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Journal of Inequalities and Applications

Erschienen in:

Open Access 01.12.2013 | Research

Refined converses of Jensen’s inequality for operators

verfasst von: Jadranka Mićić, Josip Pečarić, Jurica Perić

Erschienen in: Journal of Inequalities and Applications | Ausgabe 1/2013

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Abstract

In this paper converses of a generalized Jensen’s inequality for a continuous field of self-adjoint operators, a unital field of positive linear mappings and real-valued continuous convex functions are studied. New refined converses are presented by using the Mond-Pečarić method improvement. Obtained results are applied to refine selected inequalities with power functions.

MSC:47A63, 47A64.

Authors’ original file for figure 1

Authors’ original file for figure 2

Electronic supplementary material

The online version of this article (doi:10.1186/1029-242X-2013-353) contains supplementary material, which is available to authorized users.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

1 Introduction

Let T be a locally compact Hausdorff space and let

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-353/MediaObjects/13660_2012_Article_798_IEq1_HTML.gif

be a

C^{*}

-algebra of operators on some Hilbert space H. We say that a field

{(x_{t})}_{t \in T}

of operators in

is continuous if the function

t \mapsto x_{t}

is norm continuous on T. If in addition μ is a Radon measure on T and the function

t \mapsto ∥ x_{t} ∥

is integrable, then we can form the Bochner integral

\int_{T} x_{t} d μ (t)

, which is the unique element in

such that

φ (\int_{T} x_{t} d μ (t)) = \int_{T} φ (x_{t}) d μ (t)

for every linear functional φ in the norm dual

A^{*}

Assume further that there is a field

{(ϕ_{t})}_{t \in T}

of positive linear mappings

ϕ_{t} : A \to B

from

to another

C^{*}

-algebra ℬ of operators on a Hilbert space K. We recall that a linear mapping

ϕ : A \to B

is said to be positive if

ϕ (x) \geq 0

for all

x \geq 0

. We say that such a field

{(ϕ_{t})}_{t \in T}

is continuous if the function

t \mapsto ϕ_{t} (x)

is continuous for every

x \in A

. Let the

C^{*}

-algebras include the identity operators and let the function

t \mapsto ϕ_{t} (1_{H})

be integrable with

\int_{T} ϕ_{t} (1_{H}) d μ (t) = k 1_{K}

for some positive scalar k. If

\int_{T} ϕ_{t} (1_{H}) d μ (t) = 1_{K}

, we say that a field

{(ϕ_{t})}_{t \in T}

is unital.

Let

B (H)

be the

C^{*}

-algebra of all bounded linear operators on a Hilbert space H. We define bounds of a self-adjoint operator

x \in B (H)

m_{x} : = inf_{∥ ξ ∥ = 1} 〈 x ξ, ξ 〉 and M_{x} : = sup_{∥ ξ ∥ = 1} 〈 x ξ, ξ 〉

(1)

for

ξ \in H

. If

Sp (x)

denotes the spectrum of x, then

Sp (x) \subseteq [m_{x}, M_{x}]

For an operator

x \in B (H)

, we define the operator

| x | : = {(x^{*} x)}^{1 / 2}

. Obviously, if x is self-adjoint, then

| x | = {(x^{2})}^{1 / 2}

Jensen’s inequality is one of the most important inequalities. It has many applications in mathematics and statistics and some other well-known inequalities are its special cases.

Let f be an operator convex function defined on an interval I. Davis [1] proved the so-called Jensen operator inequality

f (ϕ (x)) \leq ϕ (f (x)),

(2)

where

ϕ : A \to B (K)

is a unital completely positive linear mapping from a

C^{*}

-algebra

to linear operators on a Hilbert space K, and x is a self-adjoint element in

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-353/MediaObjects/13660_2012_Article_798_IEq29_HTML.gif

with spectrum in I. Subsequently, Choi [2] noted that it is enough to assume that ϕ is unital and positive.

Mond, Pečarić, Hansen, Pedersen et al. in [3‐6] studied another generalization of (2) for operator convex functions. Moreover, Hansen et al. [7] presented a general formulation of Jensen’s operator inequality for a bounded continuous field of self-adjoint operators and a unital field of positive linear mappings:

f (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \leq \int_{T} ϕ_{t} (f (x_{t})) d μ (t),

(3)

where f is an operator convex function.

There is an extensive literature devoted to Jensen’s inequality concerning different refinements and extensive results, e.g., see [8‐20]. Mićić et al. [21] proved that the discrete version of (3) stands without operator convexity of f under a condition on the spectra of operators. Recently, Mićić et al. [22] presented a discrete version of refined Jensen’s inequality for real-valued continuous convex functions. A continuous version is given below.

Theorem 1 Let

{(x_{t})}_{t \in T}

be a bounded continuous field of self-adjoint elements in a unital

C^{*}

-algebra

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-353/MediaObjects/13660_2012_Article_798_IEq31_HTML.gif

defined on a locally compact Hausdorff space T equipped with a bounded Radon measure μ. Let

m_{t}

and

M_{t}

m_{t} \leq M_{t}

, be the bounds of

x_{t}

t \in T

. Let

{(ϕ_{t})}_{t \in T}

be a unital field of positive linear mappings

ϕ_{t} : A \to B

from

to another unital

C^{*}

-algebra ℬ. Let

(m_{x}, M_{x}) \cap [m_{t}, M_{t}] = \emptyset, t \in T, and a < b,

where

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, are the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

and

a = sup {M_{t} : M_{t} \leq m_{x}, t \in T}, b = inf {m_{t} : m_{t} \geq M_{x}, t \in T} .

f : I \to R

is a continuous convex (resp. concave) function provided that the interval I contains all

m_{t}

M_{t}

, then

f (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \leq \int_{T} ϕ_{t} (f (x_{t})) d μ (t) - δ_{f} \bar{x} \leq \int_{T} ϕ_{t} (f (x_{t})) d μ (t)

(resp.

f (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \geq \int_{T} ϕ_{t} (f (x_{t})) d μ (t) - δ_{f} \bar{x} \geq \int_{T} ϕ_{t} (f (x_{t})) d μ (t))

(4)

holds, where

\begin{matrix} δ_{f} \equiv δ_{f} (\bar{m}, \bar{M}) = f (\bar{m}) + f (\bar{M}) - 2 f (\frac{\bar{m} + \bar{M}}{2}) \\ (resp. δ_{f} \equiv δ_{f} (\bar{m}, \bar{M}) = 2 f (\frac{\bar{m} + \bar{M}}{2}) - f (\bar{m}) - f (\bar{M})), \\ \bar{x} \equiv {\bar{x}}_{x} (\bar{m}, \bar{M}) = \frac{1}{2} 1_{K} - \frac{1}{\bar{M} - \bar{m}} | x - \frac{\bar{m} + \bar{M}}{2} 1_{K} | \end{matrix}

and

\bar{m} \in [a, m_{x}]

\bar{M} \in [M_{x}, b]

\bar{m} < \bar{M}

, are arbitrary numbers.

The proof is similar to [[22], Theorem 3] and we omit it.

On the other hand, Mond, Pečarić, Furuta et al. in [6, 23‐27] investigated converses of Jensen’s inequality. For presenting these results, we introduce some abbreviations. Let

f : [m, M] \to R

m < M

. Then a linear function through

(m, f (m))

and

(M, f (M))

has the form

h (z) = k_{f} z + l_{f}

, where

k_{f} : = \frac{f (M) - f (m)}{M - m} and l_{f} : = \frac{M f (m) - m f (M)}{M - m} .

(5)

Using the Mond-Pečarić method, in [27] the following generalized converse of Jensen’s operator inequality (2) is presented

F [ϕ (f (A)), g (ϕ (A))] \leq max_{m \leq z \leq M} F [k_{f} z + l_{f}, g (z)] 1_{\tilde{n}},

(6)

for a convex function 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

, operator monotone in u,

U \supset f [m, M]

V \supset g [m, M]

ϕ : H_{n} \to H_{\tilde{n}}

is a unital positive linear mapping and A is a self-adjoint operator with spectrum contained in

[m, M]

A continuous version of (6) and in the case of

\int_{T} ϕ_{t} (1_{H}) d μ (t) = k 1_{K}

for some positive scalar k, is presented in [28]. Recently, Mićić et al. [29] obtained better bound than the one given in (6) as follows.

Theorem 2 [[29], Theorem 2.1]

Let

{(x_{t})}_{t \in T}

be a bounded continuous field of self-adjoint elements in a unital

C^{*}

-algebra

with the spectra in

[m, M]

m < M

, defined on a locally compact Hausdorff space T equipped with a bounded Radon measure μ, and let

{(ϕ_{t})}_{t \in T}

be a unital field of positive linear maps

ϕ_{t} : A \to B

from

to another unital

C^{*}

-algebra ℬ. Let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the self-adjoint operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

and

f : [m, M] \to R

g : [m_{x}, M_{x}] \to R

F : U \times V \to R

, where

f ([m, M]) \subseteq U

g ([m_{x}, M_{x}]) \subseteq V

and F is bounded.

If f is convex and F is an operator monotone in the first variable, then

F [\int_{T} ϕ_{t} (f (x_{t})) d μ (t), g (\int_{T} ϕ_{t} (x_{t}) d μ (t))] \leq C_{1} 1_{K} \leq C 1_{K},

(7)

where constants

C_{1} \equiv C_{1} (F, f, g, m, M, m_{x}, M_{x})

and

C \equiv C (F, f, g, m, M)

are

\begin{matrix} \begin{aligned} C_{1} & = sup_{m_{x} \leq z \leq M_{x}} F [k_{f} z + l_{f}, g (z)] \\ = sup_{\frac{M - M_{x}}{M - m} \leq p \leq \frac{M - m_{x}}{M - m}} F [p f (m) + (1 - p) f (M), g (p m + (1 - p) M)], \end{aligned} \\ \begin{aligned} C & = sup_{m \leq z \leq M} F [k_{f} z + l_{f}, g (z)] \\ = sup_{0 \leq p \leq 1} F [p f (m) + (1 - p) f (M), g (p m + (1 - p) M)] . \end{aligned} \end{matrix}

If f is concave, then reverse inequalities are valid in (7) with inf instead of sup in bounds

C_{1}

and C.

In this paper, we present refined converses of Jensen’s operator inequality. Applying these results, we further refine selected inequalities with power functions.

2 Main results

In the following we assume that

{(x_{t})}_{t \in T}

is a bounded continuous field of self-adjoint elements in a unital

C^{*}

-algebra

with the spectra in

[m, M]

m < M

, defined on a locally compact Hausdorff space T equipped with a bounded Radon measure μ and that

{(ϕ_{t})}_{t \in T}

is a unital field of positive linear mappings

ϕ_{t} : A \to B

between

C^{*}

-algebras.

For convenience, we introduce abbreviations

\tilde{x}

and

δ_{f}

as follows:

\tilde{x} \equiv {\tilde{x}}_{x_{t}, ϕ_{t}} (m, M) : = \frac{1}{2} 1_{K} - \frac{1}{M - m} \int_{T} ϕ_{t} (| x_{t} - \frac{m + M}{2} 1_{H} |) d μ (t),

(8)

where m, M,

m < M

, are some scalars such that the spectra of

x_{t}

t \in T

, are in

[m, M]

;

δ_{f} \equiv δ_{f} (m, M) : = f (m) + f (M) - 2 f (\frac{m + M}{2}),

(9)

where

f : [m, M] \to R

is a continuous function.

Obviously,

m 1_{H} \leq x_{t} \leq M 1_{H}

implies

- \frac{M - m}{2} 1_{H} \leq x_{t} - \frac{m + M}{2} 1_{H} \leq \frac{M - m}{2} 1_{H}

for

t \in T

and

\int_{T} ϕ_{t} (| x_{t} - \frac{m + M}{2} 1_{H} |) d μ (t) \leq \frac{M - m}{2} \int_{T} ϕ_{t} (1_{H}) d μ (t) = \frac{M - m}{2} 1_{K}

. It follows

\tilde{x} \geq 0

. Also, if f is convex (resp. concave), then

δ_{f} \geq 0

(resp.

δ_{f} \leq 0

To prove our main result related to converse Jensen’s inequality, we need the following lemma.

Lemma 3 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

\begin{aligned} min {p_{1}, p_{2}} [f (m) + f (M) - 2 f (\frac{m + M}{2})] \\ \leq p_{1} f (m) + p_{2} f (M) - f (p_{1} m + p_{2} M) \\ \leq max {p_{1}, p_{2}} [f (m) + f (M) - 2 f (\frac{m + M}{2})] . \end{aligned}

(10)

Proof These results follow from [[30], Theorem 1, p.717] for

n = 2

. For the reader’s convenience, we give an elementary proof of (10).

Let

a_{i} \leq b_{i}

i = 1, 2

, be positive real numbers such that

A = a_{1} + a_{2} < B = b_{1} + b_{2}

. Using Jensen’s inequality and its reverse, we get

\begin{aligned} B f (\frac{b_{1} m + b_{2} M}{B}) - A f (\frac{a_{1} m + a_{2} M}{A}) \\ \leq (B - A) f (\frac{(b_{1} - a_{1}) m + (b_{2} - a_{2}) M}{B - A}) \\ \leq (b_{1} - a_{1}) f (m) + (b_{2} - a_{2}) f (M) \\ = b_{1} f (m) + b_{2} f_{2} (M) - (a_{1} f (m) + a_{2} f_{2} (M)) . \end{aligned}

(11)

Suppose that

0 < p_{1} < p_{2} < 1

p_{1} + p_{2} = 1

. Replacing

a_{1}

and

a_{2}

p_{1}

and

p_{2}

, respectively, and putting

b_{1} = b_{2} = p_{2}

A = 1

and

B = 2 p_{2}

in (11), we get

2 p_{2} f (\frac{m + M}{2}) - f (p_{1} f (m) + p_{2} f (M)) \leq p_{2} f (m) + p_{2} f_{2} (M) - (p_{1} f (m) + p_{2} f_{2} (M)),

which gives the right-hand side of (10). Similarly, replacing

b_{1}

and

b_{2}

p_{1}

and

p_{2}

, respectively, and putting

a_{1} = a_{2} = p_{1}

A = 2 p_{1}

and

B = 1

in (11), we obtain the left-hand side of (10).

p_{1} = 0

p_{2} = 1

p_{1} = 1

p_{2} = 0

, then inequality (10) holds, since f is convex. If

p_{1} = p_{2} = 1 / 2

, then we have an equality in (10). □

The main result of an improvement of the Mond-Pečarić method follows.

Lemma 4 Let

{(x_{t})}_{t \in T}

{(ϕ_{t})}_{t \in T}

, m and M be as above. Then

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq k_{f} \int_{T} ϕ_{t} (x_{t}) d μ (t) + l_{f} 1_{K} - δ_{f} \tilde{x} \leq k_{f} \int_{T} ϕ_{t} (x_{t}) d μ (t) + l_{f} 1_{K}

(12)

for every continuous convex function

f : [m, M] \to R

, where

\tilde{x}

and

δ_{f}

are defined by (8) and (9), respectively.

If f is concave, then the reverse inequality is valid in (12).

Proof We prove only the convex case. By using (10) we get

f (p_{1} m + p_{2} M) \leq p_{1} f (m) + p_{2} f (M) - min {p_{1}, p_{2}} [f (m) + f (M) - 2 f (\frac{m + M}{2})]

(13)

for every

p_{1}, p_{2} \in [0, 1]

such that

p_{1} + p_{2} = 1

. Let functions

p_{1}, p_{2} : [m, M] \to [0, 1]

be defined by

p_{1} (z) = \frac{M - z}{M - m}, p_{2} (z) = \frac{z - m}{M - m} .

Then, for any

z \in [m, M]

, we can write

f (z) = f (\frac{M - z}{M - m} m + \frac{z - m}{M - m} M) = f (p_{1} (z) m + p_{2} (z) M) .

By using (13) we get

f (z) \leq \frac{M - z}{M - m} f (m) + \frac{z - m}{M - m} f (M) - \tilde{z} [f (m) + f (M) - 2 f (\frac{m + M}{2})],

(14)

where

\tilde{z} = \frac{1}{2} - \frac{1}{M - m} | z - \frac{m + M}{2} |,

since

min {\frac{M - z}{M - m}, \frac{z - m}{M - m}} = \frac{1}{2} - \frac{1}{M - m} | z - \frac{m + M}{2} | .

Now since

Sp (x_{t}) \subseteq [m, M]

, by utilizing the functional calculus to (14), we obtain

f (x_{t}) \leq \frac{M - x_{t}}{M - m} f (m) + \frac{x_{t} - m}{M - m} f (M) - {\tilde{x}}_{t} [f (m) + f (M) - 2 f (\frac{m + M}{2})],

where

{\tilde{x}}_{t} = \frac{1}{2} 1_{H} - \frac{1}{M - m} | x_{t} - \frac{m + M}{2} 1_{H} | .

Applying a positive linear mapping

ϕ_{t}

, integrating and using

\int_{T} ϕ_{t} (1_{H}) d μ (t) = 1_{K}

, we get the first inequality in (12) since

\tilde{x} = \int_{T} ϕ_{t} ({\tilde{x}}_{t}) d μ (t) = \frac{1}{2} 1_{K} - \frac{1}{M - m} \int_{T} ϕ_{t} (| x_{t} - \frac{m + M}{2} 1_{H} |) d μ (t) .

By using that

δ_{f} \tilde{x} \geq 0

, the second inequality in (12) holds. □

We can use Lemma 4 to obtain refinements of some other inequalities mentioned in the introduction. First, we present a refinement of Theorem 2.

Theorem 5 Let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

and let

m_{\tilde{x}}

be the lower bound of the operator

\tilde{x}

. Let

f : [m, M] \to R

g : [m_{x}, M_{x}] \to R

F : U \times V \to R

, where

f ([m, M]) \subseteq U

g ([m_{x}, M_{x}]) \subseteq V

and F is bounded.

If f is convex and F is operator monotone in the first variable, then

\begin{aligned} F [\int_{T} ϕ_{t} (f (x_{t})) d μ (t), g (\int_{T} ϕ_{t} (x_{t}) d μ (t))] \\ \leq F [k_{f} x + l_{f} - δ_{f} \tilde{x}, g (\int_{T} ϕ_{t} (x_{t}) d μ (t))] \\ \leq sup_{m_{x} \leq z \leq M_{x}} F [k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}, g (z)] 1_{K} \leq sup_{m_{x} \leq z \leq M_{x}} F [k_{f} z + l_{f}, g (z)] 1_{K} . \end{aligned}

(15)

If f is concave, then the reverse inequality is valid in (15) with inf instead of sup.

Proof We prove only the convex case. Then

δ_{f} \geq 0

implies

0 \leq δ_{f} m_{\tilde{x}} 1_{K} \leq δ_{f} \tilde{x}

. By using (12) it follows that

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq k_{f} x + l_{f} - δ_{f} \tilde{x} \leq k_{f} x + l_{f} - δ_{f} m_{\tilde{x}} 1_{K} \leq k_{f} x + l_{f} .

Using operator monotonicity of

F (\cdot, v)

in the first variable, we obtain (15). □

3 Difference-type converse inequalities

By using Jensen’s operator inequality, we obtain that

α g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \leq \int_{T} ϕ_{t} (f (x_{t})) d μ (t)

(16)

holds for every operator convex function f on

[m, M]

, every function g and real number α such that

α g \leq f

[m, M]

. Now, applying Theorem 5 to the function

F (u, v) = u - α v

α \in R

, we obtain the following converse of (16). It is also a refinement of [[29], Theorem 3.1].

Theorem 6 Let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

and

f : [m, M] \to R

g : [m_{x}, M_{x}] \to R

be continuous functions.

If f is convex and

α \in R

, then

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) - α g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \leq max_{m_{x} \leq z \leq M_{x}} {k_{f} z + l_{f} - α g (z)} 1_{K} - δ_{f} \tilde{x} .

(17)

If f is concave, then the reverse inequality is valid in (17) with min instead of max.

Remark 1 (1) Obviously,

\begin{matrix} \int_{T} ϕ_{t} (f (x_{t})) d μ (t) - α g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \\ \leq max_{m_{x} \leq z \leq M_{x}} {k_{f} z + l_{f} - α g (z)} 1_{K} - δ_{f} \tilde{y} \leq max_{m_{x} \leq z \leq M_{x}} {k_{f} z + l_{f} - α g (z)} 1_{K} \end{matrix}

for every convex function f, every

α \in R

, and

m_{\tilde{x}} 1_{K} \leq \tilde{y} \leq \tilde{x}

, where

m_{\tilde{x}}

is the lower bound of

\tilde{x}

(2)

According to [[29], Corollary 3.2], we can determine the constant in the RHS of (17).

(i)

Let f be convex. We can determine the value

C_{α}

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) - α g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \leq C_{α} 1_{K} - δ_{f} \tilde{x}

as follows:

if $α \leq 0$ , g is convex or $α \geq 0$ , g is concave, then
$C_{α} = max {k_{f} m_{x} + l_{f} - α g (m_{x}), k_{f} M_{x} + l_{f} - α g (M_{x})};$

(18)
if $α \leq 0$ , g is concave or $α \geq 0$ , g is convex, then
$C_{α} = {\begin{matrix} k_{f} m_{x} + l_{f} - α g (m_{x}) & if α g_{-}^{'} (z) \geq k_{f} for every z \in (m_{x}, M_{x}), \\ k_{f} z_{0} + l_{f} - α g (z_{0}) & if α g_{-}^{'} (z_{0}) \leq k_{f} \leq α g_{+}^{'} (z_{0}) \\ for some z_{0} \in (m_{x}, M_{x}), \\ k_{f} M_{x} + l_{f} - α g (M_{x}) & if α g_{+}^{'} (z) \leq k_{f} for every z \in (m_{x}, M_{x}) . \end{matrix}$

(19)

(ii)

Let f be concave. We can determine the value

c_{α}

c_{α} 1_{K} - δ_{f} \tilde{x} \leq \int_{T} ϕ_{t} (f (x_{t})) d μ (t) - α g (\int_{T} ϕ_{t} (x_{t}) d μ (t))

as follows:

if $α \leq 0$ , g is convex or $α \geq 0$ , g is concave, then $c_{α}$ is equal to the right-hand side in (19) with reverse inequality signs;
if $α \leq 0$ , g is concave or $α \geq 0$ , g is convex, then $c_{α}$ is equal to the right-hand side in (18) with min instead of max.

Theorem 6 and Remark 1(2) applied to functions

f (z) = z^{p}

and

g (z) = z^{q}

give the following corollary, which is a refinement of [[29], Corollary 3.3].

Corollary 7 Let

{(x_{t})}_{t \in T}

be a field of strictly positive operators, let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

. Let

\tilde{x}

be defined by (8).

(i)

Let

p \in (- \infty, 0] \cup [1, \infty)

. Then

\int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) - α {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{q} \leq C_{α}^{⋆} 1_{K} - (m^{p} + M^{p} - 2^{1 - p} {(m + M)}^{p}) \tilde{x},

where the constant

C_{α}^{⋆}

is determined as follows:

if $α \leq 0$ , $q \in (- \infty, 0] \cup [1, \infty)$ or $α \geq 0$ , $q \in (0, 1)$ , then
$C_{α}^{⋆} = max {k_{t^{p}} m_{x} + l_{t^{p}} - α m_{x}^{q}, k_{t^{p}} M_{x} + l_{t^{p}} - α M_{x}^{q}};$

(20)
if $α \leq 0$ , $q \in (0, 1)$ or $α \geq 0$ , $q \in (- \infty, 0] \cup [1, \infty)$ , then
$C_{α}^{⋆} = {\begin{matrix} k_{t^{p}} m_{x} + l_{t^{p}} - α m_{x}^{q} & if {(α q / k_{t^{p}})}^{1 / (1 - q)} \leq m_{x}, \\ l_{t^{p}} + α (q - 1) {(α q / k_{t^{p}})}^{q / (1 - q)} & if m_{x} \leq {(α q / k_{t^{p}})}^{1 / (1 - q)} \leq M_{x}, \\ k_{t^{p}} M_{x} + l_{t^{p}} - α M_{x}^{q} & if {(α q / k_{t^{p}})}^{1 / (1 - q)} \geq M_{x}, \end{matrix}$

(21)

where

k_{t^{p}} : = (M^{p} - m^{p}) / (M - m)

and

l_{t^{p}} : = (M m^{p} - m M^{p}) / (M - m)

(ii)

Let

p \in (0, 1)

. Then

c_{α}^{⋆} 1_{K} + (2^{1 - p} {(m + M)}^{p} - m^{p} - M^{p}) \tilde{x} \leq \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) - α {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{q},

where the constant

c_{α}^{⋆}

is determined as follows:

if $α \leq 0$ , $q \in (- \infty, 0] \cup [1, \infty)$ or $α \geq 0$ , $q \in (0, 1)$ , then $c_{α}^{⋆}$ is equal to the right-hand side in (21);
if $α \leq 0$ , $q \in (0, 1)$ or $α \geq 0$ , $q \in (- \infty, 0] \cup [1, \infty)$ , then $c_{α}^{⋆}$ is equal to the right-hand side in (20) with min instead of max.

Using Theorem 6 and Remark 1 for

g \equiv f

and

α = 1

and utilizing elementary calculations, we obtain the following converse of Jensen’s inequality.

Theorem 8 Let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

and let

f : [m, M] \to R

be a continuous function.

If f is convex, then

0 \leq \int_{T} ϕ_{t} (f (x_{t})) d μ (t) - f (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \leq \bar{C} 1_{K} - δ_{f} \tilde{x},

(22)

where

\tilde{x}

and

δ_{f}

are defined by (8) and (9), respectively, and

\bar{C} = max_{m_{x} \leq z \leq M_{x}} {k_{f} z + l_{f} - f (z)} .

(23)

Furthermore, if f is strictly convex differentiable, then the bound

\bar{C} 1_{K} - δ_{f} \tilde{x}

satisfies the following condition:

0 \leq \bar{C} 1_{K} - δ_{f} \tilde{x} \leq {f (M) - f (m) - f^{'} (m) (M - m) - δ_{f} m_{\tilde{x}}} 1_{K},

where

m_{\tilde{x}}

is the lower bound of the operator

\tilde{x}

. We can determine the value

\bar{C}

in (23) as follows:

\bar{C} = k_{f} z_{0} + l_{f} - f (z_{0}),

(24)

where

z_{0} = {\begin{matrix} m_{x} & if f^{'} (m_{x}) \geq k_{f}, \\ f^{' - 1} (k_{f}) & if f^{'} (m_{x}) \leq k_{f} \leq f^{'} (M_{x}), \\ M_{x} & if f^{'} (M_{x}) \leq k_{f} . \end{matrix}

(25)

In the dual case, when f is concave, the reverse inequality is valid in (22) with min instead of max in (23). Furthermore, if f is strictly concave differentiable, then the bound

\bar{C} 1_{K} - δ_{f} \tilde{x}

satisfies the following condition:

{f (M) - f (m) - f^{'} (m) (M - m) - δ_{f} m_{\tilde{x}}} 1_{K} \leq \bar{C} 1_{K} - δ_{f} \tilde{x} \leq 0 .

We can determine the value

\bar{C}

in (24) with

z_{0}

, which equals the right-hand side in (25) with reverse inequality signs.

Example 1 We give examples for the matrix cases and

T = {1, 2}

. We put

f (t) = t^{4}

, which is convex, but not operator convex. Also, we define mappings

Φ_{1}, Φ_{2} : M_{3} (C) \to M_{2} (C)

Φ_{1} ({(a_{i j})}_{1 \leq i, j \leq 3}) = \frac{1}{2} {(a_{i j})}_{1 \leq i, j \leq 2}

Φ_{2} = Φ_{1}

and measures by

μ ({1}) = μ ({2}) = 1

(I)

First, we observe an example without the spectra condition (see Figure 1(a)). Then we obtain a refined inequality as in (22), but do not have refined Jensen’s inequality.

If X_{1} = 2 (\begin{array}{c} 1 & 0 & 1 \\ 0 & 0 & 1 \\ 1 & 1 & 1 \end{array}) and X_{2} = 2 (\begin{array}{c} 1 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}), then X = 2 (\begin{array}{c} 1 & 0 \\ 0 & 0 \end{array})

and

m_{1} = - 1.604

M_{1} = 4.494

m_{2} = 0

M_{2} = 2

m = - 1.604

M = 4.494

(rounded to three decimal places). We have

https://static-content.springer.com/image/art%3A10.1186%2F1029-242X-2013-353/MediaObjects/13660_2012_Article_798_Equx_HTML.gif

and

\begin{array}{ll} Φ_{1} (X_{1}^{4}) + Φ_{2} (X_{2}^{4}) & = (\begin{array}{c} 80 & 40 \\ 40 & 24 \end{array}) \\ < Φ_{1} (X_{1}^{4}) + Φ_{2} (X_{2}^{4}) + \bar{C} I_{2} - δ_{f} \tilde{X} & = (\begin{array}{c} 111.742 & 39.327 \\ 39.327 & 142.858 \end{array}) \\ < {(Φ_{1} (X_{1}) + Φ_{2} (X_{2}))}^{4} + \bar{C} I_{2} & = (\begin{array}{c} 243.758 & 0 \\ 0 & 227.758 \end{array}), \end{array}

since

\bar{C} = 227.758

δ_{f} = 405.762

\tilde{X} = (\begin{array}{c} 0.325 & - 0.097 \\ - 0.097 & 0.2092 \end{array})

(II)

Next, we observe an example with the spectra condition (see Figure 1(b)). Then we obtain a series of inequalities involving refined Jensen’s inequality and its converses.

If X_{1} = (\begin{array}{c} - 4 & 1 & 1 \\ 1 & - 2 & - 1 \\ 1 & - 1 & - 1 \end{array}) and X_{2} = (\begin{array}{c} 5 & - 1 & - 1 \\ - 1 & 2 & 1 \\ - 1 & 1 & 3 \end{array}), then X = \frac{1}{2} (\begin{array}{c} 1 & 0 \\ 0 & 0 \end{array})

and

m_{1} = - 4.866

M_{1} = - 0.345

m_{2} = 1.345

M_{2} = 5.866

m = - 4.866

M = 5.866

a = - 0.345

b = 1.345

and we put

\bar{m} = a

\bar{M} = b

(rounded to three decimal places). We have

\begin{array}{ll} {(Φ_{1} (X_{1}) + Φ_{2} (X_{2}))}^{4} & = (\begin{array}{c} 0.0625 & 0 \\ 0 & 0 \end{array}) \\ < Φ_{1} (X_{1}^{4}) + Φ_{2} (X_{2}^{4}) - δ_{f} (a, b) \bar{X} & = (\begin{array}{c} 639.921 & - 255 \\ - 255 & 117.856 \end{array}) \\ < Φ_{1} (X_{1}^{4}) + Φ_{2} (X_{2}^{4}) & = (\begin{array}{c} 641.5 & - 255 \\ - 255 & 118.5 \end{array}) \\ < {(Φ_{1} (X_{1}) + Φ_{2} (X_{2}))}^{4} + \bar{C} I_{2} - δ_{f} (m, M) \tilde{X} & = (\begin{array}{c} 731.649 & - 162.575 \\ - 162.575 & 325.15 \end{array}) \\ < {(Φ_{1} (X_{1}) + Φ_{2} (X_{2}))}^{4} + \bar{C} I_{2} & = (\begin{array}{c} 872.471 & 0 \\ 0 & 872.409 \end{array}), \end{array}

since

δ_{f} (a, b) = 3.158

\bar{X} = (\begin{array}{c} 0.5 & 0 \\ 0 & 0.204 \end{array})

δ_{f} (m, M) = 1744.82

\tilde{X} = (\begin{array}{c} 0.325 & - 0.097 \\ - 0.097 & 0.2092 \end{array})

and

\bar{C} = 872.409

Applying Theorem 8 to

f (t) = t^{p}

, we obtain the following refinement of [[29], Corollary 3.6].

Corollary 9 Let

{(x_{t})}_{t \in T}

be a field of strictly positive operators, let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

. Let

\tilde{x}

be defined by (8). Then

\begin{aligned} 0 & \leq \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) - {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \\ \leq \bar{C} (m_{x}, M_{x}, m, M, p) 1_{K} - (m^{p} + M^{p} - 2^{1 - p} {(m + M)}^{p}) \tilde{x} \\ \leq \bar{C} (m_{x}, M_{x}, m, M, p) 1_{K} \leq C (m, M, p) 1_{K} \end{aligned}

for

p \notin (0, 1)

, and

\begin{aligned} C (m, M, p) 1_{K} & \leq \bar{c} (m_{x}, M_{x}, m, M, p) 1_{K} \\ \leq \bar{c} (m_{x}, M_{x}, m, M, p) 1_{K} + (2^{1 - p} {(m + M)}^{p} - m^{p} - M^{p}) \tilde{x} \\ \leq \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) - {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \leq 0 \end{aligned}

for

p \in (0, 1)

, where

\bar{C} (m_{x}, M_{x}, m, M, p) = {\begin{matrix} k_{t^{p}} m_{x} + l_{t^{p}} - m_{x}^{p} & if p m_{x}^{p - 1} \geq k_{t^{p}}, \\ C (m, M, p) & if p m_{x}^{p - 1} \leq k_{t^{p}} \leq p M_{x}^{p - 1}, \\ k_{t^{p}} M_{x} + l_{t^{p}} - M_{x}^{p} & if p M_{x}^{p - 1} \leq k_{t^{p}}, \end{matrix}

(26)

and

\bar{c} (m_{x}, M_{x}, m, M, p)

equals the right-hand side in (26) with reverse inequality signs.

C (m, M, p)

is the known Kantorovich-type constant for difference (see, i.e., [[6], §2.7]):

C (m, M, p) = (p - 1) {(\frac{M^{p} - m^{p}}{p (M - m)})}^{1 / (p - 1)} + \frac{M m^{p} - m M^{p}}{M - m} for p \in R .

4 Ratio-type converse inequalities

In [[29], Theorem 4.1] the following ratio-type converse of (16) is given:

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{g (z)}} g (\int_{T} ϕ_{t} (x_{t}) d μ (t)),

(27)

where f is convex and

g > 0

. Applying Theorem 5 and Theorem 6, we obtain the following two refinements of (27).

Theorem 10 Let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

and let

f : [m, M] \to R

g : [m_{x}, M_{x}] \to R

be continuous functions.

If f is convex and

g > 0

, then

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{g (z)}} g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) - δ_{f} \tilde{x}

(28)

and

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}}{g (z)}} g (\int_{T} ϕ_{t} (x_{t}) d μ (t)),

(29)

where

\tilde{x}

and

δ_{f}

are defined by (8) and (9), respectively, and

m_{\tilde{x}}

is the lower bound of the operator

\tilde{x}

. If f is concave, then reverse inequalities are valid in (28) and (29) with min instead of max.

Proof We prove only the convex case. Let

α_{1} = {max}_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{g (z)}}

. Then there is

z_{0} \in [m_{x}, M_{x}]

such that

α_{1} = \frac{k_{f} z_{0} + l_{f}}{g (z_{0})}

and

\frac{k_{f} z + l_{f}}{g (z)} \leq α_{1}

for all

z \in [m_{x}, M_{x}]

. It follows that

k_{f} z_{0} + l_{f} - α_{1} g (z_{0}) = 0

and

k_{f} z + l_{f} - α_{1} g (z) \leq 0

for all

z \in [m_{x}, M_{x}]

. So,

max_{m_{x} \leq z \leq M_{x}} {k_{f} z + l_{f} - α_{1} g (z)} = 0 .

By using (17), we obtain (28). Inequality (29) follows directly from Theorem 5 by putting

F (u, v) = v^{- 1 / 2} u v^{- 1 / 2}

. □

Remark 2 (1) Inequality (28) is a refinement of (27) since

δ_{f} \tilde{x} \geq 0

. Also, (29) is a refinement of (27) since

m_{\tilde{x}} \geq 0

and

g > 0

implies

max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}}{g (z)}} \leq max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{g (z)}} .

(2)

Let the assumptions of Theorem 10 hold. Generally, there is no relation between the right-hand sides of inequalities (28) and (29) under the operator order (see Example 2). But, for example, if

g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \leq g (z_{0}) 1_{K}

, where

z_{0} \in [m_{x}, M_{x}]

is the point where it achieves

{max}_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{g (z)}}

, then the following order holds:

\begin{array}{rcl} \int_{T} ϕ_{t} (f (x_{t})) d μ (t) & \leq & max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{g (z)}} g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) - δ_{f} \tilde{x} \\ \leq & max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}}{g (z)}} g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) . \end{array}

Example 2 Let

f (t) = g (t) = t^{4}

Φ_{k} ({(a_{i j})}_{1 \leq i, j \leq 3}) = \frac{1}{2} {(a_{i j})}_{1 \leq i, j \leq 2}

and

μ ({k}) = 1

k = 1, 2

If X_{1} = (\begin{array}{c} 4 & 1 & 1 \\ 1 & 2 & 0 \\ 1 & 0 & 1 \end{array}) and X_{2} = (\begin{array}{c} 5 & - 1 & - 1 \\ - 1 & 2 & 1 \\ - 1 & 1 & 3 \end{array}), then X = (\begin{array}{c} 4.5 & 0 \\ 0 & 2 \end{array})

and

m_{1} = 0.623

M_{1} = 4.651

m_{2} = 1.345

M_{2} = 5.866

m = 0.623

M = 5.866

(rounded to three decimal places). We have

\begin{array}{ll} Φ_{1} (X_{1}^{4}) + Φ_{2} (X_{2}^{4}) & = (\begin{array}{c} 629.5 & - 87.5 \\ - 87.5 & 99 \end{array}) \\ < α_{1} {(Φ_{1} (X_{1}) + Φ_{2} (X_{2}))}^{4} - δ_{f} \tilde{x} & = (\begin{array}{c} 7823.449 & - 53.737 \\ - 53.737 & 139.768 \end{array}) \\ < α_{1} {(Φ_{1} (X_{1}) + Φ_{2} (X_{2}))}^{4} & = (\begin{array}{c} 7974.38 & 0 \\ 0 & 311.148 \end{array}), \end{array}

(30)

since

α_{1} = {max}_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{g (z)}} = 19.447

δ_{f} = 962.73

\tilde{x} = (\begin{array}{c} 0.157 & 0.056 \\ 0.056 & 0.178 \end{array})

. Further,

\begin{array}{ll} Φ_{1} (X_{1}^{4}) + Φ_{2} (X_{2}^{4}) & = (\begin{array}{c} 629.5 & - 87.5 \\ - 87.5 & 99 \end{array}) \\ < α_{2} {(Φ_{1} (X_{1}) + Φ_{2} (X_{2}))}^{4} & = (\begin{array}{c} 5246.13 & 0 \\ 0 & 204.696 \end{array}) \\ < α_{1} {(Φ_{1} (X_{1}) + Φ_{2} (X_{2}))}^{4} & = (\begin{array}{c} 7974.38 & 0 \\ 0 & 311.148 \end{array}), \end{array}

(31)

since

α_{2} = {max}_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}}{g (z)}} = 12.794

. We remark that there is no relation between matrices in the right-hand sides of equalities (30) and (31).

Remark 3 Similar to [[29], Corollary 4.2], we can determine the constant in the RHS of (29).

(i)

Let f be convex. We can determine the value C in

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq C g (\int_{T} ϕ_{t} (x_{t}) d μ (t))

as follows:

if g is convex, then
$C_{α} = {\begin{matrix} \frac{k_{f} m_{x} + l_{f} - δ_{f} m_{\tilde{x}}}{g (m_{x})} & if g_{-}^{'} (z) \geq \frac{k_{f} g (z)}{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}} for every z \in (m_{x}, M_{x}), \\ \frac{k_{f} z_{0} + l_{f} - δ_{f} m_{\tilde{x}}}{g (z_{0})} & if g_{-}^{'} (z_{0}) \leq \frac{k_{f} g (z_{0})}{k_{f} z_{0} + l_{f} - δ_{f} m_{\tilde{x}}} \leq g_{+}^{'} (z_{0}) \\ for some z_{0} \in (m_{x}, M_{x}), \\ \frac{k_{f} M_{x} + l_{f} - δ_{f} m_{\tilde{x}}}{g (M_{x})} & if g_{+}^{'} (z) \leq \frac{k_{f} g (z)}{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}} for every z \in (m_{x}, M_{x}); \end{matrix}$

(32)
if g is concave, then
$C = max {\frac{k_{f} m_{x} + l_{f} - δ_{f} m_{\tilde{x}}}{g (m_{x})}, \frac{k_{f} M_{x} + l_{f} - δ_{f} m_{\tilde{x}}}{g (M_{x})}} .$

(33)

Also, we can determine the constant D in

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq D g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) - δ_{f} \tilde{x}

in the same way as the above constant C but without

m_{\tilde{x}}

(ii)

Let f be concave. We can determine the value c in

c g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) \leq \int_{T} ϕ_{t} (f (x_{t})) d μ (t)

as follows:

if g is convex, then c is equal to the right-hand side in (33) with min instead of max;
if g is concave, then c is equal to the right-hand side in (32) with reverse inequality signs.

Also, we can determine the constant d in

d g (\int_{T} ϕ_{t} (x_{t}) d μ (t)) - δ_{f} \tilde{x} \leq \int_{T} ϕ_{t} (f (x_{t})) d μ (t)

in the same way as the above constant c but without

m_{\tilde{x}}

Theorem 10 and Remark 3 applied to functions

f (z) = z^{p}

and

g (z) = z^{q}

give the following corollary, which is a refinement of [[29], Corollary 4.4].

Corollary 11 Let

{(x_{t})}_{t \in T}

be a field of strictly positive operators, let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

. Let

\tilde{x}

be defined by (8),

m_{\tilde{x}}

be the lower bound of the operator

\tilde{x}

and

δ_{p} : = m^{p} + M^{p} - 2^{1 - p} {(m + M)}^{p}

(i)

Let

p \in (- \infty, 0] \cup [1, \infty)

. Then

\int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) \leq C^{⋆} {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{q},

where the constant

C^{⋆}

is determined as follows:

if $q \in (- \infty, 0] \cup [1, \infty)$ , then
$C^{⋆} = {\begin{matrix} \frac{k_{t^{p}} m_{x} + l_{t^{p}} - δ_{p} m_{\tilde{x}}}{m_{x}^{q}} & if \frac{q}{1 - q} \frac{l_{t^{p}} - δ_{p} m_{\tilde{x}}}{k_{t^{p}}} \leq m_{x}, \\ \frac{l_{t^{p}} - δ_{p} m_{\tilde{x}}}{1 - q} {(\frac{1 - q}{q} \frac{k_{t^{p}}}{l_{t^{p}} - δ_{p} m_{\tilde{x}}})}^{q} & if m_{x} \leq \frac{q}{1 - q} \frac{l_{t^{p}} - δ_{p} m_{\tilde{x}}}{k_{t^{p}}} \leq M_{x}, \\ \frac{k_{t^{p}} M_{x} + l_{t^{p}} - δ_{p} m_{\tilde{x}}}{M_{x}^{q}} & if \frac{q}{1 - q} \frac{l_{t^{p}} - δ_{p} m_{\tilde{x}}}{k_{t^{p}}} \geq M_{x}; \end{matrix}$

(34)
if $q \in (0, 1)$ , then
$C^{⋆} = max {\frac{k_{t^{p}} m_{x} + l_{t^{p}} - δ_{p} m_{\tilde{x}}}{m_{x}^{q}}, \frac{k_{t^{p}} q, M_{x} + l_{t^{p}} - δ_{p} m_{\tilde{x}}}{M_{x}^{q}}} .$

(35)

Also,

\int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) \leq D^{⋆} {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{q} - δ_{p} \tilde{x}

holds, where

D^{⋆}

is determined in the same way as the above constant

C^{⋆}

but without

m_{\tilde{x}}

(ii)

Let

p \in (0, 1)

. Then

c^{⋆} {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{q} \leq \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t),

where the constant

c^{⋆}

is determined as follows:

if $q \in (- \infty, 0] \cup [1, \infty)$ , then $c^{⋆}$ is equal to the right-hand side in (35) with min instead of max;
if $q \in (0, 1)$ , then $c_{α}^{⋆}$ is equal to the right-hand side in (34).

Also,

d^{⋆} {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{q} - δ_{p} \tilde{x} \leq \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t)

holds, where

δ_{p} \leq 0

\tilde{x} \geq 0

and

d^{⋆}

is determined in the same way as the above constant

d^{⋆}

but without

m_{\tilde{x}}

Using Theorem 10 and Remark 3 for

g \equiv f

and utilizing elementary calculations, we obtain the following converse of Jensen’s operator inequality.

Theorem 12 Let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

f : [m, M] \to R

is a continuous convex function and strictly positive on

[m_{x}, M_{x}]

, then

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}}{f (z)}} f (\int_{T} ϕ_{t} (x_{t}) d μ (t))

(36)

and

\int_{T} ϕ_{t} (f (x_{t})) d μ (t) \leq max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{f (z)}} f (\int_{T} ϕ_{t} (x_{t}) d μ (t)) - δ_{f} \tilde{x},

(37)

where

\tilde{x}

and

δ_{f}

are defined by (8) and (9), respectively, and

m_{\tilde{x}}

is the lower bound of the operator

\tilde{x}

In the dual case, if f is concave, then the reverse inequalities are valid in (36) and (37) with min instead of max.

Furthermore, if f is convex differentiable on

[m_{x}, M_{x}]

, we can determine the constant

α_{1} \equiv α_{1} (m, M, m_{x}, M_{x}, f) = max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}}{f (z)}}

in (36) as follows:

α_{1} = {\begin{matrix} \frac{k_{f} m_{x} + l_{f} - δ_{f} m_{\tilde{x}}}{f (m_{x})} & if f^{'} (z) \geq \frac{k_{f} f (z)}{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}} for every z \in (m_{x}, M_{x}), \\ \frac{k_{f} z_{0} + l_{f} - δ_{f} m_{\tilde{x}}}{f (z_{0})} & if f^{'} (z_{0}) = \frac{k_{f} f (z_{0})}{k_{f} z_{0} + l_{f} - δ_{f} m_{\tilde{x}}} for some z_{0} \in (m_{x}, M_{x}), \\ \frac{k_{f} M_{x} + l_{f} - δ_{f} m_{\tilde{x}}}{f (M_{x})} & if f^{'} (z) \leq \frac{k_{f} f (z)}{k_{f} z + l_{f} - δ_{f} m_{\tilde{x}}} for every z \in (m_{x}, M_{x}) . \end{matrix}

(38)

Also, if f is strictly convex twice differentiable on

[m_{x}, M_{x}]

, then we can determine the constant

α_{2} \equiv α_{2} (m, M, m_{x}, M_{x}, f) = max_{m_{x} \leq z \leq M_{x}} {\frac{k_{f} z + l_{f}}{f (z)}}

in (37) as follows:

α_{2} = \frac{k_{f} z_{0} + l_{f}}{f (z_{0})},

(39)

where

z_{0} \in (m_{x}, M_{x})

is defined as the unique solution of the equation

k_{f} f (z) = (k_{f} z + l_{f}) f^{'} (z)

provided

(k_{f} m_{x} + l_{f}) f^{'} (m_{x}) / f (m_{x}) \leq k_{f} \leq (k_{f} M_{x} + l_{f}) f^{'} (M_{x}) / f (M_{x})

. Otherwise,

z_{0}

is defined as

m_{x}

M_{x}

provided

k_{f} \leq (k_{f} m_{x} + l_{f}) f^{'} (m_{x}) / f (m_{x})

k_{f} \geq (k_{f} M_{x} + l_{f}) f^{'} (M_{x}) / f (M_{x})

, respectively.

In the dual case, if f is concave differentiable, then the value

α_{1}

is equal to the right-hand side in (38) with reverse inequality signs. Also, if f is strictly concave twice differentiable, then we can determine the value

α_{2}

in (39) with

z_{0}

, which equals the right-hand side in (39) with reverse inequality signs.

Remark 4 If f is convex and strictly negative on

[m_{x}, M_{x}]

, then (36) and (37) are valid with min instead of max. If f is concave and strictly negative, then reverse inequalities are valid in (36) and (37).

Applying Theorem 12 to

f (t) = t^{p}

, we obtain the following refinement of [[29], Corollary 4.8].

Corollary 13 Let

{(x_{t})}_{t \in T}

be a field of strictly positive operators, let

m_{x}

and

M_{x}

m_{x} \leq M_{x}

, be the bounds of the operator

x = \int_{T} ϕ_{t} (x_{t}) d μ (t)

. Let

\tilde{x}

be defined by (8),

m_{\tilde{x}}

be the lower bound of the operator

\tilde{x}

and

δ_{p} : = m^{p} + M^{p} - 2^{1 - p} {(m + M)}^{p}

p \notin (0, 1)

, then

\begin{array}{rcl} 0 & \leq & \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) \leq \bar{K} (m_{x}, M_{x}, m, M, p, 0) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} - δ_{p} \\ \leq & \bar{K} (m_{x}, M_{x}, m, M, p, 0) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \\ \leq & K (m, M, p) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \end{array}

(40)

and

\begin{array}{rcl} 0 & \leq & \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) \leq \bar{K} (m_{x}, M_{x}, m, M, p, m_{\tilde{x}}) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \\ \leq & \bar{K} (m_{x}, M_{x}, m, M, p, 0) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \\ \leq & K (m, M, p) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p}, \end{array}

(41)

where

\bar{K} (m_{x}, M_{x}, m, M, p, c) = {\begin{matrix} \frac{k_{t^{p}} m_{x} + l_{t^{p}} - c δ_{p}}{m_{x}^{p}} & if \frac{p (l_{t^{p}} - c δ_{p})}{m_{x}} \geq (1 - p) k_{t^{p}}, \\ K (m, M, p, c) & if \frac{p (l_{t^{p}} - c δ_{p})}{m_{x}} < (1 - p) k_{t^{p}} < \frac{p (l_{t^{p}} - c δ_{p})}{M_{x}}, \\ \frac{k_{t^{p}} M_{x} + l_{t^{p}} - c δ_{p}}{M_{x}^{p}} & if \frac{p (l_{t^{p}} - c δ_{p})}{M_{x}} \leq (1 - p) k_{t^{p}} . \end{matrix}

(42)

K (m, M, p, c)

is a generalization of the known Kantorovich constant

K (m, M, p) \equiv K (m, M, p, 0)

(defined in [[6], §2.7]) as follows:

\begin{aligned} K (m, M, p, c) \\ : = \frac{m M^{p} - M m^{p} + c δ_{p} (M - m)}{(p - 1) (M - m)} {(\frac{p - 1}{p} \frac{M^{p} - m^{p}}{m M^{p} - M m^{p} + c δ_{p} (M - m)})}^{p}, \end{aligned}

(43)

for

p \in R

and

0 \leq c \leq 0.5

p \in (0, 1)

, then

\begin{aligned} \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) & \geq \bar{k} (m_{x}, M_{x}, m, M, p, 0) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} - δ_{p} \tilde{x} \\ \geq \bar{k} (m_{x}, M_{x}, m, M, p, 0) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \\ \geq K (m, M, p) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \geq 0 \end{aligned}

and

\begin{aligned} \int_{T} ϕ_{t} (x_{t}^{p}) d μ (t) & \geq \bar{k} (m_{x}, M_{x}, m, M, p, m_{\tilde{x}}) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \\ \geq \bar{k} (m_{x}, M_{x}, m, M, p, 0) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \\ \geq K (m, M, p) {(\int_{T} ϕ_{t} (x_{t}) d μ (t))}^{p} \geq 0, \end{aligned}

where

\bar{k} (m_{x}, M_{x}, m, M, p, c)

equals the right-hand side in (42) with reverse inequality signs.

Proof The second inequalities in (40) and (41) follow directly from (37) and (36) by using (39) and (38), respectively. The last inequality in (40) follows from

\begin{array}{rcl} \bar{K} (m_{x}, M_{x}, m, M, p, 0) & = & max_{m_{x} \leq z \leq M_{x}} {\frac{k_{t^{p}} z + l_{t^{p}}}{z^{p}}} \\ \leq & max_{m \leq z \leq M} {\frac{k_{t^{p}} z + l_{t^{p}}}{z^{p}}} = K (m, M, p) . \end{array}

The third inequality in (41) follows from

\bar{K} (m_{x}, M_{x}, m, M, p, m_{\tilde{x}}) = max_{m_{x} \leq z \leq M_{x}} {\frac{k_{t^{p}} z + l_{t^{p}} - δ_{p} m_{\tilde{x}}}{z^{p}}} \leq \bar{K} (m_{x}, M_{x}, m, M, p, 0),

since

δ_{p} m_{\tilde{x}} \geq 0

for

p \notin (0, 1)

and

M_{x} \geq m_{x} \geq 0

. □

Appendix 1: A new generalization of the Kantorovich constant

Definition 1 Let

h > 0

. Further generalization of Kantorovich constant

K (h, p)

(given in [[6], Definition 2.2]) is defined by

\begin{array}{rcl} K (h, p, c) & : = & \frac{h^{p} - h + c (h^{p} + 1 - 2^{1 - p} {(h + 1)}^{p}) (h - 1)}{(p - 1) (h - 1)} \\ \times {(\frac{p - 1}{p} \frac{h^{p} - 1}{h^{p} - h + c (h^{p} + 1 - 2^{1 - p} {(h + 1)}^{p}) (h - 1)})}^{p} \end{array}

for any real number

p \in R

and any c,

0 \leq c \leq 0.5

. The constant

K (h, p, c)

is sometimes denoted by

K (p, c)

briefly. Some of those constants are depicted in Figure 2.

By inserting

c = 0

K (h, p, c)

, we obtain the Kantorovich constant

K (h, p)

. The constant

K (m, M, p, c)

defined by (43) coincides with

K (h, p, c)

by putting

h = M / m > 1

Lemma 14 Let

h > 0

. The generalized Kantorovich constant

K (h, p, c)

has the following properties:

(i)

K (h, p, c) = K (\frac{1}{h}, p, c)

for all

p \in R

(ii)

K (h, 0, c) = K (h, 1, c) = 1

for all

0 \leq c \leq 0.5

and

K (1, p, c) = 1

for all

p \in R

(iii)

K (h, p, c)

is decreasing of c for

p \notin (0, 1)

and increasing of c for

p \in (0, 1)

(iv)

K (h, p, c) \geq 1

for all

p \notin (0, 1)

and

0 < K (h, 0.5, 0) \leq K (h, p, c) \leq 1

for all

p \in (0, 1)

(v)

K (h, p, c) \leq h^{p - 1}

for all

p \geq 1

Proof (i) We use an easy calculation:

\begin{array}{rcl} K (\frac{1}{h}, p, c) = \frac{h^{- p} - h^{- 1} + c (h^{- p} + 1 - 2^{1 - p} {(h^{- 1} + 1)}^{p}) (h^{- 1} - 1)}{(p - 1) (h^{- 1} - 1)} \\ \times {(\frac{p - 1}{p} \frac{h^{- p} - 1}{h^{- p} - h^{- 1} + c (h^{- p} + 1 - 2^{1 - p} {(h^{- 1} + 1)}^{p}) (h^{- 1} - 1)})}^{p} \\ = & \frac{h - h^{p} + c (1 + h^{p} - 2^{1 - p} {(h + 1)}^{p}) (1 - h)}{(p - 1) (1 - h)} \\ \times {(\frac{p - 1}{p} \frac{1 - h^{p}}{h - h^{p} + c (1 + h^{p} - 2^{1 - p} {(h + 1)}^{p}) (1 - h)})}^{p} \\ = & K (h, p, c) . \end{array}

(ii)

Let

h > 1

. The logarithms calculation and l’Hospital’s theorem give

K (h, p, b) \to 1

p \to 1

K (h, p, b) \to 1

p \to 0

and

K (h, p, b) \to 1

h \to 1 +

. Now using (i) we obtain (ii).

(iii)

Let

h > 0

and

0 \leq c \leq 0.5

\begin{array}{rcl} \frac{d K (h, p, c)}{d c} & = & 2 ({(\frac{h + 1}{2})}^{p} - \frac{h^{p} + 1}{2}) \\ \times {(\frac{p - 1}{p} \frac{h^{p} - 1}{h - h^{p} + c (h^{p} + 1 - 2^{1 - p} {(h + 1)}^{p}) (h - 1)})}^{p} . \end{array}

Since the function

z \to z^{p}

is convex (resp. concave) on

(0, \infty)

p \notin (0, 1)

(resp.

p \in (0, 1)

), then

{(\frac{h + 1}{2})}^{p} \leq \frac{h^{p} + 1}{2}

(resp.

{(\frac{h + 1}{2})}^{p} \geq \frac{h^{p} + 1}{2}

) for every

h > 0

. Then

\frac{d K (h, p, c)}{d c} \leq 0

p \notin (0, 1)

and

\frac{d K (h, p, c)}{d c} \geq 0

p \in (0, 1)

, which gives that

K (h, p, c)

is decreasing of c if

p \notin (0, 1)

and increasing of c if

p \in (0, 1)

(iv)

Let

h > 1

and

0 \leq c \leq 0.5

. If

p > 1

then

\begin{array}{rcl} 0 & < & \frac{(p - 1) (h - 1)}{h^{p} - h + c (h^{p} + 1 - 2^{1 - p} {(h + 1)}^{p}) (h - 1)} \\ \leq & \frac{p - 1}{p} \frac{h^{p} - 1}{h^{p} - h + c (h^{p} + 1 - 2^{1 - p} {(h + 1)}^{p}) (h - 1)} \end{array}

implies

\begin{aligned} \frac{(p - 1) (h - 1)}{h^{p} - h + c (h^{p} + 1 - 2^{1 - p} {(h + 1)}^{p}) (h - 1)} \\ \leq {(\frac{p - 1}{p} \frac{h^{p} - 1}{h^{p} - h + c (h^{p} + 1 - 2^{1 - p} {(h + 1)}^{p}) (h - 1)})}^{p}, \end{aligned}

which gives

K (h, p, c) \geq 1

. Similarly,

K (h, p, c) \geq 1

p < 0

and

K (h, p, c) \leq 1

p \in (0, 1)

. Next, using (iii) and [[6], Theorem 2.54(iv)],

K (h, p, c) \geq K (h, p, 0) \geq K (h, 0.5, 0)

for

p \in (0, 1)

(v)

Let

p \geq 1

. Using (iii) and [[6], Theorem 2.54(vi)],

K (h, p, c) \leq K (h, p, 0) \leq h^{p - 1}

. □

Open AccessThis article is distributed under the terms of the Creative Commons Attribution 2.0 International License (https://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

Competing interests

The authors declare that they have no competing interests.

Authors’ contributions

All authors contributed equally and significantly in writing this paper. All authors read and approved the final manuscript.

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Titel: Refined converses of Jensen’s inequality for operators
verfasst von: Jadranka Mićić
Josip Pečarić
Jurica Perić
Publikationsdatum: 01.12.2013
Verlag: Springer International Publishing
Erschienen in: Journal of Inequalities and Applications / Ausgabe 1/2013
Elektronische ISSN: 1029-242X
DOI: https://doi.org/10.1186/1029-242X-2013-353

Springer Professional

Refined converses of Jensen’s inequality for operators

Abstract

Electronic supplementary material

Competing interests

Authors’ contributions

1 Introduction

2 Main results

3 Difference-type converse inequalities

4 Ratio-type converse inequalities

Appendix 1: A new generalization of the Kantorovich constant

Competing interests

Authors’ contributions

Authors’ original submitted files for images

Premium Partner

Springer Professional

Abstract

Electronic supplementary material

Competing interests

Authors’ contributions

1 Introduction

2 Main results

3 Difference-type converse inequalities

4 Ratio-type converse inequalities

Appendix 1: A new generalization of the Kantorovich constant

Competing interests

Authors’ contributions

Authors’ original submitted files for images

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