Operator P-class functions

Abstract

We introduce and investigate the notion of an operator P-class function. We show that every nonnegative operator convex function is of operator P-class, but the converse is not true in general. We present some Jensen type operator inequalities involving P-class functions and some Hermite-Hadamard inequalities for operator P-class functions.

MSC:47A63, 47A60, 26D15.

1 Introduction and preliminaries

Let $\mathfrak{B}(\mathcal{H})$ denote the $C^{\ast }$-algebra of all bounded linear operators on a complex Hilbert space ℋ with its identity denoted by I. When $dim\mathcal{H}=n$, we identify $\mathfrak{B}(\mathcal{H})$ with the matrix algebra ${\mathbb{M}}_n$ of all $n\times n$ matrices with entries in the complex field ℂ. We denote by $\sigma (J)$ the set of all self-adjoint operators on ℋ whose spectra are contained in an interval J. An operator $A\in \mathfrak{B}(\mathcal{H})$ is called positive (positive semidefinite for a matrix) if $〈Ax,x〉\ge 0$ for all $x\in \mathcal{H}$ and in such a case we write $A\ge 0$. For self-adjoint operators $A,B\in \mathfrak{B}(\mathcal{H})$, we write $B\ge A$ if $B-A\ge 0$. The Gelfand map $f\mapsto f(A)$ is an isometrical ∗-isomorphism between the $C^{\ast }$-algebra $C(\sigma (A))$ of a complex-valued continuous functions on the spectrum $\sigma (A)$ of a self-adjoint operator A and the $C^{\ast }$-algebra generated by I and A. If $f,g\in C(\sigma (A))$, then $f(t)\ge g(t)$ ($t\in \sigma (A)$) implies that $f(A)\ge g(A)$. A real-valued continuous function f on an interval J is called operator increasing (operator decreasing, resp.) if $A\le B$ implies $f(A)\le f(B)$ ($f(B)\le f(A)$, resp.) for all $A,B\in \sigma (J)$. We recall that a real-valued continuous function f defined on an interval J is operator convex if $f(\lambda A+(1-\lambda )B)\le \lambda f(A)+(1-\lambda )f(B)$ for all $A,B\in \sigma (J)$ and all $\lambda \in [0,1]$.

A function $f:J⟶\mathbb{R}$ is said to be of P-class on J or is a P-class function on J if

$$f(\lambda x+(1-\lambda )y)\le f(x)+f(y),$$

(1)

where $x,y\in J$ and $\lambda \in [0,1]$; see [1]. Many properties of P-class functions can be found in [1–4]. Note that the set of all P-class functions contains all convex functions and also all nonnegative monotone functions. Every non-zero P-class function is nonnegative valued. In fact, choose $\lambda =1$ and fix $x_0\in J$. It follows from (1) that

$$f(x_0)\le f(x_0)+f(y),$$

where $y\in J$. Thus $0\le f(y)$ for all $y\in J$.

For a P-class function f on an interval $[a,b]$,

$$f\left(\frac{a+b}{2}\right)\le 2{\int }_0^{1}f(ta+(1-t)b)dt\le 2(f(a)+f(b)),$$

which is known as the Hermite-Hadamard inequality for the P-class continuous functions; see [3].

In this paper, we introduce and investigate the notion of an operator P-class function and give several examples. We show that if f is a P-class function on $(0,\mathrm{\infty })$ such that ${lim}_{t\to \mathrm{\infty }}f(t)=0$, then it is operator decreasing. We also prove that if f is an operator P-class function on an interval J, then

$$f\left(C^{\ast }AC\right)\le 2C^{\ast }f(A)C,$$

where $A\in \sigma (J)$ and $C\in \mathfrak{B}(\mathcal{H})$ is an isometry. In addition, we present a Hermite-Hadamard inequality for operator P-class functions.

2 Operator P-class functions

In this section, we investigate operator P-class functions and study some relations between the operator P-class functions and the operator monotone functions.

We start our work with the following definition.

Definition 1 Let f be a real-valued continuous function defined on an interval J. We say that f is of operator P-class on J if

$$f(\lambda A+(1-\lambda )B)\le f(A)+f(B)$$

for all $A,B\in \sigma (J)$ and all $\lambda \in [0,1]$.

Clearly every nonnegative operator convex function is of operator P-class.

Example 1 Let $f(t)=t^{-r}$ ($0\le r\le 1$) be defined on $(1,\mathrm{\infty })$. It follows from the operator concavity of $t^r$ ($0\le r\le 1$) [5] and the arithmetic-harmonic mean inequality that

$$\begin{array}{rl}{(\lambda A+(1-\lambda )B)}^{-r}& \le {(\lambda A^r+(1-\lambda )B^r)}^{-1}\left(\text{by the concavity of }{t}^r\right)\\ \le \lambda A^{-r}+(1-\lambda )B^{-r}(\text{by the arithmetic-harmonic mean inequality})\\ \le A^{-r}+B^{-r},\end{array}$$

where $A,B\in \sigma ((1,\mathrm{\infty }))$ and $\lambda \in [0,1]$. Thus f is an operator P-class function on $(1,\mathrm{\infty })$.

In addition, every operator P-class f on an interval J is of operator Q-class in the sense that

$$f(\lambda A+(1-\lambda )B)\le \frac{f(A)}{\lambda }+\frac{f(B)}{1-\lambda }$$

for all $A,B\in \sigma (J)$ and $\lambda \in (0,1)$; see [6]. In the next example, we show the converse is not true, in general.

Example 2 The function $f(t)=4-t^2$ defined on $[-\sqrt{3},\sqrt{3}]$ is of operator Q-class; see [[7], Example 2.1]. We put $\lambda =\frac{1}{2}$, $A=\left(\begin{array}{cc}\frac{3}{2}& 0\\ 0& \frac{3}{2}\end{array}\right)$ and $B=\left(\begin{array}{cc}-\frac{3}{2}& 0\\ 0& -\frac{3}{2}\end{array}\right)$. Then $f(\lambda A+(1-\lambda )B)=\left(\begin{array}{cc}4& 0\\ 0& 4\end{array}\right)\nleqq f(A)+f(B)=\left(\begin{array}{cc}\frac{14}{4}& 0\\ 0& \frac{14}{4}\end{array}\right)$. Hence f is not of operator P-class.

Example 3 Let $\alpha >0$ and f be a continuous function on the interval $[\alpha ,2\alpha ]$ into itself. It follows from

$$f(\lambda A+(1-\lambda )B)\le 2\alpha \le f(A)+f(B)(A,B\in \sigma ([\alpha ,2\alpha ]),\lambda \in [0,1])$$

that f is of operator P-class on $[\alpha ,2\alpha ]$.

Example 4 Let g be a nonnegative continuous function on an interval $[a,b]$ and $\alpha ={sup}_{x,y\in [a,b],t\in [x,y]}|g(t)-g(x)-g(y)|$. We put $f(t)=g(t)+\alpha$. Then

$$\begin{array}{rl}f(\lambda A+(1-\lambda )B)& =g(\lambda A+(1-\lambda )B)+\alpha \\ \le (g(A)+\alpha )+(g(B)+\alpha )=f(A)+f(B),\end{array}$$

where $A,B\in \sigma ([a,b])$ and $\lambda \in [0,1]$. Hence f is an operator P-class function.

Next, we explore some relations between operator P-class functions and operator monotone functions. In fact, we have the following.

Theorem 1 If f is an operator P-class function on the interval $(0,\mathrm{\infty })$ such that ${lim}_{t\to \mathrm{\infty }}f(t)=0$, then f is operator decreasing.

Proof Let $0<A\le B$. Fix $\epsilon >0$. We put $C=B-A+\epsilon$. Let $\theta >0$. It follows from ${lim}_{t\to \mathrm{\infty }}f(t)=0$ that there exists $M>0$ such that $f(t)\le \theta$ for all $t\ge M$. We may assume that the spectrum of the strictly positive operator C is contained in $[\alpha ,\beta ]$ for some $0<\alpha <\beta$. It follows from ${lim}_{\lambda \to 1^{-}}\frac{\lambda }{1-\lambda }=\mathrm{\infty }$ that there exists $\delta >0$ such that $\frac{\lambda }{1-\lambda }\ge \frac{M}{\alpha }$ for all $\lambda \in (1-\delta ,1)$. Hence $\sigma (\frac{\lambda }{1-\lambda }C)\subseteq [M,\mathrm{\infty })$ for all $\lambda \in (1-\delta ,1)$. Now, by the functional calculus for the positive operator $\frac{\lambda }{1-\lambda }C$, we have $f(\frac{\lambda }{1-\lambda }C)\le \theta$ for all $\lambda \in (1-\delta ,1)$. Thus $〈f(\frac{\lambda }{1-\lambda }C)x,x〉\le \theta {\parallel x\parallel }^2$ for all $\lambda \in (1-\delta ,1)$ and $x\in \mathcal{H}$. Since $\lambda (B+\epsilon )=\lambda A+(1-\lambda )(\frac{\lambda }{1-\lambda })C$ and f is P-class we have

$$f(\lambda (B+\epsilon ))\le f(A)+f\left(\left(\frac{\lambda }{1-\lambda }\right)C\right)$$

for all $\lambda \in (1-\delta ,1)$. Hence

$$〈f(\lambda (B+\epsilon ))x,x〉\le 〈f(A)x,x〉+〈f\left(\frac{\lambda }{1-\lambda }C\right)x,x〉\le 〈f(A)x,x〉+\theta {\parallel x\parallel }^2,$$

where $\lambda \in (1-\delta ,1)$ and $x\in \mathcal{H}$. As $\lambda \to 1^{-}$ and then $\theta \to 0^{+}$ we obtain $〈f(B+\epsilon )x,x〉\le 〈f(A)x,x〉$ for all $x\in \mathcal{H}$. As $\epsilon \to 0^{+}$, we conclude that $f(B)\le f(A)$. □

3 Jensen operator inequality for operator P-class functions

In this section, we present a Jensen operator inequality for operator P-class functions. We start with the following result in which we utilized the well-known technique of [8].

Theorem 2 Let f be an operator P-class function on an interval J, $A\in \sigma (J)$, and $C\in \mathfrak{B}(\mathcal{H})$ be an isometry. Then

$$f\left(C^{\ast }AC\right)\le 2C^{\ast }f(A)C.$$

(2)

Proof Let $X=\left(\begin{array}{cc}A& 0\\ 0& B\end{array}\right)\in \mathfrak{B}(\mathcal{H}\oplus \mathcal{H})$ for some $B\in \sigma (J)$ and let $U=\left(\begin{array}{cc}C& D\\ 0& -C^{\ast }\end{array}\right)$ and $V=\left(\begin{array}{cc}C& -D\\ 0& C^{\ast }\end{array}\right)$, where $D=\sqrt{1_{\mathcal{H}}-C{C}^{\ast }}$. Now we can easily conclude from the two facts $C^{\ast }D=\sqrt{1_{\mathcal{H}}-C{C}^{\ast }}C=0$ and $DC=C\sqrt{1_{\mathcal{H}}-C^{\ast }C}=0$ that U and V are unitary operators in $\mathfrak{B}(\mathcal{H}\oplus \mathcal{H})$. Further,

$$U^{\ast }XU=\left(\begin{array}{cc}{C}^{\ast }AC& C^{\ast }AD\\ DAC& DAD+CB{C}^{\ast }\end{array}\right)$$

and

$$V^{\ast }XV=\left(\begin{array}{cc}{C}^{\ast }AC& -C^{\ast }AD\\ -DAC& DAD+CB{C}^{\ast }\end{array}\right).$$

Using the operator P-class property of f we have

$$\begin{array}{rl}\left(\begin{array}{cc}f(C^{\ast }AC)& 0\\ 0& f(DAD+CB{C}^{\ast })\end{array}\right)& =f\left(\begin{array}{cc}{C}^{\ast }AC& 0\\ 0& DAD+CB{C}^{\ast }\end{array}\right)\\ =f\left(\frac{U^{\ast }XU+V^{\ast }XV}{2}\right)\\ \le f\left(U^{\ast }XU\right)+f\left(V^{\ast }XV\right)\\ =2\left(\begin{array}{cc}{C}^{\ast }f(A)C& 0\\ 0& Df(A)D+Cf(B)C^{\ast }\end{array}\right).\end{array}$$

Therefore

$$f\left(C^{\ast }AC\right)\le 2C^{\ast }f(A)C.$$

 □

Applying Theorem 2 we have some inequalities including the subadditivity.

Corollary 1 Let f be operator P-class on an interval J, $A_j\in \sigma (J)$ ($1\le j\le n$), and $C_j\in \mathfrak{B}(\mathcal{H})$ ($1\le j\le n$), where ${\sum }_{j=1}^n{C}_j^{\ast }{C}_j=1$. Then

$$f\left(\sum _{j=1}^n{C}_j^{\ast }{A}_j{C}_j\right)\le 2\sum _{j=1}^n{C}_j^{\ast }f(A_j)C_j.$$

Proof Let

$$\tilde{A}=\tilde{A}=\left(\begin{array}{c}{A}_1\\ A_2\\ \cdots \\ A_n\end{array}\right)\in \mathfrak{B}(\mathcal{H}\oplus \cdots \oplus \mathcal{H}),\tilde{C}=\left(\begin{array}{c}{C}_1\\ C_2\\ ⋮\\ C_n\end{array}\right)\in \mathfrak{B}(\mathcal{H}\oplus \cdots \oplus \mathcal{H}).$$

It follows from ${\tilde{C}}^{\ast }\tilde{C}=1$ and (2) that

$$f\left(\sum _{j=1}^n{C}_j^{\ast }{A}_j{C}_j\right)=f\left({\tilde{C}}^{\ast }\tilde{A}\tilde{C}\right)\le 2{\tilde{C}}^{\ast }f(\tilde{A})\tilde{C}=2\sum _{j=1}^n{C}_j^{\ast }f(A_j)C_j.$$

 □

Corollary 2 Let f be operator P-class on $[0,\mathrm{\infty })$ such that $f(0)=0$, $A\in \sigma ([0,\mathrm{\infty }))$, and $C\in \mathfrak{B}(\mathcal{H})$ be a contraction. Then

$$f\left(C^{\ast }AC\right)\le 2C^{\ast }f(A)C.$$

Proof For every contraction $C\in \mathfrak{B}(\mathcal{H})$, we put $D=\sqrt{1_{\mathcal{H}}-C^{\ast }C}$. It follows from $C^{\ast }C+D^{\ast }D=1_{\mathcal{H}}$ and (2) that

$$f\left(C^{\ast }AC\right)=f(C^{\ast }AC+D^{\ast }0D)\le 2f\left(C^{\ast }AC\right)+2f\left(D^{\ast }0D\right)=2C^{\ast }f(A)C.$$

 □

Corollary 3 Let f be operator P-class on $[0,\mathrm{\infty })$ such that $f(0)=0$ and $A,B\in \sigma ((0,\mathrm{\infty }))$ such that $A\le B$. Then

$$A^{-1}f(A)\le 2B^{-1}f(B).$$

Proof Let $A,B\in \sigma ((0,\mathrm{\infty }))$ such that $0<A\le B$. We put $C=B^{-1/2}{A}^{1/2}$. Then $C{C}^{\ast }=B^{-1/2}A{B}^{-1/2}\le 1_{\mathcal{H}}$, so C is a contraction. It follows from (2) that

$$f(A)=f\left(C^{\ast }BC\right)\le 2C^{\ast }f(B)C=2A^{1/2}{B}^{-1/2}f(B)B^{-1/2}{A}^{1/2}.$$

Therefore

$$A^{-1}f(A)\le 2B^{-1}f(B).$$

 □

In the following theorem, we obtain the Choi-Davis-Jensen type inequality for operator P-class functions.

Theorem 3 Let Φ be a unital positive linear map on $\mathfrak{B}(\mathcal{H})$, $A\in \sigma (J)$ and f be operator P-class on an interval J. Then

$$f(\mathrm{\Phi }(A))\le 2\mathrm{\Phi }(f(A)).$$

(3)

Proof Let $A\in \sigma (J)$. We put Ψ the restriction of Φ to the $C^{\ast }$-algebra ${\mathcal{C}}^{\ast }(A,I)$ generated by I and A. Then Ψ is a unital completely positive map on ${\mathcal{C}}^{\ast }(A,I)$. The celebrated Stinespring dilation theorem [[9], Theorem 1] states that there exist an isometry $V:\mathcal{H}⟶\mathcal{H}$ and a unital ∗-homomorphism $\pi :{\mathcal{C}}^{\ast }(A,I)⟶\mathfrak{B}(\mathcal{H})$ such that $\mathrm{\Psi }(A)=V^{\ast }\pi (A)V$. Hence

$$\begin{array}{rl}f(\mathrm{\Phi }(A))& =f(\mathrm{\Psi }(A))=f(V^{\ast }\pi (A)V)\le 2V^{\ast }f(\pi (A))V(\text{by }\text{(2)})\\ =2V^{\ast }\pi (f(A))V=2\mathrm{\Psi }(f(A))=2\mathrm{\Phi }(f(A)).\end{array}$$

 □

We will show that the constant 2 is the best possible such one in the following example.

Example 5 Let $f(t)=2-t^2$ for $t\in [-1,1]$. Then $1\le f(t)\le 2$ and

$$f(\lambda A+(1-\lambda )B)=2-{(\lambda A+(1-\lambda )B)}^2\le 2\le 2-A^2+2-B^2=f(A)+f(B),$$

where $A,B\in \sigma ([-1,1])$. Hence f is of operator P-class on $[-1,1]$. Now, consider that the unital positive map $\mathrm{\Phi }:{\mathbb{M}}_2\to {\mathbb{M}}_2$ is defined by $\mathrm{\Phi }(A)=\frac{tr(A)}{2}I$. Then for the Hermitian matrix $A=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$ we have $\mathrm{\Phi }(A)=0$, $f(\mathrm{\Phi }(A))=2$, $f(A)=I$, and $\mathrm{\Phi }(f(A))=I$. Therefore $f(\mathrm{\Phi }(A))=2\mathrm{\Phi }(f(A))$. This shows that the coefficient 2 in (2) and (3) is the best.

Example 6 Consider (the nonnegative increasing function and so) P-class function $f(t)=\sqrt{t}$ where $t\in (0,\mathrm{\infty })$. Let the unital positive map $\mathrm{\Psi }:{\mathbb{M}}_2(\mathbb{C})\to \mathbb{C}$ be defined by $\mathrm{\Psi }(A)=a_{22}$ with $A={(a_{ij})}_{1\le i,j\le 2}$ and let $A={\left(\begin{array}{cc}2& 2\\ 2& 1\end{array}\right)}^2$. Then $\mathrm{\Psi }(f(A))=1$ and $f(\mathrm{\Psi }(A))=\sqrt{8}$. Hence $f(\mathrm{\Psi }(A))\nleqq 2\mathrm{\Psi }(f(A))$. It follows from (3) that f is not of operator P-class.

We present a Hermite-Hadamard inequality for operator P-class functions in the next theorem.

Theorem 4 Let Φ be a unital positive linear map on $\mathfrak{B}(\mathcal{H})$ and f be operator P-class on J. Then

$$f\left(\frac{\mathrm{\Phi }(A)+\mathrm{\Phi }(B)}{2}\right)\le 2{\int }_0^{1}f(\lambda \mathrm{\Phi }(A)+(1-\lambda )\mathrm{\Phi }(B))d\lambda \le 4(\mathrm{\Phi }(f(A))+\mathrm{\Phi }(f(B))),$$

where $A,B\in \sigma (J)$ and $\lambda \in [0,1]$.

Proof Let $A,B\in \sigma (J)$ and $\lambda \in [0,1]$. Then

$$\begin{array}{rl}f\left(\frac{\mathrm{\Phi }(A)+\mathrm{\Phi }(B)}{2}\right)& =f\left(\frac{\lambda \mathrm{\Phi }(A)+(1-\lambda )\mathrm{\Phi }(B)+(1-\lambda )\mathrm{\Phi }(A)+\lambda \mathrm{\Phi }(B)}{2}\right)\\ \le f(\lambda \mathrm{\Phi }(A)+(1-\lambda )\mathrm{\Phi }(B))+f((1-\lambda )\mathrm{\Phi }(A)+\lambda \mathrm{\Phi }(B))\\ \le 2(f(\mathrm{\Phi }(A))+f(\mathrm{\Phi }(B))).\end{array}$$

(4)

Integrating both sides of (4) over $[0,1]$ we obtain

$$\begin{array}{rl}f\left(\frac{\mathrm{\Phi }(A)+\mathrm{\Phi }(B)}{2}\right)\le & {\int }_0^{1}f(\lambda \mathrm{\Phi }(A)+(1-\lambda )\mathrm{\Phi }(B))d\lambda \\ +{\int }_0^{1}f((1-\lambda )\mathrm{\Phi }(A)+\lambda \mathrm{\Phi }(B))d\lambda \\ =& 2{\int }_0^{1}f(\lambda \mathrm{\Phi }(A)+(1-\lambda )\mathrm{\Phi }(B))d\lambda \\ \le & 2(f(\mathrm{\Phi }(A))+f(\mathrm{\Phi }(B)))\\ \le & 4(\mathrm{\Phi }(f(A))+\mathrm{\Phi }(f(B)))(\text{by }\text{(3)}).\end{array}$$

 □

4 Some inequalities for P-class functions involving continuous operator fields

Let $\mathcal{A}$ be a $C^{\ast }$-algebra of operators acting on a Hilbert space and let T be a locally compact Hausdorff space. A field ${(A_t)}_{t\in T}$ of operators in $\mathcal{A}$ is called a continuous field of operators if the mapping $t\mapsto A_t$ is norm continuous on T. If $\mu (t)$ is a Radon measure on T and the function $t\mapsto \parallel A_t\parallel$ is integrable, one can form the Bochner integral ${\int }_T{A}_{t}d\mu (t)$, which is the unique element in $\mathcal{A}$ such that

$$\phi ({\int }_T{A}_{t}d\mu (t))={\int }_T\phi (A_t)d\mu (t)$$

for every linear functional φ in the norm dual ${\mathcal{A}}^{\ast }$ of $\mathcal{A}$.

Let $\mathcal{C}(T,\mathcal{A})$ denote the set of bounded continuous functions on T with values in $\mathcal{A}$. It is easy to see that the set $\mathcal{C}(T,\mathcal{A})$ is a $C^{\ast }$-algebra under the pointwise operations and the norm $\parallel {(A_t)}_{t\in T}\parallel ={sup}_{t\in T}\parallel A_t\parallel$; cf. [10].

Assume that there is a field ${({\mathrm{\Phi }}_t)}_{t\in T}$ of positive linear mappings ${\mathrm{\Phi }}_t:\mathcal{A}⟶\mathcal{B}$ from $\mathcal{A}$ to another $C^{\ast }$-algebra ℬ. We say that such a field is continuous if the mapping $t\mapsto {\mathrm{\Phi }}_t(A)$ is continuous for every $A\in \mathcal{A}$. If the $C^{\ast }$-algebras are unital and the field $t\mapsto {\mathrm{\Phi }}_t(I)$ is integrable with integral I, we say that ${({\mathrm{\Phi }}_t)}_{t\in T}$ is unital; see [10].

Theorem 5 Let $f:J⟶\mathbb{R}$ be an operator P-class function defined on an interval J, and let $\mathcal{A}$ and ℬ be unital $C^{\ast }$-algebras. If ${({\mathrm{\Phi }}_t)}_{t\in T}$ is a unital field of positive linear mappings ${\mathrm{\Phi }}_t:\mathcal{A}⟶\mathcal{B}$ defined on a locally compact Hausdorff space T with a bounded Radon measure μ, then

$$f({\int }_T{\mathrm{\Phi }}_t(A_t)d\mu (t))\le 2{\int }_T{\mathrm{\Phi }}_t(f(A_t))d\mu (t)$$

holds for every bounded continuous field ${(A_t)}_{t\in T}$ of self-adjoint elements in $\mathcal{A}$ with spectra contained in J.

Proof We consider the unital positive linear map $\mathrm{\Psi }:\mathcal{C}(T,\mathcal{A})⟶\mathcal{B}$ defined by $\mathrm{\Psi }({(A_t)}_{t\in T})={\int }_T{\mathrm{\Phi }}_t(A_t)d\mu (t)$. Let $\tilde{A}={(A_t)}_{t\in T}\in \mathcal{C}(T,\mathcal{A})$. It follows from $\sigma (\tilde{A})\subseteq J$ and (3) that

$$f\left(\mathrm{\Psi }\left({(A_t)}_{t\in T}\right)\right)=f(\mathrm{\Psi }(\tilde{A}))\le 2\mathrm{\Psi }(f(\tilde{A}))=2\mathrm{\Psi }\left(f\left({(A_t)}_{t\in T}\right)\right)=2\mathrm{\Psi }\left({(f(A_t))}_{t\in T}\right).$$

 □

In the discrete case, $T=\{1,\dots ,n\}$ in Theorem 5, we get the following result.

Corollary 4 Let $f:J⟶\mathbb{R}$ be an operator P-class function defined on an interval J, let $A_j\in \sigma (J)$ ($1\le j\le n$) and ${\mathrm{\Phi }}_j$ ($1\le j\le n$) be unital positive linear maps on $\mathfrak{B}(\mathcal{H})$. Then

$$f(\sum _{j=1}^n{\mathrm{\Phi }}_j(A_j))\le 2\sum _{j=1}^n{\mathrm{\Phi }}_j(f(A_j)).$$