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Research | Open | Published:

Refined converses of Jensen’s inequality for operators

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.

1 Introduction

Let T be a locally compact Hausdorff space and let be a C -algebra of operators on some Hilbert space H. We say that a field ( x t ) t T of operators in is continuous if the function t x t is norm continuous on T. If in addition μ is a Radon measure on T and the function t x t is integrable, then we can form the Bochner integral T x t dμ(t), which is the unique element in such that

φ ( T x t d μ ( t ) ) = T φ( x t )dμ(t)

for every linear functional φ in the norm dual A .

Assume further that there is a field ( ϕ t ) t T of positive linear mappings ϕ t :AB from to another C -algebra of operators on a Hilbert space K. We recall that a linear mapping ϕ:AB is said to be positive if ϕ(x)0 for all x0. We say that such a field ( ϕ t ) t T is continuous if the function t ϕ t (x) is continuous for every xA. Let the C -algebras include the identity operators and let the function t ϕ t ( 1 H ) be integrable with T ϕ t ( 1 H )dμ(t)=k 1 K for some positive scalar k. If T ϕ t ( 1 H )dμ(t)= 1 K , we say that a field ( ϕ t ) t 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 xB(H) by

m x := inf ξ = 1 xξ,ξand M x := sup ξ = 1 xξ,ξ
(1)

for ξH. If Sp(x) denotes the spectrum of x, then Sp(x)[ m x , M x ].

For an operator xB(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 ) ) ϕ ( f ( x ) ) ,
(2)

where ϕ:AB(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 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 [36] 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 ( T ϕ t ( x t ) d μ ( t ) ) T ϕ t ( f ( x t ) ) dμ(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 [820]. 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 T be a bounded continuous field of self-adjoint elements in a unital C -algebra defined on a locally compact Hausdorff space T equipped with a bounded Radon measure μ. Let m t and M t , m t M t , be the bounds of x t , tT. Let ( ϕ t ) t T be a unital field of positive linear mappings ϕ t :AB from to another unital C -algebra . Let

( m x , M x )[ m t , M t ]=,tT, and a<b,

where m x and M x , m x M x , are the bounds of the operator x= T ϕ t ( x t )dμ(t) and

a=sup{ M t : M t m x ,tT},b=inf{ m t : m t M x ,tT}.

If f:IR is a continuous convex (resp. concave) function provided that the interval I contains all m t , M t , then

f ( T ϕ t ( x t ) d μ ( t ) ) T ϕ t ( f ( x t ) ) dμ(t) δ f x ¯ T ϕ t ( f ( x t ) ) dμ(t)

(resp.

f ( T ϕ t ( x t ) d μ ( t ) ) T ϕ t ( f ( x t ) ) dμ(t) δ f x ¯ T ϕ t ( f ( x t ) ) dμ(t))
(4)

holds, where

δ f δ f ( m ¯ , M ¯ ) = f ( m ¯ ) + f ( M ¯ ) 2 f ( m ¯ + M ¯ 2 ) ( resp. δ f δ f ( m ¯ , M ¯ ) = 2 f ( m ¯ + M ¯ 2 ) f ( m ¯ ) f ( M ¯ ) ) , x ¯ x ¯ x ( m ¯ , M ¯ ) = 1 2 1 K 1 M ¯ m ¯ | x m ¯ + M ¯ 2 1 K |

and m ¯ [a, m x ], M ¯ [ M x ,b], m ¯ < 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, 2327] investigated converses of Jensen’s inequality. For presenting these results, we introduce some abbreviations. Let f:[m,M]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 := f ( M ) f ( m ) M m and l f := 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 ) ) ] max m z M F [ k f z + l f , g ( z ) ] 1 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×V, operator monotone in u, Uf[m,M], Vg[m,M], ϕ: H n H 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 T ϕ t ( 1 H )dμ(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 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 T be a unital field of positive linear maps ϕ t :AB from to another unital C -algebra . Let m x and M x , m x M x , be the bounds of the self-adjoint operator x= T ϕ t ( x t )dμ(t) and f:[m,M]R, g:[ m x , M x ]R, F:U×VR, where f([m,M])U, g([ m x , M x ])V and F is bounded.

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

F [ T ϕ t ( f ( x t ) ) d μ ( t ) , g ( T ϕ t ( x t ) d μ ( t ) ) ] C 1 1 K C 1 K ,
(7)

where constants C 1 C 1 (F,f,g,m,M, m x , M x ) and CC(F,f,g,m,M) are

C 1 = sup m x z M x F [ k f z + l f , g ( z ) ] = sup M M x M m p M m x M m F [ p f ( m ) + ( 1 p ) f ( M ) , g ( p m + ( 1 p ) M ) ] , C = sup m z M F [ k f z + l f , g ( z ) ] = sup 0 p 1 F [ p f ( m ) + ( 1 p ) f ( M ) , g ( p m + ( 1 p ) M ) ] .

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 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 T is a unital field of positive linear mappings ϕ t :AB between C -algebras.

For convenience, we introduce abbreviations x ˜ and δ f as follows:

x ˜ x ˜ x t , ϕ t (m,M):= 1 2 1 K 1 M m T ϕ t ( | x t m + M 2 1 H | ) dμ(t),
(8)

where m, M, m<M, are some scalars such that the spectra of x t , tT, are in [m,M];

δ f δ f (m,M):=f(m)+f(M)2f ( m + M 2 ) ,
(9)

where f:[m,M]R is a continuous function.

Obviously, m 1 H x t M 1 H implies M m 2 1 H x t m + M 2 1 H M m 2 1 H for tT and T ϕ t (| x t m + M 2 1 H |)dμ(t) M m 2 T ϕ t ( 1 H )dμ(t)= M m 2 1 K . It follows x ˜ 0. Also, if f is convex (resp. concave), then δ f 0 (resp. δ f 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,MI and p 1 , p 2 [0,1] such that p 1 + p 2 =1. Then

min { p 1 , p 2 } [ f ( m ) + f ( M ) 2 f ( m + M 2 ) ] p 1 f ( m ) + p 2 f ( M ) f ( p 1 m + p 2 M ) max { p 1 , p 2 } [ f ( m ) + f ( M ) 2 f ( m + M 2 ) ] .
(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 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

B f ( b 1 m + b 2 M B ) A f ( a 1 m + a 2 M A ) ( B A ) f ( ( b 1 a 1 ) m + ( b 2 a 2 ) M B A ) ( 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 ) ) .
(11)

Suppose that 0< p 1 < p 2 <1, p 1 + p 2 =1. Replacing a 1 and a 2 by 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 ( m + M 2 ) f ( p 1 f ( m ) + p 2 f ( M ) ) 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 by 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).

If p 1 =0, p 2 =1 or 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 T , ( ϕ t ) t T , m and M be as above. Then

T ϕ t ( f ( x t ) ) dμ(t) k f T ϕ t ( x t )dμ(t)+ l f 1 K δ f x ˜ k f T ϕ t ( x t )dμ(t)+ l f 1 K
(12)

for every continuous convex function f:[m,M]R, where 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) p 1 f(m)+ p 2 f(M)min{ p 1 , p 2 } [ f ( m ) + f ( M ) 2 f ( m + M 2 ) ]
(13)

for every p 1 , p 2 [0,1] such that p 1 + p 2 =1. Let functions p 1 , p 2 :[m,M][0,1] be defined by

p 1 (z)= M z M m , p 2 (z)= z m M m .

Then, for any z[m,M], we can write

f(z)=f ( M z M m m + z m M m M ) =f ( p 1 ( z ) m + p 2 ( z ) M ) .

By using (13) we get

f(z) M z M m f(m)+ z m M m f(M) z ˜ [ f ( m ) + f ( M ) 2 f ( m + M 2 ) ] ,
(14)

where

z ˜ = 1 2 1 M m |z m + M 2 |,

since

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

Now since Sp( x t )[m,M], by utilizing the functional calculus to (14), we obtain

f( x t ) M x t M m f(m)+ x t m M m f(M) x ˜ t [ f ( m ) + f ( M ) 2 f ( m + M 2 ) ] ,

where

x ˜ t = 1 2 1 H 1 M m | x t m + M 2 1 H |.

Applying a positive linear mapping ϕ t , integrating and using T ϕ t ( 1 H )dμ(t)= 1 K , we get the first inequality in (12) since

x ˜ = T ϕ t ( x ˜ t )dμ(t)= 1 2 1 K 1 M m T ϕ t ( | x t m + M 2 1 H | ) dμ(t).

By using that δ f x ˜ 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 M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t) and let m x ˜ be the lower bound of the operator x ˜ . Let f:[m,M]R, g:[ m x , M x ]R, F:U×VR, where f([m,M])U, g([ m x , M x ])V and F is bounded.

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

F [ T ϕ t ( f ( x t ) ) d μ ( t ) , g ( T ϕ t ( x t ) d μ ( t ) ) ] F [ k f x + l f δ f x ˜ , g ( T ϕ t ( x t ) d μ ( t ) ) ] sup m x z M x F [ k f z + l f δ f m x ˜ , g ( z ) ] 1 K sup m x z M x F [ k f z + l f , g ( z ) ] 1 K .
(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 0 implies 0 δ f m x ˜ 1 K δ f x ˜ . By using (12) it follows that

T ϕ t ( f ( x t ) ) dμ(t) k f x+ l f δ f x ˜ k f x+ l f δ f m x ˜ 1 K k f x+ l f .

Using operator monotonicity of F(,v) in the first variable, we obtain (15). □

3 Difference-type converse inequalities

By using Jensen’s operator inequality, we obtain that

αg ( T ϕ t ( x t ) d μ ( t ) ) T ϕ t ( f ( x t ) ) dμ(t)
(16)

holds for every operator convex function f on [m,M], every function g and real number α such that αgf on [m,M]. Now, applying Theorem 5 to the function F(u,v)=uαv, αR, 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 M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t) and f:[m,M]R, g:[ m x , M x ]R be continuous functions.

If f is convex and αR, then

T ϕ t ( f ( x t ) ) dμ(t)αg ( T ϕ t ( x t ) d μ ( t ) ) max m x z M x { k f z + l f α g ( z ) } 1 K δ f x ˜ .
(17)

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

Remark 1 (1) Obviously,

T ϕ t ( f ( x t ) ) d μ ( t ) α g ( T ϕ t ( x t ) d μ ( t ) ) max m x z M x { k f z + l f α g ( z ) } 1 K δ f y ˜ max m x z M x { k f z + l f α g ( z ) } 1 K

for every convex function f, every αR, and m x ˜ 1 K y ˜ x ˜ , where m x ˜ is the lower bound of x ˜ .

  1. (2)

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

  2. (i)

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

    T ϕ t ( f ( x t ) ) dμ(t)αg ( T ϕ t ( x t ) d μ ( t ) ) C α 1 K δ f x ˜

as follows:

  • if α0, g is convex or α0, 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 α0, g is concave or α0, g is convex, then

    C α ={ k f m x + l f α g ( m x ) if  α g ( z ) k f  for every  z ( m x , M x ) , k f z 0 + l f α g ( z 0 ) if  α g ( z 0 ) k f α g + ( z 0 ) for some  z 0 ( m x , M x ) , k f M x + l f α g ( M x ) if  α g + ( z ) k f  for every  z ( m x , M x ) .
    (19)
  1. (ii)

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

    c α 1 K δ f x ˜ T ϕ t ( f ( x t ) ) dμ(t)αg ( T ϕ t ( x t ) d μ ( t ) )

as follows:

  • if α0, g is convex or α0, g is concave, then c α is equal to the right-hand side in (19) with reverse inequality signs;

  • if α0, g is concave or α0, 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 T be a field of strictly positive operators, let m x and M x , m x M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t). Let x ˜ be defined by (8).

  1. (i)

    Let p(,0][1,). Then

    T ϕ t ( x t p ) dμ(t)α ( T ϕ t ( x t ) d μ ( t ) ) q C α 1 K ( m p + M p 2 1 p ( m + M ) p ) x ˜ ,

where the constant C α is determined as follows:

  • ifα0, q(,0][1,)orα0, q(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)
  • ifα0, q(0,1)orα0, q(,0][1,), then

    C α ={ k t p m x + l t p α m x q if  ( α q / k t p ) 1 / ( 1 q ) m x , l t p + α ( q 1 ) ( α q / k t p ) q / ( 1 q ) if  m x ( α q / k t p ) 1 / ( 1 q ) M x , k t p M x + l t p α M x q if  ( α q / k t p ) 1 / ( 1 q ) M x ,
    (21)

where k t p :=( M p m p )/(Mm) and l t p :=(M m p m M p )/(Mm).

  1. (ii)

    Let p(0,1). Then

    c α 1 K + ( 2 1 p ( m + M ) p m p M p ) x ˜ T ϕ t ( x t p ) dμ(t)α ( T ϕ t ( x t ) d μ ( t ) ) q ,

where the constant c α is determined as follows:

  • ifα0, q(,0][1,)orα0, q(0,1), then c α is equal to the right-hand side in (21);

  • ifα0, q(0,1)orα0, q(,0][1,), then c α is equal to the right-hand side in (20) with min instead of max.

Using Theorem 6 and Remark 1 for gf 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 M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t) and let f:[m,M]R be a continuous function.

If f is convex, then

0 T ϕ t ( f ( x t ) ) dμ(t)f ( T ϕ t ( x t ) d μ ( t ) ) C ¯ 1 K δ f x ˜ ,
(22)

where x ˜ and δ f are defined by (8) and (9), respectively, and

C ¯ = max m x z M x { k f z + l f f ( z ) } .
(23)

Furthermore, if f is strictly convex differentiable, then the bound C ¯ 1 K δ f x ˜ satisfies the following condition:

0 C ¯ 1 K δ f x ˜ { f ( M ) f ( m ) f ( m ) ( M m ) δ f m x ˜ } 1 K ,

where m x ˜ is the lower bound of the operator x ˜ . We can determine the value C ¯ in (23) as follows:

C ¯ = k f z 0 + l f f( z 0 ),
(24)

where

z 0 ={ m x if  f ( m x ) k f , f 1 ( k f ) if  f ( m x ) k f f ( M x ) , M x  if f ( M x ) k f .
(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 C ¯ 1 K δ f x ˜ satisfies the following condition:

{ f ( M ) f ( m ) f ( m ) ( M m ) δ f m x ˜ } 1 K C ¯ 1 K δ f x ˜ 0.

We can determine the value 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) M 2 (C) by Φ 1 ( ( a i j ) 1 i , j 3 )= 1 2 ( a i j ) 1 i , j 2 , Φ 2 = Φ 1 and measures by μ({1})=μ({2})=1.

  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( 1 0 1 0 0 1 1 1 1 )and X 2 =2( 1 0 0 0 0 0 0 0 0 ),then X=2( 1 0 0 0 )

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

and

Φ 1 ( X 1 4 ) + Φ 2 ( X 2 4 ) = ( 80 40 40 24 ) < Φ 1 ( X 1 4 ) + Φ 2 ( X 2 4 ) + C ¯ I 2 δ f X ˜ = ( 111.742 39.327 39.327 142.858 ) < ( Φ 1 ( X 1 ) + Φ 2 ( X 2 ) ) 4 + C ¯ I 2 = ( 243.758 0 0 227.758 ) ,

since C ¯ =227.758, δ f =405.762, X ˜ = ( 0.325 0.097 0.097 0.2092 ) .

Figure 1
figure1

Refinement for two operators and a convex function f .

  1. (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 =( 4 1 1 1 2 1 1 1 1 )and X 2 =( 5 1 1 1 2 1 1 1 3 ),then X= 1 2 ( 1 0 0 0 )

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 m ¯ =a, M ¯ =b (rounded to three decimal places). We have

( Φ 1 ( X 1 ) + Φ 2 ( X 2 ) ) 4 = ( 0.0625 0 0 0 ) < Φ 1 ( X 1 4 ) + Φ 2 ( X 2 4 ) δ f ( a , b ) X ¯ = ( 639.921 255 255 117.856 ) < Φ 1 ( X 1 4 ) + Φ 2 ( X 2 4 ) = ( 641.5 255 255 118.5 ) < ( Φ 1 ( X 1 ) + Φ 2 ( X 2 ) ) 4 + C ¯ I 2 δ f ( m , M ) X ˜ = ( 731.649 162.575 162.575 325.15 ) < ( Φ 1 ( X 1 ) + Φ 2 ( X 2 ) ) 4 + C ¯ I 2 = ( 872.471 0 0 872.409 ) ,

since δ f (a,b)=3.158, X ¯ = ( 0.5 0 0 0.204 ) , δ f (m,M)=1744.82, X ˜ = ( 0.325 0.097 0.097 0.2092 ) and 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 T be a field of strictly positive operators, let m x and M x , m x M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t). Let x ˜ be defined by (8). Then

0 T ϕ t ( x t p ) d μ ( t ) ( T ϕ t ( x t ) d μ ( t ) ) p C ¯ ( m x , M x , m , M , p ) 1 K ( m p + M p 2 1 p ( m + M ) p ) x ˜ C ¯ ( m x , M x , m , M , p ) 1 K C ( m , M , p ) 1 K

for p(0,1), and

C ( m , M , p ) 1 K c ¯ ( m x , M x , m , M , p ) 1 K c ¯ ( m x , M x , m , M , p ) 1 K + ( 2 1 p ( m + M ) p m p M p ) x ˜ T ϕ t ( x t p ) d μ ( t ) ( T ϕ t ( x t ) d μ ( t ) ) p 0

for p(0,1), where

C ¯ ( m x , M x ,m,M,p)={ k t p m x + l t p m x p if  p m x p 1 k t p , C ( m , M , p ) if  p m x p 1 k t p p M x p 1 , k t p M x + l t p M x p if  p M x p 1 k t p ,
(26)

and 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)=(p1) ( M p m p p ( M m ) ) 1 / ( p 1 ) + M m p m M p M m for pR.

4 Ratio-type converse inequalities

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

T ϕ t ( f ( x t ) ) dμ(t) max m x z M x { k f z + l f g ( z ) } g ( 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 M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t) and let f:[m,M]R, g:[ m x , M x ]R be continuous functions.

If f is convex and g>0, then

T ϕ t ( f ( x t ) ) dμ(t) max m x z M x { k f z + l f g ( z ) } g ( T ϕ t ( x t ) d μ ( t ) ) δ f x ˜
(28)

and

T ϕ t ( f ( x t ) ) dμ(t) max m x z M x { k f z + l f δ f m x ˜ g ( z ) } g ( T ϕ t ( x t ) d μ ( t ) ) ,
(29)

where x ˜ and δ f are defined by (8) and (9), respectively, and m x ˜ is the lower bound of the operator 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 z M x { k f z + l f g ( z ) }. Then there is z 0 [ m x , M x ] such that α 1 = k f z 0 + l f g ( z 0 ) and k f z + l f g ( z ) α 1 for all z[ 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)0 for all z[ m x , M x ]. So,

max m x z 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 x ˜ 0. Also, (29) is a refinement of (27) since m x ˜ 0 and g>0 implies

max m x z M x { k f z + l f δ f m x ˜ g ( z ) } max m x z M x { k f z + l f g ( z ) } .
  1. (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( T ϕ t ( x t )dμ(t))g( z 0 ) 1 K , where z 0 [ m x , M x ] is the point where it achieves max m x z M x { k f z + l f g ( z ) }, then the following order holds:

    T ϕ t ( f ( x t ) ) d μ ( t ) max m x z M x { k f z + l f g ( z ) } g ( T ϕ t ( x t ) d μ ( t ) ) δ f x ˜ max m x z M x { k f z + l f δ f m x ˜ g ( z ) } g ( T ϕ t ( x t ) d μ ( t ) ) .

Example 2 Let f(t)=g(t)= t 4 , Φ k ( ( a i j ) 1 i , j 3 )= 1 2 ( a i j ) 1 i , j 2 and μ({k})=1, k=1,2.

If  X 1 =( 4 1 1 1 2 0 1 0 1 )and X 2 =( 5 1 1 1 2 1 1 1 3 ),then X=( 4.5 0 0 2 )

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

Φ 1 ( X 1 4 ) + Φ 2 ( X 2 4 ) = ( 629.5 87.5 87.5 99 ) < α 1 ( Φ 1 ( X 1 ) + Φ 2 ( X 2 ) ) 4 δ f x ˜ = ( 7823.449 53.737 53.737 139.768 ) < α 1 ( Φ 1 ( X 1 ) + Φ 2 ( X 2 ) ) 4 = ( 7974.38 0 0 311.148 ) ,
(30)

since α 1 = max m x z M x { k f z + l f g ( z ) }=19.447, δ f =962.73, x ˜ = ( 0.157 0.056 0.056 0.178 ) . Further,

Φ 1 ( X 1 4 ) + Φ 2 ( X 2 4 ) = ( 629.5 87.5 87.5 99 ) < α 2 ( Φ 1 ( X 1 ) + Φ 2 ( X 2 ) ) 4 = ( 5246.13 0 0 204.696 ) < α 1 ( Φ 1 ( X 1 ) + Φ 2 ( X 2 ) ) 4 = ( 7974.38 0 0 311.148 ) ,
(31)

since α 2 = max m x z M x { k f z + l f δ f m 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).

  1. (i)

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

    T ϕ t ( f ( x t ) ) dμ(t)Cg ( T ϕ t ( x t ) d μ ( t ) )

as follows:

  • if g is convex, then

    C α ={ k f m x + l f δ f m x ˜ g ( m x ) if  g ( z ) k f g ( z ) k f z + l f δ f m x ˜  for every  z ( m x , M x ) , k f z 0 + l f δ f m x ˜ g ( z 0 ) if  g ( z 0 ) k f g ( z 0 ) k f z 0 + l f δ f m x ˜ g + ( z 0 ) for some  z 0 ( m x , M x ) , k f M x + l f δ f m x ˜ g ( M x ) if  g + ( z ) k f g ( z ) k f z + l f δ f m x ˜  for every  z ( m x , M x ) ;
    (32)
  • if g is concave, then

    C=max { k f m x + l f δ f m x ˜ g ( m x ) , k f M x + l f δ f m x ˜ g ( M x ) } .
    (33)

Also, we can determine the constant D in

T ϕ t ( f ( x t ) ) dμ(t)Dg ( T ϕ t ( x t ) d μ ( t ) ) δ f x ˜

in the same way as the above constant C but without m x ˜ .

  1. (ii)

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

    cg ( T ϕ t ( x t ) d μ ( t ) ) T ϕ t ( f ( x t ) ) dμ(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

dg ( T ϕ t ( x t ) d μ ( t ) ) δ f x ˜ T ϕ t ( f ( x t ) ) dμ(t)

in the same way as the above constant c but without m 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 T be a field of strictly positive operators, let m x and M x , m x M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t). Let x ˜ be defined by (8), m x ˜ be the lower bound of the operator x ˜ and δ p := m p + M p 2 1 p ( m + M ) p .

  1. (i)

    Let p(,0][1,). Then

    T ϕ t ( x t p ) dμ(t) C ( T ϕ t ( x t ) d μ ( t ) ) q ,

where the constant C is determined as follows:

  • ifq(,0][1,), then

    C ={ k t p m x + l t p δ p m x ˜ m x q if  q 1 q l t p δ p m x ˜ k t p m x , l t p δ p m x ˜ 1 q ( 1 q q k t p l t p δ p m x ˜ ) q if  m x q 1 q l t p δ p m x ˜ k t p M x , k t p M x + l t p δ p m x ˜ M x q if  q 1 q l t p δ p m x ˜ k t p M x ;
    (34)
  • ifq(0,1), then

    C =max { k t p m x + l t p δ p m x ˜ m x q , k t p q , M x + l t p δ p m x ˜ M x q } .
    (35)

Also,

T ϕ t ( x t p ) dμ(t) D ( T ϕ t ( x t ) d μ ( t ) ) q δ p x ˜

holds, where D is determined in the same way as the above constant C but without m x ˜ .

  1. (ii)

    Let p(0,1). Then

    c ( T ϕ t ( x t ) d μ ( t ) ) q T ϕ t ( x t p ) dμ(t),

where the constant c is determined as follows:

  • ifq(,0][1,), then c is equal to the right-hand side in (35) with min instead of max;

  • ifq(0,1), then c α is equal to the right-hand side in (34).

Also,

d ( T ϕ t ( x t ) d μ ( t ) ) q δ p x ˜ T ϕ t ( x t p ) dμ(t)

holds, where δ p 0, x ˜ 0 and d is determined in the same way as the above constant d but without m x ˜ .

Using Theorem 10 and Remark 3 for gf and utilizing elementary calculations, we obtain the following converse of Jensen’s operator inequality.

Theorem 12 Let m x and M x , m x M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t).

If f:[m,M]R is a continuous convex function and strictly positive on [ m x , M x ], then

T ϕ t ( f ( x t ) ) dμ(t) max m x z M x { k f z + l f δ f m x ˜ f ( z ) } f ( T ϕ t ( x t ) d μ ( t ) )
(36)

and

T ϕ t ( f ( x t ) ) dμ(t) max m x z M x { k f z + l f f ( z ) } f ( T ϕ t ( x t ) d μ ( t ) ) δ f x ˜ ,
(37)

where x ˜ and δ f are defined by (8) and (9), respectively, and m x ˜ is the lower bound of the operator 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 α 1 (m,M, m x , M x ,f)= max m x z M x { k f z + l f δ f m x ˜ f ( z ) }

in (36) as follows:

α 1 ={ k f m x + l f δ f m x ˜ f ( m x ) if  f ( z ) k f f ( z ) k f z + l f δ f m x ˜  for every  z ( m x , M x ) , k f z 0 + l f δ f m x ˜ f ( z 0 ) if  f ( z 0 ) = k f f ( z 0 ) k f z 0 + l f δ f m x ˜  for some  z 0 ( m x , M x ) , k f M x + l f δ f m x ˜ f ( M x ) if  f ( z ) k f f ( z ) k f z + l f δ f m x ˜  for every  z ( m x , M x ) .
(38)

Also, if f is strictly convex twice differentiable on [ m x , M x ], then we can determine the constant

α 2 α 2 (m,M, m x , M x ,f)= max m x z M x { k f z + l f f ( z ) }

in (37) as follows:

α 2 = k f z 0 + l f f ( z 0 ) ,
(39)

where z 0 ( 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 ) k f ( k f M x + l f ) f ( M x )/f( M x ). Otherwise, z 0 is defined as m x or M x provided k f ( k f m x + l f ) f ( m x )/f( m x ) or k f ( 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 T be a field of strictly positive operators, let m x and M x , m x M x , be the bounds of the operator x= T ϕ t ( x t )dμ(t). Let x ˜ be defined by (8), m x ˜ be the lower bound of the operator x ˜ and δ p := m p + M p 2 1 p ( m + M ) p .

If p(0,1), then

0 T ϕ t ( x t p ) d μ ( t ) K ¯ ( m x , M x , m , M , p , 0 ) ( T ϕ t ( x t ) d μ ( t ) ) p δ p K ¯ ( m x , M x , m , M , p , 0 ) ( T ϕ t ( x t ) d μ ( t ) ) p K ( m , M , p ) ( T ϕ t ( x t ) d μ ( t ) ) p
(40)

and

0 T ϕ t ( x t p ) d μ ( t ) K ¯ ( m x , M x , m , M , p , m x ˜ ) ( T ϕ t ( x t ) d μ ( t ) ) p K ¯ ( m x , M x , m , M , p , 0 ) ( T ϕ t ( x t ) d μ ( t ) ) p K ( m , M , p ) ( T ϕ t ( x t ) d μ ( t ) ) p ,
(41)

where

K ¯ ( m x , M x ,m,M,p,c)={ k t p m x + l t p c δ p m x p if  p ( l t p c δ p ) m x ( 1 p ) k t p , K ( m , M , p , c ) if  p ( l t p c δ p ) m x < ( 1 p ) k t p < p ( l t p c δ p ) M x , k t p M x + l t p c δ p M x p if  p ( l t p c δ p ) M x ( 1 p ) k t p .
(42)

K(m,M,p,c) is a generalization of the known Kantorovich constant K(m,M,p)K(m,M,p,0) (defined in [[6], §2.7]) as follows:

K ( m , M , p , c ) : = m M p M m p + c δ p ( M m ) ( p 1 ) ( M m ) ( p 1 p M p m p m M p M m p + c δ p ( M m ) ) p ,
(43)

for pR and 0c0.5.

If p(0,1), then

T ϕ t ( x t p ) d μ ( t ) k ¯ ( m x , M x , m , M , p , 0 ) ( T ϕ t ( x t ) d μ ( t ) ) p δ p x ˜ k ¯ ( m x , M x , m , M , p , 0 ) ( T ϕ t ( x t ) d μ ( t ) ) p K ( m , M , p ) ( T ϕ t ( x t ) d μ ( t ) ) p 0

and

T ϕ t ( x t p ) d μ ( t ) k ¯ ( m x , M x , m , M , p , m x ˜ ) ( T ϕ t ( x t ) d μ ( t ) ) p k ¯ ( m x , M x , m , M , p , 0 ) ( T ϕ t ( x t ) d μ ( t ) ) p K ( m , M , p ) ( T ϕ t ( x t ) d μ ( t ) ) p 0 ,

where 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

K ¯ ( m x , M x , m , M , p , 0 ) = max m x z M x { k t p z + l t p z p } max m z M { k t p z + l t p z p } = K ( m , M , p ) .

The third inequality in (41) follows from

K ¯ ( m x , M x ,m,M,p, m x ˜ )= max m x z M x { k t p z + l t p δ p m x ˜ z p } K ¯ ( m x , M x ,m,M,p,0),

since δ p m x ˜ 0 for p(0,1) and M x m x 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

K ( h , p , c ) : = h p h + c ( h p + 1 2 1 p ( h + 1 ) p ) ( h 1 ) ( p 1 ) ( h 1 ) × ( p 1 p h p 1 h p h + c ( h p + 1 2 1 p ( h + 1 ) p ) ( h 1 ) ) p

for any real number pR and any c, 0c0.5. The constant K(h,p,c) is sometimes denoted by K(p,c) briefly. Some of those constants are depicted in Figure 2.

Figure 2
figure2

Relation between K(p,c) for pR and 0c0.5 .

By inserting c=0 in 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:

  1. (i)

    K(h,p,c)=K( 1 h ,p,c) for all pR,

  2. (ii)

    K(h,0,c)=K(h,1,c)=1 for all 0c0.5 and K(1,p,c)=1 for all pR,

  3. (iii)

    K(h,p,c) is decreasing of c for p(0,1) and increasing of c for p(0,1),

  4. (iv)

    K(h,p,c)1 for all p(0,1) and 0<K(h,0.5,0)K(h,p,c)1 for all p(0,1),

  5. (v)

    K(h,p,c) h p 1 for all p1.

Proof (i) We use an easy calculation:

K ( 1 h , p , c ) = h p h 1 + c ( h p + 1 2 1 p ( h 1 + 1 ) p ) ( h 1 1 ) ( p 1 ) ( h 1 1 ) × ( p 1 p h p 1 h p h 1 + c ( h p + 1 2 1 p ( h 1 + 1 ) p ) ( h 1 1 ) ) p = h h p + c ( 1 + h p 2 1 p ( h + 1 ) p ) ( 1 h ) ( p 1 ) ( 1 h ) × ( p 1 p 1 h p h h p + c ( 1 + h p 2 1 p ( h + 1 ) p ) ( 1 h ) ) p = K ( h , p , c ) .
  1. (ii)

    Let h>1. The logarithms calculation and l’Hospital’s theorem give K(h,p,b)1 as p1, K(h,p,b)1 as p0 and K(h,p,b)1 as h1+. Now using (i) we obtain (ii).

  2. (iii)

    Let h>0 and 0c0.5.

    d K ( h , p , c ) d c = 2 ( ( h + 1 2 ) p h p + 1 2 ) × ( p 1 p h p 1 h h p + c ( h p + 1 2 1 p ( h + 1 ) p ) ( h 1 ) ) p .

Since the function z z p is convex (resp. concave) on (0,) if p(0,1) (resp. p(0,1)), then ( h + 1 2 ) p h p + 1 2 (resp. ( h + 1 2 ) p h p + 1 2 ) for every h>0. Then d K ( h , p , c ) d c 0 if p(0,1) and d K ( h , p , c ) d c 0 if p(0,1), which gives that K(h,p,c) is decreasing of c if p(0,1) and increasing of c if p(0,1).

  1. (iv)

    Let h>1 and 0c0.5. If p>1 then

    0 < ( p 1 ) ( h 1 ) h p h + c ( h p + 1 2 1 p ( h + 1 ) p ) ( h 1 ) p 1 p h p 1 h p h + c ( h p + 1 2 1 p ( h + 1 ) p ) ( h 1 )

implies

( p 1 ) ( h 1 ) h p h + c ( h p + 1 2 1 p ( h + 1 ) p ) ( h 1 ) ( p 1 p h p 1 h p h + c ( h p + 1 2 1 p ( h + 1 ) p ) ( h 1 ) ) p ,

which gives K(h,p,c)1. Similarly, K(h,p,c)1 if p<0 and K(h,p,c)1 if p(0,1). Next, using (iii) and [[6], Theorem 2.54(iv)], K(h,p,c)K(h,p,0)K(h,0.5,0) for p(0,1).

  1. (v)

    Let p1. Using (iii) and [[6], Theorem 2.54(vi)], K(h,p,c)K(h,p,0) h p 1 . □

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Keywords

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