Sector operator inequalities involving positive linear maps
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- 08-12-2025
- Research
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Abstract
1 Introduction
Let M, m be scalars and I be the identity operator. Other capital letters are used to denote the general elements of the \(C^{*}\)-algebra \(\mathcal {B}(\mathcal{H})\) of all bounded linear operators acting on a Hilbert space \((\mathcal {H}, \langle \cdot , \cdot \rangle )\). The operator norm is denoted by \(\|\cdot \|\). \(A^{*}\) stands for the adjoint of A. If A is self-adjoint (i.e., \(A=A^{*}\)) we say A is positive if \(\langle Ax, x\rangle \geq 0\) for all \(x\in \mathcal{H}\) and A is strictly positive if A is positive and invertible. Further, the class of the strictly positive operators is denoted by \(\mathcal {B}^{+}(\mathcal{H})\). For self-adjoint operators \(A, B\in \mathcal {B}(\mathcal{H})\), we write \(A-B \geq 0\) \((A \geq B)\) or \(B-A\leq 0\) \((B\leq A)\) to mean that the operator \(A-B\) is positive. We say linear map \(\Phi :\mathcal{B}\left ( \mathcal{H} \right )\to \mathcal{B}\left ( \mathcal{H} \right )\) is positive if \(\Phi \left ( A \right )\geq 0\) whenever \(A\geq 0\). It is said to be unital if \(\Phi \left ( I \right )=I\). There is a considerable amount of literature devoted to the study of operator inequalities related to unital positive linear maps; we refer to [7] and the references therein.
For an operator \(A\in \mathcal{B}( \mathcal{H} )\), the real part ℜA of A is defined by \(\frac{A+A^{*}}{2}\), the imaginary part is defined by \(\frac{A-A^{*}}{2i}\) and denoted by ℑA. For \(0\leq \alpha < \frac{\pi }{2}\), we define a sector as follows: Here, we recall that the numerical range of an \(n\times n\) matrix A is defined by The class of operators A with \(W(A)\subset S_{\alpha}\) is called sector operators. It has been the subject of many papers [2, 6, 8, 11, 13, 15]. For two Hermitian operators \(A, B \in \mathcal{B}( \mathcal{H} )\), a function f is said to be operator monotone (see, e.g. [3, p. 112]) if it is monotone with respect to A and B, i.e., if \(A\leq B\) implies \(f(A)\leq f(B)\). Now we will use the notation \(\mathfrak{m}\) to show analytic function class in any domain that avoids the negative x-axis,
$$\begin{aligned} S_{\alpha }=\left \{z\in \mathbb{C}:\Re z> 0, \left |\Im z\right | \leqslant \left (\Re z \right )\tan (\alpha )\right \}. \end{aligned}$$
$$\begin{aligned} W\left ( A \right )=\left \{ x^{*} A x:x\in \mathbb{C}^{n},x^{*} x=1 \right \}. \end{aligned}$$
$$ \mathfrak{m}=\{f: (0, \infty )\rightarrow (0,\infty ); f\; \mathrm{is}\; \mathrm{an} \;\mathrm{operator}\; \mathrm{monotone}\; \mathrm{function}\; \mathrm{with}\; f(1)=1\}. $$
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The so-called operator mean is strongly related to the class \(\mathfrak{m}\). For every invertible \(A, B\in \mathcal {B}^{+}(\mathcal{H})\) and \(f\in \mathfrak{m}\), we define \(\sigma _{f} : \;\mathcal {B}^{+}(\mathcal{H})\times \mathcal {B}^{+}( \mathcal{H})\rightarrow \mathcal {B}^{+}(\mathcal{H})\) by We may define a difference between two perspectives as for \(f, g\in \mathfrak{m}\). The theory of operator means has been extended to sector operators in [2], using the same identity as in (1.1). We refer the reader to [2] for a detailed discussion.
$$\begin{aligned} A\sigma _{f} B=A^{\frac{1}{2}}f(A^{-\frac{1}{2}}BA^{-\frac{1}{2}})A^{ \frac{1}{2}}. \end{aligned}$$
(1.1)
$$ D_{f, g}(A|B)=A\sigma _{f} B-A\sigma _{g} B, $$
Bedrani et al. [2, Theorem 5.5] presented the comparison for sector operators between different means that arise from different operators monotone functions as follows.
Theorem 1.1
Let \(A, B\in \mathcal {B}(\mathcal{H})\) be sector operators such that \(0< m\leq \Re A, \Re B\leq M\) and \(W(A), W(B)\subset S_{\alpha}\) for some \(0\leq \alpha <\frac{\pi}{2}\). If \(f, g\in \mathfrak{m}\), then for every unital positive linear map Φ, where \(K(m, M)=\frac{(M+m)^{2}}{4Mm}\).
$$\begin{aligned} \Phi ^{2}(\Re (A\sigma _{f} B)) \leq \sec ^{12}(\alpha )K^{2}(m, M) \Phi ^{2}(\Re (A\sigma _{g} B)), \end{aligned}$$
(1.2)
The operator Kantorovich inequality [12] states that if \(A\in \mathcal {B}(\mathcal{H})\) is satisfying the condition \(0< m\leq A\leq M\) and Φ is any unital positive linear map, then where \(K(m, M)=\frac{(M+m)^{2}}{4Mm}\) is Kantorovich constant.
$$\begin{aligned} \Phi \left ( A^{-1} \right )\leq K\left ( m, M \right )\Phi ^{-1} \left ( A \right ), \end{aligned}$$
(1.3)
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Theorem 1.2
Let \(A\in \mathcal {B}(\mathcal{H})\) be a sector operator such that \(0< m\leq \Re A\leq M\) and \(W(A)\subset S_{\alpha}\) for some \(0\leq \alpha <\frac{\pi}{2}\). Then for any unital positive linear map Φ, where \(K(m, M)=\frac{(M+m)^{2}}{4Mm}\).
$$\begin{aligned} \Re \Phi (A^{-1})\leq K(m, M)\sec ^{2}(\alpha ) \Re \Phi ^{-1}(A), \end{aligned}$$
(1.4)
The inequality (1.4) can be regarded as a counterpart to the following result under the condition that \(W(A)\subset S_{\alpha}\) for some \(0\leq \alpha <\frac{\pi}{2}\) and Φ is any unital positive linear map (see, e.g. [14, Theorem 14]), which says
$$\begin{aligned} \Re \Phi ( A^{-1} )\geq \cos ^{2}(\alpha )\Re \Phi ^{-1}( A ). \end{aligned}$$
(1.5)
It is well known [16, Theorem 7.10] (Löwner-Heinz inequality) that for two positive operators \(A, B\in \mathcal {B}(\mathcal{H})\), but,
$$\begin{aligned} A\ge B \Rightarrow & A^{p}\ge B^{p} \qquad \quad \text{for}\quad 0 \leq p \leq 1, \end{aligned}$$
(1.6)
$$\begin{aligned} A\ge B \nRightarrow & A^{p}\ge B^{p} \qquad \quad \text{for}\quad p > 1. \end{aligned}$$
Combining (1.2), (1.4) and (1.6), we obtain
$$\begin{aligned}& \Phi ^{p}(\Re (A\sigma _{f} B)) \leq \sec ^{6p}(\alpha )\left ( \frac{(M+m)^{2}}{4Mm}\right )^{p}\Phi ^{p}(\Re (A\sigma _{g} B)) \qquad \quad \text{for}\quad 0\leq p \leq 2, \end{aligned}$$
(1.7)
$$\begin{aligned}& \Re ^{p} (\Phi \left ( A^{-1} \right ) )\leq \left ( \frac{\left ( M+m \right )^{2}}{4Mm} \right )^{p}\sec ^{2p}(\alpha ) \Re ^{p}\left ( \Phi ^{-1}\left ( A \right ) \right ) \qquad \quad \text{for}\quad 0\leq p \leq 1. \end{aligned}$$
(1.8)
2 Main results
We need several lemmas for proving our main results.
Lemma 2.1
([1]) Let A and B be positive operators. Then for \(1\leq r<\infty \),
$$\begin{aligned} \left \|A^{r}+B^{r} \right \|\leq \left \| \left ( A+B \right )^{r} \right \|. \end{aligned}$$
(2.1)
Lemma 2.2
([2]) Let \(A, B\in \mathcal {B}(\mathcal{H})\) be such that \(W(A), W(B) \subset S_{\alpha}\) for some \(0\leq \alpha < \frac{\pi}{2}\). If \(f\in \mathfrak{m}\) is such that \(f^{\prime }(1)=t\) for some \(t \in (0,1)\), then where \(A!_{t}B=((1-t)A^{-1}+tB^{-1})^{-1}\) and \(A \nabla _{t}B=(1-t)A+tB\) are the weighted harmonic mean and weighted arithmetic mean, respectively. When \(t=\frac{1}{2}\), we drop t from the above notations, and we simply write ! and ∇.
$$\begin{aligned} \cos ^{2}(\alpha ) \Re (A!_{t}B) \leq \Re (A \sigma _{f} B) \leq \sec ^{2}(\alpha )\Re (A \nabla _{t}B), \end{aligned}$$
(2.2)
Lemma 2.3
([5]) Let A and B be positive operators. Then the following norm inequality holds:
$$\begin{aligned} \begin{aligned} \left \|AB \right \|\leq \frac{1}{4}\left \| A+B\right \|^{2}. \end{aligned} \end{aligned}$$
(2.3)
Lemma 2.4
([4, p. 41]) Let \(A\in \mathcal {B}(\mathcal{H})\) be strictly positive operator and Φ be unital positive linear map. Then we have
$$\begin{aligned} \Phi ^{-1}(A) \leq \Phi (A^{-1}). \end{aligned}$$
(2.4)
Lemma 2.5
Lemma 2.6
If Φ is a positive linear map and A is any complex square matrix, then
$$\begin{aligned} \Re \Phi \left ( A \right )= \Phi \left ( \Re A \right ). \end{aligned}$$
Proof
Since every positive linear map is adjoint preserving, we have □
$$\begin{aligned} \begin{aligned} \Re \Phi (A) &= \frac{\Phi (A)+\Phi (A)^{*}}{2} \\ &= \frac{\Phi (A)+\Phi (A^{*})}{2} \\ &= \Phi \left (\frac{A+A^{*}}{2}\right ) \\ &= \Phi (\Re A). \end{aligned} \end{aligned}$$
Now we first discuss whether the operator inequality (1.7) holds for \(p\geq 2\) or not.
Theorem 2.7
Let \(A, B\in \mathcal {B}(\mathcal{H})\) be such that \(0< mI \leq \Re A, \Re B\leq MI\) and \(W(A), W(B)\subset S_{\alpha }\) for some \(0\leq \alpha < \frac{\pi }{2}\). If \(f, g\in \mathfrak{m}\), then for every unital positive linear map Φ,
$$\begin{aligned} \Phi ^{p}(\Re (A\sigma _{f}B))\leq \left ( \frac{\left ( M+m \right )^{2}}{4^{\frac{2}{p}}Mm} \right )^{p}\sec ^{6p}( \alpha ) \Phi ^{p}(\Re (A\sigma _{g}B)), \qquad 2\leq p< \infty . \end{aligned}$$
(2.5)
Proof
The operator inequality (2.5) is equivalent to Since \(0< mI \leq \Re A\leq MI\), we have which means that By Lemma 2.5, we get Similarly Summing up these two operator inequalities (2.7) and (2.8), we have Compute That is, Thus, (2.6) holds. □
$$\begin{aligned} \|\Phi ^{\frac{p}{2}}(\Re (A\sigma _{f}B))\Phi ^{-\frac{p}{2}}(\Re (A \sigma _{g}B))\| \leq & \frac{(M+m)^{p}}{4M^{\frac{p}{2}}m^{\frac{p}{2}}}\sec ^{3p}(\alpha ). \end{aligned}$$
(2.6)
$$\begin{aligned} (M-\Re A)(m-\Re A)(\Re A)^{-1}\leq 0, \end{aligned}$$
$$\begin{aligned} \Re A+Mm(\Re A)^{-1}\leq (M+m)I. \end{aligned}$$
$$\begin{aligned} \frac{1}{2}\Re A+\frac{1}{2}Mm\Re (A^{-1})\leq \frac{1}{2}(M+m)I. \end{aligned}$$
(2.7)
$$\begin{aligned} \frac{1}{2}\Re B+\frac{1}{2}Mm\Re (B^{-1})\leq \frac{1}{2}(M+m)I. \end{aligned}$$
(2.8)
$$\begin{aligned} \Re (A\nabla B)+ Mm\Re ((A!B)^{-1}) \leq (M+m)I. \end{aligned}$$
(2.9)
$$ \begin{aligned} &\sec ^{p}(\alpha ) M^{\frac{p}{2}}m^{\frac{p}{2}}\|\Phi ^{ \frac{p}{2}}(\Re (A\sigma _{f}B))\Phi ^{-\frac{p}{2}}(\Re (A\sigma _{g}B)) \| \\ &\quad= \|\sec ^{p}(\alpha ) M^{\frac{p}{2}}m^{\frac{p}{2}}\Phi ^{ \frac{p}{2}}(\Re (A\sigma _{f}B))\Phi ^{-\frac{p}{2}}(\Re (A\sigma _{g}B)) \| \\ &\quad\leq \frac{1}{4}\|\sec ^{p}(\alpha )\Phi ^{\frac{p}{2}}(\Re (A \sigma _{f}B))+M^{\frac{p}{2}}m^{\frac{p}{2}}\Phi ^{-\frac{p}{2}}( \Re (A\sigma _{g}B))\|^{2}~\text{(by (2.3))} \\ &\quad\leq \frac{1}{4}\|(\sec ^{2}(\alpha )\Phi (\Re (A\sigma _{f}B))+Mm \Phi ^{-1} (\Re (A\sigma _{g}B)))^{\frac{p}{2}} \|^{2} ~\text{(by (2.1))} \\ &\quad\leq \frac{1}{4}\|\sec ^{2}(\alpha )\Phi (\Re (A\sigma _{f}B))+Mm \Phi ^{-1} (\Re (A\sigma _{g}B))\|^{p} \\ &\quad\leq \frac{1}{4} \|\sec ^{4}(\alpha ) \Phi (\Re (A\nabla B)) + Mm \Phi (\Re ^{-1}(A\sigma _{g} B)) \|^{p} ~\text{(by (2.4) and (2.2))} \\ &\quad\leq \frac{1}{4} \| \sec ^{4}(\alpha ) \Phi (\Re (A\nabla B)) + Mm \Phi ((\cos ^{2}(\alpha ) \Re (A!B))^{-1}) \|^{p} ~\text{(by (2.2))} \\ &\quad\leq \frac{1}{4}\| \sec ^{4}(\alpha ) \Phi (\Re (A\nabla B)) + \sec ^{4} (\alpha ) Mm \Phi (\Re ((A!B)^{-1})) \|^{p} ~\text{(by Lemma 2.5)} \\ &\quad= \frac{1}{4}\| \sec ^{4}(\alpha ) (\Phi (\Re (A\nabla B)) + Mm \Phi (\Re ((A!B)^{-1}))) \|^{p} \\ &\quad\leq \frac{1}{4}\sec ^{4p} (\alpha ) (M+m)^{p}.~\text{(by (2.9))} \end{aligned} $$
$$\begin{aligned} \|\Phi ^{\frac{p}{2}}(\Re (A\sigma _{f}B))\Phi ^{-\frac{p}{2}}(\Re (A \sigma _{g}B))\| \leq & \frac{(M+m)^{p}}{4M^{\frac{p}{2}}m^{\frac{p}{2}}}\sec ^{3p}(\alpha ). \end{aligned}$$
Next we present the pth power (\(p\geq 2\)) of the operator inequality (1.4) (which also means that (1.8) is valid for \(p\geq 2\)).
Theorem 2.8
Let \(A\in \mathcal {B}(\mathcal{H})\) be such that \(0< mI \leq \Re A\leq MI\) and \(W(A)\subset S_{\alpha }\) for some \(0\leq \alpha < \frac{\pi }{2}\). Then for every unital positive linear map Φ,
$$\begin{aligned} \Re ^{p}\left ( \Phi \left ( A^{-1} \right ) \right )\leq \left ( \frac{\left ( M+m \right )^{2}}{4^{\frac{2}{p}}Mm} \right )^{p}\sec ^{2p}( \alpha ) \Re ^{p}\left ( \Phi ^{-1}\left ( A \right ) \right ), \qquad 2\leq p< \infty . \end{aligned}$$
(2.10)
Proof
The operator inequality (2.10) is equivalent to In [9] Lin proved the following inequality under the condition \(0< m \leq A \leq M\): Replacing A by ℜA in inequality (2.12), we get By Lemma 2.5, we have and hence Compute
$$\begin{aligned} \left \| \Re ^{\frac{p}{2}}\left ( \Phi \left ( A^{-1} \right ) \right )\Re ^{-\frac{p}{2}}\left ( \Phi ^{-1}\left ( A \right ) \right )\right \|\leq \frac{\left ( M+m \right )^{p}}{4M^{\frac{p}{2}}m^{\frac{p}{2}}}\sec ^{p}( \alpha ) . \end{aligned}$$
(2.11)
$$\begin{aligned} Mm\Phi \left ( A^{-1} \right )+\Phi \left ( A \right )\leq M+m. \end{aligned}$$
(2.12)
$$\begin{aligned} Mm\Phi \left ( \Re ^{-1} (A) \right )+\Phi \left ( \Re A \right ) \leq M+m. \end{aligned}$$
$$\begin{aligned} Mm\Phi \left ( \Re \left ( A^{-1} \right ) \right )+\Phi \left ( \Re A \right ) \leq & Mm\Phi \left ( \Re ^{-1}(A) \right )+\Phi \left ( \Re A \right ). \end{aligned}$$
$$\begin{aligned} Mm\Phi \left ( \Re \left ( A^{-1} \right ) \right )+\Phi \left ( \Re A \right ) \leq & M+m. \end{aligned}$$
(2.13)
$$ \begin{aligned} M^{\frac{p}{2}}m^{\frac{p}{2}}\cos ^{p}(\alpha ) &\left \| \Re ^{ \frac{p}{2}}\left ( \Phi \left ( A^{-1} \right ) \right )\Re ^{- \frac{p}{2}} \left ( \Phi ^{-1}\left ( A \right ) \right )\right \| \\ & = \left \|M^{\frac{p}{2}}m^{\frac{p}{2}} \Re ^{\frac{p}{2}}\left ( \Phi \left ( A^{-1} \right ) \right )\cos ^{p}(\alpha )\Re ^{- \frac{p}{2}} \left ( \Phi ^{-1}\left ( A \right ) \right )\right \| \\ & \leq \frac{1}{4} \left \|M^{\frac{p}{2}}m^{\frac{p}{2}} \Re ^{ \frac{p}{2}}\left ( \Phi \left ( A^{-1} \right ) \right )+\cos ^{p}( \alpha ) \Re ^{-\frac{P}{2}} \left ( \Phi ^{-1}\left ( A \right ) \right )\right \|^{2}~~~~~\text{(by (2.3))} \\ &= \frac{1}{4} \left \|\left (Mm\Re \left ( \Phi \left ( A^{-1} \right ) \right )\right )^{\frac{p}{2}}+\left (\cos ^{2}(\alpha ) \Re ^{-1} \left ( \Phi ^{-1}\left ( A \right ) \right )\right )^{ \frac{p}{2}}\right \|^{2} \\ &\leq \frac{1}{4} \left \|\left (Mm\Re \left ( \Phi \left ( A^{-1} \right ) \right )+\cos ^{2}(\alpha ) \Re ^{-1} \left ( \Phi ^{-1} \left ( A \right ) \right )\right )^{\frac{p}{2}}\right \|^{2}~~~~~ \text{(by (2.1))} \\ &\leq \frac{1}{4} \left \|\left (Mm\Re \left ( \Phi \left ( A^{-1} \right ) \right )+ \Re \left ( \Phi \left ( A \right ) \right ) \right )^{\frac{p}{2}}\right \|^{2}~~~~~\text{(by Lemma 2.5))} \\ &= \frac{1}{4} \left \|\left (Mm\Phi \left ( \Re \left ( A^{-1} \right ) \right )+ \Phi \left ( \Re \left ( A \right ) \right ) \right )^{\frac{p}{2}}\right \|^{2}~~~~~\text{(by Lemma 2.6))} \\ &\leq \frac{1}{4}\left \|\left ( M+m \right )^{\frac{p}{2}} \right \|^{2}~~~~~ \text{(by (2.13))} \\ &= \frac{1}{4}\left ( M+m \right )^{p}. \end{aligned} $$
Hence, we get As every positive linear map is adjoint preserving, i.e., \(\Phi \left ( T^{*} \right )= \Phi \left ( T \right )^{*} \) for all \(T\in \mathcal {B}(\mathcal{H})\), to complete the proof, note that (2.14) is equivalent to and hence This completes the proof. □
$$\begin{aligned} \left \| \Re ^{\frac{p}{2}}\left ( \Phi \left ( A^{-1} \right ) \right )\Re ^{-\frac{p}{2}}\left ( \Phi ^{-1}\left ( A \right ) \right )\right \|\leq \frac{\left ( M+m \right )^{p}}{4M^{\frac{p}{2}}m^{\frac{p}{2}}}\sec ^{p}( \alpha ). \end{aligned}$$
(2.14)
$$\begin{aligned} \Re ^{-\frac{p}{2}}\left ( \Phi ^{-1}\left ( A \right ) \right ) \left [\Re ^{\frac{p}{2}} \left ( \Phi \left ( A^{-1} \right ) \right ) \right ]^{2}\Re ^{-\frac{p}{2}}\left (\Phi ^{-1}\left (A \right )\right )\leq \left ( \frac{\left ( M+m \right )^{2}}{4^{\frac{2}{p}}Mm} \right )^{p}\sec ^{2p}( \alpha ). \end{aligned}$$
$$\begin{aligned} \Re ^{p}\left ( \Phi \left ( A^{-1} \right ) \right )\leq \left ( \frac{\left ( M+m \right )^{2}}{4^{\frac{2}{p}}Mm} \right )^{p}\sec ^{2p}( \alpha ) \Re ^{p}\left ( \Phi ^{-1}\left ( A \right ) \right ). \end{aligned}$$
Remark 2
When \(p=2\) in Theorem 2.8, we have which is equivalent to
$$\begin{aligned} \Re ^{2}\left ( \Phi \left ( A^{-1} \right ) \right )\leq \left ( \frac{\left ( M+m \right )^{2}}{4Mm} \right )^{2}\sec ^{4}(\alpha ) \Re ^{2}\left ( \Phi ^{-1}\left ( A \right ) \right ), \end{aligned}$$
$$\begin{aligned} \left \| \Re \left ( \Phi \left ( A^{-1} \right ) \right )\Re ^{-1} \left ( \Phi ^{-1}\left ( A \right ) \right )\right \|\leq \frac{\left ( M+m \right )^{2}}{4Mm}\sec ^{2}(\alpha ) . \end{aligned}$$
(2.15)
Finally, we give an application of the inequality (2.15) which is also a generalization of [9, Theorem 2.10]. The next lemma is useful in our derivation of Theorem 2.10.
Lemma 2.9
([3]) For any bounded operator X,
$$\begin{aligned} \left |X \right |\leq tI\Leftrightarrow \left \| X\right \|\leq t \Leftrightarrow \begin{bmatrix} tI&X \\ X^{*}&tI \end{bmatrix} \geq 0. \end{aligned}$$
Theorem 2.10
Let \(A\in \mathcal {B}(\mathcal{H})\) be such that \(0< m \leq \Re A\leq M\). Then for every unital positive linear map Φ, and
$$\begin{aligned} \left |\Re \Phi \left ( A^{-1} \right )\Re ^{-1}\left ( \Phi ^{-1} \left ( A \right ) \right )+\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right )\Re \Phi \left ( A^{-1} \right ) \right |\leq \frac{\left ( M+m \right )^{2}}{2Mm}\sec ^{2}(\alpha ) . \end{aligned}$$
(2.16)
$$\begin{aligned} \Re \Phi \left ( A^{-1} \right )\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right )+\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right )\Re \Phi \left ( A^{-1} \right ) \leq \frac{\left ( M+m \right )^{2}}{2Mm}\sec ^{2}(\alpha ). \end{aligned}$$
(2.17)
Proof
By the inequality (2.15) and Lemma 2.9, we have and Summing up these two operator matrices, and let then we have By Lemma 2.9, we have Thus, (2.16) holds.
$$\begin{aligned} \begin{bmatrix} \displaystyle \frac{\left ( M+m \right )^{2}}{4Mm}\sec ^{2}(\alpha ) I & \Re \Phi \left ( A^{-1} \right )\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right ) \\ \Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right ) \Re \Phi \left ( A^{-1} \right ) & \displaystyle \frac{\left ( M+m \right )^{2}}{4Mm}\sec ^{2}(\alpha ) I \end{bmatrix} \geq 0, \end{aligned}$$
$$\begin{aligned} \begin{bmatrix} \displaystyle \frac{\left ( M+m \right )^{2}}{4Mm}\sec ^{2}(\alpha ) I &\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right ) \Re \Phi \left ( A^{-1} \right ) \\ \Re \Phi \left ( A^{-1} \right )\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right ) & \displaystyle \frac{\left ( M+m \right )^{2}}{4Mm}\sec ^{2}(\alpha ) I \end{bmatrix} \geq 0. \end{aligned}$$
$$\begin{aligned} T= \Re \Phi \left ( A^{-1} \right )\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right )+\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right )\Re \Phi \left ( A^{-1} \right ), \end{aligned}$$
$$\begin{aligned} \begin{bmatrix} \displaystyle \frac{\left ( M+m \right )^{2}}{2Mm}\sec ^{2}(\alpha ) I&T \\ T & \displaystyle \frac{\left ( M+m \right )^{2}}{2Mm}\sec ^{2}( \alpha ) I \end{bmatrix} \geq 0. \end{aligned}$$
$$\begin{aligned} \left |\Re \Phi \left ( A^{-1} \right )\Re ^{-1}\left ( \Phi ^{-1} \left ( A \right ) \right )+\Re ^{-1}\left ( \Phi ^{-1}\left ( A \right ) \right )\Re \Phi \left ( A^{-1} \right ) \right |\leq \frac{\left ( M+m \right )^{2}}{2Mm}\sec ^{2}(\alpha ). \end{aligned}$$
As T is self-adjoint, (2.17) follows from the maximal characterization of geometric mean. □
Acknowledgements
The work is supported by Hainan Provincial Natural Science Foundation (grant no. 125MS075), National Natural Science Foundation of China (grant no. 12261030), Hainan Provincial Natural Science Foundation for High-level Talents (grant no. 124RC503, 123RC474), the Hainan Province Academician Workstation (Changbin Yu), the Key Laboratory of Computational Science and Application of Hainan Province and Hainan Provincial Graduate Innovation Research Program (grant no. Qhys2024-397, Qhys2024-398). Xiaohui Fu wrote the main manuscript text and Jiqin Chen checked the proofs. All authors contributed equally to the manuscript. Jiqin Chen, Xiaohui Fu, Qi Song and Chuchu Tang are co-first authors.
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