Mathematics » "Linear Algebra - Theorems and Applications", book edited by Hassan Abid Yasser, ISBN 978-953-51-0669-2, Published: July 11, 2012 under CC BY 3.0 license. © The Author(s).

Chapter 9

Recent Research on Jensen's Inequality for Oparators

By Jadranka Mićić and Josip Pečarić
DOI: 10.5772/48468

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Recent Research on Jensen's Inequality for Oparators

Jadranka Mićić1 and Josip Pečarić2

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 I be a real interval of any type. A continuous function f:I is said to be operator convex if

fλx+(1-λ)yλf(x)+(1-λ)f(y)
()

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

Let f be an operator convex function defined on an interval I. Ch. Davis [1] provedThere is small typo in the proof. Davis states that φ by Stinespring's theorem can be written on the form φ(x)=Pρ(x)P where ρ is a *-homomorphism to B(H) and P is a projection on H. In fact, H may be embedded in a Hilbert space K on which ρ and P acts. 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)
()

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

for operator convex functions f defined on an interval I=[0,α) (with α and f(0)0) and self-adjoint operators x1,,xn with spectra in I assuming that i=1nai*ai=1. The restriction on the interval and the requirement f(0)0 was 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×n matrices, holds for the much richer class of 2n×2n matrix 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))
()

for operator convex functions f defined on an interval I, where φi:B(H)B(K) are unital positive linear mappings, x1,,xn are self-adjoint operators with spectra in I and w1,,wn are 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˜
()

for convex functions 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, matrix non-decreasing in u, Uf[m,M], Vg[m,M], φ:HnHn˜ is a unital positive linear mapping and A is 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 T be a locally compact Hausdorff space and let 𝒜 be a C*-algebra of operators on some Hilbert space H. We say that a field (xt)tT of operators in 𝒜 is continuous if the function txt is norm continuous on T. If in addition μ is a Radon measure on T and the function txt is integrable, then we can form the Bochner integral Txtdμ(t), which is the unique element in 𝒜 such that

ϕTxtdμ(t)=Tϕ(xt)dμ(t)
()

for every linear functional ϕ in the norm dual 𝒜*.

Assume furthermore that there is a field (φt)tT of positive linear mappings φt:𝒜 from 𝒜 to another 𝒞*-algebra of operators on a Hilbert space K. We recall that a linear mapping φt:𝒜 is said to be a positive mapping if φt(xt)0 for 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)=k1K for some positive scalar k. Specially, if Tφt(1H)dμ(t)=1K, we say that a field (φt)tT is 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,ξ
()

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 x is self-adjoint, then |x|=(x2)1/2 and 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:I be an operator convex function defined on an interval I and let 𝒜 and be unital C*-algebras acting on a Hilbert space H and K respectively. If (φt)tT is a unital field of positive linear mappings φt:𝒜 defined on a locally compact Hausdorff space T with a bounded Radon measure μ, then the inequality

fTφt(xt)dμ(t)Tφt(f(xt))dμ(t)
()

holds for every bounded continuous field (xt)tT of self-adjoint elements in 𝒜 with 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 T with values in 𝒜 as 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 T be a locally compact Hausdorff space equipped with a bounded Radon measure μ. Let (xt)tT be a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜 with spectra in [m,M], m<M. Furthermore, let (φt)tT be a field of positive linear mappings φt:𝒜 from 𝒜 to another unital C*-algebra , such that the function tφt(1H) is integrable with Tφt(1H)dμ(t)=k1K for some positive scalar k. Let mx and Mx, mxMx, be the bounds of the self-adjoint operator x=Tφt(xt)dμ(t) and f:[m,M], g:[mx,Mx], F:U×V be functions such that (kf)[m,M]U, g[mx,Mx]V and F is bounded. If F is operator monotone in the first variable, then

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

holds for every operator convex function h1 on [m,M] such that h1f and for every operator concave function h2 on [m,M] such that h2f.

We prove only RHS of (). Let h2 be 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 φt and integrating, we obtain

Tφtf(xt)dμ(t)Tφth2(xt)dμ(t)
()

Furthermore, replacing φt by 1kφt in 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)1K
()

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, Mx and (φt)tT be as in Theorem . Let f:[m,M], g:[mx,Mx], F:U×V be functions such that (kf)[m,M]U, g[mx,Mx]V and F is bounded. If F is operator monotone in the first variable and f is 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)1K
()

In the dual case (when f is concave) the opposite inequalities hold in () with inf instead of sup.

We prove only the convex case. For convex f the 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-αv and k=1, we obtain the following generalization of Theorem 2.4[15].

Corollary 4 Let (xt)tT, mx, Mx be as in Theorem  and (φt)tT be 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)+C1K
()

where

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

If furthermore αg is 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 f is concave and αg is strictly concave differentiable) the opposite inequalities hold in () with min instead of max with 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)=t4 which is convex but not operator convex. He demonstrated that it is sufficient to put dim H=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)4 and 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.

I)
  • 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)4 and Φ1X14+Φ2X24 under the operator order. We observe that in the above case the following stands X=Φ1(X1)+Φ2(X2)=2000 and [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).

media/Figure1.jpg

Figure 1.

Spectral conditions for a convex function f

II)
  • 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)=12000 and [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)tT be a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜 defined on a locally compact Hausdorff space T equipped with a bounded Radon measure μ. Let mt and Mt, mtMt, be the bounds of xt, tT. Let (φt)tT be a unital field of positive linear mappings φt:𝒜 from 𝒜 to another unital C*-algebra . If

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

where mx and 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)
()

holds for every continuous convex function f:I provided that the interval I contains all mt,Mt.

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

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

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

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

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

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

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)
()

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 φt and summing, we obtain

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

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

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

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

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

Tφtf(xt)dμ(t)f(mx)1K+l(mx)Tφt(xt)dμ(t)-mx1K
()

Since mx1K=Tφt(xt)dμ(t), it follows

Tφtf(xt)dμ(t)f(mx)1K=fTφt(xt)dμ(t)
()

which is the desired inequality (). Putting φt(y)=aty for every y𝒜, where at0 is a real number, we obtain the following obvious corollary of Theorem .

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

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

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

fTatxtdμ(t)Tatf(xt)dμ(t)
()

holds for every continuous convex function f:I provided that the interval I contains 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)tT be a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜 defined on a locally compact Hausdorff space T equipped with a bounded Radon measure μ. Furthermore, let (φt)tT be a field of positive linear mappings φt:𝒜 from 𝒜 to another unital C*-algebra , such that the function tφt(1H) is integrable with Tφt(1H)dμ(t)=k1K for some positive scalar k. Let mt and Mt, mtMt, be the bounds of xt, tT, m=inftT{mt}, M=suptT{Mt}, and mx and 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×V are functions such that (kf)[m,M]U, g[mx,Mx]V, f is convex, F is 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)1K
()

In the dual case (when f is concave) the opposite inequalities hold in () by replacing inf and sup with sup and 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-αv or F(u,v)=v-1/2uv-1/2 in Theorem , we obtain the next corollary.

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

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

and f is 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)1K
()

If additionally g>0 on [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)
()

In the dual case (when f is concave) the opposite inequalities hold in () by replacing min and max with max and min, respectively. If additionally g>0 on [mx,Mx], then the opposite inequalities also hold in () by replacing min and max with max and 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 f be a convex function on an interval I, m,MI and 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)
()

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

Lemma 11 Let x be a bounded self-adjoint elements in a unital C*-algebra 𝒜 of operators on some Hilbert space H. If the spectrum of x is 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˜)
()

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+M2
()

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-m and p2=z-mM-m we obtain

f(z)M-zM-mf(m)+z-mM-mf(M)-12-1M-mz-m+M2f(m)+f(M)-2fm+M2
()

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)I and fg on I implies f(x)g(x); and h(z)=|z| implies h(x)=|x|. Then by using () we obtain the desired inequality ().

Theorem 12 Let (xt)tT be a bounded continuous field of self-adjoint elements in a unital C*-algebra 𝒜 defined on a locally compact Hausdorff space T equipped with a bounded Radon measure μ. Let mt and Mt, mtMt, be the bounds of xt, tT. Let (φt)tT be a unital field of positive linear mappings φt:𝒜 from 𝒜 to another unital C*-algebra . Let

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

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

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

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

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

(resp.

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

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¯21K
()

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¯1K and f is 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˜
()

where δf and x˜ are defined by ().

But since f is 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˜
()

since Tφt(1H)dμ(t)=1K. Combining the two inequalities () and (), we have LHS of (). Since δf0 and x˜0, then we have RHS of ().

If m<M and 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.

media/Figure1-2.jpg

Figure 2.

Refinement for two operators and a convex function f

We put f(t)=t4 which 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=-3I3 and X2=2I3, then X=Φ1(X1)+Φ2(X2)=-0.5I2, so m=-3, M=2. We also put m¯=-3 and 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=-1/2 and 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¯=2 we obtain δf=135/8, X˜=I2/2 which 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=1.3446 and 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=7/8 and δfX˜=0.43751001, which is worse than the above improvement.

Putting Φt(y)=aty for every y𝒜, where at0 is a real number, we obtain the following obvious corollary of Theorem .

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

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

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

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

If f:I is a continuous convex (resp. concave) function provided that the interval I contains 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 δf is defined by (), x˜˜=121H-1M¯-m¯Tatxtdμ(t)-m¯+M¯21H and 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 Ai for each i=1,...,n, i.e. when (mA,MA)[mi,Mi]= for i=1,...,n, where mA and MA, mAMA, are the bounds of A, and mi and 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 mi and 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))
()

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

We prove only the case when f is 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=B or B=C or A=C implies A=B=C.

a) Let m<M. Since f is 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,...,n1
()

but since f is 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,...,n
()

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

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

Applying a positive linear mapping Φi and 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)
()

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)
()

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

1αi=1n1Φif(Ai)1βi=n1+1nΦif(Ai)
()

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)
()

which gives the desired double inequality ().

b) Let m=M. Since [mi,Mi][m,M] for i=1,...,n1, then Ai=m1H and f(Ai)=f(m)1H for i=1,...,n1. It follows

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

On the other hand, since f is convex on I, we have

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

where l is the subdifferential of f. Replacing z by Ai for i=n1+1,...,n, applying Φi and 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=γ and 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))
()

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,...,n and i=n1+1n1βΦi(1H)=1K hold. So we can apply Theorem  on operators An1+1,...,An and mappings 1βΦi and obtain the desired inequality.

2. We denote by mC and MC the bounds of C=i=1nΦi(Ai). If (mC,MC)[mi,Mi]=, i=1,...,n1 or f is 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 18 If neither assumptions (mC,MC)[mi,Mi]=, i=1,...,n1, nor f is 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=2 we 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)=t4 is 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 mi and 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)
()

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

If f:I is 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 mi and 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,...,n
()

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

fi=1nαiAii=1nαif(Ai)
()

holds for every continuous convex function f:I provided that the interval I contains 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=1nαiAi and Bi=Ai-1, i=2,...,n+1. Then mB1=mA, MB1=MA are the bounds of B1 and mBi=mi-1, MBi=Mi-1 are the ones of Bi, i=2,...,n+1. Also, we define (n+1)-tuple of non-negative numbers (p1,...,pn+1) by p1=1 and pi=αi-1, i=2,...,n+1. Then i=1n+1pi=2 and by using () we have

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

Since

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

then

p1B1=12i=1n+1piBi=i=2n+1piBi
()

Taking into account () and (), we can apply Corollary  for n1=1 and Bi, pi as 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 mi and 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>0 and 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))
()

holds for every continuous convex function f:I provided that the interval I contains 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¯21H
()

and   m¯[m,mL], M¯[MR,M], m¯<M¯,   are arbitrary numbers. If f:I is 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=B or B=C or A=C implies A=B=C.

Since f is 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¯2 and A˜i=121H-1M¯-m¯Ai-m¯+M¯21H. Applying a positive linear mapping Φi and 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˜
()

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

Additionally, since f is 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,...,n
()

It follows

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

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

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

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˜
()

Adding βδfA˜ in the above inequalities, we get

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

Now, we remark that δf0 and A˜0. (Indeed, since f is 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=2 and the bounds of matrices as in Fig. 3.

media/Figure1-3.jpg

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)
()

holds, where δf=M¯4+m¯4-(M¯+m¯)48 and

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)=I2 and 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=M3 and 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<Bf
()

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))
()

and

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

holds for every γ1,γ2 in the close interval joining α and β, where δf and 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))
()

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 Ai putting 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].

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