Improved Bounds for the Euclidean Numerical Radius of Operator Pairs in Hilbert Spaces

Abstract

This paper presents new lower and upper bounds for the Euclidean numerical radius of operator pairs in Hilbert spaces, demonstrating improvements over recent results by other authors. Additionally, we derive new inequalities for the numerical radius and the Davis–Wielandt radius as natural consequences of our findings.

1. Introduction

Let H be a complex Hilbert space equipped with an inner product $\langle ·,·\rangle$ and the corresponding norm $\parallel ·\parallel$. The $C^{*}$-algebra of bounded linear operators on H is denoted by $\mathcal{B}\left(H\right)$. For any operator $A\in \mathcal{B}\left(H\right)$, the adjoint of A is denoted by $A^{*}$, and $\left|A\right|={\left(A^{*}A\right)}^{{\displaystyle \frac{1}{2}}}$ represents the positive square root of $A^{*}A$. The numerical range of A, denoted by $W\left(A\right)$, is the set of values $\{\langle Ax,x\rangle :x\in H,\parallel x\parallel =1\}$.

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

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

while the numerical radius is given by

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

It is known that the numerical radius $w(·)$ defines a norm on $\mathcal{B}\left(H\right)$ that is equivalent to the operator norm $\parallel ·\parallel$. In particular, the following double inequality is valid:

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

(1)

for any operator $A\in \mathcal{B}\left(H\right)$. These inequalities are sharp. The first inequality becomes an equality if $A^2=0$, while the second inequality becomes an equality if and only if A is a normal operator. An improvement to these inequalities was established by Kittaneh [1], who proved that

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

(2)

For further advancements related to inequalities (1) and (2), interested readers can refer to sources [2,3,4,5,6,7,8] and the references therein.

Let $\left(C,D\right)\in \mathcal{B}{\left(H\right)}^2:=\mathcal{B}\left(H\right)\times \mathcal{B}\left(H\right)$. The Euclidean operator radius is defined by

$$w_e\left(C,D\right):=\underset{\begin{array}{c}x\in H\\ \parallel x\parallel =1\end{array}}{sup}{\left({\left|〈Cx,x〉\right|}^2+{\left|〈Dx,x〉\right|}^2\right)}^{1/2}.$$

As pointed out in [9], $w_e:\mathcal{B}{\left(H\right)}^2\to [0,\infty )$ is a norm and the following inequality holds:

$${\displaystyle \frac{\sqrt{2}}{4}}{∥{\left|C\right|}^2+{\left|D\right|}^2∥}^{1/2}\le w_e\left(C,D\right)\le {∥{\left|C\right|}^2+{\left|D\right|}^2∥}^{1/2}$$

(3)

for $\left(C,D\right)\in \mathcal{B}{\left(H\right)}^2,$ where the constants ${\displaystyle \frac{\sqrt{2}}{4}}$ and 1 are best possible in (3).

In [10], the second author obtained the following lower bound:

$${\displaystyle \frac{\sqrt{2}}{2}}{\left[w\left(B^2+C^2\right)\right]}^{{\displaystyle \frac{1}{2}}}\le w_e\left(B,C\right).$$

The constant ${\displaystyle \frac{\sqrt{2}}{2}}$ is best possible in the sense that it cannot be replaced by a larger constant. In the same paper, the following results were obtained as well:

$${\displaystyle \frac{\sqrt{2}}{2}}max\left\{w\left(B+C\right),w\left(B-C\right)\right\}\le w_e\left(B,C\right)\le {\displaystyle \frac{\sqrt{2}}{2}}{\left[w^2\left(B+C\right)+w^2\left(B-C\right)\right]}^{{\displaystyle \frac{1}{2}}},$$

with constant ${\displaystyle \frac{\sqrt{2}}{2}}$ sharp in both inequalities,

$$w_e^2\left(B,C\right)\le max\left\{{∥B∥}^2,{∥C∥}^2\right\}+w\left(C^{*}B\right),$$

with the inequality being sharp, and also

$$w_e^2\left(B,C\right)\le {\displaystyle \frac{1}{2}}\left[∥{\left|B\right|}^2+{\left|C\right|}^2∥+∥{\left|B\right|}^2-{\left|C\right|}^2∥\right]+w\left(C^{*}B\right),$$

which is sharp as well.

By taking $\left(B,C\right)=\left(A,A^{*}\right)$ or $\left(B,C\right)=\left(\mathfrak{R}\left(A\right),\mathfrak{I}\left(A\right)\right)$ for $A\in B\left(H\right)$, where we recall that

$$\mathfrak{R}\left(A\right):={\displaystyle \frac{A+A^{*}}{2}}\mathrm{and}\mathfrak{I}\left(A\right):={\displaystyle \frac{A-A^{*}}{2i}},$$

the second author obtained in [10] several norm and numerical radius inequalities of interest for one operator A. Note that $\mathfrak{R}\left(A\right)$ and $\mathfrak{I}\left(A\right)$ refer to the real and imaginary parts of A, respectively.

For some recent results involving the estimation of the Euclidean numerical radius in terms of similar upper and lower bounds, see [11,12,13]. These estimations appear to be simpler to calculate when the involved operators are self-adjoint, since in that case, for instance, ${\left|B\right|}^2=B^2$, ${\left|C\right|}^2=C^2$, and $C^{*}B=CB.$ Moreover, if B and C are projections orthogonal to each other, then $B^2=B$, $C^2=C$, and $CB=0$, which make the bounds even simpler to calculate.

The Davis–Wielandt radius of an operator $T\in \mathcal{B}\left(H\right)$, denoted by $dw\left(T\right)$ is defined as follows [14,15]:

$$dw\left(T\right)=\underset{\begin{array}{c}x\in H\\ \parallel x\parallel =1\end{array}}{sup}{\left({\left|\langle Tx,x\rangle \right|}^2+{\parallel Tx\parallel }^4\right)}^{1/2}.$$

It is evident that $dw\left(T\right)\ge 0$, and $dw\left(T\right)=0$ if and only if $T=0$. For any $\lambda \in \mathbb{C}$, the following inequalities hold: $dw\left(\lambda T\right)>\left|\lambda \right|dw\left(T\right)$ if $\left|\lambda \right|>1$, $dw\left(\lambda T\right)<\left|\lambda \right|dw\left(T\right)$ if $\left|\lambda \right|<1$, and $dw\left(\lambda T\right)=\left|\lambda \right|dw\left(T\right)$ when $\left|\lambda \right|=1$. Note that the triangle inequality $dw(T+S)\le dw\left(T\right)+dw\left(S\right)$ does not always hold for arbitrary operators $T,S\in \mathcal{B}\left(H\right)$. However, this inequality holds when $\mathfrak{R}\left(T^{*}S\right)=0$, as shown in ([16], Corollary 2.2). It is also straightforward to verify that

$$max\left\{{w\left(T\right),\parallel T\parallel }^2\right\}\le dw\left(T\right)\le {\left(w^2\left(T\right)+{\parallel T\parallel }^4\right)}^{1/2},$$

and these inequalities are sharp, as noted in ([16], Corollary 2.2).

Additionally, observe that for $C=T$ and $D={\left|T\right|}^2$, we have

$$\begin{array}{cc}\hfill w_e{(T,|T|}^2)& :=\underset{\parallel x\parallel =1}{sup}{\left({\left|\langle Tx,x\rangle \right|}^2+{|\langle |T|}^{2}x,x{\rangle |}^2\right)}^{1/2}\hfill \\ \hfill & =\underset{\parallel x\parallel =1}{sup}{\left({\left|\langle Tx,x\rangle \right|}^2+{\parallel Tx\parallel }^4\right)}^{1/2}=dw\left(T\right).\hfill \end{array}$$

If we set $C=T$ and $D={\left|T\right|}^2$ in (3), then we obtain

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

(4)

which provides the upper bound from (5) and a corresponding lower bound.

Zamani and Shebrawi ([17], Theorem 2.1) proved that

$$dw\left(T\right)\le {\left[w^2{(T-|T|}^2{)+2\parallel T\parallel }^{2}w\left(T\right)\right]}^{1/2}.$$

Furthermore, in ([17], Theorems 2.13, 2.14, and 2.17), they also established the following inequalities:

$$d{w}^2\left(T\right)\le {\parallel T\parallel }^{2}max\left\{{1,\parallel T\parallel }^2\right\}+\sqrt{2}w\left({\left|T\right|}^{2}T\right),$$

$$d{w}^2\left(T\right)\le {\displaystyle \frac{1}{2}}\left(∥{\left|T\right|}^4+{\left|T\right|}^2∥+∥{\left|T\right|}^4-{\left|T\right|}^2∥+\sqrt{2}w\left({\left|T\right|}^{2}T\right)\right),$$

and

$$d{w}^2\left(T\right)\le \parallel T\parallel max\left\{{w\left(T\right),\parallel T\parallel }^2\right\}{\left[1+{\parallel T\parallel }^2+2w\left(T\right)\right]}^{1/2}$$

for any operator $T\in \mathcal{B}\left(H\right)$.

Recently, Bhunia et al. in ([16], Theorem 2.4) obtained the following upper bound:

$$dw\left(T\right)\le {∥{\left|T\right|}^4+{\left|T\right|}^2∥}^{1/2}$$

(5)

for $T\in \mathcal{B}\left(H\right)$.

Additionally, in [18], the authors derived inequalities for the sum of operators. Specifically, we have

$$\begin{array}{ccc}\hfill dw(T+S)& \le & {\left[2\left(d{w}^2\left(T\right)+d{w}^2\left(S\right)\right)+6∥{\left|T\right|}^4+{\left|S\right|}^4∥\right]}^{1/2}\hfill \\ & \le & 2\sqrt{2}{\left[d{w}^2\left(T\right)+d{w}^2\left(S\right)\right]}^{1/2}\hfill \end{array}$$

for $T,S\in \mathcal{B}\left(H\right)$.

Motivated by these results, we present new lower and upper bounds for the Euclidean numerical radius of operator pairs in this paper, demonstrating that some of these bounds improve upon those recently established by other authors.

2. Lower Bounds for the Euclidean Numerical Radius of Operator Pairs

In this section, we derive and establish several lower bounds for the Euclidean numerical radius of pairs of operators. Specifically, we present and prove the following lower bounds for the Euclidean numerical radius.

Theorem 1. 

For any $B,C\in \mathcal{B}\left(H\right)$, we have

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left(B^2+C^2\right),w\left(BC+CB\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(B\right),w\left(C\right)\right\}\left|w\left(B+C\right)-w\left(B-C\right)\right|\hfill \end{array}$$

(6)

and

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left(B^2+C^2\right),w\left(B^2-C^2\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(B+C\right),w\left(B-C\right)\right\}\left|w\left(B\right)-w\left(C\right)\right|.\hfill \end{array}$$

(7)

Proof. 

As in the proof of Theorem 1 of [10] (see also [12], Theorem 2.2), we have

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& =\underset{∥x∥=1}{sup}\left[{\left|〈Bx,x〉\right|}^2+{\left|〈Cx,x〉\right|}^2\right]\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}\underset{∥x∥=1}{sup}{\left(\left|〈Bx,x〉\right|+\left|〈Cx,x〉\right|\right)}^2\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}\underset{∥x∥=1}{sup}{\left|〈Bx,x〉\pm 〈Cx,x〉\right|}^2\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\underset{∥x∥=1}{sup}{\left|〈\left(B\pm C\right)x,x〉\right|}^2={\displaystyle \frac{1}{2}}{w}^2\left(B\pm C\right).\hfill \end{array}$$

This implies that

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w^2\left(B+C\right),w^2\left(B-C\right)\right\}\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\left(B+C\right)}^2\right),w\left({\left(B-C\right)}^2\right)\right\}\hfill \\ \hfill & ={\displaystyle \frac{1}{4}}\left[w\left({\left(B+C\right)}^2\right)+w\left({\left(B-C\right)}^2\right)\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\left|w\left({\left(B+C\right)}^2\right)-w\left({\left(B-C\right)}^2\right)\right|\hfill \\ \hfill & ={\displaystyle \frac{1}{4}}\left[w\left({\left(B+C\right)}^2\right)+w\left({\left(B-C\right)}^2\right)\right]\hfill \\ \hfill & +{\displaystyle \frac{1}{4}}\left[w\left(B+C\right)+w\left(B-C\right)\right]\left|w\left(B+C\right)-w\left(B-C\right)\right|\hfill \end{array}$$

(8)

By the triangle inequality, we have

$$w\left({\left(B+C\right)}^2\right)+w\left({\left(B-C\right)}^2\right)\ge w\left[{\left(B+C\right)}^2\pm {\left(B-C\right)}^2\right],$$

which gives

$$w\left({\left(B+C\right)}^2\right)+w\left({\left(B-C\right)}^2\right)\ge 2max\left\{w\left(B^2+C^2\right),w\left(BC+CB\right)\right\}.$$

Also,

$$w\left(B+C\right)+w\left(B-C\right)\ge w\left[\left(B+C\right)\pm \left(B-C\right)\right],$$

which gives

$$w\left(B+C\right)+w\left(B-C\right)\ge 2max\left\{w\left(B\right),w\left(C\right)\right\}.$$

By (8), we obtain (6).

If we replace B by $B+C$ and C by $B-C$ in (6), then we obtain

$$\begin{array}{cc}\hfill & w_e^2\left(B+C,B-C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\left(B+C\right)}^2+{\left(B-C\right)}^2\right),w\left(\left(B+C\right)\left(B-C\right)+\left(B-C\right)\left(B+C\right)\right)\right\}\hfill \\ \hfill & +max\left\{w\left(B+C\right),w\left(B-C\right)\right\}\left|w\left(B\right)-w\left(C\right)\right|.\hfill \end{array}$$

(9)

Since

$${\left(B+C\right)}^2+{\left(B-C\right)}^2=2\left(B^2+C^2\right)$$

and

$$\left(B+C\right)\left(B-C\right)+\left(B-C\right)\left(B+C\right)=2\left(B^2-C^2\right),$$

we can apply (9) to obtain

$$\begin{array}{cc}\hfill w_e^2\left(B+C,B-C\right)& \ge max\left\{w\left(B^2+C^2\right),w\left(B^2-C^2\right)\right\}\hfill \\ \hfill & +max\left\{w\left(B+C\right),w\left(B-C\right)\right\}\left|w\left(B\right)-w\left(C\right)\right|,\hfill \end{array}$$

which, by noting that

$$w_e^2\left(B+C,B-C\right)=2w_e^2\left(B,C\right),$$

yields the desired result (7). □

Remark 1. 

Inequality (6) is better than the following result obtained recently in ([12], Theorem 2.2). Indeed, for any $B,C\in \mathcal{B}\left(H\right),$ we have

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}\left[w\left(B^2+C^2\right)+max\left\{w\left(B\right),w\left(C\right)\right\}\left|w\left(B+C\right)-w\left(B-C\right)\right|\right]\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}w\left(B^2+C^2\right).\hfill \end{array}$$

(10)

Notice that the second lower bound in (10) was obtained in 2006 by the second author (see [10]).

Several consequences of Theorem 1 can be drawn. We begin with the following corollary.

Corollary 1. 

For any $B,C\in \mathcal{B}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$, we have

$$\begin{array}{cc}\hfill & max\left\{{\left|\alpha \right|}^2,{\left|\beta \right|}^2\right\}{w}_e^2\left(B,C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{B}^2+{\beta }^2{C}^2\right),\left|\alpha \beta \right|w\left(BC+CB\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{\left|\alpha \right|w\left(B\right),\left|\beta \right|w\left(C\right)\right\}\left|w\left(\alpha B+\beta C\right)-w\left(\alpha B-\beta C\right)\right|\hfill \end{array}$$

(11)

and

$$\begin{array}{cc}\hfill & max\left\{{\left|\alpha \right|}^2,{\left|\beta \right|}^2\right\}{w}_e^2\left(B,C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{B}^2+{\beta }^2{C}^2\right),w\left({\alpha }^2{B}^2-{\beta }^2{C}^2\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(\alpha B+\beta C\right),w\left(\alpha B-\beta C\right)\right\}\left|\left|\alpha \right|w\left(B\right)-\left|\beta \right|w\left(C\right)\right|.\hfill \end{array}$$

(12)

Proof. 

By replacing B with $\alpha B$ and C with $\beta C$ in Theorem 1, we obtain

$$\begin{array}{cc}\hfill & w_e^2\left(\alpha B,\beta C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{B}^2+{\beta }^2{C}^2\right),w\left(\alpha \beta BC+\alpha \beta CB\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(\alpha B\right),w\left(\beta C\right)\right\}\left|w\left(\alpha B+\beta C\right)-w\left(\alpha B-\beta C\right)\right|\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & w_e^2\left(B,C\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{B}^2+{\beta }^2{C}^2\right),w\left({\alpha }^2{B}^2-{\beta }^2{C}^2\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(\alpha B+\beta C\right),w\left(\alpha B-\beta C\right)\right\}\left|w\left(\alpha B\right)-w\left(\beta C\right)\right|.\hfill \end{array}$$

Since

$$max\left\{{\left|\alpha \right|}^2,{\left|\beta \right|}^2\right\}{w}_e^2\left(B,C\right)\ge w_e^2\left(\alpha B,\beta C\right),$$

$$w\left(\alpha \beta BC+\alpha \beta CB\right)=\left|\alpha \beta \right|w\left(BC+CB\right)$$

and

$$w\left(\alpha B\right)=\left|\alpha \right|w\left(B\right),w\left(\beta C\right)=\left|\beta \right|w\left(C\right);$$

hence, the inequalities (11) and (12) are obtained. □

Remark 2. 

If $\alpha ,\beta \in \mathbb{C}$ with $\left|\alpha \right|=\left|\beta \right|=1,$ then from Corollary 1, we derive the simpler inequalities

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{B}^2+{\beta }^2{C}^2\right),w\left(BC+CB\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(B\right),w\left(C\right)\right\}\left|w\left(\alpha B+\beta C\right)-w\left(\alpha B-\beta C\right)\right|\hfill \end{array}$$

(13)

and

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{B}^2+{\beta }^2{C}^2\right),w\left({\alpha }^2{B}^2-{\beta }^2{C}^2\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(\alpha B+\beta C\right),w\left(\alpha B-\beta C\right)\right\}\left|w\left(B\right)-w\left(C\right)\right|.\hfill \end{array}$$

(14)

Moreover for $\alpha =1$ and $\beta =i,$ we obtain from (13) and (14) that

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left(B^2-C^2\right),w\left(BC+CB\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(B\right),w\left(C\right)\right\}\left|w\left(B+iC\right)-w\left(B-iC\right)\right|\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left(B^2-C^2\right),w\left(B^2+C^2\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(B+iC\right),w\left(B-iC\right)\right\}\left|w\left(B\right)-w\left(C\right)\right|.\hfill \end{array}$$

We also present the following corollary, which offers inequalities that improve upon the lower bound established in Corollary 2.4 of the recent paper [12].

Corollary 2. 

For any self-adjoint operators $B,C$ we have the inequalities

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{∥B^2+C^2∥,∥BC+CB∥\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{∥B∥,∥C∥\right\}\left|∥B+C∥-∥B-C∥\right|\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}∥B^2+C^2∥\hfill \end{array}$$

(15)

and

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{∥B^2+C^2∥,∥B^2-C^2∥\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{∥B+C∥,∥B-C∥\right\}\left|∥B∥-∥C∥\right|.\hfill \end{array}$$

Proof. 

The proof follows directly from Theorem 1, noting that the numerical radius of a self-adjoint operator is equal to its norm. □

The following numerical example demonstrates that the first lower bound in (15) is significantly better than the second one, which was initially obtained in [10].

Example 1. 

Consider the diagonal $2\times 2$ matrices

$$B=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I_2\mathit{and}C=\left(\begin{array}{cc}1& 0\\ 0& a\end{array}\right)\mathrm{with}a>0.$$

Then,

$$B^2=I_2,C^2=\left(\begin{array}{cc}1& 0\\ 0& a^2\end{array}\right),BC+CB=\left(\begin{array}{cc}2& 0\\ 0& 2a\end{array}\right),$$

$$B+C=\left(\begin{array}{cc}2& 0\\ 0& a+1\end{array}\right),B-C=\left(\begin{array}{cc}0& 0\\ 0& 1-a\end{array}\right)$$

and

$$B^2+C^2=\left(\begin{array}{cc}2& 0\\ 0& a^2+1\end{array}\right).$$

It is well known that if $A=\mathrm{diag}({\lambda }_1,\cdots ,{\lambda }_n),$ then ${\displaystyle ∥A∥=\underset{i=1,\cdots ,n}{max}\left|{\lambda }_i\right|.}$ Therefore,

$$∥B∥=1,∥C∥=max\left\{1,a\right\},∥B+C∥=max\left\{2,1+a\right\}=1+max\left\{1,a\right\},$$

$$∥B-C∥=\left|1-a\right|,∥BC+CB∥=2max\left\{1,a\right\}$$

and

$$∥B^2+C^2∥=max\left\{2,a^2+1\right\}=1+{\left(max\left\{1,a\right\}\right)}^2$$

for $a>0.$

Observe also that for $\overline{z}=\left(x,y\right)\in {\mathbb{C}}^2,$ we have that

$$〈B\overline{z},\overline{z}〉={\left|x\right|}^2+{\left|y\right|}^2,〈C\overline{z},\overline{z}〉={\left|x\right|}^2+a{\left|y\right|}^2.$$

Therefore,

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& =\underset{{\left|x\right|}^2+{\left|y\right|}^2=1}{sup}\left({\left({\left|x\right|}^2+{\left|y\right|}^2\right)}^2+{\left({\left|x\right|}^2+a{\left|y\right|}^2\right)}^2\right)\hfill \\ \hfill & =\underset{{\left|x\right|}^2+{\left|y\right|}^2=1}{sup}\left(1+{\left(1+\left(a-1\right)y^2\right)}^2\right)\hfill \\ \hfill & =\underset{{\left|x\right|}^2+{\left|y\right|}^2=1}{sup}\left(1+1+2\left(a-1\right)y^2+{\left(a-1\right)}^2{y}^4\right)\hfill \\ \hfill & =2+\underset{{\left|x\right|}^2+{\left|y\right|}^2=1}{sup}\left[\left(a-1\right)y^2\left(2+\left(a-1\right)y^2\right)\right]\hfill \end{array}$$

for $a>0.$

If we take $a=2,$ then

$$w_e^2\left(B,C\right)=5,$$

the middle term in (15)

$$\begin{array}{cc}\hfill & {\displaystyle \frac{1}{2}}\left[max\left\{∥B^2+C^2∥,∥BC+CB∥\right\}+max\left\{∥B∥,∥C∥\right\}\left|∥B+C∥-∥B-C∥\right|\right]\hfill \\ \hfill & =4.5\hfill \end{array}$$

and the right term

$${\displaystyle \frac{1}{2}}∥B^2+C^2∥=2.5.$$

This numerically shows that the middle term in (15) provides a significantly better lower bound for the Euclidean numerical radius than the previous one from [10].

We also have the following corollary.

Corollary 3. 

For any $A\in \mathcal{B}\left(H\right)$ and $\alpha ,\beta \in \mathbb{C}$ we have

$$\begin{array}{cc}\hfill & \left({\left|\alpha \right|}^2+{\left|\beta \right|}^2\right)w^2\left(A\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{A}^2+{\beta }^2{\left(A^2\right)}^{*}\right),\left|\alpha \beta \right|∥{\left|A\right|}^2+{\left|A^{*}\right|}^2∥\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{\left|\alpha \right|,\left|\beta \right|\right\}w\left(A\right)\left|w\left(\alpha A+\beta A^{*}\right)-w\left(\alpha A-\beta A^{*}\right)\right|\hfill \end{array}$$

(16)

and

$$\begin{array}{cc}\hfill & \left({\left|\alpha \right|}^2+{\left|\beta \right|}^2\right)w^2\left(A\right)\hfill \\ \hfill & \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{A}^2+{\beta }^2{\left(A^2\right)}^{*}\right),w\left({\alpha }^2{A}^2-{\beta }^2{\left(A^2\right)}^{*}\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left|\left|\alpha \right|-\left|\beta \right|\right|w\left(A\right)max\left\{w\left(\alpha A+\beta A^{*}\right),w\left(\alpha A-\beta A^{*}\right)\right\}.\hfill \end{array}$$

(17)

Proof. 

We take $B=\alpha A$ and $C=\beta A^{*}$ in Theorem 1 to obtain

$$\begin{array}{cc}\hfill w_e^2\left(\alpha A,\beta A^{*}\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{A}^2+{\beta }^2{\left(A^2\right)}^{*}\right),w\left(\alpha \beta A{A}^{*}+\alpha \beta A^{*}A\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(\alpha A\right),w\left(\beta A^{*}\right)\right\}\left|w\left(\alpha A+\beta A^{*}\right)-w\left(\alpha A-\beta A^{*}\right)\right|\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w_e^2\left(\alpha A,\beta A^{*}\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left({\alpha }^2{A}^2+{\beta }^2{\left(A^2\right)}^{*}\right),w\left({\alpha }^2{A}^2-{\beta }^2{\left(A^2\right)}^{*}\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(\alpha A+\beta A^{*}\right),w\left(\alpha A-\beta A^{*}\right)\right\}\left|w\left(\alpha A\right)-w\left(\beta A^{*}\right)\right|.\hfill \end{array}$$

Observe that

$$w_e^2\left(\alpha A,\beta A^{*}\right)=\left({\left|\alpha \right|}^2+{\left|\beta \right|}^2\right)w^2\left(A\right),$$

$$w\left(\alpha \beta A{A}^{*}+\alpha \beta A^{*}A\right)=\left|\alpha \beta \right|w\left(A{A}^{*}+A^{*}A\right)=\left|\alpha \beta \right|∥{\left|A\right|}^2+{\left|A^{*}\right|}^2∥$$

and

$$w\left(\beta A^{*}\right)=\left|\beta \right|w\left(A\right),$$

which proves the desired results (16) and (17). □

Remark 3. 

If we take $\alpha =cos\theta$ and $\beta =sin\theta ,$ $\theta \in \mathbb{R}$ in (17), we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left({cos}^2\theta A^2+{sin}^2\theta {\left(A^2\right)}^{*}\right),\left|sin\left(2\theta \right)\right|∥{\displaystyle \frac{{\left|A\right|}^2+{\left|A^{*}\right|}^2}{2}}∥\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{\left|cos\theta \right|,\left|sin\theta \right|\right\}w\left(A\right)\left|w\left(cos\theta A+sin\theta A^{*}\right)-w\left(cos\theta A-sin\theta A^{*}\right)\right|\hfill \end{array}$$

(18)

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left({cos}^2\theta A^2+{sin}^2\theta {\left(A^2\right)}^{*}\right),w\left({cos}^2\theta A^2-{sin}^2\theta {\left(A^2\right)}^{*}\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}\left|\left|cos\theta \right|-\left|sin\theta \right|\right|w\left(A\right)max\left\{w\left(cos\theta A+sin\theta A^{*}\right),w\left(cos\theta A-sin\theta A^{*}\right)\right\}.\hfill \end{array}$$

If we take $\theta =\pi /4$ in (18), we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{2}}max\left\{∥{\displaystyle \frac{A^2+{\left(A^2\right)}^{*}}{2}}∥,∥{\displaystyle \frac{{\left|A\right|}^2+{\left|A^{*}\right|}^2}{2}}∥\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}w\left(A\right)\left|∥{\displaystyle \frac{A+A^{*}}{2}}∥-∥{\displaystyle \frac{A-A^{*}}{2}}∥\right|.\hfill \end{array}$$

We now present the following corollary, which offers an improvement of the lower bound from ([12], Corollary 2.5).

Corollary 4. 

For any $A\in \mathcal{B}\left(H\right)$, we have

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{2}}max\left\{∥{\displaystyle \frac{{\left|A\right|}^2+{\left|A^{*}\right|}^2}{2}}∥,∥{\displaystyle \frac{{\left(A^{*}\right)}^2-A^2}{2}}∥\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{∥{\displaystyle \frac{A+A^{*}}{2}}∥,∥{\displaystyle \frac{A^{*}-A}{2}}∥\right\}\left|\left(∥{\displaystyle \frac{\left(1-i\right)A+\left(1+i\right)A^{*}}{2}}∥\right)-∥{\displaystyle \frac{\left(1+i\right)A+\left(1-i\right)A^{*}}{2}}∥\right|\hfill \end{array}$$

(19)

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \ge {\displaystyle \frac{1}{2}}max\left\{∥{\displaystyle \frac{{\left|A\right|}^2+{\left|A^{*}\right|}^2}{2}}∥,∥{\displaystyle \frac{A^2+{\left(A^{*}\right)}^2}{2}}∥\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{∥{\displaystyle \frac{\left(1-i\right)A+\left(1+i\right)A^{*}}{2}}∥,∥{\displaystyle \frac{\left(1+i\right)A+\left(1-i\right)A^{*}}{2}}∥\right\}\left|∥{\displaystyle \frac{A+A^{*}}{2}}∥-∥{\displaystyle \frac{A^{*}-A}{2}}∥\right|.\hfill \end{array}$$

(20)

Proof. 

Let $A=B+iC$ be the Cartesian decomposition of A, where

$$B=\mathfrak{R}\left(A\right)={\displaystyle \frac{A+A^{*}}{2}}\mathrm{and}C=\mathfrak{I}\left(A\right)={\displaystyle \frac{A-A^{*}}{2i}}={\displaystyle \frac{i}{2}}\left(A^{*}-A\right).$$

Observe that

$$w_e^2\left(B,C\right)=w^2\left(A\right),B^2+C^2={\displaystyle \frac{{\left|A\right|}^2+{\left|A^{*}\right|}^2}{2}}$$

$$\begin{array}{cc}\hfill BC+CB& ={\displaystyle \frac{i}{2}}\left[{\displaystyle \frac{\left(A+A^{*}\right)\left(A^{*}-A\right)}{2}}+{\displaystyle \frac{\left(A^{*}-A\right)\left(A+A^{*}\right)}{2}}\right]\hfill \\ \hfill & ={\displaystyle \frac{i}{4}}\left[A{A}^{*}+{\left(A^{*}\right)}^2-A^2-A^{*}A+A^{*}A-A^2+{\left(A^{*}\right)}^2-A{A}^{*}\right]\hfill \\ \hfill & ={\displaystyle \frac{i}{2}}\left({\left(A^{*}\right)}^2-A^2\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill B^2-C^2& ={\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^2-{\left({\displaystyle \frac{i}{2}}\left(A^{*}-A\right)\right)}^2\hfill \\ \hfill & ={\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^2+{\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^2\hfill \\ \hfill & ={\displaystyle \frac{A^2+A{A}^{*}+A^{*}A+{\left(A^{*}\right)}^2+A^2-A{A}^{*}-A^{*}A+{\left(A^{*}\right)}^2}{4}}\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left(A^2+{\left(A^{*}\right)}^2\right)\hfill \end{array}$$

$$B+C={\displaystyle \frac{A+A^{*}}{2}}+{\displaystyle \frac{i\left(A^{*}-A\right)}{2}}={\displaystyle \frac{\left(1-i\right)A+\left(1+i\right)A^{*}}{2}}$$

and

$$B-C={\displaystyle \frac{A+A^{*}}{2}}-{\displaystyle \frac{i\left(A^{*}-A\right)}{2}}={\displaystyle \frac{\left(1+i\right)A+\left(1-i\right)A^{*}}{2}}.$$

By making use of (6) and (7), we deduce the desired results (19) and (20). □

We have the following lower bounds for the Davis–Wielandt radius. These lower bounds are better than the one offered by the first inequality in (4).

Corollary 5. 

For any $T\in \mathcal{B}\left(H\right),$ we have

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left(T^2+{\left|T\right|}^4\right),w\left(T{\left|T\right|}^2+{\left|T\right|}^{2}T\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(T\right),{∥T∥}^2\right\}\left|w\left(T+{\left|T\right|}^2\right)-w\left(T-{\left|T\right|}^2\right)\right|\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left(T^2+{\left|T\right|}^4\right),w\left(T^2-{\left|T\right|}^4\right)\right\}\hfill \\ \hfill & +{\displaystyle \frac{1}{2}}max\left\{w\left(T+{\left|T\right|}^2\right),w\left(T-{\left|T\right|}^2\right)\right\}\left|w\left(T\right)-{∥T∥}^2\right|.\hfill \end{array}$$

Proof. 

The proof follows by taking $\left(B,C\right)=\left(T,{\left|T\right|}^2\right)$ in Theorem 1. □

Further, we recall the following representation result obtained in ([11], Theorem 2.2).

Lemma 1. 

Let $X,$$Y\in \mathcal{B}\left(H\right).$ Then,

$$w_e^2\left(X,Y\right)=\underset{\begin{array}{c}{\mu }^2+{\nu }^2=1,\\ \mu ,\nu \ge 0\end{array}}{sup}\underset{\theta \in \mathbb{R}}{sup}{w}^2\left(\mu e^{i\theta }X+\nu e^{-i\theta }Y\right).$$

By utilizing this result, we can prove the following lower bound for the Euclidean numerical radius.

Proposition 1. 

For $B,C\in \mathcal{B}\left(H\right)$, we have

$$w_e^2\left(B,C\right)\ge {\displaystyle \frac{1}{2}}{w}_e\left(B^2,C^2\right).$$

(21)

Proof. 

Let $\mu ,\nu \ge 0$ be such that ${\mu }^2+{\nu }^2=1$. Let also $\theta \in \mathbb{R}$. From (11), we obtain for $\alpha =\sqrt{\mu }{e}^{i{\displaystyle \frac{\theta }{2}}}$ and $\beta =\sqrt{\nu }{e}^{-i{\displaystyle \frac{\theta }{2}}}$ that

$$max\left\{\mu ,\nu \right\}{w}_e^2\left(B,C\right)\ge {\displaystyle \frac{1}{2}}w\left(\mu e^{i\theta }{B}^2+\nu e^{-i\theta }{C}^2\right).$$

So, we deduce that

$$\begin{array}{cc}\hfill \underset{\begin{array}{c}{\mu }^2+{\nu }^2=1,\\ \mu ,\nu \ge 0\end{array}}{sup}max\left\{\mu ,\nu \right\}{w}_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}\underset{\begin{array}{c}{\mu }^2+{\nu }^2=1,\\ \mu ,\nu \ge 0\end{array}}{sup}\underset{\theta \in \mathbb{R}}{sup}w\left(\mu e^{i\theta }{B}^2+\nu e^{-i\theta }{C}^2\right)\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}{w}_e\left(B^2,C^2\right)\hfill \end{array}$$

and since ${\displaystyle \underset{\begin{array}{c}{\mu }^2+{\nu }^2=1,\\ \mu ,\nu \ge 0\end{array}}{sup}max\left\{\mu ,\nu \right\}=1,}$ we deduce inequality (21). □

Remark 4. 

If we take $B=A$ and $C=A^{*}$ in (21), we obtain

$$w^2\left(A\right)\ge {\displaystyle \frac{1}{4}}{w}_e\left(A^2,{\left(A^{*}\right)}^2\right).$$

If we take $B=\mathfrak{R}\left(A\right)$ and $C=\mathfrak{I}\left(A\right),$ then by (21) we obtain

$$w^2\left(A\right)\ge {\displaystyle \frac{1}{2}}{∥{\mathfrak{R}}^2\left(A\right)+{\mathfrak{I}}^2\left(A\right)∥}^{1/2}.$$

Indeed, we have

$$\begin{array}{cc}\hfill w_e^2\left({\mathfrak{R}}^2\left(A\right),{\mathfrak{I}}^2\left(A\right)\right)& =\underset{∥x∥=1}{sup}\left[〈{\mathfrak{R}}^2\left(A\right)x,x〉+〈{\mathfrak{I}}^2\left(A\right)x,x〉\right]\hfill \\ \hfill & =\underset{∥x∥=1}{sup}\left[〈\left({\mathfrak{R}}^2\left(A\right)+{\mathfrak{I}}^2\left(A\right)\right)x,x〉\right]\hfill \\ \hfill & =\omega \left({\mathfrak{R}}^2\left(A\right)+{\mathfrak{I}}^2\left(A\right)\right)=∥{\mathfrak{R}}^2\left(A\right)+{\mathfrak{I}}^2\left(A\right)∥.\hfill \end{array}$$

If we take $B=T$ and $C={\left|T\right|}^2,$ then by (21), we obtain

$$d{w}^2\left(T\right)\ge {\displaystyle \frac{1}{2}}{w}_e\left(T^2,{\left|T\right|}^4\right).$$

The following representation is also known (see, for instance, [19]).

Lemma 2. 

Let $X,Y\in \mathcal{B}\left(H\right).$ Then,

$$w\left(\begin{array}{cc}0& X\\ Y^{*}& 0\end{array}\right)={\displaystyle \frac{1}{2}}\underset{\theta \in \mathbb{R}}{sup}∥e^{i\theta }X+e^{-i\theta }Y∥.$$

We are able now to prove the following proposition.

Proposition 2. 

For $B,C\in \mathcal{B}\left(H\right)$, we have

$$w_e^2\left(B,C\right)\ge {\displaystyle \frac{1}{2}}w\left(\begin{array}{cc}0& B^2\\ {\left(C^{*}\right)}^2& 0\end{array}\right).$$

(22)

Proof. 

By using (12) for $\alpha =e^{i{\displaystyle \frac{\theta }{2}}}$ and $\beta =e^{-i{\displaystyle \frac{\theta }{2}}},$ we obtain

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \ge {\displaystyle \frac{1}{2}}max\left\{w\left(e^{i\theta }{B}^2+e^{-i\theta }{C}^2\right),w\left(e^{-i\theta }{B}^2-e^{-i\theta }{C}^2\right)\right\}\hfill \\ \hfill & \ge {\displaystyle \frac{1}{4}}max\left\{∥e^{i\theta }{B}^2+e^{-i\theta }{C}^2∥,∥e^{-i\theta }{B}^2-e^{-i\theta }{C}^2∥\right\}\hfill \end{array}$$

for all $\theta \in \mathbb{R}$.

If we take the supremum over $\theta \in \mathbb{R}$, we obtain

$$w_e^2\left(B,C\right)\ge {\displaystyle \frac{1}{4}}max\left\{\underset{\theta \in \mathbb{R}}{sup}∥e^{i\theta }{B}^2+e^{-i\theta }{C}^2∥,\underset{\theta \in \mathbb{R}}{sup}∥e^{-i\theta }{B}^2-e^{-i\theta }{C}^2∥\right\}.$$

(23)

Since, by Lemma 2,

$${\displaystyle \frac{1}{2}}\underset{\theta \in \mathbb{R}}{sup}∥e^{i\theta }{B}^2+e^{-i\theta }{C}^2∥=w\left(\begin{array}{cc}0& B^2\\ {\left(C^2\right)}^{*}& 0\end{array}\right)=w\left(\begin{array}{cc}0& B^2\\ {\left(C^{*}\right)}^2& 0\end{array}\right)$$

and

$${\displaystyle \frac{1}{2}}\underset{\theta \in \mathbb{R}}{sup}∥e^{-i\theta }{B}^2-e^{-i\theta }{C}^2∥=w\left(\begin{array}{cc}0& B^2\\ -{\left(C^{*}\right)}^2& 0\end{array}\right)=w\left(\begin{array}{cc}0& B^2\\ {\left(C^{*}\right)}^2& 0\end{array}\right).$$

Then, by (23), we deduce (22) as requested. □

Remark 5. 

If we take $B=A$ and $C=A^{*}$ in (22), we obtain

$$w^2\left(A\right)\ge {\displaystyle \frac{1}{4}}w\left(\begin{array}{cc}0& A^2\\ A^2& 0\end{array}\right)={\displaystyle \frac{1}{4}}w\left(A^2\right).$$

If we take $B=\mathfrak{R}\left(A\right)$ and $C=\mathfrak{I}\left(A\right),$ then by (22), we obtain

$$w^2\left(A\right)\ge {\displaystyle \frac{1}{2}}w\left(\begin{array}{cc}0& {\mathfrak{R}}^2\left(A\right)\\ {\mathfrak{I}}^2\left(A\right)& 0\end{array}\right).$$

Moreover, if we take $B=T$ and $C={\left|T\right|}^2,$ then by (21), we obtain

$$d{w}^2\left(T\right)\ge {\displaystyle \frac{1}{2}}w\left(\begin{array}{cc}0& T^2\\ {\left|T\right|}^4& 0\end{array}\right).$$

3. Upper Bounds for the Euclidean Numerical Radius of Operator Pairs

In this section, we aim to establish several upper bounds for the Euclidean numerical radius of operator pairs in Hilbert spaces. To derive our first result, we recall two well-known inequalities. The first is the Kato inequality (see [20]), which is given by

$${\left|〈Tx,y〉\right|}^2\le 〈{\left|T\right|}^{2\lambda }x,x〉〈{\left|T^{*}\right|}^{2(1-\lambda )}y,y〉$$

(24)

for any $T\in \mathcal{B}\left(H\right)$, $\lambda \in [0,1]$, and any $x,y\in H.$

The second is the well-known McCarthy inequality (see [21]), which asserts that for any positive operator P and for $s\ge 1$, we have

$${〈Px,x〉}^s\le 〈P^{s}x,x〉,x\in H,\parallel x\parallel =1.$$

(25)

With these preliminaries, we can now present the following result.

Theorem 2. 

For any $B,C\in \mathcal{B}\left(H\right),$$\lambda \in \left[0,1\right]$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}$, we have

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \le w\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)w\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)\hfill \\ \hfill & \le {∥{\left|B\right|}^{4\lambda }+{\left|C\right|}^{4\lambda }∥}^{1/2}{∥{\left|B^{*}\right|}^{4\left(1-\lambda \right)}+{\left|C^{*}\right|}^{4\left(1-\lambda \right)}∥}^{1/2}\hfill \end{array}$$

(26)

and

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \le w\left(\alpha {\left|B\right|}^{2\lambda }+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)w\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\beta {\left|C\right|}^{2\lambda }\right)\hfill \\ \hfill & \le {∥{\left|B\right|}^{4\lambda }+{\left|C^{*}\right|}^{4\left(1-\lambda \right)}∥}^{1/2}{∥{\left|B^{*}\right|}^{4\left(1-\lambda \right)}+{\left|C\right|}^{4\lambda }∥}^{1/2}.\hfill \end{array}$$

(27)

Proof. 

By (24), we have

$$\begin{array}{cc}\hfill {\left|〈Bx,x〉\right|}^2+{\left|〈Cx,x〉\right|}^2& \le 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \end{array}$$

(28)

for any $B,C\in \mathcal{B}\left(H\right),$$x\in H$ and $\lambda \in \left[0,1\right]$.

By the Cauchy–Schwarz inequality, we have

$$\begin{array}{cc}\hfill & 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \\ \hfill & \le {\left({〈{\left|B\right|}^{2\lambda }x,x〉}^2+{〈{\left|C\right|}^{2\lambda }x,x〉}^2\right)}^{1/2}{\left({〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2+{〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2\right)}^{1/2}\hfill \\ \hfill & ={\left({〈\alpha {\left|B\right|}^{2\lambda }x,x〉}^2+{〈\beta {\left|C\right|}^{2\lambda }x,x〉}^2\right)}^{1/2}{\left({〈\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2+{〈\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2\right)}^{1/2},\hfill \end{array}$$

(29)

for $x\in H,$ where $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

Since

$$\begin{array}{cc}\hfill {〈\alpha {\left|B\right|}^{2\lambda }x,x〉}^2+{〈\beta {\left|C\right|}^{2\lambda }x,x〉}^2& ={\left|〈\alpha {\left|B\right|}^{2\lambda }x,x〉+i〈\beta {\left|C\right|}^{2\lambda }x,x〉\right|}^2\hfill \\ \hfill & ={\left|〈\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)x,x〉\right|}^2\hfill \end{array}$$

and

$$\begin{array}{c}\hfill {〈\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2+{〈\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2={\left|〈\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)x,x〉\right|}^2,\end{array}$$

then by taking (29) into account, we obtain

$$\begin{array}{cc}\hfill & 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \\ \hfill & \le \left|〈\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)x,x〉\right|\left|〈\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)x,x〉\right|\hfill \end{array}$$

(30)

for $x\in H.$

By (28) and (30), we derive

$$\begin{array}{cc}\hfill & {\left|〈Bx,x〉\right|}^2+{\left|〈Cx,x〉\right|}^2\hfill \\ \hfill & \le \left|〈\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)x,x〉\right|\left|〈\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)x,x〉\right|\hfill \end{array}$$

for $x\in H.$

By taking the supremum over $x\in H,$$∥x∥=1,$ we obtain

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& =\underset{∥x∥=1}{sup}\left({\left|〈Bx,x〉\right|}^2+{\left|〈Cx,x〉\right|}^2\right)\hfill \\ \hfill & \le \underset{∥x∥=1}{sup}\left[\left|〈\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)x,x〉\right|\right.\left.\left|〈\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)x,x〉\right|\right]\hfill \\ \hfill & \le \underset{∥x∥=1}{sup}\left|〈\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)x,x〉\right|\underset{∥x∥=1}{sup}\left|〈\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)x,x〉\right|\hfill \\ \hfill & =w\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)w\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right),\hfill \end{array}$$

which proves (26).

By the Cauchy–Schwarz inequality, we also have

$$\begin{array}{cc}\hfill & 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \\ \hfill & \le {\left({〈{\left|B\right|}^{2\lambda }x,x〉}^2+{〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2\right)}^{1/2}{\left({〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2+{〈{\left|C\right|}^{2\lambda }x,x〉}^2\right)}^{1/2}\hfill \\ \hfill & ={\left({〈\alpha {\left|B\right|}^{2\lambda }x,x〉}^2+{〈\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2\right)}^{1/2}{\left({〈\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^2+{〈\beta {\left|C\right|}^{2\lambda }x,x〉}^2\right)}^{1/2}\hfill \end{array}$$

for $x\in H,$ where $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

By making a similar argument as above, we then derive (27).

Furthermore, by McCarthy inequality (25), we have for $∥x∥=1$

$$\begin{array}{cc}\hfill {\left|〈\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)x,x〉\right|}^2& ={\left|\alpha 〈{\left|B\right|}^{2\lambda }x,x〉+i\beta 〈{\left|C\right|}^{2\lambda }x,x〉\right|}^2\hfill \\ \hfill & ={〈{\left|B\right|}^{2\lambda }x,x〉}^2+{〈{\left|C\right|}^{2\lambda }x,x〉}^2\hfill \\ \hfill & \le 〈{\left|B\right|}^{4\lambda }x,x〉+〈{\left|C\right|}^{4\lambda }x,x〉\hfill \\ \hfill & =〈\left({\left|B\right|}^{4\lambda }+{\left|C\right|}^{4\lambda }\right)x,x〉.\hfill \end{array}$$

By taking the supremum over $∥x∥=1,$ we obtain

$$\begin{array}{cc}\hfill w\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)& =\underset{∥x∥=1}{sup}\left|〈\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)x,x〉\right|\hfill \\ \hfill & \le \underset{∥x∥=1}{sup}{〈\left({\left|B\right|}^{4\lambda }+{\left|C\right|}^{4\lambda }\right)x,x〉}^{1/2}={∥{\left|B\right|}^{4\lambda }+{\left|C\right|}^{4\lambda }∥}^{1/2}.\hfill \end{array}$$

Similarly,

$$w\left(\gamma {\left|B^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)\le {∥{\left|B^{*}\right|}^{4\left(1-\lambda \right)}+{\left|C^{*}\right|}^{4\left(1-\lambda \right)}∥}^{1/2},$$

which proves the last part of (26). □

Several consequences of Theorem 2 can be presented. We start with the following corollary.

Corollary 6. 

For any $B,C\in \mathcal{B}\left(H\right)$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}$, we have

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \le w\left(\alpha \left|B\right|+i\beta \left|C\right|\right)w\left(\gamma \left|B^{*}\right|+i\delta \left|C^{*}\right|\right)\hfill \\ \hfill & \le {∥{\left|B\right|}^2+{\left|C\right|}^2∥}^{1/2}{∥{\left|B^{*}\right|}^2+{\left|C^{*}\right|}^2∥}^{1/2}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \le w\left(\alpha \left|B\right|+i\delta \left|C^{*}\right|\right)w\left(\gamma \left|B^{*}\right|+i\beta \left|C\right|\right)\hfill \\ \hfill & \le {∥{\left|B\right|}^2+{\left|C^{*}\right|}^2∥}^{1/2}{∥{\left|B^{*}\right|}^2+{\left|C\right|}^2∥}^{1/2}.\hfill \end{array}$$

Proof. 

The proof follows directly from Theorem 2 by setting $\lambda ={\displaystyle \frac{1}{2}}$. □

Remark 6. 

In ([12], Theorem 2.2), the authors proved that for any $B,C\in \mathcal{B}\left(H\right)$, the following result holds:

$$\begin{array}{cc}\hfill & w_e^2\left(B,C\right)\hfill \\ \hfill & \le min\left\{w\left(\left|B\right|+i\left|C\right|\right)w\left(\left|B^{*}\right|+i\left|C^{*}\right|\right),w\left(\left|B\right|+i\left|C^{*}\right|\right)w\left({\left|B^{*}\right|}^2+i\left|C\right|\right)\right\}\hfill \\ \hfill & \le min\left\{{∥{\left|B\right|}^2+{\left|C\right|}^2∥}^{1/2}{∥{\left|B^{*}\right|}^2+{\left|C^{*}\right|}^2∥}^{1/2},{∥{\left|B\right|}^2+{\left|C^{*}\right|}^2∥}^{1/2}{∥{\left|B^{*}\right|}^2+{\left|C\right|}^2∥}^{1/2}\right\}.\hfill \end{array}$$

This result can be derived from Corollary 6 by setting $\alpha =\beta =\gamma =\delta =1$. The interested reader can derive other similar bounds by choosing some of $\alpha ,\beta ,\gamma ,\delta$ to be $-1$.

We also present the following corollary, which, while straightforward, is of significant interest.

Corollary 7. 

For any $B,C$ self-adjoint operators, $\lambda \in \left[0,1\right]$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}$, we have

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \le w\left(\alpha {\left|B\right|}^{2\lambda }+i\beta {\left|C\right|}^{2\lambda }\right)w\left(\gamma {\left|B\right|}^{2\left(1-\lambda \right)}+i\delta {\left|C\right|}^{2\left(1-\lambda \right)}\right)\hfill \\ \hfill & \le {∥{\left|B\right|}^{4\lambda }+{\left|C\right|}^{4\lambda }∥}^{1/2}{∥{\left|B\right|}^{4\left(1-\lambda \right)}+{\left|C\right|}^{4\left(1-\lambda \right)}∥}^{1/2}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \le w\left(\alpha {\left|B\right|}^{2\lambda }+i\delta {\left|C\right|}^{2\left(1-\lambda \right)}\right)w\left(\gamma {\left|B\right|}^{2\left(1-\lambda \right)}+i\beta {\left|C\right|}^{2\lambda }\right)\hfill \\ \hfill & \le {∥{\left|B\right|}^{4\lambda }+{\left|C\right|}^{4\left(1-\lambda \right)}∥}^{1/2}{∥{\left|B\right|}^{4\left(1-\lambda \right)}+{\left|C\right|}^{4\lambda }∥}^{1/2}.\hfill \end{array}$$

Remark 7. 

For $\lambda =1/2$ in the above corollary, we obtain

$$w_e^2\left(B,C\right)\le w\left(\alpha \left|B\right|+i\beta \left|C\right|\right)w\left(\gamma \left|B\right|+i\delta \left|C\right|\right)\le ∥{\left|B\right|}^2+{\left|C\right|}^2∥$$

for $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

Another important consequence of Theorem 2 is derived in the next corollary.

Corollary 8. 

For $A\in \mathcal{B}\left(H\right),$ $\phi ,\psi \in \mathbb{C}$, $\lambda \in \left[0,1\right]$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\},$ we have the numerical radius inequalities

$$\begin{array}{cc}\hfill & \left({\left|\phi \right|}^2+{\left|\psi \right|}^2\right)w^2\left(A\right)\hfill \\ \hfill & \le w\left(\alpha {\left|\phi \right|}^{2\lambda }{\left|A\right|}^{2\lambda }+i\beta {\left|\psi \right|}^{2\lambda }{\left|A^{*}\right|}^{2\lambda }\right)w\left(\gamma {\left|\phi \right|}^{2\left(1-\lambda \right)}{\left|A^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|\psi \right|}^{2\left(1-\lambda \right)}{\left|A\right|}^{2\left(1-\lambda \right)}\right)\hfill \\ \hfill & \le {∥{\left|\phi \right|}^{4\lambda }{\left|A\right|}^{4\lambda }+{\left|\psi \right|}^{4\lambda }{\left|A^{*}\right|}^{4\lambda }∥}^{1/2}{∥{\left|\phi \right|}^{4\left(1-\lambda \right)}{\left|A^{*}\right|}^{4\left(1-\lambda \right)}+{\left|\psi \right|}^{4\left(1-\lambda \right)}{\left|A\right|}^{4\left(1-\lambda \right)}∥}^{1/2}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & \left({\left|\phi \right|}^2+{\left|\psi \right|}^2\right)w^2\left(A\right)\hfill \\ \hfill & \le w\left(\alpha {\left|\phi \right|}^{2\lambda }{\left|A\right|}^{2\lambda }+i\delta {\left|\psi \right|}^{2\left(1-\lambda \right)}{\left|A\right|}^{2\left(1-\lambda \right)}\right)w\left(\gamma {\left|\phi \right|}^{2\left(1-\lambda \right)}{\left|A^{*}\right|}^{2\left(1-\lambda \right)}+i\beta {\left|\psi \right|}^{2\lambda }{\left|A^{*}\right|}^{2\lambda }\right)\hfill \\ \hfill & \le {∥{\left|\phi \right|}^{4\lambda }{\left|A\right|}^{4\lambda }+{\left|\psi \right|}^{4\left(1-\lambda \right)}{\left|A\right|}^{4\left(1-\lambda \right)}∥}^{1/2}{∥{\left|\phi \right|}^{4\left(1-\lambda \right)}{\left|A^{*}\right|}^{4\left(1-\lambda \right)}+{\left|\psi \right|}^{4\lambda }{\left|A^{*}\right|}^{4\lambda }∥}^{1/2}.\hfill \end{array}$$

Proof. 

We take $B=\phi A$ and $C=\psi A^{*}$ in Theorem 2 to obtain the desired results. □

Remark 8. 

We notice that for $\phi =\psi =1,$ we obtain the simpler inequalities

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le {\displaystyle \frac{1}{2}}w\left(\alpha {\left|A\right|}^{2\lambda }+i\beta {\left|A^{*}\right|}^{2\lambda }\right)w\left(\gamma {\left|A^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|A\right|}^{2\left(1-\lambda \right)}\right)\hfill \\ \hfill & \le {∥{\left|A\right|}^{4\lambda }+{\left|A^{*}\right|}^{4\lambda }∥}^{1/2}{∥{\left|A^{*}\right|}^{4\left(1-\lambda \right)}+{\left|A\right|}^{4\left(1-\lambda \right)}∥}^{1/2}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le {\displaystyle \frac{1}{2}}w\left(\alpha {\left|A\right|}^{2\lambda }+i\delta {\left|A\right|}^{2\left(1-\lambda \right)}\right)w\left(\gamma {\left|A^{*}\right|}^{2\left(1-\lambda \right)}+i\beta {\left|A^{*}\right|}^{2\lambda }\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}{∥{\left|A\right|}^{4\lambda }+{\left|A\right|}^{4\left(1-\lambda \right)}∥}^{1/2}{∥{\left|A^{*}\right|}^{4\left(1-\lambda \right)}+{\left|A^{*}\right|}^{4\lambda }∥}^{1/2}.\hfill \end{array}$$

We also have in the particular case of $\lambda =1/2,$ that

Remark 9. 

For $\lambda =1/2$ in Corollary 8, we obtain

$$\begin{array}{cc}\hfill \left({\left|\phi \right|}^2+{\left|\psi \right|}^2\right)w^2\left(A\right)& \le w\left(\alpha \left|\phi \right|\left|A\right|+i\beta \left|\psi \right|\left|A^{*}\right|\right)w\left(\gamma \left|\phi \right|\left|A^{*}\right|+i\delta \left|\psi \right|\left|A\right|\right)\hfill \\ \hfill & \le {∥{\left|\phi \right|}^2{\left|A\right|}^2+{\left|\psi \right|}^2{\left|A^{*}\right|}^2∥}^{1/2}{∥{\left|\phi \right|}^2{\left|A^{*}\right|}^2+{\left|\psi \right|}^2{\left|A\right|}^2∥}^{1/2}.\hfill \end{array}$$

(31)

If ${\left|\phi \right|}^2+{\left|\psi \right|}^2=1$ in (31), then we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le w\left(\alpha \left|\phi \right|\left|A\right|+i\beta \left|\psi \right|\left|A^{*}\right|\right)w\left(\gamma \left|\phi \right|\left|A^{*}\right|+i\delta \left|\psi \right|\left|A\right|\right)\hfill \\ \hfill & \le {∥{\left|\phi \right|}^2{\left|A\right|}^2+{\left|\psi \right|}^2{\left|A^{*}\right|}^2∥}^{1/2}{∥{\left|\phi \right|}^2{\left|A^{*}\right|}^2+{\left|\psi \right|}^2{\left|A\right|}^2∥}^{1/2}.\hfill \end{array}$$

(32)

If we take $\phi =sin\theta ,$$\psi =cos\theta ,$ $\theta \in \mathbb{R}$ in (32), then we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le w\left(\alpha \left|sin\theta \right|\left|A\right|+i\beta \left|cos\theta \right|\left|A^{*}\right|\right)w\left(\gamma \left|sin\theta \right|\left|A^{*}\right|+i\delta \left|cos\theta \right|\left|A\right|\right)\hfill \\ \hfill & \le {∥{sin}^2\theta {\left|A\right|}^2+{cos}^2\theta {\left|A^{*}\right|}^2∥}^{1/2}{∥{sin}^2\theta {\left|A^{*}\right|}^2+{cos}^2{\left|A\right|}^2∥}^{1/2}.\hfill \end{array}$$

Also, if we take $\phi =\psi =1,$ then we obtain from (31) that

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}w\left(\alpha \left|A\right|+i\beta \left|A^{*}\right|\right)w\left(\gamma \left|A^{*}\right|+i\delta \left|A\right|\right)\le {\displaystyle \frac{1}{2}}∥{\left|A\right|}^2+{\left|A^{*}\right|}^2∥.$$

This is an improvement of the second Kittaneh’s inequality in (2).

Another significant consequence of Theorem 2 is presented in the following corollary.

Corollary 9. 

For any $A\in \mathcal{B}\left(H\right)$ and $\lambda \in \left[0,1\right],$ we have

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le w\left(\alpha {\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^{2\lambda }+i\beta {\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^{2\lambda }\right)w\left(\gamma {\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^{2\left(1-\lambda \right)}+i\delta {\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^{2\left(1-\lambda \right)}\right)\hfill \\ \hfill & \le {∥{\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^{4\lambda }+{\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^{4\lambda }∥}^{1/2}{∥{\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^{4\left(1-\lambda \right)}+{\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^{4\left(1-\lambda \right)}∥}^{1/2}\hfill \end{array}$$

(33)

and

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le w\left(\alpha {\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^{2\lambda }+i\delta {\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^{2\left(1-\lambda \right)}\right)w\left(\gamma {\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^{2\left(1-\lambda \right)}+i\beta {\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^{2\lambda }\right)\hfill \\ \hfill & \le {∥{\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^{4\lambda }+{\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^{4\left(1-\lambda \right)}∥}^{1/2}{∥{\left({\displaystyle \frac{A+A^{*}}{2}}\right)}^{4\left(1-\lambda \right)}+{\left({\displaystyle \frac{A^{*}-A}{2}}\right)}^{4\lambda }∥}^{1/2}.\hfill \end{array}$$

(34)

Proof. 

Let $A=B+iC$ be the Cartesian decomposition of A, where

$$B={\displaystyle \frac{A+A^{*}}{2}}=\mathfrak{R}\left(A\right)\mathrm{and}C={\displaystyle \frac{i}{2}}\left(A^{*}-A\right)=\mathfrak{I}\left(A\right).$$

By applying Theorem 2 for this choice of B and $C,$ we obtain (33) and (34). □

Remark 10. 

For $\lambda =1/2$ in Corollary 9, we obtain

$$\begin{array}{cc}\hfill w^2\left(A\right)& \le w\left(\alpha \left({\displaystyle \frac{A+A^{*}}{2}}\right)+i\beta \left({\displaystyle \frac{A^{*}-A}{2}}\right)\right)w\left(\gamma \left({\displaystyle \frac{A+A^{*}}{2}}\right)+i\delta \left({\displaystyle \frac{A^{*}-A}{2}}\right)\right)\le {\displaystyle \frac{1}{2}}∥A^2+{\left(A^{*}\right)}^2∥\hfill \end{array}$$

for $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}.$

This is another improvement of the second Kittaneh’s inequality in (2).

By taking $\left(B,C\right)=\left(T,{\left|T\right|}^2\right)$ in Theorem 2, we obtain

Corollary 10. 

For any $T\in \mathcal{B}\left(H\right),$ $\lambda \in \left[0,1\right]$ and $\alpha ,\beta ,\gamma ,\delta \in \left\{-1,1\right\}$, we have

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \le w\left(\alpha {\left|T\right|}^{2\lambda }+i\beta {\left|T\right|}^{4\lambda }\right)w\left(\gamma {\left|T^{*}\right|}^{2\left(1-\lambda \right)}+i\delta {\left|T\right|}^{4\left(1-\lambda \right)}\right)\hfill \\ \hfill & \le {∥{\left|T\right|}^{4\lambda }+{\left|T\right|}^{8\lambda }∥}^{1/2}{∥{\left|T^{*}\right|}^{4\left(1-\lambda \right)}+{\left|T\right|}^{8\left(1-\lambda \right)}∥}^{1/2}\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \le w\left(\alpha {\left|T\right|}^{2\lambda }+i\delta {\left|T\right|}^{4\left(1-\lambda \right)}\right)w\left(\gamma {\left|T^{*}\right|}^{2\left(1-\lambda \right)}+i\beta {\left|T\right|}^{4\lambda }\right)\hfill \\ \hfill & \le {∥{\left|T\right|}^{4\lambda }+{\left|T\right|}^{8\left(1-\lambda \right)}∥}^{1/2}{∥{\left|T^{*}\right|}^{4\left(1-\lambda \right)}+{\left|T\right|}^{8\lambda }∥}^{1/2}.\hfill \end{array}$$

In particular for $\lambda =1/2$, we have

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \le w\left(\alpha \left|T\right|+i\beta {\left|T\right|}^2\right)w\left(\gamma \left|T^{*}\right|+i\delta {\left|T\right|}^2\right)\hfill \\ \hfill & \le {∥{\left|T\right|}^2+{\left|T\right|}^4∥}^{1/2}{∥{\left|T^{*}\right|}^2+{\left|T\right|}^4∥}^{1/2}\hfill \end{array}$$

(35)

for all $T\in \mathcal{B}\left(H\right)$.

We observe that, if ${\left|T^{*}\right|}^2\le {\left|T\right|}^2,$ namely $T$is hyponormal operator, then

$$∥{\left|T^{*}\right|}^2+{\left|T\right|}^4∥\le ∥{\left|T\right|}^2+{\left|T\right|}^4∥$$

and by (35), we obtain

$$\begin{array}{cc}\hfill d{w}^2\left(T\right)& \le w\left(\alpha \left|T\right|+i\beta {\left|T\right|}^2\right)w\left(\gamma \left|T^{*}\right|+i\delta {\left|T\right|}^2\right)\hfill \\ \hfill & \le {∥{\left|T\right|}^2+{\left|T\right|}^4∥}^{1/2}{∥{\left|T^{*}\right|}^2+{\left|T\right|}^4∥}^{1/2}\le ∥{\left|T\right|}^2+{\left|T\right|}^4∥,\hfill \end{array}$$

which is an improvement of inequality (5).

To obtain our next result, we make also use of the well-known McCarthy inequality (25).

Theorem 3. 

For any $B,C\in \mathcal{B}\left(H\right),$ $\lambda \in \left[0,1\right]$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, we have

$$w_e^2\left(B,C\right)\le {∥{\left|B\right|}^{2\lambda p}+{\left|C\right|}^{2\lambda p}∥}^{1/p}{∥{\left|B^{*}\right|}^{2\left(1-\lambda \right)q}+{\left|C^{*}\right|}^{2\left(1-\lambda \right)q}∥}^{1/q}$$

(36)

and

$$w_e^2\left(B,C\right)\le {∥{\left|B\right|}^{2\lambda p}+{\left|C^{*}\right|}^{2\left(1-\lambda \right)p}∥}^{1/p}{∥{\left|B^{*}\right|}^{2\left(1-\lambda \right)q}+{\left|C\right|}^{2\lambda q}∥}^{1/q}.$$

(37)

Proof. 

By Hölder’s inequality, we have for $\lambda \in \left[0,1\right]$ that

$$\begin{array}{cc}\hfill & 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \\ \hfill & \le {\left({〈{\left|B\right|}^{2\lambda }x,x〉}^p+{〈{\left|C\right|}^{2\lambda }x,x〉}^p\right)}^{1/p}{\left({〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^q+{〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^q\right)}^{1/q}\hfill \end{array}$$

(38)

for $x\in H$ and $p,$$q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1.$

By utilizing McCarthy’s inequality (25), we obtain

$$\begin{array}{cc}\hfill {〈{\left|B\right|}^{2\lambda }x,x〉}^p+{〈{\left|C\right|}^{2\lambda }x,x〉}^p& \le 〈{\left|B\right|}^{2\lambda p}x,x〉+〈{\left|C\right|}^{2\lambda p}x,x〉\hfill \\ \hfill & =〈\left({\left|B\right|}^{2\lambda p}+{\left|C\right|}^{2\lambda p}\right)x,x〉\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill {〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^q+{〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^q& \le 〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)q}x,x〉+〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)q}x,x〉\hfill \\ \hfill & =〈\left({\left|B^{*}\right|}^{2\left(1-\lambda \right)q}+{\left|C^{*}\right|}^{2\left(1-\lambda \right)q}\right)x,x〉\hfill \end{array}$$

for $x\in H,$$∥x∥=1$. So, by (38), we obtain

$$\begin{array}{cc}\hfill & 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \\ \hfill & \le {〈\left({\left|B\right|}^{2\lambda p}+{\left|C\right|}^{2\lambda p}\right)x,x〉}^{1/p}{〈\left({\left|B^{*}\right|}^{2\left(1-\lambda \right)q}+{\left|C^{*}\right|}^{2\left(1-\lambda \right)q}\right)x,x〉}^{1/q}\hfill \end{array}$$

for $x\in H$, $∥x∥=1$.

By making use of (28), we get

$$\begin{array}{cc}\hfill & {\left|〈Bx,x〉\right|}^2+{\left|〈Cx,x〉\right|}^2\hfill \\ \hfill & \le {〈\left({\left|B\right|}^{2\lambda p}+{\left|C\right|}^{2\lambda p}\right)x,x〉}^{1/p}{〈\left({\left|B^{*}\right|}^{2\left(1-\lambda \right)q}+{\left|C^{*}\right|}^{2\left(1-\lambda \right)q}\right)x,x〉}^{1/q}\hfill \end{array}$$

for $x\in H$, $∥x∥=1$.

By taking the supremum over $∥x∥=1$, we obtain (36).

We also have

$$\begin{array}{cc}\hfill & 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \\ \hfill & \le {\left({〈{\left|B\right|}^{2\lambda }x,x〉}^p+{〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^p\right)}^{1/p}{\left({〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉}^q+{〈{\left|C\right|}^{2\lambda }x,x〉}^q\right)}^{1/q}\hfill \end{array}$$

for $x\in H$ and $p,$$q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1,$ which gives in a similar way (37). □

Remark 11. 

For $p=q=2$ in Theorem 3, we obtain the last upper bounds in Theorem 2, where $\lambda \in \left[0,1\right].$

For $\lambda =1/2$ in Theorem 3, we obtain

$$w_e^2\left(B,C\right)\le {∥{\left|B\right|}^p+{\left|C\right|}^p∥}^{1/p}{∥{\left|B^{*}\right|}^q+{\left|C^{*}\right|}^q∥}^{1/q}$$

and

$$w_e^2\left(B,C\right)\le {∥{\left|B\right|}^p+{\left|C^{*}\right|}^p∥}^{1/p}{∥{\left|B^{*}\right|}^q+{\left|C\right|}^q∥}^{1/q}$$

for $p,$$q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1.$

Several consequences of Theorem 3 can be derived. As a first step, we present the following corollary.

Corollary 11. 

For any $A\in \mathcal{B}\left(H\right),$ $\lambda \in \left[0,1\right]$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, we have

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥{\left|A\right|}^{2\lambda p}+{\left|A^{*}\right|}^{2\lambda p}∥}^{1/p}{∥{\left|A\right|}^{2\left(1-\lambda \right)q}+{\left|A^{*}\right|}^{2\left(1-\lambda \right)q}∥}^{1/q}$$

and

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥{\left|A\right|}^{2\lambda p}+{\left|A\right|}^{2\left(1-\lambda \right)p}∥}^{1/p}{∥{\left|A^{*}\right|}^{2\lambda q}+{\left|A^{*}\right|}^{2\left(1-\lambda \right)q}∥}^{1/q}.$$

In particular,

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}{∥{\left|A\right|}^p+{\left|A^{*}\right|}^p∥}^{1/p}{∥{\left|A\right|}^q+{\left|A^{*}\right|}^q∥}^{1/q}.$$

Proof. 

It follows by taking in Theorem 3 $B=A$ and $C=A^{*}$. □

Our next corollary follows by taking $B=\mathfrak{R}\left(A\right)$ and $C=\mathfrak{I}\left(A\right)$ in Theorem 3.

Corollary 12. 

For any $A\in \mathcal{B}\left(H\right),$ $\lambda \in \left[0,1\right]$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, we have

$$w^2\left(A\right)\le {∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\lambda p}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\lambda p}∥}^{1/p}{∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\left(1-\lambda \right)q}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\lambda \right)q}∥}^{1/q}$$

and

$$w_e^2\left(B,C\right)\le {∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\lambda p}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\lambda \right)p}∥}^{1/p}{∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\left(1-\lambda \right)q}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\lambda q}∥}^{1/q}.$$

In particular,

$$w^2\left(A\right)\le {∥{\left|\mathfrak{R}\left(A\right)\right|}^p+{\left|\mathfrak{I}\left(A\right)\right|}^p∥}^{1/p}{∥{\left|\mathfrak{R}\left(A\right)\right|}^q+{\left|\mathfrak{I}\left(A\right)\right|}^q∥}^{1/q}.$$

We derive also the next corollary.

Corollary 13. 

For any $T\in \mathcal{B}\left(H\right),$ $\lambda \in \left[0,1\right]$ and $p,q>1$ with ${\displaystyle \frac{1}{p}}+{\displaystyle \frac{1}{q}}=1$, we have

$$d{w}^2\left(T\right)\le {∥{\left|T\right|}^{2\lambda p}+{\left|T\right|}^{4\lambda p}∥}^{1/p}{∥{\left|T^{*}\right|}^{2\left(1-\lambda \right)q}+{\left|T\right|}^{4\left(1-\lambda \right)q}∥}^{1/q}$$

and

$$d{w}^2\left(T\right)\le {∥{\left|T\right|}^{2\lambda p}+{\left|T\right|}^{4\left(1-\lambda \right)p}∥}^{1/p}{∥{\left|T^{*}\right|}^{2\left(1-\lambda \right)q}+{\left|T\right|}^{4\lambda q}∥}^{1/q}.$$

In particular,

$$d{w}^2\left(T\right)\le {∥{\left|T\right|}^p+{\left|T\right|}^{2p}∥}^{1/p}{∥{\left|T^{*}\right|}^q+{\left|T\right|}^{2q}∥}^{1/q}.$$

Proof. 

The proof follows by taking $\left(B,C\right)=\left(T,{\left|T\right|}^2\right)$ in Theorem 3. □

We also have the following theorem.

Theorem 4. 

For any $B,C\in \mathcal{B}\left(H\right)$ and $\lambda \in \left[0,1\right],$ we have

$$\begin{array}{cc}\hfill w_e^2\left(B,C\right)& \le {\displaystyle \frac{1}{2}}\left[∥{\left|B\right|}^{2\lambda }+{\left|C\right|}^{2\lambda }∥+∥{\left|B\right|}^{2\lambda }-{\left|C\right|}^{2\lambda }∥\right]∥{\left|B^{*}\right|}^{2\left(1-\lambda \right)}+{\left|C^{*}\right|}^{2\left(1-\lambda \right)}∥\hfill \end{array}$$

(39)

and

$$\begin{array}{c}\hfill w_e^2\left(B,C\right)\le {\displaystyle \frac{1}{2}}\left[∥{\left|B\right|}^{2\lambda }+{\left|C^{*}\right|}^{2\left(1-\lambda \right)}∥+∥{\left|B\right|}^{2\lambda }-{\left|C^{*}\right|}^{2\left(1-\lambda \right)}∥\right]∥{\left|B^{*}\right|}^{2\left(1-\lambda \right)}+{\left|C\right|}^{2\lambda }∥.\end{array}$$

(40)

Proof. 

One can see that

$$\begin{array}{cc}\hfill & 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \\ \hfill & \le max\left\{〈{\left|B\right|}^{2\lambda }x,x〉,〈{\left|C\right|}^{2\lambda }x,x〉\right\}\left[〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[〈{\left|B\right|}^{2\lambda }x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉+\left|〈{\left|B\right|}^{2\lambda }x,x〉-〈{\left|C\right|}^{2\lambda }x,x〉\right|\right]\left[〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\right]\hfill \\ \hfill & ={\displaystyle \frac{1}{2}}\left[〈\left({\left|B\right|}^{2\lambda }+{\left|C\right|}^{2\lambda }\right)x,x〉+\left|〈\left({\left|B\right|}^{2\lambda }-{\left|C\right|}^{2\lambda }\right)x,x〉\right|\right]〈\left({\left|B^{*}\right|}^{2\left(1-\lambda \right)}+{\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)x,x〉\hfill \end{array}$$

for $x\in H.$

From (28), we obtain

$$\begin{array}{cc}\hfill & {\left|〈Bx,x〉\right|}^2+{\left|〈Cx,x〉\right|}^2\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left[〈\left({\left|B\right|}^{2\lambda }+{\left|C\right|}^{2\lambda }\right)x,x〉+\left|〈\left({\left|B\right|}^{2\lambda }-{\left|C\right|}^{2\lambda }\right)x,x〉\right|\right]〈\left({\left|B^{*}\right|}^{2\left(1-\lambda \right)}+{\left|C^{*}\right|}^{2\left(1-\lambda \right)}\right)x,x〉\hfill \end{array}$$

for $x\in H.$

By taking the supremum over $x\in H$, $∥x∥=1,$ we obtain the desired result (39).

Inequality (40) follows by the fact that

$$\begin{array}{cc}\hfill & 〈{\left|B\right|}^{2\lambda }x,x〉〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\hfill \\ \hfill & \le max\left\{〈{\left|B\right|}^{2\lambda }x,x〉,〈{\left|C^{*}\right|}^{2\left(1-\lambda \right)}x,x〉\right\}\left[〈{\left|B^{*}\right|}^{2\left(1-\lambda \right)}x,x〉+〈{\left|C\right|}^{2\lambda }x,x〉\right]\hfill \end{array}$$

for $x\in H.$ □

Remark 12. 

If we take $\lambda =1/2$ in Theorem 4, we obtain

$$w_e^2\left(B,C\right)\le {\displaystyle \frac{1}{2}}\left[∥\left|B\right|+\left|C\right|∥+∥\left|B\right|-\left|C\right|∥\right]∥\left|B^{*}\right|+\left|C^{*}\right|∥$$

and

$$w_e^2\left(B,C\right)\le {\displaystyle \frac{1}{2}}\left[∥\left|B\right|+\left|C^{*}\right|∥+∥\left|B\right|-\left|C^{*}\right|∥\right]∥\left|B^{*}\right|+\left|C\right|∥.$$

The first consequence of Theorem 4 is presented in the following corollary.

Corollary 14. 

For any $A\in \mathcal{B}\left(H\right)$ and $\lambda \in \left[0,1\right]$, we have

$$w^2\left(A\right)\le {\displaystyle \frac{1}{4}}\left[∥{\left|A\right|}^{2\lambda }+{\left|A^{*}\right|}^{2\lambda }∥+∥{\left|A\right|}^{2\lambda }-{\left|A^{*}\right|}^{2\lambda }∥\right]∥{\left|A^{*}\right|}^{2\left(1-\lambda \right)}+{\left|A\right|}^{2\left(1-\lambda \right)}∥$$

and

$$w^2\left(A\right)\le {\displaystyle \frac{1}{4}}\left[∥{\left|A\right|}^{2\lambda }+{\left|A\right|}^{2\left(1-\lambda \right)}∥+∥{\left|A\right|}^{2\lambda }-{\left|A\right|}^{2\left(1-\lambda \right)}∥\right]∥{\left|A^{*}\right|}^{2\left(1-\lambda \right)}+{\left|A^{*}\right|}^{2\lambda }∥.$$

In particular,

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

Proof. 

The proof follows by taking $B=A$ and $C=A^{*}$ in Theorem 4. □

The next corollary holds also.

Corollary 15. 

For any $A\in \mathcal{B}\left(H\right),$ $\lambda \in \left[0,1\right],$ we have

$$\begin{array}{cc}\hfill & w^2\left(A\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left[∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\lambda }+{\left|\mathfrak{I}\left(A\right)\right|}^{2\lambda }∥+∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\lambda }-{\left|\mathfrak{I}\left(A\right)\right|}^{2\lambda }∥\right]∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\left(1-\lambda \right)}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\lambda \right)}∥\hfill \end{array}$$

and

$$\begin{array}{cc}\hfill & w^2\left(A\right)\hfill \\ \hfill & \le {\displaystyle \frac{1}{2}}\left[∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\lambda }+{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\lambda \right)}∥+∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\lambda }-{\left|\mathfrak{I}\left(A\right)\right|}^{2\left(1-\lambda \right)}∥\right]∥{\left|\mathfrak{R}\left(A\right)\right|}^{2\left(1-\lambda \right)}+{\left|\mathfrak{I}\left(A\right)\right|}^{2\lambda }∥.\hfill \end{array}$$

In particular,

$$w^2\left(A\right)\le {\displaystyle \frac{1}{2}}\left[∥\left|\mathfrak{R}\left(A\right)\right|+\left|\mathfrak{I}\left(A\right)\right|∥+∥\left|\mathfrak{R}\left(A\right)\right|-\left|\mathfrak{I}\left(A\right)\right|∥\right]∥\left|\mathfrak{R}\left(A\right)\right|+\left|\mathfrak{I}\left(A\right)\right|∥.$$

Proof. 

The proof follows by taking $B=\mathfrak{R}\left(A\right)$ and $C=\mathfrak{I}\left(A\right)$ in Theorem 4. □

Finally, by selecting $B=T$ and $C={\left|T\right|}^2$ in Theorem 4, we can also state the following corollary.

Corollary 16. 

For any $T\in \mathcal{B}\left(H\right),$ $\lambda \in \left[0,1\right],$ we have

$$d{w}^2\left(T\right)\le {\displaystyle \frac{1}{2}}\left[∥{\left|T\right|}^{2\lambda }+{\left|T\right|}^{4\lambda }∥+∥{\left|T\right|}^{2\lambda }-{\left|T\right|}^{4\lambda }∥\right]∥{\left|T^{*}\right|}^{2\left(1-\lambda \right)}+{\left|T\right|}^{4\left(1-\lambda \right)}∥$$

and

$$d{w}^2\left(T\right)\le {\displaystyle \frac{1}{2}}\left[∥{\left|T\right|}^{2\lambda }+{\left|T\right|}^{4\left(1-\lambda \right)}∥+∥{\left|T\right|}^{2\lambda }-{\left|T\right|}^{4\left(1-\lambda \right)}∥\right]∥{\left|T^{*}\right|}^{2\left(1-\lambda \right)}+{\left|T\right|}^{4\lambda }∥.$$

In particular

$$d{w}^2\left(T\right)\le {\displaystyle \frac{1}{2}}\left[∥\left|T\right|+{\left|T\right|}^2∥+∥\left|T\right|-{\left|T\right|}^2∥\right]∥\left|T^{*}\right|+{\left|T\right|}^2∥.$$

4. Conclusions

In this paper, we have presented new lower and upper bounds for the Euclidean numerical radius of operator pairs in Hilbert spaces, improving upon recent results in the literature. We also derived new inequalities for the numerical radius and the Davis–Wielandt radius as natural consequences of our findings.

This work lays a strong foundation for future research in this area. In particular, extending these results to the setting of semi-Hilbert spaces would be a valuable direction to explore (see [22,23,24] for a solid background on operators in semi-Hilbert spaces). Another interesting direction for future work is to study the Berezin number of operator pairs in reproducing kernel Hilbert spaces (see [25,26] and the sources cited therein for background on reproducing kernel Hilbert spaces).

We hope this study will inspire further advancements in numerical radius inequalities and related concepts in operator theory.