Refin Mond-Pec-corr(2)

REFINED CONVERSES OF JENSEN’S INEQUALITY FOR OPERATORS

JADRANKA MICIC, JOSIP PECARIC, AND JURICA PERIC

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-Pecaricmethod improvement. Obtained results are applied to refine selected inequalities with power

functions.

1. Introduction

Let T be a locally compact Hausdorff space and let A be a C∗-algebra of operators on someHilbert space H. We say that a field (xt)t∈T of operators in A is continuous if the functiont 7→ xt is norm continuous on T. If in addition µ is a Radon measure on T and the functiont 7→ ‖xt‖ is integrable, then we can form the Bochner integral

∫Txt dµ(t), which is the unique

element in A such that

ϕ

(∫T

xt dµ(t)

)=

∫T

ϕ(xt) dµ(t)

for every linear functional ϕ in the norm dual A∗.Assume further that there is a field (φt)t∈T of positive linear mappings φt : A → B from A

to another C∗-algebra B of operators on a Hilbert space K. We recall that a linear mappingφ : A → B is said to be positive if φ(x) ≥ 0 for all x ≥ 0. We say that such a field iscontinuous if the function t 7→ φt(x) is continuous for every x ∈ A. Let the C∗-algebras includethe identity operators and the function t 7→ φt(1H) be integrable with

∫Tφt(1H) dµ(t) = k1K

for some positive scalar k. If∫Tφt(1H) dµ(t) = 1K , we say that a field (φt)t∈T is unital.

Let B(H) be the C∗-algebra of all bounded linear operators on a Hilbert space H. We definebounds of a self-adjoint operator x ∈ B(H) by

mx := inf‖ξ‖=1

〈xξ, ξ〉 and Mx := sup‖ξ‖=1

〈xξ, ξ〉 (1)

for ξ ∈ H. If Sp(x) denotes the spectrum of x, then Sp(x) ⊆ [mx,Mx].For an operator x ∈ B(H) we define operator |x| := (x∗x)1/2. Obviously, if x is self-adjoint,

then |x| = (x2)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 [3] proved the so-called

Jensen operator inequality

f (φ(x)) ≤ φ (f(x)) , (2)

where φ : A → B(K) is a unital completely positive linear mapping from a C∗-algebra A tolinear operators on a Hilbert space K, and x is a self-adjoint element in A with spectrum in I.Subsequently Choi [2] noted that it is enough to assume that φ is unital and positive.

Mond, Pecaric, Hansen, Pedersen et al. in [10, 11, 25, 8] studied another generalization of(2) for operator convex functions. Moreover, Hansen, Pecaric and Peric [9] presented a gene-ral formulation of Jensen’s operator inequality for a bounded continuous field of self-adjoint

2010 Mathematics Subject Classification. 47A63, 47A64.Key words and phrases. Mond-Pecaric method, self-adjoint operator, positive linear mapping, converse of

Jensen’s operator inequality, convex function.

1

2 JADRANKA MICIC, JOSIP PECARIC, AND JURICA PERIC

operators and a unital field of positive linear mappings

f

(∫T

φt(xt) dµ(t)

)≤∫T

φt(f(xt)) dµ(t), (3)

where f is an operator convex function.There is an extensive literature devoted to Jensen’s inequality concerning different refinementsand extensive results, e.g. see [1, 4, 5, 6, 12, 13, 15, 17], [26]-[30]. Micic, Pavic and Pecaric[14] proved that the discrete version of (3) stands without operator convexity of f under acondition on the spectra of operators. Recently, Micic, Pecaric and Peric [16] presented thediscrete version of the refined Jensen’s inequality for real valued continuous convex functions.A continuous version is given below.

Theorem 1. Let (xt)t∈T be a bounded continuous field of self-adjoint elements in a unitalC∗-algebra A defined on a locally compact Hausdorff space T equipped with a bounded Radonmeasure µ. Let mt and Mt, mt ≤Mt, be the bounds of xt, t ∈ T . Let (φt)t∈T be a unital fieldof positive linear mappings φt : A → B from A to another unital C∗−algebra B. Let

(mx,Mx) ∩ [mt,Mt] = ∅, t ∈ T, and a < b,

where mx and Mx, mx ≤Mx, be the bounds of the operator x =∫Tφt(xt) dµ(t) and

a = sup {Mt : Mt ≤ mx, t ∈ T} , b = inf {mt : mt ≥Mx, t ∈ T} .If f : I → R is a continuous convex (resp. concave) function provided that the interval I containsall mt,Mt, then

f

(∫T

φt(xt) dµ(t)

)≤∫T

φt(f(xt)) dµ(t)− δf x ≤∫T

φt(f(xt)) dµ(t) (4)(resp.

f

(∫T

φt(xt) dµ(t)

)≥∫T

φt(f(xt)) dµ(t)− δf x ≥∫T

φt(f(xt)) dµ(t))

(5)

holds, where

δf ≡ δf (m, M) = f(m) + f(M)− 2f(m+M

2

)(resp. δf ≡ δf (m, M) = 2f

(m+M

2

)− f(m)− f(M) ),

x ≡ xx(m, M) = 121K − 1

M−m

∣∣∣x− m+M2 1K

∣∣∣ (6)

and m ∈ [a,mx], M ∈ [Mx, b], m < M , are arbitrary numbers.

The proof is similar to [16, Theorem 3] and we omit it.

On the other hand, Mond, Pecaric, Furuta et al. in [23, 24, 7, 20, 19, 8] investigated conversesof Jensen’s inequality. For presenting these results, we introduce some abbreviations. Letf : [m,M ] → R, m < M . Then a linear function through (m, f(m)) and (M,f(M)) has theform h(z) = kfz + lf , where

kf :=f(M)− f(m)

M −mand lf :=

Mf(m)−mf(M)

M −m. (7)

Using the Mond-Pecaric method, in [19] the following generalized converse of Jensen’s oper-ator inequality (2) is presented

F [φ (f(A)) , g (φ(A))] ≤ maxm≤z≤M

F [kfz + lf , g(z)] 1n, (8)

for convex function f defined on an interval [m,M ], m < M , where g is a real valued continuousfunction on [m,M ], F (u, v) is a real valued function defined on U × V , operator monotone inu, U ⊃ f [m,M ], V ⊃ g[m,M ], φ : Hn → Hn is a unital positive linear mapping and A is aself-adjoint operator with spectrum contained in [m,M ].A continuous version of (8) and in the case of

∫Tφt(1H) dµ(t) = k1K for some positive scalar

REFINED CONVERSES OF JENSEN’S INEQUALITY FOR OPERATORS 3

k, is presented in [18]. Recently, Micic, Pavic and Pecaric [21] obtained better bound than theone given in (8) as follows.

Theorem 2. [21, Theorem 2.1] Let (xt)t∈T be a bounded continuous field of self-adjointelements in a unital C∗-algebra A with the spectra in [m,M ], m < M , defined on a locallycompact Hausdorff space T equipped with a bounded Radon measure µ, and let (φt)t∈T be aunital field of positive linear maps φt : A → B from A to another unital C∗−algebra B. Letmx and Mx, mx ≤ Mx, be the bounds of the self-adjoint operator x =

∫Tφt(xt) dµ(t) and

f : [m,M ] → R, g : [mx,Mx] → R, F : U × V → R, where f([m,M ]) ⊆ U , g([mx,Mx]) ⊆ Vand F be bounded.

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

F

[∫T

φt(f(xt)) dµ(t) , g

(∫T

φt(xt) dµ(t)

)]≤ C1 1K ≤ C 1K , (9)

where constants C1 ≡ C1(F, f, g,m,M,mx,Mx) and C ≡ C(F, f, g,m,M) are

C1 = supmx≤z≤Mx

F [kfz + lf , g(z)]

= supM−MxM−m ≤p≤

M−mxM−m

F [pf(m) + (1− p)f(M) , g(pm+ (1− p)M)] ,

C = supm≤z≤M

F [kfz + lf , g(z)]

= sup0≤p≤1

F [pf(m) + (1− p)f(M) , g(pm+ (1− p)M)] .

If f is concave, then reverse inequalities are valid in (9) with inf instead of sup in bounds C1

and C.

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

2. Main results

In the following we assume that (xt)t∈T is a bounded continuous field of self-adjoint elementsin a unital C∗-algebra A with the spectra in [m,M ], m < M , defined on a locally compactHausdorff space T equipped with a bounded Radon measure µ and (φt)t∈T is a unital field ofpositive linear mappings φt : A → B between C∗-algebras.

For convenience we introduce abbreviations x and δf as follows:

x ≡ xxt,φt(m,M) :=

1

21K −

1

M −m

∫T

φt

(|xt −

m+M

21H |)dµ(t), (10)

where m,M , m < M , are some scalars such that the spectra of xt, t ∈ T , are in [m,M ];

δf ≡ δf (m,M) := f(m) + f(M)− 2f

(m+M

2

), (11)

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

Obviously, m1H ≤ xt ≤ M1H implies −M−m2 1H ≤ xt − m+M2 1H ≤ M−m

2 1H for t ∈ T and∫Tφt(|xt − m+M

2 1H |)dµ(t) ≤ M−m

2

∫Tφt (1H) dµ(t) = M−m

2 1K . It follows x ≥ 0. Also, if f isconvex (resp. concave) then δf ≥ 0 (resp. δf ≤ 0).

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


Lemma 3. Let f be a convex function on an interval I, m,M ∈ I and p1, p2 ∈ [0, 1] such thatp1 + p2 = 1. Then

min{p1, p2}[f(m) + f(M)− 2f

(m+M

2

)]≤ p1f(m) + p2f(M)− f(p1m+ p2M) (12)

≤ max{p1, p2}[f(m) + f(M)− 2f

(m+M

2

)]Proof. These results follow from [22, Theorem 1, p. 717] for n = 2. For the reader’s convenience,we give an elementary proof of (12).

Let ai ≤ bi, i = 1, 2, be positive real numbers, such that A = a1 + a2 < B = b1 + b2. UsingJensen’s inequality and its reverse, we get

Bf

(b1m+ b2M

B

)−Af

(a1m+ a2M

A

)≤ (B −A)f

((b1 − a1)m+ (b2 − a2)M

B −A

)≤ (b1 − a1)f(m) + (b2 − a2)f(M)

= b1f(m) + b2f2(M)− (a1f(m) + a2f2(M)). (13)

Suppose that 0 < p1 < p2 < 1, p1 + p2 = 1. Replacing a1 and a2 by p1 and p2, respectively,and putting b1 = b2 = p2, A = 1 and B = 2p2 in (13), we get

2p2f

(f(m) + f(M)

2

)− f (p1f(m) + p2f(M)) ≤ p2f(m) + p2f2(M)− (p1f(m) + p2f2(M)),

which gives the right-hand side of (12). Similarly, replacing b1 and b2 by p1 and p2, respectively,and putting a1 = a2 = p1, A = 2p1 and B = 1 in (13), we obtain the left-hand side of (12).

If p1 = 0, q1 = 1 or p1 = 1, q1 = 0, then the inequality (12) holds, since f is convex. Ifp1 = q1 = 1/2, then we have an equality in (12). �

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

Lemma 4. Let (xt)t∈T , (φt)t∈T , m and M be as above. Then∫T

φt(f(xt)) dµ(t) ≤ kf∫T

φt(xt) dµ(t) + lf1K − δf x ≤ kf∫T

φt(xt) dµ(t) + lf1K (14)

for every continuous convex function f : [m,M ] → R, where x and δf are defined by (10) and(11), respectively.

If f is concave, then the reverse inequality is valid in (14).

Proof. We prove only the convex case. By using (12) we get

f (p1m+ p2M)

≤ p1f(m) + p2f(M)−min{p1, p2}[f(m) + f(M)− 2f

(m+M

2

)](15)

for every p1, p2 ∈ [0, 1] such that p1 + p2 = 1. Let functions p1, p2 : [m,M ] → [0, 1] be definedby

p1(z) =M − zM −m

, p2(z) =z −mM −m

.

Then for any z ∈ [m,M ] we can write

f (z) = f

(M − zM −m

m+z −mM −m

M

)= f (p1(z)m+ p2(z)M) .

By using (15) we get

f(z) ≤ M − zM −m

f(m) +z −mM −m

f(M)− z[f(m) + f(M)− 2f

(m+M

2

)], (16)


where

z =1

2− 1

M −m

∣∣∣∣z − m+M

2

∣∣∣∣ ,since

min

{M − zM −m

,z −mM −m

}=

1

2− 1

M −m

∣∣∣∣z − m+M

2

∣∣∣∣ .Now since Sp(xt) ⊆ [m,M ], by utilizing the functional calculus to (16) we obtain

f(xt) ≤M − xtM −m

f(m) +xt −mM −m

f(M)− xt[f(m) + f(M)− 2f

(m+M

2

)],

where

xt =1

21H −

1

M −m

∣∣∣∣xt − m+M

21H

∣∣∣∣ .Applying a positive linear mapping φt, integrating and using

∫Tφt(1H) dµ(t) = 1K , we get the

first inequality in (14), since

x =

∫T

φt (xt) dµ(t) =1

21K −

1

M −m

∫T

φt

(|xt −

m+M

21H |)dµ(t).

By using that δf x ≥ 0 the second inequality in (14) holds. �

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

Theorem 5. Let mx and Mx, mx ≤Mx, be the bounds of the operator x =∫Tφt(xt) dµ(t) and

mx be the lower bound of the operator x. Let f : [m,M ]→ R, g : [mx,Mx]→ R, F : U×V → R,where f([m,M ]) ⊆ U , g([mx,Mx]) ⊆ V and F is bounded.

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

F

[∫T

φt(f(xt)) dµ(t) , g

(∫T

φt(xt) dµ(t)

)]≤ F

[kf x+ lf − δf x , g

(∫T

φt(xt) dµ(t)

)]≤ sup

mx≤z≤Mx

F [kfz + lf − δfmx, g(z)] 1K ≤ supmx≤z≤Mx

F [kfz + lf , g(z)] 1K .

(17)

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

Proof. We prove only the convex case. Then δf ≥ 0 implies 0 ≤ δfmx1K ≤ δf x. By using (14)it follows that∫

T

φt(f(xt)) dµ(t) ≤ kfx+ lf − δf x ≤ kfx+ lf − δfmx1K ≤ kfx+ lf .

Using operator monotonicity of F (·, v) in the first variable, we obtain (17). �

3. Difference type converse inequalities

By using Jensen’s operator inequality we obtain that

αg

(∫T

φt(xt)dµ(t)

)≤∫T

φt (f(xt)) dµ(t) (18)

holds for every operator convex function f on [m,M ], every function g and real number α suchthat αg ≤ f on [m,M ]. Now, applying Theorem 5 to the function F (u, v) = u−αv, α ∈ R, weobtain the following converse of (18). It is also a refinement of [21, Theorem 3.1].



f : [m,M ]→ R, g : [mx,Mx]→ R be continuous functions.If f is convex and α ∈ R, then∫

T

φt(f(xt)) dµ(t)− αg(∫

T

φt(xt) dµ(t)

)≤ maxmx≤z≤Mx

{kfz + lf − αg(z)} 1K − δf x.(19)

If f is concave, then the reverse inequality with min instead of max is valid in (19).

Remark 1. 1) Similarly to Theorem 5 we obtain∫T


T

φt(xt) dµ(t)

)≤ max

mx≤z≤Mx

{kfz + lf − αg(z)} 1K − δf y ≤ maxmx≤z≤Mx

{kfz + lf − αg(z)} 1K

for every convex function f , every α ∈ R, and mx1K ≤ y ≤ x, where mx is the lower bound ofx.2) According to [21, Corollary 3.2] we can determine the constant in the RHS of (19).i) Let f be convex. We can determine the bound Cα in∫

T


T

φt(xt) dµ(t)

)≤ Cα1K − δf x

as follows:· if α ≤ 0 and g is convex, then

Cα = max {kfmx + lf − αg(mx) , kfMx + lf − αg(Mx)} ; (20)

· if α ≤ 0 and g is concave, then

Cα

=

kfmx + lf − αg(mx) if αg′−(z) ≥ kf for every z∈(mx,Mx),

kfz0 + lf − αg(z0) if αg′−(z0) ≤ kf ≤ αg′+(z0) for some z0∈(mx,Mx),

kfMx + lf − αg(Mx) if αg′+(z) ≤ kf for every z∈(mx,Mx),

(21)

· if α ≥ 0 and g is convex, then Cα is defined by (21);· if α ≥ 0 and g is concave, then Cα is defined by (20).ii) Let f be concave. We can determine the bound cα in

cα1K − δf x ≤∫T


T

φt(xt) dµ(t)

)as follows:· if α ≤ 0 and g is convex, then cα is equal to the right side in (21) with reverse inequalitysigns;· if α ≤ 0 and g is concave, then cα is equal to the right side in (20) with min instead of max;· if α ≥ 0 and g is convex, then cα is equal to the right side in (20) with min instead of max;· if α ≥ 0 and g is concave, then cα is equal to the right side in (21) with reverse inequalitysigns.

Theorem 6 and Remark 1-2) applied on functions f(z) = zp and g(z) = zq give the followingcorollary, which is a refinement of [21, Corollary 3.3].

Corollary 7. Let (xt)t∈T be a field of strictly positive operators, mx and Mx, mx ≤ Mx, bethe bounds of the operator x =

∫Tφt(xt) dµ(t). Let x be defined by (10).

(i) Let p ∈ (−∞, 0] ∪ [1,∞). Then∫T

φt(xpt ) dµ(t)− α

(∫T

φt(xt) dµ(t)

)q≤ C?α1K −

(mp +Mp − 21−p(m+M)p

)x,


where the bound C?α is determined as follows:· if α ≤ 0 and q ∈ (−∞, 0] ∪ [1,∞), then

C?α = max {ktpmx + ltp − αmqx, ktpMx + ltp − αMq

x} ; (22)

· if α ≤ 0 and q ∈ (0, 1), then

C?α =

ktpmx + ltp − αmq

x if (α q/ktp)1/(1−q) ≤ mx,

ltp + α(q − 1) (α q/ktp)q/(1−q)

if mx ≤ (α q/ktp)1/(1−q) ≤Mx,

ktpMx + ltp − αMqx if (α q/ktp)

1/(1−q) ≥Mx,

(23)

where ktp := (Mp −mp)/(M −m) and ltp := (Mmp −mMp)/(M −m) (i.e. replacing f withzp in (7));· if α ≥ 0 and q ∈ (−∞, 0] ∪ [1,∞), then C?α is defined by (23);· if α ≥ 0 and q ∈ (0, 1), then C?α is defined by (22).

(ii) Let p ∈ (0, 1). Then

c?α1K +(21−p(m+M)p −mp −Mp

)x

≤∫T

φt(xpt ) dµ(t)− α

(∫T

φt(xt) dµ(t)

)q,

where the bound c?α is determined as follows:· if α ≤ 0 and q ∈ (−∞, 0] ∪ [1,∞), then c?α is equal to the right side in (23);· if α ≤ 0 and q ∈ (0, 1), then c?α is equal to the right side in (22) with min instead of max;· if α ≥ 0 and q ∈ (−∞, 0] ∪ [1,∞), then c?α is equal to the right side in (22) with min insteadof max;· if α ≥ 0 and q ∈ (0, 1), then c?α is equal to the right side in (23).

Using Theorem 6 and Remark 1 for g ≡ f and α = 1 and utilizing elementary calculationswe obtain the following converse of Jensen’s inequality.


f : [m,M ]→ R be a continuous function.If f is convex, then

0 ≤∫T

φt(f(xt)) dµ(t)− f(∫

T

φt(xt) dµ(t)

)≤ C1K − δf x, (24)

where x and δf are defined by (10) and (11), respectively, and

C = maxmx≤z≤Mx

{kfz + lf − f(z)} . (25)

Furthermore, if f is strictly convex differentiable, then the bound C1K − δf x satisfies the fol-lowing condition

0 ≤ C1K − δf x ≤ {f(M)− f(m)− f ′(m)(M −m)− δfmx} 1K , (26)

where mx is the lower bound of the operator x. We can determine the value C ≡ C(m,M,mx,Mx, f)in (25) as follows

C = kfz0 + lf − f(z0), (27)

where

z0 =

mx if f ′(mx) ≥ kf ,f ′−1 (kf ) if f ′(mx) ≤ kf ≤ f ′(Mx),Mx if f ′(Mx) ≤ kf .

(28)

In the dual case, when f is concave, the reverse inequality is valid in (24) with min insteadof max in (25). Furthermore, if f is strictly concave differentiable, then the bound C1K − δf xsatisfies the following condition

{f(M)− f(m)− f ′(m)(M −m)− δfmx} 1K ≤ C1K − δf x ≤ 0.


We can determine the value C in (27), with z0 which equals the right side in (28) with reverseinequality signs.

Example 1. We give examples for the matrix cases and T = {1, 2}. We put f(t) = t4 whichis convex, but not operator convex. Also, we define mappings Φ1,Φ2 : M3(C) → M2(C) byΦ1((aij)1≤i,j≤3) = 1

2 (aij)1≤i,j≤2, Φ2 = Φ1 and measures by µ({1}) = µ({2}) = 1.

Figure 1. Refinement for two operators and a convex function f

I) First, we observe an example without the spectra condition (see Figure 1.a). Then we obtaina refined inequality as in (24), but don’t have a refined Jensen’s inequality.

If X1 = 2

1 0 10 0 11 1 1

and X2 = 2

1 0 00 0 00 0 0

, then X = 2

(1 00 0

)

and m1 = −1.604, M1 = 4.494, m2 = 0, M2 = 2, m = −1.604, M = 4.494 (rounded to threedecimal places). We have

(Φ1(X1) + Φ2(X2))4

=

(16 00 0

)6T(

80 4040 24

)= Φ1

(X4

1

)+ Φ2

(X4

2

)and

Φ1

(X4

1

)+ Φ2

(X4

2

)=

(80 4040 24

)<

(111.742 39.32739.327 142.858

)= Φ1

(X4

1

)+ Φ2

(X4

2

)+ CI2 − δf X

<

(243.758 0

0 227.758

)= (Φ1(X1) + Φ2(X2))

4+ CI2,

since C = 227.758, δf = 405.762, X =

(0.325 −0.097−0.097 0.2092

).

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

If X1 =

−4 1 11 −2 −11 −1 −1

and X2 =

5 −1 −1−1 2 1−1 1 3

, then X =1

2

(1 00 0

)


and m1 = −4.866, M1 = −0.345, m2 = 1.345, M2 = 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

(Φ1(X1) + Φ2(X2))4

=

(0.0625 0

0 0

)<

(639.921 −255−255 117.856

)= Φ1

(X4

1

)+ Φ2

(X4

2

)− δf (a, b)X

<

(641.5 −255−255 118.5

)= Φ1

(X4

1

)+ Φ2

(X4

2

)<

(731.649 −162.575−162.575 325.15

)= (Φ1(X1) + Φ2(X2))

4+ CI2 − δf (m,M)X

<

(872.471 0

0 872.409

)= (Φ1(X1) + Φ2(X2))

4+ CI2,

since δf (a, b) = 3.158, X =

(0.5 00 0.204

), δf (m,M) = 1744.82, X =

(0.325 −0.097−0.097 0.2092

)and

C = 872.409 .

Applying Theorem 8 to f(t) = tp, we obtain the following refinement of [21, Corollary 3.6].


∫Tφt(xt) dµ(t). Let x is defined by (10). Then

0 ≤∫T

φt(xpt ) dµ(t)−

(∫T

φt(xt) dµ(t)

)p≤ C(mx,Mx,m,M, p)1K −

(mp +Mp − 21−p(m+M)p

)x

≤ C(mx,Mx,m,M, p)1K ≤ C(m,M, p)1K

for p 6∈ (0, 1), and

C(m,M, p)1K ≤ c(mx,Mx,m,M, p)1K

≤ c(mx,Mx,m,M, p)1K +(21−p(m+M)p −mp −Mp

)x

≤∫T

φt(xpt ) dµ(t)−

(∫T

φt(xt) dµ(t)

)p≤ 0

for p ∈ (0, 1), where

C(mx,Mx,m,M, p) =

ktp mx + ltp −mpx if pmp−1

x ≥ ktp ,C(m,M, p) if pmp−1

x ≤ ktp ≤ pMp−1x ,

ktp Mx + ltp −Mpx if pMp−1

x ≤ ktp ,(29)

and c(mx,Mx,m,M, p) equals the right side in (29) with reverse inequality signs. C(m,M, p)is the known Kantorovich type constant for difference (see i.e. [8, §2.7]):

C(m,M, p) = (p− 1)

(Mp −mp

p(M −m)

)1/(p−1)

+Mmp −mMp

M −m, for p ∈ R.

4. Ratio type converse inequalities

In [21, Theorem 4.1] the following ratio type converse of (18) is given:∫T

φt(f(xt)) dµ(t) ≤ maxmx≤z≤Mx

{kfz + lfg(z)

}g

(∫T

φt(xt) dµ(t)

), (30)

where f is convex and g > 0. Applying Theorem 5 and Theorem 6, we obtain the following tworefinements of (30).


Theorem 10. Let mx and Mx, mx ≤ Mx, be the bounds of the operator x =∫Tφt(xt) dµ(t)

and f : [m,M ]→ R, g : [mx,Mx]→ R be continuous functions.If f is convex and g > 0, then∫

T


{kfz + lfg(z)

}g

(∫T

φt(xt) dµ(t)

)− δf x (31)

and ∫T


{kfz + lf − δfmx

g(z)

}g

(∫T

φt(xt) dµ(t)

), (32)

where x and δf are defined by (10) and (11), respectively, and mx is the lower bound of theoperator x. If f is concave, then reverse inequalities are valid in (31) and (32) with min insteadof max.

Proof. We prove only the convex case. Let α1 = maxmx≤z≤Mx

{kfz+lfg(z)

}. Then there is z0 ∈

[mx,Mx] such that α1 =kfz0+lfg(z0) and

kfz+lfg(z) ≤ α1 for all z ∈ [mx,Mx]. It follows that

kfz0 + lf − α1g(z0) = 0 and kfz + lf − α1g(z) ≤ 0 for all z ∈ [mx,Mx], since g > 0. So

maxmx≤z≤Mx

{kfz + lf − α1g(z)} = 0.

By using (19) we obtain (31). The inequality (32) follows directly from Theorem 5 by puttingF (u, v) = v−1/2uv−1/2. �

Remark 2. 1) The inequality (31) is a refinement of (30) since δf x ≥ 0. Also, (32) is arefinement of (30) since mx ≥ 0 and g > 0 implies

maxmx≤z≤Mx

{kfz + lf − δfmx

g(z)

}≤ maxmx≤z≤Mx

{kfz + lfg(z)

}.

2) Let the assumptions of Theorem 10 hold. Generally, there is no relation between the rightsides of inequalities (31) and (32) under the operator order (see Example 2). But, for ex-ample, if g

(∫Tφt(xt) dµ(t)

)≤ g(z0)1K , where z0 ∈ [mx,Mx] is the point where it achieves

maxmx≤z≤Mx

{kfz+lfg(z)

}, then the following order holds∫

T


{kfz + lfg(z)

}g

(∫T

φt(xt) dµ(t)

)− δf x

≤ maxmx≤z≤Mx

{kfz + lf − δfmx

g(z)

}g

(∫T

φt(xt) dµ(t)

).

Example 2. Let f(t) = g(t) = t4, Φk((aij)1≤i,j≤3) = 12 (aij)1≤i,j≤2 and µ({k}) = 1, k = 1, 2.

If X1 =

4 1 11 2 01 0 1

and X2 =

5 −1 −1−1 2 1−1 1 3

, then X =

(4.5 00 2

)and m1 = 0.623, M1 = 4.651, m2 = 1.345, M2 = 5.866, m = 0.623, M = 5.866 (rounded tothree decimal places). We have

Φ1

(X4

1

)+ Φ2

(X4

2

)=

(629.5 −87.5−87.5 99

)< α1 (Φ1(X1) + Φ2(X2))

4 − δf x =

(7823.449 −53.737−53.737 139.768

)< α1 (Φ1(X1) + Φ2(X2))

4=

(7974.38 0

0 311.148

),

(33)


since α1 = maxmx≤z≤Mx

{kfz+lfg(z)

}= 19.447, δf = 962.73, x =

(0.157 0.0560.056 0.178

). Further,

Φ1

(X4

1

)+ Φ2

(X4

2

)=

(629.5 −87.5−87.5 99

)< α2 (Φ1(X1) + Φ2(X2))

4=

(5246.13 0

0 204.696

)< α1 (Φ1(X1) + Φ2(X2))

4=

(7974.38 0

0 311.148

),

(34)

since α2 = maxmx≤z≤Mx

{kfz+lf−δfmx

g(z)

}= 12.794. We remark that there is no relation between

matrices in the right sides of the equalities (33) and (34).

Remark 3. Similar to [21, Corollary 4.2] we can determine the constant in the RHS of (32).(i) Let f be convex. We can determine the bound C in∫

T

φt(f(xt)) dµ(t) ≤ C g(∫

T

φt(xt) dµ(t)

)as follows:· if g is convex, then

Cα

=

kfmx + lf − δfmx

g(mx)if g′−(z) ≥ kfg(z)

kfz + lf − δfmxfor every z∈(mx,Mx),

kfz0 + lf − δfmx

g(z0)

if g′−(z0) ≤ kfg(z0)kfz0+lf−δfmx

≤ g′+(z0) for some z0∈(mx,Mx),

kfMx + lf − δfmx

g(Mx)if g′+(z) ≤ kfg(z)

kfz + lf − δfmxfor every z∈(mx,Mx);

(35)

· if g is concave, then

C = max

{kfmx + lf − δfmx

g(mx),kfMx + lf − δfmx

g(Mx)

}. (36)

Also, we can determine the bound D in∫T

φt(f(xt)) dµ(t) ≤ Dg

(∫T

φt(xt) dµ(t)

)− δf x

in the same way as the above bound C but without mx.

(ii) Let f be concave. We can determine the bound c in

c g

(∫T

φt(xt) dµ(t)

)≤∫T

φt(f(xt)) dµ(t)

as follows:· if g is convex, then c is equal to the right side in (36) with min instead of max;· if g is concave, then c is equal to the right side in (35) with reverse inequality signs.

Also, we can determine the bound d in

d g

(∫T

φt(xt) dµ(t)

)− δf x ≤

∫T

φt(f(xt)) dµ(t)

in the same way as the above bound c but without mx.

Theorem 10 and Remark 3 applied on functions f(z) = zp and g(z) = zq give the followingcorollary, which is a refinement of [21, Corollary 4.4].



∫Tφt(xt) dµ(t). Let x be defined by (10), δp := mp + Mp −

21−p(m+M)p and mx be the lower bound of the operator x.

(i) Let p ∈ (−∞, 0] ∪ [1,∞). Then∫T

φt(xpt ) dµ(t) ≤ C?

(∫T

φt(xt) dµ(t)

)q,

where the bound C? is determined as follows:· if q ∈ (−∞, 0] ∪ [1,∞), then

C? =

ktp mx+ltp−δpmx

mqx

if q1−q

ltp−δpmx

ktp≤ mx,

ltp−δpmx

1−q

(1−qq

ktpltp−δpmx

)qif mx ≤ q

1−qltp−δpmx

ktp≤Mx,

ktp Mx+ltp−δpmx

Mqx

if q1−q

ltp−δpmx

ktp≥Mx;

(37)

· if q ∈ (0, 1), then

C? = max

{ktp mx + ltp − δpmx

mqx

,ktpq,Mx + ltp − δpmx

Mqx

}. (38)

Also, ∫T

φt(xpt ) dµ(t) ≤ D?

(∫T

φt(xt) dµ(t)

)q− δpx

holds, where D? is determined in the same way as the above bound C? but without mx.

(ii) Let p ∈ (0, 1). Then

c?(∫

T

φt(xt) dµ(t)

)q≤∫T

φt(xpt ) dµ(t),

where the bound c? is determined as follows:· if q ∈ (−∞, 0] ∪ [1,∞), then c? is equal to the right side in (38) with min instead of max;· if q ∈ (0, 1), then c?α is equal to the right side in (37).

Also,

d?(∫

T

φt(xt) dµ(t)

)q− δpx ≤

∫T

φt(xpt ) dµ(t)

holds, where δp ≤ 0, x ≥ 0 and d? is determined in the same way as the above bound d? butwithout mx.

Using Theorem 10 and Remark 3 for g ≡ f and utilizing elementary calculations we obtainthe following converse of Jensen’s operator inequality.

Theorem 12. Let mx and Mx, mx ≤ Mx, be the bounds of the operator x =∫Tφt(xt) dµ(t)

and f : [m,M ]→ R be a continuous function.If f : [m,M ]→ R is a continuous convex function and strictly positive on [mx,Mx], then∫

T


{kfz + lf − δfmx

f(z)

}f

(∫T

φt(xt) dµ(t)

)(39)

and ∫T


{kfz + lff(z)

}f

(∫T

φt(xt) dµ(t)

)− δf x, (40)

where x and δf are defined by (10) and (11), respectively, and mx is the lower bound of theoperator x.

In the dual case, if f is concave, then the reverse inequalities are valid in (39) and (40) withmin instead of max.


Furthermore, if f is convex differentiable on [mx,Mx], we can determine the bound

α1 ≡ α1(m,M,mx,Mx, f) = maxmx≤z≤Mx

{kfz + lf − δfmx

f(z)

}in (39) as follows:

α1 =

kfmx + lf − δfmx

f(mx)if f ′(z) ≥ kff(z)

kfz + lf − δfmxfor every z∈(mx,Mx),

kfz0 + lf − δfmx

f(z0)if f ′(z0) =

kff(z0)

kfz0 + lf − δfmxfor some z0∈(mx,Mx),

kfMx + lf − δfmx

f(Mx)if f ′(z) ≤ kff(z)

kfz + lf − δfmxfor every z∈(mx,Mx).

(41)

Also, if f is strictly convex twice differentiable on [mx,Mx], then we can determine the bound

α2 ≡ α2(m,M,mx,Mx, f) = maxmx≤z≤Mx

{kfz + lff(z)

}in (40) as follows:

α2 =kfz0 + lff(z0)

, (42)

where z0 ∈ (mx,Mx) is defined as the unique solution of the equation kff(z) = (kfz + lf )f ′(z)provided (kfmx+lf )f ′(mx)/f(mx) ≤ kf ≤ (kfMx+lf )f ′(Mx) /f(Mx). Otherwise z0 is definedas mx or Mx provided kf ≤ (kfmx + lf )f ′(mx) /f(mx) or kf ≥ (kfMx + lf )f ′(Mx) /f(Mx),respectively.

In the dual case, if f is concave differentiable, then the value α1 is equal to the right sidein (41) with reverse inequality signs. Also, if f is strictly concave twice differentiable, then wecan determine the value α2 in (42), with z0 which equals the right side in (42) with reverseinequality signs.

Remark 4. If f is convex and strictly negative on [mx,Mx], then (39) and (40) are valid withmin instead of max. If f is concave and strictly negative, then reverse inequalities are valid in(39) and (40).

Applying Theorem 12 to f(t) = tp, we obtain the following refinement of [21, Corollary 4.8].


∫Tφt(xt) dµ(t). Let x be defined by (10), δp := mp + Mp −

21−p(m+M)p and mx be the lower bound of the operator x.If p 6∈ (0, 1), then

0 ≤∫T

φt(xpt ) dµ(t) ≤ K(mx,Mx,m,M, p, 0)

(∫T

φt(xt) dµ(t)

)p− δp

≤ K(mx,Mx,m,M, p, 0)

(∫T

φt(xt) dµ(t)

)p(43)

≤ K(m,M, p)

(∫T

φt(xt) dµ(t)

)pand

0 ≤∫T

φt(xpt ) dµ(t) ≤ K(mx,Mx,m,M, p,mx)

(∫T

φt(xt) dµ(t)

)p≤ K(mx,Mx,m,M, p, 0)

(∫T

φt(xt) dµ(t)

)p(44)

≤ K(m,M, p)

(∫T

φt(xt) dµ(t)

)p,

where


K(mx,Mx,m,M, p, c)

=

ktp mx+ltp−cδp

mpx

ifp (ltp−cδp)

mx≥ (1− p) ktp ,

K(m,M, p, c) ifp (ltp−cδp)

mx< (1− p) ktp < p (ltp−cδp)

Mx,

ktp Mx+ltp−cδpMp

xif

p (ltp−cδp)Mx

≤ (1− p) ktp .

(45)

K(m,M, p, c) is a generalization of the known Kantorovich constantK(m,M, p) ≡ K(m,M, p, 0) (defined in [8, §2.7]) as follows

K(m,M, p, c)

:=mMp −Mmp + cδp(M −m)

(p− 1)(M −m)

(p− 1

p

Mp −mp

mMp −Mmp + cδp(M −m)

)p,

(46)

for p ∈ R and 0 ≤ c ≤ 0.5.If p ∈ (0, 1), then∫

T

φt(xpt ) dµ(t) ≥ k(mx,Mx,m,M, p, 0)

(∫T

φt(xt) dµ(t)

)p− δpx

≥ k(mx,Mx,m,M, p, 0)

(∫T

φt(xt) dµ(t)

)p≥ K(m,M, p)

(∫T

φt(xt) dµ(t)

)p≥ 0

and ∫T

φt(xpt ) dµ(t) ≥ k(mx,Mx,m,M, p,mx)

(∫T

φt(xt) dµ(t)

)p≥ k(mx,Mx,m,M, p, 0)

(∫T

φt(xt) dµ(t)

)p≥ K(m,M, p)

(∫T

φt(xt) dµ(t)

)p≥ 0,

where k(mx,Mx,m,M, p, c) equals the right side in (45) with reverse inequality signs.

Proof. The second inequalities in (43) and (44) follow directly from (40) and (39) by using (42)and (41), respectively. The last inequality in (43) follows from

K(mx,Mx,m,M, p, 0) = maxmx≤z≤Mx

{ktp z + ltp

zp

}≤ max

m≤z≤M

{ktp z + ltp

zp

}= K(m,M, p).

The third inequality in (44) follows from

K(mx,Mx,m,M, p,mx) = maxmx≤z≤Mx

{ktp z + ltp − δpmx

zp

}≤ K(mx,Mx,m,M, p, 0),

since δpmx ≥ 0 for p 6∈ (0, 1) and Mx ≥ mx ≥ 0. �

Appendix A. A new generalization of the Kantorovich constant

Definition 1. Let h > 0. Further generalization of Kantorovich constant K(h, p) (given in [8,Definition 2.2]) is defined by

K(h, p, c) :=hp − h+ c(hp + 1− 21−p(h+ 1)p)(h− 1)

(p− 1)(h− 1)

×(p− 1

p

hp − 1

hp − h+ c(hp + 1− 21−p(h+ 1)p)(h− 1)

)p


for any real number p ∈ R and any c, 0 ≤ c ≤ 0.5. The constant K(h, p, c) is sometimes denotedby K(p, c) briefly.

Figure 2. Relation between K(p, c) for p ∈ R and 0 ≤ c ≤ 0.5

By inserting c = 0 in K(h, p, c) we obtain the Kantorovich constant K(h, p). The constantK(m,M, p, c) defined by (46) 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 followingproperties

(i) K(h, p, c) = K( 1h , p, c) for all p ∈ R,

(ii) K(h, 0, c) = K(h, 1, c) = 1 for all 0 ≤ c ≤ 0.5 and K(1, p, c) = 1 for all p ∈ R,(iii) K(h, p, c) is decreasing of c for p 6∈ (0, 1) and increasing of c for p ∈ (0, 1),(iv) K(h, p, c) ≥ 1 for all p 6∈ (0, 1) and 0 < K(h, 0.5, 0) ≤ K(h, p, c) ≤ 1 for all p ∈ (0, 1),(v) K(h, p, c) ≤ hp−1 for all p ≥ 1.

Proof. (i): We use an easy calculation:

K(1

h, p, c)

=h−p − h−1 + c(h−p + 1− 21−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− 21−p(h−1 + 1)p)(h−1 − 1)

)p=

h− hp + c(1 + hp − 21−p(h+ 1)p)(1− h)

(p− 1)(1− h)

×(p− 1

p

1− hp

h− hp + c(1 + hp − 21−p(h+ 1)p)(1− h)

)p= K(h, p, c)

(ii): Let h > 1. The logarithms calculation and l’Hospital’s theorem give K(h, p, b) → 1 asp→ 1, K(h, p, b)→ 1 as p→ 0 and K(h, p, b)→ 1 as h→ 1+. Now using (i) we obtain (ii).

(iii): Let h > 0 and 0 ≤ c ≤ 0.5.

dK(h, p, c)

dc= 2

((h+ 1

2

)p− hp + 1

2

)×

(p− 1

p

hp − 1

h− hp + c(hp + 1− 21−p(h+ 1)p)(h− 1)

)p.


Since the function z → zp is convex (resp. concave) on (0,∞) if p 6∈ (0, 1) (resp. p ∈ (0, 1)),

then (h+12 )p ≤ hp+1

2 (resp. (h+12 )p ≥ hp+1

2 ) for every h > 0. Then dK(h,p,c)dc ≤ 0 if p 6∈ (0, 1)

and dK(h,p,c)dc ≥ 0 if p ∈ (0, 1), which gives that K(h, p, c) is decreasing of c if p 6∈ (0, 1) and

increasing of c if p ∈ (0, 1).

(iv): Let h > 1 and 0 ≤ c ≤ 0.5. If p > 1 then

0 <(p− 1)(h− 1)

hp − h+ c(hp + 1− 21−p(h+ 1)p)(h− 1)

≤ p− 1

p

hp − 1

hp − h+ c(hp + 1− 21−p(h+ 1)p)(h− 1)

implies

(p− 1)(h− 1)

hp − h+ c(hp + 1− 21−p(h+ 1)p)(h− 1)

≤(p− 1

p

hp − 1

hp − h+ c(hp + 1− 21−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 [8, Theorem 2.54 (iv)], K(h, p, c) ≥ K(h, p, 0) ≥ K(h, 0.5, 0) for p ∈ (0, 1).

(v): Let p ≥ 1. Using (iii) and [8, Theorem 2.54 (vi)], K(h, p, c) ≤ K(h, p, 0) ≤ hp−1. �

References

[1] S. Abramovich, G. Jameson and G. Sinnamon, Refining Jensen’s inequality, Bull. Math. Soc. Sci. Math.Roumanie (N.S.) 47 (2004) 3–14.

[2] M.D. Choi, A Schwarz inequality for positive linear maps on C∗-algebras, Illinois J. Math. 18 (1974)565–574.

[3] C. Davis, A Schwarz inequality for convex operator functions, Proc. Amer. Math. Soc. 8 (1957) 42-44.

[4] S.S. Dragomir, A new refinement of Jensen’s inequality in linear spaces with applications, Math. Comput.Modelling 52 (2010) 1497–1505.

[5] J.I. Fujii, An external version of the Jensen operator inequality, Sci. Math. Japon. Online e-2011 (2011)

59–62.[6] J.I. Fujii, J. Pecaric and Y. Seo, The Jensen inequality in an external formula, J. Math. Inequal. 6 (2012)

473-480.

[7] T. Furuta, Operator inequalities associated with Holder-McCarthy and Kantorovich inequalities, J. In-equal. Appl. 2 (1998) 137–148.

[8] T. Furuta, J. Micic Hot, J. Pecaric and Y. Seo, Mond-Pecaric Method in Operator Inequalities, Mono-

graphs in Inequalities 1, Element, Zagreb, 2005.[9] F. Hansen, J. Pecaric and I. Peric, Jensen’s operator inequality and it’s converses Math. Scand. 100 (2007)

61–73.[10] F. Hansen and G.K. Pedersen, Jensen’s inequality for operators and Lowner’s theorem, Math. Ann. 258

(1982) 229–241.

[11] F. Hansen and G.K. Pedersen, Jensen’s operator inequality, Bull. London Math. Soc. 35 (2003) 553–564.[12] A. Ivelic, A. Matkovic, and J. E. Pecaric, On a Jensen-Mercer operator inequality, Banach J. Math. Anal.

5 (2011), 19–28.

[13] M. Khosravi, J.S. Aujla, S.S. Dragomir and M.S. Moslehian, Refinements of Choi-Davis-Jensen’s inequal-ity, Bull. Math. Anal. Appl. 3 (2011) 127–133.

[14] J. Micic, Z. Pavic and J. Pecaric, Jensen’s inequality for operators without operator convexity, Linear

Algebra Appl. 434 (2011) 1228–1237.[15] J. Micic, Z. Pavic and J. Pecaric Extension of Jensen’s operator inequality for operators without operator

convexity, Abstr. Appl. Anal. 2011 (2011) 1–14.

[16] J. Micic, J. Pecaric and J. Peric, Refined Jensen’s operator inequality with condition on spectra, Oper.Matrices. 7 (2013), 293–308.

[17] J. Micic, J. Pecaric and J. Peric, Extension of the refined Jensen’s operator inequality with condition onspectra, Ann. Funct. Anal. 3 (2012) 67–85.

[18] J. Micic, J. Pecaric and Y. Seo, Converses of Jensen’s operator inequality, Oper. Matrices 4 (2010) 385–

403.[19] J. Micic, J. Pecaric, Y. Seo and M. Tominaga, Inequalities of positive linear maps on Hermitian matrices,

Math. Inequal. Appl. 3 (2000) 559–591.


[20] J. Micic, Y. Seo, S.E. Takahasi and M. Tominaga, Inequalities of Furuta and Mond-Pecaric, Math. Inequal.

Appl. (1999) 2 83–111.

[21] J. Micic, Z. Pavic and J. Pecaric, Some better bounds in converses of the Jensen operator inequality, Oper.Matrices. 6 (2012), 589–605.

[22] D.S. Mitrinovic, J.E. Pecaric and A.M. Fink, Classical and New Inequalities in Analysis, Kluwer Acad.Publ., Dordrecht-Boston-London, 1993.

[23] B. Mond and J.E. Pecaric, Converses of Jensen’s inequality for linear maps of operators, An. Univ. Vest

Timis. Ser. Mat.-Inform. XXXI 2 (1993) 223–228.[24] B. Mond and J. Pecaric, Converses of Jensen’s inequality for several operators, Rev. Anal. Numer. Theor.

Approx. 23 (1994) 179–183.

[25] B. Mond and J. Pecaric, On Jensen’s inequality for operator convex functions, Houston J. Math. 21 (1995)739–754.

[26] M.S. Moslehian and M. Kian, Jensen type inequalities for Q-class functions, Bull. Aust. Math. Soc. 85

(2011), 128–142.[27] J. Rooin, A refinement of Jensen’s inequality, J. Ineq. Pure and Appl. Math. 6 2 Art. 38 (2005) 4 p.

[28] H.M. Srivastava, Z.G. Xia and Z.H. Zhang, Some further refinements and extensions of the Hermite-

Hadamard and Jensen inequalities in several variables, Math. Comput. Modelling 54 (2011) 2709–2717.[29] Z.G. Xiao, H.M. Srivastava and Z.H. Zhang, Further refinements of the Jensen inequalities based upon

samples with repetitions, Math. Comput. Modelling 51 (2010) 592–600.[30] L.C. Wang, X.F. Ma and L.H. Liu, A note on some new refinements of Jensen’s inequality for convex

functions, J. Inequal. Pure Appl. Math. 10 2 Art. 48 (2009) 6 p.

Jadranka Micic, Faculty of Mechanical Engineering and Naval Architecture, University of Za-

greb, Ivana Lucica 5, 10000 Zagreb, CroatiaE-mail address: [email protected]

Josip Pecaric, Faculty of Textile Technology, University of Zagreb, Prilaz baruna Filipovica30, 10000 Zagreb, Croatia

E-mail address: [email protected]

Jurica Peric, Faculty of Science, Department of Mathematics, University of Split, Teslina 12,21000 Split, Croatia

E-mail address: [email protected]

Refin Mond-Pec-corr(2)

Documents

Transcript of Refin Mond-Pec-corr(2)