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Research article

Some new Young type inequalities

  • Received: 23 December 2023 Revised: 21 January 2024 Accepted: 01 February 2024 Published: 21 February 2024
  • MSC : 15A45, 47A30

  • In this paper, we gave some generalized Young type inequalities due to Zuo and Li [J. Math. Inequal., 16 (2022), 1169-1178], and we also presented a new Young type inequality. As applications, we obtained some operator inequalities and matrix versions inequalities including the Hilbert-Schmidt norm and trace.

    Citation: Yonghui Ren. Some new Young type inequalities[J]. AIMS Mathematics, 2024, 9(3): 7414-7425. doi: 10.3934/math.2024359

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  • In this paper, we gave some generalized Young type inequalities due to Zuo and Li [J. Math. Inequal., 16 (2022), 1169-1178], and we also presented a new Young type inequality. As applications, we obtained some operator inequalities and matrix versions inequalities including the Hilbert-Schmidt norm and trace.



    Let (H,,) be a complex Hilbert space and let B(H) denote the algebra of all bounded linear operators acting on H. A self-adjoint operator A is said to be positive if Ax,x0 for all xH, while it is said to be strictly positive if A is positive and invertible. As usual, we say that A>B when AB>0 and AB when AB0, respectively. Moreover, Mn denotes the sets of all n×n complex matrices. The unitarily invariance of the |||||| on Mn means that |||UAV|||=|||A||| for any AMn and all unitary matrices U,VMn. For A=[aij]Mn, the Hilbert-Schmidt (or Frobenius) norm and the trace norm of A are defined by

    ||A||2=nj=1s2j(A)

    and

    ||A||1=nj=1sj(A),

    respectively, where sj(A) are the singular values of A, that is, the eigenvalues of the positive semi-definite matrix

    |A|=(AA)12,

    and arranged in a nonincreasing order. It is well known that 2 is unitarily invariant. In addition, we defined

    AsB=(1s)A+sB

    and

    AsB=A12(A12BA12)sA12

    for s[0,1], denoted by AB and AB when s=12, respectively. Similarly, we define the weighted means by

    asb=(1s)a+sb

    and

    asb=a1sbs

    for a,b>0 and s[0,1].

    The famous Young's inequality states that the convex combination of two positive numbers is greater than or equal to the product of powers of these numbers with corresponding coefficients as exponents, which reads

    asbasb (1)

    for a,b>0 and 0s1, with equation if and only if a=b. It extends the classical arithmetic-geometric means inequality. In 2015, Alzer et al. [1] showed the following refinements and reverses of Eq (1)

    (sτ)λ(asb)λ(asb)λ(aτb)λ(aτb)λ(1s1τ)λ (2)

    for a,b>0, 0<sτ<1 and λ1. In recent years, there have been a large number of works directly inspired by the Alzer-Fonseca-Kovaˇcec inequality, see [2,3,4,5,6,7] et al. Moreover, letting λ=1, the following results obtained by Zhao and Wu [8] can be regarded as some further refinements of (2) when s=12 and τ=12, respectively,

    (i) If 0s12, then

    sa+(1s)basb1s+(1s)(ab )2r0(4aba )2; (3)

    (ii) If 12s1, then

    sa+(1s)basb1s+s(ab )2r0(4abb )2; (4)

    where a,b>0, r0=min{2r,12r}, r=min{s,1s} for s[0,1]. Sababheh and Moslehian [9] show a nice multiple-term refinements of (3) and (4). Interested readers could refer to [10,11,12,13,14] and references therein for some other results about Young's inequality.

    Unless otherwise specified, we will default to a,b>0 and 0s1 in the rest of this article for our convenience.

    There are some results related to Young's inequality mentioned above. Ghazanfari et al. [15] presented an inequality

    (1s2+s3)a+(1s2)bss2asb1s+(ab )2. (5)

    It is easy to verify that both sides in the inequality (5) are greater than or equal to the corresponding sides in the inequalities (3) and (4), respectively. This indicates that the inequality (5) is a new Young type inequality.

    In 2020, Ren [4] showed a generalization of the inequality (5),

    (1sN+1+sN+2)a+(1s2)bssN(N+1)asb1s+(ab )2 (6)

    for NN+. Later, Yang and Li [16] gave a more generalized inequality than (6): If N1,N2N+, then

    (1sN1+1+sN1+2)a+(1sN2+2)bss(N1N2)(N1+1)asb1s+(ab )2. (7)

    Very recently, Zuo and Li [17] got a new generalization of inequality (5): If NN+, then

    (1sN+1+sN+2)a+(1sN+1)bss(N+1)asb1s+(ab )2. (8)

    In addition, the Kantorovich constant and the Specht's ratio are defined by

    K(h,2)=(h+1)24h  for  h>0

    and

    S(h)={h1h1elog(h1h1),  if  h(0,1)(1,),      1,    if  h=1.

    K(h,2) and S(h) have some common properties, for example:

    (i) K(1,2)=S(1)=1;

    (ii) K(h,2) and S(h) are decreasing on h(0,1) and increasing on h(1,+);

    (iii) K(h,2)1 and S(h)1.

    Zuo et al. [18] and Furuichi [19] showed

    S(hr)asbK(h,2)rasbasb (9)

    for r=min{s,1s} and h=ab>0.

    In this paper, we try to give some generalizations of (8), and we also present a new generalization of Young type inequality, then we refine these inequalities with the Kantorovich constant. As applications, we obtain some operator inequalities, Hilbert-Schmidt norm inequalities and trace inequalities.

    First, we show a generalization of the inequality (8).

    Theorem 1. Let N1,N2N+ and 1s0, a,b>0. Then, we have

    (1sN1+1+sN1+2)a+(1sN1+N2+1)bss(1N2)(N1+1)asb1s+(ab )2. (10)

    Proof. By computations, we can get

    ss(1N2)(N1+1)asb1s+(ab )2(1sN1+1+sN1+2)a(1sN1+N2+1)b=ss(1N2)(N1+1)asb1s2ab+(1s)sN1+1a+ssN1+N2bss(1N2)(N1+1)asb1s2ab+(sN1+1a)(1s)(sN1+N2b)s    (by (1))=(ss(1N2)(N1+1)2as2b1s2ss(1N2)(N1+1)2a1s2bs2)20.

    We now explain that Theorem 1 is a new generalization of the Young type inequality. In fact, comparing inequality (10) with (7), it is not difficult to find that

    (1sN1+N2+1)b(1sN2+2)b

    and

    ss(1N2)(N1+1)ss(N1N2)(N1+1)

    for N1,N2N, 1s0, b>0.

    Remark 1. Taking N2=0 in Theorem 1, we get (8).

    Next, we show another generalization of (8), which can be regarded as a complement of Theorem 1.

    Theorem 2. Let N1,N2N+ and 1s0, a,b>0. Then, we have

    (1sN1+1+sN1+2)a+(1sN1N2+2)bssN2(N1+1)asb1s+(ab )2.

    Proof. By computations, we can get

    ssN2(N1+1)asb1s+(ab )2(1sN1+1+sN1+2)a(1sN1N2+2)b=ssN2(N1+1)asb1s2ab+(1s)sN1+1a+ssN1N2+1bssN2(N1+1)asb1s2ab+(sN1+1a)(1s)(sN1N2+1b)s    (by (1))=(ssN2(N1+1)2as2b1s2ssN2(N1+1)2a1s2bs2)20.

    Comparing Theorem 2 with Theorem 1, it is easy to see that

    (1sN1+N2+1)b(1sN1N2+2)b

    and

    ss(1N2)(N1+1)ssN2(N1+1)

    for N2N+, 1s0, b>0. Therefore, Theorem 2 is a new generalization of the Young type inequality.

    Remark 2. Taking N2=1 in Theorem 2, we get (8).

    Next, we show a new generalization of Young type inequality.

    Theorem 3. Let N1,N2N, 1s0 and a,b>0. Then, we have

    (1sN1+sN1+1)a+(1sN2+1)bss(N1N2)N1asb1s+(ab )2.

    Proof. By computations, we obtain

    (ab )2+ss(N1N2)N1asb1s(1sN1+sN1+1)a(1sN2+1)b=(1s)(sN1a)+s(sN2b)2ab+ss(N1N2)N1asb1s(sN1a)1s(sN2b)s2ab+ss(N1N2)N1asb1s    (by (1))=(ss(N2N1)+N12a1s2bs2ss(N1N2)N12as2b1s2)20.

    Letting N2=0 in Theorem 3, we get

    (1sN+sN+1)a+(1s)bssNNasb1s+(ab )2 (11)

    for NN, 1s0 and a,b>0. As we can see that both sides in (11) are greater than or equal to the corresponding sides in the inequalities (3) and (4), respectively, this indicates that Theorem 3 is a new generalization of the Young type inequality.

    Following the ideas of Theorem 2, we can get a new Young type inequality by replacing N2 with 1N2 in Theorem 3. However, we omit it to avoid repetition of the article.

    We next improve Theorems 1–Theorem 3 with the Kantorovich constant by (9).

    Theorem 4. Let N1,N2N+ and 1s0, a,b>0,

    h=s1N2ab

    and

    r=min{s,1s}.

    We have

    (1sN1+1+sN1+2)a+(1sN1+N2+1)bK(h,2)rss(1N2)(N1+1)asb1s+(ab )2. (12)

    Proof. Compute

    K(h,2)rss(1N2)(N1+1)asb1s+(ab )2(1sN1+1+sN1+2)a(1sN1+N2+1)b=K(h,2)rss(1N2)(N1+1)asb1s2ab+(1s)sN1+1a+ssN1+N2bK(h,2)rss(1N2)(N1+1)asb1s2ab+K(h,2)r(sN1+1a)(1s)(sN1+N2b)s=(K(h,2)r2ss(1N2)(N1+1)2as2b1s2K(h,2)r2ss(1N2)(N1+1)2a1s2bs2)20.

    Theorem 5. Let N1,N2N+, 1s0, a,b>0 and r=min{s,1s}.

    (i) If

    h=sN2ab,

    then, we have

    (1sN1+1+sN1+2)a+(1sN1N2+2)bK(h,2)rssN2(N1+1)asb1s+(ab )2.

    (ii) If

    h=sN1N2ab,

    then, we have

    (1sN1+sN1+1)a+(1sN2+1)bK(h,2)rss(N1N2)N1asb1s+(ab )2.

    Proof. Using the same technique as in Theorem 4, we complete the proof of Theorem 5.

    Replacing a by a2 and b by b2 in Theorems 4 and 5, respectively, we get the following corollary:

    Corollary 1. Let N1,N2N+, a,b>0 and r=min{s,1s} for 1s0.

    (i) If

    h=s1N2a2b2,

    then, we have

    (1sN1+1+sN1+2)a2+(1sN1+N2+1)b2K(h,2)rss(1N2)(N1+1)(asb1s)2+(ab)2.

    (ii) If

    h=sN2a2b2,

    then, we have

    (1sN1+1+sN1+2)a2+(1sN1N2+2)b2K(h,2)rssN2(N1+1)(asb1s)2+(ab)2.

    (iii) If

    h=sN1N2a2b2,

    then, we have

    (1sN1+sN1+1)a2+(1sN2+1)b2K(h,2)rss(N1N2)N1(asb1s)2+(ab)2.

    Based on the scalars results mentioned above, we next present some operator inequalities, Hilbert-Schmidt norm inequalities and trace inequalities as promised.

    Lemma 6. [20] Let XMn be self-adjoint and f and g be continuous real functions such that f(t)g(t) for all tSp(X) (the spectrum of X). Then f(X)g(X).

    Theorem 7. Let A,BB(H) be strictly positive operators and let positive real numbers m,m and M,M satisfy either of the following conditions:

    (i) 0<s1N2mIAs1N2mI<MIBMI;

    (ii) 0<sN21mIBsN21mI<MIAMI.

    Then, we have

    (1sN1+1+sN1+2)B+(1sN1+N2+1)AK(h,2)rss(1N2)(N1+1)AsB+2(ABAB),

    where

    h=Ms1N2m,r=min{s,1s}

    and N1,N2N+.

    Proof. Let b=1 in (12), we have

    (1sN1+1+sN1+2)a+(1sN1+N2+1)K(s1N2a,2)rss(1N2)(N1+1)as+(a+12a ). (13)

    Under the conditions (i), we have

    IhI=Ms1N2mIX=A12BA12hI=Ms1N2mI,

    and then

    Sp(X)[h,h](1,+).

    By Lemma 6 and (13), we obtain

    (1sN1+1+sN1+2)X+(1sN1+N2+1)ImaxhxhK(x,2)rss(1N2)(N1+1)Xs+(X+I2X12).

    Since the Kantorovich constant K(h,2) is an increasing function on h(1,+), then

    (1sN1+1+sN1+2)A12BA12+(1sN1+N2+1)IK(h,2)rss(1N2)(N1+1)(A12BA12)s+A12BA12+I2(A12BA12)12. (14)

    Under the conditions (ii), we have

    1hI=sN21mMIX=A12BA121hI=sN21mMII,

    and then

    Sp(X)[1h,1h](0,1).

    Since the Kantorovich constant K(h,2) is a decreasing function on h(0,1), we can similarly get

    (1sN1+1+sN1+2)A12BA12+(1sN1+N2+1)IK(1h,2)rss(1N2)(N1+1)(A12BA12)s+A12BA12+I2(A12BA12)12. (15)

    In fact, inequality (14) is equal to (15) with the property

    K(t,2)=K(1t,2).

    Multiplying A12 from the both sides to the inequalities (14) or (15), we get the required results.

    Theorem 8. Let A,BB(H) be strictly positive operators and let positive real numbers m,m and M,M satisfy either of the following conditions

    (i) 0<sN2mIAsN2mI<MIBMI;

    (ii) 0<sN2mIBsN2mI<MIAMI;

    Then, we have

    (1sN1+1+sN1+2)B+(1sN1N2+2)AK(h,2)rssN2(N1+1)AsB+2(ABAB),

    where

    h=MsN2m,r=min{s,1s}

    and N1,N2N+.

    Proof. Using the same method as in Theorem 7, we can get Theorem 8 by Theorem 5 (i).

    Theorem 9. Let A,BB(H) be strictly positive operators and let positive real numbers m,m and M,M satisfy either of the following conditions

    (i) 0<sN1N2mIAsN1N2mI<MIBMI;

    (ii) 0<sN2N1mIBsN2N1mI<MIAMI;

    Then, we have

    (1sN1+sN1+1)B+(1sN2+1)AK(h,2)rss(N1N2)N1AsB+2(ABAB).

    where

    h=MsN1N2m,r=min{s,1s}

    and N1,N2N+.

    Proof. Using the same method as in Theorem 7, we can obtain Theorem 9 by Theorem 5 (ii).

    Theorem 10. Let A,BMn be positive definite and XMn. Then, we have

    (1sN1+1+sN1+2)||A||22+(1sN1+N2+1)||B||22Krss(1N2)(N1+1)||AsXB1s||22+||AXXB||22,

    where N1,N2N+, r=min{s,1s} for 1s0 and

    K=min{K(s1N2λ2iμ2l,2),1i,ln}

    for λi,μl are eigenvalues of A,B, respectively.

    Proof. Since A,BMn are positive definite, by the spectral theorem, there exist unitary matrices U,VMn such that

    A=UΛ1U,B=VΛ2V,

    where

    Λ1=diag(λ1,λ2,,λn),Λ2=diag(μ1,μ2,,μn),λi,μi>0,i=1,2,,n.

    Let

    Y=UXV=[yil],

    then,

    ||AXXB||22=||U[(λiμl)yil]V||22=ni,l=1(λiμl)2|yil|2.

    Similarly, we have

    ||AsXB1s||22=ni,l=1(λsiμ1sl)2|yil|2

    and

    ||A||22=ni,l=1λ2i|yil|2.

    By the unitarily invariance of the Hilbert-Schmidt norm and Corollary 1 (i), we get the following results

    (1sN1+1+sN1+2)||A||22+(1sN1+N2+1)||B||22=(1sN1+1+sN1+2)ni,l=1λ2i|yil|2+(1sN1+N2+1)ni,l=1μ2l|yil|2=ni,l=1[(1sN1+1+sN1+2)λ2i+(1sN1+N2+1)μ2l]|yil|2ni,l=1[Krss(1N2)(N1+1)(λsiμ1sl)2+(λiμl)2]|yil|2=Krss(1N2)(N1+1)||AsXB1s||22+||AXXB||22.

    Theorem 11. Let A,BMn be positive definite and XMn. Then, we have

    (1sN1+1+sN1+2)||A||22+(1sN1N2+2)||B||22KrssN2(N1+1)||AsXB1s||22+||AXXB||22,

    where N1,N2N+, r=min{s,1s} for 1s0 and

    K=min{K(sN2λ2iμ2l,2),1i,ln}

    for λi,μl are eigenvalues of A,B, respectively.

    Proof. We get Theorem 11 by Theorem 10 and Corollary 1 (ii).

    Theorem 12. Let A,BMn be positive definite and XMn. Then, we have

    (1sN1+sN1+1)||A||22+(1sN2+1)||B||22Krss(N1N2)N1||AsXB1s||22+||AXXB||22,

    where N1,N2N+, r=min{s,1s} for 1s0 and

    K=min{K(sN1N2λ2iμ2l,2),1i,ln}

    for λi,μl are eigenvalues of A,B, respectively.

    Proof. We get Theorem 12 by Theorem 10 and Corollary 1 (iii).

    Lemma 13. [21] Let A,BMn, then,

    nj=1sj(AB)nj=1sj(A)sj(B).

    Theorem 14. Let A,BMn be positive definite and N1,N2N+, r=min{s,1s} for 1s0 and

    K=min{K(s1N2sj(A)sj(B),2),1jn}.

    Thus, we have

    tr[(1sN1+1+sN1+2)A+(1sN1+N2+1)B]Krss(1N2)(N1+1)||As||2||B1s||2+||A||1+||B||12||A12B12||1.

    Proof. By Theorem 4, Lemma 13 and the famous Cauchy-Schwarz inequality, we obtain

    tr[(1sN1+1+sN1+2)A+(1sN1+N2+1)B]=nj=1((1sN1+1+sN1+2)sj(A)+(1sN1+N2+1)sj(B))nj=1[maxK(s1N2sj(A)sj(B),2)rss(1N2)(N1+1)ssj(A)s1sj(B)+(sj(A)sj(B) )2]=Krss(1N2)(N1+1)nj=1sj(As)sj(B1s)+nj=1sj(A)+nj=1sj(B)2nj=1sj(A)sj(B)Krss(1N2)(N1+1)(nj=1s2j(As))12(nj=1s2j(B1s))12+nj=1sj(A)+nj=1sj(B)2nj=1sj(A12B12)=Krss(1N2)(N1+1)||As||2||B1s||2+||A||1+||B||12||A12B12||1.

    Theorem 15. Let A,BMn be positive definite and N1,N2N+, r=min{s,1s} for 1s0 and

    K=min{K(sN2sj(A)sj(B),2),1jn}.

    We have

    tr[(1sN1+1+sN1+2)A+(1sN1N2+2)B]KrssN2(N1+1)||As||2||B1s||2+||A||1+||B||12||A12B12||1.

    Proof. Following the line of Theorem 14, we can get the Theorem 15 by Theorem 5 (i).

    Theorem 16. Let A,BMn be positive definite and N1,N2N+, r=min{s,1s} for 1s0 and

    K=min{K(sN1N2sj(A)sj(B),2),1jn}.

    We have

    tr[(1sN1+sN1+1)A+(1sN2+1)B]Krss(N1N2)N1||As||2||B1s||2+||A||1+||B||12||A12B12||1.

    Proof. Following the line of Theorem 14, we can get the Theorem 16 by Theorem 5 (ii).

    In this paper, we give two generalized Young type inequalities of Zuo and Li [17], and we also present a new Young type inequality by comparing with the results obtained by Zhao and Wu [8]. As applications, we obtain some inequalities including operator, Hilbert-Schmidt norm and trace using our scalars results.

    The author declares he has not used Artificial Intelligence (AI) tools in the creation of this article.

    The author wishes to express his sincere thanks to the three referees for their detailed and helpful suggestions for revising the manuscript.

    The author declares that he has no competing interest.



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