# Multiple-Term Refinements of Young Type Inequalities

Authors: Daeshik Choi
Journal: Journal of Mathematics
 (2016)

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## Abstract

Recently, a multiple-term refinement of Young’s inequality has been proved. In this paper, we show its reverse refinement. Moreover, we will present multiple-term refinements of Young’s inequality involving Kantorovich constants. Finally, we will apply scalar inequalities to operators.


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## Body

## 1. Introduction

The classical Young inequality states that ifa,b>0 and 0≤v≤1, then(1)1-va+vb≥a1-vbv.For0≤v≤1, we define three functions r0(v), r1(v), and R0(v) by (2)r0v=min⁡v,1-v,r1v=min⁡2r0v,1-2r0v,R0v=1-r0v.In [1] and the references there, the following improvements of Young inequality and its reverse are discussed:(3)1-va+vb≥a1-vbv+r0va-b2,1-va+vb≤a1-vbv+1-r0va-b2,(4)1-va+vb≥a1-vbv+r0va-b2+r1va-ab42χ0,1/2v+b-ab42χ1/2,1v,1-va+vb≤a1-vbv+R0va-b2-r1vb-ab42χ0,1/2v+a-ab42χ1/2,1v,where χI is the characteristic function defined by(5)χIx=1,ifx∈I0,ifx∉I.Another form of Young type inequalities discussed in [1] is as follows:(6)1-va+vb2≥a1-vbv2+r02va-b2+r1va-ab2χ0,1/2v+b-ab2χ1/2,1v,1-va+vb2≤a1-vbv2+R02va-b2-r1vb-ab2χ0,1/2v+a-ab2χ1/2,1v.Other types of improvements of the Young inequality is to use Kantorovich constants. Wu and Zhao [2] showed(7)1-va+vb≥K1a,br1va1-vbv+r0va-b2,1-va+vb≤K1a,b-r1va1-vbv+R0va-b2,where(8)K1a,b=a+b24ab.Note that (7) improves (3), since K1(a,b)≥1 for all a,b>0. In [3], Liao and Wu improved (4) as follows:(9)1-va+vb≥K2a,br2va1-vbv+r0va-b2+r1va-ab2χ0,1/2v+b-ab2χ1/2,1v,1-va+vb≤K2a,b-r2va1-vbv+R0va-b2-r1vb-ab2χ0,1/2v+a-ab2χ1/2,1v,where(10)K2a,b=a4+b424ab4,r2v=min⁡2r1v,1-2r1v.The constants of the form m+M2/4mM are called Kantorovich constants.Throughout the paper, we will use the following functions.Definition 1.
One defines the sequencernv of functions on [0,1] as follows: (11)r0v=min⁡v,1-v,rnv=min⁡2rn-1v,1-2rn-1v,
forn∈N.Definition 2.
Forl,k∈N and a,b>0, we define the functions gl,k(a,b) by (12)gl,ka,b=a1-k-1/2lbk-1/2l-a1-k/2lbk/2l2=a1-k-1/2lbk-1/2l+a1-k/2lbk/2l-2a1-2k-1/2l+1b2k-1/2l+1.
The following multiple-term refinement of Young inequality has been recently proved in [1]. In the next section, we will present a different and simpler proof of it.Theorem 3.
Leta,b>0 and N be any positive integer. Then(13)1-va+vb≥a1-vbv+r0va-b2+∑l=1N-1rlv∑k=12lgl,ka,bχk-1/2l,k/2lv,for 0≤v≤1.The above is a simplified expression of the original one in [1] which is written in rather complicated notation, but they are essentially identical. Note that the first inequalities of (3) and (4) are obtained from (13) with N=1 and 2, respectively.The object of this paper is to show(1)
a reverse of (13) which generalizes the second inequalities of (3) and (4),(2)
multiple-term refinements of Young inequality involving Kantorovich constants which generalize (7) and (9),(3)
operator inequalities related to Young inequality.


## 2. Multiple-Term Reverse of Young Inequality

From now on, we will fixa,b>0 and use the following functions: (14)μv=1-va+vb-a1-vbv,αNv=∑l=0N-1rlv∑k=12lgl,ka,bχk-1/2l,k/2lv,βNv=a-b2-∑l=0N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv,for v∈[0,1] and an integer N>0. As we will see (Lemma 4), rl(k/2l)=0 for any integer k with 0≤k≤2l. Thus the interval of the characteristic function in αN(v) or βN(v) can include boundary points. For example, χ((k-1)/2l,k/2l) may be replaced by χ((k-1)/2l,k/2l] or χ[(k-1)/2l,k/2l].We can expressr0(v) and r1(v) as multipart functions as follows:(15)r0v=v,0≤v≤121-v,12<v≤1,r1v=2v,0≤v≤141-2v,14<v≤122v-1,12<v≤342-2v,34<v≤1.For any l≥0, we can formulate rl(v) explicitly.Lemma 4.
Letl≥0 and 1≤k≤2l be integers. If (k-1)/2l≤v≤k/2l, then(16)rlv=2lv-k+1,k-12l≤v≤2k-12l+1k-2lv,2k-12l+1<v≤k2l.Proof.
We prove it by induction onl. The case l=0 is obvious. Assume that (k-1)/2l+1≤v≤k-1/2l+1+1/2l+2. If k=2m-1 is odd, then (m-1)/2l≤v≤(m-1)/2l+1/2l+2<(2m-1)/2l+1 and rl(v)=2lv-m+1 by induction. Since v≤(2k-1)/2l+2, one has(17)rlv=2lv-k-12≤2k-14-k-12=14,rl+1v=min⁡2rlv,1-2rlv=2rlv=2l+1v-k+1.If k=2m is even, then (2m-1)/2l+1≤v≤m/2l-1/2l+2<m/2l and rl(v)=m-2lv by induction. Since v≤(2k-1)/2l+2, we have(18)rlv=k2-2lv≥k2-2k-14=14,rl+1v=1-2rlv=2l+1v-k+1.Using the same argument, we can show that if (k-1)/2l+1+1/2l+2<v≤k/2l+1, then rl+1(v)=k-2l+1v. We omit the detailed proof.Lemma 5.
For a positive integerN, αN(v) is the linear interpolation of μ(v) at v=k/2N for k=0,1,…,2N.Proof.
Sincerl(v) is a line segment on each interval (k-1)/2l+1,k/2l+1 for 1≤k≤2l+1, αN(v) is a line segment on (k-1)/2N,k/2N for 1≤k≤2N. Thus it suffices to show(19)αNk2N=μk2N,1≤k<2N.Note that since αN(v)=μ(v)=0 at v=0,1, (19) holds for k=0,2N. We will prove (19) by induction on N. Since α1(v)=r0(v)a-b2, one has(20)α112=12a-b2=μ12.Assume that (19) holds and 1≤k<2N+1. If k=2m is even, then (21)αN+1k2N+1=αN+1m2N=αNm2N+rNm2N∑k=12NgN,ka,bχk-1/2N,k/2Nm2N=αNm2N,sincerNm2N=0=μk2N+1by  induction.If k=2m-1 is odd, then(22)αN+1k2N+1=αNm2N-12N+1+rNm2N-12N+1gN,ma,b.Since αN(v) is the line segment joining ((m-1)/2N,μ((m-1)/2N)) and (m/2N,μ(m/2N)) on (m-1)/2N,m/2N, we have(23)αNm2N-12N+1=12μm-12N+12μm2N.By Lemma 4, rN(m/2N-1/2N+1)=1/2. Noting that a1-vbv=(1-v)a+vb-μ(v), we can write gN,m(a,b) by (24)gN,ma,b=a1-m-1/2Nbm-1/2N+a1-m/2Nbm/2N-2a1-2m-1/2N+1b2m-1/2N+1=-μm-12N-μm2N+2μ2m-12N+1.Thus, from (22), we deduce that(25)αN+1k2N+1=μ2m-12N+1=μk2N+1.Remark 6.
Sinceμ(v) is concave, μ(v)≥αN(v) by Lemma 5, which proves Theorem 3 in a much simpler way, where the original proof is done by mathematical induction on N.Lemma 7.
Letλ(v) be the reflection of μ(v) about the point (1/2,μ(1/2)); that is, (26)λv=2μ12-μ1-v,=a-b2-va-1-vb+avb1-v,=1-va+vb+avb1-v-2ab.Then each of the following is true. (1)
λ(v)≥μ(v) for all 0≤v≤1.(2)
βN(v) is the linear interpolation of λ(v) at v=k/2N at k=0,…,2N.Proof.
Letf(v)=λ(v)-μ(v). Since(27)f′v=avb1-v1-ab1-2vln⁡ab,one derives that f′(v)<0 for v∈(0,1/2), f′(v)>0 for v∈(1/2,1), and f′(1/2)=0. Thus we have λ(v)-μ(v)≥λ(1/2)-μ(1/2)=0 for all 0≤v≤1. So part (1) is proved.
For the second part, it suffices to show thatβN(v) is the reflection of αN(v) about the point (1/2,αN(1/2)); that is,(28)2αN12-αN1-v=βNv.Noting that rl(v)=rl(1-v) and gl,2l-k+1(a,b)=gl,k(b,a), one gets that (29)2αN12-αN1-v=2μ12-∑l=0N-1rlv∑k=12lgl,ka,bχk-1/2l,k/2l1-v=a-b2-∑l=0N-1rlv∑k=12lgl,ka,bχ1-k/2l,1-k-1/2lv=a-b2-∑l=0N-1rlv∑k=12lgl,2l-k+1a,bχk-1/2l,k/2lv=βNv.Theorem 8.
For any integerN≥1 and 0≤v≤1, one has(30)αNv≤μv≤βNv.That is, (31)1-va+vb≥a1-vbv+∑l=0N-1rlv∑k=12lgl,ka,bχk-1/2l,k/2lv=a1-vbv+r0va-b2+∑l=1N-1rlv∑k=12lgl,ka,bχk-1/2l,k/2lv,1-va+vb≤a1-vbv+a-b2-∑l=0N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv=a1-vbv+R0va-b2-∑l=1N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv.Proof.
As mentioned in Remark6, αN(v)≤μ(v) follows from the concavity of μ(v). Since μ(v)≤λ(v) and βN(v) is the linear interpolation of λ(v) which is convex on [0,1], μ(v)≤λ(v)≤βN(v).Note that the second inequalities of (3) and (4) can be obtained from μ(v)≤βN(v) with N=1 and 2, respectively. Moreover, (6) follows from the following result with N=2.Corollary 9.
For any integerN≥1 and 0≤v≤1, one has(32)1-va+vb2≥a1-vbv2+r02va-b2+∑l=1N-1rlv∑k=12lgl,ka2,b2χk-1/2l,k/2lv,1-va+vb2≤a1-vbv2+R02va-b2-∑l=1N-1rlv∑k=12lgl,kb2,a2χk-1/2l,k/2lv.Proof.
Replacinga and b by their squares in Theorem 8, we obtain that (33)1-va2+vb2≥a1-vbv2+r0va-b2+∑l=1N-1rlv∑k=12lgl,ka2,b2χk-1/2l,k/2lv,1-va2+vb2≤a1-vbv2+R0va-b2-∑l=1N-1rlv∑k=12lgl,kb2,a2χk-1/2l,k/2lv.Since r0(v)-r02(v)=R0(v)-R02(v)=v(1-v) for all v, we derive from the above that (34)1-va2+vb2≥v1-va-b2+a1-vbv2+r02va-b2+∑l=1N-1rlv∑k=12lgl,ka2,b2χk-1/2l,k/2lv,1-va2+vb2≤v1-va-b2+a1-vbv2+R02va-b2-∑l=1N-1rlv∑k=12lgl,kb2,a2χk-1/2l,k/2lv.Hence (32) follows from the identity(35)1-va+vb2=1-va2+vb2-v1-va-b2.


## 3. Young Inequalities Involving Kantorovich Constants

In this section, we will discuss multiple-term improvements of Young inequality involving Kantorovich constants. For a nonnegative integern, we define Kn(a,b) by(36)Kna,b=a1/2n+b1/2n24ab1/2n.Lemma 10.
For0≤v≤1, one has(37)1-va+vb≥K0a,br0va1-vbv.Proof.
Replacinga-1b by t, the inequality is equivalent to(38)1-v+vt≥K01,tr0vtv,for t>0. Taking the natural logarithm, it suffices to show that(39)ft≡ln⁡1-v+vt-r0vln⁡K01,t-vln⁡t≥0.A direct computation shows that(40)t+11-v+vtf′t=1-t-1v1-vt+1-r0v1-v+vt=v1-2vt-1,0≤v≤121-v2v-11-t-1,12<v≤1.Thus f(t)≥f(1)=0 for any t>0.Lemma 11.
For a positive integerN and 0≤v≤1, define sN(t) by(41)sNt=1-v+vt-∑l=0N-1rlv∑k=12ltk-1/2l-tk/2l2χk-1/2l,k/2lv,for t>0. Then one has(42)sNt=∑k=12Nk-2Nvtk-1/2N+2Nv-k+1tk/2Nχk-1/2N,k/2Nv.Proof.
As mentioned in the previous section, sincerl(k/2l)=0 for 0≤k≤2l, χ((k-1)/2l,k/2l) in the definition of sN(t) may be replaced by χ(k-1)/2l,k/2l. We prove (42) by induction on N. We have (43)s1t=1-v+vt-r0v1-t2=1-2v+2vtχ0,1/2+2-2vt+2v-1tχ1/2,1.Suppose that (42) holds. Then (44)sN+1t=sNt-rNv∑k=12Ntk-1/2N-tk/2N2χk-1/2N,k/2Nv=∑k=12Nlk-2Nvtk-1/2N+2Nv-k+1tk/2N-rNvtk-1/2N-tk/2N2χk-1/2N,k/2Nv=∑k=12N2k-2N+1v-1tk-1/2N+2N+1v-2k+2t2k-1/2N+1χk-1/2N,2k-1/2N+1v+∑k=12N2k-2N+1vt2k-1/2N+1+2N+1v-2k+1tk/2Nχ2k-1/2N+1,k/2Nv.Replacing 2k-1 in the first summation and 2k in the second summation, respectively, by m, one has (45)sN+1t=∑m=1m,odd2N+1-1m-2N+1vtm-1/2N+1+2N+1v-m+1tm/2N+1χm-1/2N+1,m/2N+1v+∑m=2m,even2N+1m-2N+1vtm-1/2N+1+2N+1v-m+1tm/2N+1χm-1/2N+1,m/2N+1v=∑k=12N+1k-2N+1vtk-1/2N+1+2N+1v-k+1tk/2N+1χk-1/2N+1,k/2N+1v.Thus (42) holds for all positive integers N.The following shows a multiple-term refinement of Young inequality involving Kantorovich constants.Theorem 12.
Fora,b>0, 0≤v≤1, and N∈N, one has(46)1-va+vb≥KNa,brNva1-vbv+∑l=0N-1rlv∑k=12lgl,ka,bχk-1/2l,k/2lv.Proof.
Puttingt=a-1b, (46) can be rewritten as(47)1-v+vt≥KN1,trNvtv+∑l=0N-1rlv∑k=12ltk-1/2l+tk/2l-2t2k-1/2l+1χk-1/2l,k/2lv.By Lemma 11, the above can be expressed by(48)∑k=12Nk-2Nvtk-1/2N+2Nv-k+1tk/2Nχk-1/2N,k/2Nv≥KN1,trNvtv.Thus it suffices to show that if (k-1)/2N<v≤k/2N, then(49)k-2Nvtk-1/2N+2Nv-k+1tk/2N≥KN1,trNvtv,for t>0. Replacing t by t2N and letting μ=2Nv, the above is equivalent to(50)k-μtk-1+μ-k+1tk≥K01,trNvtμ.Let s=min⁡{k-μ,μ-k+1}. By Lemma 10, one gets that (51)k-μtk-1+μ-k+1tk≥K0tk-1,tkstk-μk-1+μ-k+1k=K01,tstμ.Since(52)rNv=2Nv-k+1,k-12N≤v≤2k-12N+1k-2Nv,2k-12N+1<v≤k2Nby Lemma 4, it follows that if (k-1)/2N<v≤k/2N, then s=min⁡{k-2Nv,2Nv-k+1}=rN(v) and therefore (50) holds.Note that (46) can be written as(53)1-va+vb≥KNa,brNva1-vbv+r0va-b2+∑l=1N-1rl∑k=12lgl,ka,bχk-1/2l,k/2lv,which gives the first inequalities of (7) and (9) with N=1 and 2, respectively.Now we consider a reverse inequality corresponding to Theorem12. A given inequality of the form (1-v)a+vb≥ξ(v,a,b) can be utilized to derive its reverse in many cases. For example, replacing v by 1-v in(54)1-va+vb≥a1-vbv+r0va-b2,which is the first inequality of (3), we obtain(55)1-va+vb≤a+b-avb1-v-r0va-b2=2ab-avb1-v+R0va-b2.Since 2ab≤avb1-v+a1-vbv, the above implies the second inequality of (3). Similarly, replacing v by 1-v in (56)1-va+vb≥a1-vbv+r0va-b2+r1va-ab42χ0,1/2v+b-ab42χ1/2,1v,which is the first inequality of (4), we get (57)1-va+vb≤a+b-avb1-v-r0va-b2-r1vb-ab42χ0,1/2v+a-ab42χ1/2,1v=2ab-avb1-v+R0va-b2-r1va-ab42χ1/2,1v+b-ab42χ0,1/2v.Since 2ab≤avb1-v+a1-vbv, the above implies the second inequality of (4). In the same way, the first inequality in Theorem 8 can be used to derive (58)1-va+vb≤2ab-avb1-v+R0va-b2-∑l=1N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv,which is stronger than the second inequality in the theorem. Based on such an observation, we can show a reverse inequality corresponding to Theorem 12 as follows.Theorem 13.
Fora,b>0, 0≤v≤1, and N∈N, one has (59)1-va+vb≤2ab-KNa,brNvavb1-v+R0va-b2-∑l=1N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv≤KNa,b-rNva1-vbv+R0va-b2-∑l=1N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv.Proof.
Replacingv by 1-v and k by 2l-k+1 in (46), we have (60)1-va+vb≤a+b-KNa,brNvavb1-v-∑l=0N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv=a+b-KNa,brNvavb1-v-r0va-b2-∑l=1N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv=2ab-KNa,brNvavb1-v+R0va-b2-∑l=1N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv≤KNa,b-rNva1-vbv+R0va-b2-∑l=1N-1rlv∑k=12lgl,kb,aχk-1/2l,k/2lv,where the last inequality results from(61)2ab≤KNa,brNvavb1-v+KNa,b-rNva1-vbv.Note that the second inequalities of (7) and (9) follow from the above theorem with N=1 and 2, respectively.


## 4. Operator Inequalities

From now on, we use uppercase letters for invertible positive operators on a Hilbert space and lowercase letters for real numbers. The following notations will be used:(i)
A≥B (A>B) denotes that A-B is a positive (invertible positive) operator.(ii)
A≥0(A>0) denotes that A is a positive (invertible positive) operator.For A,B>0 and 0≤v≤1, the v-arithmetic and v-geometric means of A and B are defined, respectively, by(62)A∇vB=1-vA+vB,A♯vB=A1/2A-1/2BA-1/2vA1/2.In the case v=1/2, we will omit the v-value in them. For example, A∇B denotes A∇1/2B.The operator version of (1) is well known as follows:(63)A∇vB≥A♯vB,for A and B positive invertible operators and 0≤v≤1 (see [4, 5] for more matrix Young inequalities). To show operator inequalities corresponding to their scalar versions, we will use the operator monotonicity of continuous functions; that is, if f is a real valued continuous function defined on the spectrum of a self-adjoint operator A, then f(t)≥0 for every t in the spectrum of A implies that f(A) is a positive operator.From now on,gl,k(A,B) will denote the operator version of gl,k(a,b) defined in Definition 2. That is, (64)gl,kA,B=A♯k-1/2lB+A♯k/2lB-2A♯2k-1/2l+1B,for A,B>0.Theorem 14.
LetA,B>0 and 0≤v≤1. Then(65)A∇vB≥A♯vB+2r0vA∇B-A♯B+∑l=1N-1rlv∑k=12lgl,kA,Bχk-1/2l,k/2lv,A∇vB≤A♯vB+2R0vA∇B-A♯B-∑l=1N-1rlv∑k=12lgl,kB,Aχk-1/2l,k/2lv.Proof.
For anyx>0, we have (66)1-v+vx≥xv+r0v1-x2+∑l=1N-1rlv∑k=12lgl,k1,xχk-1/2l,k/2lv,1-v+vx≤xv+R0v1-x2-∑l=1N-1rlv∑k=12lgl,kx,1χk-1/2l,k/2lv,by Theorem 8. Thus, for any positive operator X, we have(67)1-vI+vX≥Xv+r0vI+X-2X+∑l=1N-1rlv∑k=12lgl,kI,Xχk-1/2l,k/2lv,1-vI+vX≤Xv+R0vI+X-2X-∑l=1N-1rlv∑k=12lgl,kX,Iχk-1/2l,k/2lv,by the operator monotonicity of continuous functions, where I is the identity operator. Note that since A♯vB=B♯1-vA, we can express gl,k(I,X) and gl,k(X,I) by (68)gl,kI,X=I♯k-1/2lX+I♯k/2lX-2I♯2k-1/2l+1X=Xk-1/2l+Xk/2l-2X2k-1/2l+1,gl,kX,I=X♯k-1/2lI+X♯k/2lI-2X♯2k-1/2l+1I=I♯1-k-1/2lX+I♯1-k/2lX-2I♯1-2k-1/2l+1X=X1-k-1/2l+X1-k/2l-2X1-2k-1/2l+1.Letting X=A-1/2BA-1/2 and then multiplying all terms by A on both sides, (67) yields (65), where gl,k(B,A) can be obtained as follows: (69)A1/2gl,kA-1/2BA-1/2,IA1/2=A♯1-k-1/2lB+A♯1-k/2lB-2A♯1-2k-1/2l+1B=B♯k-1/2lA+B♯k/2lA-2B♯2k-1/2l+1A=gl,kB,A.The following shows matrix inequalities corresponding to Corollary9.Theorem 15.
LetA,B,X be n×n complex matrices. If A and B are positive semidefinite, then (70)1-vAX+vXB22≥A1-vXBv22+r02vAX-XB22+∑l=1N-1rlv∑k=12lhl,kA,B;Xχk-1/2l,k/2lv,1-vAX+vXB22≤A1-vXBv22+R02vAX-XB22-∑l=1N-1rlv∑k=12lhl,2l-k+1A,B;Xχk-1/2l,k/2lv,where ·2 denotes the Frobenius norm and (71)hl,kA,B;X=A1-k-1/2lXBk-1/2l-A1-k/2lXBk/2l22.Proof.
SinceA and B are positive semidefinite, there exist unitary matrices U and V such that A=Udiag⁡(a1,…,an)U∗ and B=Vdiag⁡(b1,…,bn)V∗, where diag⁡(c1,…,cn) denotes the n×n diagonal matrix whose ith diagonal entry is ci. Letting Y=U∗XV=yijn, we can show the following directly: (72)AX-XB=Uai-bjyijV∗,1-vAX+vXB=U1-vai+vbjyijV∗,A1-vXBv=Uai1-vbjvyijV∗,A1-k-1/2lXBk-1/2l-A1-k/2lXBk/2l=Uai1-k-1/2lbjk-1/2l-ai1-k/2lbjk/2lyijV∗.Since ·2 is unitarily invariant, one obtains that (73)1-vAX+vXB22-A1-vXBv22=∑i,j=1n1-vai+vbjyij2-∑i,j=1nai1-vbjvyij2=∑i,j=1n1-vai+vbj2-ai1-vbjv2yij2.Thus, by (32), we have(74)1-vAX+vXB22≥A1-vXBv22+r02vAX-XB22+∑l=1N-1rlv∑k=12l∑i,j=1ngl,kai2,bj2yij2χk-1/2l,k/2lv,1-vAX+vXB22≤A1-vXBv22+R02vAX-XB22-∑l=1N-1rlv∑k=12l∑i,j=1ngl,kbj2,ai2yij2χk-1/2l,k/2lv.Since(75)gl,kai2,bj2=ai1-k-1/2lbjk-1/2l-ai1-k/2lbjk/2l2,gl,kbj2,ai2=aik-1/2lbj1-k-1/2l-aik/2lbj1-k/2l2,we have(76)∑i,j=1ngl,kai2,bj2yij2=hl,kA,B;X,∑i,j=1ngl,kbj2,ai2yij2=hl,2l-k+1A,B;X.Next we consider operator inequalities related to Kantorovich constants.Theorem 16.
LetA,B>0 and h be a real number with h≥1. If either hA≤B or hB≤A, then (77)A∇vB≥KN1,hrNvA♯vB+r0vA+B-2A♯B+∑l=1N-1rlv∑k=12lA♯k-1/2lB+A♯k/2lB-2A♯2k-1/2l+1Bχk-1/2l,k/2lv,A∇vB≤2A♯B-KN1,hrNvB♯vA+R0vA+B-2A♯B-∑l=1N-1rlv∑k=12lB♯k-1/2lA+B♯k/2lA+B♯2k-1/2l+1Aχχk-1/2l,k/2lv≤KN1,h-rNvA♯vB+R0vA+B-2A♯B-∑l=1N-1rlv∑k=12lB♯k-1/2lA+B♯k/2lA+B♯2k-1/2l+1Aχχk-1/2l,k/2lv.Proof.
Lettingt=a-1b in Theorem 12, we have(78)1-v+vt≥KN1,trNvtv+∑l=0N-1rlv∑k=12lgl,k1,tχk-1/2l,k/2lv,for any t>0. Since KN(1,t) is an increasing function on (1,∞) and KN(1,t-1)=KN(1,t) for t>0, KN(1,h)≤KN(1,t) for t≥h or t≤1/h. Thus, from (78), we have(79)1-v+vt≥KN1,hrNvtv+∑l=0N-1rlv∑k=12lgl,k1,tχk-1/2l,k/2lv,for t≥h or t≤1/h. Replacing t by X=A-1/2BA-1/2 which satisfies hI≤X or X≤1/hI, we get that(80)1-vI+vX≥KN1,hrNvXv+∑l=0N-1rlv∑k=12lgl,kI,Xχk-1/2l,k/2lv.Multiplying the inequality by A1/2 on both sides, we have (81)A∇vB≥KN1,hrNvA♯vB+∑l=0N-1rlv∑k=12lA♯k-1/2lB+A♯k/2lB-2A♯2k-1/2l+1Bχk-1/2l,k/2lv.Since the zeroth term (l=0) in the summation is r0(v)(A+B-2A♯B), we finish the proof of the first inequality of this theorem.
We can prove the second one in the same way. Lettingt=a-1b in Theorem 13, we have(82)1-v+vt≤2t-KN1,trNvt1-v+R0vt+1-2t-∑l=1N-1rlv∑k=12lgl,kt,1χk-1/2l,k/2lv≤KN1,t-rNvtv+R0vt+1-2t-∑l=1N-1rlv∑k=12lgl,kt,1χk-1/2l,k/2lv,for any t>0. In particular, if t≥h or t≤1/h, then (83)1-v+vt≤2t-KN1,hrNvt1-v+R0vt+1-2t-∑l=1N-1rlv∑k=12lgl,kt,1χk-1/2l,k/2lv≤KN1,h-rNvtv+R0vt+1-2t-∑l=1N-1rlv∑k=12lgl,kt,1χk-1/2l,k/2lv.Replacing t by X=A-1/2BA-1/2, one has (84)1-vI+vX≤2X-KN1,hrNvX1-v+R0vX+I-2X-∑l=1N-1rlv∑k=12lgl,kX,Iχk-1/2l,k/2lv≤KN1,h-rNvXv+R0vX+I-2X-∑l=1N-1rlv∑k=12lgl,kX,Iχk-1/2l,k/2lv.Finally, multiplying each term by A1/2 on both sides, we get that (85)A∇vB≤2A♯B-KN1,hrNvA♯1-vB+R0vA+B-2A♯B-∑l=1N-1rlv∑k=12lA♯1-k-1/2lB+A♯1-k/2lB+A♯1-2k-1/2l+1Bχk-1/2l,k/2lv≤KN1,h-rNvA♯vB+R0vA+B-2A♯B-∑l=1N-1rlv∑k=12lA♯1-k-1/2lB+A♯1-k/2lB+A♯1-2k-1/2l+1Bχk-1/2l,k/2lv.Since A♯1-μB=B♯μA, the above shows the second inequality of this theorem.In the following, we consider special values ofv.Corollary 17.
LetA,B>0 and 0≤v≤1. If v=k0/2N for some 1≤k0<2N, then(86)A∇vB≥A♯vB+∑l=0N-1rlv∑k=12lA♯k-1/2lB+A♯k/2lB-2A♯2k-1/2l+1Bχk-1/2l,k/2lv,A∇vB≤2A♯B-B♯vA+R0vA+B-2A♯B-∑l=1N-1rlv∑k=12lB♯k-1/2lA+B♯k/2lA+B♯2k-1/2l+1Aχk-1/2l,k/2lv≤A♯vB+R0vA+B-2A♯B-∑l=1N-1rlv∑k=12lB♯k-1/2lA+B♯k/2lA+B♯2k-1/2l+1Aχk-1/2l,k/2lv.If v=(2k0-1)/2N+1 for some 1≤k0≤2N, then (87)A∇vB≥14A♯v+2-NB+A♯v-2-NB+2A♯vB+∑l=0N-1rlv∑k=12lA♯k-1/2lB+A♯k/2lB-2A♯2k-1/2l+1Bχk-1/2l,k/2lv,A∇vB≤2A♯B-14B♯v+2-NA+B♯v-2-NA+2B♯vA+R0vA+B-2A♯B-∑l=1N-1rlv∑k=12lB♯k-1/2lA+B♯k/2lA+B♯2k-1/2l+1Aχk-1/2l,k/2lv.Proof.
Ifv=k0/2N, then rN(v)=0. Thus from (78) and (82), we have (88)1-v+vt≥tv+∑l=0N-1rl∑k=12lgl,k1,tχk-1/2l,k/2lv,1-v+vt≤2t-t1-v+R0vt+1-2t-∑l=1N-1rlv∑k=12lgl,kt,1χk-1/2l,k/2lv≤tv+R0vt+1-2t-∑l=1N-1rlv∑k=12lgl,kt,1χk-1/2l,k/2lv.Using the same argument as in the proof of Theorem 16, it is easy to derive the desired operator inequalities from the above.
Meanwhile, ifv=(2k0-1)/2N+1, then rN(v)=1. Thus, from (78) and the first inequality of (82), we have (89)1-v+vt≥14tv+1/2N+tv-1/2N+2tv+∑l=0N-1rlv∑k=12lgl,k1,tχk-1/2l,k/2lv,1-v+vt≤2t-14t1-v+1/2N+t1-v-1/2N+2t1-v+R0vt+1-2t-∑l=1N-1rlv∑k=12lgl,kt,1χk-1/2l,k/2lv.Letting t=X=A-1/2BA-1/2, we can obtain the desired operator inequalities.



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*Source: 4346712-2016-11-16.xml*