1 Introduction and preliminaries
Let \(\mathcal{B}(\mathcal{H})\) be the \(C^{*}\)-algebra of all bounded linear operators on a complex Hilbert space \(\mathcal{H}\) with the operator norm \(\Vert \cdot \Vert \) and the identity I; also \(\mathcal{M} _{n}(\mathcal{C})\) denotes the space of all \(n\times n\) complex matrices. For an operator \(A \in \mathcal{B}(\mathcal{H})\), we write \(A\geq 0\) if A is positive, and \(A>0\) if A is positive invertible. For \(A, B \in \mathcal{B}(\mathcal{H})\), we say \(A\geq B\) if \(A-B\geq 0\). The Gelfand map \(f(t)\mapsto f(A)\) is an isometrical ∗-isomorphism between the \(C^{*}\)-algebra \(C(\operatorname {sp}(A))\) of continuous functions on the spectrum \(\operatorname {sp}(A)\) of a selfadjoint operator A and the \(C^{*}\)-algebra generated by A and I. A linear map Φ on \(\mathcal{B}(\mathcal{H})\) is positive if \(\Phi (A) \geq 0\) whenever \(A\geq 0\). It is said to be unital if \(\Phi (I)=I\). A continuous function \(f:J\rightarrow \mathcal{R}\) is operator concave if
for all selfadjoint operators \(A, B\in \mathcal{B}(\mathcal{H})\) with spectra in J and all \(\alpha \in [0,1]\).
$$\begin{aligned} f\bigl(\alpha A+(1-\alpha )B\bigr) \geq \alpha f(A)+(1-\alpha )f(B) \end{aligned}$$
The well-known Young inequality says that, for positive real numbers a, b and \(0\leq t \leq 1\), we have \(a^{t}b^{1-t}\leq t a+(1-t)b\). Refinements and reverses of this inequality are proven in [2, 9, 14‐16] and the references therein. Also Kittaneh et al. in [10] obtained the following improvement of the Young inequality for any positive definite matrices \(A,B\in \mathcal{M}_{n}(\mathcal{C})\):
where \(t\in [0,1]\), \(r=\min \{t,1-t\}\) and \(R=\max \{t,1-t\}\).
$$\begin{aligned} A^{1-t}B^{t}+r(A+B-2A\sharp B)\leq (1-t)A+t B \leq A^{1-t}B^{t}+R(A+B-2A \sharp B), \end{aligned}$$
(1)
Anzeige
Zhao et al. [19] obtained a refinement of Young’s inequality and its reverse as follows: where \(r=\min \{t,1-t\}\), \(R=\max \{t,1-t\}\) and \(r_{0}=\min \{2r,1-2r \}\).
(i)
for \(0< t\leq \frac{1}{2}\),
$$\begin{aligned} &r_{0}\bigl(\sqrt[4]{ab}-\sqrt{a}\bigr)^{2}+r( \sqrt{a}-\sqrt{b})^{2} +a ^{1-t}b^{t} \\ &\quad \leq (1-t)a+tb \\ &\quad \leq R(\sqrt{a}-\sqrt{b})^{2}-r_{0}\bigl(\sqrt[4]{ab}- \sqrt{b}\bigr)^{2}+a ^{1-t}b^{t}; \end{aligned}$$
(2)
(ii)
for \(\frac{1}{2}< t<1\),
$$\begin{aligned} &r_{0}\bigl(\sqrt[4]{ab}-\sqrt{b}\bigr)^{2}+R(\sqrt{a}- \sqrt{b})^{2} +a ^{1-t}b^{t}\\ &\quad \leq (1-t)a+tb \\ &\quad \leq r(\sqrt{a}-\sqrt{b})^{2}-r_{0}\bigl(\sqrt[4]{ab}- \sqrt{a}\bigr)^{2}+a ^{1-t}b^{t}, \end{aligned}$$
Sababheh et al. [15, 16] established some refinements and reverses of Young’s inequality as follows: where
\(s_{j}(t)= ((-1)^{r_{j}}2^{j-1}t+(-1)^{r_{j}+1}[ \frac{r_{j}+1}{2}] )\), \(r_{j}=[2^{j}t]\) and \(k_{j}=[2^{j-1}t]\). Here \([x]\) is the greatest integer less than or equal to x.
(i)
for \(0\leq t\leq \frac{1}{2}\),
$$\begin{aligned} S_{N}(t;a,b)\leq t a+(1-t)b-a^{t} b^{1-t}\leq (1-t) (\sqrt{a}- \sqrt{b})^{2}-S_{N}(2t; \sqrt{ab},a); \end{aligned}$$
(3)
(ii)
for \(\frac{1}{2}\leq t\leq 1\),
$$\begin{aligned} S_{N}(t;a,b)\leq t a+(1-t)b-a^{t} b^{1-t}\leq t( \sqrt{a}-\sqrt{b})^{2}-S _{N}(2-2t;\sqrt{ab},b), \end{aligned}$$
$$\begin{aligned} S_{N}(t;a,b)=\sum_{j=1}^{N}s_{j}(t) \bigl(\sqrt[2^{j}]{b^{2^{j-1}-k_{j}}a ^{k_{j}}}- \sqrt[2^{j}]{a^{k_{j}+1}b^{2^{j-1}-k_{j}-1}} \bigr)^{2}, \end{aligned}$$
Let \(A, B\in \mathcal{B}(\mathcal{H})\) be positive. The operator t-weighted arithmetic, geometric, and harmonic means of operators A, B are defined by \(A\nabla_{t}B=(1-t)A+tB\), \(A\sharp_{t}B=A^{ \frac{1}{2}}(A^{-\frac{1}{2}}B A^{-\frac{1}{2}})^{t}A^{\frac{1}{2}}\) and \(A!_{t}B=((1-t)A^{-1}+tB^{-1})^{-1}\), respectively. In particular, for \(t =\frac{1}{2}\) we get the usual operator arithmetic mean ∇, the geometric mean ♯ and the harmonic mean !.
2 Results and discussion
For positive real numbers \(a_{i}\) and \(b_{i} \) (\(i=1,2,\ldots,n\)) the Hölder inequality states that
for \(p,q>1\) such that \(\frac{1}{p}+\frac{1}{q}=1\). If \(p=q=2\) in (4), then we get the Cauchy–Schwarz inequality. The Hölder inequality for positive operators \(A_{i}\) and \(B_{i} \) (\(i=1,2,\ldots,n\)) is
where \(0\leq t\leq 1\). In the case \(t=\frac{1}{2}\), we get the operator Cauchy–Schwarz inequality. For further information as regards the Hölder and Cauchy–Schwarz inequalities we refer the reader to [3‐5, 11, 12, 20] and the references therein. Ando [1] proved that if Φ is a positive linear map, then for positive operators \(A, B\in \mathcal{B}(\mathcal{H})\) and \(t\in [0,1]\), we have
Recently, some authors presented several reverse-types of Ando’s inequality (see [13, 17]).
$$\begin{aligned} \sum_{i=1}^{n}a_{i}^{1/p}b_{i}^{1/q} \leq \Biggl( \sum_{i=1}^{n}a_{i} \Biggr) ^{1/p} \Biggl( \sum_{i=1}^{n}b_{i} \Biggr) ^{1/q} \end{aligned}$$
(4)
$$\begin{aligned} \sum_{i=1}^{n}A_{i} \sharp_{t} B_{i}\leq \Biggl( \sum _{i=1}^{n}A_{i} \Biggr) \sharp_{t} \Biggl( \sum_{i=1}^{n}B_{i} \Biggr) , \end{aligned}$$
$$\begin{aligned} \Phi (A\sharp_{t} B)\leq \Phi (A)\sharp_{t}\Phi (B). \end{aligned}$$
Anzeige
The Hölder–McCarthy’s inequality says that for any positive operator A and any unit vector \(x\in \mathcal{H}\), we have
Furuta [8] showed that this inequality is equivalent to Young’s inequality.
$$\begin{aligned} \bigl\langle A^{t}x,x\bigr\rangle \leq \langle Ax,x \rangle^{t}, \quad 0\leq t\leq 1. \end{aligned}$$
(5)
3 Conclusions
In this paper, we establish a reverse of Ando’s inequality for positive (non-unital) linear maps and positive definite matrices by using an inequality due to Sababheh. We obtain some reverses of the matrix Hölder and Cauchy–Schwarz inequalities and a reverse of inequality (5) for \(t\in (0,\frac{1}{2}]\) as follows:
$$\begin{aligned} &\langle Tx,x \rangle^{t}-\bigl\langle T^{t}x,x\bigr\rangle \\ & \quad \leq 2R \bigl( \langle Tx,x\rangle^{\frac{1}{2}}-\bigl\langle T^{ \frac{1}{2}}x,x\bigr\rangle \bigr) -r_{0} \bigl( \bigl\langle T^{\frac{1}{2}}x,x \bigr\rangle + \langle Tx,x\rangle^{\frac{1}{2}}-2\bigl\langle T^{\frac{1}{4}}x,x \bigr\rangle \langle Tx,x\rangle^{\frac{1}{4}} \bigr) . \end{aligned}$$
5 Main results
To prove our first result, we need the following lemmas.
Lemma 1
([16])
Let
\(A, B\in \mathcal{M}_{n}(\mathcal{C})\)
be positive definite matrices and
\(t\in [0,1]\). Then
where
\(\alpha_{j}(t)=\frac{k_{j}(t)}{2^{j-1}}\).
$$\begin{aligned} \sum_{j=1}^{N}s_{j}(t) ( A\sharp_{\alpha_{j}(t)}B+A \sharp_{2^{1-j}+\alpha_{j}(t)}B-2A\sharp_{2^{-j}+\alpha_{j}(t)}B ) +A \sharp_{t}B\leq A\nabla_{t}B, \end{aligned}$$
(6)
For \(N=2\), we have the following lemma, which is shown in [19] for positive invertible operators.
Lemma 2
Let
\(A, B\in \mathcal{B}(\mathcal{H})\)
be positive invertible operators and
\(t\in [0,1]\).
(i)
If
\(0< t\leq \frac{1}{2}\), then
$$\begin{aligned} r_{0}(A\sharp B-2A\sharp_{\frac{1}{4}}B+A)+2t(A\nabla B-A\sharp B)+A \sharp_{t}B\leq A\nabla_{t}B. \end{aligned}$$
(7)
(ii)
If
\(\frac{1}{2}< t<1\), then
where
\(r=\min \{\nu ,1-\nu \}\)
and
\(r_{0}=\min \{2r,1-2r\}\).
$$\begin{aligned} r_{0}(A\sharp B-2A\sharp_{\frac{3}{4}}B+B)+2(1-t) (A \nabla B-A\sharp B)+A \sharp_{t}B\leq A\nabla_{t}B, \end{aligned}$$
(8)
Lemma 3
([16])
Let
\(A, B\in \mathcal{M}_{n}(\mathcal{C})\)
be positive definite matrices and
\(t\in [0,1]\).
(i)
If
\(0\leq t\leq \frac{1}{2}\), then
$$\begin{aligned} A\nabla_{t}B&\leq A\sharp_{t}B+2(1-t) (A \nabla B-A\sharp B) \\ &\quad {} -\sum_{j=1}^{N}s_{j}(2t) ( A \sharp_{1-\beta_{j}(t)}B+A \sharp_{1+2^{-j}-\beta_{j}(t)}B-2A\sharp_{1-2^{-j-1}-\beta_{j}(t)}B ) . \end{aligned}$$
(9)
(ii)
If
\(\frac{1}{2}\leq t\leq 1\), then
where
\(\beta_{j}(t)=2^{-j}k_{j}(2t)\)
and
\(\gamma_{j}(t)=2^{1-j}k _{j}(2-2t)\).
$$\begin{aligned} A\nabla_{t}B &\leq A\sharp_{t}B+2t(A\nabla B-A\sharp B) \\ & \quad {} -\sum_{j=1}^{N}s_{j}(2-2t) ( A \sharp_{\gamma_{j}(t)}B+A \sharp_{\gamma_{j}(t)+2^{1-j}}B-2A\sharp_{\gamma_{j}(t)+2^{-j}}B ) , \end{aligned}$$
Remark 4
For \(N=2\), we have the following lemma, which is shown in [19] for positive invertible operators.
Lemma 5
Let
\(A, B\in \mathcal{B}(\mathcal{H})\)
be positive invertible operators and
\(t\in [0,1]\).
(i)
If
\(0< t\leq \frac{1}{2}\), then
$$\begin{aligned} A\nabla_{t}B \leq A\sharp_{t}B+2(1-t) (A\nabla B-A\sharp B)-r_{0}(A \sharp B-2A\sharp_{\frac{3}{4}}B+B). \end{aligned}$$
(10)
(ii)
If
\(\frac{1}{2}< t<1\), then
where
\(r=\min \{\nu ,1-\nu \}\)
and
\(r_{0}=\min \{2r,1-2r\}\).
$$\begin{aligned} A\nabla_{t}B \leq A\sharp_{t}B+2t(A\nabla B-A \sharp B)-r_{0}(A\sharp B-2A\sharp_{\frac{1}{4}}B+A), \end{aligned}$$
(11)
Now, we obtain a reverse of Ando’s inequality for positive invertible operators as follows.
Theorem 6
Let
\(A, B\in \mathcal{B}(\mathcal{H})\)
be positive invertible operators, Φ be a positive linear map and
\(t\in [0,1]\).
(i)
If
\(0\leq t\leq \frac{1}{2}\), then
$$\begin{aligned} & \Phi (A) \sharp_{t}\Phi (B)-\Phi (A \sharp_{t}B) \\ &\quad \leq 2R\biggl(\Phi (A)\sharp \Phi (B)-\Phi (A\sharp B)+ \frac{1}{2}\bigl(\Phi (A)+\Phi (B)-2\Phi (A) \sharp \Phi (B)\bigr)\biggr) \\ &\quad \quad {} -\sum_{j=1}^{N}s_{j}(2t) \bigl( \Phi (A\sharp_{1-\beta_{j}(t)}B)+ \Phi (A\sharp_{1+2^{-j}-\beta_{j}(t)}B)- 2\Phi (A \sharp_{1-2^{-j-1}-\beta_{j}(t)}B) \bigr) \\ &\quad \quad {} -\sum_{j=1}^{N}s_{j}(t) \bigl( \Phi (A)\sharp_{\alpha_{j}(t)}\Phi (B)+ \Phi (A)\sharp_{2^{1-j}+\alpha_{j}(t)}\Phi (B) \\ &\quad \quad {}-2\Phi (A) \sharp_{2^{-j}+\alpha_{j}(t)}\Phi (B) \bigr) . \end{aligned}$$
(12)
(ii)
If
\(\frac{1}{2}\leq t\leq 1\), then
where
\(\alpha_{j}(t)=\frac{k_{j}(t)}{2^{j-1}}\), \(\beta_{j}(t)=2^{-j}k _{j}(2t)\), \(\gamma_{j}(t)=2^{1-j}k_{j}(2-2t)\)
and
\(R=\max \{t,1-t\}\).
$$\begin{aligned} &\Phi (A) \sharp_{t}\Phi (B)-\Phi (A \sharp_{t}B) \\ &\quad \leq 2R\biggl(\Phi (A)\sharp \Phi (B)-\Phi (A\sharp B)+ \frac{1}{2}\bigl(\Phi (A)+\Phi (B)-2\Phi (A) \sharp \Phi (B)\bigr)\biggr) \\ &\quad \quad {} -\sum_{j=1}^{N}s_{j}(2-2t) \bigl( \Phi (A\sharp_{\gamma_{j}(t)}B)+ \Phi (A\sharp_{\gamma_{j}(t)+2^{1-j}}B)-2\Phi (A \sharp_{\gamma_{j}(t)+2^{-j}}B) \bigr) \\ &\quad \quad {} -\sum_{j=1}^{N}s_{j}(t) \bigl( \Phi (A)\sharp_{\alpha_{j}(t)}\Phi (B)+ \Phi (A)\sharp_{2^{1-j}+\alpha_{j}(t)}\Phi (B) \\ &\quad \quad {}-2\Phi (A) \sharp_{2^{-j}+\alpha_{j}(t)}\Phi (B) \bigr), \end{aligned}$$
(13)
Proof
The proof of inequality (13) is similar to the proof of inequality (12). Thus, we only prove inequality (12).
Let \(0\leq t\leq \frac{1}{2}\). Applying inequalities (10) and (9), we have
Now, using the positive linear map Φ on (14), we get
Moreover, if we replace A and B by \(\Phi (A)\) and \(\Phi (B)\) in inequality (14), respectively, then
From the first inequality of (16) and the second inequality of (15), we have
which implies that
Therefore, applying inequality (1), we get
□
$$\begin{aligned} &\sum_{j=1}^{N}s_{j}(t) (A\sharp_{\alpha_{j}(t)}B+A \sharp_{2^{1-j}+\alpha_{j}(t)}B-2A\sharp_{2^{-j}+\alpha_{j}(t)}B) \\ &\quad \leq A\nabla_{t} B-A\sharp_{t}B \\ &\quad \leq 2R(A\nabla B-A\sharp B) -\sum_{j=1}^{N}s_{j}(2t) ( A \sharp_{1-\beta_{j}(t)}B+A\sharp_{1+2^{-j}-\beta_{j}(t)}B-2A \sharp_{1-2^{-j-1}-\beta_{j}(t)}B ) . \end{aligned}$$
(14)
$$\begin{aligned} &\sum_{j=1}^{N} s_{j}(t) \bigl( \Phi (A\sharp_{\alpha_{j}(t)}B)+\Phi (A \sharp_{2^{1-j}+\alpha_{j}(t)}B)-2\Phi (A\sharp_{2^{-j}+\alpha_{j}(t)}B) \bigr) +\Phi (A \sharp_{t}B) \\ &\quad \leq \Phi (A)\nabla_{t} \Phi (B) \\ &\quad \leq 2R \bigl( \Phi (A)\nabla \Phi (B)-\Phi (A\sharp B) \bigr) +\Phi (A \sharp_{t}B) \\ & \quad\quad {} -\sum_{j=1}^{N}s_{j}(2t) \bigl( \Phi (A\sharp_{1-\beta_{j}(t)}B)+ \Phi (A\sharp_{1+2^{-j}-\beta_{j}(t)}B)-2\Phi (A \sharp_{1-2^{-j-1}-\beta_{j}(t)}B) \bigr) . \end{aligned}$$
(15)
$$\begin{aligned} &\sum_{j=1}^{N}s_{j}(t) \bigl( \Phi (A)\sharp_{\alpha_{j}(t)}\Phi (B)+ \Phi (A)\sharp_{2^{1-j}+ \alpha_{j}(t)}\Phi (B)-2\Phi (A) \sharp_{2^{-j}+\alpha_{j}(t)}\Phi (B) \bigr) \\ &\quad\quad {} +\Phi (A)\sharp_{t}\Phi (B) \\ &\quad \leq \Phi (A)\nabla_{t}\Phi (B) \\ &\quad \leq 2R \bigl( \Phi (A)\nabla \Phi (B)-\Phi (A\sharp B) \bigr) +\Phi (A) \sharp_{t}\Phi (B) \\ &\quad\quad {} -\sum_{j=1}^{N}s_{j}(2t) \bigl( \Phi (A)\sharp_{1-\beta_{j}(t)}\Phi (B)+ \Phi (A)\sharp_{1+2^{-j}-\beta_{j}(t)}\Phi (B) \\ &\quad\quad {}-2\Phi (A) \sharp_{1-2^{-j-1}-\beta_{j}(t)}\Phi (B) \bigr) . \end{aligned}$$
(16)
$$\begin{aligned} &\sum_{j=1}^{N} s_{j}(t) \bigl( \Phi (A)\sharp_{\alpha_{j}(t)}\Phi (B)+ \Phi (A)\sharp_{2^{1-j}+ \alpha_{j}(t)}\Phi (B)-2\Phi (A) \sharp_{2^{-j}+\alpha_{j}(t)}\Phi (B) \bigr) \\ & \quad\quad {} +\Phi (A)\sharp_{t}\Phi (B) \\ & \quad \leq \Phi (A)\nabla_{t}\Phi (B) \\ &\quad \leq 2R\bigl(\Phi (A)\nabla \Phi (B)-\Phi (A\sharp B)\bigr)+\Phi (A \sharp_{t}B) \\ & \quad\quad {} -\sum_{j=1}^{N}s_{j}(2t) \bigl( \Phi (A\sharp_{1-\beta_{j}(t)}B)+ \Phi (A\sharp_{1+2^{-j}-\beta_{j}(t)}B)-2\Phi (A \sharp_{1-2^{-j-1}-\beta_{j}(t)}B) \bigr) , \end{aligned}$$
$$\begin{aligned} &\sum_{j=1}^{N} s_{j}(t) \bigl( \Phi (A)\sharp_{\alpha_{j}(t)}\Phi (B)+ \Phi (A)\sharp_{2^{1-j}+ \alpha_{j}(t)}\Phi (B)-2\Phi (A) \sharp_{2^{-j}+\alpha_{j}(t)}\Phi (B) \bigr) \\ & \quad\quad {} +\Phi (A)\sharp_{t}\Phi (B) \\ & \quad \leq 2R \bigl( \Phi (A)\nabla \Phi (B)-\Phi (A\sharp B) \bigr) +\Phi (A \sharp_{t}B) \\ &\quad\quad {} -\sum_{j=1}^{N}s_{j}(2t) \bigl( \Phi (A\sharp_{1-\beta_{j}(t)}B)+ \Phi (A\sharp_{1+2^{-j}-\beta_{j}(t)}B)-2\Phi (A \sharp_{1-2^{-j-1}-\beta_{j}(t)}B) \bigr) . \end{aligned}$$
$$\begin{aligned} &\Phi (A) \sharp_{t}\Phi (B)-\Phi (A\sharp_{t}B) \\ &\quad \leq 2R \bigl(\Phi (A)\nabla \Phi (B)-\Phi (A\sharp B)\bigr) \\ & \quad \quad {} -\sum_{j=1}^{N}s_{j}(2t) \bigl( \Phi (A\sharp_{1-\beta_{j}(t)}B)+ \Phi (A\sharp_{1+2^{-j}-\beta_{j}(t)}B)- 2\Phi (A \sharp_{1-2^{-j-1}-\beta_{j}(t)}B) \bigr) \\ &\quad \quad {} -\sum_{j=1}^{N}s_{j}(t) \bigl( \Phi (A)\sharp_{\alpha_{j}(t)}\Phi (B)+ \Phi (A)\sharp_{2^{1-j}+\alpha_{j}(t)}\Phi (B)-2\Phi (A) \sharp_{2^{-j}+\alpha_{j}(t)}\Phi (B) \bigr) \\ & \quad \leq 2R\biggl(\Phi (A)\sharp \Phi (B)-\Phi (A\sharp B)+\frac{1}{2}\bigl( \Phi (A)+ \Phi (B)-2\Phi (A)\sharp \Phi (B)\bigr)\biggr) \\ & \quad \quad {} -\sum_{j=1}^{N}s_{j}(2t) \bigl( \Phi (A\sharp_{1-\beta_{j}(t)}B)+ \Phi (A\sharp_{1+2^{-j}-\beta_{j}(t)}B)- 2\Phi (A \sharp_{1-2^{-j-1}-\beta_{j}(t)}B) \bigr) \\ &\quad \quad {} -\sum_{j=1}^{N}s_{j}(t) \bigl( \Phi (A)\sharp_{\alpha_{j}(t)}\Phi (B)+ \Phi (A)\sharp_{2^{1-j}+\alpha_{j}(t)}\Phi (B)-2\Phi (A) \sharp_{2^{-j}+\alpha_{j}(t)}\Phi (B) \bigr) . \end{aligned}$$
Similarly for \(N=2\) by applying Lemma 2 and Lemma 5, we can obtain a reverse of Ando’s inequality for positive invertible operators.
Corollary 7
Let
\(A, B\in \mathcal{B}(\mathcal{H})\)
be positive invertible operators, Φ be a positive linear map and
\(t\in [0,1]\).
(i)
If
\(0< t\leq \frac{1}{2}\), then
$$\begin{aligned} & \Phi (A)\sharp_{t}\Phi (B)-\Phi (A\sharp_{t}B) \\ &\quad \leq 2R\biggl(\Phi (A)\sharp \Phi (B)-\Phi (A\sharp B)+\frac{1}{2}\bigl( \Phi (A)+\Phi (B)-2\Phi (A) \sharp \Phi (B)\bigr)\biggr) \\ &\quad \quad {} -r_{0}\bigl(\Phi (A\sharp B)+\Phi (B)-2\Phi (A\sharp_{\frac{3}{4}}B) \bigr) \\ &\quad \quad {} -r_{0}\bigl(\Phi (A)\sharp \Phi (B)+\Phi (A)-2\bigl(\Phi (A) \sharp_{ \frac{1}{4}}\Phi (B)\bigr)\bigr) \\ & \quad \leq 2R\biggl(\Phi (A)\sharp \Phi (B)-\Phi (A\sharp B)+\frac{1}{2}\bigl( \Phi (A)+ \Phi (B)-2\Phi (A)\sharp \Phi (B)\bigr)\biggr); \end{aligned}$$
(17)
(ii)
if
\(\frac{1}{2}< t< 1\), then
where
\(r=\min \{t,1-t\}\), \(R=\max \{t,1-t\}\)
and
\(r_{0}=\min \{2r,1-2r \}\).
$$\begin{aligned} &\Phi (A)\sharp_{t}\Phi (B)-\Phi (A\sharp_{t}B) \\ &\quad \leq 2R\biggl(\Phi (A)\sharp \Phi (B)-\Phi (A\sharp B)+\frac{1}{2}\bigl( \Phi (A)+\Phi (B)-2\Phi (A) \sharp \Phi (B)\bigr)\biggr) \\ &\quad \quad {} -r_{0}\bigl(\Phi (A)\sharp \Phi (B)+\Phi (B)-2\bigl(\Phi (A) \sharp_{ \frac{3}{4}}\Phi (B)\bigr)\bigr) \\ &\quad \quad {} -r_{0}\bigl(\Phi (A\sharp B)+\Phi (A)-2\Phi (A\sharp_{\frac{1}{4}} B)\bigr) \\ &\quad \leq 2R\biggl(\Phi (A)\sharp \Phi (B)-\Phi (A\sharp B)+\frac{1}{2}\bigl( \Phi (A)+ \Phi (B)-2\Phi (A)\sharp \Phi (B)\bigr)\biggr), \end{aligned}$$
(18)
We want to establish some inequalities for positive invertible operators.
Theorem 8
Let
\(A, B\in \mathcal{B}(\mathcal{H})\)
be positive invertible. If
\(t\in [0,1]\)
and Φ, Ψ are two unital positive linear maps, then for any unit vector
\(x\in \mathcal{H}\)
where
\(r=\min \{t,1-t\}\), \(R=\max \{t,1-t\}\), \(r_{0}=\min \{2r,1-2r\}\).
(i)
for
\(0< t\leq \frac{1}{2}\),
$$\begin{aligned} &2r \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle \nabla \bigl\langle \Psi (B)x,x\bigr\rangle -\bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x \bigr\rangle \bigl\langle \Psi \bigl(B^{1/2}\bigr)x,x \bigr\rangle \bigr) \\ &\quad\quad {} + r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x \bigr\rangle \bigr)\bigl\langle \Psi \bigl(B^{\frac{1}{2}}\bigr)x,x\bigr\rangle + \bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{\frac{3}{4}} \bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{\frac{1}{4}}\bigr)x,x \bigr\rangle \bigr) \\ &\quad \leq (1-t)\bigl\langle \Phi (A)x,x\bigr\rangle +t\bigl\langle \Psi (B)x,x\bigr\rangle - \bigl\langle \Psi \bigl(B^{t}\bigr)x,x\bigr\rangle \bigl\langle \Phi \bigl(A^{1-t}\bigr)x,x\bigr\rangle \\ &\quad \leq 2R \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle \nabla \bigl\langle \Psi (B)x,x \bigr\rangle -\bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{ \frac{1}{2}}\bigr)x,x\bigr\rangle \bigr) \\ &\quad\quad {} -r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B ^{\frac{1}{2}}\bigr)x,x\bigr\rangle +\bigl\langle \Psi (B)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{\frac{1}{4}} \bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{\frac{3}{4}}\bigr)x,x \bigr\rangle \bigr) ; \end{aligned}$$
(19)
(ii)
for
\(\frac{1}{2}< t<1\),
$$\begin{aligned} &R \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +\bigl\langle \Psi (B)x,x\bigr\rangle -2 \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{1/2}\bigr)x,x \bigr\rangle \bigr) \\ & \quad\quad {} + r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x \bigr\rangle \bigr)\bigl\langle \Psi \bigl(B^{\frac{1}{2}}\bigr)x,x\bigr\rangle + \bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{\frac{1}{4}} \bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{\frac{3}{4}}\bigr)x,x \bigr\rangle \bigr) \\ & \quad \leq (1-t)\bigl\langle \Phi (A)x,x\bigr\rangle +t\bigl\langle \Psi (B)x,x\bigr\rangle - \bigl\langle \Psi \bigl(B^{t}\bigr)x,x\bigr\rangle \bigl\langle \Phi \bigl(A^{1-t}\bigr)x,x\bigr\rangle \\ & \quad \leq r \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +\bigl\langle \Psi (B)x,x \bigr\rangle -2\bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{ \frac{1}{2}}\bigr)x,x\bigr\rangle \bigr) \\ &\quad\quad {} -r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B ^{\frac{1}{2}}\bigr)x,x\bigr\rangle +\bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{\frac{3}{4}} \bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{\frac{1}{4}}\bigr)x,x \bigr\rangle \bigr) , \end{aligned}$$
Proof
The proof of part (ii) is similar to the proof of part (i). Thus we only prove (i).
Applying inequality (2) for any positive real numbers k, s, we have
Fix s in (20). Then applying functional calculus to the operator A, we have
If we apply the positive linear map Φ and inner product for \(x\in {\mathcal{H}}\) with \(\Vert x \Vert =1\) in inequality (21), we have
Now, using the functional calculus to the operator B, we have
Taking the positive linear map Ψ and the inner product for \(y\in {\mathcal{H}}\) with \(\Vert y \Vert =1\), we get
Now, if we put \(x=y\), then we get the desired result. □
$$\begin{aligned} &r(k+s-2\sqrt{ks})+ r_{0}\bigl(k^{\frac{1}{2}}s^{\frac{1}{2}}+k-2k^{ \frac{3}{4}}s^{\frac{1}{4}} \bigr) \\ &\quad \leq (1-t)k+ts-k^{1-t}s^{t} \\ &\quad \leq R(k+s-2\sqrt{ks})-r_{0}\bigl(k^{\frac{1}{2}}s^{\frac{1}{2}}+s-2k ^{\frac{1}{4}}s^{\frac{3}{4}}\bigr). \end{aligned}$$
(20)
$$\begin{aligned} & r\bigl(A+sI-2\sqrt{s}A^{\frac{1}{2}}\bigr)+ r_{0} \bigl(A^{\frac{1}{2}}s^{ \frac{1}{2}}+A-2A^{\frac{3}{4}}s^{\frac{1}{4}}\bigr) \\ & \quad \leq (1-t)A+tsI-s^{t}A^{1-t} \\ & \quad \leq R\bigl(A+sI-2\sqrt{s}A^{\frac{1}{2}}\bigr)-r_{0} \bigl(A^{\frac{1}{2}}s^{ \frac{1}{2}}+sI-2A^{\frac{1}{4}}s^{\frac{3}{4}}\bigr). \end{aligned}$$
(21)
$$\begin{aligned} &r \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +s-2\sqrt{s}\bigl\langle \Phi \bigl(A^{ \frac{1}{2}}\bigr)x,x\bigr\rangle \bigr) \\ &\quad\quad {} + r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x \bigr\rangle \bigr)s^{ \frac{1}{2}}+\bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{ \frac{3}{4}}\bigr)x,x\bigr\rangle s^{\frac{1}{4}} \bigr) \\ &\quad \leq (1-t)\bigl\langle \Phi (A)x,x\bigr\rangle +ts-s^{t}\bigl\langle \Phi \bigl(A^{1-t}\bigr)x,x \bigr\rangle \\ &\quad \leq R \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +s-2\sqrt{s}\bigl\langle \Phi \bigl(A ^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigr) -r_{0} \bigl( \bigl\langle \Phi \bigl(A^{ \frac{1}{2}}\bigr)x,x\bigr\rangle s^{\frac{1}{2}}+s-2 \bigl\langle \Phi \bigl(A^{ \frac{1}{4}}\bigr)x,x\bigr\rangle s^{\frac{3}{4}} \bigr) . \end{aligned}$$
$$\begin{aligned} &r \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +B-2\bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x \bigr\rangle B^{1/2} \bigr) \\ & \quad\quad {} + r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x \bigr\rangle \bigr)B^{ \frac{1}{2}}+\bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{ \frac{3}{4}}\bigr)x,x\bigr\rangle B^{\frac{1}{4}} ) \\ & \quad \leq (1-t)\bigl\langle \Phi (A)x,x\bigr\rangle +tB-B^{t}\bigl\langle \Phi \bigl(A^{1-t}\bigr)x,x \bigr\rangle \\ &\quad \leq R \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +B-2\bigl\langle \Phi \bigl(A^{ \frac{1}{2}}\bigr)x,x\bigr\rangle B^{\frac{1}{2}} \bigr) -r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle B^{\frac{1}{2}}+B-2\bigl\langle \Phi \bigl(A ^{\frac{1}{4}}\bigr)x,x \bigr\rangle B^{\frac{3}{4}} \bigr) . \end{aligned}$$
$$\begin{aligned} &r \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +\bigl\langle \Psi (B)y,y\bigr\rangle -2 \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{1/2}\bigr)x,x \bigr\rangle \bigr) \\ & \quad\quad {} + r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x \bigr\rangle \bigr)\bigl\langle \Psi \bigl(B^{\frac{1}{2}}\bigr)x,x\bigr\rangle + \bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{\frac{3}{4}} \bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{\frac{1}{4}}\bigr)x,x \bigr\rangle \bigr) \\ & \quad \leq (1-t)\bigl\langle \Phi (A)x,x\bigr\rangle +t\bigl\langle \Psi (B)x,x\bigr\rangle - \bigl\langle \Psi \bigl(B^{t}\bigr)x,x\bigr\rangle \bigl\langle \Phi \bigl(A^{1-t}\bigr)x,x\bigr\rangle \\ & \quad \leq R \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +\bigl\langle \Psi (B)x,x \bigr\rangle -2\bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigl\langle \bigl(B^{\frac{1}{2}}\bigr)x,x \bigr\rangle \bigr) \\ & \quad\quad {} -r_{0} \bigl( \bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x \bigr\rangle \bigl\langle \Psi \bigl(B ^{\frac{1}{2}}\bigr)x,x\bigr\rangle +\bigl\langle \Psi (B)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{\frac{1}{4}} \bigr)x,x\bigr\rangle \bigl\langle \Psi \bigl(B^{\frac{3}{4}}\bigr)x,x \bigr\rangle \bigr) . \end{aligned}$$
Theorem 9
Let
\(A, B\in \mathcal{B}(\mathcal{H})\)
be positive invertible. If
\(t\in [0,1]\)
and Φ, Ψ are two unital positive linear maps, then for any unit vector
\(x\in \mathcal{H}\)
where
\(r=\min \{t,1-t\}\), \(R=\max \{t,1-t\}\), \(r_{0}=\min \{2r,1-2r \}\).
(i)
for
\(0< t\leq \frac{1}{2}\),
$$\begin{aligned} &2r \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle \nabla \bigl\langle \Psi (B)x,x \bigr\rangle -\bigl\langle \Phi \bigl(A^{\frac{1}{2}}\bigr)x,x\bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2} \bigr) \\ & \quad\quad {} + r_{0} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x \bigr\rangle \bigl\langle \Psi (B)x,x \bigr\rangle ^{1/2}+\bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{3/4}\bigr)x,x \bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/4} \bigr) \\ &\quad \leq (1-t)\bigl\langle \Phi (A)x,x\bigr\rangle +t\bigl\langle \Psi (B)x,x\bigr\rangle - \bigl\langle \Psi (B)x,x\bigr\rangle ^{t}\bigl\langle \Phi \bigl(A^{1-t}\bigr)x,x\bigr\rangle \\ & \quad \leq 2R \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle \nabla \bigl\langle \Psi (B)x,x \bigr\rangle -\bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2} \bigr) \\ & \quad\quad {} -r_{0} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle \bigl\langle \Psi (B)x,x \bigr\rangle ^{1/2}+\bigl\langle \Psi (B)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{1/4}\bigr)x,x \bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{3/4} \bigr) ; \end{aligned}$$
(ii)
for
\(\frac{1}{2}< t<1\),
$$\begin{aligned} &R \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +\bigl\langle \Psi (B)x,x\bigr\rangle -2 \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2} \bigr) \\ &\quad\quad {} + r_{0} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x \bigr\rangle \bigr)\bigl\langle \Psi (B)x,x \bigr\rangle ^{1/2}+\bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{1/4} \bigr)x,x \bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{3/4} \bigr) \\ & \quad \leq (1-t)\bigl\langle \Phi (A)x,x\bigr\rangle +t\bigl\langle \Psi (B)x,x\bigr\rangle - \bigl\langle \Psi (B)x,x\bigr\rangle ^{t}\bigl\langle \Phi \bigl(A^{1-t}\bigr)x,x\bigr\rangle \\ & \quad \leq r \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle +\bigl\langle \Psi (B)x,x \bigr\rangle -2\bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2} \bigr) \\ &\quad\quad {} -r_{0} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle \bigl\langle \Psi (B)x,x \bigr\rangle ^{1/2}+\bigl\langle \Phi (A)x,x\bigr\rangle -2\bigl\langle \Phi \bigl(A^{3/4}\bigr)x,x \bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/4} \bigr) , \end{aligned}$$
Proof
The proof of part (ii) is similar to the proof of part (i). Thus we just prove (i). For any positive real number k and any unit vector \(x\in \mathcal{H}\), we have
Applying inequality (22) and the functional calculus for the operator A, we have
Now, using the unital positive operator Φ and the inner product for \(y\in \mathcal{H}\) with \(\Vert y \Vert =1\) in inequality (23), we get
Now, putting \(y=x\), we get the desired result. □
$$\begin{aligned} &r \bigl( k+\bigl\langle \Psi (B)x,x\bigr\rangle -2\sqrt{k}\bigl\langle \Psi (B)x,x \bigr\rangle \bigr) +r_{0} \bigl( k^{1/2} \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2} +k-2k ^{3/4}\bigl\langle \Psi (B)x,x\bigr\rangle ^{1/4} \bigr) \\ &\quad \leq (1-t)k+t\bigl\langle \Psi (B)x,x\bigr\rangle -k^{1-t}\bigl\langle \Psi (B)x,x \bigr\rangle ^{t} \\ & \quad \leq R \bigl( k+\bigl\langle \Psi (B)x,x\bigr\rangle -2\sqrt{k}\bigl\langle \Psi (B)x,x \bigr\rangle ^{1/2} \bigr) \\ & \quad\quad {} -r_{0} \bigl( k^{1/2}\bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2}+ \bigl\langle \Psi (B)x,x \bigr\rangle -2k^{1/4} \bigl\langle \Psi (B)x,x\bigr\rangle ^{3/4} \bigr) . \end{aligned}$$
(22)
$$\begin{aligned} &r \bigl(A+\bigl\langle \Psi (B)x,x\bigr\rangle I_{\mathcal{H}} -2\sqrt{A}\bigl\langle \Psi (B)x,x\bigr\rangle \bigr) \end{aligned}$$
(23)
$$\begin{aligned} &\quad\quad {} +r_{0} \bigl(A^{1/2}\bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2} +A-2A^{3/4}\bigl\langle \Psi (B)x,x\bigr\rangle ^{1/4} \bigr) \\ &\quad \leq (1-t)A+t\bigl\langle \Psi (B)x,x\bigr\rangle I_{\mathcal{H}}-A^{1-t} \bigl\langle \Psi (B)x,x\bigr\rangle ^{t} \\ & \quad \leq R \bigl(A+\bigl\langle \Psi (B)x,x\bigr\rangle I_{\mathcal{H}}-2\sqrt{A} \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2} \bigr) \\ &\quad\quad {} -r_{0} \bigl(A^{1/2}\bigl\langle \Psi (B)x,x\bigr\rangle ^{1/2}+ \bigl\langle \Psi (B)x,x \bigr\rangle I_{\mathcal{H}}-2A^{1/4} \bigl\langle \Psi (B)x,x\bigr\rangle ^{3/4} \bigr). \end{aligned}$$
(24)
$$\begin{aligned} &r \bigl( \bigl\langle \Phi (A)y,y\bigr\rangle +\bigl\langle \Psi (B)x,x\bigr\rangle I_{ \mathcal{H}}-2\bigl\langle \Phi (A)y,y\bigr\rangle ^{1/2}\bigl\langle \Psi (B)x,x \bigr\rangle \bigr) \\ & \quad\quad {} +r_{0} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)y,y\bigr\rangle \bigl\langle \Psi (B)x,x \bigr\rangle ^{1/2} +\bigl\langle \Phi (A)y,y\bigr\rangle -2\bigl\langle \Phi \bigl(A^{3/4}\bigr)y,y \bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{1/4} \bigr) \\ &\quad \leq (1-t)\bigl\langle \Phi (A)y,y\bigr\rangle +t\bigl\langle \Psi (B)x,x\bigr\rangle I _{\mathcal{H}}-\bigl\langle \Phi \bigl(A^{1-t}\bigr)y,y\bigr\rangle \bigl\langle \Psi (B)x,x \bigr\rangle ^{t} \\ & \quad \leq R \bigl( \bigl\langle \Phi (A)y,y\bigr\rangle +\bigl\langle \Psi (B)x,x \bigr\rangle I_{\mathcal{H}}-2\bigl\langle \Phi (A)y,y\bigr\rangle ^{1/2} \bigl\langle \Psi (B)x,x \bigr\rangle ^{1/2} \bigr) \\ & \quad\quad {} -r_{0} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)y,y\bigr\rangle \bigl\langle \Psi (B)x,x \bigr\rangle ^{1/2}+ \bigl\langle \Psi (B)x,x\bigr\rangle I_{\mathcal{H}}-2\bigl\langle \Phi \bigl(A^{1/4} \bigr)y,y\bigr\rangle \bigl\langle \Psi (B)x,x\bigr\rangle ^{3/4} \bigr) . \end{aligned}$$
Corollary 10
Let
\(A\in \mathcal{B}(\mathcal{H})\)
be positive, Φ be a unital positive linear map and
\(t\in [0,1]\). Then for any unit vector
\(x\in \mathcal{H}\)
where
\(r=\min \{t,1-t\}\), \(R=\max \{t,1-t\}\), \(r_{0}=\min \{2r,1-2r \}\).
(i)
for
\(0< t\leq \frac{1}{2}\),
$$\begin{aligned} &2r\bigl\langle \Phi (A)x,x\bigr\rangle ^{t-\frac{1}{2}} \bigl( \bigl\langle \Phi (A)x,x \bigr\rangle ^{\frac{1}{2}}-\bigl\langle \Phi \bigl(A^{\frac{1}{2}} \bigr)x,x\bigr\rangle \bigr) \\ & \quad\quad {} + r_{0}\bigl\langle \Phi (A)x,x\bigr\rangle ^{t-\frac{1}{2}} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle + \bigl\langle \Phi (A)x,x\bigr\rangle ^{1/2} \\ & \quad\quad {}-2\bigl\langle \Phi \bigl(A^{3/4} \bigr)x,x\bigr\rangle \bigl\langle \Phi (A)x,x\bigr\rangle ^{-1/4} \bigr) \\ & \quad \leq \bigl\langle \Phi (A)x,x\bigr\rangle ^{t}-\bigl\langle \Phi \bigl(A^{t}\bigr)x,x\bigr\rangle \\ & \quad \leq 2R\bigl\langle \Phi (A)x,x\bigr\rangle ^{t-\frac{1}{2}} \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle ^{\frac{1}{2}}-\bigl\langle \Phi \bigl(A^{\frac{1}{2}} \bigr)x,x \bigr\rangle \bigr) \\ & \quad\quad {} - r_{0}\bigl\langle \Phi (A)x,x\bigr\rangle ^{t-\frac{1}{2}} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle + \bigl\langle \Phi (A)x,x\bigr\rangle ^{1/2} \\ & \quad\quad {}-2\bigl\langle \Phi \bigl(A^{1/4} \bigr)x,x\bigr\rangle \bigl\langle \Phi (A)x,x\bigr\rangle ^{1/4} \bigr) ; \end{aligned}$$
(ii)
for
\(\frac{1}{2}< t<1\),
$$\begin{aligned} &2R\bigl\langle \Phi (A)x,x\bigr\rangle ^{t-\frac{1}{2}} \bigl( \bigl\langle \Phi (A)x,x \bigr\rangle ^{\frac{1}{2}}-\bigl\langle \Phi \bigl(A^{\frac{1}{2}} \bigr)x,x\bigr\rangle \bigr) \\ & \quad\quad {} + r_{0}\bigl\langle \Phi (A)x,x\bigr\rangle ^{t-\frac{1}{2}} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle + \bigl\langle \Phi (A)x,x\bigr\rangle ^{1/2} \\ & \quad\quad {}-2\bigl\langle \Phi \bigl(A^{1/4} \bigr)x,x\bigr\rangle \bigl\langle \Phi (A)x,x\bigr\rangle ^{1/4} \bigr) \\ & \quad \leq \bigl\langle \Phi (A)x,x\bigr\rangle ^{t}-\bigl\langle \Phi \bigl(A^{t}\bigr)x,x\bigr\rangle \\ & \quad \leq 2r\bigl\langle \Phi (A)x,x\bigr\rangle ^{t-\frac{1}{2}} \bigl( \bigl\langle \Phi (A)x,x\bigr\rangle ^{\frac{1}{2}}-\bigl\langle \Phi \bigl(A^{\frac{1}{2}} \bigr)x,x \bigr\rangle \bigr) \\ & \quad\quad {} - r_{0}\bigl\langle \Phi (A)x,x\bigr\rangle ^{t-\frac{1}{2}} \bigl( \bigl\langle \Phi \bigl(A^{1/2}\bigr)x,x\bigr\rangle + \bigl\langle \Phi (A)x,x\bigr\rangle ^{1/2}\\ & \quad\quad {}-2\bigl\langle \Phi \bigl(A^{3/4} \bigr)x,x\bigr\rangle \bigl\langle \Phi (A)x,x\bigr\rangle ^{-1/4} \bigr) , \end{aligned}$$
Proof
In the next result, we obtain a refinement of inequality (5) for \(t\in (0,\frac{1}{2}]\).
Corollary 11
Let
\(T\in {\mathcal{B}}(\mathcal{H})\)
be positive operator and
\(x\in {\mathcal{H}}\)
be a unit vector. Then, for
\(0< t\leq \frac{1}{2}\), we have
where
\(r=\min \{t,1-t\}\), \(R=\max \{t,1-t\}\), \(r_{0}=\min \{2r,1-2r \}\).
$$\begin{aligned} &\langle Tx,x \rangle^{t}-\bigl\langle T^{t}x,x\bigr\rangle \\ &\quad \leq 2R\langle Tx,x\rangle^{t-\frac{1}{2}} \bigl( \langle Tx,x \rangle^{\frac{1}{2}}-\bigl\langle T^{\frac{1}{2}}x,x\bigr\rangle \bigr) \\ & \quad\quad {} - r_{0}\langle Tx,x\rangle^{t-\frac{1}{2}} \bigl( \bigl\langle T^{ \frac{1}{2}}x,x\bigr\rangle + \langle Tx,x\rangle^{\frac{1}{2}}-2\bigl\langle T ^{\frac{1}{4}}x,x\bigr\rangle \langle Tx,x\rangle^{\frac{1}{4}} \bigr) \\ & \quad \leq 2R \bigl( \langle Tx,x\rangle^{\frac{1}{2}}-\bigl\langle T^{ \frac{1}{2}}x,x\bigr\rangle \bigr) -r_{0} \bigl( \bigl\langle T^{\frac{1}{2}}x,x \bigr\rangle + \langle Tx,x\rangle^{\frac{1}{2}}-2\bigl\langle T^{\frac{1}{4}}x,x \bigr\rangle \langle Tx,x\rangle^{\frac{1}{4}} \bigr) , \end{aligned}$$
Proof
If we replace \(\Phi (A)=A\), \(A\in \mathcal{B}(\mathcal{H})\) and t with \(1-t\) in Corollary 10, then we get the desired result. □
6 Some new results
In this section, we prove some difference reverse-types of the Hölder and Cauchy–Schwarz inequalities.
Theorem 12
Let
\(A_{i}, B_{i}\in \mathcal{B}(\mathcal{H})\) (\(1\leq i \leq n\)) be positive invertible and
\(t\in [0,1]\).
(i)
If
\(0< t\leq \frac{1}{2}\), then
$$\begin{aligned} & \Biggl( \sum_{i=1}^{n}A_{i} \Biggr) \sharp_{t} \Biggl( \sum_{i=1}^{n}B_{i} \Biggr) - \Biggl( \sum_{i=1}^{n}A_{i} \sharp_{t}B_{i} \Biggr) \\ &\quad \leq R \Biggl( \sum_{i=1}^{n}A_{i}+ \sum_{i=1}^{n}B_{i}-2\sum _{i=1}^{n}(A _{i}\sharp B_{i}) \Biggr) \\ &\quad\quad {}-r_{0} \Biggl( \sum _{i=1}^{n}(A_{i}\sharp B_{i})+ \sum_{i=1}^{n}B_{i}-2\sum _{i=1}^{n}(A_{i} \sharp_{\frac{3}{4}}B_{i}) \Biggr) \\ &\quad\quad {} -r_{0} \Biggl( \sum_{i=1}^{n}A_{i} \sharp \sum_{i=1}^{n}B_{i}+\sum _{i=1} ^{n}A_{i}-2 \Biggl( \sum _{i=1}^{n}A_{i} \sharp_{\frac{1}{4}}\sum_{i=1} ^{n}B_{i} \Biggr) \Biggr) . \end{aligned}$$
(ii)
If
\(\frac{1}{2}< t< 1\), then
where
\(r=\min \{t,1-t\}\), \(R=\max \{t,1-t\}\)
and
\(r_{0}=\min \{2r,1-2r \}\).
$$\begin{aligned} & \Biggl( \sum_{i=1}^{n}A_{i} \Biggr) \sharp_{t} \Biggl( \sum_{i=1}^{n}B_{i} \Biggr) - \Biggl(\sum_{i=1}^{n}A_{i} \sharp_{t}B_{i} \Biggr)\\ &\quad \leq R \Biggl( \sum _{i=1} ^{n}A_{i}+\sum _{i=1}^{n}B_{i}-2 \Biggl( \sum _{i=1}^{n}A_{i}\sharp B_{i} \Biggr) \Biggr) \\ &\quad \quad {} -r_{0} \Biggl( \Biggl(\sum_{i=1}^{n}A_{i} \Biggr)\sharp \Biggl(\sum_{i=1}^{n}B_{i} \Biggr)+\sum_{i=1}^{n}B_{i}-2 \Biggl( \Biggl(\sum_{i=1}^{n}A_{i} \Biggr)\sharp_{\frac{3}{4}}\Biggl( \sum_{i=1}^{n}B_{i} \Biggr) \Biggr) \Biggr) \\ & \quad \quad {} -r_{0} \Biggl( \sum_{i=1}^{n}(A_{i} \sharp B_{i})+\sum_{i=1}^{n}A_{i}-2 \sum_{i=1}^{n}(A_{i} \sharp_{\frac{1}{4}} B_{i}) \Biggr) , \end{aligned}$$
Proof
Since the function \(f(x)=x^{t}\) (\(t\in [0,1]\)) is an operator concave function, \(\sum_{i=1}^{n}w_{i}T_{i}^{t}\leq (\sum_{i=1}^{n}w_{i}T _{i} )^{t}\) for positive operators \(T_{i}\) and positive real numbers \(w_{i}\) such that \(\sum_{i=1}^{n}w_{i}=1\). Now, Theorem 12 yields a reverse of this inequality as follows.
Example 13
If for positive operators \(T_{i}\) (\(1\leq i \leq n\)), we take \(A_{i}=w_{i}I\) and \(B_{i}=w_{i}T_{i}\) (\(1\leq i \leq n\)), in Theorem 12, where \(w_{i}\)’s are positive real numbers such that \(\sum_{i=1}^{n}w_{i}=1\), we obtain the following inequalities:
(i)
If \(0\leq t\leq \frac{1}{2}\), then
$$\begin{aligned} \Biggl(\sum_{i=1}^{n}w_{i}T_{i} \Biggr)^{t}-\sum_{i=1}^{n}w_{i}T_{i}^{t} &\leq R \Biggl(I+\sum_{i=1}^{n}w_{i}T_{i}-2 \sum_{i=1}^{n}w_{i}T_{i}^{1/2} \Biggr) \\ & \quad {} -r_{0} \Biggl(\sum_{i=1}^{n}w_{i}T_{i}^{1/2}+ \sum_{i=1}^{n}w_{i}T_{i}-2 \sum_{i=1}^{n}w_{i}T_{i}^{3/4} \Biggr) \\ & \quad {} -r_{0} \Biggl(\Biggl(\sum_{i=1}^{n}w_{i}T_{i} \Biggr)^{1/2}+I-2\Biggl(\sum_{i=1}^{n}w_{i}T _{i}\Biggr)^{1/4} \Biggr). \end{aligned}$$
(ii)
If \(\frac{1}{2}< t\leq 1\), then
$$\begin{aligned} \Biggl(\sum_{i=1}^{n}w_{i}T_{i} \Biggr)^{t}-\sum_{i=1}^{n}w_{i}T_{i}^{t} &\leq R \Biggl(I+\sum_{i=1}^{n}w_{i}T_{i}-2 \sum_{i=1}^{n}w_{i}T_{i}^{1/2} \Biggr) \\ &\quad {} -r_{0} \Biggl(\Biggl(\sum_{i=1}^{n}w_{i}T_{i} \Biggr)^{1/2}+ \sum_{i=1}^{n}w_{i}T _{i}-2\Biggl(\sum_{i=1}^{n}w_{i}T_{i} \Biggr)^{3/4} \Biggr) \\ & \quad {} -r_{0} \Biggl(\sum_{i=1}^{n}w_{i}T_{i}^{1/2}+I -2\sum_{i=1}^{n}w_{i}T _{i}^{1/4} \Biggr). \end{aligned}$$
In [18], the Tsallis relative operator entropy \(T_{t}(A|B)\) for positive invertible operators A, B and \(0< t\leq 1\) is defined as follows:
For further information as regards the Tsallis relative operator entropy see [6] and the references therein. In [7, Proposition 2.3], it is shown that for any unital positive linear map Φ the following inequality holds:
In (25), by similar techniques of Theorem 12, for positive operators \(A_{i}\), \(B_{i}\) (\(i=1,2,\ldots,n\)), we have
In the next theorem, we show a reverse of inequality (26).
$$\begin{aligned} T_{t}(A,B)=\frac{A\sharp_{t}B-A}{t}. \end{aligned}$$
$$\begin{aligned} \Phi \bigl(T_{t}(A|B)\bigr)\leq T_{t}\bigl( \Phi (A)|\Phi (B)\bigr). \end{aligned}$$
(25)
$$\begin{aligned} \sum_{i=1}^{n} \bigl( T_{t}(A_{i}|B_{i}) \bigr) \leq T_{t} \Biggl( \sum_{i=1} ^{n}A_{i}\Big|\sum _{i=1}^{n}B_{i} \Biggr) . \end{aligned}$$
(26)
Theorem 14
Let
\(A_{i}, B_{i}\in \mathcal{B}(\mathcal{H})\) (\(1\leq i\leq n\)) be positive invertible and
\(t\in (0,1)\).
(i)
If
\(0< t\leq \frac{1}{2}\), then
$$\begin{aligned} &T_{t} \Biggl( \sum_{i=1}^{n}A_{i}\Big| \sum_{i=1}^{n}B_{i} \Biggr) -\sum _{i=1} ^{n}\bigl(T_{t}(A_{i}|B_{i}) \bigr) \\ & \quad \leq \frac{1}{t} \Biggl[R \Biggl(\sum_{i=1}^{n}A_{i}+ \sum_{i=1}^{n}B_{i}-2 \sum _{i=1}^{n}(A_{i}\sharp B_{i}) \Biggr) \\ &\quad\quad {}-r_{0} \Biggl(\sum _{i=1}^{n}(A_{i} \sharp B_{i})+ \sum_{i=1}^{n}B_{i}-2\sum _{i=1}^{n}(A_{i} \sharp_{\frac{3}{4}}B_{i}) \Biggr) \\ & \quad\quad {} -r_{0} \Biggl(\sum_{i=1}^{n}A_{i} \sharp \sum_{i=1}^{n}B_{i}+\sum _{i=1} ^{n}A_{i}-2\Biggl(\sum _{i=1}^{n}A_{i} \sharp_{\frac{1}{4}}\sum_{i=1}^{n}B _{i}\Biggr) \Biggr) \Biggr]. \end{aligned}$$
(ii)
If
\(\frac{1}{2}< t<1\), then
$$\begin{aligned} &T_{t} \Biggl( \sum_{i=1}^{n}A_{i}\Big| \sum_{i=1}^{n}B_{i} \Biggr) - \sum_{i=1} ^{n}\bigl(T_{t}(A_{i}|B_{i}) \bigr)\\ &\quad \leq \frac{1}{t} \Biggl[R \Biggl(\sum_{i=1}^{n}A _{i}+\sum_{i=1}^{n}B_{i}-2 \Biggl(\sum_{i=1}^{n}A_{i}\sharp B_{i}\Biggr) \Biggr) \\ & \quad \quad {} -r_{0} \Biggl(\Biggl(\sum_{i=1}^{n}A_{i} \Biggr)\sharp \Biggl(\sum_{i=1}^{n}B_{i} \Biggr)+\sum_{i=1}^{n}B_{i}-2 \Biggl(\sum_{i=1}^{n}A_{i} \sharp_{\frac{3}{4}}\sum_{i=1} ^{n}B_{i} \Biggr) \Biggr) \\ &\quad \quad {} -r_{0} \Biggl(\sum_{i=1}^{n}(A_{i} \sharp B_{i})+\sum_{i=1}^{n}A_{i}-2 \sum_{i=1}^{n}(A_{i} \sharp_{\frac{1}{4}} B_{i}) \Biggr) \Biggr]. \end{aligned}$$
Proof
Applying Theorem 12 for \(0< t\leq \frac{1}{2}\), we have
whence we get the first inequality. The proof of the second inequality is similar. □
$$\begin{aligned} &T_{t} \Biggl( \sum_{i=1}^{n}A_{i}\Big| \sum_{i=1}^{n}B_{i} \Biggr) -\sum _{i=1} ^{n}\bigl(T_{t}(A_{i}|B_{i}) \bigr) \\ &\quad =\frac{ ( \sum_{i=1}^{n}A_{i} ) \sharp_{t} ( \sum_{i=1} ^{n}B_{i} ) -\sum_{i=1}^{n}A_{i}}{t}-\sum_{i=1}^{n} \frac{A_{i} \sharp_{t}B_{i}-A_{i}}{t} \\ &\quad \leq \frac{1}{t} \Biggl[R \Biggl(\sum_{i=1}^{n}A_{i}+ \sum_{i=1}^{n}B_{i}-2 \sum _{i=1}^{n}(A_{i}\sharp B_{i}) \Biggr)-r_{0} \Biggl(\sum _{i=1}^{n}(A_{i} \sharp B_{i})+ \sum_{i=1}^{n}B_{i}-2\sum _{i=1}^{n}(A_{i} \sharp_{\frac{3}{4}}B_{i}) \Biggr) \\ &\quad\quad {} -r_{0} \Biggl(\sum_{i=1}^{n}A_{i} \sharp \sum_{i=1}^{n}B_{i}+\sum _{i=1} ^{n}A_{i}-2\Biggl(\sum _{i=1}^{n}A_{i} \sharp_{\frac{1}{4}}\sum_{i=1}^{n}B _{i}\Biggr) \Biggr) \Biggr], \end{aligned}$$
Remark 15
We can present our results for non-invertible operators; see [6]. It is a direct consequence of the definition of the mean in the sense of Kubo–Ando [11] that \(A\sharp_{t} (B+ \varepsilon )\) is a monotone increasing net. Let B be a non-invertible operator and \(\epsilon >0\). It follows from the set \(\{A\sharp_{t}(B+ \epsilon ): \epsilon >0\}\) being bounded above for \(0 < \varepsilon <1\) that the limit
exists in the strong operator topology. So by (27), \(A\sharp _{t}B\) exists.
$$\begin{aligned} A\sharp_{t}B=\lim_{\epsilon \downarrow 0}A \sharp_{t}(B+\epsilon ) \end{aligned}$$
(27)
Acknowledgements
The third author would like to thank the Tusi Mathematical Research Group (TMRG).
Competing interests
The authors declare that they have no competing interests.
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