Recent Research on Jensen's Inequality for Oparators

1. Introduction

The self-adjoint operators on Hilbert spaces with their numerous applications play an important part in the operator theory. The bounds research for self-adjoint operators is a very useful area of this theory. There is no better inequality in bounds examination than Jensen's inequality. It is an extensively used inequality in various fields of mathematics.

Let $I$ be a real interval of any type. A continuous function $f:I\to ℝ$ is said to be operator convex if

$$f\left(\lambda x+(1-\lambda )y\right)\le \lambda f\left(x\right)+(1-\lambda )f\left(y\right)$$uid1

holds for each $\lambda \in [0,1]$ and every pair of self-adjoint operators $x$ and $y$ (acting) on an infinite dimensional Hilbert space $H$ with spectra in $I$ (the ordering is defined by setting $x\le y$ if $y-x$ is positive semi-definite).

Let $f$ be an operator convex function defined on an interval $I.$ Ch. Davis [1] provedThere is small typo in the proof. Davis states that $\phi$ by Stinespring's theorem can be written on the form $\phi \left(x\right)=P\rho \left(x\right)P$ where $\rho$ is a $*$-homomorphism to $B\left(H\right)$ and $P$ is a projection on $H.$ In fact, $H$ may be embedded in a Hilbert space $K$ on which $\rho$ and $P$ acts. The theorem then follows by the calculation $f\left(\phi \right(x\left)\right)=f\left(P\rho \right(x\left)P\right)\le Pf\left(\rho \right(x\left)\right)P=P\rho \left(f\right(x)P=\phi (f\left(x\right)),$ where the pinching inequality, proved by Davis in the same paper, is applied. a Schwarz inequality

$$f\left(\phi \left(x\right)\right)\le \phi \left(f\left(x\right)\right)$$uid3

where $\phi :𝒜\to B\left(K\right)$ is a unital completely positive linear mapping from a $C^{*}$-algebra $𝒜$ to linear operators on a Hilbert space $K,$ and $x$ is a self-adjoint element in $𝒜$ with spectrum in $I.$ Subsequently M. D. Choi [2] noted that it is enough to assume that $\phi$ is unital and positive. In fact, the restriction of $\phi$ to the commutative $C^{*}$-algebra generated by $x$ is automatically completely positive by a theorem of Stinespring.

F. Hansen and G. K. Pedersen [3] proved a Jensen type inequality

$$f\left(\sum _{i=1}^n{a}_i^{*}{x}_i{a}_i\right)\le \sum _{i=1}^n{a}_i^{*}f\left(x_i\right)a_i$$uid4

for operator convex functions $f$ defined on an interval $I=[0,\alpha )$ (with $\alpha \le \infty$ and $f\left(0\right)\le 0)$ and self-adjoint operators $x_1,\cdots ,x_n$ with spectra in $I$ assuming that ${\sum }_{i=1}^n{a}_i^{*}{a}_i=\mathbf{1}.$ The restriction on the interval and the requirement $f\left(0\right)\le 0$ was subsequently removed by B. Mond and J. Pečarić in [4], cf. also [5].

The inequality (▭) is in fact just a reformulation of (▭) although this was not noticed at the time. It is nevertheless important to note that the proof given in [3] and thus the statement of the theorem, when restricted to $n\times n$ matrices, holds for the much richer class of $2n\times 2n$ matrix convex functions. Hansen and Pedersen used (▭) to obtain elementary operations on functions, which leave invariant the class of operator monotone functions. These results then served as the basis for a new proof of Löwner's theorem applying convexity theory and Krein-Milman's theorem.

B. Mond and J. Pečarić [6] proved the inequality

$$f\left(\sum _{i=1}^n{w}_i{\phi }_i\left(x_i\right)\right)\le \sum _{i=1}^n{w}_i{\phi }_i\left(f\left(x_i\right)\right)$$uid5

for operator convex functions $f$ defined on an interval $I,$ where ${\phi }_i:B\left(H\right)\to B\left(K\right)$ are unital positive linear mappings, $x_1,\cdots ,x_n$ are self-adjoint operators with spectra in $I$ and $w_1,\cdots ,w_n$ are are non-negative real numbers with sum one.

Also, B. Mond, J. Pečarić, T. Furuta et al. [6], [7], [8], [9], [10], [11] observed conversed of some special case of Jensen's inequality. So in [10] presented the following generalized converse of a Schwarz inequality (▭)

$$F\left[\phi \left(f\left(A\right)\right),g\left(\phi \left(A\right)\right)\right]\le \underset{m\le t\le M}{max}F\left[f\left(m\right)+\frac{f\left(M\right)-f\left(m\right)}{M-m}(t-m),g\left(t\right)\right]1_{\tilde{n}}$$uid6

for convex functions $f$ defined on an interval $[m,M]$, $m<M$, where $g$ is a real valued continuous function on $[m,M]$, $F(u,v)$ is a real valued function defined on $U\times V$, matrix non-decreasing in $u$, $U\supset f[m,M]$, $V\supset g[m,M]$, $\phi :H_n\to H_{\tilde{n}}$ is a unital positive linear mapping and $A$ is a Hermitian matrix with spectrum contained in $[m,M]$.

There are a lot of new research on the classical Jensen inequality (▭) and its reverse inequalities. For example, J.I. Fujii et all. in [12], [13] expressed these inequalities by externally dividing points.

2. Classic results

In this section we present a form of Jensen's inequality which contains (▭), (▭) and (▭) as special cases. Since the inequality in (▭) was the motivating step for obtaining converses of Jensen's inequality using the so-called Mond-Pečarić method, we also give some results pertaining to converse inequalities in the new formulation.

We recall some definitions. Let $T$ be a locally compact Hausdorff space and let $𝒜$ be a $C^{*}$-algebra of operators on some Hilbert space $H.$ We say that a field ${\left(x_t\right)}_{t\in T}$ of operators in $𝒜$ is continuous if the function $t\mapsto x_t$ is norm continuous on $T.$ If in addition $\mu$ is a Radon measure on $T$ and the function $t\mapsto \parallel x_t\parallel$ is integrable, then we can form the Bochner integral ${\int }_T{x}_{t}d\mu \left(t\right)$, which is the unique element in $𝒜$ such that

$$\varphi \left({\int }_T{x}_{t}d\mu \left(t\right)\right)={\int }_T\varphi \left(x_t\right)d\mu \left(t\right)$$

for every linear functional $\varphi$ in the norm dual ${𝒜}^{*}$.

Assume furthermore that there is a field ${\left({\phi }_t\right)}_{t\in T}$ of positive linear mappings ${\phi }_t:𝒜\to ℬ$ from $𝒜$ to another ${𝒞}^{*}$-algebra $ℬ$ of operators on a Hilbert space $K$. We recall that a linear mapping ${\phi }_t:𝒜\to ℬ$ is said to be a positive mapping if ${\phi }_t\left(x_t\right)\ge 0$ for all $x_t\ge 0$. We say that such a field is continuous if the function $t\mapsto {\phi }_t\left(x\right)$ is continuous for every $x\in 𝒜.$ Let the ${𝒞}^{*}$-algebras include the identity operators and the function $t\mapsto {\phi }_t\left(1_H\right)$ be integrable with ${\int }_T{\phi }_t\left(1_H\right)d\mu \left(t\right)=k{1}_K$ for some positive scalar $k$. Specially, if ${\int }_T{\phi }_t\left(1_H\right)d\mu \left(t\right)=1_K,$ we say that a field ${\left({\phi }_t\right)}_{t\in T}$ is unital.

Let $B\left(H\right)$ be the $C^{*}$-algebra of all bounded linear operators on a Hilbert space $H$. We define bounds of an operator $x\in B\left(H\right)$ by

$$m_x=\underset{\parallel \xi \parallel =1}{inf}⟨x\xi ,\xi ⟩\text{and}{M}_x=\underset{\parallel \xi \parallel =1}{sup}⟨x\xi ,\xi ⟩$$uid7

for $\xi \in H$. If $\mathrm{𝖲𝗉}\left(x\right)$ denotes the spectrum of $x$, then $\mathrm{𝖲𝗉}\left(x\right)\subseteq [m_x,M_x]$.

For an operator $x\in B\left(H\right)$ we define operators $\left|x\right|$, $x^{+}$, $x^{-}$ by

$$\left|x\right|={\left(x^{*}x\right)}^{1/2},x^{+}=\left(\right|x|+x)/2,x^{-}=\left(\right|x|-x)/2$$

Obviously, if $x$ is self-adjoint, then $\left|x\right|={\left(x^2\right)}^{1/2}$ and $x^{+},x^{-}\ge 0$ (called positive and negative parts of $x=x^{+}-x^{-}$).

2.1. Jensen's inequality with operator convexity

Firstly, we give a general formulation of Jensen's operator inequality for a unital field of positive linear mappings (see [14]).

Theorem 1 Let $f:I\to ℝ$ be an operator convex function defined on an interval $I$ and let $𝒜$ and $ℬ$ be unital $C^{*}$-algebras acting on a Hilbert space $H$ and $K$ respectively. If ${\left({\phi }_t\right)}_{t\in T}$ is a unital field of positive linear mappings ${\phi }_t:𝒜\to ℬ$ defined on a locally compact Hausdorff space $T$ with a bounded Radon measure $\mu ,$ then the inequality

$$f\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\le {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)$$uid10

holds for every bounded continuous field ${\left(x_t\right)}_{t\in T}$ of self-adjoint elements in $𝒜$ with spectra contained in $I.$

We first note that the function $t\mapsto {\phi }_t\left(x_t\right)\in ℬ$ is continuous and bounded, hence integrable with respect to the bounded Radon measure $\mu .$ Furthermore, the integral is an element in the multiplier algebra $M\left(ℬ\right)$ acting on $K.$ We may organize the set $CB(T,𝒜)$ of bounded continuous functions on $T$ with values in $𝒜$ as a normed involutive algebra by applying the point-wise operations and setting

$$\parallel {\left(y_t\right)}_{t\in T}\parallel =\underset{t\in T}{sup}\parallel y_t\parallel {\left(y_t\right)}_{t\in T}\in CB(T,𝒜)$$

and it is not difficult to verify that the norm is already complete and satisfy the $C^{*}$-identity. In fact, this is a standard construction in $C^{*}$-algebra theory. It follows that $f\left({\left(x_t\right)}_{t\in T}\right)={\left(f\left(x_t\right)\right)}_{t\in T}$. We then consider the mapping

$$\pi :CB(T,𝒜)\to M\left(ℬ\right)\subseteq B\left(K\right)$$

defined by setting

$$\pi \left({\left(x_t\right)}_{t\in T}\right)={\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)$$

and note that it is a unital positive linear map. Setting $x={\left(x_t\right)}_{t\in T}\in CB(T,𝒜),$ we use inequality (▭) to obtain

$$f\left(\pi \left({\left(x_t\right)}_{t\in T}\right)\right)=f\left(\pi \left(x\right)\right)\le \pi \left(f\left(x\right)\right)=\pi \left(f\left({\left(x_t\right)}_{t\in T}\right)\right)=\pi \left({\left(f\left(x_t\right)\right)}_{t\in T}\right)$$

but this is just the statement of the theorem.

2.2. Converses of Jensen's inequality

In the present context we may obtain results of the Li-Mathias type cf. Chapter 3[15] and [16], [17].

Theorem 2 Let $T$ be a locally compact Hausdorff space equipped with a bounded Radon measure $\mu$. Let ${\left(x_t\right)}_{t\in T}$ be a bounded continuous field of self-adjoint elements in a unital $C^{*}$-algebra $𝒜$ with spectra in $[m,M]$, $m<M$. Furthermore, let ${\left({\phi }_t\right)}_{t\in T}$ be a field of positive linear mappings ${\phi }_t:𝒜\to ℬ$ from $𝒜$ to another unital $C^{*}-$algebra $ℬ$, such that the function $t\mapsto {\phi }_t\left(1_H\right)$ is integrable with ${\int }_T{\phi }_t\left(1_H\right)d\mu \left(t\right)=k{1}_K$ for some positive scalar $k$. Let $m_x$ and $M_x$, $m_x\le M_x$, be the bounds of the self-adjoint operator $x={\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)$ and $f:[m,M]\to ℝ$, $g:[m_x,M_x]\to ℝ$, $F:U\times V\to ℝ$ be functions such that $\left(kf\right)\left([m,M]\right)\subset U,$ $g\left([m_x,M_x]\right)\subset V$ and $F$ is bounded. If $F$ is operator monotone in the first variable, then

$$\begin{array}{c}{\displaystyle \underset{m_x\le z\le M_x}{inf}F\left[k·h_1\left(\frac{1}{k}z\right),g\left(z\right)\right]1_K\le F\left[{\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right),g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\right]}\\ {\displaystyle \le \underset{m_x\le z\le M_x}{sup}F\left[k·h_2\left(\frac{1}{k}z\right),g\left(z\right)\right]1_K}\end{array}$$uid13

holds for every operator convex function $h_1$ on $[m,M]$ such that $h_1\le f$ and for every operator concave function $h_2$ on $[m,M]$ such that $h_2\ge f$.

We prove only RHS of (▭). Let $h_2$ be operator concave function on $[m,M]$ such that $f\left(z\right)\le h_2\left(z\right)$ for every $z\in [m,M]$. By using the functional calculus, it follows that $f\left(x_t\right)\le h_2\left(x_t\right)$ for every $t\in T$. Applying the positive linear mappings ${\phi }_t$ and integrating, we obtain

$${\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)\le {\int }_T{\phi }_t\left(h_2\left(x_t\right)\right)d\mu \left(t\right)$$

Furthermore, replacing ${\phi }_t$ by $\frac{1}{k}{\phi }_t$ in Theorem ▭, we obtain ${\displaystyle \frac{1}{k}{\int }_T{\phi }_t\left(h_2\left(x_t\right)\right)d\mu \left(t\right)\le h_2\left(\frac{1}{k}{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)}$, which gives ${\displaystyle {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)\le k·h_2\left(\frac{1}{k}{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)}$. Since $m_x{1}_K\le {\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\le M_x{1}_K$, then using operator monotonicity of $F(·,v)$ we obtain

$$\begin{array}{cc}& {\displaystyle F\left[{\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right),g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\right]}\\ & {\displaystyle \le F\left[k·h_2\left(\frac{1}{k}{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right),g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\right]\le \underset{m_x\le z\le M_x}{sup}F\left[k·h_2\left(\frac{1}{k}z\right),g\left(z\right)\right]1_K}\end{array}$$uid14

Applying RHS of (▭) for a convex function $f$ (or LHS of (▭) for a concave function $f$) we obtain the following generalization of (▭).

Theorem 3 Let ${\left(x_t\right)}_{t\in T}$, $m_x$, $M_x$ and ${\left({\phi }_t\right)}_{t\in T}$ be as in Theorem ▭. Let $f:[m,M]\to ℝ$, $g:[m_x,M_x]\to ℝ$, $F:U\times V\to ℝ$ be functions such that $\left(kf\right)\left([m,M]\right)\subset U,$ $g\left([m_x,M_x]\right)\subset V$ and $F$ is bounded. If $F$ is operator monotone in the first variable and $f$ is convex on the interval $[m,M]$, then

$$\begin{array}{c}{\displaystyle F\left[{\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right),g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\right]}\\ {\displaystyle \le \underset{m_x\le z\le M_x}{sup}F\left[\frac{Mk-z}{M-m}f\left(m\right)+\frac{z-km}{M-m}f\left(M\right),g\left(z\right)\right]1_K}\end{array}$$uid16

In the dual case (when $f$ is concave) the opposite inequalities hold in (▭) with $inf$ instead of $sup$.

We prove only the convex case. For convex $f$ the inequality $f\left(z\right)\le \frac{M-z}{M-m}f\left(m\right)+\frac{z-m}{M-m}f\left(M\right)$ holds for every $z\in [m,M]$. Thus, by putting $h_2\left(z\right)=\frac{M-z}{M-m}f\left(m\right)+\frac{z-m}{M-m}f\left(M\right)$ in (▭) we obtain (▭). Numerous applications of the previous theorem can be given (see [15]). Applying Theorem ▭ for the function $F(u,v)=u-\alpha v$ and $k=1$, we obtain the following generalization of Theorem 2.4[15].

Corollary 4 Let ${\left(x_t\right)}_{t\in T}$, $m_x$, $M_x$ be as in Theorem ▭ and ${\left({\phi }_t\right)}_{t\in T}$ be a unital field of positive linear mappings ${\phi }_t:𝒜\to ℬ$. If $f:[m,M]\to ℝ$ is convex on the interval $[m,M]$, $m<M$, and $g:[m,M]\to ℝ$, then for any $\alpha \in ℝ$

$${\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)\le \alpha g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)+C{1}_K$$uid18

where

$$\begin{array}{ccc}\hfill C& =& \underset{m_x\le z\le M_x}{max}\left\{\frac{M-z}{M-m}f\left(m\right)+\frac{z-m}{M-m}f\left(M\right)-\alpha g\left(z\right)\right\}\hfill \\ & \le & \underset{m\le z\le M}{max}\left\{\frac{M-z}{M-m}f\left(m\right)+\frac{z-m}{M-m}f\left(M\right)-\alpha g\left(z\right)\right\}\hfill \end{array}$$

If furthermore $\alpha g$ is strictly convex differentiable, then the constant $C\equiv C(m,M,f,g,\alpha )$ can be written more precisely as

$$C=\frac{M-z_0}{M-m}f\left(m\right)+\frac{z_0-m}{M-m}f\left(M\right)-\alpha g\left(z_0\right)$$

where

$$z_0=\left\{\begin{array}{cc}{g}^{\text{'}-1}\left(\frac{f\left(M\right)-f\left(m\right)}{\alpha (M-m)}\right)\hfill & \text{if}\alpha g^{\text{'}}\left(m_x\right)\le \frac{f\left(M\right)-f\left(m\right)}{M-m}\le \alpha g^{\text{'}}\left(M_x\right)\hfill \\ m_x\hfill & \text{if}\alpha g^{\text{'}}\left(m_x\right)\ge \frac{f\left(M\right)-f\left(m\right)}{M-m}\hfill \\ M_x\hfill & \text{if}\alpha g^{\text{'}}\left(M_x\right)\le \frac{f\left(M\right)-f\left(m\right)}{M-m}\hfill \end{array}\right.$$

In the dual case (when $f$ is concave and $\alpha g$ is strictly concave differentiable) the opposite inequalities hold in (▭) with $min$ instead of $max$ with the opposite condition while determining $z_0$.

3. Inequalities with conditions on spectra

In this section we present Jensens's operator inequality for real valued continuous convex functions with conditions on the spectra of the operators. A discrete version of this result is given in [18]. Also, we obtain generalized converses of Jensen's inequality under the same conditions.

Operator convexity plays an essential role in (▭). In fact, the inequality (▭) will be false if we replace an operator convex function by a general convex function. For example, M.D. Choi in Remark 2.6[2] considered the function $f\left(t\right)=t^4$ which is convex but not operator convex. He demonstrated that it is sufficient to put $\dim H=3$, so we have the matrix case as follows. Let $\Phi :M_3\left(ℂ\right)\to M_2\left(ℂ\right)$ be the contraction mapping $\Phi \left({\left(a_{ij}\right)}_{1\le i,j\le 3}\right)={\left(a_{ij}\right)}_{1\le i,j\le 2}$. If $A=\left(\begin{array}{ccc}1& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right),$ then $\Phi {\left(A\right)}^4=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\neg \le \left(\begin{array}{cc}9& 5\\ 5& 3\end{array}\right)=\Phi \left(A^4\right)$ and no relation between $\Phi {\left(A\right)}^4$ and $\Phi \left(A^4\right)$ under the operator order.

Example 5 It appears that the inequality (▭) will be false if we replace the operator convex function by a general convex function. We give a small example for the matrix cases and $T=\{1,2\}$. We define mappings ${\Phi }_1,{\Phi }_2:M_3\left(ℂ\right)\to M_2\left(ℂ\right)$ by ${\Phi }_1\left({\left(a_{ij}\right)}_{1\le i,j\le 3}\right)=\frac{1}{2}{\left(a_{ij}\right)}_{1\le i,j\le 2}$, ${\Phi }_2={\Phi }_1$. Then ${\Phi }_1\left(I_3\right)+{\Phi }_2\left(I_3\right)=I_2$.

1. If

   $$X_1=2\left(\begin{array}{ccc}1& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right)\text{and}{X}_2=2\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$$

   then

   $${\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4=\left(\begin{array}{cc}16& 0\\ 0& 0\end{array}\right)\neg \le \left(\begin{array}{cc}80& 40\\ 40& 24\end{array}\right)={\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)$$

   Given the above, there is no relation between ${\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4$ and ${\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)$ under the operator order. We observe that in the above case the following stands $X={\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)=\left(\begin{array}{cc}2& 0\\ 0& 0\end{array}\right)$ and $[m_x,M_x]=[0,2]$, $[m_1,M_1]\subset [-1.60388,4.49396]$, $[m_2,M_2]=[0,2]$, i.e.

   $$(m_x,M_x)\subset [m_1,M_1]\cup [m_2,M_2]$$

   (see Fig. 1.a).

Figure 1.

Spectral conditions for a convex function f

1. If

   $$X_1=\left(\begin{array}{ccc}-14& 0& 1\\ 0& -2& -1\\ 1& -1& -1\end{array}\right)\text{and}{X}_2=\left(\begin{array}{ccc}15& 0& 0\\ 0& 2& 0\\ 0& 0& 15\end{array}\right)$$

   then

   $${\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4=\left(\begin{array}{cc}\frac{1}{16}& 0\\ 0& 0\end{array}\right)<\left(\begin{array}{cc}89660& -247\\ -247& 51\end{array}\right)={\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)$$

   So we have that an inequality of type (▭) now is valid. In the above case the following stands $X={\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)=\left(\begin{array}{cc}\frac{1}{2}& 0\\ 0& 0\end{array}\right)$ and $[m_x,M_x]=[0,0.5]$, $[m_1,M_1]\subset [-14.077,-0.328566]$, $[m_2,M_2]=[2,15]$, i.e.

   $$(m_x,M_x)\cap [m_1,M_1]=\varnothing \text{and}(m_x,M_x)\cap [m_2,M_2]=\varnothing$$

   (see Fig. 1.b).

3.1. Jensen's inequality without operator convexity

It is no coincidence that the inequality (▭) is valid in Example ▭-II). In the following theorem we prove a general result when Jensen's operator inequality (▭) holds for convex functions.

Theorem 6 Let ${\left(x_t\right)}_{t\in T}$ be a bounded continuous field of self-adjoint elements in a unital $C^{*}$-algebra $𝒜$ defined on a locally compact Hausdorff space $T$ equipped with a bounded Radon measure $\mu$. Let $m_t$ and $M_t$, $m_t\le M_t$, be the bounds of $x_t$, $t\in T$. Let ${\left({\phi }_t\right)}_{t\in T}$ be a unital field of positive linear mappings ${\phi }_t:𝒜\to ℬ$ from $𝒜$ to another unital $C^{*}-$algebra $ℬ$. If

$$(m_x,M_x)\cap [m_t,M_t]=\varnothing ,t\in T$$

where $m_x$ and $M_x$, $m_x\le M_x$, are the bounds of the self-adjoint operator $x={\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)$, then

$$f\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\le {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)$$uid25

holds for every continuous convex function $f:I\to ℝ$ provided that the interval $I$ contains all $m_t,M_t$.

If $f:I\to ℝ$ is concave, then the reverse inequality is valid in (▭).

We prove only the case when $f$ is a convex function. If we denote $m=\underset{t\in T}{inf}\left\{m_t\right\}$ and $M=\underset{t\in T}{sup}\left\{M_t\right\}$, then $[m,M]\subseteq I$ and $m{1}_H\le A_t\le M{1}_H$, $t\in T$. It follows $m{1}_K\le {\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\le M{1}_K$. Therefore $[m_x,M_x]\subseteq [m,M]\subseteq I$.

a) Let $m_x<M_x$. Since $f$ is convex on $[m_x,M_x]$, then

$$f\left(z\right)\le \frac{M_x-z}{M_x-m_x}f\left(m_x\right)+\frac{z-m_x}{M_x-m_x}f\left(M_x\right),z\in [m_x,M_x]$$uid26

but since $f$ is convex on $[m_t,M_t]$ and since $(m_x,M_x)\cap [m_t,M_t]=\varnothing$, then

$$f\left(z\right)\ge \frac{M_x-z}{M_x-m_x}f\left(m_x\right)+\frac{z-m_x}{M_x-m_x}f\left(M_x\right),z\in [m_t,M_t],t\in T$$uid27

Since $m_x{1}_K\le {\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\le M_x{1}_K,$ then by using functional calculus, it follows from (▭)

$$f\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\le \frac{M_x{1}_K-{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)}{M_x-m_x}f\left(m_x\right)+\frac{{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)-m_x{1}_K}{M_x-m_x}f\left(M_x\right)$$uid28

On the other hand, since $m_t{1}_H\le x_t\le M_t{1}_H$, $t\in T$, then by using functional calculus, it follows from (▭)

$$f\left(x_t\right)\ge \frac{M_x{1}_H-x_t}{M_x-m_x}f\left(m_x\right)+\frac{x_t-m_x{1}_H}{M_x-m_x}f\left(M_x\right),t\in T$$

Applying a positive linear mapping ${\phi }_t$ and summing, we obtain

$${\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)\ge \frac{M_x{1}_K-{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)}{M_x-m_x}f\left(m_x\right)+\frac{{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)-m_x{1}_K}{M_x-m_x}f\left(M_x\right)$$uid29

since ${\int }_T{\phi }_t\left(1_H\right)d\mu \left(t\right)=1_K$. Combining the two inequalities (▭) and (▭), we have the desired inequality (▭).

b) Let $m_x=M_x$. Since $f$ is convex on $[m,M]$, we have

$$f\left(z\right)\ge f\left(m_x\right)+l\left(m_x\right)(z-m_x)\text{for}\text{every}z\in [m,M]$$uid30

where $l$ is the subdifferential of $f$. Since $m{1}_H\le x_t\le M{1}_H$, $t\in T$, then by using functional calculus, applying a positive linear mapping ${\phi }_t$ and summing, we obtain from (▭)

$${\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)\ge f\left(m_x\right)1_K+l\left(m_x\right)\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)-m_x{1}_K\right)$$uid31

Since $m_x{1}_K={\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right),$ it follows

$${\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)\ge f\left(m_x\right)1_K=f\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)$$uid32

which is the desired inequality (▭). Putting ${\phi }_t\left(y\right)=a_{t}y$ for every $y\in 𝒜$, where $a_t\ge 0$ is a real number, we obtain the following obvious corollary of Theorem ▭.

Corollary 7 Let ${\left(x_t\right)}_{t\in T}$ be a bounded continuous field of self-adjoint elements in a unital $C^{*}$-algebra $𝒜$ defined on a locally compact Hausdorff space $T$ equipped with a bounded Radon measure $\mu$. Let $m_t$ and $M_t$, $m_t\le M_t$, be the bounds of $x_t$, $t\in T$. Let ${\left(a_t\right)}_{t\in T}$ be a continuous field of nonnegative real numbers such that ${\int }_T{a}_{t}d\mu \left(t\right)=1$. If

$$(m_x,M_x)\cap [m_t,M_t]=\varnothing ,t\in T$$

where $m_x$ and $M_x$, $m_x\le M_x$, are the bounds of the self-adjoint operator $x={\int }_T{a}_t{x}_{t}d\mu \left(t\right)$, then

$$f\left({\int }_T{a}_t{x}_{t}d\mu \left(t\right)\right)\le {\int }_T{a}_{t}f\left(x_t\right)d\mu \left(t\right)$$uid34

holds for every continuous convex function $f:I\to ℝ$ provided that the interval $I$ contains all $m_t,M_t$.

3.2. Converses of Jensen's inequality with conditions on spectra

Using the condition on spectra we obtain the following extension of Theorem ▭.

Theorem 8 Let ${\left(x_t\right)}_{t\in T}$ be a bounded continuous field of self-adjoint elements in a unital $C^{*}$-algebra $𝒜$ defined on a locally compact Hausdorff space $T$ equipped with a bounded Radon measure $\mu$. Furthermore, let ${\left({\phi }_t\right)}_{t\in T}$ be a field of positive linear mappings ${\phi }_t:𝒜\to ℬ$ from $𝒜$ to another unital $C^{*}-$algebra $ℬ$, such that the function $t\mapsto {\phi }_t\left(1_H\right)$ is integrable with ${\int }_T{\phi }_t\left(1_H\right)d\mu \left(t\right)=k{1}_K$ for some positive scalar $k$. Let $m_t$ and $M_t$, $m_t\le M_t$, be the bounds of $x_t$, $t\in T$, $m=\underset{t\in T}{inf}\left\{m_t\right\}$, $M=\underset{t\in T}{sup}\left\{M_t\right\}$, and $m_x$ and $M_x$, $m_x<M_x$, be the bounds of $x={\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)$. If

$$(m_x,M_x)\cap [m_t,M_t]=\varnothing ,t\in T$$

and $f:[m,M]\to ℝ$, $g:[m_x,M_x]\to ℝ$, $F:U\times V\to ℝ$ are functions such that $\left(kf\right)\left([m,M]\right)\subset U,$ $g\left([m_x,M_x]\right)\subset V$, $f$ is convex, $F$ is bounded and operator monotone in the first variable, then

$$\begin{array}{c}{\displaystyle \underset{m_x\le z\le M_x}{inf}F\left[\frac{M_{x}k-z}{M_x-m_x}f\left(m_x\right)+\frac{z-k{m}_x}{M_x-m_x}f\left(M_x\right),g\left(z\right)\right]1_K}\\ {\displaystyle F\left[{\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right),g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\right]}\\ {\displaystyle \le \underset{m_x\le z\le M_x}{sup}F\left[\frac{Mk-z}{M-m}f\left(m\right)+\frac{z-km}{M-m}f\left(M\right),g\left(z\right)\right]1_K}\end{array}$$uid37

In the dual case (when $f$ is concave) the opposite inequalities hold in (▭) by replacing $inf$ and $sup$ with $sup$ and $inf$, respectively.

We prove only LHS of (▭). It follows from (▭) (compare it to (▭))

$${\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)\ge \frac{M_{x}k{1}_K-{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)}{M_x-m_x}f\left(m_x\right)+\frac{{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)-m_{x}k{1}_K}{M_x-m_x}f\left(M_x\right)$$

since ${\int }_T{\phi }_t\left(1_H\right)d\mu \left(t\right)=k{1}_K$. By using operator monotonicity of $F(·,v)$ we obtain

$$\begin{array}{cc}& {\displaystyle F\left[{\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right),g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\right]}\hfill \end{array}\ge F\left[\frac{M_{x}k{1}_K-{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)}{M_x-m_x}f\left(m_x\right)+\frac{{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)-m_{x}k{1}_K}{M_x-m_x}f\left(M_x\right),g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\right]$$

mxzMx F[Mx k-zMx-mxf(mx)+z-kmxMx-mxf(Mx),g(z)] 1K

$$$$

Putting $F(u,v)=u-\alpha v$ or $F(u,v)=v^{-1/2}u{v}^{-1/2}$ in Theorem ▭, we obtain the next corollary.

Corollary 9 Let ${\left(x_t\right)}_{t\in T}$, $m_t$, $M_t$, $m_x$, $M_x$, $m$, $M$, ${\left({\phi }_t\right)}_{t\in T}$ be as in Theorem ▭ and $f:[m,M]\to ℝ$, $g:[m_x,M_x]\to ℝ$ be continuous functions. If

$$(m_x,M_x)\cap [m_t,M_t]=\varnothing ,t\in T$$

and $f$ is convex, then for any $\alpha \in ℝ$

$$\begin{array}{c}{\displaystyle \underset{m_x\le z\le M_x}{min}\left\{\frac{M_{x}k-z}{M_x-m_x}f\left(m_x\right)+\frac{z-k{m}_x}{M_x-m_x}f\left(M_x\right)-g\left(z\right)\right\}{1}_K+\alpha g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)}\\ {\displaystyle \le {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)}\\ {\displaystyle \le \alpha g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)+\underset{m_x\le z\le M_x}{max}\left\{\frac{Mk-z}{M-m}f\left(m\right)+\frac{z-km}{M-m}f\left(M\right)-g\left(z\right)\right\}{1}_K}\end{array}$$uid39

If additionally $g>0$ on $[m_x,M_x]$, then

$$\begin{array}{c}{\displaystyle \underset{m_x\le z\le M_x}{min}\left\{\frac{\frac{M_{x}k-z}{M_x-m_x}f\left(m_x\right)+\frac{z-k{m}_x}{M_x-m_x}f\left(M_x\right)}{g\left(z\right)}\right\}g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)}\\ {\displaystyle \le {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)\le \underset{m_x\le z\le M_x}{max}\left\{\frac{\frac{Mk-z}{M-m}f\left(m\right)+\frac{z-km}{M-m}f\left(M\right)}{g\left(z\right)}\right\}g\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)}\end{array}$$uid40

In the dual case (when $f$ is concave) the opposite inequalities hold in (▭) by replacing $min$ and $max$ with $max$ and $min$, respectively. If additionally $g>0$ on $[m_x,M_x]$, then the opposite inequalities also hold in (▭) by replacing $min$ and $max$ with $max$ and $min$, respectively.

4. Refined Jensen's inequality

In this section we present a refinement of Jensen's inequality for real valued continuous convex functions given in Theorem ▭. A discrete version of this result is given in [19].

To obtain our result we need the following two lemmas.

Lemma 10 Let $f$ be a convex function on an interval $I$, $m,M\in I$ and $p_1,p_2\in [0,1]$ such that $p_1+p_2=1$. Then

$$min\{p_1,p_2\}\left[f\left(m\right)+f\left(M\right)-2f\left(\frac{m+M}{2}\right)\right]\le p_{1}f\left(m\right)+p_{2}f\left(M\right)-f(p_{1}m+p_{2}M)$$uid42

These results follows from Theorem 1, p. 717[20].

Lemma 11 Let $x$ be a bounded self-adjoint elements in a unital $C^{*}$-algebra $𝒜$ of operators on some Hilbert space $H$. If the spectrum of $x$ is in $[m,M]$, for some scalars $m<M$, then

$$\begin{array}{ccc}\hfill f\left(x\right)& \le & \frac{M{1}_H-x}{M-m}f\left(m\right)+\frac{x-m{1}_H}{M-m}f\left(M\right)-{\delta }_f\tilde{x}\hfill \\ \hfill (\text{resp.}f\left(x\right)& \ge & \frac{M{1}_H-x}{M-m}f\left(m\right)+\frac{x-m{1}_H}{M-m}f\left(M\right)+{\delta }_f\tilde{x})\hfill \end{array}$$uid44

holds for every continuous convex $($resp. concave$)$ function $f:[m,M]\to ℝ$, where

$$\begin{array}{cc}\hfill & {\delta }_f=f\left(m\right)+f\left(M\right)-2f\left(\frac{m+M}{2}\right)(\text{resp.}{\delta }_f=2f\left(\frac{m+M}{2}\right)-f\left(m\right)-f\left(M\right))\\ & \text{and}\tilde{x}=\frac{1}{2}{1}_H-\frac{1}{M-m}\left|x-\frac{m+M}{2}{1}_H\right|\end{array}$$

We prove only the convex case. It follows from (▭) that

$$\begin{array}{ccc}\hfill f\left(p_{1}m+p_{2}M\right)& \le & p_{1}f\left(m\right)+p_{2}f\left(M\right)\hfill \\ & -& min\{p_1,p_2\}\left(f\left(m\right)+f\left(M\right)-2f\left(\frac{m+M}{2}\right)\right)\hfill \end{array}$$uid45

for every $p_1,p_2\in [0,1]$ such that $p_1+p_2=1$ . For any $z\in [m,M]$ we can write

$$f\left(z\right)=f\left(\frac{M-z}{M-m}m+\frac{z-m}{M-m}M\right)$$

Then by using (▭) for $p_1=\frac{M-z}{M-m}$ and $p_2=\frac{z-m}{M-m}$ we obtain

$$\begin{array}{ccc}\hfill f\left(z\right)& \le & {\displaystyle \frac{M-z}{M-m}f\left(m\right)+\frac{z-m}{M-m}f\left(M\right)}\hfill \\ & -& {\displaystyle \left(\frac{1}{2}-\frac{1}{M-m}\left|z-\frac{m+M}{2}\right|\right)\left(f\left(m\right)+f\left(M\right)-2f\left(\frac{m+M}{2}\right)\right)}\hfill \end{array}$$uid46

since

$$min\left\{\frac{M-z}{M-m},\frac{z-m}{M-m}\right\}=\frac{1}{2}-\frac{1}{M-m}\left|z-\frac{m+M}{2}\right|$$

Finally we use the continuous functional calculus for a self-adjoint operator $x$: $f,g\in 𝒞\left(I\right),Sp\left(x\right)\subseteq I$ and $f\le g$ on $I$ implies $f\left(x\right)\le g\left(x\right)$; and $h\left(z\right)=\left|z\right|$ implies $h\left(x\right)=\left|x\right|$. Then by using (▭) we obtain the desired inequality (▭).

Theorem 12 Let ${\left(x_t\right)}_{t\in T}$ be a bounded continuous field of self-adjoint elements in a unital $C^{*}$-algebra $𝒜$ defined on a locally compact Hausdorff space $T$ equipped with a bounded Radon measure $\mu$. Let $m_t$ and $M_t$, $m_t\le M_t$, be the bounds of $x_t$, $t\in T$. Let ${\left({\phi }_t\right)}_{t\in T}$ be a unital field of positive linear mappings ${\phi }_t:𝒜\to ℬ$ from $𝒜$ to another unital $C^{*}-$algebra $ℬ$. Let

$$(m_x,M_x)\cap [m_t,M_t]=\varnothing ,t\in T,\text{and}m<M$$

where $m_x$ and $M_x$, $m_x\le M_x$, be the bounds of the operator $x={\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)$ and

$$m=sup\left\{M_t:M_t\le m_x,t\in T\right\},M=inf\left\{m_t:m_t\ge M_x,t\in T\right\}$$

If $f:I\to ℝ$ is a continuous convex $($resp. concave$)$ function provided that the interval $I$ contains all $m_t,M_t$, then

$$f\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\le {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)-{\delta }_f\tilde{x}\le {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)$$uid48

$($resp.

$$f\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\ge {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)-{\delta }_f\tilde{x}\ge {\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right))$$uid49

holds, where

$$\begin{array}{ccc}\hfill {\delta }_f\equiv {\delta }_f(\overline{m},\overline{M})& =& f\left(\overline{m}\right)+f\left(\overline{M}\right)-2f\left(\frac{\overline{m}+\overline{M}}{2}\right)\hfill \\ \hfill (\text{resp.}{\delta }_f\equiv {\delta }_f(\overline{m},\overline{M})& =& 2f\left(\frac{\overline{m}+\overline{M}}{2}\right)-f\left(\overline{m}\right)-f\left(\overline{M}\right))\hfill \\ \hfill \tilde{x}\equiv {\tilde{x}}_x(\overline{m},\overline{M})& =& \frac{1}{2}{1}_K-\frac{1}{\overline{M}-\overline{m}}\left|x-\frac{\overline{m}+\overline{M}}{2}{1}_K\right|\hfill \end{array}$$uid50

and  $\overline{m}\in [m,m_A]$, $\overline{M}\in [M_A,M]$, $\overline{m}<\overline{M}$,   are arbitrary numbers.

We prove only the convex case. Since $x={\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\in ℬ$ is the self-adjoint elements such that $\overline{m}{1}_K\le m_x{1}_K\le {\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\le M_x{1}_K\le \overline{M}{1}_K$ and $f$ is convex on $[\overline{m},\overline{M}]\subseteq I$, then by Lemma ▭ we obtain

$$f\left({\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)\right)\le \frac{\overline{M}{1}_K-{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)}{\overline{M}-\overline{m}}f\left(\overline{m}\right)+\frac{{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)-\overline{m}{1}_K}{\overline{M}-\overline{m}}f\left(\overline{M}\right)-{\delta }_f\tilde{x}$$uid51

where ${\delta }_f$ and $\tilde{x}$ are defined by (▭).

But since $f$ is convex on $[m_t,M_t]$ and $(m_x,M_x)\cap [m_t,M_t]=\varnothing$ implies $(\overline{m},\overline{M})\cap [m_t,M_t]=\varnothing$, then

$$f\left(x_t\right)\ge \frac{\overline{M}{1}_H-x_t}{\overline{M}-\overline{m}}f\left(\overline{m}\right)+\frac{x_t-\overline{m}{1}_H}{\overline{M}-\overline{m}}f\left(\overline{M}\right),t\in T$$

Applying a positive linear mapping ${\phi }_t$, integrating and adding $-{\delta }_f\tilde{x}$, we obtain

$${\int }_T{\phi }_t\left(f\left(x_t\right)\right)d\mu \left(t\right)-{\delta }_f\tilde{x}\ge \frac{\overline{M}{1}_K-{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)}{\overline{M}-\overline{m}}f\left(\overline{m}\right)+\frac{{\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)-\overline{m}{1}_K}{\overline{M}-\overline{m}}f\left(\overline{M}\right)-{\delta }_f\tilde{x}$$uid52

since ${\int }_T{\phi }_t\left(1_H\right)d\mu \left(t\right)=1_K$. Combining the two inequalities (▭) and (▭), we have LHS of (▭). Since ${\delta }_f\ge 0$ and $\tilde{x}\ge 0$, then we have RHS of (▭).

If $m<M$ and $m_x=M_x$, then the inequality (▭) holds, but ${\delta }_f(m_x,M_x)\tilde{x}(m_x,M_x)$ is not defined (see Example ▭ I) and II)).

Example 13 We give examples for the matrix cases and $T=\{1,2\}$. Then we have refined inequalities given in Fig. 2.

Figure 2.

Refinement for two operators and a convex function f

We put $f\left(t\right)=t^4$ which is convex but not operator convex in (▭). Also, we define mappings ${\Phi }_1,{\Phi }_2:M_3\left(ℂ\right)\to M_2\left(ℂ\right)$ as follows: ${\Phi }_1\left({\left(a_{ij}\right)}_{1\le i,j\le 3}\right)=\frac{1}{2}{\left(a_{ij}\right)}_{1\le i,j\le 2}$, ${\Phi }_2={\Phi }_1$ (then ${\Phi }_1\left(I_3\right)+{\Phi }_2\left(I_3\right)=I_2$).

I) First, we observe an example when ${\delta }_f\tilde{X}$ is equal to the difference RHS and LHS of Jensen's inequality. If $X_1=-3I_3$ and $X_2=2I_3$, then $X={\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)=-0.5I_2$, so $m=-3$, $M=2$. We also put $\overline{m}=-3$ and $\overline{M}=2$. We obtain

$${\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4=0.0625I_2<48.5I_2={\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)$$

and its improvement

$${\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4=0.0625I_2={\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)-48.4375I_2$$

since ${\delta }_f=96.875$, $\tilde{X}=0.5I_2.$ We remark that in this case $m_x=M_x=-1/2$ and $\tilde{X}(m_x,M_x)$ is not defined.

II) Next, we observe an example when ${\delta }_f\tilde{X}$ is not equal to the difference RHS and LHS of Jensen's inequality and $m_x=M_x$. If

$$X_1=\left(\begin{array}{ccc}-1& 0& 0\\ 0& -2& 0\\ 0& 0& -1\end{array}\right),X_2=\left(\begin{array}{ccc}2& 0& 0\\ 0& 3& 0\\ 0& 0& 4\end{array}\right),\text{then}X=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\text{and}m=-1,M=2$$

In this case $\tilde{x}(m_x,M_x)$ is not defined, since $m_x=M_x=1/2$. We have

$${\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4=\frac{1}{16}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)<\left(\begin{array}{cc}\frac{17}{2}& 0\\ 0& \frac{97}{2}\end{array}\right)={\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)$$

and putting $\overline{m}=-1$, $\overline{M}=2$ we obtain ${\delta }_f=135/8$, $\tilde{X}=I_2/2$ which give the following improvement

$${\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4=\frac{1}{16}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)<\frac{1}{16}\left(\begin{array}{cc}1& 0\\ 0& 641\end{array}\right)={\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)-\frac{135}{16}\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$$

III) Next, we observe an example with matrices that are not special. If

$$X_1=\left(\begin{array}{ccc}-4& 1& 1\\ 1& -2& -1\\ 1& -1& -1\end{array}\right)\text{and}{X}_2=\left(\begin{array}{ccc}5& -1& -1\\ -1& 2& 1\\ -1& 1& 3\end{array}\right),\text{then}X=\frac{1}{2}\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)$$

so $m_1=-4.8662$, $M_1=-0.3446$, $m_2=1.3446$, $M_2=5.8662$, $m=-0.3446$, $M=1.3446$ and we put $\overline{m}=m$, $\overline{M}=M$ (rounded to four decimal places). We have

$${\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4=\frac{1}{16}\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)<\left(\begin{array}{cc}\frac{1283}{2}& -255\\ -255& \frac{237}{2}\end{array}\right)={\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)$$

and its improvement

$$\begin{array}{ccc}& & {\left({\Phi }_1\left(X_1\right)+{\Phi }_2\left(X_2\right)\right)}^4=\frac{1}{16}\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right)\hfill \\ & <& \left(\begin{array}{cc}639.9213& -255\\ -255& 117.8559\end{array}\right)={\Phi }_1\left(X_1^4\right)+{\Phi }_2\left(X_2^4\right)-\left(\begin{array}{cc}1.5787& 0\\ 0& 0.6441\end{array}\right)\hfill \end{array}$$

(rounded to four decimal places), since ${\delta }_f=3.1574$, $\tilde{X}=\left(\begin{array}{cc}0.5& 0\\ 0& 0.2040\end{array}\right)$. But, if we put $\overline{m}=m_x=0$, $\overline{M}=M_x=0.5$, then $\tilde{X}=\mathbf{0}$, so we do not have an improvement of Jensen's inequality. Also, if we put $\overline{m}=0$, $\overline{M}=1$, then $\tilde{X}=0.5\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, ${\delta }_f=7/8$ and ${\delta }_f\tilde{X}=0.4375\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$, which is worse than the above improvement.

Putting ${\Phi }_t\left(y\right)=a_{t}y$ for every $y\in 𝒜$, where $a_t\ge 0$ is a real number, we obtain the following obvious corollary of Theorem ▭.

Corollary 14 Let ${\left(x_t\right)}_{t\in T}$ be a bounded continuous field of self-adjoint elements in a unital $C^{*}$-algebra $𝒜$ defined on a locally compact Hausdorff space $T$ equipped with a bounded Radon measure $\mu$. Let $m_t$ and $M_t$, $m_t\le M_t$, be the bounds of $x_t$, $t\in T$. Let ${\left(a_t\right)}_{t\in T}$ be a continuous field of nonnegative real numbers such that ${\int }_T{a}_{t}d\mu \left(t\right)=1$. Let

$$(m_x,M_x)\cap [m_t,M_t]=\varnothing ,t\in T,\text{and}m<M$$

where $m_x$ and $M_x$, $m_x\le M_x$, are the bounds of the operator $x={\int }_T{\phi }_t\left(x_t\right)d\mu \left(t\right)$ and

$$m=sup\left\{M_t:M_t\le m_x,t\in T\right\},M=inf\left\{m_t:m_t\ge M_x,t\in T\right\}$$

If $f:I\to ℝ$ is a continuous convex $($resp. concave$)$ function provided that the interval $I$ contains all $m_t,M_t$, then

$$\begin{array}{ccc}\hfill {\displaystyle f\left({\int }_T{a}_t{x}_{t}d\mu \left(t\right)\right)}& \le & {\int }_T{a}_{t}f\left(x_t\right)d\mu \left(t\right)-{\delta }_f\tilde{\tilde{x}}\le {\int }_T{a}_{t}f\left(x_t\right)d\mu \left(t\right)\hfill \\ \hfill {\displaystyle (\text{resp.}f\left({\int }_T{a}_t{x}_{t}d\mu \left(t\right)\right)}& \ge & {\int }_T{a}_{t}f\left(x_t\right)d\mu \left(t\right)+{\delta }_f\tilde{\tilde{x}}\ge {\int }_T{a}_{t}f\left(x_t\right)d\mu \left(t\right))\hfill \end{array}$$

holds, where ${\delta }_f$ is defined by (▭), $\tilde{\tilde{x}}=\frac{1}{2}{1}_H-\frac{1}{\overline{M}-\overline{m}}\left|{\int }_T{a}_t{x}_{t}d\mu \left(t\right)-\frac{\overline{m}+\overline{M}}{2}{1}_H\right|$ and $\overline{m}\in [m,m_A]$, $\overline{M}\in [M_A,M]$, $\overline{m}<\overline{M}$, are arbitrary numbers.

5. Extension Jensen's inequality

In this section we present an extension of Jensen's operator inequality for $n-$tuples of self-adjoint operators, unital $n-$tuples of positive linear mappings and real valued continuous convex functions with conditions on the spectra of the operators.

In a discrete version of Theorem ▭ we prove that Jensen's operator inequality holds for every continuous convex function and for every $n-$tuple of self-adjoint operators $(A_1,...,A_n)$, for every $n-$tuple of positive linear mappings $({\Phi }_1,...,{\Phi }_n)$ in the case when the interval with bounds of the operator $A={\sum }_{i=1}^n{\Phi }_i\left(A_i\right)$ has no intersection points with the interval with bounds of the operator $A_i$ for each $i=1,...,n$, i.e. when $(m_A,M_A)\cap [m_i,M_i]=\varnothing$ for $i=1,...,n,$ where $m_A$ and $M_A$, $m_A\le M_A$, are the bounds of $A$, and $m_i$ and $M_i$, $m_i\le M_i$, are the bounds of $A_i$, $i=1,...,n$. It is interesting to consider the case when $(m_A,M_A)\cap [m_i,M_i]=\varnothing$ is valid for several $i\in \{1,...,n\}$, but not for all $i=1,...,n$. We study it in the following theorem (see [21]).

Theorem 15 Let $(A_1,...,A_n)$ be an $n-$tuple of self-adjoint operators $A_i\in B\left(H\right)$ with the bounds $m_i$ and $M_i$, $m_i\le M_i$, $i=1,...,n$. Let $({\Phi }_1,...,{\Phi }_n)$ be an $n-$tuple of positive linear mappings ${\Phi }_i:B\left(H\right)\to B\left(K\right)$, such that ${\sum }_{i=1}^n{\Phi }_i\left(1_H\right)=1_K$. For $1\le n_1<n$, we denote $m=min\{m_1,...,m_{n_1}\}$, $M=max\{M_1,...,M_{n_1}\}$ and ${\sum }_{i=1}^{n_1}{\Phi }_i\left(1_H\right)=\alpha 1_K$, ${\sum }_{i=n_1+1}^n{\Phi }_i\left(1_H\right)=\beta 1_K$, where $\alpha ,\beta >0$, $\alpha +\beta =1$. If

$$(m,M)\cap [m_i,M_i]=\varnothing ,i=n_1+1,...,n$$

and one of two equalities

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(A_i\right)=\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(A_i\right)=\sum _{i=1}^n{\Phi }_i\left(A_i\right)$$

is valid, then

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)$$uid57

holds for every continuous convex function $f:I\to ℝ$ provided that the interval $I$ contains all $m_i,M_i$, $i=1,...,n$. If $f:I\to ℝ$ is concave, then the reverse inequality is valid in (▭).

We prove only the case when $f$ is a convex function. Let us denote

$$A=\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(A_i\right),B=\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(A_i\right),C=\sum _{i=1}^n{\Phi }_i\left(A_i\right)$$

It is easy to verify that $A=B$ or $B=C$ or $A=C$ implies $A=B=C$.

a) Let $m<M$. Since $f$ is convex on $[m,M]$ and $[m_i,M_i]\subseteq [m,M]$ for $i=1,...,n_1$, then

$$f\left(z\right)\le \frac{M-z}{M-m}f\left(m\right)+\frac{z-m}{M-m}f\left(M\right),z\in [m_i,M_i]\text{for}i=1,...,n_1$$uid58

but since $f$ is convex on all $[m_i,M_i]$ and $(m,M)\cap [m_i,M_i]=\varnothing$ for $i=n_1+1,...,n$, then

$$f\left(z\right)\ge \frac{M-z}{M-m}f\left(m\right)+\frac{z-m}{M-m}f\left(M\right),z\in [m_i,M_i]\text{for}i=n_1+1,...,n$$uid59

Since $m_i{1}_H\le A_i\le M_i{1}_H$, $i=1,...,n_1$, it follows from (▭)

$$f\left(A_i\right)\le \frac{M{1}_H-A_i}{M-m}f\left(m\right)+\frac{A_i-m{1}_H}{M-m}f\left(M\right),i=1,...,n_1$$

Applying a positive linear mapping ${\Phi }_i$ and summing, we obtain

$$\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{M\alpha 1_K-{\sum }_{i=1}^{n_1}{\Phi }_i\left(A_i\right)}{M-m}f\left(m\right)+\frac{{\sum }_{i=1}^{n_1}{\Phi }_i\left(A_i\right)-m\alpha 1_K}{M-m}f\left(M\right)$$

since ${\sum }_{i=1}^{n_1}{\Phi }_i\left(1_H\right)=\alpha 1_K$. It follows

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{M{1}_K-A}{M-m}f\left(m\right)+\frac{A-m{1}_K}{M-m}f\left(M\right)$$uid60

Similarly to (▭) in the case $m_i{1}_H\le A_i\le M_i{1}_H$, $i=n_1+1,...,n$, it follows from (▭)

$$\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)\ge \frac{M{1}_K-B}{M-m}f\left(m\right)+\frac{B-m{1}_K}{M-m}f\left(M\right)$$uid61

Combining (▭) and (▭) and taking into account that $A=B$, we obtain

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)$$uid62

It follows

$$\begin{array}{ccc}\hfill \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)& =& \sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+\frac{\beta }{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)(\text{by}\alpha +\beta =1)\hfill \\ & \le & \sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)\left(\text{by}\left(\right)\right)\hfill \\ & =& \sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\hfill \\ & \le & \frac{\alpha }{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)+\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)\left(\text{by}\left(\right)\right)\hfill \\ & =& \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)(\text{by}\alpha +\beta =1)\hfill \end{array}$$uid63

which gives the desired double inequality (▭).

b) Let $m=M$. Since $[m_i,M_i]\subseteq [m,M]$ for $i=1,...,n_1$, then $A_i=m{1}_H$ and $f\left(A_i\right)=f\left(m\right)1_H$ for $i=1,...,n_1$. It follows

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(A_i\right)=m{1}_K\text{and}\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)=f\left(m\right)1_K$$uid64

On the other hand, since $f$ is convex on $I$, we have

$$f\left(z\right)\ge f\left(m\right)+l\left(m\right)(z-m)\text{for}\text{every}z\in I$$uid65

where $l$ is the subdifferential of $f$. Replacing $z$ by $A_i$ for $i=n_1+1,...,n$, applying ${\Phi }_i$ and summing, we obtain from (▭) and (▭)

$$\begin{array}{ccc}\hfill \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)& \ge & f\left(m\right)1_K+l\left(m\right)\left(\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(A_i\right)-m{1}_K\right)\hfill \\ & =& f\left(m\right)1_K=\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\hfill \end{array}$$

So (▭) holds again. The remaining part of the proof is the same as in the case a).

Remark 16 We obtain the equivalent inequality to the one in Theorem ▭ in the case when ${\sum }_{i=1}^n{\Phi }_i\left(1_H\right)=\gamma 1_K$, for some positive scalar $\gamma$. If $\alpha +\beta =\gamma$ and one of two equalities

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(A_i\right)=\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(A_i\right)=\frac{1}{\gamma }\sum _{i=1}^n{\Phi }_i\left(A_i\right)$$

is valid, then

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\gamma }\sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)$$

holds for every continuous convex function $f$.

Remark 17 Let the assumptions of Theorem ▭ be valid.

1. We observe that the following inequality

$$f\left(\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(A_i\right)\right)\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)$$uid68

holds for every continuous convex function $f:I\to ℝ$.

Indeed, by the assumptions of Theorem ▭ we have

$$m\alpha 1_H\le \sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le M\alpha 1_H\text{and}\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(A_i\right)=\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(A_i\right)$$

which implies

$$m{1}_H\le \sum _{i=n_1+1}^n\frac{1}{\beta }{\Phi }_i\left(f\left(A_i\right)\right)\le M{1}_H$$

Also $(m,M)\cap [m_i,M_i]=\varnothing$ for $i=n_1+1,...,n$ and ${\sum }_{i=n_1+1}^n\frac{1}{\beta }{\Phi }_i\left(1_H\right)=1_K$ hold. So we can apply Theorem ▭ on operators $A_{n_1+1},...,A_n$ and mappings $\frac{1}{\beta }{\Phi }_i$ and obtain the desired inequality.

2. We denote by $m_C$ and $M_C$ the bounds of $C={\sum }_{i=1}^n{\Phi }_i\left(A_i\right)$. If $(m_C,M_C)\cap [m_i,M_i]=\varnothing$, $i=1,...,n_1$ or $f$ is an operator convex function on $[m,M]$, then the double inequality (▭) can be extended from the left side if we use Jensen's operator inequality (see Theorem 2.1[16])

$$\begin{array}{ccc}\hfill f\left(\sum _{i=1}^n{\Phi }_i\left(A_i\right)\right)& =& f\left(\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(A_i\right)\right)\hfill \\ & \le & \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)\hfill \end{array}$$

Example 18 If neither assumptions $(m_C,M_C)\cap [m_i,M_i]=\varnothing$, $i=1,...,n_1$, nor $f$ is operator convex in Remark ▭ - 2. is satisfied and if $1<n_1<n$, then (▭) can not be extended by Jensen's operator inequality, since it is not valid. Indeed, for $n_1=2$ we define mappings ${\Phi }_1,{\Phi }_2:M_3\left(ℂ\right)\to M_2\left(ℂ\right)$ by ${\Phi }_1\left({\left(a_{ij}\right)}_{1\le i,j\le 3}\right)=\frac{\alpha }{2}{\left(a_{ij}\right)}_{1\le i,j\le 2}$, ${\Phi }_2={\Phi }_1$. Then ${\Phi }_1\left(I_3\right)+{\Phi }_2\left(I_3\right)=\alpha I_2$. If

$$A_1=2\left(\begin{array}{ccc}1& 0& 1\\ 0& 0& 1\\ 1& 1& 1\end{array}\right)\text{and}{A}_2=2\left(\begin{array}{ccc}1& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)$$

then

$${\left(\frac{1}{\alpha }{\Phi }_1\left(A_1\right)+\frac{1}{\alpha }{\Phi }_2\left(A_2\right)\right)}^4=\frac{1}{{\alpha }^4}\left(\begin{array}{cc}16& 0\\ 0& 0\end{array}\right)\neg \le \frac{1}{\alpha }\left(\begin{array}{cc}80& 40\\ 40& 24\end{array}\right)=\frac{1}{\alpha }{\Phi }_1\left(A_1^4\right)+\frac{1}{\alpha }{\Phi }_2\left(A_2^4\right)$$

for every $\alpha \in (0,1)$. We observe that $f\left(t\right)=t^4$ is not operator convex and $(m_C,M_C)\cap [m_i,M_i]\ne \varnothing ,$ since $C=A=\frac{1}{\alpha }{\Phi }_1\left(A_1\right)+\frac{1}{\alpha }{\Phi }_2\left(A_2\right)=\frac{1}{\alpha }\left(\begin{array}{cc}2& 0\\ 0& 0\end{array}\right),$ $[m_C,M_C]=[0,2/\alpha ]$, $[m_1,M_1]\subset [-1.60388,4.49396]$ and $[m_2,M_2]=[0,2]$.

With respect to Remark ▭, we obtain the following obvious corollary of Theorem ▭.

Corollary 19 Let $(A_1,...,A_n)$ be an $n-$tuple of self-adjoint operators $A_i\in B\left(H\right)$ with the bounds $m_i$ and $M_i$, $m_i\le M_i$, $i=1,...,n$. For some $1\le n_1<n$, we denote $m=min\{m_1,...,m_{n_1}\}$, $M=max\{M_1,...,M_{n_1}\}$. Let $(p_1,...,p_n)$ be an $n-$tuple of non-negative numbers, such that $0<{\sum }_{i=1}^{n_1}{p}_i={𝐩}_{{𝐧}_{\mathbf{1}}}<{𝐩}_{𝐧}={\sum }_{i=1}^n{p}_i$. If

$$(m,M)\cap [m_i,M_i]=\varnothing ,i=n_1+1,...,n$$

and one of two equalities

$$\frac{1}{{𝐩}_{{𝐧}_{\mathbf{1}}}}\sum _{i=1}^{n_1}{p}_i{A}_i=\frac{1}{{𝐩}_{𝐧}}\sum _{i=1}^n{p}_i{A}_i=\frac{1}{{𝐩}_{𝐧}-{𝐩}_{{𝐧}_{\mathbf{1}}}}\sum _{i=n_1+1}^n{p}_i{A}_i$$

is valid, then

$$\frac{1}{{𝐩}_{{𝐧}_{\mathbf{1}}}}\sum _{i=1}^{n_1}{p}_{i}f\left(A_i\right)\le \frac{1}{{𝐩}_{𝐧}}\sum _{i=1}^n{p}_{i}f\left(A_i\right)\le \frac{1}{{𝐩}_{𝐧}-{𝐩}_{{𝐧}_{\mathbf{1}}}}\sum _{i=n_1+1}^n{p}_{i}f\left(A_i\right)$$uid71

holds for every continuous convex function $f:I\to ℝ$ provided that the interval $I$ contains all $m_i,M_i$, $i=1,...,n$.

If $f:I\to ℝ$ is concave, then the reverse inequality is valid in (▭).

As a special case of Corollary ▭ we can obtain a discrete version of Corollary ▭ as follows.

Corollary 20 (Discrete version of Corollary ▭) Let $(A_1,...,A_n)$ be an $n-$tuple of self-adjoint operators $A_i\in B\left(H\right)$ with the bounds $m_i$ and $M_i$, $m_i\le M_i$, $i=1,...,n$. Let $({\alpha }_1,...,{\alpha }_n)$ be an $n-$tuple of nonnegative real numbers such that ${\sum }_{i=1}^n{\alpha }_i=1$. If

$$(m_A,M_A)\cap [m_i,M_i]=\varnothing ,i=1,...,n$$uid73

where $m_A$ and $M_A$, $m_A\le M_A$, are the bounds of $A={\sum }_{i=1}^n{\alpha }_i{A}_i$, then

$$f\left(\sum _{i=1}^n{\alpha }_i{A}_i\right)\le \sum _{i=1}^n{\alpha }_{i}f\left(A_i\right)$$uid74

holds for every continuous convex function $f:I\to ℝ$ provided that the interval $I$ contains all $m_i,M_i$.

We prove only the convex case. We define $(n+1)-$tuple of operators $(B_1,...,B_{n+1})$, $B_i\in B\left(H\right)$, by $B_1=A={\sum }_{i=1}^n{\alpha }_i{A}_i$ and $B_i=A_{i-1}$, $i=2,...,n+1$. Then $m_{B_1}=m_A$, $M_{B_1}=M_A$ are the bounds of $B_1$ and $m_{B_i}=m_{i-1}$, $M_{B_i}=M_{i-1}$ are the ones of $B_i$, $i=2,...,n+1$. Also, we define $(n+1)-$tuple of non-negative numbers $(p_1,...,p_{n+1})$ by $p_1=1$ and $p_i={\alpha }_{i-1}$, $i=2,...,n+1$. Then ${\sum }_{i=1}^{n+1}{p}_i=2$ and by using (▭) we have

$$(m_{B_1},M_{B_1})\cap [m_{B_i},M_{B_i}]=\varnothing ,i=2,...,n+1$$uid75

Since

$$\sum _{i=1}^{n+1}{p}_i{B}_i=B_1+\sum _{i=2}^{n+1}{p}_i{B}_i=\sum _{i=1}^n{\alpha }_i{A}_i+\sum _{i=1}^n{\alpha }_i{A}_i=2B_1$$

then

$$p_1{B}_1=\frac{1}{2}\sum _{i=1}^{n+1}{p}_i{B}_i=\sum _{i=2}^{n+1}{p}_i{B}_i$$uid76

Taking into account (▭) and (▭), we can apply Corollary ▭ for $n_1=1$ and $B_i$, $p_i$ as above, and we get

$$p_{1}f\left(B_1\right)\le \frac{1}{2}\sum _{i=1}^{n+1}{p}_{i}f\left(B_i\right)\le \sum _{i=2}^{n+1}{p}_{i}f\left(B_i\right)$$

which gives the desired inequality (▭).

6. Extension of the refined Jensen's inequality

There is an extensive literature devoted to Jensen's inequality concerning different refinements and extensive results, see, for example [22], [23], [24], [25], [26], [27], [28], [29].

In this section we present an extension of the refined Jensen's inequality obtained in Section ▭ and a refinement of the same inequality obtained in Section ▭.

Theorem 21 Let $(A_1,...,A_n)$ be an $n-$tuple of self-adjoint operators $A_i\in B\left(H\right)$ with the bounds $m_i$ and $M_i$, $m_i\le M_i$, $i=1,...,n$. Let $({\Phi }_1,...,{\Phi }_n)$ be an $n-$tuple of positive linear mappings ${\Phi }_i:B\left(H\right)\to B\left(K\right)$, such that ${\sum }_{i=1}^{n_1}{\Phi }_i\left(1_H\right)=\alpha 1_K$, ${\sum }_{i=n_1+1}^n{\Phi }_i\left(1_H\right)=\beta 1_K$, where $1\le n_1<n$, $\alpha ,\beta >0$ and $\alpha +\beta =1$. Let $m_L=min\{m_1,...,m_{n_1}\}$, $M_R=max\{M_1,...,M_{n_1}\}$ and

$$\begin{array}{ccc}\hfill m& =& max\left\{M_i:M_i\le m_L,i\in \{n_1+1,...,n\}\right\}\hfill \\ \hfill M& =& min\left\{m_i:m_i\ge M_R,i\in \{n_1+1,...,n\}\right\}\hfill \end{array}$$

If

$$(m_L,M_R)\cap [m_i,M_i]=\varnothing ,i=n_1+1,...,n,\text{and}m<M$$

and one of two equalities

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(A_i\right)=\sum _{i=1}^n{\Phi }_i\left(A_i\right)=\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(A_i\right)$$

is valid, then

$$\begin{array}{ccc}\hfill \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)& \le & \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+\beta {\delta }_f\tilde{A}\le \sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\hfill \\ & \le & \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-\alpha {\delta }_f\tilde{A}\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)\hfill \end{array}$$uid78

holds for every continuous convex function $f:I\to ℝ$ provided that the interval $I$ contains all $m_i,M_i$, $i=1,...,n$, where

$$\begin{array}{c}{\displaystyle {\delta }_f\equiv {\delta }_f(\overline{m},\overline{M})=f\left(\overline{m}\right)+f\left(\overline{M}\right)-2f\left(\frac{\overline{m}+\overline{M}}{2}\right)}\\ {\displaystyle \tilde{A}\equiv {\tilde{A}}_{A,\Phi ,n_1,\alpha }(\overline{m},\overline{M})=\frac{1}{2}{1}_K-\frac{1}{\alpha (\overline{M}-\overline{m})}\sum _{i=1}^{n_1}{\Phi }_i\left(\left|A_i-\frac{\overline{m}+\overline{M}}{2}{1}_H\right|\right)}\end{array}$$uid79

and   $\overline{m}\in [m,m_L]$, $\overline{M}\in [M_R,M]$, $\overline{m}<\overline{M}$,   are arbitrary numbers. If $f:I\to ℝ$ is concave, then the reverse inequality is valid in (▭).

We prove only the convex case. Let us denote

$$A=\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(A_i\right),B=\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(A_i\right),C=\sum _{i=1}^n{\Phi }_i\left(A_i\right)$$

It is easy to verify that $A=B$ or $B=C$ or $A=C$ implies $A=B=C$.

Since $f$ is convex on $[\overline{m},\overline{M}]$ and $\mathrm{𝖲𝗉}\left(A_i\right)\subseteq [m_i,M_i]\subseteq [\overline{m},\overline{M}]$ for $i=1,...,n_1$, it follows from Lemma ▭ that

$$f\left(A_i\right)\le \frac{\overline{M}{1}_H-A_i}{\overline{M}-\overline{m}}f\left(\overline{m}\right)+\frac{A_i-\overline{m}{1}_H}{\overline{M}-\overline{m}}f\left(\overline{M}\right)-{\delta }_f{\tilde{A}}_i,i=1,...,n_1$$

holds, where ${\delta }_f=f\left(\overline{m}\right)+f\left(\overline{M}\right)-2f\left(\frac{\overline{m}+\overline{M}}{2}\right)$ and ${\tilde{A}}_i=\frac{1}{2}{1}_H-\frac{1}{\overline{M}-\overline{m}}\left|A_i-\frac{\overline{m}+\overline{M}}{2}{1}_H\right|$. Applying a positive linear mapping ${\Phi }_i$ and summing, we obtain

$$\begin{array}{ccc}\hfill \sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)& \le & {\displaystyle \frac{\overline{M}\alpha 1_K-{\sum }_{i=1}^{n_1}{\Phi }_i\left(A_i\right)}{\overline{M}-\overline{m}}f\left(\overline{m}\right)+\frac{{\sum }_{i=1}^{n_1}{\Phi }_i\left(A_i\right)-\overline{m}\alpha 1_K}{\overline{M}-\overline{m}}f\left(\overline{M}\right)}\hfill \\ & -& {\displaystyle {\delta }_f\left(\frac{\alpha }{2}{1}_K-\frac{1}{\overline{M}-\overline{m}}\sum _{i=1}^{n_1}{\Phi }_i\left(\left|A_i-\frac{\overline{m}+\overline{M}}{2}{1}_H\right|\right)\right)}\hfill \end{array}$$

since ${\sum }_{i=1}^{n_1}{\Phi }_i\left(1_H\right)=\alpha 1_K$. It follows that

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{\overline{M}{1}_K-A}{\overline{M}-\overline{m}}f\left(\overline{m}\right)+\frac{A-\overline{m}{1}_K}{\overline{M}-\overline{m}}f\left(\overline{M}\right)-{\delta }_f\tilde{A}$$uid80

where $\tilde{A}=\frac{1}{2}{1}_K-\frac{1}{\alpha (\overline{M}-\overline{m})}{\sum }_{i=1}^{n_1}{\Phi }_i\left(\left|A_i-\frac{\overline{m}+\overline{M}}{2}{1}_H\right|\right)$.

Additionally, since $f$ is convex on all $[m_i,M_i]$ and $(\overline{m},\overline{M})\cap [m_i,M_i]=\varnothing$, $i=n_1+1,...,n$, then

$$f\left(A_i\right)\ge \frac{\overline{M}{1}_H-A_i}{\overline{M}-\overline{m}}f\left(\overline{m}\right)+\frac{A_i-\overline{m}{1}_H}{\overline{M}-\overline{m}}f\left(\overline{M}\right),i=n_1+1,...,n$$uid81

It follows

$$\frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-{\delta }_f\tilde{A}\ge \frac{\overline{M}{1}_K-B}{\overline{M}-\overline{m}}f\left(\overline{m}\right)+\frac{B-\overline{m}{1}_K}{\overline{M}-\overline{m}}f\left(\overline{M}\right)-{\delta }_f\tilde{A}$$uid82

Combining (▭) and (▭) and taking into account that $A=B$, we obtain

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-{\delta }_f\tilde{A}$$uid83

Next, we obtain

$$\begin{array}{ccc}& & \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\hfill \\ & =& \sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+\frac{\beta }{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)(\text{by}\alpha +\beta =1)\hfill \\ & \le & \sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-\beta {\delta }_f\tilde{A}\left(\text{by}\left(\right)\right)\hfill \\ & \le & \frac{\alpha }{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-\alpha {\delta }_f\tilde{A}+\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-\beta {\delta }_f\tilde{A}\left(\text{by}\left(\right)\right)\hfill \\ & =& \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-{\delta }_f\tilde{A}(\text{by}\alpha +\beta =1)\hfill \end{array}$$

which gives the following double inequality

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)-\beta {\delta }_f\tilde{A}\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-{\delta }_f\tilde{A}$$uid84

Adding $\beta {\delta }_f\tilde{A}$ in the above inequalities, we get

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+\beta {\delta }_f\tilde{A}\le \sum _{i=1}^n{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-\alpha {\delta }_f\tilde{A}$$uid85

Now, we remark that ${\delta }_f\ge 0$ and $\tilde{A}\ge 0$. (Indeed, since $f$ is convex, then $f\left((\overline{m}+\overline{M})/2\right)\le (f\left(\overline{m}\right)+f\left(\overline{M}\right))/2$, which implies that ${\delta }_f\ge 0$. Also, since

$$\mathrm{𝖲𝗉}\left(A_i\right)\subseteq [\overline{m},\overline{M}]\Rightarrow \left|A_i-\frac{\overline{M}+\overline{m}}{2}{1}_H\right|\le \frac{\overline{M}-\overline{m}}{2}{1}_H,i=1,...,n_1$$

then

$$\sum _{i=1}^{n_1}{\Phi }_i\left(\left|A_i-\frac{\overline{M}+\overline{m}}{2}{1}_H\right|\right)\le \frac{\overline{M}-\overline{m}}{2}\alpha 1_K$$

which gives

$$0\le \frac{1}{2}{1}_K-\frac{1}{\alpha (\overline{M}-\overline{m})}\sum _{i=1}^{n_1}{\Phi }_i\left(\left|A_i-\frac{\overline{M}+\overline{m}}{2}{1}_H\right|\right)=\tilde{A})$$

Consequently, the following inequalities

$$\begin{array}{ccc}& & \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+\beta {\delta }_f\tilde{A}\hfill \\ & & \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-\alpha {\delta }_f\tilde{A}\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)\hfill \end{array}$$

hold, which with (▭) proves the desired series inequalities (▭). 1.05

Example 22 We observe the matrix case of Theorem ▭ for $f\left(t\right)=t^4$, which is the convex function but not operator convex, $n=4$, $n_1=2$ and the bounds of matrices as in Fig. 3.

Figure 3.

An example a convex function and the bounds of four operators

We show an example such that

$$\begin{array}{cc}& \frac{1}{\alpha }\left({\Phi }_1\left(A_1^4\right)+{\Phi }_2\left(A_2^4\right)\right)<\frac{1}{\alpha }\left({\Phi }_1\left(A_1^4\right)+{\Phi }_2\left(A_2^4\right)\right)+\beta {\delta }_f\tilde{A}\\ & <{\Phi }_1\left(A_1^4\right)+{\Phi }_2\left(A_2^4\right)+{\Phi }_3\left(A_3^4\right)+{\Phi }_4\left(A_4^4\right)\\ & <\frac{1}{\beta }\left({\Phi }_3\left(A_3^4\right)+{\Phi }_4\left(A_4^4\right)\right)-\alpha {\delta }_f\tilde{A}<\frac{1}{\beta }\left({\Phi }_3\left(A_3^4\right)+{\Phi }_4\left(A_4^4\right)\right)\end{array}$$uid88

holds, where ${\delta }_f={\overline{M}}^4+{\overline{m}}^4-{(\overline{M}+\overline{m})}^{4}8$ and

$$\tilde{A}=\frac{1}{2}{I}_2-\frac{1}{\alpha (\overline{M}-\overline{m})}\left({\Phi }_1\left(|A_1-\frac{\overline{M}+\overline{m}}{2}{I}_h|\right)+{\Phi }_2\left(|A_2-\frac{\overline{M}+\overline{m}}{2}{I}_3|\right)\right)$$

We define mappings ${\Phi }_i:M_3\left(ℂ\right)\to M_2\left(ℂ\right)$ as follows: ${\Phi }_i\left({\left(a_{jk}\right)}_{1\le j,k\le 3}\right)=\frac{1}{4}{\left(a_{jk}\right)}_{1\le j,k\le 2}$, $i=1,...,4$. Then ${\sum }_{i=1}^4{\Phi }_i\left(I_3\right)=I_2$ and $\alpha =\beta =\frac{1}{2}$.

Let

$$A_1=2\left(\begin{array}{ccc}2& 9/8& 1\\ 9/8& 2& 0\\ 1& 0& 3\end{array}\right),A_2=3\left(\begin{array}{ccc}2& 9/8& 0\\ 9/8& 1& 0\\ 0& 0& 2\end{array}\right),A_3=-3\left(\begin{array}{ccc}4& 1/2& 1\\ 1/2& 4& 0\\ 1& 0& 2\end{array}\right),A_4=12\left(\begin{array}{ccc}5/3& 1/2& 0\\ 1/2& 3/2& 0\\ 0& 0& 3\end{array}\right)$$

Then $m_1=1.28607$, $M_1=7.70771$, $m_2=0.53777$, $M_2=5.46221$, $m_3=-14.15050$, $M_3=-4.71071$, $m_4=12.91724$, $M_4=36.$, so $m_L=m_2$, $M_R=M_1$, $m=M_3$ and $M=m_4$ (rounded to five decimal places). Also,

$$\frac{1}{\alpha }\left({\Phi }_1\left(A_1\right)+{\Phi }_2\left(A_2\right)\right)=\frac{1}{\beta }\left({\Phi }_3\left(A_3\right)+{\Phi }_4\left(A_4\right)\right)=\left(\begin{array}{cc}4& 9/4\\ 9/4& 3\end{array}\right)$$

and

$$\begin{array}{ccc}\hfill A_f\equiv \frac{1}{\alpha }\left({\Phi }_1\left(A_1^4\right)+{\Phi }_2\left(A_2^4\right)\right)& =& \left(\begin{array}{cc}989.00391& 663.46875\\ 663.46875& 526.12891\end{array}\right)\hfill \\ \hfill C_f\equiv {\Phi }_1\left(A_1^4\right)+{\Phi }_2\left(A_2^4\right)+{\Phi }_3\left(A_3^4\right)+{\Phi }_4\left(A_4^4\right)& =& \left(\begin{array}{cc}68093.14258& 48477.98437\\ 48477.98437& 51335.39258\end{array}\right)\hfill \\ \hfill B_f\equiv \frac{1}{\beta }\left({\Phi }_3\left(A_3^4\right)+{\Phi }_4\left(A_4^4\right)\right)& =& \left(\begin{array}{cc}135197.28125& 96292.5\\ 96292.5& 102144.65625\end{array}\right)\hfill \end{array}$$

Then

$$A_f<C_f<B_f$$uid89

holds (which is consistent with (▭)).

We will choose three pairs of numbers $(\overline{m},\overline{M})$, $\overline{m}\in [-4.71071,0.53777]$, $\overline{M}\in [7.70771,12.91724]$ as follows

i) $\overline{m}=m_L=0.53777$, $\overline{M}=M_R=7.70771$, then

${\tilde{\Delta }}_1=\beta {\delta }_f\tilde{A}=0.5·2951.69249·\left(\begin{array}{cc}0.15678& 0.09030\\ 0.09030& 0.15943\end{array}\right)=\left(\begin{array}{cc}231.38908& 133.26139\\ 133.26139& 235.29515\end{array}\right)$

ii) $\overline{m}=m=-4.71071$, $\overline{M}=M=12.91724$, then

${\tilde{\Delta }}_2=\beta {\delta }_f\tilde{A}=0.5·27766.07963·\left(\begin{array}{cc}0.36022& 0.03573\\ 0.03573& 0.36155\end{array}\right)=\left(\begin{array}{cc}5000.89860& 496.04498\\ 496.04498& 5019.50711\end{array}\right)$

iii) $\overline{m}=-1$, $\overline{M}=10$, then

${\tilde{\Delta }}_3=\beta {\delta }_f\tilde{A}=0.5·9180.875·\left(\begin{array}{cc}0.28203& 0.08975\\ 0.08975& 0.27557\end{array}\right)=\left(\begin{array}{cc}1294.66& 411.999\\ 411.999& 1265.\end{array}\right)$

New, we obtain the following improvement of (▭) (see (▭))

Table 1.

Using Theorem ▭ we get the following result.

Corollary 23 Let the assumptions of Theorem ▭ hold. Then

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+{\gamma }_1{\delta }_f\tilde{A}\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)$$uid91

and

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)-{\gamma }_2{\delta }_f\tilde{A}\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)$$uid92

holds for every ${\gamma }_1,{\gamma }_2$ in the close interval joining $\alpha$ and $\beta$, where ${\delta }_f$ and $\tilde{A}$ are defined by (▭).

Adding $\alpha {\delta }_f\tilde{A}$ in (▭) and noticing ${\delta }_f\tilde{A}\ge 0$, we obtain

$$\frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)\le \frac{1}{\alpha }\sum _{i=1}^{n_1}{\Phi }_i\left(f\left(A_i\right)\right)+\alpha {\delta }_f\tilde{A}\le \frac{1}{\beta }\sum _{i=n_1+1}^n{\Phi }_i\left(f\left(A_i\right)\right)$$uid93

Taking into account the above inequality and the left hand side of (▭) we obtain (▭).

Similarly, subtracting $\beta {\delta }_f\tilde{A}$ in (▭) we obtain (▭).

Remark 24 We can obtain extensions of inequalities which are given in Remark ▭ and ▭. Also, we can obtain a special case of Theorem ▭ with the convex combination of operators $A_i$ putting ${\Phi }_i\left(B\right)={\alpha }_{i}B$, for $i=1,...,n$, similarly as in Corollary ▭. Finally, applying this result, we can give another proof of Corollary ▭. The interested reader can see the details in [30].