In this section, we present our results. First, we start with the following lemmas that will be used to develop new results in this paper.
Lemma 1
[13] Let \(a_{i}\) (\(i=1,2,...,n\)) be apositive real number and \(r\ge 1\). Then
$$\begin{aligned} \left( {\textstyle \sum \limits _{i=1}^{n}} a_{i}\right) ^{r}\le n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} a_{i}^{r}\text {.} \end{aligned}$$
Lemma 2
[11] Let \(T\in B\left( {{\mathcal {H}}}\right)\) be positive operator and let \(x\in {{\mathcal {H}}}\) such that \(\left\| x\right\| =1\). Then
$$\begin{aligned} \left\langle Tx,x\right\rangle ^{r}\le \left\langle T^{r}x,x\right\rangle \text {for }r\ge 1\text {.} \end{aligned}$$
Lemma 3
[11] Let \(T\in B\left( {{\mathcal {H}}}\right)\) and let f and g be non-negative continuous functions on \(\left[ 0,+\infty \right)\) such that \(f\left( t\right) g\left( t\right) =t\) for all \(t\in \left[ 0,+\infty \right)\). Then
$$\begin{aligned} \left| \left\langle Tx,y\right\rangle \right| ^{2}\le \left\langle f^{2}\left( \left| T\right| \right) x,x\right\rangle \left\langle g^{2}\left( \left| T^{*}\right| \right) y,y\right\rangle \text {,} \end{aligned}$$
for all \(x,y\in {{\mathcal {H}}}\).
The following Theorem is proved in [12].
Theorem 1
Let \(A_{i},X_{i},B_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\)), \(r\ge 1\) and let f and g be non-negative continuous functions on \(\left[ 0,+\infty \right)\) satisfying the relation \(f\left( t\right) g\left( t\right) =t\) (\(t\in \left[ 0,+\infty \right)\)). Then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}X_{i}B_{i}\right) \le \frac{n^{r-1}}{2}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}+\left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}\right) \right) \text {.} \end{aligned}$$
The first main result in this section is refinement of the above Theorem.
Theorem 2
Let \(A_{i},X_{i},B_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\)) and let f and g be non-negative continuous functions on \(\left[ 0,+\infty \right)\) satisfying the relation \(f\left( t\right) g\left( t\right) =t\) (\(t\in \left[ 0,+\infty \right)\)). Then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}X_{i}B_{i}\right) \quad & \quad \le \frac{n^{r-1}}{2}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}+\left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}\right) \right) \\& -\frac{n^{r-1}}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( {\hat{k}}_{\lambda }\right), \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}{\hat{k}}_{\lambda } ,{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {,} \end{aligned}$$
for \(r\ge 1\).
Proof
Let \({\hat{k}}_{\lambda }\) be the normalized reproducing kernel of \({{\mathcal {H}} }\), then
$$\begin{aligned}&\left| \left\langle \left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}X_{i}B_{i}\right) {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right| ^{r}\\&\quad =\left| {\textstyle \sum \limits _{i=1}^{n}} \left\langle \left( A_{i}^{*}X_{i}B_{i}\right) {\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle \right| ^{r}\\&\quad \le {\textstyle \sum \limits _{i=1}^{n}} \left| \left\langle A_{i}^{*}X_{i}B_{i}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle \right| ^{r}\\&\quad = {\textstyle \sum \limits _{i=1}^{n}} \left| \left\langle X_{i}B_{i}{\hat{k}}_{\lambda },A_{i}{\hat{k}}_{\lambda }\right\rangle \right| ^{r}\\&\quad \le \left( {\textstyle \sum \limits _{i=1}^{n}} \left\langle f^{2}\left( \left| X_{i}\right| \right) B_{i}\hat{k}_{\lambda },B_{i}{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\left\langle g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}{\hat{k}} _{\lambda },A_{i}{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{r}\\&\qquad (\text {by using Lemma }2)\\&\quad \le n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} \left\langle f^{2}\left( \left| X_{i}\right| \right) B_{i}\hat{k}_{\lambda },B_{i}{\hat{k}}_{\lambda }\right\rangle ^{\frac{r}{2}}\left\langle g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}{\hat{k}} _{\lambda },A_{i}{\hat{k}}_{\lambda }\right\rangle ^{\frac{r}{2}}\\&\qquad (\text {by using Lemma }2)\\&\quad \le n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\left\langle \left( A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}\\&\qquad (\text {by using Lemma }1)\\&\quad \,=\,\frac{n^{r-1}}{2}\left[ {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}k_{\lambda },k_{\lambda }\right\rangle +\left\langle \left( A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}k_{\lambda },k_{\lambda }\right\rangle \right) \right] \\&\qquad -\frac{n^{r-1}}{2}\left[ {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}k_{\lambda },k_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left( A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}k_{\lambda },k_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\right] \\ \text {(since }\sqrt{a}\sqrt{b}&\,=\,\frac{a+b}{2}-\frac{\left( \sqrt{a} -\sqrt{b}\right) ^{2}}{2}\text { where }a,b\ge 0\text {)}\\&\quad \,=\,\frac{n^{r-1}}{2}\left[ {\textstyle \sum \limits _{i=1}^{n}} \left\langle \left( \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}+\left( A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}\right) \hat{k}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right] \\&\qquad -\frac{n^{r-1}}{2}\left[ {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}} _{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left( A_{i}^{*} g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\right] \\&\quad \le \frac{n^{r-1}}{2}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}+\left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}\right) \right) \\&\qquad -\frac{n^{r-1}}{2}\left[ {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}} _{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left( A_{i}^{*} g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\right] . \end{aligned}$$
Now, by taking supremum over \(\lambda \in \Omega\), we get the desired inequality. \(\square\)
Corollary 1
Let \(A_{i},X_{i},B_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\)), \(r\ge 1\) and \(0\le p\le 1\). Then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}X_{i}B_{i}\right) & \le \frac{n^{r-1}}{2}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left[ A_{i}^{*}\left| X_{i}^{*}\right| ^{2\left( 1-p\right) }A_{i}\right] ^{r}+\left[ B_{i}^{*}\left| X_{i} \right| ^{2p}B_{i}\right] ^{r}\right) \right) \\& -\frac{n^{r-1}}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( {\hat{k}}_{\lambda }\right), \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left[ A_{i}^{*}\left| X_{i}^{*}\right| ^{^{2\left( 1-p\right) }}A_{i}\right] ^{r}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left[ B_{i}^{*}\left| X_{i}\right| ^{2p}B_{i}\right] ^{r}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {.} \end{aligned}$$
Proof
The result follows immediately from Theorem 2 for \(f\left( t\right) =t^{p}\) and \(g\left( t\right) =t^{1-p}\) \(\left( 0\le p\le 1\right)\). \(\square\)
For \(A_{i}=B_{i}=I\) (\(i=1,2,...,n\)) in Theorem 2 we get the following result.
Corollary 2
Let \(X_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\) ) and let f and g as in Theorem 2. Then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} X_{i}\right)&\le \frac{n^{r-1}}{2}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} g^{2r}\left( \left| X_{i}^{*}\right| \right) +f^{2r}\left( \left| X_{i}\right| \right) \right) \\\quad & \quad -\frac{n^{r-1}}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( {\hat{k}}_{\lambda }\right) \text {,} \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle g^{2r}\left( \left| X_{i}^{*}\right| \right) {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle f^{2r}\left( \left| X_{i}\right| \right) {\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {,} \end{aligned}$$
for \(r\ge 1\).
Remark 1
(1) If we take \(f\left( t\right) =t^{p}\) and \(g\left( t\right) =t^{1-p}\) \(\left( 0\le p\le 1\right)\) in Corollary 2, then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} X_{i}\right)&\le \frac{n^{r-1}}{2}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left| X_{i}\right| ^{2rp}+\left| X_{i}^{*}\right| ^{2r\left( p-1\right) }\right) \\\quad & \quad -\frac{n^{r-1}}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( {\hat{k}}_{\lambda }\right) \text {,} \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left| X_{i}^{*}\right| ^{2r\left( p-1\right) }{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2} }-\left\langle \left| X_{i}\right| ^{2rp}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {,} \end{aligned}$$
for \(r\ge 1\).
(2) Taking \(f\left( t\right) =\) \(g\left( t\right) =t^{\frac{1}{2}}\) \(\left( t\in \left[ 0,+\infty \right) \right)\) and \(r=1\)in Corollary 2, we get
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} X_{i}\right)&\le \frac{1}{2}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left| X_{i}\right| +\left| X_{i}^{*}\right| \right) \\\quad & \quad -\frac{1}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1} \eta \left( {\hat{k}}_{\lambda }\right) \text {,} \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left| X_{i}^{*}\right| {\hat{k}}_{\lambda } ,{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left| X_{i}\right| {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {.} \end{aligned}$$
For \(X_{i}=I\) (\(i=1,2,...,n\)), \(r\ge 1\) and \(f\left( t\right) =\) \(g\left( t\right) =t^{\frac{1}{2}}\) \(\left( t\in \left[ 0,+\infty \right) \right)\) in Theorem 2 we get the following inequality.
Corollary 3
Let \(A_{i},B_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\)) and \(r\ge 1\). Then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}B_{i}\right)&\le \frac{n^{r-1}}{2}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left| A_{i}\right| ^{2r}+\left| B_{i}\right| ^{2r}\right) \right) \\\quad & \quad -\frac{n^{r-1}}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( {\hat{k}}_{\lambda }\right) \text {,} \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left| A_{i}\right| ^{2r}{\hat{k}}_{\lambda } ,{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left| B_{i}\right| ^{2r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {.}\ \ \end{aligned}$$
Remark 2
For \(n=1\) in Corollary 2, we get
$$\begin{aligned} {\mathbf {ber}}^{r}\left( A^{*}B\right) \le \frac{1}{2}{\mathbf {ber}}\left( \left( A^{*}A\right) ^{r}+\left( B^{*}B\right) ^{r}\right) -\frac{1}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1} \eta \left( {\hat{k}}_{\lambda }\right) \text {,} \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) =\left( \left\langle \left( A^{*}A\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2} }-\left\langle \left( B^{*}B\right) ^{r}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {.}\ \end{aligned}$$
In [5] the authors proved the following theorem.
Theorem 3
Let \(A_{i},X_{i},B_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\)), \(r\ge 1\) and let f and g be non-negative continuous functions on \(\left[ 0,+\infty \right)\) satisfying the relation \(f\left( t\right) g\left( t\right) =t\) (\(t\in \left[ 0,+\infty \right)\)). Then for all \(r\ge 1\), we have.
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}X_{i}B_{i}\right) \le \frac{n^{r-1}}{\sqrt{2}}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}+i\left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}\right) \right) \text {.} \end{aligned}$$
In the following theorem we introduce a refinement of the above Theorem.
Theorem 4
Let \(A_{i},X_{i},B_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\)) and let f and g be non-negative continuous functions on \(\left[ 0,+\infty \right)\) satisfying the relation \(f\left( t\right) g\left( t\right) =t\) (\(t\in \left[ 0,+\infty \right)\)). Then for all \(r\ge 1\), we have.
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}X_{i}B_{i}\right)&\le \frac{n^{r-1}}{\sqrt{2}}{\mathbf {ber}} \left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}+i\left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}\right) \right) \\\quad & \quad -\frac{n^{r-1}}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( {\hat{k}}_{\lambda }\right) \text {,} \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}{\hat{k}}_{\lambda } ,{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {.} \end{aligned}$$
Proof
Let \({\hat{k}}_{\lambda }\) be the normalized reproducing kernel of \({{\mathcal {H}} }\), then
$$\begin{aligned}&\left| \left\langle \left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}X_{i}B_{i}\right) {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right| ^{r}\\&=\left| {\textstyle \sum \limits _{i=1}^{n}} \left\langle \left( A_{i}^{*}X_{i}B_{i}\right) {\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle \right| ^{r}\\&\quad \le {\textstyle \sum \limits _{i=1}^{n}} \left| \left\langle A_{i}^{*}X_{i}B_{i}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle \right| ^{r}\\&\quad = {\textstyle \sum \limits _{i=1}^{n}} \left| \left\langle X_{i}B_{i}{\hat{k}}_{\lambda },A_{i}{\hat{k}}_{\lambda }\right\rangle \right| ^{r}\\&\quad \le \left( {\textstyle \sum \limits _{i=1}^{n}} \left\langle f^{2}\left( \left| X_{i}\right| \right) B_{i}\hat{k}_{\lambda },B_{i}{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\left\langle g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}{\hat{k}} _{\lambda },A_{i}{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{r}\\&\qquad (\text {by using Lemma }3)\\&\quad \le n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} \left\langle f^{2}\left( \left| X_{i}\right| \right) B_{i}\hat{k}_{\lambda },B_{i}{\hat{k}}_{\lambda }\right\rangle ^{\frac{r}{2}}\left\langle g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}{\hat{k}} _{\lambda },A_{i}{\hat{k}}_{\lambda }\right\rangle ^{\frac{r}{2}}\\&\qquad (\text {by using Lemma }1)\\&\quad \le n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\left\langle \left( A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}\\&\qquad (\text {by using Lemma }2)\\&\quad \,=\,\frac{n^{r-1}}{2}\left[ {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle +\left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}} _{\lambda }\right\rangle \right) \right] \\&\qquad -\frac{n^{r-1}}{2} {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}} _{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left( A_{i}^{*} g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\\&\quad \le \frac{n^{r-1}}{\sqrt{2}}\left[ {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}} _{\lambda }\right\rangle +i\left\langle \left( A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}} _{\lambda },{\hat{k}}_{\lambda }\right\rangle \right) \right] \\&\qquad -\frac{n^{r-1}}{2} {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}} _{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left( A_{i}^{*} g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\\&\quad =\frac{n^{r-1}}{\sqrt{2}}\left[ {\textstyle \sum \limits _{i=1}^{n}} \left\langle \left( \left( A_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) A_{i}\right) ^{r}+i\left( B_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) B_{i}\right) ^{r}\right) \hat{k}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right] \\&\qquad -\frac{n^{r-1}}{2} {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left( B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}} _{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left( A_{i}^{*} g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right) ^{r}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\\ \text {(as }\left| a+b\right|&\le \sqrt{2}\left| a+ib\right| \text {, }\forall a,b\in {\mathbb {R}} \text {)}\\&\quad \le \frac{n^{r-1}}{\sqrt{2}}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}+i\left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}\right) \right) -\frac{n^{r-1} }{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( {\hat{k}}_{\lambda }\right) \text {,}\\&\text {.} \end{aligned}$$
where
$$\begin{aligned} \eta \left( k_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}k_{\lambda },k_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}k_{\lambda },k_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {.} \end{aligned}$$
By taking supremum over \(\lambda \in \Omega\), we get
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} A_{i}^{*}X_{i}B_{i}\right)&\le \frac{n^{r-1}}{\sqrt{2}}{\mathbf {ber}} \left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left[ A_{i}^{*}g^{2}\left( \left| X_{i}^{*}\right| \right) A_{i}\right] ^{r}+i\left[ B_{i}^{*}f^{2}\left( \left| X_{i}\right| \right) B_{i}\right] ^{r}\right) \right) \\ \quad & \quad -\frac{n^{r-1}}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( {\hat{k}}_{\lambda }\right) . \end{aligned}$$
This completes the proof. \(\square\)
Putting \(f\left( t\right) =t^{p}\) and \(g\left( t\right) =t^{1-p}\) \(\left( 0\le p\le 1\right)\) in Theorem 4 we get the following corollary.
Corollary 4
Let \(A_{i},X_{i},B_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\)) and let f and g be non-negative continuous functions on \(\left[ 0,+\infty \right)\) satisfying the relation \(f\left( t\right) g\left( t\right) =t\) (\(t\in \left[ 0,+\infty \right)\)). Then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} X_{i}\right) \le \frac{n^{r-1}}{\sqrt{2}}{\mathbf {ber}}\left( {\textstyle \sum \limits _{i=1}^{n}} \left( g^{2r}\left( \left| X_{i}^{*}\right| \right) +if^{2r}\left( \left| X_{i}\right| \right) \right) \right) -\frac{n^{r-1}}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1} \eta \left( {\hat{k}}_{\lambda }\right) \text {,} \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle g^{2r}\left( \left| X_{i}^{*}\right| \right) {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle f^{2r}\left( \left| X_{i}\right| \right) {\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {,} \end{aligned}$$
for all \(r\ge 1\).
Proof
Taking \(A_{i}=B_{i}=I\) (\(i=1,2,...,n\)) in Theorem 4 we deduce the desired result. \(\square\)
Next, taking \(n=1\), \(r=1\) and \(f\left( t\right) =\) \(g\left( t\right) =t^{\frac{1}{2}}\) in Corollary 4 we get the
following inequality
Corollary 5
Let \(T\in B\left( {{\mathcal {H}}}\right)\). Then
$$\begin{aligned} {\mathbf {ber}}\left( T\right) \le \frac{1}{\sqrt{2}}{\mathbf {ber}}\left( \left| T^{*}\right| +i\left| T\right| \right) -\frac{1}{2}\inf \limits _{\left\| {\hat{k}}_{\lambda }\right\| =1}\eta \left( \hat{k}_{\lambda }\right) \text {,} \end{aligned}$$
where
$$\begin{aligned} \eta \left( {\hat{k}}_{\lambda }\right) = {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left| T^{*}\right| {\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{\frac{1}{2}}-\left\langle \left| T\right| {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{\frac{1}{2}}\right) ^{2}\text {,} \end{aligned}$$
Now, we are in a position to prove the following result.
Theorem 5
Let \(T_{i}\in B\left( {{\mathcal {H}}}\right)\) (\(i=1,2,...,n\)) with the Cartesian decomposition \(T_{i}=A_{i}+iB_{i}\) for \(i=1,2,...,n\). Then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( {\textstyle \sum \limits _{i=1}^{n}} T_{i}\right) \le 2^{\frac{r}{2}-1}n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} {\mathbf {ber}}\left( \left( \left| A_{i}\right| ^{r}+\left| B_{i}\right| ^{r}\right) \right) \text {.} \end{aligned}$$
Proof
Let \({\hat{k}}_{\lambda }\) be the normalized reproducing kernel of \({{\mathcal {H}} }\), then we have
$$\begin{aligned}&\left| \left\langle \left( {\textstyle \sum \limits _{i=1}^{n}} T_{i}\right) {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right| ^{r}\\&\quad =\left| {\textstyle \sum \limits _{i=1}^{n}} \left\langle T_{i}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right| ^{r}\\&\quad \le \left( {\textstyle \sum \limits _{i=1}^{n}} \left| \left\langle T_{i}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right| \right) ^{r}\\&\quad =\left( {\textstyle \sum \limits _{i=1}^{n}} \left| \left\langle A_{i}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle +i\left\langle B_{i}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right| \right) ^{r}\\&\quad =\left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left| \left\langle A_{i}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right| ^{2}+\left| \left\langle B_{i}{\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right| ^{2}\right) ^{\frac{1}{2} }\right) ^{r}\\&\quad \le \left( {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left| A_{i}\right| {\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{2}+\left\langle \left| B_{i}\right| {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{2}\right) ^{\frac{1}{2} }\right) ^{r}\\&\qquad (\text {by using Lemma }3)\\&\quad \le n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left| A_{i}\right| {\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{2}+\left\langle \left| B_{i}\right| {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{2}\right) ^{\frac{r}{2}}\\&\qquad (\text {by using Lemma }1)\\&\quad \le 2^{\frac{r}{2}-1}n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left| A_{i}\right| {\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle ^{r}+\left\langle \left| B_{i}\right| {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle ^{r}\right) \\&\qquad (\text {by using Lemma }1)\\&\quad \le 2^{\frac{r}{2}-1}n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} \left( \left\langle \left| A_{i}\right| ^{r}{\hat{k}}_{\lambda },\hat{k}_{\lambda }\right\rangle +\left\langle \left| B_{i}\right| ^{r} {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \right) \\&\quad =2^{\frac{r}{2}-1}n^{r-1} {\textstyle \sum \limits _{i=1}^{n}} \left\langle \left( \left| A_{i}\right| ^{r}+\left| B_{i}\right| ^{r}\right) {\hat{k}}_{\lambda },{\hat{k}}_{\lambda }\right\rangle \text {.} \end{aligned}$$
Now, taking the supremum over all \(\lambda \in \Omega\) we get the desired result. \(\square\)
For \(n=1\) we get the following inequality.
Corollary 6
Let \(T\in B\left( {{\mathcal {H}}}\right)\) with the Cartesian decomposition \(T=A+iB\) and \(r\ge 1\). Then
$$\begin{aligned} {\mathbf {ber}}^{r}\left( T\right) \le 2^{\frac{r}{2}-1}{\mathbf {ber}}\left( \left( \left| A\right| ^{r}+\left| B\right| ^{r}\right) \right) . \end{aligned}$$