln (\, \beta +\delta +1) }{(\, \beta +\delta ) ^{2}}. \end{align*} $$

4 Some examples

We define the upper incomplete Gamma function

$$ \begin{align*} \Gamma (a,z):=\int_{z}^{\infty }t^{a-1}e^{-t}\,dt, \end{align*} $$

which for $z=0$ , gives the Gamma function

$$ \begin{align*} \Gamma (a):=\int_{0}^{\infty }t^{a-1}e^{-t}\,dt \quad \text{for }\mbox{Re}\, a>0. \end{align*} $$

We have the integral representation [8]

(4.1) $$ \begin{align} \Gamma (a,z)=\frac{z^{a}e^{-z}}{\Gamma (1-a)}\int_{0}^{\infty }\frac{ t^{-a}e^{-t}}{z+t}\,dt \end{align} $$

for $\mbox {Re\,}a<1$ and $\vert \mbox {arg\,}z\vert <\pi .$

Now, we consider the weight $w_{\cdot ^{-a}e^{-\cdot }}( \lambda ) :=\lambda ^{-a}e^{-\lambda }$ for $\lambda>0.$ Then by (4.1),

(4.2) $$ \begin{align} \mathcal{D}( w_{\cdot ^{-a}e^{-\cdot }}) ( t) =\int_{0}^{\infty }\frac{\lambda ^{-a}e^{-\lambda }}{t+\lambda }\,d\lambda =\Gamma (1-a)t^{-a}e^{t}\Gamma (a,t) \end{align} $$

for $a<1$ and $t>0.$ For $a=0$ in (4.2),

$$ \begin{align*} \mathcal{D}( w_{e^{-\cdot }}) ( t) =\int_{0}^{\infty } \frac{e^{-\lambda }}{t+\lambda }\,d\lambda =\Gamma (1)e^{t}\Gamma (0,t)=e^{t}E_{1}( t) \end{align*} $$

for $t>0,$ where

$$ \begin{align*} E_{1}( t) :=\int_{t}^{\infty }\frac{e^{-u}}{u}\,du. \end{align*} $$

From Theorem 2.1, we conclude that $\mathcal {D}( w_{\cdot ^{-a}e^{-\cdot }}) $ and, in particular, $\mathcal {D}( w_{e^{-\cdot }}) $ are operator subadditive on $( 0,\infty ) .$

We can also consider the weight $w_{( \cdot ^{2}+a^{2}) ^{-1}}( \lambda ) :={1}/{(\lambda ^{2}+a^{2})}$ for $\lambda>0$ and $a>0.$ Then, by simple calculations,

$$ \begin{align*} \mathcal{D}( w_{( \cdot ^{2}+a^{2}) ^{-1}}) (t) & :=\int_{0}^{\infty }\frac{1}{( \lambda +t) (\lambda ^{2}+a^{2}) }\,d\lambda =\frac{\pi t}{2a( t^{2}+a^{2}) }-\frac{\ln t-\ln a}{t^{2}+a^{2}} \end{align*} $$

for $t>0$ and $a>0.$ For $a=1$ , we also have

$$ \begin{align*} \mathcal{D}( w_{( \cdot ^{2}+1) ^{-1}}) ( t) :=\int_{0}^{\infty }\frac{1}{( \lambda +t) ( \lambda ^{2}+1) }\,d\lambda =\frac{\pi t}{2( t^{2}+1) }-\frac{ \ln t}{t^{2}+1} \end{align*} $$

for $t>0.$ If $T>0$ and $a>0,$ then

$$ \begin{align*} ( T^{2}+a^{2}) ^{-1}\bigg( \frac{\pi }{2a}T-\ln T+\ln a\bigg) =\int_{0}^{\infty }\frac{1}{( \lambda ^{2}+a^{2}) }( \lambda +T) ^{-1}\,d\lambda \end{align*} $$

and, in particular,

$$ \begin{align*} ( T^{2}+1) ^{-1}\bigg( \frac{\pi }{2}T-\ln T\bigg) =\int_{0}^{\infty }\frac{1}{( \lambda ^{2}+1) }( \lambda +T) ^{-1}\,d\lambda. \end{align*} $$

From Theorem 2.1, we conclude that $\mathcal {D}( w_{( \cdot ^{2}+a^{2}) ^{-1}}) $ and, in particular, $\mathcal {D} ( w_{( \cdot ^{2}+1) ^{-1}}) $ are operator subadditive on $( 0,\infty ) .$

Other similar inequalities can be obtained by using the examples of operator monotone functions provided in [2–4, 6, 7].