New Bounds for the Davis–Wielandt Radius via the Moore–Penrose Inverse of Bounded Linear Operators

Abstract

In this paper, we obtain some new upper bounds involving powers of the Davis–Wielandt radius of bounded linear operators with closed ranges by using the Moore–Penrose inverse. Moreover, by providing some examples, we show that the upper bounds obtained here are better than the existing ones in some situations.

1. Introduction

The inequalities of operators on Hilbert space are interesting and useful for applying operator theory to any field of natural science. Numerical radius inequalities play an important role in various fields of operator theory and matrix analysis [1,2], and there have been many generalizations of the numerical radius. The numerical radius is not only a powerful tool for characterizing the numerical range but also provides a key tool for stability, convergence and error estimation in practical applications [3]. The close relationship between the numerical radius, norm, and spectral radius makes it an indispensable tool in functional analysis and numerical calculations. In research on numerical radius inequalities, researchers mainly focus on improving and generalizing the existing inequalities. The purpose of this paper is to study one of the generalizations of the numerical radius, that is, the Davis–Wielandt radius. By using the Moore–Penrose inverse of bounded linear operators with closed ranges, we will establish some new Davis–Wielandt radius inequalities. Furthermore, by providing numerical examples, we will show that these inequalities are better than the existing inequalities. Let us recall the following necessary definitions and related symbols.

Let $(H,\langle ·,·\rangle )$ be a complex Hilbert space and let $\mathcal{B}\left(H\right)$ be the $C^{*}$-algebra of all bounded linear operators on H. For $A\in \mathcal{B}\left(H\right)$, the Moore–Penrose inverse of A is the operator $X\in \mathcal{B}\left(H\right)$ that satisfies the following Penrose equations [4]:

$$AXA=A,XAX=X,{\left(AX\right)}^{*}=AXand{\left(XA\right)}^{*}=XA,$$

where ${(·)}^{*}$ denotes the adjoint operator. Note that the Moore–Penrose inverse of A exists if and only if the range of A is closed and that it is unique and denoted by $A^{†}$ in this case. Throughout the paper, $\mathcal{CR}\left(H\right)$ denotes the subset of $\mathcal{B}\left(H\right)$ of all operators with closed ranges. Obviously, when H is a finite dimensional space, all operators in $\mathcal{B}\left(H\right)$ have closed ranges. Therefore, each matrix has a unique Moore–Penrose inverse. It is easy to see that, for $A\in \mathcal{CR}\left(H\right)$,

$${\left(A^{*}A\right)}^{†}=A^{†}{A}^{*†},$$

$$A^{*}=A^{*}A{A}^{†}=A^{†}A{A}^{*},$$

$$A=A{A}^{†}A={\left(A{A}^{†}\right)}^{*}A={\left(A^{†}\right)}^{*}{\left|A\right|}^2,$$

and

$$A=A{A}^{†}A=A{\left(A^{†}A\right)}^{*}={\left|A^{*}\right|}^2{\left(A^{†}\right)}^{*},$$

where $\left|A\right|={\left(A^{*}A\right)}^{{\displaystyle \frac{1}{2}}}$ and $|A^{*}|={\left(A{A}^{*}\right)}^{{\displaystyle \frac{1}{2}}}$. Recently, Sababheh et al. [5] and Bhunia et al. [6] studied the numerical radius of bounded linear operators by using the Moore–Penrose inverse. For further discussion of the Moore–Penrose inverse of bounded linear operators, we direct readers to [7,8].

For $A\in \mathcal{B}\left(H\right)$, the numerical range of A is defined as [9]

$$W\left(A\right)=\{\langle Ax,x\rangle :x\in H,\parallel x\parallel =1\}.$$

The numerical radius and the operator norm of A, denoted as $w\left(A\right)$ and $\parallel A\parallel$, respectively, are defined as

$$w\left(A\right)=sup\left\{\right|\langle Ax,x\rangle |:x\in H,\parallel x\parallel =1\}$$

and

$$\parallel A\parallel =sup\{\parallel Ax\parallel :x\in H,\parallel x\parallel =1\}.$$

It is well known that $w(·)$ defines a norm on $\mathcal{B}\left(H\right)$, which is equivalent to the operator norm $\parallel ·\parallel$. Namely, for every $A\in \mathcal{B}\left(H\right)$, the following inequalities hold:

$${\displaystyle \frac{1}{2}}\parallel A\parallel \le w\left(A\right)\le \parallel A\parallel .$$

The concept of numerical radius is useful in studying the bounded linear operators, and there have many generalizations of the numerical radius, such as in, e.g., [10,11,12,13,14,15,16] and the references therein. One of the most interesting generalizations is the Davis–Wielandt radius, which is defined as [17,18]

$$dw\left(A\right)=sup\{\sqrt{{\left|\langle Ax,x\rangle \right|}^2+{\parallel Ax\parallel }^4}:x\in H,\parallel x\parallel =1\},$$

for $A\in \mathcal{B}\left(H\right)$. It is clear that the Davis–Wielandt radius $dw(·)$ satisfies the following inequality:

$${max\{w\left(A\right),\parallel A\parallel }^2\}\le dw\left(A\right)\le \sqrt{w^2\left(A\right)+{\parallel A\parallel }^4}.$$

(1)

The second inequality in (1) becomes equality if and only if A is normaloid, i.e., $w\left(A\right)=\parallel A\parallel$ (see [19], Corollary 3.2). In recent years, the Davis–Wielandt radius inequalities have been studied by many researchers. For instance, in [20,21], the authors proved that if $A\in \mathcal{B}\left(H\right)$, then

$$\begin{array}{c}\hfill d{w}^2\left(A\right)\le {\parallel \left|A\right|}^2+{\left|A\right|}^4{\parallel }^{{\displaystyle \frac{1}{2}}}\parallel {|A^{*}|}^2+{\left|A\right|}^4{\parallel }^{{\displaystyle \frac{1}{2}}}.\end{array}$$

(2)

In [22], Zamani and Shebrawi proved that

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le w^2{\left(\right|A|}^2{-A)+2\parallel A\parallel }^{2}w\left(A\right),\hfill \end{array}$$

(3)

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le {\displaystyle \frac{1}{2}}w\left(A^2\right)+{\displaystyle \frac{1}{4}}{w\left(\right|A|}^2+|A^{*}{|}^2)\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}(2w^2\left(A\right)-c^2\left(A\right)+2w\left(A\right)\sqrt{w^2\left(A\right)-c^2\left(A\right)}),\hfill \end{array}$$

(4)

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le {max\{\parallel A\parallel }^2{,\parallel A\parallel }^4\}+\sqrt{2}{w\left(\right|A|}^{2}A),\hfill \end{array}$$

(5)

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le {\displaystyle \frac{1}{2}}{\left(w\right(\left|A\right|}^4+{\left|A\right|}^2{)+w(\left|A\right|}^4-{\left|A\right|}^2\left)\right)+\sqrt{2}{w\left(\right|A|}^{2}A),\hfill \end{array}$$

(6)

where $c\left(A\right)=inf\left\{\right|\langle Ax,x\rangle |:x\in H,\parallel x\parallel =1\}.$

In [23], Bhunia et al. proved that

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le {\displaystyle \frac{1}{2}}\left\{w^2(A+A^{*}A)+w^2(A-A^{*}A)\right\},\hfill \end{array}$$

(7)

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le {\displaystyle \frac{1}{2}}{\parallel \left|A\right|}^2+|A^{*}{|}^2+{2\left|A\right|}^4\parallel ,\hfill \end{array}$$

(8)

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le {\parallel \left|A\right|}^2+{\left|A\right|}^4\parallel ,\hfill \end{array}$$

(9)

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le {\displaystyle \frac{1}{2}}(w\left(A^2\right)+{\parallel A\parallel }^2{)+\parallel A\parallel }^4.\hfill \end{array}$$

(10)

For more results on the Davis–Wielandt radius inequalities, the reader may be referred to [24,25,26,27,28].

In this paper, using the Moore–Penrose inverse of bounded linear operators, we obtain some new upper bounds for the Davis–Wielandt radius of bounded linear operators with closed ranges and show that these upper bounds are better than the existing ones mentioned above by numerical examples.

2. Main Results

In order to prove our results, we need the following lemmas:

Lemma 1

([29]). Let A be a non-negative bounded linear operator in Hilbert space H, and let $x\in H$ be any unit vector. Then, for $r\ge 1$, we have

$${\langle Ax,x\rangle }^r\le \langle A^{r}x,x\rangle .$$

Lemma 2

([30]). Let $x,y,e\in H$ with $\parallel e\parallel =1$ and $\alpha \in \mathbb{C}\setminus \left\{0\right\}$. Then,

$$\left|\langle x,e\rangle \langle e,y\rangle \right|\le {\displaystyle \frac{1}{\left|\alpha \right|}}\left(max\{1,|\alpha -1\left|\right\}\parallel x\parallel \parallel y\parallel +|\langle x,y\rangle |\right).$$

In particular, if $\alpha =2$, then the above inequality becomes the Buzano inequality ([31])

$$\left|\langle x,e\rangle \langle e,y\rangle \right|\le {\displaystyle \frac{1}{2}}\left(\right|\langle x,y\rangle |+\parallel x\parallel \parallel y\parallel ).$$

Lemma 3

([32]). Let $x,y,e\in H$ with $\parallel e\parallel =1$. Then,

$${\left|\langle x,e\rangle \langle e,y\rangle \right|}^r\le {\displaystyle \frac{1+\alpha }{2}}{\parallel x\parallel }^r{\parallel y\parallel }^r+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle x,y\rangle \right|}^r.$$

for every $0\le \alpha \le 1$ and $r\ge 1$.

Lemma 4

([5]). Let $A\in \mathcal{CR}\left(H\right)$. Then,

$${\left|\langle Ax,y\rangle \right|}^2\le {\langle \left|A\right|}^{2}x,x\rangle \langle A{A}^{†}y,y\rangle ,$$

for any $x,y\in H$.

We know that $\mathcal{R}\left(A\right)$ is closed if and only if $\mathcal{R}\left(A^{*}\right)$ is closed, while $\mathcal{R}\left(A^{*}\right)=\mathcal{R}\left(A^{*}A\right)$. So, when $A\in \mathcal{CR}\left(H\right)$, then $A^{*}A\in \mathcal{CR}\left(H\right)$. Now applying Lemma 4, we obtain the following inequality:

$$|\langle A^{*}Ax,y\rangle {|}^2\le {\langle \left|A\right|}^{4}x,x\rangle \langle A^{†}Ay,y\rangle ,$$

for $A\in \mathcal{CR}\left(H\right)$, $x,y\in H$. Indeed,

$$\begin{array}{cc}\hfill |\langle A^{*}Ax,y\rangle {|}^2& \le \langle |A^{*}A{|}^{2}x,x\rangle \langle A^{*}A{\left(A^{*}A\right)}^{†}y,y\rangle \hfill \\ \hfill & ={\langle \left|A\right|}^{4}x,x\rangle \langle A^{*}A{A}^{†}{A}^{*†}y,y\rangle \hfill \\ \hfill & ={\langle \left|A\right|}^{4}x,x\rangle \langle A^{*}{A}^{*†}y,y\rangle \hfill \\ \hfill & ={\langle \left|A\right|}^{4}x,x\rangle \langle {\left(A^{†}A\right)}^{*}y,y\rangle \hfill \\ \hfill & ={\langle \left|A\right|}^{4}x,x\rangle \langle A^{†}Ay,y\rangle .\hfill \end{array}$$

Now, we are in a position to state our main results.

Theorem 1.

Let $A\in \mathcal{CR}\left(H\right)$. Then, for $r\ge 1$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-1}{∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}},$$

(11)

$$d{w}^{2r}\left(A\right)\le 2^{r-1}{∥{\left|A\right|}^{4r}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}{∥{\left|A\right|}^{8r}+A{A}^{†}∥}^{{\displaystyle \frac{1}{2}}},$$

(12)

$$d{w}^{2r}\left(A\right)\le 2^{r-1}{∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}$$

(13)

and

$$d{w}^{2r}\left(A\right)\le 2^{r-1}{∥|A^{*}{|}^{4r}+A{A}^{†}∥}^{{\displaystyle \frac{1}{2}}}{∥|A^{*}{|}^{8r}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}.$$

(14)

Proof. 

Let $x\in H$ be a unit vector. Then,

$$\begin{array}{cc}& {\left(\right|\langle Ax,x\rangle |}^2+{\parallel Ax\parallel }^4{)}^r\hfill \\ \hfill & \le 2^{r-1}\left({\left|\langle Ax,x\rangle \right|}^{2r}+{\langle A^{*}Ax,x\rangle }^{2r}\right)\hfill \\ \hfill & \le 2^{r-1}\left({\langle \left|A\right|}^2{x,x\rangle }^r{\langle A{A}^{†}x,x\rangle }^r+{\langle \left|A\right|}^4{x,x\rangle }^r{\langle A^{†}Ax,x\rangle }^r\right)\hfill \\ \hfill & \phantom{=}\left(byLemma4\right)\hfill \\ \hfill & \le 2^{r-1}{\left({\langle \left|A\right|}^2{x,x\rangle }^{2r}+{\langle \left|A\right|}^4{x,x\rangle }^{2r}\right)}^{{\displaystyle \frac{1}{2}}}{\left({\langle A{A}^{†}x,x\rangle }^{2r}+{\langle A^{†}Ax,x\rangle }^{2r}\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \phantom{=}(bytheCauchy--Schwarzinequality)\hfill \\ \hfill & \le 2^{r-1}{\left({\langle \left|A\right|}^{4r}{x,x\rangle +\langle \left|A\right|}^{8r}x,x\rangle \right)}^{{\displaystyle \frac{1}{2}}}{\left(\langle {\left(A{A}^{†}\right)}^{2r}x,x\rangle +\langle {\left(A^{†}A\right)}^{2r}x,x\rangle \right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \phantom{=}\left(byLemma1\right)\hfill \\ \hfill & =2^{r-1}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain

$$d{w}^{2r}\left(A\right)\le 2^{r-1}{∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}.$$

Similarly, we can show that

$$d{w}^{2r}\left(A\right)\le 2^{r-1}{∥{\left|A\right|}^{4r}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}{∥{\left|A\right|}^{8r}+A{A}^{†}∥}^{{\displaystyle \frac{1}{2}}}.$$

From ([33], Theorem 3.3), we have $dw\left(A\right)=dw\left(A^{*}\right)$, and so we obtain the inequalities (13) and (14). □

In particular, by considering $r=1$ in Theorem 1, we obtain the following corollary:

Corollary 1.

Let $A\in \mathcal{CR}\left(H\right)$. Then

$$\begin{array}{c}\hfill d{w}^2\left(A\right)\le min\{\alpha ,\beta ,\gamma ,\delta \},\end{array}$$

(15)

where

$$\alpha ={∥{\left|A\right|}^4+{\left|A\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}},$$

$$\beta ={∥{\left|A\right|}^4+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}{∥{\left|A\right|}^8+A{A}^{†}∥}^{{\displaystyle \frac{1}{2}}},$$

$$\gamma ={∥|A^{*}{|}^4+{\left|A^{*}\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}$$

and

$$\delta ={∥|A^{*}{|}^4+A{A}^{†}∥}^{{\displaystyle \frac{1}{2}}}{∥|A^{*}{|}^8+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}.$$

Remark 1.

The inequality in Corollary 1 is better than inequalities (1), (2), (3), (4), (8), (9), and (10); the inequalities in [22], which are Theorems 2.2, 2.5, 2.16, and 2.17; and that in [21], which is Theorem 2.10. The inequalities in [22], namely Theorems 2.2, 2.5, 2.16, and 2.17 and that in [21], namely Theorem 2.10, are shown, respectively, as follows:

$$\begin{array}{cc}\hfill & d{w}^2\left(A\right)\le {\displaystyle \frac{1}{2}}{w\left(\right|A|}^2+{2\left|A\right|}^4+{|A^{*}|}^2)-{\displaystyle \frac{1}{2}}\underset{\parallel x\parallel =1}{inf}(\parallel Ax\parallel -\parallel A^{*}x{\parallel )}^2,\hfill \\ \hfill & d{w}^2\left(A\right)\le {\displaystyle \frac{1}{4}}\left({w\left(\right(\left|A\right|}^2{+A)}^2{)+w(\left(\right|A|}^2{-A)}^2{)+w(\left|A\right|}^2+{2\left|A\right|}^4+{|A^{*}|}^2)\right),\hfill \\ \hfill & d{w}^2\left(A\right)\le {max\{w\left(A\right),w(\left|A\right|}^2{\left)\right\}\left(w\right(\left|A\right|}^4+{\left|A\right|}^2{)+2w(\left|A\right|}^{2}A{\left)\right)}^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill & d{w}^2\left(A\right)\le {\parallel A\parallel max\{w\left(A\right),w(\left|A\right|}^2\left)\right\}(1+{\parallel A\parallel }^2{+2w\left(A\right))}^{{\displaystyle \frac{1}{2}}},\hfill \\ \hfill & d{w}^2\left(A\right)\le {\displaystyle \frac{1}{4}}{w}^2\left(\right|A|+i|A^{*}\left|\right)+{\displaystyle \frac{1}{4}}w(A|A^{*}|)\hfill \\ \hfill & +{\displaystyle \frac{1}{8}}{min\{\parallel |A|}^2+{\left|A^{*}\right|}^2+{8\left|A\right|}^4{\parallel ,\parallel \left|A\right|}^2+{|A^{*}|}^2+8{|A^{*}|}^4\parallel \}.\hfill \end{array}$$

If we take $A=\left[\begin{array}{cc}0& 2\\ 0& 0\end{array}\right]$. Then, by calculations, we obtain $\alpha =\gamma =\sqrt{272}\approx 16.49$ and $\beta =\delta =\sqrt{17}\times \sqrt{256}\approx 65.97$, and so Corollary 1 gives $d{w}^2\left(A\right)\le 16.49$, whereas inequalities (1), (2), (3), (4), (8), (9), and (10), and Theorems 2.2, 2.5, 2.16, 2.17 in [22], and Theorem 2.10 in [21] give $d{w}^2\left(A\right)\le 17$, $d{w}^2\left(A\right)\le 17.89$, $d{w}^2\left(A\right)\le 25.94$, $d{w}^2\left(A\right)\le 17.5$, $d{w}^2\left(A\right)\le 18$, $d{w}^2\left(A\right)\le 20$, $d{w}^2\left(A\right)\le 18$, $d{w}^2\left(A\right)\le 18$, $d{w}^2\left(A\right)\le 17.47$, $d{w}^2\left(A\right)\le 17.89$, $d{w}^2\left(A\right)\le 21.17$ and $d{w}^2\left(A\right)\le 17.5$, respectively. Therefore, for this example, the upper bound of $d{w}^2\left(A\right)$ in Corollary 1 is better than the existing bounds mentioned above.

Theorem 2.

Let $A\in \mathcal{CR}\left(H\right)$. Then, for $r\ge 1$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}+A{A}^{†}+A^{†}A∥$$

(16)

and

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}+A{A}^{†}+A^{†}A∥.$$

(17)

Proof. 

Let $x\in H$ be a unit vector. Then,

$$\begin{array}{cc}& {\left(\right|\langle Ax,x\rangle |}^2+{\parallel Ax\parallel }^4{)}^r\hfill \\ \hfill & \le 2^{r-1}\left({\left|\langle Ax,x\rangle \right|}^{2r}+{\langle A^{*}Ax,x\rangle }^{2r}\right)\hfill \\ \hfill & \le 2^{r-1}\left({\langle \left|A\right|}^2{x,x\rangle }^r{\langle A{A}^{†}x,x\rangle }^r+{\langle \left|A\right|}^4{x,x\rangle }^r{\langle A^{†}Ax,x\rangle }^r\right)\hfill \\ \hfill & \phantom{=}\left(byLemma4\right)\hfill \\ \hfill & \le 2^{r-2}\left({\langle \left|A\right|}^2{x,x\rangle }^{2r}+{\langle \left|A\right|}^4{x,x\rangle }^{2r}+{\langle A{A}^{†}x,x\rangle }^{2r}+{\langle A^{†}Ax,x\rangle }^{2r}\right)\hfill \\ \hfill & \le 2^{r-2}\left({\langle \left|A\right|}^{4r}{x,x\rangle +\langle \left|A\right|}^{8r}x,x\rangle +\langle {\left(A{A}^{†}\right)}^{2r}x,x\rangle +\langle {\left(A^{†}A\right)}^{2r}x,x\rangle \right)\hfill \\ \hfill & \phantom{=}\left(byLemma1\right)\hfill \\ \hfill & =2^{r-2}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}+A{A}^{†}+A^{†}A)x,x\rangle .\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}+A{A}^{†}+A^{†}A∥.$$

Similarly, we can show that

$$d{w}^{2r}\left(A^{*}\right)\le 2^{r-2}∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}+A{A}^{†}+A^{†}A∥.$$

Since $dw\left(A\right)=dw\left(A^{*}\right)$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}+A{A}^{†}+A^{†}A∥.$$

□

Theorem 3.

Let $A\in \mathcal{CR}\left(H\right)$, $0\le \alpha \le 1$. Then, for $r\ge 1$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥\alpha X+(1-\alpha )Z+{\left|A\right|}^{8r}+A^{†}A∥$$

(18)

and

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥\alpha X+(1-\alpha )Z+|A^{*}{|}^{8r}+A{A}^{†}∥,$$

(19)

where

$$X={\left|A\right|}^{4r}+A{A}^{†},Z={\left|A^{*}\right|}^{4r}+A^{†}A.$$

Proof. 

Let $x\in H$ be a unit vector. Then,

$$\begin{array}{cc}& {\left(\right|\langle Ax,x\rangle |}^2+{\parallel Ax\parallel }^4{)}^r\hfill \\ \hfill & \le 2^{r-1}\left({\left|\langle Ax,x\rangle \right|}^{2r}+{|\langle |A|}^{2}x,x{\rangle |}^{2r}\right)\hfill \\ \hfill & =2^{r-1}\left({\alpha \left|\langle Ax,x\rangle \right|}^{2r}+(1-\alpha )|\langle A^{*}x,x\rangle {|}^{2r}+{|\langle |A|}^{2}x,x{\rangle |}^{2r}\right)\hfill \\ \hfill & \le 2^{r-1}\left({\alpha \langle \left|A\right|}^2{x,x\rangle }^r{\langle A{A}^{†}x,x\rangle }^r+(1-\alpha )\langle |A^{*}{{|}^{2}x,x\rangle }^r{\langle A^{†}Ax,x\rangle }^r\right.\hfill \\ \hfill & \phantom{=}\left.+{\langle \left|A\right|}^4{x,x\rangle }^r{\langle A^{†}Ax,x\rangle }^r\right)\phantom{=}\left(byLemma4\right)\hfill \\ \hfill & \le 2^{r-1}\left[{\displaystyle \frac{\alpha }{2}}\left({\langle \left|A\right|}^2{x,x\rangle }^{2r}+{\langle A{A}^{†}x,x\rangle }^{2r}\right)+{\displaystyle \frac{1-\alpha }{2}}\left(\langle |A^{*}{{|}^{2}x,x\rangle }^{2r}+{\langle A^{†}Ax,x\rangle }^{2r}\right)\right.\hfill \\ \hfill & \phantom{=}\left.+{\displaystyle \frac{1}{2}}\left({\langle \left|A\right|}^4{x,x\rangle }^{2r}+{\langle A^{†}Ax,x\rangle }^{2r}\right)\right]\hfill \\ \hfill & \phantom{=}\le 2^{r-2}\left[{\alpha \langle \left(\right|A|}^{4r}+{\left(A{A}^{†}\right)}^{2r})x,x\rangle +(1-\alpha )\langle \left(\right|A^{*}{|}^{4r}+{\left(A^{†}A\right)}^{2r})x,x\rangle \right.\hfill \\ \hfill & \phantom{=}\phantom{=}\left.+{\langle \left(\right|A|}^{8r}+{\left(A^{†}A\right)}^{2r})x,x\rangle \right]\phantom{=}\left(byLemma1\right)\hfill \\ \hfill & \phantom{=}=2^{r-2}\left[{\alpha \langle \left(\right|A|}^{4r}+A{A}^{†})x,x\rangle +(1-\alpha )\langle \left(\right|A^{*}{|}^{4r}+A^{†}A)x,x\rangle \right.\hfill \\ \hfill & \phantom{=}\phantom{=}\left.+{\langle \left(\right|A|}^{8r}+A^{†}A)x,x\rangle \right]\hfill \\ \hfill & \phantom{=}=2^{r-2}〈{\left(\alpha \right(\left|A\right|}^{4r}+A{A}^{†})+(1-\alpha )(|A^{*}{|}^{4r}+A^{†}{A)+|A|}^{8r}+A^{†}A)x,x〉\hfill \\ \hfill & \phantom{=}\le 2^{r-2}∥\alpha X+(1-\alpha )Z+{\left|A\right|}^{8r}+A^{†}A∥.\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain inequality (18). Similarly, we can show that

$$d{w}^{2r}\left(A^{*}\right)\le 2^{r-2}∥|\alpha X+(1-\alpha )Z+|A^{*}{|}^{8r}+A{A}^{†}∥.$$

Since $dw\left(A\right)=dw\left(A^{*}\right)$, and so we obtain the inequality (19). □

Remark 2.

In particular, if we take $\alpha =1$ in inequality (18), then we obtain the inequality (16). If we take $\alpha =0$ in inequality (19), we obtain inequality (17).

Now, we observe the following inequality:

$$|a+b|\le \sqrt{2}|a+ib|,a,b\in \mathbb{R}.$$

(20)

By employing the inequality (20), we prove the following theorem:

Theorem 4.

Let $A\in \mathcal{CR}\left(H\right)$. Then,

$$d{w}^2\left(A\right)\le {\displaystyle \frac{1}{\sqrt{2}}}w\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A)\right)$$

(21)

and

$$d{w}^2\left(A\right)\le {\displaystyle \frac{1}{\sqrt{2}}}w\left(|A^{*}{|}^4+{\left|A^{*}\right|}^8+i(A{A}^{†}+A^{†}A)\right).$$

(22)

Proof. 

Let $x\in H$ be a unit vector. Then,

$$\begin{array}{cc}\hfill {\left|\langle Ax,x\rangle \right|}^2+{\parallel Ax\parallel }^4& ={\left|\langle Ax,x\rangle \right|}^2+{\langle A^{*}Ax,x\rangle }^2\hfill \\ \hfill & \le {\langle \left|A\right|}^{2}x,x\rangle \langle A{A}^{†}x,x\rangle +{\langle \left|A\right|}^{4}x,x\rangle \langle A^{†}Ax,x\rangle \hfill \\ \hfill & \phantom{=}\left(byLemma4\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({\langle \left|A\right|}^2{x,x\rangle }^2+{\langle A{A}^{†}x,x\rangle }^2\right)+{\displaystyle \frac{1}{2}}\left({\langle \left|A\right|}^4{x,x\rangle }^2+{\langle A^{†}Ax,x\rangle }^2\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({\langle \left|A\right|}^{4}x,x\rangle +\langle A{A}^{†}x,x\rangle \right)+{\displaystyle \frac{1}{2}}\left({\langle \left|A\right|}^{8}x,x\rangle +\langle A^{†}Ax,x\rangle \right)\hfill \\ \hfill & \phantom{=}\left(byLemma1\right)\hfill \\ \hfill & =〈\left({\displaystyle \frac{{\left|A\right|}^4+{\left|A\right|}^8}{2}}\right)x,x〉+〈\left({\displaystyle \frac{A{A}^{†}+A^{†}A}{2}}\right)x,x〉\hfill \\ \hfill & \le \sqrt{2}\left|〈\left({\displaystyle \frac{{\left|A\right|}^4+{\left|A\right|}^8}{2}}\right)x,x〉+i〈\left({\displaystyle \frac{A{A}^{†}+A^{†}A}{2}}\right)x,x〉\right|\hfill \\ \hfill & \phantom{=}\left(byinequality\right(20\left)\right)\hfill \\ \hfill & =\sqrt{2}\left|〈\left({\displaystyle \frac{{\left|A\right|}^4+{\left|A\right|}^8}{2}}+i{\displaystyle \frac{A{A}^{†}+A^{†}A}{2}}\right)x,x〉\right|\hfill \\ \hfill & ={\displaystyle \frac{1}{\sqrt{2}}}\left|〈\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A)\right)x,x〉\right|.\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain inequality (21). Similar to the proof of inequality (21), and using $dw\left(A\right)=dw\left(A^{*}\right)$, we obtain inequality (22). □

In [34], for Equation (4), the authors show that if $H,K\in \mathcal{B}\left(H\right)$ are self-adjointed, then

$${\displaystyle \frac{1}{2}}\parallel H^2+K^2\parallel \le w^2(H+iK)\le \parallel H^2+K^2\parallel .$$

(23)

By using the inequality (23), we obtain the following remark, which shows that Theorem 4 improves Theorem 2 for the case $r=2$:

Remark 3.

Since $A{A}^{†}$ and $A^{†}A$ are self-adjointed, we have

$$\begin{array}{cc}\hfill d{w}^4\left(A\right)& \le {\displaystyle \frac{1}{2}}{w}^2\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A)\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}∥{\left(\right|A|}^4+{\left|A\right|}^8{)}^2+{(A{A}^{†}+A^{†}A)}^2∥\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left.∥{2\left(\right|A|}^8+{\left|A\right|}^{16})+2(A{A}^{†}+A^{†}A\right)\parallel \hfill \\ \hfill & =∥{\left|A\right|}^8+{\left|A\right|}^{16}+A{A}^{†}+A^{†}A∥.\hfill \end{array}$$

Theorem 5.

Let $A\in \mathcal{CR}\left(H\right)$. Then, for $r\ge 1$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥{3\left|A\right|}^{4r}+A{A}^{†}∥$$

(24)

and

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥3|A^{*}{|}^{4r}+A^{†}A∥.$$

(25)

Proof. 

Let $x\in H$ be a unit vector. Then,

$$\begin{array}{cc}\hfill {\left(\right|\langle Ax,x\rangle |}^2+{\parallel Ax\parallel }^4{)}^r& \le 2^{r-1}\left({\left|\langle Ax,x\rangle \right|}^{2r}+{\langle A^{*}Ax,x\rangle }^{2r}\right)\hfill \\ \hfill & \le 2^{r-1}\left({\langle \left|A\right|}^2{x,x\rangle }^r{\langle A{A}^{†}x,x\rangle }^r+{\langle \left|A\right|}^2{x,x\rangle }^{2r}\right)\hfill \\ \hfill & \phantom{=}\left(byLemma4\right)\hfill \\ \hfill & \le 2^{r-1}\left({\displaystyle \frac{{\langle \left|A\right|}^2{x,x\rangle }^{2r}+{\langle A{A}^{†}x,x\rangle }^{2r}}{2}}+{\langle \left|A\right|}^2{x,x\rangle }^{2r}\right)\hfill \\ \hfill & \le 2^{r-1}\left({\displaystyle \frac{{\langle \left|A\right|}^{4r}x,x\rangle +\langle {\left(A{A}^{†}\right)}^{2r}x,x\rangle }{2}}+{\langle \left|A\right|}^{4r}x,x\rangle \right)\hfill \\ \hfill & \phantom{=}\left(byLemma1\right)\hfill \\ \hfill & =2^{r-2}〈\left(3\right|A{|}^{4r}+A{A}^{†})x,x〉\hfill \\ \hfill & \le 2^{r-2}∥{3\left|A\right|}^{4r}+A{A}^{†}∥.\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain

$$d{w}^{2r}\left(A\right)\le 2^{r-2}∥{3\left|A\right|}^{4r}+A{A}^{†}∥.$$

(26)

Similar to the proof of inequality (26), and using $dw\left(A\right)=dw\left(A^{*}\right)$, we obtain inequality (25). □

Remark 4.

In particular, if we take $r=1$ in Theorem 5, then we obtain following inequalities:

$$d{w}^2\left(A\right)\le {\displaystyle \frac{1}{2}}∥{3\left|A\right|}^4+A{A}^{†}∥$$

(27)

and

$$d{w}^2\left(A\right)\le {\displaystyle \frac{1}{2}}∥3|A^{*}{|}^4+A^{†}A∥.$$

(28)

Next, we consider an example to show that inequalities (27) and (28) are better than existing inequalities (3), (4), and (23). If we consider $A=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$, then from (27) and (28), we obtain $d{w}^2\left(A\right)\le 2$, whereas the inequalities in (3), (4), and (23) give $d{w}^2\left(A\right)\le 6$, $d{w}^2\left(A\right)\le 17$, and $d{w}^2\left(A\right)\le 4$, respectively. Thus, for this example, the upper bound of $d{w}^2\left(A\right)$ in inequalities (27) and (28) are better than the existing bounds.

Next, based on Lemmas 2 and 4, we obtain the following result:

Theorem 6.

Let $A\in \mathcal{CR}\left(H\right)$, $\alpha \in \mathbb{C}\setminus \left\{0\right\}$. Then, for $r\ge 1$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-1}\left({\displaystyle \frac{1}{2\left|\alpha \right|}}∥{3\left|A\right|}^{4r}+A{A}^{†}∥+{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{∥X∥}^{{\displaystyle \frac{1}{2}}}{∥Z∥}^{{\displaystyle \frac{1}{2}}}\right)$$

(29)

and

$$d{w}^{2r}\left(A\right)\le 2^{r-1}\left({\displaystyle \frac{1}{2\left|\alpha \right|}}∥3|A^{*}{|}^{4r}+A^{†}A∥+{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{∥Y∥}^{{\displaystyle \frac{1}{2}}}{∥Z∥}^{{\displaystyle \frac{1}{2}}}\right),$$

(30)

where

$$X={\left|A\right|}^{4r}+{\left|A\right|}^{8r},Y=|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r},Z=A{A}^{†}+A^{†}A.$$

Proof. 

Let $x\in H$ be a unit vector. Then,

$$\begin{array}{cc}& {\left|\langle Ax,x\rangle \right|}^{2r}+{\parallel Ax\parallel }^{4r}\hfill \\ \hfill & ={\left|\langle Ax,x\rangle \right|}^{2r}+{\langle A^{*}Ax,x\rangle }^{2r}\hfill \\ \hfill & \le {\langle \left|A\right|}^2{x,x\rangle }^r{\langle A{A}^{†}x,x\rangle }^r+{\langle \left|A\right|}^4{x,x\rangle }^r{\langle A^{†}Ax,x\rangle }^r\hfill \\ \hfill & \phantom{=}\left(byLemma4\right)\hfill \\ \hfill & \le {\langle \left|A\right|}^{2r}x,x\rangle \langle x,A{A}^{†}x\rangle +{\langle \left|A\right|}^{4r}x,x\rangle \langle x,A^{†}Ax\rangle \hfill \\ \hfill & \phantom{=}\left(byLemma1\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{\left|\alpha \right|}}\left(\left|\langle |A{|}^{2r}x,A{A}^{†}x\rangle \right|+max\{1,|\alpha -1\left|\right\}∥{\left|A\right|}^{2r}x∥∥A{A}^{†}x∥\right)\hfill \\ \hfill & +{\displaystyle \frac{1}{\left|\alpha \right|}}\left(\left|\langle |A{|}^{4r}x,A^{†}Ax\rangle \right|+max\{1,|\alpha -1\left|\right\}∥{\left|A\right|}^{4r}x∥∥A^{†}Ax∥\right)\hfill \\ \hfill & \phantom{=}\left(byLemma2\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{\left|\alpha \right|}}\left(\left|\langle |A{|}^{2r}x,A{A}^{†}x\rangle \right|+\left|\langle |A{|}^{4r}x,A^{†}Ax\rangle \right|\right)\hfill \\ \hfill & +{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}\left({\langle \left|A\right|}^{4r}{x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle A{A}^{†}x,x\rangle }^{{\displaystyle \frac{1}{2}}}+{\langle \left|A\right|}^{8r}{x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle A^{†}Ax,x\rangle }^{{\displaystyle \frac{1}{2}}}\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{\left|\alpha \right|}}\left(\left|\langle |A{|}^{2r}x,A{A}^{†}x\rangle \right|+\left|\langle {\left(A^{†}A\right)}^{*}|A{|}^{4r}x,x\rangle \right|\right)\hfill \\ \hfill & +{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{\left({\langle \left|A\right|}^{4r}{x,x\rangle +\langle \left|A\right|}^{8r}x,x\rangle \right)}^{{\displaystyle \frac{1}{2}}}{\left(\langle A{A}^{†}x,x\rangle +\langle A^{†}Ax,x\rangle \right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{\left|\alpha \right|}}\left(\left|\langle |A{|}^{2r}x,A{A}^{†}x\rangle \right|+\left|\langle |A{|}^{4r}x,x\rangle \right|\right)\hfill \\ \hfill & +{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{\left|\alpha \right|}}\left(∥{\left|A\right|}^{2r}x∥∥A{A}^{†}x∥+{\langle \left|A\right|}^{4r}x,x\rangle \right)\hfill \\ \hfill & +{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & ={\displaystyle \frac{1}{\left|\alpha \right|}}\left({\langle \left|A\right|}^{4r}{x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle A{A}^{†}x,x\rangle }^{{\displaystyle \frac{1}{2}}}+{\langle \left|A\right|}^{4r}x,x\rangle \right)\hfill \\ \hfill & +{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{\left|\alpha \right|}}\left({\displaystyle \frac{{\langle \left(\right|A|}^{4r}+A{A}^{†})x,x\rangle }{2}}+{\langle \left|A\right|}^{4r}x,x\rangle \right)\hfill \\ \hfill & +{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2\left|\alpha \right|}}∥{3\left|A\right|}^{4r}+A{A}^{†}∥+{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}& {\left({\left|\langle Ax,x\rangle \right|}^2+{\parallel Ax\parallel }^4\right)}^r\hfill \\ \hfill & \le 2^{r-1}\left({\left|\langle Ax,x\rangle \right|}^{2r}+{\parallel Ax\parallel }^{4r}\right)\hfill \\ \hfill & \le 2^{r-1}\left({\displaystyle \frac{1}{2\left|\alpha \right|}}∥{3\left|A\right|}^{4r}+A{A}^{†}∥+{\displaystyle \frac{max\{1,|\alpha -1\left|\right\}}{\left|\alpha \right|}}{∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}\right).\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain the inequality (29). Similar to the proof of inequality (29), and using $dw\left(T\right)=dw\left(T^{*}\right)$, we obtain inequality (30). □

As a special case for $\alpha =2$ and $r=1$ in Theorem 6, we have the following corollary:

Corollary 2.

Let $A\in \mathcal{CR}\left(H\right)$. Then,

$$d{w}^2\left(A\right)\le {\displaystyle \frac{1}{4}}∥{3\left|A\right|}^4+A{A}^{†}∥+{\displaystyle \frac{1}{2}}{∥{\left|A\right|}^4+{\left|A\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}$$

(31)

and

$$d{w}^2\left(A\right)\le {\displaystyle \frac{1}{4}}∥3|A^{*}{|}^4+A^{†}A∥+{\displaystyle \frac{1}{2}}{∥|A^{*}{|}^4+{\left|A^{*}\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}.$$

(32)

Remark 5.

We note that, if the inequalities

$${\displaystyle \frac{1}{2}}{∥{\left|A\right|}^4+{\left|A\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}\le {\displaystyle \frac{1}{4}}∥{3\left|A\right|}^4+A{A}^{†}∥$$

(33)

and

$${\displaystyle \frac{1}{2}}{∥|A^{*}{|}^4+{\left|A^{*}\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}\le {\displaystyle \frac{1}{4}}∥3|A^{*}{|}^4+A^{†}A∥$$

(34)

hold, then inequalities (31) and (32) are the improvements of inequalities (24) and (25) for the case $r=1$ in Theorem 5, respectively. Next, we give an example to show that there is $A\in \mathcal{CR}\left(H\right)$ such that inequalities (33) and (34) hold. Consider $A=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$. Then,

$${\displaystyle \frac{1}{2}}{∥{\left|A\right|}^4+{\left|A\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{\sqrt{2}}{2}}\approx 0.71<0.75\approx {\displaystyle \frac{3}{4}}={\displaystyle \frac{1}{4}}∥{3\left|A\right|}^4+A{A}^{†}∥$$

and

$${\displaystyle \frac{1}{2}}{∥|A^{*}{|}^4+{\left|A^{*}\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}={\displaystyle \frac{\sqrt{2}}{2}}\approx 0.71<0.75\approx {\displaystyle \frac{3}{4}}={\displaystyle \frac{1}{4}}∥3|A^{*}{|}^4+A^{†}A∥.$$

Next, using Lemma 3, we obtain the following Davis–Wielandt radius of the bounded linear operators:

Theorem 7.

Let $A\in \mathcal{B}\left(H\right)$, $0\le \alpha \le 1$. Then, for $r\ge 1$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-1}\left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^{2r}+{\left|A^{*}\right|}^{2r}∥+{\displaystyle \frac{1-\alpha }{2}}{w}^r\left(A^2\right)+{\parallel A\parallel }^{4r}\right)$$

(35)

and

$$d{w}^{2r}\left(A\right)\le 2^{r-1}\left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^{2r}+{\left|A^{*}\right|}^{2r}∥+{\displaystyle \frac{1-\alpha }{2}}{w}^r\left(A^{*2}\right)+{\parallel A\parallel }^{4r}\right).$$

(36)

Proof. 

Let $x\in H$ be a unit vector. Then, from the Lemma 3, we obtain

$$\begin{array}{cc}\hfill {\left|\langle Ax,x\rangle \right|}^{2r}& =|\langle Ax,x\rangle \langle x,A^{*}x\rangle {|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{\parallel Ax\parallel }^r{\parallel A^{*}x\parallel }^r+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle Ax,A^{*}x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{\langle Ax,Ax\rangle }^{{\displaystyle \frac{r}{2}}}{\langle A^{*}x,A^{*}x\rangle }^{{\displaystyle \frac{r}{2}}}+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle A^{2}x,x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{4}}\left({\langle \left|A\right|}^2{x,x\rangle }^r+\langle |A^{*}{{|}^{2}x,x\rangle }^r\right)+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle A^{2}x,x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{4}}{\langle \left(\right|A|}^{2r}+|A^{*}{|}^{2r})x,x\rangle +{\displaystyle \frac{1-\alpha }{2}}{\left|\langle A^{2}x,x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^{2r}+{\left|A^{*}\right|}^{2r}∥+{\displaystyle \frac{1-\alpha }{2}}{w}^r\left(A^2\right).\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}& {\left({\left|\langle Ax,x\rangle \right|}^2+{\parallel Ax\parallel }^4\right)}^r\hfill \\ \hfill & \le 2^{r-1}\left({\left|\langle Ax,x\rangle \right|}^{2r}+{\parallel Ax\parallel }^{4r}\right)\hfill \\ \hfill & \le 2^{r-1}\left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^{2r}+\langle |A^{*}{|}^{2r}∥+{\displaystyle \frac{1-\alpha }{2}}{w}^r\left(A^2\right)+{\parallel A\parallel }^{4r}\right).\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain the inequality (35). Similar to the proof of inequality (35), and using $dw\left(A\right)=dw\left(A^{*}\right)$, we obtain inequality (36). □

Remark 6.

If we take $\alpha =0$ in inequality (35), then we have

$$d{w}^{2r}\left(A\right)\le 2^{r-2}\left({\displaystyle \frac{1}{2}}∥{\left|A\right|}^{2r}+{\left|A^{*}\right|}^{2r}∥+w^r\left(A^2\right)+2{\parallel A\parallel }^{4r}\right).$$

(37)

It is obvious that inequality (37) improves the inequality in [20], i.e., Theorem 3.1.

In particular, if we consider $r=1$ in Theorem 7, then we have the following corollary:

Corollary 3.

Let $A\in \mathcal{B}\left(H\right)$. Then

$$d{w}^2\left(A\right)\le min\{X,Y\},$$

where

$$X=\underset{0\le \alpha \le 1}{min}\left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^2+{\left|A^{*}\right|}^2∥+{\displaystyle \frac{1-\alpha }{2}}w\left(A^2\right)+{\parallel A\parallel }^4\right)$$

and

$$Y=\underset{0\le \alpha \le 1}{min}\left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^2+{\left|A^{*}\right|}^2∥+{\displaystyle \frac{1-\alpha }{2}}w\left(A^{*2}\right)+{\parallel A\parallel }^4\right).$$

Theorem 8.

Let $A\in \mathcal{CR}\left(H\right)$, $0\le \alpha \le 1$. Then, for $r\ge 1$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-1}\left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^{4r}+A{A}^{†}∥+{\displaystyle \frac{1-\alpha }{2}}{w}^r(A{A}^{†}{\left|A\right|}^2{)+\parallel A\parallel }^{4r}\right)$$

(38)

and

$$d{w}^{2r}\left(A\right)\le 2^{r-1}\left({\displaystyle \frac{1+\alpha }{4}}∥|A^{*}{|}^{4r}+A^{†}A∥+{\displaystyle \frac{1-\alpha }{2}}{w}^r(A^{†}A|A^{*}{|}^2{)+\parallel A\parallel }^{4r}\right).$$

(39)

Proof. 

Let $x\in H$ be a unit vector. Then, from the Lemmas 3 and 4, we obtain

$$\begin{array}{cc}\hfill {\left|\langle Ax,x\rangle \right|}^{2r}& \le {\langle \left|A\right|}^2{x,x\rangle }^r{\langle A{A}^{†}x,x\rangle }^r\hfill \\ \hfill & ={\left|{\langle \left|A\right|}^{2}x,x\rangle \langle x,A{A}^{†}x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{∥{\left|A\right|}^{2}x∥}^r{\parallel A{A}^{†}x\parallel }^r+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle |A{|}^{2}x,A{A}^{†}x\rangle \right|}^r\hfill \\ \hfill & ={\displaystyle \frac{1+\alpha }{2}}{\langle \left|A\right|}^4{x,x\rangle }^{{\displaystyle \frac{r}{2}}}{\langle A{A}^{†}x,x\rangle }^{{\displaystyle \frac{r}{2}}}+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle A{A}^{†}|A{|}^{2}x,x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{4}}\left({\langle \left|A\right|}^4{x,x\rangle }^r+{\langle A{A}^{†}x,x\rangle }^r\right)+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle A{A}^{†}|A{|}^{2}x,x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{4}}{\langle \left(\right|A|}^{4r}+A{A}^{†})x,x\rangle +{\displaystyle \frac{1-\alpha }{2}}{\left|\langle A{A}^{†}|A{|}^{2}x,x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^{4r}+A{A}^{†}∥+{\displaystyle \frac{1-\alpha }{2}}{w}^r(A{A}^{†}{\left|A\right|}^2).\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}& {\left({\left|\langle Ax,x\rangle \right|}^2+{\parallel Ax\parallel }^4\right)}^r\hfill \\ \hfill & \le 2^{r-1}\left({\left|\langle Ax,x\rangle \right|}^{2r}+{\parallel Ax\parallel }^{4r}\right)\hfill \\ \hfill & \le 2^{r-1}\left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^{4r}+A{A}^{†}∥+{\displaystyle \frac{1-\alpha }{2}}{w}^r(A{A}^{†}{\left|A\right|}^2{)+\parallel A\parallel }^{4r}\right).\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain inequality (38). Similar to the proof of inequality (38), and using $dw\left(A\right)=dw\left(A^{*}\right)$, we obtain inequality (39). □

In particular, if we take $r=1$ in Theorem 8, then we obtain the following corollary:

Corollary 4.

Let $A\in \mathcal{CR}\left(H\right)$. Then,

$$d{w}^2\left(A\right)\le min\{M,N\},$$

where

$$M=\underset{0\le \alpha \le 1}{min}\left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^4+A{A}^{†}∥+{\displaystyle \frac{1-\alpha }{2}}w(A{A}^{†}{\left|A\right|}^2{)+\parallel A\parallel }^4\right)$$

and

$$N=\underset{0\le \alpha \le 1}{min}\left({\displaystyle \frac{1+\alpha }{4}}∥|A^{*}{|}^4+A^{†}A∥+{\displaystyle \frac{1-\alpha }{2}}w(A^{†}A|A^{*}{|}^2{)+\parallel A\parallel }^4\right).$$

Remark 7.

Since $w(A{A}^{†}{\left|A\right|}^2)\le {\displaystyle \frac{1}{2}}∥{\left|A\right|}^4+A{A}^{†}∥$. Then,

$$\begin{array}{cc}\hfill d{w}^2\left(A\right)& \le \left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^4+A{A}^{†}∥+{\displaystyle \frac{1-\alpha }{2}}w(A{A}^{†}{\left|A\right|}^2{)+\parallel A\parallel }^4\right)\hfill \\ \hfill & \le \left({\displaystyle \frac{1+\alpha }{4}}∥{\left|A\right|}^4+A{A}^{†}∥+{\displaystyle \frac{1-\alpha }{4}}∥{\left|A\right|}^4+A{A}^{†}∥+{\parallel A\parallel }^4\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}∥{\left|A\right|}^4+A{A}^{†}∥+{\parallel A\parallel }^4.\hfill \end{array}$$

(40)

Inequality (40) refines the inequalities in (4), (5), and (6) for some operators. Let $A=\left[\begin{array}{cc}1& 0\\ 0& 0\end{array}\right]$, then (40) gives $d{w}^2\left(A\right)\le 2$, whereas the inequalities in (4), (5), and (6) give $d{w}^2\left(A\right)\le 17$, $d{w}^2\left(A\right)\le 2.41$, and $d{w}^2\left(A\right)\le 2.41$, respectively. Thus, for this example, the inequalities in Theorem 8 for $r=1$ are better than the existing inequalities.

Theorem 9.

Let $A\in \mathcal{CR}\left(H\right)$, $0\le \alpha \le 1$. Then, for $r\ge 1$, we have

$$d{w}^{2r}\left(A\right)\le 2^{r-1}\left({\displaystyle \frac{1+\alpha }{2}}{∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}+{\displaystyle \frac{1-\alpha }{4}}∥{3\left|A\right|}^{4r}+A{A}^{†}∥\right)$$

(41)

and

$$d{w}^{2r}\left(A\right)\le 2^{r-1}\left({\displaystyle \frac{1+\alpha }{2}}{∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}+{\displaystyle \frac{1-\alpha }{4}}∥3|A^{*}{|}^{4r}+A^{†}A∥\right).$$

(42)

Proof. 

Let $x\in H$ be a unit vector. Then, from the Lemma 3, we have

$$\begin{array}{cc}& {\left|\langle Ax,x\rangle \right|}^{2r}+{\langle \left|A\right|}^2{x,x\rangle }^{2r}\hfill \\ \hfill & \le {\langle \left|A\right|}^2{x,x\rangle }^r{\langle A{A}^{†}x,x\rangle }^r+{\langle \left|A\right|}^4{x,x\rangle }^r{\langle A^{†}Ax,x\rangle }^r\hfill \\ \hfill & ={\left|{\langle \left|A\right|}^{2}x,x\rangle \langle x,A{A}^{†}x\rangle \right|}^r+{\left|{\langle \left|A\right|}^{4}x,x\rangle \langle x,A^{†}Ax\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{∥{\left|A\right|}^{2}x∥}^r{\parallel A{A}^{†}x\parallel }^r+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle |A{|}^{2}x,A{A}^{†}x\rangle \right|}^r\hfill \\ \hfill & +{\displaystyle \frac{1+\alpha }{2}}{∥{\left|A\right|}^{4}x∥}^r{\parallel A^{†}Ax\parallel }^r+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle |A{|}^{4}x,A^{†}Ax\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{\langle \left|A\right|}^4{x,x\rangle }^{{\displaystyle \frac{r}{2}}}{\langle A{A}^{†}x,A{A}^{†}x\rangle }^{{\displaystyle \frac{r}{2}}}+{\displaystyle \frac{1-\alpha }{2}}{∥{\left|A\right|}^{2}x∥}^r{∥A{A}^{†}x∥}^r\hfill \\ \hfill & +{\displaystyle \frac{1+\alpha }{2}}{\langle \left|A\right|}^8{x,x\rangle }^{{\displaystyle \frac{r}{2}}}{\langle A^{†}Ax,A{A}^{†}x\rangle }^{{\displaystyle \frac{r}{2}}}+{\displaystyle \frac{1-\alpha }{2}}{\left|\langle A^{†}A|A{|}^{4}x,x\rangle \right|}^r\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{\left({\langle \left|A\right|}^4{x,x\rangle }^r+{\langle \left|A\right|}^8{x,x\rangle }^r\right)}^{{\displaystyle \frac{1}{2}}}{\left({\langle A{A}^{†}x,x\rangle }^r+{\langle A^{†}Ax,x\rangle }^r\right)}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\displaystyle \frac{1-\alpha }{2}}\left({\langle \left|A\right|}^4{x,x\rangle }^{{\displaystyle \frac{r}{2}}}{\langle A{A}^{†}x,A{A}^{†}x\rangle }^{{\displaystyle \frac{r}{2}}}+{\left|\langle |A{|}^{4}x,x\rangle \right|}^r\right)\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\displaystyle \frac{1-\alpha }{2}}\left({\langle \left|A\right|}^4{x,x\rangle }^{{\displaystyle \frac{r}{2}}}{\langle A{A}^{†}x,x\rangle }^{{\displaystyle \frac{r}{2}}}+{\langle \left|A\right|}^{4r}x,x\rangle \right)\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & +{\displaystyle \frac{1-\alpha }{2}}\left({\displaystyle \frac{{\langle \left|A\right|}^{4r}x,x\rangle +\langle A{A}^{†}x,x\rangle }{2}}+{\langle \left|A\right|}^{4r}x,x\rangle \right)\hfill \\ \hfill & \le {\displaystyle \frac{1+\alpha }{2}}{∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}+{\displaystyle \frac{1-\alpha }{4}}∥{3\left|A\right|}^{4r}+A{A}^{†}∥.\hfill \end{array}$$

Therefore,

$$\begin{array}{cc}& {\left({\left|\langle Ax,x\rangle \right|}^2+{\parallel Ax\parallel }^4\right)}^r\hfill \\ \hfill & \le 2^{r-1}\left({\left|\langle Ax,x\rangle \right|}^{2r}+{\langle \left|A\right|}^{2}x,x{\rangle }^{2r})\right)\hfill \\ \hfill & 2^{r-1}\left({\displaystyle \frac{1+\alpha }{2}}{∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}+{\displaystyle \frac{1-\alpha }{4}}∥{3\left|A\right|}^{4r}+A{A}^{†}∥\right).\hfill \end{array}$$

Taking the supremum over all unit vectors in H, we obtain inequality (41). Similar to the proof of inequality (41), and using $dw\left(A\right)=dw\left(A^{*}\right)$, we obtain inequality (42). □

Remark 8.

If we take $\alpha =0$ and $r=1$ in Theorem 9, then Theorem 9 reduces to Corollary 2.

Finally, we compare some upper bounds of the Davis–Wieldant radius obtained by us. In the following remark, we first show that the upper bounds of the Davis–Wieldant radius in Theorem 1 are smaller than those in Theorem 2.

Remark 9.

Let $A\in \mathcal{CR}\left(H\right)$, $r\ge 1$, $x\in H$ with $\parallel x\parallel =1$. Then,

$$\begin{array}{cc}& {\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left({\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r})x,x\rangle +\langle (A{A}^{†}+A^{†}A)x,x\rangle \right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{\langle \left(\right|A|}^{4r}+{\left|A\right|}^{8r}+A{A}^{†}+A^{†}A)x,x\rangle \hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}+A{A}^{†}+A^{†}A∥.\hfill \end{array}$$

By taking the supremum over all $x\in H$ with $\parallel x\parallel =1$, we obtain

$${∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}\le {\displaystyle \frac{1}{2}}∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}+A{A}^{†}+A^{†}A∥,$$

Similarly, we obtain

$${∥{\left|A\right|}^{4r}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}{∥{\left|A\right|}^{8r}+A{A}^{†}∥}^{{\displaystyle \frac{1}{2}}}\le {\displaystyle \frac{1}{2}}∥{\left|A\right|}^{4r}+{\left|A\right|}^{8r}+A{A}^{†}+A^{†}A∥,$$

$${∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}\le {\displaystyle \frac{1}{2}}∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}+A{A}^{†}+A^{†}A∥$$

and

$${∥|A^{*}{|}^{4r}+A{A}^{†}∥}^{{\displaystyle \frac{1}{2}}}{∥|A^{*}{|}^{8r}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}\le {\displaystyle \frac{1}{2}}∥|A^{*}{|}^{4r}+{\left|A^{*}\right|}^{8r}+A{A}^{†}+A^{†}A∥.$$

Thus, the upper bounds of the Davis–Wieldant radius in Theorem 1 are smaller than those in Theorem 2.

Remark 10.

If the inequalities

$${\displaystyle \frac{1}{4}}∥{3\left|A\right|}^4+A{A}^{†}∥\le {\displaystyle \frac{\sqrt{2}}{4}}w\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A)\right)$$

(43)

and

$${\displaystyle \frac{1}{4}}∥3|A^{*}{|}^4+A^{†}A∥\le {\displaystyle \frac{\sqrt{2}}{4}}w\left(|A^{*}{|}^4+{\left|A^{*}\right|}^8+i(A{A}^{†}+A^{†}A)\right)$$

(44)

hold, then the upper bounds of the Davis–Wieldant radius in Corollary 2 are smaller than those in Theorem 4. Indeed, let $x\in H$ be a unit vector. Then,

$$\begin{array}{cc}& {\displaystyle \frac{1}{4}}{\langle \left(3\right|A|}^4+A{A}^{†})x,x\rangle +{\displaystyle \frac{1}{2}}{\langle \left(\right|A|}^4+{\left|A\right|}^8{)x,x\rangle }^{{\displaystyle \frac{1}{2}}}{\langle (A{A}^{†}+A^{†}A)x,x\rangle }^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}{\langle \left(3\right|A|}^4+A{A}^{†})x,x\rangle +{\displaystyle \frac{1}{4}}\left({\langle \left(\right|A|}^4+{\left|A\right|}^8)x,x\rangle +\langle (A{A}^{†}+A^{†}A)x,x\rangle \right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}{\langle \left(3\right|A|}^4+A{A}^{†})x,x\rangle +{\displaystyle \frac{\sqrt{2}}{4}}\left|{\langle \left(\right|A|}^4+{\left|A\right|}^8)x,x\rangle +i\langle (A{A}^{†}+A^{†}A)x,x\rangle \right|\hfill \\ \hfill & ={\displaystyle \frac{1}{4}}{\langle \left(3\right|A|}^4+A{A}^{†})x,x\rangle +{\displaystyle \frac{\sqrt{2}}{4}}\left|{\langle \left(\right|A|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A))x,x\rangle \right|\hfill \\ \hfill & \le {\displaystyle \frac{1}{4}}∥{3\left|A\right|}^4+A{A}^{†}∥+{\displaystyle \frac{\sqrt{2}}{4}}w\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A\right).\hfill \end{array}$$

Therefore, if ${\displaystyle \frac{1}{4}}∥{3\left|A\right|}^4+A{A}^{†}∥\le {\displaystyle \frac{\sqrt{2}}{4}}w\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A)\right)$, we obtain

$$\begin{array}{cc}& {\displaystyle \frac{1}{4}}∥{3\left|A\right|}^4+A{A}^{†}∥+{\displaystyle \frac{1}{2}}{∥{\left|A\right|}^4+{\left|A\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le {\displaystyle \frac{1+\sqrt{2}}{4}}w\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A\right)\hfill \\ \hfill & <{\displaystyle \frac{1}{\sqrt{2}}}w\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A\right).\hfill \end{array}$$

Similarly, if ${\displaystyle \frac{1}{4}}∥3|A^{*}{|}^4+A^{†}A∥\le {\displaystyle \frac{\sqrt{2}}{4}}w\left(|A^{*}{|}^4+{\left|A^{*}\right|}^8+i(A{A}^{†}+A^{†}A)\right)$, we obtain

$$\begin{array}{cc}& {\displaystyle \frac{1}{4}}∥3|A^{*}{|}^4+A^{†}A∥+{\displaystyle \frac{1}{2}}{∥|A^{*}{|}^4+{\left|A^{*}\right|}^8∥}^{{\displaystyle \frac{1}{2}}}{∥A{A}^{†}+A^{†}A∥}^{{\displaystyle \frac{1}{2}}}\hfill \\ \hfill & \le {\displaystyle \frac{1+\sqrt{2}}{4}}w\left(|A^{*}{|}^4+|A^{*}{|}^8+i(A{A}^{†}+A^{†}A\right)\hfill \\ \hfill & <{\displaystyle \frac{1}{\sqrt{2}}}w\left(|A^{*}{|}^4+|A^{*}{|}^8+i(A{A}^{†}+A^{†}A\right).\hfill \end{array}$$

Now, we give an example to show that there is $A\in \mathcal{CR}\left(H\right)$ such that inequalities (43) and (44) hold. Consider $A=\left[\begin{array}{cc}0& 2\\ 0& 0\end{array}\right]$. Then,

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{4}}∥{3\left|A\right|}^4+A{A}^{†}∥=12<96.167\approx {\displaystyle \frac{\sqrt{2}}{4}}w\left({\left|A\right|}^4+{\left|A\right|}^8+i(A{A}^{†}+A^{†}A\right)\end{array}$$

and

$$\begin{array}{c}\hfill {\displaystyle \frac{1}{4}}∥3|A^{*}{|}^4+A^{†}A∥=12<96.167\approx {\displaystyle \frac{\sqrt{2}}{4}}w\left(|A^{*}{|}^4+|A^{*}{|}^8+i(A{A}^{†}+A^{†}A\right).\end{array}$$

Next, we provide numerical examples to illustrate that the upper bounds of the Davis–Wieldant radius obtained by Corollary 1, Corollary 3 and Corollary 4 are generally not comparable.

Remark 11.

Let $A_1=\left[\begin{array}{cc}0& 1\\ 0& 0\end{array}\right]$, $A_2=\left[\begin{array}{cc}0& 2\\ 0& 0\end{array}\right]$, and $A_3=\left[\begin{array}{cc}1& 1\\ 0& 0\end{array}\right]$. Then, the upper bounds of the Davis–Wieldant radii in Corollary 1, Corollary 3, and Corollary 4 are as shown in

Table 1. The upper bounds of the Davis–Wieldant radii.

	${\mathit{dw}}^2\left({\mathit{A}}_1\right)$	${\mathit{dw}}^2\left({\mathit{A}}_2\right)$	${\mathit{dw}}^2\left({\mathit{A}}_3\right)$
Corollary 1	1.414	16.492	5.843
Corollary 3	1.25	17	5.457
Corollary 4	1.25	20	5.744

.