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On Levinson’s operator inequality and its converses

Abstract

We give new results on Levinson’s operator inequality and its converse for normalized positive linear mappings and some large class of ‘3-convex functions at a point c’.

Introduction and preliminary results

Let \(\mathcal{B}(H)\) be the algebra of all bounded linear operators on a complex Hilbert space H. We denote by \(\mathcal{B}_{h}(H)\) the real subspace of all self-adjoint operators on H. Bounds of \(X \in\mathcal{B}_{h}(H)\) are defined by \(m := \inf \{ \langle X \xi,\xi\rangle: \xi\in H, \| \xi\|=1 \}\) and \(M := \sup \{ \langle X \xi,\xi\rangle: \xi \in H, \| \xi\|=1 \}\).

A continuous real valued function f defined on an interval I is said to be operator convex if \(f(\lambda X+(1-\lambda)Y)\leq\lambda f(X)+(1-\lambda)f(Y)\) for all self-adjoint operators \(X,Y\) with spectra contained in I and all \(\lambda\in[0,1]\). If the function f is operator convex, then the so-called Jensen operator inequality \(f(\Phi(X))\leq\Phi(f(X))\) holds for any unital positive linear mapping Φ on \(\mathcal{B}(H)\) and any \(X \in\mathcal{B}_{h}(H)\) with spectrum contained in I. Many other versions of Jensen’s operator inequality can be found in [1, 2].

Assume furthermore that \((\Phi_{1},\ldots,\Phi_{n})\) is an n-tuple of positive linear mappings \(\Phi_{i}:\mathcal{B}(H) \rightarrow\mathcal{B}(K)\). If in addition \(\sum_{i=1}^{n}\Phi_{i} (1_{H})=1_{K}\), we say that \(\sum_{i=1}^{n}\Phi_{i} (1_{H})=1_{K}\) is unital.

Now we give the definition of classes of functions for which we observe Levinson’s operator inequality.

Definition 1

([3])

Let \(f \in\mathcal{C}(I)\) be a real valued function on an arbitrary interval I in and \(c \in I^{\circ}\), where \(I^{\circ}\) is the interior of I.

We say that \(f \in\mathcal{K}_{1}^{c}(I)\) (resp. \(f \in\mathcal {K}_{2}^{c}(I)\)) if there exists a constant α such that the function \(F(t) = f(t)- \frac{\alpha}{2} t^{2}\) is concave (resp. convex) on \(I \cap(-\infty, c]\) and convex (resp. concave) on \(I \cap[c,\infty)\). (See Figure 1.)

Figure 1
figure1

Continous 3-convex function at a point c .

Moreover, we say that \(f \in\overset{\bullet}{\mathcal{K}_{1}^{c}}(I)\) (resp. \(f \in\overset{\bullet}{\mathcal{K}_{2}^{c}}(I)\)) if there exists a constant α such that the function F is operator concave (resp. operator convex) on \(I \cap(-\infty,c]\) and operator convex (resp. operator concave) on \(I \cap[c,\infty)\).

The class of functions \(\mathcal{K}_{1}^{c}(I)\) can be interpreted as functions that are ‘3-convex at a point c’ and extends 3-convex functions in the following sense: a function is 3-convex on I if and only if it is at every \(c \in I^{\circ}\).

Next, we will review the history of research of Levison’s inequality.

Levinson [4] considered an inequality as follows:

If \(f : (0, 2c) \rightarrow \mathbb{R}\) satisfies \(f''' \geq0\) and \(p_{i}\) , \(x_{i}\) , \(y_{i}\) , \(i = 1, 2, \ldots, n\) , are such that \(p_{i} > 0\) , \(\sum_{i=1}^{n} p_{i} = 1\) , \(0 \leq x_{i} \leq c\) , and

$$ x_{1} + y_{1} = x_{2} + y_{2} = \cdots= x_{n} + y_{n} = 2c, $$
(1)

then the inequality

$$ \sum_{i=1}^{n} p_{i} f (x_{i}) - f (\bar{x}) \leq\sum _{i=1}^{n} p_{i} f (y_{i}) - f ( \bar{y}) $$
(2)

holds, where \(\bar{x} = \sum_{i=1}^{n} p_{i} x_{i}\) and \(\bar{y} = \sum_{i=1}^{n} p_{i} y_{i}\) denote the weighted arithmetic means.

Numerous papers have been devoted to generalizations and extensions of Levinson’s result. Popoviciu [5] showed that the assumptions on the differentiability of f can be weakened for (2); to hold it is enough to assume that f is 3-convex. Bullen [6] gave another proof of Popoviciu’s result rescaled to a general interval \([a, b]\).

Mercer [7] made a significant improvement by replacing (1) with the weaker condition that the variances of the two sequences are equal: \(\sum_{i=1}^{n} p_{i} (x_{i} - \bar{x})^{2} = \sum_{i=1}^{n} p_{i} (y_{i} - \bar{y})^{2}\).

Witkowski [8, 9] extended this result in several ways. Firstly, he showed that Levinson’s inequality can be stated in a more general setting with random variables. Furthermore, he showed that it is enough to assume that f is 3-convex and that the assumption of equality of the variances can be weakened to inequality in a certain direction.

Baloch et al. [10] built on and extended the methods of Witkowski [8]. They introduced a new class of functions \(\mathcal{K}_{1}^{c}((a,b))\) as in Definition 1.

Mićić et al. [3] built on the methods given in [11] on operators. We give Levinson’s operator inequality for unital fields of positive linear mappings and classes of functions given by Definition 1. Moreover, we considered order among quasi-arithmetic means under similar conditions.

Next, we give the main result in [3] for two operators and \(f \in\overset{\bullet}{\mathcal{K}_{i}^{c}}(I)\), \(i=1,2\).

Theorem 1

Let \(X,Y \in\mathcal{B}_{h}(H)\) be self-adjoint operators with spectra contained in \([m,M]\) and \([n,N]\), respectively, such that \(a< m\leq M \leq c \leq n \leq N< b\). (See Figure 2.) Let Φ, Ψ be normalized positive linear mappings \(\Phi, \Psi:\mathcal{B}(H) \rightarrow \mathcal{B}(K)\).

Figure 2
figure2

Spectra conditions in Levinson’s inequality for two operators and an operator 3-convex function.

If \(f \in\overset{\bullet}{\mathcal{K}_{1}^{c}}((a,b))\) and \(C_{1} \leq C_{2}\), then

$$ \Phi \bigl( f(X) \bigr)- f \bigl( \Phi(X) \bigr) \leq C_{1} \leq C_{2} \leq \Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr), $$
(3)

where

$$ C_{1} := \frac{\alpha}{2} \bigl[ \Phi \bigl(X^{2} \bigr) - \Phi(X)^{2} \bigr],\qquad C_{2} := \frac{\alpha}{2} \bigl[ \Psi \bigl(Y^{2} \bigr) - \Psi(Y)^{2} \bigr]. $$
(4)

But, if \(f \in\overset{\bullet}{\mathcal{K}_{2}^{c}}((a,b))\) and \(C_{1}\geq C_{2}\) holds, then the reverse inequalities are valid in (3).

Proof

This theorem is special case of [3], Theorem 1, for \(k=n=1\). For the sake of completeness, we give the proof.

Let \(f \in\overset{\bullet}{\mathcal{K}_{1}^{c}}((a,b))\). So there is a constant α such that \(F(t) = f(t)- \frac{\alpha}{2} t^{2}\) is operator concave on \([m,M] \subset(a,c]\). Jensen’s inequality for an operator concave function implies

$$0 \leq F \bigl( \Phi(X) \bigr) - \Phi \bigl( F(X) \bigr) = f \bigl( \Phi(X) \bigr) -\frac{\alpha}{2} \Phi(X)^{2} - \Phi \bigl( f(X) \bigr) + \frac{\alpha}{2} \Phi \bigl(X^{2} \bigr). $$

It follows that

$$ \Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) \leq C_{1}. $$
(5)

Similarly, since F is operator convex on \([n,N] \subset[c,b)\), then Jensen’s inequality for an operator convex function implies

$$0 \leq\Psi \bigl( F(Y) \bigr) -F \bigl(\Psi(Y) \bigr) = \Psi \bigl( f(Y) \bigr)- \frac{\alpha}{2} \Psi \bigl(Y^{2} \bigr) - f \bigl( \Psi(Y) \bigr) + \frac{\alpha}{2} \Psi(Y)^{2}. $$

It follows that

$$ C_{2} \leq\Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr). $$
(6)

Combining inequalities (5) and (6) and taking into account that \(C_{1} \leq C_{2}\) we obtain the desired inequality (3). □

Applying Theorem 1 we obtain a version of Levinson’s inequality with more operators as follows.

Corollary 2

([3], Theorem 1)

Let \((X_{1},\ldots,X_{k_{1}})\) be a \({k_{1}}\)-tuple and \((Y_{1},\ldots ,Y_{k_{2}})\) be a \({k_{2}}\)-tuple of self-adjoint operators \(X_{i},Y_{j} \in\mathcal{B}_{h}(H)\) with spectra contained in \([m,M]\) and \([n,N]\), respectively, such that \(a< m\leq M \leq c \leq n \leq N< b\). Let \((\Phi_{1},\ldots,\Phi_{k_{1}})\) be a unital \({k_{1}}\)-tuple and \((\Psi _{1},\ldots,\Psi_{k_{2}})\) be a unital \({k_{2}}\)-tuple of positive linear mappings \(\Phi_{i}, \Psi_{j}:\mathcal{B}(H) \rightarrow\mathcal{B}(K)\).

If \(f \in\overset{\bullet}{\mathcal{K}_{1}^{c}}((a,b))\) and \(D_{1} \leq D_{2}\), then

$$ \sum_{i=1}^{k_{1}} \Phi_{i} \bigl( f(X_{i}) \bigr)- f \Biggl( \sum _{i=1}^{k_{1}} \Phi_{i}(X_{i}) \Biggr) \leq D_{1} \leq D_{2} \leq \sum _{i=1}^{k_{2}} \Psi_{i} \bigl( f(Y_{i}) \bigr) - f \Biggl( \sum_{i=1}^{k_{2}} \Psi_{i}(Y_{i}) \Biggr) $$
(7)

holds, where

$$ \begin{aligned} &D_{1} := \frac{\alpha}{2} \Biggl[ \sum _{i=1}^{k_{1}} \Phi_{i} \bigl(X_{i}^{2} \bigr) - \Biggl( \sum_{i=1}^{k_{1}} \Phi_{i}(X_{i}) \Biggr)^{2} \Biggr],\\ &D_{2} := \frac{\alpha}{2} \Biggl[ \sum_{i=1}^{k_{2}} \Psi_{i} \bigl(Y_{i}^{2} \bigr) - \Biggl( \sum _{i=1}^{k_{2}} \Psi_{i}(Y_{i}) \Biggr)^{2} \Biggr]. \end{aligned} $$
(8)

If \(f \in\overset{\bullet}{\mathcal{K}_{2}^{c}}((a,b))\) and \(D_{1}\geq D_{2}\) holds, then the reverse inequalities are valid in (7).

Proof

This result is proven directly in [3], Theorem 1, using Jensen’s operator inequality on the sum of the operators. We will give the proof by applying Theorem 1. We set \(\tilde{X}= \operatorname{diag} (X_{1},\ldots, X_{k_{1}})\) and \(\tilde {Y}= \operatorname {diag} (Y_{1},\ldots, Y_{k_{2}})\). Then \(\tilde{X} \in\mathcal{B}_{h}( \underbrace{H\oplus\cdots\oplus H}_{k_{1}} )\) and \(\tilde{Y} \in\mathcal{B}_{h}( \underbrace{H\oplus \cdots\oplus H}_{k_{2}} )\), with spectra contained in \([m,M]\) and \([n,N]\), respectively. Also, we set \(\tilde{\Phi} ( \operatorname{diag} (A_{1},\ldots, A_{k_{1}}) ) = \sum_{i=1}^{k_{1}} \Phi_{i} ( A_{i})\) and \(\tilde{\Psi} ( \operatorname{diag} (B_{1},\ldots, B_{k_{2}}) )= \sum_{i=1}^{k_{2}} \Psi_{i} ( B_{i})\). Then \(\tilde{\Phi}: \mathcal{B}( \underbrace{H\oplus\cdots\oplus H}_{k_{1}} ) \rightarrow\mathcal{B}(K)\) and \(\tilde{\Psi}: \mathcal{B}( \underbrace{H\oplus\cdots\oplus H}_{k_{2}} ) \rightarrow\mathcal {B}(K)\) are normalized positive linear mappings. We have

$$\tilde{C}_{1} = \frac{\alpha}{2} \bigl[ \tilde{\Phi} \bigl(\tilde {X}^{2} \bigr) - \tilde{\Phi}(\tilde{X})^{2} \bigr] = \frac{\alpha}{2} \Biggl[ \sum_{i=1}^{k_{1}} \Phi_{i} \bigl(X_{i}^{2} \bigr) - \Biggl( \sum _{i=1}^{k_{1}} \Phi_{i}(X_{i}) \Biggr)^{2} \Biggr]=D_{1} $$

and

$$\tilde{C}_{2} = \frac{\alpha}{2} \bigl[ \tilde{\Psi } \bigl( \tilde{Y}^{2} \bigr) - \tilde{\Psi}(\tilde{Y})^{2} \bigr] = \frac {\alpha}{2} \Biggl[ \sum_{i=1}^{k_{2}} \Psi_{i} \bigl(Y_{i}^{2} \bigr) - \Biggl( \sum _{i=1}^{k_{2}} \Phi_{i}(X_{i}) \Biggr)^{2} \Biggr]=D_{2}. $$

Applying Theorem 1 on \(\tilde{X}\), \(\tilde{Y}\), \(\tilde{\Phi}\), \(\tilde {\Psi}\) and taking into account that \(D_{1} \lesseqqgtr D_{2}\) implies \(\tilde{C}_{1} \lesseqqgtr\tilde{C}_{2}\), we obtain the desired inequalities (7) or their reverse inequalities. □

In this paper, as a continuation of the above consideration, we will observe other results as regards Levinson’s operator inequality and its converse. We give a few examples for power functions.

Converse of Levinson’s operator inequality

In this section we give the converse of inequalities (3) and (7) for \(f \in {\mathcal{K}_{i}^{c}}(I)\), \(i=1,2\). First, for convenience we introduce some abbreviations.

Let \(f:[m,M]\rightarrow\mathbb{R}\), \(m< M\), such that \(F(t)=f(t)- \frac {\alpha}{2} t^{2}\), \(\alpha\in \mathbb{R}\), be a convex or a concave function. We denote a linear function through the points \((m,F(m))\) and \((M,F(M))\) by \(f_{\alpha,[m,M]}^{\mathrm{line}}\), i.e.

$$f_{\alpha,[m,M]}^{\mathrm{line}}(t)=\frac{M-t}{M-m}f(m)+\frac{t-m}{M-m}f(M)- \frac{\alpha}{2} \bigl( (M+m)t-mM \bigr), \quad t \in \mathbb{R}, $$

and the slope of the line through \((m,F(m))\) and \((M,F(M))\) by \(k_{\alpha,f[m,M]}\), i.e.

$$\begin{aligned} k_{\alpha,f[m,M]} = \frac{f(M)-f(m)}{M-m}-\frac{\alpha}{2}(M+m). \end{aligned}$$

Next, we give the converse of Levinson’s operator inequality for two operators.

Theorem 3

Let X, Y, m, M, n, N, Φ, Ψ, \(C_{1}\), \(C_{2}\) be as in Theorem 1 and \(m< M\), \(n< N\). Let \(m_{x}\), \(M_{x}\) (\(m_{x}\leq M_{x}\)), and \(n_{y}\), \(N_{y}\) (\(n_{y} \leq N_{y}\)) be bounds of the operators \(\Phi(X)\) and \(\Psi(Y)\), respectively. (See Figure 3.)

Figure 3
figure3

Spectra conditions in the converse of Levinson’s inequality for two operators and a 3-convex function.

If \(f \in{\mathcal{K}_{1}^{c}}((a,b))\) and \(C_{1} \geq C_{2}\), then

$$ \Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) + \beta_{1} 1_{K} \geq C_{1} \geq C_{2} \geq \Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr) + \beta_{2} 1_{K}, $$
(9)

where

$$\begin{aligned}& \beta_{1} = \max_{m_{x} \leq t \leq M_{x}} \biggl\{ f(t) - \frac{\alpha }{2} t^{2}- f_{\alpha,[m,M]}^{\mathrm{line}}(t) \biggr\} \geq0, \end{aligned}$$
(10)
$$\begin{aligned}& \beta_{2} = \min_{n_{y} \leq t \leq N_{y}} \biggl\{ f(t) - \frac{\alpha }{2} t^{2} - f_{\alpha,[n,N]}^{\mathrm{line}}(t) \biggr\} \leq0. \end{aligned}$$
(11)

The constants \(\beta_{1}\), \(\beta_{2}\) exist for any α, m, M, \(m_{x}\), \(M_{x}\) and n, N, \(n_{y}\), \(N_{y}\).

The value of the constant \(\beta_{1}\) is \(\beta_{1}=f(t_{0}) - \frac {\alpha }{2} t_{0}^{2}- f_{\alpha,[m,M]}^{\mathrm{line}}(t_{0})\), where \(t_{0}\) may be determined as follows:

  • if \(f'_{ {-}}(t) -\alpha t\leq k_{\alpha,f[m,M]}\) for every \(t \in (m_{x},M_{x})\), then \(t_{0}=m_{x}\),

  • if \(f'_{ {-}}(t_{1})-\alpha t_{1}\geq k_{\alpha,f[m,M]}\geq{f'_{ {+}}}(t_{1})-\alpha t_{1}\) for some \(t_{1} \in(m_{x},M_{x})\), then \(t_{0}=t_{1}\),

  • if \(f'_{ {+}}(t) -\alpha t\geq k_{\alpha,f[m,M]}\) for every \(t \in (m_{x},M_{x})\), then \(t_{0}=M_{x}\).

The value of \(\beta_{2}\) can be determined as \(\beta_{1}\) if we replace m, M, \(m_{x}\), \(M_{x}\) by n, N, \(n_{y}\), \(N_{y}\), respectively, and with reverse inequality signs.

In the dual case, if \(f \in{\mathcal{K}_{2}^{c}}((a,b))\) and \(C_{1}\leq C_{2}\) holds, then the reverse inequalities are valid in (9), where \(\beta_{1}\leq0\) with min instead of max in (10) and \(\beta_{2}\geq0\) with max instead of min in (11). The value of the constants \(\beta_{1}\) and \(\beta_{2}\) can be determined as above with reverse inequality signs.

Proof

We will give the proof for \(f \in{\mathcal{K}_{1}^{c}}((a,b))\). So there is a constant α such that \(F(t) = f(t)- \frac{\alpha}{2} t^{2}\) is concave on \([m,M] \subset(a,c]\). The converse of Jensen’s operator inequality gives (see [12], Theorem 3.4)

$$\begin{aligned} &\Phi \bigl( F(X) \bigr) - F \bigl( \Phi(X) \bigr) \geq\min _{m_{x} \leq t \leq M_{x}} \biggl\{ f_{\alpha,[m,M]}^{\mathrm{line}}(t) - f(t) - \frac {\alpha}{2} t^{2} \biggr\} 1_{K} \\ &\quad\Rightarrow\quad \Phi \bigl( f(X) \bigr) - \frac{\alpha}{2} \Phi \bigl(X^{2} \bigr)- f \bigl( \Phi(X) \bigr) + \frac{\alpha}{2} \Phi(X)^{2} + \beta_{1} 1_{K} \geq0 \\ &\quad\Rightarrow\quad \Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) + \beta _{1} 1_{K} \geq C_{1}. \end{aligned}$$
(12)

Similarly, since F is operator convex on \([n,N] \subset[c,b)\), then Jensen’s operator inequality gives (see [12], Theorem 3.4)

$$\begin{aligned} &\Psi \bigl( F(Y) \bigr) - F \bigl( \Psi(Y) \bigr) \leq\max _{n_{y} \leq t \leq N_{y}} \biggl\{ f_{\alpha,[n,N]}^{\mathrm{line}}(t) - f(t) - \frac {\alpha}{2} t^{2} \biggr\} \\ &\quad\Rightarrow\quad \Psi \bigl( f(Y) \bigr) - \frac{\alpha}{2} \Psi \bigl(Y^{2} \bigr)- f \bigl( \Psi(Y) \bigr) + \frac{\alpha}{2} \Psi(Y)^{2} + \beta_{2} 1_{K} \leq0 \\ &\quad\Rightarrow\quad C_{2} \geq\Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr) + \beta_{2} 1_{K} . \end{aligned}$$
(13)

Combining inequalities (12) and (13) and taking into account \(C_{1} \geq C_{2}\) we obtain the desired inequality (9). We obtain \(\beta_{1}=f(t_{0}) - \frac{\alpha}{2} t_{0}^{2}- f_{\alpha,[m,M]}^{\mathrm{line}}(t_{0})\), where \(t_{0}\) is determined as in the statement of Theorem 3, by applying [12], Theorem 3.4, to \(\beta_{1}=-\min_{m_{x} \leq t \leq M_{x}} \{f_{\alpha ,[m,M]}^{\mathrm{line}}(t) - f(t) +\frac{\alpha}{2} t^{2} \}\). Analogously we get \(\beta_{2}=f(t_{0}) - \frac{\alpha}{2} t_{0}^{2}- f_{\alpha,[n,N]}^{\mathrm{line}}(t_{0})\). □

Remark 1

Let the assumptions of Theorem 3 be satisfied. If \(C_{1}\geq C_{2}\), f is strictly concave differentiable on \([m,c]\) and strictly convex differentiable on \([c,N]\), then (9) holds for

$$\begin{aligned}& \beta_{1} = f(x_{0}) -\frac{\alpha}{2} x_{0}^{2}- f_{\alpha ,[m,M]}^{\mathrm{line}}(x_{0}) \leq f(\bar{x}_{0}) -\frac{\alpha}{2} \bar{x}_{0}^{2}- f_{\alpha,[m,M]}^{\mathrm{line}}(\bar{x}_{0}), \\& \beta_{2} = f(y_{0}) -\frac{\alpha}{2} y_{0}^{2} - f_{\alpha ,[n,N]}^{\mathrm{line}}(y_{0}) \geq f(\bar{y}_{0}) -\frac{\alpha}{2} \bar {y}_{0}^{2} - f_{\alpha,[n,N]}^{\mathrm{line}}(\bar{y}_{0}), \end{aligned}$$

where \(x_{0}\) may be determined as follows:

  • if \(f'(m_{x})-\alpha m_{x} \leq k_{\alpha,f[m,M]}\), then \(x_{0}=m_{x}\),

  • if \(f'(m_{x}) -\alpha m_{x} \geq k_{\alpha,f[m,M]} \geq f'(M_{x})-\alpha M_{x}\), then \(x_{0}\) is the unique solution of the equation \(f'(t) - \alpha t = k_{\alpha,f[m,M]}\),

  • if \(f'(M_{x})-\alpha M_{x} \geq k_{\alpha,f[m,M]}\), then \(x_{0}=M_{x}\),

and \(\bar{x}_{0}\) is the unique solution in \((m,M)\) of the equation \(f'(t) - \alpha t = k_{\alpha,f[m,M]}\).

The values of \(y_{0}\), \(\bar{y}_{0}\) can be determined as \(x_{0}\), \(\bar {x}_{0}\), if we replace m, M, \(m_{x}\), \(M_{x}\) by n, N, \(n_{y}\), \(N_{y}\), respectively, and with reverse inequality signs.

Example 1

Let \(\Phi, \Psi, X, Y, m,M\geq0\), \(n,N \geq0\), \(m_{x}\), \(M_{x}\), \(n_{y}\), \(N_{y}\) be as in Theorem 3.

We will apply Theorem 3 putting \(f(t)=t^{p}\) on \((0,c]\) and \(f(t)=d t^{q}\) on \([c,\infty)\), where \(c>0\) and \(d=c^{p-q}\).

(i) If \(p \in(-\infty, 0] \cup[1, \infty)\), \(q\in[0,1]\), and \(\alpha=0\), then \(f \in\mathcal{K}_{2}^{c}([0,\infty))\). So, (5) and the reverse of (9) give

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \beta_{1}^{\circ} 1_{K} \leq0 \leq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \alpha_{2} 1_{K} \leq\beta _{2}^{\circ} 1_{K}, $$

where

$$\begin{aligned}& \beta_{1}^{\circ} = \min_{m_{x} \leq t \leq M_{x}} \biggl\{ t^{p} - \frac{M-t}{M-m}m^{p}-\frac{t-m}{M-m}M^{p} \biggr\} \leq0, \\& \beta_{2}^{\circ} = d \cdot\max_{n_{y} \leq t \leq N_{y}} \biggl\{ t^{q} - \frac{N-t}{N-n}n^{q}-\frac{t-n}{N-n}N^{q} \biggr\} \geq0. \end{aligned}$$

(ii) If \(p,q \in(-\infty, 0] \cup[1, 2]\), \(p^{2}-p\geq q^{2} -q\), and \(\alpha= c^{p-2} (p^{2}-p+q^{2}-q)/2\), then

$$\begin{aligned}[b] &\frac{d^{2}}{dt^{2}} \biggl(t^{p} -\frac{\alpha}{2}t^{2} \biggr)=p(p-1)t^{p-2}- \alpha\geq p(p-1)c^{p-2}- \alpha\geq0, \quad\mbox{if }0\leq t \leq c, \\ &\frac{d^{2}}{dt^{2}} \biggl(c^{p-q}t^{q} -\frac{\alpha}{2}t^{2} \biggr)=q(q-1) c^{p-q} t^{q-2}- \alpha\leq q(q-1)c^{p-2}- \alpha\leq0,\quad \mbox{if }t \geq c. \end{aligned} $$

So, \(f \in\mathcal{K}_{2}^{c}([0,\infty))\). If

$$(0< )\quad C_{1} := \frac{\alpha}{2} \bigl[ \Phi \bigl(X^{2} \bigr) - \Phi (X)^{2} \bigr]\leq C_{2} := \frac{\alpha}{2} \bigl[ \Psi \bigl(Y^{2} \bigr) - \Psi(Y)^{2} \bigr], $$

then the reverse of (9) gives

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \beta_{1} 1_{K} \leq C_{1} \leq C_{2} \leq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \beta_{2} 1_{K}, $$

where

$$\begin{aligned}& \beta_{1} = \min_{m_{x} \leq t \leq M_{x}} \biggl\{ t^{p}- \frac {M-t}{M-m}m^{p}-\frac{t-m}{M-m}M^{p} + \frac{\alpha}{2} \bigl( (M+m)t-mM - t^{2} \bigr) \biggr\} \leq0, \\& \beta_{2} = \max_{n_{y} \leq t \leq N_{y}} \biggl\{ d \biggl( t^{q}- \frac {N-t}{N-n}n^{q}- \frac{t-n}{N-n}N^{q} \biggr) + \frac{\alpha}{2} \bigl( (N+n)t-nN - t^{2} \bigr) \biggr\} \geq0. \end{aligned}$$

(iii) If \(p,q \in[0,1] \cup[2, \infty)\), \(p^{2}-p\leq q^{2} -q\), and \(\alpha= c^{p-2} (p^{2}-p+q^{2}-q)/2\), then

$$\begin{aligned}& \frac{d^{2}}{dt^{2}} \biggl(t^{p} -\frac{\alpha}{2}t^{2} \biggr) \leq p(p-1)c^{p-2}- \alpha\leq0, \quad\mbox{if }0\leq t \leq c, \\& \frac{d^{2}}{dt^{2}} \biggl(c^{p-q}t^{q} -\frac{\alpha}{2}t^{2} \biggr) \geq q(q-1)c^{p-2}- \alpha\geq0, \quad\mbox{if }t \geq c. \end{aligned}$$

So, \(f \in\mathcal{K}_{1}^{c}([0,\infty))\). If \(C_{1} \geq C_{2}\), then (9) gives

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \gamma_{1} 1_{K} \geq C_{1} \geq C_{2} \geq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \gamma_{2} 1_{K}, $$

where \(\gamma_{1} \geq0\) is defined similar to \({\beta}_{1}\) with max instead of min and \(\gamma_{2} \leq0\) is defined similar to \({\beta }_{2}\) with min instead of max.

Remark 2

Let the assumptions of Theorem 3 be satisfied. If \(f \in \overset {\bullet}{\mathcal{K}_{1}^{c}}((a,b))\) and \(C_{1}\geq C_{2}\), we obtain the following extension of (9):

$$ \begin{aligned}[b] C_{1}+ \beta_{1} 1_{K} &\geq \Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) + \beta_{1} 1_{K} \geq C_{1} \geq C_{2} \\ &\geq \Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr) + \beta_{2} 1_{K} \geq C_{2}+\beta_{2} 1_{K}. \end{aligned} $$
(14)

In the dual case, if \(f \in\overset{\bullet}{\mathcal {K}_{2}^{c}}((a,b))\) and \(C_{1}\leq C_{2}\), then the reverse inequalities are valid in (14).

Applying Theorem 3 we obtain a version of the converse of Levinson’s inequality with more operators.

Corollary 4

Let \(X_{i}\), \(\Phi_{i}\) (\(i=1,\ldots,k_{1}\)), \(Y_{j}\), \(\Psi_{j}\) (\(j=1,\ldots,k_{2}\)), m, M, n, N, \(D_{1}\), \(D_{2}\) be as in Corollary 2. Let \(m_{x}\), \(M_{x}\) (\(m_{x}\leq M_{x}\)), and \(n_{y}\), \(N_{y}\) (\(n_{y} \leq N_{y}\)) be bounds of the operators \(X=\sum_{i=1}^{k_{1}}\Phi_{i}(X_{i})\) and \(Y=\sum_{i=1}^{k_{2}}\Psi_{i}(Y_{i})\), respectively. If \(f \in{\mathcal{K}_{1}^{c}}((a,b))\) and \(D_{1} \geq D_{2}\), then

$$\begin{aligned} & \sum_{i=1}^{k_{1}} \Phi_{i} \bigl( f(X_{i}) \bigr)- f \Biggl( \sum _{i=1}^{k_{1}} \Phi_{i}(X_{i}) \Biggr) + \beta_{1} 1_{K} \\ &\quad\geq D_{1} \geq D_{2}\geq \sum _{i=1}^{k_{2}} \Psi_{i} \bigl( f(Y_{i}) \bigr) - f \Biggl( \sum_{i=1}^{k_{2}} \Psi_{i}(Y_{i}) \Biggr) + \beta_{2} 1_{K}, \end{aligned}$$
(15)

where \(\beta_{1}\) and \(\beta_{2}\) are as in Theorem 3.

If \(f \in{\mathcal{K}_{2}^{c}}((a,b))\) and \(D_{1}\leq D_{2}\) holds, then the reverse inequalities are valid in (15) with \(\beta _{1}\) and \(\beta_{2}\) as in Theorem 3 in the dual case.

Proof

We use the same technique as in the proof of Corollary 2. We omit the details. □

Remark 3

Applying Corollary 4 to positive linear mappings \(\Phi_{i}, \Psi:\mathcal{B}(H) \rightarrow\mathcal{B}(K)\) determined by \(\Phi_{i}: B \mapsto p_{i} B\), \(i=1,\ldots,k_{1}\), and \(\Psi_{i}: B \mapsto q_{i} B\), \(i=1,\ldots,k_{2}\), we obtain the following obvious result with convex combinations of the operators \(X_{i}\), \(i=1,\ldots,k_{1}\), and \(Y_{j}\), \(j=1,\ldots,k_{2}\):

Let \(X_{i}\), \(Y_{j}\) be operators as in Corollary 4, such that \(a < m_{x} \leq M_{x} \leq c \leq m_{y} \leq M_{y} <b\) for some \(a,b,c \in \mathbb{R}\). Let \((p_{1},\ldots,p_{k_{1}})\) be a \(k_{1}\)-tuple and \((q_{1},\ldots ,q_{k_{2}})\) be a \(k_{2}\)-tuple of positive scalars such that \(\sum_{i=1}^{k_{1}}p_{i} =1\) and \(\sum_{j=1}^{k_{2}}q_{j} =1\).

If \(f \in{\mathcal{K}_{1}^{c}}((a,b))\) and \(P \leq Q\), then

$$\sum_{i=1}^{k_{1}} p_{i} f(X_{i})- f ( \bar{X} ) + \beta_{1} 1_{K} \leq P \leq Q \leq \sum_{j=1}^{k_{2}} q_{j} f(Y_{j}) - f ( \bar{Y} ) + \beta_{2} 1_{K}, $$

where \(\beta_{1}\) and \(\beta_{2}\) are as in Theorem 3,

$$P := \frac{\alpha}{2} \sum_{i=1}^{k_{1}} p_{i} (X_{i} - \bar{X} )^{2},\qquad Q := \frac{\alpha}{2} \sum_{j=1}^{k_{2}} q_{j} (Y_{j} - \bar{Y} )^{2}, $$

and \(\bar{X}:= \sum_{i=1}^{k_{1}} p_{i} X_{i}\), \(\bar{Y}:= \sum_{j=1}^{k_{1}} q_{j} Y_{j}\) denote the weighted arithmetic means of the operators.

Refined Levinson’s operator inequality

In this section we obtain a refinement of Levison’s operator inequality (7) given in Section 2 under weaker conditions.

The absolute value of \(B\in\mathcal{B}(H)\) is defined by \(|B|=(B^{*}B)^{1/2}\).

For convenience, we introduce the abbreviations \(\bar{\Delta}\) and δ as follows:

  • \(\bar{\Delta} \equiv\bar{\Delta}_{B}(m,M):= \frac{1}{2} 1_{K} - \frac {1}{{M}-{m}} |B - \frac{{m}+{M}}{2} 1_{K} |\),

where \(B \in\mathcal{B}_{h}(H)\) is a self-adjoint operator, Φ is a normalized positive linear mapping and m, M (\(m < M\)) are some scalars such that spectra \(\operatorname{Sp}(X) \subseteq[m,M]\). Since \(m 1 _{K} \leq B \leq M 1 _{K}\), we have \(- \frac{M-m}{2} 1 _{K} \leq B - \frac{m+M}{2} 1 _{K} \leq\frac{M-m}{2} 1 _{K}\) and \(0\leq |\Phi(B) - \frac {{m}+{M}}{2} 1 _{K} | \leq\frac{M-m}{2} 1 _{K}\). It follows \(\bar {\Delta}\geq0\).

  • \(\delta\equiv\delta_{f,\alpha}(m,M) := 2 f ( \frac {{m}+{M}}{2} ) - f({m})-f({M}) + \frac{\alpha}{4}(M-m)^{2}\),

where \(f:[m,M]\rightarrow\mathbb{R}\) is a continuous function and \(\alpha\in \mathbb{R}\). Obviously, if \(F(t) = f(t)- \frac{\alpha}{2} t^{2}\) is concave (resp. convex) then \(\delta\geq0\) (resp. \(\delta\leq0\)).

First, we give refined Levinson’s operator inequality for two pairs of operators.

Theorem 5

Let \(\Phi, \Psi: \mathcal{B}(H) \oplus\mathcal{B}(H) \rightarrow \mathcal{B}(K)\) be normalized mappings such that \(\Phi ( \operatorname{diag} (B_{1},B_{2}) ) = \Phi_{1} ( B_{1})+ \Phi_{2} ( B_{2})\) and \(\Psi ( \operatorname{diag}(B_{1},B_{2}) ) = \Psi_{1} ( B_{1})+ \Psi _{2} ( B_{2})\), where \(\Phi_{1}\), \(\Phi_{2}\), \(\Psi_{1}\), \(\Psi_{2}\) are positive linear mappings.

Let \(X= \operatorname{diag}(X_{1},X_{2})\), \(Y= \operatorname {diag}(Y_{1},Y_{2})\), where \(X_{1}, X_{2}, Y_{1}, Y_{2} \in\mathcal{B}_{h}(H)\) are self-adjoint operators with spectra \(\operatorname{Sp}(X_{1})\subseteq[m_{1},M_{1}]\), \(\operatorname{Sp}(X_{2}) \subseteq[m_{2},M_{2}]\), \(\operatorname{Sp}(Y_{1}) \subseteq[n_{1},N_{1}]\), \(\operatorname{Sp}(Y_{2}) \subseteq[n_{2},N_{2}]\) \(M_{1}< m_{2}\), \(N_{1}< n_{2}\). Let \(a< m_{1} \leq M_{1} \leq{m}_{x} \leq{M}_{x} \leq m_{2} \leq M_{2} \leq c \leq n_{1} \leq N_{1} \leq{n}_{y} \leq{N}_{y} \leq n_{2} \leq N_{2} < b\), where \(m_{x}\), \(M_{x}\) and \(n_{y}\), \(N_{y}\) are bounds of \(\Phi(X)\) and \(\Psi(Y)\), respectively. (See Figure  4.)

Figure 4
figure4

Spectra conditions in a refined Levinson’s inequality for two operators and a 3-convex function.

If \(f \in{\mathcal{K}_{1}^{c}}((a,b))\) and \(C_{1} \leq C_{2}\) (see (4)), then

$$\begin{aligned} \Phi \bigl( f(X) \bigr)- f \bigl( \Phi(X) \bigr) &\leq\Phi \bigl( f(X) \bigr)- f \bigl( \Phi(X) \bigr) +\delta_{1} \bar{X} \leq C_{1} \\ &\leq C_{2} \leq\Psi \bigl( f(Y) \bigr)- f \bigl( \Psi(Y) \bigr) +\delta_{2} \bar{Y} \leq\Psi \bigl( f(Y) \bigr)- f \bigl( \Psi(Y) \bigr), \end{aligned}$$
(16)

where \(\delta_{1}= \delta_{f,\alpha}(\bar{m},\bar{M}) \geq0\), \(\bar {X}=\bar{\Delta}_{\Phi(X)}(\bar{m},\bar{M})\geq0\) for arbitrary numbers \(\bar{m} \in[M_{1},m_{x}]\), \(\bar{M} \in[M_{x},m_{2}]\), \(\bar{m}< \bar{M} \) and \(\delta_{2}= \delta_{f,\alpha}(\bar{n},\bar{N}) \leq0\), \(\bar{Y}=\bar{\Delta}_{\Psi(Y)}(\bar{n},\bar{N})\geq0\) for arbitrary numbers \(\bar{n} \in[N_{1},n_{y}]\), \(\bar{N} \in[N_{y},n_{2}]\), \(\bar{n}< \bar{N}\).

But, if \(f \in{\mathcal{K}_{2}^{c}}((a,b))\) and \(C_{1}\geq C_{2}\) holds, then the reverse inequalities are valid in (16), with \(\delta_{1} \leq0\) and \(\delta_{2} \geq0\).

Proof

We will give the proof for \(f \in{\mathcal{K}_{1}^{c}}((a,b))\). Since \(F(t) = f(t)- \frac{\alpha}{2} t^{2}\) is concave on \([m_{1},c]\subset(a,c]\) for some constant α, the refined Jensen’s operator inequality for a concave function implies (see [13], Theorem 3)

$$\begin{aligned} &F \bigl( \Phi(X) \bigr) \geq\Phi \bigl( F(X) \bigr) + \widetilde {\delta}_{1} \bar{X} \geq\Phi \bigl( F(X) \bigr) \\ & \quad\Rightarrow\quad C_{1} \geq\Phi \bigl( f(X) \bigr) -f \bigl( \Phi(X) \bigr) + {\delta}_{1} \bar{X} \geq\Phi \bigl( f(X) \bigr) -f \bigl( \Phi(X) \bigr), \end{aligned}$$
(17)

since \(0 \leq\widetilde{\delta}_{1} = 2 F ( \frac{{\bar{m}}+{\bar {M}}}{2} ) - F({\bar{m}})-F({\bar{M}}) = \delta_{f,\alpha }(\bar {m},\bar{M})=\delta_{1} \) and

$$\bar{X} = \frac{1}{2} 1_{K} - \frac{1}{\bar{M}-\bar{m}} \biggl|\Phi _{1}(X_{1})+ \Phi_{2}(X_{2}) - \frac{\bar{m}+\bar{M}}{2} 1_{K} \biggr| =\bar {\Delta }_{\Phi(X)}(\bar{m}, \bar{M}). $$

Similarly, since F is convex on \([c,N_{2}]\subset[c,b)\) for some constant α, the refined Jensen’s operator inequality for a convex function implies (see [13], Theorem 3)

$$\begin{aligned} &F \bigl( \Psi(Y) \bigr) \leq\Psi \bigl( F(Y) \bigr) - \widetilde {\delta}_{2} \bar{Y} \geq\Psi \bigl( F(Y) \bigr) \\ &\quad\Rightarrow\quad C_{2} \leq\Psi \bigl( f(Y) \bigr) -f \bigl( \Psi(Y) \bigr) + {\delta}_{2} \bar{Y} \leq\Psi \bigl( f(Y) \bigr) -f \bigl( \Psi(Y) \bigr), \end{aligned}$$
(18)

since \(0 \leq\widetilde{\delta}_{2} = F(\bar{N})+ F({\bar{N}}) - 2 F ( \frac{{\bar{n}}+{\bar{N}}}{2} ) = -\delta_{f,\alpha}(\bar {n},\bar {N})=- \delta_{2} \) and

$$\bar{Y} = \frac{1}{2} 1_{K} - \frac{1}{\bar{N}-\bar{n}} \biggl|\Psi _{1}(Y_{1})+ \Psi_{2}(Y_{2}) - \frac{\bar{n}+\bar{N}}{2} 1_{K} \biggr| =\bar {\Delta }_{\Psi(Y)}(\bar{n}, \bar{N}). $$

Combining inequalities (17) and (18) we obtain the desired inequality (16). □

Example 2

Let \(\Phi_{i}, \Psi_{i}, X_{i}, Y_{i}, m_{i},M_{i}\geq0\), \(n_{i},N_{i} \geq 0\), \(i=1,2\), Φ, Ψ, X, Y, \(m_{x}\), \(M_{x}\), \(n_{y}\), \(N_{y}\) be as in Theorem 5.

We will use the same technique as in Example 1 and we will apply Theorem 5 putting \(f(t)=t^{p}\) on \((0,c]\), \(f(t)=d t^{q}\) on \([c,\infty)\), where \(c>0\) and \(d=c^{p-q}\).

(i) If \(p\in[0,1]\), \(q \in(-\infty, 0] \cup[1, \infty)\), and \(\alpha=0\), then \(f \in\mathcal{K}_{1}^{c}([0,\infty))\). So, (16) gives

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \delta_{1} \bar{X} \leq0 \leq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \delta_{2} \bar{Y}, $$

where

$$\begin{aligned}& \delta_{1} = 2^{1-p} ( \bar{m}+\bar{M} )^{p} - \bar {m}^{p}-\bar {M}^{p} \geq0, \qquad \bar{X} = \frac{1}{2} 1_{K} - \frac{1}{\bar{M}-\bar{m}} \biggl|\Phi(X) - \frac{\bar{M}+\bar{m}}{2} 1_{K} \biggr|\geq0, \\& \delta_{2} = d \bigl( 2^{1-q} ( \bar{n}+\bar{N} )^{q} - \bar {n}^{q}-\bar {N}^{q} \bigr) \leq0, \qquad \bar{Y} = \frac{1}{2} 1_{K} - \frac{1}{\bar{N}-\bar{n}} \biggl|\Psi(Y) - \frac{\bar{N}+\bar{n}}{2} 1_{K} \biggr| \geq0. \end{aligned}$$

(ii) If \(p,q \in[0,1] \cup[2, \infty)\), \(p^{2}-p\leq q^{2} -q\), and \(\alpha= c^{p-2} (p^{2}-p+q^{2}-q)/2\), then \(f \in\mathcal {K}_{1}^{c}([0,\infty))\). If

$$C_{1} := \frac{\alpha}{2} \bigl[ \Phi \bigl(X^{2} \bigr) - \Phi(X)^{2} \bigr]\leq C_{2} := \frac{\alpha}{2} \bigl[ \Psi \bigl(Y^{2} \bigr) - \Psi(Y)^{2} \bigr], $$

then (16) gives

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \delta_{1} \bar{X} \leq C_{1} \leq C_{2} \leq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \delta_{2} \bar{Y}, $$

where

$$\begin{aligned}& \delta_{1} = 2^{1-p} ( \bar{m}+\bar{M} )^{p} - \bar {m}^{p}-\bar {M}^{p} + \alpha(\bar{M}- \bar{m})^{2} /4 \geq0, \\& \delta_{2} = d \bigl( 2^{1-q} ( \bar{n}+\bar{N} )^{q} - \bar {n}^{q}-\bar{N}^{q} \bigr) + \alpha( \bar{N}-\bar{n})^{2} /4 \leq0, \end{aligned}$$

and \(\bar{X}, \bar{Y}\geq0\) as in the case (i).

(iii) If \(p,q \in(-\infty, 0] \cup[1, 2]\), \(p^{2}-p\geq q^{2} -q\), and \(\alpha= c^{p-2} (p^{2}-p+q^{2}-q)/2\), then \(f \in\mathcal{K}_{2}^{c}([0,\infty))\). If \(C_{1} \geq C_{2}\) (>0), then (16) gives

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \delta_{1} \bar{X} \geq C_{1} \geq C_{2} \geq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \delta_{2} \bar{Y}, $$

where \(\delta_{1} \geq0\), \(\delta_{2} \leq0\), and \(\bar{X}, \bar{Y} \geq 0\) as in the case (ii).

The first and the last inequality in (16) are obvious, so we omit them.

Levinson’s operator inequality (7) holds with the weaker condition: \(f \in{\mathcal{K}_{1}^{c}}(I)\) and with spectra conditions (see [3], Theorem 5). Next, applying Theorem 5 we obtain a refinement of this inequality. The proof is the same as for Corollary 2 and we omit the details.

Corollary 6

Let \((\Phi_{1},\ldots,\Phi_{k_{1}})\) be a unital \({k_{1}}\)-tuple and \((\Psi _{1},\ldots,\Psi_{k_{2}})\) be a unital \({k_{2}}\)-tuple of positive linear mappings \(\Phi_{i}, \Psi_{j}:\mathcal{B}(H) \rightarrow\mathcal{B}(K)\). Let \((X_{1},\ldots,X_{k_{1}})\) be a \({k_{1}}\)-tuple and \((Y_{1},\ldots ,Y_{k_{2}})\) be a \({k_{2}}\)-tuple of self-adjoint operators \(X_{i}\) and \(Y_{j} \in\mathcal{B}_{h}(H)\) with spectra contained in \([m_{i},M_{i}]\) and \([n_{j},N_{j}]\), respectively, such that

$$\begin{aligned}& a< m_{i}\leq M_{i} \leq c \leq n_{j} \leq N_{j}< b,\quad i=1,\ldots,k_{1}, j=1,\ldots,k_{2}, \\& (m_{x},M_{x}) \cap[m_{i},M_{i}]= \varnothing,\quad i=1,\ldots,k_{1},\qquad (m_{y},M_{y}) \cap[n_{j},N_{j}]= \varnothing,\quad j=1, \ldots,k_{2}, \\& m<M,\qquad n<N, \end{aligned}$$

where \(m_{x}\), \(M_{x}\) and \(n_{y}\), \(N_{y}\) are bounds of \(X=\sum_{i=1}^{k_{1}}\Phi _{i}(X_{i})\) and \(Y=\sum_{i=1}^{k_{2}}\Psi_{i}(Y_{i})\), respectively, and

$$\begin{aligned}& m:=\max\{M_{i}| M_{i}\leq m_{x}, i=1, \ldots,k_{1} \},\qquad M:=\min\{ m_{i}| m_{i}\geq M_{x}, i=1,\ldots,k_{1} \}, \\& n:=\max\{N_{i}| N_{i}\leq n_{y}, i=1, \ldots,k_{2} \},\qquad N:=\min\{ n_{i}| n_{i}\geq N_{y}, i=1,\ldots,k_{2} \}. \end{aligned}$$

If \(f \in{\mathcal{K}_{1}^{c}}((a,b))\) and \(D_{1} \leq D_{2}\) (see (8)), then

$$\begin{aligned} &\sum_{i=1}^{k_{1}} \Phi_{i} \bigl( f(X_{i}) \bigr)- f \Biggl( \sum _{i=1}^{k_{1}} \Phi_{i}(X_{i}) \Biggr) \\ &\quad\leq\sum_{i=1}^{k_{1}} \Phi_{i} \bigl( f(X_{i}) \bigr)- f \Biggl( \sum _{i=1}^{k_{1}} \Phi _{i}(X_{i}) \Biggr) +\delta_{1} \bar{X} \leq D_{1} \\ &\quad\leq D_{2} \leq \sum_{i=1}^{k_{2}} \Psi_{i} \bigl( f(Y_{i}) \bigr) - f \Biggl( \sum _{i=1}^{k_{2}} \Psi_{i}(Y_{i}) \Biggr) \\ &\quad\leq \sum_{i=1}^{k_{2}} \Psi_{i} \bigl( f(Y_{i}) \bigr) - f \Biggl( \sum _{i=1}^{k_{2}} \Psi_{i}(Y_{i}) \Biggr) +\delta_{2} \bar{Y}, \end{aligned}$$
(19)

where \(\delta_{1}= \delta_{f,\alpha}(\bar{m},\bar{M}) \geq0\), \(\bar {X}=\bar{\Delta}_{X}(\bar{m},\bar{M})\geq0\) for arbitrary numbers \(\bar {m} \in[m,m_{x}]\), \(\bar{M} \in[M_{x},M]\), \(\bar{m}< \bar{M} \) and \(\delta_{2}= \delta_{f,\alpha}(\bar{n},\bar{N}) \leq0\), \(\bar{Y}=\bar{\Delta}_{Y}(\bar{n},\bar{N})\geq0\) for arbitrary numbers \(\bar{n} \in[n,n_{y}]\), \(\bar{N} \in[N_{y},N]\), \(\bar{n}< \bar{N}\).

But, if \(f \in{\mathcal{K}_{2}^{c}}((a,b))\) and \(D_{1}\geq D_{2}\) holds, then the reverse inequalities are valid in (19), with \(\delta _{1} \leq0\) and \(\delta_{2} \geq0\).

Refined converse of Levinson’s operator inequality

In this section we obtain a refined converse of Levison’s operator inequality (15) given in Section 2.

For convenience, we introduce the abbreviation

$${\widetilde{\Delta}} \equiv{\widetilde{\Delta}}_{\Phi, B}(m,M):= \Phi \biggl( \frac{1}{2} 1_{H} - \frac{1}{{M}-{m}} \biggl|B - \frac{{m}+{M}}{2} 1_{H} \biggr| \biggr), $$

where \(B \in\mathcal{B}_{h}(H)\) is a self-adjoint operator, Φ is a normalized positive linear mapping and m, M (\(m < M\)) are some scalars such that spectra \(\operatorname{Sp}(X) \subseteq[m,M]\). Obviously, \({\widetilde{\Delta}} \geq0\).

First, we give a refinement of (9) for two pairs of operators.

Theorem 7

Let \(\Phi, \Psi: \mathcal{B}(H) \oplus\mathcal{B}(H) \rightarrow \mathcal{B}(K)\) be normalized mappings such that \(\Phi ( \operatorname{diag} (B_{1},B_{2}) ) = \Phi_{1} ( B_{1})+ \Phi_{2} ( B_{2})\) and \(\Psi ( \operatorname{diag}(B_{1},B_{2}) ) = \Psi _{1} ( B_{1})+ \Psi_{2} ( B_{2})\), where \(\Phi_{1}\), \(\Phi_{2}\), \(\Psi_{1}\), \(\Psi_{2}\) are positive linear mappings. Let \(X= \operatorname{diag}(X_{1},X_{2})\), \(Y= \operatorname {diag}(Y_{1},Y_{2})\), where \(X_{1}, X_{2}, Y_{1}, Y_{2} \in\mathcal{B}_{h}(H)\) are self-adjoint operators with spectra \(\operatorname{Sp}(X_{1}), \operatorname{Sp}(X_{2})\subseteq[m,M]\), \(\operatorname {Sp}(Y_{1}), \operatorname{Sp}(Y_{2}) \subseteq[n,N]\), such that \(a< m\leq M \leq c \leq n \leq N< b\). Let \(m_{x}\), \(M_{x}\) and \(n_{y}\), \(N_{y}\) be bounds of the operators \(\Phi(X)\) and \(\Psi(Y)\), respectively (see Figure 3). If \(f \in{\mathcal{K}_{1}^{c}}((a,b))\) and \(C_{1} \geq C_{2}\) (see (4)), then

$$\begin{aligned} &\Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) + \beta_{1} 1_{K} \\ &\quad\geq \Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) + \beta_{1} 1_{K} -\delta _{1} \widetilde{X} \geq C_{1} \\ &\quad\geq C_{2} \geq\Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr) + \beta_{2} 1_{K} - \delta_{2} \widetilde{Y} \geq \Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr) + \beta_{2} 1_{K}, \end{aligned}$$
(20)

where \(\beta_{1}\), \(\beta_{2}\) are defined as in Theorem 3, \(\delta_{1}= \delta_{f,\alpha}(m,M) \geq0\), \(\widetilde {X}=\widetilde {\Delta}_{\Phi,X}(m,M)\geq0\), \(\delta_{2}= \delta_{f,\alpha}(n,N) \leq 0\), and \(\widetilde{Y}=\widetilde{\Delta}_{\Psi,Y}(n,N)\geq0\).

If \(f \in{\mathcal{K}_{2}^{c}}((a,b))\) and \(C_{1}\leq C_{2}\) holds, then the reverse inequalities are valid in (20), with \(\delta_{1} \leq0\) and \(\delta_{2} \geq0\) and \(\beta_{1}\) and \(\beta _{2}\) as in Theorem 3 in the dual case.

Proof

We will give the proof for \(f \in{\mathcal{K}_{1}^{c}}((a,b))\). Since \(F(t) = f(t)- \frac{\alpha}{2} t^{2}\) is concave on \([m,c]\subset(a,c]\) for some constant α, the refined converse of Jensen’s inequality for a concave function implies (see [14], Theorem 8)

$$\begin{aligned} &\Phi \bigl( F(X) \bigr) - F \bigl( \Phi(X) \bigr) \geq\min_{m_{x} \leq t \leq M_{x}} \biggl\{ f_{\alpha,[m,M]}^{\mathrm{line}}(t) - f(t) -\frac {\alpha}{2} t^{2} \biggr\} 1_{K} - \widetilde{ \widetilde{\delta}}_{1} \widetilde{X} \\ &\quad\Rightarrow\quad \Phi \bigl( f(X) \bigr) - \frac{\alpha}{2} \Phi \bigl(X^{2} \bigr)- f \bigl( \Phi(X) \bigr) + \frac{\alpha}{2} \Phi(X)^{2} + \beta_{1} 1_{K} + \widetilde{ \widetilde{\delta}}_{1} \widetilde{X} \geq0 \\ &\quad\Rightarrow\quad \Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) + \beta _{1} 1_{K} - \widetilde{\delta}_{1} \widetilde{X} \geq C_{1}, \end{aligned}$$
(21)

since \(0 \geq\widetilde{\widetilde{\delta}}_{1} = F(m)+F(M) -2 F ( \frac {m+M}{2} )= -\delta_{f,\alpha}(m,M)= -\widetilde{\delta}_{1} \) and

$$\begin{aligned} \widetilde{X} =& \frac{1}{2} 1_{K} - \frac{1}{M-m} \biggl\{ \Phi_{1} \biggl( \biggl|X_{1}- \frac{m+M}{2} 1_{H} \biggr| \biggr) + \Phi_{2} \biggl( \biggl|X_{2}- \frac{m+M}{2} 1_{H} \biggr| \biggr) \biggr\} \\ =& \frac{1}{2} 1_{K} - \frac{1}{M-m} \Phi \biggl( \biggl|X- \frac{m+M}{2} 1_{H} \biggr| \biggr) \\ =& \Phi \biggl(\frac{1}{2} 1_{H} - \frac{1}{M-m} \biggl|X- \frac{m+M}{2} 1_{H} \biggr| \biggr) = \widetilde{\Delta}_{\Phi,X}(m,M). \end{aligned}$$

Similarly, since F is convex on \([c,N_{2}]\subset[c,b)\) for some constant α, the refined converse of Jensen’s inequality for a convex function implies (see [14], Theorem 8)

$$\begin{aligned} &\Psi \bigl( F(Y) \bigr) - F \bigl( \Psi(Y) \bigr) \leq\max_{n_{y} \leq t \leq N_{y}} \biggl\{ f_{\alpha,[n,N]}^{\mathrm{line}}(t) - f(t) -\frac {\alpha}{2} t^{2} \biggr\} - \widetilde{\widetilde{ \delta}}_{2} \widetilde {Y} \\ &\quad\Rightarrow\quad \Psi \bigl( f(Y) \bigr) - \frac{\alpha}{2} \Psi \bigl(Y^{2} \bigr)- f \bigl( \Psi(Y) \bigr) + \frac{\alpha}{2} \Psi(Y)^{2} + \beta_{2} 1_{K} + \widetilde{ \widetilde{\delta}}_{2} \widetilde{Y} \leq0 \\ &\quad\Rightarrow\quad C_{2} \geq\Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr) + \beta_{2} 1_{K} -\widetilde{ \delta}_{2} \widetilde{Y}, \end{aligned}$$
(22)

since \(0 \leq\widetilde{\widetilde{\delta}}_{2} = F(n)+F(N) -2 F ( \frac {n+N}{2} )= -\delta_{f,\alpha}(n,N)= -\widetilde{ \delta}_{2} \) and

$$\widetilde{Y} = \Psi \biggl(\frac{1}{2} 1_{H} - \frac{1}{N-n} \biggl|Y- \frac{n+N}{2} 1_{H} \biggr| \biggr) = \widetilde{ \Delta}_{\Psi ,Y}(n,N). $$

Combining inequalities (21) and (22) we obtain the desired inequality (20). □

Example 3

Let \(\Phi_{i}\), \(\Psi_{i}\), \(X_{i}\), \(Y_{i}\), \(i=1,2\), \(m,M\geq0\), \(n,N \geq 0\), Φ, Ψ, X, Y, \(m_{x}\), \(M_{x}\), \(n_{y}\), \(N_{y}\) be as in Theorem 7.

We will apply Theorem 7 putting \(f(t)=t^{p}\) on \((0,c]\), \(f(t)=d t^{q}\) on \([c,\infty)\), where \(c>0\) and \(d=c^{p-q}\).

(i) If \(p \in(-\infty, 0] \cup[1, \infty)\), \(q\in[0,1]\), and \(\alpha=0\), then reverse of (20) gives

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \beta_{1}^{\circ} 1_{K} -\delta_{1} \widetilde{X} \leq0 \leq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \beta_{2}^{\circ} 1_{K} - \delta_{2} \widetilde{Y}, $$

where \(\beta_{1}^{\circ}\), \(\beta_{2}^{\circ}\) are as in Example 1(i), and

$$\begin{aligned}& \delta_{1} = 2^{1-p} ( {m}+{M} )^{p} - {m}^{p}-{M}^{p} \geq0, \qquad \widetilde{X} = \frac{1}{2} 1_{K} - \frac{1}{{M}-{m}} \Phi \biggl( \biggl|X - \frac{{M}+{m}}{2} 1_{H} \biggr|\biggr), \\& \delta_{2} = d \bigl( 2^{1-q} ( {n}+{N} )^{q}/2 - {n}^{q}-{N}^{q} \bigr) \leq0,\qquad \widetilde{Y} = \frac{1}{2} 1_{K} - \frac{1}{{N}-{n}} \Psi \biggl( \biggl|Y - \frac{{N}+{n}}{2} 1_{H} \biggr| \biggr). \end{aligned}$$

(ii) If \(p,q \in(-\infty, 0] \cup[1, 2]\), \(p^{2}-p\geq q^{2} -q\), and \(\alpha= c^{p-2} (p^{2}-p+q^{2}-q)/2\), then \(f \in\mathcal {K}_{2}^{c}([0,\infty))\). If

$$(0< )\quad C_{1} := \frac{\alpha}{2} \bigl[ \Phi \bigl(X^{2} \bigr) - \Phi (X)^{2} \bigr] \leq C_{2} := \frac{\alpha}{2} \bigl[ \Psi \bigl(Y^{2} \bigr) - \Psi(Y)^{2} \bigr], $$

then the reverse of (20) gives

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \beta_{1} 1_{K} -\delta_{1} \widetilde {X} \leq C_{1} \leq C_{2} \leq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \beta_{2} 1_{K} - \delta_{2} \widetilde{Y}, $$

where \(\beta_{1}\), \(\beta_{2}\) are as in Example 1(ii),

$$\begin{aligned}& \delta_{1} = 2^{1-p} ( {m}+{M} )^{p} - {m}^{p}-{M}^{p} + \alpha ({M}-{m})^{2} /4 \geq0, \\& \delta_{2} = d \bigl( 2^{1-q} ( {n}+{N} )^{q} - {n}^{q}-{N}^{q} \bigr) + \alpha({N}-{n})^{2} /4 \leq0 \end{aligned}$$

and \(\widetilde{X}\), \(\widetilde{Y}\) are as in the case (i).

(iii) If \(p,q \in[0,1] \cup[2, \infty)\), \(p^{2}-p\leq q^{2} -q\), and \(\alpha= c^{p-2} (p^{2}-p+q^{2}-q)/2\), then \(f \in\mathcal{K}_{1}^{c}([0,\infty))\). If \(C_{1} \geq C_{2}\), then (20) gives

$$\Phi \bigl( X^{p} \bigr) - \Phi(X)^{p} + \gamma_{1} 1_{K} -\delta_{1} \widetilde {X} \geq C_{1} \geq C_{2} \geq d \Psi \bigl( Y^{q} \bigr) - d \Psi(Y)^{q} + \gamma_{2} 1_{K} - \delta_{2} \widetilde{Y}, $$

where \(\gamma_{1} \geq0\) is defined similar to \({\beta}_{1}\) with max instead of min and \(\gamma_{2} \leq0\) is defined similar to \({\beta }_{2}\) with min instead of max, and \(\delta_{1} \leq0\), \(\delta_{2} \geq 0\), \(\widetilde{X}\), \(\widetilde{Y}\) are as in the case (ii).

The first and the last inequality in (20) are obvious, so we omit them.

Remark 4

Let the assumptions of Theorem 5 be satisfied. If \(f \in \overset {\bullet}{\mathcal{K}_{1}^{c}}([m_{1},N_{2}])\) and \(C_{1}\geq C_{2}\), we obtain the following extension of (16):

$$\begin{aligned} C_{1}+ \beta_{1} 1_{K} &\geq\Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) + \beta_{1} 1_{K} \\ &\geq\Phi \bigl( f(X) \bigr) - f \bigl( \Phi(X) \bigr) + \beta_{1} 1_{K} - {\delta}_{1} \bar{X} \geq C_{1} \\ &\geq C_{2} \geq \Psi \bigl( f(Y) \bigr) -f \bigl( \Psi(Y) \bigr)+ \beta_{2} 1_{K} - {\delta }_{2} \bar{Y} \\ &\geq\Psi \bigl( f(Y) \bigr) - f \bigl( \Psi(Y) \bigr) + \beta_{2} 1_{K} \geq C_{2} + \beta_{2} 1_{K}. \end{aligned}$$
(23)

But, if \(f \in\overset{\bullet}{\mathcal{K}_{2}^{c}}((a,b))\) and \(C_{1}\leq C_{2}\), then the reverse inequalities are valid in (23).

Applying Theorem 7 we obtain a refinement of (15). We omit the proof.

Corollary 8

Let \((\Phi_{1},\ldots,\Phi_{k_{1}})\) be a unital \({k_{1}}\)-tuple and \((\Psi _{1},\ldots,\Psi_{k_{2}})\) be a unital \({k_{2}}\)-tuple of positive linear mappings \(\Phi_{i}, \Psi_{j}:\mathcal{B}(H) \rightarrow\mathcal{B}(K)\). Let \((X_{1},\ldots,X_{k_{1}})\) be a \({k_{1}}\)-tuple and \((Y_{1},\ldots ,Y_{k_{2}})\) be a \({k_{2}}\)-tuple of self-adjoint operators \(X_{i}\) and \(Y_{j} \in\mathcal{B}_{h}(H)\) with spectra contained in \([m,M]\) and \([n,N]\), respectively, such that \(a< m\leq M \leq c \leq n \leq N< b\). Let \(m_{x}\), \(M_{x}\) and \(n_{y}\), \(N_{y}\) be bounds of \(X=\sum_{i=1}^{k_{1}}\Phi_{i}(X_{i})\) and \(Y=\sum_{i=1}^{k_{2}}\Psi _{i}(Y_{i})\), respectively.

If \(f \in{\mathcal{K}_{1}^{c}}((a,b))\) and \(D_{1} \geq D_{2}\) (see (8)), then

$$\begin{aligned} & \sum_{i=1}^{k_{1}} \Phi_{i} \bigl( f(X_{i}) \bigr)- f \Biggl( \sum _{i=1}^{k_{1}} \Phi_{i}(X_{i}) \Biggr) + \beta_{1} 1_{K} \\ &\quad\geq \sum_{i=1}^{k_{1}} \Phi_{i} \bigl( f(X_{i}) \bigr)- f \Biggl( \sum _{i=1}^{k_{1}} \Phi_{i}(X_{i}) \Biggr) -\delta_{1} \widetilde{X} + \beta_{1} 1_{K} \geq D_{1} \\ &\quad\geq D_{2} \geq\sum_{i=1}^{k_{2}} \Psi_{i} \bigl( f(Y_{i}) \bigr) - f \Biggl( \sum _{i=1}^{k_{2}} \Psi_{i}(Y_{i}) \Biggr) + \beta_{2} 1_{K} - \delta_{2} \widetilde{Y} \\ &\quad\geq \sum_{i=1}^{k_{2}} \Psi_{i} \bigl( f(Y_{i}) \bigr) - f \Biggl( \sum _{i=1}^{k_{2}} \Psi_{i}(Y_{i}) \Biggr) + \beta_{2} 1_{K}, \end{aligned}$$
(24)

where \(\beta_{1}\) and \(\beta_{2}\) are defined as in Theorem 3, \(\delta_{1}= \delta_{f,\alpha}(m,M) \geq0\), \(\widetilde{X}=\sum_{i=1}^{k_{1}} \widetilde{\Delta}_{\Phi_{i}, X_{i}}(m,M)\geq0\), \(\delta_{2}= \delta_{f,\alpha}(n,N) \leq0\), and \(\widetilde{Y}=\sum_{i=1}^{k_{2}}\widetilde{\Delta}_{\Psi_{i}, y_{i}}(n,N)\geq0\).

If \(f \in{\mathcal{K}_{2}^{c}}((a,b))\) and \(D_{1}\leq D_{2}\) holds, then the reverse inequalities are valid in (24), with \(\delta_{1} \leq0\) and \(\delta_{2} \geq0\) and \(\beta_{1}\) and \(\beta_{2}\) as in Theorem 3 in the dual case.

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Acknowledgements

This work has been supported in part by the Croatian Science Foundation under the project 5435.

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Mićić, J., Pečarić, J. On Levinson’s operator inequality and its converses. J Inequal Appl 2015, 127 (2015). https://doi.org/10.1186/s13660-015-0638-z

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MSC

  • 47A63
  • 47B15

Keywords

  • self-adjoint operator
  • positive linear mapping
  • convex function
  • Levinson’s inequality
  • Mond-Pečarić method