Abstract
In this paper, we employ some operator techniques to establish some refinements and reverses of the Callebaut inequality involving the geometric mean and Hadamard product under some mild conditions. In particular, we show
where \(A_j, B_j\in {\mathbb {B}}({\mathscr {H}})\,\,(1\le j\le n)\) are positive operators such that \(0<m{^\prime } \le B_j\le m <M \le A_j\le M{^\prime }\,\,(1\le j\le n)\), either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\), \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \) and \(K(t,2)=\frac{(t+1)^2}{4t}\,\,(t>0)\).
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1 Introduction and Preliminaries
Let \({\mathbb B}({\mathscr {H}})\) denote the \(C^*\)-algebra of all bounded linear operators on a complex Hilbert space \({\mathscr {H}}\) with the identity I. An operator \(A\in {\mathbb B}({\mathscr {H}})\) is called positive if \(\langle Ax,x\rangle \ge 0\) for all \(x\in {\mathscr {H} }\), and we then write \(A\ge 0\). We write \(A>0\) if A is a positive invertible operator. The set of all positive invertible operators is denoted by \({\mathbb B}({\mathscr {H}})_+\). For self-adjoint operators \(A, B\in {\mathbb B}({\mathscr {H}})\), we say \(B\ge A\) if \(B-A\ge 0\). The Gelfand map f to f(A) is an isometric \(*\)-isomorphism between the \(C^*\)-algebra \(C(\mathrm {sp}(A))\) of a complex-valued continuous functions on the spectrum \(\mathrm {sp}(A)\) of a self-adjoint operator A and the \(C^*\)-algebra generated by I and A. If \(f, g\in C(\mathrm{sp}(A))\), then \(f(t)\ge g(t)\,\,(t\in \mathrm{sp}(A))\) implies that \(f(A)\ge g(A)\).
It is known that the Hadamard product can be presented by filtering the tensor product \(A \otimes B\) through a positive linear map. In fact,\(A\circ B=U^*(A\otimes B)U\), where \(U:{\mathscr {H}}\rightarrow {\mathscr {H}}\otimes {\mathscr {H}}\) is the isometry defined by \(Ue_j=e_j\otimes e_j\), where \((e_j)\) is an orthonormal basis of the Hilbert space \({\mathscr {H}}\); see [7].
For \(A, B\in {\mathbb {B}}({\mathscr {H}})_+\), the operator geometric mean \(A\sharp B\) is defined by \(A\sharp B=A^{\frac{1}{2}}\left( A^{\frac{-1}{2}}BA^{\frac{-1}{2}}\right) ^{\frac{1}{2}}A^{\frac{1}{2}}\). For \(\alpha \in (0,1)\), the operator-weighted geometric mean is defined by
Callebaut [4] showed the following refinement of the Cauchy–Schwarz inequality
where \( x_j, y_j \,\,(1\le j\le n)\) are positive real numbers and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). This is indeed an extension of the Cauchy–Schwarz inequality.
Wada [10] gave an operator version of the Callebaut inequality by showing that if \(A, B\in {\mathbb B}({\mathscr {H}})_+\), then
where \(\alpha \in [0,1]\). In [6], the authors showed another operator version of the Callebaut inequality:
where \(A_j, B_j\in {{\mathbb {B}}}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\).
In [3], the authors presented the following refinement of inequality (1.2) as follows:
in which \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\).
There have been obtained several Cauchy–Schwarz type inequalities for Hilbert space operators and matrices; see [1, 2, 5, 8] and references therein.
In this paper, we present some refinements and reverses of the Callebaut inequality involving the weighted geometric mean and Hadamard product of Hilbert space operators.
2 Further Refienements of the Callebaut Inequality Involving Hadamard Product
The Kontorovich constant is
The classical Young inequality states that
where \(a,b\ge 0\) and \(\nu \in [0,1]\). Recently, Zuo et. al. [11] showed an improvement of the Young inequality as follows:
where \(a,b>0\), \(\nu \in [0,1]\), \(r=\min \left\{ \nu ,1-\nu \right\} \). Applying this inequality, J. Wu and J. Zhao [9] showed the following refinement of the Young inequality:
where \(a,b>0\), \(\nu \in [0,1]-\left\{ \frac{1}{2}\right\} \), \(r=\min \left\{ \nu ,1-\nu \right\} \) and \(r{^\prime }=\min \left\{ 2r,1-2r\right\} \). Using (2.1), we get the following lemmas.
Lemma 2.1
Let \(a,b>0\) and \(\nu \in [0,1]-\left\{ \frac{1}{2}\right\} \). Then
where \(r=\min \left\{ \nu ,1-\nu \right\} \) and \(r{^\prime }=\min \left\{ 2r,1-2r\right\} \).
Lemma 2.2
Let \(0<m{^\prime } \le B\le m <M \le A\le M{^\prime }\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then
where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
Proof
Let \(a>0\). If we replace b by \(a^{-1}\) and take \(\nu =\frac{1-\mu }{2}\) (2.2), then we get
in which \(\mu \in (0,1]\) and \(r{^\prime }=\min \left\{ 1-\mu ,\mu \right\} \). Let us fix positive real numbers \(\alpha ,\beta \) such that \(\beta <\alpha \). It follows from \(0<m{^\prime } \le B\le m <M \le A\le M{^\prime }\) that \(I\le h=\left( \frac{M}{m}\right) ^\alpha \le A^\alpha \otimes B^{-\alpha }\le h{^\prime }=\left( \frac{M{^\prime }}{m{^\prime }}\right) ^\alpha \) and \(\mathrm {sp}(A^\alpha \otimes B^{-\alpha })\subseteq [h,h{^\prime }]\subseteq (1,+\infty )\). Since the Kontorovich constant \(\frac{(a+1)^2}{4a}\) is an increasing function on \((1,+\infty )\), by (2.4), we have
where \(\mu \in (0,1]\) and \(r{^\prime }=\min \left\{ 1-\mu ,\mu \right\} \). Using the functional calculus, if we replace a by the operator \(A^\alpha \otimes B^{-\alpha }\) and \(\mu \) by \(\frac{\beta }{\alpha }\) we have
where \(r{^\prime }=\min \left\{ 1-\frac{\beta }{\alpha },\frac{\beta }{\alpha }\right\} \). Multiplying, we both sides of (2.5) by \(A^\frac{1}{2}\otimes B^\frac{1}{2}\), we obtain
Now, if we replace \(\alpha , \beta , A, B\) by \(2t-1, 2s-1, A^\frac{1}{2}, B^\frac{1}{2}\), respectively, in (2.6), we obtain
for either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\) and \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \). \(\square \)
We are ready to prove the first result of this section.
Theorem 2.3
Let \(0<m{^\prime } \le B_j\le m <M \le A_j\le M{^\prime }\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then
where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
Proof
Put \(C_j=A_j^{-{1\over 2}}B_jA_j^{-{1\over 2}}\,\,(1\le j\le n)\). By inequality (2.3), we get
Multiplying both sides of (2.8) by \(A_j^{\frac{1}{2}}\otimes A_i^{\frac{1}{2}}\), we get
for all \(1\le i,j\le n\). Therefore
Remark 2.4
It follows from
where \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\), either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\) and \(K(t,2)=\frac{(t+1)^2}{4t}\ge 1\,\,(t>0)\) that inequality (2.7) is a refinement of the second inequality of inequalities (1.2) and (1.3).
We conclude an application of Theorem 2.3 for numerical cases which is a refinement of inequality (1.1).
Corollary 2.5
Let \(0<m{^\prime } \le y_j\le m <M \le x_j\le M{^\prime }\,\,(1\le j\le n)\) and either \(-1\le t\le s<0\) or \(1\ge t\ge s>0\). Then
where \(r{^\prime }=\min \left\{ \frac{t-s}{t},\frac{s}{t}\right\} \).
Proof
If we put \(A_j=x_j, B_j=y_j\,\,(1\le j\le n)\) in Theorem 2.3 and inequality (1.2), then we get
where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \) and either \(0\le t\le s<\frac{1}{2}\) or \(1\ge t\ge s>{\frac{1}{2}}\). Now if we replace s by \({s+1\over 2}\) and t by \({t+1\over 2}\), respectively, where either \(-1\le t\le s<0\) or \(1\ge t\ge s>0\), then we obtain the desired inequalities. \(\square \)
Lemma 2.6
Let \(a,b>0\) and \(\nu \in (0,1)\). Then
where \(r=\min \{\nu ,1-\nu \}\) and \(r{^\prime }=\min \{2r,1-2r\}\).
Proof
Let \(a,b>0\), \(\nu \in (0,1)\), \(r=\min \{\nu ,1-\nu \}\) and \(r{^\prime }=\min \{2r,1-2r\}\). Applying [12, Lemma1], we have the inequalities
where \(0<\nu \le \frac{1}{2}\) and
where \(\frac{1}{2}<\nu <1\). Summing these inequalities, we get the desired result. \(\square \)
Lemma 2.7
Let \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then
where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
Using Lemma 2.7 and the same argument in the proof of Lemma 2.2, we get another refinement of inequality (1.2).
Theorem 2.8
Let \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then
where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
Remark 2.9
If \(A_j, B_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\), then
and
are positive operators. Hence, inequality (2.10) is another refinement of the second inequality of inequalities (1.2) and (1.3).
If we put \(B_j=I\,\,(1\le j\le n)\) in Theorem 2.8, then we get the next result.
Corollary 2.10
Let \(A_j\in {\mathbb B}({\mathscr {H}})_+\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then
where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
If in Theorem 2.8 we replace \(A_j, B_j, s, t\), by \(x_j, y_j, {s+1\over 2}, {t+1\over 2}\,\,(1\le j\le n)\), respectively, then we obtain another refinement of inequality (1.1).
Corollary 2.11
Let \(x_j,y_j\,\,(1\le j\le n)\) be positive numbers and either \(-1\le t\le s<0\) or \(1\ge t\ge s>0\). Then
where \(r{^\prime }=\min \left\{ \frac{t-s}{t},\frac{s-1}{t-1}\right\} \).
3 Some Reverses of the Callebaut Type Inequality
In [9], the authors showed a reverse of the Young inequality as follows:
in which \(a,b>0\), \(\nu \in [0,1]-\left\{ \frac{1}{2}\right\} \), \(r=\min \left\{ \nu ,1-\nu \right\} \), \(r{^\prime }=\min \left\{ 2r,1-2r\right\} \) and \(s=\max \left\{ \nu ,1-\nu \right\} \). Applying (3.1), we have the next result.
Lemma 3.1
Let \(a,b>0\) and \(\nu \in [0,1]-\left\{ \frac{1}{2}\right\} \). Then
where \(r=\min \left\{ \nu ,1-\nu \right\} \), \(r{^\prime }=\min \left\{ 2r,1-2r\right\} \) and \(s=\max \left\{ \nu ,1-\nu \right\} \). In particular, If\(\nu \in [0,\frac{1}{2})\), then
Now, utilizing inequality (3.2) and the same argument in the proof of Lemma 2.2, we can accomplish the corresponding result:
Lemma 3.2
Let \(0<m{^\prime } \le B\le m <M \le A\le M{^\prime }\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then
where \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
As a consequence of Lemma 3.2, we have the following result:
Theorem 3.3
Let \(0<m{^\prime } \le B_j\le m <M \le A_j\le M{^\prime }\,\,(1\le j\le n)\). Then
for either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\) and \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
Remark 3.4
If we put \(t=1\) and \(1\ge s>{\frac{1}{2}}\) in Theorem 3.3, then we get a reverse of the third inequality of (1.2):
where \(r{^\prime }=\min \left\{ {2-2s},{2s-1}\right\} \).
If we replace \(A_j, B_j, s, t\), by \(x_j, y_j, {s+1\over 2}, {t+1\over 2}\,\,(1\le j\le n)\) in Theorem 3.3, respectively, then we obtain a reverse of the second inequality of (1.1).
Corollary 3.5
Let \(0<m{^\prime } \le y_j\le m <M \le x_j\le M{^\prime }\,\,(1\le j\le n)\). Then
for either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\) and \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
Proposition 3.6
Let \(0<m{^\prime } \le B_j\le m <M \le A_j\le M{^\prime }\,\,(1\le j\le n)\) and either \(1\ge t\ge s>{\frac{1}{2}}\) or \(0\le t\le s<\frac{1}{2}\). Then
where \(h=\frac{M}{m}\), \(h{^\prime }=\frac{M{^\prime }}{m{^\prime }}\) and \(r{^\prime }=\min \left\{ \frac{t-s}{t-1/2},\frac{s-1/2}{t-1/2}\right\} \).
Proof
Since the function \(f(a)=a-\frac{1}{a}\) is increasing on \((0,\infty )\), we have
Applying inequalities (2.2), (3.2) and the same argument in the proof of Theorem 2.3, we get the desired result.
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Acknowledgments
The author would like to sincerely thank the anonymous referee for some useful comments and suggestions. The author also would like to thank the Tusi Mathematical Research Group (TMRG).
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Communicated by Poom Kumam.
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Bakherad, M. Some Reversed and Refined Callebaut Inequalities Via Kontorovich Constant. Bull. Malays. Math. Sci. Soc. 41, 765–777 (2018). https://doi.org/10.1007/s40840-016-0364-9
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DOI: https://doi.org/10.1007/s40840-016-0364-9
Keywords
- Callebaut inequality
- Cauchy–Schwarz inequality
- Hadamard product
- Operator geometric mean
- Kontorovich constant