Open access peer-reviewed chapter

Recent Research on Jensen's Inequality for Oparators

By Jadranka Mićić and Josip Pečarić

Submitted: November 30th 2011Reviewed: April 27th 2012Published: July 11th 2012

DOI: 10.5772/48468

Downloaded: 1363

1. Introduction

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

Let Ibe a real interval of any type. A continuous function f:Iis said to be operator convex if

fλx+(1-λ)yλf(x)+(1-λ)f(y)uid1

holds for each λ[0,1]and every pair of self-adjoint operators xand y(acting) on an infinite dimensional Hilbert space Hwith spectra in I(the ordering is defined by setting xyif y-xis positive semi-definite).

Let fbe an operator convex function defined on an interval I.Ch. Davis [1] provedThere is small typo in the proof. Davis states that φby Stinespring's theorem can be written on the form φ(x)=Pρ(x)Pwhere ρis a *-homomorphism to B(H)and Pis a projection on H.In fact, Hmay be embedded in a Hilbert space Kon which ρand Pacts. The theorem then follows by the calculation f(φ(x))=f(Pρ(x)P)Pf(ρ(x))P=Pρ(f(x)P=φ(f(x)),where the pinching inequality, proved by Davis in the same paper, is applied.a Schwarz inequality

fφ(x)φf(x)uid3

where φ:𝒜B(K)is a unital completely positive linear mapping from a C*-algebra 𝒜to linear operators on a Hilbert space K,and xis a self-adjoint element in 𝒜with spectrum in I.Subsequently M. D. Choi [2] noted that it is enough to assume that φis unital and positive. In fact, the restriction of φto the commutative C*-algebra generated by xis automatically completely positive by a theorem of Stinespring.

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

fi=1nai*xiaii=1nai*f(xi)aiuid4

for operator convex functions fdefined on an interval I=[0,α)(with αand f(0)0)and self-adjoint operators x1,,xnwith spectra in Iassuming that i=1nai*ai=1.The restriction on the interval and the requirement f(0)0was subsequently removed by B. Mond and J. Pečarić in [4], cf. also [5].

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

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

fi=1nwiφi(xi)i=1nwiφi(f(xi))uid5

for operator convex functions fdefined on an interval I,where φi:B(H)B(K)are unital positive linear mappings, x1,,xnare self-adjoint operators with spectra in Iand w1,,wnare are non-negative real numbers with sum one.

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

Fφf(A),gφ(A)maxmtMFf(m)+f(M)-f(m)M-m(t-m),g(t)1n˜uid6

for convex functions fdefined on an interval [m,M], m<M, where gis a real valued continuous function on [m,M], F(u,v)is a real valued function defined on U×V, matrix non-decreasing in u, Uf[m,M], Vg[m,M], φ:HnHn˜is a unital positive linear mapping and Ais a Hermitian matrix with spectrum contained in [m,M].

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

2. Classic results

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

We recall some definitions. Let Tbe a locally compact Hausdorff space and let 𝒜be a C*-algebra of operators on some Hilbert space H.We say that a field (xt)tTof operators in 𝒜is continuous if the function txtis norm continuous on T.If in addition μis a Radon measure on Tand the function txtis integrable, then we can form the Bochner integral Txtdμ(t), which is the unique element in 𝒜such that

ϕTxtdμ(t)=Tϕ(xt)dμ(t)

for every linear functional ϕin the norm dual 𝒜*.

Assume furthermore that there is a field (φt)tTof positive linear mappings φt:𝒜from 𝒜to another 𝒞*-algebra of operators on a Hilbert space K. We recall that a linear mapping φt:𝒜is said to be a positive mapping if φt(xt)0for all xt0. We say that such a field is continuous if the function tφt(x)is continuous for every x𝒜.Let the 𝒞*-algebras include the identity operators and the function tφt(1H)be integrable with Tφt(1H)dμ(t)=k1Kfor some positive scalar k. Specially, if Tφt(1H)dμ(t)=1K,we say that a field (φt)tTis unital.

Let B(H)be the C*-algebra of all bounded linear operators on a Hilbert space H. We define bounds of an operator xB(H)by

mx=infξ=1xξ,ξandMx=supξ=1xξ,ξuid7

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

For an operator xB(H)we define operators |x|, x+, x-by

|x|=(x*x)1/2,x+=(|x|+x)/2,x-=(|x|-x)/2

Obviously, if xis self-adjoint, then |x|=(x2)1/2and x+,x-0(called positive and negative parts of x=x+-x-).

2.1. Jensen's inequality with operator convexity

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

Theorem 1 Let f:Ibe an operator convex function defined on an interval Iand let 𝒜and be unital C*-algebras acting on a Hilbert space Hand Krespectively. If (φt)tTis a unital field of positive linear mappings φt:𝒜defined on a locally compact Hausdorff space Twith a bounded Radon measure μ,then the inequality

fTφt(xt)dμ(t)Tφt(f(xt))dμ(t)uid10

holds for every bounded continuous field (xt)tTof self-adjoint elements in 𝒜with spectra contained in I.

We first note that the function tφt(xt)is continuous and bounded, hence integrable with respect to the bounded Radon measure μ.Furthermore, the integral is an element in the multiplier algebra M()acting on K.We may organize the set CB(T,𝒜)of bounded continuous functions on Twith values in 𝒜as a normed involutive algebra by applying the point-wise operations and setting

(yt)tT=suptTyt(yt)tTCB(T,𝒜)

and it is not difficult to verify that the norm is already complete and satisfy the C*-identity. In fact, this is a standard construction in C*-algebra theory. It follows that f((xt)tT)=(f(xt))tT. We then consider the mapping

π:CB(T,𝒜)M()B(K)

defined by setting

π(xt)tT=Tφt(xt)dμ(t)

and note that it is a unital positive linear map. Setting x=(xt)tTCB(T,𝒜),we use inequality () to obtain

fπ(xt)tT=f(π(x))π(f(x))=πf(xt)tT=πf(xt)tT

but this is just the statement of the theorem.

2.2. Converses of Jensen's inequality

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

Theorem 2 Let Tbe a locally compact Hausdorff space equipped with a bounded Radon measure μ. Let (xt)tTbe a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜with spectra in [m,M], m<M. Furthermore, let (φt)tTbe a field of positive linear mappings φt:𝒜from 𝒜to another unital C*-algebra , such that the function tφt(1H)is integrable with Tφt(1H)dμ(t)=k1Kfor some positive scalar k. Let mxand Mx, mxMx, be the bounds of the self-adjoint operator x=Tφt(xt)dμ(t)and f:[m,M], g:[mx,Mx], F:U×Vbe functions such that (kf)[m,M]U,g[mx,Mx]Vand Fis bounded. If Fis operator monotone in the first variable, then

infmxzMxFk·h11kz,g(z)1KFTφtf(xt)dμ(t),gTφt(xt)dμ(t)supmxzMxFk·h21kz,g(z)1Kuid13

holds for every operator convex function h1on [m,M]such that h1fand for every operator concave function h2on [m,M]such that h2f.

We prove only RHS of (). Let h2be operator concave function on [m,M]such that f(z)h2(z)for every z[m,M]. By using the functional calculus, it follows that f(xt)h2(xt)for every tT. Applying the positive linear mappings φtand integrating, we obtain

Tφtf(xt)dμ(t)Tφth2(xt)dμ(t)

Furthermore, replacing φtby 1kφtin Theorem , we obtain 1kTφth2(xt)dμ(t)h21kTφt(xt)dμ(t), which gives Tφtf(xt)dμ(t)k·h21kTφt(xt)dμ(t). Since mx1KTφt(xt)dμ(t)Mx1K, then using operator monotonicity of F(·,v)we obtain

FTφtf(xt)dμ(t),gTφt(xt)dμ(t)Fk·h21kTφt(xt)dμ(t),gTφt(xt)dμ(t)supmxzMxFk·h21kz,g(z)1Kuid14

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

Theorem 3 Let (xt)tT, mx, Mxand (φt)tTbe as in Theorem . Let f:[m,M], g:[mx,Mx], F:U×Vbe functions such that (kf)[m,M]U,g[mx,Mx]Vand Fis bounded. If Fis operator monotone in the first variable and fis convex on the interval [m,M], then

FTφtf(xt)dμ(t),gTφt(xt)dμ(t)supmxzMxFMk-zM-mf(m)+z-kmM-mf(M),g(z)1Kuid16

In the dual case (when fis concave) the opposite inequalities hold in () with infinstead of sup.

We prove only the convex case. For convex fthe inequality f(z)M-zM-mf(m)+z-mM-mf(M)holds for every z[m,M]. Thus, by putting h2(z)=M-zM-mf(m)+z-mM-mf(M)in () we obtain (). Numerous applications of the previous theorem can be given (see [15]). Applying Theorem  for the function F(u,v)=u-αvand k=1, we obtain the following generalization of Theorem 2.4[15].

Corollary 4 Let (xt)tT, mx, Mxbe as in Theorem  and (φt)tTbe a unital field of positive linear mappings φt:𝒜. If f:[m,M]is convex on the interval [m,M], m<M, and g:[m,M], then for any α

Tφtf(xt)dμ(t)αgTφt(xt)dμ(t)+C1Kuid18

where

C=maxmxzMxM-zM-mf(m)+z-mM-mf(M)-αg(z)maxmzMM-zM-mf(m)+z-mM-mf(M)-αg(z)

If furthermore αgis strictly convex differentiable, then the constant CC(m,M,f,g,α)can be written more precisely as

C=M-z0M-mf(m)+z0-mM-mf(M)-αg(z0)

where

z0=g'-1f(M)-f(m)α(M-m)ifαg'(mx)f(M)-f(m)M-mαg'(Mx)mxifαg'(mx)f(M)-f(m)M-mMxifαg'(Mx)f(M)-f(m)M-m

In the dual case (when fis concave and αgis strictly concave differentiable) the opposite inequalities hold in () with mininstead of maxwith the opposite condition while determining z0.

3. Inequalities with conditions on spectra

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

Operator convexity plays an essential role in (). In fact, the inequality () will be false if we replace an operator convex function by a general convex function. For example, M.D. Choi in Remark 2.6[2] considered the function f(t)=t4which is convex but not operator convex. He demonstrated that it is sufficient to put dimH=3, so we have the matrix case as follows. Let Φ:M3()M2()be the contraction mapping Φ((aij)1i,j3)=(aij)1i,j2. If A=101001111,then Φ(A)4=1000¬9553=Φ(A4)and no relation between Φ(A)4and Φ(A4)under the operator order.

Example 5 It appears that the inequality () will be false if we replace the operator convex function by a general convex function. We give a small example for the matrix cases and T={1,2}. We define mappings Φ1,Φ2:M3()M2()by Φ1((aij)1i,j3)=12(aij)1i,j2, Φ2=Φ1. Then Φ1(I3)+Φ2(I3)=I2.

  1. If

    X1=2101001111andX2=2100000000

    then

    Φ1(X1)+Φ2(X2)4=16000¬80404024=Φ1X14+Φ2X24

    Given the above, there is no relation between Φ1(X1)+Φ2(X2)4and Φ1X14+Φ2X24under the operator order. We observe that in the above case the following stands X=Φ1(X1)+Φ2(X2)=2000and [mx,Mx]=[0,2], [m1,M1][-1.60388,4.49396], [m2,M2]=[0,2], i.e.

    (mx,Mx)[m1,M1][m2,M2]

    (see Fig. 1.a).

Figure 1.

Spectral conditions for a convex function f

  1. If

    X1=-14010-2-11-1-1andX2=15000200015

    then

    Φ1(X1)+Φ2(X2)4=116000<89660-247-24751=Φ1X14+Φ2X24

    So we have that an inequality of type () now is valid. In the above case the following stands X=Φ1(X1)+Φ2(X2)=12000and [mx,Mx]=[0,0.5], [m1,M1][-14.077,-0.328566], [m2,M2]=[2,15], i.e.

    (mx,Mx)[m1,M1]=and(mx,Mx)[m2,M2]=

    (see Fig. 1.b).

3.1. Jensen's inequality without operator convexity

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

Theorem 6 Let (xt)tTbe a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜defined on a locally compact Hausdorff space Tequipped with a bounded Radon measure μ. Let mtand Mt, mtMt, be the bounds of xt, tT. Let (φt)tTbe a unital field of positive linear mappings φt:𝒜from 𝒜to another unital C*-algebra . If

(mx,Mx)[mt,Mt]=,tT

where mxand Mx, mxMx, are the bounds of the self-adjoint operator x=Tφt(xt)dμ(t), then

fTφt(xt)dμ(t)Tφt(f(xt))dμ(t)uid25

holds for every continuous convex function f:Iprovided that the interval Icontains all mt,Mt.

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

We prove only the case when fis a convex function. If we denote m=inftT{mt}and M=suptT{Mt}, then [m,M]Iand m1HAtM1H, tT. It follows m1KTφt(xt)dμ(t)M1K. Therefore [mx,Mx][m,M]I.

a)Let mx<Mx. Since fis convex on [mx,Mx], then

f(z)Mx-zMx-mxf(mx)+z-mxMx-mxf(Mx),z[mx,Mx]uid26

but since fis convex on [mt,Mt]and since (mx,Mx)[mt,Mt]=, then

f(z)Mx-zMx-mxf(mx)+z-mxMx-mxf(Mx),z[mt,Mt],tTuid27

Since mx1KTφt(xt)dμ(t)Mx1K,then by using functional calculus, it follows from ()

fTφt(xt)dμ(t)Mx1K-Tφt(xt)dμ(t)Mx-mxf(mx)+Tφt(xt)dμ(t)-mx1KMx-mxf(Mx)uid28

On the other hand, since mt1HxtMt1H, tT, then by using functional calculus, it follows from ()

fxtMx1H-xtMx-mxf(mx)+xt-mx1HMx-mxf(Mx),tT

Applying a positive linear mapping φtand summing, we obtain

Tφtf(xt)dμ(t)Mx1K-Tφt(xt)dμ(t)Mx-mxf(mx)+Tφt(xt)dμ(t)-mx1KMx-mxf(Mx)uid29

since Tφt(1H)dμ(t)=1K. Combining the two inequalities () and (), we have the desired inequality ().

b)Let mx=Mx. Since fis convex on [m,M], we have

f(z)f(mx)+l(mx)(z-mx)foreveryz[m,M]uid30

where lis the subdifferential of f. Since m1HxtM1H, tT, then by using functional calculus, applying a positive linear mapping φtand summing, we obtain from ()

Tφtf(xt)dμ(t)f(mx)1K+l(mx)Tφt(xt)dμ(t)-mx1Kuid31

Since mx1K=Tφt(xt)dμ(t),it follows

Tφtf(xt)dμ(t)f(mx)1K=fTφt(xt)dμ(t)uid32

which is the desired inequality (). Putting φt(y)=atyfor every y𝒜, where at0is a real number, we obtain the following obvious corollary of Theorem .

Corollary 7 Let (xt)tTbe a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜defined on a locally compact Hausdorff space Tequipped with a bounded Radon measure μ. Let mtand Mt, mtMt, be the bounds of xt, tT. Let (at)tTbe a continuous field of nonnegative real numbers such that Tatdμ(t)=1. If

(mx,Mx)[mt,Mt]=,tT

where mxand Mx, mxMx, are the bounds of the self-adjoint operator x=Tatxtdμ(t), then

fTatxtdμ(t)Tatf(xt)dμ(t)uid34

holds for every continuous convex function f:Iprovided that the interval Icontains all mt,Mt.

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

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

Theorem 8 Let (xt)tTbe a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜defined on a locally compact Hausdorff space Tequipped with a bounded Radon measure μ. Furthermore, let (φt)tTbe a field of positive linear mappings φt:𝒜from 𝒜to another unital C*-algebra , such that the function tφt(1H)is integrable with Tφt(1H)dμ(t)=k1Kfor some positive scalar k. Let mtand Mt, mtMt, be the bounds of xt, tT, m=inftT{mt}, M=suptT{Mt}, and mxand Mx, mx<Mx, be the bounds of x=Tφt(xt)dμ(t). If

(mx,Mx)[mt,Mt]=,tT

and f:[m,M], g:[mx,Mx], F:U×Vare functions such that (kf)[m,M]U,g[mx,Mx]V, fis convex, Fis bounded and operator monotone in the first variable, then

infmxzMxFMxk-zMx-mxf(mx)+z-kmxMx-mxf(Mx),g(z)1KFTφtf(xt)dμ(t),gTφt(xt)dμ(t)supmxzMxFMk-zM-mf(m)+z-kmM-mf(M),g(z)1Kuid37

In the dual case (when fis concave) the opposite inequalities hold in () by replacing infand supwith supand inf, respectively.

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

Tφtf(xt)dμ(t)Mxk1K-Tφt(xt)dμ(t)Mx-mxf(mx)+Tφt(xt)dμ(t)-mxk1KMx-mxf(Mx)

since Tφt(1H)dμ(t)=k1K. By using operator monotonicity of F(·,v)we obtain

FTφtf(xt)dμ(t),gTφt(xt)dμ(t)FMxk1K-Tφt(xt)dμ(t)Mx-mxf(mx)+Tφt(xt)dμ(t)-mxk1KMx-mxf(Mx),gTφt(xt)dμ(t)

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

Putting F(u,v)=u-αvor F(u,v)=v-1/2uv-1/2in Theorem , we obtain the next corollary.

Corollary 9 Let (xt)tT, mt, Mt, mx, Mx, m, M, (φt)tTbe as in Theorem  and f:[m,M], g:[mx,Mx]be continuous functions. If

(mx,Mx)[mt,Mt]=,tT

and fis convex, then for any α

minmxzMxMxk-zMx-mxf(mx)+z-kmxMx-mxf(Mx)-g(z)1K+αgTφt(xt)dμ(t)Tφtf(xt)dμ(t)αgTφt(xt)dμ(t)+maxmxzMxMk-zM-mf(m)+z-kmM-mf(M)-g(z)1Kuid39

If additionally g>0on [mx,Mx], then

minmxzMxMxk-zMx-mxf(mx)+z-kmxMx-mxf(Mx)g(z)gTφt(xt)dμ(t)Tφtf(xt)dμ(t)maxmxzMxMk-zM-mf(m)+z-kmM-mf(M)g(z)gTφt(xt)dμ(t)uid40

In the dual case (when fis concave) the opposite inequalities hold in () by replacing minand maxwith maxand min, respectively. If additionally g>0on [mx,Mx], then the opposite inequalities also hold in () by replacing minand maxwith maxand min, respectively.

4. Refined Jensen's inequality

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

To obtain our result we need the following two lemmas.

Lemma 10 Let fbe a convex function on an interval I, m,MIand p1,p2[0,1]such that p1+p2=1. Then

min{p1,p2}f(m)+f(M)-2fm+M2p1f(m)+p2f(M)-f(p1m+p2M)uid42

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

Lemma 11 Let xbe a bounded self-adjoint elements in a unital C*-algebra 𝒜of operators on some Hilbert space H. If the spectrum of xis in [m,M], for some scalars m<M, then

fxM1H-xM-mf(m)+x-m1HM-mf(M)-δfx˜(resp.fxM1H-xM-mf(m)+x-m1HM-mf(M)+δfx˜)uid44

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

δf=f(m)+f(M)-2fm+M2(resp.δf=2fm+M2-f(m)-f(M))andx˜=121H-1M-mx-m+M21H

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

fp1m+p2Mp1f(m)+p2f(M)-min{p1,p2}f(m)+f(M)-2fm+M2uid45

for every p1,p2[0,1]such that p1+p2=1. For any z[m,M]we can write

fz=fM-zM-mm+z-mM-mM

Then by using () for p1=M-zM-mand p2=z-mM-mwe obtain

f(z)M-zM-mf(m)+z-mM-mf(M)-12-1M-mz-m+M2f(m)+f(M)-2fm+M2uid46

since

minM-zM-m,z-mM-m=12-1M-mz-m+M2

Finally we use the continuous functional calculus for a self-adjoint operator x: f,g𝒞(I),Sp(x)Iand fgon Iimplies f(x)g(x); and h(z)=|z|implies h(x)=|x|. Then by using () we obtain the desired inequality ().

Theorem 12 Let (xt)tTbe a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜defined on a locally compact Hausdorff space Tequipped with a bounded Radon measure μ. Let mtand Mt, mtMt, be the bounds of xt, tT. Let (φt)tTbe a unital field of positive linear mappings φt:𝒜from 𝒜to another unital C*-algebra . Let

(mx,Mx)[mt,Mt]=,tT,andm<M

where mxand Mx, mxMx, be the bounds of the operator x=Tφt(xt)dμ(t)and

m=supMt:Mtmx,tT,M=infmt:mtMx,tT

If f:Iis a continuous convex (resp. concave)function provided that the interval Icontains all mt,Mt, then

fTφt(xt)dμ(t)Tφt(f(xt))dμ(t)-δfx˜Tφt(f(xt))dμ(t)uid48

(resp.

fTφt(xt)dμ(t)Tφt(f(xt))dμ(t)-δfx˜Tφt(f(xt))dμ(t))uid49

holds, where

δfδf(m¯,M¯)=f(m¯)+f(M¯)-2fm¯+M¯2(resp.δfδf(m¯,M¯)=2fm¯+M¯2-f(m¯)-f(M¯))x˜x˜x(m¯,M¯)=121K-1M¯-m¯x-m¯+M¯21Kuid50

and  m¯[m,mA], M¯[MA,M], m¯<M¯,   are arbitrary numbers.

We prove only the convex case. Since x=Tφt(xt)dμ(t)is the self-adjoint elements such that m¯1Kmx1KTφt(xt)dμ(t)Mx1KM¯1Kand fis convex on [m¯,M¯]I, then by Lemma  we obtain

fTφt(xt)dμ(t)M¯1K-Tφt(xt)dμ(t)M¯-m¯f(m¯)+Tφt(xt)dμ(t)-m¯1KM¯-m¯f(M¯)-δfx˜uid51

where δfand x˜are defined by ().

But since fis convex on [mt,Mt]and (mx,Mx)[mt,Mt]=implies (m¯,M¯)[mt,Mt]=, then

fxtM¯1H-xtM¯-m¯f(m¯)+xt-m¯1HM¯-m¯f(M¯),tT

Applying a positive linear mapping φt, integrating and adding -δfx˜, we obtain

Tφtf(xt)dμ(t)-δfx˜M¯1K-Tφt(xt)dμ(t)M¯-m¯f(m¯)+Tφt(xt)dμ(t)-m¯1KM¯-m¯f(M¯)-δfx˜uid52

since Tφt(1H)dμ(t)=1K. Combining the two inequalities () and (), we have LHS of (). Since δf0and x˜0, then we have RHS of ().

If m<Mand mx=Mx, then the inequality () holds, but δf(mx,Mx)x˜(mx,Mx)is not defined (see Example  I) and II)).

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

Figure 2.

Refinement for two operators and a convex function f

We put f(t)=t4which is convex but not operator convex in (). Also, we define mappings Φ1,Φ2:M3()M2()as follows: Φ1((aij)1i,j3)=12(aij)1i,j2, Φ2=Φ1(then Φ1(I3)+Φ2(I3)=I2).

I)First, we observe an example when δfX˜is equal to the difference RHS and LHS of Jensen's inequality. If X1=-3I3and X2=2I3, then X=Φ1(X1)+Φ2(X2)=-0.5I2, so m=-3, M=2. We also put m¯=-3and M¯=2. We obtain

Φ1(X1)+Φ2(X2)4=0.0625I2<48.5I2=Φ1X14+Φ2X24

and its improvement

Φ1(X1)+Φ2(X2)4=0.0625I2=Φ1X14+Φ2X24-48.4375I2

since δf=96.875, X˜=0.5I2.We remark that in this case mx=Mx=-1/2and X˜(mx,Mx)is not defined.

II)Next, we observe an example when δfX˜is not equal to the difference RHS and LHS of Jensen's inequality and mx=Mx. If

X1=-1000-2000-1,X2=200030004,thenX=121001andm=-1,M=2

In this case x˜(mx,Mx)is not defined, since mx=Mx=1/2. We have

Φ1(X1)+Φ2(X2)4=1161001<17200972=Φ1X14+Φ2X24

and putting m¯=-1, M¯=2we obtain δf=135/8, X˜=I2/2which give the following improvement

Φ1(X1)+Φ2(X2)4=1161001<116100641=Φ1X14+Φ2X24-135161001

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

X1=-4111-2-11-1-1andX2=5-1-1-121-113,thenX=121000

so m1=-4.8662, M1=-0.3446, m2=1.3446, M2=5.8662, m=-0.3446, M=1.3446and we put m¯=m, M¯=M(rounded to four decimal places). We have

Φ1(X1)+Φ2(X2)4=1161000<12832-255-2552372=Φ1X14+Φ2X24

and its improvement

Φ1(X1)+Φ2(X2)4=1161000<639.9213-255-255117.8559=Φ1X14+Φ2X24-1.5787000.6441

(rounded to four decimal places), since δf=3.1574, X˜=0.5000.2040. But, if we put m¯=mx=0, M¯=Mx=0.5, then X˜=0, so we do not have an improvement of Jensen's inequality. Also, if we put m¯=0, M¯=1, then X˜=0.51001, δf=7/8and δfX˜=0.43751001, which is worse than the above improvement.

Putting Φt(y)=atyfor every y𝒜, where at0is a real number, we obtain the following obvious corollary of Theorem .

Corollary 14 Let (xt)tTbe a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜defined on a locally compact Hausdorff space Tequipped with a bounded Radon measure μ. Let mtand Mt, mtMt, be the bounds of xt, tT. Let (at)tTbe a continuous field of nonnegative real numbers such that Tatdμ(t)=1. Let

(mx,Mx)[mt,Mt]=,tT,andm<M

where mxand Mx, mxMx, are the bounds of the operator x=Tφt(xt)dμ(t)and

m=supMt:Mtmx,tT,M=infmt:mtMx,tT

If f:Iis a continuous convex (resp. concave)function provided that the interval Icontains all mt,Mt, then

fTatxtdμ(t)Tatf(xt)dμ(t)-δfx˜˜Tatf(xt)dμ(t)(resp.fTatxtdμ(t)Tatf(xt)dμ(t)+δfx˜˜Tatf(xt)dμ(t))

holds, where δfis defined by (), x˜˜=121H-1M¯-m¯Tatxtdμ(t)-m¯+M¯21Hand m¯[m,mA], M¯[MA,M], m¯<M¯, are arbitrary numbers.

5. Extension Jensen's inequality

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

In a discrete version of Theorem  we prove that Jensen's operator inequality holds for every continuous convex function and for every n-tuple of self-adjoint operators (A1,...,An), for every n-tuple of positive linear mappings (Φ1,...,Φn)in the case when the interval with bounds of the operator A=i=1nΦi(Ai)has no intersection points with the interval with bounds of the operator Aifor each i=1,...,n, i.e. when (mA,MA)[mi,Mi]=for i=1,...,n,where mAand MA, mAMA, are the bounds of A, and miand Mi, miMi, are the bounds of Ai, i=1,...,n. It is interesting to consider the case when (mA,MA)[mi,Mi]=is valid for several i{1,...,n}, but not for all i=1,...,n. We study it in the following theorem (see [21]).

Theorem 15 Let (A1,...,An)be an n-tuple of self-adjoint operators AiB(H)with the bounds miand Mi, miMi, i=1,...,n. Let (Φ1,...,Φn)be an n-tuple of positive linear mappings Φi:B(H)B(K), such that i=1nΦi(1H)=1K. For 1n1<n, we denote m=min{m1,...,mn1}, M=max{M1,...,Mn1}and i=1n1Φi(1H)=α1K, i=n1+1nΦi(1H)=β1K, where α,β>0, α+β=1. If

(m,M)[mi,Mi]=,i=n1+1,...,n

and one of two equalities

1αi=1n1Φi(Ai)=1βi=n1+1nΦi(Ai)=i=1nΦi(Ai)

is valid, then

1αi=1n1Φi(f(Ai))i=1nΦi(f(Ai))1βi=n1+1nΦi(f(Ai))uid57

holds for every continuous convex function f:Iprovided that the interval Icontains all mi,Mi, i=1,...,n. If f:Iis concave, then the reverse inequality is valid in ().

We prove only the case when fis a convex function. Let us denote

A=1αi=1n1Φi(Ai),B=1βi=n1+1nΦi(Ai),C=i=1nΦi(Ai)

It is easy to verify that A=Bor B=Cor A=Cimplies A=B=C.

a)Let m<M. Since fis convex on [m,M]and [mi,Mi][m,M]for i=1,...,n1, then

f(z)M-zM-mf(m)+z-mM-mf(M),z[mi,Mi]fori=1,...,n1uid58

but since fis convex on all [mi,Mi]and (m,M)[mi,Mi]=for i=n1+1,...,n, then

f(z)M-zM-mf(m)+z-mM-mf(M),z[mi,Mi]fori=n1+1,...,nuid59

Since mi1HAiMi1H, i=1,...,n1, it follows from ()

fAiM1H-AiM-mf(m)+Ai-m1HM-mf(M),i=1,...,n1

Applying a positive linear mapping Φiand summing, we obtain

i=1n1Φif(Ai)Mα1K-i=1n1Φi(Ai)M-mf(m)+i=1n1Φi(Ai)-mα1KM-mf(M)

since i=1n1Φi(1H)=α1K. It follows

1αi=1n1Φif(Ai)M1K-AM-mf(m)+A-m1KM-mf(M)uid60

Similarly to () in the case mi1HAiMi1H, i=n1+1,...,n, it follows from ()

1βi=n1+1nΦif(Ai)M1K-BM-mf(m)+B-m1KM-mf(M)uid61

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

1αi=1n1Φif(Ai)1βi=n1+1nΦif(Ai)uid62

It follows

1αi=1n1Φi(f(Ai))=i=1n1Φi(f(Ai))+βαi=1n1Φi(f(Ai))(byα+β=1)i=1n1Φi(f(Ai))+i=n1+1nΦi(f(Ai))(by())=i=1nΦi(f(Ai))αβi=n1+1nΦi(f(Ai))+i=n1+1nΦi(f(Ai))(by())=1βi=n1+1nΦi(f(Ai))(byα+β=1)uid63

which gives the desired double inequality ().

b)Let m=M. Since [mi,Mi][m,M]for i=1,...,n1, then Ai=m1Hand f(Ai)=f(m)1Hfor i=1,...,n1. It follows

1αi=1n1Φi(Ai)=m1Kand1αi=1n1Φif(Ai)=f(m)1Kuid64

On the other hand, since fis convex on I, we have

f(z)f(m)+l(m)(z-m)foreveryzIuid65

where lis the subdifferential of f. Replacing zby Aifor i=n1+1,...,n, applying Φiand summing, we obtain from () and ()

1βi=n1+1nΦif(Ai)f(m)1K+l(m)1βi=n1+1nΦi(Ai)-m1K=f(m)1K=1αi=1n1Φif(Ai)

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

Remark 16 We obtain the equivalent inequality to the one in Theorem  in the case when i=1nΦi(1H)=γ1K, for some positive scalar γ. If α+β=γand one of two equalities

1αi=1n1Φi(Ai)=1βi=n1+1nΦi(Ai)=1γi=1nΦi(Ai)

is valid, then

1αi=1n1Φi(f(Ai))1γi=1nΦi(f(Ai))1βi=n1+1nΦi(f(Ai))

holds for every continuous convex function f.

Remark 17 Let the assumptions of Theorem  be valid.

1.We observe that the following inequality

f1βi=n1+1nΦi(Ai)1βi=n1+1nΦi(f(Ai))uid68

holds for every continuous convex function f:I.

Indeed, by the assumptions of Theorem  we have

mα1Hi=1n1Φi(f(Ai))Mα1Hand1αi=1n1Φi(Ai)=1βi=n1+1nΦi(Ai)

which implies

m1Hi=n1+1n1βΦi(f(Ai))M1H

Also (m,M)[mi,Mi]=for i=n1+1,...,nand i=n1+1n1βΦi(1H)=1Khold. So we can apply Theorem  on operators An1+1,...,Anand mappings 1βΦiand obtain the desired inequality.

2.We denote by mCand MCthe bounds of C=i=1nΦi(Ai). If (mC,MC)[mi,Mi]=, i=1,...,n1or fis an operator convex function on [m,M], then the double inequality () can be extended from the left side if we use Jensen's operator inequality (see Theorem 2.1[16])

fi=1nΦi(Ai)=f1αi=1n1Φi(Ai)1αi=1n1Φi(f(Ai))i=1nΦi(f(Ai))1βi=n1+1nΦi(f(Ai))

Example 18If neither assumptions (mC,MC)[mi,Mi]=, i=1,...,n1, nor fis operator convex in Remark  - 2. is satisfied and if 1<n1<n, then () can not be extended by Jensen's operator inequality, since it is not valid. Indeed, for n1=2we define mappings Φ1,Φ2:M3()M2()by Φ1((aij)1i,j3)=α2(aij)1i,j2, Φ2=Φ1. Then Φ1(I3)+Φ2(I3)=αI2. If

A1=2101001111andA2=2100000000

then

1αΦ1(A1)+1αΦ2(A2)4=1α416000¬1α80404024=1αΦ1A14+1αΦ2A24

for every α(0,1). We observe that f(t)=t4is not operator convex and (mC,MC)[mi,Mi],since C=A=1αΦ1(A1)+1αΦ2(A2)=1α2000,[mC,MC]=[0,2/α], [m1,M1][-1.60388,4.49396]and [m2,M2]=[0,2].

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

Corollary 19 Let (A1,...,An)be an n-tuple of self-adjoint operators AiB(H)with the bounds miand Mi, miMi, i=1,...,n. For some 1n1<n, we denote m=min{m1,...,mn1}, M=max{M1,...,Mn1}. Let (p1,...,pn)be an n-tuple of non-negative numbers, such that 0<i=1n1pi=𝐩𝐧1<𝐩𝐧=i=1npi. If

(m,M)[mi,Mi]=,i=n1+1,...,n

and one of two equalities

1𝐩𝐧1i=1n1piAi=1𝐩𝐧i=1npiAi=1𝐩𝐧-𝐩𝐧1i=n1+1npiAi

is valid, then

1𝐩𝐧1i=1n1pif(Ai)1𝐩𝐧i=1npif(Ai)1𝐩𝐧-𝐩𝐧1i=n1+1npif(Ai)uid71

holds for every continuous convex function f:Iprovided that the interval Icontains all mi,Mi, i=1,...,n.

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

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

Corollary 20 (Discrete version of Corollary ▭) Let (A1,...,An)be an n-tuple of self-adjoint operators AiB(H)with the bounds miand Mi, miMi, i=1,...,n. Let (α1,...,αn)be an n-tuple of nonnegative real numbers such that i=1nαi=1. If

(mA,MA)[mi,Mi]=,i=1,...,nuid73

where mAand MA, mAMA, are the bounds of A=i=1nαiAi, then

fi=1nαiAii=1nαif(Ai)uid74

holds for every continuous convex function f:Iprovided that the interval Icontains all mi,Mi.

We prove only the convex case. We define (n+1)-tuple of operators (B1,...,Bn+1), BiB(H), by B1=A=i=1nαiAiand Bi=Ai-1, i=2,...,n+1. Then mB1=mA, MB1=MAare the bounds of B1and mBi=mi-1, MBi=Mi-1are the ones of Bi, i=2,...,n+1. Also, we define (n+1)-tuple of non-negative numbers (p1,...,pn+1)by p1=1and pi=αi-1, i=2,...,n+1. Then i=1n+1pi=2and by using () we have

(mB1,MB1)[mBi,MBi]=,i=2,...,n+1uid75

Since

i=1n+1piBi=B1+i=2n+1piBi=i=1nαiAi+i=1nαiAi=2B1

then

p1B1=12i=1n+1piBi=i=2n+1piBiuid76

Taking into account () and (), we can apply Corollary  for n1=1and Bi, pias above, and we get

p1f(B1)12i=1n+1pif(Bi)i=2n+1pif(Bi)

which gives the desired inequality ().

6. Extension of the refined Jensen's inequality

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

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

Theorem 21 Let (A1,...,An)be an n-tuple of self-adjoint operators AiB(H)with the bounds miand Mi, miMi, i=1,...,n. Let (Φ1,...,Φn)be an n-tuple of positive linear mappings Φi:B(H)B(K), such that i=1n1Φi(1H)=α1K, i=n1+1nΦi(1H)=β1K, where 1n1<n, α,β>0and α+β=1. Let mL=min{m1,...,mn1}, MR=max{M1,...,Mn1}and

m=maxMi:MimL,i{n1+1,...,n}M=minmi:miMR,i{n1+1,...,n}

If

(mL,MR)[mi,Mi]=,i=n1+1,...,n,andm<M

and one of two equalities

1αi=1n1Φi(Ai)=i=1nΦi(Ai)=1βi=n1+1nΦi(Ai)

is valid, then

1αi=1n1Φi(f(Ai))1αi=1n1Φi(f(Ai))+βδfA˜i=1nΦi(f(Ai))1βi=n1+1nΦi(f(Ai))-αδfA˜1βi=n1+1nΦi(f(Ai))uid78

holds for every continuous convex function f:Iprovided that the interval Icontains all mi,Mi, i=1,...,n, where

δfδf(m¯,M¯)=f(m¯)+f(M¯)-2fm¯+M¯2A˜A˜A,Φ,n1,α(m¯,M¯)=121K-1α(M¯-m¯)i=1n1ΦiAi-m¯+M¯21Huid79

and   m¯[m,mL], M¯[MR,M], m¯<M¯,   are arbitrary numbers. If f:Iis concave, then the reverse inequality is valid in ().

We prove only the convex case. Let us denote

A=1αi=1n1Φi(Ai),B=1βi=n1+1nΦi(Ai),C=i=1nΦi(Ai)

It is easy to verify that A=Bor B=Cor A=Cimplies A=B=C.

Since fis convex on [m¯,M¯]and 𝖲𝗉(Ai)[mi,Mi][m¯,M¯]for i=1,...,n1, it follows from Lemma  that

fAiM¯1H-AiM¯-m¯f(m¯)+Ai-m¯1HM¯-m¯f(M¯)-δfA˜i,i=1,...,n1

holds, where δf=f(m¯)+f(M¯)-2fm¯+M¯2and A˜i=121H-1M¯-m¯Ai-m¯+M¯21H. Applying a positive linear mapping Φiand summing, we obtain

i=1n1Φif(Ai)M¯α1K-i=1n1Φi(Ai)M¯-m¯f(m¯)+i=1n1Φi(Ai)-m¯α1KM¯-m¯f(M¯)-δfα21K-1M¯-m¯i=1n1ΦiAi-m¯+M¯21H

since i=1n1Φi(1H)=α1K. It follows that

1αi=1n1Φif(Ai)M¯1K-AM¯-m¯f(m¯)+A-m¯1KM¯-m¯f(M¯)-δfA˜uid80

where A˜=121K-1α(M¯-m¯)i=1n1ΦiAi-m¯+M¯21H.

Additionally, since fis convex on all [mi,Mi]and (m¯,M¯)[mi,Mi]=, i=n1+1,...,n, then

f(Ai)M¯1H-AiM¯-m¯f(m¯)+Ai-m¯1HM¯-m¯f(M¯),i=n1+1,...,nuid81

It follows

1βi=n1+1nΦif(Ai)-δfA˜M¯1K-BM¯-m¯f(m¯)+B-m¯1KM¯-m¯f(M¯)-δfA˜uid82

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

1αi=1n1Φif(Ai)1βi=n1+1nΦif(Ai)-δfA˜uid83

Next, we obtain

1αi=1n1Φi(f(Ai))=i=1n1Φi(f(Ai))+βαi=1n1Φi(f(Ai))(byα+β=1)i=1n1Φi(f(Ai))+i=n1+1nΦi(f(Ai))-βδfA˜(by())αβi=n1+1nΦi(f(Ai))-αδfA˜+i=n1+1nΦi(f(Ai))-βδfA˜(by())=1βi=n1+1nΦi(f(Ai))-δfA˜(byα+β=1)

which gives the following double inequality

1αi=1n1Φi(f(Ai))i=1nΦi(f(Ai))-βδfA˜1βi=n1+1nΦi(f(Ai))-δfA˜uid84

Adding βδfA˜in the above inequalities, we get

1αi=1n1Φi(f(Ai))+βδfA˜i=1nΦi(f(Ai))1βi=n1+1nΦi(f(Ai))-αδfA˜uid85

Now, we remark that δf0and A˜0. (Indeed, since fis convex, then f(m¯+M¯)/2(f(m¯)+f(M¯))/2, which implies that δf0. Also, since

𝖲𝗉(Ai)[m¯,M¯]Ai-M¯+m¯21HM¯-m¯21H,i=1,...,n1

then

i=1n1ΦiAi-M¯+m¯21HM¯-m¯2α1K

which gives

0121K-1α(M¯-m¯)i=1n1ΦiAi-M¯+m¯21H=A˜)

Consequently, the following inequalities

1αi=1n1Φi(f(Ai))1αi=1n1Φi(f(Ai))+βδfA˜1βi=n1+1nΦi(f(Ai))-αδfA˜1βi=n1+1nΦi(f(Ai))

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

Example 22 We observe the matrix case of Theorem  for f(t)=t4, which is the convex function but not operator convex, n=4, n1=2and the bounds of matrices as in Fig. 3.

Figure 3.

An example a convex function and the bounds of four operators

We show an example such that

1αΦ1(A14)+Φ2(A24)<1αΦ1(A14)+Φ2(A24)+βδfA˜<Φ1(A14)+Φ2(A24)+Φ3(A34)+Φ4(A44)<1βΦ3(A34)+Φ4(A44)-αδfA˜<1βΦ3(A34)+Φ4(A44)uid88

holds, where δf=M¯4+m¯4-(M¯+m¯)48and

A˜=12I2-1α(M¯-m¯)Φ1|A1-M¯+m¯2Ih|+Φ2|A2-M¯+m¯2I3|

We define mappings Φi:M3()M2()as follows: Φi((ajk)1j,k3)=14(ajk)1j,k2, i=1,...,4. Then i=14Φi(I3)=I2and α=β=12.

Let

A1=229/819/820103,A2=329/809/810002,A3=-341/211/240102,A4=125/31/201/23/20003

Then m1=1.28607, M1=7.70771, m2=0.53777, M2=5.46221, m3=-14.15050, M3=-4.71071, m4=12.91724, M4=36., so mL=m2, MR=M1, m=M3and M=m4(rounded to five decimal places). Also,

1αΦ1(A1)+Φ2(A2)=1βΦ3(A3)+Φ4(A4)=49/49/43

and

Af1αΦ1(A14)+Φ2(A24)=989.00391663.46875663.46875526.12891CfΦ1(A14)+Φ2(A24)+Φ3(A34)+Φ4(A44)=68093.1425848477.9843748477.9843751335.39258Bf1βΦ3(A34)+Φ4(A44)=135197.2812596292.596292.5102144.65625

Then

Af<Cf<Bfuid89

holds (which is consistent with ()).

We will choose three pairs of numbers (m¯,M¯), m¯[-4.71071,0.53777], M¯[7.70771,12.91724]as follows

i)m¯=mL=0.53777, M¯=MR=7.70771, then

Δ˜1=βδfA˜=0.5·2951.69249·0.156780.090300.090300.15943=231.38908133.26139133.26139235.29515

ii)m¯=m=-4.71071, M¯=M=12.91724, then

Δ˜2=βδfA˜=0.5·27766.07963·0.360220.035730.035730.36155=5000.89860496.04498496.044985019.50711

iii)m¯=-1, M¯=10, then

Δ˜3=βδfA˜=0.5·9180.875·0.282030.089750.089750.27557=1294.66411.999411.9991265.

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

Table 1.

Using Theorem  we get the following result.

Corollary 23 Let the assumptions of Theorem  hold. Then

1αi=1n1Φi(f(Ai))1αi=1n1Φi(f(Ai))+γ1δfA˜1βi=n1+1nΦi(f(Ai))uid91

and

1αi=1n1Φi(f(Ai))1βi=n1+1nΦi(f(Ai))-γ2δfA˜1βi=n1+1nΦi(f(Ai))uid92

holds for every γ1,γ2in the close interval joining αand β, where δfand A˜are defined by ().

Adding αδfA˜in () and noticing δfA˜0, we obtain

1αi=1n1Φi(f(Ai))1αi=1n1Φi(f(Ai))+αδfA˜1βi=n1+1nΦi(f(Ai))uid93

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

Similarly, subtracting βδfA˜in () we obtain ().

Remark 24 We can obtain extensions of inequalities which are given in Remark  and . Also, we can obtain a special case of Theorem  with the convex combination of operators Aiputting Φi(B)=αiB, for i=1,...,n, similarly as in Corollary . Finally, applying this result, we can give another proof of Corollary . The interested reader can see the details in [30].

© 2012 The Author(s). Licensee IntechOpen. This chapter is distributed under the terms of the Creative Commons Attribution 3.0 License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.

How to cite and reference

Link to this chapter Copy to clipboard

Cite this chapter Copy to clipboard

Jadranka Mićić and Josip Pečarić (July 11th 2012). Recent Research on Jensen's Inequality for Oparators, Linear Algebra - Theorems and Applications, Hassan Abid Yasser, IntechOpen, DOI: 10.5772/48468. Available from:

chapter statistics

1363total chapter downloads

More statistics for editors and authors

Login to your personal dashboard for more detailed statistics on your publications.

Access personal reporting

Related Content

This Book

Next chapter

A Linear System of Both Equations and Inequalities in Max-Algebra

By Abdulhadi Aminu

Related Book

First chapter

Cramer’s Rules for the System of Two-Sided Matrix Equations and of Its Special Cases

By Ivan I. Kyrchei

We are IntechOpen, the world's leading publisher of Open Access books. Built by scientists, for scientists. Our readership spans scientists, professors, researchers, librarians, and students, as well as business professionals. We share our knowledge and peer-reveiwed research papers with libraries, scientific and engineering societies, and also work with corporate R&D departments and government entities.

More About Us